Angle of principal stress vs maximum shear stress

[fonseh]

Homework Statement

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[/B]

Homework Equations

The Attempt at a Solution

As in the picture posted

 

[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 ?

 

[Chestermiller]

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.

 

[Chestermiller]

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.

 

[fonseh]

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.

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

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

 

[fonseh]

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.
View attachment 110575
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.

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 ?

 

[fonseh]

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.
View attachment 110575
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.

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

 

[Chestermiller]

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

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

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

 

[Chestermiller]

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

I don't know what that means.

 

[fonseh]

Both are correct.

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 ?

 

[fonseh]

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.
View attachment 110575
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.

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

 

[Chestermiller]

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 ?

Here is my Mohr's circle.

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

 

[Chestermiller]

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 ?

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

 

[fonseh]

Here is my Mohr's circle.
View attachment 110580
The numbers in red are the values of $2\theta$. They run from -180 degrees to +180 degrees.

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 ?

 

[Chestermiller]

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 ?

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 $\vec{\tau}=\sigma_t \vec{i}_{\theta}$.

 

[fonseh]

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 $\vec{\tau}=\sigma_t \vec{i}_{\theta}$.

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 ?

 

[Chestermiller]

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 ?

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 $$\sigma_t=\sigma_{\theta}$$

 

[fonseh]

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 $\vec{\tau}=\sigma_t \vec{i}_{\theta}$.

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 $-30\vec{i}=+30(-1)=30$ ?

 

[Chestermiller]

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 $-30\vec{i}=+30(-1)=30$ ?

Sure. Look at the figure in the problem statement.

 

[fonseh]

Sure. Look at the figure in the problem statement.

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

 

[fonseh]

Sure. Look at the figure in the problem statement.

The problem statement in the photo in post#1 didnt stated -30i , where does it come from ?

 

[Chestermiller]

The problem statement in the photo in post#1 didnt stated -30i , where does it come from ?

Look at the direction of the arrow.

 

[fonseh]

Look at the direction of the arrow.

ya , the shear stress of arrow of normal stress of 70MPa is to the left , so the sher stress should be -30 , right ? why is (70 , 30 ) in the mohr's cicle that you sketched ?

 

[Chestermiller]

ya , the shear stress of arrow of normal stress of 70MPa is to the left , so the sher stress should be -30 , right ? why is (70 , 30 ) in the mohr's cicle that you sketched ?

It's $-30\vec{i}$ which is the same as $+30\vec{i}_{\theta}$, so $\sigma_{\theta}=+30$.

 

[fonseh]

It's $-30\vec{i}$ which is the same as $+30\vec{i}_{\theta}$, so $\sigma_{theta}=+30$.

why $-30\vec{i}$ which is the same as $+30\vec{i}_{\theta}$ ??

 

[Chestermiller]

why $-30\vec{i}$ which is the same as $+30\vec{i}_{\theta}$ ??

Because at $\theta = 90$ degrees, $\vec{i}_\theta=-\vec{i}$.

 

[fonseh]

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 $$\sigma_t=\sigma_{\theta}$$

so , do you mean at here you defined counterclockwise as positive angle ? So , the sign convention of the rotation of the element follows the sign convention of teh angle rotation in the mohr's circle ?
At the point where normal stress = 70MPa , the shear stress rotate the element anticlockwise , so in the mohr's circle , the point is (70,30) ?

At the point where normal stress = 100MPa , the shear stress at the point will rotate the element clockwise , so the shear stress at the particular point is negative ?

 

[Chestermiller]

so , do you mean at here you defined counterclockwise as positive angle ? So , the sign convention of the rotation of the element follows the sign convention of teh angle rotation in the mohr's circle ?
At the point where normal stress = 70MPa , the shear stress rotate the element anticlockwise , so in the mohr's circle , the point is (70,30) ?

At the point where normal stress = 100MPa , the shear stress at the point will rotate the element clockwise , so the shear stress at the particular point is negative ?

I don't understand this rotation business, and I don't intend to try. I already invested a ton of effort on this with you, and I would like to disengage now. Hope you can accept this.

Chet

 

[fonseh]

I don't understand this rotation business, and I don't intend to try. I already invested a ton of effort on this with you, and I would like to disengage now. Hope you can accept this.

Chet

it's ok if you don't want to deal with the rotation business ,
can you explain why at $\theta = 90$ degrees, $\vec{i}_\theta=-\vec{i}$ ??

 

[fonseh]

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 $$\sigma_t=\sigma_{\theta}$$

So , according to you , in the mohr's circle , you defined counterclockwise as positive , and the sign convention of the shear stress / force follow this also ?
So , for the point where the 100MPa tensile normal stress act , the shear stress will turn the element clockwise , so the shear stress has negative value ?

 

[Chestermiller]

it's ok if you don't want to deal with the rotation business ,
can you explain why at $\theta = 90$ degrees, $\vec{i}_\theta=-\vec{i}$ ??

Are you familiar with cylindrical polar coordinates? If so, are you familiar with the unit vectors in the radial and circumferential directions? Our focus is on the unit vector in the circumferential direction. Please describe how this circumferential unit vector varies (i.e., the direction in which it points) as a function of the circumferential coordinate, which is measured counterclockwise relative to the positive x axis. What direction does the unit vector in the circumferential direction point when the circumferential coordinate is 90 degrees?

 

[Chestermiller]

So , according to you , in the mohr's circle , you defined counterclockwise as positive , and the sign convention of the shear stress / force follow this also ?
So , for the point where the 100MPa tensile normal stress act , the shear stress will turn the element clockwise , so the shear stress has negative value ?

I told you I am no longer going to discuss the rotation business. You are using the word "turn" in place of the word "rotate." It's just, as the american idiom describes, a "wolf in sheep's clothing."

 

[fonseh]

Are you familiar with cylindrical polar coordinates? If so, are you familiar with the unit vectors in the radial and circumferential directions? Our focus is on the unit vector in the circumferential direction. Please describe how this circumferential unit vector varies (i.e., the direction in which it points) as a function of the circumferential coordinate, which is measured counterclockwise relative to the positive x axis. What direction does the unit vector in the circumferential direction point when the circumferential coordinate is 90 degrees?

Ok , i understand the cylindrical polar coordinate . But i still don't understand why you said at $\theta = 90$ degrees, $\vec{i}_\theta=-\vec{i}$ ?
In polar coordinate , the angle 0 start from positive x( horizontal axis) and rotate couterclockwise , right ?

 

[Chestermiller]

Ok , i understand the cylindrical polar coordinate . But i still don't understand why you said at $\theta = 90$ degrees, $\vec{i}_\theta=-\vec{i}$ ?
In polar coordinate , the angle 0 start from positive x( horizontal axis) and rotate couterclockwise , right ?

What direction does $\vec{i}_\theta$ point at $\theta=90$ degrees?

 

[fonseh]

What direction does $\vec{i}_\theta$ point at $\theta=90$ degrees?

Postive y-axis (upwards) am I right?

 

[Chestermiller]

Postive y-axis (upwards) am I right?

No. $i_{\theta}$ is tangent to the circumference. So, again, what direction is it pointing?

 

[fonseh]

No. $i_{\theta}$ is tangent to the circumference. So, again, what direction is it pointing?

As the angle is rotated anticlockwise , so the tangent at here pointing to the left ? Pls refer to the photo below ?

 

[Chestermiller]

As the angle is rotated anticlockwise , so the tangent at here pointing to the left ? Pls refer to the photo below ?

Yes. It is pointing in the negative x direction. It is thus equal to $-\vec{i}$, where $\vec{i}$ is the cartesian unit vector in the (positive) x direction.

 

[fonseh]

Yes. It is pointing in the negative x direction. It is thus equal to $-\vec{i}$, where $\vec{i}$ is the cartesian unit vector in the (positive) x direction.

ok , i don't understand why you are considering the vector of the tangent of the line ?

 

[Chestermiller]

ok , i don't understand why you are considering the vector of the tangent of the line ?

Because that is the direction of the shear stress on a plane tangent to the cylinder.

 

[fonseh]

Because that is the direction of the shear stress on a plane tangent to the cylinder.

Can you explain further ? I still didnt get that

 

[Chestermiller]

Can you explain further ?

We are trying to determine the normal stress and shear stress on planes of various orientations within the material. The tangent planes to a cylinder whose axis is pointing in the z direction include all possible orientations of such planes.

 

[fonseh]

We are trying to determine the normal stress and shear stress on planes of various orientations within the material. The tangent planes to a cylinder whose axis is pointing in the z direction include all possible orientations of such planes.

We are dealing with stress in 2 dimension only , right ? which are y and x axes , right ?

 

[Chestermiller]

We are dealing with stress in 2 dimension only , right ? which are y and x axes , right ?

Right. The planes tangent to the cylinder I indicated include all the planes of interest in our x-y framework. The unit vectors normal and tangent to the planes I indicated all lie within the x-y plane (a plane of constant z). Not only do these tangent planes to the cylinder include all possible orientations, but they are automatically ordered in sequence.

 

[fonseh]

We are trying to determine the normal stress and shear stress on planes of various orientations within the material. The tangent planes to a cylinder whose axis is pointing in the z direction include all possible orientations of such planes.

why are they many orientations ? there are only positive / negative x and y axes , right ?

 

[fonseh]

Right. The planes tangent to the cylinder I indicated include all the planes of interest in our x-y framework. The unit vectors normal and tangent to the planes I indicated all lie within the x-y plane (a plane of constant z). Not only do these tangent planes to the cylinder include all possible orientations, but they are automatically ordered in sequence.

why there is cylinder ? in the original question in post # 1, it's a cube , right ? or something which has square / rectangular base

 

[Chestermiller]

why there is cylinder ? in the original question in post # 1, it's a cube , right ? or something which has square / rectangular base

I doesn't have to be square or rectangular. You do realize that what we are trying to do here is determine the normal stress and the shear stress on a plane of arbitrarily specified spatial orientation, given the normal stress and shear stresses on two specific planes, one oriented perpendicular to the x-axis (normal- and shear stresses $\sigma_x$ and $\tau_{xy}$) and the other perpendicular to the y-axis (normal- and shear stresses $\sigma_y$ and $\tau_{xy}$), correct? In the problem statement, part iii, they specify a particular plane and then ask you to determine the normal- and shear stresses on this plane. That is the whole objective of what we are trying to do here. So, don't get hung up on cubes and rectangles.

The reason we are working with a cylinder is that the infinite set of planes tangent to the cylinder include all possible orientations of such planes, ordered in sequence as a function of $\theta$. The Mohr's circle presents the combination of normal stress and shear stress for the entire set of possible planes.

 

[fonseh]

I doesn't have to be square or rectangular. You do realize that what we are trying to do here is determine the normal stress and the shear stress on a plane of arbitrarily specified spatial orientation, given the normal stress and shear stresses on two specific planes, one oriented perpendicular to the x-axis (normal- and shear stresses $\sigma_x$ and $\tau_{xy}$) and the other perpendicular to the y-axis (normal- and shear stresses $\sigma_y$ and $\tau_{xy}$), correct? In the problem statement, part iii, they specify a particular plane and then ask you to determine the normal- and shear stresses on this plane. That is the whole objective of what we are trying to do here. So, don't get hung up on cubes and rectangles.

The reason we are working with a cylinder is that the infinite set of planes tangent to the cylinder include all possible orientations of such planes, ordered in sequence as a function of $\theta$. The Mohr's circle presents the combination of normal stress and shear stress for the entire set of possible planes.

ok , i understand it . Although it's not quite related to mechanics of materials . I was hoping to verify the sign convention of the stresses in mohr's circle .Although now i am still not sure why some books plot the postive shear stress axis downwards / upwards , and the shear stress that turn the element clockwise / anticlockwise is positive