Chapter 9, Problem 27P

To determine

Find the reactions and all bar forces for the truss.

Explanation of Solution

Given information:

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

The area (A) of all the bars is $1000{\text{mm}}^2$.

Identify Zero Force Members:

In a two member forces, if both the members are not parallel and the no external loads or reactions acts at that joint. Then, the force in both the members is zero.

Calculation:

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

Refer Figure 1.

Find the length of the member AB and BC as follows:

Consider triangle ABD.

$\begin{array}{c}AB=\sqrt{{\left(4+2\right)}^2+8^2}\\ =\sqrt{6^2+8^2}\\ =\sqrt{100}\\ =10\text{m}\end{array}$

By symmetry of truss,

$AB=BC=10\text{m}$

Find the length of the member AD and DC as follows:

Consider triangle ADC.

$\begin{array}{c}AD=\sqrt{2^2+8^2}\\ =\sqrt{2^2+8^2}\\ =\sqrt{68}\\ =8.246\text{m}\end{array}$

By symmetry of truss,

$AD=DC=8.246\text{m}$

Find the angle ${\theta }_1$ as follows:

$\begin{array}{l}\tan {\theta }_1=\frac{2}{8}\\ {\theta }_1={\tan }^{-1}\left(\frac{2}{8}\right)\\ {\theta }_1=14.036°\end{array}$

Find the angle ${\theta }_2$ as follows:

$\begin{array}{l}\tan {\theta }_2=\frac{6}{8}\\ {\theta }_2={\tan }^{-1}\left(\frac{6}{8}\right)\\ {\theta }_2=36.869°\end{array}$

Consider the horizontal reaction at C as the redundant.

Remove the redundant at C to get the released structure.

Show the released structure with applied load as shown in Figure 2.

Refer Figure 2.

Find the reactions at A, D, and C as follows:

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_{x1}+30=0\\ A_{x1}=-30\text{kN}\end{array}$

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ C_{y1}\times 16-\left(18\times 8\right)-\left(30\times 6\right)=0\\ C_{y1}=\left(\frac{324}{16}\right)\\ C_{y1}=20.25\text{kN}\end{array}$

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_{y1}+C_{y1}=18\\ A_{y1}=18-C_{y1}\\ A_{y1}=18-\left(20.25\right)\\ A_{y1}=-2.25\text{kN}\end{array}$

Find the forces $F_P$ in the bars of released structure due to applied load as follows:

Refer Figure 2.

Consider Joint C.

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{CD1}\cos {\theta }_1-F_{CB1}\cos {\theta }_2=0\\ -F_{CD1}\cos \left(14.036°\right)-F_{CB1}\cos \left(36.869°\right)=0\\ F_{CD1}=-F_{CB1}\left[\frac{\cos \left(36.869°\right)}{\cos \left(14.036°\right)}\right]\\ F_{CD1}=-F_{CB1}\left(\frac{0.8}{0.97}\right)\\ F_{CD1}=-0.824F_{CB1}\end{array}$

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{CD1}\sin {\theta }_1+F_{CB1}\sin {\theta }_2+C_{y1}=0\\ \left(-0.824F_{CB1}\right)\sin \left(14.036°\right)+F_{CB1}\sin \left(36.869°\right)+20.25=0\\ \left(-0.824F_{CB1}\right)\sin \left(14.036°\right)+F_{CB1}\sin \left(36.869°\right)+20.25=0\\ -0.2F_{CB1}+0.6F_{CB1}=-20.25\end{array}$

$\begin{array}{l}{F}_{CB1}=-\frac{20.25}{0.4}\\ F_{CB1}=-50.625\text{kN}\\ F_{CB1}=50.625\text{kN}\left(\text{C}\right)\end{array}$

$\begin{array}{c}{F}_{CD1}=-0.824F_{CB1}\\ =-0.824\left(-50.625\right)\\ =41.715\text{kN}\left(\text{T}\right)\end{array}$

Consider joint B.

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ 30-F_{AB1}\cos {\theta }_2+F_{BC1}\cos {\theta }_2=0\\ 30-F_{AB1}\cos \left(36.869°\right)+F_{BC1}\cos \left(36.869°\right)=0\\ 30-0.8F_{AB1}+0.8F_{BC1}=0\\ 30-0.8F_{AB1}+0.8\left(-50.625\right)=0\\ F_{AB1}=-\frac{10.5}{0.8}\\ F_{AB1}=13.125\text{kN}\left(\text{C}\right)\end{array}$

Consider joint D.

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{AD1}\cos {\theta }_2+F_{DC1}\cos {\theta }_2=0\\ -F_{AD1}\cos {\theta }_2=-F_{DC1}\cos {\theta }_2\\ F_{AD1}=F_{DC1}\\ F_{AD1}=41.75\text{kN}\left(\text{T}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{DB1}-18-F_{AD1}\sin {\theta }_1-F_{DC1}\sin {\theta }_1=0\\ F_{DB1}-18-41.75\sin \left(14.036°\right)-41.75\sin \left(14.036°\right)=0\\ F_{DB1}-18-20.2513=0\\ F_{DB1}=38.25\text{kN}\left(\text{T}\right)\end{array}$

Show the P forces in the members of the truss due to the applied load at C as shown in Table 1.

Members	$F_P$ forces
AB	$13.125\text{kN}\left(\text{C}\right)$
BC	$50.625\text{kN}\left(\text{C}\right)$
AD	$41.715\text{kN}\left(\text{T}\right)$
DC	$41.715\text{kN}\left(\text{T}\right)$
DB	$38.25\text{kN}\left(\text{T}\right)$

Table 1

Show the released structure loaded with unit horizontal load at C as shown in Figure 2.

Refer Figure 3.

By symmetry of the truss,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x+1=0\\ A_x=-1\text{kN}\end{array}$

Consider Joint D.

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{CD}\cos {\theta }_1-F_{CB}\cos {\theta }_2=-1\\ -F_{CD}\cos \left(14.036°\right)-F_{CB}\cos \left(36.869°\right)=-1\end{array}$        (1)

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{CD}\sin {\theta }_1+F_{CB}\sin {\theta }_2+C_y=0\\ F_{CD}\sin \left(14.036°\right)+F_{CB}\sin \left(36.869°\right)=0\end{array}$        (2)

Solve Equation (1) and (2).

$\begin{array}{l}{F}_{CD}=1.546\text{kN}\left(\text{T}\right)\\ F_{CB}=0.625\text{kN}\left(\text{C}\right)\end{array}$

By symmetry of the truss,

$\begin{array}{l}{F}_{AD}=F_{CD}=1.546\text{kN}\left(\text{T}\right)\\ F_{AB}=F_{CB}=0.625\text{kN}\left(\text{C}\right)\end{array}$

Consider Joint B.

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{AB}\sin {\theta }_2-F_{BC}\cos {\theta }_2-F_{BD}=0\\ -F_{BD}=F_{CD}\sin \left(36.869°\right)+F_{CB}\sin \left(36.869°\right)\\ -F_{BD}=\left(-0.625\right)\sin \left(36.869°\right)+\left(-0.625\right)\sin \left(36.869°\right)\\ -F_{BD}=-0.75\\ F_{BD}=0.75\text{kN}\left(\text{T}\right)\end{array}$

Show the Q forces in the members of the truss due to the unit load at C as shown in Table 2.

Members	$F_Q$ forces
AB	$0.625\text{kN}\left(\text{C}\right)$
BC	$0.625\text{kN}\left(\text{C}\right)$
CD	$1.546\text{kN}\left(\text{T}\right)$
AC	$1.546\text{kN}\left(\text{T}\right)$
DB	$0.75\text{kN}\left(\text{T}\right)$

Table 2

Find the deflection ${\Delta }_{CO}$ of the released structure due to the applied load and the deflection ${\delta }_{CC}$ of the released structure due to the unit load as follows:

Members	$F_P$ forces(kN)	$F_Q$ forces(kN)	$L\left(\text{m}\right)$	${\Delta }_{CX}=\frac{F_Q{F}_{P}L}{AE}$	${\delta }_{CC}=\frac{F_Q^{2}L}{AE}$
AB	$-13.125\text{kN}$	$-0.625\text{kN}$	10	$\frac{82.03}{AE}$	$\frac{3.90625}{AE}$
BC	$-50.625\text{kN}$	$-0.625\text{kN}$	10	$\frac{316.4063}{AE}$	$\frac{3.90625}{AE}$
AD	$41.715\text{kN}$	$1.546\text{kN}$	8.246	$\frac{531.796}{AE}$	$\frac{19.7089}{AE}$
DC	$41.715\text{kN}$	$1.546\text{kN}$	8.246	$\frac{531.796}{AE}$	$\frac{19.7089}{AE}$
DB	$38.25\text{kN}$	$0.75\text{kN}$	4	$\frac{114.75}{AE}$	$\frac{2.25}{AE}$

Table 3

Refer Table 3.

$\begin{array}{l}{\Delta }_{CX}={\displaystyle \sum \frac{F_Q{F}_{P}L}{AE}}\\ {\Delta }_{CX}=\frac{1576}{AE}\end{array}$

$\begin{array}{c}{\delta }_{CC}={\displaystyle \sum \frac{F_Q^{2}L}{AE}}\\ =\frac{49.48}{AE}\end{array}$

Find the horizontal reaction at C as follows:

$\begin{array}{c}{\Delta }_C={\Delta }_{CX}+{\delta }_{CC}{R}_C\\ 0=\frac{1576}{AE}+\frac{49.48}{AE}{R}_C\\ R_C=-\frac{1576}{49.48}\\ R_C=-31.89\text{kN}\end{array}$

Thus, the horizontal reaction at C is $\underset{\_}{R_C=-31.89\text{kN}}$.

Find final forces in the truss as shown in Table 4.

Members	$F_P\left(\text{kN}\right)$	$F_Q\left(\text{kN}\right)$	$R_C(\to )\left(\text{kN}\right)$	$F_i=F_P+F_Q{R}_C$ (kN)
AB	$-13.125\text{kN}$	$-0.625\text{kN}$	-31.89	6.80
BC	$-50.625\text{kN}$	$-0.625\text{kN}$	-31.89	-30.7
AD	$41.715\text{kN}$	$1.546\text{kN}$	-31.89	-7.5
DC	$41.715\text{kN}$	$1.546\text{kN}$	-31.89	-7.5
DB	$38.25\text{kN}$	$0.75\text{kN}$	-31.89	14.34

Table 4

Show the final support reaction as shown in Table 5.

Support Reactions
Members	$P_i\left(\text{kN}\right)$	$F_Q\left(\text{kN}\right)$ forces	$R_C\left(\text{kN}\right)(\to )$	$\text{Reaction}=P_i+F_Q{R}_C\text{kN}$
$H_A$	-30	-1	-31.89	1.89
$R_A$	-2.25	0	-31.89	-2.25
$R_C$	20.25	1	-31.89	20.25

Table 5

Thus, the reactions and all bar forces for the truss is tabulate in table 4 and table 5.