Mechanics of Machines Dr. Mohammad Kilani Class 3 Position Analysis

Mechanics of Machines

Dr. Mohammad Kilani

Class 3Position Analysis

TYPES OF MECHANISM ANALYSESE

Position Analysis

Position Analysis

Given θ12

find θ23

, θ34

and θ14

Find maximum and minimum values

of θ14

.

Find the location of each point on

the mechanism for a given value of

θ12

and the curve traced by the a

point during motion.

θ12

θ41

θ23

θ34

Velocity Analysis

Position Analysis

Given θ12

and ω12

find ω23

, ω34

and ω14

Find the speed of a point in the

mechanismθ

12, ω

12

θ 23

, ω23

θ 34

, ω34

θ 41

, ω41

Acceleration Analysis

Position Analysis

Given θ12

, ω12

and α12

find α23

,

α34

and α14

Find the acceleration of a point in the

mechanismθ

12, ω

12, α

12

θ 23

, ω23

, α23

θ 34

, ω34

, α34

θ 41

, ω41

, α41

Question

Why is acceleration analysis important?

θ12

, ω12

, α12

θ 23

, ω23

, α23

θ 34

, ω34

, α34

θ 41

, ω41

, α41

GRAPHICAL POSITION ANALYSIS

Graphical Position Analysis

Use graphical methods, (pen, ruler,

compass and protractors to solve

position analysis problems.

CAD drafting methods may also be used

as a convenient alternative.θ

12

θ41

θ23

θ34

VECTOR POSITION ANALYSIS

Vector Position Analysis

Use vector loop closure equation to

solve position analysis problems

Loop Closure Equation

Use vector loop

closure equation to

solve position analysis

problems

4132 rrrrrP

Planar Four Bar Loop Closure Equation

How many scalar equations can be

written from the vector loop-closure

equation?

How many unknowns can be solved for

4132 rrrrrP


How many scalar equations can be written

from the vector loop-closure equation?: Two

scalar equations could be written from a

vector equation

How many unknowns can be solved for: Two

unknowns: In a standard problem, θ1

and θ2

are given. θ3

and θ4

are to be solved for.

We will focus on planar mechanisms in this

course

4132 rrrrrP


The closure condition expresses the condition

that a loop of a linkage closes on itself.

For the four-bar linkage shown the closure

equation is

jirjirjirjir

urururur

rrrr

444111333222

41332

4132

sincossincossincossincos

412


The closure condition expresses the condition that a

loop of a linkage closes on itself.

For the four-bar linkage shown the closure equation is

The following two scalar equations are produced

jirjirjirjir

urururur

rrrr

444111333222

41332

4132

sincossincossincossincos

412

44113322

44113322

sinsinsinsin

coscoscoscos

rrrr

rrrr

Analytical Solution when θ2 is known

The above two equations could be solved for two unknowns. One approach is to

eliminate one of the unknown angles by isolating the trigonometric function involving

the angle on the left-hand side of the equation.

Squaring both sides of both equations and using the identity we

obtain

44113322

44113322

sinsinsinsin

coscoscoscos

rrrr

rrrr

22441133

22441133

sinsinsinsin

coscoscoscos

rrrr

rrrr

1cossin 22

424242212121

41414124

22

21

23

sinsincoscos2sinsincoscos2

sinsincoscos2

rrrr

rrrrrr


To obtain an explicit expression for θ4

in terms of θ2

and the constant angle θ1

,

we combine the coefficients of cos θ4

and sin θ4

in the equation above as

follows:

where

21212123

24

22

21

242141

242141

sinsincoscos2

sin2sin2

cos2cos2

rrrrrrC

rrrrB

rrrrA

424242212121

41414124

22

21

23


sinsincoscos2

rrrr

rrrrrr

0sincos 44 CBA

The above equation can be solved for θ4

by using the following half angle

identities

After substitution and simplification, we get

where

21212123

24

22

21

242141

242141

sinsincoscos2

sin2sin2

cos2cos2

rrrrrrC

rrrrB

rrrrA0sincos 44 CBA

2tan1

2tan1cos

2tan1

2tan2sin

42

42

4

424

4

0)(22 CABttAC

2tan 4t


Solving for t we obtain

21212123

24

22

21

242141

242141

sinsincoscos2

sin2sin2

cos2cos2

rrrrrrC

rrrrB

rrrrA 0)(22 CABttAC

t

CA

CBABt

AC

ACACBBt

14

222

2

tan2

2

442

and

or


The ± sign identifies the two possible assembly modes of the linkage

t

CA

CBABt

rr

rrrrC

rrrrB

rrrrA

14

222

212121

23

24

22

21

242141

242141

tan2

sinsincoscos2

sin2sin2

cos2cos2

and

Given the values of θ1

and θ2

for a four bar mechanism of known r1

,

r2

, r3

and r4

the output angle θ4

is calculated as:


The ± sign identifies the two possible assembly modes of the linkage

t

CA

CBABt

rr

rrrrC

rrrrB

rrrrA

14

222

212121

23

24

22

21

242141

242141

tan2

sinsincoscos2

sin2sin2

cos2cos2

and

Given the values of θ1

and θ2

for a four bar mechanism of known r1

,

r2

, r3

and r4

the output angle θ4

is calculated as:


t

CA

CBABt

rr

rrrrC

rrrrB

rrrrA

14

222

212121

23

24

22

21

242141

242141

tan2

sinsincoscos2

sin2sin2

cos2cos2

and

Note that –π/2 ≤ tan-1 (t) ≤ –π/2. Therefore, θ4

will have the range –π ≤ θ4

≤ –π.

Unless the linkage is a Grashof type II linkage in one of the extreme positions of its motion range, there are two valid solutions for θ4

.

These correspond to two assembly modes or branches for the linkage.


t

CA

CBABt

rr

rrrrC

rrrrB

rrrrA

14

222

212121

23

24

22

21

242141

242141

tan2

sinsincoscos2

sin2sin2

cos2cos2

and

Because of the square root in the expression for t, it can be

complex if (A2 + B2) < C2. In this case, the mechanism is a

Grashof type II linkage in one of the extreme positions of its

motion range and it cannot be assembled in specified values of

θ1

and θ2

.

The assembly configurations would then appear as shown

below


t

CA

CBABt

rr

rrrrC

rrrrB

rrrrA

14

222

212121

23

24

22

21

242141

242141

tan2

sinsincoscos2

sin2sin2

cos2cos2

and

After θ4

is known, an expression for θ3

can be obtained by

solving the loop closure equation to obtain:

Note that it is essential that the sign of the numerator and

denominator be maintained to determine the quadrant in

which the angle θ3

lies. This can be done by using the ATAN2

function. The form of this function is:

ATAN2 (sin θ3

, cos θ3

)

22441133

22441133

sinsinsinsin

coscoscoscos

rrrr

rrrr

224411

22441113 coscoscos

sinsinsintan

rrr

rrr


Once all of the angular quantities are known, it is

relatively straightforward to compute the coordinates

of any point on the mechanism.

In particular, the coordinates of Q,P, and R are given by

jirrr

jirjirrrr

jirjirrrr

jirrr

p

p

p

Q

1111

44411141

33322232

2222

sincos

sincossincos

sincossincos

sincos


If the coupler angle θ3

is given, and θ2

and θ3

are to be determined,

graphical solution will require an iterative trial and error solution.

The analytical procedure, in contrast, follows exactly the same

procedure as when θ2

is given. It starts by writing the loop closure

equations with θ2

replacing θ3

as the variable to be

eliminated.

The equations set are of exactly the same form except that the indices

2 and 3 are interchanged. Therefore, we can use directly the position

solution derived for the case of known θ2

while interchanging the

indices 2 and 3.


33441122

33441122

sinsinsinsin

coscoscoscos

rrrr

rrrr

22441133

22441133

sinsinsinsin

coscoscoscos

rrrr

rrrr

When the coupler angle θ3

is given, there is an assembly-mode

ambiguity similar to that occurine when θ2

is given.

It is necessary to know the appropriate mode of the linkage

before the analysis is begun;. The mode is determined by the +

or – sign used for the square root term when calculating t.

Once the assembly mode is determined, it remains the same for

any position of the input link unless the linkage is a class III

linkage, and passes through a singular (indeterminate) position .


33441122

33441122

sinsinsinsin

coscoscoscos

rrrr

rrrr

A four bar linkage with r1

= 1, r2

= 2, r3

= 3.5, r4

= 4, and θ1

= 0, find θ3

and θ4

for each of the solution branches when the

driving crank is in the positions θ2

= 0, π/2, π,and - -π/2.

Example: Position Analysis of a Four Bar Linkage

58.535049.0tan2

5049.075.48

75.4080

75.4)2)(1(25.3421

sinsincoscos2

0sin2sin2

8)4)(2(2)4)(1(2

cos2cos2

14

222

222

2222

212121

23

24

22

21

242141

242141

t

CA

CBABt

C

rr

rrrrC

rrrrB

A

rrrrA

A four bar linkage with r1

= 1, r2

= 2, r3

= 3.5, r4

= 4, and θ1

= 0, find θ3

and θ4

for each of the solution branches when the driving

crank is in the positions θ2

= 0, π/2, π,and - -π/2.

Example: Position Analysis of a Four Bar Linkage

POSITION ANALYSIS FOR A RIGID BODY WHEN TWO POINTS ARE KNOWN

Given the kinematic properties of one point on a rigid body and the

angular position, angular velocity, and angular acceleration of the

body, we can compute the position, velocity, and acceleration of any

defined point on the rigid body.

For the rigid body shown. Assume that A and B are two points

attached to an arbitrary link, say link 5, and a third point is defined

relative to the line between points A and B by the angle β and the

distance rC/A

, which is represented as r6

. Then the position of point

C can be computed directly if rA

and θ5

are known.

Position Analysis for a Rigid Body When Two Points are Knows

The position of point C is given as:

If θ5

If is known, the equation above can be used to calculate the location of point

C directly.

We often know the position vectors of two points A and B on the rigid body. The

value of θ5

can be calculated from the from the x and y components of the position

vectors for A and B using

Position Analysis for a Rigid Body When Two Points are Knows

56

6666 sincos

where

jirrrrr AAC

xx

yy

AB

AB

rr

rr1

5 tan

POSITION ANALYSIS FOR A SLIDER-CRANK MECHANISM

HW#2 (Prob. 4-6 and 4-7, with data in row (a)

Table P4-1).(4-10, 4-12, 4-18(f))

Next to the four-bar linkage,

the slider-crank is probably

the most commonly used

mechanism.

It appears in all internal

combustion engines and in

numerous industrial and

household devices.

Position Analysis for a Slider-Crank Mechanism

To develop the closure equations, locate vectors

r2

and r3

as was done in the regular four-bar

linkage.

One of the other two vectors is taken in the

direction of the slider velocity and the other is

taken perpendicular to the velocity direction.

The loop closure equation is


4132 rrrrrp

Offset

r3

r2

r1

r4

rp


2

sincossincos

sincossincos

14

444111

333222

41332

4132

412

wherejirjir

jirjir

urururur

rrrr

44113322

44113322

sinsinsinsin

coscoscoscos

rrrr

rrrr

Writing the loop closure equation in terms of the

vector angles, we obtain

Two scalar equations are produced. The equations

can be solved for two unknowns.

r3

r2

r1

r4

rp


44113322

44113322

sinsinsinsin

coscoscoscos

rrrr

rrrr

Unlike the four-bar linkage loop closure equations where all link

lengths are known, the piston displacement r1

is an unknown in

the slider-crank equation. The constraint resulting from a

known r1

is replaced by the constraint θ4

= θ1

+ π/2.

The following problem statements are possible

Crank angle θ2

given , find θ3

and r1

Piston displacement r1

given, find θ2

and θ3

Coupler angle θ3

given, find θ2

and r1

.

r3

r2

r1

r4

rp


44113322

44113322

sinsinsinsin

coscoscoscos

rrrr

rrrr

The analytical solution procedure follows the same major steps as

in the four-bar linkage case. To eliminate θ3

, first isolate it in the

loop closure equations as follows:

Squaring both sides of both equations and using the identity

we obtain

r3

r2

r1

r4

rp

22441133

22441133

sinsinsinsin

coscoscoscos

rrrr

rrrr

1cossin 22

424242212121

41414124

22

21

23


sinsincoscos2

rrrr

rrrrrr


The expression gives r1

in a quadratic expression

involving θ2

and the other known variables. To

obtain a solution, collect together the

coefficients of the different powers of r1

as

follows

r3

r2

r1

r4

rp

424242212121

41414124

22

21

23


sinsincoscos2

rrrr

rrrrrr

424242

23

24

22

2121241414

121

sinsincoscos2


0

rrrrrB

rrA

BArr

where


r3

r2

r1

r4

rp

424242

23

24

22

2121241414

121

sinsincoscos2

sinsincoscossinsincoscos2

0

rrrrrB

rrA

BArr

where

21212

41

4141

sinsincoscos

0)2cos()cos(

sinsincoscos

rA

giving

The expression for A could be simplified by

noting that


424242

23

24

22

21212

121

sinsincoscos2

sinsincoscos

0

rr

rrrB

rA

BArr

where

Solving for r1

gives

The ± sign indicates two possible assembly

modes for the same θ2

.

2

42

1

BAAr


424242

23

24

22

21212

sinsincoscos2

sinsincoscos

rr

rrrB

rA Because of the square root in the

expression for r1

, it becomes complex

when A2 < 4B.

If this happens, the mechanism cannot be

assembled in the position specified.

2

42

1

BAAr


424242

23

24

22

21212

sinsincoscos2

sinsincoscos

rr

rrrB

rA Once a value for r1

is determined, the closure equations can be

solved for θ3

to give

As in the case of the four-bar linkage, it is essential that the signs

of the numerator and denominator in the above expression be

maintained to determine the quadrant in which the angle θ3

lies.

2

42

1

BAAr

224411

22441113

22441133

22441133

coscoscos

sinsinsintan

sinsinsinsin

coscoscoscos

rrr

rrr

rrrr

rrrr

r3

r2

r1

r4

rp


424242

23

24

22

21212

sinsincoscos2

sinsincoscos

rr

rrrB

rA Once all of the angular quantities are known, it is relatively

straightforward to compute the coordinates of any point on the

vector loops used in the closure equations.

In particular, the coordinates of Q and P are given by

2

42

1

BAAr

r3

r2

r1

r4

rp

jirjirrrr

jirrr

p

Q

33322232

2222

sincossincos

sincos

Analytical Solution when r1 is known

The analytical solution procedure follows the same major steps as in

the previous case.

After eliminating θ3

from the loop closure equation, we simplify the

resulting equation as follows

where r3

r2 r

1

r4

rp

0sincos 22 CBA

0sinsincoscos2sinsincoscos2

sinsincoscos2

424242212121

41414123

24

22

21

rrrr

rrrrrr

41414123

24

22

21

442121

442121

sinsincoscos2

sin2sin2

cos2cos2

rrrrrrC

rrrrB

rrrrA


The trigonometric half-angle identities can be used to solve the

equation above for θ2

. Using these identities and simplifying gives

Solving for t gives

r3

r2 r

1

r4

rp

0sincos 22 CBA

414141

23

24

22

21

442121

442121

sinsincoscos2

sin2sin2

cos2cos2

rr

rrrrC

rrrrB

rrrrA

2tan1

2tan1cos

2tan1

2tan2sin

22

22

2

222

2

2tan

0)(2

2

2

t

CABttAC

where

tCA

CBABt

AC

ACACBBt

12

222

2

tan2

2

442

and


The ± sign indicates two possible assembly modes.

Typically, there are two valid solutions for θ2

.

Because tan-1 has a valid range of π/2 ≤ tan-1 (t) ≤ –π/2,

θ2

will have the range –π ≤ θ2

≤ –π.

414141

23

24

22

21

442121

442121

sinsincoscos2

sin2sin2

cos2cos2

rr

rrrrC

rrrrB

rrrrA

tCA

CBABt

AC

ACACBBt

12

222

2

tan2

2

442

and


Because of the square root in the equation for the

variable t , it becomes complex when (A2 + B2) <

C2

If this happens, the mechanism cannot be

assembled for the specified value of r1

.

414141

23

24

22

21

442121

442121

sinsincoscos2

sin2sin2

cos2cos2

rr

rrrrC

rrrrB

rrrrA

tCA

CBABt

AC

ACACBBt

12

222

2

tan2

2

442

and


Knowing θ2

, the closure equations can be

solved for θ3

. As in the previous cases, it is

essential that the signs of the numerator and

denominator be maintained to determine the

quadrant in which the angle θ3

lies.

414141

23

24

22

21

442121

442121

sinsincoscos2

sin2sin2

cos2cos2

rr

rrrrC

rrrrB

rrrrA

tCA

CBABt

AC

ACACBBt

12

222

2

tan2

2

442

and


When the coupler angle θ3

is the input link, the analytical procedure for solving

the position equations follows the same major steps as when θ2

is the input.

we can assume that θ1

, θ3

, θ3

are known and that θ2

and r1

are to be found.

The link lengths r2

and r2

and the angles θ1

, and θ4

are constants.

For the position analysis, again begin with the loop closure equations and

isolate the terms with θ2

The resulting equations of the same form obtained

when θ2

is the input except that the indices 2 and 3 are interchanged.

Therefore, we can use directly the position solution derived for the case when

θ2

is the input and interchange the indices 2 and 3.

23441132

23441132

sinsinsinsin

coscoscoscos

rrrr

rrrr

POSITION ANALYSIS FOR AN INVERTED SLIDER-CRANK MECHANISM

An inversion of the fourbar slider-crank linkage

in which the sliding joint is between links 3

and 4 is shown. The slider block has pure

rotation about point O4

.

Note that the angle between link 3 and link 4

is fixed and is equal to π/2, and that link 1 is

assumed to be aligned with the x-axis of the

reference coordinate, θ1

= 0.

Position Analysis for an Inverted Slider-Crank Mechanism

r3

r2 r

1

r4

To develop the closure equations, locate vectors r2

and r3

as was done in the original slider-crank linkage. The

magnitude of r3

is variable depending on the location of

the slider.

Vectors r1

is taken in the direction of the ground link, and

vector r4

is taken perpendicular to r3

. The loop closure

equation is


4132 rrrr

r3

r2 r

1

r4


2

sincossincos

sincossincos

34

444111

333222

41332

4132

412

wherejirjir

jirjir

urururur

rrrr

44113322

44113322

sinsinsinsin

coscoscoscos

rrrr

rrrr

Writing the loop closure equation in terms of the

vector angles, we obtain

Two scalar equations are produced. The equations

can be solved for two unknowns.

r3

r2 r

1

r4


The piston displacement r3

is an unknown in the equation. The

constraint resulting from a known r3

is replaced by the

constraint θ4

= θ3

+ π/2.

The following problem statements are possible

Crank angle θ2

given , find θ4

and r3

Piston displacement r3

given, find θ2

and θ4

Output angle θ4

given, find θ2

and r3

.

r3

r2 r

1

r4

34

34

cossin

sincos

0sin

1cos

1

1

444322

4414322

sincossin

cossincos

rrr

rrrr


The analytical solution procedure follows the same major steps as

in the non-inverted slider crank. To eliminate θ4

, first isolate it in

the loop closure equations as follows:

Squaring both sides of both equations and using the identity

we obtain

1cossin 22

22122

21

23

24 cos2 rrrrrr

r3

r2 r

1

r4

224344

1224344

sincossin

cossincos

rrr

rrrr

444322

4414322

sincossin

cossincos

rrr

rrrr


The equation can be solved for r3

, which is the one

unknown.

Substituting the resulting r3

value in the first closure

equation, we obtain an expression with only θ3

unknown

The above expression can be solve using the previous

half angle identities

22122

21

23

24 cos2 rrrrrr

r3

r2 r

1

r4

444322

4414322

sincossin

cossincos

rrr

rrrr

4414322 cossincos rrrr

Trigonometric Solution when θ2 is known

The problem could be solved using

trignonometry by solving first for the

length of link O4

A from the triangle

O2

AO4

, of which two sides and an angle

are known.

Using length of link O4

A and the

triangle ABO4

the length r3

and θ4

could be found.

r3

r2 r

1

r4

2

321

24

22122

21

24 cos2

rrAO

rrrrAO


The expression can be seen to be the same as that obtained for a

non-inverted slider-crank when r1

is known with interchanging the

indices 1 and 2 and the indices 3 and 4.

r3

r2 r

1

r4

0sinsincoscos2sinsincoscos2

sinsincoscos2

313131212121

32323224

23

22

21

rrrr

rrrrrr

TRANSMISSION ANGLES


The transmission angle μ is defined as the angle between the

output link and the coupler of a four-bar mechanism. It is

usually taken as the absolute value of the acute angle of the

pair of angles at the intersection of the two links and varies

continuously as the linkage goes through its range of motion.

The transmission angle is considered a measure of the quality

of force transmission at the joint. A transmission angle value

close to π/2 is desired, and a value of 0 or is avoided.

Documents

Mechanics of Machines Dr. Mohammad Kilani Class 3 Position Analysis