Solucionario
Páginas: 53 (13124 palabras)
Publicado: 27 de julio de 2010
Chapter 4
4-1
W 2 A RA 1 RB B A RA 1 1 RB W 2 B
(a)
1 RD 3 RA RB 2 D
(b)
1 RC C
A
B
W
(c)
W 1 RC RB RA RB 2 RA W
(d) (e)
A
RA
1
2 W RBx B RBx 1 RB RBy RBy
Scale of corner magnified
(f)
Chapter 4
51
4-2 (a)
2 kN 60° 2 30° 2 kN 60° RB RA 30° RA RB 90°
R A = 2 sin 60 = 1.732 kN R B = 2 sin 30 = 1 kN Ans.
Ans.
1
(b)
RA A
0.4 mB 45° 800 N
S = 0.6 m α = tan−1 0.6 = 30.96◦ 0.4 + 0.6
0.6 m
RO
O s RO RA 135° 30.96° 800 N 14.04 30.96°
45
30.96
800 RA = sin 135 sin 30.96 800 RO = sin 14.04 sin 30.96 RO =
⇒ ⇒
R A = 1100 N Ans. R O = 377 N Ans.
(c)
1.2 kN
30° RO RA
1.2 = 2.078 kN Ans. tan 30 1.2 = 2.4 kN Ans. RA = sin 30
60° RA RO
90° 1.2 kN
60°
(d) Step 1: Find R A and R EC 4.5 m 30° y 2 400 N
h=
4.5 = 7.794 m tan 30 MA = 0
+
h
4
B RAx RA
3
D 60°
9R E − 7.794(400 cos 30) − 4.5(400 sin 30) = 0 R E = 400 N Ans.
x
A RAy 9m
E RE
Fx = 0 Fy = 0
R Ax + 400 cos 30 = 0 RA =
⇒
R Ax = −346.4 N ⇒ R Ay = −200 N Ans.
R Ay + 400 − 400 sin 30 = 0
346.42 + 2002 = 400 N
52
Solutions Manual • Instructor’s Solution Manual toAccompany Mechanical Engineering Design
Step 2: Find components of RC on link 4 and R D
(RCy)4 C (RCx)4
+
MC = 0 R D = 305.4 N Ans. ⇒ ⇒ ( RC x ) 4 = 305.4 N ( RC y ) 4 = −400 N
400(4.5) − (7.794 − 1.9) R D = 0 ⇒ Fx = 0 Fy = 0
4 RD D
E 400 N
Step 3: Find components of RC on link 2
(RCy)2 C (RCx)2
Fx = 0 ( RC x ) 2 + 305.4 − 346.4 = 0 Fy = 0 ( RC y ) 2 = 200 N ⇒ ( RC x ) 2 =41 N
2
B 346.4 N A 200 N
305.4 N
200 N 41 N Pin C
400 N 305.4 N 30° 400 N C
200 N 41 N C
400 N 305.4 N
B 346.4 N A 200 N 400 N
305.4 N
B
D
305.4 N
D
E 400 N
Ans. 4-3 (a)
y 4" A R1 V (lbf) 60 O x 1.43 41.43 11.43 40 lbf 4" B 30 lbf 6" C R2 4" D 60 lbf x
+
M0 = 0
O
−18(60) + 14R2 + 8(30) − 4(40) = 0 R2 = 71.43 lbf Fy = 0: R1 − 40 + 30 +71.43 − 60 = 0 R1 = −1.43 lbf M1 M2 M3 M4 = −1.43(4) = −5.72 lbf · in = −5.72 − 41.43(4) = −171.44 lbf · in = −171.44 − 11.43(6) = −240 lbf · in = −240 + 60(4) = 0 checks!
M (lbf • in) O M4 M1 M2 M3 x
Chapter 4
53
(b)
MO
y
2 kN A 200 mm B
4 kN/m C 150 mm x
Fy = 0 R0 = 2 + 4(0.150) = 2.6 kN M0 = 0 M0 = 2000(0.2) + 4000(0.150)(0.425) = 655 N · m
O RO
150 mm
V (kN) 2.60.6 O M (N • m) O M1 655 M2 M3 x O x
M1 = −655 + 2600(0.2) = −135 N · m M2 = −135 + 600(0.150) = −45 N · m 1 M3 = −45 + 600(0.150) = 0 checks! 2
(c)
y 6 ft
1000 lbf 4 ft A R2 B x
M0 = 0: 10R2 − 6(1000) = 0 Fy = 0: R1 − 1000 + 600 = 0
⇒ ⇒
R2 = 600 lbf R1 = 400 lbf
O R1 V (lbf) 400 O
x
600 M (lbf • ft) O M1 M2
x
M1 = 400(6) = 2400 lbf · ft M2 = 2400 − 600(4) = 0checks!
(d)
y
1000 lbf 2 ft A R1 6 ft
2000 lbf 2 ft B C R2
+
x
MC = 0
O
−10R1 + 2(2000) + 8(1000) = 0 R1 = 1200 lbf Fy = 0: 1200 − 1000 − 2000 + R2 = 0 R2 = 1800 lbf M1 = 1200(2) = 2400 lbf · ft M2 = 2400 + 200(6) = 3600 lbf · ft M3 = 3600 − 1800(2) = 0 checks!
1200
200 1800
x
M M1 O
M2 M3 x
54
Solutions Manual • Instructor’s Solution Manual toAccompany Mechanical Engineering Design
(e)
y 4 ft
400 lbf
800 lbf x
+
MB = 0
O R1 V (lbf)
3 ft B 3 ft A R2 C
−7R1 + 3(400) − 3(800) = 0 R1 = −171.4 lbf Fy = 0: −171.4 − 400 + R2 − 800 = 0
800 x 571.4
O 171.4
R2 = 1371.4 lbf
M O M1 M2 M3 x
M1 = −171.4(4) = −685.7 lbf · ft M2 = −685.7 − 571.4(3) = −2400 lbf · ft M3 = −2400 + 800(3) = 0 checks!
(f) Break at A40 lbf/in
O R1 y 160 lbf A B 2"
8"
A VA
1 R1 = V A = 40(8) = 160 lbf 2 +
D R3
320 lbf 5" C 5"
MD = 0
R2
12(160) − 10R2 + 320(5) = 0 R2 = 352 lbf Fy = 0
x
40 lbf/in
320 lbf
160 lbf V (lbf) 160 O 160 M M1 O
352 lbf
128 lbf
−160 + 352 − 320 + R3 = 0 R3 = 128 lbf
192
x 128
M4
M5 M2 M3
x
1 M1 = 160(4) = 320 lbf · in 2 1 M2 = 320 − 160(4) =...
4-1
W 2 A RA 1 RB B A RA 1 1 RB W 2 B
(a)
1 RD 3 RA RB 2 D
(b)
1 RC C
A
B
W
(c)
W 1 RC RB RA RB 2 RA W
(d) (e)
A
RA
1
2 W RBx B RBx 1 RB RBy RBy
Scale of corner magnified
(f)
Chapter 4
51
4-2 (a)
2 kN 60° 2 30° 2 kN 60° RB RA 30° RA RB 90°
R A = 2 sin 60 = 1.732 kN R B = 2 sin 30 = 1 kN Ans.
Ans.
1
(b)
RA A
0.4 mB 45° 800 N
S = 0.6 m α = tan−1 0.6 = 30.96◦ 0.4 + 0.6
0.6 m
RO
O s RO RA 135° 30.96° 800 N 14.04 30.96°
45
30.96
800 RA = sin 135 sin 30.96 800 RO = sin 14.04 sin 30.96 RO =
⇒ ⇒
R A = 1100 N Ans. R O = 377 N Ans.
(c)
1.2 kN
30° RO RA
1.2 = 2.078 kN Ans. tan 30 1.2 = 2.4 kN Ans. RA = sin 30
60° RA RO
90° 1.2 kN
60°
(d) Step 1: Find R A and R EC 4.5 m 30° y 2 400 N
h=
4.5 = 7.794 m tan 30 MA = 0
+
h
4
B RAx RA
3
D 60°
9R E − 7.794(400 cos 30) − 4.5(400 sin 30) = 0 R E = 400 N Ans.
x
A RAy 9m
E RE
Fx = 0 Fy = 0
R Ax + 400 cos 30 = 0 RA =
⇒
R Ax = −346.4 N ⇒ R Ay = −200 N Ans.
R Ay + 400 − 400 sin 30 = 0
346.42 + 2002 = 400 N
52
Solutions Manual • Instructor’s Solution Manual toAccompany Mechanical Engineering Design
Step 2: Find components of RC on link 4 and R D
(RCy)4 C (RCx)4
+
MC = 0 R D = 305.4 N Ans. ⇒ ⇒ ( RC x ) 4 = 305.4 N ( RC y ) 4 = −400 N
400(4.5) − (7.794 − 1.9) R D = 0 ⇒ Fx = 0 Fy = 0
4 RD D
E 400 N
Step 3: Find components of RC on link 2
(RCy)2 C (RCx)2
Fx = 0 ( RC x ) 2 + 305.4 − 346.4 = 0 Fy = 0 ( RC y ) 2 = 200 N ⇒ ( RC x ) 2 =41 N
2
B 346.4 N A 200 N
305.4 N
200 N 41 N Pin C
400 N 305.4 N 30° 400 N C
200 N 41 N C
400 N 305.4 N
B 346.4 N A 200 N 400 N
305.4 N
B
D
305.4 N
D
E 400 N
Ans. 4-3 (a)
y 4" A R1 V (lbf) 60 O x 1.43 41.43 11.43 40 lbf 4" B 30 lbf 6" C R2 4" D 60 lbf x
+
M0 = 0
O
−18(60) + 14R2 + 8(30) − 4(40) = 0 R2 = 71.43 lbf Fy = 0: R1 − 40 + 30 +71.43 − 60 = 0 R1 = −1.43 lbf M1 M2 M3 M4 = −1.43(4) = −5.72 lbf · in = −5.72 − 41.43(4) = −171.44 lbf · in = −171.44 − 11.43(6) = −240 lbf · in = −240 + 60(4) = 0 checks!
M (lbf • in) O M4 M1 M2 M3 x
Chapter 4
53
(b)
MO
y
2 kN A 200 mm B
4 kN/m C 150 mm x
Fy = 0 R0 = 2 + 4(0.150) = 2.6 kN M0 = 0 M0 = 2000(0.2) + 4000(0.150)(0.425) = 655 N · m
O RO
150 mm
V (kN) 2.60.6 O M (N • m) O M1 655 M2 M3 x O x
M1 = −655 + 2600(0.2) = −135 N · m M2 = −135 + 600(0.150) = −45 N · m 1 M3 = −45 + 600(0.150) = 0 checks! 2
(c)
y 6 ft
1000 lbf 4 ft A R2 B x
M0 = 0: 10R2 − 6(1000) = 0 Fy = 0: R1 − 1000 + 600 = 0
⇒ ⇒
R2 = 600 lbf R1 = 400 lbf
O R1 V (lbf) 400 O
x
600 M (lbf • ft) O M1 M2
x
M1 = 400(6) = 2400 lbf · ft M2 = 2400 − 600(4) = 0checks!
(d)
y
1000 lbf 2 ft A R1 6 ft
2000 lbf 2 ft B C R2
+
x
MC = 0
O
−10R1 + 2(2000) + 8(1000) = 0 R1 = 1200 lbf Fy = 0: 1200 − 1000 − 2000 + R2 = 0 R2 = 1800 lbf M1 = 1200(2) = 2400 lbf · ft M2 = 2400 + 200(6) = 3600 lbf · ft M3 = 3600 − 1800(2) = 0 checks!
1200
200 1800
x
M M1 O
M2 M3 x
54
Solutions Manual • Instructor’s Solution Manual toAccompany Mechanical Engineering Design
(e)
y 4 ft
400 lbf
800 lbf x
+
MB = 0
O R1 V (lbf)
3 ft B 3 ft A R2 C
−7R1 + 3(400) − 3(800) = 0 R1 = −171.4 lbf Fy = 0: −171.4 − 400 + R2 − 800 = 0
800 x 571.4
O 171.4
R2 = 1371.4 lbf
M O M1 M2 M3 x
M1 = −171.4(4) = −685.7 lbf · ft M2 = −685.7 − 571.4(3) = −2400 lbf · ft M3 = −2400 + 800(3) = 0 checks!
(f) Break at A40 lbf/in
O R1 y 160 lbf A B 2"
8"
A VA
1 R1 = V A = 40(8) = 160 lbf 2 +
D R3
320 lbf 5" C 5"
MD = 0
R2
12(160) − 10R2 + 320(5) = 0 R2 = 352 lbf Fy = 0
x
40 lbf/in
320 lbf
160 lbf V (lbf) 160 O 160 M M1 O
352 lbf
128 lbf
−160 + 352 − 320 + R3 = 0 R3 = 128 lbf
192
x 128
M4
M5 M2 M3
x
1 M1 = 160(4) = 320 lbf · in 2 1 M2 = 320 − 160(4) =...
Leer documento completo
Regístrate para leer el documento completo.