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# HW22[1] - PROBLEM 7.5 150 mm/’f< ” V o 150 mm,«V if...

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Unformatted text preview: PROBLEM 7.5 150 mm /’f< ” V / o ,/ 150 mm ,«V' if ' Determine the internal forces at point J of the structure shown. \ \ 7}. SOLUTION FBD Frame: AB is two-force member, so A Ax : y A : 1A): 0.36 m 0.15 In y 12 Q EMC = 0: (0.3 m)Ax — (0.48 m)(390 N) = 0 AX =624N «w 5 Ayzl—ixzzsoNorAyzzsoNt MZFX=0:F—624N=0 F=624N—~»< FBDAJ: 260A, M 12Fy=0:260N—V:0 J ~ Mm» ,M %E V=260N1< . x Q ZMJ : 01M — (0.2 m)(260 N): 0 M = 52.0N-m 34 rMmm PROBLEM 7.13 ,. ‘ M25723. Two members, each consistin of strai ht and 168-mm—radius uarter— J 7 . . g g q 1339‘” c1rcle portlons, are connected as shown and support a 480-N load at D. Determine the internal forces at point .1. SOLUTION “31) Frame: Q EMA = 0: (0.336 m)[2—C] — (0.252 m)(480 N) = 0 C=375N —+ EFy =0: Ax—&C=0 Ax =—Zi(375N)=360N 25 25 X =360N~—> 12F) =0:Ay—480N+%(375N)=0 Ay :375N) Q EMU = 0: (0.324 m)(480 N) — (0.27 m)B = 0 B = 576 N —> ZFX = 0: CX —%(375N) = 0 Cx=360N —+ (>313: 0: —480N + %(375 N) + (576 N) ~ Cy = 0 CV = 201N1 \ 2ny = 0: V — (360 N)cos30° — (201 N)sin30° = 0 V : 412N \ 4 / 2F), : 0: F + (360 N)sin30° — (201N)cos30° = 0 F:—5.93N F=5.93N /< (2M0 = 0: (0.168 m)(201N + 5.93 N) — M = 0 M = 34.76 NAm M = 34.8N‘m )4 1”": r i— '1 j, , a ..> , r4 «1 fr" 3‘“? fa: .2 PROBLEM 7.21 A force P is applied to a bent rod which is supported by a roller and a pin and bracket. For each of the three cases shown, determine the intemal forces at point .1. SOLUTION (a) FBD Rod: QEMD=0:aP—2aA=o A=—«—— 2 ~+2Fx=0:V—5=0 2 § . i213:0. QEMJ=0:M—a§=0 QEMD = 0: aP—§[§A] = 0 V=——>4 2 F=04 M=%‘)< FBD AJ: F (c) FBD Rod: (3 PROBLEM 7.21 CONTINUED —»21~;=0:§£—V=0 52 V=—<k{ 2 TZF.=0:i£—F=O ’ 52 F=2Pl< M:Z P)< 2 CEMD =0:aP—2a(§A]—2a[5A]=o 5 5 21:2 14 ->2Fx:0:V—[3§1: =0 514 v=213... 14 ? y=0:i£—F= 14 [5:214 7 Q2M1=o:M—a(35—P)=o 514 3 M=—-aP‘)< ...
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HW22[1] - PROBLEM 7.5 150 mm/’f< ” V o 150 mm,«V if...

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