Dan B. Marghitu, Mechanisms and Robots Analysis with MATLAB®
Подождите немного. Документ загружается.
4.6 Slider-Crank (R-RRT) Mechanism 125
The inertia force on link 1 at C
1
is
F
in1
= −m
1
a
C
1
=
√
2
4
ı +
√
2
4
j
N.
The gravitational force on crank 1 at C
1
is
G
1
= −m
1
gj = −10j N.
The mass moment of inertia of the link 1 about C
1
is
I
C
1
= m
1
(AB
2
+ h
2
)/12 = 0.0833417 kg m
2
.
The moment of inertia on link 1 is
M
in1
= −I
C
1
α
1
= 0.
Link 2
The mass of connecting rod 2 is
m
2
= ρ
2
BC h d ,
where the density of the material of link 2 is ρ
2
. For simplicity of calculation m
2
=
1 kg. The inertia force on link 2 at C
2
is
F
in2
= −m
2
a
C
2
=
3
√
2
4
ı +
√
2
4
j
N.
The gravitational force on link 2 at C
2
is
G
2
= −m
2
gj = −10j N.
The mass moment of inertia of link 2 about C
2
is
I
C
2
= m
2
(BC
2
+ h
2
)/12 = 0.0833417 kg m
2
.
The moment of inertia on link 2 is
M
in2
= −I
C
2
α
2
= 0.
Link 3
The mass of the link 3 is
m
3
= ρ
3
h
slider
w
slider
d,
where the density of the material of link 3 is ρ
3
. For simplicity of calculation m
3
=
1 kg. The inertia force on link 3 at C
3
= C is
F
in3
= −m
3
a
C
3
=
√
2 ı N.
126 4 Dynamic Force Analysis
The gravitational force on link 3 at C
3
= C is
G
3
= −m
3
gj = −10j N.
The mass moment of inertia of slider 3 about C
3
= C is
I
C
3
= m
3
(h
2
slider
+ w
2
slider
)/12 = 0.0000166667 kg m
2
.
The moment of inertia on slider 3 is
M
in3
= −I
C
3
α
3
= 0.
The MATLAB commands for the forces and moments of inertia are:
m1=1;
IC1=m1
*
(ABˆ2+hˆ2)/12;
G1=[0-m1
*
g0];
Fin1=-m1
*
aC1;
Min1 = - IC1
*
alpha1;
m2=1;
IC2=m2
*
(BCˆ2+hˆ2)/12;
G2=[0-m2
*
g0];
Fin2=-m2
*
aC2;
Min2 = - IC2
*
alpha20;
m3=1;
IC3=m3
*
(hSliderˆ2+wSliderˆ2)/12;
G3=[0-m3
*
g0];
Fin3=-m3
*
aC3;
Min3 = - IC3
*
alpha30;
For a given value of the crank angle φ (φ = π/4) and a known driven force F
ext
find
the joint reactions and the drive (equilibrium) moment M on the crank.
4.6.2 Joint Forces and Drive Moment
4.6.2.1 Newton–Euler Equations of Motion
Figure 4.12 shows the free-body diagrams of the crank 1, the connecting rod 2, and
the slider 3. For each moving link the dynamic equilibrium equations are applied
(Newton–Euler equations of motion)
m a
C
=
∑
F and I
C
α =
∑
M
C
,
where C is the center of mass of the link.
4.6 Slider-Crank (R-RRT) Mechanism 127
1
A
B
C
2
3
M
B
C
1
C
2
C
F
21
F
12
F
32
F
23
F
01
F
23y
G
3
F
23x
F
03
F
32y
F
32x
G
2
F
12x
F
12y
F
21y
F
21x
F
01y
G
1
F
01x
x
y
F
ext
Fig. 4.12 Free-body diagrams
The force analysis starts with the link 3 because the external driven force F
ext
on
the slider is given.
The reaction joint force of the ground 0 on the slider 3, F
03
, is perpendicular
to the sliding direction, x-axis: F
03
⊥ ı (Fig. 4.13). The application point Q of the
reaction force F
03
is determined using the Euler’s moment equation
I
C
3
α
3
= r
CQ
×F
03
or 0 = r
CQ
×F
03
=⇒ r
CQ
= 0 or C = Q.
It results that the reaction force F
03
acts at C. For the slider 3 the vector sum of the
net forces (external forces F
ext
, gravitational force G
3
, joint forces F
23
, F
03
) is equal
to m
3
a
C
3
(Fig. 4.13)
m
3
a
C
3
= F
23
+ G
3
+ F
ext
+ F
03
,
3
C
m
3
a
C
3
=
= F
23
+ G
3
+ F
ext
+ F
03
= F
23
+ G
3
+ F + F
03
m
3
a
C
3
3
F
ext
C
F
23
F
23y
G
3
F
23x
F
03
Free-Body Diagram (FBD)
NEWTON
-EULER
(Kinetic Diagram)
Fig. 4.13 Newton–Euler equations for slider 3
128 4 Dynamic Force Analysis
where F
23
= F
23x
ı + F
23y
j and F
03
= F
03y
j. Projecting the previous vectorial equa-
tion onto x and y axes gives
m
3
a
C
3x
= F
23x
+ F
ext
,
m
3
a
C
3y
= F
23y
−m
3
g + F
03y
,
or numerically
(1)(−
√
2)=F
23x
+ 100, (4.30)
0 = F
23y
−(1)(10)+F
03y
. (4.31)
There are two equations Eqs. 4.30 and 4.31 and three unknowns F
03y
, F
23x
and F
23y
and that is why the analysis will continue with link 2.
The MATLAB commands for Newton-Euler equations for slider 3 are:
F03=[0sym(’F03y’,’real’) 0 ];
F23 = [ sym(’F23x’,’real’) sym(’F23y’,’real’) 0 ];
eqF3 = F03+F23+Fe+G3-m3
*
aC3;
eqF3x = eqF3(1);
eqF3y = eqF3(2);
For the connecting rod 2 (Fig. 4.14), Newton’s equation gives
m
2
a
C
2
= F
32
+ G
2
+ F
12
.
The previous equation can be projected on x and y axes
C
2
C
2
B
= F
32
+ G
2
+ F
12
=
=
m
2
a
C
2
I
C2
α
2
r
C
×F
32
C
2
r
B
×F
12
C
2
+
m
2
a
C
2
I
C2
α
2
B
C
2
C
2
F
12
F
32
F
32y
F
32x
G
2
F
12x
F
12y
FBD
NEWTON-EULER
(Kinetic Diagram)
Fig. 4.14 Newton–Euler equations for link 2
4.6 Slider-Crank (R-RRT) Mechanism 129
m
2
a
C
2x
= F
32x
+ F
12x
,
m
2
a
C
2y
= F
32y
−m
2
g + F
12y
,
or numerically
(1)(−
3
√
2
4
)=F
32x
+ F
12x
, (4.32)
(1)(−
√
2
4
)=F
32y
−1(10)+F
12y
. (4.33)
For the link 2 a moment equation can be written with respect to C
2
I
C
2
α
2
= r
C
2
C
×F
32
+ r
C
2
B
×F
12
,
or
I
C
2
α
2
k =
ıj
k
x
C
−x
C
2
y
C
−y
C
2
0
F
32x
F
32y
0
+
ıj
k
x
B
−x
C
2
y
B
−y
C
2
0
F
12x
F
12y
0
,
or
I
C
2
α
2
=(x
C
−x
C
2
)F
32y
−(y
C
−y
C
2
)F
32x
+(x
B
−x
C
2
)F
12y
−(y
B
−y
C
2
)F
12x
,
or numerically
0 =(
√
2 −
3
√
2
4
)F
32y
−(−
√
2
4
)F
32x
+(
√
2
2
−
3
√
2
4
)F
12y
−(
√
2
2
−
√
2
4
)F
12x
. (4.34)
The MATLAB commands for the Newton–Euler equations for link 2 are:
F32 = -F23;
F12 = [ sym(’F12x’,’real’) sym(’F12y’,’real’) 0 ];
eqF2 = F32+F12+G2-m2
*
aC2;
eqF2x = eqF2(1);
eqF2y = eqF2(2);
eqM2 = cross(rB-rC2,F12)+cross(rC-rC2,F32)-...
IC2
*
alpha20;
eqM2z = eqM2(3);
Equations 4.30–4.34 form a system of 5 equations with five scalar unknowns. The
system can be solved using the solve statement:
sol32 = solve(eqF3x,eqF3y,eqF2x,eqF2y,eqM2z);
F03ys = eval(sol32.F03y);
130 4 Dynamic Force Analysis
F23xs = eval(sol32.F23x);
F23ys = eval(sol32.F23y);
F12xs = eval(sol32.F12x);
F12ys = eval(sol32.F12y);
F03s=[0,F03ys, 0 ];
F23s = [ F23xs, F23ys, 0 ];
F12s = [ F12xs, F12ys, 0 ];
The numerical values for the joint forces for the links 3 and 2 are
F
03y
= −85 −
3
√
2
2
N,
F
23x
= −100 −
√
2N, F
23y
= 95 +
3
√
2
2
N,
F
12x
= −
1
4
(400 + 7
√
2) N, F
12y
=
5
4
(84 +
√
2) N,
or
F
03
= |F
03
| = 85 +
3
√
2
2
N,
F
23
= |F
23
| =
F
2
23x
+ F
2
23y
=
38063
2
+ 485
√
2N,
F
12
= |F
12
| =
F
2
12x
+ F
2
12y
=
1
2
84137 + 2450
√
2N.
For the crank 1 (Fig. 4.15), there are two vectorial equations
m
1
a
C
1
= F
21
+ G
1
+ F
01
,
I
C
1
α
1
= r
C
1
B
×F
21
+ r
C
1
A
×F
01
+ M,
where M, is the input (motor) moment on the crank, F
21
= −F
12
, and F
01
= F
01x
ı +
F
01y
j. The above vectorial equations give three scalar equations on x, y, and z
m
1
a
C
1x
= F
21x
+ F
01x
,
m
1
a
C
1y
= F
21y
−m
1
g + F
01y
,
I
C1
α
1
k =
ıj
k
x
B
−x
C
1
y
B
−y
C
1
0
F
21x
F
21y
0
+
ıj
k
x
A
−x
C
1
y
A
−y
C
1
0
F
01x
F
01y
0
+M k = 0,
or
m
1
a
C
1x
= F
21x
+ F
01x
,
m
1
a
C
1y
= F
21y
−m
1
g + F
01y
,
4.6 Slider-Crank (R-RRT) Mechanism 131
1
B
C
1
A
=
m
1
a
C
1
I
C1
α
1
= F
21
+ F
in 1
+ G
1
+ F
01
m
1
a
C
1
=
I
C1
α
1
r
A
×F
01
C
1
r
B
×F
21
C
1
+
+ M
1
A
M
B
C
1
F
21
F
01
F
21y
F
21x
F
01y
G
1
F
01x
FBD
NEWTON
-EULER
(Kinetic Diagram)
Fig. 4.15 Newton–Euler equations for link 1
I
C
1
α
1
=(x
B
−x
C
1
)F
21y
−(y
C
2
−y
C
1
)F
21x
+(x
A
−x
C
1
)F
01y
−(y
A
−y
C
1
)F
01x
+ M,
or numerically
1(−
√
2
4
)=
1
4
(400 + 7
√
2)+F
01x
, (4.35)
1(−
√
2
4
)=−
5
4
(84 +
√
2) −1(10)+F
01y
, (4.36)
0 =(
√
2
2
−
√
2
4
)
−
5
4
(84 +
√
2)
−(
√
2
2
−
√
2
4
)
1
4
(400 + 7
√
2)
−
√
2
4
F
01y
+
√
2
4
F
01y
+ M = 0. (4.37)
The MATLAB commands for the Newton–Euler equations for the crank 1 are:
F01=m1
*
aC1-G1+F12s;
Mm=-cross(rB,-F12s)-cross(rC1,G1-m1
*
aC1)-IC1
*
alpha1;
Equations 4.35–4.37 give
F
01x
= −2(50 +
√
2) N, F
01y
= 115 +
√
2N,
M = 3 + 105
√
2Nm,
or
132 4 Dynamic Force Analysis
F
01
= |F
01
| =
F
2
01x
+ F
2
01y
=
23235 + 630
√
2N,
M = |M| = 3 + 105
√
2Nm.
Another way of calculating the moment M required for dynamic equilibrium is to
write the moment equation of motion for link 1 about the fixed point A
I
A
α
1
k = r
AC
1
×G
1
+ r
AB
×F
21
+ M =⇒ M = r
B
×F
12
−r
C
1
×G
1
,
where I
A
= I
C
1
+m
1
(AB/2)
2
. The reaction force F
01
does not appear in this moment
equation.
The MATLAB program using Newton–Euler equations and the results are given
in Appendix C.1.
4.6.2.2 D’Alembert’s Principle
For each moving link the dynamic equilibrium equations are applied (d’Alembert’s
principle)
∑
F + F
in
= 0 and
∑
M
C
+ M
in
= 0,
where C is the center of mass of the link. With d’Alembert’s principle the moment
summation can be about any arbitrary point P
∑
M
P
+ M
in
+ r
PC
×F
in
= 0.
The force analysis starts with the link 3 because the external driven force F
ext
is
given. For the slider 3 the vector sum of all the forces (external forces F
ext
, gravita-
tional force G
3
, inertia forces F
in3
, joint forces F
23
, F
03
) is zero (Fig. 4.16)
∑
F
(3)
= F
23
+ F
in3
+ G
3
+ F
ext
+ F
03
= 0,
F
in 3
= −m
3
a
C
3
F
(3)
= F
23
+ F
in 3
+ G
3
+ F
ext
+ F
03
= 0
F
(3)
= F
23
+ F
3
+ G
3
+ F + F
03
= 0
D’ALEMBERT
3
F
ext
C
F
23
F
23y
G
3
F
23x
F
03
Fig. 4.16 D’Alembert’s principle for slider 3
4.6 Slider-Crank (R-RRT) Mechanism 133
where F
23
= F
23x
ı + F
23y
j and F
03
= F
03y
j. Projecting this force onto x and y axes
gives
∑
F
(3)
·ı = F
23x
+(−m
3
a
C
3x
)+F
ext
= 0,
∑
F
(3)
·j = F
23y
−m
3
g + F
03y
= 0,
or numerically
F
23x
+(−1)(−
√
2)+100 = 0, (4.38)
F
23y
−(1)(10)+F
03y
= 0. (4.39)
The MATLAB commands for slider 3 are:
F03=[0sym(’F03y’,’real’) 0 ];
F23 = [ sym(’F23x’,’real’) sym(’F23y’,’real’) 0 ];
eqF3 = F03+F23+Fe+G3+Fin3;
eqF3x = eqF3(1);
eqF3y = eqF3(2);
For the connecting rod 2 (Fig. 4.17), the sum of the forces is equal to zero
∑
F
(2)
= F
32
+ F
in2
+ G
2
+ F
12
= 0,
The previous equation can be projected on x and y axes
F
in 2
= −m
2
a
C
2
M
in 2
= −I
C2
α
2
D’ALEMBERT
F
(2)
= F
32
+ F
in 2
+ G
2
+ F
12
= 0
M
(2)
B
=(r
C
− r
B
) × F
32
+(r
C2
− r
B
) × (F
in 2
+ G
2
)+M
in 2
= 0
B
C
2
C
2
F
12
F
32
F
32y
F
32x
G
2
F
in 2
M
in 2
F
12x
F
12y
Fig. 4.17 D’Alembert’s principle for link 2
134 4 Dynamic Force Analysis
∑
F
(2)
·ı = F
32x
+(−m
2
a
C
2x
)+F
12x
= 0,
∑
F
(2)
·j = F
32y
+(−m
2
a
C
2y
) −m
2
g + F
12y
= 0,
or numerically
F
32x
+(−1)(−
3
√
2
4
)+F
12x
= 0, (4.40)
F
32y
+(−1)(−
√
2
4
) −1(10)+F
12y
= 0. (4.41)
For the link 2 a moment equation can be written with respect to C
2
∑
M
(2)
C
2
= r
C
2
C
×F
32
+ r
C
2
B
×F
12
+ M
in2
= 0.
Instead of the previous equation the sum of the moments with respect to B can be
used
∑
M
(2)
B
= r
BC
×F
32
+ r
BC
2
×(F
in2
+ G
2
)+M
in2
= 0,
or
ıj
k
x
C
−x
B
y
C
−y
B
0
F
32x
F
32y
0
+
ıj
k
x
C
2
−x
B
y
C
2
−y
B
0
−m
2
a
C
2x
−m
2
a
C
2y
−m
2
g 0
−I
C
2
α
2
k = 0,
or
(x
C
−x
B
)F
32y
−(y
C
−y
B
)F
32x
+(x
C
2
−x
B
)(−m
2
a
C
2y
−m
2
g)
−(y
C
2
−y
B
)(−m
2
a
C
2x
) −I
C
2
α
2
= 0,
or numerically
(
√
2 −
√
2
2
)F
32y
−(−
√
2
2
)F
32x
+(
3
√
2
4
−
√
2
2
)
−1(−
√
2
4
) −1(10)
−(
√
2
4
−
√
2
2
)
−1(−
3
√
2
4
)
−0 = 0. (4.42)
For the connecting rod 2 the MATLAB commands are:
F32 = -F23;
F12 = [ sym(’F12x’,’real’) sym(’F12y’,’real’) 0 ];
eqF2 = F32+F12+G2+Fin2;
eqF2x = eqF2(1);
eqF2y = eqF2(2);