Mechanical Design of
Machine
Elements/Components
Gandingan
( Coupling )
Gandingan
Tujuan:
•
Menyambung 2 aci hujung ke
hujung.
•
Sambungan boleh dilakukan dalam
tiga keadaan kegunaan:
1. paksi aci-aci adalah
sepaksi(collinear)
2. paksi aci-aci adalah
bersilang(intersect)
3. paksi aci-aci adalah
selari(parallel) tetapi tidak
sepaksi
•
Menyambung aci mesin pemacu
kepada aci mesin dipacu.
Tiga keadaan penjajaran sambungan aci
Jenis-jenis Gandingan
Gandingan Tegar (Rigid Couplings)~provide a connection between two
perfectly aligned shafts.
Frequently in use in industry ~ Steel, Copper, Paper, Petro-chemical,
Food, Wastewater, Transportation and also for applications ~ Pumps;
Compressors; Fans; Conveyors; anywhere motors, engines, and
turbines are used
cth: gandingan bebibir(flanged coupling), clamped coupling,
tapered-sleeve coupling.
Gandingan Mudah-Lentur (Flexible Coupling)
cth: rubber element Flexible Coupling, Falk Flexible
Coupling, roller chain Coupling, gear type Coupling, Oldham
Coupling, Universal Joints (~ Hooke’s joint and Constant
velocity Universal Joints ).
Jenis-jenis Gandingan

Rigid Coupling






Locks the shaft together
Alowing no relative motion between them
No provision for misalignment between them
Does not reduce shock or vibration across it from one shaft
to the other
suitable for low speeds, accurately aligned shafts
Flexible Coupling




Take care of a small amount of unintentional misalignment
Allow radial, angular and axial misalignment
Slight misalignment and vibration absorbed by coupling
Provide for end float (axial movement of a shaft)
Flanged Rigid Coupling
• Keyed Coupling
widely used
coupling halves attached to shaft ends by keys
used standard keys
key
bolt
• Compression coupling
split double cone that does not move axially
• Design consideration for flanged coupling
1.
the torque capacity of the key or wedged friction
connection with the shaft
2.
the strength of the relatively thin web portions
that are drilled to accommodate the bolts
3.
Db
the strength of the bolts.
For this course only item 1 and 3 will be discuss.
flange
shaft
Basic formula of shear stress
analysis of bolts in coupling
F
F
2T



2
As N b (d / 4) Db N b (d 2 / 4)
where
S sy
8K s P


2
Db N b (d ) N
To find the torque that can be transmited
Db
 Nb
2
Db
d 2 Db
T    As 
 Nb   

 Nb
2
4
2
d 2 Db
T  
 Nb
8
T  F  r  Nb  F 
To find the required bolts diameter
d
8T
Db N bd
T
Ks P

Where d =Ssy= 0.577Sy/N or d =
all
all = 0.18Su or 0.3Syp
(which is lower/small)
Nb= number of bolts
Db= bolt circle diameter
 = 2n/60
rad/second
P = Power, kW or hp
n = rpm
Ks= service factor for
Table of Service Factors for Machinery
Ks = c1 + c2
Driving machines
c1
Electric motors
0.5
turbomachinery
0.75
Reciprocating engines:
1.5 – 2.8
4-cylinder
6-cylinder
1-2
Driven machines
c2
Machine tools
3
Rolling mills
1.6 -2
blowers
1.5
Textile machinery
1.6 – 2.2
Rotary pumps, compressors
1.5
Rciprocating pumps, compressors
2.2 – 3.2
Paper mills
1.8 -3
Mixing machines
1.8 -3
Wood processing machines
2-3
Electric generators
1.1 – 1.3
hoists
1.5
Contoh: An electric motor is used to drive a blower at 600rpm. The shaft
of the motor is connected to the shaft of the blower by flange coupling.
Calculate the rated power of the flanged keyed coupling assembly as
shown. Given that there are 6 bolts of diameter 15mm with allowable
shear strength Sall 210MPa. The bolts circle diameter is Db = 150mm.
Area in shear for one bolt is
1
As   (15mm) 2  176.7 mm2
4
Allowable force for one bolt is
D
F  As  As S all
b
F  176.7 mm2 x 291MPa
F  51.4kN
The torque that can be
transmited by six bolts which
are at a 75mm distance from
the central axis is
T  F  r  (51.4kN)  (0.075m)  6
T  23kN.m
Therefore the
rated power is
T   T  2n
P

Ks
60 K s


2 23  103 N .m  600
P
 723kW
60 sec ond  0.5  1.5
A rigid flange coupling used 4 bolts. The bolts circle diameter is 125mm. If
the bolt has an ultimate tensile strength of 550MPa and a yield point in
tension of 345MPa, determine the bolts size if it is to transmit a torque of
2kN.m
all = 0.18Su = 0.18(550MPa) = 99MPa
or all = 0.3Syp= 0.3(345MPa) = 103.5MPa
Use all= 99MPa ; T = 2kN.m ; Nb= 4
d
8T
8(2 103 N )

Db N bd
 (0.125m)( 4)(99 106 N / m 2 )
d  0.0101m  10.1mm
Hence from Table 8-1, use M12 bolts.