Angle of principal stress vs maximum shear stress

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The discussion revolves around the calculation of the angle of principal stress and maximum shear stress, where the user initially calculated the angle as +35 degrees but was provided with -55 degrees as the answer. The calculations involve using the equations for normal and shear stress, leading to confusion regarding the angles and resulting shear stress values. The user notes that both angles yield maximum shear stress values of 90 MPa, but with different signs, prompting questions about the accuracy of their Mohr's circle representation. The thread emphasizes the importance of correctly interpreting stress distributions and angles in relation to the physical representation of shear stress vectors. Understanding these concepts is crucial for resolving discrepancies in stress analysis.
  • #31
fonseh said:
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?
 
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  • #32
fonseh said:
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."
 
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  • #33
Chestermiller said:
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 ?
 
  • #34
fonseh said:
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?
 
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  • #35
Chestermiller said:
What direction does ##\vec{i}_\theta## point at ##\theta=90## degrees?
Postive y-axis (upwards) am I right?
 
  • #36
fonseh said:
Postive y-axis (upwards) am I right?
No. ##i_{\theta}## is tangent to the circumference. So, again, what direction is it pointing?
 
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  • #37
Chestermiller said:
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 ?
 

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  • #38
fonseh said:
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.
 
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  • #39
Chestermiller said:
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 ?
 
  • #40
fonseh said:
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.
 
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  • #41
Chestermiller said:
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
 
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  • #42
fonseh said:
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.
 
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  • #43
Chestermiller said:
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 ?
 
  • #44
fonseh said:
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.
 
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  • #45
Chestermiller said:
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 ?
 
  • #46
Chestermiller said:
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
 
  • #47
fonseh said:
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.
 
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  • #48
Chestermiller said:
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
 

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