Which wall thickness gives half the stress

[Janiceleong26]

Homework Statement

Homework Equations

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

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?

 

[Chestermiller]

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

 

[Janiceleong26]

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

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?

 

[Chestermiller]

F is directly proportional to A? Oh so if stress is halved, and for the same tension, area is halved right?

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

But area of..? Is it the area of the thickness of the tube?

The annular cross sectional area of the tube.

 

[Janiceleong26]

No. Try again, getting the math correct this time.The annular cross sectional area of the tube.

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

 

[Chestermiller]

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

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

 

[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 !

 

[Chestermiller]

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 !

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

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

 

[Janiceleong26]

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

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

Ohh thanks!