Question

c) Solve for internal forces just to the left of point C 5 k/ft A В C 7' 3'

Answer 1

I Given beam. 5K/H By equilibrium equetions { fn =0 {hyao HA=0 RA+ RB = ОВ IRA TRB (1/2x9x5) = 92.5K 18.28 3.891 EMA=0 K + Mi => R8 (16) = (22.5 %) 40.5 RB z 4.99K.. K-ft 39.7 K-ft 29.54 k- RA = 18.28K.. In span BD (n from 8) : SFx = - RB = - 4.92K So, SFB - SFO = -4.29 k. BMn = 4 RB (n) = 4.22 (n). BMB (n=o) =O k-, BMD (= ) = 4.92 (7) = 29.54 k-ft In span AD (n hom o): » 5k/lt SFX = -4.29 + gxxx SFO (M=0) = -4.99 K SEC (n=3) = -1.72 K... | SF@C y = = 0.55 m. SFA (n=9) = + 18.28 K..

Srn=0 5x²/18 = 4.22 → a= 3.89 ft hom D. BMX = + 4.22 (x + 1) - $x** * * BM o (n=o) = 29.54 kaft BMA (M= q) = 0 Kaft... BM e (n= 3) = + 39.7 K-ft... | BM @C BM max (n = 3,89) = 40.5 K-ft.

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