Chapter 9, Problem 9.7.6P

A simple beam ABC has a moment of inertia 1,5 from A to B and A from B to C (see figure). A concentrated load P acts at point B.

Obtain the equations of the deflection curves for both parts of the beam. From the equations, determine the angles of rotation 0_Aand B_cat the supports and the deflection 6_Bat point B.

  

To determine

The equations of the deflection curve for both parts of the beam, the angle of rotation ${\theta }_A\text{ and }{\theta }_C$ at the supports and the deflection ${\delta }_B$ at the point for given simple beam.

Answer to Problem 9.7.6P

The equations of the deflection curve for both parts of the beam, the angle of rotation ${\theta }_A\text{ and }{\theta }_C$ at the supports and the deflection ${\delta }_B$ at the point are as below for given simple beam.

1.   $v=\frac{-2Px}{729EI}\left(19L^2-27x^2\right)\text{ When 0}\le \text{x}\le \frac{L}{3}$
2.   $v=\frac{-P}{1458}\left(-13L^3+175P{L}^{2}x-243L{x}^2+81x^2\right)\text{ When }\frac{L}{3}\le \text{x}\le L$
3.   ${\theta }_A=\frac{38P{L}^2}{729EI}$
4.   ${\theta }_C=\frac{34P{L}^2}{729EI}$
5.   ${\delta }_B=\frac{32P{L}^3}{2187EI}$

Explanation of Solution

Given Information:

We have the simple beam ABC with moment of inertia 1.5I from point A to B and I from point B to C as per shown figure with a concentrated load P acting at point B:

  

The length of beam, L

Moment of inertia between A to B, 1.5I

Moment of inertia between B to C, I

Concentrated Load P =B $AB=\frac{L}{3}\text{}\text{and BC=}\frac{2L}{3}$

For the given simple beam, we have deflective curve as shown below:

  

Now, applying differential equation,

  $\begin{array}{l}E\left(I_{AB}\right)\frac{d^{2}v}{d{x}^2}=M\\ E\left(1.5I\right)\frac{d^{2}v}{d{x}^2}=\frac{2Px}{3}\text{ 0}\le \text{x}\le \frac{L}{3}\\ EI\frac{d^{2}v}{d{x}^2}=\frac{4Px}{9}\text{ 0}\le \text{x}\le \frac{L}{3}\mathrm{.....}(1)\\ EI\frac{d^{2}v}{d{x}^2}=\frac{PL}{3}-\frac{Px}{3}\frac{L}{3}\le \text{x}\le L\mathrm{.....}(2)\end{array}$

Now doing integration of equation (1),

  $\begin{array}{l}EI\frac{d^{2}v}{d{x}^2}=\frac{4Px}{9}\text{ 0}\le \text{x}\le \frac{L}{3}\\ EI\frac{dv}{dx}=\frac{4P{x}^2}{18}{\text{+C}}_1\mathrm{......}\text{(3)}\end{array}$

Performing integration for equation (2),

  $\begin{array}{l}EI\frac{d^{2}v}{d{x}^2}=\frac{PL}{3}-\frac{Px}{3}\\ EI\frac{dv}{dx}=\frac{PL}{3}x-\frac{P{x}^2}{6}{\text{+C}}_2\mathrm{........}\text{(4) }\end{array}$

We are applying boundary conditions at $x=\frac{L}{3}$

  $\begin{array}{l}{\left(\frac{dv}{d{x}_B}\right)}_{left}={\left(\frac{dv}{d{x}_B}\right)}_{right}\\ \frac{4P{x}^2}{18}{\text{+C}}_1=\frac{PL}{3}x-\frac{P{x}^2}{2}{\text{+C}}_2\\ \frac{4P}{18}{\left(\frac{L}{3}\right)}^2{\text{+C}}_1=\frac{PL}{3}\left(\frac{L}{3}\right)-\frac{P{\left(\frac{L}{3}\right)}^2}{2}{\text{+C}}_2\\ {\text{C}}_2={\text{C}}_1+\frac{4P{L}^2}{162}-\frac{P{L}^2}{9}+\frac{P{L}^2}{18}\\ {\text{C}}_2={\text{C}}_1-\frac{11P{L}^2}{162}\mathrm{......}(5)\end{array}$

For equation (3), taking integration,

  $EIv=\frac{4P{x}^3}{54}{\text{+C}}_1{\text{x+C}}_3\text{ 0}\le \text{x}\le \frac{L}{3}$

For equation (4), taking integration,

  $\begin{array}{l}EI\frac{dv}{dx}=\frac{PL}{3}x-\frac{P{x}^2}{6}{\text{+C}}_2\\ EIv=\frac{PL}{6}{x}^4-\frac{P{x}^3}{18}{\text{+C}}_{2}x+C_4\frac{L}{3}\le x\le L\end{array}$

In the above equations, applying boundary conditions at

  $\begin{array}{l}x{\text{ =0, v=0 and C}}_3=0\\ x\text{ =L, v=0 }\\ \end{array}$

So, we will get,

  $\begin{array}{l}0=\frac{P{L}^3}{6}\text{-}\frac{P{L}^3}{18}{\text{+C}}_2{\text{L+C}}_4\\ {\text{C}}_4=-\frac{P{L}^3}{9}-C_{2}L\mathrm{........}(6)\end{array}$

For boundary conditions at $x=\frac{L}{3}$ ,

  $\begin{array}{l}{\left(v_B\right)}_{left}={\left(v_B\right)}_{right}\\ \frac{4P}{54}{\text{+C}}_1\left(\frac{L}{3}\right)=\frac{PL}{6}{\left(\frac{L}{3}\right)}^2-\frac{P{\left(\frac{L}{3}\right)}^3}{18}{\text{+C}}_2\left(\frac{L}{3}\right)+C_4\\ \frac{4P{x}^3}{54}{\text{+C}}_1\left(\frac{L}{3}\right)=\frac{P{L}^3}{54}-\frac{P{L}^3}{486}{\text{+C}}_2\left(\frac{L}{3}\right)+C_4\\ {\text{C}}_{1}L=\frac{10P{L}^3}{243}+{\text{C}}_{2}L+3{\text{C}}_4\mathrm{.........}(7)\end{array}$

With the equation (6) and (7), we will get

  $\begin{array}{l}{\text{C}}_{1}L=\frac{10P{L}^3}{243}+{\text{C}}_{2}L+3{\text{C}}_4\\ {\text{C}}_{1}L=\frac{10P{L}^3}{243}+{\text{C}}_{2}L-\frac{P{L}^3}{3}-3{\text{C}}_{2}L\\ {\text{C}}_{1}L=\frac{-71P{L}^3}{243}{\text{-2C}}_{2}L\\ {\text{C}}_1=\frac{-71P{L}^3}{243}{\text{-2C}}_2\mathrm{........}(8)\end{array}$

From equation (5) and (8), we will get

  $\begin{array}{l}{\text{C}}_1=\frac{-71P{L}^3}{243}{\text{-2C}}_2+\frac{11P{L}^2}{81}\\ {\text{3C}}_1=\frac{-71P{L}^2+33P{L}^2}{243}\\ {\text{3C}}_1=\frac{-38P{L}^2}{243}\\ {\text{C}}_1=\frac{-38P{L}^2}{729}\end{array}$

So,

  $\begin{array}{l}{\text{C}}_1=\frac{-71P{L}^2}{243}{\text{-2C}}_2\\ \frac{-38P{L}^2}{729}=\frac{-71P{L}^2}{243}{\text{-2C}}_2\\ {\text{2C}}_2=\frac{-71P{L}^2}{243}+\frac{38P{L}^2}{729}\\ {\text{2C}}_2=\frac{-213P{L}^2+P{L}^2}{729}\\ {\text{C}}_2=\frac{-175P{L}^2}{1458}\\ {\text{C}}_3=0\\ {\text{C}}_4=\frac{13P{L}^3}{1458}\end{array}$

So,

  $\begin{array}{l}\frac{dv}{dx}=\frac{1}{EI}\left[\frac{4P{x}^2}{18}+C_1\right]\\ \frac{dv}{dx}=\frac{1}{EI}\left[\frac{4P{x}^2}{18}-\frac{38P{L}^2}{729}\right]\\ \frac{dv}{dx}=\frac{1}{EI}\left[\frac{324P{x}^2-76P{L}^2}{1458}\right]\\ \frac{dv}{dx}=-\frac{4P}{1458EI}\left[-81P{x}^2+19L^2\right]\\ \frac{dv}{dx}=-\frac{2P}{729EI}\left[19L^2-81P{x}^2\right]\text{ 0}\le \text{x}\le \frac{L}{3}\end{array}$

When,

  $\begin{array}{l}\text{ 0}\le \text{x}\le \frac{L}{3}\\ \frac{dv}{dx}=\frac{1}{EI}\left[\frac{PLx}{3}-\frac{P{x}^2}{6}+C_2\right]\\ \frac{dv}{dx}=\frac{1}{EI}\left[\frac{PLx}{3}-\frac{P{x}^2}{6}-\frac{175P{L}^2}{1458}\right]\\ \frac{dv}{dx}=\frac{1}{EI}\left[\frac{486PLx-243P{x}^2-175P{L}^2}{1458}\right]\\ \frac{dv}{dx}=-\frac{P}{1458EI}\left[175L^2-486Lx+243x^2\right]\end{array}$

Angle of rotation at point A,

  $\begin{array}{l}{\theta }_A=-{\left(\frac{dv}{dx}\right)}_{x=0}\\ {\theta }_A=\frac{2P}{729EI}\left(19L^2\right)\\ {\theta }_A=\frac{38P{L}^2}{729EI}\end{array}$

Angle of rotation at point C,

  $\begin{array}{l}{\theta }_C=+{\left(\frac{dv}{dx}\right)}_{x=L}\\ {\theta }_C=\frac{P}{1458EI}\left(175L^2-486L^2+243L^2\right)\\ {\theta }_C=\frac{68P{L}^2}{1458EI}\\ {\theta }_C=\frac{34P{L}^2}{729EI}\end{array}$

Deflection of the beam is as follows:

  $\begin{array}{l}v=\frac{-2Px}{729EI}\left(19L^2-27x^2\right)\text{ When 0}\le \text{x}\le \frac{L}{3}\\ \text{When }\frac{L}{3}\le \text{x}\le L\\ v=\frac{1}{EI}\left(\frac{PL}{6}{x}^2-\frac{P{x}^3}{18}{\text{+C}}_{2}x+C_4\right)\\ v=\frac{1}{EI}\left(\frac{PL}{6}{x}^2-\frac{P{x}^3}{18}\text{+}\frac{-175P{L}^2}{1458}x+\frac{13P{L}^3}{1458}\right)\\ v=\frac{-P}{1458}\left(-13L^3+175P{L}^{2}x-243L{x}^2+81x^2\right)\text{ When }\frac{L}{3}\le \text{x}\le L\end{array}$

Deflection at point B,

  $\begin{array}{l}{\delta }_B={\left(v\right)}_{x=\frac{L}{3}}\\ {\delta }_B=\frac{2P\left(\frac{L}{3}\right)}{729EI}\left[19L^2-27\frac{L^2}{9}\right]\\ {\delta }_B=\frac{2PL}{3\left(729EI\right)}\left[19L^2-3L^2\right]\\ {\delta }_B=\frac{2PL}{3\left(729EI\right)}\left[16L^2\right]\\ {\delta }_B=\frac{32P{L}^3}{2187EI}\end{array}$

Conclusion:

The answers are calculated by integration and boundary conditions.