Angle of principal stress vs maximum shear stress

[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