Solving Design Project w/ Thermal Switch (Statics)

[cardriverx]

Homework Statement

This is for Statics and mechanics of materials 1...

In this project, we were tasked with the re-design of a cheap thermal switch that is to be a component of a low cost product. The switch consists of three metal strips that are bounded on the top and bottom by a rigid plate (the plate is not bounded tho). The two outer strips are steel, and the one in the middle is aluminum. The only loads on the switch are the loads generated by the strips expanding due to heat. Since the middle strip is aluminum, its coefficient of thermal expansion is greater than the steel outer strips. Because of this, the aluminum strip will expand faster than the steel strips, which results in an internal force being generated. When the forces in the aluminum strip reach its critical axial compressive load, it will buckle, which will cause it to contact one of the steel strips. This will cause a circuit to be completed, which will signal the product that it has reached a certain temperature. We are given dimensions for a thermo switch that completes its circuit at 180 degrees, and we are to change the dimensions of the aluminum strip so that it completes its circuit at 100 degrees instead.

Height of strips = 4 in

width of alum. strip for 180 degree switch = 1/16 in
"" "" steel strip """ """ """ """ = 1/16

length of alum. strip = 1/4 in
""" """ steel """ = 1/8 in

coef. thermal expansion for alum. (alpha) = 12.5E-6
"""" """" """" "" steel (alpha) = 6.6E-6

Youngs modulus of alum (E) = 10,000 ksi = 10,000,000 psi
"""""""""" steel (E) = 30,000 ksi

Homework Equations

Formula for critical axial load = (4*Pi^2*E*((length*width^3)/12))/Height^2


Width must be < length so the alum. strip buckles in the direction of the steel.

The Attempt at a Solution

Well what I tried doing is first using the equation:

(alpha alum. * deltaT * height) - (alpha steel * deltaT * height) = how much more the alum. expands than the steel

Then I reduced it to:

DeltaT * 2.4E-5 = how much more the alum. expands (Deformation)

By plugging in both alphas and the height of 4 in.

Then I used:

Deformation = (P*L)/(E*A) with P=force and L=height and A=cross section area.

I changed that to:

(Deformation * E * A)/L = P

Plugging in I get

(DeltaT*A*2.4E-5*1E7)/4

Which is:

DeltaT*A*60 = Force in lb

I then set that equal to the formula for critical axial load, because you want the strip to reach the critical axial load at the temperature you specify.

Now here is the problem, when I set them equal to each other, it cancels out the length of the strip, and I get:

sqrt((DeltaT*11328)/(4*Pi^2*1E7)) = width of the strip

I plugged in 180 degrees, and got .07 inches, which is 8 hundredths off the 1/16 it should be..



Please help me, I've been on this for like 4 hrs now, am I approaching this wrong? It can't be right that the length of the strip cancels out, because it affects the critical axial load! Right?

 

[Mapes]

It looks like you're assuming that the plate compresses the aluminum strip until its length matches the length of unconstrained steel strips at the higher temperature. But this isn't exactly what happens, is it? The length of the connected strips before buckling is more than an unconstrained steel strip and less than an unconstrained aluminum strip.

Try calculating the strain of each strip as a function of stress and temperature change. (Lump the steel strips together to make calculation easier.) Set these strains equal. Perform a free-body diagram on one of the plates to get the relationship between the stresses. Finally, connect the stress in the aluminum strip to the critical buckling stress.