Question 4.10: Determine the force in each member of the compound truss sho...
Determine the force in each member of the compound truss shown in Fig. 4.25(a).
Static Determinacy The truss has 11 members and 7 joints and is supported by 3 reactions. Since m + r = 2j and the reactions and the members of the truss are properly arranged, it is statically determinate.
The slopes of the inclined members, as determined from the dimensions of the truss, are shown in Fig. 4.25(a).
Reactions The reactions at supports A and B, as computed by applying the three equilibrium equations to the freebody diagram of the entire truss (Fig. 4.25(b)), are
A_x = 25 kN\longleftarrow A_y = 5 kN\uparrow B_y = 35 kN \uparrow
Section aa Since a joint with two or fewer unknown forces cannot be found to start the method of joints, we first calculate F_{AB} by using section aa, as shown in Fig. 4.25(a).
The free-body diagram of the portion of the truss on the left side of section aa is shown in Fig. 4.25(c). We determine F_{AB} by summing moments about point G, the point of intersection of the lines of action of F_{CG} and F_{DG}.
+\circlearrowleft \sum{M_G} = 0 -25(8) – 5(4) + 10(4) + F_{AB}(8) = 0
F_{AB} = 22.5 kN (T)
With F_{AB} now known, the method of joints can be started either at joint A, or at joint B, since both of these joints have only two unknowns each. We begin with joint A.
Joint A The free-body diagram of joint A is shown in Fig. 4.25(d).
+\longrightarrow \sum{F_x} = 0 -25 + 22.5 +\frac{1}{\sqrt{5}}F_{AC} + \frac{3}{5}F_{AD} = 0
+\uparrow \sum{F_y} = 0 5 + \frac{2}{\sqrt{5}}F_{AC} + \frac{4}{5}F_{AD} = 0
Solving these equations simultaneously, we obtain
F_{AC} = -27.95 kN and F_{AD} = 25 kN
F_{AC} = 27.95 kN (C) F_{AD} = 25 kN (T)
Joints C and D Focusing our attention on joints C and D in Fig. 4.25(b), and by satisfying the two equilibrium equations by inspection at each of these joints, we determine
F_{CG} = 27.95 kN (C0 F_{CD} = 10 kN (C) F_{DG} = 20.62 kN (T)Joint G Next, we consider the equilibrium of joint G (see Fig. 4.25(e)).
+\longrightarrow \sum{F_x} = 0 5 + \frac{1}{\sqrt{5}}(27.95) – \frac{1}{\sqrt{17}}(20.62) +\frac{1}{\sqrt{17}}F_{EG} +\frac{1}{\sqrt{5}}F_{FG} =0
+\uparrow \sum{F_y} = 0 -40 + \frac{2}{\sqrt{5}}(27.95) – \frac{4}{\sqrt{17}}(20.62) -\frac{4}{\sqrt{17}}F_{EG} -\frac{2}{\sqrt{5}}F_{FG} =0
Solving these equations, we obtain
F_{EG} = -20.62 kN and F_{FG} = -16.77 kN
F_{EG} = 20.62 kN (C) F_{FG} = 16.77 kN (C)
Joints E and F Finally, by considering the equilibrium, by inspection, of joints E and F (see Fig. 4.25(b)), we obtain
F_{BE} = 25 kN (C) F_{EF} = 10 kN (T) F_{BF} = 16.77 kN (C)
