29 thoughts on “HOMEWORK H28.B”

  1. We are asked to find the shear stress at pin B but are not given the diameter of the pin, D. Should we solve it in terms of this unknown diameter, or is this a typo and should the diameter be d, the same diameter as the wire (which is given)?

    1. I think yes, since we are given a radius, I think we treat it like Example 2 in lecture 28 where the professor found the area of the pin as a circle

    1. Yes, when you're calculating the shear you would multiply the area of the cross-section by 2 as the pin would break on two different parts.

  2. Since there are both x and y component the shear will happen diagonally. This means that we need to find the magnitude of x and y forces before doing the calculations. Is that correct

  3. Correct comments were made in the discussion. It is a double-shear pin so you will need to multiply the area by 2. And yes, you will first need to find the x and y reaction forces at B before calculating the shear stress.

  4. When finding the shear for a Pin, why cant we just use the force acting tangent to the pin like we did in H24 when we found the shear on a member instead of taking the magnitude of the forces?

    1. I don't exactly get your question, but Bx and By are the components of the total shear force acting on the pin, which is what you need to calculate the shear stress.

    2. Using Bx and By is an easier way to find the shear stress. However, I looked back at the homework you're talking about and I believe that your method might also work.

  5. Elliott, I believe you do need to use the moment about B to solve for the tension in CD. You can then go from there to find the reactions at B

    1. Yes, since it is a pin joint. You would have Dx and Dy that would be opposite and equal (if you find the sqrt(dx^2 +dy^2) to the tension. This happens because there boddy is in equilibrium.

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