Chapter 10, Problem 1P

To determine

Find the fixed end moments for the fixed beam.

Answer to Problem 1P

The fixed end moment AB is $\underset{\_}{-\frac{3PL}{16}}$.

The fixed end moment BA is $\underset{\_}{\frac{3PL}{16}}$.

Explanation of Solution

Apply the sign conventions for calculating reactions using the three equations of equilibrium as shown below.

• For summation of forces along x-direction is equal to zero $\left(\sum F_x=0\right)$, consider the forces acting towards right side as positive $\left(\to +\right)$ and the forces acting towards left side as negative $\left(\leftarrow -\right)$.
• For summation of forces along y-direction is equal to zero $\left(\sum F_y=0\right)$, consider the upward force as positive $\left(\uparrow +\right)$ and the downward force as negative $\left(\downarrow -\right)$.
• For summation of moment about a point is equal to zero $\left(\sum M_{\text{at}\text{a}\text{point}}=0\right)$, consider the clockwise moment as positive and the counter clockwise moment as negative.

Show the free body diagram of the beam as in Figure (1).

To draw the bending moment diagram of the given beam, the fixed beam can be converted as simply supported beam. Support A is taken as pinned support and support B is taken as roller support.

Determine vertical reaction at support A;

Take moment about point B;

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ R_A\times L-P\left(\frac{L}{2}+\frac{L}{4}\right)-P\left(\frac{L}{4}\right)=0\\ R_A\times L-\left(\frac{4PL+2PL}{8}\right)-\frac{PL}{4}=0\\ R_A\times L-\left(\frac{3PL}{4}\right)-\frac{PL}{4}=0\end{array}$

$\begin{array}{l}{R}_A\times L-\frac{3PL}{4}-\frac{PL}{4}=0\\ R_A\times L-\frac{4PL}{4}=0\\ R_A=\frac{4PL}{4L}\\ R_A=P\end{array}$

Determine the reaction at support B using the relation;

$\begin{array}{l}{\displaystyle \sum V=0}\\ R_A+R_B-P-P=0\\ R_B=2P-P\\ R_B=P\end{array}$

Determine the bending moment at A;

$\begin{array}{c}{M}_A=-P\left(L\right)+P\left(\frac{L}{2}+\frac{L}{4}\right)+P\left(\frac{L}{4}\right)\\ =-PL+\left(\frac{6PL}{8}\right)+\frac{PL}{4}\\ =-PL+\left(\frac{3PL}{4}+\frac{PL}{4}\right)\\ =-PL+PL\\ =0\end{array}$

Determine the bending moment at a distance of $\left(\frac{L}{4}\right)$ from support A;

$\begin{array}{c}{M}_{\left(\frac{L}{4}\right)}=P\left(\frac{L}{4}\right)\\ =\frac{PL}{4}\end{array}$

Determine the bending moment at $\left(\frac{3L}{4}\right)$ from support A;

$\begin{array}{c}{M}_{\left(\frac{3L}{4}\right)}=P\left(\frac{3L}{4}\right)-P\left(\frac{L}{2}\right)\\ =\frac{3PL}{4}-\frac{PL}{2}\\ =\frac{6PL-4PL}{8}\\ =\frac{2PL}{8}\\ =\frac{PL}{4}\end{array}$

Determine the bending moment at B;

$\begin{array}{c}{M}_B=P\left(L\right)-P\left(\frac{3L}{4}\right)-P\left(\frac{L}{4}\right)\\ =PL-\frac{3PL-PL}{4}\\ =PL-\left(\frac{4PL}{4}\right)\\ =0\end{array}$

Show the bending moment diagram as in Figure (2).

Refer Figure (2),

Determine the area of the moment diagram using the relation;

$\begin{array}{c}{A}_M=\text{Number}\text{of}\text{triangles}\times \text{Area}\text{of}\text{triangle}+\text{Area}\text{of}\text{rectangle}\\ =2\times \left(\frac{1}{2}\times \frac{PL}{4}\times \frac{L}{4}\right)+\left(\frac{PL}{4}\times \frac{L}{2}\right)\\ =\frac{2P{L}^2}{32}+\frac{P{L}^2}{8}\\ =\frac{P{L}^2}{16}+\frac{P{L}^2}{8}\\ =\frac{P{L}^2+2P{L}^2}{16}\\ =\frac{3P{L}^2}{16}\end{array}$

Determine the fixed end moment AB using the formula;

$FE{M}_{AB}=M_{AB}=\frac{2{\left(A_M\overline{x}\right)}_A}{L^2}-\frac{4{\left(A_M\overline{x}\right)}_B}{L^2}$

Here, $\overline{x}$ is the centroid of the area.

Substitute $\frac{3P{L}^2}{16}$ for $A_M$ and $\frac{L}{2}$ for $\overline{x}$.

$\begin{array}{c}FE{M}_{AB}=M_{AB}=\frac{2\left(\frac{3P{L}^2}{16}\left(\frac{L}{2}\right)\right)}{L^2}-\frac{4\left(\frac{3P{L}^2}{16}\left(\frac{L}{2}\right)\right)}{L^2}\\ =\frac{2}{L^2}\left(\frac{3P{L}^2}{16}\times \frac{L}{2}\right)-\frac{4}{L^2}\left(\frac{3P{L}^2}{16}\times \frac{L}{2}\right)\\ =\frac{3PL}{16}-\frac{3PL}{8}\\ =-\frac{3PL}{16}\\ =\frac{3PL}{16}\left(\text{Counterclockwise}\right)\end{array}$

Hence, the fixed end moment AB is $\underset{\_}{\frac{3PL}{16}\left(\text{Counterclockwise}\right)}$.

Determine the fixed end moment BA using the formula;

$FE{M}_{BA}=M_{BA}=\frac{4{\left(A_M\overline{x}\right)}_A}{L^2}-\frac{2{\left(A_M\overline{x}\right)}_B}{L^2}$

Here, $\overline{x}$ is the centroid of the area.

Substitute $\frac{3P{L}^2}{16}$ for $A_M$ and $\frac{L}{2}$ for $\overline{x}$.

$\begin{array}{c}FE{M}_{BA}=M_{BA}=\frac{4\left(\frac{3P{L}^2}{16}\left(\frac{L}{2}\right)\right)}{L^2}-\frac{2\left(\frac{3P{L}^2}{16}\left(\frac{L}{2}\right)\right)}{L^2}\\ =\frac{4}{L^2}\left(\frac{3P{L}^2}{16}\times \frac{L}{2}\right)-\frac{2}{L^2}\left(\frac{3P{L}^2}{16}\times \frac{L}{2}\right)\\ =\frac{3PL}{8}-\frac{3PL}{16}\\ =\frac{3PL}{16}\end{array}$

Hence, the fixed end moment BA is $\underset{\_}{\frac{3PL}{16}\left(\text{clockwise}\right)}$.