519_Mechanics Homework Mechanics of Materials Solution

519_Mechanics Homework Mechanics of Materials Solution - EM...

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Unformatted text preview: EM 5 3 6 5.36 For the beam and loading shown, determine (a) the distance a for which the PROBL ' mximmn absolute value of the bending moment in the beam is as small as possible, ([2) the corresponding mximum normal stress due to bending. (Sec hint of Prob.5.33). LE! kips 0.8 kip: SOLUTION E ‘92 M: = o - 0.80. 41.90.23 - (2.7mm + (3.9) 9 = B = ILt+~ 0.222220. 32 MB: 0 concaua.) - 3.6 C +£2.H(:.2) 4(0.9)(L2\ a o C ‘-' 1.9 + 0.223220. Beach-‘3 momen'} ¢,+ C . 92% 2 0 °" 9'“ °' M; +(O.S)(a.) g o [(-a- [+0. MC. 7: - 0.8 Q Bending momen+ 0:} D ZMD = 0 . M‘> +(0.2)(a+r.5“) _ LSC = 0 MD ' 1:5“ ‘— one“: a: Benefit-‘3 meme-n“ ¢+ E M. D in; 3 O "‘ ME 4‘ 0.9 B = 0 Ca Mg = L26 — 0.2 a. V 0.“! B 0.8a= 1.25 -.o.2a. (2:1;ch «- ME 1.008 In}. H :‘M,- 0.9m lap-H. max ”4] 2 L003 kip-H = mews {grip-in For rolfeai she! sedan. 5 3:53 8 = l. as in” Noam»? s'l'ress - '3'236 :3 7.20 Im‘ ...
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