Chapter 11.9, Problem 103P

11.103 and 11.104 Each member of the truss shown is made of steel and has a cross-sectional area of 500 mm². Using E = 200 GPa, determine the deflection indicated.

11.103 Vertical deflection of joint B.

11.104 Horizontal deflection of joint B.

Fig. P11.103 and P11.104

To determine

Calculate the vertical deflection of joint B $\left({\delta }_B\right)$.

Answer to Problem 103P

The vertical deflection of joint B $\left({\delta }_B\right)$ is $\underset{\_}{0.1459\text{mm}\downarrow }$.

Explanation of Solution

Given information:

The Young’s modulus of the steel (E) is $200\text{GPa}$.

The area of the each member (A) is $500{\text{mm}}^{\text{2}}$.

The vertical load act at the joint C (P) is $4.8\text{kN}$.

The length of the member AD $\left(L_{AD}\right)$ is $2.4\text{m}$.

The length of the member CD $\left(L_{CD}\right)$ is $2.5\text{m}$.

Calculation:

Consider the vertical force (Q) at joint B.

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

Refer to Figure 1.

Calculate the length of the member AB $\left(L_{AB}\right)$:

$\begin{array}{c}{L}_{AB}=\sqrt{{1.2}^2+{1.6}^2}\\ =2\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}\\ =2\times {10}^3\text{mm}\end{array}$

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

$\begin{array}{c}{L}_{BD}=L_{AB}\\ =2\times {10}^3\text{mm}\end{array}$

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

$\begin{array}{c}{L}_{AD}=2.4\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}\\ =2.4\times {10}^3\text{mm}\end{array}$

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

$\begin{array}{c}{L}_{CD}=2.5\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}\\ =2.5\times {10}^3\text{mm}\end{array}$

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

$\begin{array}{c}{L}_{BC}=\sqrt{{1.2}^2+{\left(2.5-1.6\right)}^2}\\ =1.5\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}\\ =1.5\times {10}^3\text{mm}\end{array}$

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.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ \frac{3}{5}{F}_{BC}+F_{CD}=0\end{array}$ (1)

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ \frac{4}{5}{F}_{BC}-4.8=0\\ \frac{4}{5}{F}_{BC}=4.8\end{array}$

$\begin{array}{l}{F}_{BC}=\frac{4.8\times 5}{4}\\ F_{BC}=6\text{kN}\end{array}$

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

Substitute $6\text{kN}$ for $F_{BC}$ in Equation (1).

$\begin{array}{l}\frac{3}{5}\left(6\right)+F_{CD}=0\\ F_{CD}=-\frac{3\times 6}{5}\\ F_{CD}=-3.6\text{kN}\end{array}$

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

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

Refer to Figure 3.

Calculate the horizontal forces by applying the equation of equilibrium:

Sum of horizontal forces is equal to 0.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ \frac{4}{5}{F}_{AB}+\frac{4}{5}{F}_{BD}-3.6=0\\ \frac{4}{5}{F}_{AB}+\frac{4}{5}{F}_{BD}=3.6\end{array}$ (2)

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ \frac{3}{5}{F}_{AB}-\frac{3}{5}{F}_{BD}-4.8-Q=0\\ \frac{3}{5}\left(F_{AB}-F_{BD}\right)=4.8+Q\end{array}$

$\begin{array}{l}{F}_{AB}-F_{BD}=\left(4.8+Q\right)\times \frac{5}{3}\\ F_{AB}=8+\frac{5}{3}Q+F_{BD}\end{array}$

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

Substitute $\left(8+\frac{5}{3}Q+F_{BD}\right)$ for $F_{AB}$ in Equation (2).

$\begin{array}{l}\frac{4}{5}\left(8+\frac{5}{3}Q+F_{BD}\right)+\frac{4}{5}{F}_{BD}=3.6\\ 6.4+\frac{4}{3}Q+\frac{4}{5}{F}_{BD}+\frac{4}{5}{F}_{BD}=3.6\\ \frac{8}{5}{F}_{BD}=3.6-6.4-\frac{4}{3}Q\\ \frac{8}{5}{F}_{BD}=-2.8-\frac{4}{3}Q\end{array}$

$\begin{array}{l}{F}_{BD}=\left(-2.8-\frac{4}{3}Q\right)\times \frac{5}{8}\\ F_{BD}=-\frac{7}{4}-\frac{5}{6}Q\end{array}$

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

Substitute $\left(-\frac{7}{4}-\frac{5}{6}Q\right)$ for $F_{BD}$ in Equation (2).

$\begin{array}{l}\frac{4}{5}{F}_{AB}+\frac{4}{5}\left(-\frac{7}{4}-\frac{5}{6}Q\right)=3.6\\ \frac{4}{5}{F}_{AB}-\frac{7}{5}-\frac{2}{3}Q=3.6\\ \frac{4}{5}{F}_{AB}=3.6+\frac{7}{5}+\frac{2}{3}Q\end{array}$

$\begin{array}{l}{F}_{AB}=\left(5+\frac{2}{3}Q\right)\times \frac{5}{4}\\ F_{AB}=\frac{25}{4}+\frac{5}{6}Q\end{array}$

Show the diagram of the joint D as in Figure 4.

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

Refer to Figure 4.

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ \frac{3}{5}{F}_{BD}+F_{AD}=0\\ F_{AD}=-\frac{3}{5}{F}_{BD}\end{array}$ (3)

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

Substitute $\left(-\frac{7}{4}-\frac{5}{6}Q\right)$ for $F_{BD}$ in Equation (3).

$\begin{array}{l}{F}_{AD}=-\frac{3}{5}\left(-\frac{7}{4}-\frac{5}{6}Q\right)\\ F_{AD}=\frac{21}{20}+\frac{1}{2}Q\end{array}$

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

$\begin{array}{c}\frac{\partial F_{AB}}{\partial Q}=0+\frac{5}{6}\\ =\frac{5}{6}\end{array}$

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

${\delta }_{AB}=\frac{F_{AB}{L}_{AB}}{AE}\times \frac{\partial F_{AB}}{\partial Q}$

Substitute $\left(\frac{25}{4}+\frac{5}{6}Q\right)$ for $F_{AB}$, $2\times {10}^3\text{mm}$ for $L_{AB}$, $500{\text{mm}}^{\text{2}}$, for $A$, $200\text{GPa}$ for E, and $\frac{5}{6}$ for $\frac{\partial F_{AB}}{\partial Q}$

$\begin{array}{c}{\delta }_{AB}=\frac{\left(\frac{25}{4}+\frac{5}{6}Q\right)\times 2\times {10}^3}{500\times 200\text{GPa}\times \frac{1{\text{kN/mm}}^2}{1\text{GPa}}}\times \frac{5}{6}\\ =\frac{5}{48}+\frac{1}{72}Q\end{array}$

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

$\begin{array}{c}\frac{\partial F_{AD}}{\partial Q}=0+\frac{1}{2}\\ =\frac{1}{2}\end{array}$

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

${\delta }_{AD}=\frac{F_{AD}{L}_{AD}}{AE}\times \frac{\partial F_{AD}}{\partial Q}$

Substitute $\left(\frac{21}{20}+\frac{1}{2}Q\right)$ for $F_{AD}$, $2.4\times {10}^3\text{mm}$ for $L_{AD}$, $500{\text{mm}}^{\text{2}}$, for $A$, $200\text{GPa}$ for E, and $\frac{1}{2}$ for $\frac{\partial F_{AD}}{\partial Q}$.

$\begin{array}{c}{\delta }_{AD}=\frac{\left(\frac{21}{20}+\frac{1}{2}Q\right)\times 2.4\times {10}^3}{500\times 200\text{GPa}\times \frac{1{\text{kN/mm}}^2}{1\text{GPa}}}\times \frac{1}{2}\\ =\frac{63}{5,000}+\frac{3}{500}Q\end{array}$

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

$\begin{array}{c}\frac{\partial F_{BD}}{\partial Q}=0-\frac{5}{6}\\ =-\frac{5}{6}\end{array}$

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

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

Substitute $\left(-\frac{7}{4}-\frac{5}{6}Q\right)$ for $F_{BD}$, $2\times {10}^3\text{mm}$ for $L_{BD}$, $500{\text{mm}}^{\text{2}}$, for $A$, $200\text{GPa}$ for E, and $\left(-\frac{5}{6}\right)$ for $\frac{\partial F_{BD}}{\partial Q}$.

$\begin{array}{c}{\delta }_{BD}=\frac{\left(-\frac{7}{4}-\frac{5}{6}Q\right)\times 2\times {10}^3}{500\times 200\text{GPa}\times \frac{1{\text{kN/mm}}^2}{1\text{GPa}}}\times \left(-\frac{5}{6}\right)\\ =\frac{7}{240}+\frac{1}{72}Q\end{array}$

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

$\frac{\partial F_{BC}}{\partial Q}=0$

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

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

Substitute $6\text{kN}$ for $F_{BC}$, $1.5\times {10}^3\text{mm}$ for $L_{BC}$, $500{\text{mm}}^{\text{2}}$, for $A$, $200\text{GPa}$ for E, and 0 for $\frac{\partial F_{BC}}{\partial Q}$.

$\begin{array}{c}{\delta }_{BC}=\frac{6\times 1.5\times {10}^3}{500\times 200\text{GPa}\times \frac{1{\text{kN/mm}}^2}{1\text{GPa}}}\times 0\\ =0\end{array}$

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

$\frac{\partial F_{CD}}{\partial Q}=0$

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

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

Substitute $\left(-3.6\text{kN}\right)$ for $F_{BC}$, $2.5\times {10}^3\text{mm}$ for $L_{BC}$, $500{\text{mm}}^{\text{2}}$, for $A$, $200\text{GPa}$ for E, and 0 for $\frac{\partial F_{BC}}{\partial Q}$.

$\begin{array}{c}{\delta }_{BC}=\frac{-3.6\times 2.5\times {10}^3}{500\times 200\text{GPa}\times \frac{1{\text{kN/mm}}^2}{1\text{GPa}}}\times 0\\ =0\end{array}$

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

${\delta }_B={\delta }_{AB}+{\delta }_{AD}+{\delta }_{BD}+{\delta }_{BC}+{\delta }_{CD}$

Substitute $\left(\frac{5}{48}+\frac{1}{72}Q\right)$ for ${\delta }_{AB}$, $\left(\frac{63}{5,000}+\frac{3}{500}Q\right)$ for ${\delta }_{AD}$, $\left(\frac{7}{240}+\frac{1}{72}Q\right)$ for ${\delta }_{BD}$, 0 for ${\delta }_{BC}$, and 0 for ${\delta }_{CD}$.

$\begin{array}{c}{\delta }_B=\left[\begin{array}{l}\left(\frac{5}{48}+\frac{1}{72}Q\right)+\left(\frac{63}{5,000}+\frac{3}{500}Q\right)+\\ \left(\frac{7}{240}+\frac{1}{72}Q\right)+0+0\end{array}\right]\\ =0.1459-0.0338Q\end{array}$

Substitute 0 for Q.

$\begin{array}{c}{\delta }_B=0.1459-0.0338\left(0\right)\\ =0.1459\text{mm}\downarrow \end{array}$

Hence the vertical deflection of joint B $\left({\delta }_B\right)$ is $\underset{\_}{0.1459\text{mm}\downarrow }$.