Machine Elements: Design Project

Drag Chain Conveyor

Matthew Gray, Michael McClain,

Pete White, Kyle Gilliam,

Alejandro Moncada and Matthew Gonzalez

Executive Summary

The logistics of many industries use drag chain conveyors to move heavy loads short distances.

The purpose of this project is to analytically define the characteristics and design parameters for

our own drag chain conveyor system.

The starting parameters used for our general design are listed below:

Maximum load of 3000 pounds (treated as a 4 feet x 4 feet wooden pallet)

Maximum speed of 35 feet per minute

Load travel distance of 10 linear feet

Conveyor frame is rigid

Neglect weight of wooden pallet

Chain idealized

Drive Sprocket parameters are taken from manufacturer data

The following components were the focus of our design:

Pinion driven by the motor

Main drive gear

Shaft connecting main drive gear

Bearings supporting the shaft

Drag chain conveyors operate by rotating a shaft that drives two or more sprockets, which in

turn, drive the chains. The chains are connected around a drive sprocket and a free rotating

sprocket (an idler sprocket) at the opposite end of the conveyor. A load to be moved rests on top

of the chains and as the chains rotate around the prescribed loop, the load is moved from one

point to another.

Our design approach was to work backwards from our starting parameters to determine the

forces necessary for our conveyor to operate as prescribed. These forces were then used to

analyze the key components of our conveyor.

From our force analysis, we were able to determine that it should take the conveyor no longer

than 17 seconds to move the 3000 pound load 10 feet. The total force required for our conveyor

was calculated to be 776 pounds.

Our sprocket design was based on our decision to use an ANSI No. 50 chain for the conveyor.

With the properties of this chain, we were able to calculate the pitch diameter of the sprocket by

using the American Chain Association Sprocket Design Guidelines. This diameter was then used

in selecting an actual sprocket to be used in our design. Finally, we determined that a torque of

81 pound-foot on each sprocket is required to satisfy our initial parameters.

The initial parameters used for our gear design are listed below:

20 degree pressure angle

Standard full depth gear

21 teeth for pinion

55 teeth for gear

2.62:1 gear ratio

3 inch pitch diameter for pinion

These parameters were used to calculate the geometries of both the gear and pinion. After

determining the geometries of the gear and pinion, a thorough stress analysis was conducted on

the gear set. Notable calculations from this analysis include the rotational speed of the pinion and

the number of cycles of both pinion and gear. The pinion was calculated to have a rotational

speed of 45 revolutions per minute, while the number of cycles of the pinion and gear in a 5 year

span were determined to be 27 million and 10 million respectively.

Other calculations that comprised this stress analysis include: bending stress, surface stress,

bending fatigue strength, and surface fatigue strength. After conducting the stress analysis of the

gear set, the safety factors of the pinion and gear were determined. We then re-evaluated the

design and re-iterated the process.

The final step in our gear design was selecting the material of the gear set. Our choice of material

is shown below:

Gear Material (Table 12-20 Norton)

o AGMA Grade 2 (Premium Grade)

o Class A5

o Carburized and case hardened

o Brinell Hardness = 710 (corresponds to 64HRC)

o Qv = 5

With this material, the final safety factors of the gear and pinion are as follows:

Gear

o Bending Safety Factor = 4.76

Pinion

o Bending Safety Factor = 3.98

o Surface Fatigue Safety Factor = 2.38

Shaft Design Parameters:

1.75 inch diameter

2 – 3 inch overhang on the outside of each sprocket

SAE/AISI 1010 Cold rolled steel

As with the gear set, a stress analysis was conducted on the shaft. The values we calculated for

the maximum stresses on the shaft are as follows: 7188 psi (bending stress), 219 psi (shear

stress), and 1850 psi (torsional stress). After calculating the shaft’s maximum stresses, we

proceeded to determine its safety factor. With the shaft’s material properties and maximum

stresses, we used the maximum shear stress theory to get a safety factor of 5.4.

In designing the bearings, we first had to determine the total force being subjected on each of the

two bearings. The bearing closest to the drive gear was calculated to have a total force of 379

pounds, while the other bearing was calculated to have a total force of 85 pounds. Using the

higher total force, we determined the dynamic load rating. The value of this rating was calculated

to be 1296 pounds. Finally, using the dynamic load rating we were able to determine the safety

factor of the bearing, which was calculated to be 15.

Report Outline

I. Introduction & Overview

A. Identification of Need

B. Definition of Problem

C. Drag Chain Conveyor System

II. Analysis

A. Nomenclature

B. Preliminary Design

C. Force Analysis

D. Sprocket Design

E. Gear Design

1. Gear and Pinion Stress Analysis

a. Bending Stress Calculations

b. Surface Stress Calculations

c. Bending Fatigue Strengths

d. Surface Fatigue Strengths

2. Safety Factor Calculation

3. Material Selection

F. Shaft Design

1. Shaft Stress Analysis

a. Bending Stress Calculation

b. Shear Stress Calculation

c. Torsional Stress Calculation

2. Safety Factor Calculations

3. Shaft Key Stress Calculations

G. Bearing Design

1. Force Calculations

2. Dynamic Load Rating Calculation

3. Safety Factor Calculation

III. Summary of Safety Factors

IV. Conclusion

V. References

VI. Appendix

VII. Engineering Drawings

Introduction & Overview

Identification of Need

The logistics of many industries require the use of human operated forklifts to transport heavy

loads. Implementation of automated conveyor systems has proven to be an exceptional solution

to minimizing human interaction in moving heavy loads short distances. These conveyor systems

are commonly used to move heavy pallets of finished products or raw materials from one area of

a manufacturing facility to another.

Definition of Problem

The purpose of this project is to analytically define the characteristics and design parameters for

a drag chain conveyor system. The criteria we chose for our design process came from our

research on existing conveyor systems. The assumed maximum load of 3000 pounds was based

on an average of maximum loads for these conveyor systems. In addition, these loads are

typically placed on 4 feet × 4 feet (l × w) wooden pallets. For our design, we will also assume

that the 3000 pounds maximum load takes the form of this pallet. We defined 10 linear feet as

the distance that this load must be moved. This distance keeps the conveyor system small enough

to be shipped as a finished product via motor freight. It also allows simple installation in a

manufacturing facility with no need for additional setup or assembly. Again, based on average

values found for existing systems, we chose our translational speed for the system to be 35 feet

per minute.

The frame is assumed rigid as it is a non-moving part. In addition, it is typically overdesigned

compared to the moving elements of the conveyor system. The weight of the wooden pallet is

taken to be negligible with respect to the 3000 pounds load, and any flexure of the pallet is

ignored. The travel distance of 10 linear feet is measured from the center of the drive sprockets.

Since the design of the chains is outside the scope of this project, all chains used are assumed to

be idealized. This means that any flexure, stretching, or stress in the chain is ignored. The

sprockets that drive the chains are taken directly from manufacturer’s specifications to match the

drive chains.

All our initial parameters are listed below:

Maximum load of 3000 pounds (treated as a 4 feet x 4 feet wooden pallet)

Maximum speed of 35 feet per minute

Load travel distance of 10 linear feet

Conveyor frame is rigid

Neglect weight of wooden pallet

Chain idealized

Sprockets that drive chains are taken from manufacture’s information

The main elements to be designed are the pinion driven by the motor, the main drive gear, the

shaft connecting the main drive gear, and the bearings supporting the shaft. Many assumptions

and idealizations have been made in order to focus on the main elements required for this report:

a shaft, gear and bearing.

Drag Chain Conveyor System

Drag chain conveyors operate by rotating a shaft that drives two or more sprockets, which in

turn, drive the chains. The chains are connected around a drive sprocket and a free rotating

sprocket (an idler sprocket) at the opposite end of the conveyor. See the Engineering Drawings

section for a better illustration of the final design. These types of conveyors have many

variations and are used in many different industries around the globe. A common application of

these types of conveyors is in the area of material handling. A load to be moved rests on top of

the chains and as the chains rotate around the prescribed loop, the load is moved from one point

to another. Many different companies manufacture drag chain conveyors, however the design

principles are similar industry wide. The parameters that do differ among designs are usually the

component materials, load ratings, and speed of travel.

Analysis

Nomenclature

𝑊𝑝 = 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑙𝑜𝑎𝑑 𝑝𝑒𝑟 𝑠𝑖𝑑𝑒

𝑊𝑐 = 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐ℎ𝑎𝑖𝑛 𝑝𝑒𝑟 𝑠𝑖𝑑𝑒

𝑊𝐿 = 𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑤𝑒𝑖𝑔ℎ𝑡 𝑝𝑒𝑟 𝑠𝑖𝑑𝑒

𝑊𝑇 = 𝑡𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑 (𝑝𝑒𝑟 𝑠𝑖𝑑𝑒) 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝑡𝑜 𝑐ℎ𝑎𝑖𝑛

𝑄𝑇 = 𝑁𝑜𝑟𝑚𝑎𝑙 𝐹𝑜𝑟𝑐𝑒

𝐹𝑓 = 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑝𝑒𝑟 𝑠𝑖𝑑𝑒

𝜇𝑘 = 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑒𝑓𝑓. 𝑜𝑓 𝑈𝐻𝑀𝑊 𝑎𝑛𝑑 𝑆𝑡𝑒𝑒𝑙

𝑉𝐶 = 𝑙𝑖𝑛𝑒𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑐ℎ𝑎𝑖𝑛 𝑎𝑛𝑑 𝑙𝑜𝑎𝑑

𝐹𝐶 = 𝑓𝑜𝑟𝑐𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑝𝑒𝑟 𝑠𝑖𝑑𝑒 𝑡𝑜 𝑚𝑜𝑣𝑒 𝑙𝑜𝑎𝑑

𝐹𝑇 = 𝑓𝑜𝑟𝑐𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 (𝑡𝑜𝑡𝑎𝑙)𝑡𝑜 𝑚𝑜𝑣𝑒 𝑙𝑜𝑎𝑑

𝑃𝑑𝑠1 = 𝑝𝑖𝑡𝑐ℎ 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑠𝑝𝑟𝑜𝑐𝑘𝑒𝑡 #1

𝑃𝑑𝑠2 = 𝑝𝑖𝑡𝑐ℎ 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑠𝑝𝑟𝑜𝑐𝑘𝑒𝑡 #2

𝑇𝑠𝑝1 = 𝑡𝑜𝑟𝑞𝑢𝑒 𝑎𝑡 𝑠𝑝𝑟𝑜𝑐𝑘𝑒𝑡 #1

𝑇𝑠𝑝2 = 𝑡𝑜𝑟𝑞𝑢𝑒 𝑎𝑡 𝑠𝑝𝑟𝑜𝑐𝑘𝑒𝑡 #2

𝜔𝑠𝑝 = 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑝𝑟𝑜𝑐𝑘𝑒𝑡𝑠

𝜔𝑠ℎ𝑎𝑓𝑡 = 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠ℎ𝑎𝑓𝑡

𝑃𝑑𝑝 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑟𝑎𝑙 𝑝𝑖𝑡𝑐ℎ 𝑜𝑓 𝑝𝑖𝑛𝑖𝑜𝑛

𝑟𝑝 = 𝑝𝑖𝑡𝑐ℎ 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑝𝑖𝑛𝑖𝑜𝑛

𝑃𝑑𝑔 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑟𝑎𝑙 𝑝𝑖𝑡𝑐ℎ 𝑜𝑓 𝑔𝑒𝑎𝑟

𝑟𝑔 = 𝑝𝑖𝑡𝑐ℎ 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑔𝑒𝑎𝑟

𝑁𝑡𝑝 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ 𝑜𝑛 𝑝𝑖𝑛𝑖𝑜𝑛

𝑁𝑡𝑔 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ 𝑜𝑛 𝑔𝑒𝑎𝑟

𝑑𝑝 = 𝑝𝑖𝑡𝑐ℎ 𝑑𝑖𝑎𝑚𝑡𝑒𝑟 𝑜𝑓 𝑝𝑖𝑛𝑖𝑜𝑛

𝑑𝑔 = 𝑝𝑖𝑡𝑐ℎ 𝑑𝑖𝑎𝑚𝑡𝑒𝑟 𝑜𝑓 𝑔𝑒𝑎𝑟

𝑃𝑐 = 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑝𝑖𝑡𝑐ℎ

𝑃𝑏 = 𝑏𝑎𝑠𝑒 𝑝𝑖𝑡𝑐ℎ

𝑎 = 𝑎𝑑𝑑𝑒𝑛𝑑𝑢𝑚

𝑏 = 𝑑𝑒𝑑𝑒𝑛𝑑𝑢𝑚

𝐷𝑊 = 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑑𝑒𝑝𝑡ℎ

𝐷𝑊ℎ = 𝑤ℎ𝑜𝑙𝑒 𝑑𝑒𝑝𝑡ℎ

𝑡𝑐𝑡 = 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑡𝑜𝑜𝑡ℎ 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠

𝑟𝑓 = 𝑓𝑖𝑙𝑙𝑒𝑡 𝑟𝑎𝑑𝑖𝑢𝑠

𝐶𝑏𝑚𝑖𝑛 = 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑏𝑎𝑠𝑖𝑐 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒

𝑊𝑡𝑙𝑚𝑖𝑛 = 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑡𝑜𝑝 𝑙𝑎𝑛𝑑

𝐶𝑠𝑔 = 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒

𝑊𝑓 = 𝑓𝑎𝑐𝑒 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑡𝑜𝑜𝑡ℎ

𝑍 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑐𝑡𝑖𝑜𝑛

𝐶 = 𝑐𝑒𝑛𝑡𝑒𝑟 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑚𝑝 = 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 𝑔𝑒𝑎𝑟 𝑎𝑛𝑑 𝑝𝑖𝑛𝑖𝑜𝑛

𝜔𝑝 = 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑝𝑖𝑛𝑖𝑜𝑛

𝜔𝑔 = 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑔𝑒𝑎𝑟

𝑁𝑐𝑦𝑐−𝑝 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑦𝑐𝑙𝑒𝑠 𝑜𝑓 𝑝𝑖𝑛𝑖𝑜𝑛

𝑁𝑐𝑦𝑐−𝑔 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑦𝑐𝑙𝑒𝑠 𝑜𝑓 𝑔𝑒𝑎𝑟

𝑊𝑡 = 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑖𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑜𝑛 𝑡𝑜𝑜𝑡ℎ

𝑇𝑔 = 𝑡𝑜𝑟𝑞𝑢𝑒 𝑜𝑛 𝑔𝑒𝑎𝑟

𝐾𝑣 = 𝐷𝑦𝑛𝑎𝑚𝑖𝑐 𝐹𝑎𝑐𝑡𝑜𝑟

𝑄𝑣 = 𝑄𝑢𝑎𝑙𝑖𝑡𝑦 𝐹𝑎𝑐𝑡𝑜𝑟

𝐾𝑚 = 𝐿𝑜𝑎𝑑 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝐹𝑎𝑐𝑡𝑜𝑟

𝐾𝐵 = 𝑅𝑖𝑚 𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝐹𝑎𝑐𝑡𝑜𝑟

𝐾𝑎 = 𝐴𝑝𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝐹𝑎𝑐𝑡𝑜𝑟

𝐾𝐼 = 𝐼𝑑𝑙𝑒𝑟 𝐹𝑎𝑐𝑡𝑜𝑟

𝐾𝑆 = 𝑆𝑖𝑧𝑒 𝐹𝑎𝑐𝑡𝑜𝑟

𝐽 = 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝐺𝑒𝑜𝑚𝑒𝑡𝑟𝑦 𝐹𝑎𝑐𝑡𝑜𝑟

𝐼 = 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐺𝑒𝑜𝑚𝑒𝑡𝑟𝑦 𝐹𝑎𝑐𝑡𝑜𝑟

𝜎𝑏𝑝 = 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑜𝑛 𝑝𝑖𝑛𝑖𝑜𝑛

𝜎𝑏𝑔 = 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑜𝑛 𝑔𝑒𝑎𝑟

𝜎𝑐𝑝 = 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑜𝑛 𝑝𝑖𝑛𝑖𝑜𝑛

𝐶𝑣 = 𝐷𝑦𝑎𝑛𝑚𝑖𝑐 𝐹𝑎𝑐𝑡𝑜𝑟

𝐶𝑚 = 𝐿𝑜𝑎𝑑 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝐹𝑎𝑐𝑡𝑜𝑟

𝐶𝑎 = 𝐴𝑝𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝐹𝑎𝑐𝑡𝑜𝑟

𝐶𝑆 = 𝑆𝑖𝑧𝑒 𝐹𝑎𝑐𝑡𝑜𝑟

𝐶𝑝 = 𝐸𝑙𝑎𝑠𝑡𝑖𝑐 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡

𝐶𝑓 = 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐹𝑖𝑛𝑖𝑠ℎ 𝐹𝑎𝑐𝑡𝑜𝑟

𝐸𝑝 = 𝑌𝑜𝑢𝑛𝑔′𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑃𝑖𝑛𝑖𝑜𝑛

𝐸𝑔 = 𝑌𝑜𝑢𝑛𝑔′𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝐺𝑒𝑎𝑟

𝑣𝑝 = 𝑃𝑜𝑖𝑠𝑠𝑜𝑛′𝑠 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 𝑃𝑖𝑛𝑖𝑜𝑛

𝑣𝑔 = 𝑃𝑜𝑖𝑠𝑠𝑜𝑛′𝑠 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 𝑔𝑒𝑎𝑟

𝑆𝑓𝑏𝑔= corrected AGMA bending strength for gear

𝑆𝑓𝑏𝑝= corrected AGMA bending strength for pinion

𝑆′𝑓𝑏= uncorrected AGMA bending strength

𝐾𝐿𝑝 = 𝐿𝑖𝑓𝑒 𝐹𝑎𝑐𝑡𝑜𝑟 𝑓𝑜𝑟 𝑝𝑖𝑛𝑖𝑜𝑛

𝐾𝐿𝑝 = 𝐿𝑖𝑓𝑒 𝐹𝑎𝑐𝑡𝑜𝑟 𝑓𝑜𝑟 𝑔𝑒𝑎𝑟

𝐾𝑇 = 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝐹𝑎𝑐𝑡𝑜𝑟

𝐾𝑅 = 𝑅𝑒𝑙𝑖𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝐹𝑎𝑐𝑡𝑜𝑟

𝑆𝑓𝑐𝑝= 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑓𝑎𝑡𝑖𝑞𝑢𝑒 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ − 𝑝𝑖𝑛𝑖𝑜𝑛

𝑆′𝑓𝑐𝑝= 𝑢𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑓𝑎𝑡𝑖𝑞𝑢𝑒 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ − 𝑝𝑖𝑛𝑖𝑜𝑛

𝐶𝑇 = 𝑇𝑒𝑚𝑒𝑟𝑝𝑎𝑡𝑢𝑟𝑒 𝐹𝑎𝑐𝑡𝑜𝑟

𝐶𝑅 = 𝑅𝑒𝑙𝑖𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝐹𝑎𝑐𝑡𝑜𝑟

𝐶𝐿 = 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐿𝑖𝑓𝑒 𝐹𝑎𝑐𝑡𝑜𝑟

𝐶𝐻 = 𝐻𝑎𝑟𝑑𝑛𝑒𝑠𝑠 𝑅𝑎𝑡𝑖𝑜 𝐹𝑎𝑐𝑡𝑜𝑟

𝑁𝑏𝑝 = 𝑆𝑎𝑓𝑒𝑡𝑦 𝑆𝑎𝑓𝑡𝑜𝑟 𝑓𝑜𝑟 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑜𝑛 𝑝𝑖𝑛𝑖𝑜𝑛

𝑁𝑏𝑔 = 𝑆𝑎𝑓𝑒𝑡𝑦 𝑆𝑎𝑓𝑡𝑜𝑟 𝑓𝑜𝑟 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑜𝑛 𝑔𝑒𝑎𝑟

𝑁𝐶𝑝 = 𝑆𝑎𝑓𝑒𝑡𝑦 𝑆𝑎𝑓𝑡𝑜𝑟 𝑓𝑜𝑟 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐹𝑎𝑡𝑖𝑔𝑢𝑒 𝑜𝑛 𝑝𝑖𝑛𝑖𝑜𝑛

𝜎𝑏𝑚𝑎𝑥= 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑜𝑛 𝑠ℎ𝑎𝑓𝑡

𝑀 = 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑛 𝑠ℎ𝑎𝑓𝑡

𝑐 = 𝑑𝑖𝑠𝑡. 𝑓𝑟𝑜𝑚 𝑛𝑢𝑒𝑡𝑟𝑎𝑙 𝑎𝑥𝑖𝑠 𝑡𝑜 𝑜𝑢𝑡𝑒𝑟 𝑒𝑑𝑔𝑒 𝑜𝑓 𝑠ℎ𝑎𝑓𝑡

𝐼𝑠 = 𝑚𝑎𝑠𝑠 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑜𝑓 𝑠ℎ𝑎𝑓𝑡

𝑑𝑠= diameter of shaft

𝜏𝑏𝑚𝑎𝑥= max shear bending stress

𝑉𝑚𝑎𝑥= max shear force

𝐴𝑠ℎ𝑒𝑎𝑟= area under shear force

𝜏𝑡𝑜𝑟𝑠𝑖𝑜𝑛 = max 𝑡𝑜𝑟𝑠𝑖𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠

𝑇𝑠ℎ𝑎𝑓𝑡 = 𝑡𝑜𝑟𝑞𝑢𝑒 𝑜𝑛 𝑠ℎ𝑎𝑓𝑡

𝐽𝑠 = 𝑝𝑜𝑙𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑜𝑓 𝑠ℎ𝑎𝑓𝑡

𝜏𝑚𝑎𝑥 = 𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑙𝑒 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠

𝑁𝑀𝑆𝑆𝑇 = 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟, 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑆ℎ𝑒𝑎𝑟 𝑇ℎ𝑒𝑜𝑟𝑦

𝑁𝐷𝐸 = 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟, 𝐷𝑖𝑠𝑡𝑜𝑟𝑡𝑖𝑜𝑛 𝐸𝑛𝑒𝑟𝑔𝑦 𝑇ℎ𝑒𝑜𝑟𝑦

𝜎1 = 𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 1

𝜎3 = 𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 3

𝜎′ = 𝑣𝑜𝑛 − 𝑀𝑖𝑠𝑒𝑠 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑠𝑡𝑟𝑒𝑠𝑠

Preliminary Design

Our initial design was based on existing conveyor technology. After reviewing existing conveyor

systems, we were able to make some justifiable preliminary design decisions. Many drag chain

conveyors contain tensioning devices. These devices put an extremely high tension on the chains

in order to minimize the flexure and keep the chain from riding on the metal frame. This aim at

reducing metal on metal wear has its flaws. As a load is applied, the tension in the chain

increases, and more force is transmitted to the sprockets that drive and guide the chain. This

causes increased fatigue on the sprockets and connecting shafts. Another issue with tensioned

conveyors is that when a load is applied on the chain, it may flex so much that there is contact

between the chain and the conveyor’s metal frame. This can cause extreme friction and

premature failure of the frame and/or chain. We opted for a design that utilized Ultra High

Molecular Weight (UHMW) Polyethylene wear blocks below the chains. These blocks have low

coefficients of friction and wear extremely well. This decision reduces the need for extremely

high tension in the chain, which in turn, reduces the overall fatigue on the system. The UHMW

wear blocks can be easily replaced at regular intervals to prevent excessive wear on the chain and

conveyor frame.

In order to simplify the amount of analysis required, we chose a chain used by existing drag

chain conveyors with sprockets that match the chain’s properties. We will analyze the force

acting on the chain and UHMW blocks to determine the amount of torque required at the shaft.

The shaft diameter is determined by the sprockets that match the drive chains. However, all other

design aspects of the shaft will be analyzed and a suitable material selection will be made.

Finally, the drive gear and bearings will also be designed from scratch based on the forces

required to transmit the load at the prescribed 35 feet per minute.

Our approach to analyzing the assembly was to work backwards. From defining our load and

desired travel speed, we calculated the force required to move the load. This force calculation

was the foundation in designing the components of the conveyor.

Force Analysis

The first step in our design was to use the given information to find the forces transmitted

throughout the system. By analyzing the 3000 pound load acting downward on the chains and

using the coefficient of kinetic friction of the UHMW block, we were able to determine the force

needed in the chains to move the load at 35 feet per minute. The 3000 pound load is applied

evenly between the 2 chains at 1500 pounds per chain. Each 1500 pound load is distributed

over 4 feet of chain. The detailed calculations of our force analysis are shown below:

𝐖𝐩 = (1500 𝑙𝑏

4 𝑓𝑡) ∗ (

1 𝑓𝑡

12 𝑖𝑛) = 𝟑𝟏. 𝟐𝟓

𝒍𝒃

𝒊𝒏 (Eq. 1)

𝐖𝐂 = (4𝑙𝑏

𝑓𝑡) ∗ (

1 𝑓𝑡

12 𝑖𝑛) = . 𝟑𝟑𝟑

𝒍𝒃

𝒊𝒏 - Data from ANSI Chain # 50 (Eq. 2)

𝐖𝐋 = 𝐿𝑜𝑎𝑑 + 𝐶ℎ𝑎𝑖𝑛 𝑤𝑒𝑖𝑔ℎ𝑡 = 31.25𝑙𝑏

𝑖𝑛+ .33

𝑙𝑏

𝑖𝑛= 𝟑𝟏. 𝟓𝟖

𝒍𝒃

𝒊𝒏 (Eq. 3)

𝐖𝐓 = (31.58𝑙𝑏

𝑖𝑛) ∗ (4𝑓𝑡 ∗ (

12 𝑖𝑛

1 𝑓𝑡)) + (

.33 𝑙𝑏

𝑖𝑛) ∗ (6 𝑓𝑡 ∗ (

12 𝑖𝑛

1 𝑓𝑡)) = 𝟏𝟓𝟑𝟗. 𝟔 𝒍𝒃 (Eq. 4)

𝜇𝑘 = 𝟎. 𝟐𝟓 𝑓𝑜𝑟 𝑈𝐻𝑀𝑊

𝑄𝑇 = WT = 𝟏𝟓𝟑𝟗. 𝟔 𝒍𝒃

𝑭𝒇 = 𝜇𝑄𝑇 = (0.25) ∗ (1539.6 𝑙𝑏) = 𝟑𝟖𝟒. 𝟗 𝒍𝒃 (Eq. 5)

𝑉𝑐 = (35𝑓𝑡

𝑚𝑖𝑛) ∗ (

1 𝑚𝑖𝑛

60 𝑠) ∗ (

12 𝑖𝑛

1 𝑓𝑡) = 7

𝑖𝑛

𝑠 (Eq.6)

→ 10 𝑓𝑡 ∗ (12 𝑖𝑛

1 𝑓𝑡) = 120 𝑖𝑛 ∗ (

1

(7𝑖𝑛

𝑠)) = 𝟏𝟕. 𝟏𝟒 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 𝒕𝒐 𝒕𝒓𝒂𝒗𝒆𝒍 𝟏𝟎 𝒇𝒆𝒆𝒕

∑ 𝐹 = 𝑚𝑎 → 𝐹𝐶 − 384.9 𝑙𝑏𝑠 = [3000 𝑙𝑏+(

.33 𝑙𝑏

𝑖𝑛)∗(10𝑓𝑡)∗(

12 𝑖𝑛

1 𝑓𝑡)]

32.2 𝑓𝑡

𝑠2

∗ (𝑎) (Eq. 7.a)

𝐹𝐶 − 384.9 𝑙𝑏 = 94.41 𝑠𝑙𝑢𝑔𝑠 ∗ (𝑎) → 𝑎 = 𝑑𝑉

𝑑𝑡 (Eq. 7.b)

𝐹𝐶 − 384.9 𝑙𝑏 = 94.41 𝑠𝑙𝑢𝑔𝑠 ∗ (𝑑𝑉

𝑑𝑡) → ∫ (𝐹𝐶 − 384.9 𝑙𝑏

𝑡

0)𝑑𝑡 = ∫ (94.41 𝑠𝑙𝑢𝑔𝑠)𝑑𝑉

𝑉

0 (Eq. 7.c)

∫ (𝐹𝐶 − 384.9 𝑙𝑏17.14 𝑠

0)𝑑𝑡 = ∫ (94.41 𝑠𝑙𝑢𝑔𝑠)𝑑𝑉

7𝑖𝑛

𝑠0

(Eq. 7.d)

→ 𝐹𝐶(17.14 𝑠) − 384.9 𝑙𝑏(17.14 𝑠) = 94.41 𝑠𝑙𝑢𝑔𝑠 ∗ 7𝑖𝑛

𝑠 (Eq. 7.e)

→ 𝐹𝐶 =1

17.14 𝑠∗ [(660.87

𝑠𝑙𝑢𝑔∗𝑖𝑛

𝑠) ∗ (

𝑓𝑡

12 𝑖𝑛) + 6597.186 𝑙𝑏]=388.11 lbs (Eq. 7.f)

→ 𝐹𝑇 = 388.113 𝑙𝑏 𝑝𝑒𝑟 𝑠𝑖𝑑𝑒 ∗ 2 𝑠𝑖𝑑𝑒𝑠 = 776.226 𝑙𝑏 (Eq. 8)

Sprocket Design

Once the force to move the maximum load of 3000 pounds was calculated, we were able to use

this information along with idealization of the chains to calculate the torque and angular velocity

required at the sprockets and the shaft. The chains were idealized to have no stretch and it was

assumed that any force on the chains was transmitted by the sprockets. We began with an ANSI

No. 50 chain, which has a pitch of 5/8 inch.

By using the properties of the ANSI No. 50 chain, the AGMA equations and the assumption that

each sprocket has 40 teeth we were able to calculate a pitch diameter of the sprockets. The bore

size was assumed to be 1.75 inches. This starting value would determine the diameter of the shaft

and its feasibility is found later in the calculations. The calculation for finding the pitch diameter

of a sprocket was found by using the following equation:

𝑃𝑑𝑠1 = 𝑃𝑑𝑠2 = 𝑐ℎ𝑎𝑖𝑛 𝑝𝑖𝑡𝑐ℎ

sin (180

𝑁)

=(

5

8𝑖𝑛)

sin(180

40)

= 7.966 𝑖𝑛 (Eq. 8)

Based on this pitch diameter, the assumed number of teeth and bore size, we chose part

#6236K744 from McMaster-Carr, a large material and component supplier. From this pitch

diameter and our calculated total force, we were able to find the torque required at each sprocket

to move the load at 35 feet per minute.

𝑇𝑠𝑝1 = 𝑇𝑠𝑝2 = 𝐹𝑐 (𝑃𝑑

2) = (244.268 𝑙𝑏) (

7.966 𝑖𝑛

2) = 972.9 𝑙𝑏 ∗ 𝑖𝑛 (

1 𝑓𝑡

12 𝑖𝑛) = 81.076 𝑙𝑏𝑓𝑡 (Eq.9)

Thus,

Tsp1 = Tsp2 = 81.076 lb * ft

𝑻𝒔𝒉𝒂𝒇𝒕 = 𝑻𝒔𝒑𝟏 + 𝑻𝒔𝒑𝟐 = 𝟏𝟔𝟐. 𝟏𝟓 𝐥𝐛𝐟𝐭 (Eq.10)

The rotational speed (angular velocity) of the sprockets is easily found through basic dynamic

relationships. The chain must move at 35 feet per minute, and this speed corresponds with the

tangential velocity of the sprockets at the pitch point or pitch diameter distance. We calculated

the angular velocity of the sprocket and used the same value for the angular velocity of the shaft.

Since both components are connected to each other, they share the same rotational speed.

𝑉𝑐 = (𝑃𝑑𝑠1

2) ∗ 𝜔𝑠𝑝 → 𝜔𝑠𝑝 = (

2∗35 𝑓𝑡

7.966 𝑖𝑛∗𝑚𝑖𝑛) ∗ (

12 𝑖𝑛

𝑓𝑡) (

1 𝑟𝑒𝑣

2𝜋 𝑟𝑎𝑑) = 16.78 𝑅𝑃𝑀 =̃ 17 𝑅𝑃𝑀 (Eq.11)

Thus; 𝛚𝐬𝐩 = 𝛚𝐬𝐡𝐚𝐟𝐭 = 𝟏𝟕 𝐫𝐩𝐦

Gear Design

In designing our gear and pinion mesh, we referred heavily to Machine Design – An Integrated

Approach by Dr. Robert L. Norton. Our starting point was to define the pressure angle for the

gear teeth. We chose a 20 degree pressure angle based on the fact that is the most widely used

pressure angle in industry. Table 12-5 (Norton) shows the minimum number of pinion teeth to

avoid interference between a full depth pinion and gear with a 20 degree pressure angle. We

wanted to avoid interference so that we could eliminate the need to undercut the teeth. This saves

time in drafting and manufacturing. We chose the range of 16 teeth minimum for the pinion to

101 teeth maximum for the gear; this range would give us more gear ratio options. We chose a

standard full depth gear teeth because the equations for the geometry of this gear with a 20

degree pressure angle are readily available in Table 12-1 (Norton). Other standard tables have

also been established for several factors, including the AGMA bending geometry factor J and I.

We referred to these tables to narrow down our gear ratios. In the end, we chose the number of

teeth to correspond with a pre-defined J and I values from the AGMA tables to further ease

calculations. Because we also know that the motor will be at a lower torque and higher speed

than the shaft requires, we estimated a required gear ratio of at least 2 to 1.

Referring to the AGMA tables for bending and surface geometry factors and our teeth range to

avoid interference, we were able to choose a desirable number of teeth for both the gear and

pinion. We decided on 21 teeth for the pinion and 55 teeth for the gear, giving us a 2.62:1 gear

ratio. We assumed a pinion pitch diameter of 3 inches. This would roughly make the gear pitch

diameter very close to the pitch diameter of the sprockets. Once these factors were established

we could now find the geometry for the gear and pinion.

Gear and Pinion Dimensional Calculations

One of the most useful properties for calculating the geometry of a gear is the diametral pitch.

Once this diametral pitch is defined, all remaining gear and pinion geometry can be found based

on this value. To find the diametral pitch, we used our assumed starting pinion pitch diameter

and number of teeth on the pinion.

𝑃𝑑𝑝 =𝑁𝑡𝑝

𝑑𝑝=

21 𝑡𝑒𝑒𝑡ℎ

3 𝑖𝑛= 7

𝑡𝑒𝑒𝑡ℎ

𝑖𝑛 (Eq.12)

In order for gears to mesh properly, they must have the same pitch angle and diametral pitch.

Using this information with our assumed 55 teeth on the gear, the pitch diameter of the gear can

be calculated.

𝑃𝑑𝑔 = 7𝑡𝑒𝑒𝑡ℎ

𝑖𝑛=

𝑁𝑡𝑔

𝑑𝑔=

55 𝑡𝑒𝑒𝑡ℎ

𝑑𝑔→ 𝑑𝑔 =

55 𝑡𝑒𝑒𝑡ℎ

(7𝑡𝑒𝑒𝑡ℎ

𝑖𝑛)

= 7.857 𝑖𝑛 =̃ 7.86 𝑖𝑛 (Eq.13)

With the known pitch diameters and diametral pitches of both gear and pinion, we can now

calculate the remaining dimensions of both the gear and pinion.

For quick and simple calculations for the gear geometries, we created a Microsoft Excel

spreadsheet (See appendix) that included all of the formulas for calculating gear geometry from

Table 12-1 (Norton). These values apply to both the gear and pinion in order to ensure the gears

mesh and transmit power properly. The calculated gear geometries are listed below:

Pc = 0.446 in

Pb = 0.4215 in

Pd = 7 teeth/in

a = 0.1429 in

b = 0.1786 in

𝑫𝑾=0.2857 in

𝑫𝑾𝒉 = 0.3214 in

𝒕𝒄𝒕 = 0.2244 in

𝒓𝒇 = 0.0429 in

𝑪𝒃𝒎𝒊𝒏 = 0.0357 in

𝑾𝒕𝒍𝒎𝒊𝒏 = 0.0357 in

𝑪𝒔𝒈= 0.0500 in

Another critical property needed for gear design is the face width. Face width (𝑊𝑓 ) is roughly

expressed as a range function of diametral pitch (8/Pd < 𝑊𝑓 < 16/Pd) (Norton). For our design, we

used the following to find a starting face width:

8

𝑃𝑑< 𝑊𝑓 <

16

𝑃𝑑 →

87 𝑡𝑒𝑒𝑡ℎ

𝑖𝑛

< 𝑊𝑓 < 16

7 𝑡𝑒𝑒𝑡ℎ

𝑖𝑛

→ 𝟏. 𝟏𝟒 𝒊𝒏

𝒕𝒐𝒐𝒕𝒉< 𝑊𝑓 < 2.29

𝒊𝒏

𝒕𝒐𝒐𝒕𝒉 (Eq. 14)

We took the average of this range and defined our starting face width (𝑊𝑓) of both gear and

pinion to be 1.714 inches.

𝑾𝒇 = 1.714 in

At this point, we have calculated all of the dimensions of the gear and pinion. We can now find

the length of action of the gear mesh. Note that “𝑎” represents the addendum length and is the same

for both the gear and the pinion.

𝑍 = √[(𝑟𝑝 + 𝑎)2

− (𝑟𝑝𝑐𝑜𝑠∅)2

] + √[(𝑟𝑔 + 𝑎)2

− (𝑟𝑔𝑐𝑜𝑠∅)2

] − 𝐶𝑠𝑖𝑛(∅) (Eq.15)

𝑍 = √[(1.5 + .1429)2 − (1.5𝑐𝑜𝑠(20))2] + √[(3.93 + .1429)2 − (3.93cos (20))2] − 5.43 sin(20)

Thus, Z = .7037

With the length of action, we can calculate the contact ratio (mp) and ensure that it agrees with

typical values for spur gears. According to Norton, the contact ratio is calculated using the length

of action (Z), the diametral pitch (Pd), and the pressure angle (Ø).

𝑚𝑝 =𝑝𝑑𝑍

𝜋cos (∅)→ 𝑚𝑝 =

7∗.7037

𝜋cos (20)= 1.67 Thus; mp = 1.67 (Eq.16)

This contact ratio falls within the acceptable range for spur gear sets of 1.4 to 2 (Norton). The

gear contact ratio ensures that more than one tooth of the gear carries the load at any given time.

Although true in theory, in reality, there will be times when one tooth carries the entire load. This

occurs at the center of the mesh where the load is applied at a lower position on the tooth. This

position is referred to as the highest point of single tooth contact (HPSTC) (Norton). We

assumed this information when we chose our number of teeth based on the geometrical bending

factors (J and I). This assumption will provide a more accurate stress analysis.

Gear and Pinion Stress Analysis

The first step in beginning a stress analysis is to gather the required values for the calculations.

We must first find the number of cycles for the gear and pinion. We already found that the gear

and shaft will spin at 17 revolutions per minute and we have the gear ratio, thus we can find the

rotational speed of the pinion and in turn the number of cycles. After finding the rotational speed

of the pinion, we will find the number of cycles for 5 years at 8 hour days and 250 working days

in each year. We decided this was a reasonable period of working time for such a system.

𝜔𝑝 = 𝜔𝑔 ∗ 𝑔𝑒𝑎𝑟 𝑟𝑎𝑡𝑖𝑜 → 𝜔𝑝 = 17 𝑟𝑝𝑚 ∗ (2.62) = 44.54 =̃ 45 𝑟𝑝𝑚 (Eq.17)

𝑁𝑐𝑦𝑐−𝑝 = 45𝑟𝑝𝑚 (60𝑚𝑖𝑛

ℎ𝑟) (

8ℎ𝑟

𝑑𝑎𝑦) (

250 𝑑𝑎𝑦

1 𝑦𝑟) (

5 𝑦𝑟

1) = 2.7𝑥107 𝑐𝑦𝑐𝑙𝑒𝑠 𝑖𝑛 5 𝑦𝑒𝑎𝑟𝑠 (Eq.18)

𝑁𝑐𝑦𝑐−𝑔 = 17𝑟𝑝𝑚 (60𝑚𝑖𝑛

ℎ𝑟) (

8ℎ𝑟

𝑑𝑎𝑦) (

250 𝑑𝑎𝑦

1 𝑦𝑟) (

5 𝑦𝑟

1) = 1.02𝑥107 𝑐𝑦𝑐𝑙𝑒𝑠 𝑖𝑛 5 𝑦𝑒𝑎𝑟𝑠 (Eq.19)

Thus;

ωp = 45 rpm

𝑵𝒄𝒚𝒄−𝒑 =2.7x107 cycles in 5 years

𝑵𝒄𝒚𝒄−𝒈 =1.02x107 cycles in 5 years

We assumed a gear quality value of Qv = 5, based on typical values found in Table 12-6 for

mining conveyor systems (Norton). With this assumption and the data calculated so far we can

now begin the calculation of various correction factors for the stress analysis of the gear and

pinion.

Gear and Pinion Bending Stress Calculations

The bending stress for a spur gear is defined by the AGMA Bending Stress Equation (Norton).

The equation is as follows:

𝜎𝑏 = (𝑊𝑡𝑃𝑑

𝑊𝑓𝐽) (

𝐾𝑎𝐾𝑚

𝐾𝑣) 𝐾𝑠𝐾𝐵𝐾𝐼 (Eq.20)

The Wt variable represents the tangential force on the tooth of the pinion and gear; other known

values include the face width and diametral pitch. The remaining variables used in this equation

are defined and justified below for both the gear and the pinion:

Tangential Forces (Wt) – Same for both pinion and gear

This force is equal and opposite for both gear and pinion because they are in mesh. Because we

already found the required torque on the shaft we can use this value and the pitch diameter of the

gear to find the tangential force that acts on the gear and pinion teeth.

𝑊𝑡 =𝑇𝑔

𝑟𝑔=

162.15 𝑙𝑏𝑓𝑡

3.93 𝑖𝑛(

12 𝑖𝑛

𝑓𝑡) = 495.11 𝑙𝑏𝑠 (Eq.21)

𝑾𝒕 = 𝟒𝟗𝟓. 𝟏𝟏 𝒍𝒃𝒔

Bending Strength Geometry Factor (J) – Table 12-9 (Norton)

Jg = 0.40 Jp = 0.34

Dynamic Factor (Kv) - Same for both pinion and gear

For gears with a Qv factor less than or equal to 5 this factor is defined as:

𝐾𝑣 =50

50+√𝑉𝑡 in our case, the Vt term (pitch line velocity) is equal to 35 ft/min for both gears.

𝐾𝑣 =50

50+√35𝑓𝑡/𝑚𝑖𝑛= 0.89419 (Eq.22)

𝑲𝒗 = 𝟎. 𝟖𝟗𝟒𝟏𝟗

Load Distribution Factor (Km) - Table 12-6 (Norton) - Same for both pinion and gear

For face widths less than 2 inches, a Km factor is equal to 1.6.

Km = 1.6

Rim Thickness Factor (KB = 1) - Same for both pinion and gear

For solid-disk gears this value is always set to 1(Norton).

Idler Factor (KI = 1) - Same for both pinion and gear

This value is always set to 1 for a non-idler gear (Norton).

Application Factor (Ka) – Table 12-17 (Norton) - Same for both pinion and gear

In our situation, we will be using an electric motor as the driving machine, which transmits a

uniform load. Torque at start up may cause slight shock; therefore the decision was made to

overdesign the component with a Ka value of 1.25.

Ka = 1.25

Size Factor (Ks) - Same for both pinion and gear

In order to be conservative we chose a value of 1.25, typically this would be set to 1 unless we

wanted to account for special situations (Norton)

Ks = 1.25

Now that all of the necessary factors have been found, the bending stress for the gear and pinion

teeth can be calculated. We employed Microsoft excel to speed these calculations as well as

those calculations for the surface stresses. The obtained values for bending stress on the gear and

pinion are found as:

𝝈𝒃𝒑 = 𝟏𝟔. 𝟔 𝒌𝒔𝒊 𝝈𝒃𝒈 = 𝟏𝟒. 𝟏𝟒 𝒌𝒔𝒊

Gear and Pinion Surface Stress Calculations

Because the gear teeth experience not only bending stresses but also surface stresses or pitting

stresses, we must also take this into account for our design. The AGMA defines the pitting

resistance formula as follows (Norton):

𝜎𝑐 = 𝐶𝑝√(𝑊𝑡

𝑊𝑓𝐼𝑑𝑝) (

𝐶𝑎𝐶𝑚

𝐶𝑣) 𝐶𝑠𝐶𝑓 (Eq.23)

This equation is only applied to the pinion because it experience more revolutions and thus will

fail before the gear. The “C” variables used in this equation correspond to the “K” variables in

the bending stress equation with exception of (Cp), and (Cf). The value of (𝑑𝑝) is the pitch

diameter of the pinion, for reasons mentioned above.

Surface Geometry Factor (I)

The variable (I) represents the surface geometry factor, which takes into account the radii of

curvature of the gear teeth and pressure angle. This value is readily available from AGMA 908-

B89.

I = .102

Elastic Coefficient (Cp)

This value accounts for differences in tooth material and is found by (Norton):

𝐶𝑝 = √(1

𝜋(1−𝑣𝑝

2

𝐸𝑝)+(

1−𝑣𝑔2

𝐸𝑔)

) = √(1

𝜋(1−0.282

30𝑥106 )+(1−0.282

30𝑥106 )) = 2276.72 (Eq.24)

Cp= 2276.72

Surface Finish Factor (Cf)

This is usually set to 1 for gears made with standard manufacturing methods (Norton).

Now that all of the factors have been defined, we may calculate the pitting resistance or surface

stress of the pinion.

𝜎𝑐𝑝 = 2276.72√(495.11

1.714∗.102∗3) (

1.25∗1.6

.89419) 1.25 ∗ 1 (Eq.25)

𝝈𝒄 = 𝟏𝟏𝟔. 𝟗 𝒌𝒔𝒊

Now that the bending and surface stresses have been calculated, we may choose a gear and

pinion material and determine the actual bending and surface strengths of that material. Once

determined, a safety factor will be calculated and re-evaluation made if required.

Bending Fatigue Strengths for Gear Materials

In order to accurately design a gear set, one must ensure an acceptable safety factor. In order to

check the safety of our design we must find the associated AGMA bending fatigue strengths

associated with the gear materials. The AGMA publishes values; however they must be corrected

to account for the situational use of a particular gear set. This corrected Bending fatigue

strength(𝑆𝑓𝑏) is found by:

𝑆𝑓𝑏 = 𝐾𝐿

𝐾𝑇𝐾𝑅𝑆𝑓𝑏

′ (Eq.26)

where 𝑆𝑓𝑏′ is the uncorrected bending fatigue strength given by AGMA. This is found for Grade

2 gear materials by:

𝑆𝑓𝑏′ =6235 + 174*HB -0.126 HB2 (Eq.27)

Grade 2 implies a premium quality by the AGMA, ensuring a good homogenous microstructure

and purity of the metal used to make the gear. HB designates the Brinell hardness of the metal to

be used. We first decided on a steel that had HB = 300, this made our 𝑺𝒇𝒃′ = 47,095 psi. Other

factors in this equation are defined and justified below:

Life Factor (KL)

This factor is based on the life required by the part. This will be different for the pinion and gear

because they have a different number of cycles (N). An acceptable calculation of this factor for

commercial uses is given by Figure 12-24 as (See Norton):

𝐾𝐿 = 1.3558(𝑁)−0.0178 (Eq.28)

𝑲𝑳𝒑 = 𝟏. 𝟑𝟓𝟓𝟖(𝟐. 𝟕𝒙𝟏𝟎𝟕)−𝟎.𝟎𝟏𝟕𝟖 = . 𝟗𝟗𝟗𝟖

𝑲𝑳𝒈 = 𝟏. 𝟑𝟓𝟓𝟖(𝟏. 𝟎𝟐𝒙𝟏𝟎𝟕)−𝟎.𝟎𝟏𝟕𝟖 = 𝟏. 𝟎𝟏𝟕

Temperature Factor (KT =1) – Same for both gear and pinion

For temperatures less than 250°F, this value is set to 1(Norton). We are designing this for use at

no more than this temperature.

Reliability Factor (KR = 1.50) – Same for both gear and pinion (Table 12-19)(Norton)

We initially chose a reliability of 99.99%, which corresponds to a value of KR = 1.50.

Now that all factors have been identified we can calculate the actual bending strength of the gear

and pinion.

𝑆𝑓𝑏𝑔=

𝐾𝐿𝑔

𝐾𝑇𝐾𝑅𝑆𝑓𝑏

′ =1.017

1∗1.5∗ 47,095 = 31.93 𝑘𝑠𝑖 → 𝑺𝒇𝒃𝒈

= 𝟑𝟏. 𝟗𝟑 𝒌𝒔𝒊 (Eq.26a)

𝑆𝑓𝑏𝑝=

𝐾𝐿𝑝

𝐾𝑇𝐾𝑅𝑆𝑓𝑏

′ =.9998

1∗1.5∗ 47,095 = 31.39 𝑘𝑠𝑖 → 𝑺𝒇𝒃𝒑

= 𝟑𝟏. 𝟑𝟗 𝒌𝒔𝒊 (Eq.26b)

Surface -Fatigue Strengths for Gear Materials

We previously calculated the surface-fatigue stress for the pinion and now must calculate the

actual strength of the pinion material in order to have a comparative analysis and formulate a

safety factor. This calculation will not be done for the gear because it relies on the number of

cycles of the gear or pinion being analyzed; in our case the pinion has significantly more cycles

than the gear and thus will fail before the gear does. The AGMA publishes surface-fatigue

strength values; however, like the bending strength values they must be corrected to account for

operational situations. The AGMA surface- fatigue strength calculation is given by:

𝑆𝑓𝑐𝑝=

𝐶𝐿𝐶𝐻

𝐶𝑇𝐶𝑅𝑆𝑓𝑐𝑝

′ (Eq.29)

where 𝑆𝑓𝑐′ is the uncorrected surface-fatigue strength given by AGMA. This is found for Grade 2

gear materials by: 𝑆𝑓𝑐′ =27000 +364HB. (Eq.30)

HB designates the Brinell hardness of the metal to be used. We first decided on a steel that had

HB = 300, this made our 𝑺𝒇𝒄′ = 136200 psi. The factors CT and CR are identical in value to the K

values with the same subscript. Other factors in this equation are defined and justified below:

Surface-Life Factor (CL)

The publish AGMA values for surface – fatigue strength are based on 1 x107 cycles. So this

factor allows for correction based on the actual number of cycles of the pinion (2.7 x107). From

Table 12-26 (Norton) and classifying our design for commercial use, we can calculate this value

using the equation: 𝐶𝐿 = 1.448𝑁−0.023 (Eq.31)

This makes our CL = .977

Hardness Ratio Factor (CH)

This factor takes into account the difference in hardness of the gear and pinion teeth, if one

exists. Because we are making both gear and pinion out of the same material with the same heat

treatment, this value is set to CH = 1 (Norton – page 728).

Now that the factors have been identified and calculated, we may calculate the actual surface-

fatigue strength of the pinion.

𝑆𝑓𝑐𝑝=

𝐶𝐿𝐶𝐻

𝐶𝑇𝐶𝑅𝑆𝑓𝑐𝑝

′ = .977∗1

1∗1.5∗ 136200 = = 𝟖𝟖. 𝟕 𝒌𝒔𝒊 (Eq.29a)

Gear and Pinion Safety Factor Calculation

Now that the bending and surface-fatigue stresses and strengths have been computed, we can

calculate the safety factor for both the gear and pinion. These are basic calculations found by:

𝑁𝑏 = 𝑠𝑓𝑏

𝜎𝑏 and 𝑁𝑐 =

𝑠𝑓𝑐

𝜎𝑐 (Eq.32 and 33)

The safety factors for the pinion are:

𝑁𝑏𝑝 = 31.39 𝑘𝑠𝑖

16.6 𝑘𝑠𝑖= 𝟏. 𝟖𝟗 𝑁𝑐𝑝 =

88.7 𝑘𝑠𝑖

116.9 𝑘𝑠𝑖=. 𝟕𝟓𝟒

The safety factor for the gear is:

𝑁𝑏𝑔 = 31.93 𝑘𝑠𝑖

14.14 𝑘𝑠𝑖= 𝟐. 𝟐𝟔

It is clear from the safety factors that our pinion will fail due to surface-fatigue stresses. There

are a number of values that we could change to increase the resistance to these stresses. These

values include: the reliability, the material hardness, the number of cycles and face width. In

order to speed the calculation process for changes we might make, we created a Microsoft Excel

spreadsheet (See appendix) that did all calculations based on the input parameters.

Gear and Pinion Material Selection and Final Values

After much iteration, we decided to reduce the reliability of our gear and pinion from 99.99 % to

99% and make the components from the following material, keeping the initial dimensions the

same, including the face width.

Gear Material (Table 12-20 Norton)

o AGMA Grade 2 (Premium Grade)

o Class A5 Steel

o Carburized and case hardened

o Brinell Hardness = 710 (corresponds to 64HRC)

o Qv = 5

Once this material was chosen, the calculations were easily repeated in the Excel spreadsheet and

the safety factors are as follows:

Gear

o Bending Safety Factor (𝑁𝑏𝑔) = 4.76

Pinion

o Bending Safety Factor (𝑁𝑏𝑝) = 3.98

o Surface Fatigue Safety Factor(𝑁𝑐𝑝) = 2.38

These safety factors are more than adequate for our design and from this point we moved on to

designing the shaft and bearings.

Shaft Design for Conveyor System In order to design an acceptable shaft to transmit power from the gear to the sprockets some

assumptions needed to be made. We determined earlier that the chain sprockets have an inside

bore of 1.75 inches. To accommodate these sprockets we decided to choose a shaft that had an

outside diameter of 1.75 inches. The length of the shaft was determined by close inspection of

existing conveyor systems and standard dimensions of the pallets that the conveyor would move.

The width of a wooden pallet is standardized at 48 inches. We decided on roughly a 2 to 3 inch

overhang on the outside of each sprocket. This allows the main supports of the pallet to be

directly over the chain. The remaining distances were found through iteration of the design and

placement of the different components.

We already had the dimensions of the sprockets and the width of their hubs; we also found the

dimensions of the gear. The bearings dimensions were found from a standard manufacturer’s

catalog. This particular manufacturer had one pillow block bearing housing with a very wide

range of dynamic load ratings available for the inserted bearing. So we used this housing’s

dimensions to finalize our locations and would choose our actual bearing based on the dynamic

load rating later in our analysis. Referring to Norton and Mott we tried to adhere to common

design guidelines for locating our power transmitting devices that would be attached to the shaft.

Bearings were placed close to these power transmitting devices to minimize bending forces. The

shaft material was initially chosen to be SAE/AISI 1010 Cold rolled steel. The final dimensions

of the shaft and location of all of the power transmitting elements and bearings is shown in the

appendix.

Once we established the tentative location of all of the elements that will be attached to the shaft

we were able to develop a loading diagram as well as the corresponding shear and moment

diagrams. We used Autodesk® Inventor® to confirm our hand calculations and model the shaft

and location of power transmitting components. Inventor® has a very useful feature called

“Design Accelerator©” which allows the user to input the force and support bearing locations on

the shaft. With this information and several dimensions required from the user, Inventor®

automatically calculates the shear and moment diagrams and maximum values for: Shear Stress,

Bending Stress, Torsional Stress and maximum shaft deflection. One can even define the

material to be used in creating the shaft, as well as the modulus of elasticity, Poisson’s ratio and

other metallurgical characteristics of the component. The plots of the maximum forces, moments

and stresses generated by this software are shown in the appendix.

We calculated all of these values by hand and compared them with the computer generated

model. They agreed fairly well, discrepancies were attributed to rounding differences. The hand

calculations for the bending, shear and torsional stresses are calculated below:

Maximum Bending Stress

𝜎𝑏𝑚𝑎𝑥=

𝑀𝑐

𝐼𝑠=

32𝑀

𝜋𝑑𝑠3 =

32(315.17 𝑙𝑏∗𝑓𝑡)

𝜋(1.75 𝑖𝑛)3 ∗ (12𝑖𝑛

𝑓𝑡) = 7188.1 𝑝𝑠𝑖 (Eq.34)

Maximum Shear Stress

𝜏𝑏𝑚𝑎𝑥=

𝑉𝑚𝑎𝑥

𝐴𝑠ℎ𝑒𝑎𝑟=

526.88 𝑙𝑏

𝜋(1.75 𝑖𝑛)2

4

= 219.05 𝑝𝑠𝑖 (Eq.35)

Maximum Torsional Stress

𝜏𝑡𝑜𝑟𝑠𝑖𝑜𝑛 =𝑇𝑠ℎ𝑎𝑓𝑡𝑐

𝐽𝑠=

(162.15 𝑙𝑏∗𝑓𝑡)(1.75 𝑖𝑛

2)

.92 𝑖𝑛4∗

12 𝑖𝑛

𝑓𝑡= 1850.625 𝑝𝑠𝑖 (Eq.36)

With this information and the assumption of SAE/AISI 1010 Cold rolled steel we were able to

calculate several safety factors for bending, shear and torsional effects on the shaft. It is

important to note that we assumed a constant loading condition; thus, eliminating any alternating

or mean stresses. This idealization would prove to be feasible after we found our safety factors

for bending, shear and torsional stresses to be relatively high. After the actual stresses were

defined we used the Maximum Shear Stress Theory and Distortion Energy Theory to calculate

the safety factor for the maximum shear stress and maximum bending stress respectively. The

information related to the SAE 1010 Cold Rolled Steel is:

o Sy = 44,000 psi

o Sut = 53,000 psi

o Se = 26,500 psi

o Density = .284 lb/in3

o Poisson’s Ratio = 0.3

o Young’s Modulus = 30x106 psi

The calculations for the safety factors of the shaft under bending and shear are as follows:

Maximum Shear Stress Theory for Static Failure

𝜏𝑚𝑎𝑥 = √(𝜎𝑥−𝜎𝑦

2)

2

+ 𝜏𝑥𝑦2 = √(

7188.1𝑝𝑠𝑖−0

2)

2

+ (1850.625 𝑝𝑠𝑖)2 = 4042.525 𝑝𝑠𝑖

(Eq.37)

𝑁𝑀𝑆𝑆𝑇 =0.5𝑆𝑦

𝜏𝑚𝑎𝑥= 0.5∗44,000 𝑝𝑠𝑖

4042.525 𝑝𝑠𝑖= 5.442 (Eq.38)

Distortion Energy /Von Mises-Hencky Theory for Ductile Static Loading Failures

𝜎1 = 𝜎𝑥−𝜎𝑧

2+ 𝜏𝑚𝑎𝑥 =

7188.1 𝑝𝑠𝑖−0

2+ 4042.525 𝑝𝑠𝑖 = 7636.575 𝑝𝑠𝑖 (Eq.39)

𝜎3 = 𝜎𝑥+𝜎𝑧

2− 𝜏𝑚𝑎𝑥 =

7188.1 𝑝𝑠𝑖+0

2− 4042.525 𝑝𝑠𝑖 = −448.475 𝑝𝑠𝑖 (Eq.40)

𝜎′ = √𝜎12 + 𝜎3

2 − 𝜎1𝜎3 = √(7636.57 𝑝𝑠𝑖)2 + (448.47 𝑝𝑠𝑖)2 − (7636.57)(−448.47)

= 7870.39 𝑝𝑠𝑖 (Eq.41)

𝑁𝐷𝐸 = 𝑆𝑦

𝜎′ = 44,000 𝑝𝑠𝑖

7870.39 𝑝𝑠𝑖= 5.59 (Eq.42)

Using the fact that the maximum shear theory is more conservative, our safety factor comes out

to be 5.442. This is more than reasonable for the level of precision required in this type of

machine.

Stress calculations on Shaft Keys

We will be attaching the gear and sprockets to the shaft via profile parallel keys and keyways.

The keys will be made from the same SAE 1010 Steel; however, the ultimate strength will be

used to define the safety factor. This will be done such that the key will fracture and prevent the

other components from spinning in the case of failure, protecting the remaining system from

damage.

From Table 10-2 (Norton) for a shaft having a diameter equal to 1.75 inches the suggested key

width is 0.375 inches. From Table 10-3 (Norton), the suggested height for a .375 inch by 1 inch

long key is .437 inches. The length of our keys will be greater than or equal to the width of the

element they are connecting to the shaft (i.e. – gear width = 1.714, key length = 1.75 inch.).

The force experienced by the shaft at the different locations along the shaft is equal to the torque

at that location over the radius of the shaft at that location:

𝐹𝑠ℎ𝑎𝑓𝑡 @𝑔𝑒𝑎𝑟 =𝑇

𝑟=

162.15 𝑙𝑏𝑓𝑡1.75 𝑖𝑛

2

∗ (12𝑖𝑛

𝑓𝑡) = 2223.17 𝑙𝑏𝑠 (Eq.43)

𝐹𝑠ℎ𝑎𝑓𝑡 @𝑠𝑝𝑟𝑜𝑐𝑘𝑒𝑡 1 =𝑇

𝑟=

81.075 𝑙𝑏𝑓𝑡1.75 𝑖𝑛

2

∗ (12𝑖𝑛

𝑓𝑡) = 1111.89 𝑙𝑏𝑠 (Eq.44)

𝐹𝑠ℎ𝑎𝑓𝑡 @𝑠𝑝𝑟𝑜𝑐𝑘𝑒𝑡 2 =𝑇

𝑟=

81.075 𝑙𝑏𝑓𝑡1.75 𝑖𝑛

2

∗ (12𝑖𝑛

𝑓𝑡) = 1111.89 𝑙𝑏𝑠 (Eq.45)

From these forces, the corresponding shear stresses can be computed for the keys at these

locations. The forces on each sprocket are equal, thus simplifying the calculations. Once the

actual shear stresses are calculated, the Von-Mises equivalent stresses can be computed and a

safety factor developed. The width of the sprocket and key interaction is 1.25 inches, the width

of the gear and key interaction is 1.714 inches. All keys are 0.375 inches in width.

𝜏𝑘𝑒𝑦@𝑔𝑒𝑎𝑟 =𝐹

𝐴𝑠ℎ𝑒𝑎𝑟=

2223.17 𝑙𝑏𝑠

(0.375 𝑖𝑛)(1.714 𝑖𝑛)= 3458.84 𝑝𝑠𝑖 (Eq.46)

𝜏𝑘𝑒𝑦@𝑠𝑝𝑟𝑜𝑐𝑘𝑒𝑡𝑠 =𝐹

𝐴𝑠ℎ𝑒𝑎𝑟=

1111.89 𝑙𝑏𝑠

(0.375 𝑖𝑛)(1.25 𝑖𝑛)= 2372.03 𝑝𝑠𝑖 (Eq.47)

*Note that the shear stress on the key for the sprockets is equal.

𝜎′𝑘𝑒𝑦@𝑔𝑒𝑎𝑟 = √3𝜏𝑥𝑦

2 = √3(3458.84𝑝𝑠𝑖)2 = 5990.89 𝑝𝑠𝑖 (Eq.48)

𝜎′𝑘𝑒𝑦@𝑠𝑝𝑟𝑜𝑐𝑘𝑒𝑡𝑠 = √3𝜏𝑥𝑦

2 = √3(2372.03𝑝𝑠𝑖)2 = 4108.47 𝑝𝑠𝑖 (Eq.49)

*Note that the von-Mises equivalent stresses on the key for the sprockets is equal.

Calculation of the safety factors for the gear and sprocket keys. (Eq 6.18e pg 368 Norton)

𝑁𝑓 =1

𝜎′𝑎𝑆𝑒

+𝜎′𝑚𝑆𝑢𝑡

𝑤ℎ𝑒𝑟𝑒 𝜎′𝑎 = 𝜎′𝑚 (Eq.50)

𝑁𝑓𝑘𝑒𝑦 @ 𝑔𝑒𝑎𝑟=

15990.89𝑝𝑠𝑖

26500𝑝𝑠𝑖+

5990.89𝑝𝑠𝑖

53000𝑝𝑠𝑖

= 2.95 (Eq.51)

𝑁𝑓𝑘𝑒𝑦 @ 𝑠𝑝𝑟𝑜𝑐𝑘𝑒𝑡𝑠=

14108.47𝑝𝑠𝑖

26500𝑝𝑠𝑖+

4108.47𝑝𝑠𝑖

53000𝑝𝑠𝑖

= 4.30 (Eq.52)

Bearing Calculations and Design

In order to find an acceptable bearing for our application we first needed to find the forces acting

at the bearings. These forces are reactions to the radial and tangential components of the forces

acting at the gear and sprockets on the shaft. The locations of the bearings, gear and sprockets is

shown in the torque diagram in the appendix. We assumed no axial movement of the shaft

because the shaft is locked in position at the bearings by clamp collars. Also note that the

sprockets have no y-component of force because the only force experienced is that of the chain

acting in the x direction. Calculations of the forces at the gear and sprockets are shown below:

𝐹𝑔𝑥 =𝑇𝑔

𝑟𝑔=

162.15 𝑙𝑏𝑓𝑡

(7.86 𝑖𝑛

2)

∗ (12 𝑖𝑛

𝑓𝑡) = 495.11 𝑙𝑏𝑠 (Positive x-direction)

𝐹𝑔𝑦 = 𝐹𝑔𝑥𝑡𝑎𝑛(∅) = 495.11𝑙𝑏 ∗ tan(20°) = 180.21 𝑙𝑏𝑠 (Positive y-direction)

𝐹𝑠1𝑥 = 𝐹𝑔𝑥 −𝑇𝑠1

𝑟𝑠1= 495.11𝑙𝑏𝑠 −

81.075 𝑙𝑏𝑓𝑡

(7.966 𝑖𝑛

2)

∗ (12 𝑖𝑛

𝑓𝑡) = −250.85 𝑙𝑏𝑠 (Negative x-direction)

𝐹𝑠1𝑥 = 0

𝐹𝑠2𝑥 = 𝐹𝑠1𝑥 −𝑇𝑠2

𝑟𝑠2= 250.85 𝑙𝑏𝑠 −

81.075 𝑙𝑏𝑓𝑡

(7.966 𝑖𝑛

2)

∗ (12 𝑖𝑛

𝑓𝑡) = −6.58 𝑙𝑏𝑠 (Negative x-direction)

𝐹𝑠2𝑥 = 0

It should be noted that the gear supplies a torque and the first sprocket removes roughly half of

this torque to drive the chain. The remaining torque is transmitted to the second sprocket in order

to drive the second chain. A minimal amount of force remains after the second sprocket. The

bearing reaction forces are then found by taking the sum of the forces about the shaft and the

sum of the moments about one of the bearings. The magnitude of the total bearing forces is

shown below:

|𝑅𝐵1| = 379.15 𝑙𝑏𝑠

|𝑅𝐵2| = 85.62 𝑙𝑏𝑠

From these forces, specifically the higher valued force at bearing one, we can calculate the

required dynamic load rating of the bearings to be used.

The bearing housing we assumed earlier in our design is manufactured by Rexnord Inc. We

referred to their lengthy product catalog and found they had a step by step process for

determining the dynamic load rating based on the materials and tolerances they use in their

bearing design. The number of cycles for the shaft is based on the 17 rpm required to maintain a

35 feet per minute velocity of the pallet. We designed the gear to withstand 5 years of service,

and the total number of cycles in this period was calculated previously to be 1.02 x 107 cycles.

Using this value and the instructions from Rexnord, we were able to choose a bearing. The

procedure provided by Rexnord is shown below:

1. Determine the number of hours the bearing is to be used over a 10 year period.

a. ℎ𝑜𝑢𝑟𝑠 = 8 ℎ𝑟

𝑑𝑎𝑦∗

250 𝑑𝑎𝑦

𝑦𝑟∗ 10𝑦𝑟 = 20000 ℎ𝑟𝑠

2. Refer to Rexnord Table that correlates the number of hours in 10 years, and the RPM to

the C/P ratio. Choose a corresponding C/P ratio.

a. The lowest RPM that Rexnord has values for was 50 RPM, where we only

required 17 RPM. We chose the corresponding C/P ratio for the 50 RPM

b. The C/P ratio was shown to be 3.42

3. Use the C/P ratio from the table and the actual applied load (P) to determine the required

Dynamic Load Rating (C).

𝐶

𝑃= 3.42 →

𝐶

379.15= 3.42 → 𝐶 = (379.15𝑙𝑏)(3.42) = 1296.7 𝑙𝑏𝑠

The dynamic load rating required is only 1296.7 lbs. Of the bearings available to fit in our

housing, the lowest dynamic load rating was 20,200 lbs. We chose this double row roller ball

bearing, though expensive, it would last must longer than all other components and require

minimal maintenance. The part number and other information about this bearing are found in the

appendix. We chose the method of calculating the bearing dynamic load rating from the

manufacturer because the typical formula 𝐿10 = (𝐶

𝑃)

3

does not account for the bearing material

makeup, hardness, revolutions per minute and manufacturing methods. The safety factor of our

bearing is as follows:

𝑁𝐵 =𝐶𝑏𝑒𝑎𝑟𝑖𝑛𝑔

𝐶𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑=

20200

129.67= 15.57 (Eq.53)

Summary of Safety Factors and Allowable Stresses

The design of this drag chain conveyor involves many complex parts that each must be analyzed

with respect to the loads they are asked to transmit. Each component was analyzed and

prescribed a safety factor. This safety factor is a simple ratio of the maximum stress allowed on

that component before it fails to the actual stress it experiences. The higher the ratio the safer, or

more overdesigned, that component will be. Below is a summary of the safety factors for each of

our components. We list the lowest safety factor that was calculated for each component to

remain consistent to a conservative design approach.

Gear

o Bending Safety Factor (𝑁𝑏𝑔) = 4.76

Pinion

o Bending Safety Factor (𝑁𝑏𝑝) = 3.98

o Surface Fatigue Safety Factor(𝑁𝑐𝑝) = 2.38

Shaft

o Shearing Safety Factor (𝑁𝑀𝑆𝑆𝑇) = 5.442

o Bending Safety Factor (𝑁𝐷𝐸) = 5.59

Bearing

o Dynamic Load Safety Factor (𝑁𝐵) = 15.57

Shaft Keys

o Safety Factor of Gear Key (𝑁𝑓𝑘𝑒𝑦 ) = 2.95

o Safety Factor of Sprocket Keys (𝑁𝑓𝑘𝑒𝑦𝑠) = 4.30

It is evident from these values that the first component to fail will be the keys. This is consistent

with the purpose of shaft keys. Keys are designed to lock a power transmitting device to the shaft

and help transmit any rotational forces. If these forces become so much as to exceed the strength

of the key, the key will fail prior to the shaft or transmitting device fails. This is optimal because

the failure of the key keeps the main components of the system from experiencing forces that are

over their design stresses. Keys are cheap to produce and provide a means of protection to the

more expensive components.

Conclusion

In conclusion, the drive shaft conveyor was designed to move a maximum load of 3000 pounds

at a speed of 35 feet per minute. The root of all calculations was the force required to accomplish

this task. Rather than defining our starting torque at the motor and working towards a feasible

load that could be moved, we started with a typical load required in industry and worked

backwards. This process proved successful, as we were able to design components specifically

for the applied load.

The lowest safety factor was 2.38 for the pinion under surface fatigue stresses. This is

understandable because of the high number of cycles experienced by the pinion, relative to the

remaining components. A safety factor of 2.38 is an acceptable conservative number in relation

to the low speed operation of the conveyor. When the pinion “fails” according to this safety

factor, it is due to pitting on the pinion teeth. Pitting is not a catastrophic failure and the

component may still be used beyond this point for some time.

As for the remaining components, the keys connecting the gear and sprockets would be the most

likely to fail. This is a desirable result, as the keys act to protect the other components from

overloading. The bearings were purposely overdesigned to accommodate a much longer life of

the conveyor. The shaft had a very high safety factor in order to avoid bending stresses from

repeated rotation under constant torque.

Our initial assumption for the life expectancy of the system was 5 years of 250 days per year and

8 work hours per day. The calculated safety factors are all based on this assumption, with

exception of the bearing which is rated for 10 years. Our final results show that the design should

be successful under normal operating conditions implied during calculations.

References

1. Norton, Robert L., “Machine Design: An Integrated Approach”, 4th edition, 2011,

Chapters 5,6,10-12

2. Mott, Robert L., “Machine Elements In Mechanical Design”,1985, Chapters 9-14

3. Taylor et al, “Standard Handbook of Chains: Chains for Power Transmission and

Material Handling”, 2nd edition, 2005

4. AGMA, “Geometry Factors for Determining Pitting Resistance and Bending Strength of

Spur, Helical and Herringbone Gear Teeth”, AGMA Information Sheet 226.01, Rev. 908-

B89, 1989

5. Rexnord, “Link-belt, MB and Rexnord Bearings Product Catalog and Selection Guide”,

Rexnord Spherical Roller Bearings-Engineering Section, pp5-7, 2012

6. Kana Chain, www.kana-chain.com/chain/ansi-chain.htm, ANSI Chain Designations

7. ASM Handbook, Characterization and Failure Analysis of Plastics, ASM International,

2003

APPENDIX

Gear and Pinion Dimension Calculations in Microsoft Excel

Bending and Surface Stresses on Pinion and Gear Excel Screen Shot

Bending and Surface Stresses on Pinion and Gear Excel Screen Shot

Shaft Torque Diagram

Shaft Shear Diagram

Shaft Shear Stress Diagram

Sprocket 1

Gear

Bearing 1

Sprocket 2 Bearing 2

Shaft Bending Moment Diagram

Shaft Bending Stress Diagram

Shaft Torsional Stress Diagram