Chapter 6.2, Problem 6.56P

A Pratt roof truss is loaded as shown. Determine the force members FH, FI, and GI.

Fig. P6.55 and P6.56

To determine

The force in the members $FH,FI\text{and }GI$.

Answer to Problem 6.56P

The force in the members $FH,FI\text{and }GI$ are, $10.0\text{ kN}$ which is a compressive force, $4.92\text{kN}$ and $6.00\text{kN}$ both of which are tension forces respectively.

Explanation of Solution

The Pratt roof truss is loaded as shown in the figure. There is load at all the point on the top. The free body diagram of the given arrangement is given by Figure 1.

 

Figure 1

The total load on the truss is given from the figure as,

$\begin{array}{c}\text{Total load}=5\left(3\text{kN}\right)+2\left(1.5\text{kN}\right)\\ =18\text{kN}\end{array}$

By symmetry the $y$ component of the reaction at $A$ will be half the total load.

$\begin{array}{c}{A}_y=\frac{1}{2}\left(18\text{kN}\right)\\ =9\text{kN}\end{array}$

The section’s free body diagram is given as,

The tangent of the angle $\alpha$ can be written from the diagram as,

$\tan \alpha =\frac{6.75}{3}$

The angle is derived from the tangent value as,

$\begin{array}{c}\alpha ={\tan }^{-1}\frac{6.75}{3}\\ =66.04°\end{array}$

From the diagram the sum of the moments in the counter clockwise direction about $I$ is given as,

$\frac{4}{5}{F}_{FH}\left(\frac{2}{3}\times 6.75\text{m}\right)+A_y\left(6\text{m}\right)-\left(1.5\text{kN}\right)\left(6\text{m}\right)-\left(3\text{kN}\right)\left(3\text{m}\right)=0$ (I)

From the diagram the sum of the moments in the counter clockwise direction about $L$ is given as,

$F_{FI}\left(\sin \alpha \right)\left(6\text{m}\right)-\left(3\text{kN}\right)\left(6\text{m}\right)-\left(3\text{kN}\right)\left(3\text{m}\right)=0$ (II)

From the diagram the sum of the moments in the counter clockwise direction about $H$ is given as,

$F_{GI}\left(6.75\text{m}\right)+\left(1.5\text{kN}\right)\left(9\text{m}\right)+\left(3\text{kN}\right)\left(3\text{m}\right)+\left(3\text{kN}\right)\left(6\text{m}\right)-A_y\left(9\text{m}\right)=0$ (III)

Conclusion:

Solve for $F_{FH}$ from (I) by substituting $9\text{kN}$ for $A_y$.

$\begin{array}{c}\left\{\begin{array}{l}\frac{4}{5}{F}_{FH}\left(\frac{2}{3}\times 6.75\text{m}\right)+\left(9\text{kN}\right)\left(6\text{m}\right)\\ -\left(1.5\text{kN}\right)\left(6\text{m}\right)-\left(3\text{kN}\right)\left(3\text{m}\right)\end{array}\right\}=0\\ F_{FH}=\frac{\left\{\begin{array}{l}-\left(9\text{kN}\right)\left(6\text{m}\right)+\left(1.5\text{kN}\right)\left(6\text{m}\right)\\ +\left(3\text{kN}\right)\left(3\text{m}\right)\end{array}\right\}}{\frac{4}{5}\left(\frac{2}{3}\times 6.75\text{m}\right)}\\ =-10.0\text{ kN}\end{array}$

Solve for $F_{DE}$ from (II) by substituting $66.04°$ for $\alpha$.

$\begin{array}{c}\left\{\begin{array}{l}{F}_{FI}\left(\sin \left(66.04°\right)\right)\left(6\text{m}\right)-\left(3\text{kN}\right)\left(6\text{m}\right)\\ -\left(3\text{kN}\right)\left(3\text{m}\right)\end{array}\right\}=0\\ F_{FI}=\frac{\left(3\text{kN}\right)\left(6\text{m}\right)+\left(3\text{kN}\right)\left(3\text{m}\right)}{\left(\sin \left(66.04°\right)\right)\left(6\text{m}\right)}\\ =+4.92\text{kN}\end{array}$

Solve for $F_{DF}$ from (III) by substituting $9\text{kN}$ for $A_y$.

$\begin{array}{c}\left\{\begin{array}{l}{F}_{GI}\left(6.75\text{m}\right)+\left(1.5\text{kN}\right)\left(9\text{m}\right)\\ +\left(3\text{kN}\right)\left(3\text{m}\right)+\left(3\text{kN}\right)\left(6\text{m}\right)\\ -\left(9\text{kN}\right)\left(9\text{m}\right)\end{array}\right\}=0\\ F_{GI}=\frac{\left\{\begin{array}{l}-\left(1.5\text{kN}\right)\left(9\text{m}\right)-\left(3\text{kN}\right)\left(3\text{m}\right)\\ -\left(3\text{kN}\right)\left(6\text{m}\right)+\left(9\text{kN}\right)\left(9\text{m}\right)\end{array}\right\}}{\left(6.75\text{m}\right)}\\ =+6.00\text{kN}\end{array}$

Therefore, the force in the members $FH,FI\text{and }GI$ are, $10.0\text{ kN}$ which is a compressive force, $4.92\text{kN}$ and $6.00\text{kN}$ both of which are tension forces respectively.