Theory of Machines

LECTURE NOTES- MECE 303 Theory of

Machines

5- Analytical Position Analysis

Fall Semester 2010/2011

Halil Orhan YILDIRAN, MS

1

Analytical Position Analysis

There are various methods to be applied in analytical solution of

mechanism:

*Solution by successive triangles

*Solution of Loop Closure Equations

-Solution by using scalar component equation-Solution by using scalar component equation

-Solution by using complex equation and its conjugate

2

Analytical Position Analysis

.

Example: Stepwise solution of Four bar linkage using successive triangles

θ13

Ø ψ

μ

γ c

3

θ12 θ14

Ø

-μ

Ø’

Analytical Position Analysis-Four Bar

.Joint variables: θ12, θ13, θ14

θ12 is the input

Find θ13 and θ14

4

µ is transmission angle

in the figure µ=open configuration assembly of mechanism

in the figure -µ=cross configuration assembly of mechanism

Known mechanism: a1, a2, a3, a4 is known (c and γ are also known)

Analytical Position Analysis-Four Bar

.

Solution of successive triangles yields:

a) 2/1

1221

2

2

2

1 )cos2( θaaaas −+= (from triangle A0AB0 using cosine law)

b)

−+= −

sa

asa 2

2

22

11

2cos'φ (from triangle A0AB0)

5

sa12

0 0

c) 'φπφ −=

d)

+−±= −

43

2

3

22

41

2cos

aa

asaµ (from triangle AB0B)

Analytical Position Analysis-Four Bar

.

e)

−+= −

sa

asa

4

2

3

22

41

2cosψ

f) θ14=Ø -ψ

g) θ13= θ14 –µ

6

If position of point C is required:

Xc=a2cos θ12+c cos(θ13+γ)

Yc=a2sin θ12+c sin(θ13+γ)

Analytical Position Analysis

.Stepwise solution of off set slider crank mechanism

a1, a2, a3 and b3 and γ3 are known. θ12 is the input. From loop closure equation, writing i and j

components

γ

7

θ12

θ13

γ3

Analytical Position Analysis-Off-set slider crank

.

a2cosθ12=s14+a3cosθ13 (1)

a2sinθ12=a1+a3sinθ13 (2)

)sin(1

sin 1122

3

13 aaa

−= θθ (3) (obtained from (2)) )sin(1

sin 1122

3

1

13 aaa

−= − θθ

8

13312214 coscos θθ aas −= (4) (obtained from (1))

Xc=s14+b3cos(θ13-γ3) (5)

yc=a1+b3sin(θ13-γ3) (6)

Equations (5) and (6) can only be solved after equations (3) and (4) are solved

Analytical Position Analysis-Inverted slider crank

Inverted Slider crank mechanism

a1, a2, a4 , are known and θ12 is the input. We will find θ14 and s43.

If we write loop equation;

θ12 θ14

Ø ψ

9

Analytical Position Analysis-Inverted slider crank

.(1) psinØ=a2sinθ12

(2) pcosØ=a2cosθ12- a1

from the above two equation

(3)

−= −

1122

1221

cos

sintan

aa

a

θ

θφ

(4) θθ cossin 1122122 aaa

p−

==

10

(4) φ

θ

φ

θ

cos

cos

sin

sin 1122122 aaap

−==

From triangle B0AB

(5) 2

4

2

43 aps −=

(6)

=

= −−

p

a

p

s 41431 cossinψ

(7) θ14=رΨ

Analytical Position Analysis

.

Example: Quick return mechanism

Known a1, a2, a4, a5 , input θ12

This is a mechanism formed by in line inverted slider crank and off set slider crank

11

This is a mechanism formed by in line inverted slider crank and off set slider crank

mechanism

For this analysis you must be careful in selecting reference axis.

1-Select in line inverted slider crank (links 1,2,3,4) and find θ14

2-Select off set slider crank (links 1,4,5,6), and θ14 as input find θ15 and s16

Analytical Position Analysis-Quick Return Mechanism

.s43sin θ14=a2sinθ12

s43cos θ14=a2cosθ12 -a1

−= −

1122

122114

cos

sintan

aa

a

θ

θθ

12

Since θ14 is found, we can use this value to be an input to the second

mechanism

b1=-a4cos θ14+a5sinθ15

+= −

5

1441115

cossin

a

ab θθ

s16=a4sin θ14+a5cos θ15

Analytical Position Analysis

AThe figure

θ15

13

θ14

θ12