# Homework Help: Angle of principal stress vs maximum shear stress

1. Dec 15, 2016

### fonseh

1. The problem statement, all variables and given/known data
I'm having problem with the angle of $$theta_s1$$ . My ans is +35 (as in my working) , but the ans provided is $$theta_s1$$ = -55

2. Relevant equations

3. The attempt at a solution
As in the picture posted

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2. Dec 18, 2016

### fonseh

Update :
When i use the
$$\tan{2\theta_p}=\frac{2\tau_{xy}}{(\sigma_x-\sigma_y)}=-\frac{(2)(-30)}{(100+70)}=$$

,$$\theta = 35degree$$

Instead of -55 degree . But , when i plug in -55 degree , i gt the $$\tau_(max)$$ = 90MPa , when i use 35 degree , i gt -90MPa , how can this occur ?

In the digram , it's clear that at (100,30) , we only need to rotate the (100,30) to upper vertical axis to get the max shear stress , so how could the angle 2 theta > 90 degree ?

3. Dec 19, 2016

### Staff: Mentor

The equations for the normal stress and the shear stress are:

$$\sigma_n=\left(\frac{\sigma_x+\sigma_y}{2}\right)+\frac{(\sigma_x-\sigma_y)}{2}\cos{2\theta}+\tau_{xy}\sin(2\theta)$$
$$\sigma_t=-\frac{(\sigma_x-\sigma_y)}{2}\sin{2\theta}+\tau_{xy}\cos(2\theta)$$
From the figure, the Cartesian coordinate system stresses are:
$\sigma_x=$+100 MPa
$\sigma_y=$-70 MPa
$\tau_{xy}=-30$ MPa
The principal stresses occur at the angles where $\tau_{xy}=0$. So,
$$\tan{2\theta_p}=\frac{2\tau_{xy}}{(\sigma_x-\sigma_y)}=\frac{-60}{170}=-0.353$$
So, $$2\theta_p=-19.44\ degrees$$ and $$\theta_p=-9.72\ degrees$$
Using the equation above, the norml stress at this angle is 105.14 MPa.
The angle for the other principal stress is $$2\theta_p=-19.44+180=160.56\ degrees$$ so $$\theta_p=80.28\ degrees$$
Using the equation above, the normal stress at this angle is -75.14 MPa.

The shear stress is maximum when $$2\tan{\theta_s}=-\frac{(\sigma_x-\sigma_y)}{2\tau{xy}}=\frac{170}{60}=2.833$$
So, $$2\theta_s=70.56\ degrees$$ and $$\theta_p=35.3\ degrees$$
Using the equation above, the shear stress at this angle is -90.13 MPa.
The unit vector in the direction of this shear stress is $\vec{i}_s=-\sin{\theta_p}\vec{i}+\cos{\theta_p}\vec{j}=-0.578\vec{i}+0.816\vec{j}$
So the vectorial maximum shear stress component of the stress vector is given by:
$$\vec{\sigma}_{t,max}=+90.13(0.578\vec{i}-0.816\vec{j})\ MPa$$
The direction of this shear stress component is consistent with the direction of the arrows in the figure accompanying the problem statement.

Another angle at which the maximum shear stress occurs is at $35.3-90=-54.7$ degrees. Using the equation above, the shear stress value at this angle is +90.13 MPa. The unit vector in the direction of this shear stress is $\vec{i}_s=-\sin{\theta_p}\vec{i}+\cos{\theta_p}\vec{j}=0.816\vec{i}+0.578\vec{j}$.
So the vectorial maximum shear stress component of the stress vector is given by:
$$\vec{\sigma}_{t,max}=+90.13(0.816\vec{i}+0.578\vec{j})\ MPa$$
The direction of this shear stress component is also consistent with the direction of the arrows in the figure accompanying the problem statement.

4. Dec 19, 2016

### Staff: Mentor

Perhaps this will help. For the state of stress shown in the present problem (assumed to also be a homogeneous state of stress), here is a sketch of the variation of the actual shear stress vectors acting on a cylinder of the material (i.e., exerted by the material outside the cylinder on the surface of the cylinder). The small arrows correspond to a shear stress of magnitude 30 MPa. The large arrows correspond to a shear stress magnitude of 90 MPa. The dots correspond to a shear stress of magnitude 0 MPa, and are situated at the directions corresponding to the principal stresses.

The large arrow in the lower right quadrant corresponds to the maximum shear stress vector at -55 degrees (i.e., it is tangent to the cylinder at -55 degrees). The large arrow in the upper right quadrant corresponds to the maximum shear stress vector at +35 degrees (i.e., it is tangent to the cylinder at +35 degrees). Note that the shear stress vector is varying in magnitude and direction over the surface of the cylinder.

Please study the figure carefully and see what you think. This is what the real shear stress distribution looks like, not the confusing picture provided by the Mohr circle.

5. Dec 19, 2016

### fonseh

This should be $$\theta_p =35.3\ degrees$$ ??

So , the maxiumum shear stress occur at $$\theta_p =-55\ degrees$$ ??

6. Dec 19, 2016

### fonseh

Can you plot the point of normal stress and shear stress on mohr's circle ? I want to compare ..
I have done in post #1 , but what i get from the mohr's circle is when i rotate the element 35 clockwise , i will get positive max shear stress of 90MPa , instead by calculation , the max positive shear stress at the angle of -55 degree.
So , what's wrogn with the mohr's circle i sketched ?

7. Dec 19, 2016

### fonseh

In this diagram , you assume the positive max shear stress is downwards ?

8. Dec 19, 2016

### Staff: Mentor

Both are correct. See my figure of the actual stress distribution.

Last edited: Dec 19, 2016
9. Dec 19, 2016

### Staff: Mentor

I don't know what that means.

10. Dec 19, 2016

### fonseh

But , when i plot in the mohr's circle , i found that by rotation of angle of +35 degree , i will get the positive max shear stress , not -55 degree , so , is my mohr's circle wrong ?

11. Dec 19, 2016

### fonseh

Can you draw the angle in the diagram where is 35 degree and where is -55 degree , it's hard to imagine it . And it may not be correct when i imagine it

12. Dec 19, 2016

### Staff: Mentor

Here is my Mohr's circle.

The numbers in red are the values of $2\theta$. They run from -180 degrees to +180 degrees.

13. Dec 19, 2016

### Staff: Mentor

The figure clearly shows that $\tau_{xy}=-30 MPa$, not +30 MPa. Redo the problem with that value and see what you get.

14. Dec 19, 2016

### fonseh

well , thanks for the diagram , can you explain why for the 100 , you plot the shear stress as negative 30 ? why for the 70 , the shear stress is positive 30 ?

15. Dec 19, 2016

### Staff: Mentor

See my post #13. For 70, the shear stress is $-30\vec{i}=+30(\vec{-i})$ (see my figure). The polar angle $\vec{i}_{\theta}$ is always oriented counterclockwise. The vectorial shear stress component in this framework is represented by $\vec{\tau}=\sigma_t \vec{i}_{\theta}$.

Last edited: Dec 19, 2016
16. Dec 19, 2016

### fonseh

do you mean if the shear stress turn the element counter clockwise , then the shear stress is positive , if the shear stress turn the element clockwise , then the shear stress is negative ? Like at 100MPa normal stress , the shear stress at that part 0f 30MPa will turn the element clockwise , so the shear stress is negative ?

17. Dec 19, 2016

### Staff: Mentor

No. What I'm saying is that, when we apply this specific framework, we are essentially expressing the stress vector in terms of its components in cylindrical coordinates. The normal stress is essentially $\sigma_r$ and the shear stress is essentially $\sigma_{\theta}$. At zero degrees, the unit vector in the theta direction points in the y direction. At 90 degrees, the unit vector in the theta direction points in the negative x direction. At 180 degrees, the unit vector in the theta direction points in the negative y direction. At 270 degrees, the unit vector in the theta direction points in the positive x direction. So the unit vector in the theta direction is always pointing the the counterclockwise direction, and the sign of the shear stress value adjusts to this. So,

$$\vec{\sigma}=\sigma_n\vec{i}_n+\sigma_t\vec{i}_t=\sigma_r\vec{i}_r+\sigma_{\theta}\vec{i}_{\theta}$$
with $$\vec{i}_n=\vec{i}_r$$ $$\vec{i}_t=\vec{i}_{\theta}$$ $$\sigma_n=\sigma_r$$ and $$\sigma_t=\sigma_{\theta}$$

18. Dec 19, 2016

### fonseh

why the shear stress is $-30\vec{i}=+30(\vec{-i})$??

So , the shear stress at that point where normal stress = 70 , so the shear stress is $-30\vec{i}=+30(-1)=30$ ???

19. Dec 19, 2016

### Staff: Mentor

Sure. Look at the figure in the problem statement.

20. Dec 19, 2016

### fonseh

From where , you know that the shear stress at normal stress = 70 MPa is -30i ?

Last edited: Dec 19, 2016