myinventions.pl · 12.11.2019 TM&AC, Lecture 5, Sebastian Korczak, only for educational purposes 3...

Theory of Machines and Automatic ControlWinter 2019/2020

Lecturer: Sebastian Korczak, PhD Eng.

12.11.2019 TM&AC, Lecture 5, Sebastian Korczak, only for educational purposes 2

Lecture 5 cont.

Dynamics of planar mechanisms.

Inverse dynamics problem

Dynamics of planar mechanisms

Calculation of forces and torques that cause given motion of a mechanism(kinetostatics)

0. Mechanism and it's geometry, driving and operating forces/torques, displacement, velocity and acceleration functions are given.

1. Calculation of inertia forces and torques acting moving members of the mechanism.

2. Decomposition of the mechanism with reaction disclosure.

3. Write down vector sums of external forces, reactions and inertia forces (d'Alembert equations).

4. Solve the equations with graphical and/or analytical method.

Inverse dynamic problem – example

Given:geometry, mass, center of a mass locations, mass moments of inertia for all mechanism members, constant angular velocity ω of a driven element, operating force P.

velocity scheme vB

acceleration scheme

aDaC 2

inertial forces and torques

inertial forces and torques + external and operating forces and torques

Structural decomposition

Unknownreaction forces

RA 0RA 1

RE 3 RE 0

RB 2+BC 2+ BC3+ P+ RE3=0

+ RB 2t

+ BC2+BC 3+ P+ RE 3n

+ RE 3t

+ RB 2t

+ RE 3t

part BD : ∑ M D : ... .

part ED : ∑ M D : ... .

+ RB 2t

+ RE 3t

+ RB 2t

+ RE 3t

RE 3 RE 0

RA 0 RA 1

part AB : R A 1+ BC 1+ RB1=0

M R=.. .

∑ M A : ... .

Lecture 6

Machine dynamics.Reduction of masses and forces.

Machine equation of motion.

Overview

Machine dynamics

Machine – a tool containing one or more parts that uses energy to perform an intended action. Machines are assembled from components.

source: wikipedia.org, The Boulton & Watt Steam Engine, 1784

Overview

Machine dynamics

starting steady-state motion

stopping

Idea of reduction

Reduction of masses and forces

n-DoFSystem

x1(t )=F1(x1 , x2 , ... , t )x2(t )=F2(x1 , x2 , ... , t )

...xn(t )=Fn(x1 , x2 , ... , t )+constraints+ limitations

complicated

Source: James Albert Bonsack (1859 – 1924) - U.S. patent 238,640cigarette rolling machine, invented in 1880 and patented in 1881

Idea of reduction

n-DoFSystem

simpler, but not always possible

mr(t)F r(t )

xr( t)

I r(t)

M r(t)

φr (t)

1-DoFSystem

Source: James Albert Bonsack (1859 – 1924) - U.S. patent 238,640cigarette rolling machine, invented in 1880 and patented in 1881

Kinetic energy

Reduction of masses

Total kinetic energy

T=∑i=1

(12mi v i

12I iωi

Kinetic energy

Reduction of masses

mr(t)F r(t )

xr( t)

T =12mr v r

reduced mass

v r=dxr (t )

T=∑i=1

(12mi v i

12I iωi

Kinetic energy

Reduction of masses

mr(t)F r(t )

xr( t)

I r(t)

M r(t)

φr (t)

T =12I r ωr

reduced moment of inertia

T =12mr v r

reduced mass

v r=dxr (t )

ωr=d φr (t)

T=∑i=1

(12mi v i

12I iωi

System power

Reduction of forces

Total system's power

P (F i , M i ,ωi , v i , ...)

System power

Reduction of forces

P (F i , M i ,ωi , v i , ...)

P=Fr vr

reduced force

mr(t)F r(t )

xr( t)

v r=dxr (t )

System power

Reduction of forces

P=M rωr

reduced torque

P (F i , M i ,ωi , v i , ...)

P=Fr vr

reduced force

mr(t)F r(t )

xr( t)

I r(t)

M r(t)

φr (t)

v r=dxr (t )

ωr=d φr (t)

Kinetic energy

Reduction of masses – details

T=∑i=1

n12mi v i

2+∑j=1

k12I jω j

n – translating elementsk – rotating elements

Kinetic energy

T=∑i=1

n12mi v i

2+∑j=1

k12I jω j

12mr vr

n12mi vi

2+∑j=1

k12I jω j

Kinetic energy

T=∑i=1

n12mi v i

2+∑j=1

k12I jω j

12mr vr

n12mi vi

2+∑j=1

k12I jω j

mr=∑i=1

vr2+∑j=1

I jω j

Kinetic energy

T=∑i=1

n12mi v i

2+∑j=1

k12I jω j

12mr vr

n12mi vi

2+∑j=1

k12I jω j

mr=∑i=1

vr2+∑j=1

I jω j

12I rωr

n12mi v i

2+∑j=1

k12I jω j

Kinetic energy

T=∑i=1

n12mi v i

2+∑j=1

k12I jω j

12mr vr

n12mi vi

2+∑j=1

k12I jω j

mr=∑i=1

vr2+∑j=1

I jω j

12I rωr

n12mi v i

2+∑j=1

k12I jω j

I r=∑i=1

ωr2+∑j=1

I jω j

Kinetic energy

T=∑i=1

n12mi v i

2+∑j=1

k12I jω j

12mr vr

n12mi vi

2+∑j=1

k12I jω j

mr=∑i=1

vr2+∑j=1

I jω j

12I rωr

n12mi v i

2+∑j=1

k12I jω j

I r=∑i=1

ωr2+∑j=1

I jω j

– arbitrary chosen velocitiesvr , ωr

Reduction of forces – details

dW=∑i=1

Pidsi cosαi+∑j=1

M jd φ j

αi=∢(Pi , dsi)

dW=∑i=1

Pidsi cosαi+∑j=1

M jd φ j

Pr dsr=∑i=1

Pidsicosα i+∑j=1

M jd φ j

dW=∑i=1

Pidsi cosαi+∑j=1

M jd φ j

Pr dsr=∑i=1

Pidsicosα i+∑j=1

M jd φ j

Pr=∑i=1

Pidsidsr

cosα i+∑j=1

d φ j

dW=∑i=1

Pidsi cosαi+∑j=1

M jd φ j

Pr dsr=∑i=1

Pidsicosα i+∑j=1

M jd φ j

Pr=∑i=1

Pidsidsr

cosα i+∑j=1

d φ j

Pr=∑i=1

Piv idt

v rdtcosα i+∑

ω jdt

dW=∑i=1

Pidsi cosαi+∑j=1

M jd φ j

Pr dsr=∑i=1

Pidsicosα i+∑j=1

M jd φ j

Pr=∑i=1

Pidsidsr

cosα i+∑j=1

d φ j

Pr=∑i=1

Piv idt

v rdtcosα i+∑

ω jdt

Pr=∑i=1

Piv iv r

cosαi+∑j=1

dW=∑i=1

Pidsi cosαi+∑j=1

M jd φ j

Pr dsr=∑i=1

Pidsicosα i+∑j=1

M jd φ j M r dφr=∑i=1

Pidsicosα i+∑j=1

M jd φ j

Pr=∑i=1

Pidsidsr

cosα i+∑j=1

d φ j

Pr=∑i=1

Piv idt

v rdtcosα i+∑

ω jdt

Pr=∑i=1

Piv iv r

cosαi+∑j=1

M r=∑i=1

Pidsid φr

cosαi+∑j=1

d φ j

M r=∑i=1

Pividt

ωrdtcosα i+∑

ω jdt

M r=∑i=1

Piviωr

cosαi+∑j=1

ω jωr

Reduction of forces/torques

Pr=∑i=1

Piv ivr

cosαi+∑j=1

v rM r=∑

Piviωr

cosαi+∑j=1

ω jωr

Reduction of masses/moments of inertia

mr=∑i=1

v r2+∑j=1

I jω j

v r2 I r=∑

ωr2+∑j=1

I jω j

Linear motion

Machine equation of motion

m(t)F (t)

Linear motion

d (12m(t) v (t)2)=F ( t)dx

12dm(t )v (t)2

+m( t)v ( t)dv (t)=F (t)dx

12dm( t)v ( t)2

+m(t )dx (t )dt

dv (t )=F (t )dx

dm(t)dx

v (t)2

dv (t)dt

=F (t )

dm(t )dt

v (t )2

+mdv (t )dt

=F (t)

if m=const . ⇒ mdv (t)dt

=P(t) o r m x ( t )=F (t )

m(t)F (t)

complete differential of kinetic energy

elementary work

elementary change of kinetic

energy

Angular motion

d (I ω(t)2

2 )=M ( t)d φ

...dI (t)d φ

ω(t)2

2+ I (t)

dω(t )dt

=M (t )

dI ( t)dt

ω( t)2

+ I ( t)dω(t)dt

=M ( t)

if I=const . ⇒ Idω( t)dt

=M (t ) o r I φ(t)=M (t )

φ( t)

Rolling wheel (without a slip)

Given: m – wheel's mass,IO – wheel's mass moment of inertia in point O,

r – wheel's radius,M – torque.

Assume: v – velocity of the wheel's center,ω – angular velocity of the wheel.

T=12mv2

+12IOω

2 but v=ω r

T=12mv2

+12IOv2

r2 =12 (m+

IOr2 )v2

=12mr v

mr=m+IOr2 =const.

P=Mvr=Mrv=F r v

F r=Mr

mrdvdt

(m+I 0

r2 ) dvdt =Mr

m1 – total mass

mr1 – reduced mass

m2 – total mass

mr2 – reduced mass

m1 = m

mr1 ? m

Reduction of masses and forces – quiz

Example 1

In this example a drum winch is analyzed. It consist of:● electric motor EM, which generates torque as a function of angular

velocity ω: M=A-Bω where A and B are given constants; rotor moment of

inertia is equal to Im

● two stage gearbox GB (reducer) with given gears moments of inertia I1, I

4; shafts moments of inertia are equal to I

s; gears ratio are given as

1 and i

● winch's drum has diameter D and moment of inertia Id; drum is set on two

ball bearing that generates resistance Mf assumed to be constant;

● inclined plane with angle α related to the horizon line;● box of weight G pulled up with winch. Friction between object and inclined

plane is represented as dry friction with μ coefficient.

Example 1

T=12( Im+ I1+ I s)ω1

12( I2+ I3+I s)ω2

12( I 4+ Id+I s)ω3

12Ggv2

Example 1

T=12( Im+ I1+ I s)ω1

12( I2+ I3+I s)ω2

12( I 4+ Id+I s)ω3

12Ggv2

ω2ω1

ω3ω2

Example 1

T=12( Im+ I1+ I s)ω1

12( I2+ I3+I s)ω2

12( I 4+ Id+I s)ω3

12Ggv2

ω2ω1

=i1→ω2=ω1 i1

ω3ω2

=i2→ω3=ω2i2=ω1i1i2

ω3=D2

ω1i1i2

Example 1

T=12( Im+ I1+ I s)ω1

12( I2+ I3+I s)ω1

2 i12+

12( I4+ Id+ I s)ω1

2 i12 i2

12GgD2

2 i12 i2

ω2ω1

=i1→ω2=ω1 i1

ω3ω2

=i2→ω3=ω2i2=ω1i1i2

ω3=D2

ω1i1i2

Example 1

T=12 [( Im+ I1+ I s)+(I 2+I 3+ I s)i1

2+( I 4+ Id+I s)i1

2 i22+GgD2

2 i22]ω1

ω2ω1

=i1→ω2=ω1 i1

ω3ω2

=i2→ω3=ω2i2=ω1i1i2

ω3=D2

ω1i1i2

Example 1

I R=Im+ I 1+ I s+( I 2+ I 3+ I s)i12+( I 4+ I d+ I s)i1

2i22+GgD2

ω2ω1

=i1→ω2=ω1 i1

ω3ω2

=i2→ω3=ω2i2=ω1i1i2

ω3=D2

ω1i1i2

reduced moment of inertia

Example 1

N=M sω1−M f ω3−P v

Reduction of masses and forcesExample 1

P=T+G sinα=μN+G sinα=μG cosα+G sinα

Example 1

N=M sω1−M f ω3−(μGcosα+G sinα) v

Example 1

N=M sω1−M f ω1 i1i2−(μGcosα+G sin α)D2

ω1i1i2

Example 1

N=[M s−M f i1i2−(μGcosα+G sinα)D2i1i2]ω1

Example 1

M R=M s−M f i1 i2−(μG cosα+G sinα)D2i1i2

reduced torque

Example 1

M R=M s −(M f i1 i2+(μGcosα+G sinα)D2i1i2)

Reduced torque

Example 1

Reduced torque

driving torque MD(electric motor torque)

passive torque MP

Example 1

Reduced torque

M R = M D−M P = (A−Bω1)−MP

driving torque MD(electric motor torque)

passive torque MP

I Rdω1

dt=M R

IR (t)

MR (t)

ω1(t )

M R=A−Bω1−M P

Start-up process

I Rdω1

dt=M R

IR (t)

MR (t)

ω1(t )

M R=A−Bω1−M P

Start-up process

I Rdω1

dt=M R

IR (t)

MR (t)

ω1(t )

M R=A−Bω1−M P

Start-up process

dt+BI R

ω1=A−M P

I RA ,M P ,B , I R - constants

general solution particular solution

non-homogeneous 1st order ODE

with const. coef.

initial condition

Start-up process

dt+BI R

ω1=A−M P

with const. coef.

initial condition

Start-up process

dt+BI R

ω1=A−M P

with const. coef.

ω1g(t)=Ee−BIRt

ω1 p(t)=F

initial condition

ω1(t=0)=0

ω1(t)=A−M P

B(1−e

−BI Rt)

Start-up process

ω1(t)=A−M P

B(1−e

−BI Rt)

ω1(t )

Start-up process

ωmax=A−M P

ω1(t )

steady state value (maximum)

ω1(t)=A−M P

B(1−e

−BI Rt)

Start-up process

ω1(t)=A−M P

B(1−e

−BI Rt)

0,95A−M P

B=A−M P

B(1−e

−BI Rt95)

ωmax=A−M P

ω1(t )

START-UP TIME(95% of maximum value)

Start-up process

ω1(t)=A−M P

B(1−e

−BI Rt)

ωmax=A−M P

ω1(t )

START-UP TIME(95% of maximum value)

t 95≈3I RB

0,95A−M P

B=A−M P

B(1−e

−BI Rt95)

Start-up process

ω1(t)=A−M P

B(1−e

−BI Rt)

v(t)=D2

ω1(t)i1 i2

electric motor angular velocity

box linear velocity

air resistance proportional to velocity

I sI 1

myinventions.pl · 12.11.2019 TM&AC, Lecture 5, Sebastian Korczak, only for educational purposes 3...

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