# !!Shear Diagram!!

Tags: diagram, shear
 P: 2,993 1. The problem statement, all variables and given/known data I have posted the problem below. I am looking to find P max. My professor wants us to draw a shear diagram to find out where it will occur. From equilibrium equations I have found that Ay=By=P+0.6 kip I am having trouble drawing the diagram. I know it starts with -P and is constant until it hits Ay. Then I add -P+(P+.6) So it is +.6 Now there is a distributed load. So it should decrease linearly. But that is where I am stuck. Where does it decrease to? Does it cross the x-axis at all? Sorry, but I just can't see this one straight....
 PF Patron HW Helper Sci Advisor P: 5,802 It will decrease linearly at the rate of the distributed load of .2k/ft, until you reach B. Your equilibrium equation and shear value at A looks real good. Now where's P_max?
 P: 2,993 Well I know that it decreases linearly at .2k/ft until B. But What value does my diagram take on at B? It is going to be By+ something....but what is that something?
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## !!Shear Diagram!!

 Quote by Saladsamurai Well I know that it decreases linearly at .2k/ft until B. But What value does my diagram take on at B? It is going to be By+ something....but what is that something?
Well, if you know it decreases linearly at .2k/ft, you should be able to find out what that 'something' is.
 P: 2,993 Okay. I added By to -1.2 since that is the area under dist load. My diagram now works out to equilibrium. Great! Now my next conceptual hurdle is this: I have a shear diagram that goes like this: Starts at -P (below x-axis) 'Jumps' to +.6 Decrease linearly to -.6 'Jumps' to +P (above axis) I don't know anything about P itself. I know that V max is either .6 kip or P kip. how does the diagram tell me anything about whether P is greater or less than .6?
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P: 5,802
 Quote by Saladsamurai Okay. I added By to -1.2 since that is the area under dist load. My diagram now works out to equilibrium. Great! Now my next conceptual hurdle is this: I have a shear diagram that goes like this: Starts at -P (below x-axis) 'Jumps' to +.6 Decrease linearly to -.6 'Jumps' to +P (above axis) I don't know anything about P itself. I know that V max is either .6 kip or P kip. how does the diagram tell me anything about whether P is greater or less than .6?
your diagram is now correct, and you are also correct in that it does not tell you anything about the value of P. But you are given that the max shear stress allowed is 10ksi, and you are given the cross section of the member.
 P: 2,993 I am not following you. So I should evaluate at both P and .6?
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P: 5,802
 Quote by Saladsamurai I am not following you. So I should evaluate at both P and .6?
I don't know why you are looking at shear stresses and not bending stresses. Shear seldom controls member design, but I haven't looked too closely at the numbers. Anyway, based on shear alone, the max (average) shear stress allowed is 10ksi, the area of the section is 16.5in^2, so the max shear allowed is 165 kips, it seems, which sure dwarfs the distributed load. But I'd check bending first, shear later, although there is no metion in the problem of bending stresses..
P: 2,993
 Quote by PhanthomJay I don't know why you are looking at shear stresses and not bending stresses. Shear seldom controls member design, but I haven't looked too closely at the numbers. Anyway, based on shear alone, the max (average) shear stress allowed is 10ksi, the area of the section is 16.5in^2, so the max shear allowed is 165 kips, it seems, which sure dwarfs the distributed load. But I'd check bending first, shear later, although there is no metion in the problem of bending stresses..
Don't mean to sound rude here. But now I have no idea what you are saying.

I am looking to find the maximum P allowed here. I know that $\tau =10 ksi$

I know that $$\tau =\frac{QV}{It}\Rightarrow V=\frac{\tau It}{Q}$$

I am not sure how your last post helps me solve this problem.

If I plug in the numbers I get 165 kip. But the answer in the text is 80.1 kip.

What am I not seeing?