Chapter 9, Problem 9.7.4P

-4 A simple beam ABCD has moment of inertia I near the supports and moment of iertia 2I in the middle region, as shown in the figure. A uniform load of intensity q acts over the entire length of the beam.

Determine the quations of the deflection curve for the left-hand half of the beam. Also, find the angle of rotation ${\theta }_A$ at the left-hand support and the deflection

   ${\delta }_{\max }$ at the midpoint.

To determine

The equations of the deflection curve for the left-hand half of the beam, the angle of rotation ${\theta }_A$ at the left-hand support and the deflection ${\delta }_{\max }$ at the midpoint should be determined for given simple beam.

Answer to Problem 9.7.4P

The equations of the deflection curve for the left-hand half of the beam, the angle of rotation ${\theta }_A$ at the left-hand support and the deflection ${\delta }_{\max }$ at the midpoint are as below for given simple beam.

1. $v=\frac{-qx}{768EI}\left(32x^3-64L{x}^2+21L^3\right)\left(0\le x\le \frac{L}{4}\right)$
2. $v=\frac{-q}{12288EI}\left(13L^4+256L^{3}x-512L{x}^3+256x^4\right)\left(\frac{L}{4}\le x\le \frac{L}{2}\right)$
3. ${\theta }_A{}^{\text{'}}=\frac{7q{L}^3}{256EI}(Clockwise)$
4. ${\delta }_{\max }=\frac{31q{L}^4}{4096EI}(\downarrow )$

Explanation of Solution

Given Information:

We have the simple beam ABCD with moment of inertia I near supports and moment of inertia 2I in the middle region as per below figure:

  

For the given simple beam below is the free body diagram.

  

For the above diagram, applying the force equilibrium in horizontal direction, we will get,

  $\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x=0\end{array}$

At the point A, we are applying moment equilibrium,

  $\begin{array}{l}{\displaystyle \sum M_A=0;}\\ D_y\times L-qL\times \frac{L}{2}=0\\ D_y=\frac{qL}{2}\end{array}$

Again we are applying the force equilibrium but in vertical direction,

  $\begin{array}{l}{{\displaystyle \sum F}}_y=0\\ A_y-qL+D_y=0\\ A_y-qL+\frac{qL}{2}=0\\ A_y=\frac{qL}{2}\end{array}$

Now we are taking a section which is at distance of x from point A where $\left(0\le x\le \frac{L}{4}\right)$ , we get below figure:

  

In above section, we are applying moment equilibrium,

  $\begin{array}{l}M+qx\left(\frac{x}{2}\right)-\frac{qL}{2}\left(x\right)=0\\ M=\frac{qL}{2}x-\frac{q}{2}{x}^2\\ EI{v}^{\text{'}\text{'}}=\frac{qL}{2}x-\frac{q}{2}{x}^2(\text{ From moment-curvature relationship)}\end{array}$

We can get the slope equation after integrating the above equation.

  $\begin{array}{l}EI{v}^{\text{'}}=\frac{qL}{2}\frac{x^2}{2}-\frac{q}{2}\frac{x^3}{3}+C_1\\ v^{\text{'}}=\frac{1}{EI}\left(\frac{qL}{4}{x}^2-\frac{q}{6}{x}^3+C_1\right)\end{array}$

Once again taking integration on both sides, we can determine the deflection equation,

  $\begin{array}{l}{v}^{\text{'}}=\frac{1}{EI}\left(\frac{qL}{2}\frac{x^2}{2}-\frac{q}{2}\frac{x^3}{3}+C_1\right)\\ v=\frac{1}{EI}\left(\frac{qL}{4}\frac{x^3}{3}-\frac{q}{6}\frac{x^4}{4}+C_{1}x+C_2\right)\\ v=\frac{1}{EI}\left(\frac{qL}{12}{x}^3-\frac{q}{24}{x}^4+C_{1}x+C_2\right)\end{array}$

Now we are taking a section which is at distance of x from point A where $\left(\frac{L}{4}\le x\le \frac{L}{2}\right)$ , we get below figure:

  

In above section, we are applying moment equilibrium,

  $\begin{array}{l}M+qx\left(\frac{x}{2}\right)-\frac{qL}{2}\left(x\right)=0\\ M=\frac{qL}{2}x-\frac{q}{2}{x}^2\\ E\left(2I\right)v^{\text{'}\text{'}}=\frac{qL}{2}x-\frac{q}{2}{x}^2(\text{ From moment-curvature relationship)}\end{array}$

We can get the slope equation after integrating the above equation.

  $\begin{array}{l}E\left(2I\right)v^{\text{'}\text{'}}=\frac{qL}{2}x-\frac{q}{2}{x}^2\\ E\left(2I\right)v^{\text{'}}=\frac{qL}{2}\frac{x^2}{2}-\frac{q}{2}\frac{x^3}{3}+C_3\\ v^{\text{'}}=\frac{1}{2EI}\left(\frac{qL}{4}{x}^2-\frac{q}{6}{x}^3+C_3\right)\end{array}$

Once again taking integration on both sides, we can determine the deflection equation,

  $\begin{array}{l}{v}^{\text{'}}=\frac{1}{2EI}\left(\frac{qL}{4}{x}^2-\frac{q}{6}{x}^3+C_3\right)\\ v=\frac{1}{2EI}\left(\frac{qL}{4}\frac{x^3}{3}-\frac{q}{6}\frac{x^4}{4}+C_{3}x+C_4\right)\\ v=\frac{1}{2EI}\left(\frac{qL}{12}{x}^3-\frac{q}{24}{x}^4+C_{3}x+C_4\right)\end{array}$

The exaggerated elastic curve for the simple beam ABCD can be represented as below diagram:

  

The boundary condition is applied for deflection equation for $\left(0\le x\le \frac{L}{4}\right)$ where deflection is zero at point A. We will get,

  $\begin{array}{l}v=\frac{1}{EI}\left(\frac{qL}{12}{x}^3-\frac{q}{24}{x}^4+C_{1}x+C_2\right)\\ \text{Putting values of x =0 and v =0}\\ 0=\frac{1}{EI}\left(\frac{qL}{12}{0}^3-\frac{q}{24}{0}^4+C_1\times 0+C_2\right)\\ C_2=0\end{array}$

This concludes that the slope contains value as zero when the maximum deflection of the beam ${\delta }_{\max }$ occurs at $x=\frac{L}{2}$ due to symmetry of loading.

Now applying boundary condition to deflection equation $\left(\frac{L}{4}\le x\le \frac{L}{2}\right)$ where $v^{\text{'}}\left(\frac{L}{2}\right)=0$ ,

  $\begin{array}{l}E\left(2I\right)v^{\text{'}}=\frac{qL}{2}\frac{x^2}{2}-\frac{q}{2}\frac{x^3}{3}+C_3\\ E\left(2I\right)\left(0\right)=\frac{qL}{4}{\left(\frac{L}{2}\right)}^2-\frac{q}{6}{\left(\frac{L}{2}\right)}^3+C_3\\ 0={\frac{qL}{16}}^3-{\frac{qL}{48}}^3+C_3\\ C_3=-{\frac{qL}{24}}^3\end{array}$

We are applying the continuity of slope condition at point B by evaluating the slope equations for segments $\left(0\le x\le \frac{L}{4}\right)$ and $\left(\frac{L}{4}\le x\le \frac{L}{2}\right)$ as follow.

  $\begin{array}{l}{\left(v_B^{\text{'}}\right)}_{left}={\left(v_B^{\text{'}}\right)}_{right}\\ \frac{1}{EI}\left(\frac{qL}{4}{x}^2-\frac{q}{6}{x}^3+C_1\right)=\frac{1}{2EI}\left(\frac{qL}{4}{x}^2-\frac{q}{6}{x}^3+C_3\right)\\ \text{Putting values of x =}\frac{L}{4}and{\text{ C}}_3=\frac{-q{L}^3}{24}\\ \frac{1}{EI}\left(\frac{qL}{4}{\left(\frac{L}{4}\right)}^2-\frac{q}{6}{\left(\frac{L}{4}\right)}^3+C_1\right)=\frac{1}{2EI}\left(\frac{qL}{4}{\left(\frac{L}{4}\right)}^2-\frac{q}{6}{\left(\frac{L}{4}\right)}^3+C_3\right)\\ \left(\frac{q{L}^3}{64}-\frac{q{L}^3}{384}+C_1\right)=\frac{1}{2}\left(\frac{q{L}^3}{64}-\frac{q{L}^3}{384}-\frac{q{L}^3}{24}\right)\\ \frac{5q{L}^3}{384}+C_1=-\frac{11q{L}^3}{768}\\ C_1=-\frac{7q{L}^3}{256}\end{array}$

Now we are applying the continuity of deflection condition at point B by evaluating the slope equations for segments $\left(0\le x\le \frac{L}{4}\right)$ and $\left(\frac{L}{4}\le x\le \frac{L}{2}\right)$ as follow.

  $\begin{array}{l}{\left(v_B^{\text{'}}\right)}_{left}={\left(v_B^{\text{'}}\right)}_{right}\\ \frac{1}{EI}\left(\frac{qL}{12}{x}^3-\frac{q}{24}{x}^4+C_{1}x+C_2\right)=\frac{1}{2EI}\left(\frac{qL}{12}{x}^3-\frac{q}{24}{x}^4+C_{3}x+C_4\right)\\ \text{Putting values of x =}\left(\frac{L}{4}\right){\text{, C}}_2=0,{\text{ C}}_3\text{=}\frac{-q{L}^3}{24}{\text{ and C}}_1=\frac{-7q{L}^3}{256}\\ \frac{1}{EI}\left(\frac{qL}{12}{\left(\frac{L}{4}\right)}^3-\frac{q}{24}{\left(\frac{L}{4}\right)}^4-\frac{7q{L}^3}{256}\left(\frac{L}{4}\right)+0\right)=\frac{1}{2EI}\left(\frac{qL}{12}{\left(\frac{L}{4}\right)}^3-\frac{q}{24}{\left(\frac{L}{4}\right)}^4-\frac{q{L}^3}{24}\left(\frac{L}{4}\right)+C_4\right)\\ \left(\frac{q{L}^4}{768}-\frac{q{L}^4}{6144}-\frac{7q{L}^4}{1024}+0\right)=\frac{1}{2}\left(\frac{q{L}^4}{768}-\frac{q{L}^4}{6144}-\frac{q{L}^4}{96}+C_4\right)\\ -\frac{35q{L}^4}{6144}=\frac{1}{2}\left(-\frac{19q{L}^4}{2048}+C_4\right)\\ C_4=-\frac{13q{L}^4}{6144}\end{array}$

For determination of deflection equation at segment $\left(0\le x\le \frac{L}{4}\right)$ ,

  $\begin{array}{l}v=\frac{1}{EI}\left(\frac{qL}{12}{x}^3-\frac{q}{24}{x}^4+C_{1}x+C_2\right)\\ C_2=0\text{}{\text{ and C}}_1=\frac{-7q{L}^3}{256}\\ v=\frac{1}{EI}\left(\frac{qL}{12}{x}^3-\frac{q}{24}{x}^4-\frac{7q{L}^3}{256}x+0\right)\\ v=\frac{qx}{EI}\left(\frac{L}{12}{x}^2-\frac{1}{24}{x}^3+\frac{-7L^3}{256}x\right)\\ v=\frac{-qx}{768EI}\left(32x^3-64L{x}^2+21L^3\right)\left(0\le x\le \frac{L}{4}\right)\end{array}$

For determination of deflection equation at segment $\left(\frac{L}{4}\le x\le \frac{L}{2}\right)$ ,

  $\begin{array}{l}v=\frac{1}{2EI}\left(\frac{qL}{12}{x}^3-\frac{q}{24}{x}^4+C_{3}x+C_4\right)\\ {\text{C}}_3\text{=}\frac{-q{L}^3}{24}\text{}\text{ and }{C}_4=-\frac{13q{L}^4}{6144}\\ v=\frac{1}{2EI}\left(\frac{qL}{12}{x}^3-\frac{q}{24}{x}^4-\frac{q{L}^3}{24}x-\frac{13q{L}^4}{6144}\right)\\ v=\frac{q}{EI}\left(\frac{L}{24}{x}^3-\frac{1}{48}{x}^4-\frac{L^3}{48}x-\frac{13L^4}{12288}\right)\\ v=\frac{-q}{12288EI}\left(13L^4+256L^{3}x-512L{x}^3+256x^4\right)\left(\frac{L}{4}\le x\le \frac{L}{2}\right)\end{array}$

For angle of rotation at point A using a slope equation for segment $\left(0\le x\le \frac{L}{4}\right)$

  $\begin{array}{l}{v}^{\text{'}}=\frac{1}{EI}\left(\frac{qL}{4}{x}^2-\frac{q}{6}{x}^3+C_1\right)\\ x=0\text{}{\text{ and C}}_1=\frac{-7q{L}^3}{256}\\ {\theta }_A{}^{\text{'}}=\frac{1}{EI}\left(\frac{qL}{4}\times 0^2-\frac{q}{6}\times 0^3-\frac{7q{L}^3}{256}\right)\\ {\theta }_A{}^{\text{'}}=-\frac{7q{L}^3}{256EI}\\ {\theta }_A{}^{\text{'}}=\frac{7q{L}^3}{256EI}(Clockwise)\end{array}$

Finally, the deflection at midpoint of the beam considering the deflection equation for segment $\left(\frac{L}{4}\le x\le \frac{L}{2}\right)$ .

  $\begin{array}{l}v=\frac{-q}{12288EI}\left(13L^4+256L^{3}x-512L{x}^3+256x^4\right)\\ x=\frac{L}{2}\\ {\delta }_{\max }=\frac{-q}{12288EI}\left(13L^4+256L^3\left(\frac{L}{2}\right)-512L{\left(\frac{L}{2}\right)}^3+256{\left(\frac{L}{2}\right)}^4\right)\\ {\delta }_{\max }=\frac{-q{L}^4}{12288EI}\left(93\right)\\ {\delta }_{\max }=\frac{-31q{L}^4}{4096EI}\\ {\delta }_{\max }=\frac{31q{L}^4}{4096EI}(\downarrow )\end{array}$

Conclusion:

The equations of the deflection curve for the left-hand half of the beam, the angle of rotation ${\theta }_A$ at the left-hand support and the deflection ${\delta }_{\max }$ at the midpoint are as below for given simple beam.

  $v=\frac{-qx}{768EI}\left(32x^3-64L{x}^2+21L^3\right)\left(0\le x\le \frac{L}{4}\right)$

  $v=\frac{-q}{12288EI}\left(13L^4+256L^{3}x-512L{x}^3+256x^4\right)\left(\frac{L}{4}\le x\le \frac{L}{2}\right)$

  ${\theta }_A{}^{\text{'}}=\frac{7q{L}^3}{256EI}(Clockwise)$

  ${\delta }_{\max }=\frac{31q{L}^4}{4096EI}(\downarrow )$