Chapter 6.2, Problem 6.47P

A floor truss is loaded as shown. Determine the force members CF, EF, and EG.

Fig. P6.47 and P6.48

To determine

The force in the members $CF,EF\text{and }EG$.

Answer to Problem 6.47P

The force in the members $CF,EF\text{and }EG$ are $26.0\text{kN}$ the member of $CF$ is in Tension, $1.118\text{kN}$ the member of the CF is in tension and $27.0\text{kN}$, the member  of CF is in compression.

Explanation of Solution

The floor truss is loaded as shown in the figure. The free body diagram of the given arrangement is given by Figure 1.

 

Figure 1

For the entire truss, the force along the positive $x$ axis is zero. This gives,

$k_x=0$

The moment of force in the counter clockwise direction about $A$ will also be zero. From the diagram,

$\begin{array}{l}{k}_y\left(4.8\text{m}\right)-\left(4\text{kN}\right)\left(0.8\text{m}\right)-\left(4\text{kN}\right)\left(1.6\text{m}\right)\\ -\left(3\text{kN}\right)\left(2.4\text{m}\right)-\left(2\text{kN}\right)\left(3.2\text{m}\right)-\left(2\text{kN}\right)\left(4\text{m}\right)-\left(1\text{kN}\right)\left(4.8\text{m}\right)=0\end{array}$ (I)

Solve for $k_y$ from (I).

$\begin{array}{c}{k}_y=\frac{\left(4\text{kN}\right)\left(0.8\text{m}\right)+\left(4\text{kN}\right)\left(1.6\text{m}\right)+\left(3\text{kN}\right)\left(2.4\text{m}\right)+\left(2\text{kN}\right)\left(3.2\text{m}\right)+\left(2\text{kN}\right)\left(4\text{m}\right)+\left(1\text{kN}\right)\left(4.8\text{m}\right)}{\left(4.8\text{m}\right)}\\ =7.5\text{kN}\end{array}$

This implies that $k=\left(7.5\text{kN}\right)j$.

The force in the positive $y$ axis sums up to zero. This gives,

$A+k-18\text{kN}=0$

Substitute $7.5\text{kN}$ for $k$ to get $A$.

$\begin{array}{c}A+7.5\text{kN}-18\text{kN}=0\\ A=10.5\text{kN}\end{array}$

This implies $A=\left(10.5\text{kN}\right)j$.

For the section $CF,EF\text{and }EG$ the sum of the moments in the counter clockwise direction about $E$ is given as,

$F_{CF}\left(0.4\text{m}\right)-A\left(1.6\text{m}\right)+\left(2\text{kN}\right)\left(1.6\text{m}\right)+\left(4\text{kN}\right)\left(0.8\text{m}\right)=0$ (II)

The force in the positive $y$ axis is given as,

$A-2\text{kN}-4\text{kN}-4\text{kN}-\frac{1}{\sqrt{5}}{F}_{EF}=0$ (III)

For the section $CF,EF\text{and }EG$ the sum of the moments in the counter clockwise direction about $F$ is given as,

$-F_{EG}\left(0.4\text{m}\right)-A\left(2.4\text{m}\right)+\left(2\text{kN}\right)\left(2.4\text{m}\right)+\left(4\text{kN}\right)\left(0.8\text{m}\right)+\left(4\text{kN}\right)\left(1.6\text{m}\right)=0$ (IV)

Conclusion:

Substitute $10.5\text{kN}$ for $A$ in (II) and solve for $F_{CF}$.

$\begin{array}{c}{F}_{CF}\left(0.4\text{m}\right)-\left(10.5\text{kN}\right)\left(1.6\text{m}\right)+\left(2\text{kN}\right)\left(1.6\text{m}\right)+\left(4\text{kN}\right)\left(0.8\text{m}\right)=0\\ F_{CF}=26.0\text{kN}\end{array}$ (V)

Substitute $10.5\text{kN}$ for $A$ in (III) and solve for $F_{EF}$.

$\begin{array}{c}10.5\text{kN}-2\text{kN}-4\text{kN}-4\text{kN}-\frac{1}{\sqrt{5}}{F}_{EF}=0\\ F_{EF}=1.118\text{kN}\end{array}$ (VI)

Substitute $10.5\text{kN}$ for $A$ in (IV) and solve for $F_{EG}$.

$\begin{array}{c}-F_{EG}\left(0.4\text{m}\right)-\left(10.5\text{kN}\right)\left(2.4\text{m}\right)+\left(2\text{kN}\right)\left(2.4\text{m}\right)+\left(4\text{kN}\right)\left(0.8\text{m}\right)+\left(4\text{kN}\right)\left(1.6\text{m}\right)=0\\ F_{EG}=27.0\text{kN}\end{array}$ (VII)

Therefore, the force in the members $CF,EF\text{and }EG$ are $26.0\text{kN}$ the member of $CF$ is in Tension, $1.118\text{kN}$ the member of the CF is in tension and $27.0\text{kN}$, the member  of CF is in compression.