An airplane has a half wingspan of 33 m Determine the change

An airplane has a half wingspan of 33 m. Determine the change in length of the aluminum alloy (alpha = 22.5 Times 10^6/degree C) wing spar if the plane leaves the ground at a temperature of 15 degree C and climbs to an altitude where the temperature is -55 degree C. The above assembly is at an initial temperature of 40 degree C. Bar 1 is aluminum alloy (alpha = 12.5 Times 10^-6/degree C) and bar 2 is stainless steel (alpha = 9.6 Times 10^-6/degree C). The supports at A and C are rigid. Determine the lowest temperature at which the two bars touch each other.

Solution

Problem 7.

Given the length of half wing span(L) = 33m.

Coefficient of thermal expansion of aluminum alloy() = 22 X 10^-6 /⁰C

let the temperature on ground T_g= 15⁰C

the temperature at altitude, T_a = -55⁰C

Change in the length of aluminum alloy half wing span is given by *L*(T_a ^_T_g)

= 22 * 10^-6 * 33 * (-55-15))

= -22 * 33 * (70) * 10^-6

= -50820 * 10^-6 m (- indicates contraction)

change in length of half wing span = 0.05082 m or 50.82 mm (contraction)

change in length of full wing span = 2 * 0.05082 m = 0.10164m or 101.64mm (contraction)

Problem 8.

Given, initial temperature T_i = 40⁰C

Let Final or minimum temperature when the two bars touch = T_f

Change in temperature dT = T_f – T_i

Let the length of bar1 be L_a = 40 in.

     The length of bar2 be L_s = 55 in.

Coefficient of expansion of bar 1 _a = 12.5 X 10^-6 /⁰C

Coefficient of expansion of bar 2 _s = 9.6 X 10^-6 /⁰C

Gap in between the bars d= 0.08 in.

The bars touch each other when the sum of expansion of two bars = d= 0.08 in.

So, (Expansion of bar1 + Expansion of bar2) = d= 0.08in.

(_a * L_a* dT) + (_s * L_s* dT) = 0.08

(12.5 X 10^-6 * 40 * dT) + (9.6 X 10^-6 * 55 * dT) = 0.08

     dT * 1028 * 10^-6 = 0.08

dT = 77.82⁰C

T_f – T_i = 77.82

T_f = 77.82 + 40

T_f= 117.82⁰C

Minimum final temperature when the two bars touch = 117.82⁰C