8/11/2019 MET_214_-_Lecture_10

1/22

Lecture #10

TYPES OF BELTS and Belt

Selection

Course Name : DESIGN OF MACHINE ELEMENTS

Course Number: MET 214

8/11/2019 MET_214_-_Lecture_10

2/22

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.

8/11/2019 MET_214_-_Lecture_10

3/22

Examples of various types of belts are shown below.

8/11/2019 MET_214_-_Lecture_10

4/22

Insert diagrams

8/11/2019 MET_214_-_Lecture_10

5/22

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.

8/11/2019 MET_214_-_Lecture_10

6/22

Standard V Belt Cross Sections are shown in the figures below.

8/11/2019 MET_214_-_Lecture_10

7/22

Insert Images

8/11/2019 MET_214_-_Lecture_10

8/22

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 andnumber 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.

8/11/2019 MET_214_-_Lecture_10

9/22

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 electricmotor (normal torque) noted at 50 hp at 1160 rpm full load speed. The driveis to a bucket elevator in a potash plant that is to be used 12 hours (h) daily atapproximately 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 tomatch speed of bucket elevator.

Procedure:

Step 1: Modify the normal rating by the V-belt service factor to produce thedesign power. The design power is related to the normal by as shown below.

Design Power = (normal power)(service factor)

8/11/2019 MET_214_-_Lecture_10

10/22

The value of the service factor depends on the applications as shown in table providedbelow.

For the example under consideration:

Service factor: 1.3

Design power =(1.3)(50)=65 hp

8/11/2019 MET_214_-_Lecture_10

11/22

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.

8/11/2019 MET_214_-_Lecture_10

12/22

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 thehorizontal axis. After locating the design power level on the X axis, move vertically until thespeed 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 sectionrecommended from the graph is a size 5V belt.

Q: Given a particular design power, why does recommend cross section decrease as speedincreases? i.e.

@ (65 hp, 110 rpm)-> 8V

@(65 hp, 500 rpm)-> 5V

@(65 hp, 4000 rpm)-> 3Vx

Step 3: Compute the nominal speed ratio

Step 4: Compute diameter of drive sheave to produce a belt speed of V b=4000 ft/min

where belt speed ft/min

diameter of sheave in inches

speed of rotation of sheave in rpm

72.1

675

1160

B

A

w

n

nm

DnDn

vb

262.12

n

D

vb

8/11/2019 MET_214_-_Lecture_10

13/22

Rearranging for D

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 diametercombinations may needed to be performed until a ratio of stock diameters produces aratio very nearly equal to the speed ratio calculated from the shaft speeds. The degree ofmismatch that is acceptable depends on the application. Table 7-3 from the book by Motthas been provided to illustrate the process of searching for stock components to producean acceptable speed ratio.

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

inchesn

vD

A

b

A 17.13

116014.3

40001212

72.1A

B

B

A

BBAA D

D

n

nDnDn

8/11/2019 MET_214_-_Lecture_10

14/22

Step 6: Determine the rated power (for the belt size selected previously) using figures 7-

10, 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.

8/11/2019 MET_214_-_Lecture_10

15/22

Images 7-10 and 7-11

8/11/2019 MET_214_-_Lecture_10

16/22

8/11/2019 MET_214_-_Lecture_10

17/22

Figure 7-12 and Figure 7-13

8/11/2019 MET_214_-_Lecture_10

18/22

Viewing graph 7-11 for the small sheave pitch diameter of 12.4 at 1160 rpm. The ratedpower using the solid curve for 1160 rpm is 26.4 hp. The dotted curve lying immediatelyabove the solid curve is an additional horse power added rating that can be added to thebasic power rating. A given belt can carry a greater power as the speed ratio increases, upto 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 1160rpm, 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 designprocess can proceed. To assist in specifying a 1st estimate on center distance, use therelationship provided below.

In the interest of conserving space, assume C = 24.0 in.

Step 8: Compute the required belt length using the equation provided below.

2(24.0) + 1.57(21.1 + 12.4) + = 101.4 in

inCin

C

DDCD

5.1001.21

4.121.2131.21

3 122

2

12

12

457.12

C

DDDDCL

L2

12

4

C

DD

8/11/2019 MET_214_-_Lecture_10

19/22

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

2) Calculate center to center spacing using the following formula.

12

28.64 DDLB

16

32 2

12

2DDBB

C

8/11/2019 MET_214_-_Lecture_10

20/22

Step 9 continued: L = 100.0 in = 189.6 = 23.3

Step 10: compute the angle of wrap of the belt around the sheaves from the following

equations.

= 158

Q: Is 1or 2associated with the larger diameter pulley and why?

Step 11: Determine the correction factors from figures 7-14 and 7-15 provided below.

C

DD

C

DD

2sin2180

2

sin2180

1210

2

1210

1

8/11/2019 MET_214_-_Lecture_10

21/22

= 0.94

= 0.96

Step 12: Compute the corrected rated power per belt and the number of belts required to carrythe design power:

Corrected power = = 24.86

Number of belts = 65/24.86 = 2.61 belts 3

8/11/2019 MET_214_-_Lecture_10

22/22

Summary of design

Input: Electric motor, 50.0 hp at 1160 rpm

Service factor: 1.3

Design power: 65.hpBelt: 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