ME211solution_4

# ME211solution_4 - The distributed load 75lb/ft can be...

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The distributed load 75lb/ft can be considered as a resultant force 75lb/ft×12ft=900lb at 6ft away from the point A. From the entire structure, normal forces at the point of A and B can be calculated. 0; (14) 2500(20) 900(8) 3000(2) 0 By MA =− + + + = , 4514 y A lb = 0 xx FB == 4514 2500 900 3000 0 yy + = 1886 y B lb = Section at the point C, and consider the F.B.D. of the left part. 2500(6) 0 CC MM =+ = 15000 15.0 C M lb ft kip ft 0 xC FN k i p = 2500 4514 0 yC FV + = , Vl 2014 2.01 C b Section at the point D, and consider the F.B.D. of the right part. 1886(2) 0 DD 3771 3.77 D M lb ft kip ft =⋅ = 0 xD k i p 3000 1886 0 yD + + = , 1114 1.11 D b

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The distributed load can be considered as one resultant force, 1 (2 ) 2 abw + , acting at the centroid of the triangle. From the entire structure, find reaction forces at the point C 11 0; (2 ) ( ) ( ) 0 23 By Ma b w b a A ⎡⎤ =+ ⎢⎥ ⎣⎦ b = )( ) 6 y w A abba b This problem requires 0 C V = From the F.B.D. of the left part of the sectioned structure at C 1 (2 )( ) 0 62 2 y wb Fa b b a a b ⎛⎞ + ⎜⎟ ⎝⎠ 2 w = )( ) ) 68 ww ab b
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## This homework help was uploaded on 04/04/2008 for the course ME 211 taught by Professor Thouless during the Fall '08 term at University of Michigan.

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ME211solution_4 - The distributed load 75lb/ft can be...

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