Feel really stupid. Need help. Thermal switch design problem.

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konto77
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Homework Statement


There are two rigid plates on the ends.

In between them are two steel plates and one aluminum plate. The aluminum one is in the center. When the temperature increases enough the aluminum will buckle and touch one of the steel plates to close the circuit.

alpha(alum) = 12.5 (10^-6) /degF
alpha(steel) = 6.6 (10^-6) /degF

E(al) = 10,000 ksi
E(s) = 30,000 ksi

t(a) = 1/16 in w(a) = 1/4 in These are thickness and width
t(a)<w(a) or else the aluminum will not buckle towards the steel

t(s) = 1/16 in w(s) = 1/8 in


Length = 4 in

With these conditions the aluminum buckles at 180 deg F

Find dimensions that will allow the circuit to close at 100 deg F

Homework Equations


critical axial compressive load: P(cr) = 4 pi^2 E I/L^2

I = w(a)t(a)^3/12



alpha*deltaT*L + PL/E/A = 0 when there are rigid plates right?

and the problem also says "As the temperature increases, the lengths of both aluminum and stell straps remain equal because of the rigid end pieces." Does this mean that length of stell and aluminum will change and remain equal? Or that their change in length equals 0.




The Attempt at a Solution



I used this equation
alpha*deltaT*L + PL/E/A = 0

with
this equation to get the square root of a negative number for thickness of a
P(cr) = 4 pi^2 E I/L^2
 
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konto77 said:
Does this mean the length of stell [sic] and aluminum will change and remain equal?

konto77: Yes, that is correct. The change in length of the aluminum and steel does not equal zero.
konto77 said:
alpha*deltaT*L + P*L/(E*A) = 0 when there are rigid plates, right?

Incorrect. The right-hand side should be displacement, x, not zero. It should be xa for aluminum, or xs for steel.

Does the problem say anything like, hold ws, ts, and wa constant? I think it perhaps should (or could). And let the unknown for the second part of the problem be ta.

Hint 1: alphaa*deltaT*L + Pa*L/(Ea*Aa) = xa.
Hint 2: Ps = -0.5*Pa.
Hint 3: Pa = Pcr = -4*Ea*Ia*(pi/L)^2.

Keep trying.