# Homework Help: Shear Diagram

1. Apr 4, 2008

!!Shear Diagram!!

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....

2. Apr 4, 2008

### PhanthomJay

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?

3. Apr 4, 2008

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?

4. Apr 4, 2008

### PhanthomJay

Well, if you know it decreases linearly at .2k/ft, you should be able to find out what that 'something' is.

5. Apr 4, 2008

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?

6. Apr 4, 2008

### PhanthomJay

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.

7. Apr 4, 2008

I am not following you. So I should evaluate at both P and .6?

8. Apr 4, 2008

### 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..

9. Apr 4, 2008

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?

Last edited: Apr 4, 2008
10. Apr 4, 2008

### PhanthomJay

Sorry for the confusion, but it has nothing to do with your work; After looking at this problem, I don't understand why it asks you to determine the maximum P based on shear allowables, without asking you to look at the maximum P based on bending stress allowables.

Assuming that P = 80.1K is the correct answer, a quick look at the max bending stress, which i assume you have studied, at A or B, yields huge values, in the order of several hundreds ksi, so surely the member will fail in bending before it ever reaches its allowable shear stress, in which case P must be much less than 80K.

But if we choose to ignore this fact, and somehow just pretend that shear governs, then the difference between the book answer of 80 vs. 165 probably comes about in the calculation of VQ/It. This value of max shear stress often occurs at the neutral axis, but because you have thin webs , the value you use for t may be 3 inches instead of 6 inches, I don't know what values you used for Q, I, or T in your calcs. So the max P may be based on max shear stress of 10ksi rather tha max average shear stress of 10 ksi. Now i know you must still be confused more than this problem confuses me.

EDIT: Ok, you forced me to painstakingly run the numbers, and i get the neutral axis at 2.34 in. from the bottom, Q=8.22, I=21.95, and t=2(1.5) =3, so V works out to 80.1K. So when the problem said the alloeable shear stress is 10ksi, it indeed did refer to the max allowable shear stress at the neutral axis (VQ/It), not the average allowable shear stress across the entire section (V/A). Thus, from your shear diagram, max shear is the greater of P or .6, thus P =80.1K, based on the given allowable shear stress. However, the problem still remains that the max bending stress, Mc/I, at A or B, is
(80.1) ( 3) (12) (2.34)/21.95 = 307 ksi, and I don't know what materail would have that high an allowable bending stress and with such a low shear stress allowable, so I am not comfortabel saying that Pmax= 80.1K without knowing more about the bending stress allowable. Clear?

Last edited: Apr 5, 2008