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anonymous
 4 years ago
Draw the shear and moment diagrams
anonymous
 4 years ago
Draw the shear and moment diagrams

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328658787358:dw FR (triangle) = (1/2)(18m)(300N/m) = 2700N centroid = 2/3(18) = 12m from A FR (triangle) = 3600N 9 meters from rectangle Sum of forces in y direction = FR+FB FR(rectangle) = 0 Then I solve for FB about A using sum of the moments Is this the right procedure thus far?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328659569886:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328659825566:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First, let's find the resultant force of the distribution. It will have a magnitude and location. You've done that right. Now we need to start at A and cut a section to the right of A. Let the length of the section be \(x\). The equation for sheer is simple enough. \[V = 0 = F_{R,A}  w_R \cdot x w_T \cdot xV\] Let's solve for \(V\), then integrate the following expression to find the moment. \[M = \int\limits V dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1329285118055:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Looks right. At least judging from the fact that you got maximum moment at minimum shear.
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