HW20 - PROBLEM 6.159...

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Unformatted text preview: PROBLEM 6.159 Astadiummofh‘ussisloadndasshown.Dctcrmincthcfirmcin -- .21.” members BC. BH» and GH’ :4 H SOLUTION I . Q Ma = n: (6.3 EJFM - (14 fi)(1.3kips) — (23 Russ kip)=o FBI) Mull: 'FGH.=3.00kiIB (3* 1 ( $51,, = o: (3.15 tt)[%Fm]—(14 ft)(0.9 kip) - 0 FM = 4.10kips T4 9 . . 9 BF =u:—F —1-81nps—0.9h +—F =0 t " 41”" P 21.93 3” F”. =' $[2 7 lu'ps — % (4-10 593)] = 4.386 kips Fm = 4.39 kips T < PROBLEM 6.161 - :=“.' For the frame and loading shown, detennine the components of ihe forces acting on m-ba DABC at B and at D. Q EMG - 0: (0.6 mm, —(0.5 m)6 kN — (1.0 m)(12 kN) - 0 .. Hy=25kNT L EME - 0: (0.5 m)B: —(0.2 m)(25 kN) - 0 a; = 10 kN 0n DABC B, = 10.00 kN—-- dl —-I.F,, =0: -Dr+B,+12kN=0. D,=(10kN+12kN)=22kN D,=22.0kN--—4 LEM, = 0: (0.81:00’, -(0.5 01);}, = 0 D,=13.75kN Dy “3.7510011 Tm; =0:B,,-D, =0 B, =13-75kN . n,=13.75kN14 PROBLEM 6.163 ' ' .¢,,-.:;..: - Forthefi'amemidloading shawndetemahethecomponentsof the forces aotingonmcmbcr CFEatCand atF. Q 2MB = n: (13 in.)(401b) —(10in.)A, = D A, = 52 lb —-- Q m, = o: (4 my; - (6 in.)(521b) = u F,=731b—»onA3F onCFEfimalz-ove F,=7&nIb«—d . ( 2M, = o: (9 in.)(40 1b)—(4 imp", —(4in.)(781b) =0 Fy=l'2.flfllbt" '(‘zFfmcffi =0 cx -78lb Ic,,-=7s.otb —»4 In; =D:—4I}1b+F,,+Cy =0 Cy=401b—121b=281b C,-23.01b11 ...
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