Sunday, January 27, 2013

MECHANICAL VIBRATION Assignment No I



CTIEMT, SHAHPUR
MECHANICAL ENGINEERING DEPARTMENT

MECHANICAL VIBRATION 
Assignment No # 1

Date of Issue: 28th Jan, 2013                                                    Date of Submission: 4th Feb, 2013


1.      Represent the periodic motion given in figure 1 by harmonic motion

Figure 1
2.      Represent the periodic motions given in figure 2 by harmonic motion.
Figure 2     

3.      Find the sum of harmonic motion of equal amplitude but of slightly different frequencies. Discuss the beat phenomenon that results from this sum.

4.      Split up the harmonic motion 8 sin (ωt + π/4) into two harmonic motions one of which has amplitude of 10 and phase difference zero.

5.      A steel shaft 6 cm diameter and 50 cm long fixed at one end carries a flywheel of weight of 1000 kgf and radius of gyration 30cm at its free end. Find the frequency of free longitudinal transverse and torsional vibration E= 2 x 106 kgf/ cm2, C = 3.8 x 106 kgf/ cm2.

6.      A shock absorber is to be designed so that its overshoot is 10% of the initial displacement when released. Determine the damping factor. If the damping factor is reduced to one half this values, what will be the overshoot?

7.      Why do we consider natural frequency in vibration?

8.      A simple U tube manometer filled with liquid is ahown in figure 3. Calculate the frequency of resulting motion if the minimum length of a manometer tube is 0.15m.



Figure 3


9.      A gun barrel having mass 560kg is designed with the following data: Initial recoil velocity 36 m/sec, recoil distance on firing 1.5 m
Calculate:
(a)    Spring constant
(b)   Damping coefficient
(c)    Time required for the barrel to return to a position 0.12 m from its initial position


Total Quality Management (TQM) Assignment No I



CTIEMT, SHAHPUR
MECHANICAL ENGINEERING DEPARTMENT

Total Quality Management (TQM)

Assignment No # 1

Date of Issue:26th Jan, 2013                                                           Date of Submission: 4th Feb, 2013


1.      Differentiate between the following:
a)      TQM and TQC
b)      JIT and MRP
c)      Traditional quality management and TQM
d)      Internal and external customers

2.      Describe the need of quality in the present day context. What are the impacts of quality?

3.      How can Indian companies attain world class excellence? Which organizations are world class companies? Why?

4.      Describe any 15 (fifteen) Models of TQM in detail. Which one of these models do you think is more suited to Indian companies?

5.      Define TQM. What are its major elements? What are the principles and barriers in TQM implementation? How can you remove these barriers?

6.      Who is customer? Explain the various strategies and ways to get customer feedback. Also explain how you will use customer feedback and complaints.

7.      “JIT approach helps in waste elimination” comment.

8.      write a short note on the following:
a)      JIT and its elements
b)      Kanban system
c)      Quality dimensions
d)      TWE and its types

Tuesday, March 13, 2012

Assignmnet No III. MP-II


Assignment No III

Manufacturing Processes- II



Date of issue: 9th March, 2012                                                 Last date of Submission: 23rd March, 2012





1.    Explain in detail of constructional feature of shaper, planner slotting machine, milling and sawing machine with diagram.

2.    Differentiate the following:
a.    Up-milling and down- milling
b.    Shaper and planner

3.    Write a short note on the following
a.    Mechanism of shaper
b.    Mechanism of planner
c.    Simple, compound and differential indexing

4.    Draw and explain with neat sketch of milling cutter and drilling tool with all its angles.

5.    What are grinders? Explain cylindrical surface and centre less grinders.

6.    Write a short note on:
a.    Broaching machine
b.    Drilling
c.    Operation of boring
d.    Drilling operation
e.    Nomenclature of grinding wheel.

Wednesday, March 7, 2012

Assignmnet No III ---- Machine Design II


 

Assignment No III
Machine Design - II
Topic: Design of Belt, Rope, Chain, Clutch, Brakes, Bearing, lubrication system and CAD


Date of issue: 9th March, 2012                                       Last date of Submission: 23rd March, 2012


1.      It is required to select a flat-belt drive for a compressor running at 720 rpm, which is driven by a 25 kW, 1440 rpm motor. Space is available for a centre distance of 3m. The belt is open type.

2.      Design a dry single plate friction clutch for following data:
P = 25 kW, N = 1440 rpm, Mean radius = 5 x face.

3.      Select a suitable chain drive to transmit 30 kW from an electric motor running at 1200 rpm to a line shaft running at 250 rpm. Motor shaft diameter is 60 mm and the centre distance is approximately 600 mm. Service is l0 hr/day, 6 days a week. Good lubrication condition is expected.

4.      A band brake has to absorb 4.5 k W at 150 rpm. For long life, the maximum pressure between lining and drum is 0.7MPa. The width of band is 50 mm and μ = 0.12. Find the angle of wrap if the brake drum diameter is 300 mm.

5.      A tentative design of a journal bearing result in a diameter of 75 mm and a length of 125 mm for supporting a load of 20 kN. The shaft runs at 1000 r.p.m. The bearing surface temperature is not to exceed75°C in a room temperature of 35°C. The oil used has an absolute viscosity of 0.01kg/m-s at the operating temperature. Determine the amount of artificial cooling required in watts. Assume d/c = 1000.

6.      Discuss the design procedure of lubrication system in transmission system.

7.      Describe the detail design process for computer aided design. Discuss how use of computers assist a designer in each stage of the design process with respect to reducing the time and providing wider spectrum to design solutions.



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Tuesday, January 31, 2012

Short questions with answers on Design of Flywheel


1.      What is flywheel?
A flywheel is a heavy body rotating about its axis. It acts as a reservoir of energy which is stored in the form of kinetic energy. The extra energy is stored during the idle stroke of the driven machinery and released during the working stroke. Thus flywheel controls the fluctuations of speed during each cycle of the driven machinery.
2.      What are the functions of flywheel in a machine?
The primary function of a flywheel is:
a.       To absorb energy when demand of energy id less than the supply
b.      To give out energy when demand of energy is more than the supply.
3.      What types of stresses are set up in the flywheel rims?
a.       Tensile stress due to the centrifugal force
b.      Tensile bending stress due to restraint of the arms
c.       Shrinkage stresses due to the unequal rate of cooling of casting.
4.      What are the various types of flywheel?
a.       Solid disc type
b.      Rimmed type with either arms or solid web
Solid disc type flywheel is rarely used because they have less capacity of storing energy.
Rimmed type flywheels with arms are preferred because they can store more energy. Small rimmed type flywheels are manufactured with solid web or holes drilled in the web.
5.      Why flywheels are used in punching machines?
Use of flywheel in punching machine is due to the following reasons:
a.       It decreases the variation of speed during each cycle of punching machine.
b.      It decreases the fluctuation of speed due to difference in output and input
c.       It stores energy during idle stroke and releases during working stroke.
6.      Why flywheel is used in IC engines?
In IC engine or stem engine the energy is developed during the power stroke, no energy is developed during suction, compression and exhaust strokes in 4 stroke engine. It helps the crank shaft to run at uniform speed by performing its primary function
7.      What is the difference in the function of governor and a flywheel?
Governor regulates the mean speed of an engine when there are variations in load by changing the supply of working fluid. Flywheel does not maintain a constant speed. It reduces the fluctuations.
8.      Coefficient of fluctuation of speed is ------------ of maximum fluctuation of speed and the mean speed
Ratio
9.      Due to centrifugal forces acting on the rim, the flywheel arms will be subjected to ------------- stresses
Tensile
10.  Why flywheel arm are usually elliptical?
This shape helps in more section modulus for the dame weight. This results in more strength than circular section
11.  Under what consideration the shaft for a flywheel is designed?
It is designed under shear stresses produced due to the combined action of torsion and bending moment.
12.   In a flywheel, the major axis of the elliptical section of the arm is the plane of rotation. Write done the reason for this arrangement.
The arms may have to carry the full torque load due to high inertia of the flywheel when the energy input to its shaft is cut off. The arm may be assumed as a cantilever fixed at the hub and carrying the load at the rim end. This bending moment lies in the plane of rotation of the flywheel. Therefore, the major axis of the arms must be parallel to the tangential force F acting on the flywheel.

Bending moment on the arm, M = F(R-dh/2)
                                    F = t/(nR)
Where n = number of arms
Section modulus of arm, Z = (π/32 )b1a12
Bending stress σb = M/Z
13.   On what basis the material of flywheel is selected?
a.       High tensile strength
b.      High fatigue strength
c.       Low shrinkage
14.  What are the advantages of having elliptical section of flywheel arm?
The flywheel arms are made of elliptical with major axis twice the minor axis. The major axis lies in the plane of rotation and provides double the resistance against bending moment.
15.   Difference between flywheel and governor
The function of a governor in engine is entirely different from that of a flywheel. It regulates the mean speed of an engine when there are variations in the load, e.g. when the load on the engine increases, it becomes necessary to increase the supply of working fluid. On the other hand, when the load decreases, less working fluid is required. The governor automatically controls the supply of working fluid to the engine with the varying load condition and keeps the mean speed within certain limits.
As discussed above, the flywheel does not maintain a constant speed; it simply reduces the fluctuation of speed. In other words, a flywheel controls the speed variations caused by the fluctuation of the engine turning moment during each cycle of operation. It does not control the speed variations caused by the varying load.
16.  Define the following terms
Coefficient of Fluctuation of Speed, coefficient of steadiness, fluctuation of energy, maximum fluctuation of energy, Coefficient of Fluctuation of Energy


Coefficient of fluctuation of speed: The difference between the maximum and minimum speeds during a cycle is called the maximum fluctuation of speed. The ratio of the maximum fluctuation of speed to the mean speed is called coefficient of fluctuation of speed.

Coefficient of steadiness: The reciprocal of coefficient of fluctuation of speed is known as coefficient of steadiness and it is denoted by m.

Fluctuation of energy, maximum fluctuation of energy: 
The fluctuation of energy may be determined by the turning moment diagram for one complete cycle of operation. Consider a turning moment diagram for a single cylinder double acting steam engine as shown in Fig.  The vertical ordinate represents the turning moment and the horizontal ordinate (abscissa) represents the crank angle.
 
A little consideration will show that the turning moment is zero when the crank angle is zero. It rises to a maximum value when crank angle reaches 90º and it is again zero when crank angle is 180º. This is shown by the curve abc in Fig. and it represents the turning moment diagram for outstroke. The curve cde is the turning moment diagram for instroke and is somewhat similar to the curve abc. Since the work done is the product of the turning moment and the angle turned, therefore the area of the turning moment diagram represents the work done per revolution. In actual practice, the engine is assumed to work against the mean resisting torque, as shown by a horizontal line AF. The height of the ordinate aA represents the mean height of the turning moment diagram. Since it is assumed that the work done by the turning moment per revolution is equal to the work done against the mean resisting torque, therefore the area of the rectangle aA Fe is proportional to the work done against the mean resisting torque. We see in Fig.  that the mean resisting torque line AF cuts the turning moment diagram at points B, C, D and E. When the crank moves from ‘a’ to ‘p’ the work done by the engine is equal to the area aBp, whereas the energy required is represented by the area aABp. In other words, the engine has done less work (equal to the area aAB) than the requirement. This amount of energy is taken fromthe flywheel and hence the speed of the flywheel decreases. Now the crank moves from p to q, the work done by the engine is equal to the area pBbCq, whereas the requirement of energy is represented by the area pBCq. Therefore the engine has done more work than the requirement. This excess work (equal to the area BbC) is stored in the flywheel and hence the speed of the flywheel increases while the crank moves from p to q.
Similarly when the crank moves from q to r, more work is taken from the engine than is developed. This loss of work is represented by the area CcD. To supply this loss, the flywheel gives up some of its energy and thus the speed decreases while the crank moves from q to r. As the crank moves from r to s, excess energy is again developed given by the area DdE and the speed again increases. As the piston moves from s to e, again there is a loss of work and the speed decreases. The variations of energy above and below the mean resisting torque line are called fluctuation of energy. The areas BbC, CcD, DdE etc. represent fluctuations of energy. A little consideration will show that the engine has a maximum speed either at q or at s. This is due to the fact that the flywheel absorbs energy while the crank moves from p to q and from r to s. On the other hand, the engine has a minimum speed either at p or at r. The reason is that the flywheel gives out some of its energy when the crank moves from a to p and from q to r. The difference between the maximum and the minimum energies is known as maximum fluctuation of energy 

Coefficient of Fluctuation of Energy 
It is defined as the ratio of the maximum fluctuation of energy to the work done per cycle. It is usually denoted by CE