1. FRICTION MECHANISMS Belt Drives & Brakes

2. Friction Friction – When one body slides on another, the tangential force resisting the motion is proportional to the force between the surfaces in contact. The coefficient of friction, μ = F f /N, Where F f = frictional force N = normal force between the two bodies. μ is constant for any given pair of surfaces and is approximately independent of the area of contact, sliding velocity and intensity of pressure. (It is suspected that friction is due to molecular attraction and formation of local small pressure welds.) The friction angle φ is given by φ = tan -1 μ F/N = tan φ = μ

3. Types of Friction Basically three types: Solid friction (a.k.a. Coulomb friction) - between surfaces in rubbing contact. Rolling friction - when one member rolls without slip over another. Fluid friction - where layers of fluid move over each other at different speeds. Limiting Friction – the friction that exists before surfaces begin sliding over each other. (After that it tends to decrease to another almost constant value, called the sliding friction.) Work done against friction is converted into heat energy => there is loss of mechanical efficiency.

4. Belt Drives Power can be transmitted from one shaft to another by means of belt or rope drives running between pulleys attached to the shafts. Belt drives are chosen because of: -Low cost over other means of power transmission -Their ability to absorb shock that might otherwise be transmitted from the driven shaft back to the driver. -They can prevent vibrations in the driver from being transmitted to the driver shaft. A belt may act as a clutch, disengaging driver and driven shafts when loose and engaging them when tight, e.g. the method used in power engagement in petrol/diesel lawn mowers. Operation of belt drives depends entirely on the development of friction forces round the arc of contact with the pulleys. Also important is the elasticity of the belt.

5. Ratio of Driving Tensions (Flat Belts)

6. Consider forces acting on a small length of belt subtending an angle of contact δθ at the centre. Centripetal acceleration of belt towards centre of pullel, a = V 2 /r. This is due to the centrifugal tension, mV 2 which acts at a tangent to the pulley. This gives rise to a normal reaction δN on the belt by the pulley.

7. Ratio of Driving Tensions (Flat Belts) If m = mass per unit length of the belt, then resolving radially: (T+ δT).sin(δθ/2) + T.sin(δθ/2) – δN = (mrδθ)V 2 /r But for small δθ and δT, sinθ→θ and δθδT→0 2Tδθ/2 – δN = (mrδθ)V 2 /r (T – mV 2 ) δθ = δN ……………………………………………. (1) Resolving tangentially: (T+ δT).cos(δθ/2) – T. cos(δθ/2) – μδN = 0 But as θ→0, cos θ→1 δT = μδN …………………………………………………………. (2) Eliminating δN between (1) & (2) [i.e. δN = δT/μ]: δT/(T – mV 2 ) = μδθ ∫ [δT/(T – mV 2 )] = ∫ μδθ ln (T 1 - mV 2 ) - ln (T 2 – mV 2 ) = μθ ln [(T 1 - mV 2 )/ (T 2 – mV 2 )] = μθ (T 1 - mV 2 )/ (T 2 – mV 2 ) = e μθ Ignoring centrifugal tension under static or low-speed conditions (i.e. taking initial conditions) T 1 /T 2 = e μθ.

8. Ratio of the Driving Tensions: V- Pulleys In diagram (a), if the groove angle of the pulley (i.e. the V-angle) = 2α, and the normal reaction on each face of a belt element is δN´, inclined at 90° - α to the plane of the symmetry of the pulley, => the component of the reactions in the plane of the symmetry of the pulley is given by: δN = 2 δN´sinα The frictional force, F f = 2μδN´ between the pulley groove and the belt, i.e.: F f = 2μδN´ = 2μδN/2sinα = μδN/sinα

9. Ratio of the Driving Tensions: V- Pulleys Resolving radially: (T + δT).sin(δθ/2) + T. sin(δθ/2) – δN = (mrδθ).(V 2 /r) 2T δθ - mδθV 2 (T - mV 2 ) δθ = δN …………………………………….. (1) Resolving tangentially: (T + δT).cos(δθ/2) - T.cos(δθ/2) - μδN/sinα = 0 Reducing for small angles: δT = μδN/sinα  δN = δT.sinα/μ ………………………………………….. (2)  (T - mV 2 ) δθ = δT.sinα/μ  δT/(T - mV 2 ) = μδθ/ sinα  ∫ [δT/(T - mV 2 )] = (μ/ sinα) ∫ δθ  ln[(T 1 - m V 2 )/ (T 2 - mV 2 )] = (μθ)/sinα  [(T 1 - m V 2 )/ (T 2 - mV 2 )] = e (μθ)/sinα Ignoring centrifugal tension => T 1 /T 2 = e (μθ)/sinα

10. Power Transmitted Power(P) transmitted by the system is P = (T 1 – T 2 )V i.e. force × velocity But T 1 -mV 2 /T 2 – mv 2 = e μθ T 2 = (T 1 - mV 2 ) e -μθ + mV 2 = T 1 e -μθ - mV 2 e -μθ + mV 2 Thus P = (T 1 – T 2 )V = (T 1 e -μθ - mV 2 e -μθ + mV 2 )V Factorising => P = (T 1 - mV 2 )(1 - e -μθ )V Putting mV 2 = T c, the centrifugal tension: P = (T 1 – T c ) (1 - e -μθ )V

11. Power and Initial Tension Let T 0 = the initial tension (a.k.a. static tension). When power is being transmitted, the tension in the tight side increases from T 0 to T 1 and on the slack side decreases from T 0 to T 2. Assuming that the belt has linear elasticity and constant overall length, then: T 1 –T 0 =T 0 – T 2 T 1 + T 2 = 2T 0 Incorporating the centrifugal tension, mV 2 T 1 - mV 2 + T 2 - mV 2 = 2(T 0 - mV 2 ) But T 2 = (T 1 - mV 2 ) e -μθ + mV 2 T 1 - mV 2 + (T 1 - mV 2 ) e -μθ + mV 2 - mV 2 = 2(T 0 - mV 2 ) So (T 1 - mV 2 )(1 + e -μθ ) = 2(T 0 - mV 2 ) T 1 - mV 2 = 2(T 0 - mV 2 )/ (1 + e -μθ ) From P = (T 1 - mV 2 )(1 - e -μθ )V P = 2(T 0 - mV 2 )(1 - e -μθ )V/(1 +e -μθ ) i.e. Power in terms of T 1 and the initial tension, T 0.

12. Maximum Power Maximum power occurs when dP/dV = 0, i.e. when (d/dV). 2(T 0 V – mV 3 )(1 - e -μθ ) /(1 +e -μθ ) = 0 (d/dV). 2(T 0 V – mV 3 ) β = 0 where β = (1 - e -μθ ) /(1 +e -μθ ) 2βT 0 - 6βmV 2 = 0 2βT 0 = 6βmV 2 T 0 = 3mV 2 V = √(T 0 /3m) For maximum power transmission, the initial tension must be 3× the centrifugal tension.

13. Efficiency & Durability NB. T 0 is not a constant, and has its greatest value at V = 0. As V increases, T 0 increases to T 1 on the tight side (and decreases to T2 on the slack side), the power rising from zero to maximum and back to zero Efficiency of Belt Drives – This is the percentage loss of speed between the input and output. It is given by η = [(T 1 – T 2 )100]/[Belt section × elastic modulus] The main energy losses in belt drives arise from Creep – Permanent deformation of a material with time. This is due to change in tension of the belt round each pulley. The belt contracts as it passes over the driving pulley and extends as it passes over the driven pulley. This causes a ‘creep’ backwards round the driving pulley and forward round the driven pulley. Hysterisis – Loss of energy due to internal friction in the material and is caused by continuous flexing of the belt between the straight and curved sections of the drive, coupled with fluctuations in tensions (continuous exchange from T 1 to T 2 toT 1 over all sections of the belt).

14. Band Brakes

15. A flexible band (backed with friction material) is wrapped round a drum. The ends of the band are anchored to positions on a lever which is pivoted to a fixed fulcrum. The braking force is applied at one end (C) of the lever. Tension Ratio round the Drum (T 1 /T 2 ) = e μθ The breaking torque T q = (T 1 – T 2 )d/2, where d = diameter of the drum.

16. Band and Block Brakes

17. Modification of the band brake. The flexible steel band has a number of wood blocks fixed to the inside surface and the friction of the blocks on the drum provides the braking action. Each block embraces (wraps) a short arc of the drum. Wear is confined the wooden blocks, which are cheap to replace. The tension ratio is given by T 1 /T 2 = [(1 + μtanθ)/(1 – μtanθ)] n, where θ= ½ the angle subtended by each block assuming they are all of equal size n = number of blocks μ = coefficient of friction

18. Block Brakes These are also known as shoe brakes, and there are two principal types: (a) the rigidly-mounted shoe on the lever (b) the pivot-mounted shoe The pivot mounted shoe ‘wraps’ itself around the drum and therefore produces a uniform pressure distribution between the shoe and the drum. This results in higher torque and longer wear life, unlike in the case where the shoe is rigidly fixed to the lever. The braking force (and therefore the torque) is entirely due to the friction between the shoe and the drum.