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Mechanical Shafts

Shafts Practice Problems

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Mechanical Engineering (BSME 2020)

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Eastern Visayas State University

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D-

SHAFTS

Definitions:

Shaft – a rotating member transmitting power

Axle – a stationary member carrying rotating wheels, pulleys, etc

Spindle – a short shaft or axle on machines

Machine Shaft – shaft which is used to transmit power between the source and the machine absorbing the power

Lineshaft or Mainshaft – transmission shaft driven by the prime mover

Countershaft, jackshaft, headshaft, shortshaft – transmission shaft intermediate between the lineshaft and the driven shaft

Shafting design diagram

Commercial Sizes of shaft, inches (Faires p. 269; Vallance p. 181)

####### ,

2,

####### ,

Materials for Transmission shaft: cold rolled, hot rolled, forged carbon steel

Relation of Power, Torque and Speed

P = 2πTN

Where: - D-

Commercial Sizes of shaft, inches (Faires p. 269; Vallance p. 181)

2, - - - , - - - , - - - ,3, - - - , - - - ,4, - - - , - -

,5, -

Ssmax = maximum shear stress St max = maximum tensile or compressive stress M =bending moment T = torsional moment

####### D-

Strength of Shaft with assumed allowable stresses (PSME Code p)

For Main Power transmitting shafts:

4 =

(. 5 80

# ( = 7

804 5

For Lineshafts carrying pulleys:

4 =

(. 5 53.

# ( = 7

For Small, Short shafts:

4 =

(. 5 38

# ( = 7

384 5

Where: P = power transmitted in HP D = diameter of shaft in inches \N = speed in RPM

Shafting formulas from machinery’s handbook Diameter of shaft: A. for allowable twist not exceeding 0 deg per ft length

( = 0:√ # ( = 4 7

; 5

Where: D = shaft diameter, inches T = torque, lb-in HP = horsepower N = speed, rpm In S units (allowable twist 0 deg per meter length)

( = 2√ : # ( = 125 7

4 5

Where: D = shaft diameter, mm T = torque, N-mm HP = Power kW N = speed, rpm

B. for allowable twist not exceeding 1 deg per20D length

( = 0√ : # ( = 4 7

; 5

D- Where: D = shaft diameter, inches T = torque, lb-in HP = horsepower N = speed, rpm

C. For short, solid shaft subjected only to heavy transverse shear

( = 7

Where: V = maximum transverse shearing loads, lbs Ss = maximum torsional shearing stress, lb/sq. in

Linear Deflection of shafting

For steel lineshafting, it is considered good practice to limit the linear deflection to a maximum of 0 inch per foot length

Maximum Distance:

A. For shafting subjected to no bending action except its own weight:

= 8 1 ( 2

B. For shafting subjected to bending action pulleys, etc.

= 5 1 ( 2 !

Where: L = maximum distance between bearings, ft D = diameter of shaft, inches

Note:

Pulleys should be placed as close to the bearings as possible
In general, shafting up to three inches in diameter is almost always made from cold-rolled steel

D-

= =

) 32

>(* "# #$ %$ ℎ"'

G = 12 000 000 psi for steel

) 180

> =

4201(12)

) 32 >(

*(12,000,000)

D = 2 in

Compute the nominal shear stress at the surface in MPa for a 40 mm diameter shaft that transmit 750 kw at 1500 rpm. Axial and bending loads are assumed negligible. Torsional shearing stress is 28 MPa A. 218 C. 232 B. 312 D. 380

Solution: P = 2πTN 750 = 2πT(1500/60) T = 4 kN-m

16 )(.

16 )(0). Ss = 379 982 kPa = 380 MPa

A hollow shaft has an inner diameter of 0 m and an outer diameter of 0 m. Compute for the torque if the shear stress is not to exceed 120 MPa in N-m (ME Bd. Oct 95) A. 4500 C. 4300 B. 4100 D. 4150

Solution:

= ? ( ?:@ A:)

120,000,000 =

16(0) )(0−0)

B = CDEE F− G

Design the size of a solid steel shaft to be used for a 500 hp, 250 rpm application if the allowable torsional deflection is 1° and the allowable stress is 10,000 psi and modulus of rigidity is 13 x 10^6 psi (ME Bd. Oct 95) A. 5 in dia C. 4 7/8 in dia B. 4 5/8 in dia D. 4 ¾ in dia

Solution:

Solving for the shaft diameter D based on stress: P = 2πTN 500(33,000) = 2πT(250) T = 10,504 ft-lb = 126,051 in-lb D-

16 )(.

10,000 =

16(126. )(. D = 4 in

Solving for the shaft diameter D based on torsional deflection:

The shaft length is not given. The common practice for torsional deflection is 1° per 20D

Θ = 1° x (180/π) rad T = 126,051 in-lbs L = 20D J = (π/32)D^ G = 13,000,000 psi

= H

#######

1( KL) =

2,LM2L =!ON>( :)(.,LLL,LLL)

D = 4 in Therefore: use D = 4 7/8 in

A 2 in solid shaft is driven by a 36 in gear and transmits power at 120 rpm. If allowable shearing stress is 12 ksi, what horsepower can be transmitted? (ME Bd. Oct 95) A. 29 C. 39. B. 35 D. 34.

Solution:

= ! 12,000 =

#######

T = 18,850 in-lbs = 1,570 ft-lbs P = 2πTN = 2π(1570)(120/60)(hp/550) = 35 HP

A short 61 mm shaft transmit 120 HP. Compute the linear speed of a pulley 55 cm mounted on the shaft. (ME Bd. Oct 95) A. 1796 fpm C. 1856 fpm B. 1766 fpm D. 2106 fpm

Solution: For short shaft (PSME Code)

Solution: P = 2πTN 7 = 2πT(1200/60) T = 0 kN-m = 59,365 N-mm

! D-

30 =

16(59,365) )(.

( = SR GG

Determine the torque received by the motor shaft running at 4250 rpm, transmiting 11 hp, through a 10 in diameter, 20° incolute gear. The shaft is supported by ball bearings at both ends and the gear is fixed at the middle of 8 in shaft length (ME Bd Apr 2006) A. 163 in-lb C in-lb B. 167 in-lb D. 138 in-lb

Solution:

P = 2πTN 11(33,000) = 2πT(4250) T = 13 ft-lb = 163 in-lb

Compute for the diameter in inches of a conveyor head pulley SAE 4130 solid steel shaft being driven by a 11HP drive motor through a gear reduces with 180 rpm output, the torsional deflection is 0 degree/foot of shaft length. (ME Bd, Apr 2006)

Solution: P = 2πTN 11(33,000) = 2πT(180) T = 321 ft-lbs = 3852 in-lbs

= H

#######

0( KL) =

.KM2(2) =!ON>( :)(,MLL,LLL)

D = 2 in, say 2 3/8 inches pulley

The power that can be transmitted by a spindle of 55 mm diameter running at 1200 rpm and allowable stress of 5 MPa is equal to : (ME Bd Apr 2006) A. 10 kw C. 23 kw B. 20 kw D. 35 kw

Solution:

= !

5000 =

16 )(0). T = 0 KN-m

D- P = 2πTN = 2π(0)(1200/60) = 20 kw

Determine the diameter in inches of a steel countershaft that delivers 13 HP at a speed of 15 rad/sec, the allowable material shear stress is 8 ksi. (ME Bd. Oct 2005) A. 1 in C. 2 in B. 1 ½ in D. 1 ¼ in

Solution: N = 15 rad/sec = 149 rpm

;4 =

(. 5 38

"# #%'Zℎ"'

13 =

(.(149) 38

[ = R]

Compute for the torsional deflection in degrees of 110 mm diameter, 1 meter long shaft subjected to a twist moment 3 x 10^6 N-mm. the torsional modulus of elasticity is 83 000 MPa. (ME Bd. Apr 2005) A. 0 C. 0. B. 0 D. 0.

Solution:

(30105 −///)(1400//)

= ) 32 > ( 110 )//(83,000 5 //

2 )

180 Z^ )

= E _`a

Two parallel shaft connected by pure rolling turn in the same direction and having a speed ratio of 2. what is the distance of the two shaft if the smaller cylinder is 22 cm in diameter? (ME Bd. Apr 2005) A. 16 cm C. 25 cm B. 30 cm D. 19 cm

Solution:

D1N1 = D2N 2 = (N1/N2) = (D2/D1) 2 = (D2/D1) D2 = 60 cm

Cylinder 2 is an internal cylinder, therefore, the center distance will be:

b =

(2 − ( 2

60 − 22 2

= RW cG

B. What is the total shear force in the pin? A. 371 lbf C. 117 lbf B. 711 lbf D. 311 lbf

Solution: T = F x 1/2 F = 711. 11 lbf

C. What pin diameter is needed? A. 0 in C. 0 in B. 0 in D. 0 in

Solution: Ss = F/ (π/4 x d^2) 17,056 = 711. 11/ (π/4 x d^2) d = 0 in

D. If the pin has a tensile yield strength of 50 MPa, manufactured from steel, what will be the yield strength in shear? A. 71 ksi C. 60 ksi B. 117 x 10^6 Pa D. 294 x 10^6 Pa

Solution: Sys = 0 Syt = 0 x (510 x 10^6) = 294 x 10^ Ss = Sys/FS = 294 x 10^6/2 = 117 x 10^6 Pa

E. Assume that the shaft torque becomes 45 N-m, compute for the total shear t the pin. A. 9641 N C. 3149 N B. 117 lbf D. 571 lbf

Solution: D = 1. 125 in x 2 cm/in x m/100 cm = 0 m T = F x D/ 45 = F x 0/2 F = 3149 N

A marine turbine developing 15,000 hp and it turns the shaft at 300 rpm. The propeller which is attached the shaft develops a thrust of 150,000 lbs. A hollow steel shaft with an outside diameter of 14 in will be used. (ME Bd. Apr 2004) A. Compute for the torque. A. 622,000 ft-lb C. 486,600 ft-lb B. 826,000 ft-lb D. 262,600 ft-lb

Solution: P = 2πTN = 15,000 = 2πT (300) T = 262,605 ft-lb

B. determine the inside diameter of the shaft if the maximum shearing stress based on torsion alone is not to exceed 7,500 psi.

Solution: Ss = 16TDo / π [Do^4 - Di^4] = 7500 = 16 x 262,605 x 14 / π [14^4 – Di^4] Di = 9 in

C. what will be the propeller’s thrust that is develop in kilogram? A. 68, 180 kg C. 86,880 kg B. 26,810 kg D. 186,000 kg

Solution: Ft = 150,000 lbs x kg/2 lbs = 68, 18182 kg.

D. what is the diameter of the solid shaft? A. 1 in C. 6. 128 in B. 12 D. 8 in

Solution: Ss = 16T/πD^3 = 7,500 = 16 x 262,605 x 12 / π D^ 3 D = 1 2 in

E. What is the percentage saving in weight over the solid shaft? A. 92 % C. 37 % B. 73 % D. 67 %

Solution: Ws = weight of solid shaft = volume x density = A x L x ρ = π/4 x 12^2 x L ρ = 130 L ρ Wh = weight of hollow shaft = π/4 [14^2 -9^2] L ρ = 81 ρ % savings = Ws-Wh / Ws x 100 = 130 ρ – 81 ρ / 130 ρ = 37 %

A shell with an outside diameter of 406 mm and a wall thickness of 2 mm is subjected to 280 MPa tensile load and 45 KN-m. What is the maximum shear stress in MPa? A. 69 MPa C. 296 MPa B. 156 MPa D. 16 MPa

Solution: Do = 0 m Di = 0 – (2x0) = 0 m Ss = 16TDo / π [Do^4 – Di^4] = 16 x 45 x 0 / π [0^4 – 0^4] = 69,721 KPa = 69 MPa

A torque of 3142 foot-pounds is applied to one end of a solid cylindrical shaft of 4 inches in diameter fastened at the other end. The maximum shearing stresses closest to? A. 2500 psi C. 3,000 psi B. 3500 psi D. 3142 psi
A taper pin with a minimum mean diameter of 6 mm will be applied to fix a lever to a 2 inches shaft with an allowable design stress of 15 ksi. Compute for the transmitted maximum torque in in-lbs. use a factor of safety of 2. A. 600 C. 700 B. 800 D. 780

Solution: Empirical formula from Machinery Handbook p. 186 d = 1(T/DS) ^1/2 d = mean diameter of taper pin (in) S = safe unit stress (psi) = 6 mm = 0 in T = torque (in-lbs) D = shaft diameter (in) 0 = 1 (T/2 x 15,000) ^1/2 T = 1,398 in-lbs Applying the factor of safety 2; T = 1,398/2 = 699 in-lbs

Determine the power transmitted by main power transmitting steel shaft with 2 7/8 inches in outside diameter, SAE 1040 driving conveyor head pulleys at a shaft speed of 150 rpm. A. 40 C. 45 B. 50 D. 38

Solution: P = D^3 x N / 80 = 2^3 x 150 / 80 = 44 HP

Find the diameter of a main power transmitting shaft in mm to transmit 100 KW at 400 rpm. A. 48 C. 76 B. 67 D. 85

Solution: P = 100 KW = 134 HP N = 400 rpm P = D^3 x N / 80 = 134 = D^3 x 400 / 80 D = 2 in = 76 mm

Design a safe shaft diameter for a power transmitting shaft to transmit 30 HP at a shaft speed of 180 rpm. A. 2 C. 3. B. 2 D. 2 3/

Solution: P = D^3 x N / 80 = 134 = D^3 x 180 / 80 D = 2 in = 2 3/8 in 30. What would be the diameter in mm of a main power transmitting steel shaft SAE 1040 to transmit 100 KW at 600 rpm? A. 67 C. 76 B. 47 D. 56

Solution: P = D^3 x N / 80 = 134 = D^3 x 600 / 80 D 2 in = 66 mm

Compute the power in HP of a line shaft having a diameter of 1 7/8 inches with a speed of 200 rm. A. 25 C. 32 B. 27 D. 35

Solution: Power = D^3 x N / 53 = 1^3 x 200 / 53 = 24 HP

A short 40 mm diameter shaft having a rotational speed of 300 rpm. Determine the power transmitted by the shaft. A. 31 HP C. 45 HP B. 38 HP D. 49 HP

Solution: Power = D^3 x N / 38 = (40/25) ^3 x 300 / 38 = 30 HP

Determine the diameter of a line shaft transmitting 25 HP and having a speed of 225 rpm. The shear stress of the shaft is 6 ksi. A. 1 7/8 in C. 1 1/5 in B. 1 ¾ in D. 1 ¼ in

Solution: Power = D^3 x N / 53 = D^3 x 225 / 53 D = 1 .811 in say 1 7/8 in

Compute the angular deflection of a shaft in degrees having a diameter of 4 5/8 inches, length 60 inches, transmit 12 HP, at 180 rpm. Use G = 11^6 psi.

A. 0 degree C. 0 degree B. 0 degree D. 0 degree

Solution: P = 2πTN = 12 x 33,000 = 2πT x 180 T = 350 ft-lbs = 4,201 in-lbs Θ = TL/JG = [(4,201 x 60)/ (π/32 x 4^4 x 11^60)] x 180/π = 0 degree

Compute for the diameter in inches of SAE 1030 steel shaft to transmit 12 HP at 120 rpm with torsional deflection below 0 degree/foot length as required. A. 2 7/8 in C. 2 ¼ in B. 2 5/8 in D. 2 3/8 in

Solution: P = 2πTN

A short 61 mm diameter shaft transmits 120 HP. Compute the linear speed of a pulley 55 cm diameter mounted on the shaft. A. 1,796 fpm C. 1,870 fpm B. 1,766 fpm D. 2,016 fpm

Solution: For short shaft: P = D^3 x N / 38 = 120 = (61/25) ^3 x N / 38 N = 330 rpm V = πDN = π x 0 x 330 = 1870 fpm

Find the power in watts transmitted by a main power transmitting shaft with a diameter of 55 mm and speed of 200 rpm. A. 15 C. 25. B. 18 D. 30.

Solution: For main power transmitting shaft: P = D^3 x N / 80 D = 55/25 = 2. P = 2^3 x 200 / 80 = 25 x 746 = 18 watts

Compute for the torsional deflection in degrees of a 3 5/8 inches diameter, 1 m long shaft subjected to a twist moment of 3x10^6 N-mm. the torsional modulus of elasticity is 80,000 MPa. A. 0 C. 0. B. 0 D. 1.

Solution: D = 3 5/8 in = 92 mm Θ = TL/JG = [(3x10^6 x 1200) / (π/32 x 92^4 x 80,000)] x 180/π = 0 degree

Compute for the torsional deflection in degrees of a 110 mm diameter, 1 m long shaft subjected to twist moment of 3x10^6 N-mm. The torsional modulus of elasticity is 83, N/mm^2. A. 0 C. 0. B. 0 D. 0.

Solution: Θ = TL/JG = [(3x10^6 x 1400) / (π/32 x 110^4 x 83,000)] x 180/π = 0 degree

Find the torsional moment in Newton-millimeter developed when the shaft delivers 20 KW at 200 rpm. A. 0^6 C. 1^ B. 1^6 D. 2^

Solution: P = 2πTN = 20 = 2πT x 200/ T = 0 KN-m = 0^6 N-mm

A 102 mm diameter shaft is driven at 3000 rpm by a 300 HP prime mover. The shaft drives a 121 cm diameter chain sprocket having 86% output efficiency. Compute the torque in in-lb develop in the shaft. A. 5,600 C. 8, B. 7,100 D. 6,

Solution: P = 2πTN = 300 x 33,000 = 2πT x 3,000 = 525 ft-lb = 6,302 in-lb

Compute for the twisting moment in in-lb developed when the shaft delivers 20 HP at 1, rpm. A. 1,166 C. 1, B. 915 D. 945

Solution: P = 2πTN = 20 x 33,000 = 2πT x 1,200 = 87 ft-lb = 1,050 in-lb

Determine the diameter in inches of a steel countershaft that delivers 13 HP at a speed of 15 rad/s, the allowable material shear stress is 8 ksi. A. 1 in C. 2 in B. 1 ½ in D. 1 ¼ in

Solution: N = 15 rad/s x rev/2πrad x 60s/min = 149 rpm HP = D^3N / 38 for countershaft = 13 = (D^3 x 149)/38 = 1 in

In SAE 1030 steel 2 inches diameter solid shaft with a deflection not to exceed 0 degree/ft length. Modulus of rigidity of 12,000,000 psi. Determine the power transmitted if the shaft rotates at 280 rpm. A. 7 HP C. 8 HP B. 3 HP D. 3 HP

Solution: Θ = TL/JG = 0 x π/180 = T x12 / (π/32 x 2^4 x 12x10^6) T = 1644 in-lbs = 137 ft-lbs P = 2πTN = 2π x 137 x 280 x HP/33,000 = 7 HP

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