Chapter 9, Problem 29P

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 of all the bars is $1000{\text{mm}}^2$.

Calculation:

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

Refer Figure 1.

Find the length of the inclined members of the truss as follows:

Consider the length of the inclined members AE, EB, BD, DC is l.

$\begin{array}{c}l=\sqrt{3^2+4^2}\\ =\sqrt{25}\\ =5\text{m}\end{array}$

By symmetry of truss,

$AE=EB=BD=DC=5\text{m}$

Find the angle $\theta$ as follows:

$\begin{array}{l}\tan \theta =\frac{4}{3}\\ \theta ={\tan }^{-1}\left(\frac{4}{3}\right)\\ \theta =53.130°\end{array}$

Consider the horizontal reaction at D as the redundant.

Remove the redundant at D 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 and D as follows:

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_{x1}=0\end{array}$

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ \left(D_{y1}\times 9\right)-\left(100\times 3\right)-\left(60\times 6\right)-\left(60\times 12\right)=0\\ D_{y1}=\left(\frac{1380}{9}\right)\\ D_{y1}=153.3\text{kN}\end{array}$

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_{y1}+D_{y1}=60+60+100\\ A_{y1}=220-D_{y1}\\ A_{y1}=220-153.3\\ A_{y1}=66.7\text{kN}\end{array}$

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

Refer Figure 2.

Consider Joint A.

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_{y1}-F_{AE1}\sin \theta =0\\ 66.7-F_{AE1}\sin \left(53.130°\right)=0\\ 66.7-0.8F_{AE1}=0\\ F_{AE1}=83.3\text{kN}\left(\text{T}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_{x_1}+F_{AB1}+F_{AE1}\cos \theta =0\\ 0+F_{AB1}+F_{AE1}\cos \left(53.130°\right)=0\\ F_{AB1}+0.6\left(83.3\right)=0\\ F_{AB1}=-50\text{kN}\\ F_{AB1}=50\text{kN}\left(\text{C}\right)\end{array}$

Consider joint C.

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{CD1}\sin \theta -60=0\\ -F_{CD1}\sin \left(53.130°\right)-60=0\\ -0.8F_{CD1}-60=0\\ F_{CD1}=75\text{kN}\left(\text{C}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{BC1}-F_{CD1}\cos \theta =0\\ F_{BC1}=-\left(-75\right)\cos \left(53.130°\right)\\ F_{BC1}=45\text{kN}\left(\text{T}\right)\end{array}$

Consider joint B.

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -60-F_{BE1}\sin \left(53.130°\right)-F_{BD1}\sin \left(53.130°\right)=0\\ 0.8F_{BE1}+0.8F_{BD1}=-60\end{array}$        (1)

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{BA1}-F_{BE1}\cos \left(53.130°\right)+F_{BD1}\cos \left(53.130°\right)+F_{BC1}=0\\ -\left(-50\right)-0.6F_{BE1}+0.6F_{BD1}+45=0\\ -0.6F_{BE1}+0.6F_{BD1}=95\end{array}$        (2)

Solve Equation (1) and (2).

$\begin{array}{l}{F}_{BE1}=41.7\text{kN}\left(\text{T}\right)\\ F_{BD1}=116.7\text{kN}\left(\text{C}\right)\end{array}$

Consider joint D.

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{EA1}\cos \theta +F_{ED1}+F_{EB1}\cos \theta =0\\ -\left(83.3\right)\cos \left(\text{53}\text{.130}°\right)+F_{ED1}+\left(41.7\right)\cos \left(\text{53}\text{.130}°\right)=0\\ -50+F_{ED1}+25=0\\ F_{ED1}=25\text{kN}\left(\text{T}\right)\end{array}$

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

Members	$F_P$ forces (kN)
AB	50 (C)
BC	45 (T)
CD	75 (C)
DE	25 (T)
EA	83.3 (T)
BE	41.7 (T)
BD	116.7 (C)

Table 1

Show the released structure loaded with unit horizontal load at D as shown in Figure 3.

Refer Figure 3.

By symmetry of the truss,

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

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ D_y\times 9-\left(1\times 4\right)=0\\ D_y=0.444\text{kN}\end{array}$

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y+D_y=0\\ A_y+0.444\text{kN=0}\\ A_y=-0.444\text{kN}\end{array}$

Find the $F_Q$ forces in the members of the released structure due to unit load as follows:

Refer Figure 3.

Consider Joint A.

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y-F_{AE}\sin \theta =0\\ -0.444-F_{AE}\sin \left(53.130°\right)=0\\ -0.444-0.8F_{AE}=0\\ F_{AE}=0.556\text{kN}\left(\text{C}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x+F_{AB}+F_{AE}\cos \theta =0\\ 1+F_{AB}+F_{AE}\cos \left(53.130°\right)=0\\ F_{AB}+0.6\left(-0.556\right)=-1\\ F_{AB}=-0.664\text{kN}\\ F_{AB}=0.664\text{kN}\left(\text{C}\right)\end{array}$

Consider joint C.

In a two member forces BC and CD, 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.

The forces in the member BC and CD is $F_{BC}=F_{CD}=0$.

Consider joint B.

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{BE}\sin \left(53.130°\right)-F_{BD}\sin \left(53.130°\right)=0\\ 0.8F_{BE}+0.8F_{BD}=0\\ F_{BE}=-F_{BD}\end{array}$        (3)

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{BA}-F_{BE}\cos \left(53.130°\right)+F_{BD}\cos \left(53.130°\right)+F_{BC}=0\\ -\left(-50\right)-0.6F_{BE1}+0.6F_{BD1}+45=0\\ -0.6F_{BE1}+0.6F_{BD1}=95\end{array}$        (4)

Solve Equation (1) and (2).

$\begin{array}{l}{F}_{BE1}=41.7\text{kN}\left(\text{T}\right)\\ F_{BD1}=116.7\text{kN}\left(\text{C}\right)\end{array}$

Consider joint D.

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{EA}\cos \theta +F_{ED}+F_{EB}\cos \theta =0\\ -\left(-0.556\right)\cos \left(\text{53}\text{.130}°\right)+F_{ED}+\left(0.556\right)\cos \left(\text{53}\text{.130}°\right)=0\\ -\left(-0.556\right)\cos \left(\text{53}\text{.130}°\right)+F_{ED}+\left(0.556\right)\cos \left(\text{53}\text{.130}°\right)=0\\ 0.3336+F_{ED}+0.3336=0\\ F_{ED}=0.667\text{kN}\left(\text{C}\right)\end{array}$

Show the Q forces in the members of the truss due to the unit load at D as shown in Table 1.

Members	$F_Q$ forces (kN)
AB	0.667 (C)
BC	0
CD	0
DE	0.667 (C)
EA	0.556 (C)
BE	0.556 (T)
BD	0.556 (C)

Table 2

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

Members	$F_P$ forces(kN)	$F_Q$ forces (kN)	$L\left(\text{m}\right)$	${\Delta }_{DO}=\frac{F_Q{F}_{P}L}{AE}$	${\delta }_{DD}=\frac{F_Q^{2}L}{AE}$
AB	-50	-0.667	6	$\frac{200.1}{AE}$	$\frac{2.669}{AE}$
BC	45	0	6	0	$0$
CD	-75	0	5	0	0
DE	25	-0.667	6	$\frac{-100.05}{AE}$	$\frac{2.669}{AE}$
EA	83.3	-0.556	5	$\frac{-231.574}{AE}$	$\frac{1.5456}{AE}$
BE	41.7	0.556	5	$\frac{115.926}{AE}$	$\frac{1.5456}{AE}$
BD	-116.7	-0.556	5	$\frac{324.426}{AE}$	$\frac{1.5456}{AE}$

Table 3

Refer Table 3.

$\begin{array}{l}{\Delta }_{DO}={\displaystyle \sum \frac{F_Q{F}_{P}L}{AE}}\\ {\Delta }_{DO}=\frac{309}{AE}\end{array}$

$\begin{array}{c}{\delta }_{DD}={\displaystyle \sum \frac{F_Q^{2}L}{AE}}\\ =\frac{10}{AE}\end{array}$

Find the horizontal reaction at D as follows:

$\begin{array}{c}{\Delta }_D={\Delta }_{DD}+{\delta }_{DD}{R}_D\\ 0=\frac{309}{AE}+\frac{10}{AE}{R}_D\\ R_D=-\frac{309}{10}\\ R_D=30.9\text{kN}(\to )\end{array}$

Thus, the horizontal reaction at D is $\underset{\_}{R_D=30.9\text{kN}(\to )}$.

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

Members	$F_P\left(\text{kN}\right)$	$F_Q\left(\text{kN}\right)$	$R_D(\to )\left(\text{kN}\right)$	$F_i=F_P+F_Q{R}_D$(kN)
AB	-50	-0.667	-30.9	-29.4
BC	45	0	-30.9	45
CD	-75	0	-30.9	-75
DE	25	-0.667	-30.9	45
EA	83.3	-0.556	-30.9	100.5
BE	41.7	0.556	-30.9	24.5
BD	-116.7	-0.556	-30.9	-99.5

Table 4

Show the final support reaction as shown in Table 5.

Support Reactions
Members	$P_i\left(\text{kN}\right)$	$P_Q\left(\text{kN}\right)$ forces	$R_D\left(\text{kN}\right)(\to )$	$\text{Reaction}=P_i+P_Q{R}_D\text{kN}$
$H_A$	0	1	-30.9	-30.9
$R_A$	66.7	-0.444	-30.9	80.4
$R_D$	153.3	0.444	-30.9	139.6

Table 5

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