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

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

2 COSMOS: Complete Online Solutions Manual Organization Sstem Chapter 6, Solution 10. FBD Truss: Σ : H F x x B smmetr: A = H = 4 kips b inspection of joints C and G : F = F and F AC CE BC F = F and F EG GH FG Joint FBDs: Joint A: also, b smmetr F = F, F = F, F = F and F = F AB FH BD DF CE EG BE EF FAB FAC 3kips = = 4 3 so F =.00 kips C AB F = 4.00 kips T AC and, from above, F =.00 kips C FH Joint B: Joint E: and FCE = FEG = FGH = 4.00 kips T Σ Fx : ( kips) FBE FBD= Σ F 0: ( kips) = 2 F BD + F BE = so FBD = kips, FBE kips or F BD = 3.98 kips C F BE.238 kips C and, from above, F DF = 3.98 kips C F EF.238 kips C 3 Σ F : FDE 2 ( kips) F.286 kips T DE

3 COSMOS: Complete Online Solutions Manual Organization Sstem Chapter 6, Solution 29. Σ : F F x x Then, b inspection of joint F, F FG Then, b inspection of joint G, F GH B inspection of joint J, F IJ Then, b inspection of joint I, F HI F Then, b inspection of joint E, F BE EI

4 \ COSMOS: Complete Online Solutions Manual Organization Sstem Chapter 6, Solution 31. B inspection of joint C: F Then, b inspection of joint B : F Then, b inspection of joint E : F BC BE DE B inspection of joint H : F FH and F HI B inspection of joint Q : F OQ and F QR B inspection of joint J : F LJ

5 COSMOS: Complete Online Solutions Manual Organization Sstem Chapter 6, Solution 42. FBD Truss: Σ : 0 F x K ( ) ( ) x = Σ M : 6a K 12 lb a 20 lb A ( ) a( ) ( ) a( ) 4a 20 lb 3 37 lb 2a 00 lb 00 lb ( ) ( ) Σ F : A 320lb 200lb 37 lb 12 lb lb K = 937. lb FBD Section ABEC: A = lb ( ) FCF ( )( ) + ( 8 ft)( 20 lb lb) Σ M : 2ft + 4ft 00lb E F = 320 lb, F = 3.2 kips T CF 1 Σ F: lb 20 lb 2( 00 lb) FEF F = 62. lb, F = lb T EF ( ) 2 Σ Fx: 320 lb lb FEG= 0 F = 337 lb, F = 3.38 kips C EG CF EG EF

6 COSMOS: Complete Online Solutions Manual Organization Sstem Chapter 6, Solution 0. FBD Truss: Distance between loads = 1. m Σ : A F x x FBD Section ABC: B smmetr, A = K = 18 kn FBD Section ABC: ( ) F ( )( ) ( )( ) Σ M : 1. m + 1. m 6 kn 3 m 18 kn 3 kn D CE 4 Σ MA : ( 1.8 m) FCD ( 1. m)( 6 kn) F = 22. kn T! CE F = 6.2 kn T! 8 4 Σ F : 18 kn 3 kn 6 kn FBD + ( 6.2 kn) 17 CD F = 29.8 kn C! BD

7 COSMOS: Complete Online Solutions Manual Organization Sstem Chapter 6, Solution 2. FBD Truss: Σ : A + 4 kips 4 kips F X x A x = B smmetr, A = N 0 FBD Section ABDC: ( )( ) ( ) 4 Σ MD : 18 ft 4 kips 9 ft FCE F = kips C! CE ( )( ) ( ) 3 Σ ME : 18 ft 4 kips 12 ft FDF F DF = kips C 4 Σ Fx : 4 kips + ( 10 kips 10 kips) FDE FBD Joint E: F = 4.00 kips C! DE 4 3 Σ F : FEF ( 4 kips) F = 3.00 kips T! EF

8 COSMOS: Complete Online Solutions Manual Organization Sstem Chapter 6, Solution 62. FBD Section above a-a: 27 Σ MG : (.9 m) FIK ( 2.7 m)( 40 kn) (.4 m)( 40 kn) 793 F = 7.27 kn, F = 7.3 kn C IK 27 Σ F : ( 7.27 kn FGJ) 793 F GJ = 7.27 kn T IK FBD Section ACIG: 27 Σ F : 7.27 kn 7.27 kn 793 ( ) 18 + ( F F ), F = F 949 HK HJ HJ HK 2 Σ Fx : 3( 40 kn) 2 FHK 949 F = kn F = 3.9 kn C HK HK

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.

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. 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 =