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

Info icon This preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 2
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
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 4
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
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 6
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
Image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 8
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 .
Image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 10
This is the end of the preview. Sign up to access the rest of the document.