LECTURE #4 REVIEW OF TORSION Course Name : DESIGN OF MACHINE ELEMENTS Course Number: MET 214
LECTURE #27 CONVEYER SYSTEMS Course Name : DESIGN OF MACHINE ELEMENTS Course Number: MET 214.
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Transcript of LECTURE #27 CONVEYER SYSTEMS Course Name : DESIGN OF MACHINE ELEMENTS Course Number: MET 214.
LECTURE #27 CONVEYER SYSTEMS
Course Name : DESIGN OF MACHINE ELEMENTSCourse Number: MET 214
Package handling requirements established from supply chain management strategies dictate transfer rates of packages.Package specifications (weight, volume, etc) and transfer rate requirements influence the conveyer system configuration selected for a particular application.Transfer rates, package specifications and conveyor system configuration influences drive requirements for the conveyer system.Conveyer configuration influences how drive requirements for the conveyor system are related to design requirements for sub system components.Various types of conveyor systems are available
To determine the drive requirements for a conveyor system, torque requirements associated with conveyor motion must be reflected back to the drive shaft of the motor. Information provided in previous lectures identified how to reflect mass moment of inertias between shafts of pulley systems and will be utilized to assist in the development of the drive requirements for a conveyor system. In addition to reflecting mass moment of inertia and/or torque loads as illustrated in the previous lecture, conveyor systems transport packages having specified weights. The torque requirements for a conveyor system are influenced by the movement of the packages. To identify how to relate motion of the packages to torque requirements for the drive of a conveyor system, consider the system shown below.
The total amount of mass being transported by the conveyor system involves the mass of the individual packages in addition to the mass of the belt.mT = m1 +m2 + m3 + m4 + m belt
Using Newton’s 2nd law we can relate mT to the force necessary to accelerate mT.
F=mT a where F =Force applied to conveyor belt in order to accelerate belt and packages on belt
a = acceleration of conveyor belt
The linear acceleration a may be belt expressed in terms of angular measures as follows.
where r1 = radius of drive pulley of conveyor system
α1 = angular acceleration of drive pulley rads/sec2
The force F necessary to accelerate mT can be related to torque about the drive shaft as follows
Recall Newton’s 2nd law for angular measures
Accordingly, the total mass mT being accelerated by an acceleration of a may be expressed in terms of an equivalent mass moment of inertia as shown below
11rmF T
1211111 rmrmrFrT TTMT
JT
The total mass mT multiply by establishes an equivalent mass moment of inertia which is to be included in the analysis of the drive requirements for a conveyor system.Prior to identifying how all of the torque requirement of a conveyor system are reflected to the shaft of the motor, let us consider how friction effects due to linear motion of the belt impact the torque requirements for a conveyor system.To relate how friction accompanying linear motion of the belt impacts the torque requirements of a conveyor system, consider the conveyor system illustrated below.
21
121
rmJJTrmT
Tm
m
T
T
21r
Belt motion creates a friction force between the conveyor belt and the support platform which supports the conveyor belt. The magnitude of the friction force is related to the normal force and the coefficient of friction existing between the belt and the platform. The relationship is presented below.
where Ff = Force due to friction between conveyor belt and support platform u = coefficient of friction between belt and platform N = Normal force acting to press belt against support platform
To relate the friction force Ff to a torque requirement for the motor of the conveyor system, note that Ff can be related to a torque acting about shaft #1.
After accounting for rotational acceleration effects, the total torque required from a motor for a conveyor system can be determined.To identify the torque required for a motor, all motion effects relating to torque must be transformed to the motor shaft.To systematically transform all motion effects relating to torque to motor shaft, proceed as follows:
uNFf
gumruNrFrT Tff 111
1) Determine mass moment of inertia of all components undergoing rotational motion and reflect to drive shaft of conveyor system using scale factors consistent with reflected impedances.
2) Determine resisting torques accompanying rotational motion of rotating components and reflect the resisting torques to drive shaft of conveyor system using scaling factors consistent with reflected torques.
3) Transform all masses experiencing linear motion to equivalent mass moment of inertias and reflect equivalent mass moment of inertias to drive shaft of conveyor system.
4) Transform all forces associated with linear motion of conveyor system to equivalent torques and reflect to drive shaft of conveyor system.
5) Combine the torques associated with steps 1-4.
6) Reflect torque in steps 1-5 to shaft of motor if motor is not directly connected to conveyor system.
As an example of how to determine the drive requirements for a conveyor system, consider the system shown below
For angular motion:1) Determine mass moment of inertia of rotating components and reflect to drive shaft of conveyor system using scaling factors consisted with reflected impedances.
A) Jm2 = mass moment of inertia about shaft #2 includes mass moment of inertia of shaft #2, and pulley #2 and any component rotating about shaft #2 . (from lecture #14) Shaft #1 is destination. reflect Jm2 to shaft #1
22
2
2
121
2
2
122
1
1
1
mR
mR
JJr
rrr
r
rJJ
B) Jm1 = mass moment of inertia of about shaft #1: includes mass moment of inertia of shaft #1, and pulley #1, and any component rotating about shaft #1.
C)
2) Determine resisting torques accompanying rotation of rotating components and reflect back to drive shaft of conveyor system using scaling factors consisted with torques.
C) Assume rotation of shaft 2 is accompanied by a viscous damping torque due to bearing behavior that depends on speed of conveyor.
B) Assume rotation of shaft #1 is accompanied by a viscous damping torque due to bearing behavior that depends on speed of conveyor.
2121 11 mmRmmT JJJJJ
122222
222
122
1
222
1
222
222
1
1
nnTT
nD
Dn
n
nn
n
nTT
nT
R
R
111 nT
C) Combine the torque loads acting on shaft #1 with all torque loads reflected to shaft #1
D) Alternatively, the torque loading on shaft #1 and/or shaft #2 could be independent of speed, and modeled simply as a drag torque.
where =drag torque existing on shaft #1=drag torque existing on shaft #2
E) Reflecting T2 to shaft #1 and combining the two torques results in the following expression:
F) Combination of torque types may exist depending on circumstances
3) Transform all masses experiencing linear motion to equivalent mass moment of inertias and reflect equivalent mass moment of inertias to drive shaft of conveyor system:
Utilizing relationships presented earlier:
211
1211
21
1
1
11
nTnnT
TTT
T
T
RT
22
11
d
d
TT
TT
1dT
2dT
211 ddT TTTd
beltT
Teq
mmmmmmrmJ
4321
211
4) Transform all forces associated with linear motion of conveyor system to equivalent torques and reflect to drive shaft of conveyor system.Utilizing relationship presented earlier:
5) Combine the torques associated with steps 1-4
or if viscous dragging effects
6) Transform torque in step 5 to shaft of motor. Recall basic relation governing transformation of torques in pulley/ chain drive systems.
where Torque supplied by motorPitch diameter of sprocket attached to motor
Note:
gumrFrT Tff 11
gumrTTrmJJT
TTJJT
TddTmmS
fTmS eqT
121121211
11 111
gumrnrmJJT TTmmS 1121121211
1
111
11
11
D
D
n
nDnDn
n
nTT
nTnT
m
mmm
mm
mm
m
m
D
T
Accordingly:
Applying the scalar factor above to the equations for T1 results in the following torque requirements for the motor
or if viscous effects are considered
Substituting
Or if viscous effects are considered
11 D
DTT mSm
1
11
2111
2121 D
Dgumr
D
DTT
D
DrmJJT m
Tm
ddm
TmmM
1
11
12111
2121 D
Dgumr
D
Dn
D
DrmJJT m
Tmm
TmmM
11
11
D
Dnn
D
D
mm
mm
21
21
2
1
2121
mT
mddm
mTmmM
Dgum
D
DTT
D
DrmJJT
2
2
121
2
1
2121
mTm
mm
mTmmM
Dgumn
D
D
D
DrmJJT
CBSASTM 2
Note:
If αm = 0 then
Or if viscous effects are considered
2
1
2121 D
DrmJJJ m
Tmmeqs
22 1
211
21m
Tm
ddm
Tm
ddm
Dgum
D
DTT
Dgum
D
DTTT
2
2
121
mTm
mm
Dgumn
D
DT