Young's Modulus for Two Materials connected

[thursdaytbs]

If Aluminum and steel are connected as one rod, and are attached to the wall and is pulled upon. How would the change in length for each material of the rod be calculated? I'm given that the Aluminum section is twice as long as the steel section, and the total change in length of the whole rod is 1mm.

So far, I've said:
F = Y(delta-L / L-naught)A + Y(delta-L / L-naught)A
*Where, the first part is for Aluminum, and the Second part is for Steel.

Manipulating this equation I got:

(2Fx)/A = (6.9E10)(delta-Laluminum) + (4E11)(delta-Lsteel)
and, delta-Aluminum + delta-steel = 0.001m
*where x = the length of steel, therefore aluminum = 2x.

This is where I'm stuck. I dont' know the Force applied, or the Area of the rod. What can i do? Thanks in advance for any help.

 

[thursdaytbs]

OK another thing I've thought up is saying...

F = Y(delta-Laluminum / L-naught)A
and
F = Y(delta-Lsteel / L-naught)A

since force is equal throughout?

Then set the two equations equal to one another, where A cancels out?

Can anyone confirm this is a way of doing it?

 

[Doc Al]

Looks good to me. The stress (F/A) is the same throughout the rod. You'll also need to use the other facts given regarding total change in length and the ratio of the two original lengths.

 

[thursdaytbs]

awsome, thanks a lot.