Different materials - Elastic modulus

[flower76]

Hi

This looks like an easy question but I'm stumped and would appreciate some help.

A rod is made of two sections joined end to end. The sections are identical, except that one is steel and the other is brass. While one end is held fixed, the other is pulled to result in a change in length of 1.20 mm. By how much does the length of each section increase?

Any ideas?

 

[Hootenanny]

What is the equation of young's modulus?

~H

 

[flower76]

I believe it to be:

change in L = (1/E)(F/A)(Lo)

I'm not sure what to do with the fact that you have two E values, add them together? Then do you ignore F and A, and is Lo actually 2Lo in this case?

Thanks

 

[Astronuc]

There would be a total length L, and one material has length x and the other one has length L-x.

Think of Hooke's Law, and the relationship between stress and strain.
See - http://en.wikipedia.org/wiki/Hooke's_law

Stress is load (force)/area.

 

[flower76]

I'm sorry now I think I'm even more confused, there are no values for force or area and I can't see how you would get them. Am I overlooking something?

 

[HallsofIvy]

Do the two parts separately. How much does each increase separately?

 

[flower76]

The question is to figure them out separately and I don't know how to do it. I keep going around in circles and getting answers that don't make sense.

Any other suggestions?

Thanks

 

[flower76]

Ok I'm still trying to figure out this question and getting nowhere, does anyone have any other suggestions. It would be greatly greatly appreciated.

Thanks

 

[Integral]

Please post some of your work.

 

[flower76]

Ok this is what i have so far:

change in L = 1/E x L

so 0.0012 = 1/100x10^9 x X
X= 1.2 x 10^9

and the other is 0.0012 = 1/200x10^9 x L-x
L-x= 2.4x10^8
therefore L = 1.44x10^9

This doesn't look right to me but I don't know what I'm missing.

 

[civil_dude]

First, you should probably assume that each section is the same length. That said, if the modulus of Elasticity of one material is twice the other, that means that it will stretch half as much. Thus one section stretches 0.0008 m and the other 0.0004 m.

 

[Astronuc]

flower76 - Sorry for the confusion, I seem to have made it more complicated than necessary.

In series, i.e. with the two sections (rods/bars) end-to-end, they are subject to the 'same' force, and assuming they have the same cross-sectional area, each develops the same stress.

However, the elastic (Young's) modulus of each is different, so the strain of each will be different.

The strain is simply \epsilon = \sigma/E, where \epsilon is the strain, \sigma is the axial stress, and E is the elastic modulus.

If one section is length L₁ and the other L₂, then the initial length is simply L = L₁ + L₂. Now when the sections strain, one obtains a combined length given by (1+\epsilon₁) L₁ + (1+\epsilon₂) L₂.

 

[flower76]

Thanks for the help I finally get it, and of course its much simpler then it originally looked.

 

[supafly]

flower76 - Sorry for the confusion, I seem to have made it more complicated than necessary.

In series, i.e. with the two sections (rods/bars) end-to-end, they are subject to the 'same' force, and assuming they have the same cross-sectional area, each develops the same stress.

However, the elastic (Young's) modulus of each is different, so the strain of each will be different.

The strain is simply \epsilon = \sigma/E, where \epsilon is the strain, \sigma is the axial stress, and E is the elastic modulus.

If one section is length L₁ and the other L₂, then the initial length is simply L = L₁ + L₂. Now when the sections strain, one obtains a combined length given by (1+\epsilon₁) L₁ + (1+\epsilon₂) L₂.

Can you try to explain this again to me? I am really confused!