# Chapter 6: Hydraulic Cylinders and Cushioning Devices¶

## Example 6.1 pgno:219¶

In :
# Aim:Refer Example 6-1 for Problem Description
# Given:
# Flow rate of pump:
Q_in=20.0; #gpm
# Bore diameter of Cylinder:
D=2.0; #in
# Load during extending and retracting:
F_ext=1000.0; #lb
F_ret=1000.0; #lb
# Rod diameter of cylinder:
d=1.0; #in
import math
from math import pi
from math import floor
from math import ceil

# Solution:
# Area of blank end of piston,
Ap=(pi/4)*(D**2); #in**2
# Area of rod end of piston,
Ar=(pi/4)*(d**2); #in**2
# hydraulic pressure during the extending stroke,
p_ext=F_ext/Ap; #psi
# piston velocity during the extending stroke,
v_ext=(Q_in/449)/(Ap/144); #ft/s
# rounding off the above answer
v_ext=round(v_ext)+(round(ceil((v_ext-round(v_ext))*100))/100); #ft/s
# cylinder horsepower during the extending stroke,
HP_ext=(v_ext*F_ext)/550; #HP
# rounding off the above answer
HP_ext=round(HP_ext)+(round(floor((HP_ext-round(HP_ext))*100))/100); #HP
# hydraulic pressure during the retraction stroke,
p_ret=ceil(F_ret/(Ap-Ar)); #psi
# piston velocity during the retraction stroke,
v_ret=(Q_in/449)/((Ap-Ar)/144); #ft/s;
# rounding off the above answer
v_ret=round(v_ret)+(round(ceil((v_ret-round(v_ret))*100))/100); #ft/s
# cylinder horsepower during the retraction stroke,
HP_ret=(v_ret*F_ret)/550; #HP

# Results:
print"\n  Results:  "
print"\n The hydraulic pressure during the extending stroke is  psi.",round(p_ext)
print"\n The piston velocity during the extending stroke is  ft/s.",v_ext
print"\n The cylinder horsepower during the extending stroke is  HP.",HP_ext
print"\n The hydraulic pressure during the retraction stroke is  psi.",p_ret
print"\n The piston velocity during the retraction stroke is  ft/s.",v_ret
print"\n The cylinder horsepower during the retraction stroke is  HP.",round(HP_ret,2)

  Results:

The hydraulic pressure during the extending stroke is  psi. 318.0

The piston velocity during the extending stroke is  ft/s. 2.05

The cylinder horsepower during the extending stroke is  HP. 3.72

The hydraulic pressure during the retraction stroke is  psi. 425.0

The piston velocity during the retraction stroke is  ft/s. 2.73

The cylinder horsepower during the retraction stroke is  HP. 4.96


## Example 6.2 pgno:221¶

In :
# Aim:Refer Example 6-2 for Problem Description
# Given:
# Weight of Body:
W=6000; #lb
# coefficient of friction between weight and horizontal support:
CF=0.14;

# Solution:
# Cylinder Force,
F=CF*W; #lb

# Results:
print"\n  Results:  "
print"\n The Cylinder Force at constant velocity is  lb.",F

  Results:

The Cylinder Force at constant velocity is  lb. 840.0


## Example 6.3 pgno:221¶

In :
# Aim:Refer Example 6-3 for Problem Description
# Given:
# Weight of Body:
W=6000; #lb
# Inclination of Weight:
theta=30; #deg
import math
from math import pi
from math import sin

# Solution:
# Inclination of Weight,
# Cylinder Force,
F=W*sin(theta); #lb

# Results:
print"\n  Results:  "
print"\n The Cylinder Force at constant velocity is  lb.",F

  Results:

The Cylinder Force at constant velocity is  lb. 3000.0


## Example6.4 pgno:222¶

In :
# Aim:Refer Example 6-4 for Problem Description
# Given:
# Weight of Body:
W=6000.0; #lb
# initial velocity:
u=0; #ft/s
# final velocity:
v=8.0; #ft/s
# Time taken:
t=0.5; #s

# Solution:
# For constant velocity,Cylinder Force,
F=W; #lb
# Rate of change of velocity,
a=(v-u)/t; #ft/s^2
# Force required to accelerate the weight,
F_acc=(F/32.2)*a; #lb
# Therefore, Cylinder Force,
F_cyl=(F+F_acc); #lb

# Results:
print"\n  Results:  "
print"\n The Cylinder Force at constant velocity is  lb.",F
print"\n The Cylinder Force required to accelerate the Body is  lb.",round(F_cyl)
print"\n The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation"

  Results:

The Cylinder Force at constant velocity is  lb. 6000.0

The Cylinder Force required to accelerate the Body is  lb. 8981.0

The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation


## Example 6.5 pgno:227¶

In :
# Aim:Refer Example 6-5 for Problem Description
# Given:
L1=10; #in
L2=10; #in
# Inclination of cylinder axis with vertical axis:
phi=0; #deg
import math
from math import pi
from math import cos

# Solution:
# Inclination of cylinder axis with vertical axis,
# cylinder force required to overcome load using First Class Lever Sytem,
# cylinder force required to overcome load using Second Class Lever Sytem,
# cylinder force required to overcome load using Third Class Lever Sytem,

# Results:
print"\n  Results:  "
print"\n The Cylinder Force using First Class lever System is  lb.",F_cyl_1
print"\n The Cylinder Force using Second Class lever System is  lb.",F_cyl_2
print"\n The Cylinder Force using Third Class lever System is  lb.",F_cyl_3

  Results:

The Cylinder Force using First Class lever System is  lb. 1000.0

The Cylinder Force using Second Class lever System is  lb. 500.0

The Cylinder Force using Third Class lever System is  lb. 2000.0


## Example 6.6 pgno:230¶

In :
# Aim:Refer Example 6-6 for Problem Description
# Given:
# Flow rate of pump:
Q_pump=18.2; #gpm
# Diameter of blank end of piston:
D=3.0; #in
# Diameter of cushion plunger:
D_cush=1.0; #in
# Stroke of cushion plunger:
L_cush=0.75; #in
# Distance Piston decelerates at the end of extending stroke:
L=0.75; #in
# Weight of Body:
W=1500.0; #lb
# coefficient of friction:
CF=0.12;
# Pressure relief valve settings:
p_relf=750.0; #psi
# maximum pressure at the Blank end:
p1=750.0; #psi

from math import pi

# Solution:
# Area of blank end of piston,
A_piston=(pi/4)*(D**2); #in**2
# piston velocity prior to deceleration,
v=(Q_pump/449)/(A_piston/144); #ft/s
# deceleration of piston at the end of extending stroke,
a=(v**2)/(2*(L/12)); #ft/s**2
# Area of cushion plunger,
A_cush=(pi/4)*(D_cush**2); #in**2
# maximum pressure developed by the cushion,
p2=(((W*a)/32.2)+(p1*A_piston)-(CF*W))/(A_piston-A_cush); #psi

# Results:
print"\n  Results:  "
print"\n The maximum pressure developed by the cushion is  psi.",round(p2)

  Results:

The maximum pressure developed by the cushion is  psi. 856.0

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