17. Shaft Design Introduction

Objectives „ Shaft must have adequate torsional strength to • Compute forces acting on shafts from gears, pulleys, and transmit torque and not be over stressed. sprockets. „ Shafts are mounted in bearings and transmit power • Find bending moments from gears, pulleys, or sprockets that are transmitting loads to or from other devices. through devices such as gears, pulleys, cams and clutches. • Determine torque in shafts from gears, pulleys, sprockets, clutches, and couplings. „ Components such as gears are mounted on shafts

• Compare combined stresses to suitable allowable stresses, using keys. including any required stress reduction factors such as stress „ Shaft must sustain a combination of bending and concentration factors and factors of safety. torsional loads. • Determine suitability of shaft design and/or necessary size of shafting.

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Standard diameters of shafts Torsion of circular shafts

Diameter (in.) Diameter increments (in.) Upto 3 1/16

3 to 5 1/8

5 to 8 1/4

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Torsion of circular shafts

T L Angle of twist, θ = G J „ θ = the angle of twist (radians) „ T = the applied torque (in-lb.) „ L = shaft length (in.) „ J = polar moment on inertia of the shaft cross section (in4) „ G = shear modulus of elasticity of the shaft material (lb/in2)

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1 Torsional Shear Stresses Shear Stress in a shaft

Torque Torque 16 T T c „ Shear stress, SS = 3 „ Torsional shear stress, S = π D S J Where π × d4 „ T = torque „ J = Polar moment of inertia = 16 T 32 „ D = diameter of the shaft = 3 „ c = radius of the shaft π SS „ T = Torque

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Forces on spur gear teeth Forces on spur gear teeth T n 63,000 P „ Ft = Transmitted force „ Power, P = or T = 63,000 n „ Fn = Normal force or separating force „ Torque, T = Ft r and r = Dp /2 „ Fr = Resultant force „ Combining the above we can write „ θ = pressure angle

„ F = F tan θ n t 2 T 2 P × 63,000 Ft = = F Dp Dp n F = t r cos θ

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Loads from Bevel gears Loads from Bevel gears

„ An additional axial force will be acting on the shaft „ Force transmitted, Fn = Ft tan θ cos γ because of the bevel angle „ θ = Pressure angle „ γ = Cone angle „ For the pinion it is relatively small, and can be γ = Cone angle neglected. „ Axial Force, Fa = Ft tan θ sin γ „ Resultant Force, F = 2 2 „ For the larger gear it will be significant and will be r Ft + F „ F = F or F depending on whichever is larger larger than the radial separating force. F = Fn or Fa depending on whichever is larger

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2 Loads from Worm gears Loads from Worm gears

To „ Driving force on the worm gear, Ft = Axial rwg „ To = Output torque Driving Ft sin φ „ Separating force, F = s cosφ cosλ - f sin λ where

„ λ = lead angle Separating „ ϕ = normal pressure angle

„ f = coefficient of friction

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Loads from Worm gears Loads from Worm gears

„ Axial force on the worm gear ⎛ cosφ sin λ + f cosλ ⎞ Fa(gear) = Ft(gear) ⎜ ⎟ ⎝ cosφ cosλ - f sin λ ⎠ where

„ λ = lead angle

„ ϕ = normal pressure angle

„ f = coefficient of friction

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Loads from Belts and Chains Bending of circular shafts

„ For a belt, Total load, Ft = Ff + Fb „ Shafts transmit power through gears and pulleys „ Net driving force, Fd = Ff – Fb „ These produce bending load in addition to „ Driving torque, T = Fd r torsion „ r = effective radius of pulley or sprocket „ Use strength of material approach to calculate „ For a chain F = 0 b the reaction forces and bending moments

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3 Bending of circular shafts Bending of circular shafts

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Shaft Design Problems Example Problem 17-1: Design Stresses in Shafts • Shaft shown drives a gear set that is transmitting 5 „ Step 1: Calculate the torque on the shaft from power hp at 1750 rpm.

„ Step 2: Find the torsional stress in the shaft • Shaft is supported in self-aligning ball bearings and gears are both 10 pitch, 40 tooth, 20° spur „ Step 3: Calculate the loads coming from gears, belts gears. or chains • Find torsional and bending stresses in shaft. „ Step 4: Calculate the bending moment due to the acting forces. If necessary combine the forces.

„ Step 5: Calculate the bending stress in the shaft

„ Step 6: Combine the bending stress and the torsional stress using the theories discussed in chapter 4

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Example Problem 17-1: Design Stresses in Shafts (cont’d.) Example Problem 17-1: Design Stresses in Shafts (cont’d.)

• Find the torsional stress in the shaft.

− First find Z': • Find the torsion in the shaft: (Appendix 3) (2-6) π D3 Z' = Tn 16 hp = 63,000 π (.75 in)3 Z' = then: 16

3 (17-1) Z' = .083 in

63,000 hp (3-6) T = T n S = s Z' 63,000 (5) T = 180 in-lb 1750 S = s .083 in3 T = 180 in-lb 2 Ss = 2170 lb/in

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4 Example Problem 17-1: Design Stresses in Shafts (cont’d.) Example Problem 17-1: Design Stresses in Shafts (cont’d.)

• Find the resultant force on the shaft:

• Find the load at the gear pitch circle: (12-2) F F = t (11-4) r cos θ N D = T p P d 90 lb F = r cos 20° 40 D = p 10 Fr = 96 lb

Dp = 4 inches • Find the maximum moment:

(12-3) (Appendix 2) 2 T FL Ft = Mm = DP 4

2 (180 in-lb) 96 lb (15 in) F = Mm = t 4 in 4

Ft = 90 lb Mm = 360 in-lb

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Example Problem 17-1: Design Stresses in Shafts (cont’d.) Combined Stresses in Shafts

• Find the stress: „ M „ As seen in Chap 4 S = Z

(Appendix 3) π D3 Z = 32

π (.75 in)3 Z = 32

Z = .041in3

M S = Z

360 in-lb S = .041 in3

S = 8780 lb/in2

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Combined maximum shear stress Example Problem 17-2: Combined Stresses in Shafts

„ τ = Maximum combined shear stress • From previous example problem, find the combined stress using the maximum shear stress theorem: „ S = normal stress 2 1/2 ⎡ 2 ⎛ S ⎞ ⎤ τ = S + ⎜ ⎟ „ S = shear stress ⎢ S ⎥ (4-5) S ⎝ 2 ⎠ ⎛ ⎛ S ⎞2 ⎞ ½ ⎣⎢ ⎦⎥ = S 2 + τ ⎜ s ⎜ 2 ⎟ ⎟ „ This can be rewritten as ⎝ ⎝ ⎠ ⎠ − Substituting stresses from previous example problem: 5.1 2 2 1/2 τ = T + M ⎛ 2 ⎛ 8780 ⎞2 ⎞ ½ 3 () τ = ⎜(2170 lb/in2) + ⎜ lb/in2⎟ ⎟ D ⎝ ⎝ 2 ⎠ ⎠ „ T = Torque in the shaft τ = 4900 lb/in2

„ M = Maximum moment − This should be compared to shear stress allowables.

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5 Maximum Normal Stress Theory Example Problem 17-3: Combined Stresses in Maximum Normal Stress Theory Shafts • From Example Problem 17-1, find the combined stress using the „ σ = equivalent combined normal stress maximum normal stress theory: 2 S ⎛ ⎛ S ⎞ ⎞ 1 „ S = normal stress from bending or axial loads σ = ± ⎜S 2 + ⎜ ⎟ ⎟ 2 2 ⎜ s 2 ⎟ ⎝ ⎝ ⎠ ⎠ „ SS = shear or torsional stress 2 1/2 – Substituting stresses from Example Problem 17-1: S ⎡ 2 ⎛ S ⎞ ⎤ σ = ± ⎢SS + ⎜ ⎟ ⎥ 2 2 2 2 ⎛ 2 ⎞ 1 ⎢ ⎝ ⎠ ⎥ 8780 lb / in ⎜ 2 2 ⎛ 8780 in ⎞ ⎟ ⎣ ⎦ (2170 lb / in ) ⎜ ⎟ 2 σ = + ⎜ + ⎜ ⎟ ⎟ 2 ⎜ ⎝ 2 ⎠ ⎟ „ This can be written as ⎝ ⎠ σ = 9300 lb / in2 5.1 σ = []M + (T2 + M2 )1/2 – This should be compared to the normal stress allowable. D3

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Solid Circular shaft Critical speeds of shafts

5.1 D = 3 ()T2 + M2 1/2 τ

5.1 2 2 1/2 D = 3 []M + (T + M ) σ

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Critical speeds of shafts

„ Operating speed should be 20% away from the critical speed.

„ Vibration frequency, f is given by 1 k g f = 2 π W

„ f = frequency in cycles per second, Hz

„ k = force constant, force per inch of deflection 2 „ g = acceleration due to gravity, 386.4 in./s

„ W = weight in pounds, lb

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6 „ Change the frequency to rpm Shaft with n concentrated loads

„ Critical speed, Nc = 60 × f „ Rayleigh’s equation is used. „ Also k is weight divided by deflection W k = δ W1 δ1 + W2 δ2 + W3 δ3 + ... + Wn δn 60 W g Nc = 187.7 2 2 2 2 N = W δ + W δ + W δ + ... + W δ c 2 π W δ 1 1 2 2 3 3 n n

1 N = 187.7 c δ

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Example Problem 17-5: Critical Speed Example Problem 17-5: Critical Speed (cont’d.)

• Find the estimated critical speed for the shaft in Example Problem 17-1 (assume the entire shaft diameter is ¾ inch). 188 − First, find deflection: Nc = (Appendix 2) FL3 ∂ δ = – 48 EI 188 (17-14) (Appendix 3) Nc = π D4 .21 I = 64 Nc = 410 rpm π (.75 in)4 I = 64

I = .016 in4

96 lb (15 in)4 δ = – 48 (30 x 106 lb/in2) (.016 in4) • This is approximate, and additional multiples would exist at 820, 1230, and 1640 rpm. δ = .21 inch

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