PROBLEM 6.1 SOLUTION. Free body: Entire truss: (3.2 m) (48 kn)(7.2 m) = 0 = = = BC. 60 kn. Free body: Joint B: = kn T. = 144.

Size: px

Start display at page:

Download "PROBLEM 6.1 SOLUTION. Free body: Entire truss: (3.2 m) (48 kn)(7.2 m) = 0 = = = BC. 60 kn. Free body: Joint B: = kn T. = 144."

Garey Green
6 years ago
Views:

shown. State whether each member is in tension or compression.

2 m) = 0 C B = 108 kn B = 108 kn Σ F = 0: C 108 kn + 48 kn = 0 C = 60 kn C = 60

1 PROBLEM 6.1 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. Free bod: Entire truss: Σ F = 0: B = 0 B = 0 Σ M = 0: B (3.2 m) (48 kn)(7.2 m) = 0 C B = 108 kn B = 108 kn Σ F = 0: C 108 kn + 48 kn = 0 C = 60 kn C = 60 kn Free bod: Joint B: F AB F 108 kn = BC = FAB = kn T FBC = kn T Free bod: Joint C: FAC FBC 60 kn = = F = 144 kn (checks) BC FAC = kn C 743

2 PROBLEM 6.11 Determine the force in each member of the Pratt roof truss shown. State whether each member is in tension or compression. Free bod: Truss: Due to smmetr of truss and load, Σ F = 0: A = 0 Free bod: Joint A: A 1 = H = total load = 21 kn 2 F AB F 15.3 kn = AC = F AB = kn F = kn F = 47.2 kn C AC AB Free bod: Joint B: FAC = 44.6 kn T From force polgon: F = kn, F = 10.5 kn F = kn C BD BC BC FBD = 47.2 kn C 758

3 PROBLEM 6.11 (Continued) Free bod: Joint C: Σ F = 0: 3 FCD 10.5 = 0 5 FCD = kn T Σ F = 0: 4 F + (17.50) = 0 5 F = kn F = 30.6 kn T Free bod: Joint E: DE is a zero-force member. F DE = 0 Truss and loading smmetrical about cl. 759

Therefore, this truss is a simple truss. Truss of Problem 6.

4 PROBLEM 6.30 Determine whether the trusses of Problems 6.31b, 6.32b, and 6.33b are simple trusses. Truss of Problem 6.31b: Starting with triangle CGM and adding two members at a time, we obtain successivel joints B, L, F, A, K, J, then H, D, N, I, E, O, and P, thus completing the truss. Therefore, this truss is a simple truss. Truss of Problem 6.32b: Starting with triangle ABC and adding two members at a time, we obtain successivel joints E, D, F, G, and H, but cannot go further. Thus, this truss is not a simple truss. Truss of Problem 6.33b: Starting with triangle GFH and adding two members at a time, we obtain successivel joints D, E, C, A, and B, thus completing the truss. Therefore, this is a simple truss. 797

5 ft) A (6000 lb)(25 ft) = 0 Σ F = 0: A+ 3600 lb 6000 lb 6000 lb = 0 k = k = 3600 lb A = 8400 lb We pass a section through members, DE, and and

5 PROBLEM 6.45 A Warren bridge truss is loaded as shown. Determine the force in members, DE, and. Free bod: Truss: Σ = 0: k = 0 F Σ M = 0: k (62.5 ft) (6000 lb)(12.5 ft) A (6000 lb)(25 ft) = 0 Σ F = 0: A lb 6000 lb 6000 lb = 0 k = k = 3600 lb A = 8400 lb We pass a section through members, DE, and and use the free bod shown. Σ MD = 0: F(15 ft) (8400 lb)(18.75 ft) + (6000 lb)(6.25 ft) = 0 F = lb F = 8000 lb T 15 Σ F = 0: 8400 lb 6000 lb FDE = F = lb F = 2600 lb T DE Σ M E = 0: 6000 lb(12.5 ft) (8400 lb)(25 ft) F (15 ft) = 0 DE F = 9000 lb F = 9000 lb C 821

We pass a section through, EF, and EG, and use the free bod shown. (We appl F at F.) Σ M = 0: (200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) (1500 lb)(18 ft) E 18 F (4.5 ft) = 0 2 2 18 + 4.

6 PROBLEM 6.57 A Polnesian, or duopitch, roof truss is loaded as shown. Determine the force in members, EF, and EG. Free bod: Truss: Σ F = 0: N = 0 Σ M = 0: (200 lb)(8 a) + (400 lb)(7a+ 6a+ 5 a)+(350 lb)(4 a) + (300 lb)(3a+ 2 a+ a) A(8 a) = 0 N Σ F = 0: 1500 lb 200 lb 3(400 lb) 350 lb 3(300 lb) 150 lb + N = 0 We pass a section through, EF, and EG, and use the free bod shown. (We appl F at F.) Σ M = 0: (200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) (1500 lb)(18 ft) E 18 F (4.5 ft) = Σ M = 0: F (18 ft) (400 lb)(6 ft) (400 lb)(12 ft) = 0 A EF A =1500 lb N = 1300 lb N = 1300 lb F = 3711 lb F = 3710 lb C F =+ 400 lb F = 400 lb T Σ M = 0: F (4.5 ft) (1500 lb)(18 ft)+(200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) = 0 F EG EF EF F = lb F = 3600 lb T EG EG 833

Chapter 6, Solution 1. Joint B: Joint C: Joint FBDs: F = 800 lb T. F = 1700 lb C lb lb F

Chapter 6, Solution 1. Joint B: Joint C: Joint FBDs: F = 800 lb T. F = 1700 lb C lb lb F \ COSMOS: Complete Online Solutions Manual Organization Sstem Chapter 6, Solution 1. Joint FBDs: Joint B: FAB 800 lb F = = 1 8 17 BC so F = 100 lb T AB F = 1700 lb C BC Joint C: FAC Cx 1700 lb = = 8 1