Lecture 13
HYDRAULIC ACTUATORS[CONTINUED]
1.5Acceleration and Deceleration of Cylinder Loads
Cylinders are subjected to acceleration and deceleration during their operation. Cylinders are decelerated
to provide cushioning and cylinders are accelerated to reduce the cycle time of the operation.
1.5.1 Acceleration
To calculate the acceleration of cylinder loads, the equations of motion must be understood.Let u be the
initial velocity, v the velocity after a time t,s the distance moved during the time t anda the acceleration
during the time t.The standard equations of motion are as follows:
v  u  at
v2  u 2  2as
1
s  ut  at 2
2
and
1
s  (u  v)t The force F to accelerate a weight W horizontally
2
with an acceleration a is given by
Force = Mass × Acceleration
F=
W
a
g
where g is the acceleration due to gravity and is 9.81 m/s2. The force P required to overcome friction is
given by P = µW, where µ is the coefficient of friction.
Note: Dynamic cylinder thrust
In dynamic applications, the load inertia, seal friction, load friction, etc., must be allowed for calculating
the dynamic thrust.As a first approximation, the dynamic thrust can be taken as 0.9 times the static thrust.
Cylinder seal friction varies with seal and cylinder design. The pressure required to overcome seal friction
is not readily available from the majority of cylinder manufacturers. The seal friction breakout pressure
can be taken as 5 bar for calculation purposes. It reduces when the piston starts to move. The pressure
required to overcome seal friction reduces as the cylinder bore size increases and varies according to the
seal design.
Example 1.3
A cylinder is required to move a 10 kN load 150 mm in 0.5 s. What is the output power?
Solution: The velocity is given by
v
d 0.15
=0.3 m/s

t
0.5
Power is given by
P  F  v  10 103  0.3 =3 kW
Example 1.4
A cylinder is required to extend at a minimum speed of 0.75 m/s in a system with a flow rate of 60 LPM.
What cylinder size is required?
1
Solution:
Let us first convert the LPM to
Q  60 LPM 
:
60
60
m3 / min 
 103 m3 / s
1000
1000  60
Now we know flow rate and velocity, so we can calculate the diameter,
Q  Ap  v
 103 

4
Dp2  0.75
Therefore,
Dp 
4  103
 41.2 mm
  0.75
Example 1.5
An 8 cm diameter hydraulic cylinder has a 4 cm diameter rod. If the cylinder receives flow at 100 LPM
and 12 MPa, find the (a) extension and retraction speeds and (b) extension and retraction load carrying
capacities.
Solution:
Let us first convert the flow in LPM to m3/s before we calculate forward velocityQin=100
LPM = 100/(1000 × 60) =1/600 m3/s
Now
DC = Diameter of cylinder = 8 cm = 8 × 10−2 m
dr = Diameter of piston rod = 4 cm = 4 × 10−2 m
p = 12 MPa = 12 × 106 N/m2 or Pa
(a) Forward velocity is given by
vext=
Qin 1/ 600
=
= 0.3315 m/s
Ap  d 2 / 4
Return velocity is given by
1
Qin
600
vret =

 0.442 m/s
( Ap  Ar ) π(dC2  d r2 )
4
(b) Force during extension is given by
 (8  102 )2
Fext  p  ap  12  106
 60318.57 N
4
Force during retraction is given by
2
Fret  p  ( Ap  Ar )
106   [(8 102 ) 2  (4 102 ) 2 ]
 12 
4
 42238.9 N  45.24 kN
Example 1.6
A pump supplies oil at 0.0016 m3/s to a 40 mm diameter double-acting hydraulic cylinder. If the load is
5000 N (extending and retracting) and the rod diameter is 20 mm, find the
(a) Hydraulic pressure during the extending stroke.
(b) Piston velocity during the extending stroke.
(c) Cylinder kW power during the extending stroke.
(d) Hydraulic pressure during the retracting stroke,
(e) Piston velocity during the retracting stroke.
(f ) Cylinder kW power during the retracting stroke.
Solution: We have Qin = 0.016 m3/s, Fext = Fret = 5000 N, dC = 40 mm = 0.04 m, dr = 20 mm = 0.02 m.
(a) Hydraulic pressure during the extending stroke is
pext 
Fext 5000
5000


 3978.8 kPa
2
 dC  (0.04)2
Ap
4
4
(b) Piston velocity during the extending stroke is
vext 
Qin Qin
0.016


 1.273 m/s
Ap  dC2  (0.04) 2
4
4
(c) Cylinder kW power during the extending stroke is
Pext  Fext  vext  5000 1.273  6.366 kW
(d) Hydraulic pressure during the retracting stroke is
pret 
Fret
5000
5000
=
= 5305.16 kPa

2
2
Ap  Ar  (dC  d r )  [(0.04)2  (0.02)2 ]
4
4
(e) Piston velocity during the retracting stroke is
vret 
Qin
0.016
=1.697 m/s

Ap  Ar  (0.042  0.022 )
4
(f) Cylinder kW power during the retracting stroke is
Pret  Fret  vret  5000 1.697  8.488 kW
Example 1.7
A hydraulic cylinder has a rod diameter equal to one half the piston diameter. Determine the difference in
load-carrying capacity between extension and retraction stroke if pressure is constant. What would
happen if the pressure were applied to both sides of the cylinder at the same time?
Solution: Forward or extending stroke is
Fext  p  Ap
3
Retracting stroke is
Fret  p  ( Ap – Ar )
Also, given that
dp = 2dr
Therefore
Ap = 4Ar
Now
p  Ap
Fext

Fret p  ( Ap – Ar )


Ap
Ap – Ar
Ap
Ap –0.25 Ap
Fext
Fret
Therefore,

4
3
Again
Fext  Fret  pAp  p( Ap –0.25 Ap )
Ap 

 pAp  p  Ap  
4

pAp
 Fext  Fret 
4
Therefore,
Difference =
Pr essure  Piston area
4
If the pressure were applied to both sides of the cylinder at the same time, there would be a net force to
extend the cylinder. This net force will be the same as obtained above:
Pr essure  Piston area
Fnet-extending =
4
Example 1.8
A cylinder with a bore of 150 mm and a piston rod diameter of 105 mm, has to extend with a speed of 7
m/s, pressure applied is 150 bar. Calculate
(a) The flow rate in LPM of oil to extend the cylinder
(b) The flow rate in LPM from annulus side to extend the cylinder.
(c) The retract speed in m/min using (a).
(d) The flow rate from full bore end on retract.
Solution: Area of piston is given by
Ap 
Area of rod is
Ar 

4

4
 Dp2 
 Dr2 

4

4
 (0.15)2  0.01767 m2
 (0.105)2  0.008659 m2
4
Hence, Ap  Ar  0.00901 m2 .
Given, vext = 7 m/min and p = 150 bar.
(a) Flow rate of oil to extend in LPM:
(Qext ) p  Ap  vext
 0.01767  7
 0.12369 m3 /min  123.69 LPM
(b) Flow rate of oil to extend from annulus in LPM:
(Qext )a  ( Ap  Ar )  vext
 0.00901  7
 0.06307 m3 /min  63 LPM
(c) Retract speed in m/min using (a):
Qret  (Qext )p  ( Ap  Ar )  vret
 0.12369  0.00901 vret
 vret  13.728 m/min
(d) Flow rate from full bore end on retract:
Qret  Ap  vret
 0.01767  13.728
 0.06587 m3 /min  65.87 LPM
1.6Various Methods of Applying Linear Motion Using Hydraulic Cylinders
A cylinder must produce a force equal to the load the cylinder is required to overcome. A cylinder may be
placed with its axis vertical, horizontal or inclined depending on the load to be actuated.
1. Vertical cylinder: Ina vertical cylinder, the load to be actuated is in the vertical direction as
shown in Fig. 1.12. Then the cylinder load F is equal to the weight W of the object, acting in the
vertical direction.
Load
= Cylinder force
Figure 1.12Cylinder load– vertical cylinder
2. Horizontal cylinder:The schematic diagram of horizontal cylinder is shown in Fig. 1.13. Ina
horizontal cylinder, the cylinder load is theoretically zero,because no component of the object’s
weight acts along the axis of the cylinder. However, when the object slides across the horizontal
surface, the cylinder must overcome the frictional force created between the object and the
horizontal surface.
5
Load
Figure 1.13Cylinder load –horizontal cylinder
3. Inclined cylinder:In an inclined cylinder as shown in Fig. 1.14, the cylinder load equals the
component of the object’s weight acting along the axis of the cylinder and frictional force
=W
= Cylinder force
Figure 1.14Cylinder load –inclined cylinder
For an inclined cylinder, the load the cylinder must overcome is less than the weight of the object to be
moved if the object does not slide on an inclined surface. The cylinder loads calculated as above are based
on moving an object at a constant velocity. But when the object has to be accelerated from zero velocity
to a steady-state velocity, an additional force called inertia force must be added to the weight component
and any frictional force involved.
LetFload = W = weight or load acting vertically downward, Fcyl = load acting on the cylinder, Fbear = force
on the bearings and α= angle between the load W and the axis of the cylinder.Then,
Fcyl = Floadcos(a)
Fbear = Floadsin(a)
Example 1.9
Find the cylinder force required to move a 6000 N weight along a horizontal surface at a constant velocity
(Fig. 1.15). The coefficient of friction between the weight and horizontal support surface is 0.14.
6
Load
Figure 1.15
Solution: The cylinder force is given by
Fcyl  Frictional force ( Ff )  W
 Fcyl  0.14  6000  840 N
Example 1.10
Find the cylinder force required to lift a 6000 N weight along a direction that is 30° from the horizontal
direction as shown in Fig. 1.16. The weight is moved at a constant velocity.
= Cylinder force
Figure 1.16
Solution: Let
Fload = W = weight or load acting vertically downward
Fcyl = load acting on the cylinder
Fbear = force on the bearings
= angle between the load W and the axis of the cylinder
Then
Fcyl = Floadcos
Fbear = Floadsin
Hence,
7
Fcyl  6000  cos60  3000 N
Example 1.11
A 6000 N weight is to be lifted upward in a vertical direction for the system shown in Fig. 1.17. Find the
cylinder force required to
(a) Move the weight at a constant velocity of 1.75 m/s.
(b) Accelerate the weight from zero velocity to 1.75 m/s in 0.5 s.
Load
= Cylinder force
Figure 1.17
Solution:
(a) For a constant velocity, the cylinder force to move weight at a constant velocity of 1.75 m/s
Fcyl  Fa  W  6000 N
(b) Force required to accelerate the weight:
First we shall calculate acceleration
a
1.75  0
 3.5 m/s2
0.5
Force required to accelerate the weight is
Fa  m  a 
6000  3.5
 2140.67 N
9.81
The cylinder force Fcyl required is equal to the sum of the weight and acceleration force
Fcyl  6000  2140.67  8140.67 N
Example 1.12
A 10000 N weight is to be lowered by a vertical cylinder as shown in Fig. 1.18. The cylinder has a 75 mm
diameter piston and 50 mm diameter rod. The weight is to decelerate from 100 m/min to a stop in 0.5 s.
Determine the required pressure in the rod end of the cylinder during the deceleration motion.
8
10000 N
Figure 1.18
Solution: As per Newton’s law of motion, we have
F  m a
where
a
100
m 1min

min 60 s  1.67 m/s  3.34 m/s 2
0.5 s
0.5 s
Summing forces on the 10000 N weight, we have
p( Ap  Ar )  10000 N 
 p (N/m2 ) 

4
10000 N
 3.34 m/s 2
2
9.81m/s
(0.0752  0.0502 ) m2  10000 N  3408 N
Solving we get
p  5450000 (N/m2 )  5450000 Pa  5450 kPa
Example 1.13
A 27000 N weight is being pushed up on an inclined surface at a constant speed by a cylinder, as shown
in Fig. 1.19. The coefficient of friction between the weight and the inclined surface equals 0.15.
(a) Determine the required cylinder piston diameter for the pressure of 6894 kPa,
(b) Determine the required cylinder piston diameter, if the weight is to accelerate from a 0 mm/s to a 1524
mm/s in 0.5 s.
9
27000 N
Figure 1.19
Solution:
(a) Cylinder piston diameter for the pressure of 6894 kPa
The component of the weight W acting along the axis of the cylinder is W sin 30. The component of
weight W acting along normal to the incline surface is W cos30. The frictional force equals the
coefficient of friction times the force normal to sliding surfaces.
Therefore, the frictional force f acting along the axis of the cylinder is
f    W cos30
 0.15  27000  cos30
 3507 N
Total force on the cylinder is frictional force and vertical component:
Fcylinder  f  W sin 30
 3507  27000 sin 30
 17007 N
Fcylinder  pAp =17007 N
Diameter to resist 17007 N is given by
Fcylinder  6894000 

4
D2 =17007 N
 D = 0.05605 m = 56 mm
(b) Cylinder piston diameter if the weight is to accelerate from a 0 mm/s to a 1524 mm/s in 0.5 s
As per Newton’s law of motion, we have calculate the acceleration
F  m a
where
10
a
1524mm/s
 3048 mm/s2  3.048 m/s 2
0.5 s
Summing forces on the 27000 N weight using values determined in (a), we have
27000  3.048
9.81
 2
27000  3.048
6894000  D  17007 
4
9.81

5414535 D2  17007  8389
 D = 0.0685 m = 68.5 mm
pAp  17007 

Example 1.14
A hydraulic cylinder has a bore of 200 mm and a piston rod diameter of 140 mm. For an extend speed of
5 m/min, calculate
(a) The supply flow rate.
(b) The flow rate from the annulus side on extend.
(c) The retract speed using QE.
(d) The flow rate from the full bore end on retract.
Also, if the maximum pressure applied to the cylinder is 100 bar, calculate the (e) dynamic extend
thrust and the (f) dynamic retract thrust assuming that dynamic thrust = 0.9  static thrust.
Moreover, the hydraulic cylinder having a bore of 200 mm diameter and a rod of 140 mm diameter
are connected regeneratively. (g) If the same flow rate of 157 L/min is used, calculate the extend
speed. (h) If the maximum system pressure is 100 bar, calculate the dynamic extend thrust.
Solution:
(a) Flow rate of oil to extend cylinder at 5 m/min:
QE = Area of piston × Velocity
5

= × (200/1000)2×
4
60
=0.00262 m3/min
=0.00262 ×60 ×1000
= 157 L/min
(b) Flow of oil leaving cylinder QE is given by
QE = Annulus area × Velocity
5

= × [(200/1000)2− (140/1000)2] ×
4
60
= 80 L/min
(c) The same fluid flow rate used to extend the cylinder (157 LPM) is used to retract the
cylinder. The retract cylinder velocity v is given by
v
QE
Aa
Now
QE= 157 L/min = 0.00262 m2
(A − a) = Annulus area= 0.01602 m2
11
v=
0.00262
= 0.614 m/s = 9.8 m/min
0.01602
(d) Flow from the full bore end of cylinder QR is given by
Q = A×v= 0.03142×0.164 = 0.00515 m3/s = 309 LPM
(e) We have
Full bore area =
  0.22
= 0.0314 m 2
4
Dynamic extend thrust = 0.9 × Pressure × Full bore area
N
= 0.9 × 100 × 105 × 0.0314  2  m2 
m
(f) Annulus area =

= 283 kN
  (0.22  0.142 )
4
= 0.016 m 2
 N

 N

Dynamic retract thrust = 0.9 × 100 × 105 × 0.016  2  m2 
m

= 144 kN
(g) Piston rod area =
  0.142
= 0.0154 m 2
4
Flow rate
Extend speed =
Piston rod area
157  103  L
m3 
=



0.0154  min L×m2 
= 10.2 m/min
This compares with 5 m/min when connected conventionally.
(h) For a regenerative system
Dynamic extend thrust = 0.9 ×100 × 105 × 0.0154  2  m2 
m

= 138.6 kN
As the area of the annulus is almost equal to that of the rod, the regenerative extend and conventional
retract thrusts and speeds are almost the same.
Example 1.15
A mass of 2000 kg is to be accelerated horizontally up to a velocity of 1 m/s from the rest over a distance
of 50 mm (Fig. 1.20). The coefficient of friction between the load and guide is 0.15. Calculate the bore of
the cylinder required to accelerate this load if the maximum allowable pressure at the full bore end is 100
bar (take seal friction to be equivalent to a pressure drop of 5 bar). Assume that the back pressure at the
annulus end of the cylinder is zero.
12
Load
P1=100 bar P2 = 0
Figure 1.20
Solution: In this case, u = 0, v = 1 m/s, s = 0.05 m and a is unknown. Using the equation
v2  u 2  2as
We have,
12  02  2a  0.05  a  10 m / s2
The force to accelerate the load is given by
 w
 2000  9.81 
F  a 
10  20 000 N
 g
9.81


The force P to overcome the load friction is given by
P  W  0.15  2000  9.81  2943 N
The total force to accelerate the load and overcome friction is
F + P = 20000 + 2943 = 22943 N
The cylinder area required for a given thrust is calculated from
Thrust = Force × Area
The pressure available is pressure at the full bore end of the cylinder minus the equivalent seal break-out
pressure.
Pressure available = 100−5 = 95 bar = 95 ×105 bar
Area is given by
22943
Area 
195  105
 0.002415 m 2  2415 mm 2
Now area is also given by
Area 
 D2
4
where D is the diameter. Comparing the two equations , we get D = 55.4 mm.
The cylinder diameter is thus 55.4 mm. This neglects the effect of any back pressure. The nearest standard
cylinder above has a 63 mm diameter bore.
13