Question 4.6: Determine the force in each member of the three-hinged truss...
Determine the force in each member of the three-hinged trussed arch shown in Fig. 4.21(a) by the method of joints.
Static Determinacy The truss contains 10 members and 7 joints and is supported by 4 reactions. Since m + r = 2j and the reactions and the members of the truss are properly arranged, it is statically determinate. Note that since m < 2j – 3, the truss is not internally stable, and it will not remain a rigid body when it is detached from its supports. However, when attached to the supports, the truss will maintain its shape and can be treated as a rigid body.
Zero-Force Members It can be seen from Fig. 4.21(a) that at joint C, three members, AC, CE, and CF, are connected, of which members AC and CF are collinear. Since joint C does not have any external load applied to it, the non-collinear member CE is a zero-force member.
F_{CE} = 0Similar reasoning can be used for joint D to identify member DG as a zero-force member.
F_{DG} = 0The slopes of the non-zero-force inclined members are shown in Fig. 4.21(a). The free-body diagram of the entire truss is shown in Fig. 4.21(b). The method of joints can be started either at joint E, or at joint G, since both of these joints have only two unknowns each.
Joint E Beginning with joint E, we observe from Fig. 4.21(b) that in order for \sum{F_x}=0 to be satisfied, the force in member EF must be compressive with a magnitude of 15 kN.
F_{EF} = 15 kN (C)Similarly, from \sum{F_y}=0, we obtain F_{AE}.
F_{AE} = 10 kN (C)Joint G By considering the equilibrium of joint G in the horizontal direction (\sum{F_x}=0), we observe that the force in member FG is zero.
F_{FG} = 0Similarly, by applying \sum{F_y}=0, we obtain F_{BG}.
F_{BG} = 10 kN (C)Joint F Next, we consider joint F. Both of the unknown forces, F_{CF} and F_{DF} , acting at this joint have inclined directions, so we draw the free-body diagram of this joint as shown in Fig. 4.21(c) and determine the unknowns by solving the equilibrium equations simultaneously:
+\longrightarrow \sum{F_x} = 0 15 – \frac{1}{\sqrt{2}} F_{CF} + \frac{4}{5}F_{DF} = 0
+\uparrow \sum{F_y} = 0 -20- \frac{1}{\sqrt{2}} F_{CF} – \frac{3}{5}F_{DF} = 0
Solving these equations, we obtain
F_{DF} = -25 kN and F_{CF} = -7.07 kN
F_{DF} = 25 kN (C)
F_{CF} = 7.07 kN (C)
Joint C (See Fig. 4.21(b).) In order for joint C to be in equilibrium, the two nonzero collinear forces acting at it must be equal and opposite.
F_{AC} = 7.07 kN (C)Joint D Using a similar reasoning at joint D, we obtain F_{BD}.
F_{BD} = 25 kN (C)Joint A Having determined all the member forces, we apply the two equilibrium equations at joint A to calculate the support reactions, A_x and A_y. By applying \sum{F_x}=0, we obtain A_x.
A_x = 5 kN \longrightarrowBy applying \sum{F_y}=0, we find that A_y is equal to 10 + 5 = 15 kN.
A_y = 15 kN \uparrowJoint B By applying \sum{F_x} =0, we obtain B_x.
B_x = 20 kN \longleftarrowFrom \sum{F_y} = 0, we find that B_y = 15 + 10 = 25 kN.
B_y = 25 kN \uparrowEquilibrium Check of Entire Truss Finally, to check our computations, we consider the equilibrium of the entire truss. Applying the three equations of equilibrium to the free body of the entire truss shown in Fig. 4.21(b), we have
+ \longrightarrow \sum{F_x} = 5 + 15 – 20 = 0 Checks
+ \uparrow \sum{F_y} = 15 – 10 – 20 – 10 + 25 = 0 Checks
+ \circlearrowleft \sum{M_B} = 5(2) – 15(16) – 15(6) + 10(16) + 20(8) = 0 Checks
