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Solutions Keys

The document contains a series of engineering problems related to the design and analysis of keyed shafts and keys, focusing on torque, shear stress, and allowable stress values. Each problem provides specific parameters and calculations to determine key dimensions, forces, and torques. The problems involve various key shapes and sizes, as well as different materials and stress limits.

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Fidel John Carlo
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84 views4 pages

Solutions Keys

The document contains a series of engineering problems related to the design and analysis of keyed shafts and keys, focusing on torque, shear stress, and allowable stress values. Each problem provides specific parameters and calculations to determine key dimensions, forces, and torques. The problems involve various key shapes and sizes, as well as different materials and stress limits.

Uploaded by

Fidel John Carlo
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF or read online on Scribd
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C. 279.20 ksi By 42.130 ksi ‘D. 47.120 ksi For keg ShaFt pected “Les” elopmatomnn ain poh ei 3a” Sy = 52000 = 25,000 psi = s sae 23,561.94x2 = 47,120 psi u ‘A keyed sprocket deliver a torque of 778.8 N.m thru the shaft of 84mm OD. The key thickness Is 1.5875 cm and the width is 1.11 em. Compute the length of the same Key. Thepeiissble sess value cf 60 Mpa or shear ‘and 90 Mpa for tension. Ze Fe = Tosti = 28M44N For shearing of key: 4.5875 e em| . 28 aAtem seb T7738 60,000,000 = 288444 ‘OortKLy L = 0.0433 m = 4,33¢m For compression of key: oF i Sear 34 mm 2(28,844.4) 0.015875(L) L = 0.040377 m = 4.0377 cm ‘Therefore choose the longer length of key. L = 4.33¢m 90,000,000 = (Oct. 1998) Same as a4 ting of SAE 1040 grade, cold rolled, C. 47.12 ksi D. 47.21 ksi A. 47.52 ksi B. 47.25 ksi Sen Sy _ 50000 _ osc psi 3/4” ot ‘Scanned with @camscanner T=Fr 192,595.94 = F(9/2) F = 88,357.29 Ibs For shearing of key: s, = F. 2 8885729 wl ~ ”0.75(5) © S)= Sx FS Sy = 23,561.94 x2 = 47,120 psi = 47.20ksi = 23,561.94 psi PROBLEM 4 (ME Bd. Apr. 98) ==" A keyed gears delivers a torque of 912.4 Nem thru its shaft of 63.5 mm outside diameter. Ifthe key has thickness of 15.875 mm and width of 11.1125, find the length of the key. Assume the permissible stress values of 61.2 Mpa for shear and tension at 99.8 Mpa, A. 47.42 mm ©. 42.22 mm B. 39.72 mm D. 46.92 mm Te Fxr 912.4 = F (0.0695/2) i F = 26,737 N 415.875 mi 1W11em Based on shearing of key: 7912.4 Nm &= wo 28,737 0.0111125L L = 0.04225 m = 42.25 mm 661,200,000 = f key: aneconjcompresavn othe asa s- = ’ aL 228737) 0.015875L 0.03628m = 36.28 mm 199,800,000 = L ‘Therefore use the longer length to be safe: L = 42.25mm (PROBLEM 5 ‘A metric M4 x 15 square key is used with a 16 mm shaft. Ifthe allowable shearine 8 is 60 Mpa. How much torque can the assembly handle? The rotational 9p. str of the shaft is 600 rpm. A. 20N-m C.24N-m B. 12N-m D.34Nm An M4 x 15 square key, means w= 4 mm, h=4mmandL= 15mm For shearing of key: F Sac ‘Ss = 50 Mpa or 50. Nimm? so= (4)(15) F =3000N T =F (d2) = 3000(0.016/2) = 24 Nem » PROBLEM 6 A flat key is to be designed for 75 mm diameter shaft which will transmit 150 llowable shearing stress is 200 Mpa and key width is 15 mm, deter 400 rpm. tha length of key. ‘A. 3065 mm C. 33.75 mm B. 31.83 mm D. 32.85 mm P=2nTN 150 = 2x T (400/60) T = 3.581 KN-m T=Fxr 3.581 = Fx (0.075/2) F = 95.493 KN For shearing of key: 5 ie Siw 95.493 (0.015) L = 0,03183m = 31.83mm 200,000 = 75 mim ‘4mm 16mm 15mm ‘Scanned with @ camscanner A rectengir key 1s used In a.puiley to transmit 400 KW at 1000 rpm on S0:mm shal iar A, 33.24 KN we Determine the force required to remove the shi B. 36.85 KN Ear C. 28.35 KN D. 30.55 KN P=2nTN 100 = 2x7 (1000/60) 0.9549 KN-m Fxr 0.9549 = Fx (0.050/2) F = 38.197 KN 0 t i gun ‘The friction on key experience both upper half on one side and lower half on the other side. = 2(0.4)(38.197) = 30.55 KN PROBLEM 8 j ‘A 7/6 in height x 3 in length flat key is keyed to a 2 inches diameter shaft. Determine the torque in the key if bearing stress allowabl En 4 w= 7/16 = 0.4375 in m6" oa F © RL 25,000 = ——F_ ; 3[0.437572] F = 16,406.25 Ibs r T= Fxr Zin T = 16,406.25 x (2/2) = 16,406.25 in-lbs PROBLEM 9 ‘A rectangular key is used in a pulley connected to a line shalt at 15 KW It shearing of key Is 230 Mpa, determine the force acting on key lengt ‘one-fourth of shaft diameter and key length is 1 inch. A. 43 KN B. 48 KN For: ©. 48 KN 1D. 48 KN. 1 shaft: D°N Ree 53.5 18/0.748 = 7AM D = 1.2148 in = 0.08085 m L = tin = 0.025398. m w= D/4 = 0,03085/4 = 0.007714 m (007714)(0.025398) F = 45,061 KN (PROBLEM 10 . ‘A square Key is to be used in a 40 mm diameter shaft and that m torque. If bearing stress of the key is 400 Mpa, deter dimension of square key to be used if key length is 30 mm. developed a 2 the cross secti ‘A. 324.80 mm C. 446.80 mm? B. 246.80 mm? D. 27.77 mm** nny Ts Fxr 2 = Fx (0.042) ‘30 mm F = 100KN w Ww fa: 2 KN-my wae 400,000 = —100___ (0.030/2)(h) h = 0.01666 m = 16.6667 mm w= h = 16.667 mm A = 16.667 x 16.667 = 277.77 mm? 40 mm ‘Scanned with @ camscanner * PROBLEM 11 7" ‘A 100 KN force is acting on a key that has a length of 4 times its height. if bearing stress of key is 400 Mpa, determine the height of key. A. 1054mm ©. 1265mm For the same shat and key material 8.11.18 mm D. 15.25 mm w=t/40 w = 1/4 (0.12) any w= 0.03m = h 4 Sia a Sint or aL wae 4 410,000 = —F__ Ww (o4y(003) Leah F = 1722KN F 100 KN Fxr=P 400,000 = —100 3 17220012)

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