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# HW18 - PROBLEM 6.73 meeﬁﬁngon member ABC(a at 3(b at C...

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Unformatted text preview: PROBLEM 6.73 meeﬁmncmdhadingshmdeteminethefmceacﬁngon member ABC(a) at 3, (b) at C. SOLUHOH ' FBI) member ABC: Nate: BB is a two-force member so Fm is through D. \$.an (a) C. me = 0: (8h)[%F‘D] -[12in.)(60 lb) =0 Fan -112.5lb7 36.9“ ‘1 (1:) —~ EE-O: 60lh+C,—%(IIZ.5]b)=O ,4”, Cx=301b 12F, -0:C,—%(112.51b)=0 c, = 67.5 lb 9., soC=733Ib £66.0'34 30L UTDN FBI) Fume: PROBLEM 6.86 components ofthe reactions atA dewhen the frame is loaded by a clockwiee couple of magIﬁtlxle 360 lb -in. applied (a) 81 B, (b) at D. N01: for analysis of the frame FBD. the location of the applied couple is - LEM4=D:(10in.)Ex—36Din-lb=0 Ex=36.olb——< (mg=o:[1oin.)A,—3min-lb=o A, = 36.011: ~— 41 Part (a): Ifcouple acts at'B, EC is a two-150m member, an E". =%E, E,=1s.01b1<l 12F,=0:Ay+131b=o A,=13.001b 14 Part (II): Ifcoupie ads atD, ACis atwo-forca member, so Afégx A,=12.oolbt< Then 12F}=0:121b+Ey =0 E, =—121h E,=12.001bl4 PROBLEM 6.100 1; For the frame and loading shown. determine the componama of the forcesactinganmembexABCatBand C. snLunou ' FBI) Frame: _¢ Hz»! 406M_ 32». C 22.1,. = o: (0.12 m)D — (0.27 m](l92 N) — (0.33 n1](80N] = o D = 652 N _. 121?, = u: A, = n FED members: ( 2M3 = o: (i.‘l.39m)(6521!‘v1]—(D.121:I:|)}JJr = u 3,, = 2119N -— «.25; =n:E,—2119N+652N=o E,=1457N —- PROBLEH 5.100 CONTINUED "—» an; =o:aoN+192N_—1457N+c, = o c, -119SN —- Q 2% = o: —(o.9 m)c,, + (0.12 m)(11bs N) —(o.5 m)(80 N) = o C, =1540N 1 meabove.u1ABC (:Jr =1.195kN-'—- 4 Cy=].540kNT( Br =2.]2kN —--‘ 12F, =0:-By+1540N=0 3, =1540N 13y = 1.54am i; ...
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HW18 - PROBLEM 6.73 meeﬁﬁngon member ABC(a at 3(b at C...

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