HW_solu_09-20-04 - TAM 210 211 Fall 2004 Homework solutions...

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TAM 210 / 211, Fall, 2004 Homework solutions for 09/20 3.143 Given: The distributed load shown. Required: Find the equipollent force/couple system at the wall. Solution: By inspection, we can easily see that the area under the curve is zero, so the resultant force should be zero. However, we could test this to get a bit of practice: () ( ) () () ( ) [] 0 0 cos 2 cos 2 / ˆ / 2 cos 2 / ˆ / 2 sin ˆ ˆ ˆ 0 0 0 = = = = = = = = = π L q L x L q dx L x q qdx F o L x o L x o L x y j j j j j F () () = = = × = L x o L x o O r dx L x x q dx L x q x 0 0 , / 2 sin ˆ ˆ / 2 sin ˆ k j i M To solve this, it helps to integrate by parts: Let u = x , so du = dx ; () dx L x dv / 2 sin = , so () ( ) L x L v / 2 cos 2 / = . Then () ( ) () [] ()( ) ()( ) [] () ( ) [] {} () 2 / ˆ / 2 sin 2 / 0 cos 0 2 cos 2 / ˆ / 2 cos 2 / / 2 cos 2 / ˆ / 2 sin ˆ 2 0 2 0 0 0 , o L x o L x L x o L x o O r q L L x L L L q dx L x L L x x L q dx L x x q k k k k M = = = = = = = = 3.147 Given: The concrete ramp shown, width = 1m. Density of concrete = 22,600 N/m 3 Find: The weight of the ramp and its line of action Solution: Weight = density x volume ( ) () () ( ) ( ) () N m m m m N W 900 , 33 1 5 . 0 6 2 / 1 / 600 , 22 3 = = The weight is distributed in proportion to the local volume distribution. Assume y is “up” and
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HW_solu_09-20-04 - TAM 210 211 Fall 2004 Homework solutions...

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