Experiment No 2
Object: Finding the support reactions of a given simply supported beam and verify analytically
Equipment: a A warren – beam apparatus
B A set of slotted masses of known values weights (1kg, 2kg, 3kg)
C measuring instrument
D spring balance
Exercise: 1 Ensure that the beam remains in a vertical plane during the
observation
2 note the initial reading on the scale of spring balance
3 place the known masses at specified distance
4 note the change reading on the scale of spring balance
5 change the loads at these distance and record the corresponding reading of spring balance scale
6 Take at least five reading
7 measures the distance of load from the left end of support of spring balance (L1, L2, L3 FOR W1W2W3 Respectively) Measure the distance L between the spring balance supports.
Considering the equilibrium of forces: ΣV=0
RA+RB = W1+W2+W3
AND, ΣMA=0
RB=W1L1+W2L2+W3L3 / L
THEN, RA= (W1+W2+W3) - RB
Presented by: Prof. Prakash Jugnake
शनिवार, 31 जुलाई 2010
Experiment no 3
Experiment no 3
Find a contact of angle by hinged pulley on 1kg of weight
A flexible rope resting over the flat rim of a stationary pulley. The tensions T1 and T2 are such that the motion is impending (just to take a place) between the belt and the pulley. Considering the impending motion to be clockwise relative to the drum, the tension T1 is more than T2. It is to be noted that only a part of the rope is in contact with the pulley. The angle subtended at the centre of the pulley by the position of belt in contact the angle of contact or the angle of lap
Angle of contact θ = angle AOB
Let the attention be focused on small element RS of the rope which subtends an angle δθ at the centre. The segment RS is acted upon by the following set of forces:
Tension T inn the rope acting tangentially at S,
Tension (T+δT) in the belt acting tangentially at R
Normal reaction R exerted F = μR which acts against the tendency to slip and is perpendicular to normal reaction R.
Using open rope drive
L open = μ (r1+r2) + (r1-r2)2 / x +2x
Angle of contact θ = 180⁰+2 α
π/180 × θ = radian
Find a contact of angle by hinged pulley on 1kg of weight
A flexible rope resting over the flat rim of a stationary pulley. The tensions T1 and T2 are such that the motion is impending (just to take a place) between the belt and the pulley. Considering the impending motion to be clockwise relative to the drum, the tension T1 is more than T2. It is to be noted that only a part of the rope is in contact with the pulley. The angle subtended at the centre of the pulley by the position of belt in contact the angle of contact or the angle of lap
Angle of contact θ = angle AOB
Let the attention be focused on small element RS of the rope which subtends an angle δθ at the centre. The segment RS is acted upon by the following set of forces:
Tension T inn the rope acting tangentially at S,
Tension (T+δT) in the belt acting tangentially at R
Normal reaction R exerted F = μR which acts against the tendency to slip and is perpendicular to normal reaction R.
Using open rope drive
L open = μ (r1+r2) + (r1-r2)2 / x +2x
Angle of contact θ = 180⁰+2 α
π/180 × θ = radian
Experiment no 4
Experiment No 4
Determine the moment of inertia of flywheel by falling weight method
Flywheel is a solid disc of significant size and weight mounted on the shaft of machines such as steam engines, diesel engines, turbine etc. Its function is to minimize the speed fluctuations that takes place when load on such machines suddenly decreases or increases. The flywheel acquires excess kinetic energy from the machines when load on the machine is less or its running idle and supplies the stored energy to the machine when it is subjected to larger loads. The capacity of storing / shedding of kinetic energy depend on the rotational inertia of the flywheel. This rotational inertia is known as moment of inertia of rotating object namely wheels. The moment of inertia about the axis of rotation can be analytically estimated as
I =∫r2dm
For the known geometry and mass density of the material used, in SI system of units the unit of moment of inertia is Kg.m2
In lab the object of the experiment is to estimate the moment of inertia of the given flywheel experimentally
A string carrying a suitable mass m at its one end having a length less than the height of the axle from the ground is wrapped completely and evenly round the axle. When the mass m is released, the string un winds itself, thus, setting the flywheel in rotation. As the mass m descends the rotation of the flywheel goes on increasing till it becomes maximum when the string leaves the axle and the mass drops off.
Let H be the distance fallen through by the mas before the string leaves the axle and the mass drops off, and let v and ω be the linear velocity of the mass and angular velocity of the flywheel respectively at the instant the mass drops off. Then, as the mass descends a distance h, it losses potential energy, mgh which is used up.
(1) Partly in providing kinetic energy of translation 1/2mv2 to the falling mass itself.
(2) Partly in giving kinetic energy of rotation ½ I ω2 to the flywheel ( I = M.I of the flywheel )
(3) Partly in doing work against friction
If the steady work done against friction is F per turn, and, if the number of rotations made by the flywheel till the mass detaches is equal to n1 , the work done against friction is equal to n1 F. hence by the principle of conservation of energy, we have
mgh =1/2mv2+1/2 I ω2 + n1 F Eq-1
after the mass has detached, the flywheel continues to rotate for considerable time t before it is brought to rest by friction. If it makes n2 F and evidently it is equal to the kinetic energy of the flywheel at the instant the mass drops off. Thus
n2F = ½ I ω2
F=1/2 I ω2 / n2
And substituting this value of F in equation 1 we get mgh= 1/2mv2 + ½ I ω2 + ½ I ω2 n1/n2
When I= 2mgh –mv2/ ω2 (I+n1/n2) Eq-2
I= 2mgh –mr2 ω2 / ω2(I+n1/n2) Eq-3
If r be the radius of the flywheel , ν=rω
After the mass has detached, its angular velocity decreases on account of friction and after some time t, the flywheel finally comes to rest. At the time of detachment of the mass
The angular velocity of the wheel is ω and when it comes to rest its angular velocity is zero. Hence, if the force of friction is steady, the motion of the flywheel is uniformly retarded and the average angular velocity during this interval is equal to ω/2. Thus
ω/2 = 2πn2 / t
ω = 4πn2 / t substituting ω from equation’s we get
I = mn2/ (n1+n2) {ght2/8π2n22 –γ2}
Thus observing the time t and counting the equations n1 and n2 made by the flywheel, its moment of inertia can be calculated from equations. Wheel, its moment of inertia can be calculated from equations.
Apparatus: 1 flywheel mounted on the shaft fitted on the wall
2 known masses (1kg, 2kg, 3kg)
3 a string with a pin on one end and a hook on the other end
4 a stop watch
5 a vernier calliper
Exercise: 1 measure the radius (r) of the axle with the use of calliper
2 Push the pin of the string into the hole of axle, wind the string closely on the axle and count the number of turns (n1)
3 Attach a known mass (m) at the free end of the string
4 Release the mass. Note the time (t) of the fall from instant of release to the instant it gets detached
5 Count the total number of revolution (n) made from the instant the mass is released to the instant it comes to rest. In order to help counting revolutions a mark with red paint has been made on the wheel and a fixed pointer close to the wheel has been provided.
6 The steps (d) and (e) should be repeated thrice and an average value should be taken for ‘t’ and ‘n’
7 Enter the observations in the respective column in the observation table for a mass (m)
8 Change the mass (m) at free end of the string
(A) Repeat the steps from 4 to 8
(B) Avoid large mass (m) otherwise accuracy in measurement would be affected.
Determine the moment of inertia of flywheel by falling weight method
Flywheel is a solid disc of significant size and weight mounted on the shaft of machines such as steam engines, diesel engines, turbine etc. Its function is to minimize the speed fluctuations that takes place when load on such machines suddenly decreases or increases. The flywheel acquires excess kinetic energy from the machines when load on the machine is less or its running idle and supplies the stored energy to the machine when it is subjected to larger loads. The capacity of storing / shedding of kinetic energy depend on the rotational inertia of the flywheel. This rotational inertia is known as moment of inertia of rotating object namely wheels. The moment of inertia about the axis of rotation can be analytically estimated as
I =∫r2dm
For the known geometry and mass density of the material used, in SI system of units the unit of moment of inertia is Kg.m2
In lab the object of the experiment is to estimate the moment of inertia of the given flywheel experimentally
A string carrying a suitable mass m at its one end having a length less than the height of the axle from the ground is wrapped completely and evenly round the axle. When the mass m is released, the string un winds itself, thus, setting the flywheel in rotation. As the mass m descends the rotation of the flywheel goes on increasing till it becomes maximum when the string leaves the axle and the mass drops off.
Let H be the distance fallen through by the mas before the string leaves the axle and the mass drops off, and let v and ω be the linear velocity of the mass and angular velocity of the flywheel respectively at the instant the mass drops off. Then, as the mass descends a distance h, it losses potential energy, mgh which is used up.
(1) Partly in providing kinetic energy of translation 1/2mv2 to the falling mass itself.
(2) Partly in giving kinetic energy of rotation ½ I ω2 to the flywheel ( I = M.I of the flywheel )
(3) Partly in doing work against friction
If the steady work done against friction is F per turn, and, if the number of rotations made by the flywheel till the mass detaches is equal to n1 , the work done against friction is equal to n1 F. hence by the principle of conservation of energy, we have
mgh =1/2mv2+1/2 I ω2 + n1 F Eq-1
after the mass has detached, the flywheel continues to rotate for considerable time t before it is brought to rest by friction. If it makes n2 F and evidently it is equal to the kinetic energy of the flywheel at the instant the mass drops off. Thus
n2F = ½ I ω2
F=1/2 I ω2 / n2
And substituting this value of F in equation 1 we get mgh= 1/2mv2 + ½ I ω2 + ½ I ω2 n1/n2
When I= 2mgh –mv2/ ω2 (I+n1/n2) Eq-2
I= 2mgh –mr2 ω2 / ω2(I+n1/n2) Eq-3
If r be the radius of the flywheel , ν=rω
After the mass has detached, its angular velocity decreases on account of friction and after some time t, the flywheel finally comes to rest. At the time of detachment of the mass
The angular velocity of the wheel is ω and when it comes to rest its angular velocity is zero. Hence, if the force of friction is steady, the motion of the flywheel is uniformly retarded and the average angular velocity during this interval is equal to ω/2. Thus
ω/2 = 2πn2 / t
ω = 4πn2 / t substituting ω from equation’s we get
I = mn2/ (n1+n2) {ght2/8π2n22 –γ2}
Thus observing the time t and counting the equations n1 and n2 made by the flywheel, its moment of inertia can be calculated from equations. Wheel, its moment of inertia can be calculated from equations.
Apparatus: 1 flywheel mounted on the shaft fitted on the wall
2 known masses (1kg, 2kg, 3kg)
3 a string with a pin on one end and a hook on the other end
4 a stop watch
5 a vernier calliper
Exercise: 1 measure the radius (r) of the axle with the use of calliper
2 Push the pin of the string into the hole of axle, wind the string closely on the axle and count the number of turns (n1)
3 Attach a known mass (m) at the free end of the string
4 Release the mass. Note the time (t) of the fall from instant of release to the instant it gets detached
5 Count the total number of revolution (n) made from the instant the mass is released to the instant it comes to rest. In order to help counting revolutions a mark with red paint has been made on the wheel and a fixed pointer close to the wheel has been provided.
6 The steps (d) and (e) should be repeated thrice and an average value should be taken for ‘t’ and ‘n’
7 Enter the observations in the respective column in the observation table for a mass (m)
8 Change the mass (m) at free end of the string
(A) Repeat the steps from 4 to 8
(B) Avoid large mass (m) otherwise accuracy in measurement would be affected.
बुधवार, 28 जुलाई 2010
सोमवार, 19 जुलाई 2010
MITS Bhopal
:https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjAs5XvtsofSwvNxo8WkD8SybI7nMqsbfARSGnxoFgqn4kSKciE2ca-A1V0IJuRkncfm9xSRejfqe8BpYiSQFhZh3w5ErfKsbbQhkaXcbaiQfjzMM2O6yN-KryEwlsXy2ne0dALzsIBsppP/s1600/mits.jpg)
Presented By:- Prof. Prakash Jugnake ( Mechanical Engineering Department)
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रविवार, 18 जुलाई 2010
Basic structural learning begins with an analysing of a simply supported beam. A beam is a structural member (horizontal) that is design to support the applied load (vertical). It resists the applied loading by a combination of internal transverse shear force and bending moment. An accurate analysis required in order to make sure the beam is construct without any excessive loads which affect its strength.
Types of Load and Support
Two types of typical loadings:
Concentrated load is one which can be considered to act at a point although of course in practice it must be distributed over a small area (normally vertical or incline loads). (Unit in kN)
Distributed load is one which is spread in some manner over the length or a significant length of the beam. It is usually quoted at a weight per unit length of beam and it may either be uniform or varying loading from point to point. (Unit in kN/m)
Three types of support:
Namely as Pinned support, Roller support and Fixed or Built-in support.
Covered in previous post together the reactions explained in diagrams.
The Sign Convention
The sign convention depends on the direction of the stress resultant with respect to the material against which it acts. It is used for both shear force and bending moments in analysing the directions. Positive (+ve) shear forces always deform right hand face downward with respect to the left hand face and negative (-ve) would be the other way round. Positive (+ve) bending moments always elongate the lower section of the beam and negative (-ve) would elongate the mid-section upward of the beam.
Shear Force and Bending Moment in Simply Supported Beam
The sign convention depends on the direction of the stress resultant with respect to the material against which it acts. It is used for both shear force and bending moments in analysing the directions. Positive (+ve) shear forces always deform right hand face downward with respect to the left hand face and negative (-ve) would be the other way round. Positive (+ve) bending moments always elongate the lower section of the beam and negative (-ve) would elongate the mid-section upward of the beam.
रविवार, 4 जुलाई 2010
रविवार, 20 जून 2010
बुधवार, 2 जून 2010
POWER TRANSMISSION
POWER TRANSMISSION
BELTS
Belts are the cheapest utility for power transmission between shafts that may not be axially aligned. Power transmission is achieved by specially designed belts and pulleys. The demands on a belt drive transmission system are large and this has led to many variations on the theme. They run smoothly and with little noise, and cushion motor and bearings against load changes, albeit with less strength than gears or chains. However, improvements in belt engineering allow use of belts in systems that only formerly allowed chains or gears
Pros and cons
Belt drive, moreover, is simple, inexpensive, and does not require axially aligned shafts. It helps protect the machinery from overload and jam, and damps and isolates noise and vibration. Load fluctuations are shock-absorbed (cushioned). They need no lubrication and minimal maintenance. They have high efficiency (90-98%, usually 95%), high tolerance for misalignment, and are inexpensive if the shafts are far apart. Clutch action is activated by releasing belt tension. Different speeds can be obtained by step or tapered pulleys.
The angular-velocity ratio may not be constant or equal to that of the pulley diameters, due to slip and stretch. However, this problem has been largely solved by the use of toothed belts. Temperatures ranges from −31 °F (−35.0 °C) to 185 °F (85 °C). Adjustment of center distance or addition of an idler pulley is crucial to compensate for wear and stretch.
Flat belts
The drive belt: used to transfer power from the engine's flywheel. Here shown driving a threshing machine.
Flat belts were used early in line shafting to transmit power in factories.[1] It is a simple system of power transmission that was well suited for its day. It delivered high power for high speeds (500 hp for 10,000 ft/min), in cases of wide belts and large pulleys. These drives are bulky, requiring high tension leading to high loads, so vee belts have mainly replaced the flat-belts except when high speed is needed over power. The Industrial Revolution soon demanded more from the system, and flat belt pulleys needed to be carefully aligned to prevent the belt from slipping off. Because flat belts tend to climb towards the higher side of the pulley, pulleys were made with a slightly convex or "crowned" surface (rather than flat) to keep the belts centered. Flat belts also tend to slip on the pulley face when heavy loads are applied and many proprietary dressings were available that could be applied to the belts to increase friction, and so power transmission. Grip was better if the belt was assembled with the hair (i.e. outer) side of the leather against the pulley although belts were also often given a half-twist before joining the ends (forming a Möbius strip), so that wear was evenly distributed on both sides of the belt (DB). Belts were joined by lacing the ends together with leather thonging,[2][3] or later by steel comb fasteners.[4] A good modern use for a flat belt is with smaller pulleys and large central distances. They can connect inside and outside pulleys, and can come in both endless and jointed construction.
Round belts
Round belts are a circular cross section belt designed to run in a pulley with a circular (or near circular) groove. They are for use in low torque situations and may be purchased in various lengths or cut to length and joined, either by a staple, gluing or welding (in the case of polyurethane). Early sewing machines utilized a leather belt, joined either by a metal staple or glued, to a great effect.
Vee belts
Belts on a Yanmar 2GM20 marine diesel engine.
Vee belts (also known as V-belt or wedge rope) solved the slippage and alignment problem. It is now the basic belt for power transmission. They provide the best combination of traction, speed of movement, load of the bearings, and long service life. The V-belt was developed in 1917 by John Gates of the Gates Rubber Company. They are generally endless, and their general cross-section shape is trapezoidal. The "V" shape of the belt tracks in a mating groove in the pulley (or sheave), with the result that the belt cannot slip off. The belt also tends to wedge into the groove as the load increases — the greater the load, the greater the wedging action — improving torque transmission and making the vee belt an effective solution, needing less width and tension than flat belts. V-belts trump flat belts with their small center distances and high reduction ratios. The preferred center distance is larger than the largest pulley diameter, but less than three times the sum of both pulleys. Optimal speed range is 1000–7000 ft/min. V-belts need larger pulleys for their larger thickness than flat belts. They can be supplied at various fixed lengths or as a segmented section, where the segments are linked (spliced) to form a belt of the required length. For high-power requirements, two or more vee belts can be joined side-by-side in an arrangement called a multi-V, running on matching multi-groove sheaves. The strength of these belts is obtained by reinforcements with fibers like steel, polyester or aramid (e.g. Twaron or Kevlar). This is known as a multiple-belt drive. When an endless belt does not fit the need, jointed and link vee-belts may be employed. However they are weaker and only usable at speeds up to 4000 ft/min. A link v-belt is a number of rubberized fabric links held together by metal fasteners. They are length adjustable by dissasembling and removing links when needed.
Multi-Groove belts
Used in modern automotive applications to drive many or all accessories on the engine, more commonly known as a serpentine belt. Belt is made up of usually 5 or 6 "V" shapes along side each other.
Ribbed belt
A ribbed belt is a power transmission belt featuring lengthwise grooves. It operates from contact between the ribs of the belt and the grooves in the pulley. Its single-piece structure it reported to offer an even distribution of tension across the width of the pulley where the belt is in contact, a power range up to 600 kW, a high speed ratio, serpentine drives (possibility to drive off the back of the belt), long life, stability and homogeneity of the drive tension, and reduced vibration. The ribbed belt may be fitted on various applications : compressors, fitness bikes, agricultural machinery, food mixers, washing machines, lawn mowers, etc.[5]
Film belts
Though often grouped with flat belts, they are actually a different kind. They consist of a very thin belt (0.5-15 millimeters or 100-4000 micrometres) strip of plastic and occasionally rubber. They are generally intended for low-power (10 hp or 7 kW), high-speed uses, allowing high efficiency (up to 98%) and long life. These are seen in business machines, printers, tape recorders, and other light-duty operations.
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