Simple Mechanics of Materials: Deformation and Torsion

In summary, the homework statement is that a shaft is subjected to a torque and an axial force. The actual shaft specifications (type of material, prismatic / changing cross sections are left up to us and is NOT the point of my question). The Attempt at a Solution is treating this as 1 deformation problem and 1 torsion problem. For the deformation / axial part: RA = P and RB = P also. And for the torsion part: here is where I am very unsure of what to do. I know there will be a reaction torque at A, but will point C have one also? I am basically treating this as 1 deformation problem and 1 torsion problem.
  • #1
Seraph042
29
0

Homework Statement



Shaft AB is fixed at point A and is subjected to a torque T
= 2.5 kN·m applied at point B and an axial force P = 30 kN applied at
point C.

The actual shaft specifications (type of material, prismatic / changing cross sections are left up to us and is NOT the point of my question)




Homework Equations



[tex]\delta[/tex] = PL/AE
P = force, L = length, A = cross-sectional area, E = elastic modulus

[tex]\phi[/tex] = TL/GIp
T = torque, L = length, G = shear modulus, Ip = polar moment of inertia


The Attempt at a Solution



I am basically treating this as 1 deformation problem and 1 torsion problem

For the deformation / axial part:
After cutting the sections, I find that RA = P and RB = P also

And thus,

[tex]\delta[/tex]total = [tex]\delta[/tex]1 + [tex]\delta[/tex]2

Where
[tex]\delta[/tex]1 = PLAB/AABEAB
And
[tex]\delta[/tex]2 = PLBC/ABCEBC

For the torsion part:
Here is where I am very unsure of what to do. I know there will be a reaction torque at A, but will point C have one also?
Right now I've got this -
To + TC + TA = 0
=> To = TA - TC

And

[tex]\phi[/tex]1 = [tex]\phi[/tex]2

Such that

[tex]\phi[/tex]1 = TABLAB/GABIP,AB

And

[tex]\phi[/tex]1 = (TAB-To)LBC/GBCIP,BC

Am I on the right track? Where is my analysis flawed ?
 

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  • #2
1) No, there won't be a reaction torque at point C, because there's no constraint there.

2) I don't agree with To = TA - TC or the idea that there's an additional angular displacement between B and C.
 
  • #3
Ok, I understand that my torsion analysis is all off (I've definitely had problems with this part of the class)

A few things:
- Is my axial deformation analysis correct?
- How should I go about the torsion analysis? We have to find the angle of twist at point C.
We can't assume that there's only the reaction torque at A because otherwise there would be no angular displacement?

I'm pretty confused with the whole torsion part overall
 
  • #4
Your axial calculations look fine. I'd suggest getting something compliant, like an eraser or strip of rubber, and twisting and holding it in different spots to build up your intuition on angle of twist and reaction torque. If you hold the sample at one end and twist it in the middle, what's the angle of twist at the other end? Does the angle keep increasing?
 
  • #5
Yes the angle does keep increasing, but I'm still having difficulty in figuring out how to calculate the angle of twist directly at that end based on the single reaction torque and the external torque... how would you, quantitatively, do this?
 
  • #6
Did you really try it? The angle doesn't keep increasing from B to C (theoretically or experimentally), because there's no torque on this section.

Try plotting the torque as a function of distance along the rod.
 
  • #7
Sorry, that isn't what I meant, I just meant that there will be an angle of twist from B to C due to the applied torque at point B.

What I'm having trouble figuring out though is exactly how to analyze this situation, because (as you said) there is no reaction torque at point C.

Does that mean that the reaction torque at point A is equal to the applied torque? [TA = To]

As far as the torque variation along the rod, would it be To from point A to point B, and then decrease linearly until it gets to zero at point C ?
 
  • #8
Seraph042 said:
Does that mean that the reaction torque at point A is equal to the applied torque? [TA = To]

Sure, you can show this by adding up the torques and setting the sum to zero (since the rod isn't accelerating).

Seraph042 said:
As far as the torque variation along the rod, would it be To from point A to point B, and then decrease linearly until it gets to zero at point C ?

It would be To on the left side, but it wouldn't decrease gradually. Try drawing a free-body diagram from just right of B to C. The lack of reaction torque at C (we know this because there's nothing marked there) means that no positive torque can exist in this section.
 
  • #9
If I do a FBD from just after B to C, I just have the reaction torque at point B (TB) and since there is no other torque for that part, TB = 0.

Would that mean [tex]\phi[/tex]C = ToLBC / GBCIP,BC?
 
  • #10
Section BC?
 
  • #11
The shaft section from point B to point C, I get that TB = 0

Any of this sound correct?
 
  • #12
I like it up to the point where you calculate the angle between A and B by using the properties from B to C!
 
  • #13
Well I'm supposed to calculate the angle of twist at point C, so I figure to use the applied torque at point B and use the properties of the shaft from B to C (shear rigidity, polar moment of inertia, and length)

Where did I go wrong?
 
  • #14
But the angle increases from A to B; the section from B to C is just along for the ride.
 
  • #15
I'm taking my eraser out again :confused:


Ok I think I see what you mean, the angle of twist from point A to point B is the same as the total angle of twist at point C (...?)

This is what I'm seeing with the eraser right now, and that would mean:
[tex]\phi[/tex]AB = ToLAB/GABIP,AB = [tex]\phi[/tex]C

But if this is true, then the total rotational properties of the shaft part from point B to point C are irrelevant.

Would all of this be true?
 
  • #16
Exactly. Did playing with a physical model help? It always helps me.
 
  • #17
Absolutely. I never would have figured that out without playing with my eraser; thank you so much for your help! I'm just assuming that the twist on point C has to be influenced by the material of the shaft part B to part C.

Thank you so much for your help! This is for part of a bigger project for my class and the book we're using seems to be over complicating everything. I'm glad I know how this works now... It's always easier to relate to things when you can 'see' them somehow (like you were talking about)
 
  • #18
Glad to help. Good luck!
 

1. What is deformation in mechanics of materials?

Deformation refers to the change in shape or size of a material when a force is applied to it. This can include stretching, compressing, bending, or twisting.

2. How is deformation measured?

Deformation is typically measured using strain, which is the ratio of the change in length to the original length of the material. It can be expressed as either a percentage or a decimal value.

3. What is Hooke's Law and how does it relate to deformation?

Hooke's Law states that the amount of deformation of a material is directly proportional to the amount of force applied to it. This means that for small deformations, the relationship between stress (force) and strain (deformation) is linear.

4. What is torsion in mechanics of materials?

Torsion refers to the twisting of a material when a torque or twisting force is applied to it. This can cause shear stress and strain in the material, which can lead to deformation and failure.

5. How are deformation and torsion related?

Deformation and torsion are both types of mechanical stress that can affect a material. They can occur separately or together, and can lead to changes in the shape and properties of a material. Understanding the relationship between these two factors is important in predicting how a material will behave under different loading conditions.

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