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

Friday, January 27, 2012

TWO MARKS QUESTIONS WITH ANSWERS ON DESIGN OF SPRINGS


SPRINGS
1.      What is a spring?
A spring is an elastic member, which deflects, or distorts under the action of load and regains its original shape after the load is removed.
2.      State any two functions of springs
a.       To measure forces in spring balance, meters and engine indicators.
b.      To store energy.
3.      What are the various types of springs?
a.       Helical springs
b.      Spiral springs
c.       Leaf springs
d.      Disc spring or Belleville springs
4.      Classify the helical springs
a.       Close – coiled or tension helical spring.
b.      Open –coiled or compression helical spring.
5.      Define : Leaf springs
A leaf spring consists of flat bars of varying lengths clamped together and supported at both ends, thus acting as a simply supported beam.
6.      Define : Belleville Springs
They are made in the form of a cone disc to carry a high compressive force. In order to improve their load carrying capacity, they may be stacked up together. The major stresses are tensile and compressive.
7.      What is spring index (C)?
The ratio of mean or pitch diameter to the diameter of wire for the spring is called the spring index.
8.      What is pitch?
The axial distance between adjacent coils in uncompressed state.
9.      What is solid length?
The length of a spring under the maximum compression is called its solid length. It is the product of total number of coils and the diameter of wire.
Ls = nt x d 
Where, nt = total number of coils.
10.  What are the requirements of spring while designing?
Spring must carry the service load without the stress exceeding the safe value. 
The spring rate must be satisfactory for the given application.
11.  What are the end conditions of spring?
a.      Plain end.
b.      Plain and ground end
c.       Squared end
d.      Squared and ground end.
12.  What is buckling of springs?
The helical compression spring behaves like a column and buckles at a comparative small load when the length of the spring is more than 4 times the mean coil diameter.
13.  What is surge in springs?
The material is subjected to higher stresses, which may cause early fatigue failure. This effect is called as spring surge.
14.  What is a laminated leaf spring?
In order to increase, the load carrying capacity, number of flat plates are placed and below the other.
15.  What semi – elliptical leaf springs?
The spring consists of number of leaves, which are held together by U- clips. The longleaf fastened to the supported is called master leaf. Remaining leaves are called as graduated leaves.
16.  What is nipping of laminated leaf spring?
Prestressing of leaf springs is obtained by a difference of radii of curvature known as nipping.
17.  What are the various applications of springs?
a.      The springs are used in various applications, they are
b.      Used to absorb energy or shocks (e.g. shock absorbers, buffers, etc.) 
c.       To apply forces as in brakes clutches, spring-loaded valves, etc.
d.      To measure forces as in spring balances and engine indicators
e.       To store energy as in watch
18.  Define free length.
Free length of the spring is the length of the spring when it is free or unloaded condition. It is equal to the solid length plus the maximum deflection or compression plus clash allowance. 
Lf = solid length + Ymax + 0.15 YMax
19.  Define spring index.
Spring index (C) is defined as the ratio of the mean diameter of the coil to the diameter of the wire. C =D/d
20.  Define spring rate (stiffness).
The spring stiffness or spring constant is defined as the load required per unit deflection of the spring. 
K= W/y 
Where 
W-load 
y-deflection
21.  Define pitch.
Pitch of the spring is defined as the axial distance between the adjacent coils in uncompressed state. 
Mathematically 
Pitch = free length/ n-1
22.  What are the points to be taken into consideration while selecting the pitch of the spring?
The points taken into consideration of selecting the pitch of the springs are
a.       The pitch of the coil should be such that if the spring is accidentally compressed the stress does not increase the yield point stress in torsion. 
b.      The spring should not be close up before the maximum service load is reached.
23.  Define active turns.
Active turns of the spring are defined as the number of turns, which impart spring action while loaded. As load increases the no of active coils decreases.
24.  Define inactive turns.
Inactive turns of the spring are defined as the number of turns which does not contribute to the spring action while loaded. As load increases number of inactive coils increases from0.5 to 1 turn.25.
25.  What are the different kinds of end connections for compression helical springs?
The different kinds of end connection for compression helical spring area.
a.       Plain ends
b.      Ground ends
c.       Squared ends
d.      Ground & square ends
26.  Write about the eccentric loading of springs?
If the load acting on the spring does not coincide with the axis of the spring, then spring is said to be have eccentric load. In eccentric loading the safe load of the spring decreases and the stiffness of the spring are also affected.
27.  Explain about surge in springs?
When one end of the spring is resting on a rigid support and the other end is loaded suddenly, all the coils of spring does not deflect equally, because some time is required for the propagation of stress along the wire. Thus a wave of compression propagates to the fixed end from where it is reflected back to the deflected end this wave passes through the spring indefinitely. If the time interval between the load application and that of the wave to propagate are equal, then resonance will occur. This will result in very high stresses and cause failure. This phenomenon is called surge.
28.  What are the methods used for eliminating surge in springs?
The methods used for eliminating surge are
a.       By using dampers on the center coil so that the wave propagation dies out
b.      By using springs having high natural frequency.
29.  What are the disadvantages of using helical spring of non-circular wires
a.       The quality of the spring is not good 
b.      The shape of the wire does not remain constant while forming the helix. It reduces the energy absorbing capacity of the spring.
c.       The stress distribution is not favorable as in circular wires. But this effect is negligible where loading is of static nature.
30.  Why concentric springs are used?
a.       To get greater spring force with in a given space 
b.      To insure the operation of a mechanism in the event of failure of one of the spring
31.  What is the advantage of leaf spring over helical spring?
The advantage of leaf spring over helical spring is that the end of the spring may be guided along a definite path as it deflects to act a structural member in addition to energy absorbing device.
32.  Write notes on the master leaf & graduated leaf?
The longest leaf of the spring is known as main leaf or master leaf has its ends in the form of an eye through which bolts are passed to secure the spring. The leaf of the spring other than master leaf is called the graduated leaves.
33.  What is meant by nip in leaf springs?
By giving greater radius of curvature to the full length leaves than the graduated leaves, before the leaves are assembled to form a spring thus a gap or clearance will be left between the leaves. This initial gap is called nip.
34.  What is the application of leaf spring?
The leaf springs are used in automobiles as shock absorbers for giving suspension to the automobile and it gives support to the structure.
35.  Define flat spiral spring.
A flat spiral spring is a long thin strip of elastic material wound like a spiral. These springs are frequently used in watch springs, gramophones, etc.
36.  What are the differences between helical torsion spring and tension helical springs?
Helical torsion springs are wound similar to that of tension springs but the ends are shaped to transmit torque. The primary stress in helical torsion spring is bending stress whereas in tension springs the stresses are tensional shear stresses.
37.  Define helical springs
The helical springs are made up of a wire coiled in the form of a helix and are primarily intended for compressive or tensile load.
38.  What are the different types of helical springs?
The different types of helical springs are.
a.       Open coil helical spring 
b.      Closed coil helical spring
39.  The extension springs are in considerably less use than the compression springs
OR
Why extension springs are in less use than compression springs
a.       Extension springs are subjected to additional stress- concentration
b.      Extension springs are wound with initial tension and this cannot be cold set without losing their tension
c.       They may be overstresses by being stretched too far.
40.  What condition must be satisfied for concentric spring?
a.       Maximum shear stress developed in the concentric springs is same for same material and free length
b.      Deflection is same for both the springs
c.       Solid length of both the springs is same
41.  Why pre stressing is done in laminated leaf spring? Explain.
By giving a greater radius of curvature to the full length leaves than graduated leaves, before the leaves are assembled to form a spring. By doing so, a gap or clearance will be left between the leaves. This initial gap is called nip. When the central bolt, holding the various leaves together, is tightened, the full length leaf will bend back and have an initial stress in a direction opposite to that of the normal load. The graduated leaves will have an initial stress in the same direction as that of the normal load. When the load is gradually applied to the spring, the full length leaf is first relieved of this initial stress and then stressed in opposite direction. Consequently, the full length leaf will be stressed less than the graduated leaf. The initial gap between the leaves may be adjusted so that under maximum load condition the stress in all the leaves is equal, or if desired, the full length leaves may have the lower stress. This is desirable in automobile springs in which full length leaves are designed for lower stress because the full length leaves carry additional loads caused by the swaying of the car, twisting and in some cases due to driving the car through the rear springs
42.  Define strain energy. How is it used to find number of coils in helical springs?
When an elastic material is deformed, work is done. This work is stored as elastic strain energy in the material. The close coiled helical spring when subjected to axial load is subjected to torsion. This gives rise to strain energy stroed in the spring. Strain energy stored in a close soiled helical spring under axial load

      U = (4W2 D3 n) / Gd4

Deflection, δ = (8 W D3 n) / Gd4

Then number of coils, n = (δ Gd4 ) / (8 W D3)

43.  What are the desirable properties of spring materials?
a.       High fatigue strength
b.      High ductility
c.       High resilience
d.      High creep
44.  What is curvature effect in helical springs?
When the wire is bent in the form of a helical coil, the length of the inside fibre is less than the length of outside fibre. This results in stress-concentration at the inside fibre of the coil. Due to the greater curvature on the inside surface the resultant stress is more on the inside fibres. This take care of by the Wahl’s stress concentration factor which accounts for curvature and direct shear stress effects.
45.  Explain the utility of centre bolt and rebound clip in leaf spring.
The centre bolt holds the leaves together. The U-clips reduces the bending stresses in the central part of the spring. The rebound clips serve to distribute some of the load to the shorter leaves which otherwise would be taken by the master leaf alone.
46.  Explain surging in spring
When one end of a helical spring is resting on a rigid support and the other end is loaded suddenly, then all the coils of the spring will not suddenly deflect equally, because some time is required for the propagation of stress along the spring wire. A little consideration will show that in the beginning, the end coils of the spring in contact with the applied load takes up whole of the deflection and then it transmits a large part of its deflection to the adjacent coils. In this way, a wave of compression propagates through the coils to the supported end from where it is reflected back to the deflected end. This wave of compression travels along the spring indefinitely. If the applied load is of fluctuating type as in the case of valve spring in internal combustion engines and if the time interval between the load applications is equal to the time required for the wave to travel from one end to the other end, then resonance will occur. This results in very large deflections of the coils and correspondingly very high stresses. Under these conditions, it is just possible that the spring may fail. This phenomenon is called surge.
It has been found that the natural frequency of spring should be atleast twenty times the frequency of application of a periodic load in order to avoid resonance with all harmonic frequencies upto twentieth order. The natural frequency for springs clamped between two plates is given by



where d = Diameter of the wire,
D = Mean diameter of the spring,
n = Number of active turns,
G = Modulus of rigidity,
g = Acceleration due to gravity, and
ρ = Density of the material of the spring.

The surge in springs may be eliminated by using the following methods:
By using friction dampers on the centre coils so that the wave propagation dies out.
By using springs of high natural frequency.
By using springs having pitch of the coils near the ends different than at the centre to have different natural frequencies
47.  What are the differences between closed coil & open coil helical springs
Closed coil helical spring                                          Open coil helical spring
The spring wires are coiled very                                 The wires are coiled such that there
Closely, each turn is nearly at right                             is a gap between the two consecutive
Angles to the axis of helix                                          turns.

Helix angle is less than 10o                              Helix angle is large (>10o)