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R
D cylinder diameter [mm]
C
d rod internal diameter [mm]
R
D rod external diameter [mm]
R
E modulus of elasticity of steel [MPa]
ST
F force requested from the actuator [N]
F maximum value of F [N]
MAX
h lever arm of the actuator on the spoiler [mm]
I moment of inertia [Kgm ]
2
K distributed loss of pressure coefficient []
k torque coefficient for the spoiler [Nm/rad]
T
M hinge moment for the spoiler [Nm]
M minimum hinge moment for the spoiler [Nm]
0
m mass of the actuator [Kg]
p pressure in the lower chamber [Pa]
1
p pressure in the upper chamber [Pa]
2
p high pressure (from the pump) [Pa]
P
p low pressure (to the tank) [Pa]
T
Δp distributed loss of hydraulic pressure [Pa]
Q volumetric flow to the lower chamber [m /s]
3
1
Q volumetric flow to the upper chamber [m /s]
3
2
R radial distance of L from the origin of the axis [mm]
s linear extension of the actuator [mm]
s maximum extension of the actuator [mm]
MAX
t (estimated) time of extraction [s]
EXTR
x abscissa of actuator-wing hinge [mm]
B
x abscissa of actuator-spoiler hinge [mm]
L
y ordinate of actuator-wing hinge [mm]
B
y ordinate of actuator-spoiler hinge [mm]
L 3
4. Problem description and solving method
It is needed defining positioning and dimensions of an hydraulic actuator, in order to allow the
extension of the spoiler from the retracted position; significant diagrams will be enclosed. It is also
requested to write the dynamic equation of the actuator and to calculate an estimated time of
extraction.
4.1 Positioning of the actuator
The hinge moment of the spoiler is considered linear, as in the following:
M =M +k ⋅θ
0 T
The actuator and the spoiler are inserted in a Cartesian system of coordinates, which origin
coincides with the hinge that links the spoiler with the wing, as in Fig.3; the hinge between
actuator and wing structure and the one between the actuator and the spoiler are related with
geometric points which position is fully defined by using problem data and geometric or
trigonometric formulas. With appropriate calculations (shown in 6.1) some graphs may be plotted:
extension of the actuator s(θ), lever arm h(θ), requested force F(θ)
Fig.3
4.2 Dimensioning of the actuator
The length of the actuator is related to its maximum extension, so it can be defined as
plus 10÷20 mm. The dimensioning condition for the diameter of cylinder and rod is
max s(θ
( ))
the maximum force requested, with a 20% margin. The area of the rod can be calculated from a
4 4
D d
π
2 R R
(π ⋅E ⋅( ⋅(( ) −( ) )))
ST 4 2 2
structural point of view, supposing a tubular steel rod, with >F MAX
2
Λ
The area of the cylinder, then, can be calculated from the following: 4
p p A A
⋅A − ⋅( − )=1.2⋅F
1 C 2 C R MAX
4.3 Dynamic equation of the actuator and estimated time of extraction
By Newton's second law of motion, corrected with the apparent inertia, the dynamic equation of
the actuator is the following: I (1)
s̈= p p A s)
(m+ )⋅ ⋅A − ⋅( −A )−F (
1 C 2 C R
2
R
Assuming turbulent flow, the following can be inserted in (1), obtaining (2):
2 2
,
p p p p
= −K⋅Q = −K⋅Q
1 P 1 2 T 2
I ˙
3 3 2 (2)
s̈+ K⋅(A A s s)= p p
(m+ )⋅ +( −A ) )⋅ +F ( ⋅A − ⋅(A −A )
C C R P C T C R
2
R , and .
A numeric solution can be found for the equation (2), plotting s̈( t) ṡ( t) s(t)
The time of extraction then can be estimated from the plot or calculated with the following:
s( t )=s
EXTR MAX
5. Problem data
5.1 Spoiler data
Minimum hinge moment for the spoiler M = 150 Nm
0
Torque coefficient for the spoiler k = 1800 Nm/rad
T
Extension angle for the spoiler 0° ≤ θ ≤ 65°
Moment of Inertia I = 0.04 Kgm 2
Spoiler length Λ = 400 mm
5.2 Hydraulic plant data
Distributed loss of pressure @ 10l/min flow Δp = 6.5 Mpa
High pressure (from the pump) p = 21 Mpa
P
Low pressure (to the tank) p = 0.1 MPa
T
5.3 Chosen data 5
Radial distance of L from the origin of the axis R = 120 mm
Abscissa of the actuator-wing hinge x = 60 mm
B
Ordinate of the actuator-wing hinge y = -90 mm
B
Modulus of elasticity of steel E = 200 Gpa
ST
Steel density ρ = 7850 Kg/m 3
ST
6. Calculations and results θ
6.1 Calculation of positions depending on
x R⋅cos(θ−5)=120⋅cos(θ−5)
=
L
y sin
=R⋅ (θ−5)=120⋅sin(θ−5)
L √ √
2 2 2 2
s=s x y y x y y
(θ)= (( −x ( θ)) +( − (θ)) )− (( −x (0)) +( − (0)) )
B L B L B L B L
√
√ 2 2 2 2
s(θ)= ((60−x (θ)) +(−90−y (θ)) )− ((60−120⋅cos (−5)) +(−90−120⋅sin (−5)) )
L L
y
( −y ) c
∣ ∣
B L
y−y ax+by+c=0 h=h(θ)=
= ⋅(x−x ) ⇒ ⇒
B B √
(x −x ) 2 2
(a +b )
B L (M +k ⋅θ)
(M (θ)) 0 T
F =F ( θ)= = h(θ))
(h (θ)) (
Each one of the formulas above are implemented in the Matlab® script, pasted below, in order to
get their plots and bound values
R = 120;
theta = (0:0.001:(pi/180*65));
M_0=150;
k_T=1800;
M_x=@(x) M_0 + k_T.*x;
x_L = @(x) R.* cos(x-5*pi/180);
y_L = @(x) R.* sin(x-5*pi/180);
s_x= @(x) sqrt((60-x_L(x)).^2 +(-90-y_L(x)).^2 )-sqrt((60-120.*cos(-
5*pi/180)).^2+(-90-120.*cos(-5*pi/180).^2));
a= @(x) (-90-y_L(x))./(60-x_L(x));
b=-1;
c= @(x) -90-((-90-y_L(x))./(60-x_L(x))).*60;
h_x= @(x) abs(c(x))./sqrt(a(x).^2+b^2);
F_x= @(x) M_x(x)./(h_x(x).*0.001);
s=s_x(theta);
h=h_x(theta);
F=F_x(theta);
M=M_x(theta);
smax=s(end);
hi=h(1);
hf=h(end);
Fmax=F(end);
fprintf('The actuator has a maximum extension of %f mm, its lever arm is between
%f mm and %f mm and the maximum force requested is %f N',smax,hi,hf,Fmax)
figure(1)
plot(theta,M)
title('hinge moment of the spoiler')
xlabel('theta (rad)')
ylabel('M(theta) (Nm)')
figure(2)
plot(theta,s) 6
title('extension of the actuator')
xlabel('theta (rad)')
ylabel('s(theta) (Nm)')
figure(3)
plot(theta,h)
title('lever arm of the actuator on the spoiler')
xlabel('theta (rad)')
ylabel('h(theta) (Nm)')
figure(4)
plot(theta,F)
title('requested force')
xlabel('theta (rad)')
ylabel('F(theta) (Nm)')
The actuator has a maximum extension of 136.134284 mm, its lever arm is between
101.967314 mm and 60.022378 mm and the maximum force requested is 36506.384249 N
Fig.4
6.2 Dimensioning of the actuator
From 6.1 the length of the actuator can be chosen as 150 mm. 7