Question

A steel bar 100 mm (4.0 in.) long and having a square cross section 20 mm (0.8 in.) on an edge is pulled in tension with a load of 89,000 N (20,000 lbf), and experiences an elongation of 0.10 mm (4.0 × 10−3 in.). Assuming that the deformation is entirely elastic, calculate the elastic modulus of the steel.

Answer 1

To calculate the elastic modulus of steel, the tensile stress is computed using the force and cross-sectional area, and strain is calculated with the change and original lengths. Using these, Young's modulus is found to be 222.5 x 10¹ N/m².

Step-by-step explanation:

To calculate the elastic modulus of the steel given the deformation is entirely elastic, we need to use Hooke's Law, where tensile stress is equal to Young's modulus times the tensile strain.

First, calculate the tensile stress using the formula σ = F / A, where F is the force applied and A is the cross-sectional area of the bar.

Given:
F = 89,000 N
A= 20 mm x 20 mm = 400 mm² or 400 x 10⁻⁶ m²

Tensile stress, σ = 89,000 N / (400 x 10⁻⁶ m²) = 222.5 x 10¶ N/m²

Next, calculate the strain using the formula ε = ΔL / L0, where ΔL is the change in length and L0 is the original length.

Given:
ΔL = 0.10 mm = 0.10 x 10⁻3 m
L0 = 100 mm = 100 x 10⁻3 m

Strain, ε = 0.10 x 10⁻3 m / 100 x 10⁻3 m = 0.001

Finally, we use Hooke's Law to find Young's modulus:

Young's modulus, E = σ / ε = 222.5 x 10¶ N/m² / 0.001 = 222.5 x 10¹ N/m²

The calculated value for the Young's modulus of the steel is 222.5 x 10¹ N/m².

Answer 2

This problem asks us to compute the elastic modulus of steel. For a square cross-section, A0 = , where b0 is the edge length. Combining Equations 6.1, 6.2, and 6.5 and solving for E, leads to

E = σε = FA0Δll0= Fl0b02Δl= (89, 000 N)(100×10−3m)(20×10−3m)2(0.10×10−3m)= 223 × 109 N/m2 = 223 GPa (31.3 × 106 psi)