Response: The width decreases by 2.18 × 10^(-6) m
Clarification:
Given data;
Shear Modulus; E = 207 GPa = 207 × 10^(9) N/m²
Force; F = 60000 N.
Poisson’s ratio; υ = 0.30
The initial width is 20 mm, and the thickness is 40 mm.
Area = 20 × 10^(-3) × 40 × 10^(-3)
Area = 8 × 10^(-4) m²
The formula for shear modulus is;
E = σ/ε_z
where σ represents stress calculated as Force(F)/Area(A)
while ε_z stands for longitudinal strain.
Thus;
E = (F/A)/ε_z
ε_z = (F/A)/E
ε_z = (60,000/(8 × 10^(-4)))/(207 × 10^(9))
ε_z = 3.62 × 10^(-4)
Next, the lateral strain is given by;
ε_x = - υ × ε_z
ε_x = -0.3 × 3.62 × 10^(-4)
ε_x = -1.09 × 10^(-4)
The change in width can be determined as;
Δw = w_o × ε_x
Where w_o denotes the original width = 20 × 10^(-3) m
So; Δw = 20 × 10^(-3) × -1.09 × 10^(-4)
Δw = -2.18 × 10^(-6) m
A negative sign indicates a reduction in width.
Therefore, the width decreases by 2.18 × 10^(-6) m