Answer to Question #158306 in Material Science Engineering for Thapelo

Question

  QUESTION 1: [Temperature Stresses]    [14] A link in a jig that experiences changes in temperature is comprise of two bars in series. The steel component has a Young’s modulus, coefficient of thermal expansion, diameter, and length of 200 GPa, 12×10−6/�, 0.06 m and 10 meters, respectively, and is fixed rigidly on the end that is further from its joint with the other bar. The aluminium component has a Young’s modulus, coefficient of thermal expansion, inner (i) and outer diameters (o), and length of 80 GPa, 23×10−6/�, 0.03 and 0.08 m, and 15 meters, respectively, and has a gap of 15 mm between its free end and the jig housing. Determine: 1.1. The maximum temperature change that is allowed on the jig before the aluminium bar touches the housing. 1.2. The type and magnitude of stresses induced in the two bars if the jig housing is exposed to a temperature rise of 800C

Answer 1

1.1 When total deformation caused by temperature increment for both the components reaches 15 mm, they both touches, for that,

$\alpha_sl_s(\Delta T) +\alpha_al_a(\Delta T) =0.015m$

$(12\times10^{-6} \times10+23\times 10^{-6}\times 15)\Delta T=0.015m$

$\Delta T=32.26^{o}C$

1.2 After this much temperature rise, the bar will be subjected to stress, virtual deformation by thermal expansion is given by,

$(12\times10^{-6} \times10+23\times 10^{-6}\times 15)(80-32.26)=\delta$

$\delta=0.0222m$

Now $\delta=(\frac{Fl}{AE})_s+(\frac{Fl}{AE})_a$

$0.0222=F(\frac{10}{\frac{\pi}{4}\times0.06^2\times 200 \times10^9})+F(\frac{15}{\frac{\pi}{4}\times(0.08^2-0.03^2)\times 80 \times10^9})$

$F=363.39 kN$

Stress in steel

$\sigma_s=\frac{363.39\times10^3}{\frac{\pi}{4}\times 0.06^2}=128.52MPa \space (Compressive)$

$\sigma_a=\frac{363.39\times10^3}{\frac{\pi}{4}\times (0.08^2-0.03^2)}=84.12MPa \space (Compressive)$