Question 4.7: Determine the forces in members CD, DG, and GH of the truss ...
Determine the forces in members CD, DG, and GH of the truss shown in Fig. 4.22(a) by the method of sections.
Section aa As shown in Fig. 4.22(a), a section aa is passed through the three members of interest, CD, DG, and GH, cutting the truss into two portions, ACGE and DHI. To avoid the calculation of support reactions, we will use the right-hand portion, DHI, to calculate the member forces.
Member Forces The free-body diagram of the portion DHI of the truss is shown in Fig. 4.22(b). All three unknown forces F_{CD}, F_{DG}, and F_{GH}, are assumed to be tensile and are indicated by arrows pulling away from the corresponding joints on the diagram. The slope of the inclined force, F_{DG}, is also shown on the free-body diagram. The desired member forces are calculated by applying the equilibrium equations as follows (see Fig. 4.22(b)).
+\circlearrowleft \sum{M_D} =0 -60(4) + F_{GH}(3) = 0
F_{GH} = 80 kN (T)
+\uparrow \sum{F_y} =0 -120 – 60 + \frac{3}{5}F_{DG} = 0
F_{DG} = 300 kN (T)
+\longrightarrow \sum{F_x} =0 -80 – \frac{4}{5}(300) – F_{CD} = 0
F_{CD} = -320 kN
The negative answer for F_{CD} indicates that our initial assumption about this force being tensile was incorrect, and F_{CD} is actually a compressive force.
F_{CD} = 320 kN (C)Checking Computations (See Fig. 4.22(b).)
+\circlearrowleft \sum{M_I} = 120(4) – (-320) 3 – \frac{4}{5} (300)(3) – \frac{3}{5} (300)(4) = 0 Checks
