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Stress Analysis

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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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