Tractor mounted fork lift

[chandra007]

I am trying to build a forklift attached to tractor.

My specifications are,4000 lbs of weight to be lifted to 8ft. with in 5 seconds.

what should be my engine horse power and pump specifications for the same.

What should be the cylinder specifications.

Learned members may give some calculation relevant to it.

Thanks in advance.

chandra007

 

[Leejohn]

I would look for a old forklift boom and fit to your tractor. 4000lbs to 8 feet is a task. Or do you have one and just wanting to know what cylinder to get?

 

[swick1]

This should get you started...

Power is work/time. Work is force * distance. Since you want to raise 4000lbs in 5 seconds to a height of 8ft, we can calculate power after calculating work.

The work to be done is force * distance = 4000 lbs * 8 ft = 32000 lb-ft

The power necessary to do the work in 5 seconds is work / time = 32000 lb-ft / 5 sec = 6400 lb-ft/sec

To convert this to horsepower, we divide by 550 lb-ft/sec = 1hp

This gives us a needed horsepower of 6400/550 = 11.63 hp

An external gear pump has an efficiency of 85%. Divide the horsepower by efficiency to get the actual driven horsepower needed = 11.63/0.85 = 13.68 hp. This is your hp requirement at the pump. If your drive is 95% efficient then your hp requirement for the engine is 13.68/0.95 = 14.4 hp

J_J, It's your turn!

 

[J_J]

I have already replied to him via PM.

He is wanting someone to do all the leg work and designed a forklift for him.

How fast something can lift is based on the cyl, GPM and the mechanical, chains and pulley's, etc

You can operate a cyl with just about any GPM the pump can produce with the pressure required.

The more GPM you feed the cyl, the faster it will move.

The other thing is how safe can he build it . 4000 lbs, 8 ft in the air.

Once he has the GPM requirement for the GPM, he can figure out the pump size, and pressure, and the HP to run it.

He should go and look at some forklifts and get a working knowledge of the setup.

They make forklifts that will work on the back of a tractor.

http://www.edwards-equip.com/SiteDocuments/tt36-416.101ab.pdf

http://www.edwards-equip.com/SiteDocuments/tf79-426.1011ab.pdf

 

[swick1]

Also I saw one for a mini skid steer that mounted to the loader arms. It had its own wheels too.

You are right about the safety aspect. Hydraulically speaking he should look into pilot operated check valves to keep it from falling if someone grabs a lever while the load is raised.

Mechanically he should rig some failsafes and overbuild

 

[chandra007]

Thank you swick and JJ,

As per your calculation , total power required to lift 4000 lb is 14.4 hp(if 95% efficient)

power = pressure*flow/1714.

what should be the flow? and cylinder specification?

chandra007

 

[masseyR1327]

There is a forklift on the front of my tractor. It is designed to lift 1 metric tonne. (1000 kilograms). It will lift almost 3 metres. It has three rams. Two for tilt and 1 for lift.

Not a very clear photo but I can take more if you are interested.

 

[chandra007]

hi,

what is the specification of your hydraulic pump.if possible please write and what is the stroke length of the lifting cylinder and bore size?do you have chain arrangement in your forklift?


chandra007

 

[swick1]

Thank you swick and JJ,

As per your calculation , total power required to lift 4000 lb is 14.4 hp(if 95% efficient)

power = pressure*flow/1714.

what should be the flow? and cylinder specification?

chandra007

Use algebra:
Flowrate=(power*1714)/pressure

You will want to build in a factor of safety for each part. The numbers we have given you are theoretical minimums. In practice you will want more power, flowrate, etc.

 

[chandra007]

swick,the formula is flowrate =power*1714/pressure

pressure=Force/area

4000lb/3.14*d*d/4,if i select 2 inch bore dia.then pressure=4000/3.14=1273psi.

14HP=(1273*Q)/1714 >>Q=(14*1714)/1273=18.84 GPM

flowrate=RPM*displacement/231

If RPM is 1500,

18.8 GPM=1500*displacement/231

Displacement=18.8*231/1500=2.88 cu.in
So a pump with 2.88 Cu.in of displacement and flowrate of 18.8 GPM running at 1500 rpm and rated pressure of 1273 psi,will be suitable for lifting 4000 lbs.

For timing 8ft=8*12=96 inches

Q=AV>v=Q/A=18.8/3.14*2*2/4=18.8/3.14=5.98 nearly 6 inches/second

for 1 second it can move 6 inches for moving 96 inches it takes 12 seconds,but i wish to lift the weight in 5 seconds,whether i have to change flow requirements?

For lifting power required is 14.HP,for drive how much power is required?

when 4000lb 's of weight lifted to 8 ft ,counterbalance valve is required to make it stay at top even power is off?

 

[swick1]

Your pressure needs to go up to lift it faster. Basically, you're using full flow, but only building about half of what a normal hydraulic system pressure would be. A smaller cylinder would lift the load faster and still be under the normal operating pressure of 2000 to 2500 psi.

Nobody can tell you what your driving horsepower should be because we don't know your tractor weight or what kind of terrain you are operating on. For just driving around on flat land moving a 4000lb load, i'd guess 5hp assuming you have a hydrostatic transmission and can creep very slowly. I would not make a forklift attachment for a geared tractor. 8ft is too high to lift without making it really unstable from popping the clutch.

 

[wdchyd]

I'd be concerned about the sizing and construction of the counterbalance weight to make it safe too.....hp would have to factor in the additional weight to move the tractor (Tractor, counterbalance & load)

 

[J_J]

If this is a first time endeavor for chandra007 , not knowing anything about his mechanical skills, welding skills, electrical, etc, I believe he is in for a lot of trial and error, expense, and frustration.

It is one thing to just lift a load of 4000 lbs, and another to lift and move it over what kind of terrain.

How much does the machine weigh that the forklift will be attached to. Is counter balance requires, should the tires be solid, what size axle to handle that kind of load, is side shift necessary?

Will it be powered by battery, propane, gas, diesel. Lots of decision to make.

Do it yourself is fine as long as you have the time, money, and expertise in the areas necessary to build something like this.

If he builds something like he plans, and it doesn't go right, is he going to blame us?

 

[swick1]

Well somebody that can't do algebra is not qualified to design a forklift. I assume every thread like this is hypothetical because if he really wanted to do it he would hire a licensed engineer to do the design. He may just be using us for back of the envelope calculations. Anyone who would build a contraption like this without having a real engineer check their math, apply factors of safety, specify material types and treatments, define joint construction methods, and inspect the final product before use is taking their life and maybe others lives into their own hands.

 

[chandra007]

Swick and JJ,

Both of you thinking in a different way,i am not using anybody for my calculations.I just mentioned give some idea ,if you know something.even though i am at starting stage,my interpretation must be useful for all.

sharing knowledge is allways good.

your words must be useful for all the members like me.

chandra007

 

[J_J]

Excerpt:

Cyl and roller chain.

Lifting Mechanism: Roller Chain Pulley
The hydraulic pistons are attached to the two main vertical structures called the "masts." However, the actual forks that carry the load are attached to the main body of the forklift through a pair of roller chain pulleys whose fulcrum is a gear at the top of the mast.

Thus, when the hydraulic pistons push the masts up, the gears on the masts push against the roller chains. Because one side of the chains are attached to the immobile frame of the forklift, the only way the masts can move up is if the gears rotate clockwise and pull the forks up.

The importance of this mechanism is that it allows the forks go far beyond the reach of the cylinders alone. If it weren't for the roller chain pulleys, forklifts would need much taller cylinders to lift loads to a comparable height. Taller cylinders would mean more building material, which would shift the vehicle's center of gravity forward and increase the risk of tipping. Likewise, taller cylinders would demand stronger pumps and higher pressure thresholds.

With a 4 in cyl and 2 in shaft, 48 in stroke, and if you have a 33 GPM pump at 3000 psi, the cyl will extend at 10.1 ips, and take about 4.752 sec to do it.

Cyl can push 37,699, or 18.84 T

A 2 in bore 1 in shaft, at 3000 psi can push about 9,425, or 2.71T

Now to power that 33 GPM pump, at 3000 psi will take about 68 HP.

 

[masseyR1327]

As you can see from my photo the tractor is a good size and yet loaded with 1 tonne on the front flattens the tyres to some degree and even with power steering, turning is heavy. The tractor becomes light on the rear wheels and I used to put a heavy rotary hoe on the back to counter balance it. 4000 lbs would be better with solid tyres. You will need an equivalent counter weight on the other side of the front axle. This needs serious engineering with this weight.

 

[JJZ 109]

A search of the forum will show that somewhere, some time ago someone bought an old fork lift carriage/mast and adapted it to his 3 point hitch.
As I recall it didn't work out too bad for him. That may be the way to go instead of asking all the algebra questions and as JJ points out, probably hit & miss fabrication.
I don't know where the OP is located, but I've seen old forklifts up for auction at Government Liquidation.

 

[J_J]

He says he is located in chennai, in south India.

 

[JJZ 109]

He says he is located in chennai, in south India.

Then I guess the GL website won't be doing him much good! :laughing: