Temperature of a bar to produce certain force on wall

[songoku]

Homework Statement

A brass bar and a steel bar, each 0.8 m long are at a temperature of 20^oC. Each bar is placed at that temperature between rigid walls 0.8 m apart. The cross - sectional areas for the brass and steel bars are 0.005 m² and 0.003 m², respectively. The coefficient of linear expansion and Young modulus of the two materials are given in previous page. The temperature is raised until the combined force exerted by the two bars is 1.8 MN. The temperature at which this occurs, in ^oC is closest to
a. 100
b. 110
c. 120
d. 130
e. 140

Homework Equations

P = F/A
Young Modulus = stress / strain
stress = force / area
strain = extension / original length
ΔL = α . L_o . ΔT

The Attempt at a Solution

I am not sure I understand the set - up explained by the question. I imagine the two bars are placed parallel to each other between rigid walls. Is this correct?

Then the two bars are heated so they expand and push the walls, creating force. The total force of 1.8 MN is force by brass + force by steel.

If the bars are heated to same final temperature, the extension of each bar will be different even though they are placed between rigid walls. Is this possible?

Thanks

 

[haruspex]

If the bars are heated to same final temperature, the extension of each bar will be different even though they are placed between rigid walls.

There is (for the purpose of the question) no actual extension. You are supposed to assume that shortening of each bar due to its compression exactly compensates the extension due to its raised temperature.

 

[songoku]

There is (for the purpose of the question) no actual extension. You are supposed to assume that shortening of each bar due to its compression exactly compensates the extension due to its raised temperature.

The bar is compressed? How can it undergo compression? The length of the bar and separation of rigid walls are the same so I don't understand how the bar is compressed.

Thanks

 

[haruspex]

The bar is compressed? How can it undergo compression? The length of the bar and separation of rigid walls are the same so I don't understand how the bar is compressed.

Thanks

Because it tries to expand but can't.

 

[songoku]

Because it tries to expand but can't.

Oh I see. So what formula can I used to calculate the change in temperature? I can not use ΔL = α . Lo . ΔT because ΔL = 0 and strain is also zero

 

[haruspex]

Oh I see. So what formula can I used to calculate the change in temperature? I can not use ΔL = α . Lo . ΔT because ΔL = 0 and strain is also zero

Treat it as two separate processes. The bars are heated by some amount ΔT, so they expand by two different amounts. Then each is compressed by whatever force is necessary to restore it to its original length.

 

[songoku]

Thank you very much haruspex