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)
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.
To calculate the elongation of the rod, we can use the formula for axial deformation:
ΔL = (F * L) / (E * A)
Where:
ΔL = Axial deformation (elongation) of the rod
F = Applied axial force on the rod (50 kN = 50,000 N)
L = Length of the rod (2 m = 2000 mm)
E = Young's modulus of elasticity for the alloy (140 GN/m² = 140 x 10^9 N/m²)
A = Cross-sectional area of the rod at the tapered end (30 mm diameter = 15 mm radius, A = π * r^2)
First, let's find the cross-sectional area of the rod at the tapered end:
A = π * r^2
A = π * (15 mm)^2
A ≈ π * 225 mm²
A ≈ 706.858 mm²
Now, we can calculate the elongation of the rod using the formula:
ΔL = (F * L) / (E * A)
ΔL = (50,000 N * 2000 mm) / (140 x 10^9 N/m² * 706.858 mm²)
ΔL = (100,000,000 mm² N) / (98.90 x 10^9 N.mm²)
ΔL ≈ 1.01 mm
Therefore, the elongation of the rod under an axial force of 50 kN is approximately 1.01 mm.
To calculate the elongation of the rod, we need to use the equation for axial deformation:
ΔL = (F * L) / (E * A)
where:
ΔL is the elongation or change in length,
F is the axial force applied to the rod,
L is the original length of the rod,
E is the Young's modulus or elasticity modulus of the material, and
A is the cross-sectional area of the rod.
First, let's calculate the change in cross-sectional area at each end of the rod:
Area1 = π * (diameter1/2)^2
Area1 = π * (30 mm / 2)^2
Area1 = π * 15^2
Area1 = 225π mm^2
Area2 = π * (diameter2/2)^2
Area2 = π * (20 mm / 2)^2
Area2 = π * 10^2
Area2 = 100π mm^2
Next, we need to find the average area along the length of the rod. Since the rod tapers uniformly, we can use the mean diameter:
diameter_mean = (diameter1 + diameter2) / 2
diameter_mean = (30 mm + 20 mm) / 2
diameter_mean = 25 mm
Now, we can calculate the average area:
Area_mean = π * (diameter_mean/2)^2
Area_mean = π * (25 mm / 2)^2
Area_mean = π * 12.5^2
Area_mean = 156.25π mm^2
We also need to convert the length to meters:
L = 2 m
Now, we have all the values to calculate the elongation:
ΔL = (F * L) / (E * A)
Plugging in the values:
ΔL = (50 kN * 2 m) / (140 GN/m^2 * 156.25π mm^2)
ΔL = (100,000 N * 2 m) / (140 * 10^9 N/m^2 * 156.25π * 10^-6 m^2)
Now, calculate the final value to get the elongation in meters.