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CALCULATIONS AND ANALYSIS See Stress Calculation Spreadsheet for sources of equations, sources of constants and material properties, and additional calculations Impact Analysis Direct wheel impact at max speed Fl 3 By using the deflection equation, s (based upon two fully constrained 192 EI rod ends), solving for F, and using a basic kinematic equation ( v f vo 2a(s) 2 2 ) to solve for s in terms of F, the force of impact can be determined (227505 N) Utilizing shaft stress equations shown below the stress can be determined (400 MPa) When comparing this to the shaft’s yield strength, a factor of safety of 1.33 is calculated Direct pulley impact at max speed Utilizing this same force and finding the stress on the shaft due to bending. Mc =8510 MPa I This means the shaft will permanently bend due to the moment applied on it The way to avoid this catastrophic failure is to ensure the chassis protects these open gears by extending past its edges or enclosing it completely. While this may not completely ensure the module’s safety, it will fix nearly every probable scenario. Shaft Stress Calculations Shaft 1 (Diameter=3/8”) Material: 1045 Steel, Yield Strength (Sy)= 530 MPa, Ultimate Strength= 625MPa Max Stress o The shaft is keyed for a 3/32” key, thus a close approximation for the actual yield strength is ¾ the materials yield strength (Keyed Yield Strength=398 MPa) o Loading is comprised of three components Moment-Based on cantilevered distance from bearing and radial load exerted on shaft from the miter gear (2.1 N-m) Force- Based on axial load exerted on shaft from miter gear (156.12 N) Torque- Exerted by the stall torque of the motor, through a gear ratio of 2:1 (9.64 N-m) o Stress Calculation- 4 max [(8M Fd ) 2 48T 2 ]1/ 2 =102 MPa d 2 max [(8M Fd ) 2 64T 2 ]1/ 2 = 58.4 MPa d 3 o Factors of Safety- Sy n = 3.9 max Sy n = 3.4 2 max Fatigue Life o Infinite Life- 2000RPM (Average operating speed)=33.3 cycles/second 5 year life @ 1 hour operating time (2 hr per week)-approximately 1,908,000 seconds of use 33.3*1,908,000=6.4E7 cycles to failure for infinite life o The endurance strength can be calculated using the stress concentration factors from the keyway (197 MPa) o ’F=Sut+345MPa= 970 MPa log( ' F / S e ) o b =-0.109915548 log( 2 N e ) ' o f F (2 103 ) b =.673 Sut ( f S ut ) 2 o a =900 MPa Se o Loads are based on typical operating conditions, not max conditions Moment-Based on cantilevered distance from bearing and radial load exerted on shaft from the miter gear (2.1 N-m) Force- Based on axial load exerted on shaft from miter gear (156.12 N) Torque- Exerted by the operating torque of the motor, through a gear ratio of 2:1 (2.82 N-m) 4 o a [(8M Fd ) 2 48T 2 ]1/ 2 = 39.4 MPa d 3 1 a b o N = 2.25E12 cycles to failure a Shaft 2 (Diameter=1/2”) Material: 1045 Steel, Yield Strength= 530 MPa, Ultimate Strength= 625MPa Max Stress o The shaft is keyed for a 1/8” key, thus the actual yield strength can be equated to ¾ the materials yield strength (Keyed Yield Strength=398 MPa) o Loading is comprised of three components Moment-Based on the axle length between bearings and radial load exerted on shaft from the miter gear (4.28 N-m) Force- Based on axial load exerted on shaft from miter gear (156.12 N) Torque- Exerted by the stall torque of the motor, through a gear ratio of 2:1 (9.64 N-m) o Stress Calculation- 4 max [(8M Fd ) 2 48T 2 ]1/ 2 = 47.2 MPa d 2 max [(8M Fd ) 2 64T 2 ]1/ 2 = 26.5 MPa d 3 o Factors of Safety- Sy n = 8.4 max Sy n = 7.5 2 max Fatigue Life o Infinite Life- 1000RPM=16.67 cycles/second 5 year life @ 1 hour operating time (2 hr per week)-approximately 1,908,000 seconds of use 16.67*1,908,000=3.2E7 cycles to failure for infinite life o The endurance strength can be calculated using the stress concentration factors from the keyway (197 MPa) o ’F=Sut+345MPa= 970 MPa log( ' F / S e ) o b =-0.109915548 log( 2 N e ) ' o f F (2 103 ) b =.673 Sut ( f S ut ) 2 o a =900 MPa Se o Loads are based on typical operating conditions, not max conditions Moment-Based on the axle length between bearings and radial load exerted on shaft from the miter gear (4.28 N-m) Force- Based on axial load exerted on shaft from miter gear (156.12 N) Torque- Exerted by the operating torque of the motor, through a gear ratio of 2:1 (2.82 N-m) 4 o a [(8M Fd ) 2 48T 2 ]1/ 2 = 25.6 MPa d 3 1 b o N a = 1.15E14 cycles to failure a Shaft 3 (Diameter=3/4”) Material: 1045 Steel, Yield Strength= 530 MPa, Ultimate Strength= 625MPa Max Stress o The shaft is keyed for a 3/16” key, thus the actual yield strength can be equated to ¾ the materials yield strength (Keyed Yield Strength=398 MPa) o Loading is comprised of three components Moment-Based on the axle length between bearings and the force exerted by the weight of the system (21.53 N-m) Force- Based on axial load exerted on the shaft from turning forces (235.44 N) Torque- Exerted by the stall torque of the motor, through a gear ratio of 8:1 (38.56 N-m) o Stress Calculation- 4 max [(8M Fd ) 2 48T 2 ]1/ 2 = 59.0 MPa d 2 max [(8M Fd ) 2 64T 2 ]1/ 2 = 32.7 MPa d 3 o Factors of Safety- Sy n = 6.7 max Sy n = 6.1 2 max Fatigue Life o Infinite Life- 500RPM=8.34 cycles/second 5 year life @ 1 hour operating time (2 hr per week)-apprx 1,908,000 seconds of use 8.34*1,908,000=1.6E7 cycles to failure for infinite life o The endurance strength can be calculated using the stress concentration factors from the keyway (197 MPa) o ’F=Sut+345MPa= 970 MPa log( ' F / S e ) o b =-0.109915548 log( 2 N e ) ' o f F (2 103 ) b =.673 Sut ( f S ut ) 2 o a =900 MPa Se o Loads are based on typical operating conditions, not max conditions Moment-Based on the axle length between bearings and the force exerted by the weight of the system (21.53 N-m) Force- Based on axial load exerted on the shaft from turning forces (235.4 N) Torque- Exerted by the operating torque of the motor, through a gear ratio of 8:1 (11.28 N-m) 4 o a [(8M Fd ) 2 48T 2 ]1/ 2 = 35.6 MPa d 3 1 a b o N = 5.7E12 cycles to failure a Steering Shaft (Diameter=1/4”) Material: 303 Stainless Steel, Yield Strength= 240 MPa, Ultimate Strength= 620 MPa Max Stress o Loading is based on torque alone (0.745 N-m) o Stress Calculation- 4 max [(8M Fd ) 2 48T 2 ]1/ 2 = 25.7 MPa d 2 max [(8M Fd ) 2 64T 2 ]1/ 2 = 14.8 MPa d 3 o Factors of Safety- Sy n = 9.4 max Sy n = 8.1 2 max Fatigue Life o ’F=Sut+345MPa=965E6 MPa log( ' F / S e ) o b =-0.07772 log( 2 N e ) ' o f F (2 103 ) b =.862 Sut ( f S ut ) 2 o a =914 MPa Se o Load is comprised of torque alone (.745 N-m) 4 o a [(8M Fd ) 2 48T 2 ]1/ 2 = 25.7 MPa d 3 1 b o N a = 9.3E19 cycles to failure a Spur Gears (Calculated using ANSI standards) Driving Spur Material- Carbon Steel, Yield Strength=76900 psi, Modulus of Elasticity=30E6 psi, Poisson’s Ratio=.29, Brunell Hardness 179 Max Bending Stress 33000 H o Wt = 306.8 lbf V o Ko= 1.25 - Overload Factor, based on light shocks encountered o Kv= 1.15 - Dynamic Factor, based on quality and velocity of gears o Ks= 1 - Size Factor o Pd= .833” – Pitch diameter o F= .25” – face width o Km= 1.20 – Load-Distribution factor, based on geometry o KB= 1 – Rim Thickness factor, based on geometry o J= .325- Geometry factor, based on number of teeth of gears P K K o W t K o K v K s d m B =5357.1 psi F J Sy o n = 9.0 max Endurance Stress 1/ 2 1 o Cp =2284.7 lbf/in2 2 (1 v p 1 vG ) 2 Ep EG o Cf=1 o I=0.08- Geometry Factor K C f 1/ 2 o C p (W t K o K s m ) =56972.4 psi Pd F I o Sc= 180000 psi- Repeatedly applied contact strength @ 107 cycles, material property o Zn=.59 - Stress cycle life factor, based on hardness and number of cycles o CH=1 -Hardness ratio factor o KT= 1- Temperature factor o KR= 1 – Reliability factor S Z C /( KT K R ) o SH c N H =1.9 o Comparable factor of safety= SH2=3.5 Driven Spur Material- Carbon Steel, Yield Strength=76900 psi, Modulus of Elasticity=30E6 psi, Poisson’s Ratio=.29, Brunell Hardness 179 Max Bending Stress 33000 H o Wt = 306.7 lbf V o Ko= 1.25 - Overload Factor, based on light shocks encountered o Kv= 1.15 - Dynamic Factor, based on quality and velocity of gears o Ks= 1 - Size Factor o Pd= 1.667” – Pitch diameter o F= .25” – face width o Km= 1.19 – Load-Distribution factor, based on geometry o KB= 1 – Rim Thickness factor, based on geometry o J= .389- Geometry factor, based on number of teeth of gears P K K o W t K o K v K s d m B =9011.0 psi F J Sy o n = 5.4 max Endurance Stress 1/ 2 1 o Cp =2284.7 lbf/in2 2 (1 v p 1 vG ) 2 Ep EG o Cf=1 o I=0.08- Geometry Factor K C o C p (W t K o K s m f )1/ 2 = 40010.7 psi Pd F I o Sc= 180000 psi- Repeatedly applied contact strength @ 107 cycles, material property o Zn=.60 - Stress cycle life factor, based on hardness and number of cycles o CH=1 -Hardness ratio factor o KT= 1- Temperature factor o KR= 1 – Reliability factor S Z C /( KT K R ) o SH c N H = 2.70 o Comparable factor of safety= SH2=7.2 Ring/Pinion Gears (Calculated using ANSI standards) Steering Spur Material- 2024-T4 Aluminum, Yield Strength=47000 psi, Modulus of Elasticity=10.4E6 psi, Poisson’s Ratio=.333 Max Bending Stress 33000 H o Wt = 39.3 lbf V o Ko= 1.25 - Overload Factor, based on light shocks encountered o Kv= 1.10 - Dynamic Factor, based on quality and velocity of gears o Ks= 1 - Size Factor o Pd= .4375” – Pitch diameter o F= .125” – face width o Km= 1.20 – Load-Distribution factor, based on geometry o KB= 1 – Rim Thickness factor, based on geometry o J= .24- Geometry factor, based on number of teeth of gears P K K o W t K o K v K s d m B =951.8 psi F J Sy o n = 44.1 max Steering Ring Material- 2024-T4 Aluminum, Yield Strength=47000 psi, Modulus of Elasticity=10.4E6 psi, Poisson’s Ratio=.333 Max Bending Stress 33000 H o Wt = 39.3 lbf V o Ko= 1.25 - Overload Factor, based on light shocks encountered o Kv= 1.10 - Dynamic Factor, based on quality and velocity of gears o Ks= 1 - Size Factor o Pd= 3.125” – Pitch diameter o F= .125” – face width o Km= 1.8 – Load-Distribution factor, based on geometry o KB= 1 – Rim Thickness factor, based on geometry o J= .4- Geometry factor, based on number of teeth of gears Pd K m K B o W t Ko Kv K s = 3996.0 psi F J Sy o n = 10.5 max Miter Gears (Calculated using ANSI standards) Both Miters (At max torque) Material- Medium Carbon Steel, Yield Strength=76900 psi Max Bending Stress o Pd= 1.25” – Pitch diameter o 2 T o Wt = 84.7 lbf Pd o Ko= 1.25 - Overload Factor, based on light shocks encountered o Kv= 1 - Dynamic Factor, based on quality and velocity of gears o Ks= .5 - Size Factor o F= .27” – face width o Km= 1.10 – Load-Distribution factor, based on geometry o J= 0.175- Geometry factor, based on number of teeth of gears o Kx= 1, Lengthwise curvature factor Wt K K o Pd K o K v s m =14922.2 psi F KxJ Sy o n = 5.15 max Both Miters (At max speed) Material- Medium Carbon Steel, Yield Strength=76900 psi Max Bending Stress o Pd= 1.25” – Pitch diameter o 2 T o Wt = 4 lbf Pd o Ko= 1.25 - Overload Factor, based on light shocks encountered o Kv= 1.28 - Dynamic Factor, based on quality and velocity of gears o Ks= .5 - Size Factor o F= .27” – face width o Km= 1.10 – Load-Distribution factor, based on geometry o J= 0.175- Geometry factor, based on number of teeth of gears o Kx= 1, Lengthwise curvature factor Wt K K o Pd K o K v s m =901.6 psi F KxJ S o n y = 85.3 max Forces o Knowing max torque on miter (9.63 N-m), we can find the max tangential force by dividing by half the pitch diameter=> Ftan=606.6 N o = 20 degress -pressure angle o d= 45 degrees F o Fn tan =645.5 N cos F o F1 n =220.8 N sin o Faxial Fradial F1 sin d =156.1 N Retaining Rings On 3/8” shaft o Ring can withstand 542.7 N of axial force o Miter gear provides axial load= 156.1 N o Factor of safety= 3.48 On 1/2” shaft o Ring can withstand 542.7 N of axial force o Miter gear provides axial load= 156.1 N o Factor of safety= 3.48 On 3/4” shaft o Ring can withstand 631.6 N of axial force o Axial load is from turning Assume wheel instantaneously turns 90 degrees, the max force that can be applied axially would be equivalent to the frictional force Faxial Ffriction W =235.4 N (assuming =.6) o Factor of safety= 2.68 Mechanical Brake Max Temperature Assuming all kinetic energy is converted directly into heat energy, 1 / 2mvehvveh m plateCv T 2 Assume emergency brake will not be used continuously, but rather for one cycle during the emergency Assume initial temperature of 23 ° Celcius Solving the above equation for Tfinal we find it to be 38.9 °Celcius Heat Dissipation Assuming Free Convection, the time required for heat dissipation can be calculated Utilizing the properties of air at room temperature, the Rayleigh number, Nusselt number, and convection heat transfer coefficient can be calculated Using this information the heat transfer rate is determined E t gives the time to dissipate the heat (4.8 minutes) q This resultant was later verified by the manufacturer of the brake Keys On 3/8” Shaft Key is High carbon steel, Yield Strength 427 MPa, 3/32” square Knowing the diameter of and torque on the shaft, the shear force on the key can be calculated (2024.1 N) Assuming a factor of safety of 4, the required length of the key is calculated (.63”) On 1/2” Shaft Key is High carbon steel, Yield Strength 427 MPa, 1/8” square Knowing the diameter of and torque on the shaft, the shear force on the key can be calculated (1518.1 N) Assuming a factor of safety of 4, the required length of the key is calculated (.35”) On 3/4” Shaft Key is High carbon steel, Yield Strength 427 MPa, 3/16” square Knowing the diameter of and torque on the shaft, the shear force on the key can be calculated (4048.3 N) Assuming a factor of safety of 4, the required length of the key is calculated (.63”) Keyways Keyway analysis was done using Cosmos FEA software By utilizing shaft diameters and torques, forces on keyway surfaces were calculated and input into the program Factor of Safety o Driving Miter=16 o Driven Miter=15 o Driving Pulley=3.4 o Driven Pulley=8.4 o Driven Spur=8.1 o Wheel=1.8, but in reality, failure would result in the slip of a pressed insert, rather than physical failure of the wheel Set Screws To connect spur gear to 5/16” drive motor shaft By choosing a screw size and quantity (2- #8’s), the maximum force at the shaft surface can be calculated (3425.1 N) The torque and diameter of the shaft is used to determine the actual force seen at this shaft surface (1214.5 N) By comparing these two values the factor of safety is determined (2.82) To connect spur gear to 8mm steering motor shaft By choosing a screw size and quantity (2-#6’s), the maximum force at the shaft surface can be calculated (2224.1 N) The torque and diameter of the shaft is used to determine the actual force seen at this shaft surface (234.6 N) By comparing these two values the factor of safety is determined (9.5) Timing Belt and Pulleys Utilizing MITCalc simulation software and inputting various parameters including distance between centers, power applied to belt, operating speeds, and other operating conditions a belt type and specific model was selected From this the 5M Powergrip GT2 belt was chosen and matched with pulleys of 18 and 72 teeth The selection of these parts was also verified with an engineer at the supplier sdp-si.com Bearings C10 = Catalog Load Rating (lbf) LR = Rating Life (hrs) nR = Rating Speed (RPM) FD = Desired Radial Load (lbf) LD = Desired Life (hrs) nD = Desired Speed (RPM) FR = Radial Force (lbf) FA = Axial Force (lbf) Fe = Equivalent Radial Load (lbf) C0 = Static Load Rating (lbf) X2 = Factor dependent on bearing geometry Y2 = Factor dependent on bearing geometry V = Rotation Factor a = 3 (ball bearing) e = abscissa Lower Drive Bearing C10 (lbs) 1171 Upper Drive Bearing Center Bearing LR * nR 1.0E+06 C10 (lbs) 1187 C10 (lbs) 691 FD (lbs) 44.125 LR * nR 1.0E+06 LR * nR 1.0E+06 nD (RPM) 5.0E+02 Fe (lbs) 84.6 Fe (lbs) 70.6 a 3 nD (RPM) 2000 nD (RPM) 4000 LD (hrs) 3.74E+07 FA (lbs) 35.1 FA (lbs) 35.1 LD (years) 4267 FR (lbs) 35.1 FR (lbs) 35.1 a 3 a 3 Steering Bearing V 1 V 1 C10 (lbs) 300 e 0.24 e 0.3 LR * nR 1.0E+06 X2 0.56 X2 0.56 FD (lbs) 10 Y2 1.85 Y2 1.45 nD (RPM) 340 LD (hrs) 1.38E+06 LD (hrs) 2.34E+05 a 3 LD (years) 158 LD (years) 27 LD (hrs) 7.94E+07 a LD (years) 9065 C LR * nR LD 10 F n D D Fe X 2VFR Y2 FA Screws Bolts connecting Yoke to Turntable Bolt type: 4 * 10-32 (SAE) Torque applied to the turntable T= 2.38 N-m Converted ASTM T = 21 lb-in Resultant load on each bolt T 1 V 21 lb in * 8.108 lb r 2.59 in M T 21 lb in Primary Shear Load per Bolt is V 8.108 F' 2.027 lb n 4 Since the secondary shear Forces are equal we have Mr M 21 F'' 2 2.027 lb 4r 4r 4 * 2.59 The resultant force is Fr 2.867 lb Fr = Fa = Fb = Fc = Fd = 2.867lb Maximum Shear Stress As = 0.155 Fr 2.867 18.497 psi As .155 Bolts connecting Brake to Brake plate Bolt type: 4 * 8-32 (SAE) Torque applied to the turntable T = 15 lb-in Resultant load on each bolt T 1 V 15 lb in * 13.33 lb r 1.125 in M T 15 lb in Primary Shear Load per Bolt is V 13.33 F' 3.33 lb n 4 Since the secondary shear Forces are equal we have Mr M 15 F'' 2 3.33 lb 4r 4r 4 *1.125 The resultant force is Fr 4.714 lb Fr = Fa = Fb = Fc = Fd = 4.714 lb Maximum Shear Stress As = .0992 Fr 2.867 28.9 psi As .0992 Yoke Yoke stress analysis was done using Cosmos FEA software Loading o Weight vertically loads lower bearing holes (196.2 N each) o Turning force loads inside wall (158.3 N) o Driven Miter axial force loads inside wall (156 N) o Driven Miter radial force loads upper bearing holes o Driving Miter axial force loads upward on top plate Minimum factor of safety= 20 Brake Plate Brake plate stress analysis was done using Cosmos FEA software Loading o Outside edge was fixed, as it is welded to the motor mount assembly o Each brake mounting hole was loaded with a force corresponding to the brake’s torque output and the holes distance from center Minimum factor of safety= 200 Turntable Capable of withstanding 750 lbs, or 340 kg Actual weight is about 40 kg per module Factor of Safety= 8.5 Figure X: Shaft Stress and Fatigue Strength Calculations Figure X: Drive Motor Spur Gear Stress Calculation Figure X: Miter Gear Stress and Force Calculation Figure X: Retaining Ring Calculations Figure X: Impact Calculations Figure X: Brake Temperature and Heat Dissipation Calculations Figure X: Key and Set Screw Analysis Figure X: Timing Belt and Pulley Analysis Figure X: Driven Miter Gear Keyway Analysis Figure X: Driving Miter Gear Keyway Analysis Figure X: Driven Pulley Keyway Analysis Figure X: Driving Pulley Keyway Analysis (Representation) Figure X: Driven Spur Gear Keyway Analysis Figure X: Wheel Keyway Analysis (Representation) Note: Actual wheel utilizes press fit keyway insert, thus failure during stall will result in slip of this insert, rather than the physical failure of a mechanical part Figure X: Yoke Displacement

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posted: | 11/28/2011 |

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