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ME315 Transmission Design Project

Published: at 12:35 PM (15 min read)

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Introduction

This past spring quarter at Northwestern, I took ME 315: Theory of Machines—Design of Elements, a fast-paced course covering fundamental mechanical components, failure theories, design criteria, and the structure of mechanical systems. The course focused on developing the skills needed to design and analyze basic machine elements and assemblies.

I had a great time in the class and especially enjoyed its rapid pace and breadth. It was a lot of material to absorb in just ten weeks, so and I definitely do not remember every equation or design rule off the top of my head. However, I thikn I came away with a strong foundation and the confidence to revisit and understand the material when needed.

For our final project, my partner, Nathan Shi, and I were tasked with designing a power transmission system that met the following specifications:

ParameterValue
Input power70 kW70~\mathrm{kW}
Input speed2600 rpm2600~\mathrm{rpm}
Required output speed300±3 rpm300 \pm 3~\mathrm{rpm}
Transmission typeTwo-stage gear reduction
Number of shaftsThree shafts
Gear typeSpur or helical gears permitted
Minimum shaft diameter20 mm20~\mathrm{mm}
Operating schedule8 hours/day, 52 weeks/year
Required service lifeMinimum 5 years
Reliability requirement99%99\% excluding bearings
Efficiency assumption100%100\% per gear mesh and bearings
Design objectiveCompact, cost-effective design

Other than that, we weren’t given much more guidelines (both fun and painful), which enabled us to be creative with our final design choice. We settled on a compact, two-stage reverted countershaft gearbox with coaxial input and output shafts. This (very late lol) post attempts to explain our design process.

Transmission Schematic and Design Description

transmission Schematic

The final transmission design utilizes a compact reverted countershaft gearbox arrangement with a coaxial input and output configuration. Unlike a traditional parallel shaft gearbox, the selected layout minimizes the distance between the input and the output shafts by aligning them along the same centerline while maintaining independent shaft rotation. This reduces our footprint significantly.

In our configuration, the first-stage reduction occurs between shafts 1 and 2, and the second stage reduction occurs between shafts 2 and 3. This allowed us to make use of the principle of component standardization to reduce costs. Both gear stages utilize identical 17-tooth pinions and 50-tooth gears. The shafts were also intentionally designed with geometric similarity - shafts 1 and 3 share similar structural layouts bearing arrangements despite operating at different torque levels. This can help to simplify assembly, machining, and maintenance required while reducing the number of unique components.

The gearbox also utilizes herringbone gears rather than traditional spur or helical gears. Herringbone gears were selected because they have the smooth transmission and strong load capacity of helical gears while avoiding their pitfall of having axial loads. In traditional helical gear arrangements, axial loads require larger bearings that can handle these loads. However, by using opposite-hand helices within each herringbone, the axial forces generated by one helix are to be balanced by the other.

schematic

full assembly

Overall Performance Metrics

Performance MetricFinal Design Value
Input Power70 kW
Input Speed2600 rpm
Output Speed300.56 rpm
Required Output Speed Range300 ± 3 rpm
Overall Transmission Ratio8.651:1
Transmission Error0.19%
Gear TypeHerringbone gears
Gear MaterialAISI 4140 Grade 2 steel, 450 HB
Shaft MaterialAISI 1045 steel
Number of Gear Stages2
Gear Tooth Combination17:50 per stage
Center Distance per Stage154.73 mm
Overall LayoutCoaxial reverted countershaft gearbox
Net Axial Gear ForceApproximately 0 N
Minimum Gear Bending FoS1.90, Gear 2B
Minimum Gear Contact FoS1.35
Minimum Shaft Static FoS2.14, Shaft 3
Minimum Shaft Fatigue FoS1.92, Shaft 3
Bearing Type SelectedSKF 32310 J tapered roller bearings
Maximum Required Bearing Rating194.24 kN
Selected Bearing Dynamic Rating211 kN
Bearing Maintenance RequirementNo bearing change service required
Expected Shaft LifeInfinite life, (N > 10^6) cycles
Critical ShaftShaft 2, due to highest moment and highest torque
Overall Limiting Design Factor1.35, gear contact safety factor

Geartrain and Gear Parameters

ItemGear 1Gear 2AGear 2BGear 3
TypeHerringboneHerringboneHerringboneHerringbone
Teeth, (N)17501750
Helix Angle (°)30303030
Normal Module, (m_n) (mm)4444
Transverse Module, (m_t) (mm)4.61884.61884.61884.6188
Pitch Diameter (mm)78.52230.9478.52230.94
Normal Pressure Angle (°)20202020
Transverse Pressure Angle (°)22.8022.8022.8022.80
Face Width, (b) (mm)50.2750.2750.2750.27
Speed (rpm)2600884884300.56
Power (kW)70707070
Total Torque (N·m)257.12756.22756.222224.18
Tangential Force on Gear (N)6549.076549.0719261.9819261.98
Radial Force on Gear (N)2752.422752.428095.368095.36
Axial Force (N)0000
Material4140 G2 steel4140 G2 steel4140 G2 steel4140 G2 steel
Hardness (HB)450450450450
Hardness Ratio1.001.001.001.00
Bending FoS5.587.521.902.56
Contact FoS2.202.201.351.35
Contact Ratio3.503.503.503.50
Stage Transmission Ratio2.9412.9412.9412.941
Overall Transmission Ratio8.6518.6518.6518.651
Center Distance (mm)154.73154.73154.73154.73
Total Center Distance (mm)0, coaxial reverted layout0, coaxial reverted layout0, coaxial reverted layout0, coaxial reverted layout
Transmission Error0.19%0.19%0.19%0.19%

Shaft Parameters

The number of cycles was calculated using N=60nLhN=60nL_h, where Lh=10400L_h=10400 hours. Since all shafts experience more than 10610^6 cycles, the shafts are treated as infinite-life fatigue components.

ItemShaft 1Shaft 2Shaft 3
MaterialAISI 1045 steelAISI 1045 steelAISI 1045 steel
Speed (rpm)2600884300.56
Maximum Torque, (T) (N·m)257.12756.222224.18
Radial Bending Moment (N·m)75.87386.29223.16
Tangential Bending Moment (N·m)180.53771.96530.98
Maximum Bending Moment, (M) (N·m)195.83863.21575.97
Minimum Diameter, (d_{\min}) (m)0.02000.03280.0287
Minimum Diameter with Keyway (m)0.02100.03440.0301
Number of Cycles (million cycles)1622.4551.6187.5
Location of Critical Cross SectionCenter of Gear 1Center of Gears 2A/2BCenter of Gear 3
Static FoS14.77Not calculated2.14
Fatigue FoS8.20Not calculated1.92
Expected LifeInfiniteInfiniteInfinite

Bearing Parameters

ItemBearings 1A & 1BBearings 2A & 2BBearings 3A & 3B
Radial Force on Bearing (N)1376.213841.28 / 7006.504047.68
Tangential Force on Bearing (N)3274.54-1288.92 / 14001.839630.99
Total Bearing Load (N)3551.984051.76 / 15657.0210446.99
Required Dynamic Rating, (C) (N)63136.3150266.57 / 194242.5290459.24
SKF Bearing Number32310 J32310 J32310 J
Bore Diameter (mm)505050
Width (mm)42.2542.2542.25
Dynamic Load Rating, (C_1) (kN)211211211
Outside Diameter (mm)110110110
Expected Life (million cycles)1622.4551.6187.5
Bearing Change ServicesNot neededNot neededNot needed

Shaft 1 Assembly

Shaft 3 Assembly

Shafts 1 and 3 Part Drawing

Shaft 2 Part Drawing

Gears 1/2B & 2A/3 Part Drawings

Appendix A: Shaft 1 Analysis

Given Values

P=70 kWP = 70~\mathrm{kW}, n1=2600 rpmn_1 = 2600~\mathrm{rpm}, T1=257.12 NmT_1 = 257.12~\mathrm{N\cdot m}, Wt=6549.07 NW_t = 6549.07~\mathrm{N}, Wr=2752.42 NW_r = 2752.42~\mathrm{N}, a=b=55 mma=b=55~\mathrm{mm}, and Wa0 NW_a \approx 0~\mathrm{N}.

Bearing Reactions

Ay=By=Wr2=2752.422=1376.21 NA_y = B_y = \frac{W_r}{2} = \frac{2752.42}{2} = 1376.21~\mathrm{N} Az=Bz=Wt2=6549.072=3274.54 NA_z = B_z = \frac{W_t}{2} = \frac{6549.07}{2} = 3274.54~\mathrm{N}

Resultant Bearing Forces

RA=Ay2+Az2=(1376.21)2+(3274.54)2R_A = \sqrt{A_y^2 + A_z^2} = \sqrt{(1376.21)^2 + (3274.54)^2} RA=RB=3551.98 NR_A = R_B = 3551.98~\mathrm{N}

Maximum Bending Moments

Mr=Ay(a)=By(b)=(1276.21)(0.055)=75.87 NmM_r = A_y(a) = B_y(b) = (1276.21)(0.055) = 75.87~\mathrm{N\cdot m} Mt=Az(a)=Bz(b)=(3274.54)(0.055)=180.53 NmM_t = A_z(a) = B_z(b) = (3274.54)(0.055) = 180.53~\mathrm{N\cdot m} M=Mr2+Mt2=(75.87)2+(180.53)2=195.83 NmM = \sqrt{M_r^2 + M_t^2} = \sqrt{(75.87)^2 + (180.53)^2} = 195.83~\mathrm{N\cdot m}

Minimum Shaft Diameter

Me=M2+3T2=(195.83)2+3(257.12)2=488.14 NmM_e = \sqrt{M^2 + 3T^2} = \sqrt{(195.83)^2 + 3(257.12)^2} = 488.14~\mathrm{N\cdot m} dmin=32Meπσallow3=32(488.14×103)π(350)3=20.0 mmd_{\min} = \sqrt[3]{\frac{32M_e}{\pi \sigma_{\mathrm{allow}}}} = \sqrt[3]{\frac{32(488.14 \times 10^3)}{\pi(350)}} = 20.0~\mathrm{mm}

Keyway Correction

dcorr=1.05dmin=1.05(20.0)=21.0 mmd_{\mathrm{corr}} = 1.05d_{\min} = 1.05(20.0) = 21.0~\mathrm{mm}

Bending and Torsional Stresses

Due to the required load bearing and subsequent size of the bearings, a diameter of 52.5 mm was chosen at the location of the highest stress and moment. Using:

d=52.5 mmd = 52.5~\mathrm{mm} σb=32Mπd3=32(195.83×103)π(52.5)3=15.96 MPa\sigma_b = \frac{32M}{\pi d^3} = \frac{32(195.83 \times 10^3)}{\pi(52.5)^3} = 15.96~\mathrm{MPa} τ=16Tπd3=16(257.12×103)π(52.5)3=10.48 MPa\tau = \frac{16T}{\pi d^3} = \frac{16(257.12 \times 10^3)}{\pi(52.5)^3} = 10.48~\mathrm{MPa}

Von Mises Stresses

σvm=σb2+3τ2=(15.96)2+3(10.48)2=24.16 MPa\sigma_{\mathrm{vm}} = \sqrt{\sigma_b^2 + 3\tau^2} = \sqrt{(15.96)^2 + 3(10.48)^2} = 24.16~\mathrm{MPa}

Static FoS

ny=Syσvm=35024.16=14.77n_y = \frac{S_y}{\sigma_{\mathrm{vm}}} = \frac{350}{24.16} = 14.77

Fatigue Analysis

The fatigue life of Shaft 1 was checked using the modified Goodman relation. Since the shaft rotates while the transverse gear load remains fixed in space, the bending stress is treated as fully reversed. The transmitted torque is treated as steady.

σa=σb,τm=τ,σm=0,τa=0\sigma_a = \sigma_b, \qquad \tau_m = \tau, \qquad \sigma_m = 0, \qquad \tau_a = 0

The shaft material is AISI 1045 steel with:

Sut=633 MPaS_{ut} = 633~\mathrm{MPa}

The uncorrected endurance limit for steel is estimated as:

Se=0.5Sut=0.5(633)=316.5 MPaS_e' = 0.5S_{ut} = 0.5(633) = 316.5~\mathrm{MPa}

The corrected endurance limit is:

Se=kakbkckdkeSeS_e = k_a k_b k_c k_d k_e S_e'

The surface finish factor for machined steel is:

ka=aSutbk_a = aS_{ut}^{b}

Using machined surface constants:

a=4.51,b=0.265a = 4.51, \qquad b = -0.265 ka=4.51(633)0.265=0.817k_a = 4.51(633)^{-0.265} = 0.817

The size factor is:

kb=(d7.62)0.107k_b = \left(\frac{d}{7.62}\right)^{-0.107}

Using:

d=52.5 mmd = 52.5~\mathrm{mm} kb=(52.57.62)0.107k_b = \left(\frac{52.5}{7.62}\right)^{-0.107} kb=0.814k_b = 0.814

The load factor for bending is:

kc=1k_c = 1

The temperature factor is:

kd=1k_d = 1

The reliability factor for 99% reliability is:

ke=0.814k_e = 0.814

Therefore:

Se=(0.817)(0.814)(1)(1)(0.814)(316.5)S_e = (0.817)(0.814)(1)(1)(0.814)(316.5) Se=171.2 MPaS_e = 171.2~\mathrm{MPa}

Using the previously calculated Shaft 1 stresses:

σb=15.96 MPa\sigma_b = 15.96~\mathrm{MPa} τ=10.48 MPa\tau = 10.48~\mathrm{MPa}

The equivalent alternating stress is:

σa,eq=σb=15.96 MPa\sigma_{a,\mathrm{eq}} = \sigma_b = 15.96~\mathrm{MPa}

The equivalent mean stress due to steady torsion is:

σm,eq=3τ2=3(10.48)2=18.15 MPa\sigma_{m,\mathrm{eq}} = \sqrt{3\tau^2} = \sqrt{3(10.48)^2} = 18.15~\mathrm{MPa}

The modified Goodman fatigue criterion is:

1nf=σa,eqSe+σm,eqSut=15.96171.2+18.15633=0.0932+0.0287\frac{1}{n_f} = \frac{\sigma_{a,\mathrm{eq}}}{S_e} + \frac{\sigma_{m,\mathrm{eq}}}{S_{ut}} = \frac{15.96}{171.2} + \frac{18.15}{633} = 0.0932 + 0.0287 1nf=0.1219\frac{1}{n_f} = 0.1219 nf=8.20n_f = 8.20

Since:

nf>1n_f > 1

Shaft 1 satisfies the fatigue requirement. Since the cycle count exceeds 10610^6, Shaft 1 is treated as an infinite-life shaft.

N=60n1Lh=60(2600)(10400)=1.62×109 cycles>106N = 60n_1L_h = 60(2600)(10400) = 1.62 \times 10^9~\mathrm{cycles} > 10^6

Appendix B: Gear 1 Calculations

Input Requirements

The gearbox input conditions were specified by the project requirements as:

P=70 kWP = 70~\mathrm{kW} n1=2600 rpmn_1 = 2600~\mathrm{rpm}

The desired output speed was:

n3=300±3 rpmn_3 = 300 \pm 3~\mathrm{rpm}

This required an overall transmission ratio near:

iT=26003008.67i_T = \frac{2600}{300} \approx 8.67

Transmission Ratio Selection

A two-stage reduction was selected to avoid excessively large gears and maintain a compact gearbox design. The selected tooth combination was:

17:5017:50

for both stages. This produced:

i1=i2=5017=2.941i_1 = i_2 = \frac{50}{17} = 2.941

and therefore:

iT=(2.941)2=8.651i_T = (2.941)^2 = 8.651

This satisfies the required output-speed range.

Helix Angle

A helix angle of:

β=30\beta = 30^\circ

was selected because larger helix angles improve tooth overlap and transmission smoothness while remaining manufacturable.

Gear Module and Pitch Diameter

A normal module of:

mn=4 mmm_n = 4~\mathrm{mm}

was selected. A smaller module would reduce gearbox size but significantly increase tooth stress, while a larger module would unnecessarily increase the gear size and center distance.

The transverse module is:

mt=mncosβm_t = \frac{m_n}{\cos\beta} mt=4cos30m_t = \frac{4}{\cos 30^\circ} mt=4.6188 mmm_t = 4.6188~\mathrm{mm}

The pitch diameter of Gear 1 is:

d1=mtN1=(4.6188)(17)=78.52 mmd_1 = m_tN_1 = (4.6188)(17) = 78.52~\mathrm{mm}

Pressure Angles

A standard normal pressure angle of:

ϕn=20\phi_n = 20^\circ

was selected because it provides a good balance between bending strength, contact stress, and manufacturability.

The transverse pressure angle becomes:

ϕt=tan1(tanϕncosβ)=22.80\phi_t = \tan^{-1}\left( \frac{\tan\phi_n}{\cos\beta} \right) = 22.80^\circ

Gear Forces

The tangential gear force was calculated from:

Ft=2Td=2(257.12)0.07852=6549.07 NF_t = \frac{2T}{d} = \frac{2(257.12)}{0.07852} = 6549.07~\mathrm{N}

The radial force was calculated using the transverse pressure angle:

Fr=Fttanϕt=6549.07tan(22.80)=2752.42 NF_r = F_t\tan\phi_t = 6549.07\tan(22.80^\circ) = 2752.42~\mathrm{N}

Since herringbone gears were selected, the net axial force is approximately zero.

Face Width

The face width was selected using a helical-gear overlap estimate:

b>2πmnsinβ=2π(4)sin30b > \frac{2\pi m_n}{\sin\beta} = \frac{2\pi(4)}{\sin 30^\circ} b>50.27 mmb > 50.27~\mathrm{mm}

The selected face width improves tooth-load distribution and reduces local bending stress.

Gear Material Selection

AISI 4140 Grade 2 steel was selected for all gears because of its high strength, good fatigue resistance, and suitability for high-power gearbox applications.

The smaller 17-tooth pinions experience the highest bending stresses because of their lower tooth counts and smaller pitch diameters. Using a higher-strength alloy steel improves reliability under long-term cyclic loading.

Appendix B: Gear 1 Calculations

Input Requirements

The gearbox input conditions were specified by the project requirements as:

P=70 kWP = 70~\mathrm{kW} n1=2600 rpmn_1 = 2600~\mathrm{rpm}

The desired output speed was:

n3=300±3 rpmn_3 = 300 \pm 3~\mathrm{rpm}

This required an overall transmission ratio near:

iT=26003008.67i_T = \frac{2600}{300} \approx 8.67

Transmission Ratio Selection

A two-stage reduction was selected to avoid excessively large gears and maintain a compact gearbox design. The selected tooth combination was:

17:5017:50

for both stages. This produced:

i1=i2=5017=2.941i_1 = i_2 = \frac{50}{17} = 2.941

and therefore:

iT=(2.941)2=8.651i_T = (2.941)^2 = 8.651

This satisfies the required output-speed range.

Helix Angle

A helix angle of:

β=30\beta = 30^\circ

was selected because larger helix angles improve tooth overlap and transmission smoothness while remaining manufacturable.

Gear Module and Pitch Diameter

A normal module of:

mn=4 mmm_n = 4~\mathrm{mm}

was selected. A smaller module would reduce gearbox size but significantly increase tooth stress, while a larger module would unnecessarily increase the gear size and center distance.

The transverse module is:

mt=mncosβm_t = \frac{m_n}{\cos\beta} mt=4cos30m_t = \frac{4}{\cos 30^\circ} mt=4.6188 mmm_t = 4.6188~\mathrm{mm}

The pitch diameter of Gear 1 is:

d1=mtN1=(4.6188)(17)=78.52 mmd_1 = m_tN_1 = (4.6188)(17) = 78.52~\mathrm{mm}

Pressure Angles

A standard normal pressure angle of:

ϕn=20\phi_n = 20^\circ

was selected because it provides a good balance between bending strength, contact stress, and manufacturability.

The transverse pressure angle becomes:

ϕt=tan1(tanϕncosβ)=22.80\phi_t = \tan^{-1}\left( \frac{\tan\phi_n}{\cos\beta} \right) = 22.80^\circ

Gear Forces

The tangential gear force was calculated from:

Ft=2Td=2(257.12)0.07852=6549.07 NF_t = \frac{2T}{d} = \frac{2(257.12)}{0.07852} = 6549.07~\mathrm{N}

The radial force was calculated using the transverse pressure angle:

Fr=Fttanϕt=6549.07tan(22.80)=2752.42 NF_r = F_t\tan\phi_t = 6549.07\tan(22.80^\circ) = 2752.42~\mathrm{N}

Since herringbone gears were selected, the net axial force is approximately zero.

Face Width

The face width was selected using a helical-gear overlap estimate:

b>2πmnsinβ=2π(4)sin30b > \frac{2\pi m_n}{\sin\beta} = \frac{2\pi(4)}{\sin 30^\circ} b>50.27 mmb > 50.27~\mathrm{mm}

The selected face width improves tooth-load distribution and reduces local bending stress.

Gear Material Selection

AISI 4140 Grade 2 steel was selected for all gears because of its high strength, good fatigue resistance, and suitability for high-power gearbox applications.

The smaller 17-tooth pinions experience the highest bending stresses because of their reduced tooth counts and smaller pitch diameters. Using a higher-strength alloy steel improves reliability under long-term cyclic loading.

Appendix C: Shaft 3 Analysis

Given Values

For Shaft 3, the input values are:

P=70 kW,n3=300.56 rpm,T3=2224.18 Nm,P = 70~\mathrm{kW}, \qquad n_3 = 300.56~\mathrm{rpm}, \qquad T_3 = 2224.18~\mathrm{N\cdot m}, Wt=19261.98 N,Wr=8095.36 N,Wa0 NW_t = 19261.98~\mathrm{N}, \qquad W_r = 8095.36~\mathrm{N}, \qquad W_a \approx 0~\mathrm{N}

The gear is centered between the bearings, so:

a=b=55 mma = b = 55~\mathrm{mm}

Bearing Reactions

Gear 3 is positioned between Bearings 3A and 3B. Since the gear is centered between the bearings in the simplified model, the bearing reactions are equal.

For the radial-force plane:

Ay=By=Wr2=8095.362=4047.68 NA_y = B_y = \frac{W_r}{2} = \frac{8095.36}{2} = 4047.68~\mathrm{N}

For the tangential-force plane:

Az=Bz=Wt2=19261.982=9630.99 NA_z = B_z = \frac{W_t}{2} = \frac{19261.98}{2} = 9630.99~\mathrm{N}

The resultant bearing force is:

RA=RB=Ay2+Az2R_A = R_B = \sqrt{A_y^2 + A_z^2} RA=RB=(4047.68)2+(9630.99)2=10446.99 NR_A = R_B = \sqrt{(4047.68)^2 + (9630.99)^2} = 10446.99~\mathrm{N}

Shear Force and Bending Moment

The shear-force diagrams are drawn separately in the radial and tangential planes.

For the radial-force plane:

Vy=+4047.68 NV_y = +4047.68~\mathrm{N}

from Bearing 3A to Gear 3. At Gear 3:

Vy=4047.688095.36=4047.68 NV_y = 4047.68 - 8095.36 = -4047.68~\mathrm{N}

from Gear 3 to Bearing 3B.

For the tangential-force plane:

Vz=+9630.99 NV_z = +9630.99~\mathrm{N}

from Bearing 3A to Gear 3. At Gear 3:

Vz=9630.9919261.98=9630.99 NV_z = 9630.99 - 19261.98 = -9630.99~\mathrm{N}

from Gear 3 to Bearing 3B.

The bending-moment diagrams are triangular, with the maximum bending moment occurring at Gear 3. From the force diagram:

Mr=223.16 NmM_r = 223.16~\mathrm{N\cdot m} Mt=530.98 NmM_t = 530.98~\mathrm{N\cdot m}

The resultant maximum bending moment is:

M=Mr2+Mt2=(223.16)2+(530.98)2=575.97 NmM = \sqrt{M_r^2 + M_t^2} = \sqrt{(223.16)^2 + (530.98)^2} = 575.97~\mathrm{N\cdot m}

Minimum Shaft Diameter

The shaft is checked under combined bending and torsion. The equivalent moment is:

Me=M2+3T2M_e = \sqrt{M^2 + 3T^2} Me=(575.97)2+3(2224.18)2=3901.12 NmM_e = \sqrt{(575.97)^2 + 3(2224.18)^2} = 3901.12~\mathrm{N\cdot m}

Using:

σallow=350 MPa\sigma_{\mathrm{allow}} = 350~\mathrm{MPa}

the theoretical minimum shaft diameter is:

dcalc=32Meπσallow3d_{\mathrm{calc}} = \sqrt[3]{ \frac{32M_e} {\pi\sigma_{\mathrm{allow}}} } dcalc=32(3901.12×103)π(350)3=28.7 mmd_{\mathrm{calc}} = \sqrt[3]{ \frac{32(3901.12 \times 10^3)} {\pi(350)} } = 28.7~\mathrm{mm}

A keyway correction factor of 1.05 is applied:

dkey=1.05dcalc=1.05(28.7)=30.1 mmd_{\mathrm{key}} = 1.05d_{\mathrm{calc}} = 1.05(28.7) = 30.1~\mathrm{mm}

However, this value is only a theoretical lower bound. Since Shaft 3 carries the largest torque and must support the output coupling, bearing seats, shoulders, and keyways, the final minimum design diameter was selected as:

d=52.5 mmd = 52.5~\mathrm{mm}

Static Stress Analysis

Using:

d=52.5 mmd = 52.5~\mathrm{mm}

the bending stress is:

σb=32Mπd3=32(575.97×103)π(52.5)3=46.93 MPa\sigma_b = \frac{32M}{\pi d^3} = \frac{32(575.97 \times 10^3)} {\pi(52.5)^3} = 46.93~\mathrm{MPa}

The torsional shear stress is:

τ=16Tπd3=16(2224.18×103)π(52.5)3=90.62 MPa\tau = \frac{16T}{\pi d^3} = \frac{16(2224.18 \times 10^3)} {\pi(52.5)^3} = 90.62~\mathrm{MPa}

The von Mises stress is:

σvm=σb2+3τ2=(46.93)2+3(90.62)2=163.83 MPa\sigma_{\mathrm{vm}} = \sqrt{\sigma_b^2 + 3\tau^2} = \sqrt{(46.93)^2 + 3(90.62)^2} = 163.83~\mathrm{MPa}

Using AISI 1045 steel with:

Sy=350 MPaS_y = 350~\mathrm{MPa}

the static factor of safety is:

ny=Syσvm=350163.83=2.14n_y = \frac{S_y}{\sigma_{\mathrm{vm}}} = \frac{350}{163.83} = 2.14

Fatigue Analysis

The operating-life requirement is:

Lh=10400 hrL_h = 10400~\mathrm{hr}

The number of cycles is:

N=60n3Lh=60(300.56)(10400)=1.875×108 cyclesN = 60n_3L_h = 60(300.56)(10400) = 1.875 \times 10^8~\mathrm{cycles}

Since:

N>106N > 10^6

Shaft 3 is treated as an infinite-life fatigue shaft.

The alternating and mean stress components are:

σa=σb=46.93 MPa\sigma_a = \sigma_b = 46.93~\mathrm{MPa} τm=τ=90.62 MPa\tau_m = \tau = 90.62~\mathrm{MPa} σm=0,τa=0\sigma_m = 0, \qquad \tau_a = 0

The material ultimate tensile strength is:

Sut=633 MPaS_{ut} = 633~\mathrm{MPa}

The uncorrected endurance limit is:

Se=0.5Sut=0.5(633)=316.5 MPaS_e' = 0.5S_{ut} = 0.5(633) = 316.5~\mathrm{MPa}

The corrected endurance limit and fatigue factor of safety are found using the Marin factors and modified Goodman criterion:

Se=kakbkckdkeSeS_e = k_a k_b k_c k_d k_e S_e' Se=(0.817)(0.814)(1)(1)(0.814)(316.5)=171.2 MPaS_e = (0.817)(0.814)(1)(1)(0.814)(316.5) = 171.2~\mathrm{MPa}

The equivalent alternating stress is:

σa,eq=σa=46.93 MPa\sigma_{a,\mathrm{eq}} = \sigma_a = 46.93~\mathrm{MPa}

The equivalent mean stress is:

σm,eq=3τm2=3(90.62)2=156.96 MPa\sigma_{m,\mathrm{eq}} = \sqrt{3\tau_m^2} = \sqrt{3(90.62)^2} = 156.96~\mathrm{MPa}

The modified Goodman fatigue criterion is:

1nf=σa,eqSe+σm,eqSut\frac{1}{n_f} = \frac{\sigma_{a,\mathrm{eq}}}{S_e} + \frac{\sigma_{m,\mathrm{eq}}}{S_{ut}} 1nf=46.93171.2+156.96633=0.522\frac{1}{n_f} = \frac{46.93}{171.2} + \frac{156.96}{633} = 0.522 nf=1.92>1n_f = 1.92 > 1

Therefore, Shaft 3 satisfies the fatigue requirement. Since it carries the highest torque, Shaft 3 is the most critical output shaft, but the selected diameter of 52.5 mm52.5~\mathrm{mm} provides an acceptable fatigue factor of safety.

Appendix D: Gear 3 Calculations

Gear 3 is the final output gear mounted on Shaft 3, so it carries the highest torque in the gearbox. The selected gear parameters are N3=50N_3 = 50, mn=4 mmm_n = 4~\mathrm{mm}, β=30\beta = 30^\circ, ϕn=20\phi_n = 20^\circ, P=70 kWP = 70~\mathrm{kW}, and n3=300.56 rpmn_3 = 300.56~\mathrm{rpm}.

Transmission Ratio

i2=5017=2.941,iT=(2.941)2=8.651i_2 = \frac{50}{17} = 2.941, \qquad i_T = (2.941)^2 = 8.651

Transverse Module and Pitch Diameter

mt=mncosβ=4cos30=4.6188 mmm_t = \frac{m_n}{\cos\beta} = \frac{4}{\cos 30^\circ} = 4.6188~\mathrm{mm} d3=mtN3=(4.6188)(50)=230.94 mmd_3 = m_tN_3 = (4.6188)(50) = 230.94~\mathrm{mm}

Transverse Pressure Angle

ϕt=tan1(tanϕncosβ)=tan1(tan20cos30)=22.80\phi_t = \tan^{-1}\left( \frac{\tan\phi_n}{\cos\beta} \right) = \tan^{-1}\left( \frac{\tan 20^\circ}{\cos 30^\circ} \right) = 22.80^\circ

Torque and Gear Forces

T3=9550Pn3=9550(70)300.56=2224.18 NmT_3 = \frac{9550P}{n_3} = \frac{9550(70)}{300.56} = 2224.18~\mathrm{N\cdot m} Ft=2T3d3=2(2224.18)0.23094=19261.98 NF_t = \frac{2T_3}{d_3} = \frac{2(2224.18)}{0.23094} = 19261.98~\mathrm{N} Fr=Fttanϕt=19261.98tan(22.80)=8095.36 NF_r = F_t\tan\phi_t = 19261.98\tan(22.80^\circ) = 8095.36~\mathrm{N}

Since Gear 3 is a herringbone gear, the axial forces from the two opposite helix directions cancel in the shaft direction:

Fa,net0 NF_{a,\mathrm{net}} \approx 0~\mathrm{N}

Face Width

The selected face width is based on the helical-overlap estimate:

b>2πmnsinβ=2π(4)sin30=50.27 mmb > \frac{2\pi m_n}{\sin\beta} = \frac{2\pi(4)}{\sin 30^\circ} = 50.27~\mathrm{mm}

Therefore, the selected face width is:

b=50.27 mmb = 50.27~\mathrm{mm}

Bending Stress and Factor of Safety

Using the Lewis bending equation with Y=0.408Y = 0.408 for the 50-tooth gear:

σb=FtbmnY=19261.98(50.27)(4)(0.408)=234.81 MPa\sigma_b = \frac{F_t}{bm_nY} = \frac{19261.98}{(50.27)(4)(0.408)} = 234.81~\mathrm{MPa}

Using an allowable bending stress of Sallow=600 MPaS_{\mathrm{allow}} = 600~\mathrm{MPa}, the bending factor of safety is:

nb=Sallowσb=600234.81=2.56n_b = \frac{S_{\mathrm{allow}}}{\sigma_b} = \frac{600}{234.81} = 2.56

Center Distance

The second-stage center distance is:

a=d2+d32=78.52+230.942=154.73 mma = \frac{d_2 + d_3}{2} = \frac{78.52 + 230.94}{2} = 154.73~\mathrm{mm}

Appendix E: Bearing Calculations

The same SKF 32310 J bearing was selected for all bearing locations to simplify part standardization. The required dynamic load rating was checked using:

C=Pb(60nLh106)1/3C = P_b \left( \frac{60nL_h}{10^6} \right)^{1/3}

where PbP_b is the bearing load in kN and:

Lh=36000 hrL_h = 36000~\mathrm{hr}

Shaft 1 Bearings

C1A=C1B=(3.55198)(60(2600)(36000)106)1/3=63.136 kNC_{1A} = C_{1B} = (3.55198) \left( \frac{60(2600)(36000)}{10^6} \right)^{1/3} = 63.136~\mathrm{kN}

Shaft 3 Bearings

C3A=C3B=(10.447)(60(300.56)(36000)106)1/3=90.459 kNC_{3A} = C_{3B} = (10.447) \left( \frac{60(300.56)(36000)}{10^6} \right)^{1/3} = 90.459~\mathrm{kN}

Shaft 2 Critical Bearing

C2B=(15.657)(60(884)(36000)106)1/3=194.242 kNC_{2B} = (15.657) \left( \frac{60(884)(36000)}{10^6} \right)^{1/3} = 194.242~\mathrm{kN}

The largest required bearing rating is therefore:

Cmax=194.242 kNC_{\max} = 194.242~\mathrm{kN}

The selected SKF 32310 J bearing has:

C1=211 kNC_1 = 211~\mathrm{kN}

Since:

C1=211 kN>Cmax=194.242 kNC_1 = 211~\mathrm{kN} > C_{\max} = 194.242~\mathrm{kN}

the selected bearing satisfies the required dynamic load rating for every bearing location in the gearbox. Therefore, the same bearing can be used throughout the design, and no bearing-change service is required.

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