MECE_303_5_10 [Uyumluluk Modu]
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Transcript of MECE_303_5_10 [Uyumluluk Modu]
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Theory of Machines
LECTURE NOTES- MECE 303 Theory of
Machines
5- Analytical Position Analysis
Fall Semester 2010/2011
Halil Orhan YILDIRAN, MS
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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
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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’
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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)
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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)
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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)
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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
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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
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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
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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
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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
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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
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Analytical Position Analysis
AThe figure
θ15
13
θ14
θ12