[SOLVED] Thermal Stress Problem + Tensions Problem

bossmombo

I actually have two problems in mind, so let me lay them out clearly:

1.

2. Relevant equations
For Steel:
d=1 in
Ls= 6 + 2 + 2 = 10 in
thermal expansion coefficient (alphaS): 6.3*10^(-6)/F
Elastic Modulus: 30 Mpsi

For Aluminum:
area= 3 in^2
La= 6 in
thermal expansion coefficient (alphaA): 12.9*10^(-6)/F
Elastic Modulus: 10.4 Mpsi

ΔT=60F

Answers given at the back of the book
Stress(steel) = 1.784 ksi
Stress(alum.) = -467 psi

3. Attempt at a solution
I tried finding the first equation needed to start the problem but each time as I go through with it I don't get the answer at the back of the book.
I found the thermal expansion for:
Steel= alphaS*ΔT*Ls = 6.3*10^(-6)/F * 60 F * 10 in = 0.00378 in
Aluminum= alphaA*ΔT*La = 12.9*10^(-6)/F * 60 * 6 = 0.004644 in

I assumed there was going to be a reaction force for the aluminum equal to:
[PaLa/EaA]
and I know that the stress of steel or aluminum in the end is P/A

my problem in the end is: finding the first equation, and how to get to P. Can anyone help out?
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1.

2. Attempt at a solution
I started with the sum of the forces in the y direction so:
T1+T2+T3+T4-P=0

Then did the sum of the moments about A after the deformation was done:
(L)T2+(2L)T3-(2L)P+(3L)T4=0

Then considered the deformation with similar triangles:
def.2/L=def.3/(2L)=def.4/(3L)

After which I'm stuck because there are too many unknowns..so, did I go about it all wrong or am I close but don't see it?

bossmombo

I managed to solve the first problem, still working on the second one though. If anyone is actually curious about the first one I can post my work.

Pyrrhus

For the second problem, i recommend expressing the displacements of B, C and D as function of the displacement A and the angle the bar makes with the horizontal times the space between the wires, then use Hooke's Law to relate the loads with the displacements, and finish working it out with static equilibrium, it'll end as 2 unknowns (displacement of A and the angle times the space between the wires) 2 equations.

AlephZero

In problem 2 you have to assume the wires are flexible, otherwise you won't get a solution (as you already discovered).

One way of looking at it is to see that because the horizontal bars are rigid, the wire lengths will be in an arithmetic progression, so by Hooke's Law the tensions will be

TA = x
TB = x + y
TC = x + 2y
TD = x + 3y

Then find x and y by resolving and taking moments.

bossmombo

Thanks you two, I got the solutions for that second one.