"It is required to design a bushedpin type flexible coupling to connect the output shaft of an electric motor to the shaft of a centrifugal pump. The motor delivers 26 kW power at 720 rpm. The starting torque of the motor can be assumed to be 150% of the rated torque. Design the coupling and specify the dimensions of its components."
Answers
Answer:
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Answer:
Selection Of F.O.S:
As the shaft is subjected to torsional shear stress and also there is a stress concentration due to keyway in shaft 50 selecting higher F.O.S n=4 based on σγ .
Permissible stresses for C-30 :-
σt=σyn=300/4=75N/mm2
By max. shear stress theory
t=0.56yF.O.S=37.5N/mm2
Selection of material for hub and flanges :-
Select C.I. as material for hub and flange because the flanges are having complex shape. So, they can be manufactured easily with C.I. without involving costly machining operations.
C.I. is having ability to damp the vibrations so they can damp the input vibrations.
C.I. is cheap and it reduces the overall cost of coupling.
Slecting GCI - 25-------- PSG 1.4
σu = 250 N/mm2
Selecting of F.O.S :
As C.I. material is brittle and also there is stress concentration due to keyways in hub. So consider higher F.O.S n=5 based on σu
Permissible Stress :
Consider τ=σt=σuF.O.S=50N/mmm2
Selection of material of bush:
Considering rubber as material for bush so that flexibility is maintained and no shock take place.
considering bearing pressure
Pb=1N/mm2
Selection of standard coupling :-
From PSG 7.108 -> at last
Rated power = KW power application * service factor * 100RPM of application
In this problem consider motor as a prime move to drive the centrifugal pump. Service factor for considered drive and driven combination is selecred from PSG 7.109
Consider service factor for above case = 1.5
Rated power at 100 rpm = 20×1.5×100720=4.16KW
From PSG 7.108 ->
Selecting coupling No.6 whose dimensions are as follows.
A= dia. of shaft d=75mm
B= outer dia of flange D2=200mm
C= dia of hub =D=100mm
E = Length of hub = 56 mm
G= Length of flange = l=40mm
H = protected length of flange tf=15mm
D= PCD of bolt = D1=150mm
n= no of bolts, n=4
F= dia of bolt = d1=12mm
db = dia of rubber bush =30 mm
t = clearance between two flanges = 4 mm
Let us check the indirect stresses in different component of coupling
Let T be the mean torque of coupling
P=2πNT60
20×103=2×π×720×T60
T=65.25N−m
As starting torque of the motor is 150% of rated torque.
Tmax=1.5×Tmean=1.5×265.25=397.87N−m
Tmax=397.87×103N−mm
I) Shear failure at shaft :
Tmax = π16
zind=4.80N/mm2<τ=37.5N/mm2
∴ Shaft is safe
II) Shear failure of hub
Tmax=π16τind×[D4−d4D]
397.87×103=π16τind×(1004−754100)
397.87×103=π16τind×683.59×103
τind22.96N/mm2<τ=50N/mm2
∴ Hub is safe is shear