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**Unformatted text preview: **Problem 6.1 Determine the axial forces in the mem- bers of the truss and indicate whether they are in tension (T) or compression (C). Strategy: Draw free-body diagram of joint A . By writing the equilibrium equations for the joint, you can determine the axial forces in the two members. B A C 30 ° 60 ° 800 N Solution: 30 ° F AB F AC 60 ° A 800 N y x Assume the forces are in the directions shown (both in tension). If a force turns up negative, that force will be in compression. Equilibrium Eqns. X F x : − F AB cos 30 ◦ + F AC cos 60 ◦ + 800 = 0 X F y : − F AB sin 30 ◦ − F AC sin 60 ◦ = 0 B A C 30 ° 60 ° 800 N Solving: We get F AB = 693 N (tension) F AC = − 400 N (compression) Problem 6.2 The truss supports a 10-kN load at C . (a) Draw the free-body diagram of the entire truss, and determine the reactions at its supports. (b) Determine the axial forces in the members. Indicate whether they are in tension (T) or compression (C). 10 kN C B A 3 m 4 m Solution: (a) The free-body diagram of the system is shown. The sum of the moments about B is: M B = 3 A x − 4(10) = 0 , from which A x = 13 . 33 kN. The sums of the forces: X F x = A x + B x = 0 , from which B x = − A x = − 13 . 33 kN . X F y = B y − 10 = 0 , from which B y = 10 kN . (b) The interior angle ACB is α = tan − 1 (0 . 75) = 36 . 87 ◦ . (b) Assume that the unknown forces act away from the joint. Denote the axial force in the member I , K by IK . The axial forces are F CB = BC ( − i cos α + j sin α ) , and F CA = − AC i . Summing the forces: X F y = BC sin α − 10 = 0 , from which BC = 16 . 67 kN ( T ) . X F x = − BC cos α − AC = 0 , from which AC = − 13 . 33 kN ( C ) . For the joint A , 10 kN C B A 3 m 4 m B X A X B Y 3 m 4 m W X F y = AB = 0 , from which AB = 0 Problem 6.3 In Example 6.1, suppose that the 2-kN load is applied at D in the horizontal direction, point- ing from D toward B . What are the axial forces in the members? Solution: First, solve for the support forces and then use the method of joints at each joint to solve for the forces. B X A X A Y B θ θ θ 6 m tan θ = 10 m D x y 2 kN 6 10 = 30.96 X F x : B x + A x − 2 kN = 0 X F y : A y = 0 X M B : − 6 A x = 0 Solving, we get A x = A y = 0 , B x = 2 kN Joint A : A X = A Y = AB AC θ θ = 30.96 θ = 30 . 96 ◦ X F x : AC cos θ = 0 X F y : − AB − AC sin θ = 0 Solving, we get AB = AC = 0 Joint C : AC BC CD θ θ θ C AC = 0 from above θ = 30.96 ° X F x : AC / cos θ − BC cos θ + CD cos θ = 0 − BC + CD = 0 C 5 m 5 m A D B 2 kN 3 m 3 m X F y : AC / sin θ − BC sin θ − CD sin θ = 0 − BC − CD = 0 Solving, we get BC = CD = 0 Joint B : AB BC BD B X x θ We already know AB = BC = 0 and B x = 2 kN X F x : BC / cos θ + BD + B x = 0 BD + 2 kN = 0 BD = − 2 kN ( compression ) X F y : ( all forces zero ) = 0 we have AB = AC = BC = CD = 0 BD = − 2 kN ( c ) Note that we did not have to use joint D as we had already solved for the forces there. The F BD at D is BD , with the (...

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