Design II (Brakes) (MEDE138/MEDS202/MNGD202) PDF

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Summary

This document contain worked examples as well with few tutorials to test yourself after, of which this document basically summarized the whole chapter for Brakes....

Description

Design of Brakes A brake is a mechanical device that uses friction to absorb kinetic energy from a moving system in order to slow down or stop motion of a moving part. The design of brakes depends on the following factors: 1. 2. 3. 4. 5.

Pressure between the braking surfaces. Coefficient of friction between braking surfaces. Velocity of moving parts. Projected area of friction surfaces. Ability of the brake to dissipate heat.

This work will analyse the design of externally contracting shoe brakes, internally expanding shoe brakes, simple band brakes, differential brakes and band and block brakes. Externally contracting shoes This type of brake is frequently used in the design of trains. When designing these brakes, it is important to take all related forces into account. When the brake shoe comes into contact with the brake wheel, a normal force due to the reaction of the wheel and a tangential force due to the rotational friction are experienced and both have an impact on the required load, ‘P’.

If the angle of contact , 2𝜃, is less than 60°, then it may be assumed that the normal pressure between the block and the wheel is uniform and the following formula may be used: 𝐹𝑡 : 𝜇: 𝑅𝑁 :

Tangential braking force (N) Coefficient of friction of brake Normal force on brake shoe (N)

𝑇𝐵 : 𝐹𝑡 : 𝑟:

Braking torque (Nm) Tangential braking force (N) Radius of wheel (m)

𝑭𝒕 = 𝝁 × 𝑹𝑵

𝑻𝑩 = 𝑭𝒕 × 𝒓

When the angle of contact is greater than 60°, then the unit pressure normal to the surface of contact is less at the ends than in the centre. In this case, the shoe is pivoted to the lever instead of being rigidly attached to the lever. This gives uniform wear of the brake lining in the direction of the applied force. In order to account for this, the following formula can be used to calculate an equivalent coefficient of friction.

𝝁′ = 𝜇′: 𝜇: 𝜃:

𝟒𝝁 𝐬𝐢𝐧 𝜽 𝟐𝜽 + 𝐬𝐢𝐧 𝟐𝜽

Equivalent coefficient of friction Actual coefficient of friction Half the total contact angle (degree)

Geometry of lever externally contracting shoe brakes The geometry of the externally contracting shoe brakes plays an important role in their potential braking power. The following cases explain how to calculate a value for the resulting normal force for the three different geometry styles. Single block brake – Case 1 This case allows the force and moment due to friction to be neglected because the frictional force passes through the pivot point and thus does not contribute to the moment.

𝑅𝑁 : 𝑥 𝑃: 𝑙:

Normal force on brake shoe (N) Braking lever distance (m) Applied force (N) Applied force lever distance (m)

𝑹𝑵 × 𝒙 = 𝑷 × 𝒍

Single block brake – Case 2 This case takes the frictional force into account because the frictional force acts at a distance, ‘a’ below the pivot point and therefore contributes to the moment about the pivot.

𝑅𝑁 : 𝑥 𝐹𝑡 : 𝑎: 𝑃: 𝑙:

Clockwise rotation:

(𝑹𝑵 × 𝒙) + (𝑭𝒕 × 𝒂) = 𝑷 × 𝒍

Anticlockwise Rotation:

(𝑹𝑵 × 𝒙) − (𝑭𝒕 × 𝒂) = 𝑷 × 𝒍

Normal force on brake shoe (N) Braking lever distance (m) Tangential braking friction force (N) Distance of shoe below pivot (m) Applied force (N) Applied force lever distance (m)

Single block brake – Case 3 This case takes the frictional force into account because the frictional force acts at a distance, ‘a’ above the pivot point and therefore contributes to the moment about the pivot.

Clockwise rotation: Anticlockwise Rotation: 𝑅𝑁 : 𝑥 𝐹𝑡 : 𝑎: 𝑃: 𝑙:

Normal force on brake shoe (N) Braking lever distance (m) Tangential braking friction force (N) Distance of shoe above pivot (m) Applied force (N) Applied force lever distance (m)

(𝑹𝑵 × 𝒙) − (𝑭𝒕 × 𝒂) = 𝑷 × 𝒍

(𝑹𝑵 × 𝒙) + (𝑭𝒕 × 𝒂) = 𝑷 × 𝒍

Band brakes This type of brake consists of a flexible band which embraces a part of the circumference of a drum. The ends of this band are attached to a lever which is used to create tension in the band by applying a force ‘P’ which creates a moment pivoting about ‘O’. This tightens the band on the drum, increasing the friction of the band on the drum, creating a braking force. These types of brakes can either be simple band brakes or differential band brakes. Both work in a similar way, however, they use different geometry.

In order to design a band brake, the following equations should be used: 𝑻𝟏 = 𝒆𝝁𝜽 𝑻𝟐

𝑻𝟏 𝟐. 𝟑 𝐥𝐨𝐠 ( ) = 𝝁𝜽 𝑻𝟐 𝐥𝐧 ( 𝑇1 : 𝑇2 : 𝜇: 𝜃:

𝑻𝟏

𝑻𝟐

) = 𝝁𝜽

Tension in the tight side of the band (N) Tension in the slack side of the band (N) Coefficient of friction between band and drum Angle of contact of band on drum (radians) 𝑭 = (𝑻𝟏 − 𝑻𝟐 )

𝐹: 𝑇1 : 𝑇2 :

Braking force on drum (N) Tension in the tight side of the band (N) Tension in the slack side of the band (N)

𝑇𝐵 : 𝑇1 : 𝑇2 : 𝑟:

𝑻𝑩 = (𝑻𝟏 − 𝑻𝟐 )𝒓 Braking torque on drum (Nm) Tension in the tight side of the band (N) Tension in the slack side of the band (N) Radius of drum (m) 𝑻𝟏 = 𝝈𝒕 × 𝒘 × 𝒕

𝑇1 : 𝜎𝑡 : 𝑤: 𝑡:

Tension in the tight side of the band (N) Maximum tensile stress in band (Pa) Width of band (m) Thickness of band (m)

Energy Equations The energy absorbed by the brake is equal to the energy lost by the drum and by a moving mass. 𝝎 = 𝑵× 𝝎= 𝜔: 𝑁: 𝑣: 𝑟:

𝒗 𝒓

𝝅

𝟑𝟎

Rotational speed (Rads per sec) Rotational speed (Rev per min) Linear velocity at ‘r’ (m/s) Any distance from the centre of rotation (m) ∆𝑲𝑬𝑩𝒓𝒂𝒌𝒆 + ∆𝑲𝑬𝑫𝒓𝒖𝒎 + ∆𝑲𝑬𝑴𝒂𝒔𝒔 + ∆𝑷𝑬𝑴𝒂𝒔𝒔 = 𝟎

∆𝐾𝐸𝐵𝑟𝑎𝑘𝑒 : ∆𝐾𝐸𝐷𝑟𝑢𝑚 : ∆𝐾𝐸𝑀𝑎𝑠𝑠 : ∆𝑃𝐸𝑀𝑎𝑠𝑠 :

Kinetic energy absorbed by the brake (J) Kinetic energy lost by the drum (J) Kinetic energy lost by the mass (J) Gravitational potential energy lost by the mass (J) ∆𝑲𝑬𝑩𝒓𝒂𝒌𝒆 = 𝑻𝑩𝒓𝒂𝒌𝒆 × 𝟐𝝅𝒏

∆𝐾𝐸𝐵𝑟𝑎𝑘𝑒 : 𝑇𝐵𝑟𝑎𝑘𝑒 : 𝑛:

Kinetic energy absorbed by the brake (J) Torque applied by the brake (Nm) Number of revolutions during braking ∆𝑷𝑬𝑴𝒂𝒔𝒔 = 𝒎𝒈(𝒉𝒇 − 𝒉𝒊 )

∆𝑃𝐸𝑀𝑎𝑠𝑠 : 𝑚: 𝑔: ℎ:

Change in gravitational potential energy of the mass (J) Mass of the mass (kg) Acceleration due to gravity = 9.81 (𝑚/𝑠2 ) Height of the mass (m) ∆𝑲𝑬𝑫𝒓𝒖𝒎 =

𝟏 𝒎𝒌𝟐 (𝝎𝟐𝒇 − 𝝎𝒊𝟐 ) 𝟐

∆𝑲𝑬𝑫𝒓𝒖𝒎 = ∆𝐾𝐸𝐷𝑟𝑢𝑚 : 𝑚: 𝑘: 𝐽: 𝜔𝑓 : 𝜔𝑖 :

Change in Kinetic energy of the drum (J) Mass of drum (kg) Radius of gyration of drum (m) Mass moment of inertia of the drum (𝑘𝑔 ∙ 𝑚 2) Final rotational speed (rads per sec) Initial rotational speed (rads per sec) 𝑲𝑬𝑴𝒂𝒔𝒔 =

∆𝐾𝐸𝑀𝑎𝑠𝑠 : 𝑚: 𝑣𝑓 : 𝑣𝑖 :

𝟏 𝑱(𝝎𝟐𝒇 − 𝝎𝒊𝟐 ) 𝟐

𝟏

𝟐

𝒎(𝒗𝟐𝒇 − 𝒗𝒊𝟐 )

Change in Kinetic energy of the mass attached to the drum (J) Mass of drum (kg) Final velocity of the mass (m/s) Initial velocity of the mass (m/s)

𝑻= 𝑇: 𝑛: 𝑁:

Time required to stop (min) Number of rotations until drum stops Starting rotational speed (Rev per min) ∆𝒗 =

∆𝑣: ∆𝑑: 𝑡:

𝒏 𝑵

∆𝒅 𝒕

Change in velocity during braking (𝑚/𝑠) Total distance travelled by a point on the drum during braking (m) Total time that brake is applied (sec)

Simple band brakes – Geometry

For clockwise rotation of the drum

𝑃: 𝑙: 𝑇1 : 𝑏:

𝑷 × 𝒍 = 𝑻𝟏 × 𝒃 External load (N) Distance from pivot to load (m) Tension in the tight side of the band (N) distance from pivot to band (m)

For anti-clockwise rotation of the drum 𝑷 × 𝒍 = 𝑻𝟐 × 𝒃 𝑃: 𝑙: 𝑇2 : 𝑏:

External load (N) Distance from pivot to load (m) Tension in the tight side of the band (N) distance from pivot to band (m)

Differential band brake – Geometry

For clockwise rotation of the drum

𝑃: 𝑙: 𝑇1 : 𝑏: 𝑇2 : 𝑎:

(𝑷 × 𝒍) + (𝑻𝟏 × 𝒃) = 𝑻𝟐 × 𝒂 External load (N) Distance from pivot to load (m) Tension in the tight side of the band (N) Distance from pivot to band (m) Tension in the slack side of the band (N) Distance from pivot to band (m)

For anti-clockwise rotation of the drum

𝑃: 𝑙: 𝑇1 : 𝑏: 𝑇2 : 𝑎:

(𝑷 × 𝒍) + (𝑻𝟐 × 𝒃) = 𝑻𝟏 × 𝒂 External load (N) Distance from pivot to load (m) Tension in the tight side of the band (N) Distance from pivot to band (m) Tension in the slack side of the band (N) Distance from pivot to band (m)

Under certain conditions, a differential band brake may be self-locking. Conditions for self-locking with clockwise rotation: 𝑻𝟐 × 𝒂 ≤ 𝑻𝟏 × 𝒃 𝑇1 : 𝑏: 𝑇2 : 𝑎:

Tension in the tight side of the band (N) Distance from pivot to band (m) Tension in the slack side of the band (N) Distance from pivot to band (m)

Conditions for self-locking with anti-clockwise rotation: 𝑻𝟏 × 𝒂 ≤ 𝑻𝟐 × 𝒃 𝑇1 : 𝑏: 𝑇2 : 𝑎:

Tension in the tight side of the band (N) Distance from pivot to band (m) Tension in the slack side of the band (N) Distance from pivot to band (m)

Tutorial Part 1: Literature Research 1. 2. 3. 4. 5. 6.

List five important factors to consider when designing brakes. List five important material characteristics to be used as brake linings. List three common brake material pairs. How does the function of a brake differ from a clutch? List three different types of brakes giving an application of each Explain how a differential brake can become self-locking using equations and words.

Part 2: Calculations 1. A single block brake has a drum diameter of 250 mm and the angle of contact is 90°. If the operating force of 700 N is applied at the end of the lever and the coefficient of friction between the drum and the lining is 0.35; determine the torque that is transmitted by the block brake.

[Ans: 𝑇𝑏 = 83.75 𝑁𝑚] 2. A brake shoe applied to a drum by a lever AB which is pivoted at a fixed point A and rigidly fixed to the shoe. The radius of the drum is 160 mm. The coefficient of friction of the brake lining is 0.3. If the drum rotates clockwise, find the braking torque due to the horizontal force of 600 N applied at B.

[Ans: 𝑇𝑏 = 59.65 𝑁𝑚]

3. The simple band brake, as shown in the figure below, is applied to a shaft carrying a flywheel of mass 400 kg. The radius of gyration of the flywheel is 450 mm and runs at 300 rpm. If the coefficient of friction is 0.2 and the brake drum diameter is 240 mm, find: a. The torque applied due to a hand load of 100 N. b. The number of turns of the wheel before it is brought to rest. c. The time required to bring it to rest, from the moment of the application of the brake.

[Ans: 𝑇𝐵 = 32.4 𝑁𝑚; 𝑛 = 196.5; 𝑇 = 39.3 𝑠𝑒𝑐] 4. A simple band brake operates on a drum of 600 mm in diameter, running at 200 rpm. The coefficient of friction is 0,25. The band has a contact angle of 270°. One end of the band is connected to a fixed pin and the other end to the brake arm, 125 mm from the fixed pin. The straight brake arm is 750 mm long and placed so the diameter bisects the angle of contact.

a. Calculate the pull necessary on the end of the brake arm to stop the wheel if 35 kW is being absorbed. b. Calculate the width of the steel band of 2.5 mm thick required if the maximum tensile stress is not to exceed 50 MPa. [Ans: 𝐹 = 292.03; 𝑤 = 0.0644]

5. A mass of 2000 kg is being lowered at a velocity of 2 𝑚/𝑠. On applying the brake, the mass is brought to rest in a distance of 0.5 m. The diameter of the drum is 2 meters.

a. Calculate the energy absorbed by the brake. The mass moment of inertia of the drum is 𝐽 = 150 𝑘𝑔 ∙ 𝑚 2 . b. Calculate the angle in degrees that the drum rotates before coming to a rest. c. Calculate the time taken for the drum to come to a stop. [Ans: ∆𝐾𝐸𝐵𝑟𝑎𝑘𝑒 = 14110 𝐽; 𝜃 = 28.65°; 𝑡 = 0. 25 𝑠𝑒𝑐𝑜𝑛𝑑𝑠] 6. The rope in the figure supports a load W and is wound around a barrel 450 mm diameter. A differential band brake acts on a drum 800 mm diameter which is keyed to the same shaft as the barrel. The two ends of the bands are attached to pins on opposite sides of the fulcrum of the brake lever and at distances of 25 mm and 100 mm from the fulcrum. The angle of lap of the brake band is 250° and the coefficient of friction is 0.25. What is the maximum load W which can be supported by the brake when a force of 750 N is applied to the lever at a distance of 3000 mm from the fulcrum?

[Ans: 𝑊 = 309𝑘𝑁]...