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Chapter 4

Q. 4.4

Determine the force in each member of the Warren truss shown in Fig. 4.19(a) by the method of joints.


Verified Solution

Static Determinacy The truss has 13 members and 8 joints and is supported by 3 reactions. Because m + r = 2j and the reactions and the members of the truss are properly arranged, it is statically determinate.

Zero-Force Members It can be seen from Fig. 4.19(a) that at joint G, three members, CG , FG, and GH, are connected, of which FG and GH are collinear and CG is not. Since no external load is applied at joint G, member CG is a zero-force member.

F_{CG} = 0                                              Ans.

From the dimensions of the truss, we find that all inclined members have slopes of 3:4, as shown in Fig. 4.19(a). The free-body diagram of the entire truss is shown in Fig. 4.19(b). As a joint with two or fewer unknowns—which should not be collinear—cannot be found, we calculate the support reactions. (Although joint G has only two unknown forces, F_{FG} and F_{GH}, acting on it, these forces are collinear, so they cannot be determined from the joint equilibrium equation,\sum{F_x}=0.)

Reactions By using proportions,



\sum{F_y}=0                      E_y = (24 + 30 + 12) – 36 = 30 k


\sum{F_x}=0                     A_x=0

Joint A Focusing our attention on joint A in Fig. 4.19(b), we observe that in order to satisfy \sum{F_y}=0, the vertical component of F_{AF} must push downward into the joint with a magnitude of 36 k to balance the upward reaction of 36 k. The slope of member AF is 3:4, so the magnitude of the horizontal component of F_{AF} is (4=3)(36), or 48 k. Thus, the force in member AF is compressive, with a magnitude of F_{AF}=\sqrt{(48)^2+(36)^2}=60k.

F_{AF} = 60 k (C)                                 Ans.

With the horizontal component of F_{AF} now known, we can see from the figure that in order for \sum{F_x}=0 to be satisfied, F_{AB} must pull to the right with a magnitude of 48 k to balance the horizontal component of F_{AF} of 48 k acting to the left. Therefore, member AB is in tension with a force of 48 k.

F_{AB} = 48 k (T)                                   Ans.

Joint B Next, we consider the equilibrium of joint B. Applying \sum{F_x}=0 , we obtain F_{BC}.

F_{BC} = 48 k (T)                                       Ans.

From \sum{F_y}=0 , we obtain F_{BF}.

F_{BF} = 24 k (T)                                     Ans.

Joint F This joint now has two unknowns, F_{CF} and F_{FG}, so they can be determined by applying the equations of equilibrium as follows. We can see from Fig. 4.19(b) that in order to satisfy \sum{F_y}=0 , the vertical component of F_{CF} must pull downward on joint F with a magnitude of 36 – 24 = 12 k. Using the 3:4 slope of member CF, we obtain the magnitude of the horizontal component as (4/3)(12) = 16 k and the magnitude of F_{CF} itself as 20 k.

F_{CF} = 20 k (T)                           Ans.

Considering the equilibrium of joint F in the horizontal direction (\sum{F_x} = 0), it should be obvious from Fig. 4.19(b) that F_{FG} must push to the left on the joint with a magnitude of 48 + 16 = 64 k.

F_{FG} = 64 k (C)                           Ans.

Joint G Similarly, by applying \sum{F_x}=0 , we obtain F_{GH}.

F_{GH} = 64 k (C)                        Ans.

Note that the second equilibrium equation, \sum{F_y}=0, at this joint has already been utilized in the identification of member CG as a zero-force member.

Joint C By considering equilibrium in the vertical direction, \sum{F_y}=0 , we observe (from Fig. 4.19(b)) that member CH should be in tension and that the magnitude of the vertical component of its force must be equal to 30 – 12 = 18 k. Therefore, the magnitudes of the horizontal component of F_{CH} and of F_{CH} itself are 24 k and 30 k, respectively, as shown in Fig. 4.19(b).

F_{CH} = 30 k (T)                          Ans.

By considering equilibrium in the horizontal direction, \sum{F_x}=0, we observe that member CD must be in tension and that the magnitude of its force should be equal to 48 + 16 – 24 = 40 k.

F_{CD} = 40 k (T)                            Ans.

Joint D By applying \sum{F_x}=0, we obtain F_{DE}.

F_{DE} = 40 k (T)                           Ans.

From \sum{F_y}=0 , we determine F_{DH}.

F_{DH} = 12 k (T)                             Ans.

Joint E Considering the vertical components of all the forces acting at joint E, we find that in order to satisfy \sum{F_y}=0 , the vertical component of F_{EH} must push downward into joint E with a magnitude of 30 k to balance the upward reaction E_y = 30 k. The magnitude of the horizontal component of F_{EH} is equal to (4/3)(30), or 40 k. Thus, F_{EH} is a compressive force with a magnitude of 50 k.

F_{EH} = 50 k (C)                       Ans.

Checking Computations To check our computations, we apply the following remaining joint equilibrium equations (see Fig. 4.19(b)). At joint E,

+\longrightarrow \sum{F_x} = -40 + 40 = 0               Checks

At joint H,

+\longrightarrow  \sum{F_x} = 64 – 24 – 40 = 0               Checks

+\uparrow \sum{F_y} = -18 – 12 + 30 = 0                     Checks