The steel rod will elongate by 2.86 mm and its diameter will decrease by 0.715 mm under the applied load.
Finding the change in length and diameter of the steel rod:
Given:
Length (L) = 4 meters
Initial diameter (d) = 20 millimeters
Axial tensile load (F) = 45 kN
Young's modulus (E) = 2 x 10^5 N/mm^2
Poisson's ratio (v) = 1/4
Formula:
Change in length (ΔL): ΔL = FL / (AE)
Change in diameter (Δd): Δd = -v * ΔL
Calculations:
Convert units:
Length (L): 4 m = 4000 mm
Load (F): 45 kN = 45000 N
Diameter (d): 20 mm = 0.02 m
Calculate the cross-sectional area (A):
A = πd^2 / 4 = π(0.02)^2 / 4 = 3.14 x 10^-4 m^2
Calculate the change in length (ΔL):
ΔL = FL / (AE) = (45000 N)(4000 mm) / ((2 x 10^5 N/mm^2)(3.14 x 10^-4 m^2)) = 2.86 mm
Calculate the change in diameter (Δd):
Δd = -v * ΔL = - (1/4) * (2.86 mm) = -0.715 mm
Results:
Change in length: 2.86 mm (increases)
Change in diameter: 0.715 mm (decreases)
Therefore, the steel rod will elongate by 2.86 mm and its diameter will decrease by 0.715 mm under the applied load.
Complete question:
A steel rod 4 m long and 20 mm diameter is subjected to an axial tensile load of 45 kN. Find the change in length and diameter of the rod. Es = 2 X 105 N/mm2. Poisson’s ratio =1/4