Chapter 11.9, Problem 102P

11.101 and 11.102 Each member of the truss shown is made of steel and has the cross-sectional area shown. Using E = 29 × 10⁶ psi, determine the deflection indicated.

11.101 Vertical deflection of joint C.

11.102 Horizontal deflection of joint C.

Fig. P11.101 and P11.102

To determine

Calculate the horizontal deflection of joint C $\left({\delta }_Q\right)$.

Answer to Problem 102P

The horizontal deflection of joint C $\left({\delta }_Q\right)$ is $\underset{\_}{0.0103\text{in}\text{.}\leftarrow }$.

Explanation of Solution

Given information:

The Young’s modulus of the steel (E) is $29\times {10}^6\text{psi}$.

The area of the member BC $\left(A_{BC}\right)$ is $4\text{in}{\text{.}}^{\text{2}}$.

The area of the member BD $\left(A_{BD}\right)$ is $3\text{in}{\text{.}}^{\text{2}}$.

The area of the member CD $\left(A_{BC}\right)$ is $6\text{in}{\text{.}}^{\text{2}}$.

The vertical load act at the joint C (P) is $80\text{kips}$.

The horizontal load act at the joint C (G) is $48\text{kips}$.

The length of the member BD $\left(L_{BD}\right)$ is $5\text{ft}$.

The length of the member (L) is $6\text{ft}$.

Calculation:

Show the free body diagram of the truss members as in Figure 1.

Refer to Figure 1.

The length of the member BC $\left(L_{BC}\right)$:

$\begin{array}{c}{L}_{BC}=\sqrt{{2.5}^2+6^2}\\ =6.5\text{ft}\times \frac{12\text{in}\text{.}}{1\text{ft}}\\ =78\text{in}\text{.}\end{array}$

The length of the member CD $\left(L_{CD}\right)$:

$\begin{array}{c}{L}_{CD}=L_{BC}\\ =78\text{in}\text{.}\end{array}$

The length of the member BD $\left(L_{BD}\right)$:

$\begin{array}{c}{L}_{BD}=5\text{ft}\times \frac{12\text{in}\text{.}}{1\text{ft}}\\ =60\text{in}\text{.}\end{array}$

Show the diagram of the joint C as in Figure 2.

Here, $F_{BC}$ is the force act at the member BC and $F_{CD}$ is the force act at the member CD.

Refer to Figure 2.

Calculate the horizontal forces by applying the equation of equilibrium:

Sum of horizontal forces is equal to 0.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -\frac{12}{13}{F}_{BC}-\frac{12}{13}{F}_{CD}-Q=0\\ \frac{12}{13}\left(F_{BC}+F_{CD}\right)=-Q\\ F_{BC}+F_{CD}=-\frac{13}{12}Q\end{array}$ (1)

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ \frac{5}{13}{F}_{BC}-\frac{5}{13}{F}_{CD}-P=0\\ \frac{5}{13}\left(F_{BC}-F_{CD}\right)=P\\ F_{BC}-F_{CD}=\frac{13}{5}P\end{array}$

$F_{BC}=\frac{13}{5}P+F_{CD}$

Calculate the force act at the member CD $\left(F_{CD}\right)$:

Substitute $\left(\frac{13}{5}P+F_{CD}\right)$ for $F_{BC}$ in Equation (1).

$\begin{array}{l}\frac{13}{5}P+F_{CD}+F_{CD}=-\frac{13}{12}Q\\ 2F_{CD}=-\frac{13}{12}Q-\frac{13}{5}P\\ F_{CD}=-\frac{13}{24}Q-\frac{13}{10}P\end{array}$

Calculate the force act at the member BC $\left(F_{BC}\right)$:

Substitute $\left(-\frac{13}{24}Q-\frac{13}{10}P\right)$ for $F_{CD}$ in Equation (1).

$\begin{array}{l}{F}_{BC}-\frac{13}{24}Q-\frac{13}{10}P=-\frac{13}{12}Q\\ F_{BC}=-\frac{13}{12}Q+\frac{13}{24}Q+\frac{13}{10}P\\ F_{BC}=-\frac{13}{24}Q+\frac{13}{10}P\end{array}$

Show the diagram of the joint D as in Figure 3.

Here, $F_{BD}$ is the force act at the member BD.

Refer to Figure 3.

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ \frac{5}{13}{F}_{CD}+F_{BD}=0\end{array}$

Substitute $\left(-\frac{13}{24}Q-\frac{13}{10}P\right)$ for $F_{CD}$.

$\begin{array}{l}\frac{5}{13}\left(-\frac{13}{24}Q-\frac{13}{10}P\right)+F_{BD}=0\\ -\frac{5}{24}Q-\frac{1}{2}P+F_{BD}=0\\ F_{BD}=\frac{5}{24}Q+\frac{1}{2}P\end{array}$

Partial differentiate the force act at the member BC $\left(F_{BD}\right)$ with respect to Q.

$\begin{array}{c}\frac{\partial F_{BC}}{\partial Q}=-\frac{13}{24}+0\\ =-\frac{13}{24}\end{array}$

Calculate the deflection of the member BC $\left({\delta }_{BC}\right)$ using the formula:

${\delta }_{BC}=\frac{F_{BC}{L}_{BC}}{A_{BC}E}\times \frac{\partial F_{BC}}{\partial Q}$

Substitute $\left(-\frac{13}{24}Q+\frac{13}{10}P\right)$ for $F_{BC}$, $78\text{in}\text{.}$ for $L_{BC}$, $4\text{in}{\text{.}}^{\text{2}}$, for $A_{BC}$, $29\times {10}^6\text{psi}$ for E, and $\left(-\frac{13}{24}\right)$ for $\frac{\partial F_{BC}}{\partial Q}$

$\begin{array}{c}{\delta }_{BC}=\frac{\left(-\frac{13}{24}Q+\frac{13}{10}P\right)\times 78}{4\times 29\times {10}^6\text{psi}\times \frac{{10}^{-3}\text{psi}}{1\text{ksi}}}\times \left(-\frac{13}{24}\right)\\ =0.197\times {10}^{-3}Q-0.473\times {10}^{-3}P\end{array}$

Partial differentiate the force act at the member CD $\left(F_{CD}\right)$ with respect to Q.

$\begin{array}{c}\frac{\partial F_{CD}}{\partial Q}=-\frac{13}{24}-0\\ =-\frac{13}{24}\end{array}$

Calculate the deflection of the member CD $\left({\delta }_{CD}\right)$ using the formula:

${\delta }_{CD}=\frac{F_{CD}{L}_{CD}}{A_{CD}E}\times \frac{\partial F_{CD}}{\partial Q}$

Substitute $\left(-\frac{13}{24}Q-\frac{13}{10}P\right)$ for $F_{CD}$, $78\text{in}\text{.}$ for $L_{CD}$, $6\text{in}{\text{.}}^{\text{2}}$, for $A_{CD}$, $29\times {10}^6\text{psi}$ for E, and $\left(-\frac{13}{24}\right)$ for $\frac{\partial F_{CD}}{\partial Q}$.

$\begin{array}{c}{\delta }_{CD}=\frac{\left(-\frac{13}{24}Q-\frac{13}{10}P\right)\times 78}{6\times 29\times {10}^6\text{psi}\times \frac{{10}^{-3}\text{psi}}{1\text{ksi}}}\times \left(-\frac{13}{24}\right)\\ =0.131\times {10}^{-3}Q+0.315\times {10}^{-3}P\end{array}$

Partial differentiate the force act at the member BD $\left(F_{BD}\right)$ with respect to Q.

$\begin{array}{c}\frac{\partial F_{BD}}{\partial Q}=\frac{5}{24}+0\\ =\frac{5}{24}\end{array}$

Calculate the strain energy of the member BD $\left({\delta }_{BD}\right)$ using the formula:

${\delta }_{BD}=\frac{F_{BD}{L}_{BD}}{A_{BD}E}\times \frac{\partial F_{BD}}{\partial Q}$

Substitute $\left(\frac{5}{24}Q+\frac{1}{2}P\right)$ for $F_{BD}$, $60\text{in}\text{.}$ for $L_{BD}$, $3\text{in}{\text{.}}^{\text{2}}$, for $A_{BD}$, $29\times {10}^6\text{psi}$ for E, and $\frac{5}{24}$ for $\frac{\partial F_{BD}}{\partial Q}$.

$\begin{array}{c}{\delta }_{BD}=\frac{\left(\frac{5}{24}Q+\frac{1}{2}P\right)\times 60}{3\times 29\times {10}^6\text{psi}\times \frac{{10}^{-3}\text{psi}}{1\text{ksi}}}\times \frac{5}{24}\\ =0.0299\times {10}^{-3}Q+0.0718\times {10}^{-3}P\end{array}$

Calculate the vertical deflection of joint C $\left({\delta }_P\right)$:

${\delta }_P={\delta }_{BC}+{\delta }_{CD}+{\delta }_{BD}$

Substitute $\left(0.197\times {10}^{-3}Q-0.473\times {10}^{-3}P\right)$ for ${\delta }_{BC}$, $\left(0.131\times {10}^{-3}Q+0.315\times {10}^{-3}P\right)$ for ${\delta }_{CD}$, and $\left(0.0299\times {10}^{-3}Q+0.0718\times {10}^{-3}P\right)$ for ${\delta }_{BD}$.

$\begin{array}{c}{\delta }_Q=\left[\begin{array}{l}\left(0.197\times {10}^{-3}Q-0.473\times {10}^{-3}P\right)+\\ \left(0.131\times {10}^{-3}Q+0.315\times {10}^{-3}P\right)+\\ \left(0.0299\times {10}^{-3}Q+0.0718\times {10}^{-3}P\right)\end{array}\right]\\ =0.3579\times {10}^{-3}Q-0.0862\times {10}^{-3}P\end{array}$

Substitute $80\text{kips}$ for P and $48\text{kips}$ for Q.

$\begin{array}{c}{\delta }_Q=0.3579\times {10}^{-3}\left(48\right)-0.0862\times {10}^{-3}\left(80\right)\\ =0.0103\text{in}\text{.}\leftarrow \end{array}$

Hence the horizontal deflection of joint C $\left({\delta }_Q\right)$ is $\underset{\_}{0.0103\text{in}\text{.}\leftarrow }$.