# Calculate the total change in length of a brass bar

1. Apr 23, 2017

### DevonZA

1. The problem statement, all variables and given/known data

2. Relevant equations

Stress = force/ cross sectional area

elongation = force x length / modulus of elasticity x cross sectional area

3. The attempt at a solution

The given solution is 0.149mm

I assumed that the bar needs to be solved in two sections because of the hole of diameter 20mm in the first 70mm of the bar. I will call this part A and the 90mm section part B.

Part A:

Stress = force/ cross sectional area
Force = stress x cross sectional area
= 125x10^6 x pi/4(.03-.02)^2
= 9.8kN

elongation = force x length / modulus of elasticity x cross sectional area
= 9.8x10^3 x 0.07 / 100x10^9 x pi/4(.03-.02)^2
= 8.73x10^-5mm

Part B:

Force = stress x cross sectional area
= 125x10^6 x pi/4(.03)^2
= 88.4kN

elongation = force x length / modulus of elasticity x cross sectional area
= 88.4x10^3 x 0.09 / 100x10^9 x pi/4(.03)^2
= 1.126x10^-4mm

now if I add the elongation from part A & B together I do not get 0.149mm.
Please show me where I am going wrong.

2. Apr 23, 2017

### TSny

You did not calculate the cross-sectional area of the left section correctly.

You have assumed that the stress is the same in both sections. But is that correct? Imagine that the stress is created by applying an overall force on the left end and an overall force on the right end. How do these two forces compare?

3. Apr 23, 2017

### DevonZA

Hi TSny

I understand what you are trying to get me to imagine I just don't know how to apply it.
I guess I need to find a ratio between the diameters and/or lengths? Then multiply this by the stress of 125MPa to find the stress induced on each end?

4. Apr 24, 2017

### haruspex

Did you understand from TSny's post that this is wrong:
Having fixed that, consider the two lengths as separate bodies. What is the interaction between them?

5. Apr 24, 2017

### DevonZA

I have found the solution. I will post it when I get a minute.
I realize that I need to use pi/4(.03^2-.02^2)

6. Apr 24, 2017

### DevonZA

#### Attached Files:

• ###### example question elongation.pdf
File size:
148.1 KB
Views:
14
7. Apr 24, 2017

### haruspex

You calculate the force as about 49,000kN. Are you sure about the units?

8. Apr 24, 2017

### DevonZA

That is 49kN not 49000kN. I should have used a period not a comma

9. Apr 24, 2017

### haruspex

Sorry, I should have noticed you were using the decimal comma in the radii. All looks good,then.