CHAPTER 7

# CHAPTER 7 - PROBLEM 7.1 Determine the internal forces(axial...

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PROBLEM 7.1 Determine the internal forces (axial force, shearing force, and bending moment) at point J of the structure indicated: Frame and loading of Prob. 6.77 . SOLUTION FBD JD: 0: 0 x F F Σ = = 0 = F W 0: 20 lb 20 lb 0 y F V Σ = = 40.0 lb = V W ( )( ) ( )( ) 0: 2 in. 20 lb 6 in. 20 lb 0 J M M Σ = = 160.0 lb in. = M W

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PROBLEM 7.2 Determine the internal forces (axial force, shearing force, and bending moment) at point J of the structure indicated: Frame and loading of Prob. 6.76. SOLUTION FBD AJ: 0: 60 lb 0 x F V Σ = = 60.0 lb = V W 0: 0 y F F Σ = = 0 = F W ( )( ) 0: 1 in. 60 lb 0 J M M Σ = = 60.0 lb in. = M W
PROBLEM 7.3 For the frame and loading of Prob. 6.80, determine the internal forces at a point J located halfway between points A and B . SOLUTION FBD Frame: FBD AJ: 0: 80 kN 0 y y F A Σ = = 80 kN y = A ( ) ( )( ) 0: 1.2 m 1.5 m 80 kN 0 E x M A Σ = = 100 kN x = A 1 0.3 m tan 21.801 0.75 m θ = = ° ( ) ( ) 0: 80 kN sin 21.801 100 kN cos21.801 0 x F F Σ = ° − ° = 122.6 kN = F W ( ) ( ) 0: 80 kN cos21.801 100 kN sin 21.801 0 y F V Σ = + ° − ° = 37.1 kN = V W ( )( ) ( )( ) 0: .3 m 100 kN .75 m 80 kN 0 J M M Σ = + = 30.0 kN m = M W

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PROBLEM 7.4 For the frame and loading of Prob. 6.101, determine the internal forces at a point J located halfway between points A and B . SOLUTION FBD Frame: FBD AJ: 0: 100 N 0 y y F A Σ = = 100 N y = A ( ) ( )( ) 0: 2 0.32 m cos30 0.48 m 100 N 0 F x M A Σ = ° = 86.603 N x = A ( ) ( ) 0: 100 N cos30 86.603 N sin30 0 x F F Σ = ° − ° = 129.9 N = F W ( ) ( ) 0: 100 N sin30 86.603 N cos30 0 y F V Σ = + ° − ° = 25.0 N = V W ( ) ( ) ( ) ( ) 0: 0.16 m cos30 86.603 N 0.16 m sin30 100 N 0 J M M Σ = ° ° = 4.00 N m = M W
PROBLEM 7.5 Determine the internal forces at point J of the structure shown. SOLUTION FBD Frame: FBD AJ: AB is two-force member, so 5 0.36 m 0.15 m 12 y x y x A A A A = = ( ) ( )( ) 0: 0.3 m 0.48 m 390 N 0 C x M A Σ = = 624 N x = A 5 260 N 12 y x A A = = or 260 N y = A 0: 624 N 0 x F F Σ = = 624 N = F W 0: 260 N 0 y F V Σ = = 260 N = V W ( )( ) 0: 0.2 m 260 N 0 J M M Σ = = 52.0 N m = M W

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PROBLEM 7.6 Determine the internal forces at point K of the structure shown. SOLUTION FBD Frame: FBD CK: ( ) ( )( ) 0: 0.3 m 0.48 m 390 N 0 C x M A Σ = = 624 N x = A AB is two-force member, so 5 0.36 m 0.15 m 12 y x y x A A A A = = 260 N y = A 0: 0 x x x F A C Σ = + = 624 N x x = = C A 0: 390 N 0 y y y F A C Σ = + = 390 N 260 N 130 N y C = = or 130 N y = C ( ) ( ) 12 5 0: 624 N 130 N 0 13 13 x F F Σ = + + = 626 N F = − 626 N = F W ( ) ( ) 12 5 0: 130 N 624 N 0 13 13 y F V Σ = = 120 N V = − 120.0 N = V W ( )( ) ( )( ) 0: 0.1 m 624 N 0.24 m 130 N 0 K M M Σ = = 31.2 N m = M W
PROBLEM 7.7 A semicircular rod is loaded as shown. Determine the internal forces at point J . SOLUTION FBD Rod: FBD AJ: ( ) 0: 2 0 B x M A r Σ = = 0 x = A ( ) 0: 30 lb cos 60 0 x F V Σ = ° = 15.00 lb = V W ( ) 0: 30 lb sin60 0 y F F Σ = + ° = 25.98 lb F = − 26.0 lb = F W [ ] ( ) 0: (9 in.) sin60 30 lb 0 J M M Σ = ° = 233.8 lb in. M = − 234 lb in. = M W

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PROBLEM 7.8 A semicircular rod is loaded as shown. Determine the internal forces at point K . SOLUTION FBD Rod: FBD BK: 0: 30 lb 0 y y F B Σ = = 30 lb y = B 0: 2 0 A x M rB Σ = = 0 x = B ( ) 0: 30 lb cos30 0 x F V Σ = ° = 25.98 lb V = 26.0 lb = V W ( ) 0: 30 lb sin30 0 y F F Σ = + ° = 15 lb F = − 15.00 lb = F W ( ) ( ) 0: 9 in. sin30 30 lb 0 K M M Σ = ° = 135.0 lb in. = M W
PROBLEM 7.9 An archer aiming at a target is pulling with a 210-N force on the bowstring. Assuming that the shape of the bow can be approximated by a parabola, determine the internal forces at point J .

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