1 2 De nition Belt, Rope and Chain Drives - Weebly · 2018. 9. 10. · Belt, Rope and Chain Drives. M Arshad Zahangir Chowdhury. B.Sc. (Hons.) in Mechanical Engineering , BUET Lecturer,

De�nition Classi�cation Speed Ratio Slip T1=T2 Tc Max-Power Chain Problems

Belt, Rope and Chain Drives

M Arshad Zahangir Chowdhury

B.Sc. (Hons.) in Mechanical Engineering , BUETLecturer, Department of Mechanical & Production Engineering

Ahsanullah University of Science and TechnologyDhaka-1208, Bangladesh

[email protected]://arshadzahangir.weebly.com/me-201.html

ME 201

M Arshad Zahangir Chowdhury (AUST) Theory of M/C & M/C Design ME 201 1 / 45


De�nition



De�nition

Belts or ropes are used to transmit power from one shaft to another by means of pulleyswhich rotate at the same speed or at different speeds.

The amount of power transmitted depends upon

1 Belt velocity2 Belt Tension3 Contact/lap angle4 Operating conditions

If the distance between shafts are too close, gears or gear trains are preferred for powertransmission.

Limitation: If the distance between shafts are greater than approximately 8m or 25 feet,these drives become poor transmission element.

Belt Materials.

1 Leather2 Cotton/Fabric3 Rubber



Example



Example



Example



Types of Belt

Figure: Types of Belts.

Flat belts are used for distance of 8m or less.V-belts are used for very close distance pulleys.Ropes or circular belts are used for 10m or more distance.



Classi�cation of Belt Drive

The power from one pulley to another one or more pulleys may be transmitted by any of thefollowing types of belt drives:

Open Belt Drive

The open belt drive is used with shafts arranged parallel and rotating in the same direction.Advantages: Simple & spaciousDisadvantages: Doesn’t provide enough belt tension.

Cross Belt Drive

The cross belt drive is used with shafts arranged parallel and rotating in the opposite direction.Advantages: Compact, provides enough tensions.Disadvantages: Less spacious, rubbing causes excessive wear.

Quarter Turn Belt Drive

The quarter turn belt drive also known as right angle belt drive is used with shafts arranged atright angles and rotating in one definite direction.Advantages: Compact, provides enough tensions.Disadvantages: Less spacious, rubbing causes excessive wear.



Classi�cation of Belt Drive (contd.)

Belt Drives with idler pulleys

A belt drive with an idler pulley is used with shafts arranged parallel and when open belt drivecan’t be used due to small angle of contact on the smaller pulley.Advantages: Transmission over large distance ,Desired velocity ratio & Flexible design.Disadvantages: Lots of pulleys & belts required and subsequent complexities.

Compound Belt Drive

If there is more than one pulley in a transmission shaft the arrangement is said to be acompound belt drive.Advantages: Speed variation, Flexible Design.Disadvantages: Spacious, Expensive.

Stepped or Coned Pulley Drive

A stepped or cone pulley drive is used for changing the speed of the driven shaft while the mainor driving shaft runs at constant speed. This is accomplished by shifting the belt from one part ofthe steps to the other.Advantages: Multi-speed capability.Disadvantages: Wear and tear of belt.



Open Belt Drive

Figure: Open Belt Drive.

Open Belt Drive

The open belt drive is used with shafts arranged parallel and rotating in the same direction.Advantages: Simple & spaciousDisadvantages: Doesn’t provide enough belt tension.



Cross Belt Drive

Figure: Cross Belt Drive.

Cross Belt Drive

The cross belt drive is used with shafts arranged parallel and rotating in the opposite direction.Advantages: Compact, provides enough tensions.Disadvantages: Less spacious, rubbing causes excessive wear.




Figure: Quarter Turn Belt Drive.


The quarter turn belt drive also known as right angle belt drive is used with shafts arranged atright angles and rotating in one definite direction.Advantages: Power transmission at right angle.Disadvantages: Pulley width must be large.



Belt Drive with idler pulleys

Figure: Belt Drive with idler pulleys.

Belt Drives with idler pulleys

A belt drive with an idler pulley is used with shafts arranged parallel and when open belt drivecan’t be used due to small angle of contact on the smaller pulley.Advantages: Transmission over large distance ,Desired velocity ratio & Flexible design.Disadvantages: Lots of pulleys & belts required and subsequent complexities.



Compound Belt Drive

Figure: Compound Belt Drive.

Compound Belt Drive

If there is more than one pulley in a transmission shaft the arrangement is said to be acompound belt drive.Advantages: Speed variation, Flexible Design.Disadvantages: Spacious, Expensive.



Stepped or Cone Pulley Drive

Figure: Stepped or Cone Pulley Drive.

Stepped or Coned Pulley Drive

A stepped or cone pulley drive is used for changing the speed of the driven shaft while the mainor driving shaft runs at constant speed. This is accomplished by shifting the belt from one part ofthe steps to the other.Advantages: Multi-speed capability.Disadvantages: Wear and tear of belt.



Speed Ratio

Figure: Speed ratio of belt drives.

De�nition

It is the ratio between the velocities of thedriver and the follower or driven.

Derivation

Consider two points on the circularportion of the belt.

If v1 6= v2 then the belt will fail due totearing or crushing.

) v1 = v2

)�d1N1

60 = �d2N260

[* v = !r ;! = 2�N60 ]

N2N1

= d1d2

If thickness of the belt is consideredthen r = d

2 + t2

N2N1

= d1+td2+t



Speed Ratio of Compound Belt Drives

Figure: Compound belt drive.

Features

) More than one pulley per shaft.) Intermediate shafts are available forchanging speed & direction of motion.

Derivation

Pulleys Speed Ratio Equality1,2 N1

N2= D2

D1N2 = N3

3,4 N3N4

= D4D3

N3 = N4

5,6 N5N6

=D6D5

N5 = N6

. . . . . . . . .

k-1,k Nk�1Nk

= DkDk�1

Nk�1 = Nk

N1

Nn=

D2 � D4 � D6 � : : :Dn

D1 � D3 � D5 � : : :Dn�1(1)

speed of �rst drive

speed of last drive= speed ratio (2)

Product of no. of diameters on drivens

Product of no. of diameters on drivers= speed ratio

(3)



Speed Ratio of Compound Belt Drives

speed of �rst drive

speed of last drive=

Product of no. of diameters on drivens

Product of no. of diameters on drivers(4)



Slip of Belt

Figure: Slip of belt.

De�nition

Slip refers to loss of speed of belt.If thefrictional grip becomes insufficient, the drivermay move forward without carrying the drivenpulley forward. This is called slip of belt.

Derivation:Let,s = s1 + s2 = Total Slips1 = % slip between driver and belts2 = % slip between belt and driven/follower) velocity of belt passing over driver,

v =�D1N1

60(1�

s1

100) (5)

Again, velocity of belt passing over follower,

�D2N2

60= v(1�

s2

100) (6)

Substituting v in the later equation we get,

�D2N260 = �D1N1

60 (1� s1100 )(1� s2

100 )

)N2N1

= D1D2

(1� s1100 �

s2100 ) [* s1s2

1000 � 0]

)N2N1

= D1D2

(1� s100 ) [* s = s1 + s2]



Belt Length (L) & Lap Angle (�)

Figure: Length of open belt drive.

Length of Open Belt Drive

L = �(r1 + r2) + 2x +(r1�r2)2

x

Lap/Contact Angle of Open Belt Drive

� = [180o � 2sin�1( r1�r2x )]� �

180

Figure: Length of cross belt drive.

Length of Cross Belt Drive

L = �(r1 + r2) + 2x +(r1+r2)2

x

Lap/Contact Angle of Cross Belt Drive

� = [180o + 2sin�1( r1+r2x )]� �

180



Power Transmitted

Figure: Power Transmitted by a belt Drive.

Power Transmitted, P = (T1 � T2)v

T1 = Tension in tight side of belt (N)T2 = Tension in slack side of belt (N)v = Belt velocity (m/s)



Ratio of Belt Tensions

Figure: Driving tensions.

Derivation:Consider a driven pulley with,T1 = Tension in tight side of belt (N)T2 = Tension in slack side of belt (N)� = Angle of contact in radians

Also consider a small portion of the belt PQ,subtending an angle �� at the centre of thepulley. The belt PQ is in equilibrium under thefollowing forces :

1 Tension T in the belt at P2 Tension T + �T in the belt at Q3 Normal reaction RN

4 Friction force F = �RN where � is thecoefficient of friction between the beltand the pulley.



Ratio of Belt Tensions(contd.)

Figure: PQ portion.

A

B Cx

Y?

Figure: PQ portion.




Figure: PQ portion.

Resolving forces in the horizontal direction,

RN = (T + �T )sin ��2 + Tsin ��

2) RN = (T + �T ) ��2 + T ��

2[* �� is very small ) sin ��

2 = ��2 ]

) RN = T��2 + �T��

2 + T��2

) RN = T�� [* �T��2 � 0]

Resolving forces in the vertical direction,�RN = (T + �T )cos ��

2 � Tcos ��2

�� is very small ) cos ��2 = 1,

) �RN = T + �T � T

) RN = �T�

Equating the red and blue equations,T�� = �T

�

) �TT = ��

)R T1

T2

�TT = �

R �

0��

) [lnT ]T1T2

= ��

) ln T1T2

= ��

)T1T2

= e��



Ratio of Belt Tensions for V-belts

Figure: V-belt(grooved pulley)

Derivation:Consider a grooved pulley with,R1 = Normal Reaction between the belt andsides of the groove (N)R = Total Reaction in the plane of the groove(N)� = Coefficient of friction between the beltand sides of the groove.2� = Angle of the groove (degrees or radians)

Resolving the reactions vertically to thegroove,

R = R1sin� + R1sin� = 2R1sin�

) R = 2R1sin�

) R1 = R2sin�

Friction force F = 2�RN [RN = NormalReaction to the pulley = R]

F = 2� R2sin� = �R

sin� = �Rcosec�

Rest of the proof is same as before.

Resolving forces in horizontal direction

R = T��

Resolving forces in vertical

F = �T

) �Rcosec� = �T

) �(T��)cosec� = �T

Integrate.




Figure: V-belts

For V-belts and Ropes,T1T2

= e��cosec� ; 2� = Angle of groove




Figure: Ropes

T1T2

= e��cosec� ; 2� = Angle of groove



Centrifugal Tension(Tc)

Figure: Centrifugal Tension.

Since the belt continuously runs over thepulleys, therefore some centrifugal force iscaused that increase tension on both tight andslack side. However , below 10 m/s beltvelocity Tc is negligible.

Consider, a small portion PQ of the belt ,subtending angle d� andm = mass of belt per unit length (kg/m)v= belt velocity (m/s)r = pulley radius (m)Tc = Tangential Centrifugal Tension at P,Q (N)

) Length of belt, PQ = rd�) Mass of belt, PQ = m(rd�)

) Centrifugal Force acting on belt PQ,Fc = mrd�v2

r = mv2d�

Resolving forces horizontally,

2Tcsin( d�2 ) = Fc = mv2d�

) 2Tc(d�2 ) = mv2d�

[ d�2 is very small ) sin d�

2 �d�2 ]

) Tc = mv2



Centrifugal Tension(contd.)



Centrifugal Tension (contd.)

) Tension in,

Tight side , Tt1 = T1 + Tc

Slack side , Tt2 = T2 + Tc

) Power Transmitted, P = (Tt1 � Tt2)v = (T1 � T2)v

) Ratio of belt tension ,T1T2

= e��

)Tt1�Tc

Tt2�Tc= e��



Initial Tension, T0

The belt is assembled with an initial tension, T0.

When power is being to transmitted, the tension in the tight side increases fro T0 to T1and slack side decreases from T0 to T2.

If the belt is assumed to obey Hooke’s law and its length remain constant, then theincrease in length of the tight side is equal to the decrease in length of the slack side, i.e.,

T1 � T0 = T0 � T2

) T0 = T1+T22

If centrifugal tension is considered,

) T0 = T1+T2+2Tc2



Condition for Maximum Power Transmission

Power Transmitted by a belt, P = (T1 � T2)v

) P = (T1 �T1

e��)v [* T1

T2= e��]

) P = T1(1� 1e��

)v

) P = T1Cv [* �; � = Const. C = 1� 1e��

]

) P = (Tt1 � Tc)Cv [* Tt1 = T1 + Tc ]

) P = (Tt1 �mv2)Cv [* Tc = mv2]

For maximum power transmission, dPdv = 0

) ddv [(Tt1 v �mv3)C] = 0

) Tt1 � 3mv2 = 0

) Tt1 = 3mv2 = 3Tc

This is the required condition.

) To transmit maximum power, themaximum tension in tight side must equalto three times the centrifugal tension.

Again,) Tt1 = 3mv2

) vmax =

qTt13m =

pT

3m

This is the expression for velocity when a beltis transmitting maximum power.



Chain Drives



Chain Drives

Figure: Sprocket and Chain

Classi�cation

1 Hauling or Hoisting Chain: Cranes.2 Tractive Chain: Conveyor Belts.3 Power transmitting chain: Bicycles.

Features

) Made of rigid links.

) Links are flexible for warping.

)Wheels have projecting teeth. (calledsprockets)

)Wheel and chain are constrained to movetogether ensuring no slip and perfect velocityratio.



Problem 1

Open belt drive

An open belt drive connects two pulleys 1.2m & 0.5m in diameter on shafts 4m apart.Mass of belt is 0.9 kg per metre length & maximum tension should be checked not toexceed 2000 N. Coefficient of friction is 0.3. The 1.2m driver pulley runs at 200 rpm. Due tobelt slip on one of the pulleys , the speed in the driven shaft is only 450 rpm. Determine,torque on the two shafts, power transmitted, power lost in friction & drive efficiency.

Solution: Here,

Belt velocity; v =�d1N1

60=

� � 1:2� 20060

� 12:57 m/s

Centrifugal Tension;Tc = mv2 = 0:9� 12:572

� 142 N

Now,Maximum (Allowable) Tension,

T = T1 + Tc

) T1 = T � Tc

= 2000� 142

) T1 � 1858 N



Problem 1 (contd.)

Now, for open belt drives ,angle of lap,

� = 180� � 2sin�1(r1 � r2

x)

= 180� � 2sin�1(0:6� 0:25

4)

= 180� � 2(5:02�)

) � = 169:96�

For a flat belt,

T1

T2= e��

) T2 =T1

e��=

1858

e0:3�169:96� �180

=18582:438

) T2 = 762 N

Torque on shaft with larger pulley,TL = (T1 � T2)rL � 657:6 Nm (Ans.)

Torque on shaft with smaller pulley,TS = (T1 � T2)rS � 274 Nm (Ans.)

Power transmitted by belt,P = (T1 � T2)v � 13:78 kW (Ans.)

Now, Input power,PI = TL!L = 657:6� 2� 200

60 � 13:78 kW

Again, Output power,PO = TS!S = 274� 2� 450

60 � 12:91 kW

) Power lost in frictionPLOST = PI � PO = 13:78� 12:91 = 0:87kW (Ans.)

Now, Drive efficiency,� =

POPI

= 12:9113:78 = 0:937 = 93:7% (Ans.)



Problem 2

V-belts

Two V-belts construct a belt drive on grooved pulleys of the same size. The angle ofgroove is 30o . The cross-sectional area of each belt is 750 mm2 and coefficient of frictionis 0.12.The density of belt material is 1:2Mg=m3 and maximum allowable stress in thematerial is 7 MPa. Determine, power transmitted between pulleys of 300mm diametersrunning at 1500 rpm and the shaft speed at which maximum power will be transmitted.

Solution: Here, Angle of groove, 2� = 30�

Belt velocity; v =�dN60

=� � 0:3� 1500

60� 23:56 m/s

Mass of belt per unit length;m = Area� Length� Density

= 750� 10�6 � 1� 1200

� 0:9 kg/m

Now,

Centrifugal Tension;Tc = mv2

= 0:9� 23:562

� 500 N



Problem 2 (contd.)

Maximum Tension in belt;T = Max. Stress� Cross-sectional Area

= 7� 10�6 � 750� 10�6

= 5250 N

Again, T1 = T � Tc = 5250� 500 � 4750 N

Now, for same sized pulleys ,angle of lap, � = 180� = � radians

For v-belts,

T1

T2= e��cosec�

)T1

T2= e0:12��cosec15�

) T2 � 1105 N



Problem 2 (contd.)

Power transmitted, P = (T1 � T2)� v � 2 [* 2 V-belts]

) P = (4750� 1105)� 23:56� 2

) P = 171:75 KW (Ans.)

For maximum power transmission,

vmax =q

Tmax3m =

q5250

3�0:9 = 44:1 m/s

Again,vmax = �DNshaft

60) 44:1 = ��0:3Nshaft

60) Nshaft � 2809 rpm (Ans.)



Problem 3

Ropes

A rope drive transmits 600 KW from a 4m pulley running at 90 rpm. Lap angle is 160o;groove angle is 45o; coefficient of friction is 0.28;mass of rope is 1.5 kg/m & allowabletension in each rope is 2400 N. Find, how many ropes are required for the drive.

Solution: Here, Angle of groove, 2� = 45�

Rope velocity; v =�dN60

=� � 4� 90

60� 18:84 m/s

Centrifugal Tension;Tc = mv2 = 1:5� 18:842

� 533 N

Now,Maximum (Allowable) Tension,

T = T1 + Tc

) T1 = T � Tc

= 2400� 533

) T1 � 1867 N



Problem 3 (contd.)

For ropes,

T1

T2= e��cosec�

)T1

T2= e0:28�160� �

180�cosec22:5� = 2:05

) T2 � 240 N

Power transmitted by a single rope, P = (T1 � T2)� v � 30:67 kW

If number of ropes required is n & total power transmitted is PTotal , then

PTotal = P � n

) n =PTotal

P=

60030:67

) n � 19:56

(Ans. : 20 ropes are required)M Arshad Zahangir Chowdhury (AUST) Theory of M/C & M/C Design ME 201 41 / 45


Problem 4

Design Problem

A flat belt is to be designed to transmit 110kW at a belt speed of 25 m/s between twopulleys of diameters 250mm and 400mm, having a centre distance of 1m. The allowablebelt stress is 8.5 MN=m2 and belts are available having a thickness-to-width ratio of 0.1and a material density of 1100 kg=m3. Given that the coefficient of friction is 0.3.Determine the required belt width.[Hint: Find Angle of lap,Centrifugal Tension, Total (Maximum) Tension and Powertransmitted then solve for width]

Try Yourself



Problem 5

Special Problem

Two parallel horizontal shafts, whose center lines are 4.8m apart, one being verticallyabove the other, are connected by an open belt drive. The pulley on the upper shaft is1.05m diameter, that on the lower shaft is 1.5m diameter. The belt is 150mm wide and theinitial tension in it when stationary and when no torque is being transmitted is 3kN. Thebelt has a mass of 1.5 kg/m length; the gravitational force on it may be neglected butcentrifugal force must be taken into account. The material of the belt may be assumed toobey Hooke’s Law, and the free lengths of the belt between pulleys may be assumed to bestraight. The coefficient of friction between the belt and either pulley is 0.3.

Calculate:(a) the pressure in N=m2 between the belt and the upper pulley when the beltand pulleys are stationary and no torque is being transmitted,

(b) the tension in the belt and the pressure between the belt and the upper pulley if theupper shaft rotates at 400 rev/min and there is no resisting torque on the lower shaft,hence no power being transmitted, and

(c)the greatest tension in the belt if upper shaft rotates at 400 rev/min and the maximumpossible power is being transmitted to the lower shaft.



Problem 5 (contd.)

Solution:(a): Here, width of belt = 150 mm = 0.15mand initial tension T0 = 3000 N.Consider anelement of the belt subtending an angle d� atthe center. Let, the pressure acting on theelement be p. Now by resolving forces(initialtension) in the radial (perpendicular to theelement) we have,

p = ForceArea

= Forcelength�width

=2�3000�sin d�

20:525d��0:15

=2�3000� d�

20:525d��0:15

) p � 38100 Nm2 (Ans.)



Problem 5 (contd.)

Solution: (b):

Belt velocity; v =�dN60

=� � 1:05� 400

60� 21:99 m/s

Centrifugal Tension;Tc = mv2 = 1:5� 222

� 725 N

The total length of the belt remains constantand the material obeys Hooke’s law howeverno power is transmitted, therefore the totaltension remains constant at 3000 N but someof this tension are now absorbed ascentrifugal tension.

) Effective tension = T � Tc =3000� 725 = 2275N

Since the area remains the same,p = 2275

3000 � 38100 � 28900 Nm2 (Ans.)

(c):Angle of Lap,� = 180� � 2sin�1( r1�r2

x )

= 180� � 2sin�1( 0:75�0:5254:8 )

) � = 174:6� = 3:05 rad Now,

Tt1 + Tt2 = 2T0 = 6000 : : : (1)

Again,Tt1�TcTt2�Tc

= e��

)Tt1�725Tt2�725 = e0:3�3:05 = 2:5 : : : (2)

Solving equations (1) and (2) we get,Tt1 = 3970 N (Ans.)


Documents

1 2 De nition Belt, Rope and Chain Drives - Weebly · 2018. 9. 10. · Belt, Rope and Chain Drives. M Arshad Zahangir Chowdhury. B.Sc. (Hons.) in Mechanical Engineering , BUET Lecturer,