# Homework Help: Strain Energy in a Rod Question

1. Jan 3, 2016

### smr101

Unsure which dimensions to use to work out the force and strain energy here.

I know that Force = yield strength * area

Yield strength = 310 MPa

Area, I'm unsure which radius to use. I assume 12 mm.

For strain energy = (Force^2 * Length)/(2 * Area * Youngs Modulus)

Unsure which length and area to use here.

Best attempt I've made so far gives 42.41 J

Working:

F = 310 x 10^6 * pi*0.012^2
= 140,240N

U = 140240^2 * 0.4 / 2 * pi*0.012^2 * 205 x 10^9

Thanks, question is below:

Last edited: Jan 3, 2016
2. Jan 3, 2016

### SteamKing

Staff Emeritus
This is true only if the axial stress on the rod = yield stress.

In general, Force = axial stress on the rod * cross sectional area of the rod.
If you have a non-prismatic rod, you'll have to look at the properties of each section of the rod.
If you have a non-prismatic rod, you'll have to look at the properties of each section of the rod.

For a given load P, the axial stress in the rod in section BC is much different from the axial stress in section AB. You cannot ignore this in your calculation of the axial strain energy.

3. Jan 4, 2016

### smr101

Ok, thanks.

So obviously I need to work out the force on each section and then the strain energy on each section.

For section A-B the strain energy and calculation are stated in the OP. Is this correct? If so, this doesn't correlate with the solution. Adding this number plus the strain energy for B-C will be too large a value.

For section B-C how is the area calculated for the equation Force = Area * yield strength? Is the outer or inner diameter used?

Also, same question for the strain energy equation - which diameter is used for the area?

If I could get some specific answers to this that'd be great. I've done many questions like this but never with a hollow tube using these equations.

4. Jan 4, 2016

### SteamKing

Staff Emeritus
The value of P must be selected so that the axial stress in any part of the rod, i.e., sections A-B and B-C, is below yield. If you pick P based on the properties of section A-B, then the axial stress in section B-C will exceed yield, and the rod will deform permanently.
The area in the formula is the area of the material in the rod to which the axial load is being applied. The inner portion of the rod in Section B-C is hollow, therefore, there is no material there to take any load.
See comment above.
You've never calculated the axial stress for a hollow tube or rod??? What sort of course is this?

5. Jan 4, 2016

### smr101

The value of P for the 500 mm section is 24,800N and the 400 mm section is 140,240N.

Are you saying to use the 24,800 in the calculations for strain energy in both cases?

Have I got any calculations correct here so far? Have you been able to calculate the correct answer for this?

This is revision and not homework, if you could give me the worked solution that would be a massive help.

Thanks.

6. Jan 4, 2016

### Staff: Mentor

Here's a geometry problem for you: What is the area of an annular cross section in terms of the inner and outer radii?

7. Jan 4, 2016

8. Jan 4, 2016

### SteamKing

Staff Emeritus
Please show your calculations and how you arrived at P = 24,800 N.
I reserve comment until you have posted your latest calculations.
Doesn't matter. If the thread is in the HW forums, it's against the rules to post complete solutions.

9. Jan 4, 2016

### smr101

Area = pi*r^2
= pi*(0.012^2 - 0.008^2)
= 8 x 10^-5

F = A x yield strength
= 8 x 10^-5 * 310 x 10^6
= 24, 800 N

10. Jan 4, 2016

### SteamKing

Staff Emeritus
Always show units in your calculations.

Did you check your arithmetic here? Arithmetic mistakes lose points on exams just as easily as not studying the material you're being tested on.
You have an error in the calculation of the area above.

11. Jan 4, 2016

### smr101

Yes, you're correct.

Correct area is 2.513 x 10^-4, correct force is 77,911.5.

From this, strain energy is

U = P^2*L / 2*A*E
= 77911.5^2 * 0.5 / 2 * 2.513 x 10^-4 * 205 x 10^9
= 29.458 J

That is for the 500 mm section, calculation for the 400 mm section is in the OP.

Again, these two numbers don't see to help me get the desired solution, I assume I've made a mistake but can't see where.

12. Jan 4, 2016

### SteamKing

Staff Emeritus
You don't seem to realize that if you pull on one end of the rod with force P, this force does not change along the length of the rod.

You can't have two different internal forces in the rod if it is in static equilibrium.

Is your strain energy calculation in the OP based on the P you calculated for section B-C?
You don't seem to have taken any courses in statics at all. You are making very basic mistakes with your calculations, and you apparently are unaware of the concept of static equilibrium.

13. Jan 4, 2016

### smr101

P^2/2 * E * (L1/A1 + L2/A2)

77911.5^2/2 * 205 x 10^9 * (0.5/ 2.513 x 10^-4 + 0.4/4.52 x 10^-4)

= 42.56 J

That's it, isn't it?

If I used exact values the answer would be 42.54 as the solution states.

14. Jan 4, 2016

### SteamKing

Staff Emeritus
Correct. Now on to part ii).

15. Jan 4, 2016

### smr101

Thank goodness... I've already done part (ii), thanks for the help!