# Homework Help: Finding the diameter and thickness using allowable normal and shear stress equations

1. Jun 3, 2012

### savva

1. The problem statement, all variables and given/known data
Given in uploaded file View attachment Q8.37.pdf

2. Relevant equations
σallow=$\frac{P}{A}$
τallow=$\frac{V}{A}$

3. The attempt at a solution
Given in uploaded file View attachment Q8.37 WORKING.pdf

I tried using methods used in previous questions but it didn't work out for me here, if anyone could give a hand, would be greatly appreciated.

I would also like to clarify a couple of other things related to the theory that i'm not sure about.
- I'm not sure what the relationship is between normal force (P) and shear force (V), particularly if given a normal force (P) for example and the question asks to find shear stress. Would Shear force (V) = P/2. I have basically used this principle and it has been generally successful but I am not sure if i'm getting answers correct because double shear applies.

-Secondly, I was doing a question that involved a couple of beams that had pin-joints on each side. Subsequently, I found the external force applied on each pin joint. I have attached a file and drawn an arrow on the diagram. I would like to verify if the arrow acting on the member is a normal force. Because that's what I assumed and when solving the normal stress it gave me the correct answer. The attachment of the problem is below
View attachment Q8.44.pdf

2. Jun 3, 2012

### PhanthomJay

Re: Finding the diameter and thickness using allowable normal and shear stress equati

The area of a circle is pi d^2/4. You forgot the 4. Otherwise your calcs look good for tensile stress in the rod and 'punching' shear stress in the supporting material.
Well, as in your example at the eye, 25 kN vert load is 12.5 kN shear load in the curved rod either side of the load (double shear).
Yes it is the normal axial force in the member, causing shear stress in the bolt (double shear). At the left support, however, you have both axial and shear forces in the member. Their resultant is the shear force on the pin...single shear.