Maximum shear stress in Mohr's circle

[fonseh]

Homework Statement

I am not sure how to get the maximum shear force in mohr's circle [/B]

Homework Equations

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 ?

 

[Chestermiller]

Homework Statement

I am not sure how to get the maximum shear force in mohr's circle [/B]

Homework Equations

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 ?

What figure?

 

[fonseh]

What figure?

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

 

[Chestermiller]

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.

 

[fonseh]

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.

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

 

[fonseh]

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.

Can you explainfurther ?

 

[Chestermiller]

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

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

 

[fonseh]

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

Can you verify my concept in post #1 ?

 

[Chestermiller]

Can you verify my concept in post #1 ?

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.

 

[fonseh]

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.

https://en.wikipedia.org/wiki/Mohr's_circle#/media/File:Mohr_circle_sign_convetion.svg

I think i use the sign convention in the 3rd case

 

[fonseh]

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.

https://en.wikipedia.org/wiki/Mohr's_circle#/media/File:Mohr_circle_sign_convetion.svg
I don't really understand the explanation 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
?

 

[Chestermiller]

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.

 

[Chestermiller]

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.

 

[fonseh]

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.

I still don't understand how you relate to the question that i asked ?

 

[Chestermiller]

I still don't understand how you relate to the question that i asked ?

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

 

[fonseh]

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

Huh ? I am still don't understand what you are trying to say

 

[Chestermiller]

Huh ? I am still don't understand what you are trying to say

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

 

[fonseh]

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

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 .

 

[fonseh]

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

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

 

[Chestermiller]

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 $$\sigma_{t,max}=\tau_{max}$$

 

[Chestermiller]

According to the Wikipedia article, the $\tau_{xy}$ in your figure should be -25, not + 25. This is the sign I would have assigned to it also.

 

[fonseh]

According to the Wikipedia article, the $\tau_{xy}$ in your figure should be -25, not + 25. This is the sign I would have assigned to it also.

so , the author 's coordinate is correct , right ? He lets's A = (-80, 25) because at the surface where the -80 normal stress act perpendicularly , the shear stress 25 will turn the plane clockwise , so , according to the wiki (no3) , when the shear stress turn the palne clockwise , it's positive shear stress , right ?

 

[Chestermiller]

so , the author 's coordinate is correct , right ? He lets's A = (-80, 25) because at the surface where the -80 normal stress act perpendicularly , the shear stress 25 will turn the plane clockwise , so , according to the wiki (no3) , when the shear stress turn the palne clockwise , it's positive shear stress , right ?

What kind of crazy sign convention is this. By the usual convention, $\tau_{xy}$ is defined as the shear stress on a plane of constant x (for which the outwardly directed normal to the plane is pointing in the + x direction) in the + y direction. According to the figure, that would make $\tau_{xy}$ equal to -25 on surface A.

So, according to what you're saying, in Figs 5 and 6 in the Wikipedia article you cited, since $\tau_{xy}$ is pointing upward on the right hand face of the block in the figure, its value is actually negative rather than positive (since it is pointing in the opposite direction of the shear stress shown in your homework problem).

 

[fonseh]

What kind of crazy sign convention is this. By the usual convention, $\tau_{xy}$ is defined as the shear stress on a plane of constant x (for which the outwardly directed normal to the plane is pointing in the + x direction) in the + y direction. According to the figure, that would make $\tau_{xy}$ equal to -25 on surface A.

So, according to what you're saying, in Figs 5 and 6 in the Wikipedia article you cited, since $\tau_{xy}$ is pointing upward on the right hand face of the block in the figure, its value is actually negative rather than positive (since it is pointing in the opposite direction of the shear stress shown in your homework problem).

yes , in that figure , we could see that the author put A at (-80, -25) in the mohr's circle although he stated wrongly that A = (-80,25 0 in the first figure . so , what are you trying to say ?

 

[Chestermiller]

yes , in that figure , we could see that the author put A at (-80, -25) in the mohr's circle although he stated wrongly that A = (-80,25 0 in the first figure . so , what are you trying to say ?

Which version of the problem do you want to solve?

 

[fonseh]

Which version of the problem do you want to solve?

My question is in post #1 ,
To get the θsmax" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.08px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; font-family: "PT Sans", san-serif; position: relative;">θsmaxθsmax , we have to 'turn' the B to τmax" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.08px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; font-family: "PT Sans", san-serif; position: relative;">τmaxτmax , 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 τmin" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.08px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; font-family: "PT Sans", san-serif; position: relative;">τminτmin ? It 's not possible to get the maximum shear stress like the author do at point A , right ?

 

[fonseh]

Which version of the problem do you want to solve?

Anyway , can you help me in this thread first ?
After solving this thread , then we will get back to this thread .
This thread is more about the basic things
https://www.physicsforums.com/forums/engineering-and-computer-science-homework.158/create-thread

 

[Chestermiller]

What I am going to do is solve this problem using the method that I am familiar with and that I know works. I will solve it for our two cases, one with $\tau_{xy}=+25$ and the other with $\tau_{xy}=-25$.

Case of $\tau_{xy}=+25$
First calculate $\tan{2\theta_p}$:
$$\tan{2\theta_p}=\frac{2\tau_{xy}}{(\sigma_x-\sigma_y)}=\frac{(2)(25)}{(-80-50}=-0.38462$$So, $$2\theta_p=-21.04\ degrees$$

Next calculate the average of $\sigma_x$ and $\sigma_y$:

$$\frac{\sigma_x+\sigma_y}{2}=\frac{-80+50}{2}=-15$$

Next calculate radius of Mohr's circle:$$R=\tau_{max}=\sqrt{\left(\frac{(\sigma_x-\sigma_y)}{2}\right)^2+\tau_{xy}^2}=\sqrt{(-65)^2+(25)^2}=69.6$$

Next, use the following equations to calculate the normal and shear stress components on a plane normal to a unit vector pointing in the direction of a the angle $\theta$ measured counterclockwise relative to the x axis:
$$\sigma_n=\left(\frac{\sigma_x+\sigma_y}{2}\right)+\frac{(\sigma_x-\sigma_y)}{2}\cos{2\theta}+\tau_{xy}\sin(2\theta)$$
$$\sigma_t=-\frac{(\sigma_x-\sigma_y)}{2}\sin{2\theta}+\tau_{xy}\cos(2\theta)$$

The minimum principal stress occurs at $2\theta=2\theta_p=$ = -21.04 degrees. The value is -84.6.

The maximum principal stress occurs at $2\theta=2\theta_p+180$ =158.96 degrees. The value is +54.63.

The following is a table of values for $2\theta$ (degrees), $\sigma_n$, and $\sigma_t$:

The sequence of values is in the clockwise direction.

I changed my mind about doing tau=-25. This has gotten too time-consuming. I leave that case for you.

 

[fonseh]

What I am going to do is solve this problem using the method that I am familiar with and that I know works. I will solve it for our two cases, one with $\tau_{xy}=+25$ and the other with $\tau_{xy}=-25$.

Case of $\tau_{xy}=+25$
First calculate $\tan{2\theta_p}$:
$$\tan{2\theta_p}=\frac{2\tau_{xy}}{(\sigma_x-\sigma_y)}=\frac{(2)(25)}{(-80-50}=-0.38462$$So, $$2\theta_p=-21.04\ degrees$$

Next calculate the average of $\sigma_x$ and $\sigma_y$:

$$\frac{\sigma_x+\sigma_y}{2}=\frac{-80+50}{2}=-15$$

Next calculate radius of Mohr's circle:$$R=\tau_{max}=\sqrt{\left(\frac{(\sigma_x-\sigma_y)}{2}\right)^2+\tau_{xy}^2}=\sqrt{(-65)^2+(25)^2}=69.6$$

Next, use the following equations to calculate the normal and shear stress components on a plane normal to a unit vector pointing in the direction of a the angle $\theta$ measured counterclockwise relative to the x axis:
$$\sigma_n=\left(\frac{\sigma_x+\sigma_y}{2}\right)+\frac{(\sigma_x-\sigma_y)}{2}\cos{2\theta}+\tau_{xy}\sin(2\theta)$$
$$\sigma_t=-\frac{(\sigma_x-\sigma_y)}{2}\sin{2\theta}+\tau_{xy}\cos(2\theta)$$

The minimum principal stress occurs at $2\theta=2\theta_p=$ = -21.04 degrees. The value is -84.6.

The maximum principal stress occurs at $2\theta=2\theta_p+180$ =158.96 degrees. The value is +54.63.

The following is a table of values for $2\theta$ (degrees), $\sigma_n$, and $\sigma_t$:

View attachment 110530
The sequence of values is in the clockwise direction.

I changed my mind about doing tau=-25. This has gotten too time-consuming. I leave that case for you.

What i am concerned about is should i use $$\tau$$ = -25Mpa or not...

Nvm , can you check out my new thread first , perhaps by solving the question in the new thread first , then i can understand this question better

 

[Chestermiller]

What i am concerned about is should i use $$\tau$$ = -25Mpa or not...

Nvm , can you check out my new thread first , perhaps by solving the question in the new thread first , then i can understand this question better

Well, there's something wrong with the solution in the pdf file you presented. The diagram clearly shows that tauxy is negative, and yet they treat it as positive. So there's now way I can answer this question.

I have some other stuff to do today, so I'll get back to that other thread you cited tomorrow.

 

[fonseh]

Well, there's something wrong with the solution in the pdf file you presented. The diagram clearly shows that tauxy is negative, and yet they treat it as positive. So there's now way I can answer this question.

I have some other stuff to do today, so I'll get back to that other thread you cited tomorrow.

Hope you can help me with the other thread . My lecturer asked me to do self -study . He is on leave . I have been thinking of this for the whole week . I have read a lots of online notes and also books, but stlil don't understand

 

[Chestermiller]

Hope you can help me with the other thread . My lecturer asked me to do self -study . He is on leave . I have been thinking of this for the whole week . I have read a lots of online notes and also books, but stlil don't understand

Are you familiar with matrix operations, such as multiplying a matrix by a column vector?

 

[fonseh]

What I am going to do is solve this problem using the method that I am familiar with and that I know works. I will solve it for our two cases, one with $\tau_{xy}=+25$ and the other with $\tau_{xy}=-25$.

Case of $\tau_{xy}=+25$
First calculate $\tan{2\theta_p}$:
$$\tan{2\theta_p}=\frac{2\tau_{xy}}{(\sigma_x-\sigma_y)}=\frac{(2)(25)}{(-80-50}=-0.38462$$So, $$2\theta_p=-21.04\ degrees$$

Next calculate the average of $\sigma_x$ and $\sigma_y$:

$$\frac{\sigma_x+\sigma_y}{2}=\frac{-80+50}{2}=-15$$

Next calculate radius of Mohr's circle:$$R=\tau_{max}=\sqrt{\left(\frac{(\sigma_x-\sigma_y)}{2}\right)^2+\tau_{xy}^2}=\sqrt{(-65)^2+(25)^2}=69.6$$

Next, use the following equations to calculate the normal and shear stress components on a plane normal to a unit vector pointing in the direction of a the angle $\theta$ measured counterclockwise relative to the x axis:
$$\sigma_n=\left(\frac{\sigma_x+\sigma_y}{2}\right)+\frac{(\sigma_x-\sigma_y)}{2}\cos{2\theta}+\tau_{xy}\sin(2\theta)$$
$$\sigma_t=-\frac{(\sigma_x-\sigma_y)}{2}\sin{2\theta}+\tau_{xy}\cos(2\theta)$$

The minimum principal stress occurs at $2\theta=2\theta_p=$ = -21.04 degrees. The value is -84.6.

The maximum principal stress occurs at $2\theta=2\theta_p+180$ =158.96 degrees. The value is +54.63.

The following is a table of values for $2\theta$ (degrees), $\sigma_n$, and $\sigma_t$:

View attachment 110530
The sequence of values is in the clockwise direction.

I changed my mind about doing tau=-25. This has gotten too time-consuming. I leave that case for you.

Are you familiar with matrix operations, such as multiplying a matrix by a column vector?

No , i learned it , but i didnt master it well. so , i am having difficuties of understanding it

 

[fonseh]

Are you familiar with matrix operations, such as multiplying a matrix by a column vector?

Is there any other way than the matrix method to check my sign convention ??

 

[Chestermiller]

Is there any other way than the matrix method to check my sign convention ??

I'm trying to think of another way, but have not been successful yet? Are you saying that you could not multiply a 2x2 matrix by a 2X1 column vector? By any chance, are you familiar with dyadic tensor notation?

Not having the most appropriate mathematical tools to analyze this stuff makes it much more difficult to learn. It's like trying to do mechanics without knowing calculus.