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A Torque of T_c = 460 N-m actions on gear C of the assembly shown below. 
Shafts (1) and (2) are solid 35 mm diameter alumin shafts, and shaft (3) is a 25 mm diameter solid shaft. 
Assume that L = 200 mm and G = 28 GPa. 

Determine:
        A) The maximum shear stress magnitude in shaft (1)
        B) The maximum shear stress magnitude in shaft (3)
        C) The rotation angle of gear E.
        D) The rotation angle of gear C.


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Added Thu, 11 Jun '15
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Section Properties: 
The polar moments of inertia for the shafts are :
J_1 = J_2 = Pi/32 * (35mm)^4 = 147,323.5 mm^4
J_3 = Pi/32*(25 mm)^4 = 38,349.5 mm^4

Equilibrium:
Consider a free-body diagram cut through shaft (2) and around gear C:
M_x = -T_2 + T_c = 0
>> T_2 = T_c = 460 N-m ----- (eq. a)

Next, consider a free-body diagram cut around gear B through shafts (1) and (2). 
The teeth of the gear E exert a force F on the teeth of gear B. This force F opposes the rotation of gear B. The radius of gear B will be denoted by R_b for now (even though the gear radius is not given explicitly).
M_x = T_2 - T_1 - F * R_b = 0 ----- (eq. b)



Finally, consider a Free Body Diagram cut around gear E through shaft (3).
The teeth of gear B exert an equal magnitude force F on the teeth of gear C, acting opposite to the direction assumed in the previous Free Body Diagram. The radius of gear E will be denoted by R_e for now.
M_x = -T_3 - F * R_e = 0
F = -(T_3 / R_e) ----- (eq. C)

The results of Equations (a) and (c) can be substituted into Equation (b) to give :
T_1 = 460 N-m + T_3 * R_b / R_e

The ration R_b / R_e is simply the gear ratio between gears B and E, which can also be expressed in terms of gear teeth N_b and N_e:
T_1 = 460 N-m + T_3 * R_b / R_e 
= 460 N-m + T_3(54 teeth / 42 teeth)
= 460 N-m + 1.285714 * T_3 ----- (eq. D)

Equation (d) summarizes the results of the equilibrium considerations, but there are still two unknowns in this equation: T_1 and T_3. 

Consequently, this problem is statically indeterminate. To solve this problem, and additional equation must be developed. 

This second equation will be derived from the relationship between the angles of twist in shafts (1) and (3).

Geometry of Deformation Relationship:
The rotation of gear B is equal to the angle of twist in shaft (1):
    Phi_b = Phi_1
  
and the rotation of gear E is equal to the angle of twist in shaft (3):
    Phi_e = Phi_3 
   

However, since the gear teeth mesh, the rotation angles for gears B and E are not independent. The arclengths associated with the respective rotations must be equal, but the gears turn in opposite directions. 
The relationship between the gear rotations can be stated as :
R_b * Phi_b = - R_e * Phi_e

Where R_b and R_e are the radii of gears B and E, respectively.
Since the gear rotation angles are related to the shaft angles of twist, this relationship can be expressed as:
R_b * Phi_1 = - R_e * Phi_3

Torque-Twist Relationships:
Phi_1 = (T_1 * L_1) / ( J_1 * G_1)
Phi_3 = (T_3 * L_3) / ( J_3 * G_3)

Compatibility Equation:
Substitute the torque-twist relationships into the geometry of deformation relationship to obtain:
R_b * (T_1 * L_1) / ( J_1 * G_1) = -R_e * (T_3 * L_3) / ( J_3 * G_3)

Which can be rearranged and expressed in terms of the gear ratio N_b/N_e:

(N_b / N_e) *  (T_1 * L_1) / ( J_1 * G_1)  =  (T_3 * L_3) / ( J_3 * G_3)

Note that the compatibility equation has two unknowns : T_1 and T_3. 
This equation can be solved simultaneously with Equation (d) to calculate the internal torques in shafts (1) and (3).

Solve the Equations:
Solve for internal torque T_3 in the compatibility equation.
T_3 = -T_1 * (N_b/N_e) * (L_1/L_3) * ( J_3 / J_1) * (G_3 / G_1)

And substitute this result into Equation (d) :
T_1 = 460 N-m + 1.285714 * T_3
       = 460 N-m + 1.285714 * (-T_1 * (N_b/N_e) * (L_1/L_3) * ( J_3 / J_1) * (G_3 / G_1))
       = 460 N-m - (T_1 * (54 Teeth / 42 Teeth) * (38,349.5 mm^4 / 147,323.5 mm^4) )
       = 460 N-m - 0.430305 T_1

Group the T_1 terms to Obtain : 
T_1 = 460 N-m / 1.430305 = 321.610 N-m
T_1 = 321.610 N-m

Back Substitute this result into equation d to find the internal torque in shaft (3).
T_1 = 460 N-m _ 1.285714*T_3

T_3 = ( T_1 - 460 N-m ) / 1.285714 
      = (321.610 N-m - 460 N-m) / 1.285714
      = - 107.637 N-m
T_3 = - 107.637 N-m


(a) Maximum Shear Stress Magnitude in Shaft (1):
t_1 = T_1 * c_1 / J_1 
      = (321.610 N-m) (35 mm / 2) (1,000 mm/m) / (147,323.5 mm^4)
      = 38.2 MPa

(b) Maximum Shear Stress Magnitude in Shaft (3):
t_3 = T_3 * c_3 / J_3 
      = (107.637 N-m)(25 mm/ 2)(1,000 mm/m) / (38,349.5 mm^4)
      = 35.1 MPa

(c) Rotation Angle of Gear E:
Phi_3 = T_3 * L_3 / ( J_3 * G_3)
      = (-107.637 N-m)(2)(200mm)(1,000 mm/m) / ( (38,349.5 mm^4) * (28,000 N/mm^2) )
      = -0.040096 rad

(d) Rotation Angle of Gear C:
Phi_1 = T_1 * L_1 / ( J_1 * G_1)
          = (321.610 N-m)(2)(200mm)(1,000 mm/m) / ( (147,323.5 mm^4) * (28,000 N/mm^2) )
          = 0.031186 rad

Phi_2 = T_2 * L_2 / ( J_2 * G_2)
          = (460 N-m)(200 mm)(1,000 mm/m) / ( ( 147,323.5 mm^4) *(28,000 N/mm^2) )
          = 0.022303 rad

Phi_c = Phi_b + Phi_2 = Phi_1 + Phi_2 
          = 0.031186 rad + 0.022303 rad 
          = 0.0535 rad


Content courtesy of Nishoka from Chegg - Made S.E. Friendly By ToughSTEM team.
Edit
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