Chapter 2, Problem 2.3.7P

-7 Repeat Problem 2.3-5, but n include the weight of the bar. See Table I-I in Appendix I for the weight density of steel.

To determine

The displacements at point $B$ , $C$ and $D$ .

Answer to Problem 2.3.7P

The displacements at point $B$ $1.135\times {10}^{-6}\text{in}$ .

The displacements at point $C$ $2.54\times {10}^{-6}\text{in}$ .

The displacements at point $D$ $-23.1\times {10}^{-6}\text{in}$ .

Explanation of Solution

Given information:

The length from point $A$ to $B$ is $20\text{in}$ , the length from point $B$ to $C$ is $20\text{in}$ , the length from point $C$ to $D$ is $40\text{in}$ , modulus of elasticity is $200\text{GPa}$ , area of bar from point $A$ to $B$ is $6000{\text{mm}}^2$ , area of bar from point $B$ to $C$ is $5000{\text{mm}}^2$ , area of bar from point $C$ to $D$ is $4000{\text{mm}}^2$ .

Write the expression for the elongation at the point $B$ .

  $\begin{array}{c}{\delta }_B=-\frac{P_B{L}_1}{AE}+\frac{P_C{L}_1}{AE}-\frac{P_D{L}_1}{AE}+\frac{\gamma A{L}_1{}^2}{2AE}+\frac{\left(\gamma A{L}_2\right)L_1}{AE}+\frac{\left(\gamma A{L}_3\right)L_1}{AE}\\ =\frac{L_1}{AE}\left(-P_B+P_C-P_D\right)+\frac{\gamma L_1}{E}\left(\frac{L_1}{2}+L_2+L_3\right)\end{array}$  .........(I)

Here, the elongation at the point $B$ is ${\delta }_B$ , the load acting at $B$ is $P_B$ . The load acting at $C$ is $P_C$ , the load acting at $D$ is $P_D$ and the modulus of elasticity is $E$ , the length of segment $AB$ is $L_1$ , the length of segment $BC$ is $L_2,$ and the length of segment $CD$ is $L_3$ .

Write the expression for the elongation at the point C.

  $\begin{array}{c}{\delta }_C={\delta }_B+\frac{P_C{L}_2}{AE}-\frac{P_D{L}_2}{AE}+\frac{\gamma A{L}_2{}^2}{2AE}+\frac{\left(\gamma A{L}_3\right)L_2}{AE}\\ ={\delta }_B+\frac{L_2}{AE}\left(P_C-P_D\right)+\frac{\gamma L_2}{E}\left(\frac{L_2}{2}+L_3\right)\end{array}$   .........(II)

Here, the elongation at the point $C$ is ${\delta }_C$ .

Write the expression for the elongation at the point $D$ .

  ${\delta }_D={\delta }_C-\frac{P_D{L}_3}{AE}+\frac{\gamma A{L}_3{}^2}{2AE}$  .........(III)

Here, the elongation at the point $D$ is ${\delta }_D$ .

Calculation:

Refer to appendix $\text{I}$ table $1.1$ “Weight and mass densities” to obtain weight density of steel.

  $\begin{array}{c}\gamma =490{\text{lb/ft}}^{\text{3}}\\ =490{\text{lb/ft}}^{\text{3}}\times {\left(\frac{1\text{ft}}{12\text{in}}\right)}^3\\ =0.2836{\text{lb/in}}^{\text{3}}\end{array}$

Substitute $20\text{in}$ for $L_1$ , $20\text{in}$ for $L_2$ , $40\text{in}$ for $L_3$ , $50\text{lb}$ for $P_B$ , $100\text{lb}$ for $P_C$ and $200\text{lb}$ for $P_D$ , $8.24{\text{in}}^{\text{2}}$ for $A$ , $29000\text{ksi}$ for $E$ and $0.2836{\text{lb/in}}^{\text{3}}$ for $\gamma$ in Equation (I).

  $\begin{array}{c}{\delta }_B=\left[\begin{array}{l}\frac{20\text{in}}{\left(8.24{\text{in}}^{\text{2}}\right)\left(29000\text{ksi}\right)}\left(-50\text{lb}+100\text{lb}-200\text{lb}\right)+\\ \frac{\left(0.2836{\text{lb/in}}^{\text{3}}\right)20\text{in}}{\left(29000\text{ksi}\right)}\left(\frac{20}{2}\text{in}+20\text{in}+40\text{in}\right)\end{array}\right]\\ =\left[\begin{array}{l}\frac{20\text{in}}{\left(8.24{\text{in}}^{\text{2}}\right)\left(29000\text{ksi}\right)\left(\frac{1000{\text{lb/in}}^{\text{2}}}{\text{1}\text{ksi}}\right)}\left(-50\text{lb}+100\text{lb}-200\text{lb}\right)+\\ \frac{\left(0.2836{\text{lb/in}}^{\text{3}}\right)20\text{in}}{\left(29000\text{ksi}\right)\left(\frac{1000{\text{lb/in}}^{\text{2}}}{\text{1}\text{ksi}}\right)}\left(\frac{20}{2}\text{in}+20\text{in}+40\text{in}\right)\end{array}\right]\\ =\left[\begin{array}{l}\frac{-3000\text{in}\cdot \text{lb}}{\left(8.24{\text{in}}^{\text{2}}\right)\left(29000\times {10}^3\text{lb}\right)}+\\ \frac{397.04\text{lb/in}}{\left(29000\times {10}^3\text{lb}\right)}\end{array}\right]\\ =1.135\times {10}^{-6}\text{in}\end{array}$

Substitute $20\text{in}$ for $L_2$ , $40\text{in}$ for $L_3$ , $100\text{lb}$ for $P_C$ and $200\text{lb}$ for $P_D$ , $8.24{\text{in}}^{\text{2}}$ for A, $29000\text{ksi}$ for E and $0.2836{\text{lb/in}}^{\text{3}}$ for $\gamma$ and $1.135\times {10}^{-6}\text{in}$ for ${\delta }_B$ in Equation (II).

  $\begin{array}{c}{\delta }_C=\left[\begin{array}{l}1.135\times {10}^{-6}\text{in}+\left\{\begin{array}{l}\frac{20\text{in}}{\left(8.24{\text{in}}^{\text{2}}\right)\left(29000\text{ksi}\right)}\\ \times \left(100\text{lb}-200\text{lb}\right)\end{array}\right\}+\\ \left\{\frac{\left(0.2836\text{lb}/{\text{in}}^{\text{2}}\right)20\text{in}}{\left(29000\text{ksi}\right)}\left(\frac{20}{2}\text{in}+40\text{in}\right)\right\}\end{array}\right]\\ =\left[\begin{array}{l}1.135\times {10}^{-6}\text{in}+\left\{\begin{array}{l}\frac{20\text{in}}{\left(8.24{\text{in}}^{\text{2}}\right)\left(29000\text{ksi}\times \frac{\text{1}\text{lb}/{\text{in}}^{\text{2}}}{\text{1}\text{ksi}}\right)}\\ \times \left(100\text{lb}-200\text{lb}\right)\end{array}\right\}+\\ \left\{\frac{\left(0.2836\text{lb}/{\text{in}}^{\text{2}}\right)20\text{in}}{\left(29000\text{ksi}\times \frac{\text{1}\text{lb}/{\text{in}}^{\text{2}}}{\text{1}\text{ksi}}\right)}\left(\frac{20}{2}\text{in}+40\text{in}\right)\right\}\end{array}\right]\\ =\left[\begin{array}{l}1.135\times {10}^{-6}\text{in}+\left\{\frac{20\text{in}}{238960\text{lb}}\times \left(-100\text{lb}\right)\right\}+\\ \left\{\frac{\left(283.6\text{lb}/{\text{in}}^{\text{2}}\right)}{\left(29000\text{lb}/{\text{in}}^{\text{2}}\right)}\right\}\end{array}\right]\\ =2.54\times {10}^{-6}\text{in}\end{array}$

Substitute $40\text{in}$ for $L_3$ , $200\text{lb}$ for $P_D$ , $8.24{\text{in}}^{\text{2}}$ for A, $29000\text{ksi}$ for E and $0.2836{\text{lb/in}}^{\text{3}}$ for $\gamma$ and and $2.54\times {10}^{-6}\text{in}$ for ${\delta }_C$ in Equation (III).

  $\begin{array}{c}{\delta }_D=\left[\begin{array}{l}2.54\times {10}^{-6}\text{in}-\frac{200\text{lb}\times 40\text{in}}{\left(8.24{\text{in}}^{\text{2}}\right)\left(29000\text{ksi}\right)}+\\ \frac{\left(0.2836\text{lb}/{\text{in}}^{\text{2}}\right){\left(40\text{in}\right)}^2}{2{\text{in}}^{\text{2}}\left(29000\text{ksi}\right)}\end{array}\right]\\ =\left[\begin{array}{l}2.54\times {10}^{-6}\text{in}-\frac{200\text{lb}\times 40\text{in}}{\left(8.24{\text{in}}^{\text{2}}\right)\left(29000\text{ksi}\times \frac{\text{1}\text{lb}/{\text{in}}^{\text{2}}}{\text{1}\text{ksi}}\right)}+\\ \frac{\left(0.2836\text{lb}/{\text{in}}^3\right){\left(40\text{in}\right)}^2}{2{\text{in}}^{\text{2}}\left(29000\text{ksi}\times \frac{\text{1}\text{lb}/{\text{in}}^{\text{2}}}{\text{1}\text{ksi}}\right)}\end{array}\right]\\ =\left[\begin{array}{l}2.54\times {10}^{-6}\text{in}-\frac{\text{8000}\text{lb}\cdot \text{in}}{238960\text{lb}}+\\ \frac{11.344\text{lb}\cdot \text{in}}{58000\text{lb}}\end{array}\right]\\ =-23.1\times {10}^{-6}\text{in}\end{array}$

Conclusion:

The displacements at point $B$ is $1.135\times {10}^{-6}\text{in}$ .

The displacements at point $C$ is $2.54\times {10}^{-6}\text{in}$ .

The displacements at point $D$ is $-23.1\times {10}^{-6}\text{in}$ .