Tài liệu shi20396 chương 16 pdf

Chapter 16

16-1

(a) θ1 = 0°, θ2 = 120°, θa = 90°, sin θa = 1, a = 5 in

Eq. (16-2): Mf = 0.28pa(1.5)(6)

1

120°

0°

sin θ(6 − 5 cos θ) dθ

= 17.96pa lbf· in

Eq. (16-3): MN = pa(1.5)(6)(5)

1

120°

0°

sin2 θ dθ = 56.87pa lbf· in

c = 2(5 cos 30◦) = 8.66 in

Eq. (16-4): F = 56.87pa − 17.96pa

8.66 = 4.49pa

pa = F/4.49 = 500/4.49 = 111.4 psi for cw rotation

Eq. (16-7): 500 = 56.87pa + 17.96pa

8.66

pa = 57.9 psi for ccw rotation

A maximum pressure of 111.4 psioccurs on the RH shoe for cw rotation. Ans.

(b) RH shoe:

Eq. (16-6): TR = 0.28(111.4)(1.5)(6)2(cos 0◦ − cos 120◦)

1 = 2530 lbf· in Ans.

LH shoe:

Eq. (16-6): TL = 0.28(57.9)(1.5)(6)2(cos 0◦ − cos 120◦)

1 = 1310 lbf· in Ans.

Ttotal = 2530 + 1310 = 3840 lbf· in Ans.

(c) Force vectors not to scale

x

y

Fy

Ry

Rx

R

Fx

F

Secondary

shoe

30
y

x

Rx

Fy

Fx

F

Ry

R

Primary

shoe

30

shi20396_ch16.qxd 8/28/03 4:01 PM Page 407

408 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

RH shoe: Fx = 500 sin 30° = 250 lbf, Fy = 500 cos 30° = 433 lbf

Eqs. (16-8): A =

1

2

sin2 θ

120◦

0◦

= 0.375, B =

θ

2 − 1

4

sin 2θ

2π/3 rad

0

= 1.264

Eqs. (16-9): Rx = 111.4(1.5)(6)

1 [0.375 − 0.28(1.264)] − 250 = −229 lbf

Ry = 111.4(1.5)(6)

1 [1.264 + 0.28(0.375)] − 433 = 940 lbf

R = [(−229)2 + (940)2

]

1/2 = 967 lbf Ans.

LH shoe: Fx = 250 lbf, Fy = 433 lbf

Eqs. (16-10): Rx = 57.9(1.5)(6)

1 [0.375 + 0.28(1.264)] − 250 = 130 lbf

Ry = 57.9(1.5)(6)

1 [1.264 − 0.28(0.375)] − 433 = 171 lbf

R = [(130)2 + (171)2

]

1/2 = 215 lbf Ans.

16-2 θ1 = 15°, θ2 = 105°, θa = 90°, sin θa = 1, a = 5 in

Eq. (16-2): Mf = 0.28pa(1.5)(6)

1

105°

15°

sin θ(6 − 5 cos θ) dθ = 13.06pa

Eq. (16-3): MN = pa(1.5)(6)(5)

1

105°

15°

sin2 θ dθ = 46.59pa

c = 2(5 cos 30°) = 8.66 in

Eq. (16-4): F = 46.59pa − 13.06pa

8.66 = 3.872pa

RH shoe:

pa = 500/3.872 = 129.1 psi on RH shoe for cw rotation Ans.

Eq. (16-6): TR = 0.28(129.1)(1.5)(62)(cos 15° − cos 105°)

1 = 2391 lbf· in

LH shoe:

500 = 46.59pa + 13.06pa

8.66 ⇒ pa = 72.59 psi on LH shoe for ccw rotation Ans.

TL = 0.28(72.59)(1.5)(62)(cos 15° − cos 105°)

1 = 1344 lbf · in

Ttotal = 2391 + 1344 = 3735 lbf · in Ans.

Comparing this result with that of Prob. 16-1, a 2.7% reduction in torque is achieved by

using 25% less braking material.

shi20396_ch16.qxd 8/28/03 4:01 PM Page 408