Principal Stress: Get Help with Mechanical Systems Module

In summary: The axle is loaded in such a way that on a surface plane aligned circumferentially a direct compressive stress of 100MN/m2 is applied. The shear stress is applied from the driving torque equivalent to -45MN/m2. The compressive stress and shear stress are both complementary. Therefore, the principal stresses are σy = 100MN/m2 and σx = -45MN/m2. The Maximum stress is found to be P = (σx + σy) / 2 = 135MN/m2.
  • #1
Jase harwood
10
1
HI everybody

I am in dire need of help. I have started a mechanical systems module as part of a distance learning foundation degree. The notes seem quite sparse to me and I can't seem to get to grips with it from the start. I have attached a photo of the tutorial questions but the are no workings to help me along can anyone give me some pointers. This is not an assignment it is just practice questions.

Any help would be appreciated.
 

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  • #2
What is the specific difficulty you are encountering?

Chet
 
  • #3
Hi Chet

Rather than me say everything I had better take it a step at a time, so here goes.

Question 1.

when it comes to determining the principal stress I haven't got any real clear guidance in the notes I have (or I just can't see it). can I assume that σy = 100MN/m2 and that σx = -45MN/m2. Then to get the principal stress I use:-

upload_2015-2-15_10-32-40.png


now if I was right with my previous statement then I should be able to use this formula, but the stumbling block comes with "where do I get
upload_2015-2-15_10-34-28.png
from."

Then once I have this I just use the formula with + then -.

Once again thanks
 

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  • #4
It depends on what information you have available. The equation you presented assumes that you know the stress components with respect to the x and y coordinate directions.

If you know all the components of the stress, then you know τxy. So, no problem.

If you don't know all the components of the stress, then you might know all the components of the strain. If you know all the components of the strain, then you can use Hooke's law to get all the components of the stress.

If you don't know all the components of the strain, then you might know the three displacements. If you know the three displacements, then you can use the strain displacement equations to get all the components of the strain.

If you don't know any of these things, then you have to solve a deformational mechanics problem to determine them.

Chet
 
  • #5
Ok all the questions I have seen online all specify the 3 values required but in the question I have been given I only had the 2 values. could you possibly tell me if in the question I am missing some information I require as I can't believe my tutor would make it this difficult at the start of the module. I believe I have the shear stress and the compressive stress can I get Txy from this.
 
  • #6
Jase harwood said:
Ok all the questions I have seen online all specify the 3 values required but in the question I have been given I only had the 2 values. could you possibly tell me if in the question I am missing some information I require as I can't believe my tutor would make it this difficult at the start of the module. I believe I have the shear stress and the compressive stress can I get Txy from this.

The shear stress is Txy.

Chet
 
  • #7
ok I'm starting to see where things fit in. if I may clarify then σy is compressive stress 100MN/m2 and sheer stress is τxy -45MN/m2, but what value is σx. the question states "on a plane perpendicular to the one carrying the compressive stress only complementary shear exists" this I am assuming is σx but have no idea what the value is. once I know this I think I may have cracked it.
 
  • #8
Jase harwood said:
ok I'm starting to see where things fit in. if I may clarify then σy is compressive stress 100MN/m2 and sheer stress is τxy -45MN/m2, but what value is σx. the question states "on a plane perpendicular to the one carrying the compressive stress only complementary shear exists" this I am assuming is σx but have no idea what the value is. once I know this I think I may have cracked it.
I'm having trouble understanding what you are asking. Let's work a specific problem and see how it plays out. But the photo you sent is unreadable(by me), so please type out the question. Thanks.

Chet
 
  • #9
ah sorry I didn't know it was unreadable. here goes.

An axle of a go kart is loaded in such a way that on a surface plane aligned circumferentially a direct compressive stress of 100MN/m2 is applied. In association with the compressive stress is a shear stress applied from the driving torque equivalent to -45MN/m2. On a plane perpendicular to the one carrying the compressive stress only complementary shear exits.

a) Sketch a portion of the axle highlighting the stress system.
b) Determine the principal stresses.
c) Determine the Maximum Stress.
d) Sketch a Mohrs stress circle for the system and compare the estimate of the orientation of the principal planes from the circle with that obtained exactly.

I have the answers which are (17.3, -117.3 & 67.3MN/m2) but I do not know how to obtain them.

Hope this helps explain.
 
  • #10
Ah. OK. I think I understand the problem.

The axle is a cylinder, so we're going to be using a cylindrical polar coordinate system with the axis of the cylinder being the z direction.

Do you use the criterion that compressive stresses are positive, or do you use the criterion that tensile stresses are positive?

In this problem, you are looking at the state of stress at the surface of the cylinder, r = R. According to the problem description, the only non-zero components of stress at r = R are σzz, σθθ, and σ. First you look at a plane perpendicular to the axis of the cylinder (i.e., "circumferentially aligned"), and examine the normal and shear stresses on this plane at r = R. Which of the three stresses that I listed are these? What are the values of these stresses?

To be continued.

Chet
 
  • #11
I may be barking up the wrong tree here but if we are using the z axis then isn't this a 3D analysis. I am currently only doing 2D analysis so do the Compressive stress of 100 MN/m2 and the shear stress of -45MN/m2 only apply to the x and y-axis with 100MN/m2 being on the y axis. The notes i have been given by my tutor seem very inadequate to me and don't lead into the questions very well.
 
  • #12
This is definitely 2D. Not all problems use xy coordinates. Some use θz coordinates. Are you able to visualize these stresses on the surface of the cylinder?

Chet
 
  • #13
I believe I do. Is theta going around the surface with z going through the axle. I had a look online and think this is what I am seeing.
 
  • #14
Jase harwood said:
I believe I do. Is theta going around the surface with z going through the axle. I had a look online and think this is what I am seeing.
Yes. Yes. Yes.

So now what are the values of the zz, θθ, and zθ stresses?

Chet
 
  • #15
This is so frustrating I am only seeing the 2 values in the question so I'm assuming one value is 0. Is zz 100MN/m2 and theta theta -45MN/m2 and z theta 0. I'm pulling my hair out trying to get my head round This.
 
  • #16
Jase harwood said:
This is so frustrating I am only seeing the 2 values in the question so I'm assuming one value is 0. Is zz 100MN/m2 and theta theta -45MN/m2 and z theta 0. I'm pulling my hair out trying to get my head round This.
The zz is -100 (compressive).
The zθ is -45.
The θθ is 0.

Chet
 
  • #17
Damn I thought I had it then. So all I need to do know is find the right formula to input these values into.
 
  • #18
Jase harwood said:
Damn I thought I had it then. So all I need to do know is find the right formula to input these values into.
You should try to figure out how these stresses are related to your original problem description. As far as the equation for the principal stresses are concerned, you already had that in post #3.
 
  • #19
So I have been trying to look at this again today. I have the principal stress formula but as we are not using x and y I take it we are using the zz, theta z & theta theta values. I'm sorry but I just can't see how these values factor into the formula.
 
  • #20
Replace xx by zz, xy by zθ, and yy by θθ.

Chet
 

1. What is the concept of principal stress in mechanical systems?

The concept of principal stress is an important aspect in understanding the behavior and failure of materials in mechanical systems. Principal stress refers to the maximum and minimum stress values experienced by a given point in a material under loading. These stresses are perpendicular to each other and are known as the principal stresses.

2. How is principal stress calculated?

The calculation of principal stress involves finding the eigenvalues and eigenvectors of the stress tensor at a given point in a material. The eigenvalues represent the maximum and minimum principal stresses, while the eigenvectors indicate the direction of these stresses. There are various mathematical methods and software programs available to perform these calculations.

3. How does principal stress affect the strength and failure of materials?

The principal stress values at a given point in a material can determine its strength and susceptibility to failure. If the maximum principal stress exceeds the material's ultimate strength, it can lead to fracture or failure. The orientation of the principal stresses also plays a crucial role, as materials are typically stronger in tension than in compression.

4. Can principal stress be controlled or altered in mechanical systems?

Yes, principal stress can be controlled or altered in mechanical systems through various methods such as stress relieving, heat treatment, or changing the material properties. Design considerations, such as avoiding sharp corners and changes in cross-section, can also help distribute stress and reduce the likelihood of failure.

5. How is principal stress used in the design and analysis of mechanical systems?

Principal stress is an essential factor in the design and analysis of mechanical systems. Engineers use the knowledge of principal stress to determine the appropriate material, shape, and size of components to ensure they can withstand the expected loads and stresses. It also helps identify potential failure points and allows for the optimization of designs to improve overall performance and safety.

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