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6–41.

Determine the force developed in members FE, EB, and BC of the truss and state if these members are in tension or compression.

11 kN

B

A D

C

F E

22 kN

2 m 1.5 m

2 m

2 m

SOLUTION Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, ND can be determined directly by writing the moment equation of equilibrium about point!A.

a+!MA = 0; ND(5.5) - 11(2) - 22(3.5) = 0 ND = 18.0 kN

Method of Sections. Referring to the FBD of the right portion of the truss sectioned through a–a shown in Fig. b, FBC and FFE can be determined directly by writing the moment equations of equilibrium about point E and B, respectively.

a+!ME = 0; 18.0(2) - FBC(2) = 0 FBC = 18.0 kN (T) Ans.

a+!MB = 0; 18.0(3.5) - 22(1.5) - FFE(2) = 0 FFE = 15.0 kN (C) Ans.

Also, FEB can be obtained directly by writing force equation of equilibrium along the y axis

+ c !Fy = 0; FEB a45 b + 18.0 - 22 = 0 FEB = 5.00 kN (C) Ans.

Ans: FBC = 18.0 kN (T) FFE = 15.0 kN (C) FEB = 5.00 kN (C)