Question 4.4: Determine the force in each member of the Warren truss shown...

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 = (100 + 125 + 50) – 150 = 125 kN

\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 150 kN to balance the upward reaction of 150 kN. The slope of member AF is 3:4, so the magnitude of the horizontal component of F_{AF} is (4/3)(150 kN), or 200 kN. Thus, the force in member AF is compressive, with a magnitude of F_{AF} = \sqrt{(200)^2 + (150)^2} = 250 kN.

F_{AF} = 250 kN (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 200 kN to balance the horizontal component of F_{AF} of 200 kN acting to the left. Therefore, member AB is in tension with a force of 200 kN.

F_{AB} = 200 kN (T)                                              Ans.

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

F_{BC} = 200 kN (T)                                        Ans.

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

F_{BF} = 100 kN (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 150 – 100 = 50 kN. Using the 3:4 slope of member CF, we obtain the magnitude of the horizontal component as (4/3)(50) = 66.7 kN and the magnitude of F_{CF} itself as 83.4 kN.

F_{CF} = 83.4 kN (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 200 + 66.7 = 266.7 kN.

F_{FG} = 266.7 kN (C)                                              Ans.

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

F_{GH} = 266.7 kN (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 125 – 50 = 75 kN. Therefore, the magnitudes of the horizontal component of F_{CH} and of F_{CH} itself are 100 kN and 125 kN, respectively, as shown in Fig. 4.19(b).

F_{CH} = 125 kN (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 200 + 66.7 – 100 = 166.7 kN.

F_{CD} = 166.7 kN (T)                                            Ans.

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

F_{DE} = 166.7 kN (T)                                           Ans.

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

F_{DH} = 50 kN (T)                                      Ans.

Joint E Considering P 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 125 kN to balance the upward reaction E_y = 125 kN. The magnitude of the horizontal component of F_{EH} is equal to (4/3)(125), or 166.7 kN. Thus, F_{EH} is a compressive force with a magnitude of 208.5 kN.

F_{EH} = 208.4 kN (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} = -166.7 + 166.7 = 0            Checks

At joint H,

+\longrightarrow  \sum{F_x} = 266.7 – 100 – 166.7 = 0            Checks

+\uparrow  \sum{F_y} = -75 – 50 + 125 = 0                Checks