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

1. anonymous

|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?

2. anonymous

|dw:1328659569886:dw|

3. anonymous

|dw:1328659825566:dw|

4. anonymous

First, 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. $V = 0 = F_{R,A} - w_R \cdot x -w_T \cdot x-V$ Let's solve for $$V$$, then integrate the following expression to find the moment. $M = \int\limits V dx$

5. anonymous

|dw:1329285118055:dw|

6. anonymous

Looks right. At least judging from the fact that you got maximum moment at minimum shear.