Chapter 9, Problem 9.7.5P

A beam ABC has a rigid segment from A to B and a flexible segment with moment of inertia / from B to C(see figure). A concentrated load P acts at point B.

Determine the angle of rotation S_Aof the rigidsegment, the deflection 8_Bat point ß, and the maximum deflection 8.

  

To determine

The angle of rotation ${\theta }_A$ of the rigid segment, the deflection ${\delta }_B$ at point B and the maximum deflection ${\delta }_{\max }$ should be determined for given beam.

Answer to Problem 9.7.5P

The angle of rotation ${\theta }_A$ of the rigid segment, the deflection ${\delta }_B$ at point B and the maximum deflection ${\delta }_{\max }$ are ${\theta }_A=\frac{8P{L}^2}{243EI}$ , ${\delta }_B=\frac{8P{L}^3}{729EI}$ and ${\delta }_{\max }=0.01363\frac{P{L}^3}{EI}$ for given beam.

Explanation of Solution

Given Information:

We have the beam ABC with rigid segment from point A to B and flexible segment with moment of inertia I from point B to C. A concentrated load P acts at Point B as shown in below figure.

  

We have,

Length of the beam as L

Moment of inertia as $BC=I$

Concentrated load at Point B = P

  $AB=\frac{L}{3}\text{ and BC=}\frac{2L}{3}$

The shear force acting over beam can be shown below diagram.

  

The shear force working from point A to B is $v=-\frac{3{\delta }_{B}x}{L}\text{ 0}\le \text{x}\le \frac{L}{3}$

We are taking derivative on both sides, we will get

  $\begin{array}{l}v=-\frac{3{\delta }_{B}x}{L}\\ \frac{dv}{dx}=-\frac{3{\delta }_B}{L}\end{array}$

Now for shear force working from point B to C is below.

  $\begin{array}{l}EI\frac{d^{2}v}{d{x}^2}=\frac{PL}{3}-\frac{Px}{3}\\ \text{ Taking integration ,}\\ \text{EI}\frac{dv}{dx}=\frac{PL}{3}x-\frac{P{x}^2}{6}+C_1\\ \text{Again taking integration on both sides,}\\ \text{EIv}=\frac{PL{x}^2}{6}-\frac{P{x}^2}{18}+C_{1}x+C_2\end{array}$

According to boundary condition at $x=\frac{L}{3}$ ,

  $\begin{array}{l}\frac{dv}{dx}=-\frac{3{\delta }_B}{L}\\ So,\\ \text{EI}\left(-\frac{3{\delta }_B}{L}\right)=\frac{PL\left(\frac{L}{3}\right)}{3}-\frac{P{L}^2/9}{6}+C_1\\ \left(-\frac{3\text{EI}{\delta }_B}{L}\right)=\frac{P{L}^2}{9}-\frac{P{L}^2}{54}+C_1\\ C_1=\frac{P{L}^2}{54}-\frac{P{L}^2}{9}-3\left(\frac{\text{EI}{\delta }_B}{L}\right)\\ C_1=-\frac{5P{L}^2}{54}-3\left(\frac{\text{EI}{\delta }_B}{L}\right)\\ So,\\ \text{EI}\frac{dv}{dx}=\frac{PL}{3}x-\frac{P{x}^2}{6}-\frac{5P{L}^2}{54}-\frac{\text{3EI}{\delta }_B}{L}\end{array}$

In above equation taking integration again,

  $\text{EI}\frac{dv}{dx}=\frac{PL{x}^2}{6}-\frac{P{x}^3}{18}-\frac{5P{L}^2}{54}x-\frac{\text{3EI}{\delta }_B}{L}+C_2$

And at 2^nd Boundary conditions, x = L and v = 0.

  $\begin{array}{l}0=\frac{P{L}^3}{6}-\frac{P{L}^3}{18}-\frac{5P{L}^2}{54}-\text{3EI}{\delta }_B+C_2\\ 0=\frac{9P{L}^3-3P{L}^3-5P{L}^3}{54}-\text{3EI}{\delta }_B+C_2\\ C_2=\frac{-P{L}^2}{54}+\text{3EI}{\delta }_B\\ So,\\ \text{EI}\frac{dv}{dx}=\frac{PL{x}^2}{6}-\frac{P{x}^3}{18}-\frac{5P{L}^2}{54}x-\frac{\text{3EI}{\delta }_B}{L}-\frac{P{L}^2}{54}+\text{3EI}{\delta }_B\end{array}$

And at 3^rd Boundary conditions, $x=\frac{L}{3}$

  $\begin{array}{l}{\left(v_B\right)}_{left}={\left(v_B\right)}_{right}\\ -\frac{3{\delta }_{B}x}{L}=\frac{1}{EI}\left(\frac{PL{x}^2}{6}-\frac{P{x}^3}{18}-\frac{5P{L}^2}{54}x-\frac{\text{3EI}{\delta }_B}{L}-\frac{P{L}^2}{54}+\text{3EI}{\delta }_B\right)\\ -\frac{3{\delta }_{B}x}{L}=\frac{\frac{PL{x}^2}{6}-\frac{P{x}^3}{18}-\frac{5P{L}^2}{54}x-\frac{P{L}^2}{54}}{EI}--\frac{\text{3}{\delta }_{B}x}{L}++\text{3}{\delta }_B\\ \text{3}{\delta }_B=\frac{1}{EI}\left(-PL\frac{{\left(\frac{L}{3}\right)}^2}{6}+\frac{P{\left(\frac{L}{3}\right)}^3}{18}+\frac{5P{L}^2}{54}\left(\frac{L}{3}\right)+\frac{P{L}^2}{54}\right)\\ \text{3}{\delta }_B=\frac{1}{EI}\left(\frac{-P{L}^3}{54}+\frac{P{L}^3}{468}+\frac{5P{L}^3}{162}+\frac{P{L}^3}{54}\right)\\ \text{3}{\delta }_B=\frac{1}{EI}\left(\frac{P{L}^3+15P{L}^3}{486}\right)\\ \text{3}{\delta }_B=\frac{16P{L}^3}{486EI}\\ {\delta }_B=\frac{16P{L}^3}{1458EI}\\ {\delta }_B=\frac{8P{L}^3}{729EI}\end{array}$

  $\begin{array}{l}EIv=\frac{P}{486}\left(7L^3-61L^{2}x+81L{x}^2-27x^3\right)\\ v=\frac{P}{486EI}\left(7L^3-61L^{2}x+81L{x}^2-27x^3\right)\\ \frac{dv}{dx}=\frac{P}{486EI}\left(-61L^2+162Lx+81x^2\right)\end{array}$

For maximum deflection,

  $\begin{array}{l}\frac{dv}{dx}=0\\ -61L^2+162Lx-81x^2=0\\ 81x^2+162LX+61L^2=0\\ x=\frac{162L\pm \sqrt{{\left(162L\right)}^2-4\left(81\right)\left(61L^2\right)}}{2\left(81\right)}\\ x=\frac{162L\pm \sqrt{\left(6480L^2\right)}}{162}\\ x=\frac{162L\pm 36\sqrt{5L}}{162}\\ x=\frac{L}{9}\left(9-2\sqrt{5}\right)\\ x=0.5030962\end{array}$

So we will calculate maximum deflection as below,

  $\begin{array}{l}v=\frac{P}{486EI}\left(C^3-61L^2\left(0.503096L\right)+81L{\left(0.503096L\right)}^2-27{\left(0.503096L\right)}^3\right)\\ v=-0.01363\frac{P{L}^3}{EI}\\ {\delta }_B=\frac{8P{L}^3}{729EI}\end{array}$

The angle of rotation,

  $\begin{array}{l}{A}_1{\theta }_A=\frac{{\delta }_B}{\frac{L}{3}}\\ =\frac{\frac{8P{L}^3}{729EI}}{\frac{L}{3}}\\ =\frac{24P{L}^2}{729EI}\\ {\theta }_A=\frac{8P{L}^2}{243EI}\end{array}$

Deflection,

  $\begin{array}{l}\text{EI}v=\frac{PL{x}^2}{6}-\frac{P{x}^3}{18}-\frac{5P{L}^2}{54}x-\frac{\text{3EI}{\delta }_{B}x}{L}-\frac{P{L}^2}{54}+\text{3EI}{\delta }_B\\ \text{EI}v=\frac{PL{x}^2}{6}-\frac{P{x}^3}{18}-\frac{5P{L}^2}{54}x-\frac{\text{3EI}\left(\frac{8P{L}^3}{729EI}\right)x}{L}-\frac{P{L}^2}{54}+\text{3EI}\left(\frac{8P{L}^3}{729EI}\right)\\ \text{EI}v=\frac{PL{x}^2}{6}-\frac{P{x}^3}{18}-\frac{5P{L}^2}{54}x-\frac{8P{L}^2}{243}x-\frac{P{L}^2}{54}+\frac{P{L}^3}{243}\end{array}$

But the maximum deflection would be,

  $\begin{array}{l}{\delta }_{\max }=-v_{\max }\\ {\delta }_{\max }=0.01363\frac{P{L}^3}{EI}\end{array}$

Conclusion:

The angle of rotation ${\theta }_A$ of the rigid segment, the deflection ${\delta }_B$ at point B and the maximum deflection are calculated by diagram and integration equation.