Chapter 13, Problem 41P

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 41P

The vertical reaction at A is $\underset{\_}{A_y=23\text{k}\uparrow }$.

The vertical reaction at B is $\underset{\_}{B_y=63\text{k}\uparrow }$.

The vertical reaction at D is $\underset{\_}{D_y=48\text{k}\uparrow }$.

The vertical reaction at F is $\underset{\_}{F_y=63\text{k}\uparrow }$.

The vertical reaction at G is $\underset{\_}{G_y=23\text{k}\uparrow }$.

The moment at A is $\underset{\_}{M_A=0}$.

The moment at B is $\underset{\_}{M_B=-210\text{k-ft}(\circlearrowleft )}$.

The moment at C is $\underset{\_}{M_C=180\text{k-ft}(\circlearrowright )}$.

The moment at D is $\underset{\_}{M_D=180\text{k-ft}(\circlearrowleft )}$.

The moment at F is $\underset{\_}{M_F=210\text{k-ft}(\circlearrowleft )}$.

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:

The beam and the loading are symmetric since only analyze only the right half DG of the beam.

Sketch the free body diagram of the right half DG of the beam as shown in Figure 1.

Refer Figure 1.

Find the degree of indeterminacy of the structure:

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

The beam is supported by 5 support reactions and the number of equilibrium equations is 3.

Therefore, the degree of indeterminacy of the beam is $i=2$.

Select the vertical reaction $F_y$ and $G_y$ at the supports F and G as redundant.

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

Sketch the free body diagram of primary beam subjected to external loading without support F and G as shown in Figure 2.

Refer Figure 2.

Find the reactions at the supports without considering support F and G using equilibrium equations:

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

$\begin{array}{c}\sum M_D=0\\ M_{DO}=-50\left(15\right)-2\left(30\right)\left(\frac{30}{2}+30\right)\\ =-3,450\text{k-ft}\\ M_{DO}=3,450\text{k-ft}(\circlearrowleft )\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_x=0\\ D_{yO}-50-2\left(30\right)=0\\ D_{yO}=110\text{k}(\uparrow )\end{array}$

For unit value of the unknown redundant $F_y$:

Consider the notation of moments due to external load as $m_F$.

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

Refer Figure 3.

Find the support reaction and moment at D when 1 k vertical load applied at C.

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

$\begin{array}{l}\sum M_D=0\\ M_{DF}=1\left(30\right)\\ M_{DF}=30\text{k-ft}(\circlearrowright )\end{array}$

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

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

For unit value of the unknown redundant $G_y$:

Consider the notation of moments due to external load as $m_G$.

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

Refer Figure 4.

Find the support reaction and moment at D when 1 k vertical load applied at C.

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

$\begin{array}{l}\sum M_D=0\\ M_{DG}=1\left(60\right)\\ M_{DF}=60\text{k-ft}(\circlearrowright )\end{array}$

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

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

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

Consider a section XX in the portion DE of the primary structure at a distance of x from D.

Refer Figure 1.

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

Refer Figure 5.

Portion AB.

Consider origin as A. $\left(0\le x\le 15\text{ft}\right)$.

Find the moment at section XX in the portion DE as shown in Figure 5.

$\begin{array}{c}{M}_O=-3,450+110\left(x\right)\\ =-3,450+110x\end{array}$

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

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

Segment	x-coordinate		$M_O$ $\left(\text{k-ft}\right)$	$m_{FY}$ $\left(\text{k-ft/k}\right)$	$m_{GY}$ $\left(\text{k-ft/k}\right)$
	Origin	Limits (ft)
DE	A	$0-15$	$-3,450+110x$	$30-1\left(x\right)$	$60-1\left(x\right)$
EF	E	$15-30$	$-3,450+110\left(x\right)-50\left(x-15\right)$	$30-1\left(x\right)$	$60-1\left(x\right)$
FG	G	$0-30$	$-2\frac{x^2}{2}=-x^2$	$0$	$x$

The vertical deflection at point F due to external loading is ${\Delta }_{FO}$, vertical deflection at point G due to external loading is ${\Delta }_{GO}$,  and the flexibility coefficient representing the deflection at F due to unit value of redundant $F_y$ is $f_{FF}$, the redundant $G_y$ is $f_{GG}$, and the both redundant $F_y$ and $G_y$ is $f_{FG}=f_{GF}$.

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

$\begin{array}{c}{\Delta }_{FO}=\Sigma {\displaystyle \int \frac{M_O{m}_{FY}}{EI}dx}\\ =\frac{1}{EI}\left\{\begin{array}{l}{\displaystyle {\int }_0^{15}\left(-3,450+110x\right)\left(30-x\right)dx}\\ +{\displaystyle {\int }_{15}^{30}\left[-3,450+110\left(x\right)-50\left(x-15\right)\right]\left[30-x\right]dx+{\displaystyle {\int }_0^{30}\left(-x^2\right)\left(0\right)dx}}\end{array}\right\}\\ =\frac{1}{EI}\left[-916,875-168,750+0\right]\\ =-\frac{1,085,625}{EI}{\text{k-ft}}^3\end{array}$

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

$\begin{array}{c}{\Delta }_{GO}=\Sigma {\displaystyle \int \frac{M_O{m}_{GY}}{EI}dx}\\ =\frac{1}{EI}\left\{\begin{array}{l}{\displaystyle {\int }_0^{15}\left(-3,450+110x\right)\left(60-x\right)dx}\\ +{\displaystyle {\int }_{15}^{30}\left[-3,450+110\left(x\right)-50\left(x-15\right)\right]\left[60-x\right]dx+{\displaystyle {\int }_0^{30}\left(-x^2\right)\left(x\right)dx}}\end{array}\right\}\\ =\frac{1}{EI}\left[-2,098,125-776,250-2,02,500\right]\\ =-\frac{3,076,875}{EI}{\text{k-ft}}^3\end{array}$

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

$\begin{array}{c}{f}_{FF}=\Sigma {\displaystyle \int \frac{m_{FY}^2}{EI}dx}\\ =\frac{1}{EI}\left\{{\displaystyle {\int }_0^{15}{\left(30-x\right)}^{2}dx}+{\displaystyle {\int }_{15}^{30}{\left(30-x\right)}^{2}dx+{\displaystyle {\int }_0^{30}{\left(0\right)}^{2}dx}}\right\}\\ =\frac{1}{EI}\left[7,875+1,125+0\right]\\ =\frac{9,000}{EI}{\text{k-ft}}^3/k\end{array}$

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

$\begin{array}{c}{f}_{GG}=\Sigma {\displaystyle \int \frac{m_{GY}^2}{EI}dx}\\ =\frac{1}{EI}\left\{\begin{array}{l}{\displaystyle {\int }_0^{15}{\left(60-x\right)}^{2}dx}\\ +{\displaystyle {\int }_{15}^{30}{\left(60-x\right)}^{2}dx+{\displaystyle {\int }_0^{30}{\left(x\right)}^{2}dx}}\end{array}\right\}\\ =\frac{1}{EI}\left[41,625+21,375+9,000\right]\\ =\frac{72,000}{EI}{\text{k-ft}}^3/k\end{array}$

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

$\begin{array}{c}{f}_{FG}=f_{GF}\\ =\Sigma {\displaystyle \int \frac{m_{FY}{m}_{GY}}{EI}dx}\\ =\frac{1}{EI}\left[{\displaystyle {\int }_0^{15}\left(30-x\right)\left(60-x\right)dx+{\displaystyle {\int }_{15}^{30}\left(30-x\right)\left(60-x\right)dx+{\displaystyle {\int }_0^{30}\left(0\right)\left(x\right)dx}}}\right]\\ =\frac{1}{EI}\left[18,000+4,500+0\right]\\ =\frac{22,500}{EI}{\text{k-ft}}^{\text{3}}\text{/k}\end{array}$

Find the reactions and moment:

Find the horizontal and vertical reaction at F and G.

Show the first compatibility Equation as follows:

${\Delta }_{FO}+f_{FF}{F}_y+f_{FG}{G}_y=0$

Substitute $-\frac{1,085,625}{EI}{\text{k-ft}}^3$ for ${\Delta }_{FO}$, $\frac{9,000}{EI}{\text{k-ft}}^3$ for $f_{FF}$, and $\frac{22,500}{EI}{\text{k-ft}}^{\text{3}}\text{/k}$ for $f_{FG}$.

$\begin{array}{l}-\frac{1,085,625}{EI}+\left(\frac{9,000}{EI}\right)F_y+\left(\frac{22,500}{EI}\right)G_y=0\\ 9,000F_y+22,500G_y=1,085,625\end{array}$        (1)

Show the second compatibility Equation as follows:

${\Delta }_{GO}+f_{GF}{F}_y+f_{GG}{G}_y=0$

Substitute $-\frac{3,076,875}{EI}{\text{k-ft}}^3$ for ${\Delta }_{GO}$, $\frac{22,500}{EI}{\text{k-ft}}^{\text{3}}\text{/k}$ for $f_{GF}$, and $\frac{72,000}{EI}{\text{k-ft}}^{\text{3}}\text{/k}$ for $f_{GG}$.

$\begin{array}{l}-\frac{3,076,875}{EI}+\left(\frac{22,500}{EI}\right)F_y+\left(\frac{72,000}{EI}\right)G_y=0\\ 22,500F_y+72000G_y=3,076,875\end{array}$        (2)

Solve Equation (1) and (2).

$\begin{array}{l}{F}_y=63\text{k}(\uparrow )\\ G_y=23\text{k}(\uparrow )\end{array}$

The beam is symmetrical. Hence, $F_y=B_y\text{and}{G}_y=A_y$.

Find the vertical reaction at D.

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y+B_y+F_y+G_y+D_y=50+2\left(30\right)+50+2\left(30\right)\end{array}$

Substitute 23 k for $A_y$, $63\text{k}$ for $B_y$, 63 k for $F_y$ and 23 k for $G_y$.

$\begin{array}{l}23+63+63+23+D_y=220\\ D_y=48\text{k}\uparrow \end{array}$

Therefore, the vertical reaction at A is $\underset{\_}{A_y=23\text{k}\uparrow }$.

Therefore, the vertical reaction at B is $\underset{\_}{B_y=63\text{k}\uparrow }$.

Therefore, the vertical reaction at D is $\underset{\_}{D_y=48\text{k}\uparrow }$.

Therefore, the vertical reaction at F is $\underset{\_}{F_y=63\text{k}\uparrow }$.

Therefore, the vertical reaction at G is $\underset{\_}{G_y=23\text{k}\uparrow }$.

Sketch the free body diagram with support reactions of the beam as shown in Figure 6.

Refer Figure 6,

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

For span AB,

At point A.

$S_A=23\text{k}$

At point B, (negative side):

$\begin{array}{c}{S}_B^{-}=23-2\left(30\right)\\ =-37\text{k}\end{array}$

For span BD,

At point B, (positive side):

$\begin{array}{c}{S}_B^{+}=-37+63\\ =26\text{k}\end{array}$

At point C, (positive side):

$S_C^{+}=26\text{k}$

At point C, (negative side):

$\begin{array}{c}{S}_C^{-}=26-50\\ =-24\text{k}\end{array}$

At point D, (negative side):

$S_D^{-}=-24\text{k}$

For span DF,

At point D, (positive side):

$\begin{array}{c}{S}_D^{+}=-24+48\\ =24\text{k}\end{array}$

At point E, (positive side):

$S_E^{+}=24\text{k}$

At point E, (negative side):

$\begin{array}{c}{S}_E^{-}=24-50\\ =-26\text{k}\end{array}$

At point F, (negative side):

$S_F^{-}=-26\text{k}$

For span FG,

At point F, (positive side):

$\begin{array}{c}{S}_F^{+}=-26+63\\ =37\text{k}\end{array}$

At point G,

$\begin{array}{c}{S}_G=37-2\left(30\right)\\ =-23\text{k}\end{array}$

Sketch the shear diagram for the given beam as shown in Figure 7.

Refer Figure 7.

The points of zero shear as H and I.

Find the point of zero shear force from A and G.

$\begin{array}{l}23-2\left(x_1\right)=S{F}_0\\ 2x_1=23\\ x_1=11.5\text{ft}\end{array}$

Refer Figure 6.

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

For span AB,

At point A, (hinge end)

$M_A=0$

At point H,

$\begin{array}{c}{M}_H=23\left(11.5\right)-2\left(11.5\right)\left(\frac{11.5}{2}\right)\\ =264.5-132.25\\ \simeq 132\text{k-ft}\end{array}$

At point B,

$\begin{array}{c}{M}_B=23\left(30\right)-2\left(30\right)\left(\frac{30}{2}\right)\\ =690-900\\ =-210\text{k-ft}\end{array}$

For span BD,

At point C,

$\begin{array}{c}{M}_C=23\left(45\right)-2\left(30\right)\left(\frac{30}{2}+15\right)+63\left(15\right)\\ =1,035-1,800+945\\ =180\text{k-ft}\end{array}$

At point D,

$\begin{array}{c}{M}_D=23\left(60\right)-2\left(30\right)\left(\frac{30}{2}+30\right)+63\left(30\right)-50\left(15\right)\\ =1,380-2,700+1,890-750\\ =-180\text{k-ft}\end{array}$

The moment at E, F, I, and G is equal to the moment at C, B, H, and A respectively due to symmetry of the beam.

Sketch the bending moment diagram for the given beam as shown in Figure 8.