Question

A cylindrical rod 120 mm long and having a diameter of 15.0 mm is to be deformed using a tensile load of 35,000 N. It must not experience either plastic deformation or a diameter reduction of more than 1.2 × 10–2 mm. Of the following materials listed, which are possible candidates? Justify your choice(s).

Answer 1

To determine possible candidates for the cylindrical rod, consider the tensile stress and diameter reduction requirements.

Step-by-step explanation:

In order to determine which materials are possible candidates for the cylindrical rod, we need to consider the tensile stress and the diameter reduction requirements.

Tensile stress is calculated by dividing the force applied by the cross-sectional area of the rod. The maximum tensile stress should not exceed the breaking stress of the material.

Diameter reduction is determined by the elongation of the rod, which is given by L = (F * L0) / (A * Y), where F is the tensile force, L0 is the original length of the rod, A is the cross-sectional area of the rod, and Y is the Young's modulus of the material.

Answer 2

Answer: Magnesium Alloy

Step-by-step 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.