Structural Engineering Formulas

SnnucrrlRAr', ExcrxnERrNG tr'onuul-,AS

. IIORSION . IMPACT . COMPRESSION . TENSION BENDING

. . BNAMS ' FRAMES . ARCIIES IIRUSSES PLA.IES . . RE||ATNING WALLS PIPES AND TUNNELS

FOUNDATIONS

ILYA MrriHBr,soN, PII.D' ILLL]SITRATIONS BY LIA MIKISLSON, M.S.

MCGRAW.HILL LONDON MADRID NE$,YORI< CIIICAGO SNF'RANCISCO LISBON SNOUL JUAN SAN NES'DEI,HI MILAN CITY MEXICO SINGAPORT SYDNEY ITORONIIO

.'..

Library of Congress Catatoging-in-publication

Data

To my wif e and son

Mikhelson, llya. Structural engineering formulas / ilya Mikhelson. cm.

p.

tsBN 0-07-14391 1-0 1. Structural engine€ring-Mathematics. TA636.M55 2004

2. Mathematics_Formulae. L Tifle.

624.1'02'12-4c22

2004044803

@ 2004 by llya lvlikhelson. AII rights reserued. printed in the united states of America. Except as permitted under the united states copyright Act of 1 976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval svstem.

copyright

without the prior written permission of the publisher. 1

234567 890

DOC/DOC 01

0987654

lsBN 0-07-14391 1-0 The sponsoring editor for this book was Larry s. Hager, the editing superyisor was stephen M. smith, and the production supervisor was sherri souffrance. The art director for the cover was Maroaret Webster-Shapiro. Printed and bound by RR Donnelley McGraw-Hill books are available at special quantity discounts to use as premiums and sales promotions, or for use in corporate training programs. For more information, please write to the Direator of special sales, Mccraw-Hlll Professional, Two p enn plaza, New york, Ny 1 0121-2298. or contact vour local bookstore.

This book is printed on recycled, acid-free paper containing a minimum of 50% recycled, de-inked fiber. Information contained in this work has been obtained by The McGraw-Hill companies, Inc. (,,MccrawHill") from sources believed to be reliable. However, neither McGraw-Hill nor its authors guarantee the accuracy or completeness of any information published herein and neither McGraw-Hill nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that Mccraw-Hill and its authors are supplying information but are not attempting lo render engineering or other professional services. lf such services are reouired. the assistance of an appropriate professional should be sought.

:-

CONTE

N

s

Preface

Acknowledgments Introduction Basis of Structural Analysis

1. Stress and Strain.

Tables

Methods ofAnalysis

Tension and compression

't.1

1.2,1.3

Bending Combination of compression (tension) and bending

'1.4

Torsion

1.5

Curved beams

1.6

Continuous deep beams Dynamics, transverse oscillations of the beams Dynamics, impact

1.7

1.8-1.10 1.1',t,1.12

2. Properties of Geometric Seciions For tension, compression, and bending structures For toFion structures

2.'t-2.5 2.6

Statics 3. Beams.

Diagrams and Formulas forVarious Loading Conditions

Simple beams Simple beams and beams overhanging one support Cantilever beams

3.1-3.3 3.4 3.5

Beams fixed at one end, supported at other

3.6,3.7

Beams fixed at both ends

3.8, 3.9

Continuous beams

3.10

Continuous beams: settlement of support

3.11

Simple beams: moving concentrated loads (general rules)

3.'12

CONTENTS

CONTENTS Bsams: influencs lines (exampl€s)

3.13,3.14

Stress distribution in soil

Eeams: computation of bending moment and shear using influene lines (examples)

3.15,3.16

4, Framos.

Diagrams and Formulas for Various Static Loadlng Conditions

5. Archos.

Diagrams and Formulas for Various Loading Conditions

8.3

8.4,8.5

Settlement of soil Shear strength of soil; slope stability analysis

8.6

Bearing capacity analysis

8.7

4.14.5

9. Foundations Three-hinged arches: support reactions, bending moment, and axial force

5.1

Symmetrical three-hinged arches of any shape: formulas for various static loading conditions

5.2,5.3

Two-hinged parabolic arches: formulas for various static loading mnditions

5.4, 5.5

Fixed parabolic arches: formulas for various static loading @nditions

5.6,5.7

Three-hinged arches: influen@ lines

5.8

Fixed parabolic arches: influence lines

9.1

Direct foundation stability; pile foundations

9.2

Pile group capacity

9.3

9.4

Pile capacity Rigid continuous beam elasti€lly supported

IV

Retaining

9.5-9.7

Structures

10. Retaining Structures

5.9

Steel rope

Direct foundations

Lateral earth pressure on retaining walls

5.10

10.1-10.5

Lateral eadh pressure on braced sheetings; lateral earth pressure on

6, Trusses.

Method of Joints and Method of Section Analysis

Method ofjoints and method of section analysis: examples Infl

uence lines (examples)

7. Plates.

6.3,6.4

moments

| | Rectangular plates: bending moments and defloctions (uniformly distributed load) | Rectangular plales: bending moments (uniformly varying load) | Circular plates: bending moments, shear and defloction (uniformly distributed load) | Rectangular plates: bending moments (uniformly distributed

ilt.

11. Retaining Structures

Bending Moments for Various Support and Loading Conditions

Rectangular plates: bending

10.6

basement walls

6.1,6.2

load)

Cantilever retaining walls; stability analysis

11.1

Cantilever sheet pilings

11.2

Anchored sheet pile walls

11.3

2.1

7.2-T.s 2.0

7,7,1.9

V.

Pipes 1

and Tunnels

2. Pipes and Tunnels.

Bending Moments for Various Static Loading

Conditions

7,g

Rectangular cross-section

Soils and Foundations 8. Soils

1

Engineering properties of soils

8.1

Weighumass and volume relationships; flow of water in soil

8.2

3. Pipes and Tunnels.

12.1-12.5

Bending Moments for Various Static Loading

Conditions Circular cross-section

13.1-13.3

ONTENTS Appendix Units: conversion between Anglo-American and metric systems

u.1, u.2

Mathematical formulas: algebra

M.1, M.2

l\4athematjcal formulas: geometry, solid bodies

M.3, M.4

Mathematical formulas; trigonometry

M.5, M.6

Symbols

PFIEFACE

s.1 This reference book is intended for those engaged in an occupation as important as lt is

interesting--{esign and analysis ofengineering structures. Engineering problems are diverse, and so are the analyses they require. Some are performed with sophisticated computer programs; others call only for a thoughtful application of ready-to-use formulas. ln any situation, the

informationinthiscompilationshouldbehelpful.

ltwillalsoaidengineeringandarchitectural

students and those studying for licensing examinations.

llya N.4ikhelson, Ph.D

INTFIODUCTION

ACI(NO-wLEDGNIENTS Deep appreciation goes to Mikhail Bromblin for his unwavering help in preparing the book's illustrations for publication.

Analysis of structures, regardless of its purpose or complexity, is generally performed in the following order:

. The author would also like to express his gratitude to colleagues Nick Ayoub, Tom Sweeney, and Davidas Neghandi for sharing their extensive engineering experience.

Loads, both permanent (dead loads) and temporary (live loads), acting upon the structure are computed.

.

Forces(axisforces,bendingmoments,sheare,torsionmoments,etc.)resultinginthestructure are determined.

Special thanks is given to Larry Hager for his valuable editorial advice.

o

.

Stresses in the cross-sections of structure elem€nts are found. Depending on the analysis method used, the obtained results are compared with allowable or ultimate forces and stresses allowed by norms.

The norms of structuEl design do not remain constant, but change with the evolving methods of analysis and increasing strength of materials. Furthermore, the norms fof design of various struclures, such as bridges and buildings, are different. Therefore, the analysis methods provided in this book are limited to determination of forces and stresses. Likewise, the included properties

of

materials and soils are approximations and may differ from those

accepted in the norms. AII the formulas provided in the book for analysis of structures are based on the elastic theory.

About the Author llya Mikhelson, Ph.D., has over 30 years'experience in design, research, and teaching design of bridges, tunnels, subway stations, and buildings. He is the author of numerous other publications, including: Precast concrete for Underyround Construction, Tunnels and Subways, and Bu ild i ng Stru cture s.

L.

STFIESS and

STFIAIN Methods of Analysis

STRESS and STRAIN TENSION and COMPRESSION eight

1.1

Axial force: N.

= YA(L_x), Y = unit volume weight,

Tables 1.1-1.1 2 provide formulas for determination of stresses in structural elements for various loading

A = cross - sectional area. N Stresses:o.=;!=f( t__x), o,,=TL. o. r=0.

conditions. To evaluate the results, it is necessary to comparc the computed stresses with existing norm reouirements.

Detbmation:

Diogroms

a =11(zr-*). a ^=0. ' 2E'

=-'}L.=IY-L

^

2E

2FA W = yAL = weight of the beam E = Modulus of elasticity

Axial force : tension, compressi Stresses

PP

o..A'"A - l.

:

6- =

r.

Defomation:

A.

=L-L, (along), Ao =b-b,

ta, e.=-i,

_-_4*-*

Poisson's

e"= ratio:

b

.

u=litl L€.

I

6

6=t€-.F,€=-: o. P uo- uPa, =€, L=-L=-L. A^ =c^b=-b:-b. EEAEEA

nooKes law

Temperature

(cross),

+4.

Case a/

Keacrron:

^ 0.at'EA ^-O*L!.

A, "=i.

L, k=--.

n Axial

force N =-R (compression),

-Rcr'AtoERo.atoE A, lrl-k'"'

nA

k(n-l)+l

n

For A, =d, b) -.

Where

T"0

md

i

o=6.r =62 =-61.4t08, T"0

re

lt'=4-lj

original and considered temperatures.

*ato

0 = coefficient of linear expansion Ato >

0

tension stress. At0
10

1.789

2.4s6

3.123

0.'743

0.742

0.742

0.740

Profile consisting of re cta n g u I a r c ro s s - s e c t i o n s i=n

,-I'

. g =-l' . Assumed: \.t0, lrrto b.;;

Ceometricproperties:

Dr

b,>b, t,,

-crbi .

b:

b,

I

=

I, =Ir, +I,,

Stress: t,,. Argle of

b3

>b, (i.e. h,bl

,

b, =b,""")

r,. =c,b1

+I,,,

I,

' = b,'

S.

.

=!!r linpoint sr

twist:

go

-

180'

rc

'

t.1.

M'L. GI,

.

n=3

STRESS and STRAIN CURVED BEAMS Cu

rved bea m (lransverse bending

)

Stresses:

M y-R,,

-

o. =-.-, '

A.c

y

"" _le

h"_

tA'

c=R_Ro hl

lf :'

-dJ 6

17. TH|N R|NG (t\

^l TI

o

^^

Iil-

o

tt

A^

o

.. r r{} F:b .ts69

v, lu

I

o

oa zu

<J -E

ll

E

oo

_= -


: }r>

*N

.t T i(lr - ?E.E.I "i ^-l ElJl =B €l- rl F -v,l drf ll't .'. il d ^_l -"-LT +Eirl | N*g* ts Tl+

+" Bil^i

>ll

F

- Tlc

!lr

>-

'l l

->

+

o U'

=

t

o

:o

H E -f-FfTtTtTn -77 -

lla

FRAMES

NOTES Example.

o ina x2'54a =10322 cma

Girder W14x82, lz=882inax2'54a =36712

cma

Load P=20kN, a=4m' b=8m moments Required. Compute support reactions and bending a 36712x3 I,h _n

, sotution. K=iI=

.. n=,3 R

_

pab.

" 2L

o

o

=7.8t3 kN.m

kN m =R"L+M" -Pb=13'5'1x12+7 8i3-20x8=10'653 m M. =-Hh+M" =-923x3+7 8!3 =-19'877 kN kN m =-Hh+Mu =-923x3+10'653=-17'037

Bending moment at Point of load

,

_

=

I

5k-l+28(k+2)

n, =r"-[M.-Morr--a)+M".l. "l r '

roil;*

_[

p.877

.

all

J

F

Mo

Mo

ll

1, vi

qt= or =lJ ll

:ln

=: ? ? g vl^ ? J slr +l ,1.: s f 6l

>'l= f j -i.""tt.+>> r& t

Tl!

i

il"

j

+

=

o

:o

o o o

il tl]l

o

|

r-

"

rl

o tl

6

t

i-l o!

til

L

o

|

'' >

t+ 4

F


"6

U

4.4

='

-1-= r6 allo€

L ? >

o 7 et

{

T

E

o

? ts

{

|=

n o F

2

o

t

! a.

-la F

t,

{ I

I

8

f

".

>

. I

o.l^ a

l-

g

tl tr H

5 l

{

E

It

{

-ITTTITITN

a

;.--

iltl

I

-82-

olll

>15

E -i-

-19

lN

tl

tt

-83-



FRAMES

NOTES

DIAGRAMS and FORMULAS for VARIOUS STATIC LOADTNG CONDTTTONS

M"

4.5

=-+, oo

ru,=*9

Ph

Mo=-]

M"=+ -, oo

ph

M"=-d! t,=*+

PB

t"=.+, Steady

M.

=-+

(+Ato)

fll'k . o.o,ot "=-h'(l+k)

tr.t.

lr^=-1r,,r.. k=I,h '2"I,L d = coefficient of linear expansion

22

steady heat

(+Ato)

M" =

3E!

I'k*,')[I-*!)o 2/

h'(l+k) \

at'

M.=ffl'k fr+L)a.at' " h'(l+k)\ 2) v ' , =-Iv.. y. =-$r[L]"a* 2

M, Cx,

h'\2)

=-M-,,

k=

I'h I'L

= coefficient of linear expansion

NOTES

5. AFICfIES Diagrams and Formulas for

Various Loading Gonditions

THREE-HINGED ARCHES SUPPORT REACTIONS, BENDING MOMENT ANd AXIAL FORCE

NOTES

IM" =RoL-P(L-x")=0,

bending moments in elaslic arches Tables 5.1-5.9 are provided for determining support reactions and

IMn

wilh constant or variable cross-sections Table 5.'l includes formulas for computing in any cross-soction

k

the axis force

Nk

and the shear

\

Axial

moment

force

Shear Where

Mr = Re xr

-Ha

Nr = Ra sinQ+ Ha Vt

yo

+Mo

-IPr

5-rrl. 4 \ =n.

-lP,

=rf;

Ru

=PlL.

!-s"i=0.

H^

!; =R.^2f

Left

'ar

Len

cosQ

=-nut,+Px, =0,

n^

Horizontal reactions:

in Tables 5 2-5 9 These formulas can also be applied in analysis of arches shown

Bending

5.1

Vertical reactions:

Ix=Ha-HB=0,

sinQ

HB=HA=H.

Section

=Racos0-Hermq-iq"otq hft

k (x.,yo) lM = Rox* -Hyu,

Bending moment: Mo =

ai =distance from load P to point k

Shear:

v-diogro.

-IP

=l R'^ ^l^al \Letl

Vu

or I

or Mk=Ml -Hvk

/\

lcosQu

Vk =Vk'cos0k-HsinQu'

or and

Vou

+ ucosq,

kftl

\

Ml

-HsinQ*

*- =[*^ -lr]s;n4o

Axial force:

'

Nk = Vk"sinQk +HcosQ*'

= bending noment and shear

in

simple beam

for section x,

Tied c

IM" =RoL-p(L-x")=s, IMo

=-R"L+pxn

=0,

n^

=r!b;

ft" =p.5-.

Horizontal reaction:

lx=-H" Force N,

rrra-

:

= n^

1.ft

=0.

!-^-o-r[!-*"]=0. l? ' )

rI !--" )-rt^;j ' =][pf dL \2 ' L

N"

I t,t^ L

Rishr

=N,c-n"!=g.

l.l, =P-"2d

.

SYMMETRICAL THREE-HINGED ARCHES OF ANY SHAPE

NOTES

FORMULAS for VARIOUS STATIC LOADING CONDITIONS

5.2

Table 5.2

Example.

Symmetrical three-hinged arch

Given.

circulararch 2 inTable52'

L=20q f=4m, 4x4'z+20'1 4f'+t m, x. =) radius R=-=-l;;-=la.)

A

\ +Ra

m,

-(14.s-4)=3.11m II

tanq. = | i-x.

ltln-r+

\r

)

y.

;=

Q-=20.170, sinQ. =9.345' Distribution

SUPPORT

LOADING9

\

1)

1t0-5)/(14.s-4+3.1

Ro

cosQ. =9'939

load w=2kN/m

Requirod. compute support reactions RA and bending moment

I

M. ,

axial force

1

sorution. n^ =iwl-=ixzx2O=15 kN e-

=I-=a=o.zs, L20'"f4

Ho,

N.

support bending moment MA

and shea

wli

3

=

N.

= Ro sinQ. + Ho cos Q.

- w. x.

sin Q,

=

=IL

lvrm =

wLl,ly E2\ _ g L+\gm-!./-'tm..l

|

*r^=""=*

,

=12'5 kN

2

Ra

ll=o.zrs

28. ='1?o' ;t{o.,t-o'2s'z )-2x0'25-0 #Ls(€. -Ei )- -n. ]

M,

=R"

\ ?.x2o2

, n^=;;=ffi n- =J-=

BENDING MOMENTS

REACTIONS

= 6.367

15x0.345+ 12.5x0.939

778] = | r' I

kN

=:wL

=aw|,,

Rs

s^ =H"

=iF

m

',

=

f

t'te,

-€'. )-26.

-n.1

r- =S{16.-n-).

- 2x5x0.345 = 13.46 kN

V.=Rocos0._HosinQ.-w.x.cosQ.=15x0.939_|2'510.345_2x5x0.939=0.38kN

#5t 4

J{ ,o lP o,

A

-90-

..f2

*f2

zL

=-lwf, ^4

H.

H"

*^ ,

R"=

*^=-;.

-,12t r ^\ t. =-?[8.-ir.+ni

J

Mk

=?(28,k-nk)'

=f*f

=t;

=r?,

*"

H^ =H"

=P*

B

-91 -

r',r^ rtlu

=r;(zle,-n,) =e|{26,* *ttu)'

SYMMETRICAL THREE-HINGED ARCHES OF ANY SHAPE FORMULAS for VARIOUS STATIC LOADING CONDITIONS SUPPORT

LOAbINGS

REACTIONS

u.

R, =:wL ^ ai w

JIIIItrnn*

,.,r 2 _

=ft

[:€. +s(8, -Ei. -8, +El )-'r.]

'|

R-

=:wL

n- =n^=-

wlllllnhn-,

5.3

BENDING MOMENTS

r

R- =R^ =

n-

=*(zB*-n,).

---r2 WL

48f

*'

v^ =ffPe^*

a(si. -Ei.

-e, * *) -n.1,

H^=n"=fr

*^=-$, *"=$ H.

=-awf" ^12 Hu=l*t

r.

=S1r{e,. -6i.)+n. -zq.]

r.=$ir€*-n*).

M- =Mol-

TWO_HINGED PARABOLIC ARCHES

o

N

FORMULAS for VARIOUS STATIC LOADING CONDITIONS

s

T E

5.4

Equation of parabola:

4ffL-x)x

Table 5.4

'

Example. Two-hingedparabolicarch

civen.

Parabolic arch

3

in Table

5

L=20m. l=3m. x=a=5m. E=i=--=O.ZS 4x3(20-2v5)

tano\=-I_=--.tr-_".' Q- =16.70, sinQ"

Coefficients: Forregulararch:

t5 B Fortledarch: D=- ._.

-^,

8 f'.

=0287, cos0. =0958

load

Concentrated

axialforce

solution. n^ -PL, ll

N- and u

SUPPORT REACTIONS

LOADINGS

P = 20 kN

M" and M" Required. compute suppon reacrions Ro and H^, bending moments ll

shear

=zo2?-

l, = | ,coso*

-.-- dy= 4f(L-2x) t-q=d* L'

5.4 q

4f(L-2x)

't:

'

NMIIITIIIITITITTIIITIIIITIT!

l+1).

5=tsttl

Er_ ' E,A,

BENDING MOI\4ENTS

n^=R"=* tt1 Ho =Hu =

W

V" (atpointofload)

t=0, k:l

| ^D= K=-.

rr.r.=$tr-t) 15

8

20

B1

k f2' ^-=- l+r)

5P!u [r rr, *E.l :]4j l9 x r^fo.zs - z-o.zs) +0.25"] = 10.7s kN .f,A=-KL9-zr, >.1 = g>l -

:ii

L

It riii ill

v

=

PL

lae - :r -"\.(a - ra,- * q* ' lt11 = 8 L:

20I

8

20

fL +ro.zs

-:(o.zs -

2x0.251 + 0.2sa) =-e.s

kN m]

3l R.=:wI_. R-= ^8"8

(L-x)x _4x3(20- 5)xs - r, _t=---F--' 4f

iiir il

Hiil

iii

M, =Roa-Hny.

=l5x5-10.75x2 25 = 50 81 kN m

N.

+ l0'75x0 958 = 14 6 kN = Ra sin 0" + HA cosq- = 15x0 287

\

287=I =Re cosO. -Hosin0* =15x0'958-l0'75x0

1 3

=H^=-K

wL

M.

==(r-k). lo tt 1 \ lwl" \16 64 i

M-'" =l--:k

l6f

kN

R..L"L =P" -q -" R-=P: I

!ii

Ho

rra- = IL[+e - -sr. (E ' 8 L

-

2E' + E" t) ' ' ).1

.L

=H"

r. ^". ".r gf L' ' ')

5PL.

4 RA

HA =

5wL -

wL

24"24

H. = 0.0228 "f

"'"

k

M" =Rof -Hof

TWO-HINGED PARABOLIC ARCHES F'ORMULAS for VARIOUS $TATIC' LOADING CONDITlONS SUPPORT

LOADINGS

BENDING I,|OMENTS

REACttOI'tS

5

R^

=-;, Hr

=

R" =-Ra

-

Ulc =

-

0.0357wf'?

0'714wf

Hs = 0'286wf 6

R^=-;, Hn

Ru=-R^

=-

M"=-0.0159wf'?

0'401wf

Hs = 0'099wf

R^=-;,

Ru=-Rn

u"=$-N,r

H=wf 2.286wf3 ,, =Iii+r50"'

..,f2

. R"=-R^

R,=-*' ^6L

M"

=rl--yr1

wf ., n=2

0.792wf1

"' = fF +15F

Ro

Mc

=Ru:Q

15 EI.A, -_ Ft=- -----X 8 f'L

.

=- Hf

505

FIXED PARABOLIC ARCHES

o

N

FORMULAS for VARIOUS STATIC LOADING CONDITIONS

T E s

Equation of parabola:

y= Table 5.6

Example.

Fixed Parabolic arch

Given.

Fixed parabolic arch

+MA

2

€,=t. L

t=20=;t=oO

=fe[r+e -,r )

(l+E(,)]=]?90.+[t+oo1t+o+x06)]=13'es )x)O2 rJl,zY

l1l

I

itl

M"

l!

iir

e' [r :q +

( I + 2E'

=

- L-x -'L

LOADINGS

SUPPORT REACTIONS

BENOING MOMENTS

1

M^=Mu=Mc=0 s +'

x[l

)] " ti" l,t. =-*t' E,E: =-2x?0' xg.+'x0.63 =-13.82 2 2"' H^ =

1l

M. kN

= r

*Rl

20

Distribution load w = 2 kN/m Mo and Requhed. Compute support reactions RA and HA, bending moments

4f(L-x)x rx=lc/cos9x L, ' . dy 4f (L-zx\ 'una= dx

6

in Table 5 6

q=]=o+, L=20m, f =3m, x=€L=8-, -....20 -

sorurion. n^

5.6

+

3x0

6( i +

2x0'o )] = I 0's8 kN

kN.m

T

=Roi-wx8x6-HAf -MA = 13.95x10-2x8x6-10.58x3-13'82

= -2'06

kN m

*u

r,

Y)9j

^ - 2'*, wt' rtpz rvro \'

ll

lrii

-

lir

-\

lii

Iil llr

3

il

li

wt_

wi_

5l

^14

^ =- 280 *f' 19 M- = *f' "

u."14 =awf

3 M^=wft '

Ru= R^=-;, lt -n^ =--WI .

+L

M.

280

140

RA

=qi(1+2q)P

le-rll n.^ =p*eif " t)-

RB

=6'?(1+28,)P

Mu =PLB'q,

[;E

-tJ

For 03(S0.5:

H=P15Lt'e:

4f"

rr.r-=I!e,lr-lell ' t - t ,- l

FIXED PARABOLIC ARCHES FORMULAS for VARIOUS STATIC LOADING CONDITIONS SUPPORT REACTIONS

LOADINGS

P

RA

"2

H=

fillnnrn*

?DI

1

'64

64f

WL

#,-f\),

B

., )wu l28f

184

OFI

6EI.

_D

=- "192

wl- = -'

^L

^L"

^ =* oEt. r" ti -. 15 EI" 2fL

D

-N

__ 45 Er,

-/\

M^ =M-

w

_

PT,

=-JL

M^ =M-

sPT :::-:

R. =R^=-

,a\

w

BENDING MOMENTS

-- 4 f)L

]FT

"L tvl- =--.-

'

tvt =

'

tvt

-

1EI

2L

ZIL 1{ FI -" "_c

4f

L

5.7

THREE-HINGED ARCHES

NOTES

INFLUENCE LINES

5.8

1," ll

I

Infl. Line R4

l1 llr

lli

ttit

lir

li

Infl. Line

Rsl

Infl. Line

H

I Infl.

uine vk

'[l,

ll iil

lI flr

fii

dl

L.f .x, yk.D+xL.r

u. =---=----*-, a- u-

^ =JslnQk, b! ' -un

u"

L.tanB

=-----;lanp-cotor

FIXED PARABOLIC ARCHES

T E s

o

N

INFLUENCE LINES

5.9

. Equation of pa€bola:

Table 5.9 Example. Fixed Parabolic arch

l\=lc,icosor,

civen. L=40nr" f=10m, xr=8m Concentrated load in point

k

Pr = 12 kN

Required, Usinginfluencelines,compute

Sotution. i1

M" and M*

Mo,

x, 8 a=l=0.2.

4xl0(40-8)8 ,,yr =-----=b.+

4f(L-2x) 4xl0(40-2x8) -^ tanQ=-t-=--FQr

lL,

bending moments

bending moment

ll

L

supportreactions Ro and

=30.960, sin0k =0

5i4,

cosQu

axialforce No' and shear

=9'357

tt^'

S,

x-l x R = 0.0960x-l:x l2 = 4.608 kN lu

r40 t

= -0.0640x40x12 = 30 72

kN m

M.

= SixLxPu = -0.0120x40x12 = -5.76

kN m

Mr

=Ra xr-He Yr-Ma =10'752x8-4'608x6'4-30'72=25

Nr

514+4 608x0'857 =9'475 kN = Ra sin0r +He cosQr = 10 752x0 sinQ, = 19.752*0 857-4 608x0'514= 6'745 kN cosQ*

Si

oRDTNATES oF tNFLUENcE LtNEs (Sr)

a

S,

Me =

\

x L

Ro =

xlxPt

I

i I

Irl

Vt =Rr

Infl. Line 805

H

kN'm

-Ho

Itr i,,

RA

H

MA

Mc

0.0

0.0

0.0

1.000

0.0

0.05

0.993

0.0085

-0.0395

-0.0016

0.1 0

0.972

0.0305

-0.0625

-0.0052

0.15

0.939

0.061 0

-0.0678

-0.0090

0.20

0.896

0.0960

-0.0M0

-0.0120

0.25

o.444

0.'1320

-0.0528

0.30

o.784

0.1655

-0.0368

0.35

0.718

0.1

940

-0.0184

0.40

0.648

0.2160

0.0

,,li

,ii

tanQ= . =---:, oy L'

Rn=S,xP, H=S x!"P, M=SixLxp

suPPort

m,

xPo = 0'896x12 = 10 752 kN

=

H^'

4f(L-x)x y= ---- L-dx af(L-2x)

-o.o127 *0.0'102

-0.0034 0.0080

0.45

0.575

0.2295

o.o'174

o.0246

0.50

0.500

o.2344

0.031 2

0.0468

0.55

o.425

0.2295

0.0418

0.0246

0.60

0.352

0.21 60

0.0480

0.0080

0.65

o.242

0.1940

0.0498

-0.0034

0.70

0.216

0.1

655

0.0473

*0.0102

0.75

0.1

56

o.1320

0.04'10

-0.0127

0.80

0.1

04

0.0960

0.0320

-0.0120

0.85

0.061

0.0610

0.0215

-0.0090

0.90

0.o28

0.0305

0.0118

-0.0052

0.95

0.007

0.0085

0.0032

-0.001 6

1,00

0.0

0.0

0.0

l]jl

{ii Jill

fir ,illl

illll

Infl.1Line.M"

SrxL

104 -

0.0

STEEL ROPE

,o.oot-o-/+\r* Rope deflection w = uniformly distributed load, f = rope sag due to natural

s

=

length of

weight, (f

= I / 20.

L)

ltT YJ

rope, s =./tJ +lf '

Forces and deflection: ll

Jo2nrl' -n=

(elastic deformations are not included)

il

-t =

(elastic deformations are included)

{ii ,lr

E = modulus of

) lit

+f

GI,'EA

i/

*

elasticity, A

=area of rope cross-section

tii !iL

tit

N.*

=fi'+ni

R=reaction, R=wLl2

,iil

,ulrr

Bending moment

M,*

=

*Ij

/8

,

Deflection

ytu

=M.* H

Tsmperature:

N,

=d.Ato.EA,

Att

=T,t-Tj, if:

Ato

>0 (tension),

2

7.1

CASE

B:

h a

a

The calculations are performed for plates of 1 meter width loads' The plates are analyzed in two directions for various support conditions and acting Units of

measurement:

Bending

a2

Plateshouldbecomputedinone (short) directionasabeamof length L=a

CaseB !,3.6x20=1409kN/m'? = q,,, . B. L. R. I P = | 409 x3.6x2.8x0.7

ri ili:,;:

/+7-

0=300, c=0, Y=130 Lblft3 =130x0.1571=20.42 kN/m3 kN, M=500kN.m, e=500/2500=0.2m, elB=0.213.6=0.06 Bearingcapacityfactors R" =0.78, N" =20.1, N,=20 Granularsoil,

p.$.

flg i

aeF ?: j" li

Table 8"4

8

p

I 2500 = 4.43 > 3

2 o

g


€iij ,--O

IY

Pi = 12.0 + 48.0 +

!

cr->o

U

m

Y|.. !

=-''^:*u s=;

sii ot

rl

:z a:l; r rl

rl

.\;

S^=

il

-i\

o

= 1.33 m

h4 d-33 =_:!=,=1.33 P" =

- 9.81)x4'

>-l > Tl1 Ttf ,,

r

a

|

P,

o

.rl

10.3, H = 6 m

Angle of friction Q =

10.3

I IY

.{ ul I

.t' I( ill I

4. I

177 -

l

*l o-:-l

NOTES

RETAINING STRUCTURES

llll

+

o

T =l; ;"?ll

6

FFc

!2. !2"

o

hh

;:x

E

ilil=

! I

:I

o o o

ai

-a E}F -F XME llll; o'd3

>-l >-

o-i r

E 6

Er! El !4 i, pil

5

--l

g

EE o

d ori oil o

10.4

-eF

Er6 rr o!-;:;il

le+;-+10." ;F6Riiul oL;+o.-l o€:,i"1

L.;€z--\!?e V sv].-f+gr ;oYo'83 Lilll d -o*--;-+€

o.E-9E '-: :ia+t.e--6d*6

-

.sl

=rt

+l N \_l-

E

!?:g ,s,

3

6.

s

-lN

g

1.>-

Edld

T

> " slH f ^-: r rr 3 Il: '"] '-

c

ni

o E

x t! il:

N

q e o P 6lE o'+ r^r E a EIE * * .1 I3 :ole" hl E F. e ii EIE _t a ,.t H 5tE I *f E EIT A --,1,-

:

F.

a

"P Frl

o

9

',

;3

2 =

: F

U

F

t

g.

o

IJJ

o

lu J

N o

a

'l VIla

.9

;z

b

(J

o

9)

q

I

(

o

e

'il Hlr

€l + >lx '^lL

htlN Ft4 Nl.r .rlN I

E

o E.

o

o o o o

E

a

o

;

E o

o

I

o o

c

ll

q

o E U

E .E

o

-185-

RETAINING STRUCTURES

NOTE

11.2

CANTILEVER SHEET PILINGS Equation to determine the embedment (D0 )

Table 11.2

Given.

properties:

Soil

=0, Q, =340, c, =0, Q,

=32",

c,

Tr =18

kN/ml

% =16 kN/m3

/

B=0, o(=0, 6=o

|

P,

=

tm' (+5' n Q,

)

*fj

="',(or,

For single Pile

r- =.f"n?

o-(K,-K-)YdDi ' 3{4H

=r.rrt,

+3D")

3

[

where d=pilediameter

= 0.5K,,T,H' = 0.5x0.307x18x10'? = 276.3

K,.Y,H -r--

I

4H+3D -' ^-'oH -\ +4o,

'*'(*" -!)=,ro, r", = tm'[+s' -9.) = = o," ""'[or' -1L]

r",

:

p

1t

M."=P[H+;iG;fi]

="''(or'-!)=

K'

6(4H+lD[

Maximum bending moment

Required. Compute depth D and maximum bending moment M,."* per unit length of sheet piling

sorution.

(r,-r,)vri

-r

Example. Cantilever sheet piling 2 in Table 11.2, H = l0 m

:

(K" - K.

kN,

D = (1.2 to I.4) Do for factor of safety at 1 5 to 2 0

0.283\lgyl0 3.254x t6

)Yr

Earth pressure:

P, = 0.5K"^1,H2, = 0.5x0.283x1 8x 10x0.98 = 24.96

kN, R = 0.5K"^YH'

P

-0.5(Ke K,.)y,(Do-2.

IVto ',

P: =o.5K,,YrH )':

= O (condition of equitibrium)

(f .o,).',

(o,

P, = 0.5(Ko.

-]) -* Jro, -,,) = o

zzo.:[f

+r.]+z4.e6Do *26.031(Do -2, )'

8.68(Dn

z,

-

)' =

zt=

z'

=0.5x3.254yt6(Do-2,)

-r",

)T, (Dn

Equation to determine

-,,)'

Do: lMo

") -\ o.-11-e 3l \3{*o,l*e,f

=0

ef

921.0 + 301.26D o

D =(1.2

=0

jro,-,

to 1.4)Do for factor of safetyat

t=o

1.5 to 2 0

Using method of trial and error:

assume Do =8.3

m,

(8.3-0.98)' =t96.16*rO..'tx8.3,

m = point of zero shear and maximum bending moment 394.19 =393.18

D = l.2Do =1.2x8.3 =9.96 m

Maximum bending moment

r.- = t [+., *,,)* v,(? ", *,, ) - o.s 1r,, - "^.l r*Z(?) = zre.:

($

+ o.rs

* t.+)*

z+.0

e(l"o.n, * r.o) - o. r',

186 -

zs+ xr e xs +,

'(Kp;Ko)T2Qo-74

(!l

='

rn r.o

u*

*.-

=R

(H [;.

z,+

zz

\,(2 )+P,l:z'

\ + zz

(, \

-os(r,-r")rzi [5J

)

w-

NOTES

RETAINING STRUCTURES

Tablo 11.3

Example, Anchored sheetpilewall in Table 11.3, H=15 m Given, Soil propedies: 0, =30', cr =0, ^L=20 kN/m3, Requrred. compute deprh D ano Solution.

f +s'

\

"' =r*'f\

K,.

02

=320,

cz

'*,'r.lJ"oo,"n"rl"l""i;t:* ,:,;fi"Tn-

-92)l-,un' \

r","'= m' f +s'

Nl^

',''

-4)=o.rrr, 2)

K.

= tan' | +s'

(

+so+91=,un'[+:"*14

2)

2 )]=r.rro,

l.

=0,

vv't =lJ ;l vl . rr ll4 oiil

T, =18 kN/ml

I

",",

_Nl F,z > -l .- it

I = o.ro, '- i +r" -14 -92l =,-, z)--" [-

K",

-K,.

=2.s48

Forces per unit length of wall P,

= 0.5K",T,dt = 0.5x0.333x20x1.2':

P, = 0.5K", y, (H + d) (H

. ' p:

+d) 3(H+d)

(H

-d)(2H

- d) = 0.5

= 4.8 kN

x20x (rs + r.2)

x 0.333

{r5 - t.2 )(2x t5 +

1.2

Q2

=32'

1.2) = 744.4

o

kN

,,,'rl or-dx

)

kN, ,

=#HE

=o'r!#frirt

: x = 0.059H = 0.059x15 = 0.885

Ivr, R(r5 -1.2+0.885 )+ T = (Pr + P, + &

o"=r,

-

3(ts+t.2)

=0.5K",rrHz, =0.5x0.307x20x15x1.74=80.13

Fot

(r s

)

-

; u

(H-a-1)

+sx! -t ++.+x8.86-80.13fl5-r.r* t tol = o, 3 z ) -' \ -

=r.r,

lll

=o

l:'

o

e X

T

g

ts

ci

o B

.-

I

a il': N.'",

o

N

-a

N

Nilo

il x+ il

- - E ----

d I N o

o

er^.| r .. E (d,b or{ c

'Ft,. E^-O h':o o: !xE

*+o :F

E

u

bT. ^i E B qP"i; :!+a Gl,

^l a

> ; E

o

9+c

J

F

= o, R(H-d+x )+r, 4-p,a, -p,

R = 4.8 + 744.4 +80. l3

= ul

O o

F

EII

*d VE v]

*-

o

I

V Jld Nl+ ^^ Yl; it

1

s' slo + \i +.1 - ^\i

I

ffsL

E

J9

ll

g -

!V

ll.,

9

lo< | Fa

Itl^ l>

lel lli

N

E

)z

NlNl

l

l'"ll

oi

: l+lr

+

6El^^li^-lTlv..

E

*t

E F

N

l'-|tr; I +F +

I

I lv+

X I 19 :E tFt-j-- ., tit

--l-

-oNl'":.

,9b;o,^IEE-

xc

"nxE n-o< Htr>d

F+ .!6

a

R = 527.46 kN

U

527.46 = 301.87 kN

* i|,,E =, ro*.@=7.58m. Y2c48xl8 V(K' -K..)Y,,

o

{assumed

z

D= l.2Do =1.2x7.58=9.1 m

-

r.*

= =

(q {

!

i e,+e,+e.-r , _ 0.5(Ke

- K"

/4J+i44l+sot3-jot-87

)y, Y

0.5v 2.948x 18

N

*r,)(4*,, *,,) *r,(2,,*".)*r(+r-d+2,+2.)-0s1r," -r", )i,*

+. s

+

za.+;

(ll

+

t.t

a+

+.+6)+

Bo

x(}t.t

+

* +.+al -

ro,.r, r, -,.,

-0.5x2.g48xt8x4

1

46'

(f)

+ 1.7 4 + 4. 46)

oo .N I :a

if,

= 2019.4 kN .m/m

Y

NOTES

Lzr L3. P

and

TUNN

IP rS ELS

Bending Moments for Various Static Loading Conditions

TUNNELS RECTANGULAR CROSS-SECTION

12.1

This chapter provides formulas for computation of bending moments in various structures with rectangular or circular cross-sections, including underground pipes and tunnels. The formulas for structures with circular cross-sections can also be used to compute axial forces and shea6.

,

The formulas provided are applicable to analysis of elastic systems only. The tables contain the most common

€ses

I,h I,L

of loading conditions.

+M =tension on inside of section

q*w

ri w(2k+3)-ak ': ' M- =M. =-12 k'+4k+3 t? c{2k+3)-wk M =M.=--.", 12 k'+4k+3 For q=Y

ffio

.- r2 k+3 M =M.=M^=M,=-ia. , 12 K'+4k+3

PT

4k +g

M =M. =-:-1. : - 24 k'+4k+3 lYl =lvlj

=--.

PL

:

4k+6

24 k'+4k+3

FOr K=l 1? M =M. =- -- PL

192

7 Pt v-=v,=- t92

I-

NOTEST

TUNNELS REC

TA

N GU

LA

R

CROS

S-

SECT

IO

12.2

N

llh-i(

l2(k+l)

rt FI

=M" =lfo =\{"

Mo

FI F F H f

For k=l and h=L M" =M. =M^ =M, Mo = o.l25ph'

M.,

-

0.5

=_I::

24

(M" + Md

)

-M" '"b-- Ph't(2k+7) o0(k,

M -M,

-- P!'!(lk+S)

60(k':'4k+3)

k=l

For

+ak+l)

and h=L

M'=M,=-lPh'. M-=M --llph/ 160 480 uo

= o.o64ph'?

-[M"

+0.s77(Md

-L2t.

a = P" I ltotr' 60h'\

(. ^ D=:-n-_a-_b-_l 2h' \ pbak

-M" )]

-:u'I 45a

-2b \

270a

)

PIPES AND TUNNELS REC

TA

N GUL

AR

CROS

S_S

TI

EC

12.3

ON

Th Table 12.3

I'L

Example.

Rectangular pipe

7

Given.

concreteframe,

L=4m, H=25 m,

in Table 12.3

Ir

b =1 m (unit length ofpipe)

bhi

looxlor -l.= bhi J=--EJJJCm

lt=-=-

'1212"1212

Uniformly distributed

r2

r2

cm, h, =lQ s6

hr =10

loox2or

Ir 4

l.

r =2k+1 I1

m=zo(k+z)(6k'+6k+1)

l2

=OOOO/Cm

+M =tension on inside of section

load w = 120 kN/m

Required. Compule bending moments

Sotution.

I.H 66667x2.5 -k=I='--'" -I,L 8333x4 =5.0, m=

20(k+2)

m=

r=2k+1=2x5+1=11

zo(k+2)(6k' +6k+l)

=

20(5+2)(oxs' +oxs +t)

= zs:+O

ffi

tv

,^=-iwti ;,t

or = 138k2 +265k+43 = 138x5'] +265x5+43 = 4818

o, oc:

= 78k'? + 205k + 33 =78x52 +205x5 + 33 = 3008

..,t2

r, =-li

= 81k2 + l48k + 37 = 81x5'? + I 48x 5+37 =2802

t. l* o' *-1!lL]= r.' = - 9l 24 I\ll 25340] 24[r m)l= ''o-10'

,

M.

-,

Ml

-22.56 kN.m

M.=_:fr1*s.l=rro_10'r+._rgol)=*,u.rr^ ' 24\r m) 24 !ll 251401

=

kN.m =-*f24 1,1m) \r "'J=+2.24 M"

y", =_wL'f :t