Chapter 10, Problem 10.1P

A $\frac{9}{16}$ - in. - diameter steel rod is tested in tension and elongates 0.00715 in. in a length of 8 in. under a tensile load of 6500 lb. Compute (a) the stress, (b) the strain, and (c) the modulus of elasticity, E, based on this one reading. The proportional limit of this steel is 34,000 psi.

To determine

The stress in the steel rod.

Answer to Problem 10.1P

  $26156.43\text{ Psi}$

Explanation of Solution

Given information:

Diameter of steel rod $(D)=\frac{9}{16}\text{ in}$

Length of rod $(L)=8\text{ in}$

Elongation of rod $(\Delta L)=0.00715\text{ in}$

Tensile load $(P)=6500\text{ lb}$

Proportional limit of steel $(E)=34000\text{ Psi}$

Formula used:

We know that stress is given by,

  $\begin{array}{l}(\sigma )=\frac{Force}{Area}\\ \end{array}$

Calculation:

The stress of the steel rod is follows,

  $\begin{array}{l}\sigma =\frac{Tensile\text{ load(P)}}{Area\text{ of rod(A)}}\\ \text{Area of the rod (A)}=\frac{\pi }{4}{d}^2\\ \Rightarrow \sigma =\frac{6500}{\frac{\pi }{4}\times {\left(\frac{9}{16}\right)}^2}=26156.43\text{ Psi}\end{array}$

Conclusion:

Therefore, the stress in the steel rod is $26156.43\text{ Psi}$

To determine

The strain of the rod.

Answer to Problem 10.1P

  $8.94\times {10}^{-4}$

  $$

Explanation of Solution

Given information:

Diameter of steel rod $(D)=\frac{9}{16}\text{ in}$

Length of rod $(L)=8\text{ in}$

Elongation of rod $(\Delta L)=0.00715\text{ in}$

Tensile load $(P)=6500\text{ lb}$

Proportional limit of steel $(E)=34000\text{ Psi}$

Formula used:

Strain of the rod is given by,

  $(\epsilon )=\frac{\Delta L}{L}$

Calculation:

  $\begin{array}{l}Strain\text{ (}\epsilon \text{)}=\frac{0.00715}{8}\\ \epsilon =8.94\times {10}^{-4}\end{array}$

Conclusion:

Therefore, the strain of the rod is given by $8.94\times {10}^{-4}$ .

To determine

The modulus of Elasticity of the steel rod.

Answer to Problem 10.1P

  $29.29\times {10}^6\text{ Psi}$

  $$

Explanation of Solution

Given information:

Diameter of steel rod $(D)=\frac{9}{16}\text{ in}$

Length of rod $(L)=8\text{ in}$

Elongation of rod $(\Delta L)=0.00715\text{ in}$

Tensile load $(P)=6500\text{ lb}$

Proportional limit of steel $(E)=34000\text{ Psi}$

Formula used:

Modulus of Elasticity is given by,

  $\begin{array}{l}E=\frac{stress}{strain}\\ \end{array}$

Calculation:

The modulus of elasticity of the rod is follows,

  $\begin{array}{l}E=\frac{26156.43}{8.94\times {10}^{-4}}\\ E=29.26\times {10}^6\text{ Psi}\end{array}$

Conclusion:

Therefore, the modulus of elasticity of the steel rod is $29.29\times {10}^6\text{ Psi}$ .