Tài liệu shi20396 chương 15 docx

Chapter 15 15-1 Given: Uncrowned, through-hardened 300 Brinell core and case, Grade 1, N C = 10 9 rev of pinion at R = 0.999 , N P = 20 teeth, N G = 60 teeth, Q v = 6, P d = 6 teeth/in, shaft angle 90°, n p = 900 rev/min, J P = 0.249 and J G = 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 =  59.77 + √ 785.3 59.77  0.8255 = 1.374 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 = √ 1.25 = 1.118 Bending Fig. 15-13: 0.99 S t = s at = 44(300) + 2100 = 15 300 psi Eq. (15-4): (σ all ) P = s wt = s at K L S F K T K R = 15 300(0.862) 1(1)(1.25) = 10 551 psi Eq. (15-3): W t P = (σ all ) P FK x J P P d K o K v K s K m = 10 551(1.25)(1)(0.249) 6(1)(1.374)(0.5222)(1.106) = 690 lbf H 1 = 690(785.3) 33 000 = 16.4hp Eq. (15-4): (σ all ) G = 15 300(0.893) 1(1)(1.25) = 10 930 psi shi20396_ch15.qxd 8/28/03 3:25 PM Page 390 Chapter 15 391 W t G = 10 930(1.25)(1)(0.216) 6(1)(1.374)(0.5222)(1.106) = 620 lbf H 2 = 620(785.3) 33 000 = 14.8hp Ans. The gear controls the bending rating. 15-2 Refer to Prob. 15-1 for the gearset specifications. Wear Fig. 15-12: s ac = 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 = s ac (C L ) P C H S H K T C R (σ c,all ) P = 125 920(1)(1) 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 = s ac (C L ) G C H S H K T C R (σ c,all ) G = 125 920(1.0685)(1.001 38) 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 t P =  (σ c,all ) P C p  2 Fd P I K o K v K m C s C xc =  112 630 2290  2  1.25(3.333)(0.083) 1(1.374)(1.106)(0.5937)(2)  = 464 lbf H 3 = 464(785.3) 33 000 = 11.0hp W t G =  120 511 2290  2  1.25(3.333)(0.083) 1(1.374)(1.106)(0.593 75)(2)  = 531 lbf H 4 = 531(785.3) 33 000 = 12.6hp shi20396_ch15.qxd 8/28/03 3:25 PM Page 391 392 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The pinion controls wear: H = 11.0hp Ans. The power rating of the mesh, considering the power ratings found in Prob. 15-1, is H = min(16.4, 14.8, 11.0, 12.6) = 11.0hp Ans. 15-3 AGMA 2003-B97 does not fully address cast iron gears, however, approximate compar- isons can be useful. This problem is similar to Prob. 15-1, but not identical. We will orga- nize the method. A follow-up could consist of completing Probs. 15-1 and 15-2 with identical pinions, and cast iron gears. Given: Uncrowned, straight teeth, P d = 6 teeth/in, N P = 30 teeth, N G = 60 teeth , ASTM 30 cast iron, material Grade 1, shaft angle 90°, F = 1.25 , n P = 900 rev/min , φ n = 20 ◦ , one gear straddle-mounted, K o = 1 , J P = 0.268 , J G = 0.228 , S F = 2 , S H = √ 2. Mesh d P = 30/6 = 5.000 in d G = 60/6 = 10.000 in v t = π(5)(900/12) = 1178 ft/min Set N L = 10 7 cycles for the pinion. For R = 0.99 , Table 15-7: s at = 4500 psi Table 15-5: s ac = 50 000 psi Eq. (15-4): s wt = s at K L S F K T K R = 4500(1) 2(1)(1) = 2250 psi The velocity factor K v represents stress augmentation due to mislocation of tooth profiles along the pitch surface and the resulting “falling” of teeth into engagement. Equation (5-67) shows that the induced bending moment in a cantilever (tooth) varies directly with √ E of the tooth material. If only the material varies (cast iron vs. steel) in the same geometry, I is the same. From the Lewis equation of Section 14-1, σ = M I /c = K v W t P FY We expect the ratio σ CI /σ steel to be σ CI σ steel = (K v ) CI (K v ) steel =  E CI E steel In the case of ASTM class 30, from Table A-24(a) (E CI ) av = (13 + 16.2)/2 = 14.7 kpsi Then (K v ) CI =  14.7 30 (K v ) steel = 0.7(K v ) steel shi20396_ch15.qxd 8/28/03 3:25 PM Page 392 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 =  59.77 + √ 1178 59.77  0.8255 = 1.454 Pinion bending (σ all ) P = s wt = 2250 psi From Prob. 15-1, K x = 1, K m = 1.106 , K s = 0.5222 Eq. (15-3): W t P = (σ all ) P FK x J P P d K o K v K s K m = 2250(1.25)(1)(0.268) 6(1)(1.454)(0.5222)(1.106) = 149.6 lbf H 1 = 149.6(1178) 33 000 = 5.34 hp Gear bending W t G = W t P J G J P = 149.6  0.228 0.268  = 127.3 lbf H 2 = 127.3(1178) 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: s ac = 50 000 psi s wt = σ c,all = 50 000 √ 2 = 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 =  35 355 1960  2  1.25(5.000)(0.086) 1(1.454)(1.106)(0.59375)(2)  = 91.6 lbf H 3 = H 4 = 91.6(1178) 33 000 = 3.27 hp shi20396_ch15.qxd 8/28/03 3:25 PM Page 393 394 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 = 4mm, shaft angle 90°, n 1 = 1800 rev/min, S F = 1, S H =  S F = √ 1 , J P = Y J 1 = 0.23, J G = 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 = d e1 = 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 =  54.77 + √ 200(8.29) 54.77  0.9148 = 1.663 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 Hβ = 1 + 5.6(10 −6 )(25 2 ) = 1.0035 From Fig. 15-8, (C L ) P = (Z NT ) P = 3.4822(10 9 ) −0.0602 = 1.00 (C L ) G = (Z NT ) 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 = √ 1.25 = 1.118 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 = 2.35H B + 162.89 = 2.35(180) + 162.89 = 585.9MPa Fig. 15-6: I = Z I = 0.066 Eq. (15-2): (σ H ) P = (σ H lim ) P (Z NT ) P Z W S H K θ Z Z = 585.9(1)(1) √ 1(1)(1.118) = 524.1MPa W t P =  σ H C p  2 bd e1 Z I 1000K A K v K Hβ Z x Z xc shi20396_ch15.qxd 8/28/03 3:25 PM Page 394 Chapter 15 395 The constant 1000 expresses W t in kN W t P =  524.1 190  2  25(88)(0.066) 1000(1)(1.663)(1.0035)(0.56)(2)  = 0.591 kN H 3 = W t rn 1 9.55 = 0.591(88/2)(1800) 9.55(10) 3 = 4.90 kW Wear of Gear σ H lim = 585.9 MPa (σ H ) G = 585.9(1.0054) √ 1(1)(1.118) = 526.9 MPa W t G = W t P (σ H ) G (σ H ) P = 0.591  526.9 524.1  = 0.594 kN H 4 = W t rn 9.55 = 0.594(88/2)(1800) 9.55(10 3 ) = 4.93 kW Thus in wear, the pinion controls the power rating; H = 4.90 kW Ans. We will rate the gear set after solving Prob. 15-6. 15-6 Refer to Prob. 15-5 for terms not defined below. Bending of Pinion (K L ) P = (Y NT ) P = 1.6831(10 9 ) −0.0323 = 0.862 (K L ) G = (Y NT ) G = 1.6831[10 9 (22/24)] −0.0323 = 0.864 Fig. 15-13: σ F lim = 0.30H B + 14.48 = 0.30(180) + 14.48 = 68.5 MPa Eq. (15-13): K x = Y β = 1 From Prob. 15-5: Y Z = 1.25, v et = 8.29 m/s K v = 1.663, K θ = 1, Y x = 0.56, K Hβ = 1.0035 (σ F ) P = σ F lim Y NT S F K θ Y Z = 68.5(0.862) 1(1)(1.25) = 47.2MPa W t p = (σ F ) P bm et Y β Y J 1 1000K A K v Y x K Hβ = 47.2(25)(4)(1)(0.23) 1000(1)(1.663)(0.56)(1.0035) = 1.16 kN H 1 = 1.16(88/2)(1800) 9.55(10 3 ) = 9.62 kW Bending of Gear σ F lim = 68.5 MPa (σ F ) G = 68.5(0.864) 1(1)(1.25) = 47.3MPa W t G = 47.3(25)(4)(1)(0.205) 1000(1)(1.663)(0.56)(1.0035) = 1.04 kN H 2 = 1.04(88/2)(1800) 9.55(10 3 ) = 8.62 kW shi20396_ch15.qxd 8/28/03 3:25 PM Page 395 396 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. 15-7 (a) (S F ) P =  σ all σ  P = (S F ) G =  σ all σ  G (s at K L /K T K R ) P (W t P d K o K v K s K m /FK x J ) P = (s at K L /K T K R ) G (W t P d K o K v K s K m /FK x J ) G All terms cancel except for s at , K L , and J, (s at ) P (K L ) P J P = (s at ) G (K L ) G J G From which (s at ) G = (s at ) P (K L ) P J P (K L ) G J G = (s at ) P J P J G m 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 FK x J P d K o K v K s K m  11 =  s at S F K L K T K R FK x J P d K o K v K s K m  11 (1) In wear  s ac 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  1/2 22 Squaring and solving for W t gives W t =  s 2 ac C 2 L C 2 H S 2 H K 2 T C 2 R C 2 P  22  Fd P I K o K v K m C s C xc  22 (2) Equating the right-hand sides of Eqs. (1) and (2) and canceling terms, and recognizing that C R =  K R and P d d P = N P , we obtain ( s ac ) 22 = C p (C L ) 22  S 2 H S F (s at ) 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  2 S F = 1 So we get (s ac ) G = C p (C L ) G C H  (s at ) P (K L ) P J P K x K T C s C xc N P IK s Ans. shi20396_ch15.qxd 8/28/03 3:25 PM Page 396 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 [s ac 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 = [s ac 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 (s ac ) P (C L ) P = (s ac ) G (C L ) G C H Solving for (s ac ) P gives, with C H . = 1 (s ac ) P = (s ac ) G (C L ) G (C L ) P C H . = (s ac ) G  1 m G  −0.0602 (1) (s ac ) P . = (s ac ) G m 0.0602 G Ans. This equation is the transpose of Eq. (14-45). 15-8 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): (s at ) P = 44(300) + 2100 = 15 300 psi (s at ) G (s at ) P J P J G m −0.0323 G = 15 300  0.249 0.216   3 −0.0323  = 17 023 psi ( H B ) 21 = 17 023 − 2100 44 = 339 Brinell Ans. (s ac ) G = 2290 1.0685(1)  15 300(0.862)(0.249)(1)(0.593 25)(2) 20(0.086)(0.5222) = 141 160 psi ( H B ) 22 = 141 160 − 23 600 341 = 345 Brinell Ans. (s ac ) P = (s ac ) G m 0.0602 G 1 C H . = 141 160(3 0.0602 )  1 1  = 150 811 psi ( H B ) 12 = 150 811 − 23 600 341 = 373 Brinell Ans. Care Case Pinion 300 373 Ans. Gear 399 345 shi20396_ch15.qxd 8/28/03 3:25 PM Page 397 398 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 15-9 Pinion core (s at ) P = 44(300) + 2100 = 15 300 psi (σ all ) P = 15 300(0.862) 1(1)(1.25) = 10 551 psi W t = 10 551(1.25)(0.249) 6(1)(1.374)(0.5222)(1.106) = 689.7 lbf Gear core (s at ) G = 44(352) + 2100 = 17 588 psi (σ all ) G = 17 588(0.893) 1(1)(1.25) = 12 565 psi W t = 12 565(1.25)(0.216) 6(1)(1.374)(0.5222)(1.106) = 712.5 lbf Pinion case (s ac ) P = 341(372) + 23 620 = 150 472 psi (σ c,all ) P = 150 472(1) 1(1)(1.118) = 134 590 psi W t =  134 590 2290  2  1.25(3.333)(0.086) 1(1.374)(1.106)(0.593 75)(2)  = 685.8 lbf Gear case (s ac ) G = 341(344) + 23 620 = 140 924 psi (σ c,all ) G = 140 924(1.0685)(1) 1(1)(1.118) = 134 685 psi W t =  134 685 2290  2  1.25(3.333)(0.086) 1(1.374)(1.106)(0.593 75)(2)  = 686.8 lbf The rating load would be W t rated = min(689.7, 712.5, 685.8, 686.8) = 685.8 lbf which is slightly less than intended. Pinion core (s at ) P = 15 300 psi (as before) (σ all ) P = 10 551 (as before) W t = 689.7 (as before) Gear core (s at ) G = 44(339) + 2100 = 17 016 psi (σ all ) G = 17 016(0.893) 1(1)(1.25) = 12 156 psi W t = 12 156(1.25)(0.216) 6(1)(1.374)(0.5222)(1.106) = 689.3 lbf shi20396_ch15.qxd 8/28/03 3:25 PM Page 398 Chapter 15 399 Pinion case (s ac ) P = 341(373) + 23 620 = 150 813 psi (σ c,all ) P = 150 813(1) 1(1)(1.118) = 134 895 psi W t =  134 895 2290  2  1.25(3.333)(0.086) 1(1.374)(1.106)(0.593 75)(2)  = 689.0 lbf Gear case (s ac ) G = 341(345) + 23 620 = 141 265 psi (σ c,all ) G = 141 265(1.0685)(1) 1(1)(1.118) = 135 010 psi W t =  135 010 2290  2  1.25(3.333)(0.086) 1(1.1374)(1.106)(0.593 75)(2)  = 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, J G = 0.201 , P d = 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 = πd 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 =  54.77 + √ 628.3 54.77  0.9148 = 1.412 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 shi20396_ch15.qxd 8/28/03 3:25 PM Page 399 [...].. .shi20396_ ch15.qxd 400 8/28/03 3:25 PM Page 400 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Analyze for 109 pinion cycles at 0.999 reliability Eq (15- 19): Bending Pinion: Eq (15- 23): K R = 0.50 − 0.25 log(1 − 0.999) = 1.25 √ C R = K R = 1.25 = 1.118 (sat ) P = 44(300) + 2100 = 15 300 psi Eq (15- 4): (σall ) P = Eq (15- 3): Wt = = H1 = 15 300(0.862)... = 8.5 in Eq (15- 39): a = px /π = 0.3927/π = 0.125 in Eq (15- 40): h = 0.3683 px = 0.1446 in Eq (15- 41): h t = 0.6866 px = 0.2696 in Eq (15- 42): do = 1.5 + 2(0.125) = 1.75 in Eq (15- 43): dr = 3 − 2(0.1446) = 2.711 in Eq (15- 44): Dt = 7 + 2(0.125) = 7.25 in Eq (15- 45): Dr = 7 − 2(0.1446) = 6.711 in c = 0.1446 − 0.125 = 0.0196 in Eq (15- 46): Eq (15- 47): ( FW ) max = 2 7.25 2 2 7 − 0.125 2 − 2 = 2.646 in... 56.45(0.7563) t t Eq (15- 57): WW = WG = 966 cos φn sin λ + f cos λ cos φn cos λ − f sin λ cos 20° sin 4.764° + 0.025 cos 4.764° cos 20° cos 4.764° − 0.025 sin 4.764° = 106.4 lbf Ans shi20396_ ch15.qxd 8/28/03 3:25 PM Page 405 Chapter 15 Cs = 1190 − 477 log 7.0 = 787 (c) Eq (15- 33): Eq (15- 36): Cm = 0.0107 −562 + 56(56) + 5145 = 0.767 Eq (15- 37): Cv = 0.659 exp[−0.0011(679.8)] = 0.312 Eq (15- 38): (W t ) all... 0.948 1.795 1.979 10 .156 177 5.25 24.9 5103 16.71 38.2 36.2 1.97 3 30 80 1.75 3.60 2.40 2000 15- 19 563 2120 1038 0.0183 0.951 1.571 1.732 11.6 171 6.0 24.98 4158 19.099 38.0 36.1 1.85 3 30 50 1.75 4.10 2.25 2000 15- 20 492 2524 1284 0.034A 0.913A 1.795 1.979 10 .156 179.6 5.25 24.9 5301 16.7 1.75 3.60 2.4 2500 FAN 41.2 37.7 3.59 3 30 115 15-21 492 2524 1284 0.034A 0.913A 1.795 1.979 10 .156 179.6 5.25 24.9... 1.795 1.979 10 .156 177 5.25 24.9 5103 16.71 HW HG Hf NW NG KW Cs Cm Cv VG t WG t WW f e ( Pt ) G Pn C-to-C ts L λ σG dG 1.75 3.60 1.68 2000 1.75 3.60 2.40 2000 15- 16 px dW FG A 15- 15 1000 0.759 0.236 492 2430 1189 0.0193 0.948 1.795 1.979 10 .156 177 5.25 24.9 8565 16.71 492 2430 1189 0.0193 0.948 1.795 1.979 10 .156 177 5.25 24.9 7247 16.71 38.2 36.2 1.97 3 30 125 1.75 3.60 1.69 2000 15- 18 492 2430... 179.6 5.25 24.9 5301 16.71 1.75 3.60 2.4 2600 FAN 41.2 37.7 3.59 3 30 185 15- 22 3:25 PM 38.2 36.2 1.97 3 30 1.75 3.60 1.43 2000 15- 17 8/28/03 #1 #2 #3 #4 Parameters Selected 15- 15 to 15- 22 Problem statement values of 25 hp, 1125 rev/min, m G = 10, K a = 1.25, n d = 1.1, φn = 20°, ta = 70°F are not referenced in the table shi20396_ ch15.qxd Page 406 ... shi20396_ ch15.qxd Page 403 403 shi20396_ ch15.qxd 404 15- 14 8/28/03 3:25 PM Page 404 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design N W = 1, NG = 56, Pt = 8 teeth/in, d P = 1.5 in, Ho = 1hp, φn = 20◦ , ta = 70◦ F, K a = 1.25, n d = 1, Fe = 2 in, A = 850 in2 m G = NG /N W = 56, (a) dG = NG /Pt = 56/8 = 7.0 in px = π/8 = 0.3927 in, C = 1.5 + 7 = 8.5 in Eq (15- 39):... 000 Gear: (sat ) G = 15 300 psi 15 300(0.881) = 10 783 psi 1(1)(1.25) Eq (15- 4): (σall ) G = Eq (15- 3): Wt = 10 783(0.71)(1)(0.201) = 171.4 lbf 10(1)(1.412)(0.508)(1.252) H2 = 171.4(628.3) = 3.3 hp 33 000 Wear Pinion: (C H ) G = 1, I = 0.078, C p = 2290 psi, C xc = 2 Cs = 0.125(0.71) + 0.4375 = 0.526 25 Eq (15- 22): (sac ) P = 341(300) + 23 620 = 125 920 psi (σc, all ) P = Eq (15- 1): 125 920(1)(1) =... acceptable 1.65 (d) Eq (15- 52): Amin = 43.2(8.5) 1.7 = 1642 in2 < 1700 in2 Eq (15- 49): Hloss = 33 000(1 − 0.7563)(2.18) = 17 530 ft · lbf/min Assuming a fan exists on the worm shaft, Eq (15- 50): Eq (15- 51): hC R = ¯ 1725 + 0.13 = 0.568 ft · lbf/(min · in2 · ◦ F) 3939 ts = 70 + 17 530 = 88.2◦ F Ans 0.568(1700) 405 406 38.2 36.2 1.47 3 30 854 0.759 0.236 492 2430 1189 0.0193 0.948 1.795 1.979 10 .156 177 5.25 24.9... 0.71(2.000)(0.078) 1(1.412)(1.252)(0.526 25)(2) = 144.0 lbf H3 = 144(628.3) = 2.7 hp 33 000 shi20396_ ch15.qxd 8/28/03 3:25 PM Page 401 Chapter 15 401 Gear: (sac ) G = 125 920 psi 125 920(1.043)(1) = 117 473 psi 1(1)(1.118) (σc, all ) = 117 473 2290 Wt = 2 0.71(2.000)(0.078) = 156 .6 lbf 1(1.412)(1.252)(0.526 25)(2) 156 .6(628.3) = 3.0 hp 33 000 H4 = Rating: H = min(3.8, 3.3, 2.7, 3.0) = 2.7 hp Pinion wear . 70°F are not referenced in the table. Parameters Selected 15- 15 15- 16 15- 17 15- 18 15- 19 15- 20 15- 21 15- 22 #1 p x 1.75 1.75 1.75 1.75 1.75 1.75 1.75 1.75 #2 d W 3.60. 1.979 1.979 1.979 1.979 1.732 1.979 1.979 C-to-C 10 .156 10 .156 10 .156 10 .156 10 .156 11.6 10 .156 10 .156 t s 177 177 177 177 177 171 179.6 179.6 L 5.25 5.25