Chapter 16 16-1 (a) θ 1 = 0°, θ 2 = 120°, θ a = 90°, sin θ a = 1, a = 5 in Eq. (16-2): M f = 0.28 p a (1.5)(6) 1 120° 0° sin θ (6 − 5 cos θ ) d θ = 17.96 p a lbf · in Eq. (16-3): M N = p a (1.5)(6)(5) 1 120° 0° sin 2 θ d θ = 56.87 p a lbf · in c = 2(5 cos 30 ◦ ) = 8.66 in Eq. (16-4): F = 56.87 p a − 17.96 p a 8.66 = 4.49 p a p a = F /4.49 = 500/4.49 = 111.4 psi for cw rotation Eq. (16-7): 500 = 56.87 p a + 17.96 p a 8.66 p a = 57.9 psi for ccw rotation A maximum pressure of 111.4 psi occurs on the RH shoe for cw rotation. Ans. (b) RH shoe: Eq. (16-6): T R = 0.28(111.4)(1.5)(6) 2 (cos 0 ◦ − cos 120 ◦ ) 1 = 2530 lbf · in Ans. LH shoe: Eq. (16-6): T L = 0.28(57.9)(1.5)(6) 2 (cos 0 ◦ − cos 120 ◦ ) 1 = 1310 lbf · in Ans. T total = 2530 + 1310 = 3840 lbf · in Ans. (c) Force vectors not to scale x y F y R y R x R F x F Secondary shoe 30y x R x F y F x F R y R Primary shoe 30
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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◦)
(a) Counter-clockwise rotation, θ2 = π/4 rad, r = 13.5/2 = 6.75 in
a = 4r sin θ2
2θ2 + sin 2θ2= 4(6.75) sin(π/4)
2π/4 + sin(2π/4)= 7.426 in
e = 2(7.426) = 14.85 in Ans.
(b)
α = tan−1(3/14.85) = 11.4°∑MR = 0 = 3Fx − 6.375P
Fx = 2.125P∑Fx = 0 = −Fx + Rx
Rx = Fx = 2.125P
F y = Fx tan 11.4◦ = 0.428P
∑Fy = −P − F y + Ry
Ry = P + 0.428P = 1.428P
Left shoe lever.∑MR = 0 = 7.78Sx − 15.28Fx
Sx = 15.28
7.78(2.125P) = 4.174P
Sy = f Sx = 0.30(4.174P)
= 1.252P∑Fy = 0 = Ry + Sy + F y
Ry = −F y − Sy
= −0.428P − 1.252P
= −1.68P∑Fx = 0 = Rx − Sx + Fx
Rx = Sx − Fx
= 4.174P − 2.125P
= 2.049P
Rx
Sx
Sy
Ry
Fx
F y
7.78"
15.28"
1.428P
2.125P
2.125P
0.428P
P
0.428P
2.125Ptie rod2.125P
0.428P
�
6.375"
ActuationleverRx
Ry
Fx
Fy
3"P
shi20396_ch16.qxd 8/28/03 4:01 PM Page 414
Chapter 16 415
(c) The direction of brake pulley rotation affects the sense of Sy , which has no effect onthe brake shoe lever moment and hence, no effect on Sx or the brake torque.
The brake shoe levers carry identical bending moments but the left lever carries atension while the right carries compression (column loading). The right lever is de-signed and used as a left lever, producing interchangeable levers (identical levers). Butdo not infer from these identical loadings.
16-10 r = 13.5/2 = 6.75 in, b = 7.5 in, θ2 = 45°
From Table 16-3 for a rigid, molded nonasbestos use a conservative estimate ofpa = 100 psi, f = 0.31.
In Eq. (16-16):
2θ2 + sin 2θ2 = 2(π/4) + sin 2(45°) = 2.571
From Prob. 16-9 solution,
N = Sx = 4.174P = pabr
2(2.571) = 1.285pabr
P = 1.285
4.174(100)(7.5)(6.75) = 1560 lbf Ans.
Applying Eq. (16-18) for two shoes,
T = 2a f N = 2(7.426)(0.31)(4.174)(1560)
= 29 980 lbf · in Ans.
16-11 From Eq. (16-22),
P1 = pabD
2= 90(4)(14)
2= 2520 lbf Ans.
f φ = 0.25(π)(270°/180°) = 1.178
Eq. (16-19): P2 = P1 exp(− f φ) = 2520 exp(−1.178) = 776 lbf Ans.
The radial load on the bearing pair is 1803 lbf. If the bearing is straddle mounted withthe drum at center span, the bearing radial load is 1803/2 = 901 lbf.
(c) Eq. (16-22):
p = 2P
bD
p|θ=0° = 2P1
3(16)= 2(1680)
3(16)= 70 psi Ans.
As it should be
p|θ=270° = 2P2
3(16)= 2(655)
3(16)= 27.3 psi Ans.
16-15 Given: φ=270°, b=2.125 in, f =0.20, T =150 lbf · ft, D =8.25 in, c2 = 2.25 in
Notice that the pivoting rocker is not located on the vertical centerline of the drum.
(a) To have the band tighten for ccw rotation, it is necessary to have c1 < c2 . When fric-tion is fully developed,
To help visualize what is going on let’s add a force W parallel to P1, at a lever arm ofc3 . Now sum moments about the rocker pivot.∑
M = 0 = c3W + c1 P1 − c2 P2
From which
W = c2 P2 − c1 P1
c3
The device is self locking for ccw rotation if W is no longer needed, that is, W ≤ 0.It follows from the equation above
P1
P2≥ c2
c1
When friction is fully developed
2.566 = 2.25/c1
c1 = 2.25
2.566= 0.877 in
When P1/P2 is less than 2.566, friction is not fully developed. Suppose P1/P2 = 2.25,then
c1 = 2.25
2.25= 1 in
We don’t want to be at the point of slip, and we need the band to tighten.
c2
P1/P2≤ c1 ≤ c2
When the developed friction is very small, P1/P2 → 1 and c1 → c2 Ans.
(b) Rocker has c1 = 1 in
P1
P2= c2
c1= 2.25
1= 2.25
f = ln( P1/P2)
φ= ln 2.25
3π/2= 0.172
Friction is not fully developed, no slip.
T = ( P1 − P2)D
2= P2
(P1
P2− 1
)D
2
Solve for P2
P2 = 2T
[( P1/P2) − 1]D= 2(150)(12)
(2.25 − 1)(8.25)= 349 lbf
P1 = 2.25P2 = 2.25(349) = 785 lbf
p = 2P1
bD= 2(785)
2.125(8.25)= 89.6 psi Ans.
shi20396_ch16.qxd 8/28/03 4:01 PM Page 418
Chapter 16 419
(c) The torque ratio is 150(12)/100 or 18-fold.
P2 = 349
18= 19.4 lbf
P1 = 2.25P2 = 2.25(19.4) = 43.6 lbf
p = 89.6
18= 4.98 psi Ans.
Comment:
As the torque opposed by the locked brake increases, P2 and P1 increase (althoughratio is still 2.25), then p follows. The brake can self-destruct. Protection could beprovided by a shear key.
Optimizing the partitioning of a double reduction lowered the gear-train inertia to20.9/112 = 0.187, or to 19% of that of a single reduction. This includes the two addi-tional gears.
16-29 Figure 16-29 applies,
t2 = 10 s, t1 = 0.5 s
t2 − t1t1
= 10 − 0.5
0.5= 19
Ie
n0 1 2
2.43
4 6 8 10
100
20.9
shi20396_ch16.qxd 8/28/03 4:01 PM Page 426
Chapter 16 427
The load torque, as seen by the motor shaft (Rule 1, Prob. 16-26), is
TL =∣∣∣∣1300(12)
10
∣∣∣∣ = 1560 lbf · in
The rated motor torque Tr is
Tr = 63 025(3)
1125= 168.07 lbf · in
For Eqs. (16-65):
ωr = 2π
60(1125) = 117.81 rad/s
ωs = 2π
60(1200) = 125.66 rad/s
a = −Tr
ωs − ωr= − 168.07
125.66 − 117.81= −21.41
b = Trωs
ωs − ωr= 168.07(125.66)
125.66 − 117.81
= 2690.4 lbf · in
The linear portion of the squirrel-cage motor characteristic can now be expressed as
TM = −21.41ω + 2690.4 lbf · in
Eq. (16-68):
T2 = 168.07
(1560 − 168.07
1560 − T2
)19
One root is 168.07 which is for infinite time. The root for 10 s is wanted. Use a successivesubstitution method
Scaling will affect do and di , but the gear ratio changed I. Scale up the flywheel in theProb. 16-29 solution by a factor of 2.5. Thickness becomes 4(2.5) = 10 in.
d̄ = 30(2.5) = 75 in
do = 75 + (10/2) = 80 in
di = 75 − (10/2) = 70 in
W = 8(386)(11 072)
802 + 702= 3026 lbf
v = 3026
0.26= 11 638 in3
V = π
4l(802 − 702) = 1178 l
l = 11 638
1178= 9.88 in
Proportions can be varied. The weight has increased 3026/189.1 or about 16-fold whilethe moment of inertia I increased 100-fold. The gear train transmits a steady 3 hp. But themotor armature has its inertia magnified 100-fold, and during the punch there are decel-eration stresses in the train. With no motor armature information, we cannot comment.
16-31 This can be the basis for a class discussion.