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
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