Question 6.SP.3: Determine the forces in members FH, GH, and GI of the roof t......
Determine the forces in members FH, GH, and GI of the roof truss shown.
STRATEGY: You are asked to determine the forces in only three members of the truss, so use the method of sections. Determine the reactions by treating the entire truss as a free body and then isolate part of it for analysis. In this case, you can use the same smaller part of the truss to determine all three desired forces.
MODELING and ANALYSIS: Your reasoning and computation should go something like the sequence given here.
Free Body: Entire Truss. From the free-body diagram of the entire truss (Fig. 1), find the reactions at A and L:
\mathbf{A}=12.50 ~\mathrm{kN} \uparrow \quad \mathbf{L}=7.50 ~\mathrm{kN} \uparrow
Note that
\tan \alpha=\frac{F G}{G L}=\frac{8 \mathrm{~m}}{15 \mathrm{~m}}=0.5333 \quad \alpha=28.07^{\circ}
Force in Member GI. Pass section nn vertically through the truss (Fig. 1). Using the portion HLI of the truss as a free body (Fig. 2), obtain the value of F_{G I}:
+↺ \Sigma M_H=0: \quad(7.50 ~\mathrm{kN})(10 \mathrm{~m})-(1 ~\mathrm{kN})(5 \mathrm{~m})-F_{G I}(5.33 \mathrm{~m})=0
F_{G I}=+13.13 ~\mathrm{kN} \quad F_{G I}=13.13 ~\mathrm{kN} ~T
Force in Member FH. Determine the value of F_{F H} from the equation \Sigma M_G=0. To do this, move F_{F H} along its line of action until it acts at point F, where you can resolve it into its x and y components (Fig. 3). The moment of F_{F H} with respect to point G is now \left(F_{F H} \cos \alpha\right)(8 \mathrm{~m}).
+↺ \Sigma M_G=0:
(7.50 ~\mathrm{kN})(15 \mathrm{~m})-(1 ~\mathrm{kN})(10 \mathrm{~m})-(1 ~\mathrm{kN})(5 \mathrm{~m})+\left(F_{F H} \cos \alpha\right)(8 \mathrm{~m})=0
F_{F H}=-13.81 ~\mathrm{kN} \quad F_{F H}=13.81 ~\mathrm{kN} ~C
Force in Member GH. First note that
\tan \beta=\frac{G I}{H I}=\frac{5 \mathrm{~m}}{\frac{2}{3}(8 \mathrm{~m})}=0.9375 \quad \beta=43.15^{\circ}
Then determine the value of F_{G H} by resolving the force F_{G H} into x and y components at point G (Fig. 4) and solving the equation \Sigma M_L=0.
+↺ \Sigma M_L=0: \quad(1 ~\mathrm{kN})(10 \mathrm{~m})+(1 ~\mathrm{kN})(5 \mathrm{~m})+\left(F_{G H} \cos \beta\right)(15 \mathrm{~m})=0
F_{G H}=-1.371 ~\mathrm{kN} \quad F_{G H}=1.371 ~\mathrm{kN} ~C
REFLECT and THINK: Sometimes you should resolve a force into components to include it in the equilibrium equations. By first sliding this force along its line of action to a more strategic point, you might eliminate one of its components from a moment equilibrium equation.
