Key and Coupling Solutions

1 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

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kips in D s T= ys = =8765 − 16 2 8 55 16 3 3 . . π π

(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

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in bD s T L s 43 4 2 2 1 6 39 65 87 2 2 . . . = = =

By compression, sc = sy. Key has lowest yield strength.

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in tD s T L c 32 5 2 2 1 66 65 87 4 4 . . = = = Use L = 5.32 in

For Flat key, b = ½ in , t = 3/8 in. By shear, ss = sys

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in bD s T L s 43 4 2 2 1 6 39 65 87 2 2 . . . = = =

By compression, sc = sy. Key has lowest yield strength.

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in tD s T L c 09 7 2 8 3 66 65 87 4 4 . . = = = 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

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in bD s T L s 12 4 2 2 1 6 42 65 87 2 2 . . . = = =

2

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in tD s T L c 94 4 2 2 1 71 65 87 4 4 . . = = = Use L = 4.94 in

For Flat key, b = ½ in , t = 3/8 in. By shear, ss = sys

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in bD s T L s 12 4 2 2 1 6 42 65 87 2 2 . . . = = =

By compression, sc = sy. Key has lowest yield strength.

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in tD s T L c 59 6 2 8 3 71 65 87 4 4 . . = = = 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

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kips in lb in n hp T= = =2358 − =2358 − 1750 5 65 000 63 000 63 . . , ,

Table AT 19, use b = 3/8 in, t = 1/4 in for D = 1 ¾ in Assume smooth load, N = 1.5

For Flat key, b = 3/8 in, t = 1/4 in. By shear, sys = 0.6sy = 0.6(66) = 39.6 ksi ksi N s s_s ys 264 5 1 6 39 . . . = = =

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in bD s T L s 272 0 75 1 8 3 4 26 358 2 2 2 . . . . = = =

By compression, use sy of shaft the lowest. Pulley has the highest compressive strength (Cast iron).

ksi N s s_c y 393 5 1 59 . . = = =

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in tD s T L c 549 0 75 1 4 1 3 39 358 2 4 4 . . . . = = = Use L = 0.549 in - answer

3 For Square key, b = 3/8 in, t = 3/8 in.

By shear, ss = sys sys = 0.6sy = 0.6(66) = 39.6 ksi ksi N s s_s ys 264 5 1 6 39 . . . = = =

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in bD s T L s 272 0 75 1 8 3 4 26 358 2 2 2 . . . . = = =

By compression, use sy of shaft the lowest. Pulley has the highest compressive strength (Cast iron).

ksi N s s_c y 393 5 1 59 . . = = =

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in tD s T L c 366 0 75 1 8 3 3 39 358 2 4 4 . . . . = = = 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

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kips in lb in n hp T= = =53550 − =5355 − 100 85 000 63 000 63 . , , ,

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 ksi N s s_s ys 192 25 2 2 43 . . . = = =

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in bD s T L s 29 2 25 3 4 3 2 19 55 53 2 2 . . . . = = =

By compression, use sy of sprocket the lowest.

ksi N s s_c y 244 25 2 55 . . = = =

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in tD s T L c 60 3 25 3 4 3 4 24 55 53 4 4 . . . . = = =

4 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 ksi N s s_s ys 192 25 2 2 43 . . . = = =

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in bD s T L s 29 2 25 3 4 3 2 19 55 53 2 2 . . . . = = =

By compression, use sy of sprocket the lowest.

ksi N s s_c y 244 25 2 55 . . = = =

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in tD s T L c 40 5 25 3 2 1 4 24 55 53 4 4 . . . . = = =

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

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kips in lb in n hp T= = =22500 − =225 − 210 75 000 63 000 63 . , , ,

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 ksi N s s_s ys 99 4 6 39 . . = = =

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in bD s T L s 02 2 3 4 3 9 9 5 22 2 2 . . . = = =

By compression, use sy of cast-steel gear the lowest.

ksi N s s_c y 875 4 35 . = = =

5

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in tD s T L c 57 4 3 4 3 75 8 5 22 4 4 . . . = = =

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 ksi N s s_s ys 99 4 6 39 . . = = =

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in bD s T L s 02 2 3 4 3 9 9 5 22 2 2 . . . = = =

By compression, use sy of cast-steel gear the lowest.

ksi N s s_c y 875 4 35 . = = =

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in tD s T L c 86 6 3 2 1 75 8 5 22 4 4 . . . = = =

Use L = 6.86 in. - answer

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

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kips in lb in n hp T= = =2716 − =2716 − 1160 50 000 63 000 63 . , , ,

For flat key: b = 3/8 in, t = ¼ in Check for compression:

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ksi LtD T s_c 1337 625 1 4 1 2 716 2 4 4 . . . = = =

Based on pulley material, 299 225 37 13 40 . . . = > = = c y s s N

Therefore safe in compression. Determine the yield stress on the key

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ksi LbD T s_s 4457 625 1 8 3 2 716 2 2 2 . . . = = =

6 25 2_. = N s y ys s Ns s =0_.6 =

(

225

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4457

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6 0_{. = . .} y s ksi s_y=16_.7

Min. s_y=

(

13_.37

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2_.25

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₌30ksi _{- 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:

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ksi LtD T s_c 8914 625 1 8 3 2 716 2 4 4 . . . = = =

Based on pulley material, 449 225 914 8 40 . . . = > = = c y s s N

Therefore safe in compression. Determine the yield stress on the key

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ksi LbD T s_s 4457 625 1 8 3 2 716 2 2 2 . . . = = = 25 2_. = N s y ys s Ns s =0_.6 =

(

225

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4457

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6 0_{. = . .} y s ksi s_y=16_.7

Min. s_y=

(

8_.914

)(

2_.25

)

₌20ksi _{- 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

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kips in lb in n hp T= = =12600 − =126 − 200 40 000 63 000 63 . , , , Table AT 19, D = 2 in b = ½ in, t = 3/8 in

7 a) Square key, b = ½ in, t = ½ in

By shear:

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ksi LbD T s_s 84 2 2 1 3 6 12 2 2 . . = = =

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50 4 4 8 63 6 0 6 0 . . . . = = = = s y s ys s s s s N < 6.21 By compression:

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ksi LtD T s_c 168 2 2 1 3 6 12 4 4 . . = = =

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38 2 8 16 40 . . = = = c y s pulley s N < 6.21 Answer N = 2.38

b) Flat key, b = ½ in, t = 3/8 in By shear:

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ksi LbD T s_s 84 2 2 1 3 6 12 2 2 . . = = =

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50 4 4 8 63 6 0 6 0 . . . . = = = = s y s ys s s s s N < 6.21 By compression:

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ksi LtD T s_c 224 2 8 3 3 6 12 4 4 . . = = =

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78 1 4 22 40 . . = = = c y s pulley s N < 6.21 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)

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in kips N D s D s T₌ s ₌ y _{= =10431 −} 16 5 1 5 2 85 6 0 16 6 0 16 3 3 3 . . . . . π π π Key: By shear: 2 LbD s T= s

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ksi N s N s s_s ys y 252 5 1 63 6 0 6 0 . . . . = = = =

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kips in T= =689 − 2 5 2 8 5 5 3 2 25 . . . . < 104.31 in-kips By compression: 4 LtD s T= c

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ksi N pulley s s_c y 2667 5 1 40 . . = = =

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kips in T= =3646 − 4 5 2 8 5 5 3 67 26 . . . . < 104.31 in-kips Use T=36_.46in−kips=36_,460in−kips

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hp Tn hp 243 000 63 420 460 36 000 63 = = = , , ,

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:

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in kips N D s D s T= s = y = =18025 − 16 5 1 3 85 6 0 16 6 0 16 3 3 3 . . . . π π π Key: By shear: 2 LbD s T₌ s

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ksi N s N s s_s ys y 252 5 1 63 6 0 6 0 . . . . = = = =

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kips in T= =14175 − 2 3 4 3 5 2 25 . . < 180.25 in-kips By compression: 4 LtD s T= c

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ksi N pulley s s_c y 2667 5 1 40 . . = = =

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kips in T= =75 − 4 3 4 3 5 67 26. < 180.25 in-kips

Use T=75in−kips=75_,000in−kips

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hp Tn hp 500 000 63 420 000 75 000 63 = = = , , , 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: a) 2 ND T = D T N=2 Axial force = D fT fN F= =2 b) T= ND=ND 2 2 D T N = Axial force = D fT fN F= =

10

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

s sA s D T F=2 = 2 D A s T= s s

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

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ksi N s N s s_s ys y 225 2 75 6 0 6 0 . . . = = = = < 24.06 ksi

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lb in kips in T = − = −       = 562 5620 2 16 11 1 296 0 5 22 . . .

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hp Tn hp 1784 000 63 200 5620 000 63, = , = . =

(b) Key, SAE 2317, OQT 1000F, Table AT 7, sy = 79 ksi

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ksi N s N s s_s ys y 237 2 79 6 0 6 0 . . . = = = = < 24.06 ksi

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lb in kips in T = − = −       = 592 5920 2 16 11 1 296 0 7 23 . . .

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hp Tn hp 1879 000 63 200 5920 000 63, = , = . =

(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.

11 Solution:

Woodruff key, 3/16 x 1 in.

D = 1 3/16 in

Shaft: Cold drawn, SAE 1045 (Table AT 8) sy = 85 ksi

2

D A s

T= s s

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

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kips in D A s D A s T s s y s = −       = = = 539 2 16 3 1 178 0 85 6 0 2 6 0 2 . . . . For shaft:

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kips in D s D s T s y _{= −}       = = = 1677 16 16 3 1 85 6 0 16 6 0 16 3 3 3 . . . π π π

Therefore the key is weaker.

(b) Key material = SAE 4130, WQT 1100, Table AT 7, sy = 114 ksi

In yield: For key

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kips in D A s D A s T s s y s _{= −}       = = = 723 2 16 3 1 178 0 114 6 0 2 6 0 2 . . . . For shaft:

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kips in D s D s T s y = −       = = = 1677 16 16 3 1 85 6 0 16 6 0 16 3 3 3 . . . π π π

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

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_      =       = = 2 4 2 2 2 2 d s d s A s F _{s s s} π _s π D d s D d s FD T _{s s} 2 2 4 1 2 2 2 π π =             = =

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ksi s s s_s= _us =0.75 _u =0.75193.5 =145.1

12

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in kips in lb T  = − = −      = 2 3205 32050 8 3 1 145 4 1 2 , . . π

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hp Tn hp 9157 000 63 1800 050 32 000 63 _, . , , = = =

(b) For shaft, cold-finished, SAE 1137, Table AT 8, su = 103 ksi

ksi s s s_{ns n} u 3096 2 103 6 0 2 6 0 6 0. . . = .      =       = = _′ But,

(

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psi ksi ksi

D T s_s 20404 204 3096 2 050 32 16 16 3 3 , . . , < = = = = π π 52 1 4 20 96 30 . . . = = = s ns s s

N > 1.5, therefore safe from fatigue at this stress.

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.

ksi s_y=90 8 1_. = n y s s ksi s_n 50 8 1 90 = = .

Pin Material, C1095, OQT 800 F, Table AT 9.

ksi s_u=176

(

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ksi

s

s_{n =}0_.5 _u=0_.5176 ₌88 (a) For the shaft.

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ksi N s s_s y 15 2 50 6 0 6 0 = = = . .

( )(

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lbs in kips in D s T₌ s _{= =19414 − =19414 −} 16 875 1 15 16 3 3 , . . π π FL T =

( )

20 414 19, _=F lb F=970.7 (b) For the pin.

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ksi N s s_s y 264 2 88 6 0 6 0 . . . = = =

13

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3 2 4 4 2 2 d T d d T A d T s s s π π = = =

( )(

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lbs in kips in d s T= s = =2592 − =2592 − 4 5 0 4 26 4 3 3 . . . π π FL T =

( )

20 2592_=F lb F=129_.6

(c) For the combination. Use F₌129_.6lb

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:

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ksi N s s_s y 96 3 48 6 0 6 0 . . . = = =

( )

3 2 4 4 2 2 d T d d T A d T s s s π π = = =

(

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3 4 4 4 6 9 d π . . = in d=0.836 Use d in 0875in 8 7 . = =

(b) For the bending stress. As cantilever beam let

(

D D

)

L= _h− 2 1

(

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in L 5 25 125 2 1 . . = − = From Table AT 2. 2 2 2 FL wL M= =

14

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lb D D T D T F h m 2347 5 2 5 4400 4 4 2 2 ₌ + = + = = . lb F=1174

(

)(

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lb FL M 734 2 25 1 1174 2 = = = . Bending stress

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(

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psi ksi d M s_b 11160 1116 875 0 734 32 32 3 3 , . . = = = = π π

If the loading varies and factor of safety of 3.

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ksi

Ns

s_{n = b=}311_.16 ₌33_.48

Pin material, SAE 1020, as rolled, Table AT 7, sy = 48 ksi, su = 65 ksu.

( )

ksi

s

s_n′=0.5 _u =0.565 =32.5

The bending stress is nearly safe as the load varies.

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

( )

lb in n hp T= = =6300 − 100 10 000 63 000 63_{, ,} lb E T F 1800 5 3 6300 = = = .

From text, Page 290. Breaking stress = 50,000 psi

(

)

psi s_s=0_.8550_,000 ₌42_,500 2 4 d F s_s π =

(

)

2 1800 4 500 42 d π = , in d=0.2322 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

lb E T F 664 3125 3 22 . . = = =

15

(

)

(

)

ksi s_s 6012 8 3 64 6 4 2 . . = = π

(b) Select material. s_{us =}1_.1

(

60_.12

)

₌66_.13ksi

From appendix, Table AT 7, select Cold drawn, C1020 with sus = 66 ksi

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

(

)( )( )

hL r

( )

N in lb T= 1000 _{m t} − 4 d D r_m = +

Table 10.2 for 10 splines, sliding connection

D h=0.095 Then

(

)

in h₌0_.0951_.25 ₌0_.11875 But in d D h 00815 2 087 1 25 1 2 . . . = − = − = actual in r_m 058425 4 087 1 25 1 . . . = + =

(

)(

)(

)(

)( )

in lb T= 1000 0.0815 1.000 0.58425 10 =476.2 − - ans

(

)(

)

hp Tn hp 272 63000 3600 2 476 63000 . . = = = - ans

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

(

)( )

hL

( )( )

r N in lb T= 1000 m t − 4 d D r_m = +

Table 10.2 for 10 splines, sliding connection

D h=0_.095 Then

(

)

in

16 But in d D h 0094 2 122 1 31 1 2 . . . = − = − = (actual) in r_m 0608 4 122 1 31 1 . . . = + =

(a) Safe Torque

(

)(

)(

)(

)( )

in lb

T= 1000 0_.094 1_.9375 0_.608 10 =1107_.32 − _{- 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

( )( )(

)(

)

kips in D DL s T s _{= −}          =             = 5876 2 31 1 8 9375 1 31 1 90 2 8 . . . . π π lbs in T=58,760 −

(c) Torsional stress in the shaft From safe torque of 1107.32 in-lb

(

)

(

)

psi D T s r s 3368 1875 1 32 1107 16 16 3 3 = = = . . π π - ans

From torque at the point of yield

(

)

(

)

psi D T s r s 178711 1875 1 760 58 16 16 3 3 , . , = = = π

π - ans (too high)

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 in P N D d t ₀₈₃₃₃ 12 10 . = = =

(b) Length of spline (same torsional strength as the shaft when one fourth the splines carry the load (Page 288).

(

)

in D D L r 02136 8333 0 5625 0 3 3 . . . = = =

Check for compression.

Failure in compression is not likely (Page 289) and can be checked by using the projected contact area.

Projected contact area:

(

)(

)

2 0356 0 8333 0 2136 0 2 0 2 0 LD in

A_c = . = . . . = . 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, cold-drawn AISI 1113; shaft, cold-rolled, AISI 1045; bolts, SAE grade 5 (§5.8). Using the static

17

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: ksi N s s_s ys 1473 3 3 6 48 . . . = = = s bs N h F 4 2 π = 8 2 2 H s N h FH T_{= =}π b s

( ) ( )(

)(

)

kips in T= =425158 − 8 25 12 73 14 6 12 . . . π lbs in T=425,158 −

(

)(

)

hp Tn hp 4252 000 63 630 158 425 000 63 = = = , , , Bolts in compression: ksi N s s_c y 2455 3 3 81 . . = = = c bhgs N F =

18 2 2 H hgs N FH T= = b c

( )( )( )(

)(

)

kips in T= =1353319 − 2 25 12 55 24 5 1 1 6 . . . . lbs in T=1_,353_,319 −

(

)(

)

hp Tn hp 13533 000 63 630 319 353 1 000 63 _, , , , , = = = Key in shear: ksi N s s_s ys 1309 3 3 2 43 . . . = = =

(

)(

)( )(

)

kips in bdL s T= s = =296570 − 2 25 7 5 25 1 09 13 2 . . . . lbs in T=296,570 −

(

)(

)

hp Tn hp 2966 000 63 630 570 296 000 63 = = = , , , Key in compression: ksi N s s_c y 2182 3 3 72 . . = = =

(

)(

)( )(

)

kips in tdL s T= c = =247180 − 4 25 7 5 25 1 82 21 4 . . . . lbs in T=247_,180 −

(

)(

)

hp Tn hp 2472 000 63 630 180 247 000 63 = = = , , , Shaft in shear: ksi N s s_s ys 1545 3 3 51 . . = = =

( ) (

)

kips in s d T₌ s _{= =379200 −} 16 45 15 5 16 3 3 . . π π lbs in T=379_,200 −

(

)(

)

hp Tn hp 3792 000 63 630 200 379 000 63 = = = , , ,

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.

19 Solution:

For ASTM 25, suc = 97 ksi, in shear sus = 35 ksi (Table AT 6)

( )

lbs in n hp T= = =52500 − 60 50 000 63 000 63 , , , kips in T=52_.5 −

(a) The outside diameter of the jaws

ksi N s s_s us 4375 8 35 . = = =

Assume Dm as the average diameter, t = thickness = Do – Di , Nj = number of jaws

Shear area,

( )

      −       +         =         = 2 2 2 1 2 1 _{o i o i} j j m s D D D D N t N D A π π

(

2 2

)

2 2 32 4 4 2 1 i o i o s D D D D A _= −       −       = π π i o m D D T D T F + = = 2 4

(

2 2

)

(

)

(

2 2

)

128 32 4 i o i o i o i o s s D D D D T D D D D T A F s − + = − ⋅ + = = π π

(

)

(

3

)

(

9

)

5 52 128 375 4 2 − + = o o D D π . .

By trial and error.

in

D_o =7.466 or D_o=7.5in

(b) The length of jaws h.

ksi N s s uc c 12125 8 97 . = = =

(

)

2 i o j c D D h N A = −

(

o i

)

(

o i

)

(

o i

)

j c c D D h F D D h F D D h N F A F s − = − = − = = 2 4 2 2

(

)

(

)

(

2 2

)

2 2 4 i o i o i o c D D h T D D h D D T s − = − + =

20

(

)

(

)

( )

[

2 2

]

3 5 7 5 52 2 125 12 − = . . . h in h=0_.1833 or h in 16 3 =

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.

ksi N s s us s 226 5 113 . = = =

Each shear area

(

)( )(

)

kips D s F _s 5616 4 5625 0 6 22 4 2 2 . . . = =       = π π

(

)(

)

in kips Fa Fa T= = = 5616 24375 =13687 − 2 2 _{. . .} lbs in T=13,687 −

(b) Torque transmitted for shear of the pin (simply supported beam) 3 Fa M T= =

(

)(

)

kips in T= =4563 − 3 4375 2 616 5 . . . lbs in T=4,563 −

21 (c) Torque transmitted for shear of the pin (pb = 3 ksi)

(

)

Da F a D F A F p b b 2 2 = = =

(

05625

)(

24375

)

2 3 . . F = ksi F=8.23 3 Fa M T= =

(

)(

)

kips in T= =6687 − 3 4375 2 23 8 . . . lbs in T=6,687 − (d) Safe power

(

)(

)

hp Tn hp 2897 000 63 400 563 4 000 63 _, . , , = = =

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.

ksi N s s_s us 415 4 166 . = = =

Each shear area

(

)( )(

)

kips D s F _s 1833 4 75 0 5 41 4 2 2 . . . = =       = π π

(

)(

)

in kips Fa M T= = = 18_.33 0_.6875 =12_.602 − lbs in T=12_,602 −

22

(b) Torque transmitted for shear of the pin (simply supported beam) 3 2Fa M T= =

(

)(

)

kips in T= =8401 − 3 6875 0 33 18 2 . . . lbs in T= 4018_, −

(c) Torque transmitted for shear of the pin (pb = 4 ksi)

Da F A F p b b = =

(

075

)(

06875

)

4 . . F = ksi F=2_.063 3 2Fa M T= =

(

)(

)

kips in T= =09455 − 3 6875 0 063 2 2 . . . lbs in T=945_.5 − (d) Safe power

(

)(

)

hp Tn hp 3602 000 63 2400 5 945 000 63 _, . . , = = = - End -