Moment Shear

8/10/2019 Moment Shear

http://slidepdf.com/reader/full/moment-shear 1/40

6/26/08 4:12 PM Page 89 l c04.qxd

89

4-1. Determi11e the i11ternal shear, axial load. and 8k the support at B is a roller. Poinl C is localed jusl Lo the l be11di11g mome11t in the beam at poi11ts C and D.Assume

right of the 8-k load.

c 8 ft 8 ft 8 ft

Reactions;

:..u; = 0; A, = 0

( +LMo = O: 40000 + 8000(16)-24A.. = 0 40 k·ft

B, = lOOO l bS' +s·=f )

A, = 7000 lb ..- TU, = O; :: 0

7000-8000+8,

By

For C :

+il:F , = 0: 7000 -8000-1{ = 0

Ve = - IOOO l b = - I k Ans

:_ i..f. = O; N e= 0 Ans

(+LMc = 0; M c - 7000(8) = 0

Mc= 56000 Lb ·ft = 56 k·ft Ans 11.

For D: ..- T Ur = 0: 7000- 8000- = 0 OJ:

Vo = - lOOOlb= -lk Ans

:.r..F , = 0; ND= 0 Ans

l===t,' ====' =B ' =i-¢ .Jp

(..+ I.Mo= 0; Mo +8000(8)-7000( 16) = 0 v Mo = 48000 lb. ft = 48 k·ft An.

l

i11 Prob. 4-1.

4-3. Determine the i11ternal shear, axial load, and bending moment in the beam at point B.

2.6.8 r<.

-

Moll

N ( f I

s L 4 ' J.. - s· -'1 Vgl I I

• :,L.F , = O; Na = 0 Ans

+ I.F , = 0; lfi -28.8 = O; \fl = 28.8 k Ans

._ I ( +kMs = O; M s +4(28.8) = 0

M 8 = -ll5k·ft Ans

© 2009 by R.C. Hibbeler. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all

copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

·.. - k/ft

·-.

T.TTlri:. B c

,. ::; ft-I ------12 fL--------<



6/26/08 4:12 PM Page 90 l c04.qxd

90

© 2009 by R.C. Hibbeler. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Tbis material is protected under all

copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without pennission in writing from the publisher.

*4-4. l)ra\.V the shear and 11101nent diagranis for the

beam in Prob. 4-3.

6 k/ft

A

Tl

3 ft

c

12 ft

4-5. l)etennine the internal shear, axial force, and

bending 11101nent in the bea111 at points C' and D.Assun1e

Lhc support al Bis a roller. Point Dis lot:alcd just to Lhc right of the 10-k load.

10 k

(+W. = (); B, (30) + 25 - 25 -10(20) = 0

B, = 6.667 k 1'5r: f t 1 I

z:;;: Jt ,J

+ t:EF, =0; A, + 6.667 - 10 = 0

A, = 3.333 k

A, ' .za '

.ft ' IO}t

SegmentA.C:

_:, IFx -= O;

Scgm<ntDB:

_:. IFz == 0;

A. = 0

Ne= 0

-It:+ 3.333= 0

If: = 3.33 k Ans

Mc - 25 - 3.333(10) = 0

Mc= SIU k· ft

No= 0

Ans

Am

Aas

"'1 /Jd"

+ t:EF, = O;

(+ = O;

lb+ 6.6'167= 0

If, = -6.67 k Am

-M,, + 2S + 6.667(10) = 0

M,, 91.7 k· ft A•



6/26/08 4:12 PM Page 91 l c04.qxd

91



4-6. i)etennine the internal shear_ axial force, and

bending n101nent in the bean1 at points (.'and 1).Assu1ne

the support al A is a roller and B is a pin.

250 lb/ft

Eollmilcln>:

(+t.w. = O;

+ru;. = o:

-•tIF, = 0;

4.50(9) + 2 - A,.(10) = 0

A,. = 4.25 k

B, + 4.25 - 4.SO = 0

B, = 0.25 k

B, = 0

Ne= 0 Am

+tIF, = O;

SegmentDB:

+tIF, = 0:

-Ii: - 1.()0= 0

If: = - 1.00 k Am

Mc + UXl(2) + 2

=0

Mc =-4.00 k· ft Am

Np = 0 Ans

1£, + 0.25 - 1.00= 0

1£, = 0.750 k Aas

-M,, + 0.25(4) -UJ0(2) = 0

M,, = -1.00 i:· fl Am



6 /2 6 /08 4:12 PM Page 92 lc04.qxd

92

l I



4-7. l)eternllne the internal shear, axial load, and

bending momcnl at point C, \Vhit:h is just Lo the right of

the roller at il.and point TJ, \Vhich is ju.;;t to the left of the

3000-lh concentrated force.

EnlireBeam :

2500lb

r,fj :.I..fx =0:

(+l:M.t = O; Bx =O 2500(20) - A, (14) + 900(8) + 3000(2)= 0

A, = 4514.29 lb

+tu;= O; -2500 -900 -3000 + 4514.29 + B, = 0

B, = 1885.71 lb

Seg!ll""t to the kft of C :

.4l:F, = O; Ne= 0 Am

+il:F, = O; -2500-1!: +4514.29 = 0

\!: = 2014.3 lb= 2.01 k Am 2. 5 1>/b

(+l:J<k = O: Mc + 2500(6) = 0 " Mc= -150001b· ft=-15.0k· ft Am <

SegmentDB : :.= O;

... il:F, =0; If,...1885.71-3000=0

\b= 1114.3 lb= l.llk Am

c+I.Mo =0-. 1835.71(2)-M,, =O

.+5!4·Z91o

'" "•

'it

188S·711/,

MD =3771.4=3.7711:. ft

*4-8. Detern1i11e the internal shear. axial force. and

bending 11101nent in the bean1 at point B.

8 k/rt

2 k/fl

Segment BC: -j;t XIZ)• Z8.t>I:. -•t f;; = Q; N, = 0 Am 2.0Z)=M-.()1<

-- + itF, = O: \i - 28.8 - 24.0 = 0 -::_ ... _ _"'::,.,_

\) = 52.8 k Am

(+I:M. = O; -M.. - 28.8(4) - 24.0(6) = 0

M,, = -259 k· ft Am "'ft J< G/t



6 /2 6 /08 4:12 PM Page 93 lc04.qxd

93

+



4-9. i)etennine the internal shear_ axial force, and

bending moment at point C. Assume the support at A is

a pin and Risa roller.

200lb 50 lb /ft 200lb

Entire llcam :

(+IM.. = 0: B,. (18) + 200(4) - 4 9) - 200(22) = 0

B,.= 425 lb

+ i:tF, = O;

Segm1:111AC:

-.I:F, = 0:

A, +425-20 -450-200=0

A, = 425 lb

A. = 0

Am

-It - 225 - 200 + 425 = 0

l(, = 0 Ans

Mc + 225(3) + 200(13) - 425(9) = 0

Mc = s.so lb· ft Alli

i!/)CJ/,

I

+4-10. Detern1i11e the shear and n10111ent in the floor

girder as a function of x, ':\"here 12 ft < x <::: 24 ft.Assu1ne

the support at A is a pin and Bis a roller.

800lb 800 lb 800lb 800 lb

Segment.

+t l:F, = O;

(+IM.. = O;

-A,.(48) + 800(36 + 24 + 12} = 0

A,. = 1200 lb

1200-800-V= 0

V = 400lb Ans

-1200r + SOO(.r- 12) + M -= 0

M = (400.t + 9600) lb· ft Ans

1;;:,jt i.i!/' ; /:?,

' ti-



6 /2 6 /08 4:12 PM Page 94 lc04.qxd

94

1 i



4-11. i)ra\v the shear and n101nent diagran1 of the floor

girder in Prob. 4-10. Assun1e there is a pin at A and a

roller al B.

8001b 8001b 8001b 8001b

A

,_I_

12 fl 12 rt 12 fl 12 ft

M{ l<·ft)

Vm,_-=1200 lbAns

114,,,a,= 19.2 k.ft .\ns

*4-12. Detern1ine the internal shear. a'-ial load and

bending n101nent in the bea111 at points C' and D.Assu1ne A is a pin and 1J is a roller.

Reactions:

2.4( )(6)-(06)( )(6)+20-6By = 0 5.. .5

By= 4.89kN

A, -m(6) = O; A,. = 3.60 kN

-A,-( )(6)+4.89= O;.<, = 00933kN

For C:

'...:EF,= 0;

+ID',= O; (+Ll1c 0;

For D:

:.IF.i: = 0;

+ rrr; = o: (+ "LMJJ "° O;

36+Nc= O;Nc= -3.6kN

- o.0933 - If· = O; 10 = -o.0933 kN

Mc+(l)(0.0933)= Q;Mc -0.0933kN·m

iVv-= 0

lf,+489 O;\f,• -4.89kN

Mo -2(4.89) + 20 = 0

Mn= -l0.2kN·m

Ans

Ans

Ans

Ans

Ans

An.o;,:



6 /2 6 /08 4:12 PM Page 95 lc04.qxd

95

t+:



4-13. Dra\v the shear and n10111e11t diagran1s for the

bean1 in Prob.4-12.

kN

\/.,,," =-4.89 KN Ans

,t/m"' = -20 K_'\. In Ans

v (j:jj)

4-14. Dclcrminc Lhc internal shear, axial load, and

bending n10111ent at point C'. Assun1e the support at A is

+ 10 k 0.8 k/ft 8k

A c

... 6 ft 12 ft 12ft + 6ft

Re.:.u:rions:

\'+ :r.1<1, = °'· 24Ay-30(!0)-l2(l9.2)+6(8) 0

A,. = 20.11<.

20.l-l0-!9.2-&+B, = 0

B>=l7.lk

A, = 0

For C +!I.F,. Vc +9.6-20. l+ 10 = -0; Ye= 05 k

Ao.s

lo.

f'l,t[

1'1' ri· fl'

A,

Ay a,

lo«: '-"k l !.' ,. l ,.-1, He

(':rMc = O: - Mc-6!9.6)+12(20.l)-l&(H))m 0

Mc 3.60k·ft

Ne= 0

Ans

Aos20.l I<. v,



6 /2 6 /08 4:12 PM Page 96 lc04.qxd

96

==t=l =!=[ T'=. l \ *=I

=·r·

-1,



4-15. l)ra\.V the shear and n101nent diagran1s of the

bean1 in Prob. 4-14.

•• •I.ff •· '

16=.

Yitl I lo./

.I k n.1 k

6 ft + 12 ft 12 ft + 6ft111 I I

vm,L\ = ---10 k

A1""" --60 k.ft

Ans Ans

f(<·fll

I. 3.16

*4-16. Detern1ine the internal 11orn1al force. shear force. and n10111ent at point.;; F: and TJ of the con1pound hean1.

• • •

A ID B IE

200N·m

2 111--l-2 111---4111---2 m-+-2 m

-200-+ 50{2) - ,'rft ::r:. 0

Me:::: -iC(JN- m An:t.

S""""tD8 :

+ t IF, :::; 0: 1,£, &00 + $0 """ o

tMp "" (r, - SQ:) (2} + s (50) - , r,, " 0

Mo"' 13CON· in" -L}JkN· m At11



6 /2 6 /08 4:12 PM Page 97 lc04.qxd

97



4-17. The cont'.rctc girder supports the tvvo column

loads. If the soil pressure under the girder is assun1ed to

be unifor111, detennine its intensity wand the place1nent d

of the colunm at B. i)ravv the shear and n101nent diagranis

for the girder.

30 ft

60 + 30 " --- = 3.00 kfft

30

Arn

90(1.SJ - 30{3) - 60;3 + dl = 0

d...:; 18 ft Ans

F·--·1j _[_._1_ 1 .. 3 <ftt

+4-18. Detern1ine the internal non11al force, shear force,

and n10111e11t at point C of the bean1.

!!cam:

('l:M• = O: 600 (2) + 1200(3) - A, (6) • 0

SegmaitAC:

800 - 600 - !SO - If: ()

1£: = Sil ti Am

- 800(3) + 600(L5J + ISO(I) + • 0



6 /2 6 /08 4:12 PM Page 98 lc04.qxd

98

4-19. l)eternllne the distance a betvveen the supports in

Lcrms of the beam's length L so Lhat the bending moment

in the synnnetric shaft is zero at the center. The intensity

-----==:

Momt:nts Function :

*4-20. Determine the shear and momcnl in Lhc beam 400 lh, IL

as a function of x. Assun1e the support at R is a roller.

l-rrn "" O; 1.5(60CL0)-15B., +600; O; 81 31Wllb

+t!F, 0; A. +3040- 6COO = 0; A, = 2960 lb

Sc:f!:mer.t:

+Tu = O: 196()- \'-41)(k = 0

V.,:; 2%0-400x Ans

M+ i.OOX)-.<(2960)-600= o 2

.\I= -·2(Xli+29'60x+600 Ans


bean1 in Prob. 4-20.

(;lJwt!4001b, fl

!

---x

-------15ft--------

c ; I

v (i,)

n.. I I 1

" j '· 11-•1.h..





6 /2 6 /08 4:12 PM Page 99 lc04.qxd

99



4-22. l)eternllne the shear and n101nent in the function

of x. \\'here 2 m < x <: 4 m.

7

l k'!

12 k'!·m

I .

= --i-2mj_4m A,

Reaction at A :

=0:

( IM,= (l;

Segment:

+iIF, 0:

{+IM= 0:

A, = 0

A,\8) • 7(6)+ l2 = O; A, = 3 75 kN

- V+ 3.75 - 7 = 0: V = - 3.25 Ans

-M+3.75x· 7(x·2) = 0

M = -3.25x+ l4

4-23. i)ravv the shear and 11101nent diagranis for

Prob. 4-22.

Ans 3.1!) l:JJ

+7 kN

V ma'\= 3.75 kl\

,tlmnx = 13 k ·1n

Ans

Ans



6 /2 6 /08 4:12 PM Page 100 lc04.qxd

100



*4-24. l)etennine the internal shear_ axial load, and bend

ing momcnl al (a) point C, \Vhit:h is just to Lhc left of the

roller at A, and (b) point TJ, \Vhich is just to the right of

3000-lb concentrated force.Assu1ne the suppo1t atB is a pin.

2500 lb 75 lb /ft 3000lb

Reactions:

- ZF.. = O;

(+Ut, = O;

+HF, = O;

For C

Bz = 0

2500(20)-A, (14)+900(8)+ 30(}( 2)= 0

.;, = 4514.29 lb

Z500+900+3000-4514,29-B.- = 0

B, = 1885.11

1-L-'-"t1"'-"1.-·-;:bl' _ + i:r: = 0; - 4500- Vc+45t4-29 = 0

!(' = 2014.3 lb = 2.01 k Ans

+ 25()()(6)= 0

Mc= -t50C()Jb.ft= -l5k·ft Ans

i 1 B,

6y

,A,

For D:

Ne 0 Ans

+f ""' 0:

'+Lllo = 0:

:.LP:.= O;

- 25-00-900- \!, +4514.29 • Oo \!, • 1114.3 lb = 1.1l k

900(6}+2500(18) 4514.29(l2)+M<>= 0

Mti"" 377L41b·ft = 3.77 k·ft Am:

·0 AM

Ans

+Prob. 4--24.

2500 lb

4-2..'j:, i)ravv the shear and 11101nent diagrarns for bea111

in 75 lh/fl

3000\b

Vmax = -2.5 k

AflililX -15 k·ft

n n

--tH-;+------12 rt ----- 2ft

A.ns

,-\ns



6 /2 6 /08 4:13 PM Page 101 lc04.qxd

101

6xax•2.x

- 1

0

F11>e.1 M::r J wdx=W --ojL 1



4-26. l)eternllne the shear and 11101nent in the bea111 as

a funt:tion of x. 600 lb /ft

H

Reaction at /\!

F,= I IQ 1. , l I(!

= 20001b

' I

10 ft

From rhe uble on the in.side back cover for .a para.bola the centrQi.d Is at-( lO ft) = 1.5 fi. 4

"-· ii. .(+Lli, = 0: 2.5(2()(1())- IOA,. • O: A,= 5001b

= o "' = 0 A·•

Seg nl'. Ai

F-;=:jl7fr; F =5:n.Jdr::::: 2.xlZ•'

+ TI.f"v""' O; 500-U-V= o pr$ ,, V= 500-'.lJ Ans

(+lM= U: M+2.x ](X -' )-500x= 0·4· .

M = 500x -fj.5.i' Am s... ii

4-27. Dra\V the shear and moment diagrams for Lhc

hean1.

< w,,L ..-dz;-- , 0 L'• 3

+i:LF 1 0; woL _ wox1 'l;O

12 3£l

x= ( )11

'L=0.630L4

woL w0x3 1 -(x)--(-<)-M=Q

12 3L' 4

M = wo.L:c _ wo..r.4 !2 I2L'

I- l.

Sub>tinJle > =0.6JQL

M = 0.0394 w0L'

V nm;:;:;:-w 0U4 2

Ans

,lyfmax:;:;: 0.0394 w 0L 1\ns



6 /2 6 /08 4:13 PM Page 102 lc04.qxd

102

= 0:



*4-28. Dra\v the shear and n10111e11t diagran1s for the

cantilever beani.

"Irrrrrrll1I'

L

... 't = 0: -v- -P=O

Y= """ -P ""' , H+ Pr= (! ·JJ

M= - "'-"' - Px A•

beam, and dclcrminc the shear and moment throughout

the hean1 as functions of x.

SuppiJrt R1.ac1icns: As shown on FBD,

Sh ar 4nd Moment F11,T1,cttqn:

30.0-2.l:-V = 0

v = (:J0.0-lx) k

-----6ft-----+--4 rt---

kip. ft

rl:.'>fNA =0; M +216+zx(i)-3().0."'0

M = {-x1+20.0.- 2t6} I::·ft

For' Ct <x !:10 ft!

+T:tF, =O;

-M-8(10-.x)-40w:O

M "{UXlx- lW} ·ft

Ans

Aru

Ans

-=j-------j·--"--''°f----Y(j't..)'

---·-10·0 -fkO



6 /2 6 /08 4:13 PM Page 103 lc04.qxd

103



4-30. Dra\v Lhc shear and momcnl diagrams for the

bean1, and detern1ine the shear and n10111e11t throughout

the hean1 as functions of x.

800 lb/ft 1200lb

x _j :J----8ft-------8ft

S "PF'"rl 81:41;/4,nu ; Ju hown on FBD,

S IJ.-llt' an4 Mv•11U FunctiDn:

Ft>r -0 S:t < 8 ft!

7.6-0- 0.800:< -V=O

v = (7.60-0.800<) k Ans

(+LY., o; M +44.S+0.800x(i)-?.60x=O

M =HlAoo..!+7.00x-44.8} kit Ans

For S Ct <x S 16 ft!

V-1.20 =O

+-M -t'.lil(l6-x) =O

M = {l.:l!lr-19.2) k fi

V(<>

M(l<jtJ



6 /2 6 /08 4:13 PM Page 104 lc04.qxd

104



4-31. i)ravv the shear and n101nent diagra1ns for the

tapered t:anlilcvcr beam.

0

V(.:) (K

7;:K.

0

n

A, = 0

+ire' =O-. M(X)

(K·ft)

A, - 48 k - 24 k = 0

A, = 72 k 1'

-48 {6 ft) - 24 l: (l6 f1) - M, = 0

*4-32. i)etennine the shear and 11101nent in the floor

girder as a function of x. \\'here 4 ft <: x < 8 ft. Assume

the support at A is a roller and Risa pin. The floor hoardsare sin1ply supported on the joists at C', TJ, F:_, F. and Ci.

200 lb/Ct

'io••h 80011. B.. H.

R.ea4.\iOn at A :

+IM, 0: -A,{161+800(12+8+4) ... 400(!6); I}

A, 1600 tb

LL_L A

J 4'

S.grm:m;

+iU =O; 1600-800-4')() -V;O

V= 400 lb

ol.. It !loot

t'M <::r.M ;0. M -800\4)-400.:x) = O

M= {400x+3200) lb·ft Ans li.oo



6 /2 6 /08 4:13 PM Page 105 lc04.qxd

105



4-33. l)ra\v the shear and 11101nent diagra1ns for the

iloor girder in Prob. 4-32.

2001b/ft

1"" It 8""11. S..IJ. !\boll. /loo;(,

=L =L_l=-===i 11"1<> lb il""•IL

v !l>l

vmax::::; ± !200 lh Ans

Afm,1x = 6400 lb·ft

Ans

a funt:tion of x.

8001b

200 lb/Ct

-\1-8002-0-0 x·

'= 0

61}

v = (- 3J3x'-800!1b

M+J '(2600,·')+800x+ 1200= ()

M = (-!.I Ix' - 800x- l200J lb·ft

Ans



6 /2 6 /08 4:13 PM Page 106 lc04.qxd

106



4-3..'j:, i)ra\v the shear and n101nent diagra1ns for the

beam in Prob. 4-34. 200 lb/ft

x

800lb

(IL)

- ... r=-I =.,--::_-::_, I lllTIJ l / _-,f,oo ,,

Vmax=-3.80 k

+ *4-36. Detern1ine the shear and n10111ent in the tapered

bean1 as a function of x.

8 k /tn

+ i!F, ;0; 36 - l 8 -(-xJ

8(x) - -(9-.r)x - V = 0

2 9 9 4 8 .,

V:;:;:. 36 -x- - &:T -x 9 9

V; 0.444x' - Bx + 36 Ans

108 + l 8 2 8 x )(x) ( ) + (9- x)(x) { ) - 3fu M :O

-l -x -x - - 2 ·9 3 9 2

M ;-108-_!x'- 4..'+ .!.}+ Jfu: 27 18

M ;0.14&' - 4x' + 36': - 108



6 /2 6 /08 4:13 PM Page 107 lc04.qxd

107



4-37. l)ra\.V the shear and 11101nent diagranis for the

bea111 in Prob. 4-36.

8.kK/111

A

!<---------9 m ---------

Vmax =36 kN

Mmax=-10.8kN·m

Ans

Ans


bean1, and detennine the shear and 11101nent in the bea111

as funt:Lions of x.

Supporl R,r;at;tiolf¥: As ,:hown o:; FBD. A n

3w 0l

Fero s x <.Ln:

---wi>x-V"'O 4

Fort.fl<><:>;!,,:

V

•(3L-4x)

1 Arn>

- ,\1- ( ;4 \L-x) j(l-x)(L; A)= 0 M.s:: {l .x) 3 Ans

3L



6 /2 6 /08 4:13 PM Page 108 lc04.qxd

108

••• i! !i2rlb;lt

"



4-39. A reinforced concrete pier is used to support the

stringers for a bridge deck. Dra\v the shear and moment

diagran1s for the pier \Vhen it is suhjected to the stringer

loads shoV\.'IL Assun1e the colunms at A and B exert only

vertical reactions on the pier.

60kN3S kN 3S kN 35 kN

1111 1.5 111 1.5 111 1111

60kl\

Im Im 1.5,,, ;.:;,,,

112·5 k;<l

V!Il"""- =±60kN

11'/max =-60 kN·m

reactions on the bean1.

beam. The bearings at A and B only exert vertical

Soo1;. 8001&-

8001b 8001b

......lJ... ·· l--2rt 2lt---2IL-

*"'lb

Vil!)

/4'Jo

/>1.(11 jl.)

(000 24<XJ

2•• .. 0

-w..

:t:(lt)!

"' • .,. ;;c(/<J

- AA>•



6 /2 6 /08 4:13 PM Page 109 lc04.qxd

109

5

0



4-41. Dra\V the shear and moment diagrams for the

bean1.

800 lh/ll

v t101

l--z:"'-- I 7 -6#0

MOb-lt)

'' p· 2-;;600 A'1ura;. = 51.2 k·ft

Ans

4-42. Detern1ine the shear and n10111ent in the bean1 as a function of x and then dra\v the shear and n10111ent

diagran1s for the beani.

a b

-------L-------

y"(,:)

-=·---- L

. . :i:k

-"' ITIJIITI l!!lli i1111!: 1! 11111I /I l x

For 0 x <a:

Mo M -x

/,

fora<x:s;L: x Mo

!.1 "" ·-;-<t .41() L



6 /2 6 /08 4:13 PM Page 110 lc04.qxd

110

4-43. The beam is subjcclcd Lo the uniformly distributed

n10111ent nz ( n10111ent/length ). Dra\v the shear and n10111e11t m

diagran1s for the bean1.

S!ippon k<11trlcn1 ;All tOOwn mt f'BO.

Slrt-or 41i4 Mru1unt F-Uht!li1u1:

Ans

:Jll!>- ,. \, ;> .,. '> Ans i ; , f , .:;&,.

>----L---.; ,,,L

µ·- =?\

- =-q v M

_JL ,j M•• "' m

1-· t v.. »v

in turn support the longitudinal sin1ply supported floor zr· 'l'1"Tr· *4--44. The Dooring system for a building t'.onsists of agirder that supports laterally running floor bean1s, \Vhich

slabs. Dravv the shear and 11101nent diagra1ns for the

girder. Assume the girder is simply supported. r5jt 1,1,I,r,,511 IC:Jt r 4k f:r 8 t:_ 7-12 k.

2 k/flv(J:..)

2ft-j_.t_

t[/t)

f'{ <J<! .. 7'"- 6' .

X{jt.j

4-45. DraV\' Lhe shear and momenl diagrams for the

bean1.12kN/111

I . j

5m-1-----Cm

Ans

111ma... = 96 K'l".111 Ans





6 /2 6 /08 4:13 PM Page 111 lc04.qxd

111



4-46. l)ra\.V the shear and n101nent diagran1s of the

bea111. Assu1ne the support at Bis a pin and A is a roller.

y(I/,)

I \/mm: = 850 Th Am

100 lb/ft

H

4-47. Dra\v the shear and n10111ent diagran1s for the

bea111. Assun1e the support at B is a pin.

,'1-f,llllx = 2.81 K.ft An!;.

+ 8kN/n1

-1--------6111-------

\l max = 24.5 KN Ans

.tf max= 34.5 Kl'\.111 Ans



6 /2 6 /08 4:13 PM Page 112 lc04.qxd

112

l

2 Ans



*4-48. l)ra\.V the shear and 11101nent diagran1s for the

beam.

----4---4.5 111 ,

From PSD(a)

fmm FBD(b)

M + (0.5556) ( 4.108l) ('1.H)S)

-9.37'5(4.!08) ,,o M "" 2 .61 KN· rn Ans

Ans

l-73m-+- +>C< C,.

'ls7;; t:..J. /!>.JZ ;:;ij

re'[ ;': t}.!J75"foi · T

{a;Jl4!')•!/ -.<.5b< r----- a 1)M (o;

M +I l.2.5{ LS) .. 9.315(4.5) "'0

M=2SJ!kN-m

9·'JJibl

MCl".N·m> .25·7-, ,,.zs.3

I "Hq r i

--,;t----"'-r-----j-X(m) ,;'-----,+,,+,-,---4,-x c...

4-49. Dra\v the shear and n10111e11t diagran1s for the bean1.1'here is a pin at C'.

p p

! !

L/4 1 L/4 L/4 L/4

1 L/+ JI '

/4

1.p· -- .J. .e.

,---

i -·--- 2.

u ..'!.. I' t'max-

-PL3:

Ans



6 /2 6 /08 4:13 PM Page 113 lc04.qxd

113

0



4-50. 'fhe concrete bean1 suppo1ts the ¥\'all, vvhich

subjct'.ls the beam to the unifonn loading shovvn.The beam

itself has cross-sectional din1e11sio11s of 12 in. hy 26 in.

and is n1ade fron1 concrete having a specific \Veight of

'Y = 150 lb/ft3. l)ravv the shear and n101nent diagranis for

Lhc beam and spct'.ify Lhc maximum and minimum

n10111e11ts in the bean1. Neglect the \Veight of the steel

reinforcen1ent in the bea111.

800 lbJt

Weig)lt of beam

A = 12(26) = 2.1667 ft'144

.,_, = 150(2.1667) = 325 lb/fl

Ymax=10.7

K Ans

Vmax=51.0K.ft Ans

r---wft 3ft-8ft

+4-51. Dra\v the shear and momcnl diagrams for Lhc

hean1.

Suppon ft#a.:d!u1s: Al ihi:-wn ;::,., f'Bn

SJt#ar rmd lltll'Jlllf DJa1ra.m: Sheu :md lnO.rncrtl J.l

r "' Lt} o.i1 be dae«nitied wi!ng d.e mcliwd of ''Xtioos.

+irr;=o:

1\1m.,,.-23w

Ans

L2/216 Ans

ct · -----+---

} 1:

"T ,



6 /2 6 /08 4:13 PM Page 114 lc04.qxd

114



*4-52. i)ra\v the shear and 11101nent diagran1s for the

beam.

,M(kH·I")

I

1-==i==""':---t-- =--ie-x(,,,.)

o


bean1.

(+ !M = -0; M - w.,L( J <> 4 3 '

H

L L 2 2

"• ,..p·"'·1..

,c-


con1pound bea111. 'lhe seg1nents are connected by pins

at Band D.

150 lb /ft

c

6 ft

V(ibJ

IOoo

-//,00



6 /2 6 /08 4:13 PM Page 115 lc04.qxd

115

-



4-55. l)rff\.V the shear and 11101nent diagranis for the

beam.

5 k/ft

'"{

"4,. I I

'" I:.

,.

I Pf')'"ft

q..(

6 ft 10 ft j_ 6 ft

Vma_,.-=25K Ans

H{tffl i .

I I "' 1

.tlmrrx =-45 K.f't Ans

- Mlfv41l'D>. "

I ir

*4-56. Dravv Lhc shear and momcnl diagrams for Lhc

con1pound bean1. It is supported by a sn1ooth plate at Jl,

\Vhich slides vvithin the groove and so it cannot supporta vcrtit:al force. allhough it can support a moment and p

+axial load.

S fpllrl R•.u:.i.i1u

Fmm !he F&DoflCJJI!ml BD

,.,.W,=O; 8 1(a)-·Pta! ..o 8 1 .,,p

-;-fI: ..o; q P-P=O c,=2P

:-t.IF,, O; B,=O

a 1. Lr r (-t-ilt..i .. j):

.i-trs<;=O: P-P=O {equilibrium is wuisficd!)

Sitar a11d" M t11t oJ)iatrlJln :

,r I I

"' p /\,

L_!J=l , -f



6 /2 6 /08 4:13 PM Page 116 lc04.qxd

116

.,

t:



4-57. 'lhe boards ABC' and BC'l) are loosely bolted

together as sho\vn. If Lhc bolts exert only vertical

reactions on the hoards. detern1ine the reactions at the

supports and dra\v the shear and n10111ent diagran1s for

each board.

t f lo tAI I1, j /.

Ll· S kN/m 10 kN

;:t,:} {tl1LLttAJ ,i T>'. .! D"-1&4 t;J l.oS;lf-"'

'""

l.., 1-. 1..

UWl' the FBDs of me:nhl:;r.; ABC and BCD·

(i.l".MA "" 0; C_,()J - /3r (J) - l ( l.5} "'Q

(+!.Yo"" 0: C 1(1) - D,(4)-.- tO{:/J) "'{l

C, "' 8.57l kN: B" "" 6.7S6 N An"'

+iLF.=O: A1 - 15 ... t.571 - 6-786 (I

... ru.=o: A,. "' lJ.1l N

D.• - !U- 8.571+6.786"' (}

Ans

D, "' !L78 kN An"'

4-58. Dra\v Lhc shear and momcnl diagrams for the

compound beam. The segments arc conncclc<l by a pin

+at R.

8kN

H

----+-l rn ------ 1111

Vmax=---18 KN Ans

Ans



6 /2 6 /08 4:13 PM Page 117 lc04.qxd

117

1



4-59. l)etennine the internal shear_ axial load, and

bending momcnl in the beam at points D and E. Point E 4k is just to the right of the 4-k load. Assun1e J\ is a roller,

the splice at R is a pin, C' is a fixed support.

S-1AB:

:.l:F, = 0:

(+l:Ma = 0:

-tu;= o;

.i!i = 0

-A,.(!2)+ 6.0(6)= O; A,.= 3.0k

3.0-6.0.+B, = 0; B, = J.Ok

•

A [ TJ

lfi lr; 6ft 6ft .. 4ft .. 4ft

Sqin<:ntBC:

:...I.Fz ;:Q;

+tu;=O: (+ =();

'=-0

-3.0-4+C, = O; c, = 7.0k

3.0(8)+4(4l-Mc = O;Jtc =<ID.Ok· fi

Scgmait..W :

= O;

.tl:F, = O;

[+ = O;

Segm<ntEC:

" = 0Am

l.0-3.0-lj, = O; Iii ., 0 Am

-3.0(6)+3.0(3}+Af,,. 0

Ab= 9.00k· fi Am

:..t ;;; O;

+ tl:F, = O;

'+Lii, = O;

= 0 Am

11:+7.0s O; II: =-7.COk Am

- .V.-«J.0 + 7.0(4) = 0 Jo = -12.0k· fi Am

+ *4-60. Dra\v the shear and n10111e11t diagran1s of the

bea111 C'DJ.c.:. Assu1ne the suppo1t at A is a roller and Bis

a pin. "!'here are fixed-connected joints at D and L'.

V ·: L.P i

M(KN. !

6kN lOkN

1.\ kN



6 /2 6 /08 4:13 PM Page 118 lc04.qxd

118

16l,11'lt>



4-61. 11ie overhanging bean1 has been fabricated vvith

projct'.tc<l arm ED on it. Dra\V Lhc shear and momcnl

diagran1s for the bean1 ARC' if it support.;; a load of 800 lb.

flint: 1'he loading in the supporting stn1t l)L' n1ust be

rcplat:cd by equivalent loads at poinl Bon Lhc axis of Lhc

beam.

8001b

H 2 fl c

5 ft

e::==='*'£:t--'1"!!6"o" 3ZOOl J!.

800(1{))-S3FDlll{-4)-jfns\°2)"o

FDE: 2000 fb

+ Vmax = -800 lb Ans

V(lb}

• +txJ1

1. <l• I

4D::;

-1----b--C-«IOt·-- --.,,,+"-x<)tJ

"

LW"max = -4000 lb.fl Ans



6 /2 6 /08 4:13 PM Page 119 lc04.qxd

119

T

1



4-62. l)ra\.V the shear and n101nent diagran1s for each

member of the frame. Assume the support at A is a pin

and D is a roller.

l/·84•

0.6 k/!t

0.8 kJt

rt-----<

Vm=--11.8K

1lJmall--87.(j K.ft

Ans

Ans

+ 4--<-i3. Dra\v the shear and n10111ent diagran1s for each

n1en1ber of the fran1e. Assu1ne A is fixed, the joint at 1J is a pin, and support C' is a roller. 0.5 k/ft B

8 ft

20k

1/5 •/11 1 '*

.ZO·OK

144t}t

6-0K

Vmllx=20.0 K Ans

An!io

,har" - +- +-I, --j,

-t>·Of(



6 /2 6 /08 4:13 PM Page 120 lc04.qxd

120

I.

Ji L



*4-64. l)ra\.V the shear and 11101nent diagranis for each

member of the frame. Assume the joinl al A is a pin and

support C is a roller. The joint at R is fixed. The \\ nd load

is transfen·ed to the n1en1bers at the girts and purlins fron1

the si1nply supported \Vall and roof seg1nents.

500 lb /ft

7ft

7 rt

Supprnt tioos;

{-1oI.\l.i. .. i); -3.$(7) - L7S(l-4} - (4;.21))(, "W'){7ros 30")

-4.20(Jin, 31}0i(J4+3.S:) + ,(:ZJ) = 0

C, = 5.LJ3 lN

l.75 + 3.5 + 1.75 + 4.20 '.11 30" -· 5.l33 - A, 0

"' 3.%'7 llN

A! 4.10 cos 30" = {)

A,. =3.64kN

'"'¥"1.-rs 1

I

71 i

o',,. c,

"'·.

MIK:fiJ . 'rr;.

3.<l

/.II<; J{ 1.1 /. t. _]

._,., 11 =1· ! 1=· =II ,..,,, A,

l.ilo33 11; Z.S,11( 1

y



6 /2 6 /08 4:13 PM Page 121 lc04.qxd

121

z . ·

4

--



4-65. l)ra\.V the shear and n101nent diagran1s for each of

Lhc three members of Lhc frame. Assume Lhc frame is pin

connected at A, C', and TJ and there is a fixed joint at R.

50kN40kN

..-n::."1

4m .... "."' f.[ <.J

4,,._

t'I>•"

+LLJZJ

··]>,.. - $'

{11.>ll fa.II '1 N

4-66. l)ra\.V the shear and n101nent diagra1ns for each

member of Lhc frame. The joints at A, B, and C arc pin

connected.

250lb/ft /,/

tifw

r I CD I

A

6 rt 6 fl



6 /2 6 /08 4:13 PM Page 122 lc04.qxd

121



6 /2 6 /08 4:13 PM Page 123 lc04.qxd

122



4-67. i)ra\v the n101nent diagran1s for the bea111 using

the method of superposition. Consider Lhc beam to be

sin1ply supported. Assun1e A is a pin and R is a roller.

""'I!. , .. ,•. 5•tl if t

1 ioJt ( l==='l'=======

Ii

800lb

"lJ)J)t Il!J 750 lb. ft

----10ft--------10 ft----

+ *4-68. Solve Prob. 4-67 by considering Lhc beam Lo be

cantilevered from the support at A. 800lb

I I I I l { 750 lb ft 1

50 lb/ft

J l L I,<J l 10 ft 10 ft



6 /2 6 /08 4:13 PM Page 124 lc04.qxd

123

+-

+·



4-69. l)ra\v the n101nent diagran1s for the bean1 using

Lhc method of superposition. Consider the beam to be

cantilevered fron1 the pin at Jl.

-zsJ sa _ r-- ::::==---·......_ x (,.,.,)

2&>'.lcr !111 !l i * i* J 11 j _=.=-==- +-·)((..)

.ffl6 K1' -t -Uf<JO

2,400 q::·m::;::::;::::;;:;;;:::::;;:;;;:=?.co;: kN 526-0 ........_ --"'11---f-X (1")

.2oo,.,i

(Mb! d1

4-70. Dra\v the n10111ent diagran1s for the hean1 using

the 1nethod of superposition.

801•/j/

II

'( !='= . . . . " I

...._ 17.. o/i -------1'

M(x)

llb)t)

feo!<i 60Dib.



6 /2 6 /08 4:13 PM Page 125 lc04.qxd

124

MA



4-71. i)ravv the 11101nent diagranis for the bea111 using

the method of superposition. Consider Lhc beam to be

cantilevered fron1 the pin at A.

50 kK/111

Ji

® +

+ +

<-;;-:,=- -& :t !lrr,---J

-r-EM, "'O:



6 /2 6 /08 4:13 PM Page 126 lc04.qxd

125

(

T

1


copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without pennission in writing from the publisher.rr· 4-lP. 'fhe balcony located on the third floor of a n1otel

is shuwn in Lhc photo. IL is t'.onstrut:Lcd using a 4-in.-Lhick

cont'.rctc (plain stone) slab ·which rests on the four simply supported floor hean1s, t\vo cantilevered side girders J\R 6 ft r·H and JJ(J, and the front and rear girders. 11ie idealized

framing plan \vith average dimensions is sho,vn in the

adjacent figure. According to local code. the balcony live

load is 45 psf. i)ra\v the shear and n101nent diagra111s for the front girder lJ(J and a side girder All. Assun1e the

front girder is a l'.hannel that has a ·weight of 25 lb/ft and

the side girders are \\ de flange sections that have a \\'eight

of 45 lb/ft. Neglect the \Veight of the floor bean1s and front railing. For this solution treat eal'.h of the five slabs

as t\\'O-\\'ay slahs.

j :I I,; n C D T

-4n4tt-.1:4 n-.J:41,--1.'- 4tt--I

Dead load = (4 in.)(12 lb/ft'. in.)= 4S p$f

Llw: load = 45 pstTotal load = 93 psf

1.5 <'. 2 Two-way slab

2R - 372(2) - 2T,-, b72

R = 7441b

)(2) = 0

+Front girder

2R' - 4(744) -5\-)<25 + 211)(4) = 0 .2.

R' = 2668 lb

Maximum moment is at center of girder

744•"'

'

"''"·6J.7ft

(...:EM.. = O;

M ... 186(0.667) ... 744{2) + 744<t\l + 372(4) + 372(8) + 250(5)-2668(10) = 0

M = 14 890 lb· ti = 14.9 k· ft Ans

Sido girder

Max.Unmn moment at: support.

(+:EM, = O; M - 1758t3)- :'336(6) = 0

M = 37 290 lb· ft = 37.3 k· ft Ans

Roof load OD inlmllcdialt; joist is (102 lblft')j..'.!_ ft)(l.5 ft) = 51 Jb/ft.12 '

--

,,.ft

,. !V - ....

R = 2[1020 + 135] = 577.5 11>

Re = 577.5 + 190 + 230 = 991.5 lb

Ro = 571.5 ... 250 + l90 = 1117.5 lb

JI, = STl.5 + 310 + 350 = 1237.5 lb

R.r = 577.5 + 370 + 410 = 1357.5 lb



6 /2 6 /08 4:13 PM Page 127 lc04.qxd

126



4-2 P. 1'he canopy sho\vn in the photo provides shelter

for the entrance of a building. Consider all members Lo

he sin1ply supported. The bar joists at C', D, F, F each have

a \Veight of 13:5 lb and are 20 ft long. The roof is 4 in. thick A

and is to be plain lighn.veight concrete having a density H

of 102 lb/ft3. Live load caused by drifting sno\V is assumed

to be trapezoidal, \Vith 60 psf at the right (against the \Vall)

and 20 psf at the left (overhang). Assun1e the concrete slab is simply supported bel\veen the joists. Dra\v Lhe

shear and moment diagrams for Lhe side girder AB. Neglect it.;; \Veight.

1.5 ft 1.5 ft 1.5 ft 1.5 ft 1.5 ft

-------- -------·--

It) j( f /'11 j{ ft., 1391; k..

+ 'k'.."'

f t I>

I I4lo"' 4;1J'"

& m.. + 190 + 230"" 997.5 lb

11o = m.s + i..'lll + m = 1111.s lh 517.5 + 310 + 350 c 1237.5 lb

fir- S/7.5 + 310 + 410 " !357.5 lb

Mm-Jo

I ! ll)l >

I r-r"-f----t- --;--\-;-;({J<)



6 /2 6 /08 4:13 PM Page 128 lc04.qxd

+

+(- j +



4-3P. The idealized framing plan for a Jloor syslcm

lot'.alcd in Lhc lobby of an offlt'.c building is shovvn in the

figure. The floor is n1ade using 4-in .-thick reinforced stone

concrete. If the \Valls of the elevator shaft are llktde fro111

4-in.-thit:k lighL\vcighL solid t'.Onl:rclc masonry, having a

height of 10 ft, detern1ine the n1axin1un1 n10111ent in hean1

AB. Neglect the \Veight of the n1e1nbers.

le 8 ft

8 ft

J

I! H

Flevalor

shaft

j A

8 ft

E GH

8£:01•/lt.

.>5..?'1:/!•

j 4

--8ft---6ft--+-6ft-

{i ' ' ;_ 1: '4J .;!,f'.

Rcinloo=f"""""' """"slab= (150 lblft') (;'.;fi) = 50 psf

Elevator lobby live x.d = 100 p.sf

T.W ""1ing = 150 pd

eoocm.blod:-:

,!.(1 ){10) = 350 lblfi

+12

From sla.b ADIB : ... = (4)(150) ;:: 600 lb/fl

&m>EF:

Fi = (350)(8)+( )((i0))(8) = 26001b

BcamHG:

Y( ) I', = I !• ( 1· 35-0X&I •50(2)

)(3) = 2'2' lb

-( -)(45-0 11·4 l 1. .2

l--i---1"'-r --+-+----t-tljt) (+l:M.. = 0; 11,(20)- 7200(4) - 2600(8)- .5700(11) - 2525(14)- 495007) = 0

II, = 11 SOO lb = 11.SH

+ t.tF,. :;::; ()'; A,+ 11590- 7200-UJOO -5700 - 2525 - 49SO .. 0

,,, = 11 31!.1 lb = ll.38l k.

f + l:.M = ll: M + 7.20( ) - U.31!.1(&) • 0

M.. - 63.t.ik· ft

Moment Shear

Documents