Solid Mechanics Question - Help

[Mapes]

I'd check the shear stress at points e and d too, in case that's where the maximum shear occurs (I don't know).

 

[dislect]

hmm I wish I could but me and my friends don't know how.. how do we check the sheer stress in a certain point?

 

[dislect]

Could you do the last part with me?
I can't figure out which point on plane AB would fail first. Its past the due date but I really want to know.. I guessed its B but I have no explanation

 

[Mapes]

Here's a possible approach:

1. Define angle $\phi$ as rotating from axis 1 towards axis 2 around axis 3, and angle $\theta$ as rotating from axis 1 towards axis 3 around axis 2. In other words, axis 2 corresponds to $\phi=\pi/2$, $\theta=0$, and axis 3 corresponds to $\phi=0$, $\theta=\pi/2$.

2. Identify the stress state at the surface from torsion to be

$$\bold{\sigma}=\tau_0\left[\begin{array}{ccc}<br /> 0& 0& \sin\phi\\<br /> 0 & 0 & -\cos\phi\\<br /> \sin\phi & -\cos\phi & 0\\<br /> \end{array}\right]$$

where $\tau_0=TR/J$.

3. To find the shear stress on a certain plane, define a direction cosine matrix $\bold{a}$ that gives the new axes when multiplied by the old axes. Double check that this is

$$\bold{a}=\left[\begin{array}{ccc}<br /> \cos\theta& 0& \sin\theta\\<br /> 0 & 1 & 0\\<br /> -\sin\theta& 0 & \cos\theta\\<br /> \end{array}\right]$$

where $\theta$ is the angle of the joint, and we know that $\sin\theta=\frac{4}{5}$ and $\cos\theta=\frac{3}{5}$. Then the stress transformed to the joint plane is

$$\bold{\sigma '}=\bold{a\sigma ' a^T}$$

The shear components of $\bold{\sigma '}$ (hint: two of them are $-\cos\phi\sin\theta$ and $\sin\phi(\cos^2\theta-\sin^2\theta)$) represent the shear stresses on the joint; by checking different values of $\phi$, you can check where the stress is highest and what its value is to solve the problem. This is a generalized technique that can be used to solve any joint problem for any loading configuration.

 

[dislect]

Wow, thanks!

So in this question there's a little dilema.
As I understand, all I really care about is a(1,3) and a(3,1) because they are the sheer stress on the direction that I am looking for, right?
$\sin\phi(\cos^2\theta-\sin^2\theta)$
with the given data leaves me with -0.28sin(phi) so the max values is when the angle phi is 90 or 270, and that means A or B on my AB plane.
Meaning, it depends on the direction of the moment T? :)

 

[Mapes]

Wow, thanks!

So in this question there's a little dilema.
As I understand, all I really care about is a(1,3) and a(3,1) because they are the sheer stress on the direction that I am looking for, right?

This approach let's you look at all the possible shear stresses to find out where the highest stress occurs. In other words, don't pick a certain location; instead, figure out where the shear stress is maximized. What's the largest shear stress term?

 

[dislect]

This approach let's you look at all the possible shear stresses to find out where the highest stress occurs. In other words, don't pick a certain location; instead, figure out where the shear stress is maximized. What's the largest shear stress term?

Got lost in the translation :) what is a shear stress term?
Do you mean that i should look at the entire matrix, for example a(2,1)= $-\cos\phi\sin\theta$
And in this question, $\sin\theta=\frac{4}{5}$ so a(2,1)=-0.8*$\cos\phi$ which is the max value I can get as long as $-\phi$
is 0 ? and so I get that the first point to break is on the the middle of the perimeter between A to B ?

 

[Mapes]

In the stress tensor (matrix) $\bold{\sigma}$, the terms that are off the diagonal are shear stress terms.

 

[dislect]

yeah i kinda guessed that 1 minute later and edited my post :S

 

[Mapes]

Great, I agree that $\bold{\sigma '_{21}}=\bold{\sigma '_{12}}=-\frac{4}{5}\tau_0\cos\phi$ has the largest magnitude over the entire joint (compare to $\bold{\sigma '_{13}}=-\frac{7}{25}\tau_0\sin\phi$ and $\bold{\sigma '_{23}}=-\frac{3}{5}\tau_0\cos\phi$). This corresponds to $\phi =0$ or $\pi$, which looks like points A and B on your diagram. (So you guessed correctly in your original solution!). But we get slightly different answers for the maximum permissable torque T; perhaps one of us has a calculation error.

EDIT: Fix typo; $\phi=\pi$, not $\pi/2$.

 

[dislect]

That's kind of far from what I got :\ Could you take a look and see if I did any critical mistakes in my calculation? Because I checked yours and its right, and I can't find where I got it wrong (plus I need to explain it to my friend)

 

[Mapes]

Actually, I can't follow your approach. I wanted you to solve in terms of the angles because that makes it easy to look at the answer and see if it works at extreme values (like joint angles of 0° and 90°). It's too difficult to follow many lines of someone else's numerical calculations.

 

[dislect]

So it's safe to say that the max tourqe is T=(5/8)*pi*(R^3)*Sigma0 (reminder:Sigma0 is the max shear stress)?

Mapes, thanks a lot for your help and supreme patience :) I know I've been a nag (and I probably will be).

 

[Mapes]

That looks good.

 

[dislect]

Oh and one last question I forgot to ask, how come there's a shear stress on a(3,1) or a(1,3), meaning in the direction of e3? I thought there was supose to be only on e2.

 

[Mapes]

I'm not sure what you mean. The matrix $\bold{a}$ is the rotation matrix, also known as the direction cosine matrix. It relates different coordinate systems. It doesn't represent stress.

 

[dislect]

I was relating to the Sigma' matrix, it has stress terms on a(3,1) and a(1,3)

 

[Mapes]

Well, doesn't the shear stress act in the 1-3 direction at points d and e in your diagram? These points correspond to $\phi=\pi/2$ and $3\pi/2$, which would correspond to $\sin\phi=1$ or $-1$.