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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--------<
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© 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•
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© 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.
4-6. 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
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l I
© 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.
4-7. 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
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+
© 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.
4-9. 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-
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1 i
© 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.
4-11. 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;,:
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t+:
© 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.
4-13. 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,
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==t=l =!=[ T'=. l \ *=I
=·r·
-1,
© 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.
4-15. 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
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© 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.
4-17. 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
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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
4-21. Dra\v the shear and n10111e11t diagran1s for the
bean1 in Prob. 4-20.
(;lJwt!4001b, fl
!
---x
-------15ft--------
c ; I
v (i,)
n.. I I 1
" j '· 11-•1.h..
© 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.
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© 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.
4-22. 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
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© 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.
*4-24. 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
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6xax•2.x
- 1
0
F11>e.1 M::r J wdx=W --ojL 1
© 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.
4-26. 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
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= 0:
© 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.
*4-28. 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
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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
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© 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.
4-31. 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
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© 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.
4-33. 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
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106
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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.
4-3..'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
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107
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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.
4-37. 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
4-38. Dra\v the shear and n10111e11t diagran1s for the
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
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108
••• i! !i2rlb;lt
"
© 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.
4-39. 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>•
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109
5
0
© 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.
4-41. 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
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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
© 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.
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111
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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
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112
l
2 Ans
© 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.
*4-48. 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
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113
0
© 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.
4-50. '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 ,
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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.
*4-52. i)ra\v the shear and 11101nent diagran1s for the
beam.
,M(kH·I")
I
1-==i==""':---t-- =--ie-x(,,,.)
o
4-53. Dra\v the shear and n10111ent diagran1s for the
bean1.
(+ !M = -0; M - w.,L( J <> 4 3 '
H
L L 2 2
"• ,..p·"'·1..
,c-
4-54. Dra\v the shear and n10111ent diagran1s for the
con1pound bea111. 'lhe seg1nents are connected by pins
at Band D.
150 lb /ft
c
6 ft
V(ibJ
IOoo
-//,00
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-
© 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.
4-55. 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
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.,
t:
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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.
4-57. '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
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1
© 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.
4-59. 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
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16l,11'lt>
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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.
4-61. 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
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119
T
1
© 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.
4-62. 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(
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I.
Ji L
© 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.
*4-64. 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
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121
z . ·
4
--
© 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.
4-65. 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
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© 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.
4-67. 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
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+-
+·
© 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.
4-69. 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.
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MA
© 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.
4-71. 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:
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(
T
1
© 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.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
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© 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.
4-2 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<)
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+
+(- j +
© 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.
4-3P. 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