Chapter 8, Problem 17P

To determine

Find the vertical and horizontal deflection of the hinge at C.

Answer to Problem 17P

The vertical deflection of the hinge at C is $\underset{\_}{8.6\text{in}\text{.}\uparrow }$.

The horizontal deflection of the hinge at C is $\underset{\_}{15.4\text{in}\text{.}\to }$.

Explanation of Solution

Given information:

A single concentrated load of 60 kips is applied at joint B.

Area of all segments of the arch is $A=20\text{in}{.}^2$.

Moment of inertia $I=600\text{in}{.}^4$ and value of E is $30,000\text{ksi}$.

Procedure to find the deflection of truss by virtual work method is shown below.

• For Real system: If the deflection of truss is determined by the external loads, then apply method of joints or method of sections to find the real axial forces in all the members of the truss.
• For virtual system: Remove all given real loads, apply a unit load at the joint where is deflection is required and also in the direction of desired deflection. Use method of joints or method of sections to find the virtual axial forces in all the member of the truss.
• Finally use the desired deflection equation.

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 negative and the counter clockwise moment as positive.

Calculation:

Find the length of the AB segment of the arch using given Figure.

$\begin{array}{c}{L}_{AB}=\sqrt{{30}^2+{38.46}^2}\\ =48.78\text{ft}\end{array}$

Find the length of the BC segment of the arch using given Figure.

$\begin{array}{c}{L}_{BC}=\sqrt{{30}^2+{\left(50-38.46\right)}^2}\\ =32.14\text{ft}\end{array}$

Find the angle made by the member AB with respect to the horizontal.

$\begin{array}{c}{\theta }_{AB}={\tan }^{-1}\left(\frac{38.46}{30}\right)\\ =52°\end{array}$

Find the angle made by the member BC with respect to the horizontal.

$\begin{array}{c}{\theta }_{BC}={\tan }^{-1}\left(\frac{50-38.46}{30}\right)\\ =21°\end{array}$

Sketch the arch with the forces produced by the P-system as shown in Figure 1.

Find the reactions at the supports using Equilibrium equations:

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ E_y\left(120\right)-60\left(30\right)=0\\ E_y=15\text{kips}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y+E_y-60=0\\ A_y+15-60=0\\ A_y=45\text{kips}\uparrow \end{array}$

Summation of moments about C is equal to 0.

$\begin{array}{l}\sum M_C=0\\ A_x\left(50\right)-60\left(30\right)+45\left(60\right)=0\\ A_x=18\text{kips}\to \end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ A_x-E_x=0\\ E_x=18\text{kips}\leftarrow \end{array}$

Find the $F_P$ forces and $M_P$ moments of each member as follows:

For AB,

$\begin{array}{c}{F}_{AB}^P=-45\sin 52°-18\cos 52°\\ =-46.55\text{kips}\end{array}$

$\begin{array}{c}{M}_{AB}^P=\left(45\cos 52°-18\sin 52°\right)x\\ =13.52x\end{array}$

For BC,

$\begin{array}{c}{F}_{BC}^P=15\sin 21°-18\cos 21°\\ =-11.23\text{kips}\end{array}$

$M_{BC}^P=659.5-20.52x$

For ED,

$\begin{array}{c}{F}_{ED}^P=-15\sin 52°-18\cos 52°\\ =-22.9\text{kips}\end{array}$

$M_{ED}^P=-4.95x$

For CD,

$\begin{array}{c}{F}_{CD}^P=-15\sin 52°-18\cos 21°\\ =-22.18\text{kips}\end{array}$

$M_{CD}^P=-241.46+7.51x$

Consider a dummy load of 1 kips directed vertically at joint C with the forces $F_Q$.

Sketch the reactions produced at the supports by the vertical dummy load at C as shown in Figure 2.

Find the reactions at the supports using Equilibrium equations:

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ E_y\left(120\right)-1\left(60\right)=0\\ E_y=0.5\text{kips}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y+E_y-1=0\\ A_y+0.5-1=0\\ A_y=0.5\text{kips}\uparrow \end{array}$

Summation of moments about C is equal to 0.

$\begin{array}{l}\sum M_C=0\\ A_x\left(50\right)-0.5\left(60\right)=0\\ A_x=0.6\text{kips}\to \end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ A_x-E_x=0\\ E_x=0.6\text{kips}\leftarrow \end{array}$

Find the $F_Q$ forces and $M_Q$ moments of each member as follows:

For AB,

$F_{AB}^Q=-0.763\text{kips}$

$M_{AB}^Q=-0.165x$

For BC,

$F_{BC}^Q=-0.739\text{kips}$

$M_{BC}^Q=-8.05+0.25x$

For ED,

$F_{ED}^Q=-0.763\text{kips}$

$M_{ED}^Q=-0.165x$

For CD,

$F_{CD}^Q=-0.739\text{kips}$

$M_{CD}^Q=-8.05+0.25x$

Find the vertical deflection at C $\left({\Delta }_{Cy}\right)$ as follows:

$\begin{array}{c}\left(1\text{kips}\right){\Delta }_{Cy}=\Sigma \frac{F_Q{F}_{P}L}{AE}+\Sigma {\displaystyle \int \frac{M_Q{M}_{P}dx}{EI}}\\ \left(1\text{kips}\right){\Delta }_{Cy}=\left[\begin{array}{l}\frac{1}{AE}\left[\begin{array}{l}\left(-46.55\right)\left(-0.763\right)\left(48.78\times 12\right)+\left(-11.23\right)\left(-0.739\right)\left(32.14\times 12\right)\\ +\left(-22.9\right)\left(-0.763\right)\left(48.78\times 12\right)+\left(-22.18\right)\left(-0.739\right)\left(32.14\times 12\right)\end{array}\right]\\ +{\displaystyle \underset{0}{\overset{48.78}{\int }}\frac{\left(13.52x\right)\left(-0.165x\right)}{EI}dx}+{\displaystyle \underset{0}{\overset{32.14}{\int }}\frac{\left(659.5-20.52x\right)\left(-8.05+0.25x\right)}{EI}dx}\\ +{\displaystyle \underset{0}{\overset{48.78}{\int }}\frac{\left(4.95x\right)\left(-0.165x\right)}{EI}dx}+{\displaystyle \underset{0}{\overset{32.14}{\int }}\frac{\left(-241.46+7.51x\right)\left(-8.05+0.25x\right)}{EI}dx}\end{array}\right]\\ {\Delta }_{Cy}=\frac{40,600}{AE}+\frac{-90,972\times {12}^3}{EI}\\ =\frac{40,600}{20\times 30,000}+\frac{-90,972\times {12}^3}{30,000\times 600}\end{array}$

$\begin{array}{c}{\Delta }_{Cy}=-8.6\text{in}\text{.}\\ =8.6\text{in}\text{.}\uparrow \end{array}$

Therefore, the vertical deflection at C is $\underset{\_}{8.6\text{in}\text{.}\uparrow }$.

Consider a dummy load of 1 kips directed horizontally at joint C with the forces $F_Q$.

Sketch the reactions produced at the supports by the horizontal dummy load at C as shown in Figure 3.

Find the reactions at the supports using Equilibrium equations:

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ E_y\left(120\right)-1\left(50\right)=0\\ E_y=0.42\text{kips}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -A_y+E_y=0\\ -A_y+0.42=0\\ A_y=0.42\text{kips}\downarrow \end{array}$

Summation of moments about C is equal to 0.

$\begin{array}{l}\sum M_C=0\\ A_x\left(50\right)-0.42\left(60\right)=0\\ A_x=\frac{1}{2}\text{kips}\leftarrow \end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ -A_x-E_x+1=0\\ -\frac{1}{2}-E_x+1=0\\ E_x=\frac{1}{2}\text{kips}\leftarrow \end{array}$

Find the $F_Q$ forces and $M_Q$ moments of each member as follows:

For AB,

$\begin{array}{c}{F}_{AB}^Q=0.42\sin 52°+0.5\cos 52°\\ =0.64\text{kips}\end{array}$

$M_{AB}^Q=0.1354x$

For BC,

$\begin{array}{c}{F}_{BC}^Q=0.42\sin 21°+0.5\cos 21°\\ =0.62\text{kips}\end{array}$

$M_{BC}^Q=6.61-0.206x$

For ED,

$\begin{array}{l}{F}_{ED}^Q=-0.42\sin 52°-0.5\cos 52°\\ =0.64\text{kips}\end{array}$

$M_{ED}^Q=-0.1354x$

For CD,

$F_{CD}^Q=-0.62\text{kips}$

$M_{CD}^Q=-6.61+0.206x$

Find the horizontal deflection at C $\left({\Delta }_{Cx}\right)$ as follows:

$\begin{array}{c}\left(1\text{kips}\right){\Delta }_{Cx}=\Sigma \frac{F_Q{F}_{P}L}{AE}+\Sigma {\displaystyle \int \frac{M_Q{M}_{P}dx}{EI}}\\ \left(1\text{kips}\right){\Delta }_{Cx}=\left[\begin{array}{l}\frac{1}{AE}\left[\begin{array}{l}\left(-46.55\right)\left(0.64\right)\left(48.78\times 12\right)+\left(-11.23\right)\left(0.62\right)\left(32.14\times 12\right)\\ +\left(-22.9\right)\left(-0.64\right)\left(48.78\times 12\right)+\left(-22.18\right)\left(-0.62\right)\left(32.14\times 12\right)\end{array}\right]\\ +{\displaystyle \underset{0}{\overset{48.78}{\int }}\frac{\left(13.52x\right)\left(0.1354x\right)}{EI}dx}+{\displaystyle \underset{0}{\overset{32.14}{\int }}\frac{\left(659.5-20.52x\right)\left(6.61-0.206x\right)}{EI}dx}\\ +{\displaystyle \underset{0}{\overset{48.78}{\int }}\frac{\left(4.95x\right)\left(-0.1354x\right)}{EI}dx}+{\displaystyle \underset{0}{\overset{32.14}{\int }}\frac{\left(-241.46+7.51x\right)\left(-6.61-0.206x\right)}{EI}dx}\end{array}\right]\\ {\Delta }_{Cx}=\frac{-6,240}{AE}+\frac{160,512.5\times {12}^3}{EI}\\ =\frac{-6,240}{20\times 30,000}+\frac{160,512.5\times {12}^3}{30,000\times 600}\end{array}$

${\Delta }_{Cx}=15.4\text{in}\text{.}\to \text{.}$

Therefore, the horizontal deflection at C is $\underset{\_}{15.4\text{in}\text{.}\to }$.