hwk15

hwk15 - ‘61; 7,490 41/4105 W $5410»; f Ila/(#5 V i I I...

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Unformatted text preview: ‘61; 7,490 41/4105 W $5410»; f Ila/(#5 V i I I KE/i/i Z? :7. ; “10+WM%,87=V 5 A6: 5%= /é HM T _ A‘Vs’ Zfit-O 2A6+AB¢543229=1> Arc): —4/§=”l3133m (I, 0 a BC $120M CL : ' 2.0 i 15 __¢ 3 ’ ' 215370 "gab-GmD-wné/sjro +3 —.ébl> +9” my,“ ’20 “hum '50-: 3% :— wmm T (0K ngé 37m Ma'fiafi Jothll D (1ng ‘ W = (3533 an! (a) ED = wow/q CT) 5 I I J." .l. ‘ ‘1 400mm I 400mm I 400mmT-400mm The truss shown is part of an airplane‘s internal structure. Determine the axial forces in members BC, BD, and BE. 0 2452c» 133 : MW I1 3 5%“ sum. Wmemwe Sam’mdwggm A; 3W wewmm " MHE—fiég 281i. M gawk: : c6 :0 130:0 AV Ac v0.53?» M5 2:5‘0 Pk»? @531333cm3637zo fi-c, \ A, 7 are_5',+(<13,33M%,67‘)+0+ BEM%.8‘I°=O -8- Mafia ,«Ls .="/533_Z§((C) V ...
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hwk15 - ‘61; 7,490 41/4105 W $5410»; f Ila/(#5 V i I I...

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