Stress Analysis of Drive Shaft

STRESS ANALYSIS OF DRIVE SHAFT The initial drive shaft design was of 6mm in diameter with a step up to 10mm for the drive pulley attachment (see Fig. F.1) Calculations were carried out for the drive shaft to ensure that high stresses were not reached in the shaft during operation which could potentially lead to a failure.

Figure F.1. Initial Drive Shaft Design

It should be noted that the track running over the pulley adds complexity to the calculation and thus the pulley/track interaction was simply treated as having one point of contact and calculations were carried out in the same way as for the gear mesh interaction. It was considered that this would provide a good approximation and would give an exaggeration of the true result rather than an underestimation. The first step was to construct free body diagrams of the shaft. All forces acting on the shaft are illustrated in Fig. F.2. Reaction forces are present at the locations of the bearings and also the gear and pulley. Maximum torque applied from the motor shaft was assumed to be 3Nm (see Appendix I). Realistically this was representative of the worst case scenario.

Figure F.2. Free Body Diagram for the Shaft in Vertical and Horizontal Planes

The forces at the gear and pulley were calculated and then used to determine the reaction forces by applying Newton’s 2nd Law. From the values calculated shear force and bending moment diagrams could be found. Finally maximum bending, shear and torsional stresses could be calculated and compared to maximum allowable stresses. Forces at gear:

T

2T

2× 3

Tangential: Ftg = r = d = 19.41 g g

= 309.14 N

1000

Where Ftg = Tangential force on gear teeth T = Torque on the shaft rg = Pitch radius of the gear dg = Pitch diameter of the gear Radial: Frg = Ftg tan φ =112 .51 N Where Frg = Radial force gear teeth φ = Pressure angle of the gear = 20° Axial: Spur gears will be used. Since a helix angle does not need to be considered for these gears there will be no axial force. Forces at pulley:

2T 2×3 F = = = 117.81N tp Tangential: d P 50.93 1000

Where Ftp = Tangential force on pulley teeth dp = Pitch diameter of the pulley

Radial: Frp = Ftp tan φ = 42 .88 N the pressure angle for the pulley will be assumed as 20°. Where Frp = Radial force at the pulley Axial: Again, since there is not helical interaction there will not be an axial force on the shaft from the pulley. Force Equilibrium: The forces in the X, Y and Z directions and the moment about the Z-axis were summed:

∑F ∑F ∑F ∑M ∑M

X

: − R BX = 0 ⇒ R BX = 0

Y

:R AY − Frg + R BY − Frp = 0

Z

: −R AZ + Ftg − R BZ + Ftp = 0 Z

| A : −Frg ( X g ) + R BY ( X B ) − Frp ( X P ) = 0

Y

| A : Ftg ( X g ) − R BZ ( X B ) + Ftp ( X P ) = 0

Equation Equation Equation Equation Equation

1 2 3 4 5

X values are illustrated in Fig. F.2. Taking Equation 4 rearranging for RBY:

and

substituting

all

known

values

and

− (112 .51 × 0.018 ) + 0.042 R BY − ( 42 .88 × 0.08 ) = 0 ⇒ R BY =

3.72372 = 88 .66 N 0.042

This can then be used in Equation 2 to find RAY which was calculated to be 66.73N. Rearranging Equation 5, RBZ can be found: R BZ =

(309 .14 × 0.018 ) + (117 .81 × 0.08 ) = 356 .89 N 0.042

This result can be used in Equation 3 to find R AZ which was calculated to be 70.06N. ΓB = Γtg + Γtp = ( Ftg × rg ) + ( Ftp × rp ) = 6.355 Nm

Diagrams: The above results were used to construct shear force and bending moment diagrams in the vertical and horizontal planes for the shaft. These diagrams can be seen in Fig. F.3.

Figure F.3. Shear Force and Bending Moment Diagrams

Gear Hole The gear was to be fixed to the shaft using a pin through the boss of the gear. Maximum bending stress, shear stress and torsional shear stress were all assessed (axial stress in the shaft was not present). Initially a 2mm diameter hole in the 6mm diameter shaft was considered. The stress concentration factors for this hole were found using appropriate stress concentration graphs1. The following stress concentration factors were calculated: Ktbending = 1.9 Kttorsion = 2.6 Bending Stress: To find the maximum bending stress at this location the bending stress equation was used: σ bending =

M max | g ×r I

Where Mmax|g = Maximum moment at the gear I = 2nd moment of area r = Shaft radius Using the bending moment diagrams it was possible to calculate the maximum moment at the location of the gear hole:

[

M max | g = ( M Y | g ) + ( M Z | g )

1

2

]

2 1/ 2

= 1.691Nm

Norton, R.,L., Machine Design: An Integrated Approach, 2nd Edition, Prentice Hall, 2000.

I=

π × rshaft 4 4

Therefore σbending = 79 .7 MPa Now multiplying by the bending stress concentration factor the following maximum stress is calculated for the gear hole: σbending × Kt beniding =151 .43 MPa

Shear Stress: The maximum shear stress due to transverse loading at the location of the gear hole was found by using the following equation: 4 Vmax τ max | g = 3 A Where Vmax = Maximum shear force A = Shear Area

[

Vmax | g = (VY | g ) + (VZ | g ) 2

]

2 1/ 2

= 243.42 N

Therefore,

τ max | g =11 .51 MPa τ max | g ×K = 29 .93 MPa

Torsional Shear Stress: The maximum torsional shear stress was found using the following equation: τ torsion | g =

Γtg × r J

Where Гtg = Torque due to tangential loading at the gear r = Shaft radius J = Polar 2nd moment of area π × rshaft 4 J = 2 Therefore, τ torsion |G = 80 .4 MPa τ torsion | g ×Kt torsion = 209 .04 MPa

A comparison of the maximum stresses obtained from these calculations to the tensile and shear yield strength of mild steel, 250MPa and 125MPa2 respectively can now be made. The torsional shear stress is definitely of concern and the maximum bending stress is approaching undesirable values. To overcome this problem the hole was increased to 4mm to reduce the stress concentration factors and thus eliminate the problem. Pulley Hole 2

Callister Jr, W.,D., Fundamentals of Materials Science and Engineering, Wiley, 2001.

It was intended to attach the pulley to the shaft using the same method as for the gear. The same method and equations outlined for the gear were followed for the pulley hole. Firstly the stress concentration factor due to the hole needed to be calculated again. Since the hole diameter was to be 4mm and the shaft diameter under examination at this point was 10mm the following stress concentration factors could be calculated3: Ktbending = 1.84 Kttorsion = 1.93 Bending Stress:

M max | P ×r = 17 .63 MPa I × Kt bending = 32 .44 MPa

σ bending = σ bending

Shear Stress: 4 Vmax τ max | P = = 2.13 MPa 3 A τ max | P ×K = 4.11 MPa

Torsional Shear Stress: τ torsion | P =

Γtp × r J

= 14 .999 MPa

τ torsion | P ×Kt torsion = 28 .94 MPa

These results show that even with the stress concentration factor considered the hole does not raise the stress in the shaft to undesirable levels. Fillet Initially a 0.2mm fillet was considered. Again, the bending, shear and torsional shear stresses were all considered following the same procedure. For the fillet the following stress concentration factors were found4: Ktbending = 2.3 Kttorsion = 1.91 It was found that the bending stress was as high as 263.97MPa at the fillet and when multiplied by the concentrating factor this was raised to 607MPa. Values for the shear and torsional shear stresses were acceptable. However the bending stresses calculated were obviously worrying therefore the fillet radius had to be 3 4

Norton, R.,L., Machine Design: An Integrated Approach, 2nd Edition, Prentice Hall, 2000. Norton, R.,L., Machine Design: An Integrated Approach, 2nd Edition, Prentice Hall, 2000.

reconsidered. To get around this problem the shaft diameter at the pulley was increased with several smaller steps and greater fillet radii. This design also provided a better means of mounting the shaft.