# Which wall thickness gives half the stress

1. Nov 2, 2015

### Janiceleong26

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

2. Relevant equations
Young's Modulus, E=σ/ξ
Stress,σ =F/A
Strain, ξ=x/L

3. The attempt at a solution
As diameter and tension are constant, so stress must be constant I assume.. So for the same Young's modulus, if stress is halved, then strain must be halved too I guess. But I don't know what is the extension and length. Do we have to consider the increase in circumference?

2. Nov 2, 2015

### Staff: Mentor

What does your second equation in your list of relevant equations tell you? Do you really need to know the extension in length to answer this question? Do you really need to know the change in strain to answer this question? They already said in the problem statement that the circumference πD stays constant, right?

Chet

3. Nov 2, 2015

### Janiceleong26

F is directly proportional to A? Oh so if stress is halved, and for the same tension, area is halved right? But area of..? Is it the area of the thickness of the tube?

4. Nov 2, 2015

### Staff: Mentor

No. Try again, getting the math correct this time.

The annular cross sectional area of the tube.

5. Nov 2, 2015

### Janiceleong26

Oh sorry, I mean area would double, not halved.
But I thought the diameter is the same?

6. Nov 2, 2015

### Staff: Mentor

The diameter is the same, but the pipe is hollow and the wall thickness doubles. The cross sectional area is $A = \pi D W$

7. Nov 2, 2015

### Janiceleong26

Oh I see it now.. The annular cross sectional area of the tube is actually the area of a rectangle, if we cut the tube and straighten it out, right? Ok thanks !

8. Nov 3, 2015

### Staff: Mentor

That's one way of looking at it. Another way is:

$$A=\frac{\pi D_{outer}^2}{4}-\frac{\pi D_{inner}^2}{4}=\pi \left(\frac{D_{outer}+D_{inner}}{2}\right)\left(\frac{D_{outer}-D_{inner}}{2}\right)=\pi D_{average}W$$

9. Nov 3, 2015

Ohh thanks!!