A circular alloy 2 m long uniformly tapers from 30 mm diameter to 20 mm diameter. Calculate the elongation of the rod under an axial force of 50 kN. (E for the alloy = 140 GN/m^2)

Answer 1

To find the elongation of the rod, we need to use the formula for elongation:

ΔL = (F * L) / (A * E)

Where:
ΔL = elongation of the rod
F = axial force
L = original length of the rod
A = cross-sectional area of the rod
E = Young's modulus

Given:
F = 50 kN = 50,000 N
L = 2 m
Diameter at one end = 30 mm = 0.03 m
Diameter at the other end = 20 mm = 0.02 m
E = 140 GN/m^2 = 140,000,000 N/m^2

First, we need to calculate the cross-sectional areas at both ends of the rod.

Area at one end: A1 = π * (d1/2)^2
Area at other end: A2 = π * (d2/2)^2

A1 = π * (0.03/2)^2 = 0.00070685 m^2
A2 = π * (0.02/2)^2 = 0.00031416 m^2

Next, we can calculate the average area:

Aavg = (A1 + A2) / 2

Aavg = (0.00070685 + 0.00031416) / 2 = 0.00051051 m^2

Now we can substitute the values into the elongation formula:

ΔL = (F * L) / (A * E)
ΔL = (50,000 * 2) / (0.00051051 * 140,000,000)
ΔL ≈ 0.553 mm

Therefore, the elongation of the rod under an axial force of 50 kN is approximately 0.553 mm.