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## 29 thoughts on “HOMEWORK H28.B”

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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)?

I think that might be a typo and that we can use the same diameter d for the wire and pin D

Yes, that's a typo. You can assume D=d.

Leaving this here for people checking this later - Prof. Jones, confirmed that we can use the same diameter d, for the pin.

Would the area for the tensile stress of wire CD be modeled as a circle?

I think it should be modeled as a circular cross section since they have given the diameter of the cable CD as d

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

Just to make sure, the pin is only going to be sheared in two spots correct? Or is there one spot where it will be sheared?

I believe so, in the problem statement it mentions that the L bracket is pinned to ground with a double -shear pin.

Yes, It says in the problem they used a double shear pin

In a double-shear pin we have to multiply the A by 2 since it has to shear at both ends. Is that right?

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.

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

Yes, you need to find the x and y forces.

Yes, find the x and y forces.

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.

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?

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.

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.

Is it necessary to take the moment about B in order to find the Tension of CD?

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

Mason is right

Is it alright if I take CB as a 2 force member and the following question with that logic?

Yes, you can consider CB to be a two-force member.

How is the area calculated if the thickness is not given?

I believe, the area will just be2( pi r^2 )and the diameter is given. You do not need the thickness.

To find the area use 2*(pi*(d/2)^2), the thickness is not needed

Would there be reaction forces at D or just a tension force?

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.