Question

Problem 7.16 (GO Tutorial) A cylindrical bar of steel 10.4 mm (0.4094 in.) in diameter is to be deformed elastically by application of a force along the bar axis. Determine the force that will produce an elastic reduction of 3.2 ×10-3 mm (1.260 ×10-4 in.) in the diameter. For steel, values for the elastic modulus (E) and Poisson's ratio (ν) are, respectively, 207 GPa and 0.30.

Answer 1

Answer:

P = 18035.25 N

Explanation:

Given

D = 10.4 mm

ΔD = 3.2 ×10⁻³ mm

E = 207 GPa

ν = 0.30

If

σ = P/A

A = 0.25*π*D²

σ = E*εx

ν = - εz / εx

εz = ΔD / D

We can get εx as follows

εz = ΔD / D = 3.2 ×10⁻³ mm / 10.4 mm = 3.0769*10⁻⁴

Now we find εx

ν = - εz / εx   ⇒   εx = - εz / ν = - 3.0769*10⁻⁴ / 0.30 = - 1.0256*10⁻³

then

σ = E*εz = (207 GPa)*(-1.0256*10⁻³) = - 2.123*10⁸ Pa

we have to obtain A:

A = 0.25*π*D² = 0.25*π*(10.4*10⁻3)² = 8.49*10⁻⁵ m²

Finally we apply the following equation in order o get P

σ = P/A   ⇒  P =  σ*A = (- 2.123*10⁸Pa)*(8.49*10⁻⁵ m²) = 18035.25 N