1

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

2 2     2v u as

21     

2s ut at

and 1

  (     )2

s u v t The force F to accelerate a weight W horizontally

with an acceleration a is given by

Force = Mass × Acceleration

F = W

ag

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

0.15

0.5

dv

t =0.3 m/s

Power is given by 310 10 0.3P F v =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?

2

Solution:

Let us first convert the LPM to :

3 3 3/ min60 60

60 LPM m 10 m1000 1

/0

s0 0 60

Q

Now we know flow rate and velocity, so we can calculate the diameter,

pQ A v

3 2

p10 0.754

D

Therefore, 3

p

4 1041.2 mm

0.75D

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/m

2 or Pa

(a) Forward velocity is given by

vext=in

p

Q

A=

2

1/ 600

/ 4d= 0.3315 m/s

Return velocity is given by

vret = in

2 2

C rp r

1

600 0.442 m/sπ( )( )

4

Q

d dA A

(b) Force during extension is given by 2 2

6

ext p

(8 10 )12 10 60318.57

4F p a

N

Force during retraction is given by

3

ret p r

6 2 2 2 210 [(8 10 ) (4 10 ) ]12

4

42238.9 N 45.24

(

k

)

N

F p A A

Example 1.6 A pump supplies oil at 0.0016 m

3/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

ext

ext 2

C

2

p

5000 50003978.8 kPa

(0.04)

44

Fp

dA

(b) Piston velocity during the extending stroke is

in inext 2 2

Cp

0.016  1.273 m/s

(0.04)

44

Q Q

dv

A

(c) Cylinder kW power during the extending stroke is

ext ext ext 5000 1.273 6.366 kWP F v

(d) Hydraulic pressure during the retracting stroke is

ret

ret

p r

Fp

A A

=

C

2 2 2 2

r

5000 5000

( ) [(0.04) (0.02) ]

44

d d

= 5305.16 kPa

(e) Piston velocity during the retracting stroke is

inret 2 2

p r

0.016

(0.04 0.02 )

4

Qv

A A

=1.697 m/s

(f) Cylinder kW power during the retracting stroke is

ret ret ret 5000 1.697 8.488 kWP F v

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

ext p   F p A

4

Retracting stroke is

ret p r   (  – )F p A A

Also, given that

dp = 2dr

Therefore

Ap = 4Ar

Now

pext

ret p r

    

 (  – )

p AF

F p A A

p

p r

  –

A

A A

p

p p

     –0.25

A

A A

Therefore, ext

ret

4      

3

F

F

Again

ext ret p p p

p

p p

  (  –0.25 )

4

F F pA p A A

ApA p A

p

ext ret  4

pAF F

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:

net-extendingF = Pr essure   Piston area

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

2 2 2

p p (0.15) 0.01767 m4 4

A D

Area of rod is

2 2 2

r r (0.105) 0.008659 m4 4

A D

5

Hence, 2

p r 0.00901mA A .

Given, vext = 7 m/min and p = 150 bar.

(a) Flow rate of oil to extend in LPM:

ext p e

3

p xt(

0.01767 7

0.12369 m /min 123.69 LPM

)Q A v

(b) Flow rate of oil to extend from annulus in LPM:

ext p r

3

a ext( ( )

0.00901 7

0.06307 m /min 63 LP

)

M

Q A A v

(c) Retract speed in m/min using (a):

pret ext p r ret( ))(Q Q A A v

ret0.12369 0.00901 v

ret 13.728 m/minv

(d) Flow rate from full bore end on retract:

ret p ret

3

0.01767 13.728

0.06587 m /min 65.87 LPM

Q A v

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.

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.

= Cylinder force

Load

6

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

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.

= W

= Cylinder force

Load

7

Figure 1.15

Solution: The cylinder force is given by

cyl fFrictional force  )(F F W

cyl 0.14 6000 840 NF

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.

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,

= Cylinder force

Load

8

cyl 6000  cos60 3000 N F

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.

Figure 1.17

Solution:

(a) For a constant velocity, the cylinder force to move weight at a constant velocity of 1.75 m/s

cyl a 6000 NF F W

(b) Force required to accelerate the weight:

First we shall calculate acceleration

21.75 03.5 m/s

0.5a

Force required to accelerate the weight is

a

6000 3.52140.67 N

9.81F m a

The cylinder force Fcyl required is equal to the sum of the weight and acceleration force

cyl 6000 2140.67 8140.67 NF

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.

= Cylinder force

Load

9

Figure 1.18

Solution: As per Newton’s law of motion, we have

  F m a

where

2

m 1 min100

1.67m/smin 60 s 3.34 m/s0.5 s 0.5 s

a

Summing forces on the 10000 N weight, we have

2

r 2p

10000 N( ) 10000 N 3.34 m/s

9.81 m/sp A A

2 2 2 2(N/m ) (0.075 0.050 ) m 10000 N 3408 N4

p

Solving we get 25450000 (N/m ) 5450000 Pa 5450 kPap

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.

10000 N

10

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 sin30 .W The component of

weight W acting along normal to the incline surface is cos30 .W 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

cos30

0.15 27000 cos30

3507 N

f W

Total force on the cylinder is frictional force and vertical component:

cylinder sin 30

3507 27000  sin 30

17007 N

F f W

cylinder pF pA =17007 N

Diameter to resist 17007 N is given by

2

cylinder 6894000 4

F D

=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

27000 N

11

2 21524mm/s3048 mm/s   3.048  m/s

0.5 sa

Summing forces on the 27000 N weight using values determined in (a), we have

p

27000 3.04817007

9.81pA

2 27000 3.048            6894000  17007

4 9.81D

2            5414535  17007 8389D

D = 0.0685 m = 68.5 mm

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

=4

× (200/1000)

2×

5

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

= 4

× [(200/1000)

2− (140/1000)

2] ×

5

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

EQv

A a

Now

QE= 157 L/min = 0.00262 m2

(A − a) = Annulus area= 0.01602 m2

12

v= 0.00262

0.01602 = 0.614 m/s = 9.8 m/min

(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 = 20.2

4

= 0.0314 2m

Dynamic extend thrust = 0.9 × Pressure × Full bore area

= 0.9 × 100 × 510 × 0.0314 2

2

Nm

m

= 283 kN

(f) Annulus area = 2 2(0.2 0.14 )

4

= 0.016 2m

Dynamic retract thrust = 0.9 × 100 × 510 × 0.016 2

2

Nm

m

= 144 kN

(g) Piston rod area = 20.14

4

= 0.0154

2m

Extend speed = Flow rate

Piston rod area

= 3157 10

0.0154

3

2

L m

min L×m

= 10.2 m/min

This compares with 5 m/min when connected conventionally.

(h) For a regenerative system

Dynamic extend thrust = 0.9 ×100 × 510 × 0.0154 2

2

Nm

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.

13

Figure 1.20

Solution: In this case, u = 0, v = 1 m/s, s = 0.05 m and a is unknown. Using the equation

2 2     2v u as

We have,

2 21    0  2 0.05a 210 m/sa

The force to accelerate the load is given by

w

F ag

2000 9.8110

9.81

20 000 N

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

5

2 2

22943Area

195 10

0.002415 m 2415 mm

Now area is also given by

2

Area4

D

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.

Load

P1=100 bar P2 = 0