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 ratio 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?

There's a decrease in width of 2.18 * 10^ (-6) m

Explanation:

We are given;

Shear Modulus; E = 207 GPa = 207 * 10^ (9) N/m²

Force; F = 60000 N.

Poisson's ratio; υ = 0.30

We are told width is 20 mm and thickness 40 mm.

Thus;

Area = 20 * 10^ (-3) * 40 * 10^ (-3)

Area = 8 * 10^ (-4) m²

Now formula for shear modulus is;

E = σ/ε_z

Where σ is stress given by the formula Force (F) / Area (A)

While ε_z is 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)

Now, formula for lateral strain is;

ε_x = - υ * ε_z

ε_x = - 0.3 * 3.62 * 10^ (-4)

ε_x = - 1.09 * 10^ (-4)

Now, change in width is given by;

Δw = w_o * ε_x

Where w_o is initial width = 20 * 10^ (-3) m

So; Δw = 20 * 10^ (-3) * - 1.09 * 10^ (-4)

Δw = - 2.18 * 10^ (-6) m

Negative means the width decreased.

So there's a decrease in width of 2.18 * 10^ (-6) m