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Problem 4 (30%) Determine the internal forces of every member of the truss in the figure. Consider that member BE undergoes a

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I ginen € = 29 Oud ksi, A= sin, a= 6.5xieb of BE → +60of , BD = .75 too short: Truss Isft ist te VA 15) E (4) 2oft - LookipCalculation of Reactions Compression (-) Tension = (+) Efyzort VA +/- 2010) 20 TVA = 200 kip +/EMA =0 Rex 30 = 200 x 20 RC =)fant IED 133-33 133.33 200 fto= 133 33 kip 1 IFBE = 200 kip joint c 133.33 kip 36-87 753.13 FCD ICB Es x=0 tcB 20536.87 +133.point D. for I 100 kip 36-87 133.33kip E&x=0 & - 133.33 = for los 36.87. For =) - 166.66 kip Ang J Compresor The cao - so app15 201 E 200 point A FAB 36.89 42 FAE fact & fore 20136.87 +4/320 (sy=0 9+ [ Jae 2) o peint & y alebo {}, со 1+ 56.300 Efrão1 IDB I foc 3687 O.SSS ху = T+ ги во *, foB los 3687 = oss foB - 0.693 (T) foc +0.416 = 1 I foc = 0.583 (T) foint & folg 536.• Member Lift) Pikip LQ OT (ft) dilo li Eufeli +4; (in) + L. 07] LAE -333.33 1.66 At 20 133.33 Bt 1 200 -832 -0.265 15x6.5x10do Member force - Pi & UiR AB = – 333.33 – 1.66*(-41.16) - 264.75 kip / Ans (C AE = 133.33 +0=1133-33 kip BE = 200 + 832 x 0149.8ke е . Р suf 76КР 264-75 41-16 за 29 kip 195-18\КР 133-33kip 156-17 Зар kiP

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