This document outlines the design and construction of a spur gear. It includes the objectives of analytically designing, CAD designing, and constructing a spur gear. It discusses applications of spur gears in increasing or decreasing power and speed. It also covers the selection of cast iron as the gear material. The problem is to design a pair of 20 degree full-depth teeth gears. The solution shows the calculation of pitch, face width, and tooth numbers. Finally, it provides steps for the CAD design and safe construction of the spur gear.

1. Department of Mechanical Engineering
Khulna University of Engineering & Technology
Khulna-9203
Supervised By
Md. Rasedul Islam
Lecturer
Department of Mechanical Engineering
Khulna University of Engineering & Technology
(KUET)
Khulna -9203, Bangladesh.
Submitted By
MOSTAQUEEM-INTESHAR (1205057)
AMITAV ROY (1205058)
K.M.MOHIUDDIN ABEER (1205060)
DESIGN AND CONSTRUCTION OF
SPUR GEAR

2. OUTLINE
 1.0 INTRODUCTION
 2.0 OBJECTIVES
 3.0 APPLICATION
 4.0 SELECTION OF MATERIAL
 5.0 PROBLEM
 6.0 SOLUTION
 7.0 CAD DESIGN
Department of Mechanical Engineering, KUET

3. 1.0 INTRODUCTION
A gear is a wheel with teeth that mesh together with other
gears. Gears change the speed, torque and direction of
rotating axles.
A spur gear is a gear having straight teeth cut on the rim
parallel to the axis of rotation.
Department of Mechanical Engineering, KUET

4. 2.0 OBJECTIVES:
1. Design analytically a spur gear.
2. CAD design of the spur gear.
3. Construction of the spur gear.
Department of Mechanical Engineering, KUET

5. 3.0 APPLICATION:
Spur gears can be used to increase or decrease the torque, or
power, of a given object. Spur gears are used to this effect in
washing machines, blenders, clothes dryers, construction
equipment, fuel pumps and mills.
2. Spur Gears For Speed
Spur gears are also used to increase or decrease the
speed of an object. For example, they are used in
mechanical clocks to adjust the relative speeds of the
second, minute and hour hands. In hand-held eggbeaters,
spur gears are used to increase the speed of it
Department of Mechanical Engineering, KUET
1. Spur Gears For Power

6. 3.0 APPLICATION (cont’d):
Department of Mechanical Engineering, KUET
3. Spur Gears and Cars :
Spur gears are not used in cars because of the loud noise they
produce at high speeds. The noise comes from the sound made when
the teeth of the gears collide. Spur gears are, however, used in
aircraft engines, where they are superior to helical gears, and where
noise is not an issue.

7. The gear material should have the following properties:
 High tensile strength to prevent failure against static loads
 High endurance strength to withstand dynamic loads
 Low coefficient of friction
 Good manufacturability
Generally cast iron, steel, brass and bronze are preferred for
manufacturing metallic gears with cut teeth. Where smooth action is
not important, cast iron gears with cut teeth may be employed.
Commercially cut gears have a pitch line velocity of about 5
metre/second. For velocities larger than this, gear sets with non-
metallic pinions as one member are used to eliminate vibration and
noise. Non-metallic materials are made of various materials such as
treated cotton pressed and moulded at high-pressure, synthetic resins
of the phenol type and rawhide.
4.0 Selection of Material:

8. 5.0 Problem:
 A pair of gears with 20 degree full-depth teeth are to
transmit 10 hp at 1750 rpm of the 3-in. pinion; velocity
ratio desired is about 3.8; intermittent service. Use a
strength reduction factor of about 1.4, with the load at the
tip and teeth commercially cut. Determine the pitch, face
width, and tooth numbers if the material is cast iron, class
20.

9. 6.0 Solution:
𝑊𝑒 𝐾𝑛𝑜𝑤,
𝜋𝐷 𝑝 𝑛 𝑝
12
=
𝜋(3)(1750)
12
= 1374 < 2000𝑓𝑝𝑚
10 hp < 20hp
Commercially cut gears,
𝑉𝑚 <2000 fpm
𝐹𝑑 =
600+ 𝑉 𝑚
600
𝐹𝑡
𝐹𝑡 =
3000 ℎ𝑝
𝑉𝑚
=
3000 10
1374
= 240 𝑙𝑏
𝐹𝑑 =
600+1374
600
𝑋 240 = 790 𝑙𝑏
𝐹𝑠 =
𝑠𝑏𝑌
𝐾𝑓 𝑃𝑑
For cast-iron , class 20 , 𝑆 𝑛 = 0.4 𝑆 𝑢
𝑆 𝑢 = .4 𝑋 20 = 8 𝑘𝑠𝑖 = 8000𝑝𝑠𝑖
𝐾𝑓 = 1.4

10. Solution (continued):
Assume, 𝑏 =
10
𝑃 𝑑
Table AT 24 , Load at tip, 20 𝑜
𝐹. 𝐷
Assume Y= 0.33
𝐹𝑠 = 𝐹𝑑
8000 10 (0.33)
1.4 𝑋 𝑃𝑑
2 = 790
So, 𝑃𝑑= 4.89
Use 𝑃𝑑 = 5
𝑁𝑝 = 𝑃𝑑 𝑋 𝐷 𝑝 = 5𝑋3 = 15
Y = 0.289
𝐹𝑠 = 𝐹𝑑
8000 𝑏 (0.33)
1.4 𝑋 5
= 790
𝑏 = 2.4 𝑖𝑛

11. 8
𝑃𝑑
< 𝑏 <
12.5
𝑃𝑑
= 2.5 𝑖𝑛
b= 2
1
2
𝑖𝑛
Summary of answers:
𝑃𝑑 = 5
b= 2
1
2
𝑖𝑛
𝑁 𝑝 = 15
𝑁𝑔 = 3.8 𝑋 15 = 57
Solution (continued):

12. 7.0 CAD Design:

13. i. Before setting up a job, be sure that the workpiece, table,
the taper in the spindle, and the arbor or cutter shank are
free from chips, nicks, or burrs.
ii. Do not select a milling cutter of larger diameter than is
necessary.
iii. Check the machine to see if it is in good running order
and properly lubricated, and that it moves freely, but not
too freely in all directions.
iv. Consider direction of rotation. Many cutters can be
reversed on the arbor, so be sure you know whether the
8.0 Construction:

14. v. Spindle is to rotate clockwise or counterclockwise.
vi. Feed the workpiece in a direction opposite the rotation of
the milling cutter (conventional milling).
vii. Do not change feeds or speeds while the milling machine
is in operation.
viii. When using clamps to secure a workpiece, be sure that
they are tight and that the piece is held so it will not
spring or vibrate under cut.
ix. Use a recommended cutting oil liberally.
x. Use good judgment and common sense in planning every
job, and profit from previous mistakes.
xi. Set up every job as close to the milling machine spindle
as circumstances will permit.
8.0 Construction (continued):