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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
ΣFx = 0: − F = 0

FBD JD:
ΣFy = 0: V − 20 lb − 20 lb = 0

F=0

V = 40.0 lb
ΣM J = 0: M − ( 2 in.)( 20 lb ) − ( 6 in.)( 20 lb ) = 0 M = 160.0 lb ⋅ in.

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:
ΣFx = 0: 60 lb − V = 0 V = 60.0 lb ΣFy = 0: − F = 0 F=0 ΣM J = 0: M − (1 in.)( 60 lb ) = 0 M = 60.0 lb ⋅ in.

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:ΣFy = 0: Ay − 80 kN = 0 A y = 80 kN

ΣM E = 0: (1.2 m ) Ax − (1.5 m )( 80 kN ) = 0

A x = 100 kN

θ = tan −1   = 21.801°  0.75 m 
FBD AJ:
ΣFx′ = 0: F − ( 80 kN ) sin 21.801° − (100 kN ) cos 21.801° = 0

 0.3 m 

F = 122.6 kN
ΣFy′ = 0: V + ( 80 kN ) cos 21.801° − (100 kN ) sin 21.801° = 0

V = 37.1 kN
ΣM J = 0: M + (.3 m )(100 kN ) − (.75 m )( 80 kN ) = 0

M = 30.0 kN ⋅ m 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:

ΣFy = 0: Ay − 100 N = 0

A y = 100 N

ΣM F = 0:  2 ( 0.32 m ) cos 30°  Ax − ( 0.48 m )(100 N ) = 0  

A x = 86.603 N

FBD AJ:

ΣFx′ = 0: F − (100 N ) cos 30° − ( 86.603 N ) sin 30° = 0

F = 129.9 N
ΣFy′ = 0: V + (100N ) sin 30° − ( 86.603 N ) cos 30° = 0

V = 25.0 N
ΣM J = 0: ( 0.16 m ) cos 30° ( 86.603 N )     − ( 0.16 m ) sin 30° (100 N ) − M = 0

M = 4.00 N ⋅ m

PROBLEM 7.5
Determine the internal forces at point J of the structure shown.

SOLUTION FBD Frame: AB is two-force member, so
Ay Ax = 0.36 m 0.15 m Ay = 5 Ax 12

ΣM C = 0: ( 0.3 m ) Ax − ( 0.48 m )( 390 N ) = 0

A x = 624 NAy = 5 Ax = 260 N or A y = 260 N 12 ΣFx = 0: F − 624 N = 0

F = 624 N

FBD AJ:
ΣFy = 0: 260 N − V = 0

V = 260 N
ΣM J = 0: M − ( 0.2 m )( 260 N ) = 0

M = 52.0 N ⋅ m

PROBLEM 7.6
Determine the internal forces at point K of the structure shown.

SOLUTION FBD Frame:
ΣM C = 0: ( 0.3 m ) Ax − ( 0.48 m )( 390 N ) = 0

A x = 624 N

AB is two-force member, so
Ay Ax 5 = → Ay = Ax0.36 m 0.15 m 12

A y = 260 N

ΣFx = 0: − Ax + C x = 0

C x = A x = 624 N

ΣFy = 0: Ay + C y − 390 N = 0 C y = 390 N − 260 N = 130 N or C y = 130 N ΣFx′ = 0: F + 12 5 ( 624 N ) + (130 N ) = 0 13 13

FBD CK:

F = −626 N
ΣFy′ = 0:

F = 626 N

12 5 (130 N ) − ( 624 N ) − V = 0 13 13

V = −120 N

V = 120.0 N

ΣM K = 0: ( 0.1 m )( 624 N ) − ( 0.24 m )(130 N ) − M = 0

M = 31.2 N⋅ m

PROBLEM 7.7
A semicircular rod is loaded as shown. Determine the internal forces at point J.

SOLUTION FBD Rod:
ΣM B = 0: Ax ( 2r ) = 0

Ax = 0

ΣFx′ = 0: V − ( 30 lb ) cos 60° = 0

V = 15.00 lb

FBD AJ:

ΣFy′ = 0: F + ( 30 lb ) sin 60° = 0

F = −25.98 lb
F = 26.0 lb
ΣM J = 0: M − [ (9 in.) sin 60°] ( 30 lb ) = 0

M = −233.8 lb ⋅ in.
M = 234 lb ⋅ in.

PROBLEM 7.8
Asemicircular rod is loaded as shown. Determine the internal forces at point K.

SOLUTION FBD Rod:

ΣFy = 0: By − 30 lb = 0 ΣM A = 0: 2rBx = 0

B y = 30 lb Bx = 0

ΣFx′ = 0: V − ( 30 lb ) cos 30° = 0

FBD BK:

V = 25.98 lb
V = 26.0 lb
ΣFy′ = 0: F + ( 30 lb ) sin 30° = 0

F = −15 lb
F = 15.00 lb
ΣM K = 0: M − ( 9 in.) sin 30°  ( 30 lb ) = 0  

M = 135.0 lb ⋅ in.

PROBLEM7.9 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.

SOLUTION FBD Point A: By symmetry T1 = T2
3  ΣFx = 0: 2  T1  − 210 N = 0 5 
T1 = T2 = 175 N

Curve CJB is parabolic: y = ax 2 FBD BJ: At B :
x = 0.64 m, y = 0.16 m
a= 0.16 m

( 0.64 m )...