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1057_Mechanics Homework Mechanics of Materials Solution

1057_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: 9.4 7 For the timber beam and loading shown, determine (a) the slope at end A, (b) the PROBLEM 9-47 deflection at the midpoint c. Use E = 1.6 x 106 psi. 2 kip: 3,5 m SOLUTION 3501M l-—>1 gin. UnRJfS: FoJ‘CQS in Ida-PS) Janefi‘ns In *1: ________ _L +35% = o ”5“ 175‘" ‘ “7%. +(2)(5.25)+(1.225'30.75) = o 1 L225 R. = 1.80625 k‘ip's woo = 0.350 (x ~ 3-5 >° RA RD 3-: = - w = —c3.3.§<x—3..5>O 9}" v= 1.05425 — I<x-L15>° _ o.35<x—3.5)' X 1} EI .. ¢' M = Leo'QZSx — 2<x- ms)’ - o.ns<x-3.5>" kaI-‘H' Er 6'“ : 0:303:25 x‘- ; <x-:.7s>‘— 0.05833 «(-3.573 +C. k}, H" E]: 3 - 0.3mon2x3- 45(x-L75)3— 0.014533 <x-3.s7" 4 C,x+cl lap-Pr? [x=°,J=0-J C1=O [157) 3:9] (o_3o|042)(7)° — 15,45,293- (Lemma (3.5 Y’ + (1.0) + o = o C, = — 7.5472? ksp. H'- DaJm: E: Leygo‘ pa; = Lzeo‘ In.- I= 1158.955)” = 43.525 m’ 512- (“mums“): 77.9w“? mph: = 5317.12 lap-H1 (a) Slope a?! A (and x=o\ El 5! = O — o -o - 7.54777 k‘P‘Ht x 9. = " "73%?" : -H.OO*!O_SMJ - “ll-OOMO':PM“q ‘ “I. (5‘ De-Lfeaflfou 05‘ C. (j a). X’ 3.5%) 213‘ = (0.30loq2)(3.s\3 - £41.75 )3 — o H-(7.5Lr27‘0(3.5) + o = —. :51297 kip-’61“ = - ’5'”? = - "‘ = .39 .‘ 1r .- j; 53%” 28.37sz H o o n ...
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