Chapter 10, Problem 11P

To determine

Find all the reactions of the beam, draw the shear force and bending moment diagram, and determine the deflection at the midspan of segment BC.

Answer to Problem 11P

The slope at B is $\underset{\_}{\frac{96}{EI}}$.

The slope at C is $\underset{\_}{-\frac{96}{EI}}$.

The end moment of member AB is $\underset{\_}{-48\text{kips-ft}}$.

The end moment of member BA is $\underset{\_}{84\text{kips-ft}}$.

The end moment of member BC is $\underset{\_}{-84\text{kips-ft}}$.

The end moment of member CB is $\underset{\_}{84\text{kips-ft}}$.

The end moment of member CD is $\underset{\_}{84\text{kips-ft}}$.

The end moment of member CD is $\underset{\_}{-48\text{kips-ft}}$.

The deflection under the load is

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.

Determine the fixed end moment of each member as shown below;

$\begin{array}{c}{\text{FEM}}_{AB}=-\frac{wL}{8}\\ =-\frac{30\left(16\right)}{8}\\ =-60\text{kips-ft}\\ {\text{FEM}}_{BA}=\frac{wL}{8}\\ =\frac{30\left(16\right)}{8}\\ =60\text{kips-ft}\end{array}$

$\begin{array}{c}{\text{FEM}}_{CD}=-\frac{wL}{8}\\ =-\frac{30\left(16\right)}{8}\\ =-60\text{kips-ft}\\ {\text{FEM}}_{DC}=\frac{wL}{8}\\ =\frac{30\left(16\right)}{8}\\ =60\text{kips-ft}\end{array}$

$\begin{array}{c}{\text{FEM}}_{BC}=-\frac{w{L}^2}{12}\\ =-\frac{2{\left(24\right)}^2}{12}\\ =-96\text{kips-ft}\\ {\text{FEM}}_{CB}=\frac{w{L}^2}{12}\\ =\frac{2{\left(24\right)}^2}{12}\\ =96\text{kips-ft}\end{array}$

Determine the end moments of each member as shown below;

$\begin{array}{c}{M}_{AB}=\frac{2EI}{L}\left(2{\theta }_A+{\theta }_B\right)+{\text{FEM}}_{AB}\\ =\frac{2EI}{16}\left(2\left(0\right)+{\theta }_B\right)-60\\ =\frac{EI}{8}\left({\theta }_B\right)-60\\ M_{BA}=\frac{2EI}{L}\left(2{\theta }_B+{\theta }_A\right)+{\text{FEM}}_{BA}\\ =\frac{2EI}{16}\left(2{\theta }_B+0\right)+60\\ =\frac{EI}{8}\left(2{\theta }_B\right)+60\end{array}$

Express the relationship between slope at C and B as follows:

${\theta }_C=-{\theta }_B$

Determine the end moments of each member as shown below;

$\begin{array}{c}{M}_{CD}=\frac{2EI}{L}\left(2{\theta }_C+{\theta }_D\right)+{\text{FEM}}_{CD}\\ =\frac{2EI}{16}\left(-2{\theta }_B+0\right)-60\\ =\frac{EI}{8}\left(-2{\theta }_B\right)-60\\ M_{DC}=\frac{2EI}{L}\left(2{\theta }_D+{\theta }_C\right)+{\text{FEM}}_{CD}\\ =\frac{2EI}{16}\left(2\left(0\right)-{\theta }_B\right)+60\\ =\frac{EI}{8}\left(-{\theta }_B\right)+60\end{array}$

$\begin{array}{c}{M}_{BC}=\frac{2EI}{L}\left(2{\theta }_B+{\theta }_C\right)+{\text{FEM}}_{BC}\\ =\frac{2EI}{24}\left(2{\theta }_B-{\theta }_B\right)-96\\ =\frac{E\left(1.5I\right)}{12}\left({\theta }_B\right)-96\\ M_{CB}=\frac{2EI}{L}\left(2{\theta }_C+{\theta }_B\right)+{\text{FEM}}_{CB}\\ =\frac{2E\left(1.5I\right)}{24}\left(-2{\theta }_B+{\theta }_B\right)+96\\ =\frac{E\left(1.5I\right)}{12}\left(-{\theta }_B\right)+96\end{array}$

Apply Equation of equilibrium at joint B;

$\begin{array}{l}{M}_{BA}+M_{BC}\\ \frac{EI}{8}\left(2{\theta }_B\right)+60+\frac{E\left(1.5I\right)}{12}\left({\theta }_B\right)-96=0\\ 0.25EI{\theta }_B+0.125EI{\theta }_B=36\\ {\theta }_B=\frac{96}{EI}\end{array}$

Determine the slope at C using the relation;

$\begin{array}{l}{\theta }_C=-{\theta }_B\\ {\theta }_C=-\frac{96}{EI}\end{array}$

Hence, the slope at B is $\underset{\_}{\frac{96}{EI}}$.

Hence, the slope at C is $\underset{\_}{-\frac{96}{EI}}$.

Determine the end moments of each member as shown below;

$\begin{array}{c}{M}_{AB}=\frac{EI}{8}\left(\frac{96}{EI}\right)-60\\ =-48\text{kips-ft}\\ M_{BA}=\frac{EI}{8}\left(2\left(\frac{96}{EI}\right)\right)+60\\ =84\text{kips-ft}\\ M_{CD}=\frac{EI}{8}\left(-2\left(\frac{96}{EI}\right)\right)-60\\ =-84\text{kips-ft}\\ M_{DC}=\frac{EI}{8}\left(-\frac{96}{EI}\right)+60\\ =48\text{kips-ft}\end{array}$

$\begin{array}{c}{M}_{BC}=\frac{E\left(1.5I\right)}{12}\left(\frac{96}{EI}\right)-96\\ =-84\text{kips-ft}\\ M_{CB}=\frac{E\left(1.5\right)}{12}\left(-\frac{96}{EI}\right)+96\\ =84\text{kips-ft}\end{array}$

Hence, the end moment of member AB is $\underset{\_}{-48\text{kips-ft}}$.

Hence, the end moment of member BA is $\underset{\_}{84\text{kips-ft}}$.

Hence, the end moment of member BC is $\underset{\_}{-84\text{kips-ft}}$.

Hence, the end moment of member CB is $\underset{\_}{84\text{kips-ft}}$.

Hence, the end moment of member CD is $\underset{\_}{84\text{kips-ft}}$.

Hence, the end moment of member CD is $\underset{\_}{-48\text{kips-ft}}$.

Show the free body diagram of support A, span AB, support B, span BC, support C, span CD, and support D as in Figure (1).

Consider span AB;

Determine the shear $V_{BA}$ using the relation;

Take moment about A;

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ -V_{BA}\times 16+\left(30\times 8\right)-48+84=0\\ V_{BA}=\frac{275}{16}\\ V_{BA}=17.3\text{kips}\end{array}$

Determine the shear $V_{AB}$ using the relation;

$\begin{array}{l}{\displaystyle \sum V=0}\\ V_{AB}+V_{BA}-30=0\\ V_{AB}=30-17.3\\ V_{AB}=12.7\text{kips}\end{array}$

Hence, the vertical reaction at A is $\underset{\_}{12.7\text{kips}}$.

Consider span BC;

Determine the shear $V_{CB}$ using the relation;

Take moment about B;

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ -V_{CB}\times 24+\left(2\times 24\times \frac{24}{2}\right)-84+84=0\\ V_{CB}=\frac{576}{24}\\ V_{CB}=24\text{kips}\end{array}$

Determine the shear $V_{BC}$ using the relation;

$\begin{array}{l}{\displaystyle \sum V=0}\\ V_{BC}+V_{CB}-\left(2\times 24\right)=0\\ V_{BC}=48-24\\ V_{BC}=24\text{kips}\end{array}$

Hence, the vertical reaction at support B is $\underset{\_}{41.3\text{kips}}$.

Consider span CD;

Determine the shear $V_{DC}$ using the relation;

Take moment about C;

$\begin{array}{l}{\displaystyle \sum M_C=0}\\ -V_{DC}\times 16+\left(30\times 8\right)-84+48=0\\ V_{DC}=\frac{204}{16}\\ V_{DC}=12.7\text{kips}\end{array}$

Hence, the vertical reaction at D is $\underset{\_}{12.7\text{kips}}$.

Determine the shear $V_{CD}$ using the relation;

$\begin{array}{l}{\displaystyle \sum V=0}\\ V_{DC}+V_{CD}-30=0\\ V_{CD}=30-12.7\\ V_{CD}=17.3\text{kips}\end{array}$

Hence, the vertical reaction at C is $\underset{\_}{41.3\text{kips}}$.

Shear force calculation;

$\begin{array}{l}{V}_A=12.7\text{kips}\\ V_{8\text{ft}\left(\text{left}\right)}=12.7\text{kips}\\ V_{8\text{ft}\left(\text{right}\right)}=12.7-30\\ V_{8\text{ft}\left(\text{right}\right)}=-17.3\text{kips}\\ V_{B\left(\text{left}\right)}=-17.3\text{kips}\\ V_{B\left(\text{right}\right)}=12.7+41.3-30\\ V_{B\left(\text{right}\right)}=24\text{kips}\end{array}$

$\begin{array}{l}{V}_{24\text{ft}}=12.7+41.3-30-\left(2\times 12\right)\\ V_{24\text{ft}}=0\\ V_{C\left(\text{left}\right)}=12.7+41.3-30-\left(2\times 24\right)\\ V_{C\left(\text{left}\right)}=-24\text{kips}\\ V_{C\left(\text{right}\right)}=12.7+41.3+41.3-30-\left(2\times 24\right)\\ V_{C\left(\text{right}\right)}=17.3\text{kips}\end{array}$

$\begin{array}{l}{V}_{48\text{ft}\left(\text{left}\right)}=12.7+41.3+41.3-30-\left(2\times 24\right)\\ V_{48\text{ft}\left(\text{left}\right)}=17.3\text{kips}\\ V_{48\text{ft}\left(\text{right}\right)}=12.7+41.3+41.3-30-\left(2\times 24\right)-30\\ V_{48\text{ft}\left(\text{right}\right)}=-12.7\text{kips}\\ V_{D\left(\text{left}\right)}=12.7+41.3+41.3-30-\left(2\times 24\right)-30\\ V_{D\left(\text{left}\right)}=-12.7\text{kips}\\ V_{D\left(\text{right}\right)}=12.7+41.3+41.3-30-\left(2\times 24\right)-30+12.7\\ V_{D\left(\text{right}\right)}=0\end{array}$

Bending moment calculations;

$\begin{array}{l}{M}_A=-48\text{kips-ft}\\ M_{\text{8}\text{ft}}=-48+\left(12.7\times 8\right)\\ M_{\text{8}\text{ft}}\simeq 54\text{kips-ft}\\ M_B=-48+\left(12.7\times 16\right)-\left(30\times 8\right)\\ M_B=-84.8\text{kips-ft}\\ M_{28\text{ft}}=-48+\left(12.7\times 28\right)-\left(30\times 20\right)-\left(2\times 12\times \frac{12}{2}\right)+\left(41.3\times 12\right)\\ M_{28\text{ft}}=59.2\text{kips-ft}\end{array}$

$\begin{array}{l}{M}_C=-48+84+\left(12.7\times 40\right)-\left(30\times 32\right)-\left(2\times 24\times \frac{24}{2}\right)+\left(41.3\times 24\right)\\ M_C=-1,584+1,499.2\\ M_C=-84.8\text{kips-ft}\\ M_{48\text{ft}}=-48+84+\left(12.7\times 48\right)-\left(30\times 40\right)-\left(2\times 24\times \left(\frac{24}{2}+8\right)\right)+\left(41.3\times 32\right)+\left(41.3\times 8\right)\\ M_{48\text{ft}}=-2,208+2,261.6\\ M_{48\text{ft}}\simeq 54\text{kips-ft}\end{array}$

$\begin{array}{l}{M}_{D\left(\text{left}\right)}=-48\text{kips-ft}\\ M_{D\left(\text{right}\right)}=\left[\begin{array}{l}\left(12.7\times 56\right)-\left(30\times 48\right)-\left(2\times 24\times \left(\frac{24}{2}+16\right)\right)\\ +\left(41.3\times 40\right)+\left(41.3\times 16\right)-\left(30\times 8\right)\end{array}\right]\\ M_{D\left(\text{right}\right)}=-3,024+3,024\\ M_{D\left(\text{right}\right)}=0\end{array}$

Consider a section at a distance of x from support A within 8 ft;

$\begin{array}{l}{M}_x=0\\ -48+12.7x=0\\ x=\frac{48}{12.7}\\ x=3.778\text{ft}\text{from}\text{support}A\end{array}$

Similarly, the bending moment is zero from support D at 3.778 ft.

Consider span AB and DC;

Consider a section at a distance of x (within 16 ft) to determine the where the bending moment is equal to zero;

$\begin{array}{l}{M}_x=0\\ 12.7x-48-30\left(x-8\right)=0\\ 12.7x-48-30x+240=0\\ x=\frac{192}{17.3}\\ x=11.09\text{ft}\text{from}\text{support}A\end{array}$

Similarly, the bending moment is zero from support D at 11.09 ft.

Consider span BC;

Consider a section at a distance of x (within 24 ft) to determine the where the bending moment is equal to zero;

$\begin{array}{l}{M}_x=0\\ 24\times x-84-2\times x\times \frac{x}{2}=0\\ -x^2+24x-84=0\\ x=4.25\text{ft}\text{from}\text{support}B\end{array}$

Similarly, the bending moment is zero from support C at 4.25 ft.

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

The maximum deflection occurs between span B and C at point E.

Determine the deflection at E under the load using the relations;

$\begin{array}{c}{\Delta }_E=\frac{-84\left(4.25\right)}{3E\left(1.5I\right)}\frac{4.25}{4}+\frac{2\left(7.75\right)60}{3E\left(1.5I\right)}\left(12-\frac{3}{8}\times 7.75\right)\\ =\frac{1,795.08}{EI}\end{array}$

Hence, the maximum deflection at E under the load is $\underset{\_}{\frac{1,795.08}{EI}}$.

Show the deflected shape of the frame as in Figure (3).