Lecture #10
TYPES OF BELTS and Belt
Selection
Course Name : DESIGN OF MACHINE ELEMENTS
Course Number: MET 214
Types of belts:
1) flat belt
2) V belt drive
3) synchronous belts (timing belts)
Flat belt: Simplest type often made from leather or rubber coated fabric. The sheave
(pulley) is also flat and smooth, and the driving force is therefore limited by pure friction
between the belt and the sheave. Some designers prefer flat belts for delicate machinery
because the flat belt will slip if the torque tends to rise to a level high enough to damage
the machine.
V-belt: The V shape causes the belt to wedge tightly into a groove formed in a pulley,
(called a sheave) increasing friction and allowing high torques to be transmitted before
slipping occurs. Most V belts have high strength cords positioned at the pitch diameter of
the belt cross section to increase the tensile strength of the belt. The cords, made from
natural fibers, synthetic strands, or steel, are embedded in a firm rubber compound to
provide the flexibility needed to allow the belt to pass around the sheave. Often an outer
fabric cover is added to give the belt good durability.
Synchronous belt drives
Synchronous belts are constructed with ribs or teeth across the underside of the belt. The
teeth mate with corresponding grooves in the driving and driven pulleys, called sprockets,
providing a position drive without slippage. Therefore, there is a fixed relationship
between the speed of the driver and the speed of the driven sprocket.
Examples of various types of belts are shown below.
Insert diagrams
Comparison of V-belts with Flat belts
Plain flat belts have been almost entirely replaced by V-belts and timing belts
(synchronous belts) due to their inherent drawbacks, such as excessive slippage, high
bearing pressure, axial space requirements, and noise. As detailed on a previous slide, a
distinction must be made between the outside diameter of the sheave and the pitch
diameter of the pitch circle of the sheave.
Standard V Belt Cross Sections are shown in the figures below.
Insert Images
The design of a V belt drive involves specifying 1) the sheave diameters
needed to implement a desired speed ratio existing between an input and
output shafts of a system, 2) Selecting the belt size and length and
number of belts required to transfer the amount of power from one shaft
to another required by the application consistent with the shaft to shaft
spacing that can exist in a particular application, and 3) selecting a
mounting technique for attaching the sheaves to the shafts of the system.
As noted in the book by Mott, page 273, several guidelines can be used to
initialize a design.
1) if belt speed is less than 1000 ft/min, an alternative drive component
such as gear type or chain should be considered.
2) Most commercially available sheaves are cast iron, which should be
limited to a belt speed of 6500 ft/min.
3) The angle of wrap on the smaller sheave should be greater than 1200.
The design process will be illustrated by example 7-1 page 278 in book by Mott.
Example 7-1
Design a V belt drive that has the input sheave on the shaft of an electric
motor (normal torque) noted at 50 hp at 1160 rpm full load speed. The drive
is to a bucket elevator in a potash plant that is to be used 12 hours (h) daily at
approximately 675 rpm.
Note:
Bucket elevator operates at 675 rpm. Motor provides 50 hp at 1160 rpm.
Accordingly pulley system must be designed to reduce speed of motor to
match speed of bucket elevator.
Procedure:
Step 1: Modify the normal rating by the V-belt service factor to produce the
design power. The design power is related to the normal by as shown below.
Design Power = (normal power)(service factor)
The value of the service factor depends on the applications as shown in table provided
below.
For the example under consideration:
Service factor: 1.3
Design power =(1.3)(50)=65 hp
Step 2: Select the belt section: Figure 7-9 from the book by Mott is provided below to assist in
identifying an appropriate cross section with the design power.
To utilize figure 7-9, the following issues need to be noted.
1) Design hp appears on the horizontal axis
2) Speed of faster shaft appears on vertical axis.
3) To identify an appropriate cross section for an application, the design power is located on the
horizontal axis. After locating the design power level on the X axis, move vertically until the
speed of the faster shaft is incurred. Cross section is determined from the region of the figure
containing the operating point , i.e. operating point = (X coordinate, Y coordinate) = (H.P.,
speed of faster shaft)
For the design power of 65 hp and the input (faster) speed of 1160 rpm, the cross section
recommended from the graph is a size 5V belt.
Q: Given a particular design power, why does recommend cross section decrease as speed
increases? i.e.
@ (65 hp, 110 rpm)-> 8V
@(65 hp, 500 rpm)-> 5V
@(65 hp, 4000 rpm)-> 3Vx
Step 3: Compute the nominal speed ratio
mw 
n A 1160

 1.72
nB 675
Step 4: Compute diameter of drive sheave to produce a belt speed of Vb=4000 ft/min
vb 
 Dn
 .262 Dn
12
where vb  belt speed ft/min
D  diameter of sheave in inches
n  speed of rotation of sheave in rpm
Rearranging for D
DA 
124000  13.17inches
12vb

 nA 3.141160
Step 5: Select trait sizes for the input sheave and compute the desired size of the output
sheave. In order to utilize stock components, a few iterations involving different diameter
combinations may needed to be performed until a ratio of stock diameters produces a
ratio very nearly equal to the speed ratio calculated from the shaft speeds. The degree of
mismatch that is acceptable depends on the application. Table 7-3 from the book by Mott
has been provided to illustrate the process of searching for stock components to produce
an acceptable speed ratio.
nA DA  nB DB 
nA DB

 1.72
nB DA
From previous step DA = 13.17 inches
Estimate DB = (1.72)(13.17) = 22.65
As is evident from table 7-3 below
DA = 12.4 and DB = 21.1
Step 6: Determine the rated power (for the belt size selected previously) using figures 710, 7-11 or 7-12 repeated below for convenience.
The figures provided below identify the power handling capability of a single belt.
Utilizing the design power and power rating of a belt along with some correction factors
to be discussed later will enable the designer to specify the member of belts required for
an application.
Figure 7-10 is used with 3V belts. Figure 7-11 is used with 5V belts and figure 7-12 is used
with 8V belts.
Images 7-10 and 7-11
Figure 7-12 and Figure 7-13
Viewing graph 7-11 for the small sheave pitch diameter of 12.4 at 1160 rpm. The rated
power using the solid curve for 1160 rpm is 26.4 hp. The dotted curve lying immediately
above the solid curve is an additional horse power added rating that can be added to the
basic power rating. A given belt can carry a greater power as the speed ratio increases, up
to a ratio of approximately 3.38. Figure 7-13 is a plot of the data for power to be added to
the basic rating as a function of speed ratio for 5 V belts. For a speed ratio of 1.72 at 1160
rpm, the added power is 1.15
Actual rated power of belt = 26.4 +1.15 =27.55 hp
Step 7: Specify a trail center distance. The trial distance is an estimate so that the design
process can proceed. To assist in specifying a 1st estimate on center distance, use the
relationship provided below.
D2  C  3D2  D1 
21.1  C  321.1  12.4
21.1in  C  100.5in
In the interest of conserving space, assume C = 24.0 in.
Step 8: Compute the required belt length using the equation provided below.
2
 D2  D1 
L  2C  1.57D2  D1   

 4C 
L
 D  D1 

2(24.0) + 1.57(21.1 + 12.4) +  2
 4C 
2
= 101.4 in
Step 9: Select a standard belt length from table 7-2 provided below that is close to the length
calculated in the previous step and compute the resulting center distance using the procedure
defined below. Use the values of D1 and D2 and the length L of the belt you are going to use in the
equations provided below to calculate the center to center spacing necessary for use with the belt
length chosen.
1) To find center to center separation, first calculate the following quantity
B  4L  6.28D2  D1 
2) Calculate center to center spacing using the following formula.
B  B 2  32D2  D1 
C
16
2
Step 9 continued: L = 100.0 in → 𝐵 = 189.6 → 𝐶 = 23.3 𝑖𝑛
Step 10: compute the angle of wrap of the belt around the sheaves from the following
equations.
 D  D1 
1  1800  2 sin 1  2

 2C 
 D  D1 
 2  1800  2 sin 1  2

 2C 
𝜃1 = 158 𝑑𝑒𝑔
Q: Is θ1 or θ2 associated with the larger diameter pulley and why?
Step 11: Determine the correction factors from figures 7-14 and 7-15 provided below.
𝐶𝜃 = 0.94
𝐶𝐿 = 0.96
Step 12: Compute the corrected rated power per belt and the number of belts required to carry
the design power:
Corrected power = 𝐶𝜃 𝐶𝐿 𝑃 = 24.86ℎ𝑝
Number of belts = 65/24.86 = 2.61 belts → 3 𝑏𝑒𝑙𝑡𝑠
Summary of design
Input: Electric motor, 50.0 hp at 1160 rpm
Service factor: 1.3
Design power: 65.hp
Belt: 5V cross section, 100 in length, 3 belts
Sheaves: Driver, 12.4 in pitch diameter, 3 grooves, 5V.
Driven, 21.1 in pitch diameter, 3 grooves, 5V
Actual output speed: 682 rpm