Download := 5.065 − 0.281

DESIGN OF MACHINERY
"
SOLUTION MANUAL 5-2-1
PROBLEM 5-2
Statement:
Design a fourbar mechanism to give the two positions shown in Figure P3-1 of coupler motion.
Given:
Coordinates of the points A1, B1, A2, and B2 with respect to A1:
A1x := 0.0
B1x := 1.721
A2x := 2.656
B2x := 5.065
A1y := 0.0
B1y := −1.750
A2y := −0.751
B2y := −0.281
Assumptions: Use the points A1 and A2 as the precision points P1 and P2. Define position vectors in the global
frame whose origin is at A1.
Solution:
See solution to Problem 3-4 and Mathcad file P0502.
1.
Note that this is a two-position motion generation (MG) problem because the output is specified as a complex
motion of the coupler, link 3. See Section 5.3 in which the equations for the two-position motion generation
problem are derived.
2.
Two solution methods are derived in Section 5.3 and are presented in equations 5.7 and 5.8 for the left dyad and
equations 5.11 and 5.12 for the right dyad.
3.
Method 1 (equations 5.7 and 5.11) requires the choosing of three angles for each dyad. Method 2 (equations 5.8
and 5.12) requires the choosing of two angles and a length for each dyad. Method 1 is used in this solution.
4.
In order to obtain the same solution as was done graphically in Problem 3-4, the necessary assumed values were
taken from that solution as shown below.
5.
Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1.
p 21 := 2.760
6.
7.
p 21 = 2.760
From the trigonometric relationships given in Figure 5-1, determine α2 and δ2.
α 2 := 56.519⋅ deg
α 2 = 56.519 deg
δ 2 := −15.789⋅ deg
δ 2 = −15.789 deg
From the graphical solution (see figure below), determine the values necessary for input to equations 5.7.
θ := 94.394⋅ deg
β 2 := −40.366⋅ deg
φ := −45.479⋅ deg
DESIGN OF MACHINERY
SOLUTION MANUAL 5-2-2
O4
93.449°
jY
54.330°
2.760
u
A1
X
45.479°
0.281
P 21
15.789°
B2
0.751
v
1.750
A2
134.521°
2.656
2.409
B1
w
40.366°
94.394°
75.124°
O2
8.
Solve for the WZ dyad using equations 5.7.
( ( )
)
( )
( ( )
)
( )
( ( )
)
( )
( ( )
)
( )
A := cos( θ ) ⋅ cos β 2 − 1 − sin( θ ) ⋅ sin β 2
B := cos( φ) ⋅ cos α 2 − 1 − sin( φ) ⋅ sin α 2
( )
C := p 21⋅ cos δ 2
D := sin( θ ) ⋅ cos β 2 − 1 + cos( θ ) ⋅ sin β 2
( )
E := sin( φ) ⋅ cos α 2 − 1 + cos( φ) ⋅ sin α 2
F := p 21⋅ sin δ 2
w := 4.000
z := 0.000
w = 4.000
z = 0.000
These are the expected values of w and z based on the design choices made in the graphical solution and the
assumptions made in this problem.
9.
From the graphical solution (see figure above), determine the values necessary for input to equations 5.11.
σ := −93.449⋅ deg
γ 2 := 54.330⋅ deg
ψ := 134.521 ⋅ deg
10. Solve for the US dyad using equations 5.11.
( ( )
)
( )
( ( )
)
( )
A' := cos( σ ) ⋅ cos γ 2 − 1 − sin( σ ) ⋅ sin γ 2
B' := cos( ψ ) ⋅ cos α 2 − 1 − sin( ψ ) ⋅ sin α 2
( )
C := p 21⋅ cos δ 2
DESIGN OF MACHINERY
SOLUTION MANUAL 5-2-3
( ( )
)
( )
( ( )
)
( )
D' := sin( σ ) ⋅ cos γ 2 − 1 + cos( σ ) ⋅ sin γ 2
( )
E' := sin( ψ ) ⋅ cos α 2 − 1 + cos( ψ ) ⋅ sin α 2
F := p 21⋅ sin δ 2
u := 4.000
s := 2.454
u = 4.000
s = 2.454
These are the expected values of u and s based on the design choices made in the graphical solution and the
assumptions made in this problem.
11. Solve for the links 3 and 1 using the vector definitions of V and G.
Link 3:
θ 3 := −45.479⋅ deg
θ 3 = −45.479 deg
v := 2.454
v = 2.454
θ 1 := 75.120⋅ deg
θ 1 = 75.120 deg
g := 6.447
g = 6.447
Link 1:
12. Determine the initial and final values of the input crank with respect to the vector G.
θ 2i := θ − θ 1
θ 2i = 19.274 deg
θ 2f := θ 2i + β 2
θ 2f = −21.092 deg
13. Define the coupler point with respect to point A and the vector V.
rp := z
δ p := φ − θ 3
rp = 0.000
δ p = 0.000 deg
which is correct for the assumption that the precision point is at A.
14. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2.
O2x := 0.306
O2x = 0.306
O2y := −3.988
O2y = −3.988
DESIGN OF MACHINERY
SOLUTION MANUAL 5-2-4
O4x := 1.961
O4x = 1.961
O4y := 2.243
O4y = 2.243
15. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to
the line O2O4.
θ rot := atan2( O4x − O2x) , ( O4y − O2y)
θ rot = 75.125 deg
16. Determine the Grashof condition.
Condition( S , L , P , Q) :=
SL ← S + L
PQ ← P + Q
return "Grashof" if SL < PQ
return "Special Grashof" if SL = PQ
return "non-Grashof" otherwise
Condition( v , g , w , u ) = "non-Grashof"
17. DESIGN SUMMARY
Link 2:
w = 4.000
θ = 94.394 deg
Link 3:
v = 2.454
θ 3 = −45.479 deg
Link 4:
u = 4.000
σ = −93.449 deg
Link 1:
g = 6.447
θ 1 = 75.120 deg
Coupler:
rp = 0.000
δ p = 0.000 deg
Crank angles:
θ 2i = 19.274 deg
θ 2f = −21.092 deg