Course in Machine Design Gears and mechanical transmissions Udvekslinger og mekaniske transmissioner Machine Design Computational Mechanics, AAU, Esbjerg
Outline 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
Product design and development [Ch. 1] Steel design [Ch. 3, App. D] Gears and mechanical transmissions [Ch. 11, 12, App. C] Mechanism and dynamics [Ch. 3, notes] Shafts, keys, and couplings [Ch. 9, App. B, C] Shafts, keys, and couplings [Ch. 9, App. B, C] Tolerances [Notes] Fatigue I [Ch. 4, 5, App. E] Fatigue II [Ch. 5, 6, App. E] Fasteners – fatigue [Ch. 14]
Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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Case study • Life saving system for beaches – Mobility: terrain, offroad – Non buoyant – Load: 2 persons + rescue equipment
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Program • Case study • Project management – – – –
• • • • •
Vision Mission Objective/goal Strategy
Power considerations Discussion of Gears Gears Spur Gears Bevel gears
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Gears • Mechanical power transmission devices between two rotating shafts Friction Gears Friction ensures contact. Since contact point velocity is constant
ω1R1 = ω 2 R2 ω1 R2 = ω 2 R1
Speed ratio
Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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Gears Spur
Bevel
Helical
Worm
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Gears: Constant Speed Ratio Two arbitrary profiles in contact.
ω2 ω1
O1
C
O2
Desired: We want the ratio of angular velocity to be constant as different points come in contact.
Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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Gears: Constant Speed Ratio ω2 ω1
O1
C
v1 = ω1O1C
O2
v2 = ω1O1C
Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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Gears: Constant Speed Ratio v2
ω1
θ2
O1
θ1 v1
C
Note: The tangential components of the velocities O 2 ω2 need not be equal … they can slip relative to each other
Key: The component of the velocities along the common normal must be equal … else the parts will lose contact or interfere
v1 cosθ1 = v2 cosθ 2 ω1O1C cos θ1 = ω2O2C cos θ 2
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Gears: Constant Speed Ratio O1
θ1
D1 θ1
D θ2 2
θ2
C
90 − θ1
From previous: ω1O1C cosθ1 = ω 2O2C cosθ 2 But
O1C cosθ1 = O1D1
ω1O1D1 = ω 2O2 D2 Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
ω1 O2 D2 = ω 2 O1D1
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Gears: Constant Speed Ratio D2
P
O1
D1 C From previous:
By similarity triangles
ω1 O2 D2 = ω 2 O1D1
ω1 O2 D2 O2 P = = ω 2 O1D1 O1P Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
P: Pitch point. 11
Gears: Constant Speed Ratio ω2
P
ω1
O1 O2
P: Pitch point. Intersection of normal at contact and line joining centers. Law of conjugate action • The ratio of angular velocities remains constant only if the pitch point remains fixed!! • There are many ways to ensure conjugate action Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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ω1 R2 = ω 2 R1
Spur gears • Spur gears achieve constant speed ratio through involute teeth. See animation at: http://auto.howstuffworks.co m/gear7.html
Pressure angle also remains constant!
• Pitch point does not move for two involute-profile teeth gears in contact
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Spur Gears rb 2πrb N= pb
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Spur Gears Pressure angle governed by intended pitch circle
cos ϕ =
rpb rp
Gears behave like two friction gears in contact except high power transmission
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Spur Gears: Definitions Involute for rack is a straight line! Tool Gear being manufactured
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Spur Gears: Definitions
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Spur Gears: Definitions
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Spur Gears: Meshing • For two gears to mesh, they must have the same: – Nominal pressure angle – Nominal circular pitch
• But can have different ω1 N 2 = ω 2 N1
– Number of teeth – Pitch diameter dp, dg
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Spur Gears • Actual center distance determines
• Smaller = Pinion • Larger = Gear
– – – – – –
Actual pressure angle Actual circular pitch Actual tooth thickness Back-lash Interference Contact ratio
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Spur Gears
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Gears: Forces Two components of force: Ft: Useful tangential force Fr: Radial force
Fr = Ft tan ϕ Torque = rFt Power = ωrFt The tangential force is determined from the power being transmitted. Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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Gears: Stresses
Contact Stresses Need to design gear against both types of stresses. Bending stresses
Fatigue stresses!
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Gears: Bending Stress
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Gears: Bending Stresses Two approaches: – Simplistic Lewis Equation – Modern AGMA method
• Lewis assumptions: – Neglect radial force – Force acts at tip – Uniform force distribution – No stress concentration – No sliding force – Zero contact velocity
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Gears: Bending Stresses Mc σ (bending ) = I M = Ft h c =t/2 3
bt I= 12 6 Ft h σ= 2 bt Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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Gears: Bending Stresses 6 Ft h
Ft σ= 2 = b bt
⎛ 6h ⎞ ⎜⎜ ⎟⎟ ⎝ t2 ⎠
Function of diametral pitch
Ft ⎛ P ⎞ σ= 2 = ⎜ ⎟ b ⎝Y ⎠ bt Y : Lewis Factor 6 Ft h
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Gears: Bending Stresses Example : Ft = 100lbf b = 0.1" ; P = 4; ϕ = 20; N = 28 Y = 0.35 Ft σ= b
⎛ P⎞ ⎜ ⎟ = 11.45ksi ⎝Y ⎠
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Gears: Bending Stresses AGMA model: • Neglect radial force • Force acts at tip • No sliding force • Non-uniform force distribution • Stress concentration • Non-zero contact velocity
Ft ⎛ P ⎞ σ = ⎜ ⎟Kv Ko Km b ⎝J⎠ J : geometry factor K v : velocity factor K o : overload factor K m : mounting factor
Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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Gears: Bending Stresses Geometry factor J: allows for 2 different assumptions
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Gears: Bending Stresses Velocity factor Kv: also depends on how gear is manufactured
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Gears: Bending Stresses • Overload factor Ko: depends on power source and driven machinery Table 15.1 • Light shock power and moderate shock machinery, Ko = 1.5 • Mounting factor Km: depends on accuracy of mounting and face width Table 15.2 very accurate and width of 4”, Km = 1.4 Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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Gears: Bending Stress σm
σa
1 + = Sut S n n
σ
σ
1 + = 2 Sut 2 S n n S n = (Other factors) * CG C LCS 0.5Sut S n ≈ 0.4 Sut
σ 0.4σ 1 + = 2S n 2S n n ... σ
1 = 1.4 S n n Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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Gears: Bending Stress For the idler
σm
σa
1 + = Sut S n n σ 1 = Sn n
Include 1.4 for regular gears as a factor in Sn
Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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Gears: Bending Stress S n = kt k r k ms CG C LCS 0.5Sut kt : temperature factor
σ : AGMA
k r : reliability factor
S n : From above
1 for idler k ms :} 1.4 for other
1 = Sn n
Conclusion
σ
or design ' b' for a given safety factor
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Gears: Bending Strength - Example Problem info: • Pinion: 20 teeth, gear: 40 teeth • 20 degrees pressure angle; P = 8 • Accurate mounting • Material: Steel, heat treated to 350 Bhn. • Standard full depth teeth • Required life: 5 years, 60 hours/week, 50 week/year, 1100 rpm (pinion) •
Find: max horsepower that can be transmitted with a safety factor of 1.5, and 99% reliability, based only on bending strength.
Assumptions: • AGMA method of analysis • Gear tooth loads transmitted at pitch point • No load sharing • Assume top quality hobbing operation Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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Gears: Bending Strength - Example Ft ⎛ P ⎞ σ = ⎜ ⎟Kv Ko Km b ⎝J⎠ V : 720rpm(pitch line velocity) From Curve C (Fig.15.24) K v = 1.54
σ
1 = Sn n
From Fig.15.23a J = 0.24 (pinion is weaker) K m = 1.6(assumption) K o = 1.0(assumption)
σ = 82.1Ft
Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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Gears: Bending Strength - Example S n = kt k r k ms CG C LCS 0.5Sut kt : 1
σ
1 = Sn n
k r : 0.814 k ms : 1.4 CG : 1; C L : 1.0; CS : 0.66 S n = 65812 psi
Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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Gears: Bending Strength - Example 82.1Ft = 65812 / 1.5 Ft = 534lb
σ
1 = Sn n
HP = 11.7 hp
Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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Gears: Hertzian Contact Stress As before Find : Eeq Req Area of contact Max stress... But also include the various factors
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Gears: Hertzian Contact Stress Buckingham model
Ft σ H = Cp Kv Ko Km bd p I C p : E and ν dependent (Table 15.4) d p : Pinion pitch diameter I : Geometry factor (Eqn 15.23)
Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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Gears: Contact Strength Recall contact stress problems have only finite life. Table 15.5 provides typical contact strength values Sfe for 107 cycles, 99 percent reliability and temp < 250 F To find strength at a different number of cycles and temperature use eqn. 15.25
S H = S feC Li C R
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Bevel Gears Note: b: Face width L: Pitch cone length γ: Pitch cone angle d: pitch dia
Bevel gears are not interchangeable! Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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Gears: Forces Fr
Ft Fa
Determined from power requirements
Fr = Ft tan ϕ cos γ Fa = Ft tan ϕ sin γ
Note: What is axial force for one bevel gear is radial force for the other.
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Gears: Stresses
Both bending and contact stresses Bending stress and strength
Ft σ= b
⎛ P⎞ ⎜ ⎟Kv Ko Km ⎝J⎠
S n = kt k r k ms CG C LCS 0.5Sut Factors similar or slightly different for Bevel gears Contact stress and strength
Ft σ H = Cp Kv Ko Km bd p I
S H = S feC Li C R
Factors similar or slightly different for Bevel gears Machine Design Gears and mechanical transmissions Computational Mechanics, AAU, Esbjerg
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