# Maximum shear stress in Mohr's circle

1. Dec 15, 2016

### fonseh

1. The problem statement, all variables and given/known data
I am not sure how to get the maximum shear force in mohr's circle

2. Relevant equations

3. The attempt at a solution
To get the $\theta_s max$ , we have to 'turn' the B to $\tau_{max}$ , right ? So , it should be like this ( in the figure , orange arrow) ? Am i right ? Since at A , the shear stress is negative , we can only get the minimum shear stress by 'rotating ' the A to $\tau_{min}$ ? It 's not possible to get the maximum shear stress like the author do at point A , right ?

Last edited by a moderator: Dec 15, 2016
2. Dec 15, 2016

### Staff: Mentor

What figure?

Last edited by a moderator: Dec 15, 2016
3. Dec 15, 2016

### fonseh

sorry , due to slow internet connection , it's here

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4. Dec 15, 2016

### Staff: Mentor

I have trouble relating to this development. I can relate much better to the development in the following Wikipedia article, for which I independently confirmed the equations for the stresses on a plane of arbitrary orientation: https://en.wikipedia.org/wiki/Mohr's_circle
The section entitled Drawing Mohr's Circle should be of particular interest to you.

5. Dec 15, 2016

### fonseh

Which part contains Drawing Mohr's Circle ? Can you point out in the link ?

6. Dec 16, 2016

### fonseh

Can you explainfurther ?

7. Dec 16, 2016

### Staff: Mentor

The entire section, but particularly the inset with the equations for the normal- and shear stresses, and the diagram.

8. Dec 16, 2016

### fonseh

Can you verify my concept in post #1 ?

9. Dec 16, 2016

### Staff: Mentor

No. As I said, I can't relate to the development in the figures you presented. If you try solving it using the development in Wikipedia that I referred to, I can comment and verify what you do.

10. Dec 16, 2016

### fonseh

Last edited: Dec 16, 2016
11. Dec 16, 2016

### fonseh

https://en.wikipedia.org/wiki/Mohr's_circle#/media/File:Mohr_circle_sign_convetion.svg
I dont really understand the explaination in circled part .
From the diagram we , we can see that the shear stress that cause the element to turn clockwise is positive , but in the word there , the author stated that the shear stress that cause the element that to turn clockwise is negative .... So , which is correct
?

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12. Dec 16, 2016

### Staff: Mentor

This clockwise and counterclockwise garbage in the figures makes my head spin, and I totally disregard it. I let the mathematics do all the work for me in determining the normal stress and shear stress components on a plane of specified orientation. The mathematics gets there results simply and infallably. Therefore, I won't be answering any questions about the counterclockwise and clockwise diagrams. Here are the simple rules:

1. To get the stress vector on a plane of specified orientation, dot the stress tensor with a unit normal to the plane.
2. To get the scalar normal stress component, dot the stress vector with the unit normal to the plane.
3. To get the vector normal stress component, multiply the scalar normal component by the unit normal to the plane.
4. To get the vector shear stress component, subtract the vector normal stress component from the stress vector.

Mathematically, the unit normal vector to a plane is given by $$\vec{n}=\cos{\theta}\vec{i}+\sin{\theta}\vec{j}$$where $\theta$ is the angle that the unit normal to the plane makes with the x axis.

Last edited: Dec 16, 2016
13. Dec 17, 2016

### Staff: Mentor

If I apply step 1 to the 2D stress pattern you are considering, I obtain for the stress vector on a plane oriented perpendicular to the unit vector of the previous post the following:
$$\vec{\sigma}=(\sigma_x\cos{\theta}+\tau_{xy}\sin{\theta})\vec{i}+(\tau_{xy}\cos{\theta}+\sigma_{y}\sin{\theta})\vec{j}$$
If I dot this stress vector with the unit normal to the plane $\vec{n}$ of the previous post, I obtain:
$$\sigma_n=\vec{\sigma}\centerdot \vec{n}=\left(\frac{\sigma_x+\sigma_y}{2}\right)+\left(\frac{\sigma_x-\sigma_y}{2}\right)\cos{2\theta}+\tau_{xy}\sin{2\theta}$$
A unit vector tangent to the plane under consideration is given by: $$\vec{t}=-\sin{\theta}\vec{i}+\cos{\theta}\vec{j}$$
If I dot the stress vector with this unit tangent vector to the plane, I obtain the tangential component of the stress on the plane:
$$\sigma_t=\vec{\sigma}\centerdot \vec{t}=-\left(\frac{\sigma_x-\sigma_y}{2}\right)\sin{2\theta}+\tau_{xy}\cos{2\theta}$$
So the stress vector acting on the plane can also be expressed in terms of the normal and tangent stress components on the plane by:
$$\vec{\sigma}=\sigma_n \vec{n}+\sigma_t \vec{t}$$
or equivalently, by
$$\vec{\sigma}=\left[\left(\frac{\sigma_x+\sigma_y}{2}\right)+\left(\frac{\sigma_x-\sigma_y}{2}\right)\cos{2\theta}+\tau_{xy}\sin{2\theta}\right] \vec{n}+\left[-\left(\frac{\sigma_x-\sigma_y}{2}\right)\sin{2\theta}+\tau_{xy}\cos{2\theta}\right]\vec{t}$$

We can check all this by calculating the stress vector on planes of various simple orientations. Thus,

At $\theta=0$, this gives $\vec{n}=\vec{i}$, $\vec{t}=\vec{j}$, and $\vec{\sigma}=\sigma_x\vec{i}+\tau_{xy}\vec{j}$
At $\theta = \pi /2$, this gives $\vec{n}=\vec{j}$, $\vec{t}=-\vec{i}$, and $\vec{\sigma}=\sigma_y\vec{j}+\tau_{xy}\vec{i}$
At $\theta = \pi$, this gives $\vec{n}=-\vec{i}$, $\vec{t}=-\vec{j}$, and $\vec{\sigma}=-\sigma_x\vec{i}-\tau_{xy}\vec{j}$
At $\theta = 3\pi /2$, this gives $\vec{n}=-\vec{j}$, $\vec{t}=\vec{i}$, and $\vec{\sigma}=-\sigma_y\vec{j}-\tau_{xy}\vec{i}$

All these result are all as expected.

14. Dec 17, 2016

### fonseh

I still dont understand how you relate to the question that i asked ?

15. Dec 17, 2016

### Staff: Mentor

What do you get if you use the equations I presented here?

16. Dec 17, 2016

### fonseh

Huh ? I am still dont understand what you are trying to say

17. Dec 17, 2016

### Staff: Mentor

I just want you to compare your results with the results you get using the equations I presented.

18. Dec 17, 2016

### fonseh

https://www.physicsforums.com/attachments/463-jpg.110469/
Do you mean you are using the sign convention in 1 ?

If so , can you modify your working and use sign convention in 3 ..I'm more comfortable with it .

19. Dec 17, 2016

### fonseh

Well , if you refer to post #1 , I just want to check my concept that to get the 'B ' to $$\tau_(max)$$ , right ? Why the author tun 'A' to $$\tau_(max)$$ ?
I'm confused

Actually , the $$\tau_(min)$$ is the $$\tau_(max)$$ for point A , right ? It's just the sign difference only

If so , then the $$\tau_(max)$$ should be = 35 degree clockwise , which is equivalent to -35 degree , right ? At here , we assign anticlockwise as positive

20. Dec 17, 2016

### Staff: Mentor

Using trigonometric identities, my equations for $\sigma_n$ and $\sigma_t$ can be reexpressed as:

$$\sigma_n=\left(\frac{\sigma_x+\sigma_y}{2}\right)+\tau_{max}\cos{[2(\theta-\theta_p)]}$$
$$\sigma_t=-\tau_{max}\sin{[2(\theta-\theta_p)]}$$
where $$\tau_{max}=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$$and $$\tan{2\theta_p}=\frac{2\tau_{xy}}{(\sigma_x-\sigma_y)}$$
From these equations, it follows that the maximum possible magnitude of the normal stress is:
$$\sigma_{n,max}=\left(\frac{\sigma_x+\sigma_y}{2}\right)+\tau_{max}$$
and the maximum possible magnitude of the shear stress is $$\sigma_{t,max}=\tau_{max}$$