Finding Increase in Temperature for Two Rods

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A steel rod and an aluminum rod, secured end to end, are heated to the same final temperature, resulting in a one-tenth percent increase in the steel rod's length. The challenge is to determine the temperature increase and mutual stress on both rods while maintaining a constant total length. Initial calculations suggest a temperature increase of 83.33 °C for the steel, but this does not account for the aluminum rod's behavior under the same conditions. The rods cannot flex or change total length, indicating that the expansion of one rod is countered by the compression of the other. The correct answers for the temperature increase and mutual stress are 375 °C and 700 MPa, respectively.
jfulky
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Homework Statement


A steel rod and an aluminum rod of equal length and diameter are placed end to end and secured so that they cannot flex. The rods are heated to the same final temperature, and the steel is found to increase in length by one-tenth of a percent. If the total length of the rods together remains constant, find the increase in temperature for the rods and the mutual stress on the rods.

Ans: 375 oC , 700 MPa

Homework Equations



(RATE OF HEAT FLOW)
H = ΔT/R R = L/KA

(LINEAR HEAT EXPANSION)
ΔL = L0αΔT[/B]

(STRESS)
F/A = -YαΔT

The Attempt at a Solution



I first solved for ΔT from the linear heat of expansion equation...

Lf(steel) - li(steel) = li(steel)*α*ΔT

1.001*Li(steel) - li(steel) = li(steel)*α*ΔT

0.001 = α*ΔT where α = 1.2 *10^-5

ΔT = 83.33 C

After I find this, I though I would then plug this into the Stress equation for a given metal since they have the same change in temperature. Though this is obviously not the correct answer.

Am I missing something here?
 
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You need to find the change in the length of aluminium rod. Do you expect it to be the same as that of steel if they are not secured together? First calculate and then visualize.
 
jfulky said:

Homework Statement


A steel rod and an aluminum rod of equal length and diameter are placed end to end and secured so that they cannot flex. The rods are heated to the same final temperature, and the steel is found to increase in length by one-tenth of a percent. If the total length of the rods together remains constant, find the increase in temperature for the rods and the mutual stress on the rods.

Ans: 375 oC , 700 MPa

Homework Equations



(RATE OF HEAT FLOW)
H = ΔT/R R = L/KA

(LINEAR HEAT EXPANSION)
ΔL = L0αΔT[/B]

(STRESS)
F/A = -YαΔT

The Attempt at a Solution



I first solved for ΔT from the linear heat of expansion equation...

Lf(steel) - li(steel) = li(steel)*α*ΔT

1.001*Li(steel) - li(steel) = li(steel)*α*ΔT

0.001 = α*ΔT where α = 1.2 *10^-5

ΔT = 83.33 C

After I find this, I though I would then plug this into the Stress equation for a given metal since they have the same change in temperature. Though this is obviously not the correct answer.

Am I missing something here?
All I see are calculations for the steel rod. What happened to the aluminum rod?

Also, you need a formula which relates mechanical stress to axial deflection in a rod.
 
jfulky said:

Homework Statement


A steel rod and an aluminum rod of equal length and diameter are placed end to end and secured so that they cannot flex. The rods are heated to the same final temperature, and the steel is found to increase in length by one-tenth of a percent. If the total length of the rods together remains constant, find the increase in temperature for the rods and the mutual stress on the rods.

Am I missing something here?
My guess is that you are missing a crucial piece of the statement of the question. I believe the idea is that the rods are clamped in such a way that overall length cannot change, as well as not being able to flex. Thus, the tendency to expand due to the raised temperature is exactly compensated by the compression of the rods due to the end loading.
You will need the Young's modulus of the materials, though it gets a bit messy since that may depend on the temperature, which you do not know yet. Perhaps it doesn't make enough difference to matter.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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