# HOMEWORK H28.B ## 29 thoughts on “HOMEWORK H28.B”

1. Jacob Pierce Harmon says:

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. tc says:

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

1. Kaushal Kamal Jain says:

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

2. Rahul Maheswaran says:

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

2. Karthik says:

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

1. Abhirakshak Raja says:

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

2. Ammar Al Nas says:

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

3. Samuel James Pike says:

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?

1. Corina Marie Capuano says:

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

2. Zaid Basil Qubain says:

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

4. Zaid Basil Qubain says:

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

1. Daniel Pereira Ventura says:

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.

5. Zaid Basil Qubain says:

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

1. Chalisa Joan Kulprathipanja says:

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

2. Jennifer Leigh Smith says:

Yes, find the x and y forces.

6. Kaushal Kamal Jain says:

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.

7. Ammar Al Nas says:

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. Kaushal Kamal Jain says:

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. Jennifer Leigh Smith says:

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.

8. Elliott Huy-Anh Eve says:

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

9. Mason Lee Trenaman says:

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. Kaushal Kamal Jain says:

Mason is right

10. Chirag Pradeep Nimani says:

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

1. Kaushal Kamal Jain says:

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

11. huttonl says:

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

1. Mansha Sunil Chimnani says:

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

2. jr says:

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

12. Kelsie Ann Chisholm says:

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

1. Ruth Ivania Guerra says:

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.