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Neko
17 days ago
10

A specimen of steel has a rectangular cross section 20 mm wide and 40 mm thick, an elastic modulus of 207 GPa, and a Poisson’s r

atio of 0.30. If this specimen is pulled in tension with a force of 60,000 N, what is the change in width if deformation is totally elastic?
Physics
1 answer:
Softa [2K]17 days ago
6 0

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

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