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Draw the shear and moment diagrams
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|dw: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?
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