Chapter 6.2, Problem 6.57P

A Howe scissors roof truss is loaded as shown. Determine force in members DF, DG, and EG.

Fig. P6.57 and P6.58

To determine

The force in the members $DF,DG\text{and }EG$.

Answer to Problem 6.57P

The force in the members $DF,DG\text{and }EG$ are, $10.48\text{kips}$, $3.35\text{kips}$ both of which are compressive forces and $13.02\text{kips}$ which is a tension force respectively.

Explanation of Solution

The Howe scissors 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.

 

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

$\begin{array}{c}\text{Total load}=5\left(1.6\text{kips}\right)+2\left(0.8\text{kips}\right)\\ =9.60\text{kips}\end{array}$

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

$\begin{array}{c}{A}_y=\frac{1}{2}\left(9.60\text{kips}\right)\\ =4.80\text{kips}\end{array}$

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

$\left(0.8\text{kips}\right)\left(24\text{ft}\right)+\left(1.6\text{kips}\right)\left(16\text{ft}\right)+\left(1.6\text{kips}\right)\left(8\text{ft}\right)-A_y\left(24\text{ft}\right)-\frac{8F_{DF}}{\sqrt{8^2+{3.5}^2}}\left(6\text{ft}\right)=0$ (I)

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

$-\left(1.6\text{kips}\right)\left(8\text{ft}\right)-\left(1.6\text{kips}\right)\left(16\text{ft}\right)-\frac{2.5F_{DG}}{\sqrt{8^2+{2.5}^2}}\left(16\text{ft}\right)-\frac{8F_{DG}}{\sqrt{8^2+{2.5}^2}}\left(7\text{ft}\right)=0$ (II)

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

$\left(0.8\text{kips}\right)\left(16\text{ft}\right)+\left(1.6\text{kips}\right)\left(8\text{ft}\right)-A_y\left(16\text{ft}\right)-\frac{8F_{EG}}{\sqrt{8^2+{1.5}^2}}\left(4\text{ft}\right)=0$ (III)

Conclusion:

Solve for $F_{DF}$ from (I) by substituting $4.80\text{kips}$ for $A_y$.

$\begin{array}{c}\left(0.8\text{kips}\right)\left(24\text{ft}\right)+\left(1.6\text{kips}\right)\left(16\text{ft}\right)+\left(1.6\text{kips}\right)\left(8\text{ft}\right)-\left(4.80\text{kips}\right)\left(24\text{ft}\right)-\frac{8F_{DF}}{\sqrt{8^2+{3.5}^2}}\left(6\text{ft}\right)=0\\ F_{DF}=-10.48\text{kips}\end{array}$

Solve for $F_{DG}$ from (II).

$\begin{array}{c}-\left(1.6\text{kips}\right)\left(8\text{ft}\right)-\left(1.6\text{kips}\right)\left(16\text{ft}\right)-\frac{2.5F_{DG}}{\sqrt{8^2+{2.5}^2}}\left(16\text{ft}\right)-\frac{8F_{DG}}{\sqrt{8^2+{2.5}^2}}\left(7\text{ft}\right)=0\\ F_{DG}=-3.35\text{kips}\end{array}$

Solve for $F_{DF}$ from (III) by substituting $4.80\text{kips}$ for $A_y$.

$\begin{array}{c}\left(0.8\text{kips}\right)\left(16\text{ft}\right)+\left(1.6\text{kips}\right)\left(8\text{ft}\right)-\left(4.80\text{kips}\right)\left(16\text{ft}\right)-\frac{8F_{EG}}{\sqrt{8^2+{1.5}^2}}\left(4\text{ft}\right)=0\\ F_{EG}=+13.02\text{kips}\end{array}$

Therefore, the force in the members $DF,DG\text{and }EG$ are, $10.48\text{kips}$, $3.35\text{kips}$ both of which are compressive forces and $13.02\text{kips}$ which is a tension force respectively.