Question 4.5: Determine the force in each member of the truss shown in Fig...
Determine the force in each member of the truss shown in Fig. 4.20(a) by the method of joints.
Static Determinacy The truss is composed of 7 members and 5 joints and is supported by 3 reactions. Thus, m + r = 2j. Since the reactions and the members of the truss are properly arranged, it is statically determinate.
From the dimensions of the truss given in Fig. 4.20(a), we find that all inclined members have slopes of 12:5. Since joint E has two unknown non-collinear forces, F_{CE} and F_{DE}, acting on it, we can begin the method of joints without first calculating the support reactions.
Joint E Focusing our attention on joint E in Fig. 4.20(b), we observe that in order to satisfy \sum{F_x} = 0, the horizontal component of F_{DE} must push to the left into the joint with a magnitude of 25 kN to balance the 25 kN external load acting to the right. The slope of member DE is 12:5 , so the magnitude of the vertical component of F_{DE} is (12/5)(25), or 60 kN. Thus, the force in member DE is compressive, with a magnitude of
F_{DE} = \sqrt{(25)^2 + (60)^2} = 65 kNF_{DE} = 65 kN (C)
With the vertical component of F_{DE} now known, we can see from the figure that in order for \sum{F_y} = 0 to be satisfied, F_{CE} must pull downward on joint E with a magnitude of 60 – 30 = 30 kN.
F_{CE} = 30 kN (T)Joint C Next, we consider the equilibrium of joint C. Applying \sum{F_x} = 0 , we obtain F_{CD}.
F_{CD} = 50 kN (C)From \sum{F_y} = 0, we obtain F_{AC}.
F_{AC} = 30 kN (T)Joint D Both of the unknown forces, F_{AD} and F_{BD}, acting at this joint have inclined directions, so we draw the free-body diagram of this joint as shown in Fig. 4.20(c) and determine the unknowns by solving the equilibrium equations simultaneously:
+\longrightarrow \sum{F_x} = 0 50 + \frac{5}{13}(65) – \frac{5}{13} F_{AD} + \frac{5}{13} F_{BD} = 0
+\uparrow \sum{F_y} = 0 -\frac{12}{13}(65) -\frac{12}{13}F_{AD} -\frac{12}{13}F_{BD} = 0
Solving these equations simultaneously, we obtain
F_{AD} = 65 kN and F_{BD} = -130 kN
F_{AD} = 65 kN (T)
F_{BD} = 130 kN (C)
Joint B (See Fig. 4.20(b).) By considering the equilibrium of joint B in the horizontal direction (\sum{F_x}=0) , we obtain F_{AB}.
F_{AB} = 50 kN (T)Having determined all the member forces, we apply the remaining equilibrium equation (\sum{F_y}=0) at joint B to calculate the support reaction B_y.
B_y = 120 kN \uparrowJoint A By applying \sum{F_x}=0, , we obtain A_x.
A_x = 75 kN \longleftarrowFrom \sum{F_y}=0, we obtain A_y.
A_y = 90 kN \downarrowChecking Computations To check our computations, we consider the equilibrium of the entire truss. Applying the three equilibrium equations to the free body of the entire truss shown in Fig. 4.20(b), we obtain
+\longrightarrow \sum{F_x} = 25 + 50 – 75 = 0 Checks
+\uparrow \sum{F_y} = -30 – 90 + 120 = 0 Checks
+\circlearrowleft \sum{M_B} = 30(5) – 25(12) – 50(6) + 90(5) = 0 Checks
