Baixe Shigley - Cap 15 e outras Notas de estudo em PDF para Engenharia Mecânica, somente na Docsity!
Chapter 15
15-1 Given: Uncrowned, through-hardened 300 Brinell core and case, Grade 1, N (^) C = 109 rev of pinion at R = 0 .999, N (^) P = 20 teeth, N (^) G = 60 teeth, Q v = 6, Pd = 6 teeth/in, shaft angle 90°, n (^) p = 900 rev/min, J (^) P = 0 .249 and JG = 0. 216 (Fig. 15-7), F = 1 .25 in, S (^) F = S (^) H = 1, K (^) o = 1. Mesh d (^) P = 20 / 6 = 3 .333 in d (^) G = 60 / 6 = 10 .000 in Eq. (15-7): v t = π(3.333)(900/12) = 785 .3 ft/min Eq. (15-6): B = 0 .25(12 − 6) 2 /^3 = 0. 8255 A = 50 + 56(1 − 0 .8255) = 59. 77
Eq. (15-5): K v =
Eq. (15-8): v t , max = [59. 77 + (6 − 3)]^2 = 3940 ft/min Since 785. 3 < 3904, K v = 1 .374 is valid. The size factor for bending is: Eq. (15-10): K (^) s = 0. 4867 + 0. 2132 / 6 = 0. 5222 For one gear straddle-mounted, the load-distribution factor is: Eq. (15-11): K (^) m = 1. 10 + 0 .0036(1.25) 2 = 1. 106 Eq. (15-15): ( K (^) L ) (^) P = 1 .6831(10^9 )−^0.^0323 = 0. 862 ( K (^) L ) G = 1 .6831(10^9 /3)−^0.^0323 = 0. 893 Eq. (15-14): ( C (^) L ) (^) P = 3 .4822(10^9 )−^0.^0602 = 1 ( C (^) L ) G = 3 .4822(10^9 /3)−^0.^0602 = 1. 069 Eq. (15-19): K (^) R = 0. 50 − 0 .25 log(1 − 0 .999) = 1. 25 (or Table 15-3) C (^) R =
K R =
Bending Fig. 15-13: (^0). 99 St = sat = 44(300) + 2100 = 15 300 psi
Eq. (15-4): (σall ) (^) P = s w t =
sat K (^) L S (^) F K (^) T K (^) R^ =^
1(1)(1.25) =^ 10 551 psi
Eq. (15-3): W (^) Pt =
(σall ) (^) P F K (^) x J (^) P Pd K (^) o K v K (^) s K (^) m
=
6(1)(1.374)(0.5222)(1.106) =^ 690 lbf
H 1 =
33 000 =^16 .4 hp
Eq. (15-4): (σall ) G =
1(1)(1.25) =^ 10 930 psi
Chapter 15 391
W (^) Gt =
6(1)(1.374)(0.5222)(1.106) =^ 620 lbf
H 2 =
33 000 =^14 .8 hp^ Ans. The gear controls the bending rating.
15-2 Refer to Prob. 15-1 for the gearset specifications. Wear Fig. 15-12: sac = 341(300) + 23 620 = 125 920 psi For the pinion, C (^) H = 1. From Prob. 15-1, C (^) R = 1 .118. Thus, from Eq. (15-2):
(σ c , all ) (^) P =
sac ( C (^) L ) (^) P C (^) H S (^) H K (^) T C (^) R
(σ c , all ) (^) P =
1(1)(1.118) =^ 112 630 psi For the gear, from Eq. (15-16), B 1 = 0 .008 98(300/300) − 0 .008 29 = 0 .000 69 C (^) H = 1 + 0 .000 69(3 − 1) = 1 .001 38 And Prob. 15-1, ( C (^) L ) G = 1. 0685. Equation (15-2) thus gives
(σ c , all ) G =
sac ( C (^) L ) G C (^) H S (^) H K (^) T C (^) R
(σ c , all ) G =
1(1)(1.118) =^ 120 511 psi For steel: C (^) p = 2290
psi Eq. (15-9): C (^) s = 0 .125(1.25) + 0. 4375 = 0 .593 75 Fig. 15-6: I = 0. 083 Eq. (15-12): C (^) xc = 2
Eq. (15-1): W (^) Pt =
((σ c , all )^ P C (^) p
) (^2) Fd P I K (^) o K v K (^) m C (^) s C (^) xc
=
) 2 [ 1 .25(3.333)(0.083)
]
= 464 lbf
H 3 =
33 000 =^11 .0 hp
W (^) Gt =
) 2 [ 1 .25(3.333)(0.083)
]
= 531 lbf
H 4 =
33 000 =^12 .6 hp
Chapter 15 393
Our modeling is rough, but it convinces us that ( K v) (^) CI < ( K v) (^) steel , but we are not sure of the value of ( K v) (^) CI. We will use K v for steel as a basis for a conservative rating. Eq. (15-6): B = 0 .25(12 − 6) 2 /^3 = 0. 8255 A = 50 + 56(1 − 0 .8255) = 59. 77
Eq. (15-5): K v =
Pinion bending (σall ) (^) P = s w t = 2250 psi From Prob. 15-1, K (^) x = 1, K (^) m = 1 .106, K (^) s = 0. 5222
Eq. (15-3): W tP =
(σall ) (^) P F K (^) x J (^) P Pd K (^) o K v K (^) s K (^) m
=
6(1)(1.454)(0.5222)(1.106) =^149 .6 lbf
H 1 =
33 000 =^5 .34 hp Gear bending
W (^) Gt = W (^) Pt
JG
J P^ =^149.^6
= 127 .3 lbf
H 2 =
33 000 =^4 .54 hp The gear controls in bending fatigue. H = 4 .54 hp Ans.
15-4 Continuing Prob. 15-3, Table 15-5: sac = 50 000 psi
s w t = σ c , all =
= 35 355 psi
Eq. (15-1): W t^ =
(σ c , all C (^) p
) (^2) Fd P I K (^) o K v K (^) m C (^) s C (^) xc Fig. 15-6: I = 0. 86 From Probs. 15-1 and 15-2: C (^) s = 0 .593 75, K (^) s = 0 .5222, K (^) m = 1 .106, C (^) xc = 2 From Table 14-8: C (^) p = 1960
psi
Thus, W t^ =
) 2 [ 1 .25(5.000)(0.086)
]
= 91 .6 lbf
H 3 = H 4 =
33 000 =^3 .27 hp
394 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Rating Based on results of Probs. 15-3 and 15-4, H = min(5.34, 4.54, 3.27, 3.27) = 3 .27 hp Ans. The mesh is weakest in wear fatigue.
15-5 Uncrowned, through-hardened to 180 Brinell (core and case), Grade 1, 10^9 rev of pinion at R = 0 .999, N (^) p = z 1 = 22 teeth, N (^) a = z 2 = 24 teeth, Q v = 5, m (^) et = 4 mm, shaft angle 90°, n 1 = 1800 rev/min, S (^) F = 1, S (^) H =
S F =
1, J P = Y J 1 = 0 .23, JG = Y J 2 =
0 .205, F = b = 25 mm, K (^) o = K (^) A = K (^) T = K θ = 1 and C (^) p = 190
MPa. Mesh d (^) P = de 1 = mz 1 = 4(22) = 88 mm d (^) G = m (^) et z 2 = 4(24) = 96 mm Eq. (15-7): v et = 5 .236(10−^5 )(88)(1800) = 8 .29 m/s Eq. (15-6): B = 0 .25(12 − 5) 2 /^3 = 0. 9148 A = 50 + 56(1 − 0 .9148) = 54. 77
Eq. (15-5): K v =
Eq. (15-10): K (^) s = Y (^) x = 0. 4867 + 0 .008 339(4) = 0. 520 Eq. (15-11) with K (^) mb = 1 (both straddle-mounted), K (^) m = K (^) H β = 1 + 5 .6(10−^6 )(25^2 ) = 1. 0035 From Fig. 15-8, ( C (^) L ) (^) P = ( Z (^) N T ) (^) P = 3 .4822(10^9 )−^0.^0602 = 1. 00 ( C (^) L ) G = ( Z (^) N T ) G = 3 .4822[10^9 (22/24)]−^0.^0602 = 1. 0054 Eq. (15-12): C (^) xc = Z (^) xc = 2 (uncrowned) Eq. (15-19): K (^) R = Y (^) Z = 0. 50 − 0 .25 log (1 − 0 .999) = 1. 25 C (^) R = Z (^) Z =
Y Z =
From Fig. 15-10, C (^) H = Z w = 1 Eq. (15-9): Z (^) x = 0 .004 92(25) + 0. 4375 = 0. 560 Wear of Pinion Fig. 15-12: σ H lim = 2. 35 H (^) B + 162. 89 = 2 .35(180) + 162. 89 = 585 .9 MPa Fig. 15-6: I = Z (^) I = 0. 066 Eq. (15-2): (σ H ) (^) P =
(σ H lim ) (^) P ( Z (^) N T ) (^) P Z (^) W S (^) H K θ Z (^) Z
=
= 524 .1 MPa
W (^) Pt =
(σ H C (^) p
) (^2) bd e 1 Z^ I 1000 K (^) A K v K (^) H β Z (^) x Z (^) xc
396 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Rating of mesh is H rating = min(9.62, 8.62, 4.90, 4.93) = 4 .90 kW Ans. with pinion wear controlling.
15- (a) ( S (^) F ) (^) P =
(σall σ
P
= ( S F ) G =
(σall σ
G ( sat K (^) L / K (^) T K (^) R ) (^) P ( W t^ Pd K (^) o K v K (^) s K (^) m / F K (^) x J ) (^) P^ =^
( sat K (^) L / K (^) T K (^) R ) G ( W t^ Pd K (^) o K v K (^) s K (^) m / F K (^) x J ) G All terms cancel except for sat , K (^) L , and J , ( sat ) (^) P ( K (^) L ) (^) P J (^) P = ( sat ) G ( K (^) L ) G JG From which
( sat ) G =
( sat ) (^) P ( K (^) L ) (^) P J (^) P ( K (^) L ) G JG^ =^ ( sat^ )^ P
J P
JGm
β G
Where β = − 0 .0178 or β = − 0 .0323 as appropriate. This equation is the same as Eq. (14-44). Ans. (b) In bending
W t^ =
(σ all S (^) F
F K (^) x J Pd K (^) o K v K (^) s K (^) m
11
( (^) s at S (^) F
K L
K T K R
F K (^) x J Pd K (^) o K v K (^) s K (^) m
11
In wear (^) ( sac C (^) L C (^) U S (^) H K (^) T C (^) R
22
= C (^) p
( (^) W t (^) K o K v K^ m C^ s C^ xc Fd (^) P I
22 Squaring and solving for W t^ gives
W t^ =
( (^) s 2 ac C^2 L C^2 H S^2 H K (^) T^2 C^2 R C^2 P
22
( (^) Fd P I K (^) o K v K (^) m C (^) s C (^) xc
22
Equating the right-hand sides of Eqs. (1) and (2) and canceling terms, and recognizing that C (^) R =
K (^) R and Pd d (^) P = N (^) P , we obtain
( sac ) 22 =
C (^) p ( C (^) L ) (^22)
S^2 H
S F
( sat ) 11 ( K (^) L ) 11 K (^) x J 11 K (^) T C (^) s C (^) xc C^2 H N (^) P K (^) s I For equal W t^ in bending and wear S^2 H S (^) F^ =
S F
S F^ =^1
So we get
( sac ) G =
C (^) p ( C (^) L ) G C (^) H
( sat ) (^) P ( K (^) L ) (^) P J (^) P K (^) x K (^) T C (^) s C (^) xc N (^) P I K (^) s Ans.
Chapter 15 397
(c) ( S (^) H ) (^) P = ( S (^) H ) G =
σ c , all σ c
P
σ c , all σ c
G Substituting in the right-hand equality gives
[ [ sac^ C^ L^ /( C^ R^ K^ T^ )] P C (^) p
W t^ K (^) o K v K (^) m C (^) s C (^) xc /( Fd (^) P I )
]
P
= [ [ sac^ C^ L^ C^ H^ /( C^ R^ K^ T^ )] G C (^) p
W t^ K (^) o K v K (^) m C (^) s C (^) xc /( Fd (^) P I )
]
G Denominators cancel leaving ( sac ) (^) P ( C (^) L ) (^) P = ( sac ) G ( C (^) L ) G C (^) H Solving for ( sac ) (^) P gives, with C (^) H =. 1
( sac ) (^) P = ( sac ) G^ ( ( CC^ L^ ) G L )^ P
C (^) H =. ( sac ) G
m (^) G
( sac ) (^) P =. ( sac ) G m^0 G.^0602 Ans. This equation is the transpose of Eq. (14-45).
15- Core Case Pinion ( H (^) B ) 11 ( H (^) B ) (^12) Gear ( H (^) B ) 21 ( H (^) B ) (^22) Given ( H (^) B ) 11 = 300 Brinell Eq. (15-23): ( sat ) (^) P = 44(300) + 2100 = 15 300 psi
( sat ) G ( sat ) (^) PJJ^ P G
m − G^0.^0323 = 15 300
3 −^0.^0323
= 17 023 psi
( H (^) B ) 21 = 17 023 44 − 2100 = 339 Brinell Ans.
( sac ) G = (^1) .0685(1)^2290
20(0.086)(0.5222) =^ 141 160 psi
( H (^) B ) 22 = 141 160 341 −^ 23 600= 345 Brinell Ans.
( sac ) (^) P = ( sac ) G m^0 G.^0602 C^1 H
=^. 141 160(3^0.^0602 )
= 150 811 psi
( H (^) B ) 12 = 150 811 341 −^ 23 600= 373 Brinell Ans.
Care Case Pinion 300 373 Ans. Gear 399 345
Chapter 15 399
Pinion case ( sac ) (^) P = 341(373) + 23 620 = 150 813 psi
(σ c , all ) (^) P =
1(1)(1.118) =^ 134 895 psi
W t^ =
) 2 [ 1 .25(3.333)(0.086)
]
= 689 .0 lbf
Gear case ( sac ) G = 341(345) + 23 620 = 141 265 psi
(σ c , all ) G =
1(1)(1.118) =^ 135 010 psi
W t^ =
) 2 [ 1 .25(3.333)(0.086)
]
= 690 .1 lbf
The equations developed within Prob. 15-7 are effective. In bevel gears, the gear tooth is weaker than the pinion so ( C (^) H ) G = 1. (See p. 784.) Thus the approximations in Prob. 15-7 with C (^) H = 1 are really still exact.
15-10 The catalog rating is 5.2 hp at 1200 rev/min for a straight bevel gearset. Also given: N (^) P = 20 teeth, N (^) G = 40 teeth, φ n = 20 ◦^ , F = 0 .71 in, J (^) P = 0 .241, JG = 0. 201 , Pd = 10 teeth/in, through-hardened to 300 Brinell-General Industrial Service, and Q v = 5 uncrowned. Mesh d (^) P = 20 / 10 = 2 .000 in, d (^) G = 40 / 10 = 4 .000 in
v t =
π d (^) P n (^) P 12 =^
π(2)(1200) 12 =^628 .3 ft/min K (^) o = 1, S (^) F = 1, S (^) H = 1 Eq. (15-6): B = 0 .25(12 − 5) 2 /^3 = 0. 9148 A = 50 + 56(1 − 0 .9148) = 54. 77
Eq. (15-5): K v =
Eq. (15-10): K (^) s = 0. 4867 + 0. 2132 / 10 = 0. 508 K (^) mb = 1. 25 Eq. (15-11): K (^) m = 1. 25 + 0 .0036(0.71) 2 = 1. 252 Eq. (15-15): ( K (^) L ) (^) P = 1 .6831(10^9 )−^0.^0323 = 0. 862 ( K (^) L ) G = 1 .6831(10^9 /2)−^0.^0323 = 0. 881 Eq. (15-14): ( C (^) L ) (^) P = 3 .4822(10^9 )−^0.^0602 = 1. 000 ( C (^) L ) G = 3 .4822(10^9 /2)−^0.^0602 = 1. 043
400 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Analyze for 10^9 pinion cycles at 0.999 reliability Eq. (15-19): K (^) R = 0. 50 − 0 .25 log(1 − 0 .999) = 1. 25 C (^) R =
K R =
Bending Pinion: Eq. (15-23): ( sat ) (^) P = 44(300) + 2100 = 15 300 psi
Eq. (15-4): (σall ) (^) P =
1(1)(1.25) =^ 10 551 psi
Eq. (15-3): W t^ =
(σall ) (^) P F K (^) x J (^) P Pd K (^) o K v K (^) s K (^) m
=
10(1)(1.412)(0.508)(1.252) =^ 201 lbf
H 1 =
33 000 =^3 .8 hp Gear: ( sat ) G = 15 300 psi
Eq. (15-4): (σall ) G =
1(1)(1.25) =^ 10 783 psi
Eq. (15-3): W t^ =
10(1)(1.412)(0.508)(1.252) =^171 .4 lbf
H 2 =
33 000 =^3 .3 hp Wear Pinion: ( C (^) H ) G = 1, I = 0 .078, C (^) p = 2290
psi, C (^) xc = 2 C (^) s = 0 .125(0.71) + 0. 4375 = 0 .526 25 Eq. (15-22): ( sac ) (^) P = 341(300) + 23 620 = 125 920 psi
(σ c , all ) (^) P =
1(1)(1.118) =^ 112 630 psi
Eq. (15-1): W t^ =
[(σ c , all )^ P C (^) p
] (^2) Fd P I K (^) o K v K (^) m C (^) s C (^) xc
=
) 2 [ 0 .71(2.000)(0.078)
]
= 144 .0 lbf
H 3 =
33 000 =^2 .7 hp
prerequisites, and how quantitative it was. The most important thing is to have the stu- dent think about it. The instructor can comment in class when students curiosity is heightened. Options that will surface may include:
- Select a through-hardening steel which will meet or exceed core hardness in the hot- rolled condition, then heat-treating to gain the additional 86 points of Brinell hardness by bath-quenching, then tempering, then generating the teeth in the blank.
- Flame or induction hardening are possibilities.
- The hardness goal for the case is sufficiently modest that carburizing and case harden- ing may be too costly. In this case the material selection will be different.
- The initial step in a nitriding process brings the core hardness to 33–38 Rockwell C-scale (about 300–350 Brinell) which is too much. Emphasize that development procedures are necessary in order to tune the “Black Art” to the occasion. Manufacturing personnel know what to do and the direction of adjust- ments, but how much is obtained by asking the gear (or gear blank). Refer your students to D. W. Dudley, Gear Handbook, library reference section, for descriptions of heat-treat- ing processes.
15-12 Computer programs will vary.
15-13 A design program would ask the user to make the a priori decisions, as indicated in Sec. 15-5, p. 794, MED7. The decision set can be organized as follows: A priori decisions
- Function: H , K (^) o , rpm, mG , temp., N (^) L , R
- Design factor: n (^) d ( S (^) F = n (^) d , S (^) H =
n (^) d )
- Tooth system: Involute, Straight Teeth, Crowning, φ n
- Straddling: K (^) mb
- Tooth count: N (^) P ( N (^) G = m (^) G N (^) P ) Design decisions
- Pitch and Face: Pd , F
- Quality number: Q v
- Pinion hardness: ( H (^) B ) 1 , ( H (^) B ) (^3)
- Gear hardness: ( H (^) B ) 2 , ( H (^) B ) (^4)
402 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
403
First gather all of the equations one needs, then arrange them before coding. Find the required hardnesses, express the consequences of thechosen hardnesses, and allow for revisions as appropriate.
Pinion Bending
Gear Bending
Pinion Wear
Gear Wear
Load-induced
σ^
=^
W
t^ P K
Ko^
K v Km^
s
F K
Jx^
P
=^
s^11
σ^
=^
W^
t^ P K
Ko^
K v Km^
s
F K
Jx^
G
=^
s^21
σ c
= C
( p W
t^ K
Ko^
C v Cs^
xc
Fd
IP^
)^1
/^2
=^
s^12
s^22
= s^12
stress (Allowablestress)Tabulated
( sat
) P
=
s^11
S^ F
K KT^
R
( K
) L^
P
( sat
) G
=
s^21
S^ F
K^
KT R
( K
) L^
G
( sac
) P^
=
s
S 12 KH^
CT^
R
( C
) L^
( P C^ H
) P
( sac
) G
=
s
S 22 KH^
CT^
R
( C
) L^
( G C^ H
) G
strengthAssociated
bhn
=
( sat
) P
− 2100 44 ( sat
) P
− 5980 48
bhn
=
( sat
) G
− 2100 44 ( sat
) G
− 5980 48
bhn
=
( sac
) P
− 23 620 341 ( sac
) P
− 29 560 363
.^6
bhn
=
( sac
) G
− 23 620 341
( sac
) G
− 29 560 363
.^6
hardnessChosen
(^ H
) B^
11
(^ H
) B^
21
(^ H
) B^
12
(^ H
) B^
22
hardnessNew tabulated
( sat
) 1 P
=
{^ 44(
H
) B^
11
+^
2100
48(
H
) B^
11
+^
5980
( sat
) 1 G
=
{^ 44(
H
) B^
21
+^
2100
48(
H
) B^
21
+^
5980
( sac
) 1
= P
{^ 341(
H^
)^ B 12
+^
23 620
363
.6(
H
)^ B^
12
+^
29 560
( sac
) 1 G
=
{^ 341(
H
) B^
22
+^
23 620
363
.6(
H
) B^
22
+^
29 560
strengthFactor of
n^11
=
σ all σ^
=^
( sat
) 1
( P K^ L
) P
s^11
K^
KT R
n^21
=
( s at^1
) G
( K
) L^
G
s^21
K^
KT R
n^12
=
[^ (
sac
) 1
( P C^ L
) P
( C
) H^
P
s^12
K^ T
C R
]^2
n^22
=
[^ (
sac
) 1 G
( C
) L^
( G C^ H
) G
s^22
K^ T
C R
]^2
safety Note: S
F^
=^
n^ d
,^ S
H^
=^
√
S^ F
Chapter 15 405
(c) Eq. (15-33): C (^) s = 1190 − 477 log 7. 0 = 787
Eq. (15-36): C (^) m = 0. 0107
Eq. (15-37): C v = 0 .659 exp[− 0 .0011(679.8)] = 0. 312 Eq. (15-38): ( W t^ ) (^) all = 787(7) 0.^8 (2)(0.767)(0.312) = 1787 lbf Since W (^) Gt < ( W t^ ) (^) all , the mesh will survive at least 25 000 h.
Eq. (15-61): W (^) f = (^0) .025 sin 4.764°^0 .025(966) − cos 20° cos 4.764° = − 29 .5 lbf
Eq. (15-63): H (^) f = 29 .33 0005(679. 8)= 0 .608 hp
HW = (^106) 33 000.4(677 .4)= 2 .18 hp
HG = 966(5633 000.45) = 1 .65 hp
The mesh is sufficient Ans. Pn = Pt /cos λ = 8 /cos 4. 764 ◦^ = 8. 028 pn = π/ 8. 028 = 0 .3913 in
σ G = (^0) .3913(0^966 .5)(0.125) = 39 500 psi
The stress is high. At the rated horsepower,
σ G = (^1).^165 39 500 = 23 940 psi acceptable
(d) Eq. (15-52): A min = 43 .2(8.5) 1.^7 = 1642 in^2 < 1700 in^2
Eq. (15-49): H loss = 33 000(1 − 0 .7563)(2.18) = 17 530 ft · lbf/min Assuming a fan exists on the worm shaft,
Eq. (15-50): ¯ h^ C R =^
- 13 = 0 .568 ft · lbf/(min · in^2 · ◦F)
Eq. (15-51): ts = 70 + (^0) .568(1700)17 530 = 88. 2 ◦F Ans.
406
15-15 to 15-22 Problem statement values of 25 hp, 1125 rev/min,
m
G^
K
a^
n
d^
φ
n^
ta
70°F are not referenced in the table.
ParametersSelected
p^ x
d^ W
FG
A^
FAN
FAN
HW
HG
H^
f^
N^ W
N^ G
K^
W^
C^ s
C^ m
C v
VG
W
t G^
W
t W^
f^
0.034A
0.034A
e^
0.913A
0.913A
(^ P
) t G
Pn
C -to-
C^
ts^
L^
λ^
σ G
d^ G
