4. Definition of Coupling : -
A coupling is a device used to connect two shafts together at their ends
for the purpose of transmitting power. Couplings do not normally allow
disconnection of shafts during operation, however there are torque limiting
couplings which can slip or disconnect when some torque limit is
exceeded.
Uses of Coupling: -
To provide for the connection of shafts of units that are manufactured
separately such as a motor and generator and to provide for disconnection
for repairs or alterations.
To provide for misalignment of the shafts or to introduce mechanical
flexibility.
To reduce the transmission of shock loads from one shaft to another.
To introduce protection against overloads.
To alter the vibration characteristics of rotating units.
5. PROPERTIES OF GOOD COUPLING: -
1. It should be easy to connect or disconnect the coupling.
2. The protected parts like bolts, nuts, keys etc. must be
protected by suitable flange.
3. It does allow some misalignment between the two adjacent
shaft rotation axes.
4. It must transmit full torque of the shaft.
5. It must keep proper alignment of the shaft.
9. APPLICATION OF COUPLING: -
1. Rigid couplings are used where there must be no relative movement
between two shaft.
2. Protected type flange couplings are used for belt pulley.
3. Flexible coupling are used for directly connecting motor to a machine.
4. Universal couplings are used to couple together two shafts whose
axes are intersect.
5. Claw couplings are used for slow speed shafts.
6. Oldham’s coupling are used where axes of the shafts are parallel but
not in the alignment.
7. Loose coupling are used where shafts may be coupled together.
8. Gear coupling are used to connect high speed rotary machines.
11. POWER(P)=12 kW
SPEED(N)=1050 rpm
SHEAR STRESS FOR SHAFT, KEY AND NUT BOLT
(fs)=45 mpa = 45N/mm2
SHEAR STRESS FOR CAST IRON
(fs`)=12 Mpa = 12N/mm2
CRUSHING STRESS FOR KEY AND BOLT
(fc)=80 MPA = 80N/mm2
SERVICE FACTOR (S.F) = 1.25
12. T = 60P/2πN
=109.13 N-m
Max. torque to be transmitted
Tmax = T x (S.F) = 136.42 x 103
It can be transmitted by a shaft,
Tmax = πfsd3/16
(putting the values of Tmax , fs we get)
Diameter of shaft (d)=25mm
13. •Dia of the hub(2d)=50mm(d1)
•Dia of the socket(1.5d)=37.5mm(d2)
•Dia of the bolt circle(3d)=75mm(d3)
•Dia of the outer flange (4d+d/4) =106.25mm(d4)
•Length of the hub(1.5d)=37.5mm(I)
•Length of the key(1.5d)=37.5mm(L)
•Thickness of the flange(d/2)=12.5mm(tf)
14. Material = Cast Iron
Diameter=106.25mm
Allowable shear stress=12MPa
Induced shear stress in C.I. flange(2.8N/mm2) is less than
permissible limit(12N/mm2).
so, the design is safe.
Thickness of flange (tf) = 12.5mm
15. Material = Mild Steel
Induced shear stress(24.1N/mm2) is less than permissible
value(45N/mm2)
so, the design is safe.
Induced crushing stress(72.75N/mm2) is less than
permissible value(90N/mm2)
so, the design is safe.
NO. OF KEY-2
Length of key (L)= 37.5mm
Width of key (W)= 6mm
Thickness of key (t)= 5mm
16. Material = Mild Steel
Nominal diameter of bolt(d`1)=6mm
Pitch circle diameter of bolts(D`1)=3d=3*25=75mm
No. of bolts = 4.
DIA OF SHAFT NO. OF BOLTS
UP TO 40MM 3
40-100 4
ABOVE 100 6
17.
18. NAME OF THE PARTS MATERIALS No. of items
PROTECTED TYPE
FLANGE COUPLING
(INCLUDING HUB
& FLANGE)
C.I 2
KEYS MS 2
NUTS & BOLTS MS 4
20. 1. A shear load is a force that tends to produce a sliding failure on a
material along a plane that is parallel to the direction of the force.
When a paper is cut with scissors , the paper fails in shear.
2. Crushing strength is the capacity of a material
or structure to withstand axially directed pushing
forces. When the limit of compressive strength is
reached, material are crushed.
21. •Torque transmitted (Tmax )=1360.418N-m=136.418x103 N-mm
•Diameter of the shaft (d)=25mm=0.025m
•Diameter of the hub(d1)=50mm=0.050m
•Diameter of the pitch circle(d2)=75mm=0.75m
•Thickness of the flange (tf )=12.5mm=0.0125m
•Length of the key,(L)=37.5mm=0.0375m
•Width of the key,(W)=10mm=0.01m
•Thickness of the key (t)=8mm=0.008m
•Diameter of bolts (d`1)=6mm=0.006m
•No. of bolts (n)=4
22. GIVEN IN PROBLEM: -
Permissible shear stress of m/s = 45 Mpa
Permissible crushing strss of m/s = 90 Mpa
Permissible shear stess of C.I = 12 MPa
23. A)SHEAR FAILURE OF KEY
SHAFT DIAMETER= d= 25mm
T max = 136.18 N-m = 136.18x103 N-mm
Shear area=(w x I) = 26205mm2
SHEAR STRESS(fs(key)) = FORCE/AREA
force=(w x I) x fs(key)
T max = FORCE X d/2
=(w x I) x fs(key) x d/2
Putting the values we get fs(key) = 41.5 Mpa [1 Mpa = 1 N/mm^2]
As the calculated value of fs(key) is less than the 45
Mpa(permissible shear stress for M/s),
hence the design is ok.
24. B) CRUSHING FAILURE OF KEY :-
SHAFT DIAMETER=d=25mm
T max =136.18 N-m = 136.18X103 N-mm
Crushing area = L x t/2 = 131.25mm^2
Crushing stress (fc(key)) = FORCE/AREA
FORCE = fc(key) X (L x t/2)
T max = FORCE x d/2
= fc(key) x (L x t/2) x d/2
Putting the values we get fc(key) = 83 Mpa [ 1MPa = 1 N/mm2]
As the calculated value of fc(key) is less than the 90 Mpa
(permissible crushing stress of m/s)
hence the design is ok.
25. AGAINST THE SHEAR FAILURE OF THE HUB
Torque transmitted, T max =136.418 X 103
N-mm
Diameter of the shaft (d) = 25mm
Diameter of the hub (d1) = 50 mm
Shear stress (fs(hub))
T max = ((π/16) x (d1
4 – d4)/d1 x fs(hub)
Putting the values we get fs(hub)=5.928 Mpa [ 1MPa = 1N/mm2]
As the calculated value of the fs(hub) is less than the 12 MPa(permisibble shear
stress of C.I),
hence the deigns is ok.
27. A pattern may be defined as a replica of the design casting when it is
packed in a suitable moulding material, produces a cavity called mould.
There are many types of materials for making pattern of protected type flange coupling.
They are Wood, Metal, Plaster, Plastic, Wax.
Here we use wooden pattern for making the casting.
The factors for choosing the wood for making the pattern of protected type flange coupling
are-
1. It is cheap.
2. It can be easily shaped into different forms.
3. It’s manipulation is easy because of lightness in weight.
4. Good surface finish can be easily obtained.
5. It can be preserved for a long time by applying proper preservatives.
PATTERN:-
28. Pattern do not made the exact same size as the desired same size or the
casting due to some allowances. The allowances are -
i) Shrinkage Allowance,
ii) Draft Allowance,
iii) Machining Allowance,
iv) Distortion Allowance,
v) Rapping Allowance.
ALLOWANCES:-
29. Foundry or casting is a process of forming metallic products by melting
the metal, pouring it into a cavity which is known as the mould.
Required materials for making moulding sand are
1. Silica sand grain(80 -90 %)
2. Clay(5 -20%)
3. Moisture(2 -8%)
4. Miscellaneous(2%)
MOULDING:-
30.
31. CASTING:-
Casting is a manufacturing process by which a liquid material is usually poured
into a mould, which contains a hollow cavity of the desired shape, and then
allowed to solidify. The solidified part is also known as a casting, which is
ejected or broken out of the mould to complete the process.
•Metal casting is one of the most common casting processes.
35. NAME OF INSTRUMENT USES
VERNIER CALIPER to measure the distance between two
opposite sides of an object.
OUTSIDE CALIPER to measure the external size of an
object.
INSIDE CALIPER to measure the internal size of an
object.
STEEL RULE to measure distances and/or to rule
straight lines.
39. I. To apply the concerned knowledge (Machine Tools, Manufacturing process,
Machine design, Engineering design, Strength of material) gained previously as a
whole and at same lime - as recap.
2. To understand how closely the cost of manufacturing an item an purpose of
items are related and to know that is a duty of an engineer to establish a balance
between them.
3. To develop our problem solving and innovative abilities.
4. To develop the feeling of "Group Work" this is a must for our industrial life.
5.To get a sense of sequential steps involved in manufacturing on item.
6. To get an idea about the various over heads taken ill to account during the
costing of the manufactured items.
7. To grasp an ideas about the current market conditions.
