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					                                                                         C. Claur, C. Bylin, T.Pham


                                                                      M.E. 314 Design Project 3

                                                                                        May 17, 2006

                                                 Crank Shaft Design Project
                                                                                     Christopher Claur

                                                                                             Chris Bylin

                                                                                      Thompson Pham




1. DESIGN REQUIREMENTS

         Design an automobile, truck, or motorcycle crankshaft associated with the camshaft you designed
         in Project #2. This requires the consideration of both dynamic loading (fatigue) due to the
         bending and torsion stresses exerted on the crankshaft as well as the design of the main and
         offset (or throw) hydrodynamic journal bearings.


2. BACKGROUND INFORMATION

         The internal combustion engine is based on the four stroke combustion cycle, also known as the
         Otto Cycle. The four strokes of the Otto cycle are the INTAKE, COMPRESSION, POWER, and
         EXHAUST stokes. The cycle continues repeatedly during engine operation. The INTAKE stroke
         draws in air/fuel mixture to the compression chamber while the piston moves downward. As the
         piston moves upward the air/fuel mixture is compressed in the chamber, this is known as the
         COMPRESSION stroke. As the piston advances to the TDC (top dead center) position, the fuel
         mixture is ignited and forces the piston in the downward direction. This is known as the POWER
         stroke. The depleted fuel mixture is pushed out of the chamber in the EXHAUST stroke and the
         cycle repeats.
         The crankshaft has a sequence of main journals that support the crank in the engine block. The
         rod journals are offset from the main centers. The offset form the main journal is half the stroke.
         A 1990 1.6L Civic DX crankshaft with 5 main journals and 4 rod journals was analyzed in this
         report.




Fig. 1 Image of modeled crank




                                              Page 1
                                                                            C. Claur, C. Bylin, T.Pham




                 Arrangement                          4 cylinder, In-line, transverse
                 Displacement                         91 ci.
                 Compression Ratio                    9.1:1
                 Bore Diameter                        2.95 in.
                 Bore Spacing                         3.3 in
                 Stroke                               3.33 in.
                 Rod Length (Center to Center)        5.28 in.
                 Throw                                1.665 in.
                 Redline                              5500 rpm
                 Idle                                 650 rpm
                 Firing Order                         1-3-4-2

Fig. 2 Specifications


     MATERIAL SELECTION

          Nodular Cast Iron 80-55-06

                 Density( ρ )                          6.9 g/m3
                 Weight Density ( γ )                  0.25 lb/in3
                 Ultimate Tensile Strength (Sut)       80 kpsi
                 Tensile Yield Strength (Sy)           53 kpsi
                 Modulus of Elasticity (E)             24.5 MPa
                 Poisson’s Ratio (ν )                  0.30

3. ANALYSIS

     ASSUMPTIONS

          •    At INTAKE, the pressure is NOT equal to atmospheric air pressure because it is a vacuum
               and has less than atmospheric air pressure. INTAKE temperature is above atmospheric
               temperature because of the heat of the engine.
          •    The max temperature is below the adiabatic flame temperature of gasoline.




     CYLINDER PRESSURE

          According to Thermodynamics by Cengel, the adiabatic flame temperature of gasoline is 2395K.
          The engine max temperature we used was based on 90% of this temperature which was 2155.5
          K. According to manufacturing specifications the compression ratio of our motor is 9.2:1. We
          used isentropic ideal gas equations to solve for the pressures and temperatures, with a polytropic
          exponent (n) of 1.3.

                n = 1.3
                K air = 1.3




                                                   Page 2
                                                      C. Claur, C. Bylin, T.Pham


Intake Stroke
    Vacuum = 15.00 Hg = 7.368 psi
     Pint ake = −7.368
     Tint ake = 320.0 K


Compression Stroke (Isentropic Compression)
                                      k
                        V
                                 
                                  
     Pcomp   = Pint ake  int ake 
                         Vcomp 
                                 
                                      k −1
                        P
                                 
                                      k
     Tcomp   = Tint ake  int ake 
                         Pcomp 
                                 

Ignition

pv = RT

vignition = v compression

                          Tmax 
Pignition = Pcompression               
                          Tcompression 
                                       


Power Stroke (Isentropic Expansion)




Exhaust Stroke

     P6 = P1 = 7.35 psi




                                             Page 3
                                                                                  C. Claur, C. Bylin, T.Pham




The following graphs show the pressure and the force of the gas vs. the crankshaft angle.




                                                    Cylinder Temperature

                      2500

                      2000
  Temperature (K)




                      1500

                      1000

                      500

                        0
                             0                180               360                       540                  720
                                                      Crankshaft Rotation (deg)




                                                    Cylinder Pressure

                       500

                       400
    Pressure (psia)




                       300

                       200

                       100

                         0
                             0                180               360                      540                   720
                      -100
                                                     Crankshaft Rotation (deg)



Forces Due to Gas Pressure

                             π
Fgas (θ ) =
                             4
                               (Bore )P(θ )
                                    2



Fc , gas (θ ) = − Fgas cos(φ ) cos(π − φ − θ )




                                                     Page 4
                                                                         C. Claur, C. Bylin, T.Pham



                                         Force of Gas

              3500.000

              3000.000

              2500.000

              2000.000
Force (lbf)




              1500.000

              1000.000

               500.000

                 0.000
                         0   100   200     300        400          500        600      700      800
              -500.000
                                                 Crank Rot (deg)




 Redline RPM and stroke were used to determine the maximum acceleration of the piston and
 connecting rod. Below are the equations we used for the motion of the piston.




                                         Page 5
                                                                   C. Claur, C. Bylin, T.Pham


Displacement

          a × sin ω 2 − c 
φ = sin −1                
                 b        
d = a × cos θ 2 − b × cos θ 3

Velocity
                         π 
                sin θ ×     
       −1               180    180 
β = sin l1 ×                      ×
                      l2         π 
                                       
                               
                               
             sin(θ + β )
VD = VB ×
            sin(90 − β )
        V  sin(90 − θ )
ω BD =  D 
        l  sin(90 − β )
        2 

Acceleration

aB = ω AB r
            2


aBD = l2ω BD
                2


           − a B sin(θ ) − (a BD ) sin( β )
α D/ B =                                    = (rad / s 2 )
                    l2 cos( β )
ad = − aB cos(θ ) − a BD cos( β ) + l2α D / B sin( β )

Force Analysis



  Component                   Weight (oz)          Weight (lb)   Mass (slug)
 Connecting Rod                  11.0                .688          .0214
    Wrist Pin                    2.46                .154          .0048
     Piston                      11.1                .694          .0215
     Throw                       36.0                2.25          .0698




                                           Page 6
                                                                                    C. Claur, C. Bylin, T.Pham




The following equations were used to determine the force acting on the cylinders. These, combined with
Fgas , resulted in Ftotal acting on the crankshaft.

                   m piston × a B
Fpiston =
                        gc
Fc, piston = Fpiston × r × cos(φ ) cos(π − φ − θ )
                   mrod × an
Fc,rod =
                      gc
Ftotal = Fc , piston + Fc ,rod + Fcentrifugal + Fc , gas




                                                           Ftotal, cylinder

                 7000.000

                 6000.000

                 5000.000
   Force (lbf)




                 4000.000


                 3000.000

                 2000.000

                 1000.000

                    0.000
                             0           180                   360            540              720               900
                                                                Crank Rot (deg)




                                                           Page 7
                                                                      C. Claur, C. Bylin, T.Pham



       Using the geometry of the crankshaft, we found the max and min moments to be at the fillet. This,
combined with Ftotal , we were able to determine the bending stress, σbend.

                  πd 4
             I=        = .329in 4
                 64
            C = .805
            M max (θ ) = 3878inlb
            M min (θ ) = 242inlb



        Stress Concentration Factors For Bending
            D
                = 1.1
            d
            A = .971
            b = −.218
            FilletRadius = .06"
                         b
                    r
            K t ≅ A  = 2.04
                    d 
                    1
            q=           = .754
                      a
                 1+
                      r
            K f = 1 + q(K t − 1) = 1.78
             K f σ max ≤ S y ⇒ K f = K fm
             K fm = K f = 1.78

        Bending Stress
                        M (θ )C
        σ bend (θ ) = K f
                           I
        σ bend max= 19.36kpsi
        σ bend min = 1.21kpsi




                                            Page 8
                                                                         C. Claur, C. Bylin, T.Pham


Next, we used throw offset in conjunction with Ftotal to determine the torque exerted on the crankshaft.
This in turn, was used to find the torsional shear, τ (θ).

        Torsional Shear

        Tmax = 8556in − lb
        Tmin = −7726in − lb
               πd 4
        J=              = .658
                32
                                r
        τ torsion max = M (θ )    = 21.637 kpsi
                                J
                                r
        τ torsion min   = M (θ ) = −19.539kpsi
                                J




        Stress Concentration Factors for Torsion

        D
          = 1.10
        d
        A = .903
        b = −.127
                      b
                 r
        K ts ≅ A  = 1.36
                 d 
        K fs = 1 + q(K ts − 1) = 1.38
        K fs σ max ≤ S y ⇒ K fs = K fsm
        K fs = K fsm = 1.38




                                                  Page 9
                                                                                         C. Claur, C. Bylin, T.Pham




                                                                 T total and shear

                                   25000.000

                                   20000.000

                                   15000.000
T (in lb) shear (psi)




                                   10000.000

                                    5000.000
                                                                                                                 T(theta)
                                        0.000
                                                 0    180       360          540         720        900          torsional shear
                                    -5000.000
                                                                                                                 stress(theta)
                                  -10000.000

                                  -15000.000

                                  -20000.000

                                  -25000.000
                                                             Crank Rotation (deg)




                                                     Sigma bend and torsional shear stress

                                  25000.000
                                  20000.000
 bending and shear stress (psi)




                                  15000.000
                                  10000.000
                                   5000.000
                                                                                                             torsional shear stress
                                       0.000
                                                                                                             sigma bend
                                   -5000.000 0         200             400         600             800
                                  -10000.000
                                  -15000.000
                                  -20000.000
                                  -25000.000
                                                               Crank Rot (deg)




                                                             Page 10
                                                                                          C. Claur, C. Bylin, T.Pham


Von Mises Stresses



               σ ' max =   (σ   2
                                bend _ max                       )
                                             + 3τ total _ max = 19,361 psi = 19.31ksi

               σ ' min =   (σ   2
                                bend _ min   + 3τ total _ min   ) = 1853 psi = 1.85ksi
                      σ ' max −σ ' min
               σa =              = 8.75kpsi
                         2
                    σ ' +σ ' min
               σ m = max         = 10.58kpsi
                         2
                           Main Journal Diameter                                         1.77 in.
                           Main Journal Length                                           0.90 in.
                           Rod Journal Diameter                                          1.49 in.
                           Rod Journal Length                                            0.86 in.
                           Throw Diameter                                                3.33in
                           Fillet Radii                                                  0.06 in.
                           Bearing Clearance                                             .002 in.

    FATIGUE SAFETY FACTOR

          For 80-55-06 nodular cast iron, the ultimate stress (Sut ) is 53 kpsi and the yield stress (Sy) is 82
          kpsi.

σ m = σ ' m = 8.75ksi
σ a = σ ' a = 10.58 psi
K t = 2.04
K f = 1.78
K f σ max ≤ S y ∴ K fm = K f
S y = 53kpsi
S ut = 82kpsi
         1
S 'e =     S ut = 41kpsi
         2
C load   =1
C size = 0.84
C surf = 0.84
C temp = 1
C reliab = .702
S e = 20.3kpsi
                 S e S ut
Nf =                            = 1.6
         σ ' a S ut + σ ' m S e



                                                                Page 11
                                                        C. Claur, C. Bylin, T.Pham



HYDRODYNAMIC JOURNAL BEARING ANALYSIS

        Main Journal Diameter                1.77 in.
        Main Journal Length                  0.90 in.
        Rod Journal Diameter                 1.49 in.
        Rod Journal Length                   0.86 in.
        Bearing Clearance                    .002 in.


  Design of Main Journal Bearing

  SAE 30W @ 120°F

   Pmax = 6357lbs
                     1           rev
   n' redline = 6000       = 100
                     60 s         s
                  1           rev
   n'idle = 700        = 11.6
                  60s          s
   ON = 30
   d = 1.77in.
   l = .90in.
  η = 7.1 × 10 −6 µ − reyns
            Pmax Pmax
   Pave =       =     = 3990 psi
             A    ld
                      2       2
          P  d   c 
   ON =  ave    d 
          ηn'  l
                 d 
   Cd
       = .001in.
    d
   C d = 2C r
   C r = .001in.




                                         Page 12
                                                          C. Claur, C. Bylin, T.Pham



    Design of Rod Journal Bearing
         Pmax = 6357lbs


                  1            rev
n' redline = 6000        = 100
                  60s           s
               1            rev
n' idle = 700        = 11.6
               60 s          s
O N = 30
d = 1.49in.
l = .86in.
η = 7.1E − 6 µ − reyns
       P      P
Pave = max = max = 331 psi
         A     ld
                    2       2
       P  d   c 
O N =  ave    d  ∴ c d = .007in.
       ηn'  l
              d 
Cd
    = .004in.
 d
C d = 2C r
C r = .0035in.


ε = 0.93( Fig.10 − 10)
                      C 
hmin = C r (1 − ε ) =  d (1 − ε ) = 70 µin.
                       2 
      h         70 µin
Req ≤ min =             = 17.5µin.
       Λ           4




                                                Page 13
                                                                       C. Claur, C. Bylin, T.Pham


4. REFERENCES
Bennett, Dick. "Flame Temperature: What Is It? ." 01 2002. Process Heating. 07 Dec. 2005
<http://www.process-heating.com/CDA/ArticleInformation/Energy_Notes_Item/0,3271,82221,00.html>.

"Component Driveline Design for a Formula SAE Car." 2003. ACTA Press. 05 Dec. 2005
<http://www.actapress.com/PaperInfo.aspx?PaperID=13113>.

"Crank-Slider Slider Position Analysis." 12 Dec. 2005
<http://www.softintegration.com/chhtml/toolkit/mechanism/crankslider/SliderPos.html>.

"Digital Library." Generic Crank Slider. Kinematics Models for Design. 8 Dec. 2005
<http://kmoddl.library.cornell.edu/tutorials/01/p1.php>.

"Engine Formula." 4-Stroke. Engineers Edge. 9 Dec. 2005
<http://www.engineersedge.com/engine_formula_automotive.htm>.

"Honda non-VTEC 1.6L SOHC D16A6." 1988-1991. Crane Cams. 14 Nov. 2005
<http://www.cranecams.com/index.php?show=browseParts&lvl=4&prt=2032>.

Norton, Robert L. Machine Design, an Integrated Approach. 2nd ed. : Worcester Polutechnic Institute,
2000.

Rollins, Mike. "What is the Speed of a Piston with an offset crankshaft?." 9 Dec. 2005
<http://www.wfu.edu/~rollins/piston/offset/>.

"Slider Crank Model." thermodynamics. 9 Dec. 2005 <http://succ.shirazu.ac.ir/~motor/page2t.htm>.




                                             Page 14

				
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