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assignment 14 solutions

# assignment 14 solutions - 1“ Wm PROBLEM [email protected],2 m'l 3.6...

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Unformatted text preview: 1“ Wm PROBLEM 5.63 @1,2 m'l 3.6 m —-—><——* Detemline the reactions at the beam supports for the given loading. 0.8 m ‘ SOLUTION R = (1800 N/m)(3.2 m) = 5.76 kN + MA = 0: —(5.76 kN)(1.2 m +1.6 m) + Bv(3.6 m) = 0, or B), = 4.48 kN B = 4.48 kN T 4 +1 ZFy = o: A), + 4.48 — 5.76 = 0, or Ay = 1.28 kN + 217, = o: AX = 0 Therefore: A=1.28kNT4 PROBLEM 5.65 Determine the reactions at the beam supports for the given loading. 2“) lb/Il 180 H m 418 f1 —>l<——3.6 ft SOLUTION 2-“(4‘ R1 = (240%)(48 ﬂ) = 11521b 1 lb R =— 180— 3.6ﬁ =324 lb 2 2( ﬁ)< 1 Equilibrium: +EFX = 0: AC = 0 +2Fv=0z Ay—11521b+324lb=0 A), = 828.001b A = 8281b 14 +ZMA = 0: MA — (2.4 ﬂ)(1152 lb) + (6 ﬁ)(324 lb) = 0 MA = 820.80 1130 MA = 8211M: )4 . PROBLEM 5.66 90 llw‘lt ‘ 8 Determine the reactions at the beam supports for the given loading. SOLUTION The distributed load given can be simpliﬁed as in the diagram below with the resultants R1 and R2. l-S' H’ I .20 mm .30 Una/u '<——-—-— 3“ The resultants are: R, = (6 ﬁ)(301b/ﬁ) = 180 1b, and R2 = 54.5 ﬁ)(1201b/ﬁ) = 2701b 1 Now, for equilibrium: ZFX = 0: A. = 0 25. = 0; Ay+l80—270=0 A), = 90.01b Therefore: A = 90.0 lb T 4 EMA = 0: MA + (3 ﬁ)(180 1b) — (1.5 + g x 4.5)ﬁ x (2701b) = 0 or M = 6751b-ft 34 ...
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