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anonymous

  • 4 years ago

Draw the shear and moment diagrams

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  1. anonymous
    • 4 years ago
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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?

  2. anonymous
    • 4 years ago
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    |dw:1328659569886:dw|

  3. anonymous
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    |dw:1328659825566:dw|

  4. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    |dw:1329285118055:dw|

  6. anonymous
    • 4 years ago
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    Looks right. At least judging from the fact that you got maximum moment at minimum shear.

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