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Shaft Design

A design project submitted to the

Department of Mechanical Engineering Technology

Engineering Technologies Division

Cincinnati State and Technical Community College

In partial fulfillment of the

Requirements for the course

MET – 7150 Machine Design I

Late Fall Term – 2009 - 2010

By

Jeremy Bryant

MET Class of 2010

TABLE OF CONTENTS:

Problem Description: pg 1

Shaft Data: pg 2

Critical Points: pg 3

Torque Diagram: pg 4

Forces Exerted on the Shaft: pg 5

Reactions at the Bearings: pg 6-7

Horizontal Shear Forces: pg 8-10

Bending Moment Forces: pg 11-14

Calculations for Horizontal Shear and Bending Moments: pg 15

Material Selection: pg 16

Design Stresses: pg 16-17

Calculating Shaft Diameters: pg 17

1st Iteration of Diameters: pg 18-19

2nd Iteration Cs Values: pg 20

2nd Iteration Sn` Values: pg 21

2nd Iteration Diameters: pg 22

3rd Iteration Cs values: pg 23

3rd Iteration Sn` Values: pg 23

3rd Iteration Diameters: pg 24

Final Calculated Diameters: pg 25

Diameter Calculations: pg 26-27

Key Selection: pg 28-29

Key Calculations: pg 30

Shaft Drawing: pg 31

1

MET – 7150 MACHINE DESIGN I SHAFT DESIGN:

The objective of this project is to design a safe shaft given certain dimensions of gears

and bearings as well as forces acting on the shaft through the gears and the effect of

fillets and keys. Taking the dimensions, forces and other determining factors, the

ultimate goal is to determine the shaft material, design stress and diameters along the

shaft.

2

SHAFT DATA:

- Final shaft surfaces are to be ground - Required design reliability of 99% - Gears 2 & 4 will experience shock loading - Gear 3 will experience normal loading - Shaft rotational speed = 2750 RPM - Shaft input horsepower = 65 HP

COMPONENT DATA:

Gear #2

- 6 inch pitch diameter - Face width = 2 inches - Profile Key - Tangential force = ��

� = -244.4 lb - Radial force = ��

� = -63.2 lb - Axial Force = 0

Gear #3


� = -297.8 lb - Radial force = ��

� = 77 lb - Axial Force = 0

Gear #4


� = -189 lb - Radial force = ��

� = -48.9 lb - Axial Force = 0

Bearing A

- Face width = 1 inch

Bearing B

- Face width = 1 inch

CRITICAL POINTS:

There are a minimum of 15 critical points on the shaft. The gears and bearings each

have 3 critical points. At these points the shaft diameter must be determined.



3



TORQUE ON THE SHAFT:

Data given earlier:

P = 65 HP

n= 2750 rpm

Overall torque on the shaft = ��

Torque at Gear 2 = ��



� � ��

�� . � ��

� � � � � � � � � � � � � �

The torque diagram show the torque applied at all parts of the shaftline. At points 7, 8 and 9 the max torque of 1489 lb

�� ! � ��

�� . ��

� "#$� % � �244.4 ()� "� *!

� % � ��. � ��

� "#+� % � �297.8 ()� " � *!

� % � ��

� "#/� % � �189 ()� "� *!

� % � �12 ��

Torque Diagram

��

� � � � �11. � ��

The torque diagram show the torque applied at all parts of the shaft displayed by the red . At points 7, 8 and 9 the max torque of 1489 lb - in is applied.

4

displayed by the red

FORCES EXERTED ON THE SHAFT:

The data provided earlier:

Forces exerted in the gears:

Force Tangential on Gear 2 =

Force Radial on Gear 2 = �� =


Force Radial on Gear 3 = �� = 77 lb (Y Direction)


Force Radial on Gear 4 = �� =

FORCES EXERTED ON THE SHAFT:

Force Tangential on Gear 2 = �� = -244.4 lb (Z Direction)

= -63.2 lb (Y Direction)

Force Tangential on Gear 3 = �� = -297.8 lb (Z Direction)

= 77 lb (Y Direction)

Force Tangential on Gear 4 = �� = -189 lb (Z Direction)

= -48.9 lb (Y Direction)

5

SOLVE FOR THE REACTIONS AT THE BEARINGS:

Solve for reaction at bearing A and B in the XY plane.

Σ�� 0 � �63.2 � �4.5 ��

�� 27.5 �

Σ�� 48.9 � �4.5 �� 77 �

�� 7.6 �

Σ�� 0 � 7.6 � � 63.2� � 77


Solve for reaction at bearing A and B in the XY plane.

� 77 � �11 �� 16 �� 48.9 � �20.5 ��

� �5 �� 63.2 � �11.5 �� 16��

77 � � 27.5 � � 48.9 �

6

��


Solve for reaction at bearing A and B in the XZ plane.

Σ�� 0 � �244.4 � �4.5 ��

�� 515.63 �

Σ�� 189 � �4.5 �� 297

�� 215.57 �

Σ�� 0 � 215.57 � � 244.4�


Solve for reaction at bearing A and B in the XZ plane.

� � 297.8 � �11 �� 16 �� 189 � �20

297.8 � �5 �� 244.4 � �11.5 �� 16��

� � 297.8 � � 515.63 � � 189 �

7

�20.5 ��

�

HORIZONTAL SHEAR FORCES

Shear force diagrams for the XY and XZ planes.

FORCES:

Shear force diagrams for the XY and XZ planes.

8

9

HORIZONTAL SHEAR FORCES:

Shear forces taken from the shear force diagrams.

Shear Forces: XY Plane Shear Forces: XZ Plane

�� = 0.00 � �� = 0.00 �

� �� = 3.80 � � �� = 107.78 �

�� = 7.61 � �� = 215.57 �

�� = 7.61 � �� = 215.57 �

�� = −23.99 � �� = 93.37 �

�� = −55.59 � �� = −28.83 �

�� = −55.59 � �� = −28.83 �

�� = −17.09 � �� = −177.73 �

�� = 21.41 � �� = −326.63 �

�� = 21.41 � �� = −326.63 �

�� = 35.15 � �� = −68.82 �

�� = 48.90 � �� = 189.00 �

�� = 48.90 � �� = 189.00 �

�� = 24.45 � �� = 94.50 �

�� = 0.00 � �� = 0.00 �

10

COMBINED HORIZONTAL SHEAR FORCES:

Using the general form equation:

� = �� + ��

The combined horizontal shear force can be calculated.

Sample problem: � �� = 3.80 �, � �� = 107.78 �

� = � �� + � �� = � = √3.80 � + 107.78 � = 107.85 lbs

Combined Horizontal Shear Force at the Critical Points:

�� = 0.00 �

� = 107.85 �

�� = 215.70 �

�� = 215.70 �

�� = 96.40�

�� = 62.62 �

�� = 62.62 �

�� = 178.55 �

�� = 327.33 �

�� = 327.33 �

�� = 77.28 �

�� = 195.22 �

�� = 195.22 �

�� = 97.61 �

�� = 0.00 �

BENDING MOMENTS XY:

Bending Moment diagram for the XY plane based on the Momentshear force diagrams.

Bending Moment diagram for the XY plane based on the Moment-Area Method and the

11

Area Method and the

BENDING MOMENTS XZ:

Bending Moment diagram for the XZshear force diagrams.

gram for the XZ plane based on the Moment-Area Method and the

12

Area Method and the

13

COMBINED BENDING MOMENTS:

Determine the combined bending moments using the Moment-Area Method and the shear force diagrams.

Bending Moments: XY Plane Bending Moments: XZ Plane

�� = 0.00 �� = 0.00 ��

�� = .950 �� = 26.95 ��

�� = 3.80 �� = 107.78 ��

�� = 26.63 �� = 754.49 ��

�� = 18.44 �� = 908.96 ��

�� = −21.35 �� = 941.23 ��

�� = −243.72 �� = 825.90 ��

�� = −298.23 �� = 670.98 ��

� �� = −295.00 �� = 292.71 ��

��!�� = −230.77 �� !�� = −687.18 ��

�� = −216.63 �� = −786.05 ��

�� = −195.61 �� = −756.00 ��

�� = −48.91 �� = −189.00 ��

�� = −12.24 �� = −47.25 ��

�� = −.01 �� = 0.00 ��

14

COMBINED BENDING MOMENTS:

Using the general form equation:

� = "�� + ��

The combined bending moments can be calculated.

Sample problem: �� = .950 �� , �� = 26.95 ��

� = "�� + �� = $ = √. 950 �� + 26.95 �� = 26.96 in lbs

Combined Bending Moments at the Critical Points:

�� = 0.00 ��

�� = 26.96 ��

�� = 107.85 ��

�� = 754.96 ��

�� = 909.15 ��

�� = 941.47 ��

�� = 861.11 ��

�� = 734.27 ��

� �� = 415.57 ��

��!�� = 724.90 ��

�� = 815.35 ��

�� = 780.90 ��

�� = 195.23 ��

�� = 48.81 ��

�� = .01 ��

Forces in the Gears (XY Plane) ΣMA Shear Forces lb (XY) Shear Forces lb (XZ) Bending Moment in lb (XY) Bending Moment in lb (XZ)

Reaction Force (lb) Distance from Ay (in) V1 0.00 V1 0.00 M1 0.00 M1 0.00

F2 radial -63.2 4.5 V2 3.80 V2 107.78 M2 0.95 M2 26.95

F3 radial 77 11 V3 7.61 V3 215.57 M3 3.80 M3 107.78

F4 radial -48.9 20.5 V4 7.61 V4 215.57 M4 26.63 M4 754.49

Ay 7.609375 0 V5 -23.99 V5 93.37 M5 18.44 M5 908.96

By 27.49063 16 V6 -55.59 V6 -28.83 M6 -21.35 M6 941.23

V7 -55.59 V7 -28.83 M7 -243.72 M7 825.90

Forces in the Gears (XZ Plane) ΣMA V8 -17.09 V8 -177.73 M8 -298.23 M8 670.98

Reaction Force (lb) Distance from Ay (in) V9 21.41 V9 -326.63 M9 -295.00 M9 292.71

F2 tang. -244.4 4.5 V10 21.41 V10 -326.63 M10 -230.77 M10 -687.18

F3 tang. -297.8 11 V11 35.15 V11 -68.82 M11 -216.63 M11 -786.05

F4 tang. -189 20.5 V12 48.90 V12 189.00 M12 -195.61 M12 -756.00

Az 215.5688 0 V13 48.90 V13 189.00 M13 -48.91 M13 -189.00

Bz 515.6313 16 V14 24.45 V14 94.50 M14 -12.24 M14 -47.25

V15 0.00 V15 0.00 M15 -0.01 M15 0.00

Combined Horizontal Shear (lb) Combined Bending Moments (in lbs)

V1 0.00 M1 0.00

V2 107.85 M2 26.96

V3 215.70 M3 107.85

V4 215.70 M4 754.96

V5 96.40 M5 909.15

Shear and Bending Moment Forces

V5 96.40 M5 909.15

V6 62.62 M6 941.47

V7 62.62 M7 861.11

V8 178.55 M8 734.27

V9 327.33 M9 415.57

V10 327.33 M10 724.90

V11 77.28 M11 815.35

V12 195.22 M12 780.90

V13 195.22 M13 195.23

V14 97.61 M14 48.81

V15 0.00 M15 0.01

16

MATERIAL SELECTION:

Based on the loads select an adequate material based on the following rules:

1. Percent elongation must be between 12% and 25%.

2. The ratio of Yield Strength (ksi) to Percent Elongation should be greater than 4.0:

��% �� ≥ 4.0

For this shaft, 4140 OQT 1300 Alloy Steel was selected. Check the material properties with the rules above to see if the material will work:

From Appendix 3:

�� = 117,000 �� = 100,000 �� % Elongation = 23%

Check:

1. % Elongation of 23% is ≥ 12%

2. �� = 4.35 ≥ 4.0

Thus, making the material suitable for this shaft design.

DESIGN STRESS:

An appropriate design stress must be selected depending on the manner of loading (smooth, shock, repeated and reversed, or other.)

In this design there is combined stress due to the combination of torsion, shear and bending in the shaft. Therefore using the formula below the value of Modified Endurance Strength for each critical point can be found:

Sn`= Sn(Cs)(Cm)(Cst)(Cr)

Since we do not know the diameter of the shaft a 1 in diameter will be assumed. Using this diameter calculate the values:

Sn = 54,000 psi (based on, �� = 117#� , from table in text)

Cs = .876 ($� = %&. '(.�� and we assume D = 1in, then $� = %�. '(.�� = .876)

Cm = .80 (Cast, value from text)

Cst = 1.0 (Bending, value from text)

17

Cr = .81(99% reliability from table in text)

Calculate the initial Sn`:

Sn`= Sn(Cs)(Cm)(Cst)(Cr)

�` = 54,000 �� ,. 876-,. 80-,1.0-,. 81- = 30651.53 psi

CALCULATING SHAFT DIAMETERS:

Using the formulas below, calculate the minimum shaft diameter for each critical point in both bending and horizontal shear:

Bending: ./ = 0 �12 345678`9 :� + < = >8?@�A� ⁄

Horizontal shear: .ℎ = D2.94G�,H- I �`⁄

The Kt’s at each point depend upon the components at each gear (fillets, keys or retaining rings). If Kt values overlap the higher Kt is used.

Sample Problem:

Bending:

Point 2: N = 2 (normal load, from the text), Kt = 2 (used for a profile key, from the text), M = 26.96 in lb, Sn`= 30651.53 psi, T = 0 in lb, Sy = 100,000 psi

./ = 0 �12 345678`9 :� + < = >8?@�A� ⁄

./� = J �,�-2 K4,�-,�L.ML �9 NO- �LP�.P Q�� :� + < R � �9 NO�� Q��S�T� ⁄

./� =. UUV WX

Horizontal Shear:

Point 2: N = 2, Kt = 2, V = 107.85 lb, Sn` = 30651.53 psi

.ℎ = D2.94G�,H- I �`⁄

.ℎ� = D2.94,2-,107.85 �Y- 2 30651.53 �� ⁄

.ℎ� =. ZVU WX

18

1st ITERATION DIAMETERS:

The first iteration of calculated diameters for shaft bending and torque, and horizontal shear at every critical point

Based on Shaft Bending and Torque: Based on Horizontal Shear:

∗ �� = 0.0000 �ℎ � �� = 0.0000 �ℎ �

∗ �� = .3297 �ℎ � �� = .2034 �ℎ �

∗ �� = .5638 �ℎ � �� = .3216 �ℎ �

∗ �� = 2.0851 �ℎ � �� = .8630 �ℎ �

∗ �� = 1.9392 �ℎ � �� = .4711 �ℎ �

∗ �� = 1.9617 �ℎ � �� = .3797 �ℎ �

∗ �� = 1.3293 �ℎ � �� = .2192 �ℎ �

∗ �� = 1.2644 �ℎ � �� = .3701 �ℎ �

∗ �� = 1.0695 �ℎ � �� = .5011 �ℎ �

∗ ��! = 1.0604 �ℎ � ��! = .3962 �ℎ �

∗ �� = 1.0272 �ℎ � �� = .1722 �ℎ �

∗ �� = 1.0125 �ℎ � �� = .2737 �ℎ �

∗ �� = 1.2052 �ℎ � �� = .6704 �ℎ �

∗ �� = .9618 �ℎ � �� = .4740 �ℎ �

∗ �� = .9283 �ℎ � �� = 0.0000 �ℎ �

Now based on the calculations for the calculated diameters the larger diameters should be selected since that is the minimum diameter required for that point to withstand the forces at each critical point. The selected diameters are labeled with a * above.

19

Selected Diameter at Each Critical Point:

�� = 0.0000 �ℎ �

�� = .3297 �ℎ �

�� = .5638 �ℎ �

�� = 2.0851 �ℎ �

�� = 1.9392 �ℎ �

�� = 1.9617 �ℎ �

�� = 1.3293 �ℎ �

�� = 1.2644 �ℎ �

�� = 1.0695 �ℎ �

��! = 1.0604 �ℎ �

�� = 1.0272 �ℎ �

�� = 1.0125 �ℎ �

�� = 1.2052 �ℎ �

�� = .9618 �ℎ �

�� = .9283 �ℎ �

Note: Now with these calculated diameters return to Design Stress and recalculate the Cs values in place of the first iteration’s assumed diameter of 1 inch. Then recalculate the Sn` with the recalculated Cs values. Continue with this same process until diameters are no longer changing and remain constant.

20

2nd

ITERATION Cs VALUES:

Recalculate Cs values for first iteration of calculated diameters:

�� = �� = .990

�� = �. 330 �� . 3⁄ �.�� = .990

�� = �. 564 �� . 3⁄ �.�� = .933

�� = .859 − .00837�2.09 �� = .857

�� = �1.94 �� . 3⁄ �.�� = .814

�� = �1.96 �� . 3⁄ �.�� = .813

�� = �1.33 �� . 3⁄ �.�� = .849

�� = �1.26 �� . 3⁄ �.�� = .854

��! = �1.07 �� . 3⁄ �.�� = .870

��" = �1.06 �� . 3⁄ �.�� = .870

�� = �1.03 �� . 3⁄ �.�� = .873

�� = �1.01 �� . 3⁄ �.�� = .875

�� = �1.21 �� . 3⁄ �.�� = .858

�� = �. 962 �� . 3⁄ �.�� = .880

�� = �. 928 �� . 3⁄ �.�� = .883

Now recalculate the Sn` values with Cs values above.

21

2nd

ITERATION Sn` VALUES:

Now recalculate the Sn` values:

Sn` = Sn (Cs)(Cm)(Cst)(Cr)

#��` = 54,000 psi (.990)(.80)(1.0)(.81) = 34,630 psi

#�` = 54,000 psi (.990)(.80)(1.0)(.81) = 34,630 psi

#��` = 54,000 psi (.933)(.80)(1.0)(.81) = 32,646 psi

#��` = 54,000 psi (.857)(.80)(1.0)(.81) = 29,997 psi

#��` = 54,000 psi (.814)(.80)(1.0)(.81) = 28,498 psi

#��` = 54,000 psi (.813)(.80)(1.0)(.81) = 28,462 psi

#��` = 54,000 psi (.849)(.80)(1.0)(.81) = 29,707 psi

#� ` = 54,000 psi (.854)(.80)(1.0)(.81) = 29,871 psi

#�!` = 54,000 psi (.870)(.80)(1.0)(.81) = 30,246 psi

#��"` = 54,000 psi (.870)(.80)(1.0)(.81) = 30,443 psi

#��` = 54,000 psi (.873)(.80)(1.0)(.81) = 30,561 psi

#��` = 54,000 psi (.875)(.80)(1.0)(.81) = 30,610 psi

#��` = 54,000 psi (.858)(.80)(1.0)(.81) = 30,029 psi

#��` = 54,000 psi (.880)(.80)(1.0)(.81) = 30,783 psi

#��` = 54,000 psi (.883)(.80)(1.0)(.81) = 30,903 psi

Now recalculate the diameters using the values of Sn` above to get the 2nd Iteration of diameters.

22

2nd

ITERATION DIAMETERS:

1st Iteration 2nd Iteration

$%� = 0.0000 ��&ℎ�� $%� = 0.0000 ��&ℎ�� ∆ = 0

$% = .3297 ��&ℎ�� $% = .3166 ��&ℎ�� ∆ = .0131

$%� = .5638 ��&ℎ�� $%� = .5521 ��&ℎ�� ∆ = .0117

$%� = 2.0851 ��&ℎ�� $%� = 2.1000 ��&ℎ�� ∆ = .0149

$%� = 1.9392 ��&ℎ�� $%� = 1.9864 ��&ℎ�� ∆ = .0472

$%� = 1.9617 ��&ℎ�� $%� = 2.0102 ��&ℎ�� ∆ = .0486

$%� = 1.3293 ��&ℎ�� $%� = 1.3425 ��&ℎ�� ∆ = .0133

$% = 1.2644 ��&ℎ�� $% = 1.2746 ��&ℎ�� ∆ = .0102

$%! = 1.0695 ��&ℎ�� $%! = 1.0717 ��&ℎ�� ∆ = .0022

$%�" = 1.0604 ��&ℎ�� $%�" = 1.0664 ��&ℎ�� ∆ = .0024

$%�� = 1.0272 ��&ℎ�� $%�� = 1.0282 ��&ℎ�� ∆ = .0010

$%� = 1.0125 ��&ℎ�� $%� = 1.0130 ��&ℎ�� ∆ = .0005

$%�� = 1.2052 ��&ℎ�� $%�� = 1.2118 ��&ℎ�� ∆ = .0066

$%�� = .9618 ��&ℎ�� $%�� = .9615 ��&ℎ�� ∆ = .0003

$%�� = .9283 ��&ℎ�� $%�� = 0.9283 ��&ℎ�� ∆ = 0

As shown above, the values changed from the 1st iteration of diameters to the second iteration. Therefore, the Sn` must be recalculated until change stops and the diameters are constant. Recalculate the Sn` value using the 2nd iteration of diameters for the Cs values.

23

3rd

ITERATION Cs VALUES:

Recalculate the Cs values using the 2nd iteration diameters:

�� = �� (�� = 0)

�� = (. 317 �� . 3⁄ )�.�� = .994

�� = (. 552 �� . 3⁄ )�.�� = .935

��" = .859 − .00837(2.10 ��) = .857

��% = (1.99 �� . 3⁄ )�.�� = .812

��& = .859 − .00837(2.01 ��) = .857

��' = (1.34 �� . 3⁄ )�.�� = .848

��( = (1.27 �� . 3⁄ )�.�� = .853

��) = (1.07 �� . 3⁄ )�.�� = .869

��+ = (1.07 �� . 3⁄ )�.�� = .870

�� = (1.03 �� . 3⁄ )�.�� = .873

�� = (1.01 �� . 3⁄ )�.�� = .875

�� = (1.21 �� . 3⁄ )�.�� = .858

��" = (. 962 �� . 3⁄ )�.�� = .880

��% = (. 928 �� . 3⁄ )�.�� = .883

Now recalculate the Sn` values with Cs values above.

3rd

ITERATION Sn` VALUES:

Now recalculate the Sn` values:

,�� ` , ,�"` ��/ ,�'` − ,��%` ∶ 1�� 2� � ∆�� = 0

,��` = 54,000 psi (.994)(.80)(1.0)(.81) = 34,786 psi

,��` = 54,000 psi (.935)(.80)(1.0)(.81) = 32,722 psi

,�%` = 54,000 psi (.812)(.80)(1.0)(.81) = 28,423 psi

,�&` = 54,000 psi (.857)(.80)(1.0)(.81) = 29,999 psi

Now recalculate the diameters using the values of Sn` above to get the 3rd Iteration of diameters.

24

3rd

ITERATION DIAMETERS:

2nd Iteration 3rd Iteration

�4� = 0.0000 ��ℎ� �4� = 0.0000 ��ℎ� ∆ = 0

�4� = .3166 ��ℎ� �4� = .3161 ��ℎ� ∆ = .0005

�4� = .5521 ��ℎ� �4� = .5516 ��ℎ� ∆ = .0005

�4" = 2.1000 ��ℎ� �4" = 2.1000 ��ℎ� ∆ = 0

�4% = 1.9864 ��ℎ� �4% = 1.9881 ��ℎ� ∆ = .0017

�4& = 2.0102 ��ℎ� �4& = 1.9756 ��ℎ� ∆ = .0346

�4' = 1.3425 ��ℎ� �4' = 1.3425 ��ℎ� ∆ = 0

�4( = 1.2746 ��ℎ� �4( = 1.2746 ��ℎ� ∆ = 0

�4) = 1.0717 ��ℎ� �4) = 1.0717 ��ℎ� ∆ = 0

�4�+ = 1.0664 ��ℎ� �4�+ = 1.0664 ��ℎ� ∆ = 0

�4�� = 1.0282 ��ℎ� �4�� = 1.0282 ��ℎ� ∆ = 0

�4�� = 1.0130 ��ℎ� �4�� = 1.0130 ��ℎ� ∆ = 0

�4�� = 1.2118 ��ℎ� �4�� = 1.2118 ��ℎ� ∆ = 0

�4�" = .9615 ��ℎ� �4�" = .9615 ��ℎ� ∆ = 0

�4�% = 0.9283 ��ℎ� �4�% = 0.9283 ��ℎ� ∆ = 0

The diameters have virtually stopped changing and have remained constant, so further recalculation is unnecessary and the final diameters are to be used for the shaft design.

25

FINAL CALCULATED DIAMETERS:

With the final calculated diameters at each critical point determine the best size diameter for each gear and bearing based on the largest minimum required diameter.

Bearing A:

�� = 0.0000 ��ℎ�

�� = .3161 ��ℎ�

�� = .5516 ��ℎ�

Gear 2:

�" = 2.1000 ��ℎ�

�% = 1.9881 ��ℎ�

�& = 1.9756 ��ℎ�

Gear 3:

�' = 1.3425 ��ℎ�

�( = 1.2746 ��ℎ�

�) = 1.0717 ��ℎ�

Bearing B:

��+ = 1.0664 ��ℎ�

�� = 1.0282 ��ℎ�

�� = 1.0130 ��ℎ�

Gear 4:

�� = 1.2118 ��ℎ�

�4" = .9615 ��ℎ�

��% = 0.9283 ��ℎ�

Size the diameters up to 1/16 inch scale so it is a standard size and can be easily manufactured. As well as tolerance the parts with an RC2 fit for the shaft dimensions.

Minimum Diameter = .552 in. Round to .625 in.

Tolerance = .&�"'% 67

.&�""% 67

Minimum Diameter = 2.10 in. Round to 2.125 in.

Tolerance = �.��"& 67

�.��"� 67


Tolerance = �.�'"& 67

�.�'"� 67


Tolerance = �.��"' 67

�.��"� 67


Tolerance = �.�")& 67

�.�")� 67

Intial Sn` assuming 1" diameter

Sn(psi) Cs Cm Cst Cr Sn`

54000 0.87595811 0.80 1.0 0.81 30651.53

Bending and Torque Horizontal Shear

Point N Kt M Sn` (psi) T(in lb) Sy(psi) Diameter (in) Point N Kt V(lb) Sn` (psi) Diameter (in)

1 2 2 0.00 30651.526 0 100000 0 1 2 2 0.00 30651.5262 0

2 2 2 26.96 30651.526 0 100000 0.329704396 2 2 2 107.85 30651.5262 0.203418625

3 2 2.5 107.85 30651.526 0 100000 0.563786586 3 2 2.5 215.70 30651.5262 0.321633086

4 12 3 754.96 30651.526 733.2 100000 2.085086739 4 12 3 215.70 30651.5262 0.863032134

5 12 2 909.15 30651.526 733.2 100000 1.939198518 5 12 2 96.40 30651.5262 0.471080812

6 12 2 941.47 30651.526 733.2 100000 1.961663592 6 12 2 62.62 30651.5262 0.379680166

7 4 2 861.11 30651.526 1489 100000 1.329274526 7 4 2 62.62 30651.5262 0.219208446

8 4 2 734.27 30651.526 1489 100000 1.264427333 8 4 2 178.55 30651.5262 0.370146667

9 4 2 415.57 30651.526 1489 100000 1.069512632 9 4 2 327.33 30651.5262 0.501172159

10 2 2.5 724.90 30651.526 0 100000 1.063975152 10 2 2.5 327.33 30651.5262 0.39621138

11 2 2 815.35 30651.526 0 100000 1.02718962 11 2 2 77.28 30651.5262 0.172185834

12 2 2 780.90 30651.526 0 100000 1.012512759 12 2 2 195.22 30651.5262 0.273680366

13 12 2 195.23 30651.526 755.8 100000 1.205197646 13 12 2 195.22 30651.5262 0.67037725

14 12 2 48.81 30651.526 755.8 100000 0.961802779 14 12 2 97.61 30651.5262 0.474028299

15 12 3 0.01 30651.526 755.8 100000 0.928338351 15 12 3 0.00 30651.5262 0

1st Iteration 2nd Iteration 2nd Iteration 2nd Iteration (BM's & T's)

Selected Diameters Recalculate Cs Values Recalculate Sn` Selected Diameters Δ Diameters 1st & 2nd

Point D(in) Point Cs Point Sn` (psi) Point D(in) Point D (in)

Diameter Calculations

Point D(in) Point Cs Point Sn` (psi) Point D(in) Point D (in)

D1 0 Cs1 0.9896682 Sn`1 34630.46975 D1 0 D1 0

D2 0.329704396 Cs2 0.9896682 Sn`2 34630.46975 D2 0.316559957 D2 0.01314444

D3 0.563786586 Cs3 0.932955 Sn`3 32645.96288 D3 0.552063387 D3 0.0117232

D4 2.085086739 Cs4 0.8572548 Sn`4 29997.05935 D4 2.100033295 D4 0.01494656

D5 1.939198518 Cs5 0.8144134 Sn`5 28497.95222 D5 1.986356375 D5 0.04715786

D6 1.961663592 Cs6 0.8133822 Sn`6 28461.86833 D6 2.010243209 D6 0.04857962

D7 1.329274526 Cs7 0.8489571 Sn`7 29706.70535 D7 1.342539433 D7 0.01326491

D8 1.264427333 Cs8 0.8536405 Sn`8 29870.58839 D8 1.274628439 D8 0.01020111

D9 1.069512632 Cs9 0.8695066 Sn`9 30425.77524 D9 1.071667046 D9 0.00215441

D10 1.063975152 Cs10 0.8700032 Sn`10 30443.1537 D10 1.066397145 D10 0.00242199

D11 1.02718962 Cs11 0.873377 Sn`11 30561.20931 D11 1.028200502 D11 0.00101088

D12 1.012512759 Cs12 0.8747607 Sn`12 30609.62776 D12 1.012974524 D12 0.00046176

D13 1.205197646 Cs13 0.8581574 Sn`13 30028.64208 D13 1.211768355 D13 0.00657071

D14 0.961802779 Cs14 0.8797188 Sn`14 30783.12038 D14 0.961540811 D14 0.00026197

D15 0.928338351 Cs15 0.8831524 Sn`15 30903.26824 D15 0.928338351 D15 6.19E-11

3rd Iteration 3rd Iteration 3rd Iteration

Recalculate Cs Values Δ Cs values Recalculate Sn` Selected Diameters Δ Diameters 2nd & 3rd

Point Cs Point Cs Point Sn`(psi) Point D(in) Point D (in)

Cs1 0.989668203 Cs1 0 Sn`1 34630.46975 D1 0 D1 0

Cs2 0.994107111 Cs2 0.0044389 Sn`2 34785.79604 D2 0.316088083 D2 0.00047187

Cs3 0.935113989 Cs3 0.0021589 Sn`3 32721.50871 D3 0.551638201 D3 0.00042519

Cs4 0.857242272 Cs4 1.251E-05 Sn`4 29997.05935 D4 2.100033295 D4 0

Cs5 0.812263708 Cs5 0.0021496 Sn`5 28422.73168 D5 1.988089991 D5 0.00173362

Cs6 0.857317426 Cs6 0.0439353 Sn`6 29999.25139 D6 1.975632676 D6 0.03461053

Cs7 0.848030286 Cs7 0.0009268 Sn`7 29706.70535 D7 1.342539433 D7 0

Cs8 0.852886306 Cs8 0.0007542 Sn`8 29870.58839 D8 1.274628439 D8 0

Cs9 0.869314156 Cs9 0.0001925 Sn`9 30425.77524 D9 1.071667046 D9 0

Cs10 0.869785675 Cs10 0.0002176 Sn`10 30443.1537 D10 1.066397145 D10 0

Cs11 0.873377038 Cs11 0 Sn`11 30561.20931 D11 1.028200502 D11 0

Cs12 0.874760738 Cs12 0 Sn`12 30609.62776 D12 1.012974524 D12 0

Cs13 0.857644252 Cs13 0.0005131 Sn`13 30028.64208 D13 1.211768355 D13 0

Cs14 0.879718804 Cs14 0 Sn`14 30783.12038 D14 0.961540811 D14 0

Cs15 0.883152385 Cs15 0 Sn`15 30903.26824 D15 0.928338351 D15 0

28

KEY SELECTION:

Select a suitable material for the keys.

Based on the length of the gears and bearings the yield strength of a suitable material can be found using the following equation for design stress in shear:

�� = ��

�� =

� �

.�

L = Length of key (use whole length of gear or bearing)

D = Diameter of shaft

W = Width of key

T = Torque in the shaft

Bearing A:

L = 1 in T = 0 in lb D = .625 in W = 3/16 in (from table in text)

�� = 2(0 �� )

(1 ��)(. 625 ��)(3 16⁄ ��)= 0 ��

� = �( !"#)

.�= $ %&' (Any material is suitable)

Material: 1020 HR Sy = 30,000 psi

Gear 2:

L = 2 in T = 733.2 in lb D = 2.125 in W = 1/2 in (from table in text)

�� = 690 ��

� = )*)$ %&'


Gear 3:

L = 3 in T = 1489 in lb D = 1.375 in W = 5/16 in (from table in text)

�� = 2310 ��

� = *+,-* %&'


29

Bearing B:

L = 1 in T = 0 in lb D = 1.125 in W = 1/4 in (from table in text)

�� = 0 ��

� = $ %&'


Gear 4:

L = 2 in T = 755.8in lb D = 1.25 in W = 1/2 in (from table in text)

�� = 2419 ��

� = *)/** %&'


Key Length(in) T(in lbs) D (in) W (in) τd (psi) Sy (psi) Material and it's Sy

A 1.00 0.00 0.6250 0.1875 0 0 1020 HR, Sy = 30,000 psi

Gear 2 2.00 733.20 2.1250 0.5000 690.070588 4140.424 1020 HR, Sy = 30,000 psi

Gear 3 3.00 1489.00 1.3750 0.3125 2310.20606 13861.24 1020 HR, Sy = 30,000 psi

B 1.00 0.00 1.1250 0.2500 0 0 1020 HR, Sy = 30,000 psi

Gear 4 2.00 755.80 1.2500 0.2500 2418.56 14511.36 1020 HR, Sy = 30,000 psi

Key Selection

shaft design - coroflots3images.coroflot.com/user_files/individual_files/174942_d9... · shaft...

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