Chapter 8, Problem 38P

To determine

Find the vertical deflection and horizontal deflection at B.

Find the vertical deflection and horizontal deflection at the mid span of the member CD.

Answer to Problem 38P

The vertical deflection at B is $\underset{\_}{8.6\text{in}\text{.}\uparrow }$.

The horizontal deflection at B is $\underset{\_}{\text{10}\text{.61}\text{in}\text{.}\to }$.

The vertical deflection at mid span of the member CD is $\underset{\_}{20.30\text{in}\text{.}\downarrow }$.

The horizontal deflection at mid span of the member CD is $\underset{\_}{\text{10}\text{.61}\text{in}\text{.}\to }$.

Explanation of Solution

Given information:

The moment of inertia of all members is $I=180\text{in}{.}^4$.

The area of the column is $6\text{in}{.}^2$ and the area of the girder is $10\text{in}{.}^2$.

The value of Young’s modulus is $29,000\text{kips/in}{\text{.}}^{\text{2}}$.

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 vertical deflection at B $\left({\delta }_{BV}\right)$:

Find the moment $M_P$ of the P-system:

For span BC,

$M_B=0$

$\begin{array}{c}{M}_C=-10\left(5\right)\\ =-50\text{kip-ft}\end{array}$

Consider the point E as midspan of span CD.

For span CED,

$M_D=0$

$\begin{array}{c}{M}_E=-2.4\left(10\right)\left(5\right)\\ =-120\text{kip-ft}\end{array}$

$\begin{array}{c}{M}_C=-2.4\left(20\right)\left(10\right)\\ =-480\text{kip-ft}\end{array}$

For span AC,

$\begin{array}{c}{M}_C=-2.4\left(20\right)\left(10\right)+10\left(5\right)\\ =-430\text{kip-ft}\end{array}$

$\begin{array}{c}{M}_{\text{at}\text{the}\text{point}\text{where}\text{6}\text{kips}\text{acts}}=-2.4\left(20\right)\left(10\right)+10\left(5\right)\\ =-430\text{kip-ft}\end{array}$

$\begin{array}{c}{M}_A=-2.4\left(20\right)\left(10\right)+10\left(5\right)-6\left(6\right)\\ =-466\text{kip-ft}\end{array}$

Find the force $F_P$ of the P-system:

$\begin{array}{c}{F}_C=-2.4\left(20\right)-10\\ =-58\text{kips}\end{array}$

$\begin{array}{c}{F}_A=-2.4\left(20\right)-10\\ =-58\text{kips}\end{array}$

Sketch the $M_P$ diagram and $F_P$ diagram for the structure as shown in Figure 1.

Apply 1 kips vertically at point B for vertical deflection $\left({\delta }_{BV}\right)$.

Find the moment $M_Q$ of the Q-system:

For span BC,

$M_B=0$

$\begin{array}{c}{M}_C=-1\left(5\right)\\ =-5\text{kip-ft}\end{array}$

For span AC,

$\begin{array}{c}{M}_C=1\left(5\right)\\ =5\text{kip-ft}\end{array}$

$\begin{array}{c}{M}_A=1\left(5\right)\\ =5\text{kip-ft}\end{array}$

Find the force $F_Q$ of the Q-system:

$F_C=-1\text{kips}$

$F_A=-1\text{kips}$

Sketch the $M_P$ diagram and $F_P$ diagram for ${\delta }_{BV}$ as shown in Figure 2.

Find the vertical deflection at B $\left({\delta }_{BV}\right)$ using Figure 1 and Figure 2 as follows:

$\begin{array}{l}\Sigma Q\cdot {\delta }_{BV}=\Sigma \frac{1}{EI}\left(C{M}_1{M}_{3}L\right)+\Sigma F_Q\frac{F_{P}L}{AE}\\ \left(1\text{kips}\right)\left({\delta }_{BV}\right)=\left[\begin{array}{l}\frac{1}{2EI}\left(\frac{1}{3}\times 50\times 5\right)\times 5-\frac{1}{EI}\left(430\times 5\right)\times 6\\ -\frac{1}{EI}\left(\frac{1}{2}\left(\left(430+466\right)\times 5\right)\right)\times 6+\frac{58\left(12\right)}{AE}\end{array}\right]\\ {\delta }_{BV}=\frac{-26,131.67}{EI}+\frac{696}{AE}\\ {\delta }_{BV}=\frac{-26,131.67\left(1,728\right)}{\left(29,000\right)\left(180\right)}+\frac{696\left(12\right)}{\left(6\right)\left(29,000\right)}\end{array}$

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

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

Apply 1 kips horizontally at point B for horizontal deflection $\left({\delta }_{BH}\right)$.

Sketch the $M_P$ diagram and $F_P$ diagram for ${\delta }_{BH}$ as shown in Figure 3.

Find the moment $M_Q$ of the Q-system:

For span AC,

$M_C=0$

$\begin{array}{c}{M}_A=-1\left(12\right)\\ =-12\text{kip-ft}\end{array}$

Find the force $F_Q$ of the Q-system:

$F_B=-1\text{kips}$

$F_C=-1\text{kips}$

Find the horizontal deflection at B $\left({\delta }_{BH}\right)$ using Figure 1 and Figure 3 as follows:

$\begin{array}{l}\Sigma Q\cdot {\delta }_{BH}=\Sigma \frac{1}{EI}\left(C{M}_1{M}_{3}L\right)\\ \left(1\text{kips}\right)\left({\delta }_{BH}\right)=\left[\begin{array}{l}\frac{1}{EI}\left(\frac{1}{2}\times 6\times 430\right)\times 6+\frac{1}{EI}\left(\frac{1}{6}\times 6\times \left(2\left(430\right)+466\right)\right)\times 6\\ +\frac{1}{EI}\left(\frac{1}{6}\left(12\left(430+2\left(466\right)\right)\right)\right)\times 6\end{array}\right]\\ {\delta }_{BH}=\frac{32,040}{EI}\\ {\delta }_{BH}=\frac{32,040\left(1,728\right)}{\left(29,000\right)\left(180\right)}\end{array}$

${\delta }_{BH}\text{=10}\text{.61}\text{in}\mathrm{..}$

Therefore, the horizontal deflection at B is $\underset{\_}{\text{10}\text{.61}\text{in}\text{.}\to }$.

Apply 1 kips vertically at point E for vertical deflection $\left({\delta }_{EV}\right)$.

Find the moment $M_Q$ of the Q-system:

For span CED,

$M_D=0$

$M_E=0$

$\begin{array}{c}{M}_C=-1\left(10\right)\\ =-10\text{kip-ft}\end{array}$

For span AC,

$\begin{array}{c}{M}_C=-1\left(10\right)\\ =-10\text{kip-ft}\end{array}$

$\begin{array}{c}{M}_A=-1\left(10\right)\\ =-10\text{kip-ft}\end{array}$

Find the force $F_Q$ of the Q-system:

$F_C=-1\text{kips}$

$F_A=-1\text{kips}$

Sketch the $M_P$ diagram and $F_P$ diagram for ${\delta }_{EV}$ as shown in Figure 4.

Find the vertical deflection at E $\left({\delta }_{EV}\right)$ using Figure 1 and Figure 4 as follows:

$\begin{array}{l}\Sigma Q\cdot {\delta }_{EV}={\displaystyle \int M_Q\frac{M_{P}dx}{EI}+}\Sigma \frac{1}{EI}\left(C{M}_1{M}_{3}L\right)+\Sigma F_Q\frac{F_{P}L}{AE}\\ \left(1\text{kips}\right)\left({\delta }_{EV}\right)=\left[\begin{array}{l}\frac{1}{2EI}{\displaystyle \underset{0}{\overset{10}{\int }}\left(-10+x\right)\left(-1.2x^2+48x-480\right)dx}+\frac{1}{EI}\left(10\times 430\right)\times 6\\ +\frac{1}{EI}\left(\frac{1}{2}\left(10\left(430+466\right)\right)\right)\times 6+\frac{58\left(12\right)}{AE}\end{array}\right]\\ \left(1\text{kips}\right)\left({\delta }_{EV}\right)=\left[\begin{array}{l}\frac{1}{2EI}{\displaystyle \underset{0}{\overset{10}{\int }}\left(12x^2-480x+4,800-1.2x^3+48x^2-480x\right)dx}\\ +\frac{25,800}{EI}+\frac{26,880}{EI}+\frac{696}{AE}\end{array}\right]\\ {\delta }_{EV}=\left[\begin{array}{l}\frac{1}{2EI}\left[\frac{12{\left(10\right)}^3}{3}-\frac{480{\left(10\right)}^2}{2}+4,800\left(10\right)-\frac{1.2{\left(10\right)}^4}{4}+\frac{48{\left(10\right)}^3}{3}-\frac{480{\left(10\right)}^2}{2}\right]+\frac{25,800}{EI}\\ +\frac{26,880}{EI}+\frac{696}{AE}\end{array}\right]\end{array}$$\begin{array}{c}{\delta }_{EV}=\frac{61,180}{EI}+\frac{696}{AE}\\ =\frac{61,180\left(1,728\right)}{\left(29,000\right)\left(180\right)}+\frac{696\left(12\right)}{\left(6\right)\left(29,000\right)}\\ =20.30\text{in}\text{.}\downarrow \end{array}$

Therefore, the vertical deflection at mid span of the member CD is $\underset{\_}{20.30\text{in}\text{.}\downarrow }$.

Apply 1 kips horizontally at point E for horizontal deflection $\left({\delta }_{EH}\right)$.

Sketch the $M_P$ diagram and $F_P$ diagram for ${\delta }_{EH}$ as shown in Figure 5.

Find the moment $M_Q$ of the Q-system:

For span AC,

$M_C=0$

$\begin{array}{c}{M}_A=-1\left(12\right)\\ =-12\text{kip-ft}\end{array}$

Find the force $F_Q$ of the Q-system:

$F_C=1\text{kips}$

$F_E=1\text{kips}$

The moment $M_Q$ for horizontal deflection at E and horizontal deflection at H are same and no axial loads are developed in the member BCED due to applied loading.

Therefore, ${\delta }_{BH}={\delta }_{EH}=\text{10}\text{.61}\text{in}\text{.}\to$.

Thus, the horizontal deflection at mid span of the member CD is $\underset{\_}{\text{10}\text{.61}\text{in}\text{.}\to }$.