Chapter 9.6, Problem 152P

To determine

The reaction $\left(R_A\right)$, $\left(R_B\right)$ and $\left(R_c\right)$ at each support.

Answer to Problem 152P

The reaction $\left(R_A\right)$ at the support A is $\underset{\_}{\frac{3P}{32}\left(\text{downward}\right)}$.

The reaction $\left(R_B\right)$ at the support B is $\underset{\_}{\frac{13}{32}P\left(\text{upward}\right)}$.

The reaction $\left(R_C\right)$ at the support C is $\underset{\_}{\frac{11}{16}P\left(\text{upward}\right)}$.

Explanation of Solution

Calculation:

Let, choose reaction $R_C$ as redundant.

Show the free body diagram of the beam in below Figure 1.

Calculate the vertical reaction $\left(R_{B1}\right)$ at point B by taking moment about A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ R_{B1}\times \left(2L\right)+R_C\left(L\right)=0\\ 2R_{B1}L+R_{C}L=0\\ 2R_{B1}L+R_{C}L=\frac{-R_C}{2}\\ 2R_{B1}L+R_{C}L=\frac{-R_C}{2}\end{array}$

Calculate the vertical reaction $\left(R_{A1}\right)$ at point A by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ R_{A1}+R_C-\frac{R_C}{2}=0\\ R_{A1}=-\frac{R_C}{2}\end{array}$

Show the free body diagram of the beam in below Figure 2.

Calculate the vertical reaction $\left(R_{B2}\right)$ at point B by taking moment about A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ R_{B2}\times \left(2L\right)-P\left(\frac{3}{2}L\right)=0\\ R_{B2}\times \left(2L\right)=P\left(\frac{3}{2}L\right)\\ R_{B2}=\frac{\left(\frac{3PL}{2}\right)}{2L}\\ R_{B2}=\frac{3P}{4}\end{array}$

Calculate the vertical reaction $\left(R_{A2}\right)$ at point A by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ R_{A2}-P+\frac{3P}{4}=0\\ R_{A2}-\frac{P}{4}=0\\ R_{A2}=\frac{P}{4}\end{array}$

Calculate the moment $\left(M_1\right)$ due to reaction $R_{A2}$:

$\begin{array}{c}{M}_1=\frac{3P}{4}\times \frac{L}{2}\\ =\frac{3PL}{8}\end{array}$

Calculate the $\left(\frac{M_1}{EI}\right)$ value:

Substitute $\frac{3PL}{8}$ for $M_1$.

$\begin{array}{c}\frac{M_1}{EI}=\frac{\left(\frac{3PL}{8}\right)}{EI}\\ =\frac{3PL}{8EI}\end{array}$

Show the $\frac{M_1}{EI}$ diagram in below Figure 3.

Calculate the area $\left(A_1\right)$ using the relation:

$A_1=\frac{1}{2}\left(L+\frac{L}{2}\right)\left(\frac{M_1}{EI}\right)$

Substitute $\frac{3PL}{8EI}$ for $\left(\frac{M_1}{EI}\right)$.

$\begin{array}{c}{A}_1=\frac{1}{2}\left(L+\frac{L}{2}\right)\left(\frac{3PL}{8EI}\right)\\ =\frac{1}{2}\left(\frac{2L}{3}\right)\left(\frac{3PL}{8EI}\right)\\ =\frac{6P{L}^2}{48EI}\\ =\frac{P{L}^2}{8EI}\end{array}$

Calculate the area $\left(A_1+A_2\right)$ using the relation:

$\left(A_1+A_2\right)=\frac{1}{2}\left(L+\frac{L}{2}\right)\left(\frac{M_2}{EI}\right)$

Substitute $\frac{3PL}{8EI}$ for $\left(\frac{M_2}{EI}\right)$.

$\begin{array}{c}\left(A_1+A_2\right)=\frac{1}{2}\left(L+\frac{L}{2}\right)\left(\frac{3PL}{8EI}\right)\\ =\frac{9P{L}^2}{32EI}\end{array}$

Calculate the moment $\left(M_2\right)$ due to reaction $R_{B2}$:

$M_2=R_{B2}\times \left(L+\frac{L}{2}\right)$

Substitute $\frac{P}{4}$ for $\left(\frac{M_1}{EI}\right)$.

$\begin{array}{c}{M}_2=\frac{P}{4}\times \left(L+\frac{L}{2}\right)\\ =\frac{3PL}{8}\end{array}$

Calculate the $\frac{M_2}{EI}$ value:

Substitute $\frac{3PL}{8}$ for $M_2$.

$\begin{array}{c}\frac{M_2}{EI}=\frac{\frac{3PL}{8}}{EI}\\ =\frac{3PL}{8EI}\end{array}$

Show the $\frac{M_2}{EI}$ diagram in below Figure 4.

Calculate the area $\left(A_3\right)$ using the relation:

$\left(A_3\right)=\frac{1}{2}\left(\frac{L}{2}\right)\left(\frac{M_2}{EI}\right)$

Substitute $\frac{3PL}{8EI}$ for $\left(\frac{M_2}{EI}\right)$.

$\begin{array}{c}\left(A_3\right)=\frac{1}{2}\left(\frac{L}{2}\right)\left(\frac{3PL}{8EI}\right)\\ =\frac{3P{L}^2}{32EI}\end{array}$

Calculate the moment $\left(M_3\right)$ to reaction $R_{A1}$:

$\begin{array}{c}{M}_3=-\frac{R_C}{2}\times L\\ =-\frac{R_{C}L}{2}\end{array}$

Calculate the $\frac{M_3}{EI}$ value:

Substitute $\left(-\frac{R_{C}L}{2}\right)$ for $M_3$.

$\begin{array}{c}\frac{M_3}{EI}=\frac{\left(-\frac{R_{C}L}{2}\right)}{EI}\\ =-\frac{R_{C}L}{2EI}\end{array}$

Show the $\frac{M_3}{EI}$ diagram in below Figure 5.

Calculate the area $\left(A_4\right)$ using the relation:

$A_4=\frac{1}{2}L\left(\frac{M_3}{EI}\right)$

Substitute $-\frac{R_{C}L}{2EI}$ for $\left(\frac{M_3}{EI}\right)$.

$\begin{array}{c}{A}_4=\frac{1}{2}L\left(-\frac{R_{C}L}{2EI}\right)\\ =-\frac{R_C{L}^2}{4EI}\end{array}$

Calculate the area $\left(A_4+A_5\right)$ using the relation:

$\left(A_4+A_5\right)=\frac{1}{2}\left(2L\right)\left(\frac{M_3}{EI}\right)$

Substitute $-\frac{R_{C}L}{2EI}$ for $\left(\frac{M_3}{EI}\right)$.

$\begin{array}{c}\left(A_4+A_5\right)=\frac{1}{2}\left(2L\right)\left(-\frac{R_{C}L}{2EI}\right)\\ =-\frac{R_C{L}^2}{2EI}\end{array}$

Calculate the tangential deviation of A with respect to B using the relation:

$t_{B/A}=\left(A_1+A_2\right)\left(\left(\frac{1}{3}\left(L+\frac{L}{2}\right)\right)+\frac{L}{2}\right)+A_3\left(\frac{2}{3}\frac{L}{2}\right)+\left(A_4+A_5\right)\left(\frac{1}{2}2L\right)$

Substitute $\frac{9P{L}^2}{32EI}$ for $\left(A_1+A_2\right)$, $\frac{3P{L}^2}{32EI}$ for $A_3$ and $\left(-\frac{R_C{L}^2}{2EI}\right)$ for $\left(A_4+A_5\right)$.

$\begin{array}{c}{t}_{B/A}=\left(\frac{9P{L}^2}{32EI}\right)\left(\left(\frac{1}{3}\left(L+\frac{L}{2}\right)\right)+\frac{L}{2}\right)+\left(\frac{3P{L}^2}{32EI}\right)\left(\frac{2}{3}\frac{L}{2}\right)+\left(-\frac{R_C{L}^2}{2EI}\right)\left(\frac{1}{2}2L\right)\\ =\frac{9P{L}^3}{32EI}+\frac{P{L}^3}{32EI}-\frac{R_C{L}^2}{2EI}\end{array}$

Calculate the tangential deviation of C with respect to A using the relation:

$t_{C/A}=A_1\left(\frac{1}{3}L\right)+A_4\left(\frac{1}{3}L\right)$

Substitute $\frac{P{L}^2}{8EI}$ for $A_1$ and $-\frac{R_C{L}^2}{4EI}$ for $A_2$.

$\begin{array}{c}{t}_{C/A}=\frac{P{L}^2}{8EI}\left(\frac{1}{3}L\right)-\frac{R_C{L}^2}{4EI}\left(\frac{1}{3}L\right)\\ =\frac{P{L}^3}{24EI}-\frac{R_C{L}^3}{12EI}\end{array}$

Express the deflection $y_C$ equation as below;

$\begin{array}{l}{y}_C=y_A+L{\theta }_A+t_{B/A}\\ 0=0+L{\theta }_A+t_{C/A}\\ {\theta }_A=\frac{-t_{C/A}}{L}\end{array}$

Express the deflection $y_C$ equation as below;

$y_B=2L{\theta }_A+t_{B/A}$

Substitute $\frac{-t_{C/A}}{L}$ for ${\theta }_A$.

$\begin{array}{l}{y}_B=2L\frac{-t_{C/A}}{L}+t_{B/A}\\ 0=-2t_{C/A}+t_{B/A}\\ -2t_{C/A}+t_{B/A}=0\end{array}$ (1).

Calculate the reaction $\left(R_B\right)$ at the point B using the Equation (1).

$-2L\frac{-t_{C/A}}{L}+t_{B/A}=0$

Substitute $\frac{9P{L}^3}{32EI}+\frac{P{L}^3}{32EI}-\frac{R_C{L}^2}{2EI}$ for $t_{B/A}$ and $\frac{P{L}^3}{24EI}-\frac{R_C{L}^3}{12EI}$ for $t_{C/A}$.

$\begin{array}{l}-2\left(\frac{P{L}^3}{24EI}-\frac{R_C{L}^3}{12EI}\right)+\left(\frac{9P{L}^3}{32EI}+\frac{P{L}^3}{32EI}-\frac{R_C{L}^2}{2EI}\right)=0\\ -\frac{P{L}^3}{12EI}+\frac{R_C{L}^3}{6EI}+\frac{9P{L}^3}{32EI}+\frac{P{L}^3}{32EI}-\frac{R_C{L}^2}{2EI}=0\\ \frac{11P{L}^3}{48EI}-\frac{R_C{L}^3}{3EI}=0\end{array}$

$\begin{array}{l}\frac{R_C{L}^3}{3EI}=\frac{11P{L}^3}{48EI}\\ R_C=\frac{11P{L}^3}{48EI}\times \frac{3EI}{L^3}\\ R_C=\frac{11}{16}P\end{array}$

Calculate the reaction $\left(R_A\right)$ at the point A by taking the moment about B.

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ -2L{R}_A-L{R}_C+\frac{L}{2}P=0\end{array}$

Substitute $\frac{11}{16}P$ for $R_C$.

$\begin{array}{l}-2L{R}_A-L\left(\frac{11}{16}P\right)+\frac{L}{2}P=0\\ -2L{R}_A-\frac{3PL}{16}=0\\ -2L{R}_A=\frac{3PL}{16}\end{array}$

$\begin{array}{l}{R}_A=\frac{\left(\frac{3PL}{16}\right)}{-2L}\\ R_A=-\frac{3}{32}P\\ R_A=\frac{3}{32}P\left(\text{downward}\right)\end{array}$

Calculate the reaction $\left(R_B\right)$ at the point B by taking the moment about A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ 2L{R}_B+L{R}_C-\frac{3L}{2}P=0\end{array}$

Substitute $\frac{11}{16}P$ for $R_C$.

$\begin{array}{l}2L{R}_B+L\left(\frac{11}{16}P\right)-\frac{3L}{2}P=0\\ 2L{R}_B=\frac{13PL}{16}\\ R_B=\frac{\left(\frac{13PL}{16}\right)}{2L}\\ R_B=\frac{13}{32}P\end{array}$

Thus, the reaction $\left(R_A\right)$ at the support A is $\underset{\_}{\frac{3P}{32}\left(\text{downward}\right)}$.

Thus, the reaction $\left(R_B\right)$ at the support B is $\underset{\_}{\frac{13}{32}P\left(\text{upward}\right)}$.

Thus, the reaction $\left(R_C\right)$ at the support C is $\underset{\_}{\frac{11}{16}P\left(\text{upward}\right)}$.