# Hydraulic jack

• 07-22-2012, 09:18 PM
SANDIP
Hydraulic jack
How to calculate the wall thickness of 500 ton jack ? How to calculate the bore dia of hydraulic jack of 500 ton ? How to calculate the working pressure of 500 ton jack ? How to calculate the buckling effect of the ram of 500 ton hydraulic jack ?

Kindly make me understand in a very simple mode.

With b'rgds,
b.sandip
• 07-23-2012, 09:53 AM
Kelly Bramble
[QUOTE]How to calculate the wall thickness of 500 ton jack ? [/QUOTE]

[url]http://www.engineeringmotion.com/videos/1077/thin-wall-pressure-vessel-stress-calculations[/url]
[url]http://www.engineeringmotion.com/videos/1078/thin-wall-pressure-vessel-stress-calculations-ii[/url]
[url]http://www.engineersedge.com/calculators/pressure-vessel-longitudinal-stress.htm[/url]
[url]http://www.engineersedge.com/material_science/torsion-calc-round.htm[/url]

[QUOTE]How to calculate the working pressure of 500 ton jack ? [/QUOTE]

[url]http://www.engineersedge.com/hydraulic/hydraulic_pneumatic_cylinder_force_10105.htm[/url]

[QUOTE]How to calculate the buckling effect of the ram of 500 ton hydraulic jack ? [/QUOTE]

[url]http://www.engineersedge.com/column_buckling/column_ideal.htm[/url]
• 07-23-2012, 03:08 PM
PinkertonD
If you are thinking of making your own 500-ton hydraulic jack, I suggest you think again and go buy one. Hydraulic pressure to move 500-tons is no small feat. There are seal designs to think of, surface finishing of rams, pistons and bores, just for a start.

Working out the strength of materials is the easiest part.

But then, if it is a homework project then you should check...
• 07-26-2012, 08:31 PM
Kelly Bramble
[url]http://www.engineersedge.com/calculators/pressure_vessel_pipe_shell_design_tool_10043.htm[/url]
• 10-23-2013, 12:06 PM
abhi kumar
plz send me the design and calculation of hydraulic jack????
• 10-23-2013, 03:35 PM
Cragyon
[QUOTE=abhi kumar;7605]plz send me the design and calculation of hydraulic jack????[/QUOTE]

Seriously?
• 11-22-2016, 01:27 AM
rai0179
Dear Esteemed Engineers,

I am in the process of finishing up my thesis for the final year of my mechanical engineering project and require your kind help.

I am trying to show that by using hydraulic jacks one can perform the shifting of hatch covers of a vessel efficiently without using a crane.

How do i show the calculations.

I have 4 jacks rated at 10 tons each shifting a hatch cover approximately 25 tons in weight a distance of about 300mm upwards.

Single Stroke with spring return, Manual hydraulic pump ( Jacking Method)

Already done the practical part but need to show calculations to show proof of research.

Could someone point me in the correct direction.

Very much appreciated.
• 11-22-2016, 07:00 AM
jboggs
You are in your final year of mechanical engineering and you don't know how to show calculations? It's a little late to ask. You could not have "done the practical part" without some calculations. Show them.
• 11-22-2016, 05:32 PM
Kelly Bramble
See: [url]http://www.engineersedge.com/hydraulic/hydraulic_pneumatic_cylinder_force_10105.htm[/url]
• 11-23-2016, 01:20 AM
webastraltraing

[COLOR=#CC0000][FONT=Arial][B]Cylinder Blind End Area (in square inches):[/B][/FONT][/COLOR]
[/B][/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Example[/B]: What is the area of a 6" diameter cylinder?[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Diameter[/B] = 6"
Radius is 1/2 of diameter = 3"
[B]Radius[/B][B][SUP]2[/SUP][/B] = 3" x 3" = 9"[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]π[/B] x [B](Cylinder Radius)[SUP]2[/SUP][/B] = 3.14 x (3)[SUP]2[/SUP] = 3.14 x 9 = 28.26 square inches[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial] [/FONT][/COLOR]
[COLOR=#CC0000][FONT=Arial][B]Cylinder Rod End Area (in square inches):[/B][/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Blind End Area - Rod Area[/B][/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Example[/B]: What is the rod end area of a 6" diameter cylinder which has a 3" diameter rod?[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Cylinder Blind End Area[/B] = 28.26 square inches
[B]Rod Diameter[/B] = 3"
Radius is 1/2 of rod diameter = 1.5"
[B]Radius[SUP]2[/SUP][/B] = 1.5" x 1.5" = 2.25"
[B]π[/B] x [B]Radius[SUP]2[/SUP][/B] = 3.14 x 2.25 = 7.07 square inches[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Blind End Area[/B] - [B]Rod Area[/B] = 28.26 - 7.07 = 21.19 square inches[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial] [/FONT][/COLOR]
[COLOR=#CC0000][FONT=Arial][B]Cylinder Output Force (in pounds):[/B][/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Pressure (in PSI) x Cylinder Area[/B][/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Example[/B]: What is the push force of a 6" diameter cylinder operating at 2,500 PSI?[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Cylinder Blind End Area[/B] = 28.26 square inches
[B]Pressure[/B] = 2,500 psi
[B]Pressure[/B] x [B]Cylinder Area[/B] = 2,500 X 28.26 = 70,650 pounds[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial]What is the pull force of a 6" diameter cylinder with a 3" diameter rod operating at 2,500 PSI?[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Cylinder Rod End Area[/B] = 21.19 square inches
[B]Pressure[/B] = 2,500 psi[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Pressure[/B] x [B]Cylinder Area[/B] = 2,500 x 21.19 = 52,975 pounds[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial] [/FONT][/COLOR]
[COLOR=#CC0000][FONT=Arial][B]Fluid Pressure in PSI Required to Lift Load (in PSI):[/B][/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Pounds of Force Needed ÷ Cylinder Area[/B][/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Example[/B]: What pressure is needed to develop 50,000 pounds of push force from a 6" diameter cylinder?[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Pounds of Force[/B] = 50,000 pounds
[B]Cylinder Blind End Area[/B] = 28.26 square inches
[B]Pounds of Force Needed[/B] ÷ [B]Cylinder Area[/B] = 50,000 ÷ 28.26 = 1,769.29 PSI[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial]What pressure is needed to develop 50,000 pounds of pull force from a 6" diameter cylinder which has a 3" diameter rod?[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Pounds of Force[/B] = 50,000 pounds
[B]Cylinder Rod End Area[/B] = 21.19 square inches[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Pounds of Force Needed[/B] ÷ [B]Cylinder Area[/B] = 50,000 ÷ 21.19 = 2,359.60 PSI[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial] [/FONT][/COLOR]
[COLOR=#CC0000][FONT=Arial][B]Cylinder Speed (in inches per second):[/B][/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B](231 x GPM) ÷ (60 x Net Cylinder Area)[/B][/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Example[/B]: How fast will a 6" diameter cylinder with a 3" diameter rod extend with 15 gpm input?[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]GPM[/B] = 6
[B]Net Cylinder Area[/B] = 28.26 square inches
([B]231[/B] x [B]GPM[/B]) ÷ ([B]60[/B] x [B]Net Cylinder Area[/B]) = (231 x 15) ÷ (60 x 28.26) = 2.04 inches per second[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial]How fast will it retract?[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Net Cylinder Area[/B] = 21.19 square inches[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial]([B]231[/B] x [B]GPM[/B]) ÷ ([B]60[/B] x [B]Net Cylinder Area[/B]) = (231 x 15) ÷ (60 x 21.19) = 2.73 inches per second[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial] [/FONT][/COLOR]
[COLOR=#CC0000][FONT=Arial][B]GPM of Flow Needed for Cylinder Speed:[/B][/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Cylinder Area x Stroke Length in Inches ÷ 231 x 60 ÷ Time in seconds for one stroke[/B][/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Example[/B]: How many GPM are needed to extend a 6" diameter cylinder 8 inches in 10 seconds?[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Cylinder Area[/B] = 28.26 square inches
[B]Stroke Length[/B] = 8 inches
[B]Time for 1 stroke[/B] = 10 seconds
[B]Area[/B] x [B]Length[/B] ÷ [B]231[/B] x [B]60[/B] ÷ [B]Time[/B] = 28.26 x 8 ÷ 231 x 60 ÷ 10 = 5.88 gpm[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial]If the cylinder has a 3" diameter rod, how many gpm is needed to retract 8 inches in 10 seconds?[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Cylinder Area[/B] = 21.19 square inches
[B]Stroke Length[/B] = 8 inches
[B]Time for 1 stroke[/B] = 10 seconds[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Area[/B] x [B]Length[/B] ÷ [B]231[/B] x [B]60[/B] ÷ [B]Time[/B] = 21.19 x 8 ÷ 231 x 60 ÷ 10 = 4.40 gpm[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial] [/FONT][/COLOR]
[COLOR=#CC0000][FONT=Arial][B]Cylinder Blind End Output (GPM):[/B][/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Blind End Area ÷ Rod End Area x GPM In[/B][/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Example[/B]: How many GPM come out the blind end of a 6" diameter cylinder with a 3" diameter rod when there is 15 gallons per minute put in the rod end?[/FONT][/COLOR]
[COLOR=#333333][FONT=Arial][B]Cylinder Blind End Area[/B] =28.26 square inches
[B]Cylinder Rod End Area[/B] = 21.19 square inches
[B]GPM Input[/B] = 15 gpm
[B]Blind End Area[/B] ÷ [B]Rod End Area[/B] x [B]GPM In[/B] = 28.26 ÷ 21.19 x 15 = 20 gpm[/FONT][/COLOR]
• 11-24-2016, 06:50 PM
rai0179
jboggs. Am not a design engineer. Thats why i asked in the forum. ANyway there are others in the forum more helpful then you.
• 11-25-2016, 02:01 PM
Cragyon
[QUOTE=rai0179;13038]jboggs. Am not a design engineer. Thats why i asked in the forum. ANyway there are others in the forum more helpful then you.[/QUOTE]

I'm confused because everything you need is here in this forum thread. Where are you going to school?