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程序代写案例-544G P
时间：2022-07-08
Homework 5 Solutions
1.
1. Continued
1. Continued
13-1 17 / 8 2.125 inPd = =
( )2
3
1120 2.125 4.375 in
544G P
Nd d
N
= = =
( )8 4.375 35 teeth .G GN Pd Ans= = =
( )2.125 4.375 / 2 3.25 in .C Ans= + =
______________________________________________________________________________
14-3 1.25(18) 22.5 mmd mN= = =
Table 14-2: Y = 0.309
3(22.5)(10 )(1800) 2.121 m/s
60 60
dnV π π
−
= = =
Eq. (14-6b): 6.1 2.121 1.348
6.1
K += =v
Eq. (13-36): 60 000 60 000(0.5) 0.2358 kN 235.8 N
(22.5)(1800)
t HW
dnπ π
= = = =
Eq. (14-8): 1.348(235.8) 68.6 MPa .
12(1.25)(0.309)
tK W Ans
FmY
σ = = =v
________________________________________________________________________
13-10 (a) The smallest pinion tooth count that will run with itself is found from Eq. (13-10).
( )
( ) ( )
2
2
2
2
2 1 1 3sin
3sin
2 1
1 1 3sin 20
3sin 20
12.32 13 teeth .
P
kN
Ans
φ
φ
≥ + +
≥ + +
≥ →
(b) The smallest pinion that will mesh with a gear ratio of mG = 2.5, from Eq. (13-11) is
( ) ( )( )
( )
( )
( ){ }
2 2
2
2 2
2
2 1 2 sin
1 2 sin
2 1
2.5 2.5 1 2 2.5 sin 20
1 2 2.5 sin 20
14.64 15 teeth .
P
kN m m m
m
Ans
φ
φ
≥ + + +
+
≥ + + + +
≥ →
The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-12) is
( )
( ) ( )
2 2 2
2
22 2
2
sin 4
4 2 sin
15 sin 20 4 1
4 1 2 15 sin 20
45.49 45 teeth .
P
G
P
N kN
k N
Ans
φ
φ
−
≤
−
−
≤
−
≤ →
(c) The smallest pinion that will mesh with a rack, from Eq. (13-13),
( )
2 2
2 12
sin sin 20
17.097 18 teeth .
P
kN
Ans
φ
≥ =
≥ →
______________________________________________________________________________
3-81 (or 3-70)
(a)
( ) ( ) ( )
( )
2 1
2 1 1 1
1 1
2
0.15
0 300 50 (4) (3) 1000 0.15 (3)
1000 2.55 0 392.16 lbf .
0.15 392.16 58.82 lbf .
T T
T T T T T
T T Ans
T Ans
=
= = − + − = + −
− = ⇒ =
= =
∑
(b)
0 450.98(16) (22)
327.99 lbf .
0 450.98 327.99
122.99 lbf .
0 350(8) (22)
127.27 lbf .
0 350 127.27
222.73 lbf .
O y C z
C z
z O z
O z
O z C y
C y
y O y
O y
M R
R Ans
F R
R Ans
M R
R Ans
F R
R Ans
= = − −
= −
= = + −
= −
= = +
= −
= = + −
= −
∑
∑
∑
∑
(c)
(d) Combine the bending moments from both planes at A and B to find the critical location.
2 2
2 2
(983.92) ( 1781.84) 2035 lbf in
(1967.84) ( 763.65) 2111 lbf in
A
B
M
M
= + − = ⋅
= + − = ⋅
The critical location is at B. The torque transmitted through the shaft from A to B is T =
(300 − 50)(4) = 1000 lbf·in. For a stress element on the outer surface where the bending
stress and the torsional stress are both maximum,
( )
3 3
32 211132 21502 psi = 21.5 kpsi .
(1)
Mc M Ans
I d
σ
π π
= = = =
3 3
16 16(1000) 5093 psi = 5.09 kpsi .
(1)
Tr T Ans
J d
τ
π π
= = = =
______________________________________________________________________________
3.83 (or 3-72) (a)
0 300(cos 20º )(10) (cos 20º )(4)
750 lbf .
B
B
T F
F Ans
= = − +
=
∑
(b)
0 300(cos 20º )(16) 750(sin 20º )(39) (30)
183.1 lbf .
0 300(cos 20º ) 183.1 750(sin 20º )
208.5 lbf .
0 300(sin 20º )(16) (30) 750(cos 20º )(39)
861.5 lbf .
0
O z C y
C y
y O y
O y
O y C z
C z
z
M R
R Ans
F R
R Ans
M R
R Ans
F
= = − +
=
= = + + −
= −
= = − −
= −
= =
∑
∑
∑
300(sin 20º ) 861.5 750(cos 20º )
259.3 lbf .
O z
O z
R
R Ans
− − +
=
∑
(c)
(d) Combine the bending moments from both planes at A and C to find the critical location.
2 2
2 2
( 3336) ( 4149) 5324 lbf in
( 2308) ( 6343) 6750 lbf in
A
C
M
M
= − + − = ⋅
= − + − = ⋅
The critical location is at C. The torque transmitted through the shaft from A to B is
( )( )300cos 20º 10 2819 lbf inT = = ⋅ . For a stress element on the outer surface where the
bending stress and the torsional stress are both maximum,
( )
3 3
32 675032 35 203 psi = 35.2 kpsi .
(1.25)
Mc M Ans
I d
σ
π π
= = = =
3 3
16 16(2819) 7351 psi = 7.35 kpsi .
(1.25)
Tr T Ans
J d
τ
π π
= = = =
13-44 (or 13-38)
(a) For 2 ,
1
o
i
ω
ω
= from Eq. (13-11), with m = 2, k = 1, 20φ =
( )
( )
( ){ }2 222 1 2 2 1 2 2 sin 20 14.161 2 2 sin 20PN = + + + = +
So min 15 .PN Ans=
(b) 15 1.875 teeth/in .
8
NP Ans
d
= = =
(c) To transmit the same power with no change in pitch diameters, the speed and
transmitted force must remain the same.
For A, with φ = 20°,
WtA = FA cos20° = 300 cos20° = 281.9 lbf
For A, with φ = 25°, same transmitted load,
FA = WtA/cos25° = 281.9/cos25° = 311.0 lbf Ans.
Summing the torque about the shaft axis,
2 2
A B
tA tB
d dW W =
( )
( ) ( )
/ 2 20281.9 704.75 lbf
/ 2 8
A A
tB tA tA
B B
d dW W W
d d
= = = =
704.75 777.6 lbf .
cos 25 cos 25
tB
B
WF Ans= = =
______________________________________________________________________________
13-17 20 8 20 4
40 17 60 51
e = =
( )4 00 47.06 rev/min cw .
51d
n Ans= 6 =
______________________________________________________________________________
13-32 (or 13-31)
(a) 2 / 60nω π=
2 / 60 ( in N m, in W)H T Tn T Hω π= = ⋅
So
( )360 10
2
9550 / ( in kW, in rev/min)
H
T
n
H n H n
π
=
=
( )9550 75 398 N m
1800a
T = = ⋅
( )2
2
5 17
42.5 mm
2 2
mNr = = =
So 32
2
398 9.36 kN
42.5
t aTF
r
= = =
( )3 3 2 9.36 18.73 kN in the positive -direction. .b bF F x Ans= − = =
(b) ( )44
5 51
127.5 mm
2 2
mNr = = =
( )4 9.36 127.5 1193 N m ccwcT = = ⋅
4 1193 N m cw .cT Ans∴ = ⋅
Note: The solution is independent of the pressure angle.
13-51 (or 13-45)
NOTE: The shaft forces exerted on the gears are not shown in the figures above.
cos 5cos30 4.330 teeth/int nP P ψ= = =
1 1tan tan 20tan tan 22.80
cos cos30
n
t
φ
φ
ψ
− −= = =
18 4.157 in
4.330P
d = =
The forces on the shafts will be equal to the forces transmitted to the gears through the meshing
teeth.
Pinion (Gear 2)
tan 800 tan 22.80 336 lbfr t tW W φ= = =
tan 800 tan 30 462 lbfa tW W ψ= = =
336 462 800 lbf .Ans= − − +W i j k
( ) ( )
1/22 2 2336 462 800 983 lbf .W Ans = − + − + =
Gear 3
336 462 800 lbf .Ans= + −W i j k
983 lbf .W Ans=
32 7.390 in
4.330G
d = =
( )800 7.390 5912 lbf intGT W r= = = ⋅
1.
1. Continued
1. Continued
13-1 17 / 8 2.125 inPd = =
( )2
3
1120 2.125 4.375 in
544G P
Nd d
N
= = =
( )8 4.375 35 teeth .G GN Pd Ans= = =
( )2.125 4.375 / 2 3.25 in .C Ans= + =
______________________________________________________________________________
14-3 1.25(18) 22.5 mmd mN= = =
Table 14-2: Y = 0.309
3(22.5)(10 )(1800) 2.121 m/s
60 60
dnV π π
−
= = =
Eq. (14-6b): 6.1 2.121 1.348
6.1
K += =v
Eq. (13-36): 60 000 60 000(0.5) 0.2358 kN 235.8 N
(22.5)(1800)
t HW
dnπ π
= = = =
Eq. (14-8): 1.348(235.8) 68.6 MPa .
12(1.25)(0.309)
tK W Ans
FmY
σ = = =v
________________________________________________________________________
13-10 (a) The smallest pinion tooth count that will run with itself is found from Eq. (13-10).
( )
( ) ( )
2
2
2
2
2 1 1 3sin
3sin
2 1
1 1 3sin 20
3sin 20
12.32 13 teeth .
P
kN
Ans
φ
φ
≥ + +
≥ + +
≥ →
(b) The smallest pinion that will mesh with a gear ratio of mG = 2.5, from Eq. (13-11) is
( ) ( )( )
( )
( )
( ){ }
2 2
2
2 2
2
2 1 2 sin
1 2 sin
2 1
2.5 2.5 1 2 2.5 sin 20
1 2 2.5 sin 20
14.64 15 teeth .
P
kN m m m
m
Ans
φ
φ
≥ + + +
+
≥ + + + +
≥ →
The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-12) is
( )
( ) ( )
2 2 2
2
22 2
2
sin 4
4 2 sin
15 sin 20 4 1
4 1 2 15 sin 20
45.49 45 teeth .
P
G
P
N kN
k N
Ans
φ
φ
−
≤
−
−
≤
−
≤ →
(c) The smallest pinion that will mesh with a rack, from Eq. (13-13),
( )
2 2
2 12
sin sin 20
17.097 18 teeth .
P
kN
Ans
φ
≥ =
≥ →
______________________________________________________________________________
3-81 (or 3-70)
(a)
( ) ( ) ( )
( )
2 1
2 1 1 1
1 1
2
0.15
0 300 50 (4) (3) 1000 0.15 (3)
1000 2.55 0 392.16 lbf .
0.15 392.16 58.82 lbf .
T T
T T T T T
T T Ans
T Ans
=
= = − + − = + −
− = ⇒ =
= =
∑
(b)
0 450.98(16) (22)
327.99 lbf .
0 450.98 327.99
122.99 lbf .
0 350(8) (22)
127.27 lbf .
0 350 127.27
222.73 lbf .
O y C z
C z
z O z
O z
O z C y
C y
y O y
O y
M R
R Ans
F R
R Ans
M R
R Ans
F R
R Ans
= = − −
= −
= = + −
= −
= = +
= −
= = + −
= −
∑
∑
∑
∑
(c)
(d) Combine the bending moments from both planes at A and B to find the critical location.
2 2
2 2
(983.92) ( 1781.84) 2035 lbf in
(1967.84) ( 763.65) 2111 lbf in
A
B
M
M
= + − = ⋅
= + − = ⋅
The critical location is at B. The torque transmitted through the shaft from A to B is T =
(300 − 50)(4) = 1000 lbf·in. For a stress element on the outer surface where the bending
stress and the torsional stress are both maximum,
( )
3 3
32 211132 21502 psi = 21.5 kpsi .
(1)
Mc M Ans
I d
σ
π π
= = = =
3 3
16 16(1000) 5093 psi = 5.09 kpsi .
(1)
Tr T Ans
J d
τ
π π
= = = =
______________________________________________________________________________
3.83 (or 3-72) (a)
0 300(cos 20º )(10) (cos 20º )(4)
750 lbf .
B
B
T F
F Ans
= = − +
=
∑
(b)
0 300(cos 20º )(16) 750(sin 20º )(39) (30)
183.1 lbf .
0 300(cos 20º ) 183.1 750(sin 20º )
208.5 lbf .
0 300(sin 20º )(16) (30) 750(cos 20º )(39)
861.5 lbf .
0
O z C y
C y
y O y
O y
O y C z
C z
z
M R
R Ans
F R
R Ans
M R
R Ans
F
= = − +
=
= = + + −
= −
= = − −
= −
= =
∑
∑
∑
300(sin 20º ) 861.5 750(cos 20º )
259.3 lbf .
O z
O z
R
R Ans
− − +
=
∑
(c)
(d) Combine the bending moments from both planes at A and C to find the critical location.
2 2
2 2
( 3336) ( 4149) 5324 lbf in
( 2308) ( 6343) 6750 lbf in
A
C
M
M
= − + − = ⋅
= − + − = ⋅
The critical location is at C. The torque transmitted through the shaft from A to B is
( )( )300cos 20º 10 2819 lbf inT = = ⋅ . For a stress element on the outer surface where the
bending stress and the torsional stress are both maximum,
( )
3 3
32 675032 35 203 psi = 35.2 kpsi .
(1.25)
Mc M Ans
I d
σ
π π
= = = =
3 3
16 16(2819) 7351 psi = 7.35 kpsi .
(1.25)
Tr T Ans
J d
τ
π π
= = = =
13-44 (or 13-38)
(a) For 2 ,
1
o
i
ω
ω
= from Eq. (13-11), with m = 2, k = 1, 20φ =
( )
( )
( ){ }2 222 1 2 2 1 2 2 sin 20 14.161 2 2 sin 20PN = + + + = +
So min 15 .PN Ans=
(b) 15 1.875 teeth/in .
8
NP Ans
d
= = =
(c) To transmit the same power with no change in pitch diameters, the speed and
transmitted force must remain the same.
For A, with φ = 20°,
WtA = FA cos20° = 300 cos20° = 281.9 lbf
For A, with φ = 25°, same transmitted load,
FA = WtA/cos25° = 281.9/cos25° = 311.0 lbf Ans.
Summing the torque about the shaft axis,
2 2
A B
tA tB
d dW W =
( )
( ) ( )
/ 2 20281.9 704.75 lbf
/ 2 8
A A
tB tA tA
B B
d dW W W
d d
= = = =
704.75 777.6 lbf .
cos 25 cos 25
tB
B
WF Ans= = =
______________________________________________________________________________
13-17 20 8 20 4
40 17 60 51
e = =
( )4 00 47.06 rev/min cw .
51d
n Ans= 6 =
______________________________________________________________________________
13-32 (or 13-31)
(a) 2 / 60nω π=
2 / 60 ( in N m, in W)H T Tn T Hω π= = ⋅
So
( )360 10
2
9550 / ( in kW, in rev/min)
H
T
n
H n H n
π
=
=
( )9550 75 398 N m
1800a
T = = ⋅
( )2
2
5 17
42.5 mm
2 2
mNr = = =
So 32
2
398 9.36 kN
42.5
t aTF
r
= = =
( )3 3 2 9.36 18.73 kN in the positive -direction. .b bF F x Ans= − = =
(b) ( )44
5 51
127.5 mm
2 2
mNr = = =
( )4 9.36 127.5 1193 N m ccwcT = = ⋅
4 1193 N m cw .cT Ans∴ = ⋅
Note: The solution is independent of the pressure angle.
13-51 (or 13-45)
NOTE: The shaft forces exerted on the gears are not shown in the figures above.
cos 5cos30 4.330 teeth/int nP P ψ= = =
1 1tan tan 20tan tan 22.80
cos cos30
n
t
φ
φ
ψ
− −= = =
18 4.157 in
4.330P
d = =
The forces on the shafts will be equal to the forces transmitted to the gears through the meshing
teeth.
Pinion (Gear 2)
tan 800 tan 22.80 336 lbfr t tW W φ= = =
tan 800 tan 30 462 lbfa tW W ψ= = =
336 462 800 lbf .Ans= − − +W i j k
( ) ( )
1/22 2 2336 462 800 983 lbf .W Ans = − + − + =
Gear 3
336 462 800 lbf .Ans= + −W i j k
983 lbf .W Ans=
32 7.390 in
4.330G
d = =
( )800 7.390 5912 lbf intGT W r= = = ⋅
