Chapter 11, Problem 11.1P

To determine

Change in length of steel rod subjected to uniaxial load of 50 lb.

Answer to Problem 11.1P

Estimated change in length is 3.397 × 10-4 in.

Explanation of Solution

Given information:

Material is Steel, Young’s Modulus E_m= 30 x 10⁶ psi, Original Length l = 10 in., Diameter D= ¼ in. = 0.25 in., ForceF = 50 lbf

The Rod Supports the mass so that uniaxial tension is developed. Let Change in length be denoted as dl, we know acceleration due to gravity g = 9.81 m/s² = 386.088 in/s²

Calculate Area of Rod

$\begin{array}{l}A=\frac{\pi \times D^2}{4}\\ \Rightarrow A=\frac{3.14159\times {(1/4)}^2}{4}\\ \Rightarrow A=0.04909i{n}^2\end{array}$

Calculate Stress as

$\begin{array}{l}\sigma =\frac{F(pounds)}{A}\\ \Rightarrow \sigma =\frac{50}{0.04909}\\ \Rightarrow \sigma =1.019\times {10}^{3}psi\end{array}$

Assuming that material is stressed within elastic limit, hence it obeys Hook’s Law. Therefor calculate strain e

$\begin{array}{l}\sigma =E_m\times \epsilon \\ \Rightarrow \epsilon =\frac{\sigma }{E_m}\\ \Rightarrow \epsilon =\frac{1.019\times {10}^3}{30\times {10}^6}\\ \Rightarrow \epsilon =3.397\times {10}^{-5}\end{array}$

Calculate change in length dl$\begin{array}{l}\epsilon =\frac{\delta l}{L}\\ \Rightarrow \delta l=\epsilon \times L\\ \Rightarrow \delta l=3.397\times {10}^{-5}\times 10\\ \Rightarrow \delta l=3.397\times {10}^{-4}in\end{array}$.