problem set 7 (key)

8/14/2019 Problem Set 7 (Key)

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I:2.';Omm

L :~.~ I

2.';Omm..j

~mm

PROBLEM 7.2

400 For the frame and loading of Prob. 6.78, detennine the internal forces at

point J.

),

L-1-250 mn.-

SOLUTION

( IMA = 0: (0.95 m)(480 N) -(0.25 m)Dx = 0,

D = 1824N --.x

Ay + Dy + 480 N = 0(1)

FBD FRAME:

II~

-f'FDN

F

P...

FBD ABC:Note: BE is a two-force member

(WB=O: (0.75m)Ay=0 Ay=O

Dy = -480 Nhen, from (1) above,

Dy==480N

F -1824 N = 0 F= 1.824kN-~

-480 N -V = 0 V = -480 N

V=480N t ~

M + (0.25m)(480 N) = 0

M=-120 N.m M = 120.0 . m ) ..

PROPRIETARY MATERIAL. ~ 2007 The McGraw-Hill Companies, Inc.. All rights reserved.. No part of his Manual may be displayed, reproduced

or distributed in any orm or by any means, without the prior written permission of he publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission.

1004



PROBLEM7.5

Detennine the internal forces at point J of the structure shown.

j

7.<; u

lOin,

T1.5 in

7.5 ill. CocK

1111.-

~~

L""-

SOLUTION

FBD Frame:

(1)

Note: AB s a two-forcemember, o

~=.1.12 5

( IMc = 0: (15 n.)Ax- (24 n.)(781b) 0

Ax = 124.8b -

From 1) above, Ay = 52.0 lb

FBDAJ:

-I.Fx =0: -124.8 Ib + F = 0 F = 124.8b ..

t rEv = 0: 52 lb -V = 0 v = 52.0 b ~ ~

(EMJ =0: M -(10 in.)(521b) = 0 M = 520 b. n. ) ..

PROPRIETARY MATERIAl- ~ 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of his Manual may be displayed. reproduced

or distributed in any ()rm or by any means, without the prior written permission of he publisher. or used beyond the limited distribution to teachers and

educators permitted by McGraM'-Hi/l for their individual course preparation./[ )'OU are a student using thi,vManual. you are using it ,,'ilhout permission.

1007



SOLUTION

(a) FBD Beam:-~=o:

f (EMs = 0:

D

B =0

aP + 2aC -3.5aP = 0

C = I.25Pt

-P+ By + I.25P- P= 0

I>

~~=~.' ~ ,~I I ~y

I

I I By ==O.75P

p

Along AD:v

--,: Pl't

-p

v=-p

M=-Px

1:Fy = 0:

(rMJ =0:

Along BC:

I'r~,.j ~~

-p-v=o

M+xP=O~

~~==~~I~~I

I I-PAw -I.~-P~

:..

~~,~o.~1> ~

-P+ 0.75P- V = 0 V =-0.25P

M + xP -(x -a)(0.75P) = 0

M = -0.75Pa -0.25Px

1:Fy= 0:

( rMK = 0:

M(3a) = -I.5Pa

Along CD: ~ p

1::1 L. y.

t Wy = 0 v-P=o v=p

( IML = 0: -M- Xl P= 0 M = -PxI

M(I.Sa) = -I.SPa

IvJ~ = P along AB and CD ~b)

IMlmI){ = I.5Pa at C ~

PROPRIETARY MA TERIAL. @ 2007 The McGraw-Hill Companies,nc. All rights reserved.No part of his Manual maybe displayed, eproducedor distributed n anyorm or by ally means,without heprior writtenpermission of he publis'her, r usedbeyond he imited distribution o teachers ndeducators ermittedby McGraw-Hill or their individualcourse reparation. f you are a student sing his Manual, ou are using t without ermission.

1036



continued

PROPRIETARY MATERIAL. ~ 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of th,s Manual may be displayed. reproducedor distributed in an}' form or by any means. without the prior written permission of the publLl'her. or used beyond the limited distrihution to teachers and

educlltor.'i permitted by McGraw-Hilifor their individual course preparation. !f}'OU are a student using this Manual. you are u.~ing t liithout permission.

1041



PROBL~

Along DE:

7.35 CONTINUED

M

t1:F 0: .810N

rMN 0: M + (X3+ 0.3 m)(540 N) (1350 N) 0

M .162N.m + (810N)X3

162N.m atD (x. 0.4)

Along EB:

I\J11

M'"-

t~ JI 540N: 0 J' 540N

WL 0 x2(540N) 0 -540 Nx2

162N.matE O.3m)

Fromdiagrams 810NonDE-'rnx

162.0 N.matD andE

PROPRIETARY MATERIAL

or distributed in any fonn or b:

educators permitted by McGral

.@2007Thy any means,

w-Hill for the

McGraw-Hili Companies, Inc. All rights reserved. No part of his Manual may be displayed, reproduce.

vithout the prior written permission of he publisher. or used beyond the limited distribution to teachers an.,individual course preparation. If you are a student using this Manual. .vou are using it without permission.

1042



continued

PROPRIETARY MATERIAL @ 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of his Manual ma.v be displayed, reproducedor distributed in anyform or by any means, without the prior ..ritten permis.sion of/he publi.,.her, or used beyond the limited distribution to teachers and

educators permitted by McGraw-Hillfor their individual course preparation. If}'ou are a student uYing thi.s Manual, you are uYing t ~'ithout permission.

1045



A

1.50 liPROBLEM 7 39)" l()()lb !ou\!.. II' .

For the beam and loading shown, (a) draw the shear and bending-momentB diagrams, (b) detennine the maximum absolute values of the shear and

.i I bendingmoment.10111. ~-10in'"

6in.6in.6hl.6ill.

continued

PROPRIETARY MATERIAL. @ 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part ofthi$ ,'vtanual may be displayed, reproducedor distributed in any orm or by any means, ~ithout the prior written permission of he publisher, or U$edbeyond the limited di$tribution to teacher$ and

educator$ permitted by McGraK'-Hill.for their individual course preparation. If.Vou are a .~tudentu.~ing his ,l.,1anual, ou are tL~ing tl\'ithout permi~~ion.

1049



PROBLEM 7.39 CONTINUED

AlongDE: ~/b/t" lei-' b

bi.. I~Mc ,'" It) ~

;"".$"/1> ~~- ---x ~

v = 75 Ib,=o:

(rMN =0:

-(121b/in.)(10 in.) + 2951b 100 lb -V = 0,

M + (x -16 in.)(I00 lb) -(x -10 in.)(295 lb)

.+ (x -5 in.)(12 lb/in.) 10 in.) = 0

M = -750 lb. in + (751b)x

Complete diagramsusing symmetry.

1J1~ = 175.0 b alongCD andFG ~

IMlmI)( = 900 lb. in. at E ..

MATERIAL. <02007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedany onn or by any means, without the prior written permission of he publisher. or used beyond the limited distribution to teachers and

permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are ILl'ing t without permission.



SOLUTION

~~:~J.QAV'I4DOIJf2DN IMA = 0:

;UIIh- ~ --e

"" .J'*.lAf "" ~

~y

~- A.

~'II

-(2 m)(420 N) = 0 Ey = 1200 N

Uy .;, 0: Ay + 1200 N -(360 N/m)(1 m) -600 N -420 N = 0

(a) Ay = 180.0N'/z.o

"::JYo

~ \

-/8'2>

v'

(AI)

ShearDiag:

V wnps to Ay = 180N atA, thendecreasesinearly

(~ = -w = -360 N/m) to 180N -(360 N/m)(1 m) = -180 N at C.

FromC, V s piecewise onstant w = 0) with wnps of -600 N at D,

+1200N atE, -420N atE.

MomentDiag:M

(11/

-5'1 M starts at zero at A with slope ~ = V = 180 N/m, decreasing to zerodx

at x = 0.5 m. There M = !(180 N)(0.5 m) = 45 N.m Mis zero again2

at C, decreasingo -(180 N)(0.3 m) = -54 N.m atD. Mthen decreases

by (780N)(O.2m) = 156N.m to -210N.m atE, and ncreases y

(420 N)(0.5 m) = 210 N.m to zeroatE.

(b) Ivlnmx 780 N alongEB ~rom the diagrams,

lM1nm= 210 N.m atE ~

PROPRIETARY MATERIAL. @ 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedor distributed in any fonn or by any means, without the prior written pennission of he publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, .YOU re using it ~'ithout permission.

1082

problem set 7 (key)

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