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641_Mechanics Homework Mechanics of Materials Solution

# 641_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: 5.167 Determine (a) the distance a for which the maximum absolute value of the PROBLEM 5'167 bending moment in the beam is as small as possible, (b) the corresponding maximum normal stress due to bending. (Hint: Draw the bending-moment diagram and then equate the absolute values of the largest positive and negative bending moments l<—20 in.—~|~— 20 in. obtained.) 120 II) 12911) 0.5 in. .l H—L SOLUTION A 40 ’2' Me = O T - 1?¢ a. + (WHIZD) +(20)(t20) = 0 0.75m. _ 72 9c — _5_°—9~ Banding momew'i' ad“ C EZMC = (9 120 M; Mc+ |20 (LID-a.) 2-. O L? 9 M6 = 4:300 + :20 a O-aa Baseline: Women'l' oilL D +DEMD = 0 MD +(20)(|20)~— R~(a_—2o) = o M D MD= RC (cu-2o) — 2400 V O P \ B = QCQ " 20 QC. ‘- 2v00 ° m2 ': 7200 - szlgﬁel — 2400 _ .. M2 - 4800 a 1 [Waco Elinor}: -M,_= M° I4300 — no a. = 9800 - q |20 of: 194000. \$34.44: in. .- M.._ = - 4800 +(lQo)(aI+.m3 v — 643.0! )5»... 44000 . Ma 1' 4800 — iéTfJ—‘(I €43.08 IL "I V ‘ I. m, .2, For redanaularseG-Hon S: é‘bk ._ 3. 2° 8 =(é o.s)(o.7s) = 0.0%275 in“ 34"“ 215 Maximum normal dress M ("H") ‘43- 6. _ Mun-w _ €93.08 - . ———---— —. 3-72am" i '“ S 0.0%375 I P‘ = 13.72 us; - ...
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