Question

cns 220 week #3 – materials

1. a tensile load of 50,000 pounds is applied to a metal bar with a 0.5-inch by 0.5-inch cross-section and a length of 2 inches. under this load, the bar elastically deforms and becomes 2.007 inches. determine the modulus of elasticity (e) of the bar in psi.

Explanation:

Step1: Calculate the cross - sectional area

The cross - section of the bar is a square with side length \(s = 0.5\) inches. The area formula for a square is \(A=s\times s\).
\(A = 0.5\times0.5=\frac{0.5\times0.5}{1}= 0.25\) square inches.

Step2: Calculate the stress (\(\sigma\))

Stress is given by the formula \(\sigma=\frac{F}{A}\), where \(F\) is the force (tensile load) and \(A\) is the cross - sectional area. Here, \(F = 50000\) pounds and \(A=0.25\) square inches.
\(\sigma=\frac{50000}{0.25}=200000\) PSI.

Step3: Calculate the strain (\(\epsilon\))

Strain is the ratio of the change in length (\(\Delta L\)) to the original length (\(L_0\)). The original length \(L_0 = 2\) inches and the new length \(L = 2.007\) inches. So, \(\Delta L=L - L_0=2.007 - 2=0.007\) inches.
The formula for strain is \(\epsilon=\frac{\Delta L}{L_0}\).
\(\epsilon=\frac{0.007}{2}=0.0035\).

Step4: Calculate the modulus of elasticity (\(E\))

The modulus of elasticity is given by the formula \(E = \frac{\sigma}{\epsilon}\). We know that \(\sigma = 200000\) PSI and \(\epsilon=0.0035\).
\(E=\frac{200000}{0.0035}\approx57142857.14\) PSI.

Answer:

The modulus of elasticity of the bar is approximately \(\boldsymbol{5.71\times10^{7}}\) PSI (or \(57142857.14\) PSI).