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DYNAMICS - CASE STUDY SOLUTION

   

Begin with a diagram at the instant the crankshaft is in the horizontal plane.

Since the principal axes of inertia are not readily known for either the center of gravity or point O, let the reference frame xyz be located at point O as shown,

     


Problem Geometry

     
   

The crankshaft only rotates about the z-axis,

     ωx = ωy = 0,       ωz = ω = 104 rad/s
     

The angular equations (Eqs. 1 in the theory section) simplify to

    ΣMx = Iyz ωz2 - Ixz αz2
    ΣMy = -Iyz αz2 - Ixz ωz2
    ΣMx = Izz αz2

The moments are due to forces F1, F2, Ay and Ax,

     -Ay (5L) + F2 (3.5L) + 9mg (2.5L) + F1 (1.5L)
                 = Iyz ωz2 - Ixz αz2
     Ax (5L) = -Iyz αz2 - Ixz ωz2
     F1 L - F2 L = Izz αz2

These equations can be rearranged in terms of the three unknowns Ax, Ay, and αz, giving

     

     


Moment and Product of Inertia
for a Slender Rod


Crankshaft split into Nine Sections


Rotating Reference Axis

 

To solve these three equations, the inertia terms, Izz, Iyz, and Ixz, need to be determined. To find the inertia terms, split the shaft into 9 straight slender rods. The total moment of inertia and product of inertia terms can be found by summing the inertia of all sections about themselves, and using the parallel axis theorem. This gives,

     Izz = Izz-1 + Izz-2 + Izz-3 + Izz-4 +
              Izz-5 + Izz-6 + Izz-7 + Izz-8 + Izz-9
         = 0 + mL2/3 + (0 + mL2) + mL2/3 +
              0 + mL2/3 + (0 + mL2) + mL2/3 + 0
         = 3.333 mL2
         = 3.333 (0.308 kg) (0.100)2
         = 0.01027 kg-m2

     Iyz = 0 (all sections in x-z plane at given instant)

     Ixz = 0 + (-L/2)(L)m + (-L)(1.5L)m + (-L/2)(2L)m +
              0 + (L/2)(3L)m + (L)(3.5L)m + (L/2)(4L)m + 0
         = 4 mL2
         = 4 (0.308 kg) (0.100)2
         = 0.01232 kg-m2

Substituting the inertia terms and known forces, F1 = 100 N, and F2 = 65 N, gives

  

     
   
 
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