Mechanical Engineering Design (SOLUTIONS MANUAL)

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shi20396_ch01.qxd 6/5/03 12:11 PM Page 1

Chapter 1 Problems 1-1 through 1-4 are for student research. 1-5

Impending motion to left E

1

1

f

f

A

B

D

G Fcr ␪cr

C

F

␪

Facc

Consider force F at G, reactions at B and D. Extend lines of action for fully-developed friction D E and B E to find the point of concurrency at E for impending motion to the left. The critical angle is θcr . Resolve force F into components Facc and Fcr . Facc is related to mass and acceleration. Pin accelerates to left for any angle 0 < θ < θcr . When θ > θcr , no magnitude of F will move the pin. Impending motion to right E⬘

1 f

⭈E

1 f B

A

G

d

⬘ Fcr D

C

⬘ ␪cr

F⬘ ␪⬘ F⬘acc

Consider force F at G, reactions at A and C. Extend lines of action for fully-developed friction AE and C E to find the point of concurrency at E for impending motion to the left. The and F . F is related to mass critical angle is θcr . Resolve force F into components Facc acc cr and acceleration. Pin accelerates to right for any angle 0 < θ < θcr . When θ > θcr , no magnitude of F will move the pin. The intent of the question is to get the student to draw and understand the free body in order to recognize what it teaches. The graphic approach accomplishes this quickly. It is important to point out that this understanding enables a mathematical model to be constructed, and that there are two of them. This is the simplest problem in mechanical engineering. Using it is a good way to begin a course. What is the role of pin diameter d? Yes, changing the sense of F changes the response.

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2

Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

1-6

(a)

y

F

T r

␪ N

Fy = −F − f N cos θ + N sin θ = 0

(1)

T =0 r Ans.

Fx = f N sin θ + N cos θ − F = N (sin θ − f cos θ) T = Nr( f sin θ + cos θ)

x

fN

Combining T = Fr

1 + f tan θ = KFr tan θ − f

(b) If T → ∞ detent self-locking tan θ − f = 0 (Friction is fully developed.) Check: If

F = 10 lbf, N=

Ans.

∴ θcr = tan−1 f

f = 0.20, θ = 45◦ ,

(2) Ans.

r = 2 in

10 = 17.68 lbf −0.20 cos 45◦ + sin 45◦

T = 17.28(0.20 sin 45◦ + cos 45◦ ) = 15 lbf r f N = 0.20(17.28) = 3.54 lbf θcr = tan−1 f = tan−1 (0.20) = 11.31◦ 11.31° < θ < 90° 1-7 (a) F = F0 + k(0) = F0 T1 = F0r Ans. (b) When teeth are about to clear F = F0 + kx2 From Prob. 1-6 T2 = Fr T2 = r

f tan θ + 1 tan θ − f

( F0 + kx2 )( f tan θ + 1) tan θ − f

Ans.

1-8 Given, F = 10 + 2.5x lbf, r = 2 in, h = 0.2 in, θ = 60◦ , f = 0.25, xi = 0, x f = 0.2 Fi = 10 lbf; Ff = 10 + 2.5(0.2) = 10.5 lbf Ans.

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3

Chapter 1

From Eq. (1) of Prob. 1-6 N=

F − f cos θ + sin θ 10 = 13.49 lbf Ni = −0.25 cos 60◦ + sin 60◦

Ans.

10.5 13.49 = 14.17 lbf Ans. 10

Nf = From Eq. (2) of Prob. 1-6 K =

1 + f tan θ 1 + 0.25 tan 60◦ = = 0.967 Ans. tan θ − f tan 60◦ − 0.25

Ti = 0.967(10)(2) = 19.33 lbf · in Tf = 0.967(10.5)(2) = 20.31 lbf · in 1-9 (a) Point vehicles v x

Q=

v 42.1v − v 2 cars = = hour x 0.324

Seek stationary point maximum 42.1 − 2v dQ =0= ∴ v* = 21.05 mph dv 0.324 Q* =

42.1(21.05) − 21.052 = 1367.6 cars/h Ans. 0.324

(b)

v l 2

v Q= = x +l

x

l 2

l 0.324 + v(42.1) − v 2 v

−1

Maximize Q with l = 10/5280 mi v 22.18 22.19 22.20 22.21 22.22 % loss of throughput

Q 1221.431 1221.433 1221.435 ← 1221.435 1221.434 1368 − 1221 = 12% 1221

Ans.

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4


22.2 − 21.05 = 5.5% 21.05 Modest change in optimal speed Ans.

(c) % increase in speed

1-10

This and the following problem may be the student’s first experience with a figure of merit. • Formulate fom to reflect larger figure of merit for larger merit. • Use a maximization optimization algorithm. When one gets into computer implementation and answers are not known, minimizing instead of maximizing is the largest error one can make. FV = F1 sin θ − W = 0 FH = −F1 cos θ − F2 = 0 From which F1 = W/sin θ F2 = −W cos θ/sin θ fom = −S = −¢γ (volume) . = −¢γ(l1 A1 + l2 A2 ) W F1 l1 = , l2 = A1 = S S sin θ cos θ F2 W cos θ A2 = = S S sin θ l2 W cos θ l2 W + fom = −¢γ cos θ S sin θ S sin θ −¢γ W l2 1 + cos2 θ = S cos θ sin θ Set leading constant to unity θ◦

fom

0 20 30 40 45 50 54.736 60

−∞ −5.86 −4.04 −3.22 −3.00 −2.87 −2.828 −2.886

θ* = 54.736◦ fom* = −2.828

Ans.

Alternative: d 1 + cos2 θ =0 dθ cos θ sin θ And solve resulting transcendental for θ*.

Check second derivative to see if a maximum, minimum, or point of inflection has been found. Or, evaluate fom on either side of θ*.

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Chapter 1

1-11 (a) x1 + x2 = X 1 + e1 + X 2 + e2 error = e = (x1 + x2 ) − ( X 1 + X 2 ) = e1 + e2 Ans. (b) x1 − x2 = X 1 + e1 − ( X 2 + e2 ) e = (x1 − x2 ) − ( X 1 − X 2 ) = e1 − e2 Ans. (c) x1 x2 = ( X 1 + e1 )( X 2 + e2 ) e = x1 x2 − X 1 X 2 = X 1 e2 + X 2 e1 + e1 e2 e2 e1 . Ans. + = X 1 e2 + X 2 e1 = X 1 X 2 X1 X2 x1 X 1 + e1 X 1 1 + e1 / X 1 = = (d) x2 X 2 + e2 X 2 1 + e2 / X 2 −1 e2 e2 e1 e2 . e1 e2 . 1+ =1− and 1+ 1− =1+ − X2 X2 X1 X2 X1 X2 x1 X 1 . X 1 e1 e2 e= − − = Ans. x2 X2 X2 X1 X2 1-12 (a)

x1 =

√ 5 = 2.236 067 977 5

X 1 = 2.23 3-correct digits √ x2 = 6 = 2.449 487 742 78 X 2 = 2.44 3-correct digits √ √ x1 + x2 = 5 + 6 = 4.685 557 720 28 √ e1 = x1 − X 1 = 5 − 2.23 = 0.006 067 977 5 √ e2 = x2 − X 2 = 6 − 2.44 = 0.009 489 742 78 √ √ e = e1 + e2 = 5 − 2.23 + 6 − 2.44 = 0.015 557 720 28 Sum = x1 + x2 = X 1 + X 2 + e = 2.23 + 2.44 + 0.015 557 720 28 = 4.685 557 720 28 (Checks) Ans. (b) X 1 = 2.24, X 2 = 2.45 √ e1 = 5 − 2.24 = −0.003 932 022 50 √ e2 = 6 − 2.45 = −0.000 510 257 22 e = e1 + e2 = −0.004 442 279 72 Sum = X 1 + X 2 + e = 2.24 + 2.45 + (−0.004 442 279 72) = 4.685 557 720 28 Ans.

5

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6


1-13 (a) (b) (c) (d) (e) (f) (g) (h) (i)

σ = 20(6.89) = 137.8 MPa F = 350(4.45) = 1558 N = 1.558 kN M = 1200 lbf · in (0.113) = 135.6 N · m A = 2.4(645) = 1548 mm2 I = 17.4 in4 (2.54) 4 = 724.2 cm4 A = 3.6(1.610) 2 = 9.332 km2 E = 21(1000)(6.89) = 144.69(103 ) MPa = 144.7 GPa v = 45 mi/h (1.61) = 72.45 km/h V = 60 in3 (2.54) 3 = 983.2 cm3 = 0.983 liter

(a) (b) (c) (d) (e) (f) (g) (h) (i)

l = 1.5/0.305 = 4.918 ft = 59.02 in σ = 600/6.89 = 86.96 kpsi p = 160/6.89 = 23.22 psi Z = 1.84(105 )/(25.4) 3 = 11.23 in3 w = 38.1/175 = 0.218 lbf/in δ = 0.05/25.4 = 0.00197 in v = 6.12/0.0051 = 1200 ft/min = 0.0021 in/in V = 30/(0.254) 3 = 1831 in3

1-14

1-15 200 = 13.1 MPa 15.3 42(103 ) = 70(106 ) N/m2 = 70 MPa (b) σ = 6(10−2 ) 2

(a) σ =

(c) y =

1200(800) 3 (10−3 ) 3 = 1.546(10−2 ) m = 15.5 mm 3(207)(6.4)(109 )(10−2 ) 4

(d) θ =

1100(250)(10−3 ) = 9.043(10−2 ) rad = 5.18◦ 4 9 − 3 4 79.3(π/32)(25) (10 )(10 )

1-16 600 = 5 MPa 20(6) 1 (b) I = 8(24) 3 = 9216 mm4 12 π (c) I = 324 (10−1 ) 4 = 5.147 cm4 64 16(16) = 5.215(106 ) N/m2 = 5.215 MPa (d) τ = π(253 )(10−3 ) 3 (a) σ =

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Chapter 1

1-17 (a) τ =

120(103 ) = 382 MPa (π/4)(202 )

32(800)(800)(10−3 ) = 198.9(106 ) N/m2 = 198.9 MPa π(32) 3 (10−3 ) 3 π (364 − 264 ) = 3334 mm3 (c) Z = 32(36) (b) σ =

(d) k =

(1.6) 4 (79.3)(10−3 ) 4 (109 ) = 286.8 N/m 8(19.2) 3 (32)(10−3 ) 3

7

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Chapter 2 2-1 (a)

12 10 8 6 4 2 0

60

70

80

90 100 110 120 130 140 150 160 170 180 190 200 210

(b) f/(Nx) = f/(69 · 10) = f/690

Eq. (2-9) Eq. (2-10)

x

f

fx

f x2

f/(Nx)

60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210

2 1 3 5 8 12 6 10 8 5 2 3 2 1 0 1 69

120 70 240 450 800 1320 720 1300 1120 750 320 510 360 190 0 210 8480

7200 4900 19 200 40 500 80 000 145 200 86 400 169 000 156 800 112 500 51 200 86 700 64 800 36 100 0 44 100 1 104 600

0.0029 0.0015 0.0043 0.0072 0.0116 0.0174 0.0087 0.0145 0.0116 0.0174 0.0029 0.0043 0.0029 0.0015 0 0.0015

8480 = 122.9 kcycles 69 1/2 1 104 600 − 84802 /69 sx = 69 − 1 x¯ =

= 30.3 kcycles

Ans.

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9

Chapter 2

2-2 Data represents a 7-class histogram with N = 197. x

f

fx

f x2

174 182 190 198 206 214 220

6 9 44 67 53 12 6 197

1044 1638 8360 13 266 10 918 2568 1320 39 114

181 656 298 116 1 588 400 2 626 688 2 249 108 549 552 290 400 7 789 900

39 114 = 198.55 kpsi Ans. 197 1/2 7 783 900 − 39 1142 /197 sx = 197 − 1 x¯ =

= 9.55 kpsi Ans. 2-3 Form a table: x

f

fx

f x2

64 68 72 76 80 84 88 92

2 6 6 9 19 10 4 2 58

128 408 432 684 1520 840 352 184 4548

8192 27 744 31 104 51 984 121 600 70 560 30 976 16 928 359 088

4548 = 78.4 kpsi 58 1/2 359 088 − 45482 /58 sx = = 6.57 kpsi 58 − 1 x¯ =

From Eq. (2-14)

1 x − 78.4 2 1 f (x) = √ exp − 2 6.57 6.57 2π

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2-4 (a) y

f

fy

f y2

y

f/(Nw)

f (y)

g(y)

5.625 5.875 6.125 6.375 6.625 6.875 7.125 7.375 7.625 7.875 8.125

1 0 0 3 3 6 14 15 10 2 1 55

5.625 0 0 19.125 19.875 41.25 99.75 110.625 76.25 15.75 8.125 396.375

31.640 63 0 0 121.9219 131.6719 283.5938 710.7188 815.8594 581.4063 124.0313 66.015 63 2866.859

5.625 5.875 6.125 6.375 6.625 6.875 7.125 7.375 7.625 7.875 8.125

0.072 727 0 0 0.218 182 0.218 182 0.436 364 1.018 182 1.090 909 0.727 273 0.145 455 0.072 727

0.001 262 0.008 586 0.042 038 0.148 106 0.375 493 0.685 057 0.899 389 0.849 697 0.577 665 0.282 608 0.099 492

0.000 295 0.004 088 0.031 194 0.140 262 0.393 667 0.725 002 0.915 128 0.822 462 0.544 251 0.273 138 0.106 72

For a normal distribution,

2866.859 − (396.3752 /55) y¯ = 396.375/55 = 7.207, sy = 55 − 1 1 x − 7.207 2 1 f ( y) = √ exp − 2 0.4358 0.4358 2π

1/2 = 0.4358

For a lognormal distribution, √ √ x¯ = ln 7.206 818 − ln 1 + 0.060 4742 = 1.9732, sx = ln 1 + 0.060 4742 = 0.0604 1 ln x − 1.9732 2 1 exp − g( y) = √ 2 0.0604 x(0.0604)( 2π) (b) Histogram f 1.2

Data N LN

1 0.8 0.6 0.4 0.2 0 5.63

5.88

6.13

6.38

6.63

6.88 7.13 log N

7.38

7.63

7.88

8.13

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Chapter 2

11

2-5 Distribution is uniform in interval 0.5000 to 0.5008 in, range numbers are a = 0.5000, b = 0.5008 in. (a) Eq. (2-22)

µx =

0.5000 + 0.5008 a+b = = 0.5004 2 2

Eq. (2-23)

σx =

0.5008 − 0.5000 b−a = 0.000 231 √ = √ 2 3 2 3

(b) PDF from Eq. (2-20) f (x) =

1250 0.5000 ≤ x ≤ 0.5008 in 0 otherwise

(c) CDF from Eq. (2-21)  x < 0.5000 0 F(x) = (x − 0.5)/0.0008 0.5000 ≤ x ≤ 0.5008  1 x > 0.5008 If all smaller diameters are removed by inspection, a = 0.5002, b = 0.5008 µx =

0.5002 + 0.5008 = 0.5005 in 2

0.5008 − 0.5002 = 0.000 173 in √ 2 3 1666.7 0.5002 ≤ x ≤ 0.5008 f (x) = 0 otherwise σˆ x =

 x < 0.5002 0 F(x) = 1666.7(x − 0.5002) 0.5002 ≤ x ≤ 0.5008  1 x > 0.5008 2-6 Dimensions produced are due to tool dulling and wear. When parts are mixed, the distribution is uniform. From Eqs. (2-22) and (2-23), √ √ a = µx − 3s = 0.6241 − 3(0.000 581) = 0.6231 in √ √ b = µx + 3s = 0.6241 + 3(0.000 581) = 0.6251 in We suspect the dimension was

0.623 in 0.625

Ans.

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2-7 F(x) = 0.555x − 33 mm (a) Since F(x) is linear, the distribution is uniform at x = a F(a) = 0 = 0.555(a) − 33 ∴ a = 59.46 mm. Therefore, at x = b F(b) = 1 = 0.555b − 33 ∴ b = 61.26 mm. Therefore,  x < 59.46 mm 0 F(x) = 0.555x − 33 59.46 ≤ x ≤ 61.26 mm  1 x > 61.26 mm The PDF is d F/dx , thus the range numbers are: 0.555 59.46 ≤ x ≤ 61.26 mm f (x) = 0 otherwise

Ans.

From the range numbers, µx =

59.46 + 61.26 = 60.36 mm Ans. 2

61.26 − 59.46 = 0.520 mm Ans. σˆ x = √ 2 3

1

(b) σ is an uncorrelated quotient F¯ = 3600 lbf, A¯ = 0.112 in2 C F = 300/3600 = 0.083 33,

C A = 0.001/0.112 = 0.008 929

From Table 2-6, for σ µF 3600 = 32 143 psi Ans. = µA 0.112 1/2 (0.083332 + 0.0089292 ) σˆ σ = 32 143 = 2694 psi Ans. (1 + 0.0089292 ) σ¯ =

Cσ = 2694/32 143 = 0.0838 Ans. Since F and A are lognormal, division is closed and σ is lognormal too. σ = LN(32 143, 2694) psi Ans.

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13

Chapter 2

2-8 Cramer’s rule

y x 2 x y x 3 yx 3 − x yx 2 = a1 = Ans. x x 2 xx 3 − (x 2 ) 2 x 2 x 3 y x 2 xx y − yx 2 x x y = a2 = Ans. x x 2 xx 3 − (x 2 ) 2 x 2 x 3 x

y

x2

x3

xy

0 0.2 0.4 0.6 0.8 1.0 3.0

0.01 0.15 0.25 0.25 0.17 −0.01 0.82

0 0.04 0.16 0.36 0.64 1.00 2.20

0 0.008 0.064 0.216 0.512 1.000 1.800

0 0.030 0.100 0.150 0.136 −0.010 0.406

a1 = 1.040 714

a2 = −1.046 43

Ans.

x

Data y

Regression y

0 0.2 0.4 0.6 0.8 1.0

0.01 0.15 0.25 0.25 0.17 −0.01

0 0.166 286 0.248 857 0.247 714 0.162 857 −0.005 71

y 0.3

Data Regression

0.25 0.2 0.15 0.1 0.05 0 0.05

x 0

0.2

0.4

0.6

0.8

1

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2-9 Su 0 60 64 65 82 101 119 120 130 134 145 180 195 205 207 210 213 225 225 227 230 238 242 265 280 295 325 325 355 5462

Data Se 30 48 29.5 45 51 50 48 67 60 64 84 78 96 87 87 75 99 87 116 105 109 106 105 96 99 114 117 122 2274.5

m = 0.312 067 Se 140

Regression Se 20.356 75 39.080 78 40.329 05 40.641 12 45.946 26 51.875 54 57.492 75 57.804 81 60.925 48 62.173 75 65.606 49 76.528 84 81.209 85 84.330 52 84.954 66 85.890 86 86.827 06 90.571 87 90.571 87 91.196 92.132 2 94.628 74 95.877 01 103.054 6 107.735 6 112.416 6 121.778 6 121.778 6 131.140 6

Su2

Su Se

3 600 4 096 4 225 6 724 10 201 14 161 14 400 16 900 17 956 21 025 32 400 38 025 42 025 42 849 44 100 45 369 50 625 50 625 51 529 52 900 56 644 58 564 70 225 78 400 87 025 105 625 105 625 126 025 1 251 868

1 800 3 072 1 917.5 3 690 5 151 5 950 5 760 8 710 8 040 9 280 15 120 15 210 19 680 18 009 18 270 15 975 22 275 19 575 26 332 24 150 25 942 25 652 27 825 26 880 29 205 37 050 38 025 43 310 501 855.5

b = 20.356 75

Ans.

Data Regression

120 100 80 60 40 20 0

0

100

200

300

400

Su

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15

Chapter 2

2-10

E

=

y − a0 − a2 x 2

2

∂E = −2 y − a0 − a2 x 2 = 0 ∂a0 y = na0 + a2 x2 = 0 ⇒ x2 y − na0 − a2 ∂E =2 y − a0 − a2 x 2 (2x) = 0 ⇒ x + a2 x3 x y = a0 ∂a2 Cramer’s rule

y x y a0 = n x n x a2 = n x

x 2 x 3 x 3 y − x 2 x y = x 2 nx 3 − xx 2 3 x y nx y − xy x y 2 = x nx 3 − xx 2 x 3

x

Data y

Regression y

x2

20 40 60 80 200

19 17 13 7 56

19.2 16.8 12.8 7.2

400 1600 3600 6400 12 000

x3

xy

8 000 380 64 000 680 216 000 780 512 000 560 800 000 2400

a0 =

800 000(56) − 12 000(2400) = 20 4(800 000) − 200(12 000)

a2 =

4(2400) − 200(56) = −0.002 4(800 000) − 200(12 000)

y 25

Data Regression

20

15

10

5

0

x 0

20

40

60

80

100

Ans.

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2-11

x

Data y

Regression y

0.2 0.4 0.6 0.8 1 2 5

7.1 10.3 12.1 13.8 16.2 25.2 84.7

7.931 803 9.884 918 11.838 032 13.791 147 15.744 262 25.509 836

x2

y2

xy

x − x¯

(x − x) ¯ 2

0.04 0.16 0.36 0.64 1.00 4.00 6.2

50.41 106.09 146.41 190.44 262.44 635.04 1390.83

1.42 4.12 7.26 11.04 16.20 50.40 90.44

−0.633 333 −0.433 333 −0.233 333 −0.033 333 0.166 666 1.166 666 0

0.401 111 111 0.187 777 778 0.054 444 444 0.001 111 111 0.027 777 778 1.361 111 111 2.033 333 333

mˆ = k =

6(90.44) − 5(84.7) = 9.7656 6(6.2) − (5) 2

84.7 − 9.7656(5) = 5.9787 bˆ = Fi = 6 F 30

Data Regression

25 20 15 10 5 0

x 0

0.5

5 x¯ = ; 6

(a)

1

y¯ =

1.5

2

2.5

84.7 = 14.117 6

Eq. (2-37) s yx =

1390.83 − 5.9787(84.7) − 9.7656(90.44) 6−2

= 0.556 Eq. (2-36) sbˆ = 0.556

1 (5/6) 2 + = 0.3964 lbf 6 2.0333

Fi = (5.9787, 0.3964) lbf

Ans.

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17

Chapter 2

(b) Eq. (2-35) 0.556 = 0.3899 lbf/in smˆ = √ 2.0333 k = (9.7656, 0.3899) lbf/in 2-12

Ans.

The expression = δ/l is of the form x/y. Now δ = (0.0015, 0.000 092) in, unspecified distribution; l = (2.000, 0.0081) in, unspecified distribution; C x = 0.000 092/0.0015 = 0.0613 C y = 0.0081/2.000 = 0.000 75 From Table 2-6, ¯ = 0.0015/2.000 = 0.000 75 1/2 0.06132 + 0.004 052 σˆ = 0.000 75 1 + 0.004 052 = 4.607(10−5 ) = 0.000 046 We can predict ¯ and σˆ but not the distribution of .

2-13

σ = E = (0.0005, 0.000 034) distribution unspecified; E = (29.5, 0.885) Mpsi, distribution unspecified; C x = 0.000 034/0.0005 = 0.068, C y = 0.0885/29.5 = 0.030 σ is of the form x, y Table 2-6 σ¯ = ¯ E¯ = 0.0005(29.5)106 = 14 750 psi σˆ σ = 14 750(0.0682 + 0.0302 + 0.0682 + 0.0302 ) 1/2 = 1096.7 psi Cσ = 1096.7/14 750 = 0.074 35

2-14 δ=

Fl AE

F = (14.7, 1.3) kip, A = (0.226, 0.003) in2 , l = (1.5, 0.004) in, E = (29.5, 0.885) Mpsi distributions unspecified. C F = 1.3/14.7 = 0.0884 ;

C A = 0.003/0.226 = 0.0133 ;

C E = 0.885/29.5 = 0.03 Mean of δ:

Fl 1 1 = Fl δ= AE A E

Cl = 0.004/1.5 = 0.00267 ;

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From Table 2-6, ¯ A)(1/ ¯ ¯ δ¯ = F¯ l(1/ E) 1 1 δ¯ = 14 700(1.5) 0.226 29.5(106 ) = 0.003 31 in

Ans.

For the standard deviation, using the first-order terms in Table 2-6,

1/2 1/2 . F¯ l¯ 2 = δ¯ C F2 + Cl2 + C 2A + C 2E σˆ δ = C F + Cl2 + C 2A + C E2 A¯ E¯ σˆ δ = 0.003 31(0.08842 + 0.002672 + 0.01332 + 0.032 ) 1/2 = 0.000 313 in

Ans.

COV Cδ = 0.000 313/0.003 31 = 0.0945

Ans.

Force COV dominates. There is no distributional information on δ. 2-15

M = (15 000, 1350) lbf · in, distribution unspecified; d = (2.00, 0.005) in distribution unspecified. σ=

32M , πd3

CM =

1350 = 0.09 , 15 000

Cd =

0.005 = 0.0025 2.00

σ is of the form x/y, Table 2-6. Mean: σ¯ =

32 M¯ . 32 M¯ 32(15 000) = = π(23 ) π d¯3 π d3

= 19 099 psi

Ans.

Standard Deviation:

From Table 2-6,

1/2 2 σˆ σ = σ¯ C M + Cd23 1 + Cd23 . Cd 3 = 3Cd = 3(0.0025) = 0.0075 1/2 2 + (3Cd ) 2 (1 + (3Cd )) 2 σˆ σ = σ¯ C M = 19 099[(0.092 + 0.00752 )/(1 + 0.00752 )]1/2 = 1725 psi

Ans.

COV: Cσ =

1725 = 0.0903 19 099

Ans.

Stress COV dominates. No information of distribution of σ.

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19

Chapter 2

2-16 f (x)

␣

␤ x1

x2

x

Fraction discarded is α + β. The area under the PDF was unity. Having discarded α + β fraction, the ordinates to the truncated PDF are multiplied by a. a=

1 1 − (α + β)

New PDF, g(x) , is given by

f (x)/[1 − (α + β)] x1 ≤ x ≤ x2 0 otherwise

g(x) =

More formal proof: g(x) has the property

x2

1=

x1

f (x) dx

x1

1=a

x2

g(x) dx = a ∞

−∞

f (x) dx −

x1

f (x) dx −

0

∞

f (x) dx

x2

1 = a {1 − F(x1 ) − [1 − F(x2 )]} a=

1 1 1 = = F(x2 ) − F(x1 ) (1 − β) − α 1 − (α + β)

2-17 (a) d = U[0.748, 0.751] µd =

0.751 + 0.748 = 0.7495 in 2

σˆ d =

0.751 − 0.748 = 0.000 866 in √ 2 3

f (x) =

1 1 = = 333.3 in−1 b−a 0.751 − 0.748

F(x) =

x − 0.748 = 333.3(x − 0.748) 0.751 − 0.748

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F(x1 ) = F(0.748) = 0

(b)

F(x2 ) = (0.750 − 0.748)333.3 = 0.6667 If g(x) is truncated, PDF becomes g(x) =

g(x) 500 f (x) 333.3

0.748

0.749

0.750

0.751

x

333.3 f (x) = = 500 in−1 F(x2 ) − F(x1 ) 0.6667 − 0

µx =

0.748 + 0.750 a + b = = 0.749 in 2 2

σˆ x =

0.750 − 0.748 b − a = 0.000 577 in √ = √ 2 3 2 3

2-18 From Table A-10, 8.1% corresponds to z 1 = −1.4 and 5.5% corresponds to z 2 = +1.6. k1 = µ + z 1 σˆ k2 = µ + z 2 σˆ From which 1.6(9) − (−1.4)11 z 2 k1 − z 1 k2 = z2 − z1 1.6 − (−1.4)

µ=

= 9.933 σˆ = =

k2 − k1 z2 − z1 11 − 9 = 0.6667 1.6 − (−1.4)

The original density function is

1 f (k) = √ exp − 2 0.6667 2π 1

k − 9.933 0.6667

2-19 From Prob. 2-1, µ = 122.9 kcycles and σˆ = 30.3 kcycles. z 10 =

x10 − 122.9 x10 − µ = σˆ 30.3

x10 = 122.9 + 30.3z 10 From Table A-10, for 10 percent failure, z 10 = −1.282 x10 = 122.9 + 30.3(−1.282) = 84.1 kcycles Ans.

2 Ans.

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21

Chapter 2

2-20 x

f

fx

f x2

x

f/(Nw)

f (x)

60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210

2 1 3 5 8 12 6 10 8 5 2 3 2 1 0 1 69

120 70 240 450 800 1320 720 1300 1120 750 320 510 360 190 0 210 8480

7200 4900 19 200 40 500 80 000 145 200 86 400 169 000 156 800 112 500 51 200 86 700 64 800 36 100 0 44 100

60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210

0.002 899 0.001 449 0.004 348 0.007 246 0.011 594 0.017 391 0.008 696 0.014 493 0.011 594 0.007 246 0.002 899 0.004 348 0.002 899 0.001 449 0 0.001 449

0.000 399 0.001 206 0.003 009 0.006 204 0.010 567 0.014 871 0.017 292 0.016 612 0.013 185 0.008 647 0.004 685 0.002 097 0.000 776 0.000 237 5.98E-05 1.25E-05

x¯ = 122.8986

sx = 22.887 19

x

f/(Nw)

f (x)

x

f/(Nw)

f (x)

55 55 65 65 75 75 85 85 95 95 105 105 115 115 125 125 135 135

0 0.002 899 0.002 899 0.001 449 0.001 449 0.004 348 0.004 348 0.007 246 0.007 246 0.011 594 0.011 594 0.017 391 0.017 391 0.008 696 0.008 696 0.014 493 0.014 493 0.011 594

0.000 214 0.000 214 0.000 711 0.000 711 0.001 951 0.001 951 0.004 425 0.004 425 0.008 292 0.008 292 0.012 839 0.012 839 0.016 423 0.016 423 0.017 357 0.017 357 0.015 157 0.015 157

145 145 155 155 165 165 175 175 185 185 195 195 205 205 215 215

0.011 594 0.007 246 0.007 246 0.002 899 0.002 899 0.004 348 0.004 348 0.002 899 0.002 899 0.001 449 0.001 449 0 0 0.001 499 0.001 499 0

0.010 935 0.010 935 0.006 518 0.006 518 0.003 21 0.003 21 0.001 306 0.001 306 0.000 439 0.000 439 0.000 122 0.000 122 2.8E-05 2.8E-05 5.31E-06 5.31E-06

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f 0.02

Histogram PDF

0.018 0.016 0.014 0.012 0.01 0.008 0.006 0.004 0.002 0

0

50

100

150

200

250

x

2-21 x

f

fx

f x2

f/(Nw)

f (x)

174 182 190 198 206 214 222 1386

6 9 44 67 53 12 6 197

1044 1638 8360 13 266 10 918 2568 1332 39 126

181 656 298 116 1 588 400 2 626 668 2 249 108 549 552 295 704 7 789 204

0.003 807 0.005 711 0.027 919 0.042 513 0.033 629 0.007 614 0.003 807

0.001 642 0.009 485 0.027 742 0.041 068 0.030 773 0.011 671 0.002 241

x¯ = 198.6091 x 170 170 178 178 186 186 194 194 202 202 210 210 218 218 226 226

f/(Nw) 0 0.003 807 0.003 807 0.005 711 0.005 711 0.027 919 0.027 919 0.042 513 0.042 513 0.033 629 0.033 629 0.007 614 0.007 614 0.003 807 0.003 807 0

sx = 9.695 071

f (x) 0.000 529 0.000 529 0.004 297 0.004 297 0.017 663 0.017 663 0.036 752 0.036 752 0.038 708 0.038 708 0.020 635 0.020 635 0.005 568 0.005 568 0.000 76 0.000 76

f 0.045 0.04

Data PDF

0.035 0.03 0.025 0.02 0.015 0.01 0.005 0 150

x 170

190

210

230

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23

Chapter 2

2-22 f

fx

f x2

f/(Nw)

2 6 6 9 19 10 4 2 58

128 408 432 684 1520 840 352 184 4548

8192 27 744 31 104 51 984 121 600 70 560 30 976 16 928 359 088

0.008 621 0.025 862 0.025 862 0.038 793 0.081 897 0.043 103 0.017 241 0.008 621

x 64 68 72 76 80 84 88 92 624

x¯ = 78.413 79

f (x) 0.005 48 0.017 299 0.037 705 0.056 742 0.058 959 0.042 298 0.020 952 0.007 165

sx = 6.572 229

x

f/(Nw)

f(x)

x

f/(Nw)

f(x)

62 62 66 66 70 70 74 74 78 78

0 0.008 621 0.008 621 0.025 862 0.025 862 0.025 862 0.025 862 0.038 793 0.038 793 0.081 897

0.002 684 0.002 684 0.010 197 0.010 197 0.026 749 0.026 749 0.048 446 0.048 446 0.060 581 0.060 581

82 82 86 86 90 90 94 94

0.081 897 0.043 103 0.043 103 0.017 241 0.017 241 0.008 621 0.008 621 0

0.052 305 0.052 305 0.031 18 0.031 18 0.012 833 0.012 833 0.003 647 0.003 647

f 0.09

Data PDF

0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01

x

0 60

70

80

90

2-23 σ¯ =

4(40) 4 P¯ = 50.93 kpsi = 2 πd π(12 )

σˆ σ =

4 σˆ P 4(8.5) = 10.82 kpsi = 2 πd π(12 )

σˆ s y = 5.9 kpsi

100

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For no yield, m = S y − σ ≥ 0 m − µm 0 − µm µm = =− σˆ m σˆ m σˆ m ¯ µm = S y − σ¯ = 27.47 kpsi, 1/2 2 2 σˆ m = σˆ σ + σˆ Sy = 12.32 kpsi z=

z=

−27.47 = −2.230 12.32

m 0

From Table A-10, p f = 0.0129 R = 1 − p f = 1 − 0.0129 = 0.987 2-24

Ans.

For a lognormal distribution,

µ y = ln µx − ln 1 + C x2 σˆ y = ln 1 + C x2

Eq. (2-18) Eq. (2-19) From Prob. (2-23)

µm = S¯ y − σ¯ = µx 2 2 µ y = ln S¯ y − ln 1 + C Sy − ln σ¯ − ln 1 + Cσ

S¯ y = ln σ¯

1 + Cσ2 1 + C S2y

1/2 σˆ y = ln 1 + C S2y + ln 1 + Cσ2 2 2 = ln 1 + C Sy 1 + Cσ S¯ y 1 + Cσ2 ln σ¯ 1 + C S2 µ y z = − = −

σˆ ln 1 + C S2y 1 + Cσ2 σ¯ =

4(30) 4 P¯ = 38.197 kpsi = 2 πd π(12 )

4 σˆ P 4(5.1) = 6.494 kpsi = 2 πd π(12 ) 6.494 = 0.1700 Cσ = 38.197 3.81 C Sy = = 0.076 81 49.6 σˆ σ =

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25

Chapter 2

 ln 

49.6 38.197



1 + 0.170  1 + 0.076 812 2

z = − = −1.470 ln (1 + 0.076 812 )(1 + 0.1702 ) From Table A-10 p f = 0.0708 R = 1 − p f = 0.929

Ans.

2-25 (a) a = 1.000 ± 0.001 in b = 2.000 ± 0.003 in c = 3.000 ± 0.005 in d = 6.020 ± 0.006 in w¯ = d − a − b − c = 6.020 − 1 − 2 − 3 = 0.020 in ! tw = tall = 0.001 + 0.003 + 0.005 + 0.006 = 0.015 in w = 0.020 ± 0.015 in

Ans.

(b) w¯ = 0.020 0.001 2 0.003 2 0.005 2 0.006 2 2 + √ + √ + √ σˆ w = σˆ all = √ 3 3 3 3 = 0.004 86 → 0.005 in (uniform) w = 0.020 ± 0.005 in Ans. 2-26 V + V = (a + a)(b + b)(c + c) V + V = abc + bca + acb + abc + small higher order terms V . a b c = + + Ans. a b c V¯ V¯ = a¯ b¯ c¯ = 1.25(1.875)(2.75) = 6.4453 in3 V 0.001 0.002 0.003 + + = 0.00296 = 1.250 1.875 2.750 V¯ V ¯ V = V = 0.00296(6.4453) = 0.0191 in3 V¯ Lower range number: V¯ − V = 6.4453 − 0.0191 = 6.4262 in3

Ans.

Upper range number: V¯ + V = 6.4453 + 0.0191 = 6.4644 in3

Ans.

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2-27 a

(a)

c

b

w

wmax = 0.014 in, wmin = 0.004 in w¯ = (0.014 + 0.004)/2 = 0.009 in w = 0.009 ± 0.005 in w¯ = x¯ − y¯ = a¯ − b¯ − c¯ 0.009 = a¯ − 0.042 − 1.000 a¯ = 1.051 in tw = tall 0.005 = ta + 0.002 + 0.002 ta = 0.005 − 0.002 − 0.002 = 0.001 in a = 1.051 ± 0.001 in Ans. 2 = σˆ a2 + σˆ b2 + σˆ c2 σˆ w = σˆ all

(b)

σˆ a2 = σˆ w2 − σˆ b2 − σˆ c2 0.005 2 0.002 2 0.002 2 = − √ − √ √ 3 3 3 2 −6 σˆ a = 5.667(10 ) " σˆ a = 5.667(10−6 ) = 0.00238 in a¯ = 1.051 in, 2-28

σˆ a = 0.00238 in Ans.

Choose 15 mm as basic size, D, d. Table 2-8: fit is designated as 15H7/h6. From Table A-11, the tolerance grades are D = 0.018 mm and d = 0.011 mm. Hole: Eq. (2-38) Dmax = D + D = 15 + 0.018 = 15.018 mm Dmin = D = 15.000 mm Ans.

Ans.

Shaft: From Table A-12, fundamental deviation δ F = 0. From Eq. (2-39) dmax = d + δ F = 15.000 + 0 = 15.000 mm Ans. dmin = d + δ R − d = 15.000 + 0 − 0.011 = 14.989 mm 2-29

Ans.

Choose 45 mm as basic size. Table 2-8 designates fit as 45H7/s6. From Table A-11, the tolerance grades are D = 0.025 mm and d = 0.016 mm Hole: Eq. (2-38) Dmax = D + D = 45.000 + 0.025 = 45.025 mm Dmin = D = 45.000 mm Ans.

Ans.

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27

Chapter 2

Shaft: From Table A-12, fundamental deviation δ F = +0.043 mm. From Eq. (2-40) dmin = d + δ F = 45.000 + 0.043 = 45.043 mm Ans. dmax = d + δ F + d = 45.000 + 0.043 + 0.016 = 45.059 mm 2-30

Ans.

Choose 50 mm as basic size. From Table 2-8 fit is 50H7/g6. From Table A-11, the tolerance grades are D = 0.025 mm and d = 0.016 mm. Hole: Dmax = D + D = 50 + 0.025 = 50.025 mm Dmin = D = 50.000 mm Ans.

Ans.

Shaft: From Table A-12 fundamental deviation = −0.009 mm dmax = d + δ F = 50.000 + (−0.009) = 49.991 mm dmin = d + δ F − d = 50.000 + (−0.009) − 0.016 = 49.975 mm 2-31

Ans.

Choose the basic size as 1.000 in. From Table 2-8, for 1.0 in, the fit is H8/f7. From Table A-13, the tolerance grades are D = 0.0013 in and d = 0.0008 in. Hole: Dmax = D + (D) hole = 1.000 + 0.0013 = 1.0013 in Dmin = D = 1.0000 in Ans.

Ans.

Shaft: From Table A-14: Fundamental deviation = −0.0008 in dmax = d + δ F = 1.0000 + (−0.0008) = 0.9992 in Ans. dmin = d + δ F − d = 1.0000 + (−0.0008) − 0.0008 = 0.9984 in Ans. Alternatively, dmin = dmax − d = 0.9992 − 0.0008 = 0.9984 in. Ans. 2-32 W

Di

W

Do

Do = W + Di + W D¯ o = W¯ + D¯ i + W¯ = 0.139 + 3.734 + 0.139 = 4.012 in tall = 0.004 + 0.028 + 0.004 t Do = = 0.036 in Do = 4.012 ± 0.036 in Ans.

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2-33 Do = Di + 2W D¯ o = D¯ i + 2W¯ = 208.92 + 2(5.33) = 219.58 mm t = t Di + 2tw t Do = all

= 1.30 + 2(0.13) = 1.56 mm Do = 219.58 ± 1.56 mm Ans. 2-34 Do = Di + 2W D¯ o = D¯ i + 2W¯ = 3.734 + 2(0.139) = 4.012 mm 2 1/2 t Do = t 2 = tD + (2 tw ) 2 o all

= [0.0282 + (2) 2 (0.004) 2 ]1/2 = 0.029 in Do = 4.012 ± 0.029 in Ans. 2-35 Do = Di + 2W D¯ o = D¯ i + 2W¯ = 208.92 + 2(5.33) = 219.58 mm t Do = t 2 = [1.302 + (2) 2 (0.13) 2 ]1/2 all

= 1.33 mm Do = 219.58 ± 1.33 mm Ans. 2-36 (a)

w F

W

w= F−W w¯ = F¯ − W¯ = 0.106 − 0.139 = −0.033 in tw = t = 0.003 + 0.004 all

tw = 0.007 in wmax = w¯ + tw = −0.033 + 0.007 = −0.026 in wmin = w¯ − tw = −0.033 − 0.007 = −0.040 in The minimum “squeeze” is 0.026 in.

Ans.

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Chapter 2

(b)

Ymax = D¯ o = 4.012 in Ymin = max[0.99 D¯ o , D¯ o − 0.06] = max[3.9719, 3.952] = 3.972 in

Y w

Do

29

Y = 3.992 ± 0.020 in Do + w − Y = 0 w = Y − D¯ o w¯ = Y¯ − D¯ o = 3.992 − 4.012 = −0.020 in tw = t = tY + t Do = 0.020 + 0.036 = 0.056 in all

wmax = 0.036 in wmin = −0.076 in

w = −0.020 ± 0.056 in O-ring is more likely compressed than free prior to assembly of the end plate.

2-37 (a) Figure defines w as gap. w=F−W w¯ = F¯ − W¯ = 4.32 − 5.33 = −1.01 mm t = t F + tW = 0.13 + 0.13 = 0.26 mm tw =

w F

W

all

wmax = w¯ + tw = −1.01 + 0.26 = −0.75 mm wmin = w¯ − tw = −1.01 − 0.26 = −1.27 mm The O-ring is “squeezed” at least 0.75 mm. (b)

Ymax = D¯ o = 219.58 mm

Y w

Do

Ymin = max[0.99 D¯ o , D¯ o − 1.52] = max[0.99(219.58, 219.58 − 1.52)] = 217.38 mm Y = 218.48 ± 1.10 mm

From the figure, the stochastic equation is: or,

Do + w = Y w = Y − Do w¯ = Y¯ − D¯ o = 218.48 − 219.58 = −1.10 mm t = tY + t Do = 1.10 + 0.34 = 1.44 mm tw = all

wmax = w¯ + tw = −1.10 + 1.44 = 0.34 mm wmin = w¯ − tw = −1.10 − 1.44 = −2.54 mm The O-ring is more likely to be circumferentially compressed than free prior to assembly of the end plate.

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2-38

wmax = −0.020 in, wmin = −0.040 in 1 w¯ = (−0.020 + (−0.040)) = −0.030 in 2 1 tw = (−0.020 − (−0.040)) = 0.010 in 2 b = 0.750 ± 0.001 in c = 0.120 ± 0.005 in d = 0.875 ± 0.001 in w¯ = a¯ − b¯ − c¯ − d¯ −0.030 = a¯ − 0.875 − 0.120 − 0.750 a¯ = 0.875 + 0.120 + 0.750 − 0.030 a¯ = 1.715 in

w b

c

d

a

Absolute: tw =

t = 0.010 = ta + 0.001 + 0.005 + 0.001

all

ta = 0.010 − 0.001 − 0.005 − 0.001 = 0.003 in a = 1.715 ± 0.003 in Ans. Statistical: For a normal distribution of dimensions tw2 = t 2 = ta2 + tb2 + tc2 + td2 all

1/2 ta = tw2 − tb2 − tc2 − td2 = (0.0102 − 0.0012 − 0.0052 − 0.0012 ) 1/2 = 0.0085 a = 1.715 ± 0.0085 in Ans. 2-39 x

n

nx

nx 2

93 95 97 99 101 103 105 107 109 111

19 25 38 17 12 10 5 4 4 2 136

1767 2375 3685 1683 1212 1030 525 428 436 222 13 364

164 311 225 625 357 542 166 617 122 412 106 090 55 125 45 796 47 524 24 624 1315 704

x¯ = 13 364/136 = 98.26 kpsi 1/2 1 315 704 − 13 3642 /136 sx = = 4.30 kpsi 135

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31

Chapter 2

Under normal hypothesis, z 0.01 = (x0.01 − 98.26)/4.30 x0.01 = 98.26 + 4.30z 0.01 = 98.26 + 4.30(−2.3267) . = 88.26 = 88.3 kpsi Ans. 2-40

From Prob. 2-39, µx = 98.26 kpsi, and σˆ x = 4.30 kpsi. C x = σˆ x /µx = 4.30/98.26 = 0.043 76 From Eqs. (2-18) and (2-19), µ y = ln(98.26) − 0.043 762 /2 = 4.587 " σˆ y = ln(1 + 0.043 762 ) = 0.043 74 For a yield strength exceeded by 99% of the population, z 0.01 = (ln x0.01 − µ y )/σˆ y

⇒

ln x0.01 = µ y + σˆ y z 0.01

From Table A-10, for 1% failure, z 0.01 = −2.326. Thus, ln x0.01 = 4.587 + 0.043 74(−2.326) = 4.485 x0.01 = 88.7 kpsi Ans. The normal PDF is given by Eq. (2-14) as

1 x − 98.26 2 1 f (x) = √ exp − 2 4.30 4.30 2π

For the lognormal distribution, from Eq. (2-17), defining g(x), 1 1 ln x − 4.587 2 g(x) = √ exp − 2 0.043 74 x(0.043 74) 2π x (kpsi)

f/(Nw)

f (x)

g (x)

x (kpsi)

f/(Nw)

f (x)

g (x)

92 92 94 94 96 96 98 98 100 100 102

0.000 00 0.069 85 0.069 85 0.091 91 0.091 91 0.139 71 0.139 71 0.062 50 0.062 50 0.044 12 0.044 12

0.032 15 0.032 15 0.056 80 0.056 80 0.080 81 0.080 81 0.092 61 0.092 61 0.085 48 0.085 48 0.063 56

0.032 63 0.032 63 0.058 90 0.058 90 0.083 08 0.083 08 0.092 97 0.092 97 0.083 67 0.083 67 0.061 34

102 104 104 106 106 108 108 110 110 112 112

0.036 76 0.036 76 0.018 38 0.018 38 0.014 71 0.014 71 0.014 71 0.014 71 0.007 35 0.007 35 0.000 00

0.063 56 0.038 06 0.038 06 0.018 36 0.018 36 0.007 13 0.007 13 0.002 23 0.002 23 0.000 56 0.000 56

0.061 34 0.037 08 0.037 08 0.018 69 0.018 69 0.007 93 0.007 93 0.002 86 0.002 86 0.000 89 0.000 89

Note: rows are repeated to draw histogram

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Histogram f (x) g(x)

0.16

Probability density

0.14 0.12 0.1 0.08 0.06 0.04 0.02 0 90

92

94

96

98

100 102 x (kpsi)

104

106

108

110

112

The normal and lognormal are almost the same. However the data is quite skewed and perhaps a Weibull distribution should be explored. For a method of establishing the Weibull parameters see Shigley, J. E., and C. R. Mischke, Mechanical Engineering Design, McGraw-Hill, 5th ed., 1989, Sec. 4-12. 2-41

Let x = (S fe ) 104 x0 = 79 kpsi, θ = 86.2 kpsi, b = 2.6 Eq. (2-28) x¯ = x0 + (θ − x0 )(1 + 1/b) x¯ = 79 + (86.2 − 79)(1 + 1/2.6) = 79 + 7.2 (1.38) From Table A-34, (1.38) = 0.888 54 x¯ = 79 + 7.2(0.888 54) = 85.4 kpsi Eq. (2-29)

Ans.

σˆ x = (θ − x0 )[(1 + 2/b) − 2 (1 + 1/b)]1/2 = (86.2 − 79)[(1 + 2/2.6) − 2 (1 + 1/2.6)]1/2 = 7.2[0.923 76 − 0.888 542 ]1/2 = 2.64 kpsi Ans. 2.64 σˆ x = = 0.031 Ans. Cx = x¯ 85.4

2-42 x = Sut x0 = 27.7, θ = 46.2, b = 4.38 µx = 27.7 + (46.2 − 27.7)(1 + 1/4.38) = 27.7 + 18.5 (1.23) = 27.7 + 18.5(0.910 75) = 44.55 kpsi Ans.

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33

Chapter 2

σˆ x = (46.2 − 27.7)[(1 + 2/4.38) − 2 (1 + 1/4.38)]1/2 = 18.5[(1.46) − 2 (1.23)]1/2 = 18.5[0.8856 − 0.910 752 ]1/2 = 4.38 kpsi Ans. 4.38 = 0.098 Ans. Cx = 44.55 From the Weibull survival equation x − x0 b R = exp − =1− p θ − x0 x40 − x0 b = 1 − p40 R40 = exp − θ − x0 40 − 27.7 4.38 = exp − = 0.846 46.2 − 27.7 p40 = 1 − R40 = 1 − 0.846 = 0.154 = 15.4%

Ans.

2-43 x = Sut

x0 = 151.9, θ = 193.6, b = 8 µx = 151.9 + (193.6 − 151.9)(1 + 1/8) = 151.9 + 41.7 (1.125) = 151.9 + 41.7(0.941 76) = 191.2 kpsi Ans. σˆ x = (193.6 − 151.9)[(1 + 2/8) − 2 (1 + 1/8)]1/2 = 41.7[(1.25) − 2 (1.125)]1/2 = 41.7[0.906 40 − 0.941 762 ]1/2 = 5.82 kpsi Ans. 5.82 = 0.030 Cx = 191.2

2-44 x = Sut

x0 = 47.6, θ = 125.6, b = 11.84 x¯ = 47.6 + (125.6 − 47.6)(1 + 1/11.84) x¯ = 47.6 + 78 (1.08) = 47.6 + 78(0.959 73) = 122.5 kpsi σˆ x = (125.6 − 47.6)[(1 + 2/11.84) − 2 (1 + 1/11.84)]1/2 = 78[(1.08) − 2 (1.17)]1/2 = 78(0.959 73 − 0.936 702 ) 1/2 = 22.4 kpsi

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From Prob. 2-42

x − x0 b p = 1 − exp − θ − θ0 100 − 47.6 11.84 = 1 − exp − 125.6 − 47.6 = 0.0090 Ans.

y = Sy y0 = 64.1, θ = 81.0, b = 3.77 y¯ = 64.1 + (81.0 − 64.1)(1 + 1/3.77) = 64.1 + 16.9 (1.27) = 64.1 + 16.9(0.902 50) = 79.35 kpsi σ y = (81 − 64.1)[(1 + 2/3.77) − (1 + 1/3.77)]1/2 σ y = 16.9[(0.887 57) − 0.902 502 ]1/2 = 4.57 kpsi y − y0 3.77 p = 1 − exp − θ − y0 70 − 64.1 3.77 p = 1 − exp − = 0.019 Ans. 81 − 64.1 2-45 x = Sut = W[122.3, 134.6, 3.64] kpsi, p(x > 120) = 1 = 100% since x0 > 120 kpsi 133 − 122.3 3.64 p(x > 133) = exp − 134.6 − 122.3 = 0.548 = 54.8% Ans. 2-46

Using Eqs. (2-28) and (2-29) and Table A-34, µn = n 0 + (θ − n 0 )(1 + 1/b) = 36.9 + (133.6 − 36.9)(1 + 1/2.66) = 122.85 kcycles σˆ n = (θ − n 0 )[(1 + 2/b) − 2 (1 + 1/b)] = 34.79 kcycles For the Weibull density function, Eq. (2-27), 2.66−1 2.66 n − 36.9 2.66 n − 36.9 exp − f W (n) = 133.6 − 36.9 133.6 − 36.9 133.6 − 36.9 For the lognormal distribution, Eqs. (2-18) and (2-19) give, µ y = ln(122.85) − (34.79/122.85) 2 /2 = 4.771 " σˆ y = [1 + (34.79/122.85) 2 ] = 0.2778

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35

Chapter 2

From Eq. (2-17), the lognormal PDF is

1 f L N (n) = √ exp − 2 0.2778 n 2π 1

ln n − 4.771 0.2778

2

We form a table of densities f W (n) and f L N (n) and plot. n (kcycles)

f W (n)

f L N (n)

40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220

9.1E-05 0.000 991 0.002 498 0.004 380 0.006 401 0.008 301 0.009 822 0.010 750 0.010 965 0.010 459 0.009 346 0.007 827 0.006 139 0.004 507 0.003 092 0.001 979 0.001 180 0.000 654 0.000 336

1.82E-05 0.000 241 0.001 233 0.003 501 0.006 739 0.009 913 0.012 022 0.012 644 0.011 947 0.010 399 0.008 492 0.006 597 0.004 926 0.003 564 0.002 515 0.001 739 0.001 184 0.000 795 0.000 529

f (n) 0.014

LN W

0.012 0.010 0.008 0.006 0.004 0.002 0

0

50

100

150

200

250

n, kcycles

The Weibull L10 life comes from Eq. (2-26) with a reliability of R = 0.90. Thus, n 0.10 = 36.9 + (133 − 36.9)[ln(1/0.90)]1/2.66 = 78.1 kcycles Ans.

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The lognormal L10 life comes from the definition of the z variable. That is, ln n 0 = µ y + σˆ y z

or n 0 = exp(µ y + σˆ y z)

From Table A-10, for R = 0.90, z = −1.282. Thus, n 0 = exp[4.771 + 0.2778(−1.282)] = 82.7 kcycles Ans. 2-47

Form a table

i

x L(10−5)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

3.05 3.55 4.05 4.55 5.05 5.55 6.05 6.55 7.05 7.55 8.05 8.55 9.05 9.55 10.05

−5

fi

f i x(10 )

3 7 11 16 21 13 13 6 2 0 4 3 0 0 1 100

9.15 24.85 44.55 72.80 106.05 72.15 78.65 39.30 14.10 0 32.20 25.65 0 0 10.05 529.50

2

−10

f i x (10

)

27.9075 88.2175 180.4275 331.24 535.5525 400.4325 475.8325 257.415 99.405 0 259.21 219.3075 0 0 101.0025 2975.95

g(x) (105 ) 0.0557 0.1474 0.2514 0.3168 0.3216 0.2789 0.2151 0.1517 0.1000 0.0625 0.0375 0.0218 0.0124 0.0069 0.0038

x¯ = 529.5(105 )/100 = 5.295(105 ) cycles Ans. 1/2 2975.95(1010 ) − [529.5(105 )]2 /100 sx = 100 − 1 = 1.319(105 ) cycles Ans. C x = s/x¯ = 1.319/5.295 = 0.249 µ y = ln 5.295(105 ) − 0.2492 /2 = 13.149 " σˆ y = ln(1 + 0.2492 ) = 0.245 1 ln x − µ y 2 1 g(x) = √ exp − 2 σˆ y x σˆ y 2π 1 ln x − 13.149 2 1.628 exp − g(x) = x 2 0.245

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37

Chapter 2

105 g(x) 0.5

Superposed histogram and PDF

0.4

0.3

0.2

0.1

0

3.05(105)

10.05(105) x, cycles

2-48 Eq. (2-28)

Eq. (2-29)

x = Su = W[70.3, 84.4, 2.01] µx = 70.3 + (84.4 − 70.3)(1 + 1/2.01) = 70.3 + (84.4 − 70.3)(1.498) = 70.3 + (84.4 − 70.3)0.886 17 = 82.8 kpsi Ans. σˆ x = (84.4 − 70.3)[(1 + 2/2.01) − 2 (1 + 1/2.01)]1/2 σˆ x = 14.1[0.997 91 − 0.886 172 ]1/2 = 6.502 kpsi Cx =

2-49

6.502 = 0.079 Ans. 82.8

Take the Weibull equation for the standard deviation σˆ x = (θ − x0 )[(1 + 2/b) − 2 (1 + 1/b)]1/2 and the mean equation solved for x¯ − x0 x¯ − x0 = (θ − x0 )(1 + 1/b) Dividing the first by the second, [(1 + 2/b) − 2 (1 + 1/b)]1/2 σˆ x = x¯ − x0 (1 + 1/b) √ 4.2 (1 + 2/b) = − 1 = R = 0.2763 49 − 33.8 2 (1 + 1/b)

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Make a table and solve for b iteratively b 3 4 4.1

1 + 2/b 1 + 1/b (1 + 2/b) (1 + 1/b) 1.67 1.5 1.49

1.33 1.25 1.24

0.903 30 0.886 23 0.885 95

0.893 38 0.906 40 0.908 52

0.363 0.280 0.271

. b = 4.068 Using MathCad Ans. 49 − 33.8 x¯ − x0 = 33.8 + θ = x0 + (1 + 1/b) (1 + 1/4.068) = 49.8 kpsi Ans. 2-50

x = S y = W[34.7, 39, 2.93] kpsi x¯ = 34.7 + (39 − 34.7)(1 + 1/2.93) = 34.7 + 4.3(1.34) = 34.7 + 4.3(0.892 22) = 38.5 kpsi σˆ x = (39 − 34.7)[(1 + 2/2.93) − 2 (1 + 1/2.93)]1/2 = 4.3[(1.68) − 2 (1.34)]1/2 = 4.3[0.905 00 − 0.892 222 ]1/2 = 1.42 kpsi Ans. C x = 1.42/38.5 = 0.037 Ans.

2-51 x (Mrev) 1 2 3 4 5 6 7 8 9 10 11 12 Sum 78

f

fx

f x2

11 22 38 57 31 19 15 12 11 9 7 5 237

11 44 114 228 155 114 105 96 99 90 77 60 1193

11 88 342 912 775 684 735 768 891 900 847 720 7673

µx = 1193(106 )/237 = 5.034(106 ) cycles 7673(1012 ) − [1193(106 )]2 /237 = 2.658(106 ) cycles σˆ x = 237 − 1 C x = 2.658/5.034 = 0.528

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39

Chapter 2

From Eqs. (2-18) and (2-19), µ y = ln[5.034(106 )] − 0.5282 /2 = 15.292 " σˆ y = ln(1 + 0.5282 ) = 0.496 From Eq. (2-17), defining g(x),

1 g (x) = √ exp − 2 x(0.496) 2π 1

x (Mrev)

f/(Nw)

g(x) · (106 )

0.5 0.5 1.5 1.5 2.5 2.5 3.5 3.5 4.5 4.5 5.5 5.5 6.5 6.5 7.5 7.5 8.5 8.5 9.5 9.5 10.5 10.5 11.5 11.5 12.5 12.5

0.000 00 0.046 41 0.046 41 0.092 83 0.092 83 0.160 34 0.160 34 0.240 51 0.240 51 0.130 80 0.130 80 0.080 17 0.080 17 0.063 29 0.063 29 0.050 63 0.050 63 0.046 41 0.046 41 0.037 97 0.037 97 0.029 54 0.029 54 0.021 10 0.021 10 0.000 00

0.000 11 0.000 11 0.052 04 0.052 04 0.169 92 0.169 92 0.207 54 0.207 54 0.178 48 0.178 48 0.131 58 0.131 58 0.090 11 0.090 11 0.059 53 0.059 53 0.038 69 0.038 69 0.025 01 0.025 01 0.016 18 0.016 18 0.010 51 0.010 51 0.006 87 0.006 87

z=

ln x − µ y σˆ y

⇒

ln x − 15.292 0.496

2

Histogram PDF

0.25

0.2

g(x)(106)

shi20396_ch02.qxd

0.15

0.1

0.05

0

0

2

4

6 x, Mrev

8

ln x = µ y + σˆ y z = 15.292 + 0.496z

L10 life, where 10% of bearings fail, from Table A-10, z = −1.282. Thus, ln x = 15.292 + 0.496(−1.282) = 14.66 ∴ x = 2.32 × 106 rev Ans.

10

12

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Page 40

Chapter 3 3-1 From Table A-20 Sut = 470 MPa (68 kpsi), S y = 390 MPa (57 kpsi)

Ans.

3-2 From Table A-20 Sut = 620 MPa (90 kpsi), S y = 340 MPa (49.5 kpsi) Ans. 3-3 Comparison of yield strengths: Sut of G10 500 HR is S yt of SAE1020 CD is

620 = 1.32 times larger than SAE1020 CD 470

Ans.

390 = 1.15 times larger than G10500 HR Ans. 340

From Table A-20, the ductilities (reduction in areas) show, SAE1020 CD is

40 = 1.14 times larger than G10500 35

Ans.

The stiffness values of these materials are identical Ans. Sut MPa (kpsi)

Sy MPa (kpsi)

Table A-20 Ductility R%

SAE1020 CD 470(68) UNS10500 HR 620(90)

390 (57) 340(495)

40 35

Table A-5 Stiffness GPa (Mpsi) 207(30) 207(30)

3-4 From Table A-21 1040 Q&T S¯ y = 593 (86) MPa (kpsi)

at 205◦C (400◦F)

Ans.

3-5 From Table A-21 1040 Q&T

R = 65%

at 650◦C (1200◦F)

Ans.

3-6 Using Table A-5, the specific strengths are: Sy 39.5(103 ) = = 1.40(105 ) in Ans. UNS G10350 HR steel: W 0.282 Sy 43(103 ) = = 4.39(105 ) in Ans. 2024 T4 aluminum: W 0.098 Ti-6Al-4V titanium:

Sy 140(103 ) = = 8.75(105 ) in W 0.16

ASTM 30 gray cast iron has no yield strength.

Ans.

Ans.

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41

Chapter 3

3-7 The specific moduli are: 30(106 ) E = = 1.06(108 ) in W 0.282

UNS G10350 HR steel:

Ans.

2024 T4 aluminum:

10.3(106 ) E = = 1.05(108 ) in W 0.098

Ans.

Ti-6Al-4V titanium:

16.5(106 ) E = = 1.03(108 ) in W 0.16

Ans.

Gray cast iron:

14.5(106 ) E = = 5.58(107 ) in W 0.26

Ans.

2G(1 + ν) = E

3-8

⇒ ν=

E − 2G 2G

From Table A-5 Steel: ν = Aluminum: ν = Beryllium copper: ν = Gray cast iron: ν =

30 − 2(11.5) = 0.304 2(11.5)

Ans.

10.4 − 2(3.90) = 0.333 2(3.90) 18 − 2(7) = 0.286 2(7) 14.5 − 2(6) = 0.208 2(6)

Ans.

Ans. Ans.

3-9 E U 80 70 60 Stress P兾A0 kpsi

shi20396_ch03.qxd

50 Y

40

Su ⫽ 85.5 kpsi Ans. Sy ⫽ 45.5 kpsi Ans.

30

E ⫽ 90兾0.003 ⫽ 30 000 kpsi Ans.

20

R⫽

A 0 ⫺ AF 0.1987 ⫺ 0.1077 (100) ⫽ 45.8% Ans. ⫽ 0.1987 A0

10

⫽

A0 l ⫺ l0 ⌬l l ⫽ ⫽ ⫺ 1⫽ ⫺1 A l0 l0 l0

0

0

0.002 0.1

0.004 0.2

0.006 0.3

0.008 0.4 Strain,

0.010 0.5

0.012 0.6

0.014 0.7

0.016 (Lower curve) 0.8 (Upper curve)

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3-10

To plot σtrue vs. ε, the following equations are applied to the data. A0 = Eq. (3-4)

π(0.503) 2 = 0.1987 in2 4

ε = ln

l l0

ε = ln

A0 A

σtrue =

for 0 ≤ L ≤ 0.0028 in for L > 0.0028 in

P A

The results are summarized in the table below and plotted on the next page. The last 5 points of data are used to plot log σ vs log ε m = 0.2306

The curve fit gives

log σ0 = 5.1852 ⇒ σ0 = 153.2 kpsi

Ans.

For 20% cold work, Eq. (3-10) and Eq. (3-13) give, A = A0 (1 − W ) = 0.1987(1 − 0.2) = 0.1590 in2 ε = ln

0.1987 A0 = ln = 0.2231 A 0.1590

Eq. (3-14): S y = σ0 εm = 153.2(0.2231) 0.2306 = 108.4 kpsi

Ans.

Eq. (3-15), with Su = 85.5 kpsi from Prob. 3-9, Su = P 0 1 000 2 000 3 000 4 000 7 000 8 400 8 800 9 200 9 100 13 200 15 200 17 000 16 400 14 800

85.5 Su = = 106.9 kpsi 1−W 1 − 0.2

Ans.

L

A

ε

σtrue

log ε

log σtrue

0 0.0004 0.0006 0.0010 0.0013 0.0023 0.0028 0.0036 0.0089

0.198 713 0.198 713 0.198 713 0.198 713 0.198 713 0.198 713 0.198 713 0.198 4 0.197 8 0.196 3 0.192 4 0.187 5 0.156 3 0.130 7 0.107 7

0 0.000 2 0.000 3 0.000 5 0.000 65 0.001 149 0.001 399 0.001 575 0.004 604 0.012 216 0.032 284 0.058 082 0.240 083 0.418 956 0.612 511

0 5032.388 10 064.78 15 097.17 20 129.55 35 226.72 42 272.06 44 354.84 46 511.63 46 357.62 68 607.07 81 066.67 108 765.2 125 478.2 137 418.8

−3.699 01 −3.522 94 −3.301 14 −3.187 23 −2.939 55 −2.854 18 −2.802 61 −2.336 85 −1.913 05 −1.491 01 −1.235 96 −0.619 64 −0.377 83 −0.212 89

3.701 774 4.002 804 4.178 895 4.303 834 4.546 872 4.626 053 4.646 941 4.667 562 4.666 121 4.836 369 4.908 842 5.036 49 5.098 568 5.138 046

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43

Chapter 3

160000 140000 120000 true (psi)

100000 80000 60000 40000 20000 0

0

0.1

0.2

0.3

0.4 true

0.5

0.6

0.7

5.2

y ⫽ 0.2306x ⫹ 5.1852 5

log

5.1

4.9

⫺1.6

3-11

⫺1.4

Tangent modulus at σ = 0 is E0 = At σ = 20 kpsi

⫺1

⫺0.8 log

⫺0.6

⫺0.4

⫺0.2

0

4.8

5000 − 0 σ . = = 25(106 ) psi ε 0.2(10−3 ) − 0

. (26 − 19)(103 ) E 20 = = 14.0(106 ) psi Ans. − 3 (1.5 − 1)(10 )

ε(10−3 )

σ (kpsi)

60

0 5 10 16 19 26 32 40 46 49 54

50

0 0.20 0.44 0.80 1.0 1.5 2.0 2.8 3.4 4.0 5.0 3-12

⫺1.2

40 (kpsi)

shi20396_ch03.qxd

(Sy)0.001 ⫽ ˙ 35 kpsi Ans.

30 20 10 0

0

1

2

3 (10⫺3)

From Prob. 2-8, for y = a1 x + a2 x 2 a1 =

yx 3 − x yx 2 xx 3 − (x 2 ) 2

a2 =

xx y − yx 2 xx 3 − (x 2 ) 2

4

5

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Let x represent ε(10−3 ) and y represent σ (kpsi), x 0 0.2 0.44 0.80 1.0 1.5 2.0 2.8 3.4 4.0 5.0 = 21.14

y

x2

x3

xy

0 5 10 16 19 26 32 40 46 49 54 297

0 0.04 0.1936 0.64 1.00 2.25 4.00 7.84 11.56 16.00 25.00 68.5236

0 0.008 0.085 184 0.512 1.000 3.375 8.000 21.952 39.304 64.000 125.000 263.2362

0 1.0 4.4 12.8 19.0 39.0 64.0 112.0 156.4 196.0 270.0 874.6

Substituting, 297(263.2362) − 874.6(68.5236) = 20.993 67 21.14(263.2362) − (68.5236) 2 21.14(874.6) − 297(68.5236) a2 = = −2.142 42 21.14(263.2362) − (68.5236) 2 The tangent modulus is a1 =

dy dσ = = 20.993 67 − 2(2.142 42)x = 20.993 67 − 4.284 83x dx dε At σ = 0, E 0 = 20.99 Mpsi Ans. At σ = 20 kpsi 20 = 20.993 67x − 2.142 42x 2 ⇒ x = 1.069, 8.73 Taking the first root, ε = 1.069 and the tangent modulus is E 20 = 20.993 67 − 4.284 83(1.069) = 16.41 Mpsi Ans. Determine the equation for the 0.1 percent offset line y = 20.99x + b at y = 0, x = 1 ∴ b = −20.99 y = 20.99x − 20.99 = 20.993 67x − 2.142 42x 2 2.142 42x 2 − 20.99 = 0 ⇒ x = 3.130 (S y ) 0.001 = 20.99(3.13) − 2.142(3.13) 2 = 44.7 kpsi 3-13

Since |εo | = |εi |

R+h R R + N ln = ln = −ln R+ N R+ N R R+N R+h = R+N R ( R + N ) 2 = R( R + h)

From which,

N 2 + 2R N − Rh = 0

Ans.

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45

Chapter 3

h 1/2 N = R −1 ± 1 + R

The roots are:

The + sign being significant,

h 1/2 N=R 1+ −1 R

Ans.

Substitute for N in εo = ln  Gives

  ε0 = ln  

R+h R+N

R+h

h 1/2 R+ R 1+ −R R

  h 1/2   = ln 1 +  R

Ans.

These constitute a useful pair of equations in cold-forming situations, allowing the surface strains to be found so that cold-working strength enhancement can be estimated. 3-14 16T 16T 10−6 = = 2.6076T MPa πd 3 π(12.5) 3 (10−3 ) 3 π π ◦ ◦ θ r (12.5) θ 180 180 = = 6.2333(10−4 )θ ◦ γ = L 350 τ=

For G, take the first 10 data points for the linear part of the curve.

T

θ (deg.) γ (10−3 )

0 0 7.7 0.38 15.3 0.80 23.0 1.24 30.7 1.64 38.3 2.01 46.0 2.40 53.7 2.85 61.4 3.25 69.0 3.80 76.7 4.50 80.0 5.10 85.0 6.48 90.0 8.01 95.0 9.58 100.0 11.18

0 0.236 865 0.498 664 0.772 929 1.022 261 1.252 893 1.495 992 1.776 491 2.025 823 2.368 654 2.804 985 3.178 983 4.039 178 4.992 873 5.971 501 6.968 829

τ (MPa) 0 20.078 52 39.896 28 59.974 8 80.053 32 99.871 08 119.949 6 140.028 1 160.106 6 179.924 4 200.002 9 = 208.608 221.646 234.684 247.722 260.76

γ (10−3 ) x

τ (MPa) y

x2

0 0.236 865 0.498 664 0.772 929 1.022 261 1.252 893 1.495 992 1.776 491 2.025 823 2.368 654 11.450 57

0 20.078 52 39.896 28 59.974 8 80.053 32 99.871 08 119.949 6 140.028 1 160.106 6 179.924 4 899.882 8

0 0.056 105 0.248 666 0.597 420 1.045 018 1.569 742 2.237 992 3.155 918 4.103 957 5.610 522 18.625 34

xy 0 4.7559 19.8948 46.3563 81.8354 125.1278 179.4436 248.7586 324.3476 426.1786 1456.6986

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300 250

(MPa)

200 150 100 50 0

0

1

2

3

4

5

6

7

(10⫺3)

y = mx + b,

τ = y,

γ = x where m is the shear modulus G,

m=

MPa N x y − xy = 77.3 −3 = 77.3 GPa 2 2 N x − (x) 10

b=

y − mx = 1.462 MPa N

Ans.

. From curve S ys = 200 MPa Ans. Note since τ is not uniform, the offset yield does not apply, so we are using the elastic limit as an approximation. 3-15 x 38.5 39.5 40.5 41.5 42.5 43.5 44.5 45.5 46.5 47.5 48.5 49.5 = 528.0

f

fx

f x2

2 9 30 65 101 112 90 54 25 9 2 1 500

77.0 355.5 1215.0 2697.5 4292.5 4872.0 4005.0 2457.0 1162.5 427.5 97.0 49.5 21 708.0

2 964.50 14 042.25 49 207.50 111 946.30 182 431.30 211 932.00 178 222.50 111793.50 54 056.25 20 306.25 4 704.50 2 450.25 944 057.00

x¯ = 21 708/500 = 43.416, σˆ x =

944 057 − (21 7082 /500) = 1.7808 500 − 1

C x = 1.7808/43.416 = 0.041 02, y¯ = ln 43.416 − ln(1 + 0.041 022 ) = 3.7691

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47

Chapter 3

ln(1 + 0.041 022 ) = 0.0410, 1 ln x − 3.7691 2 1 g(x) = √ exp − 2 0.0410 x(0.0410) 2π σˆ y =

x

f /( N w)

g(x)

x

f /( N w)

g(x)

38 38 39 39 40 40 41 41 42 42 43 43 44 44

0 0.004 0.004 0.018 0.018 0.060 0.060 0.130 0.130 0.202 0.202 0.224 0.224 0.180

0.001 488 0.001 488 0.009 057 0.009 057 0.035 793 0.035 793 0.094 704 0.094 704 0.172 538 0.172 538 0.222 074 0.222 074 0.206 748 0.206 748

45 45 46 46 47 47 48 48 49 49 50 50

0.180 0.108 0.108 0.050 0.050 0.018 0.018 0.004 0.004 0.002 0.002 0

0.142 268 0.142 268 0.073 814 0.073 814 0.029 410 0.029 410 0.009 152 0.009 152 0.002 259 0.002 259 0.000 449 0.000 449

f (x) 0.25

Histogram PDF

0.2

0.15

0.1

0.05

0 35

40

45

S y = LN(43.42, 1.781) kpsi 3-16

50

x

Ans.

From Table A-22 AISI 1212

S y = 28.0 kpsi, σ f = 106 kpsi, Sut = 61.5 kpsi σ0 = 110 kpsi,

From Eq. (3-12)

m = 0.24,

ε f = 0.85

εu = m = 0.24

Eq. (3-10)

1 1 A0 = = 1.25 = Ai 1−W 1 − 0.2

Eq. (3-13)

εi = ln 1.25 = 0.2231 ⇒ εi < εu

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S y = σ0 εim = 110(0.2231) 0.24 = 76.7 kpsi Ans.

Eq. (3-14)

Su =

Eq. (3-15) 3-17

Page 48

61.5 Su = = 76.9 kpsi Ans. 1−W 1 − 0.2

For H B = 250, Su = 0.495 (250) = 124 kpsi = 3.41 (250) = 853 MPa

Eq. (3-17)

3-18

For the data given, H B = 2530

Ans.

H B2 = 640 226 640 226 − (2530) 2 /10 2530 = 3.887 = 253 σˆ H B = H¯ B = 9 10

Eq. (3-17)

S¯u = 0.495(253) = 125.2 kpsi

Ans.

σ¯ su = 0.495(3.887) = 1.92 kpsi Ans. 3-19

From Prob. 3-18,

H¯ B = 253

and σˆ HB = 3.887

Eq. (3-18) S¯u = 0.23(253) − 12.5 = 45.7 kpsi σˆ su = 0.23(3.887) = 0.894 kpsi 3-20

. 45.52 = 34.5 in · lbf/in3 uR = 2(30)

(a)

Ans.

Ans.

Ans.

(b) P

L

0 1 000 2 000 3 000 4 000 7 000 8 400 8 800 9 200 9 100 13 200 15 200 17 000 16 400 14 800

0 0.0004 0.0006 0.0010 0.0013 0.0023 0.0028 0.0036 0.0089

A

0.1963 0.1924 0.1875 0.1563 0.1307 0.1077

A0 /A − 1

ε

0.012 291 0.032 811 0.059 802 0.271 355 0.520 373 0.845059

0 0.0002 0.0003 0.0005 0.000 65 0.001 15 0.0014 0.0018 0.004 45 0.012 291 0.032 811 0.059 802 0.271 355 0.520 373 0.845 059

σ = P/A0 0 5 032.39 10 064.78 15 097.17 20 129.55 35 226.72 42 272.06 44 285.02 46 297.97 45 794.73 66 427.53 76 492.30 85 550.60 82 531.17 74 479.35

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49

Chapter 3 90000 80000 70000 60000 50000 40000 30000 20000 10000 0

0

0.2

0.4 All data points

0.6

0.8

50000 45000 40000 35000 30000 25000 20000 A1

15000

A2

10000 5000 0

0

0.001

0.002 0.003 First 9 data points

0.004

0.005

90000 80000 70000 60000 50000 40000

A4

A5

30000 20000 A3 10000 0

0

0.2

0.4 0.6 Last 6 data points

0.8

5 1 . uT = Ai = (43 000)(0.001 5) + 45 000(0.004 45 − 0.001 5) 2 i=1 1 + (45 000 + 76 500)(0.059 8 − 0.004 45) 2 +81 000(0.4 − 0.059 8) + 80 000(0.845 − 0.4)

. = 66.7(103 )in · lbf/in3

Ans.

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Chapter 4 4-1 W

W 2

2 B

A RA

B

A

RB

RA

1

RB 1

1

(a)

(b) 1

1 RD

RC

D

2

3

C

A

RA RB

B

W

(c)

W 1

RC RB

RA RB

2 RA W

(d) (e) A

2 W RBx B RBx 1

RB

RBy

RBy

Scale of corner magnified

(f)

RA

1

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51

Chapter 4

4-2 (a)

2 kN

R A = 2 sin 60 = 1.732 kN

RB

60°

90°

R B = 2 sin 30 = 1 kN

2 2 kN

30°

60° RB

Ans.

Ans.

RA

30°

RA

1

(b)

RA

0.4 m

S = 0.6 m

B

A

α = tan−1

45° 800 N

0.6 = 30.96◦ 0.4 + 0.6

0.6 m

RO

O s

800 RA = sin 135 sin 30.96 800 RO = sin 14.04 sin 30.96

RO RA 135° 30.96°

800 N

45 30.96 14.04

⇒

R A = 1100 N Ans.

⇒

R O = 377 N Ans.

30.96°

(c)

1.2 = 2.078 kN Ans. tan 30 1.2 = 2.4 kN Ans. RA = sin 30

1.2 kN

RO = 30° RA

RO

60°

90°

60° 1.2 kN

RA RO

(d) Step 1: Find R A and R E

4.5 = 7.794 m tan 30 MA = 0 +

h=

4.5 m

C

30° y

400 N

4

2

9R E − 7.794(400 cos 30) − 4.5(400 sin 30) = 0 R E = 400 N Ans.

h B RAx RA

3

D 60°

A RAy

x

E 9m

RE

Fx = 0

R Ax + 400 cos 30 = 0

⇒

R Ax = −346.4 N

Fy = 0

R Ay + 400 − 400 sin 30 = 0 ⇒ R Ay = −200 N R A = 346.42 + 2002 = 400 N Ans.

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Step 2: Find components of RC on link 4 and R D (RCy)4 MC = 0 + (RCx)4

C

400(4.5) − (7.794 − 1.9) R D = 0 ⇒ Fx = 0 ⇒ ( RC x ) 4 = 305.4 N Fy = 0 ⇒ ( RC y ) 4 = −400 N

4 RD

D

R D = 305.4 N

Ans.

E 400 N

Step 3: Find components of RC on link 2 (RCy)2 Fx = 0 (RCx)2

C

( RC x ) 2 + 305.4 − 346.4 = 0 Fy = 0

2

⇒

( RC x ) 2 = 41 N

( RC y ) 2 = 200 N

B

305.4 N

346.4 N A 200 N 400 N

200 N 41 N

200 N C

305.4 N

400 N C

41 N

305.4 N

30° Pin C

400 N

305.4 N

B

D

B

346.4 N

305.4 N

D

E

A 200 N

400 N

400 N

Ans. 4-3 (a)

y

40 lbf 4"

O

60 lbf 4"

6"

A R1

4"

B

C

30 lbf

R2

D

+

−18(60) + 14R2 + 8(30) − 4(40) = 0 R2 = 71.43 lbf Fy = 0: R1 − 40 + 30 + 71.43 − 60 = 0

x

R1 = −1.43 lbf

60 11.43

41.43

M (lbf • in) O

M4 M1 M2

M3

M0 = 0

x

V (lbf) O 1.43

x

M1 M2 M3 M4

= −1.43(4) = −5.72 lbf · in = −5.72 − 41.43(4) = −171.44 lbf · in = −171.44 − 11.43(6) = −240 lbf · in = −240 + 60(4) = 0 checks!

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53

Chapter 4

(b)

y

A

O MO

4 kN/m

2 kN

200 mm

RO

B

C

150 mm

R0 = 2 + 4(0.150) = 2.6 kN M0 = 0

x

150 mm

M0 = 2000(0.2) + 4000(0.150)(0.425) = 655 N · m

V (kN) 2.6 0.6

O

O

x

M1 = −655 + 2600(0.2) = −135 N · m

M (N • m) M3

O

M2 = −135 + 600(0.150) = −45 N · m 1 M3 = −45 + 600(0.150) = 0 checks! 2

x

M2

M1 655

(c)

Fy = 0

y

1000 lbf 6 ft

O

4 ft

B

x

A

M0 = 0: 10R2 − 6(1000) = 0 Fy = 0: R1 − 1000 + 600 = 0

⇒

R2 = 600 lbf

⇒

R2

R1 V (lbf) 400

x

O

600 M (lbf • ft)

M1 M2

O

(d)

y

O

2 ft

+

2000 lbf

1000 lbf 6 ft

2 ft

A

C

B

x

R2

R1 1200

x

M1 = 400(6) = 2400 lbf · ft M2 = 2400 − 600(4) = 0 checks!

200

x

MC = 0

−10R1 + 2(2000) + 8(1000) = 0 R1 = 1200 lbf Fy = 0: 1200 − 1000 − 2000 + R2 = 0 R2 = 1800 lbf

1800 M M1 O

M2 M3

x

M1 = 1200(2) = 2400 lbf · ft M2 = 2400 + 200(6) = 3600 lbf · ft M3 = 3600 − 1800(2) = 0 checks!

R1 = 400 lbf

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(e)

y

400 lbf 4 ft

O

+

800 lbf

3 ft B 3 ft A

R1

R1 = −171.4 lbf Fy = 0: −171.4 − 400 + R2 − 800 = 0

R2

V (lbf)

800

R2 = 1371.4 lbf

x

O 171.4

MB = 0

−7R1 + 3(400) − 3(800) = 0

x

C

571.4

M M3

O

M1 = −171.4(4) = −685.7 lbf · ft M2 = −685.7 − 571.4(3) = −2400 lbf · ft M3 = −2400 + 800(3) = 0 checks!

x

M1 M2

(f) Break at A 40 lbf/in

O

1 R1 = V A = 40(8) = 160 lbf 2

A

8" R1

VA

y

+

320 lbf

160 lbf 5"

A B 2"

5"

D R3

40 lbf/in

320 lbf x

160 lbf

352 lbf

MD = 0

12(160) − 10R2 + 320(5) = 0 R2 = 352 lbf

C

R2

128 lbf

Fy = 0

−160 + 352 − 320 + R3 = 0 R3 = 128 lbf

V (lbf) 192

160

x

O 128

160 M

M4 M1

O

M5 M2 M3

x

1 M1 = 160(4) = 320 lbf · in 2 1 M2 = 320 − 160(4) = 0 checks! (hinge) 2 M3 = 0 − 160(2) = −320 lbf · in M4 = −320 + 192(5) = 640 lbf · in M5 = 640 − 128(5) = 0

checks!

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55

Chapter 4

4-4 (a)

q = R1 x−1 − 40x − 4−1 + 30x − 8−1 + R2 x − 14−1 − 60x − 18−1 V = R1 − 40x − 40 + 30x − 80 + R2 x − 140 − 60x − 180 M = R1 x − 40x − 41 + 30x − 81 + R2 x − 141 − 60x − 181 for x = 18+

V =0

and M = 0

0 = R1 − 40 + 30 + R2 − 60

(1) (2)

Eqs. (1) and (2) give ⇒

R1 + R2 = 70

0 = R1 (18) − 40(14) + 30(10) + 4R2

⇒

(3)

9R1 + 2R2 = 130

(4)

Solve (3) and (4) simultaneously to get R1 = −1.43 lbf, R2 = 71.43 lbf. Ans. From Eqs. (1) and (2), at x = 0+ , V = R1 = −1.43 lbf, M = 0 x = 4+ : x = 8+ :

V V M x = 14+ : V M + x = 18 : V

= −1.43 − 40 = −41.43, M = −1.43x = −1.43 − 40 + 30 = −11.43 = −1.43(8) − 40(8 − 4) 1 = −171.44 = −1.43 − 40 + 30 + 71.43 = 60 = −1.43(14) − 40(14 − 4) + 30(14 − 8) = −240 . = 0, M = 0 See curves of V and M in Prob. 4-3 solution.

(b) q = R0 x−1 − M0 x−2 − 2000x − 0.2−1 − 4000x − 0.350 + 4000x − 0.50 V = R0 − M0 x−1 − 2000x − 0.20 − 4000x − 0.351 + 4000x − 0.51 M = R0 x − M0 − 2000x − 0.21 − 2000x − 0.352 + 2000x − 0.52

(1) (2)

at x = 0.5+ m, V = M = 0, Eqs. (1) and (2) give R0 − 2000 − 4000(0.5 − 0.35) = 0 ⇒ R1 = 2600 N = 2.6 kN R0 (0.5) − M0 − 2000(0.5 − 0.2) − 2000(0.5 − 0.35) 2 = 0

Ans.

with R0 = 2600 N, M0 = 655 N · m Ans. With R0 and M0, Eqs. (1) and (2) give the same V and M curves as Prob. 4-3 (note for V, M0 x−1 has no physical meaning). q = R1 x−1 − 1000x − 6−1 + R2 x − 10−1 V = R1 − 1000x − 60 + R2 x − 100 M = R1 x − 1000x − 61 + R2 x − 101

(c)

at x = 10+ ft, V = M = 0, Eqs. (1) and (2) give R1 − 1000 + R2 = 0 10R1 − 1000(10 − 6) = 0 0 ≤ x ≤ 6: 6 ≤ x ≤ 10:

⇒ ⇒

R1 + R2 = 1000 R1 = 400 lbf , R2 = 1000 − 400 = 600 lbf

V = 400 lbf, M = 400x V = 400 − 1000(x − 6) 0 = 600 lbf M = 400x − 1000(x − 6) = 6000 − 600x

See curves of Prob. 4-3 solution.

(1) (2)

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(d)

q = R1 x−1 − 1000x − 2−1 − 2000x − 8−1 + R2 x − 10−1 V = R1 − 1000x − 20 − 2000x − 80 + R2 x − 100 M = R1 x − 1000x − 21 − 2000x − 81 + R2 x − 101

(1) (2)

At x = 10+ , V = M = 0 from Eqs. (1) and (2) R1 − 1000 − 2000 + R2 = 0

⇒

R1 + R2 = 3000

10R1 − 1000(10 − 2) − 2000(10 − 8) = 0

⇒

R1 = 1200 lbf , R2 = 3000 − 1200 = 1800 lbf

0 ≤ x ≤ 2: 2 ≤ x ≤ 8:

V = 1200 lbf, M = 1200x lbf · ft V = 1200 − 1000 = 200 lbf M = 1200x − 1000(x − 2) = 200x + 2000 lbf · ft

8 ≤ x ≤ 10:

V = 1200 − 1000 − 2000 = −1800 lbf M = 1200x − 1000(x − 2) − 2000(x − 8) = −1800x + 18 000 lbf · ft

Plots are the same as in Prob. 4-3. (e)

q = R1 x−1 − 400x − 4−1 + R2 x − 7−1 − 800x − 10−1 V = R1 − 400x − 40 + R2 x − 70 − 800x − 100 M = R1 x − 400x − 41 + R2 x − 71 − 800x − 101

(1) (2)

at x = 10+ , V = M = 0 R1 − 400 + R2 − 800 = 0 10R1 − 400(6) + R2 (3) = 0

⇒ ⇒

R1 + R2 = 1200 10R1 + 3R2 = 2400

(3) (4)

Solve Eqs. (3) and (4) simultaneously: R1 = −171.4 lbf, R2 = 1371.4 lbf 0 ≤ x ≤ 4: 4 ≤ x ≤ 7: 7 ≤ x ≤ 10:

V = −171.4 lbf, M = −171.4x lbf · ft V = −171.4 − 400 = −571.4 lbf M = −171.4x − 400(x − 4) lbf · ft = −571.4x + 1600 V = −171.4 − 400 + 1371.4 = 800 lbf M = −171.4x − 400(x − 4) + 1371.4(x − 7) = 800x − 8000 lbf · ft

Plots are the same as in Prob. 4-3. (f) q = R1 x−1 − 40x0 + 40x − 80 + R2 x − 10−1 − 320x − 15−1 + R3 x − 20 V = R1 − 40x + 40x − 81 + R2 x − 100 − 320x − 150 + R3 x − 200 (1) 2 2 1 1 1 M = R1 x − 20x + 20x − 8 + R2 x − 10 − 320x − 15 + R3 x − 20 (2) 2 M = 0 at x = 8 in ∴ 8R1 − 20(8) = 0 ⇒ R1 = 160 lbf at x = 20+ ,

V and M = 0

160 − 40(20) + 40(12) + R2 − 320 + R3 = 0 160(20) − 20(20) 2 + 20(12) 2 + 10R2 − 320(5) = 0

⇒ ⇒

R2 + R3 = 480 R2 = 352 lbf R3 = 480 − 352 = 128 lbf 2 0 ≤ x ≤ 8: V = 160 − 40x lbf, M = 160x − 20x lbf · in 8 ≤ x ≤ 10: V = 160 − 40x + 40(x − 8) = −160 lbf , M = 160x − 20x 2 + 20(x − 8) 2 = 1280 − 160x lbf · in

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57

Chapter 4

10 ≤ x ≤ 15:

V = 160 − 40x + 40(x − 8) + 352 = 192 lbf M = 160x − 20x 2 + 20(x − 8) + 352(x − 10) = 192x − 2240

15 ≤ x ≤ 20:

V = 160 − 40x + 40(x − 8) + 352 − 320 = −128 lbf M = 160x − 20x 2 − 20(x − 8) + 352(x − 10) − 320(x − 15) = −128x + 2560

Plots of V and M are the same as in Prob. 4-3. 4-5 Solution depends upon the beam selected. 4-6 (a) Moment at center, xc = (l − 2a)/2 2 w l wl l l Mc = (l − 2a) − −a = 2 2 2 2 4 At reaction, |Mr | = wa 2 /2 a = 2.25, l = 10 in, w = 100 lbf/in 100(10) Mc = 2 Mr =

10 − 2.25 = 125 lbf · in 4

100(2.252 ) = 253.1 lbf · in Ans. 2

(b) Minimum occurs when Mc = |Mr | wa 2 wl l −a = 2 4 2

⇒

a 2 + al − 0.25l 2 = 0

Taking the positive root l √

1 −l + l 2 + 4(0.25l 2 ) = a= 2 − 1 = 0.2071l 2 2 for l = 10 in and w = 100 lbf,

Mmin = (100/2)[(0.2071)(10)]2 = 214.5 lbf · in

4-7 For the ith wire from bottom, from summing forces vertically (a)

Ti xi

Ti = (i + 1)W

a W

Ans.

iW

From summing moments about point a, Ma = W (l − xi ) − i W xi = 0 Giving, xi =

l i +1

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So l l = 1+1 2 l l = x= 2+1 3 l l = y= 3+1 4 l l = z= 4+1 5

W =

(b) With straight rigid wires, the mobile is not stable. Any perturbation can lead to all wires becoming collinear. Consider a wire of length l bent at its string support: Ti il i1

l i1

iW

W

Ma = 0 Ma =

ilW iWl cos α − cos β = 0 i +1 i +1

iWl (cos α − cos β) = 0 i +1 Moment vanishes when α = β for any wire. Consider a ccw rotation angle β, which makes α → α + β and β → α − β iWl [cos(α + β) − cos(α − β)] i +1 2i W l . 2i W lβ = sin α sin β = sin α i +1 i +1

Ma =

There exists a correcting moment of opposite sense to arbitrary rotation β. An equation for an upward bend can be found by changing the sign of W . The moment will no longer be correcting. A curved, convex-upward bend of wire will produce stable equilibrium too, but the equation would change somewhat. 4-8 (a)

1

x

cw 2s 2p 2

C R

(6, 4

ccw

ccw y

) 2

D

1

12 + 6 =9 2 12 − 6 CD = =3 2 R = 32 + 42 = 5 C=

(12, 4cw)

σ1 = 5 + 9 = 14 σ2 = 9 − 5 = 4

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59

Chapter 4

1 −1 4 = 26.6◦ cw φ p = tan 2 3

4

x 26.6 14

φs = 45◦ − 26.6◦ = 18.4◦ ccw

τ1 = R = 5, 3

3 18.4

x

5

3 3

(b)

y

1

(9, 5cw) R D

2

C

1

2p

σ1 = 6.1 + 12.5 = 18.6 1 5 φ p = tan−1 = 27.5◦ ccw 2 3.5 σ2 = 12.5 − 6.1 = 6.4

2s (16, 5ccw)

ccw

9 + 16 = 12.5 2 16 − 9 CD = = 3.5 2 R = 52 + 3.52 = 6.10 C=

cw

2

x

6.4 18.6 27.5 x

τ1 = R = 6.10 , φs = 45◦ − 27.5◦ = 17.5◦ cw 12.5 6.10 x 17.5 12.5

(c)

1 cw

y

C=

(24, 6cw) R D

2

C

(10, 6 ccw

)

2s

x 2

24 − 10 =7 2 R = 72 + 62 = 9.22

CD = 1

2p ccw

24 + 10 = 17 2

σ1 = 17 + 9.22 = 26.22 σ2 = 17 − 9.22 = 7.78

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1 −1 7 90 + tan = 69.7◦ ccw φp = 2 6

26.22 69.7 7.78 x

φs = 69.7◦ − 45◦ = 24.7◦ ccw

τ1 = R = 9.22, 17

9.22 17 24.7 x

(d)

x cw

1

C=

(9, 8cw) 2s

D

1

C R

(19, 8ccw)

ccw

2

19 − 9 =5 2 R = 52 + 82 = 9.434

CD =

2p 2

σ1 = 14 + 9.43 = 23.43 σ2 = 14 − 9.43 = 4.57

y

1 −1 5 90 + tan = 61.0◦ cw φp = 2 8

4.57 x 61 23.43

τ1 = R = 9.434,

φs = 61◦ − 45◦ = 16◦ cw

14

x 16 14 9.434

9 + 19 = 14 2

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61

Chapter 4

4-9 (a)

1

cw

C=

y (12, 7cw) R

2

D

1

2p

σ1 = 4 + 10.63 = 14.63 σ2 = 4 − 10.63 = −6.63

2s

x 2

ccw

12 + 4 =8 2 R = 82 + 72 = 10.63

CD =

C

(4, 7ccw)

12 − 4 =4 2

1 −1 8 90 + tan = 69.4◦ ccw φp = 2 7

14.63 69.4 6.63 x

φs = 69.4◦ − 45◦ = 24.4◦ ccw

τ1 = R = 10.63, 4

10.63 4 24.4 x

(b)

y

cw 1

C=

(5, 8cw)

R

6−5 = 0.5 2

6+5 = 5.5 2 R = 5.52 + 82 = 9.71

CD = C

2

D

1

2p 2s (6, 8ccw) ccw

2

σ1 = 0.5 + 9.71 = 10.21 σ2 = 0.5 − 9.71 = −9.21

x

9.21

φp =

10.21 27.75 x

1 −1 8 tan = 27.75◦ ccw 2 5.5

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φs = 45◦ − 27.75◦ = 17.25◦ cw

τ1 = R = 9.71, 0.5 9.71

x 17.25 0.5

(c)

cw 1

C=

x (8, 6cw)

8+7 = 7.5 2 R = 7.52 + 62 = 9.60

2s

CD =

2p 2

D

−8 + 7 = −0.5 2

1

C

σ1 = 9.60 − 0.5 = 9.10

R (7, 6ccw)

σ2 = −0.5 − 9.6 = −10.1

y 2 ccw

1 −1 7.5 90 + tan = 70.67◦ cw φp = 2 6

10.1 x

70.67 9.1

φs = 70.67◦ − 45◦ = 25.67◦ cw

τ1 = R = 9.60, 0.5

x 25.67 0.5 9.60

(d)

cw

1

C= x

2s

2

C

R

(6, 3ccw)

D 1

9+6 = 7.5 2 R = 7.52 + 32 = 8.078

CD =

(9, 3cw) 2p

9−6 = 1.5 2

σ1 = 1.5 + 8.078 = 9.58

y

ccw

2

σ2 = 1.5 − 8.078 = −6.58

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63

Chapter 4 6.58

φp =

1 −1 3 tan = 10.9◦ cw 2 7.5

x

10.9 9.58

φs = 45◦ − 10.9◦ = 34.1◦ ccw

τ1 = R = 8.078, 1.5

1.5 34.1 x 8.08

4-10 (a)

1

cw

C= x (20, 8cw)

2s 2p 2

D 1

C

R (10, 8ccw)

20 − 10 =5 2

20 + 10 = 15 2 R = 152 + 82 = 17

CD =

σ1 = 5 + 17 = 22

y

σ2 = 5 − 17 = −12

2

ccw 12

φp =

1 −1 8 tan = 14.04◦ cw 2 15

x 14.04 22

τ1 = R = 17,

φs = 45◦ − 14.04◦ = 30.96◦ ccw

5 5 30.96 x 17

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(b)

1

cw

C= y (10, 10cw)

R D

2

C

2p

1

(30, 10ccw) x

2

12.36

φp = 32.36 13.28

τ1 = R = 22.36,

30 + 10 = 20 2 R = 202 + 102 = 22.36

CD =

2s

ccw

30 − 10 = 10 2

σ1 = 10 + 22.36 = 32.36 σ2 = 10 − 22.36 = −12.36

1 −1 10 tan = 13.28◦ ccw 2 20

x

φs = 45◦ − 13.28◦ = 31.72◦ cw

10 22.36 x 31.72 10

(c)

1

cw

2s

(10, 9cw)

2

2p

D

1

C R

(18, 9ccw) y ccw

−10 + 18 =4 2 10 + 18 CD = = 14 2 R = 142 + 92 = 16.64 C=

x

2

σ1 = 4 + 16.64 = 20.64 σ2 = 4 − 16.64 = −12.64

1 −1 14 90 + tan = 73.63◦ cw φp = 2 9

12.64 x

73.63

20.64

τ1 = R = 16.64,

φs = 73.63◦ − 45◦ = 28.63◦ cw

4

x 28.63 4 16.64

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65

Chapter 4

(d)

1

cw x 2s

(12, 12cw)

12 + 22 = 17 2 R = 172 + 122 = 20.81

CD =

2p 2

D

1

C R

σ1 = 5 + 20.81 = 25.81 σ2 = 5 − 20.81 = −15.81

(22, 12ccw) y 2

ccw

−12 + 22 =5 2

C=

1 −1 17 90 + tan = 72.39◦ cw φp = 2 12

15.81 x

72.39 25.81

φs = 72.39◦ − 45◦ = 27.39◦ cw

τ1 = R = 20.81, 5

x 27.39 5 20.81

4-11 (a)

1/3

14 7 2

1/2 5 2/3 2 4 3 y

y 2 0

1/3

(b) y

x 10 x 1

(0, 4cw)

3

R 2

C

D

1

(10, 4ccw) x

τ1/3 = R = 6.40,

0 + 10 =5 2 10 − 0 CD = =5 2 R = 52 + 42 = 6.40 C=

1/2

2/3

τ1/2 =

σ1 = 5 + 6.40 = 11.40 σ2 = 0, σ3 = 5 − 6.40 = −1.40

11.40 = 5.70, 2

τ2/3 =

1.40 = 0.70 2

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(c)

(2, 4cw) x

2 circles

Point is a circle C

3

1 2

D

−2 − 8 = −5 2

C=

8−2 =3 2 R = 32 + 42 = 5

CD =

(8, 4ccw) y

σ1 = −5 + 5 = 0, σ2 = 0 σ3 = −5 − 5 = −10 10 = 5, 2

τ1/3 =

τ1/2 = 0, 1/3

(d)

τ2/3 = 5

10 − 30 = −10 2 10 + 30 CD = = 20 2 R = 202 + 102 = 22.36 C=

2/3 y 1/2

(30, 10cw) R C

D

2

3

1

(10, 10ccw) x

σ1 = −10 + 22.36 = 12.36 σ2 = 0 σ3 = −10 − 22.36 = −32.36

τ1/3 = 22.36,

τ1/2 =

12.36 = 6.18, 2

τ2/3 =

32.36 = 16.18 2

4-12 1/3

(a)

C=

2/3 x (80, 20cw)

1/2

C 3

D

R

2

1

(30, 20ccw) y

−80 − 30 = −55 2

80 − 30 = 25 2 R = 252 + 202 = 32.02

CD =

σ1 = 0 σ2 = −55 + 32.02 = −22.98 = −23.0 σ3 = −55 − 32.0 = −87.0

τ1/2 =

23 = 11.5, 2

τ2/3 = 32.0,

τ1/3 =

87 = 43.5 2

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67

Chapter 4 1/3

(b) 2/3

C=

x (30, 30cw)

R

60 + 30 = 45 2 R = 452 + 302 = 54.1

CD =

1/2 3

C

2

30 − 60 = −15 2

1

D

(60, 30ccw)

σ1 = −15 + 54.1 = 39.1

y

σ2 = 0 σ3 = −15 − 54.1 = −69.1 τ1/3 = (c)

39.1 + 69.1 = 54.1, 2

y

τ1/2 =

1/3

C=

1/2 (0, 20cw)

2/3

τ2/3 =

69.1 = 34.6 2

40 + 0 = 20 2

40 − 0 = 20 2 R = 202 + 202 = 28.3

CD =

R

3

39.1 = 19.6, 2

D

2

1

C

(40, 20ccw)

σ1 = 20 + 28.3 = 48.3

x

σ2 = 20 − 28.3 = −8.3 σ3 = σz = −30 τ1/3 =

48.3 + 30 = 39.1, 2

(d)

1/3

τ1/2 = 28.3,

τ2/3 =

C=

50 = 25 2

x

1/2

(50, 30cw)

2/3 3

2

C

D

(0, 30ccw)

1

30 − 8.3 = 10.9 2

50 = 25 2 R = 252 + 302 = 39.1

CD =

σ1 = 25 + 39.1 = 64.1

y

σ2 = 25 − 39.1 = −14.1 σ3 = σz = −20 τ1/3 =

64.1 + 20 = 42.1, 2

τ1/2 = 39.1,

τ2/3 =

20 − 14.1 = 2.95 2

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4-13 σ =

2000 F = = 10 190 psi = 10.19 kpsi A (π/4)(0.52 )

Ans.

FL L 72 = σ = 10 190 = 0.024 46 in Ans. AE E 30(106 ) 0.024 46 δ = 340(10−6 ) = 340µ Ans. 1 = = L 72 δ=

From Table A-5, ν = 0.292 2 = −ν1 = −0.292(340) = −99.3µ

Ans.

d = 2 d = −99.3(10−6 )(0.5) = −49.6(10−6 ) in 4-14

4-15

4-16

Ans.

From Table A-5, E = 71.7 GPa L 3 = 5.65(10−3 ) m = 5.65 mm δ = σ = 135(106 ) E 71.7(109 )

Ans.

From Table 4-2, biaxial case. From Table A-5, E = 207 GPa and ν = 0.292 σx =

E(x + ν y ) 207(109 )[0.0021 + 0.292(−0.000 67)] −6 = (10 ) = 431 MPa 1 − ν2 1 − 0.2922

σy =

207(109 )[−0.000 67 + 0.292(0.0021)] −6 (10 ) = −12.9 MPa 1 − 0.2922

Ans.

Ans.

The engineer has assumed the stress to be uniform. That is, t F

Ft = −F cos θ + τ A = 0 ⇒ τ =

F cos θ A

When failure occurs in shear F cos θ A The uniform stress assumption is common practice but is not exact. If interested in the details, see p. 570 of 6th edition. Ssu =

4-17

From Eq. (4-15) σ 3 − (−2 + 6 − 4)σ 2 + [−2(6) + (−2)(−4) + 6(−4) − 32 − 22 − (−5) 2 ]σ − [−2(6)(−4) + 2(3)(2)(−5) − (−2)(2) 2 − 6(−5) 2 − (−4)(3) 2 ] = 0 σ 3 − 66σ + 118 = 0 Roots are: 7.012, 1.89, −8.903 kpsi Ans.

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69

Chapter 4

7.012 − 1.89 = 2.56 kpsi 2 8.903 + 1.89 = 5.40 kpsi = 2

1/3 (kpsi)

τ1/2 = τ2/3

τmax = τ1/3 =

2/3 1/2

8.903 + 7.012 = 7.96 kpsi Ans. 2

8.903

1.89

7.012

(kpsi)

Note: For Probs. 4-17 to 4-19, one can also find the eigenvalues of the matrix σx τx y τzx [σ ] = τx y σ y τ yz τzx τ yz σz for the principal stresses 4-18

From Eq. (4-15)

√ 2 σ 3 − (10 + 0 + 10)σ 2 + 10(0) + 10(10) + 0(10) − 202 − −10 2 − 02 σ √ 2 √

− 10(0)(10) + 2(20) −10 2 (0) − 10 −10 2 − 0(0) 2 − 10(20) 2 = 0 σ 3 − 20σ 2 − 500σ + 6000 = 0

Roots are: 30, 10, −20 MPa 30 − 10 = 10 MPa τ1/2 = 2

Ans.

10 + 20 = 15 MPa τ2/3 = 2 30 + 20 = 25 MPa τmax = τ1/3 = 2

4-19

(MPa)

1/3

2/3 1/2

20

10

Ans.

From Eq. (4-15) σ 3 − (1 + 4 + 4)σ 2 + [1(4) + 1(4) + 4(4) − 22 − (−4) 2 − (−2) 2 ]σ −[1(4)(4) + 2(2)(−4)(−2) − 1(−4) 2 − 4(−2) 2 − 4(2) 2 ] = 0 σ 3 − 9σ 2 = 0

30

(MPa)

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Roots are: 9, 0, 0 kpsi

1/2 1/3

(kpsi)

2/3 O 0

τ2/3 = 0, 4-20 (a) R1 =

τ1/2 = τ1/3 = τmax =

c F l

Mmax = R1 a = σ =

(b)

9

(kpsi)

9 = 4.5 kpsi Ans. 2

ac F l

6M 6 ac = 2 F 2 bh bh l

⇒

F=

σ bh 2l 6ac

Ans.

(σm /σ )(bm /b) (h m / h) 2 (lm /l) 1(s)(s) 2 (s) Fm = = = s2 F (am /a) (cm /c) (s)(s)

Ans.

For equal stress, the model load varies by the square of the scale factor. 4-21

wl , R1 = 2 σ =

Mmax |x=l/2

l wl 2 wl l− = = 22 2 8

6M 3W l 6 wl 2 = = 2 2 bh bh 8 4bh 2

⇒

W =

4 σ bh 2 3 l

Wm (σm /σ )(bm /b) (h m / h) 2 1(s)(s) 2 = = = s2 W lm /l s wm lm = s2 wl

⇒

s2 wm = =s w s

Ans.

Ans.

Ans.

For equal stress, the model load w varies linearily with the scale factor. 4-22 (a) Can solve by iteration or derive equations for the general case. W1

W2

W3 . . .

A a23 RA

d3

a13

WT . . .

Find maximum moment under wheel W3

WT = W at centroid of W’s

Wn B RB

x3 l

RA =

l − x3 − d3 WT l

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71

Chapter 4

Under wheel 3 M3 = R A x3 − W1 a13 − W2 a23 = For maximum,

(l − x3 − d3 ) WT x3 − W1 a13 − W2 a23 l

WT d M3 = 0 = (l − d3 − 2x3 ) dx3 l

substitute into M,

M3 =

⇒

⇒

x3 =

l − d3 2

(l − d3 ) 2 WT − W1 a13 − W2 a23 4l

This means the midpoint of d3 intersects the midpoint of the beam For wheel i

i−1 (l − di ) 2 Mi = WT − W j a ji 4l j=1

l − di , xi = 2

Note for wheel 1: W j a ji = 0 WT = 104.4,

W1 = W2 = W3 = W4 =

(1200 − 238) 2 M1 = (104.4) = 20 128 kip · in 4(1200)

476 = 238 in, Wheel 1: d1 = 2 Wheel 2:

104.4 = 26.1 kip 4

d2 = 238 − 84 = 154 in (1200 − 154) 2 M2 = (104.4) − 26.1(84) = 21 605 kip · in = Mmax 4(1200)

Check if all of the wheels are on the rail 84" 77"

84"

315"

xmax 600"

600"

(b) xmax = 600 − 77 = 523 in (c) See above sketch. (d) inner axles 4-23 (a)

D c1 0.833"

a

a

0.75"

Ga

0.083"

C

1

1" 4

y c2 0.667"

Gb

3" 8

b

0.167"

B

0.5"

A 1" 4

1 12 "

1" 4

Aa = Ab = 0.25(1.5) = 0.375 in2 A = 3(0.375) = 1.125 in2

1

1.5"

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y¯ =

2(0.375)(0.75) + 0.375(0.5) = 0.667 in 1.125

0.25(1.5) 3 Ia = = 0.0703 in4 12 Ib =

1.5(0.25) 3 = 0.001 95 in4 12

I1 = 2[0.0703 + 0.375(0.083) 2 ] + [0.001 95 + 0.375(0.167) 2 ] = 0.158 in4 σA =

10 000(0.667) = 42(10) 3 psi Ans. 0.158

σB =

10 000(0.667 − 0.375) = 18.5(10) 3 psi Ans. 0.158

σC =

10 000(0.167 − 0.125) = 2.7(10) 3 psi Ans. 0.158

σD = −

10 000(0.833) = −52.7(10) 3 psi 0.158

(b)

Ans.

Ans.

D

c1 1.155"

C

a

b

1.732" 1

1

c2 0.577"

y

B

0.982" Ga

0.327"

0.577" 0.25" A

0.577" 1.134" A

2"

Here we treat the hole as a negative area. Aa = 1.732 in2 0.982 = 0.557 in2 Ab = 1.134 2 A = 1.732 − 0.557 = 1.175 in2 y¯ =

1.732(0.577) − 0.557(0.577) = 0.577 in 1.175

Ia =

2(1.732) 3 bh 3 = = 0.289 in4 36 36

Ib =

1.134(0.982) 3 = 0.0298 in4 36

I1 = Ia − Ib = 0.289 − 0.0298 = 0.259 in4

Gb

Ans.

Ans.

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73

Chapter 4

because the centroids are coincident. 10 000(0.577) = 22.3(10) 3 psi Ans. 0.259 10 000(0.327) = 12.6(10) 3 psi Ans. σB = 0.259 10 000(0.982 − 0.327) = −25.3(10) 3 psi σC = − 0.259 10 000(1.155) = −44.6(10) 3 psi Ans. σD = − 0.259 (c) Use two negative areas. σA =

Ans.

D C

c1 1.708"

b 1

1 Gb Gc

c2 2.292"

1.5" 2"

c

B a G a

0.25"

A

Ab = 9 in2 , Ac = 16 in2 , A = 16 − 9 − 1 = 6 in2 ; Aa = 1 in2 , y¯a = 0.25 in, y¯b = 2.0 in, y¯c = 2 in 16(2) − 9(2) − 1(0.25) = 2.292 in Ans. y¯ = 6 c1 = 4 − 2.292 = 1.708 in 2(0.5) 3 = 0.020 83 in4 12 3(3) 3 = 6.75 in4 Ib = 12 4(4) 3 = 21.333 in4 Ic = 12 Ia =

I1 = [21.333 + 16(0.292) 2 ] − [6.75 + 9(0.292) 2 ] − [0.020 83 + 1(2.292 − 0.25) 2 ] = 10.99 in4 Ans. 10 000(2.292) = 2086 psi Ans. 10.99 10 000(2.292 − 0.5) σB = = 1631 psi Ans. 10.99 10 000(1.708 − 0.5) = −1099 psi Ans. σC = − 10.99 10 000(1.708) σD = − = −1554 psi Ans. 10.99 σA =

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(d) Use a as a negative area. C c1 1.355"

b

B

1

1 1.490"

a

3.464"

c2 2.645" Ga 1.155" A

Aa = 6.928 in2 , y¯a = 1.155 in,

Ab = 16 in2 , y¯b = 2 in

A = 9.072 in2 ;

2(16) − 1.155(6.928) = 2.645 in 9.072 c1 = 4 − 2.645 = 1.355 in y¯ =

Ans.

4(3.464) 3 bh 3 = = 4.618 in4 36 36 3 4(4) = 21.33 in4 Ib = 12 I1 = [21.33 + 16(0.645) 2 ] − [4.618 + 6.928(1.490) 2 ] = 7.99 in4 Ans. 10 000(2.645) = 3310 psi Ans. σA = 7.99 10 000(3.464 − 2.645) σB = − = −1025 psi Ans. 7.99 10 000(1.355) = −1696 psi Ans. σC = − 7.99 Ia =

(e)

C c1 1.422"

a B

c2 2.828"

3.625(7.5) + 1.5(4.5) = 2.828 in Ans. 12 1 1 = (6)(1.25) 3 + 7.5(3.625 − 2.828) 2 + (1.5)(3) 3 + 4.5(2.828 − 1.5) 2 12 12 = 17.05 in4 Ans. 10 000(2.828) = 1659 psi Ans. = 17.05 10 000(3 − 2.828) = −101 psi Ans. =− 17.05 10 000(1.422) = −834 psi Ans. =− 17.05 b

A

I

σA σB σC

Aa = 6(1.25) = 7.5 in2 Ab = 3(1.5) = 4.5 in2 A = Ac + Ab = 12 in2 y¯ =

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75

Chapter 4

(f)

Let a = total area

D a

c 1.5

A = 1.5(3) − 1(1.25) = 3.25 in2 1 1 I = Ia − 2Ib = (1.5)(3) 3 − (1.25)(1) 3 12 12 = 3.271 in4 Ans.

C b

b

c 1.5

B A

1.5

10 000(1.5) = 4586 psi, 3.271

σ D = −4586 psi

10 000(0.5) = 1529 psi, σB = 3.271

σC = −1529 psi

σA =

Ans.

4-24 (a) The moment is maximum and constant between A and B I =

M = −50(20) = −1000 lbf · in ,

1 (0.5)(2) 3 = 0.3333 in4 12

E I 1.6(106 )(0.3333) = 533.3 in ρ = = M 1000 (x, y) = (30, −533.3) in

Ans.

(b) The moment is maximum and constant between A and B M = 50(5) = 250 lbf · in, ρ=

I = 0.3333 in4

1.6(106 )(0.3333) = 2133 in 250

(x, y) = (20, 2133) in

Ans.

Ans.

4-25 (a)

1000 lbf 12"

O

6"

I = B

A 333 lbf

A = 0.75(1.5) = 1.125 in 667 lbf

Mmax is at A. At the bottom of the section,

V (lbf)

x 667

O

4000(0.75) Mc = = 14 225 psi Ans. I 0.2109 Due to V, τmax constant is between A and B at y = 0

σmax =

333 O

M (lbf • in)

1 (0.75)(1.5) 3 = 0.2109 in4 12

4000

τmax = x

3 667 3V = = 889 psi 2A 2 1.125

Ans.

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(b)

1000 lbf

1000 lbf

8"

A

I =

8"

x

B

O 2000 lbf V (lbf)

1 (1)(2) 3 = 0.6667 in4 12

Mmax is at A at the top of the beam σmax =

1000 x

O

|Vmax | = 1000 lbf τmax =

1000

8000(1) = 12 000 psi 0.6667

Ans.

from O to B at y = 0

3 1000 3V = = 750 psi 2A 2 (2)(1)

Ans.

M (lbf • in) x

O

8000

(c)

120 lbf/in x

O

A

5"

B

15"

1500 lbf

5" C

1500 lbf

V (lbf) 900 600 x

O 600

900

M (lbf • in)

x

O M1

M3

τmax =

100 lbf/in x O

A

6"

B

12"

1350 lbf V (lbf)

450 lbf

750 x

O 7.5" 600 M (lbf • in)

1875(1) = 3750 psi 0.5 At A and B at y = 0

σmax =

M2

(d)

1 (0.75)(2) 3 = 0.5 in4 12 1 M1 = − 600(5) = −1500 lbf · in = M3 2 1 M2 = −1500 + (900)(7.5) = 1875 lbf · in 2 Mmax is at span center. At the bottom of the beam, I =

450

M1

Ans.

1 (1)(2) 3 = 0.6667 in4 12 600 (6) = −1800 lbf · in M1 = − 2 1 M2 = −1800 + 750(7.5) = 1013 lbf · in 2 At A, top of beam

x

O

3 900 = 900 psi 2 (0.75)(2)

I =

σmax =

M2

Ans.

1800(1) = 2700 psi 0.6667

Ans.

At A, y = 0 τmax =

3 750 = 563 psi 2 (2)(1)

Ans.

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77

Chapter 4

4-26 Mmax =

wl 2 8

⇒

σmax =

wl 2 c 8I

⇒

w=

8σ I cl 2

(a) l = 12(12) = 144 in, I = (1/12)(1.5)(9.5) 3 = 107.2 in4 w=

8(1200)(107.2) = 10.4 lbf/in 4.75(1442 )

Ans.

(b) l = 48 in, I = (π/64)(24 − 1.254 ) = 0.6656 in4 w=

8(12)(103 )(0.6656) = 27.7 lbf/in 1(48) 2

Ans.

8(12)(103 )(2.051) = 57.0 lbf/in 1.5(48) 2

Ans.

. l = 48 in, I = (1/12)(2)(33 ) − (1/12)(1.625)(2.6253 ) = 2.051 in4

(c)

w=

(d) l = 72 in; Table A-6, I = 2(1.24) = 2.48 in4 0.842"

cmax = 2.158"

2.158"

8(12)(103 )(2.48) = 21.3 lbf/in 2.158(72) 2 (e) l = 72 in; Table A-7, I = 3.85 in4 w=

2

w=

8(12)(103 )(3.85) = 35.6 lbf/in 2(722 )

Ans.

Ans.

(f) l = 72 in, I = (1/12)(1)(43 ) = 5.333 in4 w= 4-27

8(12)(103 )(5.333) = 49.4 lbf/in (2)(72) 2

(a) Model (c)

π (0.54 ) = 3.068(10−3 ) in4 64 π A = (0.52 ) = 0.1963 in2 4 218.75(0.25) Mc = σ = I 3.068(10−3 ) = 17 825 psi = 17.8 kpsi Ans. I =

500 lbf 500 lbf 0.4375 1.25 in 500 lbf

Ans.

500 lbf

V (lbf) 500

O

τmax =

500 M (lbf ¥ in) O

Mmax 500(0.4375) 218.75 lbf ¥ in

4 500 4V = = 3400 psi 3A 3 0.1963

Ans.

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(b) Model (d) 1333 lbf/in 0.25" 1.25" 500 lbf

1 Mmax = 500(0.25) + (500)(0.375) 2 = 218.75 lbf · in

500 lbf

V (lbf) 500

Vmax = 500 lbf

O

Same M and V 500 M

∴ σ = 17.8 kpsi Ans. τmax = 3400 psi

Mmax

Ans.

O

4-28

If support R B is between F1 and F2 at position x = l, maximum moments occur at x = 3 and l. M B = R A l − 2000(l − 3) + 1100(7.75 − l) = 0 R A = 3100 − 14 525/l Mx=3 = 3R A = 9300 − 43 575/l M B = R A l − 2000(l − 3) = 1100 l − 8525 To minimize the moments, equate Mx=3 to −M B giving 9300 − 43 575/l = −1100l + 8525 Multiply by l and simplify to l 2 + 0.7046l − 39.61 = 0 The positive solution for l is 5.95 in and the magnitude of the moment is M = 9300 − 43 575/5.95 = 1976 lbf · in Placing the bearing to the right of F2 , the bending moment would be minimized by placing it as close as possible to F2 . If the bearing is near point B as in the original figure, then we need to equate the reaction forces. From statics, R B = 14 525/l, and R A = 3100 − R B . For R A = R B , then R A = R B = 1550 lbf, and l = 14 575/1550 = 9.37 in.

4-29 F l

p2 b

p1 a

p1 + p2 x − l1 + terms for x > l + a a p1 + p2 1 x − l2 + terms for x > l + a V = −F + p1 x − l − 2a p1 p 1 + p2 M = −F x + x − l2 − x − l3 + terms for x > l + a 2 6a q = −Fx−1 + p1 x − l0 −

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79

Chapter 4

At x = (l + a) + , V = M = 0, terms for x > l + a = 0 p1 + p2 2 a =0 2a

⇒

p1 − p2 =

p1 + p2 3 p1 a 2 − a =0 −F(l + a) + 2 6a

⇒

2 p1 − p2 =

−F + p1 a −

p1 =

From (1) and (2)

2F (3l + 2a), a2

a b = p2 p1 + p2

From similar triangles

p2 =

⇒

2F a 6F(l + a) a2

(1) (2)

2F (3l + a) a2

(3)

ap2 p1 + p2

(4)

b=

Mmax occurs where V = 0 F a 2b

l

p2 p2

p2 b

p1

xmax = l + a − 2b

b

Mmax = −F(l + a − 2b) +

p1 p1 + p2 (a − 2b) 2 − (a − 2b) 3 2 6a

= −Fl − F(a − 2b) +

p1 p1 + p2 (a − 2b) 2 − (a − 2b) 3 2 6a

Normally Mmax = −Fl The fractional increase in the magnitude is

=

F(a − 2b) − ( p1 /2)(a − 2b) 2 − [( p1 + p2 )/6a](a − 2b) 3 Fl

For example, consider F = 1500 lbf, a = 1.2 in, l = 1.5 in (3)

(4)

p1 =

2(1500) [3(1.5) + 2(1.2)] = 14 375 lbf/in 1.22

p2 =

2(1500) [3(1.5) + 1.2] = 11 875 lbf/in 1.22

b = 1.2(11 875)/(14 375 + 11 875) = 0.5429 in

Substituting into (5) yields

= 0.036 89 4-30

or

Computer program; no solution given here.

3.7% higher than −Fl

(5)

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4-31 y a

c

x

l R2

R1

σ =

c F l c M = Fx l

R1 =

F

6M 6(c/l) Fx = 2 bh bh 2

0≤x ≤a ⇒

6cFx blσmax

h=

0≤x ≤a

4-32 F a l R1

b F l b M = Fx l 32M 32 b σmax = Fx = 3 πd πd 3 l 32 bFx 1/3 d= 0≤x ≤a π lσmax R1 =

b R2

4-33

Ans.

t

t

b

Square:

b

Am = (b − t) 2 Tsq = 2Am tτall = 2(b − t) 2 tτall

Round:

Am = π(b − t) 2 /4 Trd = 2π(b − t) 2 tτall /4

Ratio of torques

Tsq 4 2(b − t) 2 tτall = = 1.27 = 2 Trd π(b − t) tτall /2 π

Twist per unit length square: 2Gθ1 t θsq = tτall Round:

L 4(b − t) L = C = C A m A m (b − t) 2

4(b − t) π(b − t) L =C =C θrd = C 2 A m π(b − t) /4 (b − t) 2

Ratio equals 1, twists are the same.

Ans.

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81

Chapter 4

Note the weight ratio is Wsq b−t ρl(b − t) 2 = = Wrd ρlπ(b − t)(t) πt

thin-walled assumes b ≥ 20t

19 = 6.04 π = 2.86 =

4-34

with b = 20 with b = 10t

l = 40 in, τall = 11 500 psi, G = 11.5(106 ) psi, t = 0.050 in rm = ri + t/2 = ri + 0.025 for ri > 0 =0 for ri = 0 π Am = (1 − 0.05) 2 − 4 rm2 − rm2 = 0.952 − (4 − π)rm2 4 L m = 4(1 − 0.05 − 2rm + 2πrm /4) = 4[0.95 − (2 − π/2)rm ] T = 2Am tτ = 2(0.05)(11 500) Am = 1150Am

Eq. (4-45): Eq. (4-46):

θ(deg) = θ1 l

T L m l 180 T L m (40) 180 180 = = π 4G A2m t π 4(11.5)(106 ) A2m (0.05) π

= 9.9645(10−4 )

T Lm A2m

Equations can then be put into a spreadsheet resulting in: ri 0 0.10 0.20 0.30 0.40 0.45

rm

Am

Lm

ri

T(lbf · in)

ri

θ(deg)

0 0.125 0.225 0.325 0.425 0.475

0.902 5 0.889 087 0.859 043 0.811 831 0.747 450 0.708 822

3.8 3.585 398 3.413 717 3.242 035 3.070 354 2.984 513

0 0.10 0.20 0.30 0.40 0.45

1037.9 1022.5 987.9 933.6 859.6 815.1

0 0.10 0.20 0.30 0.40 0.45

4.825 4.621 4.553 4.576 4.707 4.825

1200 1000 800 T (lbf • in)

shi20396_ch04.qxd

600 400 200 0

0

0.1

0.2

0.3 ri (in)

0.4

0.5

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4.85 4.80

(deg)

4.75 4.70 4.65 4.60 4.55 4.50

0

0.1

0.2

0.3

0.4

0.5

ri (in)

Torque carrying capacity reduces with ri . However, this is based on an assumption of uniform stresses which is not the case for small ri . Also note that weight also goes down with an increase in ri . 4-35

From Eq. (4-47) where θ1 is the same for each leg. 1 1 T1 = Gθ1 L 1 c13 , T2 = Gθ1 L 2 c23 3 3

1 1 T = T1 + T2 = Gθ1 L 1 c13 + L 2 c23 = Gθ1 L i ci3 3 3 τ1 = Gθ1 c1 , τ2 = Gθ1 c2 τmax = Gθ1 cmax Ans.

Ans.

4-36 (a)

τmax = Gθ1 cmax τmax 11 500 = 1.227(105 ) psi · rad = cmax 3/32 1 1 T1/16 = Gθ1 (Lc3 ) 1/16 = (1.227)(105 )(0.5)(1/16) 3 = 4.99 lbf · in Ans. 3 3 1 T3/32 = (1.227)(105 )(0.5)(3/32) 3 = 16.85 lbf · in Ans. 3 τ1/16 = 1.227(105 )1/16 = 7669 psi, τ3/32 = 1.227(105 )3/32 = 11 500 psi Ans. Gθ1 =

(b)

4-37

θ1 =

1.227(105 ) = 1.0667(10−2 ) rad/in = 0.611◦ /in 11.5(106 )

Ans.

Separate strips: For each 1/16 in thick strip, T = ∴ Tmax

Lc2 τ (1)(1/16) 2 (11 500) = = 14.97 lbf · in 3 3

= 2(14.97) = 29.95 lbf · in

Ans.

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83

Chapter 4

For each strip, 3T l 3(14.97)(12) = = 0.192 rad Ans. Lc3 G (1)(1/16) 3 (11.5)(106 )

θ=

kt = T /θ = 29.95/0.192 = 156.0 lbf · in

Ans.

Solid strip: From Example 4-12, Tmax = 59.90 lbf · in

4-38

Ans.

θ = 0.0960 rad

Ans.

kt = 624 lbf · in

Ans.

τall = 8000 psi, 50 hp (a) n = 2000 rpm 63 025(50) 63 025H = = 1575.6 lbf · in n 2000 16T 1/3 16(1575.6) 1/3 ⇒ d= = = 1.00 in Ans. πτmax π(8000)

T =

Eq. (4-40) τmax =

16T πd 3

(b) n = 200 rpm

4-39

∴

T = 15 756 lbf · in 16(15 756) 1/3 d= = 2.157 in π(8000)

Ans.

τall = 110 MPa, θ = 30◦ , d = 15 mm, l = ? τ=

16T πd 3

Tl θ= JG

⇒

180 π

T =

π 3 τd 16

π π dGθ π J Gθ π d 4 Gθ = = l= 3 180 T 180 32 (π/16) τ d 360 τ =

4-40

π (0.015)(79.3)(109 )(30) = 2.83 m Ans. 360 110(106 )

d = 70 mm, replaced by 70 mm hollow with t = 6 mm (a)

Tsolid =

π τ (703 ) 16

Thollow =

π (704 − 584 ) τ 32 35

(π/16)(703 ) − (π/32) [(704 − 584 )/35] (100) = 47.1% % T = (π/16)(703 )

Ans.

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(b) Wsolid = kd 2 = k(702 ) , Whollow = k(702 − 582 ) % W =

4-41

k(702 ) − k(702 − 582 ) (100) = 68.7% k(702 )

Ans.

T = 5400 N · m, τall = 150 MPa 4.023(104 ) 5400(d/2) Tc = ⇒ 150(106 ) = (π/32)[d 4 − (0.75d) 4 ] d3 J 1/3 4.023(104 ) d= = 6.45(10−2 ) m = 64.5 mm 150(106 ) τ=

(a)

From Table A-17, the next preferred size is d = 80 mm; ID = 60 mm π (0.084 − 0.064 ) = 2.749(10−6 ) mm4 32 5400(0.030) = 58.9(106 ) Pa = 58.9 MPa τi = 2.749(10−6 )

Ans.

J=

(b)

Ans.

4-42 63 025(1) 63 025H = = 12 605 lbf · in n 5 16T 16T 1/3 16(12 605) 1/3 τ= ⇒ dC = = = 1.66 in πτ π(14 000) πdC3

T =

(a)

Ans.

From Table A-17, select 1 3/4 in τstart =

16(2)(12 605) = 23.96(103 ) psi = 23.96 kpsi 3 π(1.75 )

(b) design activity 4-43

ω = 2πn/60 = 2π(8)/60 = 0.8378 rad/s 1000 H = = 1194 N · m ω 0.8378 16T 1/3 16(1194) 1/3 dC = = = 4.328(10−2 ) m = 43.3 mm 6 πτ π(75)(10 ) T =

From Table A-17, select 45 mm 4-44

s=

√

A, d =

Ans.

4A/π

Square: Eq. (4-43) with b = c

4.8T c3 4.8T (τmax ) sq = ( A) 3/2 τmax =

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85

Chapter 4

(τmax ) rd =

Round:

16 T 16T 3.545T = = 3 3 / 2 π d π(4A/π) ( A) 3/2 (τmax ) sq 4.8 = 1.354 = (τmax ) rd 3.545

Square stress is 1.354 times the round stress 4-45

s=

√

A,

d=

Ans.

4A/π

Square: Eq. (4-44) with b = c, β = 0.141 θsq =

Tl Tl = 4 0.141c G 0.141( A) 4/2 G

Round: θrd =

Tl Tl 6.2832T l = = JG ( A) 4/2 G (π/32) (4A/π) 4/2 G

θsq 1/0.141 = 1.129 = θrd 6.2832 Square has greater θ by a factor of 1.13 4-46

Ans.

Text Eq. (4-43) gives τmax =

T T 1 = 2· 2 αbc bc α

From in-text table, p. 139, α is a function of b/c. Arrange equation in the form 1 b2 cτmax 1 = = y = a0 + a1 = a0 + a1 x T α b/c To plot 1/α vs 1/(b/c) , first form a table.

b/c 1 1.5 1.75 2 2.5 3 4 6 8 10 ∞

α

x 1/(b/c)

y 1/α

0.208 0.231 0.239 0.246 0.258 0.267 0.282 0.299 0.307 0.313 0.333

1 0.666 667 0.571 429 0.5 0.4 0.333 333 0.25 0.166 667 0.125 0.1 0

4.807 692 4.329 004 4.184 100 4.065 041 3.875 969 3.745 318 3.546 099 3.344 482 3.257 329 3.194 888 3.003 003

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Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 1兾 5

4.5

y 1.867x 3.061 4

3.5

3

0

0.2

0.4

0.6

0.8

1

1兾(b兾c)

Plot is a gentle convex-upward curve. Roark uses a polynomial, which in our notation is 1 3T τmax = 1 + 0.6095 + ··· 8(b/2)(c/2) 2 b/c 1 . T τmax = 2 3 + 1.8285 bc b/c Linear regression on table data y = 3.06 + 1.87x 1 1 = 3.06 + 1.87 α b/c 1 T τmax = 2 3.06 + 1.87 bc b/c 1.8 T τmax = 2 3 + bc b/c

Eq. (4-43)

4-47 Gear F 1000 lbf • in

1000 = 400 lbf 2.5 Fn = 400 tan 20 = 145.6 lbf Ft =

2.5R

Shaft ABCD y

Fn

Ft

Torque at C TC = 400(5) = 2000 lbf · in

RAy 666.7 lbf A 3" z

P=

2000 lbf • in B 145.6 lbf

RAz

2000 lbf • in 10"

RDy C C 400 lbf

5" D RDz

x

2000 = 666.7 lbf 3

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87

Chapter 4

( M A)z = 0 ( M A) y = 0

⇒

18R Dy − 145.6(13) − 666.7(3) = 0

⇒

−18R Dz + 400(13) = 0

Fy = 0

⇒

R Ay + 216.3 − 666.7 − 145.6 = 0

Fz = 0

⇒

⇒

⇒

R Dy = 216.3 lbf

R Dz = 288.9 lbf ⇒

R Ay = 596.0 lbf

R Az + 288.9 − 400 = 0 ⇒ R Az = 111.1 lbf M B = 3 5962 + 111.12 = 1819 lbf · in MC = 5 216.32 + 288.92 = 1805 lbf · in

∴ Maximum stresses occur at B. Ans. 32M B 32(1819) = 9486 psi = σB = 3 πd π(1.253 ) 16TB 16(2000) = 5215 psi = πd 3 π(1.253 ) σB 9486 σB 2 9486 2 2 + + = + τB = + 52152 = 11 792 psi 2 2 2 2 σB 2 τmax = + τ B2 = 7049 psi Ans. 2 τB =

σmax

Ans.

4-48 (a) At θ = 90◦ , σr = τrθ = 0, σθ = −σ Ans. θ = 0◦ , σr = τrθ = 0, σθ = 3σ Ans. (b) r

σθ /σ

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

3.000 2.071 1.646 1.424 1.297 1.219 1.167 1.132 1.107 1.088 1.074 1.063 1.054 1.048 1.042 1.037

兾 3 2.5 2 1.5 1 0.5 0

0

5

10 r (mm)

15

20

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4-49

1.5 = 1.5 1 1/8 r/d = = 0.125 1 . K ts = 1.39 . K t = 1.60

D/d =

Fig. A-15-8: Fig. A-15-9: σ A = Kt

32K t M 32(1.6)(200)(14) Mc = = 45 630 psi = 3 I πd π(13 )

16K ts T 16(1.39)(200)(15) Tc = = 21 240 psi = 3 J πd π(13 ) 2 σA 45.63 σA 45.63 2 2 + + = + τA = + 21.242 2 2 2 2

τ A = K ts σmax

τmax

4-50

= 54.0 kpsi Ans. 45.63 2 = + 21.242 = 31.2 kpsi Ans. 2

As shown in Fig. 4-34, the maximum stresses occur at the inside fiber where r = ri . Therefore, from Eq. (4-51) ri2 pi ro2 1+ 2 σt, max = 2 ro − ri2 ri

σr, max

4-51

ro2 + ri2 Ans. = pi ro2 − ri2 ri2 pi ro2 = 2 1 − 2 = − pi ro − ri2 ri

If pi = 0, Eq. (4-50) becomes − poro2 − ri2ro2 po /r 2 ro2 − ri2 ri2 poro2 =− 2 1+ 2 r ro − ri2

σt =

The maximum tangential stress occurs at r = ri . So σt, max = −

2 poro2 ro2 − ri2

Ans.

Ans.

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89

Chapter 4

For σr , we have − poro2 + ri2ro2 po /r 2 ro2 − ri2 ri2 poro2 = 2 −1 ro − ri2 r 2

σr =

So σr = 0 at r = ri . Thus at r = ro σr, max

por 2 = 2 o2 r o − ri

ri2 − ro2 ro2

= − po

Ans.

4-52 rav t

2 F = p A = πrav p 2 prav F πrav p σ1 = σ2 = = = Awall 2πrav t 2t

p

Ans.

F

4-53

σt > σl > σr τmax = (σt − σr )/2 at r = ri where σl is intermediate in value. From Prob. 4-50 1 τmax = (σt, max − σr, max ) 2 pi ro2 + ri2 +1 τmax = 2 ro2 − ri2 Now solve for pi using ro = 3 in, ri = 2.75 in, and τmax = 4000 psi. This gives pi = 639 psi Ans.

4-54

Given ro = 120 mm, ri = 110 mm and referring to the solution of Prob. 4-53, 2.4 MPa (120) 2 + (110) 2 τmax = +1 2 (120) 2 − (110) 2 = 15.0 MPa Ans.

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4-55

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From Table A-20, S y = 57 kpsi; also, ro = 0.875 in and ri = 0.625 in From Prob. 4-51 σt, max = −

4-56

Rearranging

2

ro − ri2 (0.8S y ) po = 2ro2

Solving, gives po = 11 200 psi

Ans.

From Table A-20, S y = 57 kpsi; also ro = 1.1875 in, ri = 0.875 in. From Prob. 4-50

σt, max = pi

ro2 + ri2 ro2 − ri2

solving gives 4-57

2 poro2 ro2 − ri2

therefore

pi = 0.8S y

pi = 13 510 psi

ro2 − ri2 ro2 + ri2

Ans.

Since σt and σr are both positive and σt > σr τmax = (σt ) max /2 where σt is max at ri Eq. (4-56) for r = ri = 0.375 in 0.282 2π(7200) 2 3 + 0.292 (σt ) max = 386 60 8 (0.3752 )(52 ) 1 + 3(0.292) 2 2 2 (0.375 ) = 8556 psi × 0.375 + 5 + − 0.3752 3 + 0.292 τmax =

8556 = 4278 psi Ans. 2

Radial stress:

r 2r 2 σr = k ri2 + ro2 − i 2o − r 2 r

ri2ro2 √ dσr = k 2 3 − 2r = 0 ⇒ r = ri ro = 0.375(5) = 1.3693 in Maxima: dr r 0.282 2π(7200) 2 3 + 0.292 0.3752 (52 ) 2 2 2 0.375 + 5 − − 1.3693 (σr ) max = 386 60 8 1.36932 = 3656 psi Ans.

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91

Chapter 4

ω = 2π(2069)/60 = 216.7 rad/s,

4-58

ρ = 3320 kg/m3 , ν = 0.24, ri = 0.0125 m, ro = 0.15 m; use Eq. (4-56) σt = 3320(216.7)

2

3 + 0.24 8

(0.0125) 2 + (0.15) 2 + (0.15) 2 1 + 3(0.24) 2 (0.0125) (10) −6 − 3 + 0.24

= 2.85 MPa Ans. 4-59 ρ=

(6/16) 386(1/16)(π/4)(62 − 12 )

= 5.655(10−4 ) lbf · s2 /in

4

τmax is at bore and equals Eq. (4-56) (σt ) max

σt 2

2π(10 000) = 5.655(10 ) 60 = 4496 psi

τmax =

−4

2

3 + 0.20 8

1 + 3(0.20) 2 2 2 2 0.5 + 3 + 3 − (0.5) 3 + 0.20

4496 = 2248 psi Ans. 2 ω = 2π(3000)/60 = 314.2 rad/s

4-60

m=

0.282(1.25)(12)(0.125) 386

= 1.370(10−3 ) lbf · s2 /in F

F 6"

F = mω2r = 1.370(10−3 )(314.22 )(6) = 811.5 lbf Anom = (1.25 − 0.5)(1/8) = 0.093 75 in2 811.5 = 8656 psi Ans. 0.093 75 . Note: Stress concentration Fig. A-15-1 gives K t = 2.25 which increases σmax and fatigue. σnom =

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4-61 to 4-66 ν = 0.292,

E = 30 Mpsi (207 GPa), ri = 0

R = 0.75 in (20 mm), ro = 1.5 in (40 mm) Eq. (4-60)

30(106 )δ (1.52 − 0.752 )(0.752 − 0) = 1.5(107 )δ ppsi = 2 2 0.75 in 2(0.75 )(1.5 − 0) 207(109 )δ (0.042 − 0.022 )(0.022 − 0) = 3.881(1012 )δ pPa = 2 2 0.020 2(0.02 )(0.04 − 0)

4-61 1 δmax = [40.042 − 40.000] = 0.021 mm 2

Ans.

1 δmin = [40.026 − 40.025] = 0.0005 mm 2

Ans.

From (2) pmax = 81.5 MPa,

pmin = 1.94 MPa

Ans.

4-62 1 δmax = (1.5016 − 1.5000) = 0.0008 in 2 1 δmin = (1.5010 − 1.5010) = 0 Ans. 2 Eq. (1)

pmax = 12 000 psi,

pmin = 0

Ans.

Ans.

4-63 1 δmax = (40.059 − 40.000) = 0.0295 mm Ans. 2 1 δmin = (40.043 − 40.025) = 0.009 mm Ans. 2 Eq. (2)

pmax = 114.5 MPa,

pmin = 34.9 MPa

Ans.

4-64 1 δmax = (1.5023 − 1.5000) = 0.001 15 in Ans. 2 1 δmin = (1.5017 − 1.5010) = 0.000 35 in Ans. 2 Eq. (1)

pmax = 17 250 psi

pmin = 5250 psi Ans.

(1)

(2)

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93

Chapter 4

4-65 1 δmax = (40.076 − 40.000) = 0.038 mm Ans. 2 1 δmin = (40.060 − 40.025) = 0.0175 mm Ans. 2 pmax = 147.5 MPa

Eq. (2)

pmin = 67.9 MPa

Ans.

4-66 1 δmax = (1.5030 − 1.500) = 0.0015 in Ans. 2 1 δmin = (1.5024 − 1.5010) = 0.0007 in Ans. 2 pmax = 22 500 psi

Eq. (1)

pmin = 10 500 psi

Ans.

4-67 1 δ = (1.002 − 1.000) = 0.001 in ri = 0, R = 0.5 in, 2 ν = 0.292, E = 30 Mpsi

ro = 1 in

Eq. (4-60) 30(106 )(0.001) (12 − 0.52 )(0.52 − 0) = 2.25(104 ) psi Ans. p= 2 2 0.5 2(0.5 )(1 − 0) Eq. (4-51) for outer member at ri = 0.5 in 12 0.52 (2.25)(104 ) (σt ) o = 1+ = 37 500 psi Ans. 12 − 0.52 0.52 Inner member, from Prob. 4-51 ri2 0 poro2 2.25(104 )(0.52 ) 1+ = −22 500 psi Ans. (σt ) i = − 2 1+ 2 =− ro 0.52 − 0 0.52 ro − ri2 Eqs. (d) and (e) above Eq. (4-59) 2 2.25(104 ) 1 + 0.52 δo = 0.5 2 + 0.292 = 0.000 735 in Ans. 30(106 ) 1 − 0.52 2.25(104 )(0.5) 0.52 + 0 δi = − − 0.292 = −0.000 265 in Ans. 30(106 ) 0.52 − 0

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4-68 νi = 0.292,

E i = 30(106 ) psi,

νo = 0.211,

1 δ = (1.002 − 1.000) = 0.001 in, 2

ri = 0,

E o = 14.5(106 ) psi R = 0.5,

ro = 1

Eq. (4-59)

0.5 0.001 = 14.5(106 )

2 12 + 0.52 0.5 + 0 0.5 − 0.292 p + 0.211 + 12 − 0.52 30(106 ) 0.52 − 0

p = 13 064 psi Ans. Eq. (4-51) for outer member at ri = 0.5 in 12 0.52 (13 064) 1+ = 21 770 psi (σt ) o = 12 − 0.52 0.52

Ans.

Inner member, from Prob. 4-51

0 13 064(0.52 ) 1+ = −13 064 psi Ans. (σt )i = − 0.52 − 0 0.52

Eqs. (d ) and (e) above Eq. (4-59) 13 064(0.5) 12 + 0.52 δo = + 0.211 = 0.000 846 in Ans. 14.5(106 ) 12 − 0.52 13 064(0.5) 0.52 + 0 − 0.292 = −0.000 154 in Ans. δi = − 30(106 ) 0.52 − 0 4-69 1 δmax = (1.003 − 1.000) = 0.0015 in ri = 0, R = 0.5 in, ro = 1 in 2 1 δmin = (1.002 − 1.001) = 0.0005 in 2 Eq. (4-60) pmax

30(106 )(0.0015) (12 − 0.52 )(0.52 − 0) = 33 750 psi Ans. = 0.5 2(0.52 )(12 − 0)

Eq. (4-51) for outer member at r = 0.5 in 12 0.52 (33 750) (σt ) o = 1+ = 56 250 psi Ans. 12 − 0.52 0.52 For inner member, from Prob. 4-51, with r = 0.5 in (σt ) i = −33 750 psi Ans.

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Chapter 4

Eqs. (d) and (e) just above Eq. (4-59) 33 750(0.5) 12 + 0.52 + 0.292 = 0.001 10 in Ans. δo = 30(106 ) 12 − 0.52 33 750(0.5) 0.52 + 0 − 0.292 = −0.000 398 in Ans. δi = − 30(106 ) 0.52 − 0 For δmin all answers are 0.0005/0.0015 = 1/3 of above answers

Ans.

4-70 νi = 0.292,

E i = 30 Mpsi,

νo = 0.334,

E o = 10.4 Mpsi

1 δmax = (2.005 − 2.000) = 0.0025 in 2 1 δmin = (2.003 − 2.002) = 0.0005 in 2 2 2 1.0 2 + 12 1 +0 1.0 − 0.292 pmax 0.0025 = + 0.334 + 10.4(106 ) 22 − 12 30(106 ) 12 − 0 pmax = 11 576 psi Ans. Eq. (4-51) for outer member at r = 1 in 22 12 (11 576) 1 + 2 = 19 293 psi Ans. (σt ) o = 22 − 12 1 Inner member from Prob. 4-51 with r = 1 in (σt ) i = −11 576 psi Ans. Eqs. (d) and (e) just above Eq. (4-59) 11 576(1) 22 + 12 δo = + 0.334 = 0.002 23 in Ans. 10.4(106 ) 22 − 12 11 576(1) 12 + 0 δi = − − 0.292 = −0.000 273 in Ans. 30(106 ) 12 − 0 For δmin all above answers are 0.0005/0.0025 = 1/5 Ans. 4-71 (a) Axial resistance Normal force at fit interface N = p A = p(2π Rl) = 2π p Rl Fully-developed friction force Fax = f N = 2π f p Rl

Ans.

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(b) Torsional resistance at fully developed friction is T = f R N = 2π f p R 2l 4-72

Ans.

d = 1 in, ri = 1.5 in, ro = 2.5 in. From Table 4-5, for R = 0.5 in, rc = 1.5 + 0.5 = 2 in rn =

0.52 √

= 1.968 245 8 in 2 2 − 22 − 0.52

e = rc − rn = 2.0 − 1.968 245 8 = 0.031 754 in ci = rn − ri = 1.9682 − 1.5 = 0.4682 in co = ro − rn = 2.5 − 1.9682 = 0.5318 in A = πd 2 /4 = π(1) 2 /4 = 0.7854 in2 M = Frc = 1000(2) = 2000 lbf · in Using Eq. (4-66) Mci 2000(0.4682) F 1000 + + = 26 300 psi Ans. = A Aeri 0.7854 0.7854(0.031 754)(1.5) Mco 2000(0.5318) F 1000 σo = − − = −15 800 psi Ans. = A Aero 0.7854 0.7854(0.031 754)(2.5) σi =

4-73

Section AA: D = 0.75 in, ri = 0.75/2 = 0.375 in, ro = 0.75/2 + 0.25 = 0.625 in From Table 4-5, for R = 0.125 in, rc = (0.75 + 0.25)/2 = 0.500 in 0.1252 rn = √

= 0.492 061 5 in 2 0.5 − 0.52 − 0.1252 e = 0.5 − rn = 0.007 939 in co = ro − rn = 0.625 − 0.492 06 = 0.132 94 in ci = rn − ri = 0.492 06 − 0.375 = 0.117 06 in A = π(0.25) 2 /4 = 0.049 087 M = Frc = 100(0.5) = 50 lbf · in 50(0.117 06) 100 + = 42 100 ksi Ans. 0.049 09 0.049 09(0.007 939)(0.375) 50(0.132 94) 100 σo = − = −25 250 psi Ans. 0.049 09 0.049 09(0.007 939)(0.625) σi =

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97

Chapter 4

Section BB: Abscissa angle θ of line of radius centers is r2 + d/2 −1 θ = cos r2 + d + D/2 0.375 + 0.25/2 −1 = 60◦ = cos 0.375 + 0.25 + 0.75/2 D+d cos θ = 100(0.5) cos 60◦ = 25 lbf · in 2 ri = r2 = 0.375 in

M=F

ro = r2 + d = 0.375 + 0.25 = 0.625 in e = 0.007 939 in σi = = σo =

(as before)

Mci Fcos θ − A Aeri 25(0.117 06) 100 cos 60◦ − = −19 000 psi 0.049 09 0.049 09(0.007 939)0.375 25(0.132 94) 100 cos 60◦ + = 14 700 psi 0.049 09 0.049 09(0.007 939)0.625

Ans. Ans.

On section BB, the shear stress due to the shear force is zero at the surface. 4-74 ri = 0.125 in, ro = 0.125 + 0.1094 = 0.2344 in From Table 4-5 for h = 0.1094 rc = 0.125 + 0.1094/2 = 0.1797 in rn = 0.1094/ln(0.2344/0.125) = 0.174 006 in e = rc − rn = 0.1797 − 0.174 006 = 0.005 694 in ci = rn − ri = 0.174 006 − 0.125 = 0.049 006 in co = ro − rn = 0.2344 − 0.174 006 = 0.060 394 in A = 0.75(0.1094) = 0.082 050 in2 M = F(4 + h/2) = 3(4 + 0.1094/2) = 12.16 lbf · in 12.16(0.0490) 3 σi = − − = −10 240 psi Ans. 0.082 05 0.082 05(0.005 694)(0.125) σo = −

12.16(0.0604) 3 + = 6670 psi Ans. 0.082 05 0.082 05(0.005 694)(0.2344)

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Find the resultant of F1 and F2 . Fx = F1x + F2x = 250 cos 60◦ + 333 cos 0◦ = 458 lbf Fy = F1 y + F2 y = 250 sin 60◦ + 333 sin 0◦ = 216.5 lbf F = (4582 + 216.52 ) 1/2 = 506.6 lbf This is the pin force on the lever which acts in a direction θ = tan−1

Fy 216.5 = 25.3◦ = tan−1 Fx 458

On the 25.3◦ surface from F1 2000 lbf • in 25.3 206

142

Ft = 250 cos(60◦ − 25.3◦ ) = 206 lbf Fn = 250 sin(60◦ − 25.3◦ ) = 142 lbf

507

A = 2[0.8125(0.375) + 1.25(0.375)] = 1.546 875 in2 The denomenator of Eq. (3-67), given below, has four additive parts. A rn = (d A/r) For d A/r , add the results of the following equation for each of the four rectangles. ro ro bdr = b ln , b = width r ri ri

1.8125 2.1875 3.6875 4.5 dA = 0.375 ln + 1.25 ln + 1.25 ln + 0.375 ln r 1 1.8125 3.3125 3.6875 = 0.666 810 6 rn =

1.546 875 = 2.3198 in 0.666 810 6

e = rc − rn = 2.75 − 2.3198 = 0.4302 in ci = rn − ri = 2.320 − 1 = 1.320 in co = ro − rn = 4.5 − 2.320 = 2.180 in Shear stress due to 206 lbf force is zero at inner and outer surfaces. σi = −

2000(1.32) 142 + = 3875 psi 1.547 1.547(0.4302)(1)

σo = −

2000(2.18) 142 − = −1548 psi 1.547 1.547(0.4302)(4.5)

Ans. Ans.

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Chapter 4

4-76 A = (6 − 2 − 1)(0.75) = 2.25 in2 6+2 = 4 in rc = 2 Similar to Prob. 4-75, 3.5 6 dA = 0.75 ln + 0.75 ln = 0.635 473 4 in r 2 4.5 rn =

2.25 A = = 3.5407 in 0.635 473 4 (d A/r)

e = 4 − 3.5407 = 0.4593 in σi =

5000 20 000(3.5407 − 2) + = 17 130 psi Ans. 2.25 2.25(0.4593)(2)

σo =

5000 20 000(6 − 3.5407) − = −5710 psi Ans. 2.25 2.25(0.4593)(6)

4-77

(a)

ro

A= ri

b dr = 2

6

2 6 dr = 2 ln r 2

= 2.197 225 in2 6 2r 1 ro 1 dr rc = br dr = A ri 2.197 225 2 r =

2 (6 − 2) = 3.640 957 in 2.197 225

r n = ro ri

=

A 2.197 225 = 6 2 (b/r) dr 2 (2/r ) dr

2.197 225 = 3.295 837 in 2[1/2 − 1/6]

e = R − rn = 3.640 957 − 3.295 837 = 0.345 12 ci = rn − ri = 3.2958 − 2 = 1.2958 in co = ro − rn = 6 − 3.2958 = 2.7042 in σi =

20 000 20 000(3.641)(1.2958) + = 71 330 psi Ans. 2.197 2.197(0.345 12)(2)

σo =

20 000 20 000(3.641)(2.7042) − = −34 180 psi Ans. 2.197 2.197(0.345 12)(6)

99

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(b) For the centroid, Eq. (4-70) gives,

r(2/r) s

s 4 rb s = = = rc = b s (2/r) s

s/r

s/r Let s = 4 × 10−3 in with the following visual basic program, “cen.” Function cen(R) DS = 4 / 1000 R = R + DS / 2 Sum = 0 For I = 1 To 1000 Step 1 Sum = Sum + DS / R R = R + DS Next I cen = 4 / Sum End Function For eccentricity, Eq. (4-71) gives,

[(s/r)/(rc − s)] s [s/(rc − s)](2/r) s = e= (2/r) s/(rc − s) [(1/r)/(rc − s)] s Program the following visual basic program, “ecc.” Function ecc(RC) DS = 4 / 1000 S = −(6 − RC) + DS / 2 R = 6 − DS / 2 SUM1 = 0 SUM2 = 0 For I = 1 To 1000 Step 1 SUM1 = SUM1 + DS * (S / R) / (RC − S) SUM2 = SUM2 + DS * (1 / R) / (RC − S) S = S + DS R = R − DS Next I ecc = SUM1 / SUM2 End Function In the spreadsheet enter the following,

1 2

A

B

C

D

ri 2

rc = cen(A2)

e = ecc(B2)

rn = B2 − C2

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101

Chapter 4

which results in,

1 2

A

B

C

D

ri 2

rc 3.640 957

e 0.345 119

rn 3.295 838

which are basically the same as the analytical results of part (a) and will thus yield the same final stresses. Ans. 4-78

rc = 12

s

s 2 (b/2) 2 2 /4 = + = 1 ⇒ b = 2 1 − s 4 − s2 22 12

√ [(s 4 − s 2 )/(rc − s)] s e= √ [( 4 − s 2 )/(rc − s)] s

1"

2"

A = πab = π(2)(1) = 6.283 in2 Function ecc(rc) DS = 4 / 1000 S = −2 + DS/2 SUM1 = 0 SUM2 = 0 For I = 1 To 1000 Step 1 SUM1 = SUM1 + DS * (S * Sqr(4 − S ^ 2)) / (rc − S) SUM2 = SUM2 + DS * Sqr(4 − S ^ 2) / (rc − S) S = S + DS Next I ecc = SUM1/SUM2 End Function rc

e

rn = rc − e

12

0.083 923

11.91 608

ci = 11.916 08 − 10 = 1.9161 co = 14 − 11.916 08 = 2.0839 M = F(2 + 2) = 20(4) = 80 kip · in σi =

80(1.9161) 20 + = 32.25 kpsi Ans. 6.283 6.283(0.083 923)(10)

σo =

80(2.0839) 20 − = −19.40 kpsi Ans. 6.283 6.283(0.083 923)(14)

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4-79 0.4"

0.4"

0.4"R 1"

For circle,

1"

For rectangle,

A r2 =

, (d A/r) 2 rc − rc2 − r 2

dA = b ln ro /ri r

Ao = πr 2

dA = 2π rc − rc2 − r 2 ∴ r dA 2.6 = 1 ln − 2π 1.8 − 1.82 − 0.42 = 0.672 723 4 r 1 A = 1(1.6) − π(0.42 ) = 1.097 345 2 in2 rn =

1.097 345 2 = 1.6312 in 0.672 723 4

e = 1.8 − rn = 0.1688 in ci = 1.6312 − 1 = 0.6312 in co = 2.6 − 1.6312 = 0.9688 in M = 3000(5.8) = 17 400 lbf · in

4-80

σi =

17.4(0.6312) 3 + = 62.03 kpsi 1.0973 1.0973(0.1688)(1)

σo =

17.4(0.9688) 3 − = −32.27 kpsi Ans. 1.0973 1.0973(0.1688)(2.6)

100 and 1000 elements give virtually the same results as shown below. Visual basic program for 100 elements: Function ecc(RC) DS = 1.6 / 100 S = −0.8 + DS / 2 SUM1 = 0 SUM2 = 0 For I = 1 To 25 Step 1 SUM1 = SUM1 + DS * S / (RC − S) SUM2 = SUM2 + DS / (RC − S) S = S + DS Next I

Ans.

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103

Chapter 4

For I = 1 To 50 Step 1 SUM1 = SUM1 + DS * S * (1 − 2 * Sqr(0.4 ^ 2 − S ^ 2)) / (RC − S) SUM2 = SUM2 + DS * (1 − 2 * Sqr(0.4 ^ 2 − S ^ 2)) / (RC − S) S = S + DS Next I For I = 1 To 25 Step 1 SUM1 = SUM1 + DS * S / (RC − S) SUM2 = SUM2 + DS / (RC − S) S = S + DS Next ecc = SUM1 / SUM2 End Function 100 elements

1000 elements

rc

e

rn = rc − e

rc

e

rn = rc − e

1.8

0.168 811

1.631 189

1.8

0.168 802

1.631 198

e = 0.1688 in, rn = 1.6312 in Yields same results as Prob. 4-79. 4-81

Ans.

From Eq. (4-72)

a = KF

1/3

=F

1/3

3 2[(1 − ν 2 )/E] 8 2(1/d)

1/3

Use ν = 0.292, F in newtons, E in N/mm2 and d in mm, then 1/3 3 [(1 − 0.2922 )/207 000] K = = 0.0346 8 1/25 pmax = =

3F 3F = 2πa 2 2π(K F 1/3 ) 2 3F 1/3 3F 1/3 = 2π K 2 2π(0.0346) 2

= 399F 1/3 MPa = |σmax | τmax = 0.3 pmax = 120F 1/3 MPa 4-82

From Prob. 4-81,

3 2[(1 − 0.2922 )/207 000] K = 8 1/25 + 0 pmax =

1/3 = 0.0436

3F 1/3 3F 1/3 = = 251F 1/3 2π K 2 2π(0.0436) 2

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and so,

σz = −251F 1/3 MPa τmax = 0.3(251) F

1/3

Ans.

= 75.3F 1/3 MPa 1/3

z = 0.48a = 0.48(0.0436)18 4-83

Ans.

= 0.055 mm

Ans.

ν1 = 0.334, E 1 = 10.4 Mpsi, l = 2 in, d1 = 1 in, ν2 = 0.211, E 2 = 14.5 Mpsi, d2 = −8 in. With b = K c F 1/2 1/2 2 (1 − 0.3342 )/[10.4(106 )] + (1 − 0.2112 )/[14.5(106 )] Kc = π(2) 1 − 0.125 = 0.000 234 6 Be sure to check σx for both ν1 and ν2 . Shear stress is maximum in the aluminum roller. So, τmax = 0.3 pmax 4000 = 13 300 psi 0.3

pmax = Since pmax = 2F/(πbl) we have pmax = So,

F= =

2F 2F 1/2 = πl K c F 1/2 πl K c

πl K c pmax 2

π(2)(0.000 234 6)(13 300) 2

= 96.1 lbf 4-84

Good class problem

4-85

From Table A-5, ν = 0.211

2 2

Ans.

1 σx 1 = (1 + ν) − = (1 + 0.211) − = 0.711 pmax 2 2 σy = 0.711 pmax σz =1 pmax These are principal stresses 1 1 τmax = (σ1 − σ3 ) = (1 − 0.711) = 0.1445 pmax 2 2

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Chapter 4

4-86

105

From Table A-5: ν1 = 0.211, ν2 = 0.292, E 1 = 14.5(106 ) psi, E 2 = 30(106 ) psi, d1 = 6 in, d2 = ∞, l = 2 in 2(800) (1 − 0.2112 )/14.5(106 ) + (1 − 0.2922 )/[30(106 )] b= (a) π(2) 1/6 + 1/∞ = 0.012 135 in pmax =

2(800) = 20 984 psi π(0.012 135)(2)

For z = 0 in, σx 1 = −2ν1 pmax = −2(0.211)20 984 = −8855 psi in wheel σx 2 = −2(0.292)20 984 = −12 254 psi In plate σ y = − pmax = −20 984 psi σz = −20 984 psi These are principal stresses. (b) For z = 0.010 in, σx1 = −4177 psi in wheel σx2 = −5781 psi in plate σ y = −3604 psi σz = −16 194 psi

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Chapter 5 5-1 (a)

F k1

k2

k3

so

y

F F F + + k1 k2 k3

k=

F ; y

k=

1 (1/k1 ) + (1/k2 ) + (1/k3 )

y=

Ans.

(b) k1

F

k2

y

F = k1 y + k2 y + k3 y k = F/y = k1 + k2 + k3

Ans.

k3

(c) k2 k1

1 1 1 = + k k1 k2 + k3

k=

1 1 + k1 k2 + k3

−1

k3

5-2 For a torsion bar, k T = T /θ = Fl/θ, and so θ = Fl/k T . For a cantilever, kC = F/δ, δ = F/kC . For the assembly, k = F/y, y = F/k = lθ + δ So

y=

Fl 2 F F = + k kT kC

Or

k=

1 (l 2 /k T ) + (1/kC )

Ans.

5-3 For a torsion bar, k = T/θ = GJ/l where J = πd 4 /32. So k = πd 4 G/(32l) = K d 4 /l . The springs, 1 and 2, are in parallel so d4 d4 +K l1 l2 1 4 1 + = Kd x l−x

k = k1 + k2 = K

And

Then

θ=

T = k

K d4

T 1 1 + x l−x

K d 4θ K d4 θ+ T = kθ = x l−x

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Chapter 5

K d4 K d 4θ T1 = θ; T2 = Thus x l−x If x = l/2, then T1 = T2 . If x < l/2, then T1 > T2 Using τ = 16T /πd 3 and θ = 32T l/(Gπd 4 ) gives πd 3 τ T = 16 and so θall =

2lτall 32l πd 3 τ = · 4 Gπd 16 Gd

Thus, if x < l/2, the allowable twist is 2xτall θall = Ans. Gd 1 4 1 k = Kd + Since x l−x π Gd 4 1 1 + = 32 x l−x Then the maximum torque is found to be Tmax

πd 3 xτall = 16

Ans.

1 1 + x l−x

Ans.

5-4 Both legs have the same twist angle. From Prob. 5-3, for equal shear, d is linear in x. Thus, d1 = 0.2d2 Ans. d24 π G (0.2d2 )4 πG + 1.258d24 k= = Ans. 32 0.2l 0.8l 32l 2(0.8l)τall θall = Ans. Gd2 Tmax = kθall = 0.198d23 τall

Ans.

5-5 A = πr 2 = π(r1 + x tan α)2

F r1

Fdx Fdx = AE Eπ(r1 + x tan α) 2 l dx F δ= π E 0 (r1 + x tan α) 2 l 1 F − = πE tan α(r1 + x tan α) 0

dδ = x ␣

l dx

= F

F 1 π E r1 (r1 + l tan α)

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Then

π Er1 (r1 + l tan α) F = δ l 2l E A1 1 + tan α = Ans. l d1

k=

5-6

dT = −w dx

Enlarged free body of length dx

x

T l dx

F = (T + dT ) + w dx − T = 0

Solution is T = −wx + c T |x=0 = P + wl = c

w dx

T = −wx + P + wl T dT

T = P + w(l − x)

w is cable’s weight per foot P

The infinitesmal stretch of the free body of original length dx is T dx AE P + w(l − x) dx = AE

dδ =

Integrating,

δ= 0

δ=

l

[P + w(l − x)] dx AE

wl 2 Pl + AE 2AE

Ans.

5-7 M = wlx − EI

wl 2 wx 2 − 2 2

dy wlx 2 wl 2 wx 3 = − x− + C1 , dx 2 2 6

EIy = y=

wlx 3 wl 2 x 2 wx 4 − − + C2 , 6 4 24 wx 2 (4lx − 6l 2 − x 2 ) 24E I

Ans.

dy = 0 at x = 0, dx

C1 = 0

y = 0 at x = 0,

C2 = 0

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109

Chapter 5

5-8 M = M1 = M B EI

dy = M B x + C1 , dx

EIy = y= 5-9 y ds dy dx

␭

MB x 2 + C2 , 2 MB x 2 2E I

dy = 0 at x = 0, dx

C1 = 0

y = 0 at x = 0,

C2 = 0

Ans.

2 dy ds = dx 2 + dy 2 = dx 1 + dx

Expand right-hand term by Binomial theorem 2 1/2 1 dy 2 dy =1+ + ··· 1+ dx 2 dx Since dy/dx is small compared to 1, use only the first two terms, dλ = ds − dx

1 dy 2 = dx 1 + − dx 2 dx 1 dy 2 dx = 2 dx 1 l dy 2 dx Ans. λ= 2 0 dx This contraction becomes important in a nonlinear, non-breaking extension spring. 5-10

4a 2 4ax 4ax − 2x y = − 2 (l − x) = − l l l dy 4a 8ax =− − 2 dx l l 2 16a 2 64a 2 x 64a 2 x 2 dy = 2 − + dx l l3 l4 1 l dy 2 8 a2 Ans. λ= dx = 2 0 dx 3 l

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5-11 y = a sin

πx l

aπ πx dy = cos dx l l 2 dy a2π 2 πx = 2 cos2 dx l l 1 l dy 2 λ= dx 2 0 dx λ=

a2 π 2 a2 = 2.467 4 l l

Ans.

Compare result with that of Prob. 5-10. See Charles R. Mischke, Elements of Mechanical Analysis, Addison-Wesley, Reading, Mass., 1963, pp. 244–249, for application to a nonlinear extension spring. 5-12 I = 2(5.56) = 11.12 in4 ymax = y1 + y2 = −

Fa 2 wl 4 + (a − 3l) 8E I 6E I

Here w = 50/12 = 4.167 lbf/in, and a = 7(12) = 84 in, and l = 10(12) = 120 in.

So

y1 = −

4.167(120) 4 = −0.324 in 8(30)(106 )(11.12)

y2 = −

600(84) 2 [3(120) − 84] = −0.584 in 6(30)(106 )(11.12)

ymax = −0.324 − 0.584 = −0.908 in

Ans.

M0 = −Fa − (wl 2 /2) = −600(84) − [4.167(120) 2 /2] = −80 400 lbf · in c = 4 − 1.18 = 2.82 in (−80 400)(−2.82) −M y =− (10−3 ) I 11.12 = −20.4 kpsi Ans.

σmax =

σmax is at the bottom of the section.

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111

Chapter 5

5-13

7 (800) + 10 3 RC = (800) + 10

800 lbf 600 lbf 3 ft

O

2 ft

RO = 5 ft

A

C

B

RO

RC

V (lbf)

5 (600) = 860 lbf 10 5 (600) = 540 lbf 10

860

O

60

M1 = 860(3)(12) = 30.96(103 ) lbf · in

540 M (lbf • in)

M1

M2 = 30.96(103 ) + 60(2)(12)

M2

= 32.40(103 ) lbf · in 32.40 Mmax ⇒ 6= Z = 5.4 in3 Z Z 2 F2l 3 F1 a[l − (l/2)] l l 2 − = + a − 2l 6E I l 2 2 48E I

σmax = y|x=5ft −

800(36)(60) 600(1203 ) 1 2 2 2 = [60 + 36 − 120 ] − 16 6(30)(106 ) I (120) 48(30)(106 ) I I = 23.69 in4

⇒

I /2 = 11.84 in4

Select two 6 in-8.2 lbf/ft channels; from Table A-7, I = 2(13.1) = 26.2 in4 , Z = 2(4.38) in3 1 23.69 − = −0.0565 in ymax = 26.2 16 32.40 = 3.70 kpsi σmax = 2(4.38) 5-14 I =

π (1.54 ) = 0.2485 in4 64

Superpose beams A-9-6 and A-9-7, yA =

300(24)(16) (162 + 242 − 402 ) 6(30)(106 )(0.2485)(40) +

12(16) [2(40)(162 ) − 163 − 403 ] 24(30)(106 )(0.2485)

y A = −0.1006 in Ans. y|x=20 =

300(16)(20) [202 + 162 − 2(40)(20)] 6(30)(106 )(0.2485)(40) −

% difference =

5(12)(404 ) = −0.1043 in Ans. 384(30)(106 )(0.2485)

0.1043 − 0.1006 (100) = 3.79% Ans. 0.1006

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5-15 1 I = 12

3 (1.53 ) = 0.105 47 in4 8

From Table A-9-10 F

␪A A

E

B

a

D

C a

Fa 2 (l + a) 3E I dy AB Fa 2 = (l − 3x 2 ) dx 6E I l yC = −

Thus, Fal Fal 2 = 6E I l 6E I Fa 2l y D = −θ A a = − 6E I With both loads, θA =

yD = − =−

Fa 2 Fa 2l − (l + a) 6E I 3E I Fa 2 120(102 ) (3l + 2a) = − [3(20) + 2(10)] 6E I 6(30)(106 )(0.105 47)

= −0.050 57 in Ans. 2 2Fa(l/2) 2 3 Fal 2 l yE = = l − 6E I l 2 24 E I = 5-16

120(10)(202 ) 3 = 0.018 96 in Ans. 24 (30)(106 )(0.105 47)

a = 36 in, l = 72 in, I = 13 in4, E = 30 Mpsi y= =

F2l 3 F1 a 2 (a − 3l) − 6E I 3E I 400(72) 3 400(36) 2 (36 − 216) − 6(30)(106 )(13) 3(30)(106 )(13)

= −0.1675 in Ans. I = 2(1.85) = 3.7 in4

5-17

Adding the weight of the channels, 2(5)/12 = 0.833 lbf/in, yA = −

Fl 3 10.833(484 ) 220(483 ) wl 4 − =− − 8E I 3E I 8(30)(106 )(3.7) 3(30)(106 )(3.7)

= −0.1378 in Ans.

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113

Chapter 5

5-18 I = πd 4 /64 = π(2)4 /64 = 0.7854 in4 Tables A-9-5 and A-9-9 y =− =−

F1 a F2l 3 + (4a 2 − 3l 2 ) 48E I 24E I 120(40) 3 80(10)(400 − 4800) + = −0.0130 in Ans. 6 48(30)(10 )(0.7854) 24(30)(106 )(0.7854)

5-19 (a) Useful relations k=

F 48E I = 3 y l

I =

2400(48) 3 kl 3 = = 0.1843 in4 48E 48(30)106

From I = bh 3 /12

12(0.1843) b Form a table. First, Table A-17 gives likely available fractional sizes for b: h=

3

8 12 , 9, 9 12 , 10 in

For h:

1 9 5 11 3 , , , , 2 16 8 16 4 For available b what is necessary h for required I?

3

b 8.5 9.0 9.5 10.0

12(0.1843) b 0.638 5" 0.626 ← choose 9"× 8 0.615 0.605

Ans.

(b) I = 9(0.625)3 /12 = 0.1831 in4 48E I 48(30)(106 )(0.1831) = = 2384 lbf/in l3 483 4(90 000)(0.1831) 4σ I = = 4394 lbf F= cl (0.625/2)(48) F 4394 y= = = 1.84 in Ans. k 2384 k=

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5-20 z R1 353 lbf 12"

502 lbf 21"

A

15"

O

C

x

B

R2 175 lbf

680 lbf

Torque = (600 − 80)(9/2) = 2340 lbf · in (T2 − T1 )

12 = T2 (1 − 0.125)(6) = 2340 2 2340 = 446 lbf, T1 = 0.125(446) = 56 lbf T2 = 6(0.875) M0 = 12(680) − 33(502) + 48R2 = 0 R2 =

33(502) − 12(680) = 175 lbf 48

R1 = 680 − 502 + 175 = 353 lbf We will treat this as two separate problems and then sum the results. First, consider the 680 lbf load as acting alone. z R1 510 lbf O

12"

A

R2 170 lbf 21"

B

15"

C

x

680 lbf

Fbx 2 (x + b2 − l 2 ); here b = 36 ", 6E I l x = 12", l = 48", F = 680 lbf

zO A = −

Also, πd 4 π(1.5) 4 = = 0.2485 in4 64 64 680(36)(12)(144 + 1296 − 2304) zA = − 6(30)(106 )(0.2485)(48) = +0.1182 in I =

z AC = −

Fa(l − x) 2 (x + a 2 − 2lx) 6E I l

where a = 12" and x = 21 + 12 = 33" 680(12)(15)(1089 + 144 − 3168) 6(30)(106 )(0.2485)(48) = +0.1103 in

zB = −

Next, consider the 502 lbf load as acting alone.

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115

Chapter 5

z 502 lbf 12"

O

A

21"

15"

B

C

R1

x

R2

Fbx 2 (x + b2 − l 2 ), where b = 15 ", 6E I l x = 12", l = 48", I = 0.2485 in4

zO B =

502(15)(12)(144 + 225 − 2304) = −0.081 44 in 6(30)(106 )(0.2485)(48) For z B use x = 33" Then,

zA =

502(15)(33)(1089 + 225 − 2304) 6(30)(106 )(0.2485)(48) = −0.1146 in

zB =

Therefore, by superposition z A = +0.1182 − 0.0814 = +0.0368 in Ans. z B = +0.1103 − 0.1146 = −0.0043 in Ans. 5-21 (a) Calculate torques and moment of inertia T = (400 − 50)(16/2) = 2800 lbf · in (8T2 − T2 )(10/2) = 2800 ⇒ T2 = 80 lbf, T1 = 8(80) = 640 lbf π I = (1.254 ) = 0.1198 in4 64 y RO O

720 lbf 9"

11"

450 lbf 12"

B

A

C RB

Due to 720 lbf, flip beam A-9-6 such that y AB → b = 9, x = 0, l = 20, F = −720 lbf dy Fb θB = (3x 2 + b2 − l 2 ) =− dx x=0 6E I l =−

−720(9) (0 + 81 − 400) = −4.793(10−3 ) rad 6(30)(106 )(0.1198)(20)

yC = −12θ B = −0.057 52 in Due to 450 lbf, use beam A-9-10, 450(144)(32) Fa 2 (l + a) = − = −0.1923 in yC = − 3E I 3(30)(106 )(0.1198)

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Adding the two deflections, yC = −0.057 52 − 0.1923 = −0.2498 in Ans. (b) At O: Due to 450 lbf:

θO = −

Fa 2 Fal dy 2 (l = − 3x ) = dx x=0 6E I l 6E I x=0

450(12)(20) 720(11)(0 + 112 − 400) + = 0.010 13 rad = 0.5805◦ 6 6(30)(10 )(0.1198)(20) 6(30)(106 )(0.1198)

At B: 450(12) [202 − 3(202 )] 6(30)(106 )(0.1198)(20) = −0.014 81 rad = 0.8485◦ 0.8485◦ = 1.694 in4 I = 0.1198 0.06◦ 64I 1/4 64(1.694) 1/4 d= = = 2.424 in π π

θ B = −4.793(10−3 ) +

Use d = 2.5 in Ans. π (2.54 ) = 1.917 in4 64 0.1198 = −0.015 61 in Ans. yC = −0.2498 1.917 I =

5-22 (a) l = 36(12) = 432 in y

x l

ymax = −

5(5000/12)(432) 4 5wl 4 =− 384E I 384(30)(106 )(5450) = −1.16 in

The frame is bowed up 1.16 in with respect to the bolsters. It is fabricated upside down and then inverted. Ans. (b) The equation in xy-coordinates is for the center sill neutral surface y=

wx (2lx 2 − x 3 − l 3 ) 24E I

Ans.

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117

Chapter 5

Differentiating this equation and solving for the slope at the left bolster gives w dy = (6lx 2 − 4x 3 − l 3 ) dx 24E I

(5000/12)(432) 3 wl 3 dy = − = − dx x=0 24E I 24(30)(106 )(5450)

Thus,

= −0.008 57 The slope at the right bolster is 0.008 57, so equation at left end is y = −0.008 57x and at the right end is y = 0.008 57(x − l). Ans. 5-23

From Table A-9-6, yL =

Fbx 2 (x + b2 − l 2 ) 6E I l

yL =

Fb 3 (x + b2 x − l 2 x) 6E I l

Fb dy L = (3x 2 + b2 − l 2 ) dx 6E I l Fb(b2 − l 2 ) dy L = dx x=0 6E I l Fb(b2 − l 2 ) ξ = 6E I l

Let And set

I =

And solve for d L

πd L4 64 32Fb(b2 − l 2 ) 1/4 dL = 3π Elξ

Ans.

For the other end view, observe the figure of Table A-9-6 from the back of the page, noting that a and b interchange as do x and −x 32Fa(l 2 − a 2 ) 1/4 Ans. dR = 3π Elξ For a uniform diameter shaft the necessary diameter is the larger of d L and d R .

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5-24

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Incorporating a design factor into the solution for d L of Prob. 5-23,

1/4 32n 2 2 Fb(l − b ) d= 3π Elξ 3 3 −9 ) 1/4 kN mm (10 10 − 3 = (mm 10 ) GPa mm 109 (10−3 )

32(1.28)(3.5)(150)|(2502 − 1502 )| −12 10 d=4 3π(207)(250)(0.001)

3.5 kN 100

150

d 250

= 36.4 mm Ans. 5-25

The maximum occurs in the right section. Flip beam A-9-6 and use y=

Fbx 2 (x + b2 − l 2 ) 6E I l

where b = 100 mm

Fb dy = (3x 2 + b2 − l 2 ) = 0 dx 6E I l Solving for x,

x= y=

l 2 − b2 = 3

2502 − 1002 = 132.29 mm from right 3

3.5(103 )(0.1)(0.132 29) [0.132 292 + 0.12 − 0.252 ](103 ) 6(207)(109 )(π/64)(0.03644 )(0.25)

= −0.0606 mm Ans. 5-26 The slope at x = 0 due to F1 in the xy plane is F1 b1 b12 − l 2 θx y = 6E I l

y a1

F1

b1 z

a2 F2

b2 x

and in the xz plane due to F2 is F2 b2 b22 − l 2 θx z = 6E I l

For small angles, the slopes add as vectors. Thus 1/2 θ L = θx2y + θx2z  2 2 2 2 1/2 2 2 F2 b2 b2 − l F1 b1 b1 − l  = + 6E I l 6E I l

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Chapter 5

119

Designating the slope constraint as ξ, we then have 2 1/2 1 ξ = |θ L | = Fi bi bi2 − l 2 6E I l Setting I = πd 4 /64 and solving for d 32 2 2 1/2 1/4 2 Fi bi bi − l d = 3π Elξ For the LH bearing, E = 30 Mpsi, ξ = 0.001, b1 = 12, b2 = 6, and l = 16. The result is d L = 1.31 in. Using a similar flip beam procedure, we get d R = 1.36 in for the RH bearing. So use d = 1 3/8 in Ans. 5-27

For the xy plane, use yBC of Table A-9-6 y=

100(4)(16 − 8) 2 [8 + 42 − 2(16)8] = −1.956(10−4 ) in 6(30)(106 )(16)

For the xz plane use yAB z=

300(6)(8) [82 + 62 − 162 ] = −7.8(10−4 ) in 6(30)(106 )(16)

δ = (−1.956j − 7.8k)(10−4 ) in |δ| = 8.04(10−4 ) in Ans. 5-28 32n 2 2 1/2 1/4 2 Fi bi bi − l d L = 3π Elξ 1/4 32(1.5) 2 2 2 2 2 2 1/2 [800(6)(6 − 10 )] + [600(3)(3 − 10 )] = 3π(29.8)(106 )(10)(0.001) = 1.56 in 1/2 1/4 32(1.5) 2 2 2 2 2 2 d R = [800(4)(10 − 4 )] + [600(7)(10 − 7 )] 3π(29.8)(106 )(10)(0.001) = 1.56 in choose d ≥ 1.56 in Ans. 5-29

From Table A-9-8 we have yL =

MB x 2 (x + 3a 2 − 6al + 2l 2 ) 6E I l

MB dy L = (3x 2 + 3a 2 − 6al + 2l 2 ) dx 6E I l

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At x = 0, the LH slope is θL =

MB dy L = (3a 2 − 6al + 2l 2 ) dx 6E I l

from which ξ = |θ L | =

MB 2 (l − 3b2 ) 6E I l

Setting I = πd 4 /64 and solving for d 32M B (l 2 − 3b2 ) 1/4 d= 3π Elξ For a multiplicity of moments, the slopes add vectorially and 32 2 1/2 1/4 2 2 Mi l − 3bi dL = 3π Elξ 1/2 1/4 32 2 2 Mi 3ai − l 2 d R = 3π Elξ The greatest slope is at the LH bearing. So 32(1200)[92 − 3(42 )] 1/4 = 0.706 in d = 3π(30)(106 )(9)(0.002) So use d = 3/4 in Ans. 5-30 RO

FAC

80 lbf

6

12 B

6FAC = 18(80) FAC = 240 lbf R O = 160 lbf I =

1 (0.25)(23 ) = 0.1667 in4 12

Initially, ignore the stretch of AC. From Table A-9-10 80(122 ) Fa 2 (l + a) = − (6 + 12) = −0.041 47 in 3E I 3(10)(106 )(0.1667) FL 240(12) δ= = = 1.4668(10−3 ) in 2 6 AE AC (π/4)(1/2) (10)(10 )

yB1 = −

Stretch of AC:

Due to stretch of AC By superposition,

y B 2 = −3δ = −4.400(10−3 ) in y B = −0.041 47 − 0.0044 = −0.045 87 in Ans.

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121

Chapter 5

5-31 θ=

(0.1F)(1.5) TL = = 9.292(10−4 ) F JG (π/32)(0.0124 )(79.3)(109 )

Due to twist δ B 1 = 0.1(θ) = 9.292(10−5 ) F Due to bending δB2 =

F(0.13 ) F L3 = = 1.582(10−6 ) F 3E I 3(207)(109 )(π/64)(0.0124 )

δ B = 1.582(10−6 ) F + 9.292(10−5 ) F = 9.450(10−5 ) F k=

1 = 10.58(103 ) N/m = 10.58 kN/m Ans. 9.450(10−5 )

5-32 F A

a

B

b

C

l

Fb l

R2 =

Fa l

δ1 =

R1 k1

δ2 =

R2 k2

R2

R1 ␦1

␦2

Spring deflection

Fb Fa Fb x =− + x − k1l k1l 2 k2l 2 Fb Fbx 2 Fx b a 2 2 (x + b − l ) + 2 Ans. − = − 6E I l l k1 k2 k1l Fb Fa(l − x) 2 Fx b a 2 − (x + a − 2lx) + 2 Ans. = − 6E I l l k1 k2 k1l

δ1 − δ2 yS = −δ1 + l y AB

y BC

5-33

R1 =

See Prob. 5-32 for deflection due to springs. Replace Fb/l and Fa/l with wl/2 wl wx 1 wl wl 1 wl − x= − + + yS = − 2k1 2k1l 2k2l 2 k1 k2 2k1 wl wx wx 1 1 2 3 3 (2lx − x − l ) + − y= + Ans. 24E I 2 k1 k2 2k1

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5-34

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Let the load be at x > l/2. The maximum deflection will be in Section AB (Table A-9-10) Fbx 2 (x + b2 − l 2 ) y AB = 6E I l Fb dy AB = (3x 2 + b2 − l 2 ) = 0 ⇒ 3x 2 + b2 − l 2 = 0 dx 6E I l

2 2 l −b l2 = 0.577l Ans. x= xmax = , 3 3 xmin = l − 0.577l = 0.423l

For x < l/2 5-35

50 lbf/ft O

M O = 50(10)(60) + 600(84)

600 lbf

= 80 400 lbf · in

B

A

7'

MO

Ans.

10'

R O = 50(10) + 600 = 1100 lbf

RO

I = 11.12 in4 from Prob. 5-12 M = −80 400 + 1100x − EI

4.167x 2 − 600x − 841 2

dy = −80 400x + 550x 2 − 0.6944x 3 − 300x − 842 + C1 dx

dy = 0 at x = 0 C1 = 0 dx E I y = −402 00x 2 + 183.33x 3 − 0.1736x 4 − 100x − 843 + C2 y = 0 at x = 0

C2 = 0

yB =

1 30(106 )(11.12)

[−40 200(1202 ) + 183.33(1203 )

− 0.1736(1204 ) − 100(120 − 84) 3 ] = −0.9075 in Ans. 5-36

See Prob. 5-13 for reactions: R O = 860 lbf, RC = 540 lbf M = 860x − 800x − 361 − 600x − 601 dy = 430x 2 − 400x − 362 − 300x − 602 + C1 EI dx E I y = 143.33x 3 − 133.33x − 363 − 100x − 603 + C1 x + C2 y = 0 at x = 0 ⇒ C2 = 0 y = 0 at x = 120 in ⇒ C1 = −1.2254(106 ) lbf · in2 Substituting C1 and C2 and evaluating at x = 60, 1 6 = 143.33(603 ) − 133.33(60 − 36) 3 − 1.2254(106 )(60) E I y = 30(10 ) I − 16 I = 23.68 in4 Agrees with Prob. 5-13. The rest of the solution is the same.

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Chapter 5

5-37 I = 0.2485 in4 24 R O = 12(20) + (300) = 420 lbf 40 12 M = 420x − x 2 − 300x − 161 2 dy = 210x 2 − 2x 3 − 150x − 162 + C1 EI dx E I y = 70x 3 − 0.5x 4 − 50x − 163 + C1 x + C2 y = 0 at x = 0 ⇒ C2 = 0 y = 0 at x = 40 in ⇒ C1 = −6.272(104 ) lbf · in2 Substituting for C1 and C2 and evaluating at x = 16, 1 [70(163 ) − 0.5(164 ) − 6.272(104 )(16)] yA = 6 30(10 )(0.2485) = −0.1006 in Ans. y|x=20 =

1 [70(203 ) − 0.5(204 ) − 50(20 − 16) 3 − 6.272(104 )(20)] 6 30(10 )(0.2485)

= 0.1043 in Ans. 3.7% difference Ans. 5-38 w[(l + a)/2][(l − a)/2)] l w a la = (l 2 − a 2 ) 2 4l w w w R2 = (l + a) − (l 2 − a 2 ) = (l + a)2 2 4l 4l 2 w w wx + (l + a) 2 x − l1 M = (l 2 − a 2 )x − 4l 2 4l w dy w w = (l 2 − a 2 )x 2 − x 3 + (l + a)2 x − l2 + C1 EI dx 8l 6 8l w 2 w w (l − a 2 )x 3 − x 4 + (l + a) 2 x − l3 + C1 x + C2 EIy = 24l 24 24l

冢l 2 a冣 w

R1 =

y = 0 at x = 0 ⇒ C2 = 0 y = 0 at x = l w 2 w 4 wa 2l 2 3 0= (l − a )l − l + C1l ⇒ C1 = 24l 24 24 w [(l 2 − a 2 )x 3 − lx 4 + (l + a) 2 x − l3 + a 2l 2 x] Ans. y= 24E I l

123

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5-39

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From Prob. 5-15, R A = R B = 120 lbf, and I = 0.105 47 in4 First half of beam, M = −120x + 120x − 101 dy = −60x 2 + 60x − 102 + C1 EI dx dy/dx = 0 at x = 20 in ⇒ 0 = −60(202 ) + 60(20 − 10) 2 + C1 ⇒ C1 = 1.8(104 ) lbf · in2 E I y = −20x 3 + 20x − 103 + 1.8(104 )x + C2 y = 0 at x = 10 in ⇒ C2 = −1.6(105 ) lbf · in3 y|x=0 =

1 30(106 )(0.105 47)

(−1.6)(105 )

= −0.050 57 in Ans. y|x=20 =

1 30(106 )(0.105 47)

[−20(203 ) + 20(20 − 10) 3 + 1.8(104 )(20) − 1.6(105 )]

= 0.018 96 in Ans. 5-40

From Prob. 5-30, R O = 160 lbf ↓, FAC = 240 lbf

I = 0.1667 in4

M = −160x + 240x − 61 dy = −80x 2 + 120x − 62 + C1 EI dx E I y = −26.67x 3 + 40x − 63 + C1 x + C2 y = 0 at x = 0 ⇒ C2 = 0 240(12) FL = −1.4668(10−3 ) in =− yA = − AE AC (π/4)(1/2) 2 (10)(106 ) at x = 6 10(106 )(0.1667)(−1.4668)(10−3 ) = −26.67(63 ) + C1 (6) C1 = 552.58 lbf · in2 yB =

1 [−26.67(183 ) + 40(18 − 6) 3 + 552.58(18)] 6 10(10 )(0.1667)

= −0.045 87 in 5-41 I1 = R1 =

Ans.

π (1.54 ) = 0.2485 in4 64 200 (12) = 1200 lbf 2

For 0 ≤ x ≤ 16 in,

M = 1200x −

I2 =

π 4 (2 ) = 0.7854 in4 64 MI

200 x − 42 2

x

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Chapter 5

1200x M = − 4800 I I1

125

1 1 100 1 1 0 x − 4 − 1200 x − 41 − − − x − 42 I1 I2 I1 I2 I2

= 4829x − 13 204x − 40 − 3301.1x − 41 − 127.32x − 42 E

dy = 2414.5x 2 − 13 204x − 41 − 1651x − 42 − 42.44x − 43 + C1 dx dy = 0 at x = 10 in dx 0 = 2414.5(102 ) − 13 204(10 − 4) 1 − 1651(10 − 4) 2 − 42.44(10 − 4) 3 + C1 C1 = −9.362(104 )

Boundary Condition:

E y = 804.83x 3 − 6602x − 42 − 550.3x − 43 − 10.61x − 44 − 9.362(104 )x + C2 y=0

at x = 0 ⇒ C2 = 0

For 0 ≤ x ≤ 16 in y=

1 [804.83x 3 − 6602x − 42 − 550.3x − 43 6 30(10 ) − 10.61x − 44 − 9.362(104 )x] Ans.

at x = 10 in y|x=10 =

1 [804.83(103 ) − 6602(10 − 4) 2 − 550.3(10 − 4) 3 6 30(10 ) − 10.61(10 − 4) 4 − 9.362(104 )(10)]

= −0.016 72 in Ans. 5-42

Define δi j as the deflection in the direction of the load at station i due to a unit load at station j. If U is the potential energy of strain for a body obeying Hooke’s law, apply P1 first. Then U=

1 P1 ( P1 δ11 ) 2

When the second load is added, U becomes U=

1 1 P1 ( P1 δ11 ) + P2 ( P2 δ22 ) + P1 ( P2 δ12 ) 2 2

For loading in the reverse order U =

1 1 P2 ( P2 δ22 ) + P1 ( P1 δ11 ) + P2 ( P1 δ21 ) 2 2

Since the order of loading is immaterial U = U and P1 P2 δ12 = P2 P1 δ21

when P1 = P2 , δ12 = δ21

which states that the deflection at station 1 due to a unit load at station 2 is the same as the deflection at station 2 due to a unit load at 1. δ is sometimes called an influence coefficient.

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5-43 (a) From Table A-9-10 y AB δ12

y

Fcx(l 2 − x 2 ) = 6E I l y ca(l 2 − a 2 ) = = F x=a 6E I l

y2 = Fδ21 = Fδ12 = Substituting I =

400 lbf b

a

A

c

B 1

2 7"

9"

Fca(l 2 − a 2 ) 6E I l

x

23"

πd 4 64 y2 =

400(7)(9)(232 − 92 )(64) = 0.00347 in Ans. 6(30)(106 )(π)(2) 4 (23)

(b) The slope of the shaft at left bearing at x = 0 is θ=

Fb(b2 − l 2 ) 6E I l

Viewing the illustration in Section 6 of Table A-9 from the back of the page provides the correct view of this problem. Noting that a is to be interchanged with b and −x with x leads to θ=

Fa(l 2 − a 2 )(64) Fa(l 2 − a 2 ) = 6E I l 6Eπd 4l

θ=

400(9)(232 − 92 )(64) = 0.000 496 in/in 6(30)(106 )(π)(2) 4 (23)

So y2 = 7θ = 7(0.000 496) = 0.00347 in Ans. 5-44

Place a dummy load Q at the center. Then, wx Qx (l − x) + 2 2 l/2 2 M dx ∂U U =2 , ymax = 2E I ∂ Q Q=0 0 l/2 2M ∂ M dx ymax = 2 2E I ∂ Q 0 Q=0 l/2 wx Qx x 2 (l − x) + dx ymax = EI 2 2 2 0 Q=0 M=

Set Q = 0 and integrate

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127

Chapter 5

ymax

w = 2E I

ymax =

lx 3 x 4 − 3 4

l/2 0

4

5wl 384E I

Ans.

5-45 I = 2(1.85) = 3.7 in4 Adding weight of channels of 0.833 lbf · in, ∂M 10.833 2 x = −F x − 5.417x 2 = −x M = −F x − 2 ∂F 48 48 1 1 ∂M δB = dx = M ( F x + 5.417x 2 )(x) dx EI 0 ∂F EI 0 =

(220/3)(483 ) + (5.417/4)(484 ) = 0.1378 in 30(106 )(3.7)

in direction of 220 lbf

y B = −0.1378 in Ans. 5-46 IO B

1 = (0.25)(23 ) = 0.1667 in4 , 12

O

FAC 3F A 6"

F

right

left

B

M = −F x¯

M = −2F x

12" x¯

x

1 U= 2E I

l

2 1 = 0.196 35 in2 2

∂ FAC =3 ∂F

FAC = 3F, 2F

A AC

π = 4

M 2 dx +

0

∂M = −x¯ ∂F

∂M = −2x ∂F

2 L AC FAC 2A AC E

l 1 FAC (∂ FAC /∂ F)L AC ∂U ∂M = dx + M δB = ∂F EI 0 ∂F A AC E 12 6 3F(3)(12) 1 −F x(− ¯ x) ¯ d x¯ + (−2F x)(−2x) dx + = EI 0 A AC E 0 3 1 F 108F 6 = (123 ) + 4F + EI 3 3 A AC E =

108F 864F + EI A AC E

=

108(80) 864(80) + = 0.045 86 in Ans. 6 10(10 )(0.1667) 0.196 35(10)(106 )

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5-47 Torsion

T = 0.1F

∂T = 0.1 ∂F

Bending

M = −F x¯

∂M = −x¯ ∂F

1 U= 2E I

M 2 dx +

F x

T 2L 2J G

1 ∂M T (∂ T /∂ F)L ∂U = M dx + δB = ∂F EI ∂F JG 0.1 0.1F(0.1)(1.5) 1 −F x(− ¯ x) ¯ d x¯ + = EI 0 JG =

0.015F F (0.13 ) + 3E I JG

Where π (0.012) 4 = 1.0179(10−9 ) m4 64 J = 2I = 2.0358(10−9 ) m4 0.015 0.001 + = 9.45(10−5 ) F δB = F 9 − 9 − 9 9 3(207)(10 )(1.0179)(10 ) 2.0358(10 )(79.3)(10 ) I =

k= 5-48

1 = 10.58(103 ) N/m = 10.58 kN/m Ans. 9.45(10−5 )

From Prob. 5-41, I1 = 0.2485 in4 , I2 = 0.7854 in4 For a dummy load ↑ Q at the center 200 −x Q ∂M x − 42 , = M = 1200x − x − 0 ≤ x ≤ 10 in 2 2 ∂Q 2 ∂U y|x=10 = ∂ Q Q=0 x x 2 1 4 1 10 2 = (1200x) − dx + [1200x − 100(x − 4) ] − dx E I1 0 2 I2 4 2 200(43 ) 1.566(105 ) 2 − − = E I1 I2 2 1.28(104 ) 1.566(105 ) =− + 30(106 ) 0.2485 0.7854 = −0.016 73 in Ans.

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129

Chapter 5

5-49

O x 3 Fa 5

l A O

4 Fa 5 l

4 F 5

A A

3 F 5

a

a

3 F 5

B

B 3 F 5

x

4 F 5

4 F 5

AB M = Fx OA

1 ∂u δB = = ∂F EI

I =

π 4 d , 64

N=

3 F 5

∂N 3 = ∂F 5

T =

4 Fa 5

∂T 4 = a ∂F 5

M1 =

4 F x¯ 5

∂ M1 4 = x¯ ∂F 5

M2 =

3 Fa 5

∂ M2 3 = a ∂F 5

(3/5) F(3/5)l (4/5) Fa(4a/5)l + AE JG

F x(x) dx +

0

1 + EI 9 Fa 3 + = 3E I 25

a

l

0

Fl AE

J = 2I,

4 F x¯ 5

∂M =x ∂F

16 + 25

A=

l 3 4 1 3 x¯ d x¯ + Fa a d x¯ 5 EI 0 5 5

Fa 2l JG

16 + 75

Fl 3 EI

9 + 25

Fa 2l EI

π 2 d 4

4Fl 16 32Fa 2l 16 64Fl 3 9 64Fa 2l 64Fa 3 9 + + + δB = + 3Eπd 4 25 πd 2 E 25 πd 4 G 75 Eπd 4 25 Eπd 4 4F 3 2 2 E 3 2 400a + 27ld + 384a l + 256l + 432a l Ans. = 75π Ed 4 G

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5-50

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The force applied to the copper and steel wire assembly is Fc + Fs = 250 lbf Since δc = δs Fs L Fc L = 2 6 3(π/4)(0.0801) (17.2)(10 ) (π/4)(0.0625) 2 (30)(106 ) Fc = 2.825Fs ∴ 3.825Fs

= 250

⇒

Fs = 65.36 lbf,

Fc = 2.825Fs = 184.64 lbf

σc =

184.64 = 12 200 psi = 12.2 kpsi Ans. 3(π/4)(0.0801) 2

σs =

65.36 = 21 300 psi = 21.3 kpsi Ans. (π/4)(0.06252 )

5-51 σb = 0.9(85) = 76.5 kpsi Ans. π (0.3752 ) = 50.69 kips Fb = 6(76.5) Bolt force 4 Fb 50.69 = −15.19 kpsi Ans. σc = − =− Cylinder stress Ac (π/4)(4.52 − 42 ) (b) Force from pressure

(a) Bolt stress

P=

π D2 π(42 ) p= (600) = 7540 lbf = 7.54 kip 4 4

50.69 Pc 50.69 Pb

6 bolts

x

Fx = 0

Pb + Pc = 7.54 (1)

P 7.54 kip

Pb L Pc L = 2 2 (π/4)(4.5 − 4 ) E 6(π/4)(0.3752 ) E

Since δc = δb ,

Pc = 5.037Pb

(2)

Substituting into Eq. (1) 6.037Pb = 7.54

⇒

Pb = 1.249 kip;

and from Eq (2),

Pc = 6.291 kip

Using the results of (a) above, the total bolt and cylinder stresses are σb = 76.5 +

1.249 = 78.4 kpsi Ans. 6(π/4)(0.3752 )

σc = −15.19 +

6.291 = −13.3 kpsi Ans. (π/4)(4.52 − 42 )

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131

Chapter 5

5-52 T = Tc + Ts

and θc = θs

Ts L Tc L = (G J ) c (G J ) s

Also,

Tc =

(G J ) c Ts (G J ) s

Substituting into equation for T ,

(G J ) c Ts T = 1+ (G J ) s

%Ts =

(G J ) s Ts = T (G J ) s + (G J ) c

Ans.

5-53 RB

RO + RB = W δ O A = δ AB

B

30"

20" O RO

(2)

30R B 20R O = , AE AE

W 800 lbf

A

RO =

3 RB 2

3 R B + R B = 800 2 1600 RB = = 320 lbf Ans. 5 R O = 800 − 320 = 480 lbf Ans. 480 = −480 psi Ans. σO = − 1 σB = +

5-54

(1)

320 = 320 psi Ans. 1

Since θ O A = θ AB T AB (6) TO A (4) = , GJ GJ

3 TO A = T AB 2

Also TO A + T AB = T T AB

3 + 1 = T, 2

T AB =

3 3T TO A = T AB = 2 5

T 2.5

Ans.

Ans.

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5-55 T2 T1 T2 T1 = = ⇒ r1 r2 1.25 3 3 T2 = T1 1.25 3 4π ∴ θ1 + θ2 = rad 1.25 180 3 4π T1 (48) (3/1.25)T1 (48) + = 4 6 4 6 (π/32)(7/8) (11.5)(10 ) 1.25 (π/32)(1.25) (11.5)(10 ) 180 F1 = F2

⇒

T1 = 403.9 lbf · in 3 T1 = 969.4 lbf · in T2 = 1.25 16T1 16(403.9) = = 3071 psi Ans. τ1 = 3 πd π(7/8)3 τ2 =

16(969.4) = 2528 psi Ans. π(1.25)3

5-56 10 kip

5 kip

FA

RO

FB

x

RC

(1) Arbitrarily, choose RC as redundant reaction Fx = 0, 10(103 ) − 5(103 ) − R O − RC = 0 (2) R O + RC = 5(103 ) lbf (3)

δC =

RC (15) [10(103 ) − 5(103 ) − RC ]20 [5(103 ) + RC ] − (10) − =0 AE AE AE −45RC + 5(104 ) = 0

⇒

RC = 1111 lbf Ans.

R O = 5000 − 1111 = 3889 lbf Ans. 5-57 w A

MC

B a

C RC

RB l x

(1) Choose R B as redundant reaction (2)

R B + RC = wl

(a)

R B (l − a) −

wl 2 + MC = 0 (b) 2

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133

Chapter 5

R B (l − a) 3 w(l − a) 2 yB = + [4l(l − a) − (l − a) 2 − 6l 2 ] = 0 3E I 24E I w [6l 2 − 4l(l − a) + (l − a)2 ] RB = 8(l − a) w (3l 2 + 2al + a 2 ) Ans. = 8(l − a)

(3)

Substituting, w (5l 2 − 10al − a 2 ) Ans. 8(l − a) wl 2 w − R B (l − a) = (l 2 − 2al − a 2 ) Ans. MC = 2 8 RC = wl − R B =

Eq. (a) Eq. (b)

5-58 w MC A

a x

B

C

RB

RC

∂M wx 2 + R B x − a1 , = x − a1 M =− 2 ∂ RB l ∂U 1 ∂M = M dx ∂ RB EI 0 ∂ RB a l −wx 2 1 −wx 2 1 (0) dx + + R B (x − a) (x − a) dx = 0 = EI 0 2 EI a 2 a 3 RB w 1 4 4 3 (l − a ) − (l − a ) + (l − a) 3 − (a − a) 3 = 0 − 2 4 3 3 RB =

w w (3l 2 + 2al + a 2 ) Ans. [3(L 4 − a 4 ) − 4a(l 3 − a 3 )] = 3 (l − a) 8(l − a)

RC = wl − R B = MC =

w (5l 2 − 10al − a 2 ) Ans. 8(l − a)

wl 2 w − R B (l − a) = (l 2 − 2al − a 2 ) Ans. 2 8

5-59 FBE 500

A

FDF 500

C

500

B RA

(1)

D

A=

π (0.0122 ) = 1.131(10−4 ) m2 4

20 kN

R A + FB E + FD F = 20 kN M A = 3FD F − 2(20) + FB E = 0

(a)

FB E + 3FD F = 40 kN

(b)

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M = R A x + FB E x − 0.51 − 20(103 )x − 11

(2) EI

FB E dy x2 = RA + x − 0.52 − 10(103 )x − 12 + C1 dx 2 2

E I y = RA

FB E 10 x3 + x − 0.53 − (103 )x − 13 + C1 x + C2 6 6 3

(3) y = 0 at x = 0 C2 = 0 FB E (1) Fl = −4.2305(10−8 ) FB E =− yB = − AE B E 1.131(10−4 )209(109 ) Substituting and evaluating at x = 0.5 m 0.53 + C1 (0.5) E I y B = 209(109 )(8)(10−7 )(−4.2305)(10−8 ) FB E = R A 6 2.0833(10−2 ) R A + 7.0734(10−3 ) FB E + 0.5C1 = 0 FD F (1) Fl yD = − = −4.2305(10−8 ) FD F =− − 4 9 AE D F 1.131(10 )(209)(10 )

(c)

Substituting and evaluating at x = 1.5 m FB E 10 1.53 + (1.5 − 0.5) 3 − (103 )(1.5 − 1) 3 +1.5C1 E I y D = −7.0734(10−3 ) FD F = R A 6 6 3 0.5625R A + 0.166 67FB E + 7.0734(10−3 ) FD F + 1.5C1 = 416.67 (d )      1 1 1 0  20 000  RA             0 1 3 0 40 000 F B E   =  2.0833(10−2 ) 7.0734(10−3 ) 0 0.5   0  FD F          0.5625 0.166 67 7.0734(10−3 ) 1.5 416.67 C1 Solve simultaneously or use software R A = −3885 N, FB E = 15 830 N, σB E =

15 830 = 140 MPa Ans., (π/4)(122 )

FD F = 8058 N, σD F =

C1 = −62.045 N · m2

8058 = 71.2 MPa Ans. (π/4)(122 )

E I = 209(109 )(8)(10−7 ) = 167.2(103 ) N · m2 3885 3 15 830 1 10 3 3 3 − x + x − 0.5 − (10 )x − 1 − 62.045x y= 167.2(103 ) 6 6 3 B: x = 0.5 m, C: x = 1 m,

D: x = 1.5,

y B = −6.70(10−4 ) m = −0.670 mm Ans. 3885 3 1 15 830 3 − (1 ) + (1 − 0.5) − 62.045(1) yC = 167.2(103 ) 6 6 = −2.27(10−3 ) m = −2.27 mm Ans. 3885 1 15 830 − (1.53 ) + (1.5 − 0.5) 3 yD = 3 167.2(10 ) 6 6 10 3 3 − (10 )(1.5 − 1) − 62.045(1.5) 3 = −3.39(10−4 ) m = −0.339 mm Ans.

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135

Chapter 5

5-60 y FBE

500 lbf 3" A

3"

C

3"

D

E I = 30(106 )(0.050) = 1.5(106 ) lbf · in2

x

B FFD

RC

RC + FB E − FF D = 500 3RC + 6FB E = 9(500) = 4500 M = −500x + FB E x − 31 + RC x − 61

(1) (2)

(a) (b)

dy FB E RC = −250x 2 + x − 32 + x − 62 + C1 dx 2 2 250 3 FB E RC x + x − 33 + x − 63 + C1 x + C2 EIy = − 3 6 6 FB E (2) Fl = −8.692(10−7 ) FB E yB = =− AE B E (π/4)(5/16) 2 (30)(106 )

EI

Substituting and evaluating at x = 3 in E I y B = 1.5(106 )[−8.692(10−7 ) FB E ] = −

250 3 (3 ) + 3C1 + C2 3

1.3038FB E + 3C1 + C2 = 2250

(c)

250 3 FB E (6 ) + (6 − 3) 3 + 6C1 + C2 3 6 4.5FB E + 6C1 + C2 = 1.8(104 )

(d)

Since y = 0 at x = 6 in E I y|=0 = − yD =

Fl AE

= DF

FD F (2.5) = 1.0865(10−6 ) FD F 2 6 (π/4)(5/16) (30)(10 )

Substituting and evaluating at x = 9 in 250 3 FB E (9 ) + (9 − 3) 3 3 6 RC (9 − 6) 3 + 9C1 + C2 + 6

E I y D = 1.5(106 )[1.0865(10−6 ) FD F ] = −

4.5RC + 36FB E − 1.6297FD F  1 1 −1 0  3 6 0 0   0 1.3038 0 3   0 4.5 0 6 4.5 36 −1.6297 9 RC = −590.4 lbf,

+ 9C1 + C2 = 6.075(104 )     500 RC  0                0  FB E   4500   2250 = 1 F  D F     1.8(104 )      1  C1          4 1 C2 6.075(10 )

FB E = 1045.2 lbf,

C1 = 4136.4 lbf · in2 ,

FD F = −45.2 lbf

C2 = −11 522 lbf · in3

(e)

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σB E =

1045.2 = 13 627 psi = 13.6 kpsi Ans. (π/4)(5/16) 2

σD F = −

45.2 = −589 psi Ans. (π/4)(5/16) 2

1 (−11 522) = −0.007 68 in Ans. 1.5(106 ) 250 3 1 yB = − (3 ) + 4136.4(3) − 11 522 = −0.000 909 in Ans. 1.5(106 ) 3 250 3 1 1045.2 −590.4 3 3 − (9 ) + (9 − 3) + (9 − 6) + 4136.4(9) − 11 522 yD = 1.5(106 ) 3 6 6 yA =

= −4.93(10−5 ) in Ans. 5-61

␪

F Q (dummy load)

M = −P R sin θ + Q R(1 − cos θ)

∂M = R(1 − cos θ) ∂Q

π ∂U 1 P R3 δQ = = (−P R sin θ) R(1 − cos θ) R dθ = −2 ∂ Q Q=0 EI 0 EI Deflection is upward and equals 2( P R 3 /E I ) 5-62

Equation (4-34) becomes

U =2 0

π

M 2 R dθ 2E I

Ans.

R/ h > 10

∂M = R(1 − cos θ) ∂F π ∂U 2 ∂M δ= = R dθ M ∂F EI 0 ∂F π 2 F R 3 (1 − cos θ) 2 dθ = EI 0

where M = F R(1 − cos θ) and

=

3π F R 3 EI

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137

Chapter 5

Since I = bh 3 /12 = 4(6) 3 /12 = 72 mm4 and R = 81/2 = 40.5 mm, we have δ=

3π(40.5) 3 F = 66.4F mm Ans. 131(72)

where F is in kN. 5-63

P

R ␪ l x

M = −P x,

∂M = −x ∂P

0≤x ≤l

∂M = l + R(1 − cos θ) 0 ≤ θ ≤ l M = Pl + P R(1 − cos θ), ∂P l π/2 1 2 δP = −P x(−x) dx + P[l + R(1 − cos θ)] R dθ EI 0 0 = 5-64

P {4l 3 + 3R[2πl 2 + 4(π − 2)l R + (3π − 8) R 2 ]} Ans. 12E I

A: Dummy load Q is applied at A. Bending in AB due only to Q which is zero. Q

A

C P

B

␪

∂M π M = P R sin θ + Q R(1 + sin θ), = R(1 + sin θ) 0 ≤ θ ≤ ∂Q 2 π/2 ∂U 1 = ( P R sin θ)[R(1 + sin θ)]R dθ (δ A ) V = ∂ Q Q=0 EI 0 θ sin 2θ π/2 π P R3 P R3 −cos θ + − 1+ = = EI 2 4 EI 4 0 =

π + 4 P R3 4 EI

Ans.

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M = P R sin θ,

B:

1 ∂U = (δ B ) V = ∂P EI =

π P R3 4 EI

∂M = R sin θ ∂P π/2

( P R sin θ)( R sin θ) R dθ

0

Ans.

5-65

y T

M x 200 N

␪

z

R A 200 N

M = P R sin θ,

∂M = R sin θ ∂P

0 h mg

Now let x = y − h; then x˙ = y˙ and x¨ = y¨ . So the D.E. is x¨ + (k/m)x = g with solution ω = (k/m) 1/2 and mg x = A cos ωt + B sin ωt + k

At contact, t = 0, x = 0, and x˙ = v0 . Evaluating A and B then yields v0 mg mg cos ωt + sin ωt + x =− k ω k or W v0 W y = − cos ωt + sin ωt + +h k ω k and Wω y˙ = sin ωt + v0 cos ωt

k To find ymax set y˙ = 0. Solving gives v0 k tan ωt = − Wω v0 k

−1 (ωt )* = tan − or Wω The first value of (ωt )* is a minimum and negative. So add π radians to it to find the maximum. Numerical example: h = 1 in, W = 30 lbf, k = 100 lbf/in. Then ω = (k/m) 1/2 = [100(386)/30]1/2 = 35.87 rad/s W/k = 30/100 = 0.3 v0 = (2gh) 1/2 = [2(386)(1)]1/2 = 27.78 in/s Then y = −0.3 cos 35.87t +

27.78 sin 35.87t + 0.3 + 1 35.87

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147

Chapter 5

For ymax

27.78(100) v0 k =− = −2.58 Wω 30(35.87) (ωt )* = −1.20 rad (minimum)

tan ωt = −

(ωt )* = −1.20 + π = 1.940 (maximum) Then t * = 1.940/35.87 = 0.0541 s. This means that the spring bottoms out at t * seconds. Then (ωt )* = 35.87(0.0541) = 1.94 rad 27.78 ymax = −0.3 cos 1.94 + sin 1.94 + 0.3 + 1 = 2.130 in Ans. So 35.87 The maximum spring force is Fmax = k( ymax − h) = 100(2.130 − 1) = 113 lbf Ans. The action is illustrated by the graph below. Applications: Impact, such as a dropped package or a pogo stick with a passive rider. The idea has also been used for a one-legged robotic walking machine. y Free fall

0

Speeds agree 0.05 Time of release

0.01 Inflection point of trig curve (The maximum speed about this point is 29.8 in/s.)

0.01

1

2

0.05

Time t Equilibrium, rest deflection

During contact ymax

5-80

Choose t = 0 at the instant of impact. At this instant, v1 = (2gh) 1/2 . Using momentum, m 1 v1 = m 2 v2 . Thus W1 + W2 W1 (2gh) 1/2 = v2 g g

Therefore at t = 0, y = 0, and y˙ = v2 y

W1 W2 ky

W1 (2gh) 1/2 v2 = W1 + W2

Let W = W1 + W2 W1

Because the spring force at y = 0 includes a reaction to W2 , the D.E. is W y¨ = −ky + W1 g With ω = (kg/W ) 1/2 the solution is y = A cos ωt + B sin ωt + W1 /k y˙ = −Aω sin ωt + Bω cos ωt

At t = 0, y = 0 ⇒ A = −W1 /k At t = 0, y˙ = v2 ⇒ v2 = Bω

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Then B=

W1 (2gh) 1/2 v2 = ω (W1 + W2 )[kg/(W1 + W2 )]1/2

We now have

1/2 W1 2h W1

y=− cos ωt + W1 sin ωt + k k(W1 + W2 ) k Transforming gives 1/2 2hk W1 W1 +1 cos(ωt − φ) + y= k W1 + W2 k where φ is a phase angle. The maximum deflection of W2 and the maximum spring force are thus 1/2 2hk W1 W1 Ans. +1 + ymax = k W1 + W2 k 1/2 2hk Fmax = kymax + W2 = W1 +1 + W1 + W2 Ans. W1 + W2 5-81

Assume x > y to get a free-body diagram. y k1(x y)

W

k2 y

x

Then W y¨ = k1 (x − y) − k2 y g A particular solution for x = a is k1 a y= k1 + k2 Then the complementary plus the particular solution is y = A cos ωt + B sin ωt +

where

(k1 + k2 )g ω= W

1/2

k1 a k1 + k2

At t = 0, y = 0, and y˙ = 0. Therefore B = 0 and k1 a A=− k1 + k2 Substituting, k1 a (1 − cos ωt) y= k1 + k2 Since y is maximum when the cosine is −1 2k1 a ymax = k1 + k2

Ans.

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Chapter 6 6-1 σ1 − σ3 = S y /n

MSS:

⇒

Sy σ1 − σ3

n=

Sy σ 1/2 2 1/2 σ = σ A2 − σ A σ B + σ B2 = σx − σx σ y + σ y2 + 3τx2y n=

DE:

(a) MSS:

σ1 = 12, σ2 = 6, σ3 = 0 kpsi 50 n= = 4.17 Ans. 12

σ = (122 − 6(12) + 62 ) 1/2 = 10.39 kpsi, 12 2 12 ± + (−8) 2 = 16, −4 kpsi (b) σ A , σ B = 2 2 DE:

n=

50 = 4.81 10.39

Ans.

σ1 = 16, σ2 = 0, σ3 = −4 kpsi MSS:

n=

50 = 2.5 16 − (−4)

Ans.

50 = 2.73 Ans. σ = (122 + 3(−82 )) 1/2 = 18.33 kpsi, n = 18.33 −6 + 10 2 −6 − 10 ± + (−5) 2 = −2.615, −13.385 kpsi (c) σ A , σ B = 2 2 DE:

σ1 = 0, σ2 = −2.615, σ3 = −13.385 kpsi MSS:

50 = 3.74 n= 0 − (−13.385)

␴B

Ans.

σ = [(−6) 2 − (−6)(−10) + (−10) 2 + 3(−5) 2 ]1/2 = 12.29 kpsi 50 n= = 4.07 Ans. 12.29 12 − 4 2 12 + 4 ± + 12 = 12.123, 3.877 kpsi (d) σ A , σ B = 2 2 DE:

σ1 = 12.123, σ2 = 3.877, σ3 = 0 kpsi MSS: DE:

n=

50 = 4.12 12.123 − 0

Ans.

σ = [122 − 12(4) + 42 + 3(12 )]1/2 = 10.72 kpsi n=

50 = 4.66 10.72

Ans.

␴A

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6-2 S y = 50 kpsi Sy σ1 − σ3

MSS:

σ1 − σ3 = S y /n

DE:

1/2 2 = S y /n σ A − σ A σ B + σ B2

n=

DE:

[122

n=

[122

1/2 n = S y / σ A2 − σ A σ B + σ B2

⇒

50 = 4.17 Ans. 12 − 0

50 = 4.17 − (12)(12) + 122 ]1/2

Ans.

50 = 4.81 − (12)(6) + 62 ]1/2

Ans.

σ1 = 12 kpsi, σ3 = −12 kpsi, n =

(c) MSS:

n=

DE:

[122

n=

50 = 2.08 Ans. 12 − (−12)

50 = 2.41 − (12)(−12) + (−12) 2 ]1/3

σ1 = 0, σ3 = −12 kpsi, n =

DE:

Ans.

50 = 4.17 12

σ1 = 12 kpsi, σ3 = 0, n =

DE:

(d) MSS:

n=

σ1 = 12 kpsi, σ3 = 0, n =

(a) MSS:

(b) MSS:

⇒

[(−6) 2

Ans.

50 = 4.17 Ans. −(−12)

50 = 4.81 − (−6)(−12) + (−12) 2 ]1/2

6-3 S y = 390 MPa Sy σ1 − σ3

MSS:

σ1 − σ3 = S y /n

DE:

1/2 2 = S y /n σ A − σ A σ B + σ B2

(a) MSS:

⇒

n=

σ1 = 180 MPa, σ3 = 0, n =

⇒

1/2 n = S y / σ A2 − σ A σ B + σ B2

390 = 2.17 Ans. 180

390 = 2.50 Ans. [1802 − 180(100) + 1002 ]1/2 180 180 2 ± + 1002 = 224.5, −44.5 MPa = σ1 , σ3 (b) σ A , σ B = 2 2 DE:

n=

MSS:

n=

DE:

n=

390 = 1.45 Ans. 224.5 − (−44.5) [1802

390 = 1.56 + 3(1002 )]1/2

Ans.

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Chapter 6

160 2 160 ± − + 1002 = 48.06, −208.06 MPa = σ1 , σ3 (c) σ A , σ B = − 2 2 390 = 1.52 Ans. 48.06 − (−208.06) 390 n= = 1.65 Ans. 2 [−160 + 3(1002 )]1/2 n=

MSS: DE:

(d) σ A , σ B = 150, −150 MPa = σ1 , σ3 380 = 1.27 Ans. n= MSS: 150 − (−150) 390 n= = 1.50 Ans. DE: [3(150) 2 ]1/2 6-4 S y = 220 MPa (a) σ1 = 100, σ2 = 80, σ3 = 0 MPa MSS: DET:

220 = 2.20 Ans. 100 − 0 σ = [1002 − 100(80) + 802 ]1/2 = 91.65 MPa 220 = 2.40 Ans. n= 91.65 n=

(b) σ1 = 100, σ2 = 10, σ3 = 0 MPa MSS: DET:

n=

220 = 2.20 Ans. 100

σ = [1002 − 100(10) + 102 ]1/2 = 95.39 MPa n=

220 = 2.31 Ans. 95.39

(c) σ1 = 100, σ2 = 0, σ3 = −80 MPa MSS: DE:

n=

220 = 1.22 Ans. 100 − (−80)

σ = [1002 − 100(−80) + (−80) 2 ]1/2 = 156.2 MPa 220 = 1.41 Ans. n= 156.2

(d) σ1 = 0, σ2 = −80, σ3 = −100 MPa 220 n= = 2.20 Ans. MSS: 0 − (−100) σ = [(−80) 2 − (−80)(−100) + (−100) 2 ] = 91.65 MPa DE: n=

220 = 2.40 Ans. 91.65

151

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6-5 (a) MSS: DE: (b) MSS: DE:

n=

2.23 OB = = 2.1 OA 1.08

n=

OC 2.56 = = 2.4 OA 1.08

n=

OE 1.65 = = 1.5 OD 1.10

n=

OF 1.8 = = 1.6 OD 1.1 ␴B

(a) C B

A Scale 1" ⫽ 200 MPa ␴A

O D E F

J K

G

L (d)

H I (c)

(c) MSS: DE: (d) MSS: DE:

n=

OH 1.68 = = 1.6 OG 1.05

n=

OI 1.85 = = 1.8 OG 1.05

n=

OK 1.38 = = 1.3 OJ 1.05

n=

OL 1.62 = = 1.5 OJ 1.05

(b)

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153

Chapter 6

6-6 S y = 220 MPa OB 2.82 = = 2.2 OA 1.3 OC 3.1 n= = = 2.4 OA 1.3 OE 2.2 n= = = 2.2 OD 1

n=

(a) MSS: DE: (b) MSS:

n=

DE:

OF 2.33 = = 2.3 OD 1 ␴B

(a) C B

A

1" ⫽ 100 MPa

E D

F

O

␴A

G J

H I (c)

K

L (d)

(c) MSS: DE: (d) MSS: DE:

n=

OH 1.55 = = 1.2 OG 1.3

n=

OI 1.8 = = 1.4 OG 1.3

OK 2.82 = = 2.2 OJ 1.3 OL 3.1 n= = = 2.4 OJ 1.3

n=

(b)

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6-7 Sut = 30 kpsi, Suc = 100 kpsi; σ A = 20 kpsi, σ B = 6 kpsi n=

(a) MNS: Eq. (6-30a)

Sut 30 = 1.5 Ans. = σx 20

30 = 1.5 Ans. 20 30 n= = 1.5 Ans. 20 30 n= = 1.5 Ans. 20 n=

BCM: Eq. (6-31a) M1M: Eq. (6-32a) M2M: Eq. (6-33a) (b) σx = 12 kpsi,τx y = −8 kpsi 12 ± σ A, σB = 2

12 2

2 + (−8) 2 = 16, −4 kpsi

30 = 1.88 Ans. 16 16 (−4) 1 = − ⇒ n 30 100 30 n= = 1.88 Ans. 16 30 n= = 1.88 Ans. 16 n=

MNS: Eq. (6-30a) BCM: Eq. (6-31b) M1M: Eq. (6-32a) M2M: Eq. (6-33a)

n = 1.74 Ans.

(c) σx = −6 kpsi, σ y = −10 kpsi,τx y = −5 kpsi −6 − 10 −6 + 10 2 ± + (−5) 2 = −2.61, −13.39 kpsi σ A, σB = 2 2 MNS: Eq. (6-30b)

n=−

100 = 7.47 Ans. −13.39

BCM: Eq. (6-31c)

n=−

100 = 7.47 Ans. −13.39

M1M: Eq. (6-32c)

n=−

100 = 7.47 Ans. −13.39

M2M: Eq. (6-33c)

n=−

100 = 7.47 Ans. −13.39

(d) σx = −12 kpsi,τx y = 8 kpsi 12 σ A, σB = − ± 2 MNS: Eq. (6-30b)

n=

12 2 − + 82 = 4, −16 kpsi 2

−100 = 6.25 Ans. −16

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155

Chapter 6

4 (−16) 1 = − n 30 100

BCM: Eq. (6-31b)

n = 3.41 Ans.

(100 − 30)4 −16 1 = − ⇒ n = 3.95 Ans. n 100(30) 100 4 n(−16) + 30 2 n + =1 30 30 − 100

M1M: Eq. (6-32b) M2M: Eq. (6-33b) Reduces to

⇒

n 2 − 1.1979n − 15.625 = 0 1.1979 + 1.19792 + 4(15.625) = 4.60 n= 2

Ans.

␴B

B

1" ⫽ 20 kpsi

(a)

A O

␴A C

E D

K F

G

H

I

L (c)

J (d)

(b)

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6-8 See Prob. 6-7 for plot. (a) For all methods:

n=

1.55 OB = = 1.5 OA 1.03

(b) BCM:

n=

OD 1.4 = = 1.75 OC 0.8

n=

OE 1.55 = = 1.9 OC 0.8

(c) For all methods:

n=

OL 5.2 = = 7.6 OK 0.68

(d) MNS:

n=

OJ 5.12 = = 6.2 OF 0.82

BCM:

n=

OG 2.85 = = 3.5 OF 0.82

M1M:

n=

OH 3.3 = = 4.0 OF 0.82

M2M:

n=

3.82 OI = = 4.7 OF 0.82

All other methods:

6-9 Given: S y = 42 kpsi, Sut = 66.2 kpsi, ε f = 0.90. Since ε f > 0.05, the material is ductile and thus we may follow convention by setting S yc = S yt . Use DE theory for analytical solution. For σ , use Eq. (6-13) or (6-15) for plane stress and Eq. (6-12) or (6-14) for general 3-D. (a) σ = [92 − 9(−5) + (−5) 2 ]1/2 = 12.29 kpsi n=

42 = 3.42 Ans. 12.29

(b) σ = [122 + 3(32 )]1/2 = 13.08 kpsi n=

42 = 3.21 Ans. 13.08

(c) σ = [(−4) 2 − (−4)(−9) + (−9) 2 + 3(52 )]1/2 = 11.66 kpsi n=

42 = 3.60 Ans. 11.66

(d) σ = [112 − (11)(4) + 42 + 3(12 )]1/2 = 9.798 n=

42 = 4.29 Ans. 9.798

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157

Chapter 6 ␴B

(d) H

1 cm ⫽ 10 kpsi G O C

D

(b)

␴A

A E

B (a)

F (c)

For graphical solution, plot load lines on DE envelope as shown. σ A = 9, σ B = −5 kpsi OB 3.5 n= = = 3.5 Ans. OA 1 12 2 12 ± + 32 = 12.7, −0.708 kpsi (b) σ A , σ B = 2 2

(a)

4.2 OD = = 3.23 OC 1.3 4−9 2 −4 − 9 ± + 52 = −0.910, −12.09 kpsi (c) σ A , σ B = 2 2 n=

4.5 OF = = 3.6 Ans. OE 1.25 11 − 4 2 11 + 4 ± + 12 = 11.14, 3.86 kpsi (d) σ A , σ B = 2 2 n=

n= 6-10

5.0 OH = = 4.35 Ans. OG 1.15

This heat-treated steel exhibits S yt = 235 kpsi, S yc = 275 kpsi and ε f = 0.06. The steel is ductile (ε f > 0.05) but of unequal yield strengths. The Ductile Coulomb-Mohr hypothesis (DCM) of Fig. 6-27 applies — confine its use to first and fourth quadrants.

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(a) σx = 90 kpsi, σ y = −50 kpsi, σz = 0 σ A = 90 kpsi and σ B = −50 kpsi. For the fourth quadrant, from Eq. (6-13) 1 1 = = 1.77 Ans. n= (σ A /S yt ) − (σ B /Suc ) (90/235) − (−50/275) (b) σx = 120 kpsi, τx y = −30 kpsi ccw. σ A , σ B = 127.1, −7.08 kpsi. For the fourth quadrant 1 = 1.76 Ans. n= (127.1/235) − (−7.08/275) (c) σx = −40 kpsi, σ y = −90 kpsi, τx y = 50 kpsi . σ A , σ B = −9.10, −120.9 kpsi. Although no solution exists for the third quadrant, use S yc 275 = 2.27 Ans. n=− =− σy −120.9 (d) σx = 110 kpsi, σ y = 40 kpsi, τx y = 10 kpsi cw. σ A , σ B = 111.4, 38.6 kpsi. For the first quadrant S yt 235 = 2.11 Ans. = n= σA 111.4 Graphical Solution: ␴B

OB 1.82 = = 1.78 (a) n = OA 1.02 (b) n =

OD 2.24 = = 1.75 OC 1.28

(c) n =

OF 2.75 = = 2.22 OE 1.24

(d) n =

OH 2.46 = = 2.08 OG 1.18

(d)

H 1 in ⫽ 100 kpsi

G

␴A

O C

D

A

B (a) E

F (c)

(b)

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159

Chapter 6

6-11

The material is brittle and exhibits unequal tensile and compressive strengths. Decision: Use the Modified II-Mohr theory as shown in Fig. 6-28 which is limited to first and fourth quadrants. Sut = 22 kpsi, Suc = 83 kpsi Parabolic failure segment:

S B + 22 2 S A = 22 1 − 22 − 83 SB

SA

SB

SA

−22 −30 −40 −50

22.0 21.6 20.1 17.4

−60 −70 −80 −83

13.5 8.4 2.3 0

(a) σx = 9 kpsi, σ y = −5 kpsi. σ A , σ B = 9, −5 kpsi. For the fourth quadrant, use Eq. (6-33a) Sut 22 = 2.44 Ans. n= = σA 9 (b) σx = 12 kpsi, τx y = −3 kpsi ccw. σ A , σ B = 12.7, 0.708 kpsi. For the first quadrant, n=

Sut 22 = 1.73 Ans. = σA 12.7

(c) σx = −4 kpsi, σ y = −9 kpsi, τx y = 5 kpsi . σ A , σ B = −0.910, −12.09 kpsi. For the third quadrant, no solution exists; however, use Eq. (6-33c) −83 = 6.87 Ans. n= −12.09 (d) σx = 11 kpsi,σ y = 4 kpsi,τx y = 1 kpsi. σ A , σ B = 11.14, 3.86 kpsi. For the first quadrant n=

S yt SA 22 = 1.97 Ans. = = σA σA 11.14 ␴B 30 Sut ⫽ 22

30

–50

Sut ⫽ 83 –90

␴A

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. Since ε f < 0.05, the material is brittle. Thus, Sut = Suc and we may use M2M which is basically the same as MNS. (a) σ A , σ B = 9, −5 kpsi 35 = 3.89 Ans. 9 (b) σ A , σ B = 12.7, −0.708 kpsi n=

35 = 2.76 Ans. 12.7 (c) σ A , σ B = −0.910, −12.09 kpsi (3rd quadrant) n=

n=

36 = 2.98 Ans. 12.09

␴B

(d) σ A , σ B = 11.14, 3.86 kpsi n=

35 = 3.14 Ans. 11.14

1 cm ⫽ 10 kpsi

Graphical Solution: O

OD 3.45 = = 2.70 Ans. OC 1.28

(c) n =

OF 3.7 = = 2.85 Ans. (3rd quadrant) OE 1.3

(d) n =

OH 3.6 = = 3.13 Ans. OG 1.15

D

(b)

B

(a)

E

F (c)

Sut = 30 kpsi, Suc = 109 kpsi Use M2M: (a) σ A , σ B = 20, 20 kpsi 30 n= = 1.5 Ans. Eq. (6-33a): 20 (b) σ A , σ B = ± (15) 2 = 15, −15 kpsi n=

30 = 2 Ans. 15

(c) σ A , σ B = −80, −80 kpsi For the 3rd quadrant, there is no solution but use Eq. (6-33c). Eq. (6-33c):

C A

(b) n =

Eq. (6-33a)

(d)

G

OB 4 = = 4.0 Ans. (a) n = OA 1

6-13

H

n=−

109 = 1.36 Ans. −80

␴A

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161

Chapter 6

(d) σ A , σ B = 15, −25 kpsi n(15) −25n + 30 2 + =1 30 30 − 109

Eq. (6-33b):

n = 1.90 Ans. OB 4.25 = = 1.50 OA 2.83 OD 4.24 = = 2.00 (b) n = OC 2.12 (a) n =

(c) n =

OF 15.5 = = 1.37 (3rd quadrant) OE 11.3

(d) n =

5.3 OH = = 1.83 OG 2.9

␴B

B (a)

A

␴A

O

C 1 cm ⫽ 10 kpsi G D (b)

H (d)

E

F (c)

6-14

Given: AISI 1006 CD steel, F = 0.55 N, P = 8.0 kN, and T = 30 N · m, applying the DE theory to stress elements A and B with S y = 280 MPa A:

σx =

4(8)(103 ) 32Fl 4P 32(0.55)(103 )(0.1) + + = πd 3 πd 2 π(0.0203 ) π(0.0202 )

= 95.49(106 ) Pa = 95.49 MPa

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16T 16(30) = 19.10(106 ) Pa = 19.10 MPa = 3 πd π(0.0203 ) 1/2 σ = σx2 + 3τx2y = [95.492 + 3(19.1) 2 ]1/2 = 101.1 MPa

τx y =

n= B:

Sy 280 = 2.77 = σ 101.1

Ans.

4P 4(8)(103 ) = 25.47(106 ) Pa = 25.47 MPa = σx = 3 2 πd π(0.020 ) 16(30) 4 0.55(103 ) 16T 4V = + + τx y = πd 3 3 A π(0.0203 ) 3 (π/4)(0.0202 ) = 21.43(106 ) Pa = 21.43 MPa σ = [25.472 + 3(21.432 )]1/2 = 45.02 MPa n=

6-15

280 = 6.22 45.02

Ans.

Design decisions required: • • • •

Material and condition Design factor Failure model Diameter of pin

Using F = 416 lbf from Ex. 6-3 32M πd 3 32M 1/3 d= πσmax

σmax =

Decision 1: Select the same material and condition of Ex. 6-3 (AISI 1035 steel, S y = 81 000). Decision 2: Since we prefer the pin to yield, set n d a little larger than 1. Further explanation will follow. Decision 3: Use the Distortion Energy static failure theory. Decision 4: Initially set n d = 1 Sy Sy = 81 000 psi = nd 1 32(416)(15) 1/3 d= = 0.922 in π(81 000)

σmax =

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163

Chapter 6

Choose preferred size of d = 1.000 in F=

π(1) 3 (81 000) = 530 lbf 32(15)

n=

530 = 1.274 416

Set design factor to n d = 1.274 Adequacy Assessment: Sy 81 000 = 63 580 psi = nd 1.274 32(416)(15) 1/3 d= = 1.000 in (OK ) π(63 580)

σmax =

6-16

F=

π(1) 3 (81 000) = 530 lbf 32(15)

n=

530 = 1.274 (OK) 416

For a thin walled cylinder made of AISI 1018 steel, S y = 54 kpsi, Sut = 64 kpsi. The state of stress is p(8) pd pd = = 40 p, σl = = 20 p, 4t 4(0.05) 8t These three are all principal stresses. Therefore, σt =

σr = − p

1 σ = √ [(σ1 − σ2 ) 2 + (σ2 − σ3 ) 2 + (σ3 − σ1 ) 2 ]1/2 2 1 = √ [(40 p − 20 p) 2 + (20 p + p) 2 + (− p − 40 p) 2 ] 2 = 35.51 p = 54 ⇒ p = 1.52 kpsi (for yield) Ans. . . For rupture, 35.51 p = 64 ⇒ p = 1.80 kpsi Ans. 6-17

For hot-forged AISI steel w = 0.282 lbf/in3 , S y = 30 kpsi and ν = 0.292. Then ρ = w/g = 0.282/386 lbf · s2 /in; ri = 3 in; ro = 5 in; ri2 = 9; ro2 = 25; 3 + ν = 3.292; 1 + 3ν = 1.876. Eq. (4-56) for r = ri becomes σt = ρω

2

3+ν 8

1 + 3ν 2 2 2ro + ri 1 − 3+ν

Rearranging and substituting the above values: Sy 1.876 0.282 3.292 50 + 9 1 − = ω2 386 8 3.292 = 0.016 19

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Setting the tangential stress equal to the yield stress, 30 000 1/2 = 1361 rad/s ω= 0.016 19 or

n = 60ω/2π = 60(1361)/(2π) = 13 000 rev/min

Now check the stresses at r = (rori ) 1/2 , or r = [5(3)]1/2 = 3.873 in 2 3+ν (ro − ri ) 2 σr = ρω 8 0.282ω2 3.292 (5 − 3) 2 = 386 8 = 0.001 203ω2 Applying Eq. (4-56) for σt 9(25) 1.876(15) 3.292 2 0.282 9 + 25 + − σt = ω 386 8 15 3.292 = 0.012 16ω2 Using the Distortion-Energy theory 1/2 = 0.011 61ω2 σ = σt2 − σr σt + σr2 Solving

ω=

30 000 0.011 61

1/2

= 1607 rad/s

So the inner radius governs and n = 13 000 rev/min 6-18

Ans.

For a thin-walled pressure vessel, di = 3.5 − 2(0.065) = 3.37 in σt =

p(di + t) 2t

σt =

500(3.37 + 0.065) = 13 212 psi 2(0.065)

σl =

500(3.37) pdi = = 6481 psi 4t 4(0.065)

σr = − pi = −500 psi

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165

Chapter 6

These are all principal stresses, thus, 1 σ = √ {(13 212 − 6481) 2 + [6481 − (−500)]2 + (−500 − 13 212) 2 }1/2 2 σ = 11 876 psi n=

Sy 46 000 46 000 = = σ σ 11 876

= 3.87 Ans. 6-19

Table A-20 gives S y as 320 MPa. The maximum significant stress condition occurs at ri where σ1 = σr = 0, σ2 = 0, and σ3 = σt . From Eq. (4-50) for r = ri σt = −

2ro2 po 2(1502 ) po = − = −3.6 po 1502 − 1002 ro2 − ri2

σ = 3.6 po = S y = 320 po = 6-20

320 = 88.9 MPa Ans. 3.6

Sut = 30 kpsi, w = 0.260 lbf/in3 , ν = 0.211, 3 + ν = 3.211, 1 + 3ν = 1.633. At the inner radius, from Prob. 6-17 1 + 3ν 2 σt 3+ν 2 2 2ro + ri − r =ρ ω2 8 3+ν i Here ro2 = 25, ri2 = 9, and so σt 0.260 = ω2 386

3.211 8

1.633(9) 50 + 9 − = 0.0147 3.211

Since σr is of the same sign, we use M2M failure criteria in the first quadrant. From Table A-24, Sut = 31 kpsi, thus, 31 000 1/2 ω= = 1452 rad/s 0.0147 rpm = 60ω/(2π) = 60(1452)/(2π) = 13 866 rev/min Using the grade number of 30 for Sut = 30 000 kpsi gives a bursting speed of 13640 rev/min. 6-21

TC = (360 − 27)(3) = 1000 lbf · in , TB = (300 − 50)(4) = 1000 lbf · in y 223 lbf A

127 lbf B

8"

C 8" 350 lbf xy plane

6"

D

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In x y plane, M B = 223(8) = 1784 lbf · in and MC = 127(6) = 762 lbf · in. 387 lbf 8"

A

8" B

6" C

106 lbf

D 281 lbf

xz plane

In the x z plane, M B = 848 lbf · in and MC = 1686 lbf · in. The resultants are M B = [(1784) 2 + (848) 2 ]1/2 = 1975 lbf · in MC = [(1686) 2 + (762) 2 ]1/2 = 1850 lbf · in So point B governs and the stresses are 16T 16(1000) 5093 = = 3 psi 3 3 πd πd d 32M B 32(1975) 20 120 = = psi σx = πd 3 πd 3 d3 2 1/2 σx σx ± + τx2y σ A, σB = 2 2  1/2  2   20.12 1 20.12 ± σ A, σB = 3 + (5.09) 2  d  2 2 τx y =

Then

= Then

(10.06 ± 11.27) kpsi · in3 d3

σA =

10.06 + 11.27 21.33 = kpsi d3 d3

σB =

10.06 − 11.27 1.21 = − 3 kpsi 3 d d

and

For this state of stress, use the Brittle-Coulomb-Mohr theory for illustration. Here we use Sut (min) = 25 kpsi, Suc (min) = 97 kpsi, and Eq. (6-31b) to arrive at 1 21.33 −1.21 − = 25d 3 97d 3 2.8 Solving gives d = 1.34 in. So use d = 1 3/8 in

Ans.

Note that this has been solved as a statics problem. Fatigue will be considered in the next chapter. 6-22

As in Prob. 6-21, we will assume this to be statics problem. Since the proportions are unchanged, the bearing reactions will be the same as in Prob. 6-21. Thus x y plane: x z plane:

M B = 223(4) = 892 lbf · in M B = 106(4) = 424 lbf · in

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167

Chapter 6

So Mmax = [(892) 2 + (424) 2 ]1/2 = 988 lbf · in σx =

32M B 32(988) 10 060 = = psi πd 3 πd 3 d3

Since the torsional stress is unchanged, τx z = 5.09/d 3 kpsi 

1/2  2   10.06 10.06 1 σ A, σB = 3 ± + (5.09) 2  d  2 2 σ A = 12.19/d 3

and σ B = −2.13/d 3

Using the Brittle-Coulomb-Mohr, as was used in Prob. 6-21, gives 1 12.19 −2.13 − = 3 3 25d 97d 2.8 Solving gives d = 1 1/8 in. Now compare to Modified II-Mohr theory 6-23

Ans.

( FA ) t = 300 cos 20 = 281.9 lbf, ( FA )r = 300 sin 20 = 102.6 lbf 3383 = 676.6 lbf 5

T = 281.9(12) = 3383 lbf · in, ( FC ) t = ( FC )r = 676.6 tan 20 = 246.3 lbf y ROy = 193.7 lbf O

ROz = 233.5 lbf

246.3 lbf

A

C

B

20"

16" 281.9 lbf

10"

x

RBy = 158.1 lbf

xy plane

O

676.6 lbf

A

B

20"

16"

z 102.6 lbf

C 10" RBz = 807.5 lbf

xz plane

M A = 20 193.72 + 233.52 = 6068 lbf · in M B = 10 246.32 + 676.62 = 7200 lbf · in (maximum) σx =

32(7200) 73 340 = 3 πd d3

16(3383) 17 230 = 3 πd d3 1/2 Sy σ = σx2 + 3τx2y = n 2 1/2 79 180 60 000 73 340 17 230 2 +3 = = 3 3 3 d d d 3.5

τx y =

d = 1.665 in

so use a standard diameter size of 1.75 in Ans.

x

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6-24

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From Prob. 6-23, τmax =

73 340 2d 3

2

σx 2

1/2

2 + τx y

=

2

17 230 + d3

2 1/2 =

Sy 2n

40 516 60 000 = d3 2(3.5)

d = 1.678 in so use 1.75 in Ans. 6-25

T = (270 − 50)(0.150) = 33 N · m , S y = 370 MPa (T1 − 0.15T1 )(0.125) = 33

⇒

T1 = 310.6 N,

T2 = 0.15(310.6) = 46.6 N

(T1 + T2 ) cos 45 = 252.6 N 107.0 N

y 163.4 N O

300

252.6 N A

89.2 N

400

150

300

400

z

C

B

252.6 N 150

320 N

174.4 N

xz plane

xy plane

M A = 0.3 163.42 + 1072 = 58.59 N · m M B = 0.15 89.22 + 174.42 = 29.38 N · m

(maximum)

32(58.59) 596.8 = 3 πd d3 16(33) 168.1 τx y = = 3 πd d3 2 2 1/2 596.8 664.0 370(106 ) 168.1 2 2 1/2 σ = σx + 3τx y = +3 = = d3 d3 d3 3.0 σx =

d = 17.5(10−3 ) m = 17.5 mm, 6-26

so use 18 mm Ans.

From Prob. 6-25,

τmax =

596.8 2d 3

2

d = 17.7(10−3 ) m = 17.7 mm,

σx 2

168.1 + d3

1/2

2 + τx2y 2 1/2

=

=

Sy 2n

342.5 370(106 ) = d3 2(3.0)

so use 18 mm Ans.

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169

Chapter 6

6-27

For the loading scheme shown in Figure (c), 4.4 F a b + = (6 + 4.5) Mmax = 2 2 4 2

V

M

= 23.1 N · m y

For a stress element at A:

B C

32M 32(23.1)(103 ) σx = = = 136.2 MPa πd 3 π(12) 3

x

A

The shear at C is 4(4.4/2)(103 ) 4( F/2) = = 25.94 MPa 3πd 2 /4 3π(12) 2 /4 1/2 136.2 2 = = 68.1 MPa 2

τx y = τmax

Since S y = 220 MPa, Ssy = 220/2 = 110 MPa, and n=

Ssy 110 = 1.62 Ans. = τmax 68.1

For the loading scheme depicted in Figure (d) 2 F 1 F a+b F a b b Mmax = − + = 2 2 2 2 2 2 2 4 This result is the same as that obtained for Figure (c). At point B, we also have a surface compression of −F −4.4(103 ) −F = − = −20.4 MPa σy = A bd 18(12) With σx = −136.2 MPa. From a Mohrs circle diagram, τmax = 136.2/2 = 68.1 MPa. n= 6-28

110 = 1.62 MPa 68.1

Ans.

Based on Figure (c) and using Eq. (6-15) 1/2 σ = σx2 = (136.22 ) 1/2 = 136.2 MPa n=

Sy 220 = 1.62 Ans. = σ 136.2

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Based on Figure (d) and using Eq. (6-15) and the solution of Prob. 6-27, 1/2 σ = σx2 − σx σ y + σ y2 = [(−136.2) 2 − (−136.2)(−20.4) + (−20.4) 2 ]1/2 = 127.2 MPa n=

Sy 220 = 1.73 Ans. = σ 127.2

6-29 w dF

r

When the ring is set, the hoop tension in the ring is equal to the screw tension. ri2 pi ro2 1+ 2 σt = 2 r ro − ri2

We have the hoop tension at any radius. The differential hoop tension d F is d F = wσt dr ro ro wri2 pi ro2 F= 1 + 2 dr = wri pi wσt dr = 2 r ro − ri2 ri ri

(1)

The screw equation is Fi =

T 0.2d

(2)

From Eqs. (1) and (2) F T = wri 0.2dwri d Fx = f pi ri dθ 2π f pi wri dθ = Fx =

pi ri d␪

pi =

dFx

o

=

2π f T 0.2d

f Tw ri 0.2dwri

2π

dθ o

Ans.

6-30 (a) From Prob. 6-29,

(b) From Prob. 6-29,

T = 0.2Fi d 190 T Fi = = = 3800 lbf 0.2d 0.2(0.25)

Ans.

F = wri pi

pi =

F Fi 3800 = 15 200 psi = = wri wri 0.5(0.5)

Ans.

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171

Chapter 6

pi ri2 + ro2 ri2 pi ro2 1+ = σt = 2 r r=ri ro − ri2 ro2 − ri2

(c)

15 200(0.52 + 12 ) = = 25 333 psi Ans. 12 − 0.52 σr = − pi = −15 200 psi σt − σr σ1 − σ3 τmax = = 2 2 25 333 − (−15 200) = 20 267 psi Ans. = 2 1/2 σ = σ A2 + σ B2 − σ A σ B

(d)

= [25 3332 + (−15 200) 2 − 25 333(−15 200)]1/2 = 35 466 psi Ans. (e) Maximum Shear hypothesis n=

0.5S y Ssy 0.5(63) = 1.55 Ans. = = τmax τmax 20.267

Distortion Energy theory n=

Sy 63 = 1.78 Ans. = σ 35 466

6-31 1"R

re

1" R 2

␴t r

The moment about the center caused by force F is Fre where re is the effective radius. This is balanced by the moment about the center caused by the tangential (hoop) stress. ro rσt w dr Fre = ri

ro2 r+ dr r ri ro2 − ri2 wpi ri2 ro 2 re = 2 + ro ln 2 ri F ro − ri2 wpi r 2 = 2 i2 r o − ri

ro

From Prob. 6-29, F = wri pi . Therefore, ro2 − ri2 ri r o + ro2 ln re = 2 2 ri ro − ri2 For the conditions of Prob. 6-29, ri = 0.5 and ro = 1 in 2 1 − 0.52 0.5 1 2 + 1 ln = 0.712 in re = 2 1 − 0.52 2 0.5

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6-32

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δnom = 0.0005 in (a) From Eq. (4-60) 30(106 )(0.0005) (1.52 − 12 )(12 − 0.52 ) = 3516 psi Ans. p= 1 2(12 )(1.52 − 0.52 ) Inner member: Eq. (4-57)

Eq. (6-13)

2 R 2 + ri2 1 + 0.52 = −5860 psi (σt ) i = − p 2 = −3516 2 1 − 0.52 R − ri2

(σr ) i = − p = −3516 psi 1/2 σi = σ A2 − σ A σ B + σ B2 = [(−5860) 2 − (−5860)(−3516) + (−3516) 2 ]1/2 = 5110 psi Ans.

Outer member: Eq. (4-58)

1.52 + 12 (σt ) o = 3516 1.52 − 12

= 9142 psi

(σr ) o = − p = −3516 psi Eq. (6-13)

σo = [91422 − 9142(−3516) + (−3516) 2 ]1/2 = 11 320 psi Ans.

(b) For a solid inner tube,

30(106 )(0.0005) (1.52 − 12 )(12 ) = 4167 psi Ans. p= 1 2(12 )(1.52 )

(σt ) i = − p = −4167 psi,

(σr ) i = −4167 psi

σi = [(−4167) 2 − (−4167)(−4167) + (−4167) 2 ]1/2 = 4167 psi Ans. 2 1.5 + 12 = 10 830 psi, (σr ) o = −4167 psi (σt ) o = 4167 1.52 − 12 σo = [10 8302 − 10 830(−4167) + (−4167) 2 ]1/2 = 13 410 psi Ans. 6-33

Using Eq. (4-60) with diametral values, 207(103 )(0.02) (752 − 502 )(502 − 252 ) = 19.41 MPa Ans. p= 50 2(502 )(752 − 252 ) 2 50 + 252 (σt ) i = −19.41 = −32.35 MPa Eq. (4-57) 502 − 252 (σr ) i = −19.41 MPa Eq. (6-13)

σi = [(−32.35) 2 − (−32.35)(−19.41) + (−19.41) 2 ]1/2 = 28.20 MPa Ans.

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Chapter 6

Eq. (4-58)

752 + 502 (σt ) o = 19.41 752 − 502

173

= 50.47 MPa,

(σr ) o = −19.41 MPa σo = [50.472 − 50.47(−19.41) + (−19.41) 2 ]1/2 = 62.48 MPa Ans. 6-34 δ=

1.9998 1.999 − = 0.0004 in 2 2

Eq. (4-59)

2 2 p(1) p(1) 2 + 12 1 +0 0.0004 = − 0.292 + 0.211 + 14.5(106 ) 22 − 12 30(106 ) 12 − 0 p = 2613 psi

Applying Eq. (4-58) at R,

Eq. (6-33b)

22 + 12 = 4355 psi (σt ) o = 2613 2 2 − 12 (σr ) o = −2613 psi, Sut = 20 kpsi, Suc = 83 kpsi n(4355) −2613n + 20 000 2 + =1 20 000 20 000 − 83 000

n = 4.52 Ans. 6-35

E = 30(106 ) psi, ν = 0.292, I = (π/64)(24 − 1.54 ) = 0.5369 in4 Eq. (4-60) can be written in terms of diameters,

2 Eδd do2 − D 2 D 2 − di2 30(106 ) (2 − 1.752 )(1.752 − 1.52 ) (0.002 46) p= = D 1.75 2(1.752 )(22 − 1.52 ) 2D 2 do2 − di2 = 2997 psi = 2.997 kpsi Outer member: Outer radius: Inner radius:

1.752 (2.997) (2) = 19.58 kpsi, (σr ) o = 0 22 − 1.752 22 1.752 (2.997) 1+ = 22.58 kpsi, (σr ) i = −2.997 kpsi (σt ) i = 2 2 − 1.752 1.752

(σt ) o =

Bending: ro :

(σx ) o =

6.000(2/2) = 11.18 kpsi 0.5369

ri :

(σx ) i =

6.000(1.75/2) = 9.78 kpsi 0.5369

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J = 2I = 1.0738 in4

Torsion: ro :

(τx y ) o =

8.000(2/2) = 7.45 kpsi 1.0738

ri :

(τx y ) i =

8.000(1.75/2) = 6.52 kpsi 1.0738

Outer radius is plane stress σx = 11.18 kpsi,

σ y = 19.58 kpsi,

τx y = 7.45 kpsi

σ = [11.182 − (11.18)(19.58) + 19.582 + 3(7.452 )]1/2 =

Eq. (6-15)

21.35 =

60 no

⇒

Inner radius, 3D state of stress

Sy 60 = no no

n o = 2.81 Ans. z –2.997 kpsi 9.78 kpsi

22.58 kpsi x

6.52 kpsi

y

From Eq. (6-14) with τ yz = τzx = 0 1 60 σ = √ [(9.78 − 22.58) 2 + (22.58 + 2.997) 2 + (−2.997 − 9.78) 2 + 6(6.52) 2 ]1/2 = ni 2 60 24.86 = ⇒ n i = 2.41 Ans. ni 6-36

From Prob. 6-35: p = 2.997 kpsi, I = 0.5369 in4 , J = 1.0738 in4 Inner member: Outer radius:

(0.8752 + 0.752 ) = −19.60 kpsi (σt ) o = −2.997 (0.8752 − 0.752 )

(σr ) o = −2.997 kpsi Inner radius:

2(2.997)(0.8752 ) = −22.59 kpsi (σt ) i = − 0.8752 − 0.752 (σr ) i = 0

Bending: ro :

(σx ) o =

6(0.875) = 9.78 kpsi 0.5369

ri :

(σx ) i =

6(0.75) = 8.38 kpsi 0.5369

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175

Chapter 6

Torsion: ro :

(τx y ) o =

8(0.875) = 6.52 kpsi 1.0738

ri :

(τx y ) i =

8(0.75) = 5.59 kpsi 1.0738

The inner radius is in plane stress: σx = 8.38 kpsi, σ y = −22.59 kpsi, τx y = 5.59 kpsi σi = [8.382 − (8.38)(−22.59) + (−22.59) 2 + 3(5.592 )]1/2 = 29.4 kpsi ni =

Sy 60 = 2.04 Ans. = σi 29.4

Outer radius experiences a radial stress, σr 1/2 1 σo = √ (−19.60 + 2.997) 2 + (−2.997 − 9.78) 2 + (9.78 + 19.60) 2 + 6(6.52) 2 2 = 27.9 kpsi 60 = 2.15 Ans. no = 27.9 6-37

KI KI θ θ θ 3θ 2 1 2√ cos ± sin cos sin σp = √ 2 2 2 2 2 2πr 2πr 1/2 θ 3θ 2 θ KI sin cos cos + √ 2 2 2 2πr 1/2

KI θ 2 θ 2 θ 2 3θ 2 θ 2 θ 2 3θ + sin cos cos =√ cos ± sin cos sin 2 2 2 2 2 2 2 2πr θ θ θ KI θ θ KI cos ± cos sin =√ cos 1 ± sin =√ 2 2 2 2 2 2πr 2πr Plane stress: The third principal stress is zero and θ θ θ θ KI KI cos 1 + sin , σ2 = √ cos 1 − sin , σ1 = √ 2 2 2 2 2πr 2πr

σ3 = 0 Ans.

Plane strain: σ1 and σ2 equations still valid however, θ KI cos σ3 = ν(σx + σ y ) = 2ν √ 2 2πr 6-38

Ans.

For θ = 0 and plane strain, the principal stress equations of Prob. 6-37 give KI , σ1 = σ2 = √ 2πr

KI σ3 = 2ν √ = 2νσ1 2πr

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1 √ [(σ1 − σ1 ) 2 + (σ1 − 2νσ1 ) 2 + (2νσ1 − σ1 ) 2 ]1/2 = S y 2 σ1 − 2νσ1 = S y 1 1 1−2 σ1 = S y ⇒ σ1 = 3S y Ans. For ν = , 3 3

(a) DE:

σ1 − σ3 = S y

(b) MSS:

ν=

⇒

σ1 − 2νσ1 = S y

σ1 = 3S y

Ans.

2 σ3 = σ1 3

␶

2␴ 3 1

6-39

1 3

⇒

␴1, ␴2

␴

Radius of largest circle σ1 1 2 σ1 − σ1 = R= 2 3 6

(a) Ignoring stress concentration F = S y A = 160(4)(0.5) = 320 kips Ans. (b) From Fig. 6-36: h/b = 1, a/b = 0.625/4 = 0.1563, β = 1.3 70 = 1.3

Eq. (6-51)

F π(0.625) 4(0.5)

F = 76.9 kips Ans. 6-40

Given: a = 12.5 mm, K I c = 80 MPa · ro = a/(ro − ri ) = ri /ro = Fig. 6-40: Eq. (6-51):

√ m, S y = 1200 MPa, Sut = 1350 MPa

350 = 175 mm, 2 12.5 = 0.5 175 − 150 150 = 0.857 175

. β = 2.5

√ K I c = βσ πa 80 = 2.5σ π(0.0125) σ = 161.5 MPa

ri =

350 − 50 = 150 mm 2

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177

Chapter 6

Eq. (4-51) at r = ro : ri2 pi σ = 2 (2) ro − ri2 161.5 =

1502 pi (2) 1752 − 1502

pi = 29.2 MPa Ans. 6-41 (a) First convert the data to radial dimensions to agree with the formulations of Fig. 4-25. Thus ro = 0.5625 ± 0.001in ri = 0.1875 ± 0.001 in Ro = 0.375 ± 0.0002 in Ri = 0.376 ± 0.0002 in The stochastic nature of the dimensions affects the δ = |Ri | − |Ro | relation in Eq. (4-60) but not the others. Set R = (1/2)( Ri + Ro ) = 0.3755. From Eq. (4-60)

Eδ ro2 − R 2 R 2 − ri2 p= R 2R 2 ro2 − ri2 Substituting and solving with E = 30 Mpsi gives p = 18.70(106 ) δ

Since δ = Ri − Ro

δ¯ = R¯ i − R¯ o = 0.376 − 0.375 = 0.001 in and

σˆ δ =

0.0002 4

2

0.0002 + 4

2 1/2

= 0.000 070 7 in Then Cδ =

σˆ δ 0.000 070 7 = 0.0707 = 0.001 δ¯

The tangential inner-cylinder stress at the shrink-fit surface is given by R¯ 2 + r¯i2 σit = −p 2 R¯ − r¯ 2 i

0.37552 + 0.18752 = −18.70(10 ) δ 0.37552 − 0.18752 6

= −31.1(106 ) δ σ¯ it = −31.1(106 ) δ¯ = −31.1(106 )(0.001) = −31.1(103 ) psi

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Also σˆ σit = |Cδ σ¯ it | = 0.0707(−31.1)103 = 2899 psi σit = N(−31 100, 2899) psi Ans. (b) The tangential stress for the outer cylinder at the shrink-fit surface is given by 2 r¯o + R¯ 2 σot = p 2 r¯o − R¯ 2 0.56252 + 0.37552 6 = 18.70(10 ) δ 0.56252 − 0.37552 = 48.76(106 ) δ psi σ¯ ot = 48.76(106 )(0.001) = 48.76(103 ) psi σˆ σot = Cδ σ¯ ot = 0.0707(48.76)(103 ) = 34.45 psi σot = N(48 760, 3445) psi Ans. 6-42

From Prob. 6-41, at the fit surface σot = N(48.8, 3.45) kpsi. The radial stress is the fit pressure which was found to be p = 18.70(106 ) δ p¯ = 18.70(106 )(0.001) = 18.7(103 ) psi σˆ p = Cδ p¯ = 0.0707(18.70)(103 ) = 1322 psi and so p = N(18.7, 1.32) kpsi and σor = −N(18.7, 1.32) kpsi These represent the principal stresses. The von Mises stress is next assessed. σ¯ A = 48.8 kpsi,

σ¯ B = −18.7 kpsi

k = σ¯ B /σ¯ A = −18.7/48.8 = −0.383 σ¯ = σ¯ A (1 − k + k 2 ) 1/2 = 48.8[1 − (−0.383) + (−0.383) 2 ]1/2 = 60.4 kpsi σˆ σ = C p σ¯ = 0.0707(60.4) = 4.27 kpsi

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Chapter 6

179

Using the interference equation S¯ − σ¯ z = − 1/2 σˆ S2 + σˆ σ2 =−

95.5 − 60.4 = −4.5 [(6.59) 2 + (4.27) 2 ]1/2

p f = α = 0.000 003 40, or about 3 chances in a million.

Ans.

6-43 σt =

6000N(1, 0.083 33)(0.75) pd = 2t 2(0.125)

= 18N(1, 0.083 33) kpsi σl =

6000N(1, 0.083 33)(0.75) pd = 4t 4(0.125)

= 9N(1, 0.083 33) kpsi σr = −p = −6000N(1, 0.083 33) kpsi These three stresses are principal stresses whose variability is due to the loading. From Eq. (6-12), we find the von Mises stress to be 1/2 (18 − 9) 2 + [9 − (−6)]2 + (−6 − 18) 2 σ = 2 = 21.0 kpsi σˆ σ = C p σ¯ = 0.083 33(21.0) = 1.75 kpsi S¯ − σ¯ z = − 1/2 σˆ S2 + σˆ σ2 =

50 − 21.0 = −6.5 (4.12 + 1.752 ) 1/2

The reliability is very high . R = 1 − (6.5) = 1 − 4.02(10−11 ) = 1 Ans.

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Chapter 7 7-1 H B = 490 Eq. (3-17):

Sut = 0.495(490) = 242.6 kpsi > 212 kpsi Se = 107 kpsi

Eq. (7-8):

a = 1.34, b = −0.085

Table 7-4:

ka = 1.34(242.6) −0.085 = 0.840 3/16 −0.107 kb = = 1.05 0.3

Eq. (7-18): Eq. (7-19):

Se = ka kb Se = 0.840(1.05)(107) = 94.4 kpsi

Eq. (7-17):

Ans.

7-2 (a) Sut = 68 kpsi, Se = 0.495(68) = 33.7 kpsi

Ans.

(b) Sut = 112 kpsi, Se = 0.495(112) = 55.4 kpsi (c) 2024T3 has no endurance limit (d) Eq. (3-17): 7-3

Se

= 107 kpsi

Ans.

Ans.

Ans.

σ F = σ0 εm = 115(0.90) 0.22 = 112.4 kpsi Eq. (7-8):

Se = 0.504(66.2) = 33.4 kpsi log(112.4/33.4) = −0.083 64 log(2 · 106 )

Eq. (7-11):

b=−

Eq. (7-9):

f =

112.4 (2 · 103 ) −0.083 64 = 0.8991 66.2

Eq. (7-13):

a=

[0.8991(66.2)]2 = 106.1 kpsi 33.4

Eq. (7-12): Eq. (7-15):

S f = a N b = 106.1(12 500) −0.083 64 = 48.2 kpsi Ans. σ 1/b 36 −1/0.083 64 a N= = = 409 530 cycles Ans. a 106.1

7-4 From S f = a N b log S f = log a + b log N Substituting (1, Sut ) log Sut = log a + b log (1) From which

a = Sut

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181

Chapter 7

Substituting (103 , f Sut ) and a = Sut log f Sut = log Sut + b log 103 From which b= ∴

1 log f 3

S f = Sut N (log f )/3

1 ≤ N ≤ 103

For 500 cycles as in Prob. 7-3 500S f ≥ 66.2(500) (log 0.8991)/3 = 60.2 kpsi Ans. 7-5 Read from graph: (103, 90) and (106, 50). From S = a N b log S1 = log a + b log N1 log S2 = log a + b log N2 From which log a =

log S1 log N2 − log S2 log N1 log N2 /N1

log 90 log 106 − log 50 log 103 log 106 /103 = 2.2095

=

a = 10log a = 102.2095 = 162.0 b=

log 50/90 = −0.085 09 3

(S f ) ax = 162−0.085 09 Check:

103 ≤ N ≤ 106 in kpsi Ans.

103 (S f ) ax = 162(103 ) −0.085 09 = 90 kpsi 106 (S f ) ax = 162(106 ) −0.085 09 = 50 kpsi

The end points agree. 7-6 Eq. (7-8): Table 7-4: Eq. (7-18): Eq. (7-19): Eq. (7-17):

Se = 0.504(710) = 357.8 MPa a = 4.51,

b = −0.265

ka = 4.51(710) −0.265 = 0.792 d −0.107 32 −0.107 kb = = = 0.858 7.62 7.62 Se = ka kb Se = 0.792(0.858)(357.8) = 243 MPa Ans.

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7-7 For AISI 4340 as forged steel, Se = 107 kpsi Eq. (7-8): a = 39.9,

Table 7-4:

b = −0.995

ka = 39.9(260) −0.995 = 0.158 0.75 −0.107 kb = = 0.907 Eq. (7-19): 0.30 Each of the other Marin factors is unity. Eq. (7-18):

Se = 0.158(0.907)(107) = 15.3 kpsi For AISI 1040: Se = 0.504(113) = 57.0 kpsi ka = 39.9(113) −0.995 = 0.362 kb = 0.907 (same as 4340) Each of the other Marin factors is unity. Se = 0.362(0.907)(57.2) = 18.7 kpsi Not only is AISI 1040 steel a contender, it has a superior endurance strength. Can you see why? 7-8 (a) For an AISI 1018 CD-machined steel, the strengths are

2.5 mm

20 mm

Eq. (3-17):

25 mm

Sut = 440 MPa

⇒

HB =

S y = 370 MPa Ssu = 0.67(440) = 295 MPa Fig. A-15-15: Fig. 7-21: Eq. (7-31):

2.5 D 25 r = = 0.125, = = 1.25, d 20 d 20 qs = 0.94 K f s = 1 + 0.94(1.4 − 1) = 1.376

K ts = 1.4

For a purely reversing torque of 200 N · m K f s 16T 1.376(16)(200 × 103 N · mm) = τmax = πd 3 π(20 mm) 3 τmax = 175.2 MPa = τa Se = 0.504(440) = 222 MPa The Marin factors are

Eq. (7-17):

ka = 4.51(440) −0.265 = 0.899 20 −0.107 = 0.902 kb = 7.62 kc = 0.59, kd = 1, ke = 1 Se = 0.899(0.902)(0.59)(222) = 106.2 MPa

440 = 129 3.41

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Chapter 7

[0.9(295)]2 a= = 664 106.2

Eq. (7-13):

1 0.9(295) b = − log = −0.132 65 3 106.2 175.2 1/−0.132 65 N= 664

Eq. (7-14): Eq. (7-15):

N = 23 000 cycles Ans. (b) For an operating temperature of 450°C, the temperature modification factor, from Table 7-6, is kd = 0.843 Se = 0.899(0.902)(0.59)(0.843)(222) = 89.5 MPa Thus [0.9(295)]2 = 788 a= 89.5 1 0.9(295) b = − log = −0.157 41 3 89.5 175.2 1/−0.157 41 N= 788 N = 14 100 cycles

Ans.

7-9 F 1 kN 800 mm b b

f = 0.9 n = 1.5 N = 104 cycles

For AISI 1045 HR steel, Sut = 570 MPa and S y = 310 MPa Se = 0.504(570 MPa) = 287.3 MPa Find an initial guess based on yielding: σa = σmax =

M(b/2) 6M Mc = = 3 3 I b(b )/12 b

Mmax = (1 kN)(800 mm) = 800 N · m σmax

Sy = n

⇒

6(800 × 103 N · mm) 310 N/mm2 = b3 1.5

b = 28.5 mm Eq. (7-24): Eq. (7-19):

de = 0.808b 0.808b −0.107 kb = = 1.2714b−0.107 7.62 kb = 0.888

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The remaining Marin factors are ka = 57.7(570) −0.718 = 0.606 kc = kd = ke = k f = 1 Eq. (7-17): Eq. (7-13): Eq. (7-14): Eq. (7-12):

Se = 0.606(0.888)(287.3 MPa) = 154.6 MPa a=

[0.9(570)]2 = 1702 154.6

1 0.9(570) b = − log = −0.173 64 3 154.6 S f = a N b = 1702[(104 ) −0.173 64 ] = 343.9 MPa n=

Sf σa

Sf n

or σa =

343.9 6(800 × 103 ) ⇒ b = 27.6 mm = 3 b 1.5 Check values for kb , Se , etc. kb = 1.2714(27.6) −0.107 = 0.891 Se = 0.606(0.891)(287.3) = 155.1 MPa a=

[0.9(570)]2 = 1697 155.1

1 0.9(570) b = − log = −0.173 17 3 155.1 S f = 1697[(104 ) −0.173 17 ] = 344.4 MPa 344.4 6(800 × 103 ) = b3 1.5 b = 27.5 mm Ans. 7-10

10 12

Fa

Fa

60 1018

Table A-20:

Sut = 440 MPa, Se

Table 7-4:

S y = 370 MPa

= 0.504(440) = 221.8 MPa

ka = 4.51(440)−0.265 = 0.899 kb = 1 (axial loading)

Eq. (7-25):

kc = 0.85 Se = 0.899(1)(0.85)(221.8) = 169.5 MPa

Table A-15-1:

d/w = 12/60 = 0.2,

K t = 2.5

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Chapter 7

185

From Eq. (7-35) and Table 7-8 Kf =

Kt 2.5 √ = 2.09 √ = √ 1 + 2/ r [(K t − 1)/K t ] a 1 + 2/ 6 [(2.5 − 1)/2.5](174/440)

σa = K f

Fa A

Se 169.5 2.09Fa = = nf 10(60 − 12) 1.8

⇒

Fa = 21 630 N = 21.6 kN Ans. Sy Fa = A ny

⇒

370 Fa = 10(60 − 12) 1.8

Fa = 98 667 N = 98.7 kN Ans. Largest force amplitude is 21.6 kN. 7-11

Ans.

A priori design decisions: The design decision will be: d Material and condition: 1095 HR and from Table A-20 Sut = 120, S y = 66 kpsi. Design factor: n f = 1.6 per problem statement. Life: (1150)(3) = 3450 cycles Function: carry 10 000 lbf load Preliminaries to iterative solution: Se = 0.504(120) = 60.5 kpsi ka = 2.70(120) −0.265 = 0.759 πd 3 I = = 0.098 17d 3 c 32 6 (10 000)(12) = 30 000 lbf · in M(crit.) = 24 The critical location is in the middle of the shaft at the shoulder. From Fig. A-15-9: D/d = 1.5, r/d = 0.10, and K t = 1.68. With no direct information concerning f, use f = 0.9. For an initial trial, set d = 2.00 in 2.00 −0.107 = 0.816 kb = 0.30 Se = 0.759(0.816)(60.5) = 37.5 kpsi [0.9(120)]2 = 311.0 37.5 1 0.9(120) b = − log = −0.1531 3 37.5

a=

S f = 311.0(3450) −0.1531 = 89.3

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σ0 =

30 M 305.6 = = 3 I /c 0.098 17d d3

305.6 = 38.2 kpsi 23 2 d = = 0.2 r= 10 10 1.68 = 1.584 Kf = √ 1 + 2/ 0.2 [(1.68 − 1)/1.68](4/120) =

Eq. (7-37): (K f ) 103 = 1 − (1.584 − 1)[0.18 − 0.43(10−2 )120 + 0.45(10−5 )1202 ] = 1.158 Eq. (7-38):

1.1582 (3450) −(1/3) log(1.158/1.584) = 1.584

(K f ) N = K 3450

= 1.225 305.6 σ0 = = 38.2 kpsi 23 σa = (K f ) N σ0 = 1.225(38.2) = 46.8 kpsi nf =

(S f )3450 89.3 = 1.91 = σa 46.8

The design is satisfactory. Reducing the diameter will reduce n, but the resulting preferred size will be d = 2.00 in. 7-12 σa = 172 MPa, Yield:

172 + 178.4 =

σm =

√ √ 3τm = 3(103) = 178.4 MPa

Sy 413 = ny ny

⇒

n y = 1.18 Ans.

(a) Modified Goodman, Table 7-9 nf =

1 = 1.06 Ans. (172/276) + (178.4/551)

(b) Gerber, Table 7-10  

2  2  1 551 172 2(178.4)(276) = 1.31 Ans. −1 + 1 + nf =  2 178.4 276  551(172) (c) ASME-Elliptic, Table 7-11 1/2 1 nf = = 1.32 Ans. (172/276) 2 + (178.4/413) 2

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Chapter 7

7-13 σa = 69 MPa, 69 + 239 =

Yield:

σm =

413 ny

√ 3(138) = 239 MPa

⇒

n y = 1.34 Ans.

(a) Modified Goodman, Table 7-9 nf = (b) Gerber, Table 7-10 nf =

1 2

551 239

2

1 = 1.46 Ans. (69/276) + (239/551)

 

 2  69 2(239)(276) = 1.73 Ans. −1 + 1 +  276  551(69)

(c) ASME-Elliptic, Table 7-11 1/2 1 nf = = 1.59 Ans. (69/276) 2 + (239/413) 2 7-14 σa

= σa2 + 3τa2 = 832 + 3(692 ) = 145.5 MPa, 145.5 + 178.4 =

Yield:

413 ny

⇒

σm =

√ 3(103) = 178.4 MPa

n y = 1.28 Ans.

(a) Modified Goodman, Table 7-9 nf = (b) Gerber, Table 7-10 nf =

1 2

551 178.4

2

1 = 1.18 Ans. (145.5/276) + (178.4/551)  

 2  145.5 2(178.4)(276) −1 + 1 + = 1.47 Ans.  276  551(145.5)

(c) ASME-Elliptic, Table 7-11 1/2 1 nf = = 1.47 Ans. (145.5/276) 2 + (178.4/413) 2 7-15 σa = Yield:

√ 3(207) = 358.5 MPa,

358.5 =

413 ny

⇒

σm = 0

n y = 1.15 Ans.

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1 = 0.77 Ans. (358.5/276)

(b) Gerber criterion of Table 7-10 does not work; therefore use Eq. (7-50). nf

σa =1 Se

(c) ASME-Elliptic, Table 7-11

⇒

nf =

nf =

Se 276 = 0.77 Ans. = σa 358.5

1 358.5/276

2 = 0.77 Ans.

Let f = 0.9 to assess the cycles to failure by fatigue Eq. (7-13): Eq. (7-14): Eq. (7-15):

[0.9(551)]2 = 891.0 MPa 276

a=

1 0.9(551) b = − log = −0.084 828 3 276 358.5 −1/0.084 828 N= = 45 800 cycles Ans. 891.0

7-16 σa = Yield:

√ 3(103) = 178.4 MPa,

178.4 + 103 =

413 ny

⇒

σm = 103 MPa

n y = 1.47 Ans.


1 = 1.20 Ans. (178.4/276) + (103/551)

(b) Gerber, Table 7-10  

2  2  178.4 1 551 2(103)(276) nf = −1 + 1 + = 1.44 Ans. 2 103 276  551(178.4)  (c) ASME-Elliptic, Table 7-11 1/2 1 nf = = 1.44 Ans. (178.4/276) 2 + (103/413) 2 7-17

Table A-20:

Sut = 64 kpsi, S y = 54 kpsi A = 0.375(1 − 0.25) = 0.2813 in2 σmax =

3000 Fmax = (10−3 ) = 10.67 kpsi A 0.2813

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Chapter 7

ny =

54 = 5.06 Ans. 10.67

Se = 0.504(64) = 32.3 kpsi ka = 2.70(64) −0.265 = 0.897 kb = 1,

kc = 0.85

Se = 0.897(1)(0.85)(32.3) = 24.6 kpsi Table A-15-1: w = 1 in, d = 1/4 in, d/w = 0.25 K t = 2.45. From Eq. (7-35) and Table 7-8 2.45 = 1.94 √ 1 + 2/ 0.125 [(2.45 − 1)/2.45](5/64) Fmax − Fmin σa = K f 2A 3.000 − 0.800 = 7.59 kpsi = 1.94 2(0.2813)

Kf =

Fmax + Fmin 2A 3.000 + 0.800 = 13.1 kpsi = 1.94 2(0.2813)

σm = K f

r=

σa 7.59 = 0.579 = σm 13.1

(a) DE-Gerber, Table 7-10 Sa =

 2

2

0.579 (64 )  −1 + 2(24.6)

1+

2(24.6) 0.579(64)

2

  = 18.5 kpsi

18.5 Sa = = 32.0 kpsi r 0.579  

2 2 1 64 7.59  2(13.1)(24.6)  nf = −1 + 1 + 2 13.1 24.6 7.59(64)

Sm =

= 2.44 Ans. (b) DE-Elliptic, Table 7-11 Sa = Sm =

(0.5792 )(24.62 )(542 ) = 19.33 kpsi 24.62 + (0.5792 )(542 ) 19.33 Sa = = 33.40 kpsi r 0.579

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Table 7-16

1 = 2.55 Ans. (7.59/24.6) 2 + (13.1/54) 2

nf =

7-18

Referring to the solution of Prob. 7-17, for load fluctuations of −800 to 3000 lbf 3.000 − (−0.800) = 13.1 kpsi σa = 1.94 2(0.2813) 3.000 + (−0.800) = 7.59 kpsi σm = 1.94 2(0.2813) r= (a) Table 7-10, DE-Gerber nf =

1 2

64 7.59

2

σa 13.13 = 1.728 = σm 7.60  

2 13.1  2(7.59)(24.6)  −1 + 1 + = 1.79 Ans. 24.6 64(13.1)

(b) Table 7-11, DE-Elliptic

nf =

7-19

1 = 1.82 Ans. (13.1/24.6) 2 + (7.59/54) 2

Referring to the solution of Prob. 7-17, for load fluctuations of 800 to −3000 lbf 0.800 − (−3.000) = 13.1 kpsi σa = 1.94 2(0.2813) 0.800 + (−3.000) = −7.59 kpsi σm = 1.94 2(0.2813) r=

σa 13.1 = −1.726 = σm −7.59

(a) We have a compressive midrange stress for which the failure locus is horizontal at the Se level. nf =

Se 24.6 = 1.88 Ans. = σa 13.1

nf =

Se 24.6 = 1.88 Ans. = σa 13.1

(b) Same as (a)

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Chapter 7

191

7-20 Sut = 0.495(380) = 188.1 kpsi Se = 0.504(188.1) = 94.8 kpsi ka = 14.4(188.1) −0.718 = 0.335 For a non-rotating round bar in bending, Eq. (7-23) gives: de = 0.370d = 0.370(3/8) = 0.1388 in 0.1388 −0.107 kb = = 1.086 0.3 Se = 0.335(1.086)(94.8) = 34.49 kpsi 30 − 15 30 + 15 = 7.5 lbf, Fm = = 22.5 lbf Fa = 2 2 32Mm 32(22.5)(16) (10−3 ) = 69.54 kpsi = σm = 3 πd π(0.3753 ) σa = r=

32(7.5)(16) (10−3 ) = 23.18 kpsi 3 π(0.375 ) 23.18 = 0.333 69.54

0


1 = 0.960 (23.18/34.49) + (69.54/188.1)

Since finite failure is predicted, proceed to calculate N Eq. (7-10):

σ F = 188.1 + 50 = 238.1 kpsi log(238.1/34.49) = −0.133 13 log(2 · 106 ) 238.1 f = (2 · 103 ) −0.133 13 = 0.4601 188.1

b=−

Eq. (7-11): Eq. (7-9): Eq. (7-13): σa σm + =1 Sf Sut

[0.4601(188.1)]2 = 217.16 kpsi 34.49 23.18 σa = = 36.78 kpsi Sf = 1 − (σm /Sut ) 1 − (69.54/188.1)

a= ⇒

Eq. (7-15) with σa = S f 36.78 1/−0.133 13 N= = 620 000 cycles Ans. 217.16

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(b) Gerber, Table 7-10 nf =

1 2

188.1 69.54

= 1.20

2

 

 2  23.18 2(69.54)(34.49) −1 + 1 +  34.49  188.1(23.18)

Thus, infinite life is predicted ( N ≥ 106 cycles). Ans.

7-21 (a)

I =

1 (18)(33 ) = 40.5 mm4 12

y=

Fl 3 3E I

⇒

F=

3E I y l3

3(207)(109 )(40.5)(10−12 )(2)(10−3 ) = 50.3 N Ans. (1003 )(10−9 ) 6 = (50.3) = 150.9 N Ans. 2

Fmin = Fmax (b)

F

101.5 mm

M = 0.1015F N · m A = 3(18) = 54 mm2 M

F

Curved beam: rc = 4.5 mm, σi = − σo =

rn =

3 h = = 4.3281 mm ln(ro /ri ) ln(6/3)

e = rc − rn = 4.5 − 4.3281 = 0.1719 mm

(0.1015F)(1.5 − 0.1719) F Mci F − = −4.859F MPa − =− − 3 Aeri A 54(0.1719)(3)(10 ) 54

(0.1015F)(1.5 + 0.1719) F Mco F − = 3.028F MPa − = − 3 Aero A 54(0.1719)(6)(10 ) 54 (σi ) min = −4.859(150.9) = −733.2 MPa (σi ) max = −4.859(50.3) = −244.4 MPa (σo ) max = 3.028(150.9) = 456.9 MPa (σo ) min = 3.028(50.3) = 152.3 MPa

Eq. (3-17)

Sut = 3.41(490) = 1671 MPa

Per the problem statement, estimate the yield as S y = 0.9Sut = 0.9(1671) = 1504 MPa. Then from Eq. (7-8), Se = 740 MPa; Eq. (7-18), ka = 1.58(1671) −0.085 = 0.841; Eq. (7-24) de = 0.808[18(3)]1/2 = 5.938 mm; and Eq. (7-19), kb = (5.938/7.62) −0.107 = 1.027.

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Chapter 7

Se = 0.841(1.027)(740) = 639 MPa −733.2 + 244.4 = 244.4 MPa (σi ) a = 2

At Inner Radius

(σi ) m =

−733.2 − 244.4 = −488.8 MPa 2 ␴a 1504 MPa

639

244.4

1504

Load line: Langer (yield) line: Intersection:

488.4

␴m

σm = −244.4 − σa σm = σa − 1504 = −244.4 − σa σa = 629.8 MPa, σm = −874.2 MPa (Note that σa is less than 639 MPa)

Yield:

ny =

629.8 = 2.58 244.4

Fatigue:

nf =

639 = 2.61 244.4

Thus, the spring is not likely to fail in fatigue at the inner radius. Ans.

At Outer Radius

Yield load line: Langer line: Intersection:

Fatigue line:

(σo ) a =

456.9 − 152.3 = 152.3 MPa 2

(σo ) m =

456.9 + 152.3 = 304.6 MPa 2

σm = 152.3 + σa σm = 1504 − σa = 152.3 + σa σa = 675.9 MPa, σm = 828.2 MPa 675.9 = 4.44 ny = 152.3 σa = [1 − (σm /Sut ) 2 ]Se = σm − 152.3 2 σm 639 1 − = σm − 152.3 1671 σm2 + 4369.7σm − 3.4577(106 ) = 0

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σm =

−4369.7 +

4369.72 + 4(3.4577)(106 ) = 684.2 MPa 2

σa = 684.2 − 152.3 = 531.9 MPa nf =

531.9 = 3.49 152.3

Thus, the spring is not likely to fail in fatigue at the outer radius. Ans. 7-22

The solution at the inner radius is the same as in Prob. 7-21. At the outer radius, the yield solution is the same. σm Se = σm − 152.3 σa = 1 − Fatigue line: Sut σm 639 1 − = σm − 152.3 1671 1.382σm = 791.3

⇒

σm = 572.4 MPa

σa = 572.4 − 152.3 = 420 MPa nf = 7-23

420 = 2.76 Ans. 152.3

Preliminaries: Table A-20:

Sut = 64 kpsi,

Sy = 54 kpsi

Se = 0.504(64) = 32.3 kpsi ka = 2.70(64) −0.265 = 0.897 kb = 1 kc = 0.85 Se = 0.897(1)(0.85)(32.3) = 24.6 kpsi Fillet: Fig. A-15-5: D = 3.75 in, d = 2.5 in, D/d = 3.75/2.5 = 1.5, and r/d = 0.25/2.5 = 0.10 ∴ K t = 2.1 Kf = σmax =

2.1 = 1.86 √ 1 + 2/ 0.25 [(2.1 − 1)/2.1](4/64) 4 = 3.2 kpsi 2.5(0.5)

−16 = −12.8 kpsi 2.5(0.5) 3.2 − (−12.8) = 14.88 kpsi σa = 1.86 2

σmin =

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Chapter 7

3.2 + (−12.8) = −8.93 kpsi σm = 1.86 2 S y 54 = 4.22 = ny = σmin −12.8

Since the midrange stress is negative, Sa = Se = 24.6 kpsi nf =

Sa 24.6 = 1.65 = σa 14.88

Hole: Fig. A-15-1: d/w = 0.75/3.75 = 0.20, K t = 2.5 Kf = σmax =

2.5 √ = 2.17 1 + 2/ 0.75/2 [(2.5 − 1)/2.5](5/64) 4 = 2.67 kpsi 0.5(3.75 − 0.75)

−16 = −10.67 kpsi 0.5(3.75 − 0.75) 2.67 − (−10.67) = 14.47 kpsi σa = 2.17 2

σmin =

σm = 2.17

2.67 + (−10.67) = −8.68 kpsi 2

Since the midrange stress is negative, S y 54 = = 5.06 n y = σmin −10.67 Sa = Se = 24.6 kpsi Sa 24.6 = 1.70 = σa 14.47 Thus the design is controlled by the threat of fatigue at the fillet; the minimum factor of safety is n f = 1.65. Ans. nf =

7-24 (a)

T

T

M = −T , h = 5 mm, A = 25 mm2 rc = 20 mm, ro = 22.5 mm, ri = 17.5 mm h 5 = 19.8954 mm rn = = ln ro /ri ln (22.5/17.5) e = rc − rn = 20 − 19.8954 = 0.1046 mm co = 2.605 mm,

ci = 2.395 mm

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σi =

Mci −T (0.002 395) = −52.34(106 )T = − 6 Aeri 25(10 )(0.1046)(10−3 )(17.5)(10−3 )

−Mco T (2.605)(10−3 ) = 44.27(106 )T = σo = − 6 − 3 − 3 Aero 25(10 )(0.1046)(10 )(22.5)(10 ) For fatigue, σo is most severe as it represents a tensile stress. 1 σm = σa = (44.27)(106 )T = 22.14(106 )T 2 Se = 0.504Sut = 0.504(770) = 388.1 MPa ka = 4.51(770) −0.265 = 0.775 de = 0.808[5(5)]1/2 = 4.04 mm 4.04 −0.107 = 1.070 kb = 7.62 Se = 0.775(1.07)(388.1) = 321.8 MPa Modified Goodman, Table 7-9 σm 1 σa + = Se Sut nf

⇒

22.14T 1 22.14T + = 321.8 770 3

T = 3.42 N · m Ans. (b) Gerber, Eq. (7-50)

nσa nσm 2 + =1 Se Sut 3(22.14)T 3(22.14)T 2 + =1 321.8 770

T 2 + 27.74T − 134.40 = 0 1 T = −27.74 + 27.742 + 4(134.40) = 4.21 N · m Ans. 2 (c) To guard against yield, use T of part (b) and the inner stress. 420 = 1.91 Ans. ny = 52.34(4.21) 7-25

From Prob. 7-24, Se = 321.8 MPa, S y = 420 MPa, and Sut = 770 MPa (a) Assuming the beam is straight, 6M 6T = 48(106 )T σmax = 2 = 3 bh 5 [(10−3 ) 3 ] Goodman:

24T 1 24T + = 321.8 770 3

⇒

T = 3.15 N · m Ans.

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197

Chapter 7

3(24)T 2 3(24)T + =1 321.8 770

(b) Gerber:

T 2 + 25.59T − 114.37 = 1 1 T = −25.59 + 25.592 + 4(114.37) = 3.88 N · m Ans. 2 (c) Using σmax = 52.34(106 )T from Prob. 7-24, ny =

420 = 2.07 Ans. 52.34(3.88)

7-26 (a)

16K f s Tmax πd 3 . Fig. 7-21 for H B > 200, r = 3 mm, qs = 1 τmax =

K f s = 1 + qs (K ts − 1) K f s = 1 + 1(1.6 − 1) = 1.6 Tmax = 2000(0.05) = 100 N · m, τmax

Tmin =

500 (100) = 25 N · m 2000

16(1.6)(100)(10−6 ) = = 101.9 MPa π(0.02) 3

τmin =

500 (101.9) = 25.46 MPa 2000

1 τm = (101.9 + 25.46) = 63.68 MPa 2 1 τa = (101.9 − 25.46) = 38.22 MPa 2 Ssu = 0.67Sut = 0.67(320) = 214.4 MPa Ssy = 0.577S y = 0.577(180) = 103.9 MPa Se = 0.504(320) = 161.3 MPa ka = 57.7(320) −0.718 = 0.917 de = 0.370(20) = 7.4 mm 7.4 −0.107 kb = = 1.003 7.62 kc = 0.59 Se = 0.917(1.003)(0.59)(161.3) = 87.5 MPa

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Modified Goodman, Table 7-9, nf =

1 1 = = 1.36 Ans. (τa /Se ) + (τm /Ssu ) (38.22/87.5) + (63.68/214.4)

(b) Gerber, Table 7-10 nf =

=

7-27

1 2 1 2

Ssu τm

2

214.4 63.68



τa  −1 + Se

2

1+

2τm Se Ssu τa

2

 

 

2   38.22 2(63.68)(87.5) = 1.70 −1 + 1 + 87.5  214.4(38.22) 

Ans.

S y = 800 MPa, Sut = 1000 MPa

. (a) From Fig. 7-20, for a notch radius of 3 mm and Sut = 1 GPa, q = 0.92. K f = 1 + q(K t − 1) = 1 + 0.92(3 − 1) = 2.84 2.84(4) P 4P σmax = −K f =− = −4018P 2 πd π(0.030) 2 1 σm = σa = (−4018P) = −2009P 2 D+d T = fP 4 0.150 + 0.03 = 0.0135P Tmax = 0.3P 4 . From Fig. 7-21, qs = 0.95. Also, K ts is given as 1.8. Thus, K f s = 1 + qs (K ts − 1) = 1 + 0.95(1.8 − 1) = 1.76 16K f s T 16(1.76)(0.0135P) = = 4482P πd 3 π(0.03) 3 1 τa = τm = (4482P) = 2241P 2 1/2 2 σm = σm + 3τm2 = [(−2009P) 2 + 3(2241P) 2 ]1/2 = 4366P

τmax =

σa = σm = 4366P Se = 0.504(1000) = 504 MPa ka = 4.51(1000) −0.265 = 0.723 30 −0.107 = 0.864 kb = 7.62 kc = 0.85 (Note that torsion is accounted for in the von Mises stress.) Se = 0.723(0.864)(0.85)(504) = 267.6 MPa

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199

Chapter 7

σ 1 σa + m = Se Sut n

Modified Goodman:

4366P 1 4366P + = 6 6 267.6(10 ) 1000(10 ) 3

⇒

P = 16.1(103 ) N = 16.1 kN Ans.

1 σ + σm = a ny Sy

Yield:

ny =

800(106 ) = 5.69 Ans. 2(4366)(16.1)(103 )

(b) If the shaft is not rotating, τm = τa = 0. σm = σa = −2009P kb = 1 (axial) kc = 0.85 (Since there is no tension, kc = 1 might be more appropriate.) Se = 0.723(1)(0.85)(504) = 309.7 MPa ␴a 800

309.7

␴m

800

nf =

⇒

P=

309.7(106 ) = 51.4(103 ) N 3(2009) = 51.4 kN Ans.

800(106 ) = 3.87 Ans. ny = 2(2009)(51.4)(103 )

Yield: 7-28

309.7(106 ) 2009P

From Prob. 7-27, K f = 2.84, K f s = 1.76, Se = 267.6 MPa 4Pmax (4)(80)(10−3 ) = −321.4 MPa = −2.84 σmax = −K f πd 2 π(0.030) 2 20 (−321.4) = −80.4 MPa 80 D+d 0.150 + 0.03 3 = 0.3(80)(10 ) = 1080 N · m = f Pmax 4 4

σmin = Tmax

Tmin =

20 (1080) = 270 N · m 80

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τmax τmin

16Tmax 16(1080) −6 = Kfs = 1.76 (10 ) = 358.5 MPa πd 3 π(0.030) 3 20 = (358.5) = 89.6 MPa 80 321.4 − 80.4 = 120.5 MPa 2 −321.4 − 80.4 = −200.9 MPa = 2 358.5 − 89.6 = 134.5 MPa = 2 358.5 + 89.6 = 224.1 MPa = 2 1/2 = σa2 + 3τa2 = [120.52 + 3(134.5) 2 ]1/2 = 262.3 MPa

σa = σm τa τm σa

σm = [(−200.9) 2 + 3(224.1) 2 ]1/2 = 437.1 MPa Goodman:

σa 262.3 = 466.0 MPa (σa ) e = = 1 − σm /Sut 1 − 437.1/1000

Let f = 0.9 [0.9(1000)]2 = 2928 MPa a= 276.6 1 0.9(1000) b = − log = −0.1708 3 276.6 (σa ) e 1/b 466.0 1/−0.1708 N= = = 47 130 cycles Ans. a 2928 7-29

S y = 490 MPa, Sut = 590 MPa, Se = 200 MPa σm =

420 + 140 420 − 140 = 280 MPa , σa = = 140 MPa 2 2

Goodman: (σa ) e =

σa 140 = 266.5 MPa > Se = 1 − σm /Sut 1 − (280/590)

[0.9(590)]2 = 1409.8 MPa 200 1 0.9(590) b = − log = −0.141 355 3 200

a=

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201

Chapter 7

N=

266.5 1409.8

−1/0.143 55

= 131 200 cycles

Nremaining = 131 200 − 50 000 = 81 200 cycles 350 + (−200) = 75 MPa 2 350 − (−200) (σa ) 2 = = 275 MPa 2 275 (σa ) e2 = = 315.0 MPa 1 − (75/590) (σm ) 2 =

Second loading:

(a) Miner’s method

N2 =

n2 n1 + =1 N1 N2

315 1409.8

−1/0.141 355

⇒

= 40 200 cycles

n2 50 000 + =1 131 200 40 200

n 2 = 24 880 cycles Ans. (b) Manson’s method Two data points:

0.9(590 MPa), 103 cycles 266.5 MPa, 81 200 cycles

a2 (103 ) b2 0.9(590) = 266.5 a2 (81 200) b2 1.9925 = (0.012 315) b2 log 1.9925 = −0.156 789 b2 = log 0.012 315 266.5 a2 = = 1568.4 MPa (81 200) −0.156 789 315 1/−0.156 789 n2 = = 27 950 cycles Ans. 1568.4 7-30

(a) Miner’s method [0.9(76)]2 = 155.95 kpsi a= 30 1 0.9(76) b = − log = −0.119 31 3 30 1/−0.119 31 48 σ1 = 48 kpsi, N1 = = 19 460 cycles 155.95 1/−0.119 31 38 σ2 = 38 kpsi, N2 = = 137 880 cycles 155.95

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1/−0.119 31 32 σ3 = 32 kpsi, N3 = = 582 150 cycles 155.95 n1 n2 n3 + + =1 N1 N2 N3 60 000 n3 4000 + + = 1 ⇒ n 3 = 209 160 cycles Ans. 19 460 137 880 582 150 (b) Manson’s method

The life remaining after the first cycle is N R1 = 19 460 − 4000 = 15 460 cycles . The two data points required to define Se, 1 are [0.9(76), 103 ] and (48, 15 460) . 0.9(76) a2 (103 ) b2 = 48 a2 (15 460)

⇒

1.425 = (0.064 683) b2

log(1.425) = −0.129 342 log(0.064 683) 48 a2 = = 167.14 kpsi (15 460) −0.129 342 −1/0.129 342 38 = 94 110 cycles N2 = 167.14 b2 =

N R2 = 94 110 − 60 000 = 34 110 cycles 0.9(76) a3 (103 ) b3 = 38 a3 (34 110) b3 b3 =

7-31

⇒

1.8 = (0.029 317) b3

log 1.8 38 = −0.166 531, a3 = = 216.10 kpsi log(0.029 317) (34 110) −0.166 531 32 −1/0.166 531 N3 = = 95 740 cycles Ans. 216.1

Using Miner’s method a=

[0.9(100)]2 = 162 kpsi 50

0.9(100) 1 = −0.085 091 b = − log 3 50 70 1/−0.085 091 σ1 = 70 kpsi, N1 = = 19 170 cycles 162 55 1/−0.085 091 = 326 250 cycles σ2 = 55 kpsi, N2 = 162 σ3 = 40 kpsi,

N3 → ∞ 0.5N 0.3N 0.2N + + =1 19 170 326 250 ∞ N = 83 570 cycles Ans.

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Chapter 7

7-32

Given H B = 495LN(1, 0.03) Eq. (3-20)

Sut = 0.495 [LN(1, 0.041)]H B = 0.495 [LN(1, 0.041)][495 LN(1, 0.03)] ¯Sut = 0.495(495) = 245 kpsi

Table 2-6 for the COV of a product. . C x y = C x2 + C y2 = (0.0412 + 0.032 ) 1/2 = 0.0508 Sut = 245LN(1, 0.0508) kpsi From Table 7-13:

a = 1.34, b = −0.086, C = 0.12 −0.086 LN(1, 0.120) ka = 1.34 S¯ut

= 1.34(245) −0.086 LN(1, 0.12) = 0.835LN(1, 0.12) kb = 1.05 (as in Prob. 7-1) Se = 0.835LN(1, 0.12)(1.05)[107LN(1, 0.139)] S¯e = 0.835(1.05)(107) = 93.8 kpsi

Now

7-33

. C Se = (0.122 + 0.1392 ) 1/2 = 0.184 Se = 93.8LN(1, 0.184) kpsi Ans.

A Priori Decisions: • Material and condition: 1018 CD, Sut = 440LN(1, 0.03), and S y = 370LN(1, 0.061) MPa • Reliability goal: R = 0.999 (z = −3.09) • Function: Critical location—hole • Variabilities: Cka = 0.058 Ckc = 0.125 Cφ = 0.138 2 1/2 2 C Se = Cka + Ckc + Cφ2 = (0.0582 + 0.1252 + 0.1382 ) 1/2 = 0.195 Ckc = 0.10 C Fa = 0.20 Cσ a = (0.102 + 0.202 ) 1/2 = 0.234

2 C Se + Cσ2 a 0.1952 + 0.2342 = = 0.297 Cn = 1 + Cσ2 a 1 + 0.2342

203

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Resulting in a design factor n f of, Eq. (6-59): n f = exp[−(−3.09) ln(1 + 0.2972 ) + ln 1 + 0.2972 ] = 2.56 • Decision: Set n f = 2.56 Now proceed deterministically using the mean values: k¯a = 0.887, kb = 1, k¯c = 0.890, and from Prob. 7-10, K f = 2.09 F¯a F¯a S¯e = K¯ f = σ¯ a = K¯ f A t (60 − 12) n¯ f n¯ f K¯ f F¯a 2.56(2.09)(15.103 ) = 9.5 mm = ∴ t= (60 − 12)(175.7) (60 − 12) S¯e Decision: If 10 mm 1018 CD is available, t = 10 mm 7-34

Ans.

1.25"

1.00"

M

M

Rotation is presumed. M and Sut are given as deterministic, but notice that σ is not; therefore, a reliability estimation can be made. From Eq. (7-70): Se = 0.506(110)LN(1, 0.138) = 55.7LN(1, 0.138) kpsi Table 7-13: ka = 2.67(110) −0.265 LN(1, 0.058) = 0.768LN(1, 0.058) Based on d = 1 in, Eq. (7-19) gives kb =

1 0.30

−0.107

= 0.879

Conservatism is not necessary Se = 0.768[LN(1, 0.058)](0.879)(55.7)[LN(1, 0.138)] S¯e = 37.6 kpsi C Se = (0.0582 + 0.1382 ) 1/2 = 0.150 Se = 37.6LN(1, 0.150)

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205

Chapter 7

Fig. A-15-14: D/d = 1.25, r/d = 0.125. Thus K t = 1.70 and Eqs. (7-35), (7-78) and Table 7-8 give 1.70LN(1, 0.15) Kf = √ 1 + 2/ 0.125 [(1.70 − 1)/(1.70)](3/110) = 1.598LN(1, 0.15)

32M 32(1400) = 1.598[LN(1 − 0.15)] σ = Kf πd 3 π(1) 3 = 22.8LN(1, 0.15) kpsi From Eq. (6-57):

ln (37.6/22.8) (1 + 0.152 )/(1 + 0.152 ) z=− = −2.37 ln[(1 + 0.152 )(1 + 0.152 )]

From Table A-10, p f = 0.008 89 ∴ R = 1 − 0.008 89 = 0.991 Ans. Note: The correlation method uses only the mean of Sut ; its variability is already included in the 0.138. When a deterministic load, in this case M, is used in a reliability estimate, engineers state, “For a Design Load of M, the reliability is 0.991.” They are in fact referring to a Deterministic Design Load. 7-35

For completely reversed torsion, ka and kb of Prob. 7-34 apply, but kc must also be considered. kc = 0.328(110) 0.125 LN(1, 0.125) Eq. 7-74: = 0.590LN(1, 0.125) Note 0.590 is close to 0.577. S Se = ka kb kc Se = 0.768[LN(1, 0.058)](0.878)[0.590LN(1, 0.125)][55.7LN(1, 0.138)] S¯ Se = 0.768(0.878)(0.590)(55.7) = 22.2 kpsi C Se = (0.0582 + 0.1252 + 0.1382 ) 1/2 = 0.195 S Se = 22.2LN(1, 0.195) kpsi Fig. A-15-15: D/d = 1.25, r/d = 0.125, then K ts = 1.40. From Eqs. (7-35), (7-78) and Table 7-8 1.40LN(1, 0.15) = 1.34LN(1, 0.15) Kts = √ 1 + 2/ 0.125 [(1.4 − 1)/1.4](3/110) τ = Kts

16T πd 3

16(1.4) τ = 1.34[LN(1, 0.15)] π(1) 3 = 9.55LN(1, 0.15) kpsi

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From Eq. (6-57):

ln (22.2/9.55) (1 + 0.152 )/(1 + 0.1952 ) = −3.43 z=− ln [(1 + 0.1952 )(1 + 0.152 )]

From Table A-10, p f = 0.0003 R = 1 − p f = 1 − 0.0003 = 0.9997 Ans. For a design with completely-reversed torsion of 1400 lbf · in, the reliability is 0.9997. The improvement comes from a smaller stress-concentration factor in torsion. See the note at the end of the solution of Prob. 7-34 for the reason for the phraseology. 7-36

1" 1 D 4

1" D 8

Non-rotating

M

M

Sut = 58 kpsi Se = 0.506(58)LN(1, 0.138) = 29.3LN(1, 0.138) kpsi Table 7-13:

ka = 14.5(58) −0.719 LN(1, 0.11) = 0.782LN(1, 0.11)

Eq. (7-23): de = 0.37(1.25) = 0.463 in 0.463 −0.107 kb = = 0.955 0.30 Se = 0.782[LN(1, 0.11)](0.955)[29.3LN(1, 0.138)] S¯e = 0.782(0.955)(29.3) = 21.9 kpsi C Se = (0.112 + 0.1382 ) 1/2 = 0.150 Table A-16: d/D = 0, a/D = 0.1, A = 0.83 ∴ K t = 2.27. From Eqs. (7-35) and (7-78) and Table 7-8 2.27LN(1, 0.10) = 1.783LN(1, 0.10) Kf = √ 1 + 2/ 0.125 [(2.27 − 1)/2.27](5/58) Table A-16:

π AD 3 π(0.83)(1.253 ) = 0.159 in3 Z= = 2 3 32 M 1.6 = 1.783LN(1, 0.10) σ = Kf Z 0.159 = 17.95LN(1, 0.10) kpsi

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207

Chapter 7

σ¯ = 17.95 kpsi Cσ = 0.10 2 2 ln (21.9/17.95) (1 + 0.10 )/(1 + 0.15 ) = −1.07 z=− ln[(1 + 0.152 )(1 + 0.102 )]

Eq. (6-57):

p f = 0.1423

Table A-10:

R = 1 − p f = 1 − 0.1423 = 0.858 Ans. For a completely-reversed design load Ma of 1400 lbf · in, the reliability estimate is 0.858. 7-37

For a non-rotating bar subjected to completely reversed torsion of Ta = 2400 lbf · in From Prob. 7-36: Se = 29.3LN(1, 0.138) kpsi ka = 0.782LN(1, 0.11) kb = 0.955 For kc use Eq. (7-74): kc = 0.328(58) 0.125 LN(1, 0.125) = 0.545LN(1, 0.125) S Se = 0.782[LN(1, 0.11)](0.955)[0.545LN(1, 0.125)][29.3LN(1, 0.138)] S¯ Se = 0.782(0.955)(0.545)(29.3) = 11.9 kpsi C Se = (0.112 + 0.1252 + 0.1382 ) 1/2 = 0.216 Table A-16: d/D = 0, a/D = 0.1, A = 0.92, K ts = 1.68 From Eqs. (7-35), (7-78), Table 7-8 Kf s =

1.68LN(1, 0.10) √ 1 + 2/ 0.125 [(1.68 − 1)/1.68](5/58)

= 1.403LN(1, 0.10) Table A-16: Jnet =

π(0.92)(1.254 ) π AD 4 = = 0.2201 32 32

τa = K f s

Ta c Jnet

2.4(1.25/2) = 1.403[LN(1, 0.10)] 0.2201 = 9.56LN(1, 0.10) kpsi

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From Eq. (6-57):

ln (11.9/9.56) (1 + 0.102 )/(1 + 0.2162 ) = −0.85 z=− ln[(1 + 0.102 )(1 + 0.2162 )]

Table A-10, p f = 0.1977 R = 1 − p f = 1 − 0.1977 = 0.80 Ans. 7-38

This is a very important task for the student to attempt before starting Part 3. It illustrates the drawback of the deterministic factor of safety method. It also identifies the a priori decisions and their consequences. The range of force fluctuation in Prob. 7-23 is −16 to +4 kip, or 20 kip. Repeatedlyapplied Fa is 10 kip. The stochastic properties of this heat of AISI 1018 CD are given. Function

Consequences

Axial Fatigue load

Fa = 10 kip C Fa = 0 Ckc = 0.125 z = −3.09 C K f = 0.11

Overall reliability R ≥ 0.998; with twin fillets √ R ≥ 0.998 ≥ 0.999 Cold rolled or machined surfaces Ambient temperature Use correlation method Stress amplitude Significant strength Se

Cka = 0.058 Ckd Cφ CK f Cσ a C Se

=0 = 0.138 = 0.11 = 0.11 = (0.0582 + 0.1252 + 0.1382 ) 1/2 = 0.195

Choose the mean design factor which will meet the reliability goal

0.1952 + 0.112 = 0.223 Cn = 1 + 0.112 n¯ = exp −(−3.09) ln(1 + 0.2232 ) + ln 1 + 0.2232 n¯ = 2.02 Review the number and quantitative consequences of the designer’s a priori decisions to accomplish this. The operative equation is the definition of the design factor Se σa = n K¯ f Fa S¯e S¯e ⇒ = σ¯ a = n¯ w2 h n¯

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209

Chapter 7

Solve for thickness h. To do so we need −0.265 = 2.67(64) −0.265 = 0.887 k¯a = 2.67 S¯ut kb = 1 −0.078 k¯c = 1.23 S¯ut = 1.23(64) −0.078 = 0.889 k¯d = k¯e = 1 S¯e = 0.887(1)(0.889)(1)(1)(0.506)(64) = 25.5 kpsi

Fig. A-15-5: D = 3.75 in, d = 2.5 in, D/d = 3.75/2.5 = 1.5, r/d = 0.25/2.5 = 0.10 ∴ K t = 2.1 K¯ f = h=

2.1 = 1.857 √ 1 + 2/ 0.25 [(2.1 − 1)/(2.1)](4/64) K¯ f n¯ Fa 1.857(2.02)(10) = 0.667 Ans. = ¯ 2.5(25.5) w2 Se

This thickness separates S¯e and σ¯ a so as to realize the reliability goal of 0.999 at each shoulder. The design decision is to make t the next available thickness of 1018 CD steel strap from the same heat. This eliminates machining to the desired thickness and the extra cost of thicker work stock will be less than machining the fares. Ask your steel supplier what is available in this heat. 7-39

3" 4

1" 1 2

1200 lbf

1" 4

Fa = 1200 lbf Sut = 80 kpsi (a) Strength ka = 2.67(80) −0.265 LN(1, 0.058) = 0.836LN(1, 0.058) kb = 1 kc = 1.23(80) −0.078 LN(1, 0.125) = 0.874LN(1, 0.125) Sa = 0.506(80)LN(1, 0.138) = 40.5LN(1, 0.138) kpsi Se = 0.836[LN(1, 0.058)](1)[0.874LN(1, 0.125)][40.5LN(1, 0.138)] S¯e = 0.836(1)(0.874)(40.5) = 29.6 kpsi C Se = (0.0582 + 0.1252 + 0.1382 ) 1/2 = 0.195

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Stress: Fig. A-15-1; d/w = 0.75/1.5 = 0.5, K t = 2.17. From Eqs. (7-35), (7-78) and Table 7-8 2.17LN(1, 0.10) Kf = √ 1 + 2/ 0.375 [(2.17 − 1)/2.17](5/80) = 1.95LN(1, 0.10) K f Fa σa = , Cσ = 0.10 (w − d)t K¯ f Fa 1.95(1.2) = = 12.48 kpsi σ¯ a = (w − d)t (1.5 − 0.75)(0.25) S¯a = S¯e = 29.6 kpsi ln ( S¯a /σ¯ a ) 1 + Cσ2 1 + C S2 z=− ln 1 + Cσ2 1 + C S2 ln (29.6/12.48) (1 + 0.102 )/(1 + 0.1952 ) =− = −3.9 ln (1 + 0.102 )(1 + 0.1952 ) From Table A-20 p f = 4.481(10−5 ) R = 1 − 4.481(10−5 ) = 0.999 955 Ans. (b) All computer programs will differ in detail. 7-40

Each computer program will differ in detail. When the programs are working, the experience should reinforce that the decision regarding n¯ f is independent of mean values of strength, stress or associated geometry. The reliability goal can be realized by noting the impact of all those a priori decisions.

7-41

Such subprograms allow a simple call when the information is needed. The calling program is often named an executive routine (executives tend to delegate chores to others and only want the answers).

7-42

This task is similar to Prob. 7-41.

7-43

Again, a similar task.

7-44

The results of Probs. 7-41 to 7-44 will be the basis of a class computer aid for fatigue problems. The codes should be made available to the class through the library of the computer network or main frame available to your students.

7-45

Peterson’s notch sensitivity q has very little statistical basis. This subroutine can be used to show the variation in q, which is not apparent to those who embrace a deterministic q .

7-46

An additional program which is useful.

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Chapter 8 8-1 (a)

Thread depth = 2.5 mm

5 mm

Width = 2.5 mm Ans. dm = 25 − 1.25 − 1.25 = 22.5 mm dr = 25 − 5 = 20 mm l = p = 5 mm Ans.

2.5 2.5 mm 25 mm

(b)

Ans.

Thread depth = 2.5 mm Ans. Width at pitch line = 2.5 mm Ans. dm = 22.5 mm dr = 20 mm l = p = 5 mm Ans.

5 mm 2.5 5 mm

8-2 From Table 8-1, dr = d − 1.226 869 p dm = d − 0.649 519 p d − 1.226 869 p + d − 0.649 519 p = d − 0.938 194 p d¯ = 2 π π d¯2 = (d − 0.938 194 p) 2 Ans. At = 4 4 8-3 From Eq. (c) of Sec. 8-2, P=F T = e=

tan λ + f 1 − f tan λ

Pdm Fdm tan λ + f = 2 2 1 − f tan λ

T0 Fl/(2π) 1 − f tan λ 1 − f tan λ = = tan λ T Fdm /2 tan λ + f tan λ + f

Using f = 0.08, form a table and plot the efficiency curve. λ, deg. 0 10 20 30 40 45

1

e 0 0.678 0.796 0.838 0.8517 0.8519

e

0

50 ␭, deg.

Ans.

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8-4 Given F = 6 kN, l = 5 mm, and dm = 22.5 mm, the torque required to raise the load is found using Eqs. (8-1) and (8-6) 6(0.05)(40) 6(22.5) 5 + π(0.08)(22.5) + TR = 2 π(22.5) − 0.08(5) 2 = 10.23 + 6 = 16.23 N · m Ans. The torque required to lower the load, from Eqs. (8-2) and (8-6) is 6(0.05)(40) 6(22.5) π(0.08)22.5 − 5 + TL = 2 π(22.5) + 0.08(5) 2 = 0.622 + 6 = 6.622 N · m Ans. Since TL is positive, the thread is self-locking. The efficiency is e=

Eq. (8-4):

6(5) = 0.294 Ans. 2π(16.23)

8-5 Collar (thrust) bearings, at the bottom of the screws, must bear on the collars. The bottom segment of the screws must be in compression. Where as tension specimens and their grips must be in tension. Both screws must be of the same-hand threads. 8-6 Screws rotate at an angular rate of n=

1720 = 22.9 rev/min 75

(a) The lead is 0.5 in, so the linear speed of the press head is V = 22.9(0.5) = 11.5 in/min Ans. (b) F = 2500 lbf/screw dm = 3 − 0.25 = 2.75 in sec α = 1/cos(29/2) = 1.033 Eq. (8-5): 2500(2.75) TR = 2

0.5 + π(0.05)(2.75)(1.033) π(2.75) − 0.5(0.05)(1.033)

= 377.6 lbf · in

Eq. (8-6): Tc = 2500(0.06)(5/2) = 375 lbf · in Ttotal = 377.6 + 375 = 753 lbf · in/screw Tmotor = H=

753(2) = 21.1 lbf · in 75(0.95) Tn 21.1(1720) = = 0.58 hp Ans. 63 025 63 025

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Chapter 8

8-7 The force F is perpendicular to the paper. 1" 4 7" 16

3" D. 16 2.406"

3"

7 1 1 − − = 2.406 in 8 4 32 T = 2.406F 7 7 M= L− F = 2.406 − F = 2.188F 32 32 L =3−

S y = 41 kpsi σ = Sy =

32M 32(2.188) F = = 41 000 3 πd π(0.1875) 3

F = 12.13 lbf T = 2.406(12.13) = 29.2 lbf · in Ans. (b) Eq. (8-5), 2α = 60◦ , l = 1/14 = 0.0714 in, f = 0.075, sec α = 1.155, p = 1/14 in 7 1 − 0.649 519 = 0.3911 in dm = 16 14 Fclamp (0.3911) Num TR = 2 Den Num = 0.0714 + π(0.075)(0.3911)(1.155) Den = π(0.3911) − 0.075(0.0714)(1.155) T = 0.028 45Fclamp 29.2 T = = 1030 lbf Ans. Fclamp = 0.028 45 0.028 45 (c) The column has one end fixed and the other end pivoted. Base decision on the mean diameter column. Input: C = 1.2, D = 0.391 in, S y = 41 kpsi, E = 30(106 ) psi, L = 4.1875 in, k = D/4 = 0.097 75 in, L/k = 42.8. For this J. B. Johnson column, the critical load represents the limiting clamping force for bucking. Thus, Fclamp = Pcr = 4663 lbf. (d) This is a subject for class discussion. 8-8

T = 6(2.75) = 16.5 lbf · in 1 5 − = 0.5417 in 8 12 1 29◦ l = = 0.1667 in, α = = 14.5◦ , 6 2

dm =

sec 14.5◦ = 1.033

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0.1667 + π(0.15)(0.5417)(1.033) = 0.0696F T = 0.5417( F/2) π(0.5417) − 0.15(0.1667)(1.033)

Eq. (8-5): Eq. (8-6):

Tc = 0.15(7/16)( F/2) = 0.032 81F Ttotal = (0.0696 + 0.0328) F = 0.1024F F=

16.5 = 161 lbf Ans. 0.1024

8-9 dm = 40 − 3 = 37 mm, l = 2(6) = 12 mm From Eq. (8-1) and Eq. (8-6)

10(0.15)(60) 10(37) 12 + π(0.10)(37) + TR = 2 π(37) − 0.10(12) 2 = 38.0 + 45 = 83.0 N · m

Since n = V /l = 48/12 = 4 rev/s ω = 2πn = 2π(4) = 8π rad/s so the power is H = T ω = 83.0(8π) = 2086 W

Ans.

8-10 (a) dm = 36 − 3 = 33 mm, l = p = 6 mm From Eqs. (8-1) and (8-6) 0.09(90) F 33F 6 + π(0.14)(33) + T = 2 π(33) − 0.14(6) 2 = (3.292 + 4.050) F = 7.34F N · m ω = 2πn = 2π(1) = 2π rad/s H = Tω 3000 H = = 477 N · m T = ω 2π 477 F= = 65.0 kN Ans. 7.34 Fl 65.0(6) = = 0.130 Ans. (b) e = 2π T 2π(477) 8-11 1 = 2(0.5) + 0.25 = 1.25 in Ans. 4 (b) From Table A-32 the washer thickness is 0.109 in. Thus, (a) L T = 2D +

L G = 0.5 + 0.5 + 0.109 = 1.109 in (c) From Table A-31, H =

7 = 0.4375 in 16

Ans.

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Chapter 8

(d) L G + H = 1.109 + 0.4375 = 1.5465 in This would be rounded to 1.75 in per Table A-17. The bolt is long enough.

Ans.

(e) ld = L − L T = 1.75 − 1.25 = 0.500 in Ans. lt = L G − ld = 1.109 − 0.500 = 0.609 in Ans. These lengths are needed to estimate bolt spring rate kb . Note: In an analysis problem, you need not know the fastener’s length at the outset, although you can certainly check, if appropriate. 8-12 (a) L T = 2D + 6 = 2(14) + 6 = 34 mm

Ans.

(b) From Table A-33, the maximum washer thickness is 3.5 mm. Thus, the grip is, L G = 14 + 14 + 3.5 = 31.5 mm Ans. (c) From Table A-31, H = 12.8 mm (d) L G + H = 31.5 + 12.8 = 44.3 mm This would be rounded to L = 50 mm. The bolt is long enough. (e) ld = L − L T = 50 − 34 = 16 mm Ans.

Ans.

lt = L G − ld = 31.5 − 16 = 15.5 mm Ans. These lengths are needed to estimate the bolt spring rate kb . 8-13 (a) L T = 2D + (b) L G > h +

1 = 2(0.5) + 0.25 = 1.25 in 4

Ans.

0.5 d d = t1 + = 0.875 + = 1.125 in 2 2 2

Ans.

(c) L > h + 1.5d = t1 + 1.5d = 0.875 + 1.5(0.5) = 1.625 in From Table A-17, this rounds to 1.75 in. The cap screw is long enough. (d) ld = L − L T = 1.75 − 1.25 = 0.500 in Ans. lt = L G − ld = 1.125 − 0.5 = 0.625 in Ans. 8-14 (a) L T = 2(12) + 6 = 30 mm (b) L G = h +

Ans.

12 d d = t1 + = 20 + = 26 mm 2 2 2

Ans.

(c) L > h + 1.5d = t1 + 1.5d = 20 + 1.5(12) = 38 mm This rounds to 40 mm (Table A-17). The fastener is long enough. (d) ld = L − L T = 40 − 30 = 10 mm Ans. l T = L G − ld = 26 − 10 = 16 mm Ans.

Ans.

Ans.

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8-15 Ad = 0.7854(0.75) 2 = 0.442 in2 Atube = 0.7854(1.1252 − 0.752 ) = 0.552 in2

(a)

kb =

0.442(30)(106 ) Ad E = = 1.02(106 ) lbf/in grip 13

km =

0.552(30)(106 ) Atube E = = 1.27(106 ) lbf/in 13 13

C=

(b)

1.02 = 0.445 1.02 + 1.27

Ans. Ans.

Ans.

1 1 1 · = = 0.020 83 in 16 3 48 (13 − 0.020 83) |P|l |δb | = = |P| = 9.79(10−7 )|P| in AE 0.442(30)(106 ) δ=

Grip Original bolt A ␦

Nut advance ␦b

␦m

|δm | =

Equilibrium A

|P|(13) |P|l = = 7.85(10−7 )|P| in AE 0.552(30)(106 )

|δb | + |δm | = δ = 0.020 83 9.79(10−7 )|P| + 7.85(10−7 )|P| = 0.020 83 Fi = |P| =

0.020 83 = 11 810 lbf Ans. + 7.85(10−7 )

9.79(10−7 )

(c) At opening load P0 9.79(10−7 ) P0 = 0.020 83 P0 =

0.020 83 = 21 280 lbf Ans. 9.79(10−7 )

As a check use Fi = (1 − C) P0 P0 = 8-16

11 810 Fi = = 21 280 lbf 1−C 1 − 0.445

The movement is known at one location when the nut is free to turn δ = pt = t/N Letting Nt represent the turn of the nut from snug tight, Nt = θ/360◦ and δ = Nt /N. The elongation of the bolt δb is δb =

Fi kb

The advance of the nut along the bolt is the algebraic sum of |δb | and |δm |

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Chapter 8

Nt N Fi Fi Nt + = kb km N 1 1 kb + km Nt = N Fi = Fi N , + kb km kb km |δb | + |δm | =

θ 360◦

Ans.

As a check invert Prob. 8-15. What Turn-of-Nut will induce Fi = 11 808 lbf? 1 1 + Nt = 16(11 808) 1.02(106 ) 1.27(106 ) . = 0.334 turns = 1/3 turn (checks) The relationship between the Turn-of-Nut method and the Torque Wrench method is as follows. kb + km Fi N (Turn-of-Nut) Nt = kb km T = K Fi d Eliminate Fi

Nt =

(Torque Wrench)

kb + km kb km

θ NT = Kd 360◦

Ans.

8-17 (a) From Ex. 8-4, Fi = 14.4 kip, kb = 5.21(106 ) lbf/in, km = 8.95(106 ) lbf/in Eq. (8-27):

T = k Fi d = 0.2(14.4)(103 )(5/8) = 1800 lbf · in Ans.

From Prob. 8-16,

t = N Fi

1 1 + kb km

1 1 = 16(14.4)(10 ) + 6 5.21(10 ) 8.95(106 )

= 0.132 turns = 47.5◦

3

Ans.

Bolt group is (1.5)/(5/8) = 2.4 diameters. Answer is lower than RB&W recommendations. (b) From Ex. 8-5, Fi = 14.4 kip, kb = 6.78 Mlbf/in, and km = 17.4 Mlbf/in T = 0.2(14.4)(103 )(5/8) = 1800 lbf · in Ans. 1 1 3 t = 11(14.4)(10 ) + 6.78(106 ) 17.4(106 ) = 0.0325 = 11.7◦ 8-18

Ans.

Again lower than RB&W.

From Eq. (8-22) for the conical frusta, with d/l = 0.5 0.577π km = 1.11 = Ed (d/l)=0.5 2 ln {5[0.577 + 0.5(0.5)]/[0.577 + 2.5(0.5)]}

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Eq. (8-23), from the Wileman et al. finite element study, km = 0.787 15 exp[0.628 75(0.5)] = 1.08 Ed (d/l)=0.5 8-19

For cast iron, from Table 8-8 For gray iron: A = 0.778 71, B = 0.616 16

0.625 km = 12(10 )(0.625)(0.778 71) exp 0.616 16 = 7.55(106 ) lbf/in 1.5 6

This member’s spring rate applies to both members. We need km for the upper member which represents half of the joint. kci = 2km = 2[7.55(106 )] = 15.1(106 ) lbf/in For steel from Table 8-8: A = 0.787 15, B = 0.628 73 0.625 6 = 19.18(106 ) lbf/in km = 30(10 )(0.625)(0.787 15) exp 0.628 73 1.5 ksteel = 2km = 2(19.18)(106 ) = 38.36(106 ) lbf/in For springs in series 1 1 1 1 1 + = + = 6 km kci ksteel 15.1(10 ) 38.36(106 ) km = 10.83(106 ) lbf/in Ans. 8-20

The external tensile load per bolt is 1 π (150) 2 (6)(10−3 ) = 10.6 kN P= 10 4 Also, L G = 45 mm and from Table A-31, for d = 12 mm, H = 10.8 mm. No washer is specified. L T = 2D + 6 = 2(12) + 6 = 30 mm L G + H = 45 + 10.8 = 55.8 mm L = 60 mm ld = 60 − 30 = 30 mm lt = 45 − 30 = 15 mm

Table A-17:

Ad =

π(12) 2 = 113 mm2 4

At = 84.3 mm2

Table 8-1: Eq. (8-17): kb =

113(84.3)(207) = 466.8 MN/m 113(15) + 84.3(30)

Steel: Using Eq. (8-23) for A = 0.787 15, B = 0.628 73 and E = 207 GPa

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Chapter 8

Eq. (8-23):

km = 207(12)(0.787 15) exp[(0.628 73)(12/40)] = 2361 MN/m ks = 2km = 4722 MN/m

Cast iron: A = 0.778 71, B = 0.616 16, E = 100 GPa km = 100(12)(0.778 71) exp[(0.616 16)(12/40)] = 1124 MN/m kci = 2km = 2248 MN/m 1 1 1 = + km ks kci C=

⇒

km = 1523 MN/m

466.8 = 0.2346 466.8 + 1523

Table 8-1: At = 84.3 mm2 , Table 8-11, Sp = 600 MPa Fi = 0.75(84.3)(600)(10−3 ) = 37.9 kN Eqs. (8-30) and (8-31): Eq. (8-28): Sp At − Fi 600(10−3 )(84.3) − 37.9 = = 5.1 Ans. n= CP 0.2346(10.6) 8-21

Computer programs will vary.

8-22

D3 = 150 mm, A = 100 mm, B = 200 mm, C = 300 mm, D = 20 mm, E = 25 mm. ISO 8.8 bolts: d = 12 mm, p = 1.75 mm, coarse pitch of p = 6 MPa. 1 π (1502 )(6)(10−3 ) = 10.6 kN/bolt P= 10 4 L G = D + E = 20 + 25 = 45 mm L T = 2D + 6 = 2(12) + 6 = 30 mm Table A-31: H = 10.8 mm L G + H = 45 + 10.8 = 55.8 mm Table A-17: L = 60 mm D1 dw

20 2.5 45

22.5 25

ld = 60 − 30 = 30 mm, Table 8-1: At = 84.3 mm2

lt = 45 − 30 = 15 mm,

Ad = π(122 /4) = 113 mm2

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Eq. (8-17): kb =

113(84.3)(207) = 466.8 MN/m 113(15) + 84.3(30)

There are three frusta: dm = 1.5(12) = 18 mm D1 = (20 tan 30◦ )2 + dw = (20 tan 30◦ )2 + 18 = 41.09 mm Upper Frustum: t = 20 mm, E = 207 GPa, D = 1.5(12) = 18 mm k1 = 4470 MN/m

Eq. (8-20):

Central Frustum: t = 2.5 mm, D = 41.09 mm, E = 100 GPa (Table A-5) ⇒ k2 = 52 230 MN/m Lower Frustum: t = 22.5 mm, E = 100 GPa, D = 18 mm ⇒ k3 = 2074 MN/m km = [(1/4470) + (1/52 230) + (1/2074)]−1 = 1379 MN/m 466.8 C= = 0.253 Eq. (e), p. 421: 466.8 + 1379 Eqs. (8-30) and (8-31): From Eq. (8-18):

Fi = K Fp = K At Sp = 0.75(84.3)(600)(10−3 ) = 37.9 kN Eq. (8-28):

8-23

n=

Sp At − Fi 600(10−3 )(84.3) − 37.9 = = 4.73 Ans. Cm P 0.253(10.6)

1 π P= (1202 )(6)(10−3 ) = 8.48 kN 8 4 From Fig. 8-21, t1 = h = 20 mm and t2 = 25 mm l = 20 + 12/2 = 26 mm t = 0 (no washer), L T = 2(12) + 6 = 30 mm L > h + 1.5d = 20 + 1.5(12) = 38 mm Use 40 mm cap screws. ld = 40 − 30 = 10 mm lt = l − ld = 26 − 10 = 16 mm Ad = 113 mm , 2

12 13

At = 84.3 mm

2

h 20 l 26

Eq. (8-17):

7

13 6

113(84.3)(207) kb = 113(16) + 84.3(10)

t2 25

= 744 MN/m Ans. dw = 1.5(12) = 18 mm D = 18 + 2(6)(tan 30) = 24.9 mm

D

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Chapter 8

From Eq. (8-20): Top frustum: D = 18, t = 13, E = 207 GPa

⇒ k1 = 5316 MN/m

Mid-frustum: t = 7, E = 207 GPa, D = 24.9 mm

⇒ k2 = 15 620 MN/m

D = 18, t = 6, E = 100 GPa ⇒ k3 = 3887 MN/m 1 = 2158 MN/m Ans. km = (1/5316) + (1/55 620) + (1/3887)

Bottom frustum:

C=

744 = 0.256 Ans. 744 + 2158

From Prob. 8-22, Fi = 37.9 kN Sp At − Fi 600(0.0843) − 37.9 = = 5.84 Ans. n= CP 0.256(8.48) 8-24

Calculation of bolt stiffness: H = 7/16 in L T = 2(1/2) + 1/4 = 1 1/4 in L G = 1/2 + 5/8 + 0.095 = 1.22 in L > 1.125 + 7/16 + 0.095 = 1.66 in Use L = 1.75 in 1.454 1.327 3 4 1 2

0.61 1.22

5 8

3 4

0.095

0.860

ld = L − L T = 1.75 − 1.25 = 0.500 in lt = 1.125 + 0.095 − 0.500 = 0.72 in Ad = π(0.502 )/4 = 0.1963 in2 At = 0.1419 in2 (UNC) At E 0.1419(30) = 5.9125 Mlbf/in = lt 0.72 Ad E 0.1963(30) kd = = 11.778 Mlbf/in = ld 0.500 1 kb = = 3.936 Mlbf/in Ans. (1/5.9125) + (1/11.778) kt =

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Member stiffness for four frusta and joint constant C using Eqs. (8-20) and (e). ⇒ k1 = 33.30 Mlbf/in

Top frustum:

D = 0.75, t = 0.5, d = 0.5, E = 30

2nd frustum:

D = 1.327, t = 0.11, d = 0.5, E = 14.5

3rd frustum:

D = 0.860, t = 0.515, E = 14.5

Fourth frustum:

⇒ k2 = 173.8 Mlbf/in

⇒ k3 = 21.47 Mlbf/in

D = 0.75, t = 0.095, d = 0.5, E = 30 ⇒ k4 = 97.27 Mlbf/in −1 4 1/ki = 10.79 Mlbf/in Ans. km = i=1

C = 3.94/(3.94 + 10.79) = 0.267 Ans. 8-25 0.095"

0.4225" 0.845"

1.238" 0.5"

0.1725"

1.018"

0.595" Steel

0.25" 0.75"

kb =

0.625"

Cast iron

0.1419(30) At E = = 5.04 Mlbf/in Ans. l 0.845

From Fig. 8-21, h=

1 + 0.095 = 0.595 in 2

l=h+

0.5 d = 0.595 + = 0.845 2 2

D1 = 0.75 + 0.845 tan 30◦ = 1.238 in l/2 = 0.845/2 = 0.4225 in From Eq. (8-20): Frustum 1: D = 0.75, t = 0.4225 in, d = 0.5 in, E = 30 Mpsi ⇒ k1 = 36.14 Mlbf/in Frustum 2: Frustum 3:

D = 1.018 in, t = 0.1725 in, E = 70 Mpsi, d = 0.5 in ⇒ k2 = 134.6 Mlbf/in D = 0.75, t = 0.25 in, d = 0.5 in, E = 14.5 Mpsi ⇒ k3 = 23.49 Mlbf/in 1 km = = 12.87 Mlbf/in Ans. (1/36.14) + (1/134.6) + (1/23.49) C=

5.04 = 0.281 Ans. 5.04 + 12.87

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Chapter 8

8-26

Refer to Prob. 8-24 and its solution. Additional information: A = 3.5 in, Ds = 4.25 in, static pressure 1500 psi, Db = 6 in, C (joint constant) = 0.267, ten SAE grade 5 bolts. P=

1 π(4.252 ) (1500) = 2128 lbf 10 4

From Tables 8-2 and 8-9, At = 0.1419 in2 Sp = 85 000 psi Fi = 0.75(0.1419)(85) = 9.046 kip From Eq. (8-28), n= 8-27

Sp At − Fi 85(0.1419) − 9.046 = = 5.31 Ans. CP 0.267(2.128)

From Fig. 8-21, t1 = 0.25 in h l D1 D2

= 0.25 + 0.065 = 0.315 in = h + (d/2) = 0.315 + (3/16) = 0.5025 in = 1.5(0.375) + 0.577(0.5025) = 0.8524 in = 1.5(0.375) = 0.5625 in

l/2 = 0.5025/2 = 0.251 25 in Frustum 1: Washer 0.5625"

0.6375"

0.065"

E = 30 Mpsi, t = 0.065 in, D = 0.5625 in k = 78.57 Mlbf/in (by computer) Frustum 2: Cap portion 0.6375" 0.18625" 0.8524"

E = 14 Mpsi, t = 0.186 25 in D = 0.5625 + 2(0.065)(0.577) = 0.6375 in k = 23.46 Mlbf/in (by computer) Frustum 3: Frame and Cap 0.8524" 0.25125" 0.5625"

E = 14 Mpsi, t = 0.251 25 in, D = 0.5625 in k = 14.31 Mlbf/in (by computer) 1 = 7.99 Mlbf/in Ans. km = (1/78.57) + (1/23.46) + (1/14.31)

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For the bolt, L T = 2(3/8) + (1/4) = 1 in. So the bolt is threaded all the way. Since At = 0.0775 in2 kb =

0.0775(30) = 4.63 Mlbf/in Ans. 0.5025

8-28 (a) Fb = R Fb, max sin θ Half of the external moment is contributed by the line load in the interval 0 ≤ θ ≤ π. π π M 2 = Fb R sin θ dθ = Fb, max R 2 sin2 θ dθ 2 0 0 π M = Fb, max R 2 2 2 from which Fb, max = Fmax =

φ2

φ1

M π R2

Fb R sin θ

M dθ = π R2

φ2

φ1

R sin θ dθ =

M (cos φ1 − cos φ2 ) πR

Noting φ1 = 75◦ , φ2 = 105◦ 12 000 (cos 75◦ − cos 105◦ ) = 494 lbf Ans. π(8/2) 2M M 2π = ( R) Fmax = Fb, max R φ = 2 πR N RN 2(12 000) = 500 lbf Ans. Fmax = (8/2)(12)

Fmax = (b)

(c) F = Fmax sin θ M = 2Fmax R[(1) sin2 90◦ + 2 sin2 60◦ + 2 sin2 30◦ + (1) sin2 (0)] = 6Fmax R from which Fmax =

12 000 M = = 500 lbf Ans. 6R 6(8/2)

The simple general equation resulted from part (b) Fmax = 8-29

(a) Table 8-11: Eq. (8-30): Table (8-15): Eq. (8-27)

2M RN

Sp = 600 MPa Fi = 0.9At Sp = 0.9(245)(600)(10−3 ) = 132.3 kN K = 0.18 T = 0.18(132.3)(20) = 476 N · m Ans.

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225

Chapter 8

(b) Washers: t = 3.4 mm, d = 20 mm, D = 30 mm, E = 207 GPa ⇒ k1 = 42 175 MN/m Cast iron: t = 20 mm, d = 20 mm, D = 30 + 2(3.4) tan 30◦ = 33.93 mm, E = 135 GPa

⇒

k2 = 7885 MN/m

Steel: t = 20 mm, d = 20 mm, D = 33.93 mm, E = 207 GPa ⇒ k3 = 12 090 MN/m km = (2/42 175 + 1/7885 + 1/12 090) −1 = 3892 MN/m Bolt: L G = 46.8 mm. Nut: H = 18 mm. L > 46.8 + 18 = 64.8 mm. Use L = 80 mm. L T = 2(20) + 6 = 46 mm, ld = 80 − 46 = 34 mm, lt = 46.8 − 34 = 12.8 mm, At = 245 mm2 , kb =

Ad = π202 /4 = 314.2 mm2

Ad At E 314.2(245)(207) = 1290 MN/m = Ad lt + At ld 314.2(12.8) + 245(34)

C = 1290/(1290 + 3892) = 0.2489, n=

Sp = 600 MPa,

Fi = 132.3 kN

Sp At − Fi 600(0.245) − 132.3 = = 15.7 Ans. C( P/N ) 0.2489(15/4)

Bolts are a bit oversized for the load. 8-30

(a) ISO M 20 × 2.5 grade 8.8 coarse pitch bolts, lubricated. Table 8-2

At = 245 mm2

Table 8-11

Sp = 600 MPa Ad = π(20) 2 /4 = 314.2 mm2 Fp = 245(0.600) = 147 kN Fi = 0.90Fp = 0.90(147) = 132.3 kN T = 0.18(132.3)(20) = 476 N · m Ans.

(b) L ≥ L G + H = 48 + 18 = 66 mm . Therefore, set L = 80 mm per Table A-17. L T = 2D + 6 = 2(20) + 6 = 46 mm ld = L − L T = 80 − 46 = 34 mm lt = L G − ld = 48 − 34 = 14 mm 80 48 grip Not to scale 14 34

46

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kb =

Ad At E 314.2(245)(207) = 1251.9 MN/m = Ad lt + At ld 314.2(14) + 245(34) 30

24

24

30

Use Wileman et al. Eq. (8-23) A = 0.787 15, B = 0.628 73 km 20 Bd = 0.787 15 exp 0.628 73 = A exp = 1.0229 Ed LG 48 km = 1.0229(207)(20) = 4235 MN/m 1251.9 = 0.228 1251.9 + 4235 Bolts carry 0.228 of the external load; members carry 0.772 of the external load. Ans. Thus, the actual loads are C=

Fb = C P + Fi = 0.228(20) + 132.3 = 136.9 kN Fm = (1 − C) P − Fi = (1 − 0.228)20 − 132.3 = −116.9 kN 8-31

Given pmax = 6 MPa, pmin = 0 and from Prob. 8-20 solution, C = 0.2346, Fi = 37.9 kN, At = 84.3 mm2 . For 6 MPa, P = 10.6 kN per bolt σi =

Fi 37.9(103 ) = 450 MPa = At 84.3

Eq. (8-35): σa =

CP 0.2346(10.6)(103 ) = 14.75 MPa = 2At 2(84.3)

σm = σa + σi = 14.75 + 450 = 464.8 MPa (a) Goodman Eq. (8-40) for 8.8 bolts with Se = 129 MPa, Sut = 830 MPa Sa =

Se (Sut − σi ) 129(830 − 450) = 51.12 MPa = Sut + Se 830 + 129 nf =

Sa 51.12 = 3.47 Ans. = σa 14.75

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Chapter 8

(b) Gerber Eq. (8-42) 1

2 2 Sa = Sut Sut + 4Se (Se + σi ) − Sut − 2σi Se 2Se 1 = 830 8302 + 4(129)(129 + 450) − 8302 − 2(450)(129) 2(129) = 76.99 MPa nf =

76.99 = 5.22 Ans. 14.75

(c) ASME-elliptic Eq. (8-43) with Sp = 600 MPa Se 2 2 − σ2 − σ S Sa = 2 S + S S p i e p e i Sp + Se2

129 600 6002 + 1292 − 4502 − 450(129) = 65.87 MPa = 2 2 600 + 129 65.87 = 4.47 Ans. nf = 14.75 8-32 P= Table 8-11:

pA π D2 p π(0.92 )(550) = = = 9.72 kN/bolt N 4N 4(36)

Sp = 830 MPa, Sut = 1040 MPa, S y = 940 MPa At = 58 mm2

Table 8-1:

Ad = π(102 )/4 = 78.5 mm2 L G = D + E = 20 + 25 = 45 mm L T = 2(10) + 6 = 26 mm H = 8.4 mm

Table A-31:

L ≥ L G + H = 45 + 8.4 = 53.4 mm Choose L = 60 mm from Table A-17 ld = L − L T = 60 − 26 = 34 mm lt = L G − ld = 45 − 34 = 11 mm kb =

Ad At E 78.5(58)(207) = 332.4 MN/m = Ad lt + At ld 78.5(11) + 58(34) 15

20

22.5

2.5 25

22.5

10

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Frustum 1: Top, E = 207, t = 20 mm, d = 10 mm, D = 15 mm

k1 = ln

0.5774π(207)(10) 15 + 10 1.155(20) + 15 − 10 1.155(20) + 15 + 10 15 − 10

= 3503 MN/m Frustum 2: Middle, E = 96 GPa, D = 38.09 mm, t = 2.5 mm, d = 10 mm 0.5774π(96)(10) k2 = 38.09 + 10 1.155(2.5) + 38.09 − 10 ln 1.155(2.5) + 38.09 + 10 38.09 − 10 = 44 044 MN/m could be neglected due to its small influence on km . Frustum 3: Bottom, E = 96 GPa, t = 22.5 mm, d = 10 mm, D = 15 mm

k3 = ln

0.5774π(96)(10) 15 + 10 1.155(22.5) + 15 − 10 1.155(22.5) + 15 + 10 15 − 10

= 1567 MN/m 1 = 1057 MN/m (1/3503) + (1/44 044) + (1/1567) 332.4 C= = 0.239 332.4 + 1057

km =

Fi = 0.75At Sp = 0.75(58)(830)(10−3 ) = 36.1 kN Table 8-17: Se = 162 MPa σi =

Fi 36.1(103 ) = 622 MPa = At 58

(a) Goodman Eq. (8-40) Sa =

Se (Sut − σi ) 162(1040 − 622) = 56.34 MPa = Sut + Se 1040 + 162 nf =

56.34 = 2.82 Ans. 20

(b) Gerber Eq. (8-42) 1

2 2 Sa = + 4Se (Se + σi ) − Sut − 2σi Se Sut Sut 2Se 1 = 1040 10402 + 4(162)(162 + 622) − 10402 − 2(622)(162) 2(162) = 86.8 MPa

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Chapter 8

CP 0.239(9.72)(103 ) = 20 MPa σa = = 2At 2(58) nf =

Sa 86.8 = 4.34 Ans. = σa 20

(c) ASME elliptic Se 2 2 − σ2 − σ S S + S S p i e p e i Sp2 + Se2

162 2 + 1622 − 6222 − 622(162) = 84.90 MPa 830 830 = 8302 + 1622

Sa =

nf = 8-33

84.90 = 4.24 Ans. 20

Let the repeatedly-applied load be designated as P. From Table A-22, Sut = 93.7 kpsi. Referring to the Figure of Prob. 4-73, the following notation will be used for the radii of Section AA. ri = 1 in,

ro = 2 in,

rc = 1.5 in

From Table 4-5, with R = 0.5 in 0.52 rn = √ = 1.457 107 in 2 1.5 − 1.52 − 0.52 e = rc − rn = 1.5 − 1.457 107 = 0.042 893 in co = ro − rn = 2 − 1.457 109 = 0.542 893 in ci = rn − ri = 1.457 107 − 1 = 0.457 107 in A = π(12 )/4 = 0.7854 in2 If P is the maximum load M = Prc = 1.5P rc ci 1.5(0.457) P P 1+ 1+ = 21.62P = σi = A eri 0.7854 0.0429(1) σa = σm =

21.62P σi = = 10.81P 2 2

(a) Eye: Section AA ka = 14.4(93.7) −0.718 = 0.553 de = 0.37d = 0.37(1) = 0.37 in 0.37 −0.107 kb = = 0.978 0.30 kc = 0.85 Se = 0.504(93.7) = 47.2 kpsi Se = 0.553(0.978)(0.85)(47.2) = 21.7 kpsi

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Since no stress concentration exists, use a load line slope of 1. From Table 7-10 for Gerber   2 2 93.7  2(21.7)  −1 + 1 + = 20.65 kpsi Sa = 2(21.7) 93.7 Note the mere 5 percent degrading of Se in Sa 1910 Sa 20.65(103 ) = nf = = σa 10.81P P Thread: Die cut. Table 8-17 gives 18.6 kpsi for rolled threads. Use Table 8-16 to find Se for die cut threads Se = 18.6(3.0/3.8) = 14.7 kpsi Table 8-2: At = 0.663 in2 σ = P/At = P/0.663 = 1.51P σa = σm = σ/2 = 1.51P/2 = 0.755P From Table 7-10, Gerber Sa =



2

120  −1 + 2(14.7)

1+

2(14.7) 120

2

  = 14.5 kpsi

19 200 Sa 14 500 = = σa 0.755P P Comparing 1910/P with 19 200/P, we conclude that the eye is weaker in fatigue. Ans. nf =

(b) Strengthening steps can include heat treatment, cold forming, cross section change (a round is a poor cross section for a curved bar in bending because the bulk of the material is located where the stress is small). Ans. (c) For n f = 2 P=

8-34

1910 = 955 lbf, max. load Ans. 2

41 = 2.41 in . Use L = 2 12 in Ans. 64 (b) Four frusta: Two washers and two members

(a) L ≥ 1.5 + 2(0.134) +

1.125" 0.134" D1

1.280"

0.75"

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Chapter 8

Washer: E = 30 Mpsi, t = 0.134 in, D = 1.125 in, d = 0.75 in k1 = 153.3 Mlbf/in Eq. (8-20): Member: E = 16 Mpsi, t = 0.75 in, D = 1.280 in, d = 0.75 in k2 = 35.5 Mlbf/in Eq. (8-20): 1 = 14.41 Mlbf/in Ans. km = (2/153.3) + (2/35.5) Bolt: L T = 2(3/4) + 1/4 = 13/4 in L G = 2(0.134) + 2(0.75) = 1.768 in ld = L − L T = 2.50 − 1.75 = 0.75 in lt = L G − ld = 1.768 − 0.75 = 1.018 in At = 0.373 in2 (Table 8-2) Ad = π(0.75) 2 /4 = 0.442 in2 Ad At E 0.442(0.373)(30) = 6.78 Mlbf/in Ans. kb = = Ad lt + At ld 0.442(1.018) + 0.373(0.75) 6.78 C= = 0.320 Ans. 6.78 + 14.41 (c) From Eq. (8-40), Goodman with Se = 18.6 kpsi, Sut = 120 kpsi 18.6[120 − (25/0.373)] = 7.11 kpsi 120 + 18.6

Sa = The stress components are σa =

CP 0.320(6) = 2.574 kpsi = 2At 2(0.373)

σm = σa + nf =

Fi 25 = 69.6 kpsi = 2.574 + At 0.373 Sa 7.11 = 2.76 = σa 2.574

Ans.

(d) Eq. (8-42) for Gerber 1 25 25 Sa = − 1202 − 2 18.6 120 1202 + 4(18.6) 18.6 + 2(18.6) 0.373 0.373 = 10.78 kpsi nf = (e) n proof =

10.78 = 4.19 2.574

85 = 1.17 Ans. 2.654 + 69.8

Ans.

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8-35 At = 0.1419 in2 Sp = 85 kpsi, Sut = 120 kpsi Se = 18.6 kpsi

(a) Table 8-2: Table 8-9: Table 8-17:

Fi = 0.75At Sp = 0.75(0.1419)(85) = 9.046 kip c= σa =

4.94 = 0.236 4.94 + 15.97 CP 0.236P = 0.832P kpsi = 2At 2(0.1419)

Eq. (8-40) for Goodman criterion 18.6(120 − 9.046/0.1419) Sa = = 7.55 kpsi 120 + 18.6 nf =

Sa 7.55 =2 = σa 0.832P

⇒

P = 4.54 kip Ans.

(b) Eq. (8-42) for Gerber criterion 1 9.046 9.046 − 1202 − 2 18.6 Sa = 120 1202 + 4(18.6) 18.6 + 2(18.6) 0.1419 0.1419 = 11.32 kpsi nf =

Sa 11.32 =2 = σa 0.832P

From which P=

11.32 = 6.80 kip Ans. 2(0.832)

(c) σa = 0.832P = 0.832(6.80) = 5.66 kpsi σm = Sa + σa = 11.32 + 63.75 = 75.07 kpsi Load factor, Eq. (8-28) n=

Sp At − Fi 85(0.1419) − 9.046 = = 1.88 Ans. CP 0.236(6.80)

Separation load factor, Eq. (8-29) n= 8-36

Table 8-2: Table 8-9: Table 8-17:

9.046 Fi = = 1.74 Ans. (1 − C) P 6.80(1 − 0.236) At = 0.969 in2 (coarse) At = 1.073 in2 (fine) Sp = 74 kpsi, Sut = 105 kpsi Se = 16.3 kpsi

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Chapter 8

233

Coarse thread, UNC Fi = 0.75(0.969)(74) = 53.78 kip

Eq. (8-42):

σi =

Fi 53.78 = 55.5 kpsi = At 0.969

σa =

CP 0.30P = 0.155P kpsi = 2At 2(0.969)

Sa =

1 105 1052 + 4(16.3)(16.3 + 55.5) − 1052 − 2(55.5)(16.3) = 9.96 kpsi 2(16.3)

nf =

Sa 9.96 =2 = σa 0.155P

From which P=

9.96 = 32.13 kip Ans. 0.155(2)

Fine thread, UNF Fi = 0.75(1.073)(74) = 59.55 kip 59.55 σi = = 55.5 kpsi 1.073 0.32P σa = = 0.149P kpsi 2(1.073) Sa = 9.96 (as before) Sa 9.96 nf = =2 = σa 0.149P From which P=

9.96 = 33.42 kip Ans. 0.149(2)

Percent improvement 33.42 − 32.13 . (100) = 4% Ans. 32.13 8-37

For a M 30 × 3.5 ISO 8.8 bolt with P = 80 kN/bolt and C = 0.33 Table 8-1:

At = 561 mm2

Table 8-11:

Sp = 600 MPa Sut = 830 MPa

Table 8-17:

Se = 129 MPa

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Fi = 0.75(561)(10−3 )(600) = 252.45 kN σi =

252.45(10−3 ) = 450 MPa 561

σa =

CP 0.33(80)(103 ) = 23.53 MPa = 2At 2(561)

Eq. (8-42): Sa =

1 830 8302 + 4(129)(129 + 450) − 8302 − 2(450)(129) = 77.0 MPa 2(129)

Fatigue factor of safety nf =

Sa 77.0 = 3.27 Ans. = σa 23.53

Load factor from Eq. (8-28), n=

Sp At − Fi 600(10−3 )(561) − 252.45 = = 3.19 Ans. CP 0.33(80)

Separation load factor from Eq. (8-29), n=

252.45 Fi = = 4.71 Ans. (1 − C) P (1 − 0.33)(80)

8-38 (a) Table 8-2: Table 8-9: Table 8-17: Unthreaded grip

At = 0.0775 in2 Sp = 85 kpsi, Sut = 120 kpsi Se = 18.6 kpsi

π(0.375) 2 (30) Ad E kb = = = 0.245 Mlbf/in per bolt Ans. l 4(13.5) π π [( D + 2t) 2 − D 2 ] = (4.752 − 42 ) = 5.154 in2 4 4 5.154(30) 1 Am E = = 2.148 Mlbf/in/bolt. Ans. km = l 12 6

Am =

(b)

Fi = 0.75(0.0775)(85) = 4.94 kip σi = 0.75(85) = 63.75 kpsi 2000 π 2 P = pA = (4) = 4189 lbf/bolt 6 4 C=

0.245 = 0.102 0.245 + 2.148

σa =

CP 0.102(4.189) = 2.77 kpsi = 2At 2(0.0775)

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Chapter 8

235

Eq. (8-40) for Goodman Sa =

18.6(120 − 63.75) = 7.55 kpsi 120 + 18.6

nf =

Sa 7.55 = 2.73 Ans. = σa 2.77

(c) From Eq. (8-42) for Gerber fatigue criterion,

1 Sa = 120 1202 + 4(18.6)(18.6 + 63.75) − 1202 − 2(63.75)(18.6) 2(18.6) = 11.32 kpsi nf =

Sa 11.32 = 4.09 Ans. = σa 2.77

(d) Pressure causing joint separation from Eq. (8-29) Fi =1 (1 − C) P Fi 4.94 P= = = 5.50 kip 1−C 1 − 0.102 5500 P = 6 = 2626 psi Ans. p= A π(42 )/4 n=

8-39

This analysis is important should the initial bolt tension fail. Members: S y = 71 kpsi, Ssy = 0.577(71) = 41.0 kpsi. Bolts: SAE grade 8, S y = 130 kpsi, Ssy = 0.577(130) = 75.01 kpsi Shear in bolts

π(0.3752 ) = 0.221 in2 As = 2 4 As Ssy 0.221(75.01) = = 5.53 kip Fs = n 3

Bearing on bolts Ab = 2(0.375)(0.25) = 0.188 in2 Ab S yc 0.188(130) = = 12.2 kip Fb = n 2 Bearing on member Fb =

0.188(71) = 5.34 kip 2.5

Tension of members At = (1.25 − 0.375)(0.25) = 0.219 in2 0.219(71) = 5.18 kip 3 F = min(5.53, 12.2, 5.34, 5.18) = 5.18 kip Ans.

Ft =

The tension in the members controls the design.

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Members: S y = 32 kpsi Bolts: S y = 92 kpsi, Ssy = (0.577)92 = 53.08 kpsi Shear of bolts

π(0.375) 2 = 0.221 in2 As = 2 4

τ=

Fs 4 = 18.1 kpsi = As 0.221

n=

Ssy 53.08 = = 2.93 Ans. τ 18.1

Bearing on bolts Ab = 2(0.25)(0.375) = 0.188 in2 −4 = −21.3 kpsi σb = 0.188 n=

Sy 92 = = 4.32 Ans. |σb | |−21.3|

n=

S yc 32 = = 1.50 Ans. |σb | |−21.3|

Bearing on members

Tension of members At = (2.375 − 0.75)(1/4) = 0.406 in2 4 = 9.85 kpsi σt = 0.406 Sy 32 = 3.25 Ans. n= = At 9.85 8-41

Members: S y = 71 kpsi Bolts: S y = 92 kpsi, Ssy = 0.577(92) = 53.08 kpsi Shear of bolts F = Ssy A/n Fs =

53.08(2)(π/4)(7/8) 2 = 35.46 kip 1.8

Bearing on bolts Fb =

2(7/8)(3/4)(92) = 54.89 kip 2.2

Fb =

2(7/8)(3/4)(71) = 38.83 kip 2.4

Bearing on members

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Chapter 8

Tension in members Ft =

(3 − 0.875)(3/4)(71) = 43.52 kip 2.6

F = min(35.46, 54.89, 38.83, 43.52) = 35.46 kip 8-42

Ans.

Members: S y = 47 kpsi Bolts: S y = 92 kpsi, Ssy = 0.577(92) = 53.08 kpsi Shear of bolts π(0.75) 2 = 0.442 in2 4 20 = 15.08 kpsi τs = 3(0.442)

Ad =

n=

Ssy 53.08 = 3.52 Ans. = τs 15.08

Bearing on bolt 20 = −14.22 kpsi 3[(3/4) · (5/8)] Sy 92 = 6.47 Ans. n=− =− σb −14.22

σb = −

Bearing on members F 20 = −14.22 kpsi =− Ab 3[(3/4) · (5/8)] Sy 47 = 3.31 Ans. n=− =− σb 14.22

σb = −

Tension on members 20 F = = 6.10 kpsi A (5/8)[7.5 − 3(3/4)] Sy 47 = 7.71 Ans. n= = σt 6.10

σt =

8-43

Members: S y = 57 kpsi Bolts: S y = 92 kpsi, Ssy = 0.577(92) = 53.08 kpsi Shear of bolts

π(3/8) 2 As = 3 = 0.3313 in2 4 5.4 F = = 16.3 kpsi τs = A 0.3313 Ssy 53.08 n= = 3.26 Ans. = τs 16.3

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Bearing on bolt

5 3 = 0.3516 in2 Ab = 3 8 16 F 5.4 = −15.36 kpsi σb = − =− Ab 0.3516 Sy 92 = 5.99 Ans. n=− =− σb −15.36

Bearing on members Ab = 0.3516 in2 (From bearing on bolt calculations) σb = −15.36 kpsi (From bearing on bolt calculations) Sy 57 = 3.71 Ans. n=− =− σb −15.36 Tension in members Failure across two bolts

5 3 3 A= 2 −2 = 0.5078 in2 16 8 8 5.4 F = = 10.63 kpsi A 0.5078 Sy 57 = 5.36 Ans. n= = σt 10.63

σ =

8-44

2.8 kN

350

By symmetry, R1 = R2 = 1.4 kN

350

C

R2

R1

RA 200 1.4 kN

B

A 50

RB

M B = 0 1.4(250) − 50R A = 0 MA = 0

200(1.4) − 50R B = 0

Members: S y = 370 MPa Bolts: S y = 420 MPa, Ssy = 0.577(420) = 242.3 MPa Bolt shear:

π (102 ) = 78.54 mm2 4 7(103 ) = 89.13 MPa τ= 78.54

As =

n=

Ssy 242.3 = = 2.72 τ 89.13

⇒ ⇒

R A = 7 kN R B = 5.6 kN

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239

Chapter 8

Ab = td = 10(10) = 100 mm2

Bearing on member:

−7(103 ) = −70 MPa 100 Sy −370 = 5.29 n=− = σ −70

σb =

Strength of member M = 1.4(200) = 280 N · m At A, 1 I A = [10(503 ) − 10(103 )] = 103.3(103 ) mm4 12 Mc 280(25) (103 ) = 67.76 MPa σA = = 3 IA 103.3(10 ) Sy 370 = 5.46 n= = σA 67.76 At C, M = 1.4(350) = 490 N · m IC =

1 (10)(503 ) = 104.2(103 ) mm4 12

σC =

490(25) (103 ) = 117.56 MPa 3 104.2(10 )

n=

Sy 370 = 3.15 < 5.46 = σC 117.56

C more critical

n = min(2.72, 5.29, 3.15) = 2.72 Ans. 8-45

Fs P

Fs = 3000 lbf 3000(3) = 1286 lbf P= 7

3" 3000 lbf

7"

3"

Pivot about this point O

H= 1" 2

1 1 + + 0.095 = 1.095 in 2 2 L ≥ L G + H = 1.095 + (7/16) = 1.532 in l = LG =

l 1" 2

7 in 16

1

3" 4

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Use 1

3" bolts 4

1 = 2(0.5) + 0.25 = 1.25 in 4 = 1.75 − 1.25 = 0.5 = 1.095 − 0.5 = 0.595 π(0.5) 2 = 0.1963 in2 = 4 = 0.1419 in

L T = 2D + ld lt Ad At

Ad At E Ad lt + At ld 0.1963(0.1419)(30) = 0.1963(0.595) + 0.1419(0.5) = 4.451 Mlbf/in

kb =

0.75" t 0.5475"

0.5"

Two identical frusta A = 0.787 15, B = 0.628 73 d km = Ed A exp 0.628 73 L G 0.5 = 30(0.5)(0.787 15) exp 0.628 73 1.095 km = 15.733 Mlbf/in C=

4.451 = 0.2205 4.451 + 15.733

Sp = 85 kpsi Fi = 0.75(0.1419)(85) = 9.046 kip σi = 0.75(85) = 63.75 kpsi σb =

C P + Fi 0.2205(1.286) + 9.046 = 65.75 kpsi = At 0.1419

τs =

Fs 3 = 15.28 kpsi = As 0.1963

von Mises stress 1/2 = [65.742 + 3(15.282 )]1/2 = 70.87 kpsi σ = σb2 + 3τs2 Stress margin m = Sp − σ = 85 − 70.87 = 14.1 kpsi Ans.

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241

Chapter 8

8-46

2P(200) = 12(50)

2Fs 12 kN

2P 50

12(50) = 1.5 kN per bolt 2(200) = 6 kN/bolt = 380 MPa π = 245 mm2 , Ad = (202 ) = 314.2 mm2 4 = 0.75(245)(380)(10−3 ) = 69.83 kN

P=

200

Fs Sp

O

At Fi

69.83(103 ) σi = = 285 MPa 245 C P + Fi 0.30(1.5) + 69.83 σb = (103 ) = 287 MPa = At 245 Fs 6(103 ) = 19.1 MPa τ= = Ad 314.2 σ = [2872 + 3(19.12 )]1/2 = 289 MPa m = Sp − σ = 380 − 289 = 91 MPa Thus the bolt will not exceed the proof stress. 8-47

Ans.

Using the result of Prob. 6-29 for lubricated assembly 2π f T Fx = 0.18d n With a design factor of d gives 0.18(3)(1000)d 0.18n d Fx d = = 716d T = 2π f 2π(0.12) or T /d = 716. Also T = K (0.75Sp At ) d = 0.18(0.75)(85 000) At = 11 475At Form a table Size 1 4

28

5 16 3 8

24

24

At

T /d = 11 475At

n

0.0364

417.7

1.75

0.058

665.55

2.8

0.0878

1007.5

4.23

The factor of safety in the last column of the table comes from 2π f (T /d) 2π(0.12)(T /d) n= = 0.0042(T /d) = 0.18Fx 0.18(1000)

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Select a 38 " − 24 UNF capscrew. The setting is given by T = (11 475At )d = 1007.5(0.375) = 378 lbf · in Given the coarse scale on a torque wrench, specify a torque wrench setting of 400 lbf · in. Check the factor of safety 2π(0.12)(400) 2π f T = = 4.47 n= 0.18Fx d 0.18(1000)(0.375) 8-48

26

F'A

F'C

F'B 50

A

50 C

B M

F"A

F"C

152

Bolts: Sp = 380 MPa, S y = 420 MPa Channel: t = 6.4 mm, S y = 170 MPa Cantilever: S y = 190 MPa Nut: H = 10.8 mm

L T = 2(12) + 6 = 30 mm L > 12 + 6.4 + 10.8 = 29.2 mm Therefore, use L = 30 mm All threads, so At = 84.3 mm2 FA + FB + FC = F/3 M = (50 + 26 + 125) F = 201F 201F = 2.01F 2(50) 1 + 2.01 F = 2.343F FC = FC + FC = 3 FA = FC =

Bolts: The shear bolt area is As = At = 84.3 mm2 Ssy = 0.577(420) = 242.3 MPa Ssy 242.3(84.3)(10−3 ) As = = 3.11 kN F= n 2.343 2.8(2.343) Bearing on bolt: For a 12-mm bolt, dm = 12 − 0.649 519(1.75) = 10.86 mm Ab = tdm = (6.4)(10.86) = 69.5 mm2 Sy Ab 420 69.5(10−3 ) = = 4.45 kN F= n 2.343 2.8 2.343

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Bearing on member: Ab = 12(10.86) = 130.3 mm2 170 (130.3)(10−3 ) = 3.38 kN F= 2.8 2.343 Strength of cantilever: I =

1 (12)(503 − 123 ) = 1.233(105 ) mm4 12

I 1.233(105 ) = = 4932 c 25 4932(190) M = = 2.22 kN F= 151 2.8(151)(103 ) So F = 2.22 kN based on bending of cantilever 8-49

F = 4 kN; M = 12(200) = 2400 N · m 2400 = 37.5 kN FA = FB = 64 FA = FB = (4) 2 + (37.5) 2 = 37.7 kN FO = 4 kN

Ans. F'A 4 kN A

F"A 37.5 kN

Ans.

32

F'O 4 kN

Ans. O

Bolt shear: As =

π(12) 2 = 113 mm2 4

37.7(10) 3 = 334 MPa Ans. τ= 113

32

B

F'B 4 kN

F"B 37.5 kN

Bearing on member: Ab = 12(8) = 96 mm2 σ =−

37.7(10) 3 = −393 MPa Ans. 96

Bending stress in plate:

b

3 bh 3 bd 3 bd 2 − −2 + a bd I = 12 12 12 8(12) 3 8(136) 3 8(12) 3 2 − −2 + (32) (8)(12) = 12 12 12 = 1.48(10) 6 mm4

Ans.

M = 12(200) = 2400 N · m σ =

Mc 2400(68) = (10) 3 = 110 MPa Ans. I 1.48(10) 6

a h a

d

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M 16.5(300) 4950 lbf • in

F'A 150 lbf

300 lbf

16 1" 2

y

1 1" 2

A

F'B 150 lbf 1 1" 2

O V 300 lbf

F"A 1650 lbf

x

B

F"B 1650 lbf

4950 = 1650 lbf 3 FA = 1500 lbf, FB = 1800 lbf FA = FB =

Shear of bolt: π (0.5) 2 = 0.1963 in2 4 1800 F = = 9170 psi τ= A 0.1963

As =

Ssy = 0.577(92) = 53.08 kpsi

Bearing on bolt: 1 3 = 0.1875 in2 Ab = 2 8 σ =−

53.08 n= = 5.79 Ans. 9.17

n=

Bearing on members: S y = 54 kpsi, n =

1800 F =− = −9600 psi A 0.1875

92 = 9.58 Ans. 9.6

54 = 5.63 Ans. 9.6

Strength of members: Considering the right-hand bolt M = 300(15) = 4500 lbf · in I =

0.375(2) 3 0.375(0.5) 3 − = 0.246 in4 12 12

4500(1) Mc = = 18 300 psi I 0.246 54(10) 3 n= = 2.95 Ans. 18 300

1" 2

2"

σ =

8-51

3" 8

The direct shear load per bolt is F = 2500/6 = 417 lbf. The moment is taken only by the four outside bolts. This moment is M = 2500(5) = 12 500 lbf · in. Thus F =

12 500 = 1250 lbf and the resultant bolt load is 2(5) F = (417) 2 + (1250) 2 = 1318 lbf

Bolt strength, S y = 57 kpsi; Channel strength, S y = 46 kpsi; Plate strength, S y = 45.5 kpsi As = π(0.625) 2 /4 = 0.3068 in2

Shear of bolt: n=

Ssy (0.577)(57 000) = = 7.66 Ans. τ 1318/0.3068

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Bearing on bolt: Channel thickness is t = 3/16 in; Ab = (0.625)(3/16) = 0.117 in2 ; n = Bearing on channel: Bearing on plate:

n=

57 000 = 5.07 Ans. 1318/0.117

46 000 = 4.08 Ans. 1318/0.117

Ab = 0.625(1/4) = 0.1563 in2 n=

45 500 = 5.40 Ans. 1318/0.1563

1" 4

Strength of plate: 0.25(7.5) 3 0.25(0.625) 3 − I = 12 12 1 5 0.25(0.625) 3 2 + (2.5) = 6.821 in4 −2 12 4 8

5 "D 8

5"

7

1" 2

M = 6250 lbf · in per plate 6250(3.75) Mc = = 3436 psi I 6.821 45 500 = 13.2 Ans. n= 3436

σ =

8-52

Specifying bolts, screws, dowels and rivets is the way a student learns about such components. However, choosing an array a priori is based on experience. Here is a chance for students to build some experience.

8-53

Now that the student can put an a priori decision of an array together with the specification of fasteners.

8-54

A computer program will vary with computer language or software application.

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Chapter 9 9-1 Eq. (9-3): F = 0.707hlτ = 0.707(5/16)(4)(20) = 17.7 kip Ans. 9-2 Table 9-6: τall = 21.0 kpsi f = 14.85h kip/in = 14.85(5/16) = 4.64 kip/in F = f l = 4.64(4) = 18.56 kip Ans. 9-3 Table A-20: 1018 HR: Sut = 58 kpsi, S y = 32 kpsi 1018 CR: Sut = 64 kpsi, S y = 54 kpsi Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: τall = min(0.30Sut , 0.40S y ) = min[0.30(58), 0.40(32)] = min(17.4, 12.8) = 12.8 kpsi for both materials. Eq. (9-3):

F = 0.707hlτall F = 0.707(5/16)(4)(12.8) = 11.3 kip Ans.

9-4 Eq. (9-3)

√ √ 2F 2(32) = = 18.1 kpsi τ= hl (5/16)(4)(2)

Ans.

9-5 b = d = 2 in F 1.414 7"

(a) Primary shear

Table 9-1 τ y =

F V = = 1.13F kpsi A 1.414(5/16)(2)

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Secondary shear

Table 9-1 Ju =

2[(3)(22 ) + 22 ] d(3b2 + d 2 ) = = 5.333 in3 6 6

J = 0.707h Ju = 0.707(5/16)(5.333) = 1.18 in4 τx = τ y =

Mr y 7F(1) = = 5.93F kpsi J 1.18

Maximum shear τmax = τx2 + (τ y + τ y ) 2 = F 5.932 + (1.13 + 5.93) 2 = 9.22F kpsi F=

20 τall = = 2.17 kip Ans. 9.22 9.22

(1)

(b) For E7010 from Table 9-6, τall = 21 kpsi Table A-20: HR 1020 Bar:

Sut = 55 kpsi, S y = 30 kpsi

HR 1015 Support:

Sut = 50 kpsi, S y = 27.5 kpsi

Table 9-5, E7010 Electrode: Sut = 70 kpsi, S y = 57 kpsi Therefore, the bar controls the design. Table 9-4: τall = min[0.30(50), 0.40(27.5)] = min[15, 11] = 11 kpsi The allowable load from Eq. (1) is F=

11 τall = = 1.19 kip Ans. 9.22 9.22

9-6 b = d = 2 in F

7"

Primary shear τ y =

F V = = 0.566F A 4(0.707)(5/16)(2)

Secondary shear Table 9-1:

Ju =

(2 + 2) 3 (b + d) 3 = = 10.67 in3 6 6

J = 0.707h Ju = 0.707(5/16)(10.67) = 2.36 in4 τx = τ y =

Mr y (7F)(1) = = 2.97F J 2.36

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Maximum shear τmax

= τx2 + (τ y + τ y ) 2 = F 2.972 + (0.556 + 2.97) 2 = 4.61F kpsi

F=

τall 4.61

Ans.

which is twice τmax /9.22 of Prob. 9-5. 9-7 Weldment, subjected to alternating fatigue, has throat area of A = 0.707(6)(60 + 50 + 60) = 721 mm2 Members’ endurance limit: AISI 1010 steel Sut = 320 MPa,

Se = 0.504(320) = 161.3 MPa

ka = 272(320) −0.995 = 0.875 kb = 1 (direct shear) kc = 0.59 (shear) kd = 1 kf =

1 1 = 0.370 = Kfs 2.7

Sse = 0.875(1)(0.59)(0.37)(161.3) = 30.81 MPa Electrode’s endurance: 6010 Sut = 62(6.89) = 427 MPa Se = 0.504(427) = 215 MPa ka = 272(427) −0.995 = 0.657 kb = 1 (direct shear) kc = 0.59 (shear) kd = 1 k f = 1/K f s = 1/2.7 = 0.370 . Sse = 0.657(1)(0.59)(0.37)(215) = 30.84 MPa = 30.81 Thus, the members and the electrode are of equal strength. For a factor of safety of 1, Fa = τa A = 30.8(721)(10−3 ) = 22.2 kN Ans.

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τ = 0

9-8 Primary shear

249

(why?)

Secondary shear Ju = 2πr 3 = 2π(4) 3 = 402 cm3

Table 9-1:

J = 0.707h Ju = 0.707(0.5)(402) = 142 cm4 M = 200F N · m ( F in kN) (200F)(4) Mr = = 2.82F (2 welds) 2J 2(142) τall 140 F = = = 49.2 kN Ans. τ 2.82

τ =

9-9

2 a 3 /12 a2 a Ju = = = 0.0833 fom = lh ah 12h h 2 a2 a(3a 2 + a 2 ) a = = 0.3333 fom = 6(2a)h 3h h 2 5a 2 a (2a) 4 − 6a 2 a 2 = = 0.2083 fom = 12(a + a)2ah 24h h 2 a4 11 a 2 a 1 8a 3 + 6a 3 + a 3 − = = 0.3056 fom = 3ah 12 2a + a 36 h h 2 8a 3 a2 a (2a) 3 1 = = = 0.3333 fom = 6h 4a 24ah 3h h 2 a3 a2 a 2π(a/2) 3 = = = 0.25 fom = πah 4ah 4h h

Rank 5

1

4

2

1

3

These rankings apply to fillet weld patterns in torsion that have a square area a × a in which to place weld metal. The object is to place as much metal as possible to the border. If your area is rectangular, your goal is the same but the rankings may change. Students will be surprised that the circular weld bead does not rank first. 9-10 1 Iu = fom = lh a

fom =

1 Iu = lh 2ah

fom =

1 Iu = lh 2ah

2 1 1 a2 a = = 0.0833 h 12 h h a3 a2 = 0.0833 6 h 2 2 2 1 a a a = = 0.25 2 4 h h

a3 12

5 5 1

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1 Iu = fom = lh [2(2a)]h

2 1 a2 a2 a (3a + a) = = 0.1667 6 6 h h

d2 a2 a = = b + 2d 3a 3 2 2a 3 2a 3 a3 2d 3 a2 a 2 a Iu = − 2d + (b + 2d) = − + 3a = 3 3 9 3 3 9 3 a 3 /3 1 a2 a2 Iu = = = 0.1111 fom = lh 3ah 9 h h 3 πa Iu = πr 3 = 8 2 πa 3 /8 a2 a Iu = = = 0.125 fom = lh πah 8h h x¯ =

b a = , 2 2

2

y¯ =

4

3

The CEE-section pattern was not ranked because the deflection of the beam is out-of-plane. If you have a square area in which to place a fillet weldment pattern under bending, your objective is to place as much material as possible away from the x-axis. If your area is rectangular, your goal is the same, but the rankings may change. 9-11

Materials: Attachment (1018 HR) S y = 32 kpsi, Sut = 58 kpsi S y = 36 kpsi, Sut ranges from 58 to 80 kpsi, use 58. Member (A36) The member and attachment are weak compared to the E60XX electrode. Decision Specify E6010 electrode Controlling property:

τall = min[0.3(58), 0.4(32)] = min(16.6, 12.8) = 12.8 kpsi

For a static load the parallel and transverse fillets are the same. If n is the number of beads, F = τall n(0.707)hl F 25 = 0.921 nh = = 0.707lτall 0.707(3)(12.8) τ=

Make a table. Number of beads n

Leg size h

1 2 3 4

0.921 0.460 → 1/2" 0.307 → 5/16" 0.230 → 1/4"

Decision: Specify 1/4" leg size Decision: Weld all-around

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Weldment Specifications: Pattern: All-around square Electrode: E6010 Type: Two parallel fillets Two transverse fillets Length of bead: 12 in Leg: 1/4 in

251

Ans.

For a figure of merit of, in terms of weldbead volume, is this design optimal? 9-12

Decision: Choose a parallel fillet weldment pattern. By so-doing, we’ve chosen an optimal pattern (see Prob. 9-9) and have thus reduced a synthesis problem to an analysis problem: A = 1.414hd = 1.414(h)(3) = 4.24h in3

Table 9-1: Primary shear

τ y =

3000 707 V = = A 4.24h h

Secondary shear Table 9-1:

Ju =

3[3(32 ) + 32 ] d(3b2 + d 2 ) = = 18 in3 6 6

J = 0.707(h)(18) = 12.7h in4 Mr y 3000(7.5)(1.5) 2657 = = = τ y J 12.7h h 1 4287 = τx2 + (τ y + τ y ) 2 = 26572 + (707 + 2657) 2 = h h

τx = τmax

Attachment (1018 HR): S y = 32 kpsi, Sut = 58 kpsi Member (A36): S y = 36 kpsi The attachment is weaker Decision: Use E60XX electrode τall = min[0.3(58), 0.4(32)] = 12.8 kpsi τmax = τall = h=

4287 = 12 800 psi h

4287 = 0.335 in 12 800

Decision: Specify 3/8" leg size Weldment Specifications: Pattern: Parallel fillet welds Electrode: E6010 Type: Fillet Length of bead: 6 in Leg size: 3/8 in

Ans.

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An optimal square space (3" × 3" ) weldment pattern is or or . In Prob. 9-12, there was roundup of leg size to 3/8 in. Consider the member material to be structural A36 steel. Decision: Use a parallel horizontal weld bead pattern for welding optimization and convenience. Materials: Attachment (1018 HR): S y = 32 kpsi, Sut = 58 kpsi Member (A36): S y = 36 kpsi, Sut 58–80 kpsi; use 58 kpsi From Table 9-4 AISC welding code, τall = min[0.3(58), 0.4(32)] = min(16.6, 12.8) = 12.8 kpsi Select a stronger electrode material from Table 9-3. Decision: Specify E6010 Throat area and other properties: A = 1.414hd = 1.414(h)(3) = 4.24h in2 x¯ = b/2 = 3/2 = 1.5 in y¯ = d/2 = 3/2 = 1.5 in 3[3(32 ) + 32 ] d(3b2 + d 2 ) = = 18 in3 6 6 J = 0.707h Ju = 0.707(h)(18) = 12.73h in4

Ju =

Primary shear: τx =

3000 707.5 V = = A 4.24h h ␶x ␶y rx

r

ry

␶x

␶ y

x

Secondary shear: τ =

Mr J

Mr Mr x cos 45◦ = J J 2651 3000(6 + 1.5)(1.5) τx = = 12.73h h 2651 τ y = τx = h

τx = τ cos 45◦ =

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Chapter 9

τmax

= (τx + τx ) 2 + τ y2 1 (2651 + 707.5) 2 + 26512 h 4279 psi = h =

Relate stress and strength: τmax = τall 4279 = 12 800 h 4279 h= = 0.334 in → 3/8 in 12 800 Weldment Specifications: Pattern: Horizontal parallel weld tracks Electrode: E6010 Type of weld: Two parallel fillet welds Length of bead: 6 in Leg size: 3/8 in Additional thoughts: Since the round-up in leg size was substantial, why not investigate a backward C weld pattern. One might then expect shorter horizontal weld beads which will have the advantage of allowing a shorter member (assuming the member has not yet been designed). This will show the inter-relationship between attachment design and supporting members. 9-14

Materials: S y = 36 kpsi, Sut = 58 to 80 kpsi; use Sut = 58 kpsi Member (A36): Attachment (1018 HR): S y = 32 kpsi, Sut = 58 kpsi τall = min[0.3(58), 0.4(32)] = 12.8 kpsi Decision: Use E6010 electrode. From Table 9-3: S y = 50 kpsi, Sut = 62 kpsi, τall = min[0.3(62), 0.4(50)] = 20 kpsi Decision: Since A36 and 1018 HR are weld metals to an unknown extent, use τall = 12.8 kpsi Decision: Use the most efficient weld pattern – square, weld-all-around. Choose6" × 6" size. Attachment length: l1 = 6 + a = 6 + 6.25 = 12.25 in Throat area and other properties: A = 1.414h(b + d) = 1.414(h)(6 + 6) = 17.0h x¯ =

6 b = = 3 in, 2 2

y¯ =

d 6 = = 3 in 2 2

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Primary shear τ y =

F 20 000 1176 V = = = psi A A 17h h

Secondary shear Ju =

(6 + 6) 3 (b + d) 3 = = 288 in3 6 6

J = 0.707h(288) = 203.6h in4 Mr y 20 000(6.25 + 3)(3) 2726 = = psi J 203.6h h 1 4760 psi = τx2 + (τ y + τ y ) 2 = 27262 + (2726 + 1176) 2 = h h

τx = τ y = τmax

Relate stress to strength τmax = τall 4760 = 12 800 h 4760 h= = 0.372 in 12 800 Decision: Specify 3/8 in leg size Specifications: Pattern: All-around square weld bead track Electrode: E6010 Type of weld: Fillet Weld bead length: 24 in Leg size: 3/8 in Attachment length: 12.25 in 9-15

This is a good analysis task to test the students’ understanding (1) Solicit information related to a priori decisions. (2) Solicit design variables b and d. (3) Find h and round and output all parameters on a single screen. Allow return to Step 1 or Step 2 (4) When the iteration is complete, the final display can be the bulk of your adequacy assessment. Such a program can teach too.

9-16

The objective of this design task is to have the students teach themselves that the weld patterns of Table 9-3 can be added or subtracted to obtain the properties of a comtemplated weld pattern. The instructor can control the level of complication. I have left the

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Chapter 9

presentation of the drawing to you. Here is one possibility. Study the problem’s opportunities, then present this (or your sketch) with the problem assignment. Section AA b1

A 1" 2 8"

d

1018 HR A

A36 a

10000 lbf

8"

b

Body welds not shown

Attachment weld pattern considered

Use b1 as the design variable. Express properties as a function of b1 . From Table 9-3, category 3: A = 1.414h(b − b1 ) x¯ = b/2,

y¯ = d/2

(b − b1 )d 2 bd 2 b1 d 2 − = 2 2 2 I = 0.707h Iu F V = τ = A 1.414h(b − b1 ) Iu =

Fa(d/2) Mc = I 0.707h Iu = τ 2 + τ 2

τ = τmax Parametric study

Let a = 10 in, b = 8 in, d = 8 in, b1 = 2 in, τall = 12.8 kpsi, l = 2(8 − 2) = 12 in A = 1.414h(8 − 2) = 8.48h in2 Iu = (8 − 2)(82 /2) = 192 in3 I = 0.707(h)(192) = 135.7h in4 τ =

1179 10 000 = psi 8.48h h

2948 10 000(10)(8/2) = psi 135.7h h 1 3175 = 12 800 = 11792 + 29482 = h h

τ = τmax

from which h = 0.248 in. Do not round off the leg size – something to learn. fom =

192 Iu = = 64.5 hl 0.248(12)

A = 8.48(0.248) = 2.10 in2 I = 135.7(0.248) = 33.65 in4

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vol = I = vol τ = τ = τmax =

0.2482 h2 l= 12 = 0.369 in3 2 2 33.65 = 91.2 = eff 0.369 1179 = 4754 psi 0.248 2948 = 11 887 psi 0.248 4127 . = 12 800 psi 0.248

Now consider the case of uninterrupted welds, b1 = 0 A = 1.414(h)(8 − 0) = 11.31h Iu = (8 − 0)(82 /2) = 256 in3 I = 0.707(256)h = 181h in4 884 10 000 = 11.31h h 2210 10 000(10)(8/2) τ = = 181h h 1 2380 τmax = = τall 8842 + 22102 = h h τmax 2380 = 0.186 in h= = τall 12 800 τ =

Do not round off h. A = 11.31(0.186) = 2.10 in2 I = 181(0.186) = 33.67 0.1862 884 = 4753 psi, vol = 16 = 0.277 in3 τ = 0.186 2 2210 = 11 882 psi τ = 0.186 256 Iu fom = = = 86.0 hl 0.186(16) I 33.67 eff = 2 = = 121.7 (h /2)l (0.1862 /2)16 Conclusions: To meet allowable stress limitations, I and A do not change, nor do τ and σ . To meet the shortened bead length, h is increased proportionately. However, volume of bead laid down increases as h 2 . The uninterrupted bead is superior. In this example, we did not round h and as a result we learned something. Our measures of merit are also sensitive to rounding. When the design decision is made, rounding to the next larger standard weld fillet size will decrease the merit.

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Chapter 9

Had the weld bead gone around the corners, the situation would change. Here is a followup task analyzing an alternative weld pattern. b1

d1

d

b

9-17

From Table 9-2 A = 1.414h(b + d)

For the box

Subtracting b1 from b and d1 from d A = 1.414 h(b − b1 + d − d1 ) d 3 b1 d 2 d2 (3b + d) − 1 − 6 6 2 1 1 = (b − b1 )d 2 + d 3 − d13 2 6

Iu =

length of bead

l = 2(b − b1 + d − d1 ) fom = Iu / hl

9-18

Below is a Fortran interactive program listing which, if imitated in any computer language of convenience, will reduce to drudgery. Furthermore, the program allows synthesis by interaction or learning without fatigue. C Weld2.f for rect. fillet beads resisting bending C. Mischke Oct 98 1 print*,’weld2.f rectangular fillet weld-beads in bending,’ print*,’gaps allowed - C. Mischke Oct. 98’ print*,’ ’ print*,’Enter largest permissible shear stress tauall’ read*,tauall print*,’Enter force F and clearance a’ read*,F,a 2 print*,’Enter width b and depth d of rectangular pattern’ read*,b,d xbar=b/2. ybar=d/2. 3 print*,’Enter width of gap b1, and depth of gap d1’ print*,’both gaps central in their respective sides’ read*,b1,d1 xIu=(b-b1)*d**2/2.+(d**3-d1**3)/6. xl=2.*(d-d1)+2.*(b-b1)

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C Following calculations based on unit leg h = 1 AA=1.414*(b-b1+d-d1) xI=0.707*xIu tau2=F*a*d/2./xI tau1=F/AA taumax=sqrt(tau2**2+tau1**2) h=taumax/tauall C Adjust parameters for now-known h AA=AA*h xI=xI*h tau2=tau2/h tau1=tau1/h taumax=taumax/h fom=xIu/h/xl print*,’F=’,F,’ a=’,a,’ bead length = ’,xl print*,’b=’,b,’ b1=’,b1,’ d=’,d,’ d1=’,d1 print*,’xbar=’,xbar,’ ybar=’,ybar,’ Iu=’,xIu print*,’I=’,xI,’ tau2=’,tau2,’ tau1=’,tau1 print*,’taumax=’,taumax,’ h=’,h,’ A=’,AA print*,’fom=’,fom, ’I/(weld volume)=’,2.*xI/h**2/xl print*,’ ’ print*,’To change b1,d1: 1; b,d: 2; New: 3; Quit: 4’ read*,index go to (3,2,1,4), index 4 call exit end

9-19

τall = 12 800 psi. Use Fig. 9-17(a) for general geometry, but employ beads. Horizontal parallel weld bead pattern 6"

8"

b = 6 in d = 8 in

From Table 9-2, category 3 A = 1.414 hb = 1.414(h)(6) = 8.48 h in2 x¯ = b/2 = 6/2 = 3 in, y¯ = d/2 = 8/2 = 4 in 6(8) 2 bd 2 = = 192 in3 Iu = 2 2 I = 0.707h Iu = 0.707(h)(192) = 135.7h in4 τ =

1179 10 000 = psi 8.48h h

beads and then

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Chapter 9

10 000(10)(8/2) 2948 Mc = = psi I 135.7h h 1 3175 psi = τ 2 + τ 2 = (11792 + 29482 ) 1/2 = h h

τ = τmax

Equate the maximum and allowable shear stresses. τmax = τall =

3175 = 12 800 h

from which h = 0.248 in. It follows that I = 135.7(0.248) = 33.65 in4 The volume of the weld metal is 0.2482 (6 + 6) h 2l = = 0.369 in3 vol = 2 2 The effectiveness, (eff) H , is 33.65 I = = 91.2 in vol 0.369 192 Iu = = 64.5 in (fom ) H = hl 0.248(6 + 6) Vertical parallel weld beads (eff) H =

6" 8"

b = 6 in d = 8 in

From Table 9-2, category 2 A = 1.414hd = 1.414(h)(8) = 11.31h in2 x¯ = b/2 = 6/2 = 3 in,

y¯ = d/2 = 8/2 = 4 in

83 d3 = = 85.33 in3 Iu = 6 6 I = 0.707h Iu = 0.707(h)(85.33) = 60.3h τ =

10 000(10)(8/2) 6633 Mc = = psi I 60.3 h h 1 = τ 2 + τ 2 = (8842 + 66332 ) 1/2 h 6692 psi = h

τ = τmax

884 10 000 = psi 11.31h h

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Equating τmax to τall gives h = 0.523 in. It follows that I = 60.3(0.523) = 31.5 in4 0.5232 h 2l = (8 + 8) = 2.19 in3 2 2 31.6 I = = 14.4 in (eff) V = vol 2.19 85.33 Iu (fom ) V = = = 10.2 in hl 0.523(8 + 8) vol =

The ratio of (eff) V /(eff) H is 14.4/91.2 = 0.158. The ratio (fom ) V /(fom ) H is 10.2/64.5 = 0.158. This is not surprising since eff =

I I 0.707 h Iu Iu = 2 = = 1.414 = 1.414 fom 2 vol (h /2)l (h /2)l hl

The ratios (eff) V /(eff) H and (fom ) V /(fom ) H give the same information. 9-20

Because the loading is pure torsion, there is no primary shear. From Table 9-1, category 6: Ju = 2πr 3 = 2π(1) 3 = 6.28 in3 J = 0.707 h Ju = 0.707(0.25)(6.28) = 1.11 in4 τ=

20(1) Tr = = 18.0 kpsi Ans. J 1.11

h = 0.375 in,

9-21

d = 8 in,

b = 1 in

From Table 9-2, category 2: A = 1.414(0.375)(8) = 4.24 in2 Iu =

83 d3 = = 85.3 in3 6 6

I = 0.707h Iu = 0.707(0.375)(85.3) = 22.6 in4 τ =

5 F = = 1.18 kpsi A 4.24

M = 5(6) = 30 kip · in c = (1 + 8 + 1 − 2)/2 = 4 in 30(4) Mc = = 5.31 kpsi I 22.6 = τ 2 + τ 2 = 1.182 + 5.312

τ = τmax

= 5.44 kpsi Ans.

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Chapter 9

h = 0.6 cm,

9-22

b = 6 cm,

d = 12 cm.

Table 9-3, category 5: A = 0.707h(b + 2d)

6 B

= 0.707(0.6)[6 + 2(12)] = 12.7 cm2

4.8 G

y¯ =

7.2

Iu =

A

122 d2 = = 4.8 cm b + 2d 6 + 2(12) 2d 3 − 2d 2 y¯ + (b + 2d) y¯ 2 3

2(12)3 − 2(122 )(4.8) + [6 + 2(12)]4.82 = 3 = 461 cm3 I = 0.707h Iu = 0.707(0.6)(461) = 196 cm4 τ =

7.5(103 ) F = = 5.91 MPa A 12.7(102 )

M = 7.5(120) = 900 N · m c A = 7.2 cm,

c B = 4.8 cm

The critical location is at A. 900(7.2) Mc A = = 33.1 MPa I 196 = τ 2 + τ 2 = (5.912 + 33.12 ) 1/2 = 33.6 MPa

τ A = τmax

n=

9-23

120 τall = 3.57 Ans. = τmax 33.6

The largest possible weld size is 1/16 in. This is a small weld and thus difficult to accomplish. The bracket’s load-carrying capability is not known. There are geometry problems associated with sheet metal folding, load-placement and location of the center of twist. This is not available to us. We will identify the strongest possible weldment. Use a rectangular, weld-all-around pattern – Table 9-2, category 6: A = 1.414 h(b + d) = 1.414(1/16)(1 + 7.5) = 0.751 in2 7.5"

x¯ = b/2 = 0.5 in y¯ = 1"

7.5 d = = 3.75 in 2 2

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7.52 d2 [3(1) + 7.5] = 98.4 in3 Iu = (3b + d) = 6 6 I = 0.707h Iu = 0.707(1/16)(98.4) = 4.35 in4 M = (3.75 + 0.5)W = 4.25W W V = = 1.332W A 0.751 4.25W (7.5/2) Mc = = 3.664W τ = I 4.35 τmax = τ 2 + τ 2 = W 1.3322 + 3.6642 = 3.90W τ =

Material properties: The allowable stress given is low. Let’s demonstrate that. For the A36 structural steel member, S y = 36 kpsi and Sut = 58 kpsi. For the 1020 CD attachment, use HR properties of S y = 30 kpsi and Sut = 55. The E6010 electrode has strengths of S y = 50 and Sut = 62 kpsi. Allowable stresses: A36:

τall = min[0.3(58), 0.4(36)] = min(17.4, 14.4) = 14.4 kpsi

1020:

τall = min[0.3(55), 0.4(30)] τall = min(16.5, 12) = 12 kpsi

E6010:

τall = min[0.3(62), 0.4(50)] = min(18.6, 20) = 18.6 kpsi

Since Table 9-6 gives 18.0 kpsi as the allowable shear stress, use this lower value. Therefore, the allowable shear stress is τall = min(14.4, 12, 18.0) = 12 kpsi However, the allowable stress in the problem statement is 0.9 kpsi which is low from the weldment perspective. The load associated with this strength is τmax = τall = 3.90W = 900 W =

900 = 231 lbf 3.90

If the welding can be accomplished (1/16 leg size is a small weld), the weld strength is 12 000 psi and the load W = 3047 lbf. Can the bracket carry such a load? There are geometry problems associated with sheet metal folding. Load placement is important and the center of twist has not been identified. Also, the load-carrying capability of the top bend is unknown. These uncertainties may require the use of a different weld pattern. Our solution provides the best weldment and thus insight for comparing a welded joint to one which employs screw fasteners.

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263

Chapter 9

9-24 y x

F

FB

x

RA

60

A RyA B

F = 100 lbf,

τall = 3 kpsi

FB = 100(16/3) = 533.3 lbf FBx = −533.3 cos 60◦ = −266.7 lbf FB = −533.3 cos 30◦ = −462 lbf y

y

It follows that R A = 562 lbf and R xA = 266.7 lbf, R A = 622 lbf M = 100(16) = 1600 lbf · in 100 462 266.7 16

3

266.7

562

The OD of the tubes is 1 in. From Table 9-1, category 6: A = 1.414(π hr)(2) = 2(1.414)(πh)(1/2) = 4.44h in2 Ju = 2πr 3 = 2π(1/2)3 = 0.785 in3 J = 2(0.707)h Ju = 1.414(0.785)h = 1.11h in4 622 140 V = = A 4.44h h Mc 1600(0.5) 720.7 Tc = = = τ = J J 1.11h h τ =

The shear stresses, τ and τ , are additive algebraically 861 1 (140 + 720.7) = psi h h 861 τmax = τall = = 3000 h 861 h= = 0.287 → 5/16" 3000 τmax =

Decision: Use 5/16 in fillet welds Ans.

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9-25 y 1" 4 1" 4B

g

g G

g

g 9"

3" 8 3" 8

x

7"

For the pattern in bending shown, find the centroid G of the weld group. 6(0.707)(1/4)(3) + 6(0.707)(3/8)(13) 6(0.707)(1/4) + 6(0.707)(3/8) = 9 in = 2 IG + A2x¯ 0.707(1/4)(63 ) 2 + 0.707(1/4)(6)(6 ) =2 12

x¯ =

I1/4

= 82.7 in4 0.707(3/8)(63 ) 2 + 0.707(3/8)(6)(4 ) =2 12

I3/8

= 60.4 in4 I = I1/4 + I3/8 = 82.7 + 60.4 = 143.1 in4 The critical location is at B. From Eq. (9-3), F = 0.189F 2[6(0.707)(3/8 + 1/4)] (8F)(9) Mc τ = = = 0.503F I 143.1 τmax = τ 2 + τ 2 = F 0.1892 + 0.5032 = 0.537F τ =

Materials: A36 Member: S y = 36 kpsi 1015 HR Attachment: S y = 27.5 kpsi E6010 Electrode: S y = 50 kpsi τall = 0.577 min(36, 27.5, 50) = 15.9 kpsi F= 9-26

15.9/2 τall /n = = 14.8 kip Ans. 0.537 0.537

Figure P9-26b is a free-body diagram of the bracket. Forces and moments that act on the welds are equal, but of opposite sense. M = 1200(0.366) = 439 lbf · in Ans. (a) Fy = 1200 sin 30◦ = 600 lbf Ans. (b) Fx = 1200 cos 30◦ = 1039 lbf Ans. (c)

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265

Chapter 9

(d) From Table 9-2, category 6: A = 1.414(0.25)(0.25 + 2.5) = 0.972 in2 2.52 d2 (3b + d) = [3(0.25) + 2.5] = 3.39 in3 6 6 The second area moment about an axis through G and parallel to z is Iu =

I = 0.707h Iu = 0.707(0.25)(3.39) = 0.599 in4

Ans.

(e) Refer to Fig. P.9-26b. The shear stress due to Fy is τ1 =

Fy 600 = = 617 psi A 0.972

The shear stress along the throat due to Fx is τ2 =

1039 Fx = = 1069 psi A 0.972

The resultant of τ1 and τ2 is in the throat plane 1/2 τ = τ12 + τ22 = (6172 + 10692 ) 1/2 = 1234 psi The bending of the throat gives τ =

439(1.25) Mc = = 916 psi I 0.599

The maximum shear stress is τmax = (τ 2 + τ 2 ) 1/2 = (12342 + 9162 ) 1/2 = 1537 psi Ans. (f) Materials: 1018 HR Member: S y = 32 kpsi, Sut = 58 kpsi (Table A-20) S y = 50 kpsi (Table 9-3) E6010 Electrode: n=

Ssy 0.577S y 0.577(32) = 12.0 Ans. = = τmax τmax 1.537

(g) Bending in the attachment near the base. The cross-sectional area is approximately equal to bh. . A1 = bh = 0.25(2.5) = 0.625 in2 τx y =

Fx 1039 = 1662 psi = A1 0.625

bd 2 0.25(2.5) 2 I = = = 0.260 in3 c 6 6 At location A σy =

Fy M + A1 I /c

σy =

439 600 + = 2648 psi 0.625 0.260

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The von Mises stress σ is 1/2 σ = σ y2 + 3τx2y = [26482 + 3(1662) 2 ]1/2 = 3912 psi Thus, the factor of safety is, n=

Sy 32 = 8.18 Ans. = σ 3.912

The clip on the mooring line bears against the side of the 1/2-in hole. If the clip fills the hole −1200 F = = −9600 psi σ = td 0.25(0.50) n=−

Sy 32(103 ) = 3.33 Ans. = − σ −9600

Further investigation of this situation requires more detail than is included in the task statement. (h) In shear fatigue, the weakest constituent of the weld melt is 1018 with Sut = 58 kpsi Se = 0.504Sut = 0.504(58) = 29.2 kpsi Table 7-4:

ka = 14.4(58) −0.718 = 0.780

For the size factor estimate, we first employ Eq. (7-24) for the equivalent diameter. √ de = 0.808 0.707hb = 0.808 0.707(2.5)(0.25) = 0.537 in Eq. (7-19) is used next to find kb de −0.107 0.537 −0.107 kb = = = 0.940 0.30 0.30 The load factor for shear kc , is kc = 0.59 The endurance strength in shear is Sse = 0.780(0.940)(0.59)(29.2) = 12.6 kpsi From Table 9-5, the shear stress-concentration factor is K f s = 2.7. The loading is repeatedly-applied. τa = τm = k f

1.537 τmax = 2.7 = 2.07 kpsi 2 2

Table 7-10: Gerber factor of safety n f , adjusted for shear, with Ssu = 0.67Sut   2  2  1 0.67(58) 2.07 2(2.07)(12.6) −1 + 1 + = 5.55 Ans. nf = 2 2.07 12.6  0.67(58)(2.07)  Attachment metal should be checked for bending fatigue.

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Chapter 9

9-27

267

Use b = d = 4 in. Since h = 5/8 in, the primary shear is F = 0.283F 1.414(5/8)(4) The secondary shear calculations, for a moment arm of 14 in give τ =

Ju =

4[3(42 ) + 42 ] = 42.67 in3 6

J = 0.707h Ju = 0.707(5/8)42.67 = 18.9 in4 Mr y 14F(2) = = 1.48F J 18.9

τx = τ y =

Thus, the maximum shear and allowable load are: τmax = F 1.482 + (0.283 + 1.48) 2 = 2.30F F=

20 τall = = 8.70 kip Ans. 2.30 2.30

From Prob. 9-5b, τall = 11 kpsi 11 τall = = 4.78 kip 2.30 2.30 The allowable load has thus increased by a factor of 1.8 Ans. Fall =

9-28

Purchase the hook having the design shown in Fig. P9-28b. Referring to text Fig. 9-32a, this design reduces peel stresses.

9-29

(a) 1 τ¯ = l

l/2

−l/2

= A1

Pω cosh(ωx) dx 4b sinh(ωl/2)

l/2

−l/2

cosh(ωx) dx

=

l/2 A1 sinh(ωx) −l/2 ω

=

A1 [sinh(ωl/2) − sinh(−ωl/2)] ω

=

A1 [sinh(ωl/2) − (−sinh(ωl/2))] ω

=

2A1 sinh(ωl/2) ω

=

Pω [2 sinh(ωl/2)] 4bl sinh(ωl/2) P Ans. τ¯ = 2bl

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(b) (c)

τ (l/2) =

Pω Pω cosh(ωl/2) = 4b sinh(ωl/2) 4b tanh(ωl/2)

Pω τ (l/2) = K = τ¯ 4b sinh(ωl/2) K =

ωl/2 tanh(ωl/2)

2bl P

Ans.

Ans.

For computer programming, it can be useful to express the hyperbolic tangent in terms of exponentials: K = 9-30

ωl exp(ωl/2) − exp(−ωl/2) 2 exp(ωl/2) + exp(−ωl/2)

Ans.

This is a computer programming exercise. All programs will vary.

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Chapter 10 10-1

1" 2

1" 4"

10-2

1" 2

1" 4"

A = Sd m dim( Auscu ) = dim(S) dim(d m ) = kpsi · inm dim( ASI ) = dim(S1 ) dim d1m = MPa · mmm MPa mmm . · m Auscu = 6.894 757(25.40) m Auscu = 6.895(25.4) m Auscu ASI = kpsi in For music wire, from Table 10-4: Auscu = 201, m = 0.145; what is ASI ? ASI = 6.89(25.4) 0.145 (201) = 2214 MPa · mmm

10-3

Ans.

Given: Music wire, d = 0.105 in, OD = 1.225 in, plain ground ends, Nt = 12 coils. Na = Nt − 1 = 12 − 1 = 11 L s = d Nt = 0.105(12) = 1.26 in

Table 10-1:

A = 201, m = 0.145 201 Sut = = 278.7 kpsi (0.105) 0.145

Table 10-4: (a) Eq. (10-14):

Ssy = 0.45(278.7) = 125.4 kpsi D = 1.225 − 0.105 = 1.120 in D 1.120 C= = = 10.67 d 0.105

Table 10-6:

Eq. (10-6): Eq. (10-3): Eq. (10-9):

KB = F| Ssy

4(10.67) + 2 = 1.126 4(10.67) − 3

πd 3 Ssy π(0.105) 3 (125.4)(103 ) = = 45.2 lbf = 8K B D 8(1.126)(1.120)

(0.105) 4 (11.75)(106 ) d4G = 11.55 lbf/in = 8D 3 Na 8(1.120) 3 (11) F| Ssy 45.2 + Ls = + 1.26 = 5.17 in Ans. L0 = k 11.55 k=

Ans.

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270 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

(b) F| Ssy = 45.2 lbf Ans. (c) k = 11.55 lbf/in Ans. 2.63(1.120) 2.63D = = 5.89 in (d) (L 0 ) cr = α 0.5 Many designers provide (L 0 ) cr /L 0 ≥ 5 or more; therefore, plain ground ends are not often used in machinery due to buckling uncertainty. 10-4

Referring to Prob. 10-3 solution, C = 10.67, Na = 11, Ssy = 125.4 kpsi, (L 0 ) cr = 5.89 in and F = 45.2 lbf (at yield). 4 ≤ C ≤ 12 C = 10.67 O.K. Eq. (10-18): Na = 11 O.K. 3 ≤ Na ≤ 15 Eq. (10-19): y1

F1

ys

L0

Fs L1 Ls

L 0 = 5.17 in, y1 =

L s = 1.26 in

30 F1 = = 2.60 in k 11.55

L 1 = L 0 − y1 = 5.17 − 2.60 = 2.57 in ξ=

5.17 − 1.26 ys − 1 = 0.50 −1= y1 2.60

ξ ≥ 0.15, ξ = 0.50 O.K. Eq. (10-20): From Eq. (10-3) for static service 8F1 D 8(30)(1.120) τ1 = K B = 1.126 = 83 224 psi πd 3 π(0.105) 3 Ssy 125.4(103 ) ns = = 1.51 = τ1 83 224 n s ≥ 1.2, n s = 1.51 O.K. Eq. (10-21): 45.2 45.2 = 83 224 = 125 391 psi τs = τ1 30 30 . Ssy /τs = 125.4(103 )/125 391 = 1 Ssy /τs ≥ (n s ) d : Not solid-safe. Not O.K. L 0 ≤ (L 0 ) cr :

5.17 ≤ 5.89

Margin could be higher,

Design is unsatisfactory. Operate over a rod?

Ans.

Not O.K.

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Chapter 10

10-5

Static service spring with: HD steel wire, d = 2 mm, OD = 22 mm, Nt = 8.5 turns plain and ground ends. Preliminaries A = 1783 MPa · mmm , m = 0.190 Table 10-5: 1783 = 1563 MPa Sut = Eq. (10-14): (2) 0.190 Ssy = 0.45(1563) = 703.4 MPa Table 10-6: Then, D = OD − d = 22 − 2 = 20 mm C = 20/2 = 10 KB =

4(10) + 2 4C + 2 = = 1.135 4C − 3 4(10) − 3

Na = 8.5 − 1 = 7.5 turns L s = 2(8.5) = 17 mm Eq. (10-21): Use (n s ) d = 1.2 for solid-safe property. πd 3 Ssy /n d π(2) 3 (703.4/1.2) (10−3 ) 3 (106 ) = = 81.12 N Fs = 8K B D 8(1.135)(20) 10−3 (2) 4 (79.3) (10−3 ) 4 (109 ) d4G = 0.002 643(106 ) = 2643 N/m = k= 8D 3 Na 8(20) 3 (7.5) (10−3 ) 3 ys =

81.12 Fs = = 30.69 mm k 2643(10−3 )

(a) L 0 = y + L s = 30.69 + 17 = 47.7 mm Ans. L0 47.7 = 5.61 mm Ans. = (b) Table 10-1: p = Nt 8.5 (c) Fs = 81.12 N (from above) Ans. (d) k = 2643 N/m (from above) Ans. (e) Table 10-2 and Eq. (10-13): 2.63(20) 2.63D = = 105.2 mm α 0.5 (L 0 ) cr /L 0 = 105.2/47.7 = 2.21 (L 0 ) cr =

This is less than 5. Operate over a rod? Plain and ground ends have a poor eccentric footprint. 10-6

Ans.

Referring to Prob. 10-5 solution: C = 10, Na = 7.5, k = 2643 N/m, d = 2 mm, D = 20 mm, Fs = 81.12 N and Nt = 8.5 turns. 4 ≤ C ≤ 12, C = 10 O.K. Eq. (10-18):

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3 ≤ Na ≤ 15,

Eq. (10-19):

Na = 7.5 O.K.

75 F1 = = 28.4 mm k 2643(10−3 ) 81.12(1.2) ( y) for yield = = 36.8 mm 2643(10−3 ) 81.12 = 30.69 mm ys = 2643(10−3 ) ( y) for yield 36.8 ξ= − 1 = 0.296 −1= y1 28.4 ξ ≥ 0.15, ξ = 0.296 O.K. Ssy = 0.45Sut O.K. y1 =

Eq. (10-20): Table 10-6: As-wound τs = K B

8Fs D πd 3

8(81.12)(20) = 1.135 π(2) 3

10−3 = 586 MPa (10−3 ) 3 (106 )

Ssy 703.4 = 1.2 O.K. (Basis for Prob. 10-5 solution) = τs 586 L s = Nt d = 8.5(2) = 17 mm Table 10-1: Fs 81.12 + Ls = + 17 = 47.7 mm L0 = k 2.643 2.63(20) 2.63D = = 105.2 mm α 0.5 (L 0 ) cr 105.2 = 2.21 = L0 47.7 which is less than 5. Operate over a rod? Not O.K. Eq. (10-21):

Plain and ground ends have a poor eccentric footprint. 10-7

Ans.

Given: A228 (music wire), SQ&GRD ends, d = 0.006 in, OD = 0.036 in, L 0 = 0.63 in, Nt = 40 turns. Table 10-4:

Table 10-1:

A = 201 kpsi · inm , m = 0.145 D = OD − d = 0.036 − 0.006 = 0.030 in C = D/d = 0.030/0.006 = 5 4(5) + 2 = 1.294 KB = 4(5) − 3 Na = Nt − 2 = 40 − 2 = 38 turns Sut =

201 = 422.1 kpsi (0.006) 0.145

Ssy = 0.45(422.1) = 189.9 kpsi Gd 4 12(106 )(0.006) 4 = 1.895 lbf/in k= = 8D 3 Na 8(0.030) 3 (38)

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Chapter 10

L s = Nt d = 40(0.006) = 0.240 in

Table 10-1:

Now Fs = kys where ys = L 0 − L s = 0.390 in. Thus, 8(kys ) D 8(1.895)(0.39)(0.030) = 1.294 (10−3 ) = 338.2 kpsi τs = K B πd 3 π(0.006) 3

(1)

τs > Ssy , that is, 338.2 > 189.9 kpsi; the spring is not solid-safe. Solving Eq. (1) for ys gives ys =

(189 900/1.2)(π)(0.006) 3 (τs /n)(πd 3 ) = = 0.182 in 8K B k D 8(1.294)(1.895)(0.030)

Using a design factor of 1.2, L 0 = L s + ys = 0.240 + 0.182 = 0.422 in The spring should be wound to a free length of 0.422 in. 10-8

Ans.

Given: B159 (phosphor bronze), SQ&GRD ends, d = 0.012 in, OD = 0.120 in, L 0 = 0.81 in, Nt = 15.1 turns. m=0

Table 10-4:

A = 145 kpsi · inm ,

Table 10-5:

G = 6 Mpsi D = OD − d = 0.120 − 0.012 = 0.108 in C = D/d = 0.108/0.012 = 9 KB =

4(9) + 2 = 1.152 4(9) − 3

Na = Nt − 2 = 15.1 − 2 = 13.1 turns

Table 10-1:

Sut =

145 = 145 kpsi 0.0120

Ssy = 0.35(145) = 50.8 kpsi

Table 10-6:

k=

6(106 )(0.012) 4 Gd 4 = 0.942 lbf/in = 8D 3 Na 8(0.108) 3 (13.1)

L s = d Nt = 0.012(15.1) = 0.181 in

Table 10-1:

Now Fs = kys , ys = L 0 − L s = 0.81 − 0.181 = 0.629 in 8(kys ) D 8(0.942)(0.6)(0.108) = 1.152 (10−3 ) = 108.6 kpsi τs = K B πd 3 π(0.012) 3

(1)

τs > Ssy , that is, 108.6 > 50.8 kpsi; the spring is not solid safe. Solving Eq. (1) for ys gives ys =

(Ssy /n)πd 3 (50.8/1.2)(π)(0.012) 3 (103 ) = = 0.245 in 8K B k D 8(1.152)(0.942)(0.108)

L 0 = L s + ys = 0.181 + 0.245 = 0.426 in Wind the spring to a free length of 0.426 in.

Ans.

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10-9

Given: A313 (stainless steel), SQ&GRD ends, d = 0.040 in, OD = 0.240 in, L 0 = 0.75 in, Nt = 10.4 turns. Table 10-4:


Table 10-5:

G = 10(106 ) psi

m = 0.146

D = OD − d = 0.240 − 0.040 = 0.200 in C = D/d = 0.200/0.040 = 5 KB =

4(5) + 2 = 1.294 4(5) − 3

Table 10-6:

Na = Nt − 2 = 10.4 − 2 = 8.4 turns 169 Sut = = 270.4 kpsi (0.040) 0.146

Table 10-13:

Ssy = 0.35(270.4) = 94.6 kpsi k=

10(106 )(0.040) 4 Gd 4 = 47.62 lbf/in = 8D 3 Na 8(0.2) 3 (8.4)

L s = d Nt = 0.040(10.4) = 0.416 in

Table 10-6:

Now Fs = kys , ys = L 0 − L s = 0.75 − 0.416 = 0.334 in 8(kys ) D 8(47.62)(0.334)(0.2) τs = K B = 1.294 (10−3 ) = 163.8 kpsi 3 3 πd π(0.040)

(1)

τs > Ssy , that is, 163.8 > 94.6 kpsi; the spring is not solid-safe. Solving Eq. (1) for ys gives ys =

(Ssy /n)(πd 3 ) (94 600/1.2)(π)(0.040) 3 = = 0.161 in 8K B k D 8(1.294)(47.62)(0.2)

L 0 = L s + ys = 0.416 + 0.161 = 0.577 in Wind the spring to a free length 0.577 in. 10-10

Ans.

Given: A227 (hard drawn steel), d = 0.135 in, OD = 2.0 in, L 0 = 2.94 in, Nt = 5.25 turns. Table 10-4:


Table 10-5:

G = 11.4(106 ) psi

m = 0.190

D = OD − d = 2 − 0.135 = 1.865 in C = D/d = 1.865/0.135 = 13.81 KB =

4(13.81) + 2 = 1.096 4(13.81) − 3

Na = Nt − 2 = 5.25 − 2 = 3.25 turns Sut =

140 = 204.8 kpsi (0.135) 0.190

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Table 10-6:

Ssy = 0.45(204.8) = 92.2 kpsi k=

Gd 4 11.4(106 )(0.135) 4 = 22.45 lbf/in = 8D 3 Na 8(1.865) 3 (3.25)

L s = d Nt = 0.135(5.25) = 0.709 in

Table 10-1:

Now Fs = kys , ys = L 0 − L s = 2.94 − 0.709 = 2.231 in 8(kys ) D 8(22.45)(2.231)(1.865) = 1.096 (10−3 ) = 106.0 kpsi (1) τs = K B πd 3 π(0.135) 3 τs > Ssy , that is, 106 > 92.2 kpsi; the spring is not solid-safe. Solving Eq. (1) for ys gives ys =

(Ssy /n)(πd 3 ) (92 200/1.2)(π)(0.135) 3 = = 1.612 in 8K B k D 8(1.096)(22.45)(1.865)

L 0 = L s + ys = 0.709 + 1.612 = 2.321 in Wind the spring to a free length of 2.32 in. 10-11

Ans.

Given: A229 (OQ&T steel), SQ&GRD ends, d = 0.144 in, OD = 1.0 in, L 0 = 3.75 in, Nt = 13 turns. A = 147 kpsi · inm , m = 0.187 Table 10-4: Table 10-5:

G = 11.4(106 ) psi D = OD − d = 1.0 − 0.144 = 0.856 in C = D/d = 0.856/0.144 = 5.944 4(5.944) + 2 KB = = 1.241 4(5.944) − 3

Table 10-1:

Na = Nt − 2 = 13 − 2 = 11 turns 147 Sut = = 211.2 kpsi (0.144) 0.187

Table 10-6:

Ssy = 0.50(211.2) = 105.6 kpsi Gd 4 11.4(106 )(0.144) 4 k= = 88.8 lbf/in = 8D 3 Na 8(0.856) 3 (11)

Table 10-1:

L s = d Nt = 0.144(13) = 1.872 in

Now Fs = kys , ys = L 0 − L s = 3.75 − 1.872 = 1.878 in 8(kys ) D 8(88.8)(1.878)(0.856) = 1.241 (10−3 ) = 151.1 kpsi τs = K B 3 3 πd π(0.144)

(1)

τs > Ssy , that is,151.1 > 105.6 kpsi; the spring is not solid-safe. Solving Eq. (1) for ys gives (Ssy /n)(πd 3 ) (105 600/1.2)(π)(0.144) 3 = = 1.094 in ys = 8K B k D 8(1.241)(88.8)(0.856) L 0 = L s + ys = 1.878 + 1.094 = 2.972 in Wind the spring to a free length 2.972 in.

Ans.

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10-12

Given: A232 (Cr-V steel), SQ&GRD ends, d = 0.192 in, OD = 3 in, L 0 = 9 in, Nt = 8 turns. m = 0.168

Table 10-4:


Table 10-5:

G = 11.2(106 ) psi D = OD − d = 3 − 0.192 = 2.808 in C = D/d = 2.808/0.192 = 14.625 KB =

4(14.625) + 2 = 1.090 4(14.625) − 3

Na = Nt − 2 = 8 − 2 = 6 turns

Table 10-1:

Sut =

169 = 223.0 kpsi (0.192) 0.168

Ssy = 0.50(223.0) = 111.5 kpsi

Table 10-6:

k=

11.2(106 )(0.192) 4 Gd 4 = 14.32 lbf/in = 8D 3 Na 8(2.808) 3 (6)

L s = d Nt = 0.192(8) = 1.536 in

Table 10-1:

Now Fs = kys , ys = L 0 − L s = 9 − 1.536 = 7.464 in 8(kys ) D 8(14.32)(7.464)(2.808) = 1.090 (10−3 ) = 117.7 kpsi τs = K B πd 3 π(0.192) 3

(1)

τs > Ssy , that is,117.7 > 111.5 kpsi; the spring is not solid safe. Solving Eq. (1) for ys gives ys =

(Ssy /n)(πd 3 ) (111 500/1.2)(π)(0.192) 3 = = 5.892 in 8K B k D 8(1.090)(14.32)(2.808)

L 0 = L s + ys = 1.536 + 5.892 = 7.428 in Wind the spring to a free length of 7.428 in. 10-13

Ans.

Given: A313 (stainless steel) SQ&GRD ends, d = 0.2 mm, OD = 0.91 mm, L 0 = 15.9 mm, Nt = 40 turns. Table 10-4:

A = 1867 MPa · mmm , m = 0.146

Table 10-5:

G = 69.0 GPa D = OD − d = 0.91 − 0.2 = 0.71 mm C = D/d = 0.71/0.2 = 3.55 KB =

4(3.55) + 2 = 1.446 4(3.55) − 3

Na = Nt − 2 = 40 − 2 = 38 turns Sut =

1867 = 2361.5 MPa (0.2) 0.146

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277

Table 10-6: Ssy = 0.35(2361.5) = 826.5 MPa (0.2) 4 (69.0) (10−3 ) 4 (109 ) d4G = k= 8D 3 Na 8(0.71) 3 (38) (10−3 ) 3 = 1.0147(10−3 )(106 ) = 1014.7 N/m or 1.0147 N/mm L s = d Nt = 0.2(40) = 8 mm Fs = kys ys = L 0 − L s = 15.9 − 8 = 7.9 8(kys ) D 8(1.0147)(7.9)(0.71) 10−3 (10−3 )(10−3 ) = 1.446 τs = K B πd 3 π(0.2) 3 (10−3 ) 3 = 2620(1) = 2620 MPa

(1)

τs > Ssy , that is, 2620 > 826.5 MPa; the spring is not solid safe. Solve Eq. (1) for ys giving ys =

(Ssy /n)(πd 3 ) (826.5/1.2)(π)(0.2) 3 = = 2.08 mm 8K B k D 8(1.446)(1.0147)(0.71)

L 0 = L s + ys = 8.0 + 2.08 = 10.08 mm Wind the spring to a free length of 10.08 mm. This only addresses the solid-safe criteria. There are additional problems. Ans. 10-14

Given: A228 (music wire), SQ&GRD ends, d = 1 mm, OD = 6.10 mm, L 0 = 19.1 mm, Nt = 10.4 turns. Table 10-4:

A = 2211 MPa · mmm , m = 0.145

Table 10-5: G = 81.7 GPa D = OD − d = 6.10 − 1 = 5.1 mm C = D/d = 5.1/1 = 5.1 Na = Nt − 2 = 10.4 − 2 = 8.4 turns KB =

4(5.1) + 2 = 1.287 4(5.1) − 3

Sut =

2211 = 2211 MPa (1) 0.145

Table 10-6: Ssy = 0.45(2211) = 995 MPa

(1) 4 (81.7) (10−3 ) 4 (109 ) d4G = 0.009 165(106 ) = k= 3 3 − 3 3 8D Na 8(5.1) (8.4) (10 ) = 9165 N/m or 9.165 N/mm

L s = d Nt = 1(10.4) = 10.4 mm Fs = kys

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ys = L 0 − L s = 19.1 − 10.4 = 8.7 mm 8(kys ) D 8(9.165)(8.7)(5.1) = 1.287 = 1333 MPa τs = K B πd 3 π(1) 3

(1)

τs > Ssy , that is, 1333 > 995 MPa; the spring is not solid safe. Solve Eq. (1) for ys giving ys =

(Ssy /n)(πd 3 ) (995/1.2)(π)(1) 3 = = 5.43 mm 8K B k D 8(1.287)(9.165)(5.1)

L 0 = L s + ys = 10.4 + 5.43 = 15.83 mm Wind the spring to a free length of 15.83 mm. 10-15

Ans.

Given: A229 (OQ&T spring steel), SQ&GRD ends, d = 3.4 mm, OD = 50.8 mm, L 0 = 74.6 mm, Nt = 5.25. Table 10-4:

A = 1855 MPa · mmm ,

Table 10-5:

G = 77.2 GPa

m = 0.187

D = OD − d = 50.8 − 3.4 = 47.4 mm C = D/d = 47.4/3.4 = 13.94 Na = Nt − 2 = 5.25 − 2 = 3.25 turns

Table 10-6:

KB =

4(13.94) + 2 = 1.095 4(13.94) − 3

Sut =

1855 = 1476 MPa (3.4) 0.187

Ssy = 0.50(1476) = 737.8 MPa

(3.4) 4 (77.2) (10−3 ) 4 (109 ) d4G = 0.003 75(106 ) = k= 8D 3 Na 8(47.4) 3 (3.25) (10−3 ) 3 = 3750 N/m or 3.750 N/mm

L s = d Nt = 3.4(5.25) = 17.85 Fs = kys ys = L 0 − L s = 74.6 − 17.85 = 56.75 mm 8(kys ) D τs = K B πd 3 8(3.750)(56.75)(47.4) = 720.2 MPa = 1.095 π(3.4) 3 τs < Ssy , that is, 720.2 < 737.8 MPa

(1)

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∴ The spring is solid safe. With n s = 1.2, ys =

(Ssy /n)(πd 3 ) (737.8/1.2)(π)(3.4) 3 = = 48.76 mm 8K B k D 8(1.095)(3.75)(47.4)

L 0 = L s + ys = 17.85 + 48.76 = 66.61 mm Wind the spring to a free length of 66.61 mm. 10-16

Ans.

Given: B159 (phosphor bronze), SQ&GRD ends, d = 3.7 mm, OD = 25.4 mm, L 0 = 95.3 mm, Nt = 13 turns. Table 10-4:

A = 932 MPa · mmm , m = 0.064

Table 10-5:

G = 41.4 GPa D = OD − d = 25.4 − 3.7 = 21.7 mm C = D/d = 21.7/3.7 = 5.865 KB =

4(5.865) + 2 = 1.244 4(5.865) − 3

Na = Nt − 2 = 13 − 2 = 11 turns Sut =

932 = 857.1 MPa (3.7) 0.064

Table 10-6: Ssy = 0.35(857.1) = 300 MPa d4G (3.7) 4 (41.4) (10−3 ) 4 (109 ) = 0.008 629(106 ) k= = 3 3 − 3 3 8D Na 8(21.7) (11) (10 ) = 8629 N/m or 8.629 N/mm L s = d Nt = 3.7(13) = 48.1 mm Fs = kys ys = L 0 − L s = 95.3 − 48.1 = 47.2 mm 8(kys ) D τs = K B πd 3 8(8.629)(47.2)(21.7) = 553 MPa = 1.244 π(3.7) 3

(1)

τs > Ssy , that is, 553 > 300 MPa; the spring is not solid-safe. Solving Eq. (1) for ys gives ys =

(Ssy /n)(πd 3 ) (300/1.2)(π)(3.7) 3 = = 21.35 mm 8K B k D 8(1.244)(8.629)(21.7)

L 0 = L s + ys = 48.1 + 21.35 = 69.45 mm Wind the spring to a free length of 69.45 mm.

Ans.

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10-17

Given: A232 (Cr-V steel), SQ&GRD ends, d = 4.3 mm, OD = 76.2 mm, L 0 = 228.6 mm, Nt = 8 turns. Table 10-4:

A = 2005 MPa · mmm ,

Table 10-5:

G = 77.2 GPa

m = 0.168

D = OD − d = 76.2 − 4.3 = 71.9 mm C = D/d = 71.9/4.3 = 16.72 KB =

4(16.72) + 2 = 1.078 4(16.72) − 3

Na = Nt − 2 = 8 − 2 = 6 turns Sut =

2005 = 1569 MPa (4.3) 0.168

Table 10-6: Ssy = 0.50(1569) = 784.5 MPa d4G (4.3) 4 (77.2) (10−3 ) 4 (109 ) k= = 0.001 479(106 ) = 3 3 − 3 3 8D Na 8(71.9) (6) (10 ) = 1479 N/m or 1.479 N/mm L s = d Nt = 4.3(8) = 34.4 mm Fs = kys ys = L 0 − L s = 228.6 − 34.4 = 194.2 mm 8(kys ) D 8(1.479)(194.2)(71.9) = 1.078 = 713.0 MPa τs = K B πd 3 π(4.3) 3

(1)

τs < Ssy , that is, 713.0 < 784.5; the spring is solid safe. With n s = 1.2 Eq. (1) becomes ys =

(Ssy /n)(πd 3 ) (784.5/1.2)(π)(4.3) 3 = = 178.1 mm 8K B k D 8(1.078)(1.479)(71.9)

L 0 = L s + ys = 34.4 + 178.1 = 212.5 mm Wind the spring to a free length of L 0 = 212.5 mm. 10-18

Ans.

For the wire diameter analyzed, G = 11.75 Mpsi per Table 10-5. Use squared and ground ends. The following is a spread-sheet study using Fig. 10-3 for parts (a) and (b). For Na , k = 20/2 = 10 lbf/in.

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(a) Spring over a Rod Source

Eq. (10-2) Eq. (10-9) Table 10-1 Table 10-1 1.15y + L s Eq. (10-13) Table 10-4 Table 10-4 Eq. (10-14) Table 10-6 Eq. (10-6) Eq. (10-3) Eq. (10-22)

(b) Spring in a Hole

Parameter Values d D ID OD C Na Nt Ls L0 (L 0 ) cr A m Sut Ssy KB n fom

0.075 0.875 0.800 0.950 11.667 6.937 8.937 0.670 2.970 4.603 201.000 0.145 292.626 131.681 1.115 0.973 −0.282

0.08 0.88 0.800 0.960 11.000 8.828 10.828 0.866 3.166 4.629 201.000 0.145 289.900 130.455 1.122 1.155 −0.391

Source 0.085 0.885 0.800 0.970 10.412 11.061 13.061 1.110 3.410 4.655 201.000 0.145 287.363 129.313 1.129 1.357 −0.536

Eq. (10-2) Eq. (10-9) Table 10-1 Table 10-1 1.15y + L s Eq. (10-13) Table 10-4 Table 10-4 Eq. (10-14) Table 10-6 Eq. (10-6) Eq. (10-3) Eq. (10-22)

Parameter Values d D ID OD C Na Nt Ls L0 (L 0 ) cr A m Sut Ssy KB n fom

0.075 0.875 0.800 0.950 11.667 6.937 8.937 0.670 2.970 4.603 201.000 0.145 292.626 131.681 1.115 0.973 −0.282

0.08 0.870 0.790 0.950 10.875 9.136 11.136 0.891 3.191 4.576 201.000 0.145 289.900 130.455 1.123 1.167 −0.398

0.085 0.865 0.780 0.950 10.176 11.846 13.846 1.177 3.477 4.550 201.000 0.145 287.363 129.313 1.133 1.384 −0.555

For n s ≥ 1.2, the optimal size is d = 0.085 in for both cases. 10-19

From the figure: L 0 = 120 mm, OD = 50 mm, and d = 3.4 mm. Thus D = OD − d = 50 − 3.4 = 46.6 mm (a) By counting, Nt = 12.5 turns. Since the ends are squared along 1/4 turn on each end, Na = 12.5 − 0.5 = 12 turns Ans. p = 120/12 = 10 mm Ans. The solid stack is 13 diameters across the top and 12 across the bottom. L s = 13(3.4) = 44.2 mm Ans. (b) d = 3.4/25.4 = 0.1339 in and from Table 10-5, G = 78.6 GPa k=

d4G (3.4) 4 (78.6)(109 ) (10−3 ) = 1080 N/m Ans. = 3 3 8D Na 8(46.6) (12)

(c) Fs = k(L 0 − L s ) = 1080(120 − 44.2)(10−3 ) = 81.9 N Ans. (d) C = D/d = 46.6/3.4 = 13.71 KB = τs = 10-20

4(13.71) + 2 = 1.096 4(13.71) − 3 8K B Fs D 8(1.096)(81.9)(46.6) = = 271 MPa Ans. 3 πd π(3.4) 3

One approach is to select A227-47 HD steel for its low cost. Then, for y1 ≤ 3/8 at F1 = 10 lbf, k ≥10/ 0.375 = 26.67 lbf/in . Try d = 0.080 in #14 gauge

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For a clearance of 0.05 in: ID = (7/16) + 0.05 = 0.4875 in; OD = 0.4875 + 0.16 = 0.6475 in D = 0.4875 + 0.080 = 0.5675 in C = 0.5675/0.08 = 7.094 G = 11.5 Mpsi d4G (0.08) 4 (11.5)(106 ) = = 12.0 turns Na = 8k D 3 8(26.67)(0.5675) 3 Nt = 12 + 2 = 14 turns, L s = d Nt = 0.08(14) = 1.12 in O.K. L 0 = 1.875 in, ys = 1.875 − 1.12 = 0.755 in Fs = kys = 26.67(0.755) = 20.14 lbf 4(7.094) + 2 KB = = 1.197 4(7.094) − 3 8Fs D 8(20.14)(0.5675) τs = K B = 1.197 = 68 046 psi πd 3 π(0.08) 3 A = 140 kpsi · inm , m = 0.190 140 Ssy = 0.45 = 101.8 kpsi (0.080) 0.190 101.8 = 1.50 > 1.2 O.K. n= 68.05 F1 10 (68.05) = 33.79 kpsi, τ1 = τs = Fs 20.14 101.8 n1 = = 3.01 > 1.5 O.K. 33.79 There is much latitude for reducing the amount of material. Iterate on y1 using a spread sheet. The final results are: y1 = 0.32 in, k = 31.25 lbf/in, Na = 10.3 turns, Nt = 12.3 turns, L s = 0.985 in, L 0 = 1.820 in, ys = 0.835 in, Fs = 26.1 lbf, K B = 1.197, τs = 88 190 kpsi, n s = 1.15, and n 1 = 3.01. ID = 0.4875 in, OD = 0.6475 in, d = 0.080 in Try other sizes and/or materials. Table 10-4:

10-21

A stock spring catalog may have over two hundred pages of compression springs with up to 80 springs per page listed. • • • •

Students should be aware that such catalogs exist. Many springs are selected from catalogs rather than designed. The wire size you want may not be listed. Catalogs may also be available on disk or the web through search routines. For example, disks are available from Century Spring at 1 − (800) − 237 − 5225 www.centuryspring.com • It is better to familiarize yourself with vendor resources rather than invent them yourself. • Sample catalog pages can be given to students for study.

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10-22

For a coil radius given by: R2 − R1 θ R = R1 + 2π N The torsion of a section is T = P R where d L = R dθ 2π N 1 ∂T 1 ∂U δp = = T dL = P R 3 dθ ∂P GJ ∂P GJ 0 2π N R2 − R1 3 P θ dθ R1 + = GJ 0 2π N 2π N 2π N R2 − R1 4

P 1 θ R1 + =

GJ 4 R2 − R1 2π N 0 πPN 4 πPN 4 ( R1 + R2 ) R12 + R22 = R2 − R1 = 2G J ( R2 − R1 ) 2G J 2 π 16P N 2 J = d 4 ∴ δp = ( R + R ) R + R 1 2 1 2 32 Gd 4 k=

10-23

P d4G = δp 16N ( R1 + R2 ) R12 + R22

Ans.

For a food service machinery application select A313 Stainless wire. G = 10(106 ) psi 0.013 ≤ d ≤ 0.10 in A = 169, m = 0.146 Note that for 0.10 < d ≤ 0.20 in A = 128, m = 0.263 18 − 4 18 + 4 = 7 lbf, Fm = = 11 lbf, r = 7/11 Fa = 2 2 18 − 4 = 9.333 lbf/in k = F/y = 2.5 − 1 169 d = 0.080 in, Sut = = 244.4 kpsi Try (0.08) 0.146 Ssu = 0.67Sut = 163.7 kpsi, Ssy = 0.35Sut = 85.5 kpsi Try unpeened using Zimmerli’s endurance data: Ssa = 35 kpsi, Ssm = 55 kpsi Gerber:

Ssa 35 = = 39.5 kpsi 2 1 − (Ssm /Ssu ) 1 − (55/163.7) 2   2  2 2  (7/11) (163.7) 2(39.5) −1 + 1 + = 35.0 kpsi =  2(39.5) (7/11)(163.7) 

Sse =

Ssa

α = Ssa /n f = 35.0/1.5 = 23.3 kpsi 8Fa 8(7) −3 β= (10−3 ) = 2.785 kpsi (10 ) = 2 2 πd π(0.08 ) 3(23.3) 2(23.3) − 2.785 2(23.3) − 2.785 2 + = 6.97 − C= 4(2.785) 4(2.785) 4(2.785) D = Cd = 6.97(0.08) = 0.558 in

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4(6.97) + 2 = 1.201 4(6.97) − 3 8Fa D 8(7)(0.558) −3 τa = K B = 1.201 (10 ) = 23.3 kpsi πd 3 π(0.083 )

KB =

n f = 35/23.3 = 1.50 checks Na =

Gd 4 10(106 )(0.08) 4 = = 31.58 turns 8k D 3 8(9.333)(0.558) 3

Nt = 31.58 + 2 = 33.58 turns, L s = d Nt = 0.08(33.58) = 2.686 in ys = (1 + ξ ) ymax = (1 + 0.15)(2.5) = 2.875 in L 0 = 2.686 + 2.875 = 5.561 in 2.63(0.558) D = = 2.935 in α 0.5 τs = 1.15(18/7)τa = 1.15(18/7)(23.3) = 68.9 kpsi

(L 0 ) cr = 2.63

n s = Ssy /τs = 85.5/68.9 = 1.24 kg 9.333(386) = = 107 Hz f = π 2 d 2 D Na γ π 2 (0.082 )(0.558)(31.58)(0.283) These steps are easily implemented on a spreadsheet, as shown below, for different diameters.

d m A Sut Ssu Ssy Sse Ssa α β C D KB τa nf Na Nt Ls ymax L0 (L 0 ) cr τs ns f (Hz)

d1

d2

d3

d4

0.080 0.146 169.000 244.363 163.723 85.527 39.452 35.000 23.333 2.785 6.977 0.558 1.201 23.333 1.500 31.547 33.547 2.684 2.875 5.559 2.936 69.000 1.240 106.985

0.0915 0.146 169.000 239.618 160.544 83.866 39.654 35.000 23.333 2.129 9.603 0.879 1.141 23.333 1.500 13.836 15.836 1.449 2.875 4.324 4.622 69.000 1.215 112.568

0.1055 0.263 128.000 231.257 154.942 80.940 40.046 35.000 23.333 1.602 13.244 1.397 1.100 23.333 1.500 6.082 8.082 0.853 2.875 3.728 7.350 69.000 1.173 116.778

0.1205 0.263 128.000 223.311 149.618 78.159 40.469 35.000 23.333 1.228 17.702 2.133 1.074 23.333 1.500 2.910 4.910 0.592 2.875 3.467 11.220 69.000 1.133 119.639

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The shaded areas depict conditions outside the recommended design conditions. Thus, one spring is satisfactory–A313, as wound, unpeened, squared and ground, d = 0.0915 in, OD = 0.879 + 0.092 = 0.971 in, Nt = 15.84 turns 10-24

The steps are the same as in Prob. 10-23 except that the Gerber-Zimmerli criterion is replaced with Goodman-Zimmerli: Ssa Sse = 1 − (Ssm /Ssu ) The problem then proceeds as in Prob. 10-23. The results for the wire sizes are shown below (see solution to Prob. 10-23 for additional details). d1

d m A Sut Ssu Ssy Sse Ssa α β C D

d2

d3

Iteration of d for the first trial d4 d1

0.080 0.0915 0.1055 0.1205 0.146 0.146 0.263 0.263 169.000 169.000 128.000 128.000 244.363 239.618 231.257 223.311 163.723 160.544 154.942 149.618 85.527 83.866 80.940 78.159 52.706 53.239 54.261 55.345 43.513 43.560 43.634 43.691 29.008 29.040 29.090 29.127 2.785 2.129 1.602 1.228 9.052 12.309 16.856 22.433 0.724 1.126 1.778 2.703

d2

d3

d 0.080 0.0915 0.1055 KB 1.151 1.108 1.078 τa 29.008 29.040 29.090 nf 1.500 1.500 1.500 Na 14.444 6.572 2.951 Nt 16.444 8.572 4.951 Ls 1.316 0.784 0.522 ymax 2.875 2.875 2.875 L0 4.191 3.659 3.397 (L 0 ) cr 3.809 5.924 9.354 τs 85.782 85.876 86.022 ns 0.997 0.977 0.941 f (Hz) 138.806 144.277 148.617

d4 0.1205 1.058 29.127 1.500 1.429 3.429 0.413 2.875 3.288 14.219 86.133 0.907 151.618

Without checking all of the design conditions, it is obvious that none of the wire sizes satisfy n s ≥ 1.2. Also, the Gerber line is closer to the yield line than the Goodman. Setting n f = 1.5 for Goodman makes it impossible to reach the yield line (n s < 1) . The table below uses n f = 2. d1 d m A Sut Ssu Ssy Sse Ssa α β C D

d2

Iteration of d for the second trial d3 d4 d1

0.080 0.0915 0.1055 0.1205 0.146 0.146 0.263 0.263 169.000 169.000 128.000 128.000 244.363 239.618 231.257 223.311 163.723 160.544 154.942 149.618 85.527 83.866 80.940 78.159 52.706 53.239 54.261 55.345 43.513 43.560 43.634 43.691 21.756 21.780 21.817 21.845 2.785 2.129 1.602 1.228 6.395 8.864 12.292 16.485 0.512 0.811 1.297 1.986

d KB τa nf Na Nt Ls ymax L0 (L 0 ) cr τs ns f (Hz)

d2

d3

0.080 0.0915 0.1055 1.221 1.154 1.108 21.756 21.780 21.817 2.000 2.000 2.000 40.962 17.594 7.609 42.962 19.594 9.609 3.437 1.793 1.014 2.875 2.875 2.875 6.312 4.668 3.889 2.691 4.266 6.821 64.336 64.407 64.517 1.329 1.302 1.255 98.065 103.903 108.376

d4 0.1205 1.079 21.845 2.000 3.602 5.602 0.675 2.875 3.550 10.449 64.600 1.210 111.418

The satisfactory spring has design specifications of: A313, as wound, unpeened, squared and ground, d = 0.0915 in, OD = 0.811 + 0.092 = 0.903 in , Nt = 19.6 turns.

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10-25

This is the same as Prob. 10-23 since Sse = Ssa = 35 kpsi. Therefore, design the spring using: A313, as wound, un-peened, squared and ground, d = 0.915 in, OD = 0.971 in, Nt = 15.84 turns.

10-26

For the Gerber fatigue-failure criterion, Ssu = 0.67S ut , Sse =

Ssa , 1 − (Ssm /Ssu ) 2

Ssa =

2 2

r Ssu  −1 + 2Sse

1+

2Sse r Ssu

2

 

The equation for Ssa is the basic difference. The last 2 columns of diameters of Ex. 10-5 are presented below with additional calculations. d = 0.105

d = 0.112

278.691 186.723 38.325 125.411 34.658 23.105 1.732 12.004 1.260 1.155 1.365

276.096 184.984 38.394 124.243 34.652 23.101 1.523 13.851 1.551 1.439 1.663

Sut Ssu Sse Ssy Ssa α β C D ID OD

Na Ls L0 (L 0 ) cr KB τa nf τs ns fn fom

d = 0.105

d = 0.112

8.915 1.146 3.446 6.630 1.111 23.105 1.500 70.855 1.770 105.433 −0.973

6.190 0.917 3.217 8.160 1.095 23.101 1.500 70.844 1.754 106.922 −1.022

There are only slight changes in the results. 10-27

As in Prob. 10-26, the basic change is Ssa . Sse =

For Goodman,

Ssa 1 − (Ssm /Ssu )

Recalculate Ssa with

r Sse Ssu r Ssu + Sse Calculations for the last 2 diameters of Ex. 10-5 are given below. Ssa =

Sut Ssu Sse Ssy Ssa α β C D ID OD

d = 0.105

d = 0.112

278.691 186.723 49.614 125.411 34.386 22.924 1.732 11.899 1.249 1.144 1.354

276.096 184.984 49.810 124.243 34.380 22.920 1.523 13.732 1.538 1.426 1.650

There are only slight differences in the results.

Na Ls L0 (L 0 ) cr KB τa nf τs ns fn fom

d = 0.105

d = 0.112

9.153 1.171 3.471 6.572 1.112 22.924 1.500 70.301 1.784 104.509 −0.986

6.353 0.936 3.236 8.090 1.096 22.920 1.500 70.289 1.768 106.000 −1.034

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Chapter 10

10-28

287

Use: E = 28.6 Mpsi, G = 11.5 Mpsi, A = 140 kpsi · inm , m = 0.190, rel cost = 1. d = 0.067 in,

Try

Sut =

140 = 234.0 kpsi (0.067) 0.190

Table 10-6:

Ssy = 0.45Sut = 105.3 kpsi

Table 10-7:

S y = 0.75Sut = 175.5 kpsi

Eq. (10-34) with D/d = C and C1 = C Sy Fmax σA = [(K ) A (16C) + 4] = 2 πd ny πd 2 S y 4C 2 − C − 1 (16C) + 4 = 4C(C − 1) n y Fmax πd 2 S y 2 4C − C − 1 = (C − 1) −1 4n y Fmax πd 2 S y 1 1 πd 2 S y 2 1+ −1 C + −2 =0 C − 4 4n y Fmax 4 4n y Fmax   2 2 2 2 πd S y πd S y πd S y 1 ± − + 2  take positive root C=  2 16n y Fmax 16n y Fmax 4n y Fmax 1 π(0.0672 )(175.5)(103 ) = 2 16(1.5)(18)  2  2 3 2 3 π(0.067) (175.5)(10 ) π(0.067) (175.5)(10 ) + 2 = 4.590 + −  16(1.5)(18) 4(1.5)(18) D = Cd = 0.3075 in πd 3 C −3 πd 3 τi 33 500 = ± 1000 4 − Fi = 8D 8D exp(0.105C) 6.5 Use the lowest Fi in the preferred range. This results in the best fom. 33 500 4.590 − 3 π(0.067) 3 − 1000 4 − = 6.505 lbf Fi = 8(0.3075) exp[0.105(4.590)] 6.5 For simplicity, we will round up to the next integer or half integer; therefore, use Fi = 7 lbf 18 − 7 = 22 lbf/in 0.5 d4G (0.067) 4 (11.5)(106 ) Na = = = 45.28 turns 8k D 3 8(22)(0.3075) 3 11.5 G = 44.88 turns Nb = Na − = 45.28 − E 28.6 L 0 = (2C − 1 + Nb )d = [2(4.590) − 1 + 44.88](0.067) = 3.555 in k=

L 18 lbf = 3.555 + 0.5 = 4.055 in

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Body: K B = τmax = (n y ) body =

4(4.590) + 2 4C + 2 = = 1.326 4C − 3 4(4.590) − 3 8K B Fmax D 8(1.326)(18)(0.3075) = (10−3 ) = 62.1 kpsi 3 3 πd π(0.067) Ssy 105.3 = 1.70 = τmax 62.1

r2 = 2d = 2(0.067) = 0.134 in, C2 =

2(0.134) 2r2 = =4 d 0.067

4(4) − 1 4C2 − 1 = = 1.25 4C2 − 4 4(4) − 4 8Fmax D 8(18)(0.3075) τ B = (K ) B = 1.25 (10−3 ) = 58.58 kpsi πd 3 π(0.067) 3

(K ) B =

(n y ) B =

Ssy 105.3 = 1.80 = τB 58.58

fom = −(1)

π 2 (0.067) 2 (44.88 + 2)(0.3075) π 2 d 2 ( Nb + 2) D =− = −0.160 4 4

Several diameters, evaluated using a spreadsheet, are shown below. d:

0.067

0.072

0.076

0.081

0.085

Sut 233.977 105.290 Ssy Sy 175.483 C 4.589 D 0.307 Fi (calc) 6.505 Fi (rd) 7.0 k 22.000 Na 45.29 Nb 44.89 L0 3.556 L 18 lbf 4.056 KB 1.326 τmax 62.118 (n y ) body 1.695 τB 58.576 (n y ) B 1.797 (n y ) A 1.500 −0.160 fom

230.799 103.860 173.100 5.412 0.390 5.773 6.0 24.000 27.20 26.80 2.637 3.137 1.268 60.686 1.711 59.820 1.736 1.500 −0.144

228.441 102.798 171.331 6.099 0.463 5.257 5.5 25.000 19.27 18.86 2.285 2.785 1.234 59.707 1.722 60.495 1.699 1.500 −0.138

225.692 101.561 169.269 6.993 0.566 4.675 5.0 26.000 13.10 12.69 2.080 2.580 1.200 58.636 1.732 61.067 1.663 1.500 −0.135

223.634 100.635 167.726 7.738 0.658 4.251 4.5 27.000 9.77 9.36 2.026 2.526 1.179 57.875 1.739 61.367 1.640 1.500 −0.133

0.09 221.219 99.548 165.914 8.708 0.784 3.764 4.0 28.000 7.00 6.59 2.071 2.571 1.157 57.019 1.746 61.598 1.616 1.500 −0.135

0.095

0.104

218.958 98.531 164.218 9.721 0.923 3.320 3.5 29.000 5.13 4.72 2.201 2.701 1.139 56.249 1.752 61.712 1.597 1.500 −0.138

215.224 96.851 161.418 11.650 1.212 2.621 3.0 30.000 3.15 2.75 2.605 3.105 1.115 55.031 1.760 61.712 1.569 1.500 −0.154

Except for the 0.067 in wire, all springs satisfy the requirements of length and number of coils. The 0.085 in wire has the highest fom.

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Chapter 10

10-29

Given: Nb = 84 coils, Fi = 16 lbf, OQ&T steel, OD = 1.5 in, d = 0.162 in. D = 1.5 − 0.162 =1.338 in (a) Eq. (10-39): L 0 = 2( D − d) + ( Nb + 1)d = 2(1.338 − 0.162) + (84 + 1)(0.162) = 16.12 in Ans. 2d + L 0 = 2(0.162) + 16.12 = 16.45 in overall.

or

1.338 D = = 8.26 d 0.162 4(8.26) + 2 KB = = 1.166 4(8.26) − 3 8Fi D 8(16)(1.338) τi = K B = 1.166 = 14 950 psi Ans. πd 3 π(0.162) 3 C=

(b)

(c) From Table 10-5 use: G = 11.4(106 ) psi and E = 28.5(106 ) psi 11.4 G Na = Nb + = 84 + = 84.4 turns E 28.5 d4G (0.162) 4 (11.4)(106 ) = 4.855 lbf/in Ans. k= = 8D 3 Na 8(1.338) 3 (84.4) (d) Table 10-4:

A = 147 psi · inm , m = 0.187 147 = 207.1 kpsi Sut = (0.162) 0.187 S y = 0.75(207.1) = 155.3 kpsi Ssy = 0.50(207.1) = 103.5 kpsi

Body πd 3 Ssy π KB D π(0.162) 3 (103.5)(103 ) = 110.8 lbf = 8(1.166)(1.338) Torsional stress on hook point B F=

2(0.25 + 0.162/2) 2r2 = = 4.086 d 0.162 4(4.086) − 1 4C2 − 1 (K ) B = = = 1.243 4C2 − 4 4(4.086) − 4 π(0.162) 3 (103.5)(103 ) = 103.9 lbf F= 8(1.243)(1.338) Normal stress on hook point A C2 =

1.338 2r1 = = 8.26 d 0.162 4C12 − C1 − 1 4(8.26) 2 − 8.26 − 1 (K ) A = = = 1.099 4C1 (C1 − 1) 4(8.26)(8.26 − 1) C1 =

289

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4 16(K ) A D S yt = σ = F + πd 3 πd 2

155.3(103 ) F= = 85.8 lbf [16(1.099)(1.338)]/[π(0.162) 3 ] + {4/[π(0.162) 2 ]} = min(110.8, 103.9, 85.8) = 85.8 lbf Ans. (e) Eq. (10-48): 85.8 − 16 F − Fi = = 14.4 in Ans. k 4.855

y= 10-30

Fmin = 9 lbf,

Fmax = 18 lbf Fa =

A313 stainless:

18 − 9 = 4.5 lbf, 2

0.013 ≤ d ≤ 0.1 0.1 ≤ d ≤ 0.2 E = 28 Mpsi,

Fm =

18 + 9 = 13.5 lbf 2

A = 169 kpsi · inm , m = 0.146 A = 128 kpsi · inm , m = 0.263 G = 10 Gpsi

Try d = 0.081 in and refer to the discussion following Ex. 10-7 Sut =

169 = 243.9 kpsi (0.081) 0.146

Ssu = 0.67Sut = 163.4 kpsi Ssy = 0.35Sut = 85.4 kpsi S y = 0.55Sut = 134.2 kpsi Table 10-8:

Sr = 0.45Sut = 109.8 kpsi Se =

Sr /2 109.8/2 = = 57.8 kpsi 1 − [Sr /(2Sut )]2 1 − [(109.8/2)/243.9]2

r = Fa /Fm = 4.5/13.5 = 0.333   2 2 2 r Sut  2Se  −1 + 1 + Table 7-10: Sa = 2Se r Sut   2 2 2 (0.333) (243.9 )  2(57.8)  = 42.2 kpsi Sa = −1 + 1 + 2(57.8) 0.333(243.9) Hook bending

Sa 4 Sa 16C = (σa ) A = Fa (K ) A 2 + = πd πd 2 (n f ) A 2 Sa 4.5 (4C 2 − C − 1)16C +4 = 2 πd 4C(C − 1) 2

This equation reduces to a quadratic in C—see Prob. 10-28

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Chapter 10

291

The useable root for C is   2 2 2 2 πd Sa πd Sa πd Sa + + 2 − C = 0.5  144 144 36   2  π(0.081) 2 (42.2)(103 )  2 3 2 3 π(0.081) (42.2)(10 ) π(0.081) (42.2)(10 ) = 0.5 + +2 −   144 144 36 = 4.91

D = Cd = 0.398 in

πd 3 C −3 πd 3 τi 33 500 = ± 1000 4 − Fi = 8D 8D exp(0.105C) 6.5 Use the lowest Fi in the preferred range. 4.91 − 3 π(0.081) 3 33 500 − 1000 4 − Fi = 8(0.398) exp[0.105(4.91)] 6.5 = 8.55 lbf For simplicity we will round up to next 1/4 integer. Fi = 8.75 lbf k=

18 − 9 = 36 lbf/in 0.25

d4G (0.081) 4 (10)(106 ) = = 23.7 turns 8k D 3 8(36)(0.398) 3 10 G = 23.3 turns Nb = Na − = 23.7 − E 28 L 0 = (2C − 1 + Nb )d = [2(4.91) − 1 + 23.3](0.081) = 2.602 in L max = L 0 + ( Fmax − Fi )/k = 2.602 + (18 − 8.75)/36 = 2.859 in 4.5(4) 4C 2 − C − 1 +1 (σa ) A = πd 2 C −1 18(10−3 ) 4(4.912 ) − 4.91 − 1 + 1 = 21.1 kpsi = π(0.0812 ) 4.91 − 1 Na =

(n f ) A = Body:

Sa 42.2 = 2 checks = (σa ) A 21.1 KB =

4(4.91) + 2 4C + 2 = = 1.300 4C − 3 4(4.91) − 3

τa =

8(1.300)(4.5)(0.398) (10−3 ) = 11.16 kpsi π(0.081) 3

τm =

Fm 13.5 (11.16) = 33.47 kpsi τa = Fa 4.5

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The repeating allowable stress from Table 7-8 is Ssr = 0.30Sut = 0.30(243.9) = 73.17 kpsi The Gerber intercept is Sse = From Table 7-10, (n f ) body =

1 2

163.4 33.47

73.17/2 = 38.5 kpsi 1 − [(73.17/2)/163.4]2

2

   2  11.16 2(33.47)(38.5) = 2.53 −1 + 1 + 38.5  163.4(11.16) 

Let r2 = 2d = 2(0.081) = 0.162 2r2 = 4, C2 = d

(K ) B =

4(4) − 1 = 1.25 4(4) − 4

(τa ) B =

(K ) B 1.25 (11.16) = 10.73 kpsi τa = KB 1.30

(τm ) B =

(K ) B 1.25 (33.47) = 32.18 kpsi τm = KB 1.30

Table 10-8: (Ssr ) B = 0.28Sut = 0.28(243.9) = 68.3 kpsi 68.3/2 = 35.7 kpsi 1 − [(68.3/2)/163.4]2   2  2  10.73 1 163.4 2(32.18)(35.7) (n f ) B = −1 + 1 + = 2.51 2 32.18 35.7  163.4(10.73) 

(Sse ) B =

Yield Bending:

4Fmax (4C 2 − C − 1) +1 (σ A ) max = πd 2 C −1 4(18) 4(4.91) 2 − 4.91 − 1 + 1 (10−3 ) = 84.4 kpsi = 2 π(0.081 ) 4.91 − 1 (n y ) A =

Body:

134.2 = 1.59 84.4

τi = ( Fi /Fa )τa = (8.75/4.5)(11.16) = 21.7 kpsi r = τa /(τm − τi ) = 11.16/(33.47 − 21.7) = 0.948 r 0.948 (Ssa ) y = (Ssy − τi ) = (85.4 − 21.7) = 31.0 kpsi r +1 0.948 + 1 (Ssa ) y 31.0 = 2.78 = (n y ) body = τa 11.16

Hook shear: Ssy = 0.3Sut = 0.3(243.9) = 73.2 kpsi τmax = (τa ) B + (τm ) B = 10.73 + 32.18 = 42.9 kpsi

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Chapter 10

73.2 = 1.71 42.9 7.6π 2 d 2 ( Nb + 2) D 7.6π 2 (0.081) 2 (23.3 + 2)(0.398) fom = − =− = −1.239 4 4 A tabulation of several wire sizes follow (n y ) B =

d Sut Ssu Sr Se Sa C D OD Fi (calc) Fi (rd) k Na Nb L0 L 18 lbf (σa ) A (n f ) A KB (τa ) body (τm ) body Ssr Sse (n f ) body (K ) B (τa ) B (τm ) B (Ssr ) B (Sse ) B (n f ) B Sy (σ A ) max (n y ) A τi r (Ssy ) body (Ssa ) y (n y ) body (Ssy ) B (τ B ) max (n y ) B fom

0.081 243.920 163.427 109.764 57.809 42.136 4.903 0.397 0.478 8.572 8.75 36.000 23.86 23.50 2.617 2.874 21.068 2.000 1.301 11.141 33.424 73.176 38.519 2.531 1.250 10.705 32.114 68.298 35.708 2.519 134.156 84.273 1.592 21.663 0.945 85.372 30.958 2.779 73.176 42.819 1.709 −1.246

0.085 242.210 162.281 108.994 57.403 41.841 5.484 0.466 0.551 7.874 9.75 36.000 17.90 17.54 2.338 2.567 20.920 2.000 1.264 10.994 32.982 72.663 38.249 2.547 1.250 10.872 32.615 67.819 35.458 2.463 133.215 83.682 1.592 23.820 1.157 84.773 32.688 2.973 72.663 43.486 1.671 −1.234

0.092 239.427 160.416 107.742 56.744 41.360 6.547 0.602 0.694 6.798 10.75 36.000 11.38 11.02 2.127 2.328 20.680 2.000 1.216 10.775 32.326 71.828 37.809 2.569 1.250 11.080 33.240 67.040 35.050 2.388 131.685 82.720 1.592 25.741 1.444 83.800 34.302 3.183 71.828 44.321 1.621 −1.245 optimal fom

The shaded areas show the conditions not satisfied.

0.098 237.229 158.943 106.753 56.223 40.980 7.510 0.736 0.834 5.987 11.75 36.000 8.03 7.68 2.126 2.300 20.490 2.000 1.185 10.617 31.852 71.169 37.462 2.583 1.250 11.200 33.601 66.424 34.728 2.341 130.476 81.961 1.592 27.723 1.942 83.030 36.507 3.438 71.169 44.801 1.589 −1.283

0.105 234.851 157.350 105.683 55.659 40.570 8.693 0.913 1.018 5.141 12.75 36.000 5.55 5.19 2.266 2.412 20.285 2.000 1.157 10.457 31.372 70.455 37.087 2.596 1.250 11.294 33.883 65.758 34.380 2.298 129.168 81.139 1.592 29.629 2.906 82.198 39.109 3.740 70.455 45.177 1.560 −1.357

0.12 230.317 154.312 103.643 54.585 39.786 11.451 1.374 1.494 3.637 13.75 36.000 2.77 2.42 2.918 3.036 19.893 2.000 1.117 10.177 30.532 69.095 36.371 2.616 1.250 11.391 34.173 64.489 33.717 2.235 126.674 79.573 1.592 31.097 4.703 80.611 40.832 4.012 69.095 45.564 1.516 −1.639

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10-31

For the hook, F

M = F R sin θ, ∂ M/∂ F = R sin θ π/2 1 π P R3 δF = F R 2 sin2 R dθ = EI 0 2 EI

R D兾2 ␪

The total deflection of the body and the two hooks π F R3 8F D 3 Nb π F( D/2) 3 8F D 3 Nb + 2 = + d4G 2 EI d4G E(π/64)(d 4 ) 8F D 3 8F D 3 Na G = 4 Nb + = d G E d4G

δ=

10-32

Na = Nb +

G E

QED

Table 10-4 for A227: A = 140 kpsi · inm ,

m = 0.190

E = 28.5(106 ) psi

Table 10-5:

Sut =

140 = 197.8 kpsi (0.162) 0.190

Eq. (10-57): S y = σall = 0.78(197.8) = 154.3 kpsi D = 1.25 − 0.162 = 1.088 in C = D/d = 1.088/0.162 = 6.72 4(6.72) 2 − 6.72 − 1 4C 2 − C − 1 = = 1.125 Ki = 4C(C − 1) 4(6.72)(6.72 − 1) σ = Ki

From

32M πd 3

Solving for M for the yield condition, My =

πd 3 S y π(0.162) 3 (154 300) = 57.2 lbf · in = 32K i 32(1.125)

Count the turns when M = 0 N = 2.5 −

My d 4 E/(10.8D N )

from which N= =

2.5 1 + [10.8D M y /(d 4 E)] 2.5 = 2.417 turns 1 + {[10.8(1.088)(57.2)]/[(0.162) 4 (28.5)(106 )]}

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295

Chapter 10

This means (2.5 − 2.417)(360◦ ) or 29.9◦ from closed. Treating the hand force as in the middle of the grip 3.5 = 2.75 in r =1+ 2 My 57.2 = = 20.8 lbf Ans. F= r 2.75 10-33

The spring material and condition are unknown. Given d = 0.081 in and OD = 0.500, (a) D = 0.500 − 0.081 = 0.419 in Using E = 28.6 Mpsi for an estimate (0.081) 4 (28.6)(106 ) d4 E k = = = 24.7 lbf · in/turn 10.8D N 10.8(0.419)(11) for each spring. The moment corresponding to a force of 8 lbf Fr = (8/2)(3.3125) = 13.25 lbf · in/spring The fraction windup turn is n=

13.25 Fr = 0.536 turns = k 24.7

The arm swings through an arc of slightly less than 180◦ , say 165◦ . This uses up 165/360 or 0.458 turns. So n = 0.536 − 0.458 = 0.078 turns are left (or 0.078(360◦ ) = 28.1◦ ). The original configuration of the spring was

Ans. 28.1

(b)

C=

0.419 = 5.17 0.081

Ki =

4(5.17) 2 − 5.17 − 1 = 1.168 4(5.17)(5.17 − 1)

32M πd 3 32(13.25) = 296 623 psi Ans. = 1.168 π(0.081) 3

σ = Ki

To achieve this stress level, the spring had to have set removed. 10-34

Consider half and double results 3FR

Straight section:

L兾2

F

M = 3F R,

∂M = 3R ∂P

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Upper 180◦ section: M = F[R + R(1 − cos φ)] ␾ ␪

R

= F R(2 − cos φ),

∂M = R(2 − cos φ) ∂P

F

Lower section:

M = F R sin θ ∂M = R sin θ ∂P

Considering bending only: L/2 π π/2 2 2 2 2 2 δ= 9F R dx + F R (2 − cos φ) R dφ + F( R sin θ) R dθ EI 0 0 0

π π 2F 9 2

3 3 π R L + R 4π − 4 sin φ + +R = 0 EI 2 2 4 2F R 2 19π 9 F R2 = R+ L = (19π R + 18L) Ans. EI 4 2 2E I 10-35


10-36


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Chapter 11 11-1

For the deep-groove 02-series ball bearing with R = 0.90, the design life x D , in multiples of rating life, is xD =

30 000(300)(60) = 540 106

Ans.

The design radial load FD is FD = 1.2(1.898) = 2.278 kN From Eq. (11-6),

C10

540 = 2.278 0.02 + 4.439[ln(1/0.9)]1/1.483 = 18.59 kN

1/3

Ans.

Table 11-2: Choose a 02-30 mm with C10 = 19.5 kN. Ans. Eq. (11-18):

1.483 540(2.278/19.5) 3 − 0.02 R = exp − 4.439 = 0.919 Ans.

11-2

For the Angular-contact 02-series ball bearing as described, the rating life multiple is xD =

50 000(480)(60) = 1440 106

The design load is radial and equal to FD = 1.4(610) = 854 lbf = 3.80 kN Eq. (11-6):

C10

1440 = 854 0.02 + 4.439[ln(1/0.9)]1/1.483

1/3

= 9665 lbf = 43.0 kN Table 11-2: Select a 02-55 mm with C10 = 46.2 kN. Ans. Using Eq. (11-18),

1.483 1440(3.8/46.2) 3 − 0.02 R = exp − 4.439 = 0.927 Ans.

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11-3

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For the straight-Roller 03-series bearing selection, x D = 1440 rating lives from Prob. 11-2 solution. FD = 1.4(1650) = 2310 lbf = 10.279 kN 1440 3/10 = 91.1 kN C10 = 10.279 1 Table 11-3: Select a 03-55 mm with C10 = 102 kN. Ans. Using Eq. (11-18),

1.483 1440(10.28/102) 10/3 − 0.02 R = exp − = 0.942 Ans. 4.439

11-4

√ We can choose a reliability goal of 0.90 = 0.95 for each bearing. We make the selections, find the existing reliabilities, multiply them together, and observe that the reliability goal is exceeded due to the roundup of capacity upon table entry. Another possibility is to use the reliability of one bearing, say R1 . Then set the reliability goal of the second as R2 =

0.90 R1

or vice versa. This gives three pairs of selections to compare in terms of cost, geometry implications, etc. 11-5

Establish a reliability goal of tact ball bearing,

√ 0.90 = 0.95 for each bearing. For a 02-series angular con

C10

1440 = 854 0.02 + 4.439[ln(1/0.95)]1/1.483

1/3

= 11 315 lbf = 50.4 kN Select a 02-60 mm angular-contact bearing with C10 = 55.9 kN. 1.483 1440(3.8/55.9) 3 − 0.02 R A = exp − = 0.969 4.439 For a 03-series straight-roller bearing, 3/10 1440 = 105.2 kN C10 = 10.279 0.02 + 4.439[ln(1/0.95)]1/1.483 Select a 03-60 mm straight-roller bearing with C10 = 123 kN. 1.483 1440(10.28/123) 10/3 − 0.02 R B = exp − = 0.977 4.439

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299

Chapter 11

Form a table of existing reliabilities Rgoal

RA

RB

0.912

0.90 0.95

0.927 0.969

0.941 0.977

0.872 0.947 0.906

The possible products in the body of the table are displayed to the right of the table. One, 0.872, is predictably less than the overall reliability goal. The remaining three are the choices for a combined reliability goal of 0.90. Choices can be compared for the cost of bearings, outside diameter considerations, bore implications for shaft modifications and housing modifications. The point is that the designer has choices. Discover them before making the selection decision. Did the answer to Prob. 11-4 uncover the possibilities? To reduce the work to fill in the body of the table above, a computer program can be helpful. 11-6

Choose a 02-series ball bearing from manufacturer #2, having a service factor of 1. For Fr = 8 kN and Fa = 4 kN xD = Eq. (11-5):

5000(900)(60) = 270 106

C10

270 =8 0.02 + 4.439[ln(1/0.90)]1/1.483

1/3

= 51.8 kN

Trial #1: From Table (11-2) make a tentative selection of a deep-groove 02-70 mm with C0 = 37.5 kN. 4 Fa = 0.107 = C0 37.5 Table 11-1: Fa /(V Fr ) = 0.5 > e X 2 = 0.56,

Y2 = 1.46

Eq. (11-9): Fe = 0.56(1)(8) + 1.46(4) = 10.32 kN Eq. (11-6):

C10

270 = 10.32 1

1/3

= 66.7 kN > 61.8 kN

Trial #2: From Table 11-2 choose a 02-80 mm having C10 = 70.2 and C0 = 45.0. Check: Fa 4 = 0.089 = C0 45

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Table 11-1: X 2 = 0.56, Y2 = 1.53 Fe = 0.56(8) + 1.53(4) = 10.60 kN Eq. (11-6):

C10

270 = 10.60 1

1/3

= 68.51 kN < 70.2 kN

∴ Selection stands. Decision: Specify a 02-80 mm deep-groove ball bearing. 11-7

Ans.

From Prob. 11-6, x D = 270 and the final value of Fe is 10.60 kN. 1/3 270 = 84.47 kN C10 = 10.6 0.02 + 4.439[ln(1/0.96)]1/1.483 Table 11-2: Choose a deep-groove ball bearing, based upon C10 load ratings. Trial #1: Tentatively select a 02-90 mm. C10 = 95.6,

C0 = 62 kN

4 Fa = 0.0645 = C0 62 From Table 11-1, interpolate for Y2 . Fa /C0

Y2

0.056 0.0645 0.070

1.71 Y2 1.63

0.0645 − 0.056 Y2 − 1.71 = = 0.607 1.63 − 1.71 0.070 − 0.056 Y2 = 1.71 + 0.607(1.63 − 1.71) = 1.661 Fe = 0.56(8) + 1.661(4) = 11.12 kN 1/3 270 C10 = 11.12 0.02 + 4.439[ln(1/0.96)]1/1.483 = 88.61 kN < 95.6 kN Bearing is OK. Decision: Specify a deep-groove 02-90 mm ball bearing.

Ans.

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301

Chapter 11

11-8

For the straight cylindrical roller bearing specified with a service factor of 1, R = 0.90 and Fr = 12 kN 4000(750)(60) = 180 106 180 3/10 = 12 = 57.0 kN Ans. 1

xD = C10 11-9

y R yO

O

z

T

R zO

Pz R yA 11

1" 2

F

Py R zA

A B

2

3" 4

20⬚

x T

Assume concentrated forces as shown.

Pz = 8(24) = 192 lbf Py = 8(30) = 240 lbf T = 192(2) = 384 lbf · in T x = −384 + 1.5F cos 20◦ = 0 F=

384 = 272 lbf 1.5(0.940)

M Oz = 5.75Py + 11.5R A − 14.25F sin 20◦ = 0; y

y

5.75(240) + 11.5R A − 14.25(272)(0.342) = 0

thus y

thus

R A = −4.73 lbf M O = −5.75Pz − 11.5R zA − 14.25F cos 20◦ = 0; y

−5.75(192) − 11.5R zA − 14.25(272)(0.940) = 0

R zA = −413 lbf;

R A = [(−413) 2 + (−4.73) 2 ]1/2 = 413 lbf

z + Pz + R zA + F cos 20◦ = 0 F z = RO z + 192 − 413 + 272(0.940) = 0 RO z = −34.7 lbf RO

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F y = R O + Py + R A − F sin 20◦ = 0 y

y

y

R O + 240 − 4.73 − 272(0.342) = 0 y

R O = −142 lbf R O = [(−34.6) 2 + (−142) 2 ]1/2 = 146 lbf So the reaction at A governs. √ Reliability Goal: 0.92 = 0.96 FD = 1.2(413) = 496 lbf x D = 30 000(300)(60/106 ) = 540 1/3 540 C10 = 496 0.02 + 4.439[ln(1/0.96)]1/1.483 = 4980 lbf = 22.16 kN A 02-35 bearing will do. Decision: Specify an angular-contact 02-35 mm ball bearing for the locations at A and O. Check combined reliability. Ans. 11-10

For a combined reliability goal of 0.90, use

√ 0.90 = 0.95 for the individual bearings.

y

O z

RO A

FA

20 B

FC

RB

x0 = 20⬚

50 000(480)(60) = 1440 106

16 C

x

10

The resultant of the given forces are R O = 607 lbf and R B = 1646 lbf. At O: Fe = 1.4(607) = 850 lbf C10 = 850 Ball:

1440 0.02 + 4.439[ln(1/0.95)]1/1.483

1/3

= 11 262 lbf or 50.1 kN Select a 02-60 mm angular-contact ball bearing with a basic load rating of 55.9 kN. At B: Fe = 1.4(1646) = 2304 lbf C10 = 2304 Roller:

1440 0.02 + 4.439[ln(1/0.95)]1/1.483

3/10

= 23 576 lbf or 104.9 kN Select a 02-80 mm cylindrical roller or a 03-60 mm cylindrical roller. The 03-series roller has the same bore as the 02-series ball.

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303

Chapter 11

11-11

The reliability of the individual bearings is R =

√ 0.999 = 0.9995

y R yO

R zO

O z

A 300 F zA

FC F yA 400

R yE

C

E 150

x R zE

From statics, y

R O = −163.4 N, y RE

= −89.1 N,

z RO = 107 N,

R Ez = −174.4 N,

R O = 195 N R E = 196 N

60 000(1200)(60) = 4320 106 1/3 4340 = 0.196 0.02 + 4.439[ln(1/0.9995)]1/1.483

xD = C10

= 8.9 kN A 02-25 mm deep-groove ball bearing has a basic load rating of 14.0 kN which is ample. An extra-light bearing could also be investigated. 11-12

Given: Fr A = 560 lbf or 2.492 kN Fr B = 1095 lbf or 4.873 kN Trial #1: Use K A = K B = 1.5 and from Table 11-6 choose an indirect mounting. 0.47Fr B 0.47Fr A

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