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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) =...
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