A classical mechanics problem by Jaber Al-arbash

At a temperature of 6 0 F 60 ^\circ F , a 0.04 in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy [ α = 12.5 × 1 0 6 / F \alpha = 12.5 \times 10^{-6} / ^\circ F ] bar with a width of 3 in and a thickness of 0.75 in. Bar (2) [ α = 9.6 × 1 0 6 / F \alpha = 9.6 \times 10^{-6} / ^\circ F ] bar with a width of 2 in and a thickness of 0.75 in.

Assume the supports at A A and C C are rigid. What is the lowest temperature at which the two bars contact each other?

Equations to be used:

a) Elongation or lengthening δ = α Δ T L \delta = \alpha \Delta TL .

b) δ 1 + δ 2 = 0.04 \delta_1 + \delta_2 =0.04 .

Note : Since there is a gap of 0.04 in between the two bars, the sum of the elongations of bar (1) and (2) is given as shown in equation (b).

Symbols:

α \alpha : Thermal expansion of a bar

L L : Length

T T : Temperature

δ \delta : Elongation.

108. 6 F 108.6 ^\circ F 48. 6 F 48.6 ^\circ F 6 0 F 60 ^\circ F None of these choices

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1 solution

Jaber Al-arbash
Feb 13, 2016

𝛿= αΔTL and 𝛿1+𝛿2= 0.04

α1ΔTL1 + α2ΔTL2 = 0.04

( 12.5 x 10^-6 ) ( ΔT) (32) + ( 9.6 x 10^-6 ) (ΔT) (44) = 0.04

solve for ΔT = 48.6

Tf= ΔT + Ti

Tf= 48.6 + 60 = 108.6 °F ( answer)

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