Chapter 13, Problem 45P

To determine

Calculate the support reactions for the given structure using method of consistent deformation.

Sketch the shear and bending moment diagrams for the given structure.

Answer to Problem 45P

The horizontal reaction at B is $\underset{\_}{B_x=15.71\text{k}\leftarrow }$.

The vertical reaction at B is $\underset{\_}{B_y=36.75\text{k}\uparrow }$.

The moment at A is $\underset{\_}{M_B=222.1\text{k-ft}\circlearrowleft }$ acting in the anticlockwise direction.

The horizontal reaction at A is $\underset{\_}{A_x=4.29\text{k}\leftarrow }$.

The vertical reaction at A is $\underset{\_}{A_y=23.25\text{k}\uparrow }$.

The moment at A is $\underset{\_}{M_A=107.9\text{k-ft}\circlearrowleft }$ acting in the anticlockwise direction.

Explanation of Solution

Given information:

The structure is given in the Figure.

Apply the sign conventions for calculating reactions, forces and moments 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 when moment taken from left and the counter clockwise moment as positive when moment taken from right.
• Consider the positive sign indicates the clockwise moment the negative sign indicates the counterclockwise moment.

Calculation:

Find the degree of indeterminacy of the structure:

Degree of determinacy of the frame is equal to the number of unknown reactions minus the number of equilibrium equations.

The frame is supported by 6 support reactions and the number of equilibrium equations is 3.

Therefore, the degree of indeterminacy of the frame is $i=3$.

Select the reaction $B_x$, $B_y$ and $M_B$ at the support B as redundant.

Release the support B and consider the notation of moments due to external load as $M_O$.

Sketch the free body diagram of primary frame subjected to external loading without support B as shown in Figure 1.

Find the reactions at the supports without considering support D using equilibrium equations:

Summation of moments of all forces about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ M_{AO}=-20\left(30\right)-1.5\left(40\right)\left(\frac{40}{2}\right)\\ M_{AO}=-1,800\text{k-ft}\\ M_{AO}=1,800\text{k-ft}(\circlearrowleft )\end{array}$

Summation of forces along x-direction is equal to 0.

$\begin{array}{l}+\to \sum F_x=0\\ -A_{xO}+20=0\\ A_{xO}=20\text{k}(\leftarrow )\end{array}$

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_{yO}-1.5\left(40\right)=0\\ A_{yO}=60\text{k}(\uparrow )\end{array}$

For unit value of the unknown redundant $B_x$:

Consider the notation of moments due to external load as $m_{BX}$.

Sketch the free body diagram of primary frame Subjected to unit value of redundant $B_x$ as shown in Figure 2.

Find the support reaction at A when 1 k horizontal load applied at B.

Summation of forces along x-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_x=0\\ A_{xBX}-1=0\\ A_{xBX}=1\text{k}(\to )\end{array}$

For unit value of the unknown redundant $B_y$:

Consider the notation of moments due to external load as $m_{BY}$.

Sketch the free body diagram of primary frame Subjected to unit value of redundant $B_y$ as shown in Figure 3.

Find the support reaction and moment at A when 1 k vertical load applied at B.

Summation of moments of all forces about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ m_{ABY}=1\left(40\right)\\ m_{ABY}=40\text{k-ft}(\circlearrowright )\end{array}$

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ -A_{yAY}+1=0\\ A_{yAY}=1\text{k}(\downarrow )\end{array}$

For unit value of the unknown redundant $M_B$:

Consider the notation of moments due to external load as $m_{MB}$.

Sketch the free body diagram of primary frame Subjected to unit value of redundant $M_B$ as shown in Figure 4.

Refer Figure 4.

Find the support reaction and moment at A when 1 k-ft unit moment applied at B.

Summation of moments of all forces about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ m_{AMB}=1\\ m_{AMB}=1\text{k-ft}(\circlearrowright )\end{array}$

Find the moment equation of the frame for different sections on the frame.

Consider a section XX in the portion AB of the primary structure at a distance of x from A.

Refer Figure 1.

Draw the primary structure with section XX as shown in Figure 5.

Refer Figure 5.

Portion AC.

Consider origin as A. $\left(0\le x\le 30\text{ft}\right)$.

Find the moment at section XX in the portion AB as shown in Figure 4.

$\begin{array}{c}{M}_O=-1,800+20\left(x\right)\\ =-1,800+20x\end{array}$

Similarly calculate the moment of the remaining section in the external loading of primary structures and redundant loading structures.

Tabulate the moment equation of different segment of frame as in Table 1.

Segment	x-coordinate		Moment of Inertia	$M_O$ $\left(\text{k-ft}\right)$	$m_{BX}$ $\left(\text{k-ft/k}\right)$	$m_{BY}$ $\left(\text{k-ft/k}\right)$	$m_{MB}$ $\left(\text{k-ft/k}\right)$
	Origin	Limits (ft)
AC	A	$0-30$	I	$-1,800+20x$	$-x$	40	1
DC	D	$0-40$	2I	$1.5x\left(\frac{x}{2}\right)=0.75x^2$	$-30$	$x$	1
BD	B	$0-30$	I	0	$x$	$0$	$-1$

The horizontal deflection at point B due to external loading is ${\Delta }_{BXO}$, vertical deflection at point B due to external loading is ${\Delta }_{BYO}$, deflection at B due to moment is ${\Theta }_{BO}$ and the flexibility coefficient representing the deflection at B due to unit value of redundant $B_x,B_y,\text{and}{M}_B$ are $f_{BX,BX}$, $f_{BY,BY}$, $f_{MB,MB}$, $f_{BX,BY}$, $f_{BX,MB}$, and $f_{BY,MB}$.

Calculate the value of ${\Delta }_{BXO}$ using the equation as follows:

$\begin{array}{c}{\Delta }_{BXO}=\Sigma {\displaystyle \int \frac{M_O{m}_{BX}}{EI}dx}\\ =\left[{\displaystyle {\int }_0^{30}\frac{1}{EI}\left(-1,800+20x\right)\left(-x\right)dx+{\displaystyle {\int }_0^{40}\frac{1}{2EI}\left(-0.75x^2\right)\left(-30\right)dx+\frac{1}{EI}{\displaystyle {\int }_0^{30}\left(0\right)\left(x\right)dx}}}\right]\\ =\frac{1}{EI}\left[630,000+\frac{480,000}{2}+0\right]\\ =\frac{870,000}{EI}{\text{k-ft}}^3\end{array}$

Calculate the value of ${\Delta }_{BYO}$ using the equation as follows:

$\begin{array}{c}{\Delta }_{BYO}=\Sigma {\displaystyle \int \frac{M_O{m}_{BY}}{EI}dx}\\ =\left[{\displaystyle {\int }_0^{30}\frac{1}{EI}\left(-1,800+20x\right)\left(40\right)dx+{\displaystyle {\int }_0^{40}\frac{1}{2EI}\left(-0.75x^2\right)\left(x\right)dx+\frac{1}{EI}{\displaystyle {\int }_0^{30}\left(0\right)\left(0\right)dx}}}\right]\\ =\frac{1}{EI}\left[-1,800,000-\frac{480,000}{2}+0\right]\\ =-\frac{2,040,000}{EI}{\text{k-ft}}^3\end{array}$

Calculate the value of ${\Theta }_{BO}$ using the equation as follows:

$\begin{array}{c}{\Theta }_{BO}=\Sigma {\displaystyle \int \frac{M_O{m}_{MB}}{EI}dx}\\ =\left[{\displaystyle {\int }_0^{30}\frac{1}{EI}\left(-1,800+20x\right)\left(1\right)dx+{\displaystyle {\int }_0^{40}\frac{1}{2EI}\left(-0.75x^2\right)\left(1\right)dx+\frac{1}{EI}{\displaystyle {\int }_0^{30}\left(0\right)\left(-1\right)dx}}}\right]\\ =\frac{1}{EI}\left[-45,000-\frac{16,000}{2}+0\right]\\ =-\frac{53,000}{EI}{\text{k-ft}}^3\end{array}$

Calculate the value of $f_{BX,BX}$ using the equation as follows:

$\begin{array}{c}{f}_{BX,BX}=\Sigma {\displaystyle \int \frac{m_{BX}^2}{EI}dx}\\ =\left[\frac{1}{EI}{\displaystyle {\int }_0^{30}{\left(-x\right)}^{2}dx+\frac{1}{2EI}{\displaystyle {\int }_0^{40}{\left(-30\right)}^{2}dx+\frac{1}{EI}{\displaystyle {\int }_0^{30}{\left(x\right)}^{2}dx}}}\right]\\ =\frac{1}{EI}\left[9,000+\frac{36,000}{2}+9,000\right]\\ =\frac{36,000}{EI}{\text{k-ft}}^{\text{3}}\text{/k}\end{array}$

Calculate the value of $f_{BY,BY}$ using the equation as follows:

$\begin{array}{c}{f}_{BY,BY}=\Sigma {\displaystyle \int \frac{m_{BY}^2}{EI}dx}\\ =\left[\frac{1}{EI}{\displaystyle {\int }_0^{30}{\left(40\right)}^{2}dx+\frac{1}{2EI}{\displaystyle {\int }_0^{40}{\left(x\right)}^{2}dx+\frac{1}{EI}{\displaystyle {\int }_0^{30}{\left(0\right)}^{2}dx}}}\right]\\ =\frac{1}{EI}\left[48,000+\frac{21,333.33}{2}+0\right]\\ =\frac{58,666.67}{EI}{\text{k-ft}}^{\text{3}}\text{/k}\end{array}$

Calculate the value of $f_{MB,MB}$ using the equation as follows:

$\begin{array}{c}{f}_{MB,MB}=\Sigma {\displaystyle \int \frac{m_{MB}^2}{EI}dx}\\ =\left[\frac{1}{EI}{\displaystyle {\int }_0^{30}{\left(1\right)}^{2}dx+\frac{1}{2EI}{\displaystyle {\int }_0^{40}{\left(1\right)}^{2}dx+\frac{1}{EI}{\displaystyle {\int }_0^{30}{\left(-1\right)}^{2}dx}}}\right]\\ =\frac{1}{EI}\left[30+\frac{40}{2}+30\right]\\ =\frac{80}{EI}{\text{k-ft}}^{\text{3}}\text{/k}\end{array}$

Calculate the value of $f_{BX,BY}$ using the equation as follows:

$\begin{array}{c}{f}_{BX,BY}=\Sigma {\displaystyle \int \frac{m_{BX}{m}_{BY}}{EI}dx}\\ =\left[\frac{1}{EI}{\displaystyle {\int }_0^{30}\left(-x\right)\left(40\right)dx+\frac{1}{2EI}{\displaystyle {\int }_0^{40}\left(-30\right)\left(x\right)dx+\frac{1}{EI}{\displaystyle {\int }_0^{30}\left(x\right)\left(0\right)dx}}}\right]\\ =\frac{1}{EI}\left[-18,000-\frac{24,000}{2}+0\right]\\ =-\frac{30,000}{EI}{\text{k-ft}}^{\text{3}}\text{/k}\end{array}$

Calculate the value of $f_{BX,MB}$ using the equation as follows:

$\begin{array}{c}{f}_{BX,MB}=\Sigma {\displaystyle \int \frac{m_{BX}{m}_{MB}}{EI}dx}\\ =\left[\frac{1}{EI}{\displaystyle {\int }_0^{30}\left(-x\right)\left(1\right)dx+\frac{1}{2EI}{\displaystyle {\int }_0^{40}\left(-30\right)\left(1\right)dx+\frac{1}{EI}{\displaystyle {\int }_0^{30}\left(x\right)\left(-1\right)dx}}}\right]\\ =\frac{1}{EI}\left[-450-\frac{1,200}{2}-450\right]\\ =-\frac{1,500}{EI}{\text{k-ft}}^{\text{3}}\text{/k}\end{array}$

Calculate the value of $f_{BY,MB}$ using the equation as follows:

$\begin{array}{c}{f}_{BY,MB}=\Sigma {\displaystyle \int \frac{m_{BY}{m}_{MB}}{EI}dx}\\ =\left[\frac{1}{EI}{\displaystyle {\int }_0^{30}\left(40\right)\left(1\right)dx+\frac{1}{2EI}{\displaystyle {\int }_0^{40}\left(x\right)\left(1\right)dx+\frac{1}{EI}{\displaystyle {\int }_0^{30}\left(0\right)\left(-1\right)dx}}}\right]\\ =\frac{1}{EI}\left[1,200+\frac{800}{2}+0\right]\\ =\frac{1,600}{EI}{\text{k-ft}}^{\text{3}}\text{/k}\end{array}$

Find the reactions and moment for the given frame:

Find the horizontal, vertical reaction and moment at B.

Show the first compatibility Equation as follows:

${\Delta }_{BXO}+f_{BX,BX}{B}_x+f_{BX,BY}{B}_y+f_{BX,MB}{M}_B=0$

Substitute $-\frac{870,000}{EI}{\text{k-ft}}^3$ for ${\Delta }_{BXO}$, $\frac{36,000}{EI}{\text{k-ft}}^{\text{3}}\text{/k}$ for $f_{BX,BX}$, $-\frac{30,000}{EI}{\text{k-ft}}^{\text{3}}\text{/k}$ for $f_{BX,BY}$, and $-\frac{1,500}{EI}{\text{k-ft}}^{\text{3}}\text{/k}$ for $f_{BX,MB}$.

$\begin{array}{l}\frac{870,000}{EI}+\left(\frac{36,000}{EI}\right)B_x+\left(-\frac{30,000}{EI}\right)B_y+\left(-\frac{1,500}{EI}\right)M_B=0\\ 36,000B_X-30,000B_y-1,500M_B=-870,000\end{array}$        (1)

Show the second compatibility Equation as follows:

${\Delta }_{BYO}+f_{BY,BX}{B}_x+f_{BY,BY}{B}_y+f_{BY,MB}{M}_B=0$

Substitute $-\frac{2,040,000}{EI}{\text{k-ft}}^3$ for ${\Delta }_{BYO}$, $-\frac{30,000}{EI}{\text{k-ft}}^{\text{3}}\text{/k}$ for $f_{BY,BX}$, $\frac{58,666.67}{EI}{\text{k-ft}}^{\text{3}}\text{/k}$ for $f_{BY,BY}$, and $\frac{1,600}{EI}{\text{k-ft}}^{\text{3}}\text{/k}$ for $f_{BY,MB}$.

$\begin{array}{l}-\frac{2,040,000}{EI}+\left(-\frac{30,000}{EI}\right)B_x+\left(\frac{58,666.67}{EI}\right)B_y+\left(\frac{1,600}{EI}\right)M_B=0\\ -30,000B_X+58,666.67B_y+1,600M_B=2,040,000\end{array}$        (2)

Show the third compatibility Equation as follows:

${\Theta }_{BO}+f_{MB,BX}{B}_x+f_{MB,BY}{B}_y+f_{MB,MB}{M}_B=0$

Substitute $-\frac{53,000}{EI}{\text{k-ft}}^3$ for ${\Theta }_{BO}$, $-\frac{1,500}{EI}{\text{k-ft}}^{\text{3}}\text{/k}$ for $f_{MB,BX}$, $\frac{1,600}{EI}{\text{k-ft}}^{\text{3}}\text{/k}$ for $f_{MB,BY}$, and $\frac{80}{EI}{\text{k-ft}}^{\text{3}}\text{/k}$ for $f_{MB,MB}$.

$\begin{array}{l}-\frac{53,000}{EI}+\left(-\frac{1,500}{EI}\right)B_x+\left(\frac{1,600}{EI}\right)B_y+\left(\frac{80}{EI}\right)M_B=0\\ -1,500B_X+1,600B_y+80M_B=53,000\end{array}$        (3)

Solve Equation (1), (2), and (3).

$\begin{array}{l}{B}_x=15.71\text{k}(\leftarrow )\\ B_y=36.75\text{k}(\uparrow )\\ M_B=222.1\text{k-ft}(\circlearrowleft )\end{array}$

Therefore, the horizontal reaction at B is $\underset{\_}{B_x=15.71\text{k}\leftarrow }$.

Therefore, the vertical reaction at B is $\underset{\_}{B_y=36.75\text{k}\downarrow }$.

Therefore, the moment at A is $\underset{\_}{M_B=222.1\text{k-ft}\circlearrowleft }$ acting in the anticlockwise direction.

Sketch the free body diagram of the frame as shown in Figure 6.

Refer Figure 6.

Find the horizontal reaction at A.

Summation of forces along x-direction is equal to 0.

$\begin{array}{l}+\to \sum F_x=0\\ -A_x-15.71+20=0\\ A_x=4.29\text{k}(\leftarrow )\end{array}$

Therefore, horizontal reaction at A is $\underset{\_}{A_x=4.29\text{k}\leftarrow }$.

Find the vertical reaction at A.

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y+36.75-1.5\left(40\right)=0\\ A_y=23.25\text{k}(\uparrow )\end{array}$

Therefore, the vertical reaction at A is $\underset{\_}{A_y=23.25\text{k}\uparrow }$.

Find the moment at A.

Summation of moments of all forces about A is equal to 0.

$\begin{array}{c}\sum M_A=0\\ M_A=222.1+36.75\left(40\right)-1.5\left(40\right)\left(\frac{40}{2}\right)-20\left(30\right)\\ =222.1+1,470-1,200-600\\ =-108.9\text{k-ft}\\ M_A=108.9\text{k-ft}(\circlearrowleft )\end{array}$

Therefore, the moment at A is $\underset{\_}{M_A=108.9\text{k-ft}\circlearrowleft }$ acting in the anticlockwise direction.

Sketch the reactions and moment for the given frame as shown in Figure 7.

Refer Figure 7.

Find the shear force $\left(S\right)$ for the given frame:

For span AC,

At point A.

$S_A=4.29\text{k}$

At point B,

$S_B=4.29\text{k}$

For span CD,

At point C.

$\begin{array}{c}{S}_C=A_y\\ =23.25\text{k}\end{array}$

At point D,

$\begin{array}{c}{S}_D=23.25-1.5\left(40\right)\\ =-36.75\text{k}\end{array}$

For span BD,

At point B.

$\begin{array}{c}{S}_B=B_x\\ =15.71\text{k}\end{array}$

At point D,

$S_D=15.71\text{k}$

Sketch the shear diagram for the given frame as shown in Figure 8.

Refer Figure 8.

Find the point of zero shear force.

Take shear force at E.

$\begin{array}{l}{S}_E=0\\ 23.25-1.5\left(x_1\right)=0\\ 1.5x_1=23.25\\ x_1=15.5\text{ft}\text{from}C\end{array}$

Refer Figure 7.

Find the bending moment $\left(M\right)$ for the frame:

For span AC,

At point A,

$M_A=-107.9\text{k-ft}$

At point C,

$\begin{array}{c}{M}_C=-107.9+4.29\left(30\right)\\ =20.8\text{k-ft}\end{array}$

For span CD,

At point C,

$M_C=20.8\text{k-ft}$

At point D,

$\begin{array}{c}{M}_D=20.8-1.5\left(40\right)\left(\frac{40}{2}\right)+23.25\left(40\right)\\ =20.8-1,200+930\\ =-249.2\text{k-ft}\end{array}$

At point E,

$M_E=20.8-1.5\left(x_1\right)\left(\frac{x_1}{2}\right)+23.25\left(x_1\right)$

Substitute 15.5 ft for $x_1$.

$\begin{array}{c}{M}_E=20.8-1.5\left(15.5\right)\left(\frac{15.5}{2}\right)+23.25\left(15.5\right)\\ =20.8-180.1875+360.375\\ =201\text{k-ft}\end{array}$

For span BD,

At point B,

$M_B=-222.1\text{k-ft}$

At point D,

$\begin{array}{c}{M}_D=-222.1+15.71\left(30\right)\\ =249.2\text{k-ft}\end{array}$

Sketch the bending moment diagram for the given frame as shown in Figure 9.