Chapter four - Philadelphia University. acceleration... · Acceleration analysis For a known...
Transcript of Chapter four - Philadelphia University. acceleration... · Acceleration analysis For a known...
Chapter four
Laith Batarseh
Acceleration analysis
Acceleration analysis
For a known four-bar mechanism, in a given configuration and known velocities, and a given angular acceleration of the crank, α2 (say CCW), construct the acceleration polygon. Determine α 3 and α 4 .
Acceleration analysis
Solution
For the position vector loop equation:
RAO2+RBA − RBO4 − RO4O2 = 0 --- (1)the velocity equation is
VAO2+VBA − VBO4 = 0 --- (2)
Acceleration analysis
Solution
The acceleration equation is obtained from the time derivative of the
velocity equation as:
AA+ABA = AB
Since RAO2 , RBA , and RBO2 are moving vectors with constant lengths, their
acceleration vectors have normal and tangential components:
Now, ω2, ω3, ω4 and α2 are known, so the components
are known and can be drawn directly
n
B
n
BA
t
A
n
A AandAAA ,,
Acceleration analysis
Solution
Rearrange the loop equation to be seem as follow
Drawing procedures:
Acceleration analysis
Solution
Acceleration analysis
Solution
Acceleration analysis
Solution
Acceleration analysis
Solution
Acceleration analysis
Solution
Acceleration analysis
Using vector algebra
tUtStr
Derive with respect to time
USUStr
UUSUSUSUStr 2
Derive another time with respect to time to find the acceleration
But
Rearrange the terms
USSUSStr 22
tUtU
Acceleration analysis
4-bar mechanism
For the 4-bar mechanism, the length of links are constant and so:
And the equation for acceleration become
As shown in pervious chapters we can find the position and velocity analysis to
4-bar mech. And in this section we will find the acceleration analysis by adding
new input which is α2 and new unknown and they are α3 and α4
0 and SS
USUSUSUStr 22
Acceleration analysis
4-bar mechanism
To apply acceleration analysis on 4-bar mechanism, we derive the loop closure
equation
Dot product both sides by Uθ3 to eliminate α3
443322
2
4444
2
3333
2
2222 UdUdUdUdUdUd
43
2
444344
2
3323
2
222322
cossin
cossin
dd
ddd
Solve for α4:
434
43
2
44
2
3323
2
2223224
sin
coscossin
d
dddd
Acceleration analysis
4-bar mechanism
Dot product both sides by Uθ4 to eliminate α4
Solve for α3:
343
2
4434
2
3324
2
2224223
sin
coscossin
d
dddd
2
44
34
2
33343324
2
222422 cossincossin
d
dddd
Acceleration analysis
Slider crank mechanism
For slider crank mechanism the input is α2 and the outputs will be:
The L.C.E is
Derive twice with respect to time
3 and S
SUaUUdUd 903322
USUdUdUdUd 3322
2
3333
2
2222
Dot product both sides by Uθ3 to eliminate α3
3
2
3323
2
222322 coscossin Sddd
Acceleration analysis
Slider crank mechanism
Solve for
3
2
3323
2
222322
cos
cossin dddS
S
Dot product both sides by U.αto eliminate S
0sincossincos 3
2
333332
2
22222 dddd
33
2223
2
332
2
223
cos
cossinsin
d
ddd