Find the average shear stress at each pin

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ptguard1
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Homework Statement



View image for all details

Homework Equations



tao = V/A

The Attempt at a Solution



I went ahead and solved the force at BC and the x and y components at A, but I feel as though my solutions are incorrect. At BC I have F = 825 kN (up and to the left), A_x = 714.5 kN positive, A_y = 247.5 kN downward. I don't know how to solve the pins from here. The correct answer at each pin is 324MPa. I found the area to be 508.94mm^2 but am unable to find what force is pulling on each pin.
 
Attachments
  • Diagram of a beam supported by a pin at A and a short inclined link at BC at 30 degrees, with downward point loads of 15 kN, 60 kN, 60 kN and 30 kN at marked positions; dimensions 0.5 m, 1 m, 1.5 m, 1.5 m and 0.5 m. Beside it is a sketch of a double-shear pin.
    7-38.png
    10.8 KB · Views: 718
ptguard1: Your area is correct. Your reaction forces are currently incorrect. If you want to show your work, someone might check your math.
 
I recalculated my solutions and found my mistake. F at BC = 165 kN (up and to the right), A_x = 142.89 kN to the right, A_y = 82.5 kN up. The issue I'm having is solving for the shear stress in the pins. I don't understand why all of the pins would have the same result.

For the pin at A, I interpret it as shown in the attached image

I used the equation A_x/Area, but i know that I need the force to be 165kN
 
Attachments
  • Free-body diagram of a pin in double shear: a red rectangular pin piece between two plates is shown with a horizontal force arrow labelled A_x pointing left, and shear force arrows labelled V/2 pointing right at the top and bottom.
    7-38 2.png
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