Chapter 9, Problem 9.6.10P

-10 The simple beam AB shown in the figure supports two equal concentrated loads P: one acting downward and the other upward.

Determine the angle of rotation

${\theta }_A$ at the left-hand end, the deflection ${\delta }_1$under the downward load, and the deflection ${\delta }_2$ at the midpoint of the beam.

To determine

The equation for the angle of rotation ${\theta }_A$ at the left-hand end, the deflection ${\delta }_1$ under the downward load and the deflection ${\delta }_2$ at the midpoint for a simple beam AB.

Answer to Problem 9.6.10P

The equation for the angle of rotation ${\theta }_A$ at the left-hand end, the deflection ${\delta }_1$ under the downward load and the deflection ${\delta }_2$ at the midpoint are ${\theta }_A=\frac{Pa\left(L-2a\right)\left(L-2a\right)}{6LEI}$ , ${\delta }_1=\frac{P{a}^2{\left(L-2a\right)}^2}{6LEI}$ and ${\delta }_2=0$ for a simple beam AB.

Explanation of Solution

Given Information:

We have,

Length of the beam, AB =L

Concentrated Load, as P

  $\begin{array}{l}AD=CB=a\\ CD=L-2a\end{array}$

We have below diagram for $\frac{M}{EI}$ with respect to simple beam.

  

The diagram for deflection of center is below:

  

The bending moment:

  $\begin{array}{l}{M}_{1}L=\frac{Pa\left(L-2a\right)}{L}\\ So,\frac{M_1}{EI}=\frac{Pa\left(L-2a\right)}{LEI}\end{array}$

The total area A is determined by first and second part of area as $A_1\text{}{\text{and A}}_2$ . With respect to this centroid can be represented as ${\overline{x}}_1\text{ and }{\overline{x}}_2$ .

So, first and second parts of area have the equations below.

  $\begin{array}{l}{A}_1=\frac{1}{2}\left(\frac{M_1}{EI}\right)\left(a\right)\\ A_1=\frac{1}{2}\left(\frac{Pa\left(L-2a\right)}{LEI}\right)\left(a\right)\\ A_1=\left(\frac{P{a}^2\left(L-2a\right)}{2LEI}\right)\\ A_2=\frac{1}{2}\left(\frac{M_1}{EI}\right)\left(\frac{L}{2}-a\right)\\ A_2=\frac{1}{2}\left(\frac{Pa\left(L-2a\right)}{LEI}\right)\left(\frac{L}{2}-a\right)\\ A_2=\left(\frac{Pa{\left(L-2a\right)}^2}{4LEI}\right)\end{array}$

Now, we can determine first moment area point C :

  $C{C}_1={\displaystyle \sum A\overline{x}}$

  $C{C}_1=A_1{\overline{x}}_1+A_2{\overline{x}}_2$

  $C{C}_1=A_1\left(\frac{L}{2}-a+\frac{a}{3}\right)+A_2\left(\frac{2}{3}\right)\left(\frac{L}{2}-a\right)$

  $C{C}_1=\frac{P{a}^2\left(L-2a\right)}{2LEI}\left(\frac{L}{2}-a+\frac{a}{3}\right)+\frac{Pa{\left(L-2a\right)}^2}{4LEI}\left(\frac{2}{3}\right)\left(\frac{L}{2}-a\right)$

  $C{C}_1=\frac{P{a}^2\left(L-2a\right)}{2LEI}\left(\frac{L}{2}-\frac{2a}{3}\right)+\frac{Pa{\left(L-2a\right)}^3}{12LEI}$

  $C{C}_1=\frac{P{a}^2\left(L-2a\right)}{2LEI}\left(\frac{3L-4a}{6}\right)+\frac{Pa{\left(L-2a\right)}^3}{12LEI}$

  $C{C}_1=\frac{P{a}^2\left(L-2a\right)\left(3L-4a\right)}{12LEI}+\frac{Pa{\left(L-2a\right)}^3}{12LEI}$

  $C{C}_1=\frac{Pa\left(L-2a\right)}{12LEI}\left(\frac{a\left(3L-4a\right)}{12}+\frac{{\left(L-2a\right)}^2}{12}\right)$

  $C{C}_1=\frac{Pa\left(L-2a\right)}{12LEI}\left(\frac{3La-4a^2+L^2-4aL+4a^2}{12}\right)$

  $C{C}_1=\frac{Pa\left(L-2a\right)}{12LEI}\left(\frac{L^2-aL}{12}\right)$

  $C{C}_1=\frac{Pa\left(L-2a\right)}{12EI}\left(L-a\right)$

The angle of rotation for the point A:

  $\begin{array}{l}{\theta }_A=\frac{C{C}_1}{\frac{L}{2}}\\ {\theta }_A=\frac{\frac{Pa\left(L-2a\right)}{12EI}\left(L-a\right)}{\frac{L}{2}}\\ {\theta }_A=\frac{Pa\left(L-2a\right)\left(L-2a\right)}{6LEI}\end{array}$

The distance between D and $D_1$ :

  $\begin{array}{l}D{D}_1=\frac{a}{\left(\frac{L}{2}\right)}C{C}_1\\ D{D}_1=\frac{a}{\left(\frac{L}{2}\right)}\frac{Pa\left(L-2a\right)\left(L-2a\right)}{12EI}\\ D{D}_1=\frac{P{a}^2\left(L-a\right)\left(L-2a\right)}{6LEI}\end{array}$

For point D, first moment area would be :

:

  $\begin{array}{l}{D}_1{D}_2=A_1\left(\frac{a}{3}\right)\\ D_1{D}_2=\frac{P{a}^3\left(L-2a\right)}{6LEI}\end{array}$

The deflection at download,

  $\begin{array}{l}{\delta }_1=D{D}_1-D_2{D}_1\\ {\delta }_1=\frac{P{a}^2\left(L-a\right)\left(L-2a\right)}{6LEI}-\frac{P{a}^3\left(L-2a\right)}{6LEI}\\ {\delta }_1=\frac{P{a}^2\left(L-2a\right)}{6LEI}\left[L-a-a\right]\\ {\delta }_1=\frac{P{a}^2{\left(L-2a\right)}^2}{6LEI}\end{array}$

The symmetric deflection at the midpoint is 0.

  ${\delta }_2=0$

Conclusion:

The equation for the angle of rotation ${\theta }_A$ at the left-hand end, the deflection ${\delta }_1$ under the downward load and the deflection ${\delta }_2$ at the midpoint are calculated using the diagram and given information.