Solution chapter 2 Dynamic Systems Modeling Simulation and Control - Kluever
Short Description
Solution chapter 02....
Description
Chapter 2: Modeling Mechanical Systems
2.1
The free-body diagram (FBD) is shown below, assuming z zin (t ) and z
zin (t ) :
b1 z + z
m
k ( z z – – z z in in)
b2 z z in
Applying Newton’s Newton’s second second law (summing positive upward):
F b1z k ( z zin ) b2 ( z zin ) mz Rearrange and put all dynamic variables ( z ( z and and z ) on the left – – hand hand side and input variables ( zin and z in ) on the right-hand side.
mz (b1 b2 ) z kz b2 zin ( t) kzin ( t)
Mathematical model
1
Copyright © 2016 John Wiley & Sons
Chapter 2 2.2
The free-body diagram (FBD) is below, assuming xin (t ) x and x 0 + x
k xin x
bx
m
Applying Newton’s Newton’s second second law (positive to the right):
F k ( xin x) bx mx
Rearrange with dynamic variables on the left – left – hand hand side and input variable xin (t ) on the right-hand side.
b x kx kxin (t ) m x
Mathematical model
2
Copyright © 2016 John Wiley & Sons
Chapter 2 2.3 a) The free-body diagram (FBD) is below, assuming x 0 + x
f a (t )
bx
m
Applying Newton’s Newton’s second second law (positive to the right):
F f a (t ) bx mx Rearrange with dynamic variables on the left-hand side and input variable f a on the right-hand side.
b x f a (t ) m x
Model using x using x as as dynamic variable
b) Substitute v x and v x into the mathematical model in part (a):
mv bv f a (t )
Mathematical model using velocity v as dynamic variable
3
Copyright © 2016 John Wiley & Sons
Chapter 2 2.4
2 FBD of mass m and “massless node,” assuming z1 z 2 and z + z 1
+ z 2
k z1 z 2
bz 2
0
f a (t )
m “massless node” mnode
Applying Newton’s second law to “massless node” and mass m (positive is to the right): Massless node: Mass m:
F bz2 k ( z1 z2 ) mnode z 2 0 (node has zero mass; mnode = 0)
F k (z z ) f (t ) mz 1
2
a
1
Rearrange with dynamic variables ( z1 , z2 , z1 , z 2 ) on the left-hand side and input variable f a on the righthand side:
1 k ( z 1 z 2 ) f a (t ) m z
Mathematical model
2 k ( z 2 z 1 ) 0 b z
4
Copyright © 2016 John Wiley & Sons
Chapter 2
1 z 2 and z 2 0 : 2.5 FBD of mass m and “massless node,” assuming z + z 1
+ z 2
b z1 z 2
kz 2
f a (t ) m
“massless node” mnode
Apply Newton’s second law to “massless node” and mass m. Massless node:
Mass m:
1 z 2 ) mnode F kz 2 b( z z 2 0 F b( z1 z2 ) f a (t ) mz 1
Rearrange with dynamic variables ( z1 , z2 , z1 , z 2 ) on the left-hand side and input variable f a on the righthand sides:
1 b( z 1 z 2 ) f a (t ) m z
Mathematical model
2 z 1 ) kz 2 0 b( z
5
Copyright © 2016 John Wiley & Sons
Chapter 2 2.6 FBD of mass m when x < 0.5 m (no contact with spring): + x
m
Applying Newton’s second law: Rearrange:
bx
F bx mx
mx bx 0
( holds for x < 0.5 m )
FBD of mass m when x ≥ 0.5 m (contact with spring): + x
k ( x 0.5) m
bx Apply Newton’s second law:
F bx k ( x 0.5) mx
Rearrange: mx bx k ( x 0.5) 0 (for x ≥ 0.5 m) Collect the two equations:
bx 0 for x < 0.5 m m x b x k ( x 0.5) 0 for x > 0.5 m m x
Mathematical model
6
Copyright © 2016 John Wiley & Sons
Chapter 2 2.7
FBD of masses m1 and m2 assuming z1
0, z2 0, and z1 z 2
f a (t )
k1 z 1
m1
b z1 z 2
Friction force
k2 z 2
m2
Apply Newton’s second law to each mass (positive to the left):
F f (t ) k z b(z z ) m z Mass m : F k z b( z z ) m z Mass m1: 2
a
1 1
2 2
1
1
2
2
1 1
2 2
Rearrange with dynamic variables on the left-hand side and input variable f a on the right-hand side:
m1 z1 b( z1 z2 ) k1 z1 fa ( t)
Mathematical model
m2 z2 b( z2 z1 ) k2 z2 0
7
Copyright © 2016 John Wiley & Sons
Chapter 2 2.8
Horizontal displacements of the link at the spring connections are L1sinθ (upper) and L2sinθ (lower). For small rotation angles, sin . FBD of the link and mass m assuming L1 x
0 (all springs are in compression) and x 0 + x
k1 ( L1 x)
bx m
J
k2 x
k3 L2
Apply Newton’s second law: sum torques about pivot point (clockwise) and sum forces on mass m: Link: +
T k (L x)L k L L
Mass: +
1
1
1
3
2
2
J
F k (L x) bx k x mx 1
1
2
Rearrange with dynamic variables ( , x, x ) on the left-hand side:
( k L2 k L2 ) k L x 0 J 1 1 3 2 1 1
Mathematical model
b x (k 1 k 2 ) x k 1 L1 0 m x
8
Copyright © 2016 John Wiley & Sons
Chapter 2 2.9 a) Draw FBD with assumption x 2 x1 (spring k 2 is in tension) and x1 > 0
k 2 ( x2 x1 )
k1 x1
m1
m2
Apply Newton’s second law to each mass (positive is to the right)
F k x k (x x ) m x Mass m : F k ( x x ) m x Mass m1:
1 1
2
2
2
2
2
1
1
1 1
2 2
Rearrange with dynamic variables on the left-hand side:
m1 x1 k1 x1 k2 ( x1 x2 ) 0
Mathematical model
m2 x2 k2 ( x2 x1 ) 0 b) Use constant total energy to derive the model. Total system energy = total potential energy + total kinetic energy
total P K
1 2
k1 x12
1 2
k2 ( x1 x2 )2
1 2
total P.E. (two springs)
m1 x12
1 2
m2 x22
total K.E. (two masses)
Take the time derivative of total energy:
total constant total 0 total k1 x1x1 k 2 (x 1 x 2 ) x1 x 2 m 1x 1x 1 m 2x 2x 2 0 Factor out x1and x2 from total equation:
total k1 x1 k2 ( x1 x2 ) m1x1 x1 k 2 (x1 x 2 ) m 2x 2 x 2 0 Because x1 and x2 cannot both be equal to zero for all time 0 ≤ t ≤ ∞, the two bracket terms must be set equal total 0 . Therefore, set both bracket terms to zero. to zero so that
m1 x1 k1 x1 k2 ( x1 x2 ) 0
Mathematical model - same as part (a)
m2 x2 k2 ( x2 x1 ) 0
9
Copyright © 2016 John Wiley & Sons
Chapter 2 2.10
The given model is
m1 x1 b x1 x2 k1 k 2 x1 k1x 2 0 m2 x2 b x2 x1 k1 x2 x1 f a (t ) 1) Friction term b x1 x2 depends on relative velocity hence dashpot b
Observations:
connects masses m1 and m2. 2) Stiffness term k1 x2
x1 depends on relative displacement hence
spring k 1 connects m1 and m2. 3) Force f a (t ) acts on mass m2 only and in the positive direction. 4) Spring k 2 is only connected to mass m1 . Mechanical system: + x1
+ x2 b
k 2
m1
m2
f a (t )
k 1
FBD of both masses assuming x2 x1 (spring k 1 in tension), x2 x1 0 , and x1 0 (spring k 2 in tension):
b x2 x1 k2 x1
f a t
m1
m2
k1 ( x2 x1 )
Apply Newton’s second law:
F k x b x x k (x x ) m x Mass m : F b x x k ( x x ) f t m x Mass m1: 2
2 1
2
2
1
1
1
1
2
2
1
1
a
1 1
2 2
Rearranging:
m1 x11 b x1 x2 k1 k 2 x1 k1x 2 0
Mathematical model - matches
m2 x2 b x2 x1 k1 ( x2 x1) f a t
given modeling equations
10
Copyright © 2016 John Wiley & Sons
Chapter 2 2.11
The given mathematical model is
m1 x1 b1x1 k1 x1 x 2 f a (t ) m2 x2 b2 x2 k1 x2 x1 k 2x 2 0 1) Dashpots b1and b2 are only connected to masses m1and m2 , respectively,
Observations:
since the friction terms depend on absolute velocities. 2) Spring k 1 is connected to m1 and m2 since the stiffness force depends on relative displacement. 3) Force f a (t ) is applied directly to mass m1 in the positive direction. 4) Spring k 2 is only connected to mass m2 . Mechanical system: + x2
+ x1
f a (t )
b2
k 1
b1
m1
m2
k 2
FBD of both masses, assuming x2 x1 , x1 0 , x2 0 and x2 0 :
f a t
b2 x2
m1
b1x1
k1 ( x2 x1 )
m2
k 2 x2
Apply Newton’s second law to each mass:
F f t b x k (x Mass m : F k ( x x ) b x Mass m1: 2
a
1 1
1
2
1
1
2
2 2
x1) m1x1 k 2x 2 m2x 2
Rearranging:
m1 x1 b1x1 k1 x1 x 2 f a (t)
Mathematical model – matches given
m2 x2 b2 x2 k1 x2 x1 k 2x 2 0
modeling equations
11
Copyright © 2016 John Wiley & Sons
Chapter 2 2.12
The given mathematical model is
m1 x1 k1x1 k 2 x1 x 2 0
m2 x2 k 2 x 2 x1 0 Observations:
1) No friction terms no dashpots. 2) No applied forces. 3) Spring k 1 is only connected to mass m1 since the stiffness force only depends on x1 . 4) Spring k 2 is connected to masses m1and m2 since the stiffness force depends on relative displacement.
Mechanical system:
+ x2
+ x1
k 1
k 2 m1
m2
FBD of both masses, assuming x2 x1 and x1 0 :
k1 x1
m1
k2 x2 x1
m2
Apply Newton’s second law to each mass: Mass m1:
F k 1 x1 k 2 ( x2 x1 ) m1x1
Mass m2:
F k 2 ( x2 x1 ) m2 x2
Rearranging:
m1 x1 k1x1 k 2 x1 x 2 0
Mathematical model - matches
m2 x2 k2 x2 x1 0
given modeling equations
12
Copyright © 2016 John Wiley & Sons
Chapter 2 2.13
a) Derive an equation for rate of energy loss (power dissipated). Power
d dt
Therefore
where total energy
d d
J
d dt
1 2
J 2 (kinetic energy)
Use Newton’s second law to get the “in ertia torque” J FBD of disk J :
J
b , friction torque
Sum all the torques on disk J with sign convention as positive clockwise:
T b = J
Substitute for J from the torque equation into the power equation ( ): b 2 Power dissipated = b
b) Rotational system: Power = torque x angular velocity. Using the free-body diagram from part (a), the only torque is the friction torque. Summing torques (positive rotation in clockwise direction):
Therefore, Power b b 2
T b = J Power dissipated = b 2
The power dissipated equation must have a minus sign since friction causes energy losses, 0
13
Copyright © 2016 John Wiley & Sons
Chapter 2
2.14
Free-body diagram of the pulley-system disk J :
Input torque,
T in
b ,friction torque
mg , weight of mass m
Apply Newton’s second law to disk, J (sum torques in the counter-clockwise direction)
T T
in
b mgr J
) on the left-hand side and input variables T and mgr on Rearrange with dynamic variables ( and in the right-hand side:
J b Tin mgr
Mathematical model
Note that mgr acts as a “load torque” on the pulley.
14
Copyright © 2016 John Wiley & Sons
Chapter 2 2.15
Free-body diagram of the system, assuming 1 2 and 2
0
Torsional shaft torque,
J 1
J 2
k 1 2
+
+
r 2 f c Contact force
Friction torque, b 2
f c
Gear 1 (no inertia)
Input torque,
T in
Apply Newton’s second law to all three disks: Gear 1:
T T
in
Gear 2 + J 1:
Disk 2:
f c r1 J gear1 gear1 0
T f r
c 2
T k
1
k 1 2 J1 1
2 b2 J 22
Because Gear 1 has negligible inertia, Tin f c r 1 or f c
1
r 1
T in . Substitute f c
1
r 1
T in for the contact force
in the equation for Disk J 1 . Rearranging we obtain:
J11 k 1 2
r 2 r 1
T in
Mathematical model
J 22 b2 k 2 1 0 We can substitute the gear ratio N = r 2/r 1 into the first equation if desired.
15
Copyright © 2016 John Wiley & Sons
Chapter 2 2.16
Free-body diagram of both disks, assuming 1 2 , 1 0 and 2 0 :
Torsional shaft torque, k 1 1 2
J 1
Input torque,
+
T in
+
J 2
Spring torque, k 2θ 2
Friction torque, b 1
Apply Newton’s second law to each disk: Disk J 1:
T T in b 1 k 1 ( 1 2 ) J 1 1
Disk J 2:
T k k 1
1
2
2 2
J 22
Rearranging we obtain:
b k ( ) T J 1 1 1 1 1 2 in
Mathematical model
J 22 k1 2 1 k 2 2 0
16
Copyright © 2016 John Wiley & Sons
Chapter 2 2.17 Free-body diagram assuming 0 and 0
b d2
kd 1
d 3
+
f a (t ) Note that for small angle , the vertical deflection of the left end is d1 sin d 1 and the vertical deflection of the right end is d2 d3 sin
d 2 d3 .
Hence the vertical velocity of the right end is d 2
d 3 .
Apply Newton’s second law and sum torques about the pivot (counter-clockwise):
T kd d f d 1
1
a
2
b d 2 d 3 d 2 d 3 J
Rearranging we obtain: 2
J b d 2 d 3 kd12 f a (t )d 2
17
Copyright © 2016 John Wiley & Sons
Mathematical model
Chapter 2 2.18
Using the polyfit command in MATLAB, fit the data with the spring deflection as the independent variable and the load force as the dependent variable. The MATLAB commands are
>> F = [ -50 -45 -35 -25 -15 -10 0 10 15 25 >> x1 = [-7.2055 -6.2354 -4.6099 -3.1938 -1.8822 >> pp1 = polyfit(x1,F,3)
35 45 50 ]; -1.2482 0 1.2482
… ]
The third-order polynomial coefficients are pp1 = [-0.0213 -0.0000 8.0449 -0.0000 ] and
x 8.0449 x where x is in mm. The MATLAB command hence the spring force is F k 0.0213 3
polyval is used to evaluate the polynomial for -8 > x1_fit = linspace(-8,8,200); >> F_fit = polyval(pp1,x1_fit);
% evaluate 3rd-order polynomial for spring force
The same steps are applied to the data for Spring #2. However, Spring #2 exhibits a linear relationship with deflection (see plots below):
Spring #1 has a nonlinear relationship with deflection (cubic polynomial), while Spring #2 has a linear relationship with deflection. 80
60 Data Polynomial fit
40
60 N
N
40 ,
, inr
g
g inr
20
20 ps
ps o
n
n o
0
0 e
Data Linear fit
e cr
cr
of -20
of
d
d -20 a a o
o
L -40 L
-40
-60 -8
-60 -6
-4
-2
0
2
4
6
-80 -8
8
-6
-4
-2
0
2
Spring #2 deflection, mm
Spring #1 deflection, mm
18
Copyright © 2016 John Wiley & Sons
4
6
8
2.19
The M-file for computing the friction force is below, along with the plot. % % Problem 2.19 % % constants Fst = 10; Fc = 7; b = 70; cpts = [ 0.001 0.002 0.005 ]; xdot = linspace(-0.05,0.05,10000); for i=1:3 c = cpts(i); Ff(i,:) = sign(xdot).*(Fc + (Fst-Fc).*exp(-abs(xdot)./cpts(i))) + b.*xdot; end plot(xdot,Ff) grid xlabel('Relative velocity, m/s') ylabel('Friction force, N')
15
10
c = 0.001 m/s c = 0.002 m/s c = 0.005 m/s
5 N , e cr of
0 n oi t ci r F
-5
-10
-15 -0.05 -0.04 -0.03 -0.02 -0.01
0
0.01 0.02 0.03 0.04 0.05
Relative velocity, m/s
At very low relative velocities (near zero) the friction force instantaneously switches between +/- the static friction force F st (i.e., -10 N to +10 N) regardless of the value for the velocity coefficient c. As the magnitude of the relative velocity increases from zero the friction force decreases to a value nearly equal to the dry friction force F C = 7 N. The “rate” of decrease in friction force near zero velocity depends on the coefficient c where smaller values of c produce a sharper decrease in friction near zero relative velocity. At higher relative velocities (such as 0.02 m/s) the friction force is simply the sum of dry friction F C = 7 N and viscous friction b x regardless of coefficient c.
19
Copyright © 2016 John Wiley & Sons
2.20
The Mfile for creating the plot is below:
% % Problem 2.20 % xdot = [-1.5:0.1:1.5]; % range of xdot values (m/s) v = 0.2; % m/s Fd = 4500*xdot./sqrt(xdot.^2 + v^2); plot(xdot,Fd) grid xlabel('Relative velocity, m/s') ylabel('Damper force, N')
The plot shows that when the magnitude of relative velocity is small, the damper force is linear with velocity, but when the magnitude of relative velocity is large, the damper force is large and nearly constant at
4500 N. 5000 4000 3000 2000 e
,
N
1000 r
0 e
of
cr p
m -1000 D
a
-2000 -3000 -4000 -5000 -1.5
-1
-0.5
0
0.5
Relative velocity, m/s
20
Copyright © 2016 John Wiley & Sons
1
1.5
Chapter 2 2.21
The Mfile for creating the plot of damper force vs. relative velocity is % % Problem 2.21 % xdot = linspace(-1.5,1.5,500); v1 = 0.06; v2 = 0.19; Fd = (3389*(xdot-v1)./sqrt((xdot-v1).^2 + v2^2) ) + 1020.84; plot(xdot,Fd)
5000 4000 3000 N ,
2000 e cr of
1000 r e p m a D
0 -1000 -2000 -3000 -1.5
-1
-0.5
0
0.5
1
1.5
Relative velocity, m/s
The plot looks much like the plot from P roblem 2.20. When the relative velocity is “small” (near zero) the damper force is approximately linear with x . Note that the magnitude of the damper force for negative values of relative velocity (compression stroke) is much less than corresponding force from Problem 2.20. Large positive relative velocity exhibits a nearly constant damper force of about 4400 N (extension) while large negative relative velocity shows a nearly constant damper force of about -2300 N (compression).
21
Copyright © 2016 John Wiley & Sons
Chapter 2 2.22 where
The round-wire spring constant is k
Gd 4 8 ND3
G = modulus of elasticity in shear d = wire diameter ( = 1 mm) N = number of coils ( = 5) D = spring diameter ( = 1.5 cm = 15 mm)
For stainless steel, G = 77.2 GPa = 77.2 (109) Pa (or N/m²)
77.2 10 N/m 1 mm k 9
2
8 5 15 mm
3
4
1 m 1000 mm
571.85 N/m
22
Copyright © 2016 John Wiley & Sons
Chapter 2 2.23
For a square-wire coiled spring (cross-sectional area = 0.8 mm 2), the spring constant is
k where
Gt 4 5.6 ND3
G = modulus of elasticity in shear t = width (or height) of square wire ( =
0.8 mm2 )
N = number of coils ( = 5) D = spring diameter ( = 1.5 cm = 15 mm)
t 0.8 0.894 mm , using G = 77.2 (109) or N/m² for stainless steel:
77.2 10 N/m 0.894 mm k 9
2
5.6 5 15 mm
4
3
1 m 1000 mm
522.84 N/m
23
Copyright © 2016 John Wiley & Sons
Chapter 2 2.24 Draw the FBDs for all three disks, assuming 1 2 and 2 3
T in
Viscous friction torque,
+
b 1 2
Torsional shaft torque,
+
Disk, J 2 (Turbine)
Disk J 1 (Impeller)
k ( 2 3 )
+
T L
Disk, J 3 (Load)
Apply Newton’s second law to each disk (positive clockwise direction ): Disk J 1:
T T
Disk J 2:
T b
T k
Disk J 3:
in
b 1 2 J 1 1 2 k 2 3 J 2 2
1
2
3 T L J 3 3
Rearrange with all dynamic variables on the left-hand sides:
b
k
J11 b 1 2 Tin (t ) J 2 2
2
1
2
Mathematical model
3 0
J 33 k 3 2 T L
24
Copyright © 2016 John Wiley & Sons
Chapter 2 2.25 Draw FBDs of gears 1, 2 and robot arm assuming 1 2 :
Robot arm
d
mg
Gear 2 + Disk J 1
Torsional shaft torque,
+
f c
k 1 2
Contact force
Input torque,
d cos 2 90
f c
Gear 1
T in
Note that the moment of inertia of the robot arm about the rotation axis is J 2 J cm md (parallel axis 2
theorem) where J cm is the moment of inertia of the arm about its c.m. Sum torques for each inertia: Gear 1 (neglect inertia):
T T
in
Gear 2 + Disk J 1:
Robot Arm:
f c
fc r1 J gear1 gear1 0
1
r 1
T in
T f r k J c 2
1
2
T k mgd cos 1
2
1 1
2
Substitute for contact force f c and gear ratio N
90 J 22 r 2 r 1
and rearrange:
J11 k 1 2 NT in
Mathematical model
J 22 k 2 1 mgd cos 2 90
25
Copyright © 2016 John Wiley & Sons
Chapter 2
2.26 Draw the FBDs of masses m1 and m2 assuming xin > x2 (spring k 2 is compressed), x1 > x2 (spring k 1 is in tension), xin
x2 , and x1 x2 . + x1
+ x2 Stiffness force between cart and head
Stiffness force between cart and head
Frame stiffness force
k2 xin x2
k1 x1 x2
k1 x1 x2 m2
Frame vibrational friction
b2 xin x2
m1 Friction force between cart and head
Friction force between cart and head
b1 x1 x2
b1 x1 x2
Sum forces (positive is to the right): Mass 1:
F k1 x1 x2 b1 x1 x 2 m1x 1
Mass 2:
F k2 xin x2 b2 x in x 2 k 1 x 1 x 2 b 1 x 1 x 2 m 2x 2
Move all dynamic variables to the left-hand sides and all input variables to the right-hand sides:
m1 x1 b1 x1 x2 k1 x1 x 2 0 m2 x2 b2 x2 b1 x2 x1 k1 x 2 x1 k 2x 2 b2x in k 2x in
26
Copyright © 2016 John Wiley & Sons
Mathematical model
Chapter 2 2.27 FBD of both masses assuming z1 z w , z2 z 1 (i.e., both springs are in compression) and z2 z 1 , and z 2 0 : k1 z1 z w
z1
Only if
m1
z w
(compression only)
k2 z2 z 1
b1 z2 z 1
m2
f a (t )
b2 z 2
Apply Newton’s second law to each mass (summing positive upwards): Head:
F k1 z1 zw k 2 z2 z1 b1 z 2 z1 m1z 1
Frame:
F k2 z2 z1 b1 z2 z1 f a b2z 2 m2z 2
Rearrange the above equations:
k ( z z ) if z 1 z w 1 b1 ( z 1 z 2 ) k 2 ( z 1 z 2 ) 1 w 1 m1 z 0 if z 1 z w 2 b1 ( z 2 z 1 ) b2 z 2 k 2 ( z 2 z 1 ) f a (t ) m2 z
Note: the spring k 1 can only be in compression ( z1 z w )
27
Copyright © 2016 John Wiley & Sons
Mathematical model
Chapter 2 2.28
When θ = 0 (level), the return spring k 2 has compressive pre-load force F L . When xc 0 ,
pushrod “spring” k 1 is undeflected. Draw the FBD of the rocker arm (moment of inertia J ) and assume a small rotation angle
θ so
that the
vertical deflection of each end of the rocker arm is L1 sin L1 (left end) and L2 sin L2 (right end). Furthermore, assume θ > 0 and xc L1 (pushrod spring k 1 is compressed) friction torque,
+
b k1 xc L1
k2 L2 F L
Pushrod force (compressive only)
Sum torques about the pivot (clockwise is positive)
T k x 1
c
L1 L1 b k2 L2 FL L2 J
Rearrange to obtain the mathematical model:
k 1 ( xc L1 ) L1 F L L2 b k L2 J 2 2 F L L2
if xc L1
Mathematical model
if xc L1
Note that the pushrod force can only be compressive, which occurs when xc L1 In order to compute the cam-follower displacement xc for static equilibrium with a level rocker arm (θ = 0) use the compressive pushrod equation with 0
J b k2 L22 k1 xc L1 L1 FL L2 k 1 xc L1 F L L2 0
Cam-follower displacement to balance return-spring pre-load force:
28
Copyright © 2016 John Wiley & Sons
xc
F L L2 k1L1
Chapter 2 2.29 FBD of each mass assuming z1 z2 , z2 z 3 (springs in tension) and z1 z 2 and z2 z 3
b z2 z 3
b z1 z 2 F a
m1
m2
k ( z 1 z 2 )
br z 1 friction
br z 2
m3
k ( z 2 z 3 )
friction
br z 3 friction
Apply Newton’s second law (sum forces positive to the left):
Mass m1:
F F
a
b z1 z2 k z1 z 2 br z1 m1z 1
Mass m2:
F b z1 z2 k ( z1 z 2 ) b z 2 z 3 k z 2 z 3 br z 2 m 2z 2
Mass m3:
F b z2 z3 k z2 z3 br z3 m3 z 3
Rearrange and place all dynamic variables on the left-hand sides:
1 b( z 1 z 2 ) br z 1 k ( z 1 z 2 ) F a m1 z 2 b( z 2 z 1 ) b( z 2 z 3 ) br z 2 k ( z 2 z 1 ) k ( z 2 z 3 ) 0 m2 z
Mathematical
3 b( z 3 z 2 ) br z 3 k ( z 3 z 2 ) 0 m3 z
model
29
Copyright © 2016 John Wiley & Sons
Chapter 2 2.30
Draw the FBD of both masses assuming z 2
z 1 (spring k 1 is in compression), z 2 z 1 , and
z in z 2 (ignore weight terms since displacements are measured from static equilibrium)
m1
k 1 ( z 2 z 1 )
2 z 1 ) b( z m2
k2 zin z 2
Apply Newton’s second law and sum all forces (positive upward): ¼ car mass:
2 z 1 ) m1z 1 F k 1 ( z 2 z 1 ) b( z
Wheel/axle mass:
2 z 1 ) k 2 ( z in z 2 ) m2 z 2 F k 1 ( z 2 z 1 ) b( z
Rearrange and place all dynamic variables on the left-hand side and input (z in) on the right-hand side:
1 b( z 1 z 2 ) k 1 ( z 1 z 2 ) 0 m1 z 2 b( z 2 z 1 ) k 1 ( z 2 z 1 ) k 2 z 2 k 2 z in (t ) m2 z
30
Copyright © 2016 John Wiley & Sons
Mathematical model
Chapter 2 2.31
Draw the FBD of both masses with sliding friction (dry friction), assuming x1 x2 and x1 > 0
m2 F f
kx1
(sliding friction force)
m1
bx1 F PZT
where F f
Fdry sgn x1 x2 k Nc sgn x1 x2 and k is the kinetic friction coefficient and N c is the
normal clamp force.
Apply Newton’s second law and sum forces on all masses (positive to the right): Clamp mass:
F F f kx1 bx1 FPZT m1x1
Slide mass:
F F f m2 x2
(holds for sliding contact)
For the case of “stiction,” we have x1 x2 and there is no relative motion between mass m1 and m2 (in addition x1 x2 ). The FBD for the stiction case is below:
m2 m1
kx1 bx1
+x1
m1 m2
F PZT
Sum forces on single joined mass:
F kx1 bx1 FPZT m1 m 2 x1 Therefore the mathematical model for the stiction case is
1 b x1 kx1 F PZT (m1 m2 ) x
To compute stiction force F c , draw FBD with equal and opposite stiction force acting on m1 and m2 : m2
kx1 bx1
F st “sticking friction” force
m1
F PZT
31
Copyright © 2016 John Wiley & Sons
Chapter 2 Summing forces (positive to the right)
Slide mass:
F Fst m2 x2 m2 x1 (note that x2 x1 for stiction)
Clamp mass:
F kx1 bx1 FPZT Fst m1x1
Because x1 x2 , substitute clamp-mass equation (stiction case) for x1 and solve for F st : F st
m2 m1 m2
F PZT b x1 kx1
Stiction force (for joined masses)
For the “no-clamp” (release) case, use the first FBD with F f 0 (no friction) and F PZT = 0:
F kx b x m x F 0 m x x 0 ,
Clamp mass: Slide mass:
1
1
2 2
1 1
2
or x2 constant
Complete mathematical model (three cases):
m1 x1 bx1 kx1 k Nc sgn x1 x2 FPZT
Sliding contact case:
m2 x2 k N c sgn x1 x 2 “stiction” (joined) case:
m1 m2 x1 bx1 kx1 FPZT
No contact (release) case:
1 b x1 kx1 0 m1 x m2 x2 0
32
Copyright © 2016 John Wiley & Sons
View more...
Comments
