Chapter 9.4, Problem 83P

Use the method of superposition to solve the following problems and assume that the flexural rigidity El of each beam is constant.

9.83 and 9.84 For the beam shown, determine the reaction at B.

Fig. P9.83

To determine

Find the reaction at point B of the beam using superposition method.

Answer to Problem 83P

The vertical reaction at point B is $\underset{\_}{B_y=\frac{13wL}{32}(\uparrow )}$.

The moment at point B is $\underset{\_}{M_B=\frac{11w{L}^2}{192}\left(\text{Clockwise}\right)}$.

Explanation of Solution

The flexural rigidity of the beam is EI.

Convert the beam into two sections as in Figure 1.

Consider portion AC of the beam:

Show the free-body diagram of the superimposed beam as in Figure 2.

Loading I:

The downward reaction $C_y$ is action at point C.

Refer to case 1 in Appendix D “Beam Deflections and Slopes” in the textbook.

Write the equation for slope and deflection for point load acting in a cantilever beam as follows;

$\begin{array}{l}y=-\frac{P{L}^3}{3EI}\\ \theta =-\frac{P{L}^2}{2EI}\end{array}$

Find the deflection at point C due to vertical reaction as follows;

$\begin{array}{c}{\left(y_C\right)}_I=-\frac{C_y{\left(\frac{L}{2}\right)}^3}{3EI}\\ =-\frac{C_y{L}^3}{24EI}\end{array}$

Find the slope at point C due to vertical reaction as follows;

$\begin{array}{c}{\left({\theta }_C\right)}_I=-\frac{C_y{\left(\frac{L}{2}\right)}^2}{2EI}\\ =-\frac{C_y{L}^2}{8EI}\end{array}$

Loading II:

The counter-clockwise moment is acting at point C.

Refer to case 3 in Appendix D “Beam Deflections and Slopes” in the textbook.

Write the equation for slope and deflection for moment acting in a cantilever beam as follows;

$\begin{array}{l}y=\frac{M{L}^2}{2EI}\\ \theta =\frac{ML}{EI}\end{array}$

Find the deflection at point C due to moment as follows;

$\begin{array}{c}{\left(y_C\right)}_{II}=-\frac{M_C{\left(\frac{L}{2}\right)}^2}{2EI}\\ =-\frac{M_C{L}^2}{8EI}\end{array}$

Find the slope at point C due to moment as follows;

$\begin{array}{c}{\left({\theta }_C\right)}_{II}=\frac{M_C\left(\frac{L}{2}\right)}{EI}\\ =\frac{M_{C}L}{2EI}\end{array}$

Find the deflection at point C as follows;

${\left(y_C\right)}_{AC}={\left(y_C\right)}_I+{\left(y_C\right)}_{II}$

Substitute $-\frac{C_y{L}^3}{24EI}$ for ${\left(y_C\right)}_I$ and $\frac{M_C{L}^2}{8EI}$ for ${\left(y_C\right)}_{II}$.

${\left(y_C\right)}_{AC}=-\frac{C_y{L}^3}{24EI}+\frac{M_C{L}^2}{8EI}$ (1)

Find the slope at point C as follows;

${\left({\theta }_C\right)}_{AC}={\left({\theta }_C\right)}_I+{\left({\theta }_C\right)}_{II}$

Substitute $-\frac{C_y{L}^2}{8EI}$ for ${\left({\theta }_C\right)}_I$ and $\frac{M_{C}L}{2EI}$ for ${\left({\theta }_C\right)}_{II}$.

${\left({\theta }_C\right)}_{AC}=-\frac{C_y{L}^2}{8EI}+\frac{M_{C}L}{2EI}$ (2)

Consider portion CB of the beam:

Show the free-body diagram of the superimposed beam as in Figure 3.

Loading III:

The upward reaction $C_y$ is action at point C.

Refer to case 1 in Appendix D “Beam Deflections and Slopes” in the textbook.

Write the equation for slope and deflection for point load acting in a cantilever beam as follows;

$\begin{array}{l}y=\frac{P{L}^3}{3EI}\\ \theta =-\frac{P{L}^2}{2EI}\end{array}$

Find the deflection at point C due to vertical reaction as follows;

$\begin{array}{c}{\left(y_C\right)}_{III}=\frac{C_y{\left(\frac{L}{2}\right)}^3}{3EI}\\ =\frac{C_y{L}^3}{24EI}\end{array}$

Find the slope at point C due to vertical reaction as follows;

$\begin{array}{c}{\left({\theta }_C\right)}_{III}=-\frac{C_y{\left(\frac{L}{2}\right)}^2}{2EI}\\ =-\frac{C_y{L}^2}{8EI}\end{array}$

Loading IV:

The downward uniformly distributed load (udl) is spread throughout the portion BC.

Refer to case 2 in Appendix D “Beam Deflections and Slopes” in the textbook.

Write the equation for slope and deflection for udl spread throughout the cantilever beam as follows;

$\begin{array}{l}y=-\frac{w{L}^4}{8EI}\\ \theta =\frac{w{L}^3}{6EI}\end{array}$

Find the deflection at point C due to udl as follows;

$\begin{array}{c}{\left(y_C\right)}_{IV}=-\frac{w{\left(\frac{L}{2}\right)}^4}{8EI}\\ =-\frac{w{L}^4}{128EI}\end{array}$

Find the slope at point C due to udl as follows;

$\begin{array}{c}{\left({\theta }_C\right)}_{IV}=\frac{w{\left(\frac{L}{2}\right)}^3}{6EI}\\ =\frac{w{L}^3}{48EI}\end{array}$

Loading IV:

The clockwise moment is acting at point C.

Refer to case 3 in Appendix D “Beam Deflections and Slopes” in the textbook.

Write the equation for slope and deflection for moment acting in a cantilever beam as follows;

$\begin{array}{l}y=\frac{M{L}^2}{2EI}\\ \theta =-\frac{ML}{EI}\end{array}$

Find the deflection at point C due to moment as follows;

$\begin{array}{c}{\left(y_C\right)}_V=\frac{M_C{\left(\frac{L}{2}\right)}^2}{2EI}\\ =\frac{M_C{L}^2}{8EI}\end{array}$

Find the slope at point C due to moment as follows;

$\begin{array}{c}{\left({\theta }_C\right)}_V=-\frac{M_C\left(\frac{L}{2}\right)}{EI}\\ =-\frac{M_{C}L}{2EI}\end{array}$

Find the deflection at point C as follows;

${\left(y_C\right)}_{CB}={\left(y_C\right)}_{III}+{\left(y_C\right)}_{IV}+{\left(y_C\right)}_V$

Substitute $-\frac{C_y{L}^3}{24EI}$ for ${\left(y_C\right)}_{III}$, $-\frac{w{L}^4}{128EI}$ for ${\left(y_C\right)}_{IV}$, and $\frac{M_C{L}^2}{8EI}$ for ${\left(y_C\right)}_V$.

${\left(y_C\right)}_{CB}=\frac{C_y{L}^3}{24EI}-\frac{w{L}^4}{128EI}+\frac{M_C{L}^2}{8EI}$ (3)

Find the slope at point C as follows;

${\left({\theta }_C\right)}_{CB}={\left({\theta }_C\right)}_{III}+{\left({\theta }_C\right)}_{IV}+{\left({\theta }_C\right)}_V$

Substitute $-\frac{C_y{L}^2}{8EI}$ for ${\left({\theta }_C\right)}_{III}$, $\frac{w{L}^3}{48EI}$ for ${\left({\theta }_C\right)}_{IV}$, and $-\frac{M_{C}L}{2EI}$ for ${\left({\theta }_C\right)}_V$.

${\left({\theta }_C\right)}_{CB}=-\frac{C_y{L}^2}{8EI}+\frac{w{L}^3}{48EI}-\frac{M_{C}L}{2EI}$ (4)

The resultant deflection at point C is equal.

${\left(y_C\right)}_{AC}={\left(y_C\right)}_{CB}$

Equate the equations (1) and (3).

$\begin{array}{l}-\frac{C_y{L}^3}{24EI}+\frac{M_C{L}^2}{8EI}=\frac{C_y{L}^3}{24EI}-\frac{w{L}^4}{128EI}+\frac{M_C{L}^2}{8EI}\\ -\frac{2C_y{L}^3}{24EI}=-\frac{w{L}^4}{128EI}\\ C_y=\frac{3wL}{32}\end{array}$

The resultant slope at point C is equal.

${\left({\theta }_C\right)}_{AC}={\left({\theta }_C\right)}_{CB}$

Equate the equations (2) and (4).

$\begin{array}{l}-\frac{C_y{L}^2}{8EI}+\frac{M_{C}L}{2EI}=-\frac{C_y{L}^2}{8EI}+\frac{w{L}^3}{48EI}-\frac{M_{C}L}{2EI}\\ \frac{2M_{C}L}{2EI}=\frac{w{L}^3}{48EI}\\ M_C=\frac{w{L}^2}{48}\end{array}$

Show the free-body diagram of the portion CB as in Figure 4.

Find the vertical reaction at point B by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ C_y-w\left(\frac{L}{2}\right)+B_y=0\\ \frac{3wL}{32}-\frac{16wL}{32}+B_y=0\\ B_y=\frac{13wL}{32}(\uparrow )\end{array}$

Find the moment at point B by taking moment about point B.

$\begin{array}{c}{\displaystyle \sum M_B=0}\\ -\frac{w{L}^2}{48}-\frac{3wL}{32}\left(\frac{L}{2}\right)-M_B+w\left(\frac{L}{2}\right)\left(\frac{L}{4}\right)=0\\ -\frac{w{L}^2}{48}-\frac{3w{L}^2}{64}-M_B+\frac{w{L}^2}{8}=0\\ -\frac{4w{L}^2}{192}-\frac{9w{L}^2}{192}-M_B+\frac{24w{L}^2}{192}=0\end{array}$

$M_B=\frac{11w{L}^2}{192}\left(\text{Clockwise}\right)$

Therefore,

The vertical reaction at point B is $\underset{\_}{B_y=\frac{13wL}{32}(\uparrow )}$.

The moment at point B is $\underset{\_}{M_B=\frac{11w{L}^2}{192}\left(\text{Clockwise}\right)}$.