Chapter 2, Problem 2.3.6P

Repeat Problem 2.3-4, but now include the weight of the bar. Sec Table 1.1 in Appendix I for the weight density of steel.

To determine

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

Answer to Problem 2.3.6P

The displacements at point $B$ $\underset{\_}{4.22\times {10}^{-4}\text{mm}}$ .

The displacements at point $C$ $\underset{\_}{6.46\times {10}^{-4}\text{mm}}$ .

The displacements at point $D$ $\underset{\_}{7.84\times {10}^{-4}\text{mm}}$ .

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.

  $\begin{array}{c}{\delta }_B=\left[\begin{array}{l}-\frac{P_B{L}_1}{A_{1}E}+\frac{P_C{L}_1}{A_{1}E}+\frac{P_E{L}_1}{A_{1}E}\\ +\frac{\gamma A_1{L}_1{}^2}{2A_{1}E}+\frac{\left(\gamma A_2{L}_2\right)L_1}{A_{1}E}+\frac{\left(\gamma A_2{L}_3\right)L_1}{A_{1}E}+\frac{\left(\gamma A_3{L}_4\right)L_1}{A_{1}E}\end{array}\right]\\ =\frac{L_1}{A_{1}E}\left(-P_B+P_C+P_E\right)+\frac{\gamma L_1}{A_{1}E}\left(\frac{A_1{L}_1}{2}+A_2{L}_2+A_2{L}_3+A_3{L}_4\right)\end{array}$  .........(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 $E$ is $P_E$ and the modulus of elasticity is $E$ , the length from the point $A$ to point $B$ is $L_1$ , the length from point $B$ to $C$ is $L_2$ , the length from the point $C$ to $D$ is $L_3$ , the length from the point $D$ to $E$ is $L_4$ , the cross-section area of bar from $A$ to $B$ is $A_1$ , the cross-section of bar from point $B$ to $C$ is $A_2,$ and the cross-section area of bar from point $D$ to $E$ is $A_3$ .

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

  $\begin{array}{c}{\delta }_C={\delta }_B+\left[\begin{array}{l}\frac{P_C{L}_2}{A_{2}E}+\frac{P_E{L}_2}{A_{2}E}\\ +\frac{\gamma A_2{L}_3{}^2}{2A_{2}E}+\frac{\left(\gamma A_2{L}_3\right)L_1}{A_{2}E}+\frac{\left(\gamma A_3{L}_4\right)L_1}{A_{2}E}\end{array}\right]\\ ={\delta }_B+\frac{L_2}{A_{2}E}\left(P_C+P_E\right)+\frac{\gamma L_2}{A_{2}E}\left(\frac{A_2{L}_2}{2}+A_2{L}_3+A_3{L}_4\right)\end{array}$  .........(II)

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

  $\begin{array}{c}{\delta }_D={\delta }_C+\left[\begin{array}{l}\frac{P_E{L}_3}{A_{2}E}\\ +\frac{\gamma A_2{L}_3{}^2}{2A_{2}E}+\frac{\left(\gamma A_3{L}_4\right)L_3}{A_{2}E}\end{array}\right]\\ ={\delta }_C+\frac{P_E{L}_3}{A_{2}E}+\frac{\gamma L_3}{A_{2}E}\left(\frac{A_2{L}_3}{2}+A_3{L}_4\right)\end{array}$   .........(III)

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

  ${\delta }_E={\delta }_D+\left[\frac{P_E{L}_4}{A_{3}E}+\frac{\left(\gamma A_3{L}_4\right)L_4}{A_{3}E}\right]$   .........(IV)

Calculation:

Refer to appendix ${\rm I}$ table $1.1$ “Weight and mass densities” to obtain weight density of steel as $77{\text{kN/m}}^{\text{3}}$ .

  $\begin{array}{c}\gamma =77{\text{kN/m}}^{\text{3}}\\ =77{\text{kN/m}}^{\text{3}}\times {\left(\frac{1\text{m}}{1000\text{mm}}\right)}^3\\ =7.7\times {10}^{-5}{\text{N/mm}}^{\text{3}}\end{array}$

Substitute $500\text{mm}$ for $L_1$ , $250\text{mm}$ for $L_2$ , $250\text{mm}$ for $L_3$ , $500\text{mm}$ for $L_4$ , $50\text{N}$ for $P_B$ , $100\text{N}$ for $P_C$ and $250\text{N}$ for $P_C$ , $350\text{N}$ for $P_E$ , $200\text{GPa}$ for E, $6000{\text{mm}}^{\text{2}}$ for $A_1$ , $5000{\text{mm}}^{\text{2}}$ for $A_2$ , $4000{\text{mm}}^{\text{2}}$ for $A_3$ and $7.7\times {10}^{-5}{\text{N/mm}}^{\text{3}}$ for $\gamma$ in Equation (I).

  ${\delta }_B=\left[\begin{array}{l}\left\{\begin{array}{l}\frac{500\text{mm}}{\left(6000{\text{mm}}^{\text{2}}\right)\left(200\text{GPa}\right)}\\ \times \left(-50\text{N}+250\text{N+35}0\text{N}\right)\end{array}\right\}+\\ \left\{\begin{array}{l}\frac{\left(7.7\times {10}^{-5}{\text{N/mm}}^{\text{3}}\right)\times 500\text{mm}}{\left(6000{\text{mm}}^{\text{2}}\right)\left(200\text{GPa}\right)}\\ \times \left(\begin{array}{l}\frac{\left(6000{\text{mm}}^{\text{2}}\right)\times 500\text{mm}}{2}\\ +\left(5000{\text{mm}}^{\text{2}}\times 250\text{N}\right)\\ +\left(5000{\text{mm}}^{\text{2}}\times 250\text{N}\right)\\ +\left(4000{\text{mm}}^{\text{2}}\times 500\text{N}\right)\end{array}\right)\end{array}\right\}\end{array}\right]$

  $\begin{array}{c}=\left[\begin{array}{l}\left\{\begin{array}{l}\frac{500\text{mm}}{\left(6000{\text{mm}}^{\text{2}}\right)\left(200\text{GPa}\right)\left(\frac{1000\text{N}/{\text{mm}}^{\text{2}}}{\text{1}\text{GPa}}\right)}\\ \times \left(-50\text{N}+250\text{N+35}0\text{N}\right)\end{array}\right\}\\ +\left\{\begin{array}{l}\frac{\left(7.7\times {10}^{-5}\text{N}/{\text{mm}}^3\right)\times 500\text{mm}}{\left(6000{\text{mm}}^{\text{2}}\right)\left(200\text{GPa}\right)\left(\frac{1000\text{N}/{\text{mm}}^{\text{2}}}{\text{1}\text{GPa}}\right)}\\ \times \left(\begin{array}{l}\frac{\left(6000{\text{mm}}^{\text{2}}\right)\times 500\text{mm}}{2}\\ +\left(5000{\text{mm}}^{\text{2}}\times 250\text{mm}\right)\\ +\left(5000{\text{mm}}^{\text{2}}\times 250\text{mm}\right)\\ +\left(4000{\text{mm}}^{\text{2}}\times 500\text{mm}\right)\end{array}\right)\end{array}\right\}\end{array}\right]\\ =\left[\frac{275000\text{N}\cdot \text{mm}}{12\times {10}^8\text{N}}+\frac{0.0385\text{N}/{\text{mm}}^{\text{2}}}{12\times {10}^8\text{N}}\times \left(\begin{array}{l}1500000\text{mm+1250000}\text{mm}\\ +1250000\text{mm}+2000000\text{mm}\end{array}\right)\right]\\ =\left[2.291\times {10}^{-4}\text{mm+}\text{N}/{\text{mm}}^{\text{2}}\times \left(60\times {10}^{-5}\text{mm}\right)\right]\\ =4.22\times {10}^{-4}\text{mm}\end{array}$

Substitute $250\text{mm}$ for $L_2$ , $250\text{mm}$ for $L_3$ , $500\text{mm}$ for $L_4$ , $50\text{N}$ for $P_B$ , $100\text{lb}$ for $P_C$ and $250\text{N}$ for $P_C$ , $350\text{N}$ for $P_E$ , $200\text{GPa}$ for E, $5000{\text{mm}}^{\text{2}}$ for $A_2$ , $4000{\text{mm}}^{\text{2}}$ for $A_3$ and $7.7\times {10}^{-5}{\text{N/mm}}^{\text{3}}$ for $\gamma$ in Equation (II).

  ${\delta }_C=4.22\times {10}^{-4}\text{mm}+\left[\begin{array}{l}\frac{250\text{mm}}{\left(5000{\text{mm}}^{\text{2}}\right)\left(200\text{GPa}\right)}\left(250\text{N+35}0\text{N}\right)+\\ \left\{\begin{array}{l}\frac{\left(7.7\times {10}^{-5}\text{N}/{\text{mm}}^3\right)\times 250\text{mm}}{\left(5000{\text{mm}}^{\text{2}}\right)\left(200\text{GPa}\right)}\\ \times \left(\begin{array}{l}\frac{\left(5000{\text{mm}}^{\text{2}}\right)\times 250\text{mm}}{2}\\ +\left(5000{\text{mm}}^{\text{2}}\times 250\text{N}\right)\\ +\left(4000{\text{mm}}^{\text{2}}\times 500\text{N}\right)\end{array}\right)\end{array}\right\}\end{array}\right]$

  $=4.22\times {10}^{-4}\text{mm}+\left[\begin{array}{l}\left\{\begin{array}{l}\frac{250\text{mm}}{\left(5000{\text{mm}}^{\text{2}}\right)\left(200\text{GPa}\right)\left(\frac{1000\text{N}/{\text{mm}}^{\text{2}}}{\text{1}\text{GPa}}\right)}\\ \times \left(250\text{N+35}0\text{N}\right)\end{array}\right\}+\\ \left\{\begin{array}{l}\frac{\left(7.7\times {10}^{-5}{\text{N/mm}}^{\text{3}}\right)\times 250\text{mm}}{\left(5000{\text{mm}}^{\text{2}}\right)\left(200\text{GPa}\right)\left(\frac{1000{\text{N}/{\text{mm}}^{\text{2}}}^{\text{2}}}{\text{1}\text{GPa}}\right)}\\ \times \left(\begin{array}{l}\frac{\left(5000{\text{mm}}^{\text{2}}\right)\times 250\text{mm}}{2}\\ +\left(5000{\text{mm}}^{\text{2}}\times 250\text{N}\right)\\ +\left(4000{\text{mm}}^{\text{2}}\times 500\text{N}\right)\end{array}\right)\end{array}\right\}\end{array}\right]$

  $=4.22\times {10}^{-4}\text{mm}+\left[\frac{150000\text{N}\cdot \text{mm}}{10\times {10}^8\text{N}}+\left\{\begin{array}{l}\frac{0.01925\text{N}/{\text{mm}}^{\text{2}}}{10\times {10}^8\text{N}}\\ \times \left(\begin{array}{l}625000{\text{mm}}^3+\\ 1250000{\text{mm}}^3+\\ 2000000{\text{mm}}^3\end{array}\right)\end{array}\right\}\right]$

  $=4.22\times {10}^{-4}\text{mm}+\left[\frac{150000\text{N}\cdot \text{mm}}{10\times {10}^8\text{N}}+\left\{\begin{array}{l}\frac{0.01925\text{N}/{\text{mm}}^3}{10\times {10}^8\text{N}}\\ \times 3875000{\text{mm}}^3\end{array}\right\}\right]$

  $\begin{array}{c}=4.22\times {10}^{-4}\text{mm+}\frac{150000\text{N}\cdot \text{mm}}{10\times {10}^8\text{N}}+\frac{74593.75\text{N}\cdot \text{mm}}{10\times {10}^8\text{N}}\\ =6.46\times {10}^{-4}\text{mm}\end{array}$

Substitute $250\text{mm}$ for $L_3$ , $500\text{mm}$ for $L_4$ , $350\text{N}$ for $P_E$ , $200\text{GPa}$ for E, $5000{\text{mm}}^{\text{2}}$ for $A_2$ , $4000{\text{mm}}^{\text{2}}$ for $A_3$ and $7.7\times {10}^{-5}\text{N}/{\text{mm}}^{\text{2}}$ for $\gamma$ in Equation (II).

  ${\delta }_D=6.46\times {10}^{-4}\text{mm}+\left[\begin{array}{l}\left\{\begin{array}{l}\frac{250\text{mm}}{\left(5000{\text{mm}}^{\text{2}}\right)\left(200\text{GPa}\right)}\\ \times \left(\text{35}0\text{N}\right)\end{array}\right\}+\\ \left\{\begin{array}{l}\frac{\left(7.7\times {10}^{-5}{\text{N/mm}}^{\text{3}}\right)\times 250\text{mm}}{\left(5000{\text{mm}}^{\text{2}}\right)\left(200\text{GPa}\right)}\\ \times \left(\begin{array}{l}\frac{\left(5000{\text{mm}}^{\text{2}}\right)\times 250\text{mm}}{2}\\ +\left(4000{\text{mm}}^{\text{2}}\times 500\text{N}\right)\end{array}\right)\end{array}\right\}\end{array}\right]$

  $\begin{array}{c}=6.46\times {10}^{-4}\text{mm}+\left[\begin{array}{l}\left\{\begin{array}{l}\frac{250\text{mm}}{\left(5000{\text{mm}}^{\text{2}}\right)\left(200\text{GPa}\right)\left(\frac{1000{\text{N/mm}}^{\text{2}}}{\text{1}\text{GPa}}\right)}\\ \times \left(\text{35}0\text{N}\right)\end{array}\right\}+\\ \left\{\begin{array}{l}\frac{\left(7.7\times {10}^{-5}{\text{N/mm}}^{\text{3}}\right)\times 250\text{mm}}{\left(5000{\text{mm}}^{\text{2}}\right)\left(200\text{GPa}\right)\left(\frac{1000{\text{N/mm}}^{\text{2}}}{\text{1}\text{GPa}}\right)}\\ \times \left(\begin{array}{l}\frac{\left(5000{\text{mm}}^{\text{2}}\right)\times 250\text{mm}}{2}\\ +\left(4000{\text{mm}}^{\text{2}}\times 500\text{mm}\right)\end{array}\right)\end{array}\right\}\end{array}\right]\\ =6.46\times {10}^{-4}\text{mm}+\left[\frac{87500\text{N}\cdot \text{mm}}{10\times {10}^8\text{N}}+\left\{\begin{array}{l}\frac{0.01925{\text{N/mm}}^{\text{2}}}{10\times {10}^8\text{N}}\\ \times \left(625000{\text{mm}}^3+2000000{\text{mm}}^3\right)\end{array}\right\}\right]\\ =6.46\times {10}^{-4}\text{mm}+\left[\frac{87500\text{N}\cdot \text{mm}}{10\times {10}^8\text{N}}+\frac{50531.25\text{N}\cdot \text{mm}}{10\times {10}^8\text{N}}\right]\end{array}$

  $=7.84\times {10}^{-4}\text{mm}$

Conclusion:

The displacements at point $B$ is $\underset{\_}{4.22\times {10}^{-4}\text{mm}}$ .

The displacements at point $C$ is $\underset{\_}{6.46\times {10}^{-4}\text{mm}}$ .

The displacements at point $D$ is $\underset{\_}{7.84\times {10}^{-4}\text{mm}}$ .