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SECTION 8 – KEYS AND COUPLINGS FLAT AND SQUARE KEYS DESIGN PROBLEMS 521. A 2-in. shaft, of cold-drawn AISI 1137, has a pulley keyed to it. (a) Compute the length of square key and the length of flat key such that a key made of cold-drawn C1020 has the same yield strength as the shaft does in pure torsion. (b) The same as (a), except that the key material is AISI 2317, OQT 1000 F. (c) Would you discard either of these keys? Explain. Solution: For AISI 1137 shaft, Table AT 8, sy = 93 ksi Yield Strength of Shaft sys = 0.6sy = 0.6(93) = 55.8 ksi sysπD 3 (55.8)(π )(2)3 T= = = 87.65 in − kips 16 16

(a) Key Material, cold-drawn, C1020, Table AT 7 sy = 66 ksi sys = 0.6sy = 0.6(66) = 39.6 ksi Table AT 19, use b = ½ in , t = 3/8 in. For square key, b = t = ½ in. By shear, ss = sys 2T 2(87.65) L= = = 4.43 in ss bD 39.6(1 2 )(2 ) By compression, sc = sy. Key has lowest yield strength. 4T 4(87.65) L= = = 5.32 in sc tD 66(1 2)(2) Use L = 5.32 in For Flat key, b = ½ in , t = 3/8 in. By shear, ss = sys 2T 2(87.65) L= = = 4.43 in ss bD 39.6(1 2 )(2 ) By compression, sc = sy. Key has lowest yield strength. 4T 4(87.65) L= = = 7.09 in sc tD 66(3 8)(2) Use L = 7.09 in (b) Key Material, AISI 2317, QOT 1000 F. Table AT 8 sy = 71 ksi sys = 0.6sy = 0.6(71) = 42.6 ksi Table AT 19, use b = ½ in , t = 3/8 in. For square key, b = t = ½ in. By shear, ss = sys 2T 2(87.65) L= = = 4.12 in ss bD 42.6(1 2)(2) By compression, sc = sy. Key has lowest yield strength.

1

SECTION 8 – KEYS AND COUPLINGS 4T 4(87.65) = = 4.94 in sc tD 71(1 2 )(2 ) Use L = 4.94 in L=

For Flat key, b = ½ in , t = 3/8 in. By shear, ss = sys 2T 2(87.65) L= = = 4.12 in ss bD 42.6(1 2)(2) By compression, sc = sy. Key has lowest yield strength. 4T 4(87.65) L= = = 6.59 in sc tD 71(3 8)(2 ) Use L = 6.59 in (c) Either of the above is not to be discarded since they are designed based on yield strength with the same factor of safety. 522.

A cast-iron pulley transmits 65.5 hp at 1750 rpm. The 1045 as-rolled shaft to which it is to be keyed is 1 ¾ in. in diameter; key material, cold-drawn 1020. Compute the length of flat key and of square key needed. Solution: For shaft: 1045 as-rolled, Table AT 7, sy = 59 ksi For key: Cold-drawn 1020, sy = 66 ksi D = 1 ¾ in = 1.75 in hp = 65.5 hp, n = 1750 rpm

63,000hp 63,000(65.5) = = 2358 in − lb = 2.358 in − kips n 1750 Table AT 19, use b = 3/8 in, t = 1/4 in for D = 1 ¾ in Assume smooth load, N = 1.5

T=

For Flat key, b = 3/8 in, t = 1/4 in. By shear, sys = 0.6sy = 0.6(66) = 39.6 ksi sys 39.6 ss = = = 26.4 ksi N 1. 5 2T 2(2.358) L= = = 0.272 in ss bD 26.4(3 8)(1.75) By compression, use sy of shaft the lowest. Pulley has the highest compressive strength (Cast iron). sy 59 sc = = = 39.3 ksi N 1. 5 4T 4(2.358) L= = = 0.549 in sc tD 39.3(1 4)(1.75) Use L = 0.549 in - answer

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SECTION 8 – KEYS AND COUPLINGS For Square key, b = 3/8 in, t = 3/8 in. By shear, ss = sys sys = 0.6sy = 0.6(66) = 39.6 ksi sys 39.6 ss = = = 26.4 ksi N 1. 5 2T 2(2.358) L= = = 0.272 in ss bD 26.4(3 8)(1.75) By compression, use sy of shaft the lowest. Pulley has the highest compressive strength (Cast iron). sy 59 sc = = = 39.3 ksi N 1. 5 4T 4(2.358) L= = = 0.366 in sc tD 39.3(3 8)(1.75) Use L = 0.366 in - answer 523.

A 3 ¼-in. shaft transmits with medium shock 85 hp at 100 rpm. Power is received through a sprocket (annealed nodular iron 60-45-10) keyed to the shaft of cold-rolled AISI 1040 (10% work), with a key of cold-finished B1113. What should be the length of (a) a square key? (b) a flat key?

Solution: For sprocket, annealed nodular iron, 60-45-10, Table AT 6, sy = 55 ksi For shaft, cold-rolled AISI 1040 (10% work), Table AT 10, sy = 85 ksi For key, cold-finished B1113, Table AT 7, sy = 72 ksi D = 3 ¼ in = 3.25 in hp = 85 hp n = 100 rpm 63,000hp 63,000(85) = = 53,550 in − lb = 53.55 in − kips n 100 Table AT 19, use b = ¾ in, t = 1/2 in for D = 3 ¼ in For medium shock, N = 2.25 (a) Square key, b = ¾ in, t = ¾ in. By shear, sys = 0.6sy = 0.6(72) = 43.2 ksi sys 43.2 ss = = = 19.2 ksi N 2.25 2T 2(53.55) L= = = 2.29 in ss bD 19.2(3 4)(3.25) By compression, use sy of sprocket the lowest. sy 55 sc = = = 24.4 ksi N 2.25 4T 4(53.55) L= = = 3.60 in sc tD 24.4(3 4)(3.25) T=

3

SECTION 8 – KEYS AND COUPLINGS Use L = 3.60 in. - answer (b) Flat key, b = ¾ in, t = ½ in. By shear, ss = sys sys = 0.6sy = 0.6(72) = 43.2 ksi sys 43.2 ss = = = 19.2 ksi N 2.25 2T 2(53.55) L= = = 2.29 in ss bD 19.2(3 4)(3.25) By compression, use sy of sprocket the lowest. sy 55 sc = = = 24.4 ksi N 2.25 4T 4(53.55) L= = = 5.40 in sc tD 24.4(1 2)(3.25) Use L = 5.40 in. - answer 524.

A cast-steel gear (SAE 0030), with a pitch diameter of 36 in., is transmitting 75 hp at 210 rpm to a rock crusher, and is keyed to a 3-in. shaft (AISI 1045, as rolled); the key is made of AISI C1020, cold drawn. For a design factor of 4 based on yield strength, what should be the length of (a) a square key, (b) flat key? (c) Would either of these keys be satisfactory? Solution: For cast-steel gear (SAE 0030), Table AT 6, sy = 35 ksi For shaft, AISI 1045, as rolled, Table AT 7, sy = 59 ksi For key, AISI C1020, cold-drawn, Table AT 7, sy = 66 ksi D = 3 in hp = 75 hp n = 210 rpm 63,000hp 63,000(75) = = 22,500 in − lb = 22.5 in − kips n 210 Table AT 19, use b = ¾ in, t = 1/2 in for D = 3 in Design factor, N = 4 (a) Square key, b = ¾ in, t = ¾ in. By shear, sys = 0.6sy = 0.6(66) = 39.6 ksi sys 39.6 ss = = = 9.9 ksi N 4 2T 2(22.5) L= = = 2.02 in ss bD 9.9(3 4)(3) By compression, use sy of cast-steel gear the lowest. sy 35 sc = = = 8.75 ksi N 4 T=

4

SECTION 8 – KEYS AND COUPLINGS 4T 4(22.5) = = 4.57 in sc tD 8.75(3 4)(3) Use L = 4.57 in. - answer (b) Flat key, b = ¾ in, t = ½ in. By shear, ss = sys sys = 0.6sy = 0.6(66) = 39.6 ksi sys 39.6 ss = = = 9.9 ksi N 4 2T 2(22.5) L= = = 2.02 in ss bD 9.9(3 4)(3) By compression, use sy of cast-steel gear the lowest. sy 35 sc = = = 8.75 ksi N 4 4T 4(22.5) L= = = 6.86 in sc tD 8.75(1 2)(3) Use L = 6.86 in. - answer L=

525.

An electric motor delivers 50 hp at 1160 rpm to a 1 5/8 in. shaft (AISI 13B45, OQT 1100 F). Keyed to this shaft is a cast-steel (SAE 080, N & T) pulley whose hub is 2 in. long. The loading may be classified as mild shock. Decide upon a key for this pulley (material), investigating both flat and square keys. Solution: hp = 50 hp n = 1160 rpm D = 1 5/8 in = 1.625 in Shaft material – AISI 13B45, OQT 1100 F, Table AT 10, sy = 112 ksi Pulley material – SAE 080, N & T, Table AT 6, sy = 40 ksi L = 2 in N = 2.0 to 2.25 for mild shock From Table AT 19 for D = 1 5/8 in b = 3/8 in, t = ¼ in 63,000hp 63,000(50) T= = = 2,716 in − lb = 2.716 in − kips n 1160 For flat key: b = 3/8 in, t = ¼ in Check for compression: 4T 4(2.716) sc = = = 13.37 ksi LtD (2 )(1 4)(1.625) sy 40 = 2.99 > 2.25 Based on pulley material, N = = sc 13.37 Therefore safe in compression. Determine the yield stress on the key 2T 2(2.716) ss = = = 4.457 ksi LbD (2)(3 8)(1.625) 5

SECTION 8 – KEYS AND COUPLINGS N = 2.25 s ys = 0.6 sy = Nss 0.6 sy = (2.25 )(4.457 ) s y = 16.7 ksi

Min. s y = (13.37 )(2.25) = 30 ksi - Minimum yield strength of key material required. Select SAE 003, Table AT 6, sy = 35 ksi – Answer. For square key: b = t = 3/8 in Check for compression: 4T 4(2.716) sc = = = 8.914 ksi LtD (2 )(3 8)(1.625) sy 40 = 4.49 > 2.25 Based on pulley material, N = = sc 8.914 Therefore safe in compression. Determine the yield stress on the key 2T 2(2.716) ss = = = 4.457 ksi LbD (2)(3 8)(1.625) N = 2.25 s ys = 0.6 sy = Nss 0.6 sy = (2.25 )(4.457 ) s y = 16.7 ksi

Min. s y = (8.914)(2.25) = 20 ksi - Minimum yield strength of key material required. Select SAE 003, Table AT 6, sy = 35 ksi – Answer.

CHECK PROBLEMS 526. A cast-steel (SAE 080, N & T) pulley, attached to a 2-in. shaft, is transmitting 40 hp at 200 rpm, and is keyed by a standard square key, 3 in. long, made of SAE 1015, cold drawn; shaft material, C1144, OQT 1000 F. (a) What is the factor of safety of the key? (b) The same as (a) except a flat key is used. Solution: Pulley, Cast steel, SAE 080, N & T, Table AT 6, sy = 40 ksi Key, SAE 1015, cold drawn, Table AT 7, sy = 63 ksi Shaft, C1144, OQT 1000 F, sy = 83 ksi hp = 40 hp N = 200 rpm D = 2 in L = 3 in

63,000hp 63,000(40) = = 12,600 in − lb = 12.6 in − kips n 200 Table AT 19, D = 2 in b = ½ in, t = 3/8 in T=

6

SECTION 8 – KEYS AND COUPLINGS a) Square key, b = ½ in, t = ½ in By shear: 2T 2(12.6 ) ss = = = 8.4 ksi LbD (3)(1 2)(2) sys 0.6sy 0.6(63) N= = = = 4.50 < 6.21 ss ss 8.4 By compression: 4T 4(12.6) sc = = = 16.8 ksi LtD (3)(1 2)(2) sy (pulley ) 40 N= = = 2.38 < 6.21 sc 16.8 Answer N = 2.38 b) Flat key, b = ½ in, t = 3/8 in By shear: 2T 2(12.6 ) ss = = = 8.4 ksi LbD (3)(1 2)(2) sys 0.6sy 0.6(63) N= = = = 4.50 < 6.21 ss ss 8.4 By compression: 4T 4(12.6) sc = = = 22.4 ksi LtD (3)(3 8)(2) sy (pulley ) 40 N= = = 1.78 < 6.21 sc 22.4 Answer N = 1.78 527.

A cast-steel (SAE 080, N & T) pulley is keyed to a 2 1/2-in. shaft by means of a standard square key, 3 ½-in. long, made of cold-drawn SAE 1015. The shaft is made of cold-drawn AISI 1045. If the shaft is in virtually pure torsion, and turns at 420 rpm, what horsepower could the assembly safely transmit (steady loading)?

Solution: Pulley, Cast steel, SAE 080, N & T, Table AT 6, sy = 40 ksi Key, SAE 1015, cold drawn, Table AT 7, sy = 63 ksi Shaft, AISI 1045, cold drawn, Table AT 7, sy = 85 ksi N = 420 rpm L = 3 ½ in = 3.5 in D = 2 ½ in Table AT 19, D = 2 ½ in b = 5/8 in, t = 7/16 in Square Key, b = t = 5/8 in N = 1.5 for steady loading (smooth) For shaft:

7

SECTION 8 – KEYS AND COUPLINGS 3 ssπD 3 0.6 syπD 0.6(85)(π )(2.5)3 T= = = = 104.31 in − kips 16 N(16) 1.5(16)

Key: By shear: s LbD T= s 2 sys 0.6 sy 0.6(63) ss = = = = 25.2 ksi N N 1. 5 (25.2 )(3.5)(5 8)(2.5) T= = 68.9 in − kips < 104.31 in-kips 2 By compression: s LtD T= c 4 sy (pulley ) 40 sc = = = 26.67 ksi N 1.5 (26.67 )(3.5)(5 8)(2.5) T= = 36.46 in − kips < 104.31 in-kips 4 Use T = 36.46 in − kips = 36,460 in − kips Tn 36,460(420) hp = = = 243 hp 63,000 63,000 528.

The same as 527, except that the diameter is 3 in. and the length of the key is 5 in.

Solution: Pulley, Cast steel, SAE 080, N & T, Table AT 6, sy = 40 ksi Key, SAE 1015, cold drawn, Table AT 7, sy = 63 ksi Shaft, AISI 1045, cold drawn, Table AT 7, sy = 85 ksi N = 420 rpm L = 5 in D = 3 in Table AT 19, D = 3 in b = 3/4 in, t = 1/2 in Square Key, b = t = 3/4 in N = 1.5 for steady loading (smooth) For shaft: 3 s πD 3 0.6 sy πD 0.6(85)(π )(3)3 T= s = = = 180.25 in − kips 16 N(16 ) 1.5(16 ) Key: By shear: s LbD T= s 2

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SECTION 8 – KEYS AND COUPLINGS 0.6(63) = 25.2 ksi N N 1. 5 (25.2)(5)(3 4)(3) T= = 141.75 in − kips < 180.25 in-kips 2 By compression: s LtD T= c 4 sy (pulley ) 40 = = 26.67 ksi sc = N 1.5 (26.67 )(5)(3 4)(3) T= = 75 in − kips < 180.25 in-kips 4 ss =

sys

=

0.6 sy

=

Use T = 75 in − kips = 75,000 in − kips Tn 75,000(420) hp = = = 500 hp 63,000 63,000 MISCELLANEOUS KEYS 529. Two assemblies, one with one feather keys, are shown, with the assumed positions of the normal forces N. Each assembly is transmitting a torque T. Derive an equation for each case giving the axial force needed to slide the hub along the shaft (f = coefficient of friction). Does either have an advantage in this respect?

Solution: ND 2 2T N= D

a) T =

Axial force = F = fN =

2 fT D

2ND = ND 2 T N= D

b) T =

fT D Assembly (b) is stronger than assembly (a) which has an axial force half that of assembly (b). Axial force = F = fN =

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SECTION 8 – KEYS AND COUPLINGS

530.

A 1 11/16-in. shaft rotating at 200 rpm, carries a cast-iron gear keyed to it by a ¼ x 1 ¼-in. Woodruff key; shaft material is cold-finished SAE 1045. The power is transmitted with mild shock. What horsepower may be safely transmitted by the key, (a) if it is made of cold-drawn SAE 1118? (b) if it is made of SAE 2317, OQT 1000 F? (c) How many keys of each material are needed to give a capacity of 25 hp? Specify a choice. Solution: Only shear is used. D = 1 11/16 in n = 200 rpm Woodruff key = ¼ x 1 ¼ in N = 2 for mild shock Shear force for key 2T F= = ss As D s AD T= s s 2 Table 10.1, ¼ x 1 ¼ in Woodruff Key is Key No. 810 Shear area, As = 0.296 sq. in. (a) Key, cold-drawn 1118, Table AT 7, sy = 75 ksi sys 0.6 sy 0.6(75) = = = 22.5 ksi < 24.06 ksi ss = N N 2 (22.5)(0.296)1 11 16 = 5.62 in − kips = 5620 in − lb T= 2 (5620)(200) Tn hp = = = 17.84 hp 63,000 63,000 (b) Key, SAE 2317, OQT 1000F, Table AT 7, sy = 79 ksi sys 0.6 sy 0.6(79) ss = = = = 23.7 ksi < 24.06 ksi N N 2 (23.7 )(0.296 )1 11 16 = 5.92 in − kips = 5920 in − lb T= 2 (5920)(200) Tn hp = = = 18.79 hp 63,000 63,000 (c) Number of keys for (a) = 25 / 17.84 = 1.4 or 2 keys Number of keys for (b) = 25 / 18.79 = 1.33 or 2 keys Select (b) which is stronger. 531.

A 3/16 x 1-in. Woodruff key is used in a 1 3/16-in. shaft (cold-drawn SAE 1045). (a) If the key is made of the same material, will it be weaker or stronger than the shaft in pure torsion? (b) If the key is made of SAE 4130, WQT 1100 F, will it be weaker or stronger? For the purposes here, the weakening of the shaft by the keyway is ignored.

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SECTION 8 – KEYS AND COUPLINGS Solution: Woodruff key, 3/16 x 1 in. D = 1 3/16 in Shaft: Cold drawn, SAE 1045 (Table AT 8) sy = 85 ksi s AD T= s s 2 Table 10.1, 3/16 x 1 in., Woodruff key is Key no. 608. Shear area = As = 0.178 sq. in. (a) Key material = Shaft Material In yield: For key 3 0.6(85)(0.178)1 ss As D 0.6sy As D 16 = 5.39 in − kips T= = = 2 2 2 For shaft: 3

3 0.6(85)(π )1 3 ssπD 3 0.6 syπD 16 = 16.77 in − kips T= = = 16 16 16 Therefore the key is weaker. (b) Key material = SAE 4130, WQT 1100, Table AT 7, sy = 114 ksi In yield: For key 3 0.6(114)(0.178)1 ss As D 0.6 sy As D 16 = 7.23 in − kips T= = = 2 2 2 For shaft: 3

3 0.6(85)(π )1 3 ssπD 3 0.6 syπD 16 = 16.77 in − kips T= = = 16 16 16 Therefore the key is weaker. 532.

A 2-in. shaft (cold-finished SAE 1137) is connected to a hub by a 3/8-in. radial taper pin made of 4150, OQT 1000 F. (a) What horsepower at 1800 rpm would be transmitted when the pin is about to be sheared off? (b) For this horsepower, what peak torsional stress may be repeated in the shaft? Is the shaft safe from fatigue at this stress?

Solution: D = 2 in d = 3/8 in n = 1800 rpm (a) For pin material , 4150, OQT 1000 F, Table AT9, su = 193.5 ksi πd 2 πd 2 = ss F = ss (2 As ) = ss (2) 4 2 πd 2 D 1 FD = ssπd 2D T= = ss 2 2 2 4 s s = sus = 0.75 su = 0.75(193.5 ) = 145.1 ksi

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SECTION 8 – KEYS AND COUPLINGS 2

1 3 T = (145.1)(π ) (2 ) = 32.05 in − kips = 32,050 in − lb 4 8 (32,050 )(1800) Tn hp = = = 915.7 hp 63,000 63,000 (b) For shaft, cold-finished, SAE 1137, Table AT 8, su = 103 ksi s 103 sns = 0.6 sn′ = 0.6 u = 0.6 = 30.96 ksi 2 2 But, 16T 16(32,050) ss = 3 = = 20,404 psi = 20.4 ksi < 30.96 ksi πD π (2)3 s 30.96 N = ns = = 1.52 > 1.5, therefore safe from fatigue at this stress. ss 20.4 533.

A 20-in. lever is keyed to a 1 7/8-in. shaft (cold-finished SAE 1141) by a radial taper pin whose mean diameter is 0.5 in.; pin material, C1095, OQT 800 F. The load on the lever is repeatedly reversed; N = 2 on endurance strength. What is the safe lever load (a) for the shaft, (b) for the pin key (shear only), (c) for the combination? Solution: T = FL where F = safe lever load. L = 20 in D = 1 7/8 in = 1.875 in Shaft Material, cold finished, SAE 1141, Table AT 10. sy = 90 ksi sy sn

= 1.8

90 = 50 ksi 1.8 Pin Material, C1095, OQT 800 F, Table AT 9. su = 176 ksi sn = 0.5su = 0.5(176 ) = 88 ksi (a) For the shaft. 0.6sy 0.6(50) ss = = = 15 ksi N 2 s πD 3 15(π )(1.875)3 T= s = = 19.414 in − kips = 19,414 in − lbs 16 16 T = FL 19,414 = F (20 ) F = 970.7 lb (b) For the pin. 0.6 sy 0.6(88) ss = = = 26.4 ksi N 2 sn =

12

SECTION 8 – KEYS AND COUPLINGS 2T d T 4T = = 3 π 2 2 As d (d ) πd 4 s πd 3 26.4(π )(0.5)3 T= s = = 2.592 in − kips = 2592 in − lbs 4 4 T = FL 2592 = F (20 ) F = 129.6 lb (c) For the combination. Use F = 129.6 lb ss =

534.

A lever is keyed to a 2 ½-in. shaft made of SAE 1035, as rolled, by a radial taper pin, made of SAE 1020, as rolled. A load of 200 lb. is applied to the lever 22 in. from the center of the shaft. (a) What size pin should be used for N = 3 based on the yield strength in shear? (b) Let the hub diameter be 5 in. and assume that the part of the pin in the hub is uniformly loaded cantilever beam. Compute the bending stress and comment on the bending strength (especially if the loading varies). Solution: Shaft material, SAE 1035, as rolled, Table AT 7, sy = 55 ksi Pin material, SAE 1020, as rolled, Table AT 7, sy = 48 ksi F = 200 lb, L = 22 in, N = 3 D = 2 ½ in, Dh = 5 in T = FL = (200)(22) = 4400 in-lb = 4.4 in-kips (a) For the pin: 0.6 sy 0.6(48) ss = = = 9.6 ksi N 3 2T d T 4T ss = = = 3 π 2 2 As d (d ) πd 4 4(4.4) 9.6 = πd 3 d = 0.836 in 7 Use d = in = 0.875 in 8 (b) For the bending stress. As cantilever beam let 1 L = (Dh − D ) 2 1 L = (5 − 2.5) = 1.25 in 2 From Table AT 2. wL2 FL M= = 2 2 Where F is the uniform load.

13

SECTION 8 – KEYS AND COUPLINGS

2T 4T 4(4400) = = = 2347 lb Dm Dh + D 5 + 2.5 F = 1174 lb FL (1174)(1.25) M= = = 734 lb 2 2 Bending stress 32M 32(734) sb = 3 = = 11,160 psi = 11.16 ksi πd π (0.875)3 If the loading varies and factor of safety of 3. sn = Nsb = 3(11.16 ) = 33.48 ksi Pin material, SAE 1020, as rolled, Table AT 7, sy = 48 ksi, su = 65 ksu. sn′ = 0.5 su = 0.5(65 ) = 32.5 ksi The bending stress is nearly safe as the load varies. 2F =

535.

A sprocket, transmitting 10 hp at 100 rpm, is attached to a 1 7/16 in. shaft as shown in Fig. 10.15, p. 290., Text; E = 3-1/2 in. What should be the minimum shear pin diameter if the computed stress is 85% of the breaking stress mentioned in the Text? Solution: hp = 10 hp, n = 100 rpm, D = 1 7/16 in = 1.4375 in, E = 3 ½ in = 3.5 in 63,000hp 63,000(10) T= = = 6300 in − lb n 100 T 6300 F= = = 1800 lb E 3.5 From text, Page 290. Breaking stress = 50,000 psi s s = 0.85(50 ,000 ) = 42 ,500 psi 4F ss = 2 πd 4(1800 ) 42,500 = πd 2 d = 0.2322 in use ¼ in. 536.

A gear is attached to a 2-in. shaft somewhat as shown in Fig. 10-15, p. 290, Text; E = 3 5/16 in.; minimum shear-pin diameter = 3/8 in. with a rated torque of 22 in-kips. (a) For this torque, compute the stress in the shear pin. (b) From the ferrous metals given in the Appendix, select those that would give a resisting torque of about 110% of the rated value. Choose one, specifying its heat treatments or other conditions. Solution: D = 2 in E = 3 5/16 in = 3.3125 in d = 3/8 in T = 22 in-kips (a) Stress in shear-pin T 22 F= = = 6.64 lb E 3.3125

14

SECTION 8 – KEYS AND COUPLINGS

4(6.64) = 60.12 ksi π (3 8)2 (b) Select material. sus = 1.1(60.12 ) = 66.13 ksi From appendix, Table AT 7, select Cold drawn, C1020 with sus = 66 ksi ss =

SPLINES 537. A shaft for an automobile transmission has 10 splines with the following dimensions: D = 1.25 in., d = 1.087 in., and L = 1.000 in. (see Table 10.2, p. 287, Text). Determine the safe torque capacity and horsepower at 3600 rpm of this sliding connection. Solution: D = 1.25 in, d = 1.087 in, L = 1.000 in, Nt = 10, n = 3600 rpm T = (1000 )(hL )(rm )(Nt ) in − lb D +d rm = 4 Table 10.2 for 10 splines, sliding connection h = 0.095D Then h = 0.095(1.25 ) = 0.11875 in But D − d 1.25 − 1.087 h= = = 0.0815 in actual 2 2 1.25 + 1.087 rm = = 0.58425 in 4 T = (1000 )(0.0815)(1.000 )(0.58425 )(10 ) = 476.2 in − lb - ans Tn (476.2 )(3600) hp = = = 27.2 hp - ans 63000 63000

538.

The rear axle of an automobile has one end splined. For this fitting there are ten splines, and D = 1.31 in., d = 1.122 in., and L = 1 15/16 in. The minimum shaft diameter is 1 3/16 in. (a) Determine the safe torque capacity of the splined connection, sliding under load. (b) Determine the torque that would have the splines on the point of yielding if the shaft is AISI 8640, OQT 1000 F, if one fourth of the splines are in contact. (c) Determine the torsional stress in the shaft corresponding to each of these torques. Solution: D = 1.31 in, d = 1.122 in, L = 1 15/16 = 1.9375 in, Nt = 10 Dr = 1 3/16 in = 1.1875 in T = (1000 )(hL )(rm )(Nt ) in − lb D +d rm = 4 Table 10.2 for 10 splines, sliding connection h = 0.095D Then h = 0.095(1.31) = 0.12445 in 15

SECTION 8 – KEYS AND COUPLINGS But

D − d 1.31 − 1.122 = = 0.094 in (actual) 2 2 1.31 + 1.122 rm = = 0.608 in 4 (a) Safe Torque T = (1000 )(0.094 )(1.9375)(0.608 )(10 ) = 1107.32 in − lb - ans (b) Torque by splines required on the point of yielding with one fourth of splines in contact (Page 288). From Table AT-7, AISI 8640, OQT 1000 F. sy = 150 ksi, ss = sys = 0.6sy = 0.6(150) = 90 ksi s πDL D (90)(π )(1.31)(1.9375) 1.31 T = s = 2 = 58.76 in − kips 8 8 2 T = 58,760 in − lbs (c) Torsional stress in the shaft From safe torque of 1107.32 in-lb 16T 16(1107.32 ) ss = 3 = = 3368 psi - ans πDr π (1.1875)3 From torque at the point of yield 16T 16(58,760) ss = 3 = = 178,711 psi - ans (too high) πDr π (1.1875)3 h=

539.

An involute splined connection has 10 splines with a pitch Pd of 12/24 (a) Determine the dimension of this connection. (b) Compute the length of spline to have the same torsional strength as the shaft when one fourth the splines carry the load; minimum shaft diameter is 9/16 in. (no sliding). Check for compression. Solution: Nt = 10, Pd = 12, Dr = 9/16 in = 0.5625 in (a) Dimension D N 10 D = t = = 0.8333 in Pd 12 (b) Length of spline (same torsional strength as the shaft when one fourth the splines carry the load (Page 288). D 3 (0.5625)3 L= r = = 0.2136 in D 0.8333 Check for compression. Failure in compression is not likely (Page 289) and can be checked by using the projected contact area. Projected contact area: Ac = 0.2LD = 0.2(0.2136 )(0.8333) = 0.0356 in 2 based on one-fourth of the teeth being under load. COUPLINGS 540. A flange coupling has the following dimensions (Fig. 10.19, p. 291, Text): d = 5, D = 8 5/8, H = 12 ¼, g = 1 ½, h = 1, L = 7 ¼ in.; number of bolts = 6; 1 ¼ x 1 ¼-in. square key. Materials: key, colddrawn AISI 1113; shaft, cold-rolled, AISI 1045; bolts, SAE grade 5 (§5.8). Using the static

16

SECTION 8 – KEYS AND COUPLINGS approach with N = 3.3 on yield strengths, determine the safe horsepower that this connection may transmit at 630 rpm. Solution:

d = 5 in D = 8 5/8 in = 8.625 in H = 12 ¼ in = 12.25 in g = 1 ½ in = 1.5 in h = 1 in L = 7 ¼ in = 7.25 in N = 3.3 nb = 630 rpm Square key = 1 ¼ in x 1 ¼ in Materials: Key: cold-drawn AISI 1113, Table AT 7, sy = 72 ksi, sys = 0.6sy = 0.6(72) = 43.2 ksi Shaft: cold-rolled, AISI 1045, Table AT 8, sy = 85 ksi, sys = 0.6sy = 0.6(85) = 51 ksi Bolt: SAE Grade 5, h = 1 in. sy = 81 ksi, sys = 0.6sy = 0.6(81) = 48.6 ksi No given material for the flange. Bolts in shear: sys 48.6 ss = = = 14.73 ksi N 3. 3 πh 2 F= N b ss 4 FH πh 2 Nb ss H T= = 2 8 2 π (1) (6 )(14.73)(12.25) T= = 425.158 in − kips 8 T = 425,158 in − lbs (425,158)(630) Tn hp = = = 4252 hp 63,000 63,000 Bolts in compression: sy 81 sc = = = 24.55 ksi N 3.3 F = Nb hgs c

17

SECTION 8 – KEYS AND COUPLINGS

FH Nb hgsc H = 2 2 (6)(1)(1.5)(24.55)(12.25) T= = 1353.319 in − kips 2 T = 1,353,319 in − lbs (1,353,319)(630) Tn hp = = = 13,533 hp 63,000 63,000 T=

Key in shear: sys 43.2 ss = = = 13.09 ksi N 3. 3 s bdL (13.09 )(1.25)(5)(7.25) T= s = = 296.570 in − kips 2 2 T = 296,570 in − lbs (296,570)(630) Tn hp = = = 2966 hp 63,000 63,000 Key in compression: sy 72 sc = = = 21.82 ksi N 3.3 s tdL (21.82)(1.25)(5)(7.25) T= c = = 247.180 in − kips 4 4 T = 247 ,180 in − lbs (247,180 )(630) Tn hp = = = 2472 hp 63,000 63,000 Shaft in shear: sys 51 ss = = = 15.45 ksi N 3.3 πd 3 ss π (5)3 (15.45) T= = = 379.200 in − kips 16 16 T = 379,200 in − lbs (379,200)(630) Tn hp = = = 3792 hp 63,000 63,000 The safest horsepower is the lowest which is 2472 hp. 541.

A cast-iron (ASTM 25) jaw clutch with 4 jaws transmits 50 hp at 60 rpm. The inside diameter of the jaws is 3 in. Considering rough handling, choose N = 8 on ultimate strengths. Make reasonable and conservative assumptions and compute (a) the outside diameter of the jaws, (b) the length of jaws h.

18

SECTION 8 – KEYS AND COUPLINGS

Solution: For ASTM 25, suc = 97 ksi, in shear sus = 35 ksi (Table AT 6) 63,000hp 63,000(50) T= = = 52,500 in − lbs n 60 T = 52.5 in − kips (a) The outside diameter of the jaws s 35 ss = us = = 4.375 ksi N 8 Assume Dm as the average diameter, t = thickness = Do – Di , Nj = number of jaws Shear area, 1 πD 1 π D + Di Do − Di As = m (t ) = o 2 Nj 2 N j 2 2 1 π D 2 − Di2 π 2 = As = o Do − Di2 2 4 4 32 2T 4T F= = Dm Do + Di F 4T 32 128T ss = = ⋅ = 2 2 π (Do + Di ) Do2 − Di2 As Do + Di π Do − Di 128(52.5) 4.375 = π (Do + 3) Do2 − 9 By trial and error. Do = 7.466 in or D o = 7.5 in (b) The length of jaws h. s 97 sc = uc = = 12.125 ksi N 8 N j h(Do − Di ) Ac = 2 F 2F 2F F sc = = = = Ac N j h(Do − Di ) 4h(Do − Di ) 2h(Do − Di )

(

(

(

sc =

)

)

(

)

)

4T (Do + Di ) 2T = 2 2h(Do − Di ) h Do − Di2

(

) 19

SECTION 8 – KEYS AND COUPLINGS

2(52.5) h (7.5)2 − (3)2 3 h = 0.1833 in or h = in 16

12.125 =

542.

[

]

The universal joint shown is made of AISI 3150, OQT 1000 F; a = 2 7/16 in., D = 9/16; n = 400 rpm. (a) What torque may be transmitted for shear of the pin (N = 5 on ultimate)? (b) Considering the pin as a simply supported beam of length a with the load distributed from a maximum at the outer (triangular), compute the safe transmitted torque (Same N). (c) In order not to have excessive wear on the pin, the average bearing pressure should not excced 3 ksi. Compute this transmitted torque. (d) What is the safe power?

Solution: For AISI 3150, OQT 1000 F, Table AT 7, su = 151 ksi, sus = 113 ksi N=5 pb = 3 ksi a = 2 7/16 in = 2.4375 in D = 9/16 in = 0.5625 in n = 400 rpm (a) Torque transmitted for shear of the pin. s 113 ss = us = = 22.6 ksi N 5 Each shear area πD 2 (22.6)(π )(0.5625)2 = F = ss = 5.616 kips 4 4 Fa T = 2 = Fa = (5.616)(2.4375) = 13.687 in − kips 2 T = 13,687 in − lbs

(b) Torque transmitted for shear of the pin (simply supported beam) Fa T =M= 3 (5.616)(2.4375) T= = 4.563 in − kips 3 T = 4,563 in − lbs

20

SECTION 8 – KEYS AND COUPLINGS

(c) Torque transmitted for shear of the pin (pb = 3 ksi) F F F pb = = = Ab D(a 2) 2Da F 3= 2(0.5625)(2.4375) F = 8.23 ksi Fa T =M= 3 (8.23)(2.4375) T= = 6.687 in − kips 3 T = 6,687 in − lbs (d) Safe power (4,563)(400) Tn hp = = = 28.97 hp 63,000 63,000 544.

A diagrammatic representation of a universal joint is shown, two yoke parts, the type being similar to Figs. 10.28 and 12.10, Text. The pin extensions have a diameter D = ¾ in.; a = 11/16 in., material of all parts is 4340, OQT 800. Let N = 4 on ultimate stresses; n = 2400 rpm. Compute the safe torque for (a) shear of pins, (b) the pin extensions in bending, assuming that the load distribution is from zero at the outside pin ends to a maximum at the inside yoke surfaces, (c) an average bearing pressure on pins of 4 ksi. (d) What is the corresponding horsepower capacity?

Solution: For AISI 4340, OQT 800 F, Table AT 7, su = 221 ksi, sus = 0.75su = 0.75(221) = 166 ksi N=4 pb = 4 ksi a = 11/16 in = 0.6875 in D = 3/4 in = 0.75 in n = 2400 rpm (a) Torque transmitted for shear of the pin. s 166 ss = us = = 41.5 ksi N 4 Each shear area πD 2 (41.5)(π )(0.75)2 = F = ss = 18.33 kips 4 4 T = M = Fa = (18.33)(0.6875) = 12.602 in − kips T = 12,602 in − lbs 21

SECTION 8 – KEYS AND COUPLINGS

(b) Torque transmitted for shear of the pin (simply supported beam) 2Fa T =M= 3 2(18.33)(0.6875) T= = 8.401 in − kips 3 T = 8,401 in − lbs (c) Torque transmitted for shear of the pin (pb = 4 ksi) F F pb = = Ab Da F 4= (0.75)(0.6875) F = 2.063 ksi 2Fa T =M= 3 2(2.063)(0.6875) T= = 0.9455 in − kips 3 T = 945.5 in − lbs (d) Safe power (945.5)(2400) Tn hp = = = 36.02 hp 63,000 63,000 -

End -

22

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(a) Key Material, cold-drawn, C1020, Table AT 7 sy = 66 ksi sys = 0.6sy = 0.6(66) = 39.6 ksi Table AT 19, use b = ½ in , t = 3/8 in. For square key, b = t = ½ in. By shear, ss = sys 2T 2(87.65) L= = = 4.43 in ss bD 39.6(1 2 )(2 ) By compression, sc = sy. Key has lowest yield strength. 4T 4(87.65) L= = = 5.32 in sc tD 66(1 2)(2) Use L = 5.32 in For Flat key, b = ½ in , t = 3/8 in. By shear, ss = sys 2T 2(87.65) L= = = 4.43 in ss bD 39.6(1 2 )(2 ) By compression, sc = sy. Key has lowest yield strength. 4T 4(87.65) L= = = 7.09 in sc tD 66(3 8)(2) Use L = 7.09 in (b) Key Material, AISI 2317, QOT 1000 F. Table AT 8 sy = 71 ksi sys = 0.6sy = 0.6(71) = 42.6 ksi Table AT 19, use b = ½ in , t = 3/8 in. For square key, b = t = ½ in. By shear, ss = sys 2T 2(87.65) L= = = 4.12 in ss bD 42.6(1 2)(2) By compression, sc = sy. Key has lowest yield strength.

1

SECTION 8 – KEYS AND COUPLINGS 4T 4(87.65) = = 4.94 in sc tD 71(1 2 )(2 ) Use L = 4.94 in L=

For Flat key, b = ½ in , t = 3/8 in. By shear, ss = sys 2T 2(87.65) L= = = 4.12 in ss bD 42.6(1 2)(2) By compression, sc = sy. Key has lowest yield strength. 4T 4(87.65) L= = = 6.59 in sc tD 71(3 8)(2 ) Use L = 6.59 in (c) Either of the above is not to be discarded since they are designed based on yield strength with the same factor of safety. 522.

A cast-iron pulley transmits 65.5 hp at 1750 rpm. The 1045 as-rolled shaft to which it is to be keyed is 1 ¾ in. in diameter; key material, cold-drawn 1020. Compute the length of flat key and of square key needed. Solution: For shaft: 1045 as-rolled, Table AT 7, sy = 59 ksi For key: Cold-drawn 1020, sy = 66 ksi D = 1 ¾ in = 1.75 in hp = 65.5 hp, n = 1750 rpm

63,000hp 63,000(65.5) = = 2358 in − lb = 2.358 in − kips n 1750 Table AT 19, use b = 3/8 in, t = 1/4 in for D = 1 ¾ in Assume smooth load, N = 1.5

T=

For Flat key, b = 3/8 in, t = 1/4 in. By shear, sys = 0.6sy = 0.6(66) = 39.6 ksi sys 39.6 ss = = = 26.4 ksi N 1. 5 2T 2(2.358) L= = = 0.272 in ss bD 26.4(3 8)(1.75) By compression, use sy of shaft the lowest. Pulley has the highest compressive strength (Cast iron). sy 59 sc = = = 39.3 ksi N 1. 5 4T 4(2.358) L= = = 0.549 in sc tD 39.3(1 4)(1.75) Use L = 0.549 in - answer

2

SECTION 8 – KEYS AND COUPLINGS For Square key, b = 3/8 in, t = 3/8 in. By shear, ss = sys sys = 0.6sy = 0.6(66) = 39.6 ksi sys 39.6 ss = = = 26.4 ksi N 1. 5 2T 2(2.358) L= = = 0.272 in ss bD 26.4(3 8)(1.75) By compression, use sy of shaft the lowest. Pulley has the highest compressive strength (Cast iron). sy 59 sc = = = 39.3 ksi N 1. 5 4T 4(2.358) L= = = 0.366 in sc tD 39.3(3 8)(1.75) Use L = 0.366 in - answer 523.

A 3 ¼-in. shaft transmits with medium shock 85 hp at 100 rpm. Power is received through a sprocket (annealed nodular iron 60-45-10) keyed to the shaft of cold-rolled AISI 1040 (10% work), with a key of cold-finished B1113. What should be the length of (a) a square key? (b) a flat key?

Solution: For sprocket, annealed nodular iron, 60-45-10, Table AT 6, sy = 55 ksi For shaft, cold-rolled AISI 1040 (10% work), Table AT 10, sy = 85 ksi For key, cold-finished B1113, Table AT 7, sy = 72 ksi D = 3 ¼ in = 3.25 in hp = 85 hp n = 100 rpm 63,000hp 63,000(85) = = 53,550 in − lb = 53.55 in − kips n 100 Table AT 19, use b = ¾ in, t = 1/2 in for D = 3 ¼ in For medium shock, N = 2.25 (a) Square key, b = ¾ in, t = ¾ in. By shear, sys = 0.6sy = 0.6(72) = 43.2 ksi sys 43.2 ss = = = 19.2 ksi N 2.25 2T 2(53.55) L= = = 2.29 in ss bD 19.2(3 4)(3.25) By compression, use sy of sprocket the lowest. sy 55 sc = = = 24.4 ksi N 2.25 4T 4(53.55) L= = = 3.60 in sc tD 24.4(3 4)(3.25) T=

3

SECTION 8 – KEYS AND COUPLINGS Use L = 3.60 in. - answer (b) Flat key, b = ¾ in, t = ½ in. By shear, ss = sys sys = 0.6sy = 0.6(72) = 43.2 ksi sys 43.2 ss = = = 19.2 ksi N 2.25 2T 2(53.55) L= = = 2.29 in ss bD 19.2(3 4)(3.25) By compression, use sy of sprocket the lowest. sy 55 sc = = = 24.4 ksi N 2.25 4T 4(53.55) L= = = 5.40 in sc tD 24.4(1 2)(3.25) Use L = 5.40 in. - answer 524.

A cast-steel gear (SAE 0030), with a pitch diameter of 36 in., is transmitting 75 hp at 210 rpm to a rock crusher, and is keyed to a 3-in. shaft (AISI 1045, as rolled); the key is made of AISI C1020, cold drawn. For a design factor of 4 based on yield strength, what should be the length of (a) a square key, (b) flat key? (c) Would either of these keys be satisfactory? Solution: For cast-steel gear (SAE 0030), Table AT 6, sy = 35 ksi For shaft, AISI 1045, as rolled, Table AT 7, sy = 59 ksi For key, AISI C1020, cold-drawn, Table AT 7, sy = 66 ksi D = 3 in hp = 75 hp n = 210 rpm 63,000hp 63,000(75) = = 22,500 in − lb = 22.5 in − kips n 210 Table AT 19, use b = ¾ in, t = 1/2 in for D = 3 in Design factor, N = 4 (a) Square key, b = ¾ in, t = ¾ in. By shear, sys = 0.6sy = 0.6(66) = 39.6 ksi sys 39.6 ss = = = 9.9 ksi N 4 2T 2(22.5) L= = = 2.02 in ss bD 9.9(3 4)(3) By compression, use sy of cast-steel gear the lowest. sy 35 sc = = = 8.75 ksi N 4 T=

4

SECTION 8 – KEYS AND COUPLINGS 4T 4(22.5) = = 4.57 in sc tD 8.75(3 4)(3) Use L = 4.57 in. - answer (b) Flat key, b = ¾ in, t = ½ in. By shear, ss = sys sys = 0.6sy = 0.6(66) = 39.6 ksi sys 39.6 ss = = = 9.9 ksi N 4 2T 2(22.5) L= = = 2.02 in ss bD 9.9(3 4)(3) By compression, use sy of cast-steel gear the lowest. sy 35 sc = = = 8.75 ksi N 4 4T 4(22.5) L= = = 6.86 in sc tD 8.75(1 2)(3) Use L = 6.86 in. - answer L=

525.

An electric motor delivers 50 hp at 1160 rpm to a 1 5/8 in. shaft (AISI 13B45, OQT 1100 F). Keyed to this shaft is a cast-steel (SAE 080, N & T) pulley whose hub is 2 in. long. The loading may be classified as mild shock. Decide upon a key for this pulley (material), investigating both flat and square keys. Solution: hp = 50 hp n = 1160 rpm D = 1 5/8 in = 1.625 in Shaft material – AISI 13B45, OQT 1100 F, Table AT 10, sy = 112 ksi Pulley material – SAE 080, N & T, Table AT 6, sy = 40 ksi L = 2 in N = 2.0 to 2.25 for mild shock From Table AT 19 for D = 1 5/8 in b = 3/8 in, t = ¼ in 63,000hp 63,000(50) T= = = 2,716 in − lb = 2.716 in − kips n 1160 For flat key: b = 3/8 in, t = ¼ in Check for compression: 4T 4(2.716) sc = = = 13.37 ksi LtD (2 )(1 4)(1.625) sy 40 = 2.99 > 2.25 Based on pulley material, N = = sc 13.37 Therefore safe in compression. Determine the yield stress on the key 2T 2(2.716) ss = = = 4.457 ksi LbD (2)(3 8)(1.625) 5

SECTION 8 – KEYS AND COUPLINGS N = 2.25 s ys = 0.6 sy = Nss 0.6 sy = (2.25 )(4.457 ) s y = 16.7 ksi

Min. s y = (13.37 )(2.25) = 30 ksi - Minimum yield strength of key material required. Select SAE 003, Table AT 6, sy = 35 ksi – Answer. For square key: b = t = 3/8 in Check for compression: 4T 4(2.716) sc = = = 8.914 ksi LtD (2 )(3 8)(1.625) sy 40 = 4.49 > 2.25 Based on pulley material, N = = sc 8.914 Therefore safe in compression. Determine the yield stress on the key 2T 2(2.716) ss = = = 4.457 ksi LbD (2)(3 8)(1.625) N = 2.25 s ys = 0.6 sy = Nss 0.6 sy = (2.25 )(4.457 ) s y = 16.7 ksi

Min. s y = (8.914)(2.25) = 20 ksi - Minimum yield strength of key material required. Select SAE 003, Table AT 6, sy = 35 ksi – Answer.

CHECK PROBLEMS 526. A cast-steel (SAE 080, N & T) pulley, attached to a 2-in. shaft, is transmitting 40 hp at 200 rpm, and is keyed by a standard square key, 3 in. long, made of SAE 1015, cold drawn; shaft material, C1144, OQT 1000 F. (a) What is the factor of safety of the key? (b) The same as (a) except a flat key is used. Solution: Pulley, Cast steel, SAE 080, N & T, Table AT 6, sy = 40 ksi Key, SAE 1015, cold drawn, Table AT 7, sy = 63 ksi Shaft, C1144, OQT 1000 F, sy = 83 ksi hp = 40 hp N = 200 rpm D = 2 in L = 3 in

63,000hp 63,000(40) = = 12,600 in − lb = 12.6 in − kips n 200 Table AT 19, D = 2 in b = ½ in, t = 3/8 in T=

6

SECTION 8 – KEYS AND COUPLINGS a) Square key, b = ½ in, t = ½ in By shear: 2T 2(12.6 ) ss = = = 8.4 ksi LbD (3)(1 2)(2) sys 0.6sy 0.6(63) N= = = = 4.50 < 6.21 ss ss 8.4 By compression: 4T 4(12.6) sc = = = 16.8 ksi LtD (3)(1 2)(2) sy (pulley ) 40 N= = = 2.38 < 6.21 sc 16.8 Answer N = 2.38 b) Flat key, b = ½ in, t = 3/8 in By shear: 2T 2(12.6 ) ss = = = 8.4 ksi LbD (3)(1 2)(2) sys 0.6sy 0.6(63) N= = = = 4.50 < 6.21 ss ss 8.4 By compression: 4T 4(12.6) sc = = = 22.4 ksi LtD (3)(3 8)(2) sy (pulley ) 40 N= = = 1.78 < 6.21 sc 22.4 Answer N = 1.78 527.

A cast-steel (SAE 080, N & T) pulley is keyed to a 2 1/2-in. shaft by means of a standard square key, 3 ½-in. long, made of cold-drawn SAE 1015. The shaft is made of cold-drawn AISI 1045. If the shaft is in virtually pure torsion, and turns at 420 rpm, what horsepower could the assembly safely transmit (steady loading)?

Solution: Pulley, Cast steel, SAE 080, N & T, Table AT 6, sy = 40 ksi Key, SAE 1015, cold drawn, Table AT 7, sy = 63 ksi Shaft, AISI 1045, cold drawn, Table AT 7, sy = 85 ksi N = 420 rpm L = 3 ½ in = 3.5 in D = 2 ½ in Table AT 19, D = 2 ½ in b = 5/8 in, t = 7/16 in Square Key, b = t = 5/8 in N = 1.5 for steady loading (smooth) For shaft:

7

SECTION 8 – KEYS AND COUPLINGS 3 ssπD 3 0.6 syπD 0.6(85)(π )(2.5)3 T= = = = 104.31 in − kips 16 N(16) 1.5(16)

Key: By shear: s LbD T= s 2 sys 0.6 sy 0.6(63) ss = = = = 25.2 ksi N N 1. 5 (25.2 )(3.5)(5 8)(2.5) T= = 68.9 in − kips < 104.31 in-kips 2 By compression: s LtD T= c 4 sy (pulley ) 40 sc = = = 26.67 ksi N 1.5 (26.67 )(3.5)(5 8)(2.5) T= = 36.46 in − kips < 104.31 in-kips 4 Use T = 36.46 in − kips = 36,460 in − kips Tn 36,460(420) hp = = = 243 hp 63,000 63,000 528.

The same as 527, except that the diameter is 3 in. and the length of the key is 5 in.

Solution: Pulley, Cast steel, SAE 080, N & T, Table AT 6, sy = 40 ksi Key, SAE 1015, cold drawn, Table AT 7, sy = 63 ksi Shaft, AISI 1045, cold drawn, Table AT 7, sy = 85 ksi N = 420 rpm L = 5 in D = 3 in Table AT 19, D = 3 in b = 3/4 in, t = 1/2 in Square Key, b = t = 3/4 in N = 1.5 for steady loading (smooth) For shaft: 3 s πD 3 0.6 sy πD 0.6(85)(π )(3)3 T= s = = = 180.25 in − kips 16 N(16 ) 1.5(16 ) Key: By shear: s LbD T= s 2

8

SECTION 8 – KEYS AND COUPLINGS 0.6(63) = 25.2 ksi N N 1. 5 (25.2)(5)(3 4)(3) T= = 141.75 in − kips < 180.25 in-kips 2 By compression: s LtD T= c 4 sy (pulley ) 40 = = 26.67 ksi sc = N 1.5 (26.67 )(5)(3 4)(3) T= = 75 in − kips < 180.25 in-kips 4 ss =

sys

=

0.6 sy

=

Use T = 75 in − kips = 75,000 in − kips Tn 75,000(420) hp = = = 500 hp 63,000 63,000 MISCELLANEOUS KEYS 529. Two assemblies, one with one feather keys, are shown, with the assumed positions of the normal forces N. Each assembly is transmitting a torque T. Derive an equation for each case giving the axial force needed to slide the hub along the shaft (f = coefficient of friction). Does either have an advantage in this respect?

Solution: ND 2 2T N= D

a) T =

Axial force = F = fN =

2 fT D

2ND = ND 2 T N= D

b) T =

fT D Assembly (b) is stronger than assembly (a) which has an axial force half that of assembly (b). Axial force = F = fN =

9

SECTION 8 – KEYS AND COUPLINGS

530.

A 1 11/16-in. shaft rotating at 200 rpm, carries a cast-iron gear keyed to it by a ¼ x 1 ¼-in. Woodruff key; shaft material is cold-finished SAE 1045. The power is transmitted with mild shock. What horsepower may be safely transmitted by the key, (a) if it is made of cold-drawn SAE 1118? (b) if it is made of SAE 2317, OQT 1000 F? (c) How many keys of each material are needed to give a capacity of 25 hp? Specify a choice. Solution: Only shear is used. D = 1 11/16 in n = 200 rpm Woodruff key = ¼ x 1 ¼ in N = 2 for mild shock Shear force for key 2T F= = ss As D s AD T= s s 2 Table 10.1, ¼ x 1 ¼ in Woodruff Key is Key No. 810 Shear area, As = 0.296 sq. in. (a) Key, cold-drawn 1118, Table AT 7, sy = 75 ksi sys 0.6 sy 0.6(75) = = = 22.5 ksi < 24.06 ksi ss = N N 2 (22.5)(0.296)1 11 16 = 5.62 in − kips = 5620 in − lb T= 2 (5620)(200) Tn hp = = = 17.84 hp 63,000 63,000 (b) Key, SAE 2317, OQT 1000F, Table AT 7, sy = 79 ksi sys 0.6 sy 0.6(79) ss = = = = 23.7 ksi < 24.06 ksi N N 2 (23.7 )(0.296 )1 11 16 = 5.92 in − kips = 5920 in − lb T= 2 (5920)(200) Tn hp = = = 18.79 hp 63,000 63,000 (c) Number of keys for (a) = 25 / 17.84 = 1.4 or 2 keys Number of keys for (b) = 25 / 18.79 = 1.33 or 2 keys Select (b) which is stronger. 531.

A 3/16 x 1-in. Woodruff key is used in a 1 3/16-in. shaft (cold-drawn SAE 1045). (a) If the key is made of the same material, will it be weaker or stronger than the shaft in pure torsion? (b) If the key is made of SAE 4130, WQT 1100 F, will it be weaker or stronger? For the purposes here, the weakening of the shaft by the keyway is ignored.

10

SECTION 8 – KEYS AND COUPLINGS Solution: Woodruff key, 3/16 x 1 in. D = 1 3/16 in Shaft: Cold drawn, SAE 1045 (Table AT 8) sy = 85 ksi s AD T= s s 2 Table 10.1, 3/16 x 1 in., Woodruff key is Key no. 608. Shear area = As = 0.178 sq. in. (a) Key material = Shaft Material In yield: For key 3 0.6(85)(0.178)1 ss As D 0.6sy As D 16 = 5.39 in − kips T= = = 2 2 2 For shaft: 3

3 0.6(85)(π )1 3 ssπD 3 0.6 syπD 16 = 16.77 in − kips T= = = 16 16 16 Therefore the key is weaker. (b) Key material = SAE 4130, WQT 1100, Table AT 7, sy = 114 ksi In yield: For key 3 0.6(114)(0.178)1 ss As D 0.6 sy As D 16 = 7.23 in − kips T= = = 2 2 2 For shaft: 3

3 0.6(85)(π )1 3 ssπD 3 0.6 syπD 16 = 16.77 in − kips T= = = 16 16 16 Therefore the key is weaker. 532.

A 2-in. shaft (cold-finished SAE 1137) is connected to a hub by a 3/8-in. radial taper pin made of 4150, OQT 1000 F. (a) What horsepower at 1800 rpm would be transmitted when the pin is about to be sheared off? (b) For this horsepower, what peak torsional stress may be repeated in the shaft? Is the shaft safe from fatigue at this stress?

Solution: D = 2 in d = 3/8 in n = 1800 rpm (a) For pin material , 4150, OQT 1000 F, Table AT9, su = 193.5 ksi πd 2 πd 2 = ss F = ss (2 As ) = ss (2) 4 2 πd 2 D 1 FD = ssπd 2D T= = ss 2 2 2 4 s s = sus = 0.75 su = 0.75(193.5 ) = 145.1 ksi

11

SECTION 8 – KEYS AND COUPLINGS 2

1 3 T = (145.1)(π ) (2 ) = 32.05 in − kips = 32,050 in − lb 4 8 (32,050 )(1800) Tn hp = = = 915.7 hp 63,000 63,000 (b) For shaft, cold-finished, SAE 1137, Table AT 8, su = 103 ksi s 103 sns = 0.6 sn′ = 0.6 u = 0.6 = 30.96 ksi 2 2 But, 16T 16(32,050) ss = 3 = = 20,404 psi = 20.4 ksi < 30.96 ksi πD π (2)3 s 30.96 N = ns = = 1.52 > 1.5, therefore safe from fatigue at this stress. ss 20.4 533.

A 20-in. lever is keyed to a 1 7/8-in. shaft (cold-finished SAE 1141) by a radial taper pin whose mean diameter is 0.5 in.; pin material, C1095, OQT 800 F. The load on the lever is repeatedly reversed; N = 2 on endurance strength. What is the safe lever load (a) for the shaft, (b) for the pin key (shear only), (c) for the combination? Solution: T = FL where F = safe lever load. L = 20 in D = 1 7/8 in = 1.875 in Shaft Material, cold finished, SAE 1141, Table AT 10. sy = 90 ksi sy sn

= 1.8

90 = 50 ksi 1.8 Pin Material, C1095, OQT 800 F, Table AT 9. su = 176 ksi sn = 0.5su = 0.5(176 ) = 88 ksi (a) For the shaft. 0.6sy 0.6(50) ss = = = 15 ksi N 2 s πD 3 15(π )(1.875)3 T= s = = 19.414 in − kips = 19,414 in − lbs 16 16 T = FL 19,414 = F (20 ) F = 970.7 lb (b) For the pin. 0.6 sy 0.6(88) ss = = = 26.4 ksi N 2 sn =

12

SECTION 8 – KEYS AND COUPLINGS 2T d T 4T = = 3 π 2 2 As d (d ) πd 4 s πd 3 26.4(π )(0.5)3 T= s = = 2.592 in − kips = 2592 in − lbs 4 4 T = FL 2592 = F (20 ) F = 129.6 lb (c) For the combination. Use F = 129.6 lb ss =

534.

A lever is keyed to a 2 ½-in. shaft made of SAE 1035, as rolled, by a radial taper pin, made of SAE 1020, as rolled. A load of 200 lb. is applied to the lever 22 in. from the center of the shaft. (a) What size pin should be used for N = 3 based on the yield strength in shear? (b) Let the hub diameter be 5 in. and assume that the part of the pin in the hub is uniformly loaded cantilever beam. Compute the bending stress and comment on the bending strength (especially if the loading varies). Solution: Shaft material, SAE 1035, as rolled, Table AT 7, sy = 55 ksi Pin material, SAE 1020, as rolled, Table AT 7, sy = 48 ksi F = 200 lb, L = 22 in, N = 3 D = 2 ½ in, Dh = 5 in T = FL = (200)(22) = 4400 in-lb = 4.4 in-kips (a) For the pin: 0.6 sy 0.6(48) ss = = = 9.6 ksi N 3 2T d T 4T ss = = = 3 π 2 2 As d (d ) πd 4 4(4.4) 9.6 = πd 3 d = 0.836 in 7 Use d = in = 0.875 in 8 (b) For the bending stress. As cantilever beam let 1 L = (Dh − D ) 2 1 L = (5 − 2.5) = 1.25 in 2 From Table AT 2. wL2 FL M= = 2 2 Where F is the uniform load.

13

SECTION 8 – KEYS AND COUPLINGS

2T 4T 4(4400) = = = 2347 lb Dm Dh + D 5 + 2.5 F = 1174 lb FL (1174)(1.25) M= = = 734 lb 2 2 Bending stress 32M 32(734) sb = 3 = = 11,160 psi = 11.16 ksi πd π (0.875)3 If the loading varies and factor of safety of 3. sn = Nsb = 3(11.16 ) = 33.48 ksi Pin material, SAE 1020, as rolled, Table AT 7, sy = 48 ksi, su = 65 ksu. sn′ = 0.5 su = 0.5(65 ) = 32.5 ksi The bending stress is nearly safe as the load varies. 2F =

535.

A sprocket, transmitting 10 hp at 100 rpm, is attached to a 1 7/16 in. shaft as shown in Fig. 10.15, p. 290., Text; E = 3-1/2 in. What should be the minimum shear pin diameter if the computed stress is 85% of the breaking stress mentioned in the Text? Solution: hp = 10 hp, n = 100 rpm, D = 1 7/16 in = 1.4375 in, E = 3 ½ in = 3.5 in 63,000hp 63,000(10) T= = = 6300 in − lb n 100 T 6300 F= = = 1800 lb E 3.5 From text, Page 290. Breaking stress = 50,000 psi s s = 0.85(50 ,000 ) = 42 ,500 psi 4F ss = 2 πd 4(1800 ) 42,500 = πd 2 d = 0.2322 in use ¼ in. 536.

A gear is attached to a 2-in. shaft somewhat as shown in Fig. 10-15, p. 290, Text; E = 3 5/16 in.; minimum shear-pin diameter = 3/8 in. with a rated torque of 22 in-kips. (a) For this torque, compute the stress in the shear pin. (b) From the ferrous metals given in the Appendix, select those that would give a resisting torque of about 110% of the rated value. Choose one, specifying its heat treatments or other conditions. Solution: D = 2 in E = 3 5/16 in = 3.3125 in d = 3/8 in T = 22 in-kips (a) Stress in shear-pin T 22 F= = = 6.64 lb E 3.3125

14

SECTION 8 – KEYS AND COUPLINGS

4(6.64) = 60.12 ksi π (3 8)2 (b) Select material. sus = 1.1(60.12 ) = 66.13 ksi From appendix, Table AT 7, select Cold drawn, C1020 with sus = 66 ksi ss =

SPLINES 537. A shaft for an automobile transmission has 10 splines with the following dimensions: D = 1.25 in., d = 1.087 in., and L = 1.000 in. (see Table 10.2, p. 287, Text). Determine the safe torque capacity and horsepower at 3600 rpm of this sliding connection. Solution: D = 1.25 in, d = 1.087 in, L = 1.000 in, Nt = 10, n = 3600 rpm T = (1000 )(hL )(rm )(Nt ) in − lb D +d rm = 4 Table 10.2 for 10 splines, sliding connection h = 0.095D Then h = 0.095(1.25 ) = 0.11875 in But D − d 1.25 − 1.087 h= = = 0.0815 in actual 2 2 1.25 + 1.087 rm = = 0.58425 in 4 T = (1000 )(0.0815)(1.000 )(0.58425 )(10 ) = 476.2 in − lb - ans Tn (476.2 )(3600) hp = = = 27.2 hp - ans 63000 63000

538.

The rear axle of an automobile has one end splined. For this fitting there are ten splines, and D = 1.31 in., d = 1.122 in., and L = 1 15/16 in. The minimum shaft diameter is 1 3/16 in. (a) Determine the safe torque capacity of the splined connection, sliding under load. (b) Determine the torque that would have the splines on the point of yielding if the shaft is AISI 8640, OQT 1000 F, if one fourth of the splines are in contact. (c) Determine the torsional stress in the shaft corresponding to each of these torques. Solution: D = 1.31 in, d = 1.122 in, L = 1 15/16 = 1.9375 in, Nt = 10 Dr = 1 3/16 in = 1.1875 in T = (1000 )(hL )(rm )(Nt ) in − lb D +d rm = 4 Table 10.2 for 10 splines, sliding connection h = 0.095D Then h = 0.095(1.31) = 0.12445 in 15

SECTION 8 – KEYS AND COUPLINGS But

D − d 1.31 − 1.122 = = 0.094 in (actual) 2 2 1.31 + 1.122 rm = = 0.608 in 4 (a) Safe Torque T = (1000 )(0.094 )(1.9375)(0.608 )(10 ) = 1107.32 in − lb - ans (b) Torque by splines required on the point of yielding with one fourth of splines in contact (Page 288). From Table AT-7, AISI 8640, OQT 1000 F. sy = 150 ksi, ss = sys = 0.6sy = 0.6(150) = 90 ksi s πDL D (90)(π )(1.31)(1.9375) 1.31 T = s = 2 = 58.76 in − kips 8 8 2 T = 58,760 in − lbs (c) Torsional stress in the shaft From safe torque of 1107.32 in-lb 16T 16(1107.32 ) ss = 3 = = 3368 psi - ans πDr π (1.1875)3 From torque at the point of yield 16T 16(58,760) ss = 3 = = 178,711 psi - ans (too high) πDr π (1.1875)3 h=

539.

An involute splined connection has 10 splines with a pitch Pd of 12/24 (a) Determine the dimension of this connection. (b) Compute the length of spline to have the same torsional strength as the shaft when one fourth the splines carry the load; minimum shaft diameter is 9/16 in. (no sliding). Check for compression. Solution: Nt = 10, Pd = 12, Dr = 9/16 in = 0.5625 in (a) Dimension D N 10 D = t = = 0.8333 in Pd 12 (b) Length of spline (same torsional strength as the shaft when one fourth the splines carry the load (Page 288). D 3 (0.5625)3 L= r = = 0.2136 in D 0.8333 Check for compression. Failure in compression is not likely (Page 289) and can be checked by using the projected contact area. Projected contact area: Ac = 0.2LD = 0.2(0.2136 )(0.8333) = 0.0356 in 2 based on one-fourth of the teeth being under load. COUPLINGS 540. A flange coupling has the following dimensions (Fig. 10.19, p. 291, Text): d = 5, D = 8 5/8, H = 12 ¼, g = 1 ½, h = 1, L = 7 ¼ in.; number of bolts = 6; 1 ¼ x 1 ¼-in. square key. Materials: key, colddrawn AISI 1113; shaft, cold-rolled, AISI 1045; bolts, SAE grade 5 (§5.8). Using the static

16

SECTION 8 – KEYS AND COUPLINGS approach with N = 3.3 on yield strengths, determine the safe horsepower that this connection may transmit at 630 rpm. Solution:

d = 5 in D = 8 5/8 in = 8.625 in H = 12 ¼ in = 12.25 in g = 1 ½ in = 1.5 in h = 1 in L = 7 ¼ in = 7.25 in N = 3.3 nb = 630 rpm Square key = 1 ¼ in x 1 ¼ in Materials: Key: cold-drawn AISI 1113, Table AT 7, sy = 72 ksi, sys = 0.6sy = 0.6(72) = 43.2 ksi Shaft: cold-rolled, AISI 1045, Table AT 8, sy = 85 ksi, sys = 0.6sy = 0.6(85) = 51 ksi Bolt: SAE Grade 5, h = 1 in. sy = 81 ksi, sys = 0.6sy = 0.6(81) = 48.6 ksi No given material for the flange. Bolts in shear: sys 48.6 ss = = = 14.73 ksi N 3. 3 πh 2 F= N b ss 4 FH πh 2 Nb ss H T= = 2 8 2 π (1) (6 )(14.73)(12.25) T= = 425.158 in − kips 8 T = 425,158 in − lbs (425,158)(630) Tn hp = = = 4252 hp 63,000 63,000 Bolts in compression: sy 81 sc = = = 24.55 ksi N 3.3 F = Nb hgs c

17

SECTION 8 – KEYS AND COUPLINGS

FH Nb hgsc H = 2 2 (6)(1)(1.5)(24.55)(12.25) T= = 1353.319 in − kips 2 T = 1,353,319 in − lbs (1,353,319)(630) Tn hp = = = 13,533 hp 63,000 63,000 T=

Key in shear: sys 43.2 ss = = = 13.09 ksi N 3. 3 s bdL (13.09 )(1.25)(5)(7.25) T= s = = 296.570 in − kips 2 2 T = 296,570 in − lbs (296,570)(630) Tn hp = = = 2966 hp 63,000 63,000 Key in compression: sy 72 sc = = = 21.82 ksi N 3.3 s tdL (21.82)(1.25)(5)(7.25) T= c = = 247.180 in − kips 4 4 T = 247 ,180 in − lbs (247,180 )(630) Tn hp = = = 2472 hp 63,000 63,000 Shaft in shear: sys 51 ss = = = 15.45 ksi N 3.3 πd 3 ss π (5)3 (15.45) T= = = 379.200 in − kips 16 16 T = 379,200 in − lbs (379,200)(630) Tn hp = = = 3792 hp 63,000 63,000 The safest horsepower is the lowest which is 2472 hp. 541.

A cast-iron (ASTM 25) jaw clutch with 4 jaws transmits 50 hp at 60 rpm. The inside diameter of the jaws is 3 in. Considering rough handling, choose N = 8 on ultimate strengths. Make reasonable and conservative assumptions and compute (a) the outside diameter of the jaws, (b) the length of jaws h.

18

SECTION 8 – KEYS AND COUPLINGS

Solution: For ASTM 25, suc = 97 ksi, in shear sus = 35 ksi (Table AT 6) 63,000hp 63,000(50) T= = = 52,500 in − lbs n 60 T = 52.5 in − kips (a) The outside diameter of the jaws s 35 ss = us = = 4.375 ksi N 8 Assume Dm as the average diameter, t = thickness = Do – Di , Nj = number of jaws Shear area, 1 πD 1 π D + Di Do − Di As = m (t ) = o 2 Nj 2 N j 2 2 1 π D 2 − Di2 π 2 = As = o Do − Di2 2 4 4 32 2T 4T F= = Dm Do + Di F 4T 32 128T ss = = ⋅ = 2 2 π (Do + Di ) Do2 − Di2 As Do + Di π Do − Di 128(52.5) 4.375 = π (Do + 3) Do2 − 9 By trial and error. Do = 7.466 in or D o = 7.5 in (b) The length of jaws h. s 97 sc = uc = = 12.125 ksi N 8 N j h(Do − Di ) Ac = 2 F 2F 2F F sc = = = = Ac N j h(Do − Di ) 4h(Do − Di ) 2h(Do − Di )

(

(

(

sc =

)

)

(

)

)

4T (Do + Di ) 2T = 2 2h(Do − Di ) h Do − Di2

(

) 19

SECTION 8 – KEYS AND COUPLINGS

2(52.5) h (7.5)2 − (3)2 3 h = 0.1833 in or h = in 16

12.125 =

542.

[

]

The universal joint shown is made of AISI 3150, OQT 1000 F; a = 2 7/16 in., D = 9/16; n = 400 rpm. (a) What torque may be transmitted for shear of the pin (N = 5 on ultimate)? (b) Considering the pin as a simply supported beam of length a with the load distributed from a maximum at the outer (triangular), compute the safe transmitted torque (Same N). (c) In order not to have excessive wear on the pin, the average bearing pressure should not excced 3 ksi. Compute this transmitted torque. (d) What is the safe power?

Solution: For AISI 3150, OQT 1000 F, Table AT 7, su = 151 ksi, sus = 113 ksi N=5 pb = 3 ksi a = 2 7/16 in = 2.4375 in D = 9/16 in = 0.5625 in n = 400 rpm (a) Torque transmitted for shear of the pin. s 113 ss = us = = 22.6 ksi N 5 Each shear area πD 2 (22.6)(π )(0.5625)2 = F = ss = 5.616 kips 4 4 Fa T = 2 = Fa = (5.616)(2.4375) = 13.687 in − kips 2 T = 13,687 in − lbs

(b) Torque transmitted for shear of the pin (simply supported beam) Fa T =M= 3 (5.616)(2.4375) T= = 4.563 in − kips 3 T = 4,563 in − lbs

20

SECTION 8 – KEYS AND COUPLINGS

(c) Torque transmitted for shear of the pin (pb = 3 ksi) F F F pb = = = Ab D(a 2) 2Da F 3= 2(0.5625)(2.4375) F = 8.23 ksi Fa T =M= 3 (8.23)(2.4375) T= = 6.687 in − kips 3 T = 6,687 in − lbs (d) Safe power (4,563)(400) Tn hp = = = 28.97 hp 63,000 63,000 544.

A diagrammatic representation of a universal joint is shown, two yoke parts, the type being similar to Figs. 10.28 and 12.10, Text. The pin extensions have a diameter D = ¾ in.; a = 11/16 in., material of all parts is 4340, OQT 800. Let N = 4 on ultimate stresses; n = 2400 rpm. Compute the safe torque for (a) shear of pins, (b) the pin extensions in bending, assuming that the load distribution is from zero at the outside pin ends to a maximum at the inside yoke surfaces, (c) an average bearing pressure on pins of 4 ksi. (d) What is the corresponding horsepower capacity?

Solution: For AISI 4340, OQT 800 F, Table AT 7, su = 221 ksi, sus = 0.75su = 0.75(221) = 166 ksi N=4 pb = 4 ksi a = 11/16 in = 0.6875 in D = 3/4 in = 0.75 in n = 2400 rpm (a) Torque transmitted for shear of the pin. s 166 ss = us = = 41.5 ksi N 4 Each shear area πD 2 (41.5)(π )(0.75)2 = F = ss = 18.33 kips 4 4 T = M = Fa = (18.33)(0.6875) = 12.602 in − kips T = 12,602 in − lbs 21

SECTION 8 – KEYS AND COUPLINGS

(b) Torque transmitted for shear of the pin (simply supported beam) 2Fa T =M= 3 2(18.33)(0.6875) T= = 8.401 in − kips 3 T = 8,401 in − lbs (c) Torque transmitted for shear of the pin (pb = 4 ksi) F F pb = = Ab Da F 4= (0.75)(0.6875) F = 2.063 ksi 2Fa T =M= 3 2(2.063)(0.6875) T= = 0.9455 in − kips 3 T = 945.5 in − lbs (d) Safe power (945.5)(2400) Tn hp = = = 36.02 hp 63,000 63,000 -

End -

22

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