Answer: Magnesium Alloy
Explanation:
Cross sectional area A = (pi/4)d sqr
A = pi / 4 x 0.015 sqr
A = 1.76 x 10^-4 m2
Stressed induced in rod S = P/A = 35000/1.76x 10^-4
S= 198.06 M Pa
Reduction in diameter of the titanium alloy = -v (alpha/E) x d
= - 0.33 (( 198.06 X 10^6) / 70 x 10^9 )x 0.015 = - 1.41 x 10^-2 mm
For steel alloy
= - 0.36 (( 198.06 X 10^6) / 150 x 10^9 )x 0.015 = - 1.02 x 10^-2 mm
For magnesium
= 0.27 (( 198.06 X 10^6) / 205 x 10^9 )x 0.015 = 0.391 x 10^-2 mm
The titanium alloy and the steel alloy does not satisfy the second criteria since the reduction in diameter exceeds the allowable limit of 1.2 x 10^2mm
Considering the yield strength of the material, we find the aluminium alloy is not suitable and hence no need to check for the second criteria. By considering the both the given conditions, we find the magnesium alloy is the suitable material.
