FINITE ELEMENT ANALYSIS 3

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M2: Basic Mechanical Quantities Force

Work / Energy and Power

Types of motion

Torque Rotational/translational position, velocity and acceleration

M2.1 Force

A force (F) is basically a push or a pull on an object with a certain mass (m) that can resultin its velocity to change. Force is measured in newtons (N). If the mass remains constant,

Newtons Second Law of Motion states:

The force on a mass is proportional to the acceleration that it producesF v ma

In the SI system of units the conversion factor between mass and the product of mass andacceleration is unity, therefore

F = ma where, F force in newtons (N)m mass of the object in kilograms (kg)

a acceleration in m/s2

M2.2 Weight

The weight is a special kind of force, and is simply the force due to gravity (g). On earth,g is equal to 9.81 m/s

2

F = mg where, F force in newtons (N)m mass of the object in kilograms (kg)

g acceleration due to gravity

Example: Two people are pushing on both sides of a 6.0 kg object sitting on a table. If the

magnitudes of the two forces are 13 N and 11 N, and the angle from the horizontal are

60q and 30q respectfully. What is the acceleration of the object? What is the normal

force exerted on it by the table?

M2.3 Work

In the simplest mathematical terms, work (W) is defined as the product of force and thedistance over which the force is applied. Units of Work are N x m orjoules (J)

Work = W = F x d where, F force in newtons (N)d distance in meters (m)


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M2.4 Energy

Energy is the ability to do work and is also measured in joules (J). There are different forms

of energy, Kinetic Energy and PotentialEnergy Kinetic Energy is energy due to motion (and less obviously thermal energy).

Depending on the type of motion, kinetic energy can be further classified as

translational kinetic energy (TKE) orrotational kinetic energy (RKE).

TKE = mv2

where, v is the speed in m/s

Potential Energy is energy due to its position and this includes gravitational potential

energy. Considering that the mass is subjected to a constant gravitational field we can

derive potential energy (PE) to be,

PE=mgh where, m is mass in kg, g is in m/s2, h is altitude

The change in potential energy of the object could be given by,

PE=mg'h where, 'h is the difference in elevation.

M2.5 Power

Power is the time rate of doing work, measured in watts (W) using the SI standard ofmeasurement, where one watt is a joule per second.


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Example: Your vehicle is stuck at the bottom of a ravine that slopes down from the highway at

an angle of 30o. The distance from your vehicle to the road is 30 meters (measured

along the slope, not the horizontal distance). You would like to winch your vehicle

(which has a mass of 900 Kg) out of the ravine:

1. Neglecting friction and rolling resistance, what is the minimum rated capacity of thewinch?

2. How much work must be done by the winch to pull the vehicle up the incline?

3. If it takes 5 minutes to pull out the vehicle, how much power is the winch consuming?

4. If the cable breaks at the top of the ravine, how fast will the vehicle be traveling when ithits the bottom (neglecting friction)?

5. How long will it take to hit the bottom?

6. If the battery on your winch is rated at 20 amphours, approximately how many times canyou pull the vehicle out of the ravine before recharging? (Assuming a standard 12v carbattery)


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Example 2:

Given, a=-0.5 m/s2, V0,

S=S0 + V0 t + (1/2) ac t2

S= 0 + V0 t + (1/2) (-0.5) t2

S= V0 t - 0.25 t2

------(A)

By substituting values for t & V0 in Equation (A),

Table: Distance for V0 = 0, 1 & 2 m/s

t/(s) 0 1 2 3 4 5

S/(m) 0 0.00 0.25 1.00 2.25 4.00 6.25

1 0.00 0.75 1.00 0.75 0.00 1.25

2 0.00 1.75 3.00 3.75 4.00 3.75

The graph plotted is shown on the other page.

From the graph, slope corresponds to velocity. Therefore, final velocity V for V0=0, 1, 2 m/s

cases could be estimated as -2.5 m/s, -1.5 & -0.5 m/s respectively by calculating the slope at

t=5s.


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Example 3:

When it starts to reverse the direction velocity becomes V = 0 m/s.

V2 = V02 + 2 ac(s-s0)

If V0=2 m/s,

0 = 22 + 2 x -0.5 x s

S = 4 mSimilarly we can show that if V0=0 m/s, s = 0 m & V0=1 m/s, s = 1 m.

Velocity at t=5 s,V = V0 + ac t

V = 2 + -0.5 * 5

V = -0.5 m/s

Similarly we can show that if V0 =0 m/s, V = -2.5 m/s & V0=1 m/s, V = -1.5 m/s.


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M3: Mechanical transmission systemsThis section will cover an introduction on,

x

Gears Drivesx Belts/pulleys

x Power screws

M 3.1 Gear Drives

A gear wheel is a toothed machine part which meshes with another toothed part to transmit

motion or to change speed or direction. Gears could produce mechanical advantage through a

gear ratio and could be considered as a simple machine. Gear wheels are often used forconversion of torque and speed of a power source. These can transfer large torques to drive very

large machines. These are often used when speed changes are required. A gear setup which

increases speed is called a step up gear while a setup which decreases speed is called a step down

gear box. Some of these gears are capable of running at very low speeds. Some of the gear boxesare capable of achieving large reductions even with a small package

M3.1.1 Gear Nomenclature

Pitch circle - is a theoretical circle upon which all calculations are based. Pitch circles of mating

gears are tangent to each other.

Pinion - is the smaller of two mating gears. Gear is the larger.

Circular pitch - is equal to the sum of the tooth thickness and the width of space measured on

the pitch circle.


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Diametral pitch - is the ratio of the number of teeth to the pitch diameter. Module - is the ratio of the pitch diameter to the number of teeth (SI).

Addendum - is the radial distance from the top land to the pitch circle.

Dedendum - is the radial distance from the bottom land to the pitch circle.

Whole depth - is the sum of the addendum and dedendum.

Clearance circle - is a circle that is tangent to the addendum circle of the mating gear.

Clearance - is the amount by which the dedendum in a given gear exceeds the addendum of its

mating gear.

There are factors to be considered when selecting gears. These are,

x Pitch

x Number of teeth

x Face width

x Material used to manufacture the gear

x Style of hub, bore, etc

Pressure angle of a gear:

If a tangent is drawn to the involute profile of a tooth at any point on the curve and if a radial line

is drawn through this point of tangency, connecting this point with the centre of the gear, then

the acute angle included between this tangent and the radial is defined as the pressure angle.

The pressure angle is a constant for a given gear. Commonly used values for pressure angles are

14.5, 20 and 25 degrees. Gears with smaller pressure angles result in weaker teeth.


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M3.1.2 Gear Relationships


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Conditions for meshing two gears:

x Should have the same diametral pitch.

x Should have the same pressure angle.

M3.1.4 Problems with gears drives

1. Interference


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Interference is the contact between gear teeth other than the intended point of their surfaces. This

could be corrected by removing the interfering tooth tips or the shaded portion or the tooth flanks

of the mating gear could be undercut. Due to interference the efficiency of gear drive will reduce.Constant rubbing due to interference will also reduce the strength of the gears.

2 Backlash

Backlash is the clearance between the meshed gears. It could be described as the amount of lost

motion due to clearance or slackness when movement is reversed and contact is re-established.

For example, in a pair of gears, backlash is the amount of clearance between mated gear teeth. In

an ideal system the backlash should be zero. It is sometimes needed to leave a space between the

gear tooth and the mating gear, so that it is possible to have a film of lubricant in between forsmooth operation or to prevent it binding from heat expansion or to counteract eccentricity or

manufacturing inaccuracies. In practice backlash could reduce the gear efficiency.

3.1.5 Types of gears

Spur Gears

This is the simplest type of gear. These are used to transmit power when the shafts are parallel

with each other. These gears are very economical for single applications. These have a simple

shape and a design. These also encounter no thrust loads from tooth engagement.


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Helical Gears

These are ideal for a system which switches gear rations frequently. These ensure a gradual tooth

engagement which results in lower noise during operation. These could be used when the shafts

are at an angle. Resulting thrust loads from teeth reaction forces generated during engagement

could cause problems. Bevel Gears

These gears are often used when the two shafts are an angle of 90 degrees. However, these could

also be used when the two shafts are at other angles. The teeth of the two gears are on a conically

shaped surface. These could have either straight or helical teeth.


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Worm Gears

In this type of gears a worm, which has a form of a screw meshes with a worm gear wheel. These

have very low transmission ratios. It can obtain higher speed reductions allowing higher torques

to be transmitted. There is greater friction involved between the worm and worm-wheel

introducing higher losses reducing the efficiency.

Rack and Pinion

A rack is the toothed linear drive and pinion is the toothed wheel of the gear train. A rack and

pinion is converts the rotational motion into linear motion. These could have straight or rack

teeth. This could also be considered as a type of a linear actuator. The rack and pinion was used

in the steering mechanism of old automobiles. It provides less backlash and greater feedback, or

steering "feel" for the driver.


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3.1.6 Examples

Example 1: A simple speed reducer is composed of 2 spur gears. The pinion gear has a pitch

diameter of 0.75 and 36 teeth while the driven gear has a pitch diameter of 4.0 and 192 teeth.

1. What is the Diametral pitch of each gear?

2. If an electric motor rotating CCW at 3000 rpm is coupled to the pinion, what is the

rotational speed of the driven gear?3. If the torque delivered to the pinion is 1 N-m, what is the torque on the driven gear?

4. What is the power transmitted by the gear train?


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Example 2: Compound Gear Set Problem:

A motor rotating at 3000 rpm drives a 50 tooth gear that is coupled in turn to a 100 tooth gear.

The 100 tooth gear is mounted on a common shaft with an 80 tooth gear that is coupled to a 200

tooth gear. The 200 tooth gear is mounted on a shaft that drives the load.

1. What is the rpm of the load?

2. In which direction (relative to the motor) is the load rotating?

3. If the motor is delivering 2 Hp, what is the torque on the output shaft (assuming the losses

associated with the power transmission system are negligible)?


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M3.2 Belt Drives

M3.2.1 Introduction

Figure: A V-Belt drive

Belt drives are suitable when the power source (eg: motor) is at some distance away from the

load. These could be used for torque and speed conversion like with gears. A belt is a strip of

rubber or some other flexible material that is looped over two or more sheaves (i.e. pulleys).They are used as a simple and efficient way to transmit power between two rotating shafts.

Belts have a very high efficiency of power transmission of around 95%. Belts are inexpensiveand easy to design. The maintenance of these devices is also easy. The elasticity present in the

belts can provide damping and shock absorption which results in less vibration.

Usually the belts have of a composite structure. They have a rubber or a synthetic surface forproviding a sufficient amount of friction. In order to provide increased tensile strength the belts

are reinforced with steel wires.

M3.2.2 Types of belts

Flat belts

Flat looped strip of flexible material

Mechanically links two or more rotating shafts


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Also used as a source of motion (i.e. conveyor belt), to continually carry a load between two

points

Pros: Inexpensive and efficiently transmits power

Low Noise

Absorbs the shock force against the motor caused by load fluctuations

Cons:

Susceptible to Slipping and misalignment

V-Belts Belt is shaped in a Shaped in a V

V belt allows higher torques to be transmitted

The sheave circumference has grooves that would mate with the V-belt. These grooveswedge the belt at higher loads, allowing more torque to be placed on the belt.

The grooves solve the problem of slipping and misalignment

For higher power requirement, two or more belts can be joined side-by-side to form a

multi-V belt

When a belt cannot be specified, a linked V-belt can be used, which is made up of rubber

links held together by metal fasteners. However, these are weaker and runs at slower speeds

Timing Belts

Also known as notch, tooth, or cog Belts

These encounters no slippage

Capable for running at constant speed

Transfers direct motion for indexing and timing purposes

These are often used in vehicles as a part of the internal combustion engine to control the

timing of engine's valves.


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M3.2.3 Belt Relations

Basic Terminology

Contact Angle Geometry


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Example 1: A V-belt drive system consists of two sheaves with a pitch diameter 12.5 cm and 30cm. The smaller sheave is driven by a 2 Hp electric motor rotating at 3600 rpm.

1. What is the (linear) belt speed?

2. What is the rotational speed of the larger sheave?

3. What is the tension in the belt when the motor is operating at the rated power?


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Example 2: An electric motor rotating at 1800 rpm (counterclockwise) is couples directly to a

sheave with a pitch diameter of 80 mm. A V-belt is used to drive a load that is directly

coupled to a second sheave with a pitch diameter of 200 mm. The maximumpermissible tension in the v-belt is 200 N. Assume that the initial (i.e. no load)

pretension in the belt in negligible and the efficiency of the power transmission system

is 100%

1. Sketch the V-belt system and indicate the direction of rotation of both sheaves.

2. What is the angular velocity (measured in radians/second) of the driven sheave (i.e. of the

load)?

3. What is maximum permissible torque on the motor shaft based on a maximum permissible

tension in the V-belt of 200 N?


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4. What is the maximum permissible load (i.e. torque on the output shaft)?

5. How much horsepower can the V-belt system safely transmit? (Note: 1 Hp == 745.7 W)


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FINITE ELEMENT ANALYSIS 3

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Transcript of FINITE ELEMENT ANALYSIS 3