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 FF Σ =− = 0 = F W 0: 20 lb 20 lb 0 y FV Σ = 40.0 lb = V W ( )( ) ( )( ) 2 in. 20 lb 6 in. 20 lb 0 J MM 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 FV Σ =− = 60.0 lb = V W 0: 0 y FF Σ = 0 = F W ( )( ) 1 in. 60 lb 0 J MM Σ = 60.0 lb in. =⋅ M W
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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 yy FA Σ =− = 80 kN y = A ( ) ( )( ) 0: 1.2 m 1.5 m 80 kN 0 Ex MA 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 FF Σ ° ° = 122.6 kN = F W ( ) ( ) 0: 80 kN cos21.801 100 kN sin 21.801 0 y FV Σ =+ ° ° = 37.1 kN = V W ( )( ) ( )( ) 0: .3 m 100 kN .75 m 80 kN 0 J MM 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 yy FA Σ =− = 100 N y = A ( ) ( )( ) 0: 2 0.32 m cos30 0.48 m 100 N 0 Fx MA  Σ= °− =  86.603 N x = A ( ) ( ) 0: 100 N cos30 86.603 N sin30 0 x FF ° ° = 129.9 N = F W ( ) ( ) 0: 100 N sin30 86.603 N cos30 0 y FV + ° ° = 25.0 N = V W ( ) ( ) () ( ) 0: 0.16 m cos30 86.603 N 0.16 m sin30 100 N 0 J M M ° = 4.00 N m =⋅ M W
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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 yx A A A A == ( ) ( )( ) 0: 0.3 m 0.48 m 390 N 0 Cx MA Σ =− = 624 N x = A 5 260 N 12 AA or 260 N y = A 0: 624 N 0 x FF Σ = 624 N = F W 0: 260 N 0 y FV Σ = 260 N = V W ( )( ) 0.2 m 260 N 0 J MM 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 Cx MA Σ =− = 624 N x = A AB is two-force member, so 5 0.36 m 0.15 m 12 y x yx A A A A =→ = 260 N y = A 0: 0 xx x FA C Σ =−+ = 624 N == CA 0: 390 N 0 yy y 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 FF Σ + = 626 N F = − 626 N = F W 12 5 0: 130 N 624 N 0 13 13 y FV Σ = 120 N V = − 120.0 N = V W ( )( ) ( )( ) 0: 0.1 m 624 N 0.24 m 130 N 0 K MM Σ= = 31.2 N m =⋅ M W
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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 Bx MA r Σ == 0 x = A ( ) 30 lb cos 60 0 x FV Σ =− ° = 15.00 lb = V W ( ) 30 lb sin60 0 y FF Σ =+ ° = 25.98 lb F = − 26.0 lb = F W [ ]( ) (9 in.) sin60 30 lb 0 J MM 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 yy FB Σ =−= 30 lb y = B 0: 2 0 Ax Mr B Σ == 0 x = B ( ) 30 lb cos30 0 x FV Σ =− ° = 25.98 lb V = 26.0 lb = V W ( ) 30 lb sin30 0 y FF Σ =+ ° = 15 lb F = − 15.00 lb = F W ( ) ( ) 9 in. sin30 30 lb 0 K MM  Σ ° =  135.0 lb in. =⋅ M W
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