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bossmombo
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[SOLVED] Thermal Stress Problem + Tensions Problem
I actually have two problems in mind, so let me lay them out clearly:
1. http://www.straydreamers.com/images/hw1.jpg
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?
--------------------------------------------------------------------------------------
1.http://www.straydreamers.com/images/hw1bis.jpg
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?
I actually have two problems in mind, so let me lay them out clearly:
1. http://www.straydreamers.com/images/hw1.jpg
Homework 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?
--------------------------------------------------------------------------------------
1.http://www.straydreamers.com/images/hw1bis.jpg
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?
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