Two bars, one copper and one aluminum, have the same length L1 = L2 = 1.50 m and
ID: 1484249 • Letter: T
Question
Two bars, one copper and one aluminum, have the same length L1 = L2 = 1.50 m and cross-sectional area A = 10.0 cm2, and are fixed on immovable walls with a gap of 1.00 mm between them at 73.0 F, and the temperature is going up steadily, as shown below. The linear thermal expansion coefficient and Young’s modulus of copper are 1.70× 105 K1 and 1.10×1011 Pa and of aluminum are 2.30×105 K1 and 6.90 × 1010 Pa, respectively. What is the force (magnitude) exerted on each wall at 113.0 F? (A) 9.42 kN; (B) 8.42 kN; (C) 7.42 kN; (D) 6.42 kN; (E) 5.42 kN.
Show your work please!
Explanation / Answer
Let us first calculate the temp when the gap of 1mm is fulfilled
Let the change in length of aluminium be x mm and copper be y mm
x+y=1
x*10^-3/1.5=2.30×105*T=>x=69T/2000 where T is change in emperature
y*10^-3/1.5=1.70×105*T=>y=51T/2000
x+y=1
120T=2000
T=50/3 K=30F
new Temperature=73+30=103 F
young modulas =(F/A)/strain
strain = change in length/length
at 103 F 1 mm gap is fixed so force will start exerting after temperature increases above 103 F
change in length of aluminium at 113 F =2.30×105*(10/1.8)*1.5=23/120000 m
strain=23/180000
force=young modulas*area*strain=8.816 kN
change in length of cpooer =1.70× 105*(10/1.8)*1.5=17/120000 m
strain=17/180000
force=10.388 kN
force on wall=19.204 kN
total force on each side=19.204/2=9.6 kN
A is correct