Chapter 2, Problem 2.4.7P

A circular bar ACB of a diameter d having a cylindrical hole of length .r and diameter till from A to C is held between rigid supports at A and B. A load P acts at U₂from ends A and B. Assume E is constant.

(a) Obtain formulas for the reactions R, and R_Bat supports A and B. respectively, due to the load P (see figure part a).

(b) Obtain a formula for the displacement S at the point of load application (see figure part a).

(c) For what value of x is R_B= (6/5)?,? (See figure part a.)

(d) Repeat part (a) if the bar is now rotated to a vertical position, load P is removed, and the bar is hanging under its own weight (assume mass density = p). (See figure part b.) Assume that

x = LI2.

  

To determine

The formulas for the reactions at the point A and point B due to the load.

Answer to Problem 2.4.7P

The reaction force at point B is $\frac{-P\left(2x+3L\right)}{2\left(x+3L\right)}$ if $x\le \frac{L}{2}$ .

The reaction force at point A is $\frac{-3\left(PL\right)}{2\left(x+3L\right)}$ if $x\le \frac{L}{2}$ .

The reaction force at point B is $\frac{-2PL}{\left(x+3L\right)}$ if $x\ge \frac{L}{2}$ .

The reaction force at point A is $\frac{-2PL}{\left(x+3L\right)}$ if $x\ge \frac{L}{2}$ .

Explanation of Solution

Given information:

The Diameter of circular bar is $d$ , length of the circular hole is $x$ .

The figure below shows the free body diagram of the bar.

  

  Figure-(1)

Write the expression for the area when $x<\frac{L}{2}$ .

  $\begin{array}{c}{A}_{AC}=\frac{\pi }{4}\left(d^2-\frac{d^2}{4}\right)\\ A_{AC}=\left(\frac{3\pi d^2}{16}\right)\end{array}$   ......(I)

Here, the area of the section AC is $A_{AC}$ and diameter of the bar is $d$ .

Write the expression for the elongation of the bar at point B .

  ${\left({\delta }_B\right)}_1=\frac{Px}{E{A}_{AC}}+\frac{P\left(\frac{L}{2}-x\right)}{E{A}_{CB}}$   ......(II)

Here, load is $P$ , distance is $x$ , length is $L$ , modulus of elasticity is $E,$ and area of the bar CB is $A_{CB}$ .

Write the expression for the area of bar CB when $x<\frac{L}{2}$ .

  $A_{CB}=\frac{\pi d^2}{4}$   ......(III)

Write the expression for the elongation at point B .

  ${\left({\delta }_B\right)}_2=\frac{P\left(\frac{L}{2}\right)}{E{A}_{AC}}$   ......(IV)

Write the expression for the elongation at point B in terms of the reaction force.

  ${\left({\delta }_B\right)}_R=\frac{R_B}{E}\left(\frac{x}{A_{AC}}+\frac{L-x}{A_{CB}}\right)$   ......(V)

Here, the reaction force at point B is $R_B$ and length of the beam from point A is $x$ .

Write the compatibility equation if $x\ge \frac{L}{2}$ .

  ${\left({\delta }_B\right)}_1+{\left({\delta }_B\right)}_R=0$   ......(VI)

Write the expression for the rod held under rigid supports if $x\ge \frac{L}{2}$ .

  ${\left({\delta }_B\right)}_2+{\left({\delta }_B\right)}_R=0$   ......(VII)

Write the expression for the force balance in horizontal direction.

  $R_{A_2}=-R_{B_2}-P$   ......(VIII)

Here, the reaction force at point A is $R_{A_2}$ .

Calculation:

Substitute $\frac{3\pi d^2}{16}$ for $A_{AC}$ and $\frac{\pi d^2}{4}$ for $A_{CB}$ in Equation (II).

  $\begin{array}{c}{\left({\delta }_B\right)}_1=\frac{Px}{E\left(\frac{3\pi d^2}{16}\right)}+\frac{P\left(\frac{L}{2}-x\right)}{E\left(\frac{\pi d^2}{4}\right)}\\ =\frac{P}{E}\left[\frac{16x}{3\pi d^2}+\frac{\left(2L-4x\right)}{\pi d^2}\right]\\ {\left({\delta }_B\right)}_1=\frac{2P}{3E}\left(\frac{2x+3L}{\pi d^2}\right)\end{array}$

Substitute $\frac{3\pi d^2}{16}$ for $A_{AC}$ in Equation (IV).

  $\begin{array}{c}{\left({\delta }_B\right)}_2=\frac{P\left(\frac{L}{2}\right)}{E\left(\frac{3\pi d^2}{16}\right)}\\ {\left({\delta }_B\right)}_2=\frac{8PL}{3E\pi d^2}\end{array}$

Substitute $\frac{3\pi d^2}{16}$ for $A_{AC}$ and $\frac{\pi d^2}{4}$ for $A_{CB}$ in Equation (V).

  $\begin{array}{c}{\left({\delta }_B\right)}_R=\frac{R_B}{E}\left(\frac{x}{\left(\frac{3\pi d^2}{16}\right)}+\frac{L-x}{\left(\frac{\pi d^2}{4}\right)}\right)\\ =\frac{R_B}{E}\left(\frac{16x}{3\pi d^2}+\frac{4\left(L-x\right)}{\pi d^2}\right)\\ {\left({\delta }_B\right)}_R=\frac{4R_B}{E}\left(\frac{x+3L}{3\pi d^2}\right)\end{array}$

Substitute $\frac{4R_B}{E}\left(\frac{x+3L}{3\pi d^2}\right)$ for ${\left({\delta }_B\right)}_R$ and $\frac{2P}{3E}\left(\frac{2x+3L}{\pi d^2}\right)$ for ${\left({\delta }_B\right)}_1$ in Equation (VI).

  $\begin{array}{l}\frac{2P}{3E}\left(\frac{2x+3L}{\pi d^2}\right)+\frac{4R_{B_1}}{E}\left(\frac{x+3L}{3\pi d^2}\right)=0\\ \frac{4R_{B_1}}{E}\left(\frac{x+3L}{3\pi d^2}\right)=-\frac{2P}{3E}\left(\frac{2x+3L}{\pi d^2}\right)\\ R_{B_1}=\left(\frac{2x+3L}{x+3L}\right)\left(\frac{-P}{2}\right)\end{array}$

Substitute $\frac{4R_B}{E}\left(\frac{x+3L}{3\pi d^2}\right)$ for ${\left({\delta }_B\right)}_R$ and $\frac{8PL}{3E\pi d^2}$ for ${\left({\delta }_B\right)}_2$ in Equation (VII).

  $\begin{array}{l}\frac{8PL}{3E\pi d^2}+\frac{4R_{B_2}}{E}\left(\frac{x+3L}{3\pi d^2}\right)=0\\ \frac{4R_{B_2}}{E}\left(\frac{x+3L}{3\pi d^2}\right)=-\frac{8PL}{3E\pi d^2}\\ \left(R_{B_2}\right)=\left(\frac{-2PL}{x+3L}\right)\end{array}$

Substitute $\left(\frac{-2PL}{x+3L}\right)$ for $R_{B_2}$ in Equation (VIII).

  $\begin{array}{c}{R}_{A_2}=-\left(\frac{-2PL}{x+3L}\right)-P\\ =\left(\frac{2PL}{x+3L}\right)-P\\ =P\left(1-\frac{2L}{x+3L}\right)\\ R_{A_2}=-P\left(\frac{x+L}{x+3L}\right)\end{array}$

Conclusion:

The reaction force at point B is $\frac{-P\left(2x+3L\right)}{2\left(x+3L\right)}$ if $x\le \frac{L}{2}$ .

The reaction force at point A is $\frac{-3\left(PL\right)}{2\left(x+3L\right)}$ if $x\le \frac{L}{2}$ .

The reaction force at point B is $\frac{-2PL}{\left(x+3L\right)}$ if $x\ge \frac{L}{2}$ .

The reaction force at point A is $\frac{-2PL}{\left(x+3L\right)}$ if $x\ge \frac{L}{2}$ .

To determine

The formula for the displacement at the point of load.

Answer to Problem 2.4.7P

The displacement at the point of load is $\frac{PL}{E\pi d^2}\left(\frac{2x+3L}{x+3L}\right)$ if $x\le \frac{L}{2}$ .

The displacement at the point of load is $\frac{8P}{7}\left(\frac{L}{E\pi d^2}\right)$ if $x=\frac{L}{2}$ .

The displacement at the point of load is $\frac{8P}{3}\left(\frac{x+L}{x+3L}\right)\frac{L}{E\pi d^2}$ if $x\ge \frac{L}{2}$ .

Explanation of Solution

Write the expression for the displacement at the point of load if $x\le \frac{L}{2}$ .

  ${\left({\delta }_P\right)}_1=\frac{-R_{A_1}}{E}\left(\frac{x}{A_{AC}}+\frac{\left(\frac{L}{2}-x\right)}{A_{CB}}\right)$   ......(IX)

Here, the reaction force at point A is $R_{A_1}$ .

Write the expression for the load at point if $x\ge \frac{L}{2}$ .

  ${\left({\delta }_P\right)}_2=\frac{-R_{A_2}}{E}\left(\frac{\left(\frac{L}{2}\right)}{A_{AC}}\right)$   ......(X)

Here, the reaction force at point A is $R_{A_2}$ .

Calculation:

Substitute $\frac{3\pi d^2}{16}$ for $A_{AC}$ and $\frac{\pi d^2}{4}$ for $A_{CB}$ in Equation (IX).

  $\begin{array}{c}{\left({\delta }_P\right)}_1=\frac{-R_{A_1}}{E}\left(\frac{x}{\left(\frac{3\pi d^2}{16}\right)}+\frac{\left(\frac{L}{2}-x\right)}{\left(\frac{\pi d^2}{4}\right)}\right)\\ =\frac{\left(\frac{-3P}{2}\left(\frac{L}{x+3L}\right)\right)}{E}\left(\frac{16x}{3\pi d^2}+\frac{2L-4x}{\pi d^2}\right)\\ {\left({\delta }_P\right)}_1=\frac{PL}{E\pi d^2}\left(\frac{2x+3L}{x+3L}\right)\end{array}$   ......(XI)

Substitute $\frac{L}{2}$ for $x$ in Equation (X).

  $\begin{array}{c}{\left({\delta }_P\right)}_1=\frac{PL}{E\pi d^2}\left(\frac{2\left(\frac{L}{2}\right)+3L}{\left(\frac{L}{2}\right)+3L}\right)\\ =\frac{PL}{E\pi d^2}\left(\frac{8L}{7L}\right)\\ =\frac{8PL}{7E\pi d^2}\end{array}$

Substitute $\frac{3\pi d^2}{16}$ for $A_{AC}$ and $P\left(\frac{x+L}{x+3L}\right)$ for $R_{A_2}$ in Equation (X).

  $\begin{array}{c}{\left({\delta }_P\right)}_2=\frac{P\left(\frac{x+L}{x+3L}\right)}{E}\left(\frac{\left(\frac{L}{2}\right)}{\left(\frac{3\pi d^2}{16}\right)}\right)\\ =\frac{16P\left(\frac{x+L}{x+3L}\right)}{3\pi d^{2}E}\left(\frac{L}{2}\right)\\ {\left({\delta }_P\right)}_2=\frac{8P}{3}\left(\frac{x+L}{x+3L}\right)\frac{L}{E\pi d^2}\end{array}$   ......(XII)

Substitute $\frac{L}{2}$ for $x$ in Equation (XII).

  $\begin{array}{c}{\left({\delta }_P\right)}_2=\frac{8P}{3}\left(\frac{\left(\frac{L}{2}\right)+L}{\left(\frac{L}{2}\right)+3L}\right)\frac{L}{E\pi d^2}\\ =\frac{8P}{3}\left(\frac{\frac{3L}{2}}{\frac{7L}{2}}\right)\frac{L}{E\pi d^2}\\ {\left({\delta }_P\right)}_2=\frac{8P}{7}\left(\frac{L}{E\pi d^2}\right)\end{array}$   ......(XIII)

Conclusion:

The displacement at the point of load is $\frac{PL}{E\pi d^2}\left(\frac{2x+3L}{x+3L}\right)$ if $x\le \frac{L}{2}$ .

The displacement at the point of load is $\frac{8P}{7}\left(\frac{L}{E\pi d^2}\right)$ if $x=\frac{L}{2}$ .

The displacement at the point of load is $\frac{8P}{3}\left(\frac{x+L}{x+3L}\right)\frac{L}{E\pi d^2}$ if $x\ge \frac{L}{2}$ .

To determine

The value of $x$ .

Answer to Problem 2.4.7P

The value of $x$ is $\frac{3L}{10}$if $x\le \frac{L}{2}$ .

The value of $x$ is $\frac{2L}{3}$ if $x\ge \frac{L}{2}$

Explanation of Solution

Write the expression for the reaction force at B if $x\le \frac{L}{2}$ .

  $R_{B_1}=\left(\frac{6}{5}\right)R_{A_1}$   ......(XIV)

Write the expression for the reaction force at B case if $x\ge \frac{L}{2}$ .

  $R_{B_2}=\left(\frac{6}{5}\right)R_{A_2}$   ......(XV)

Calculation:

Substitute $\frac{-P}{2}\left(\frac{2x+3L}{x+3L}\right)$ for $R_{B_1}$ and $\frac{-3P}{2}\left(\frac{L}{x+3L}\right)$ for $R_{A_1}$ in Equation (XIV).

  $\begin{array}{l}\frac{-P}{2}\left(\frac{2x+3L}{x+3L}\right)=\left(\frac{6}{5}\right)\left(\frac{-3P}{2}\left(\frac{L}{x+3L}\right)\right)\\ \left(2x+3L\right)5=18L\\ x=\frac{3L}{10}\end{array}$

Substitute $\frac{-2PL}{x+3L}$ for $R_{B_2}$ and $P\left(\frac{x+L}{x+3L}\right)$ for $R_{A_2}$ in Equation (XV).

  $\begin{array}{l}\frac{-2PL}{x+3L}=\left(\frac{6}{5}\right)P\left(\frac{x+L}{x+3L}\right)\\ 5L=3x+3L\\ x=\frac{2L}{3}\end{array}$

Conclusion:

The value of $x$ is $\frac{3L}{10}$ if $x\le \frac{L}{2}$ .

The value of $x$ is $\frac{2L}{3}$ if $x\ge \frac{L}{2}$ .

To determine

The formulas for the reactions at the point A and point B due to the load.

Answer to Problem 2.4.7P

The reaction force at point B is $\frac{\rho gL\pi d^2}{8}$ .

The reaction force at point A is $\frac{3\pi d^{2}L\left(\rho g\right)}{32}$ .

Explanation of Solution

Given information:

The bar is placed vertically.

The below figure shows the free body diagram of the bar.

  

  Figure-(2)

Write the compatibility equation if

  ${\left({\delta }_B\right)}_1+{\left({\delta }_B\right)}_R=0$   ......(XVI)

Write the expression for the elongation at point B in terms of the reaction force.

  ${\left({\delta }_B\right)}_R=\frac{R_B}{E}\left(\frac{x}{A_{AC}}+\frac{L-x}{A_{CB}}\right)$   ......(XVII)

Write the expression for the elongation of the bar at point B .

  ${\left({\delta }_B\right)}_1={\displaystyle \underset{0}{\overset{\frac{L}{2}}{\int }}\frac{N_{AC}}{E{A}_{AC}}dh}+{\displaystyle \underset{0}{\overset{\frac{L}{2}}{\int }}\frac{N_{CB}}{E{A}_{CB}}dh}$   ......(XVIII)

Here, the axial stress in section AC is $N_{AC}$ and the axial stress in section CB is $N_{CB}$ .

Write the expression for the axial stress in section AC is $N_{AC}$ .

  $N_{AC}=-\left[\rho g{A}_{CB}\frac{L}{2}+\rho g{A}_{AC}\left(\frac{L}{2}-h\right)\right]$   ......(XIX)

Here, the density is $\rho$ , acceleration due to gravity is $g$ , length is $L,$ and the height is $h$ .

Write the expression for the axial stress in section CB is $N_{CB}$ .

  $N_{CB}=-\rho g{A}_{CB}\left(L-h\right)$   ......(XX)

Write the expression for the elongation of the bar held between rigid bars.

  ${\left({\delta }_B\right)}_R=-{\left({\delta }_B\right)}_1$   ......(XXI)

Write the expression for the reaction at point A .

  $R_A=\left(W_{AC}+W_{CB}\right)-R_B$   ......(XXII)

Here, the weight of the bar of section AC is $W_{AC}$ and the weight of the bar of section CB is $W_{CB}$ .

Write the expression for the weight of the bar of section AC .

  $W_{AC}=\rho g{A}_{AC}\frac{L}{2}$

Write the expression for the weight of the bar of section CB .

  $W_{CB}=\rho g{A}_{CB}\frac{L}{2}$

Substitute $\rho g{A}_{AC}\frac{L}{2}$ for $W_{AC}$ and $\rho g{A}_{CB}\frac{L}{2}$ for $W_{CB}$ in Equation (XXII).

  $R_A=\left(\rho g{A}_{AC}\frac{L}{2}+\rho g{A}_{CB}\frac{L}{2}\right)-R_B$   ......(XXIII)

Calculation:

Substitute $\frac{L}{2}$ for $x$ , $\frac{3\pi d^2}{16}$ for in Equation (XVII).

  $\begin{array}{c}{\left({\delta }_B\right)}_R=\frac{R_B}{E}\left(\frac{\frac{L}{2}}{\left(\frac{3\pi d^2}{16}\right)}+\frac{L-\frac{L}{2}}{\left(\frac{\pi d^2}{4}\right)}\right)\\ =\frac{R_B}{E}\left(\frac{8L}{3\pi d^2}+\frac{4\left(L-\frac{L}{2}\right)}{\pi d^2}\right)\\ {\left({\delta }_B\right)}_R=\frac{14R_{B}L}{E3\pi d^2}\end{array}$

Substitute $\frac{3\pi d^2}{16}$ for $A_{AC}$ and $\frac{\pi d^2}{4}$ for in Equation (XIX).

  $\begin{array}{c}{N}_{AC}=-\left[\rho g\left(\frac{\pi d^2}{4}\right)\frac{L}{2}+\rho g\left(\frac{3\pi d^2}{16}\right)\left(\frac{L}{2}-h\right)\right]\\ =\frac{-1}{8}\rho g\pi d^{2}L-\frac{3}{16}\rho g\pi d^2\left(\frac{L}{2}-h\right)\end{array}$

Substitute $\frac{\pi d^2}{4}$ for in Equation (XX).

  $\begin{array}{c}{N}_{CB}=-\rho g\left(\frac{\pi d^2}{4}\right)\left(L-h\right)\\ =\frac{-1}{4}\rho g\pi d^2\left(L-h\right)\end{array}$

Substitute, $\frac{3\pi d^2}{16}$ for $\frac{-1}{8}\rho g\pi d^{2}L-\frac{3}{16}\rho g\pi d^2\left(\frac{L}{2}-h\right)$ for $N_{AC}$ and $\frac{-1}{4}\rho g\pi d^2\left(L-h\right)$ for $N_{CB}$ in Equation (XVIII).

  ${\left({\delta }_B\right)}_1=\left[\begin{array}{r}{\displaystyle \underset{0}{\overset{\frac{L}{2}}{\int }}\frac{\left(\frac{-1}{8}\rho g\pi d^{2}L-\frac{3}{16}\rho g\pi d^2\left(\frac{L}{2}-h\right)\right)}{E\left(\frac{3\pi d^2}{16}\right)}dh}+\\ {\displaystyle \underset{0}{\overset{\frac{L}{2}}{\int }}\frac{\left(\frac{-1}{4}\rho g\pi d^2\left(L-h\right)\right)}{E{A}_{CB}}dh}\end{array}\right]$   ......(XXIV)

Integrate the Equation (XXIV).

  $\begin{array}{c}{\left({\delta }_B\right)}_1=\left[\begin{array}{r}{\displaystyle \underset{0}{\overset{\frac{L}{2}}{\int }}\frac{\left(\frac{-1}{8}\rho g\pi d^{2}L-\frac{3}{16}\rho g\pi d^2\left(\frac{L}{2}-h\right)\right)}{E\left(\frac{3\pi d^2}{16}\right)}dh}+\\ {\displaystyle \underset{0}{\overset{\frac{L}{2}}{\int }}\frac{\left(\frac{-1}{4}\rho g\pi d^2\left(L-h\right)\right)}{E{A}_{CB}}dh}\end{array}\right]\\ =\left[\begin{array}{c}\frac{16}{E3\pi d^2}\left[\begin{array}{l}{\left(\frac{-1}{8}\rho g\pi d^{2}Lh\right)}_0^{\frac{L}{2}}-\\ \frac{3}{16}\rho g\pi d^2{\left(\frac{Lh}{2}-\frac{h^2}{2}\right)}_0^{\frac{L}{2}}\end{array}\right]+\\ \frac{-1}{4E{A}_{CB}}\left(\rho g\pi d^2{\left(Lh-\frac{h^2}{2}\right)}_0^{\frac{L}{2}}\right)\end{array}\right]\\ =\left(\frac{-11}{24}\rho g\left(\frac{L^2}{E}\right)-\frac{1}{8}\rho g\left(\frac{L^2}{E}\right)\right)\\ =\frac{-7}{12}\rho g\left(\frac{L^2}{E}\right)\end{array}$

Substitute $\frac{14R_{B}L}{E3\pi d^2}$ for ${\left({\delta }_B\right)}_R$ and $\frac{-7}{12}\rho g\left(\frac{L^2}{E}\right)$ for ${\left({\delta }_B\right)}_1$ in Equation (XXI).

  $\begin{array}{l}\frac{14R_{B}L}{E3\pi d^2}=\frac{7}{12}\rho g\left(\frac{L^2}{E}\right)\\ \frac{2R_B}{\pi d^2}=\frac{\rho gL}{4}\\ R_B=\frac{\rho gL\pi d^2}{8}\end{array}$

Substitute $\frac{\rho gL\pi d^2}{8}$ for $R_B$ in Equation (XXIII).

  $R_A=\left(\rho g{A}_{AC}\frac{L}{2}+\rho g{A}_{CB}\frac{L}{2}\right)-\frac{\rho gL\pi d^2}{8}$   ......(XXV)

Substitute $\frac{3\pi d^2}{16}$ for in Equation ......(XXV).

  $\begin{array}{c}{R}_A=\left(\rho g\left(\frac{3\pi d^2}{16}\right)\frac{L}{2}+\rho g\left(\frac{\pi d^2}{4}\right)\frac{L}{2}\right)-\frac{\rho gL\pi d^2}{8}\\ =\left(\rho g\left(\frac{3\pi d^{2}L}{32}\right)+\rho g\left(\frac{\pi d^{2}L}{8}\right)\right)-\frac{\rho gL\pi d^2}{8}\\ =\frac{3\pi d^{2}L\left(\rho g\right)}{32}\end{array}$

Conclusion:

The reaction force at point B is $\frac{\rho gL\pi d^2}{8}$ .

The reaction force at point A is $\frac{3\pi d^{2}L\left(\rho g\right)}{32}$ .