Question

A 10 cm long bar, measuring 1 cm in diameter is loaded under tension. Assuming yield occurs at a load of 500 N, which corresponds to a 0.7% elongation, determine the modulus of resilience of this bar.

Answer 1

Answer:

$2.23*10^{4}\frac{N}{m^{2} }$

Explanation:

Modulus of resilience is the maximum amount of strain that an elastic material can support per unit volume, without deformation, and is calculated using the following equation:

μ = σ^2 ÷ 2*E

σ = yield strain= force / cross area

force = 500N; area=π*$r^{2}$= ${7.8540^{-5}m^{2}$

σ = $\frac{500N}{7.8540^{-5}m^{2} } =6.3662*10^{6} \frac{N}{m^{2} }$

E= young modulus: relation between stress and strain, measures stiffness

E=σ/∈, where

∈=(L-Lo)/Lo=7*$10^{-3}$

where

L=current length = 10 cm * 1.007 = 1.0070*$10^{-1}$ m

Lo=original lenght = 10 cm = 1.0*$10^{-1}$m

so, E=σ/∈ = $9.0094*10^{8}$

μ = modulus of resilience = $({6.3662*10^{6}})^{2} / 2*9.094*10^{8} = 2.2283*10^{4} \frac{N}{m^{2} }$