Bench Top Hydraulic Press Questions

[pirana]

I'm building a small hydraulic press & i'll probably use a 6 ton bottle jack for power. The top member is a W4 x 13 & the two uprights are C4 x 5.4. For the bed i'll probably use 2 pieces of 2-1/2 x 2-1/2 x 3/8 angle. My questions are what size & type round bar should I use to support the bed? Is 3/4" dia. rod stout enough ? And where is a good source for it, McMaster-Carr or MSC ? I want to drill the bed support holes in the uprights this weekend but need to know what size bar will suffice. Any advice would be appreciated.

[musgofasta]

I believe my hydraulic press only has 1/2'' or so rods. I'd think 3/4'' solid rods would be plenty strong. I know my Home Depot sells some pretty thick solid rod, and I've bought 3/4'' threaded rod there before.

Sorry I can't be any more help.

[t. jones]

Used spare wheel wrenches work well
Thanx Trevor

[pirana]

Originally Posted by t. jones

Used spare wheel wrenches work well
Thanx Trevor

I sure didn't think of that. I happen to have a big 4 way lug wrench laying around that would be perfect. Thanks

[akdiesel]

Why not just use 1/2" or 5/8" grade 8 bolts. Your local hardware store will carry them.

[RobSmith]

ooooh 6 tons on a 1/2" steel rod in shear ? I turned up a couple of 1" bars for my press. It may be overkill. But I have all my fingers and eyes. 3/4" would be OK. Tire / wheel wrenches are the softest steel you can buy...put one in a vise and bend it with one hand !

[balddave]

Ah the engineer in me is pulling out the calculator...

6ton * 2000lbs/ton = 12kip force

12kip / 4 = 3 kip (double shear on 2 bolts = 4)

(3/4)^2*pi/4 = .442 in^2

3kip / .442 in^2 = 6.8ksi

36ksi * .4 = 14.4ksi allowable based on A36 steel rod

6.8 / 14.4 = .472 < 1 therefore you'd be fine.

The only problem I see is with bearing... if the bearing area is too small you may see some localized yielding... which might bugger up the rods beyond use after a while...

3 kip / 14.4 ksi = .208 in^2

.208 in^2 / .75in = .277in

This means you need to support the bolt for a minimum of .277" each place it is in shear on both sides of the shear plane.

That's what I'd do..I'm sure someone could argue something different. But my 12 ton only has 3/4" bar, although that is higher strength material than A36 which is pretty much any run of the mill steel. So go with some 3/4" bar and you should be fine...or upgrade to some nice alloy steel and never think twice about it.

[z28snksknr]

Originally Posted by balddave

Ah the engineer in me is pulling out the calculator...

6ton * 2000lbs/ton = 12kip force

12kip / 4 = 3 kip (double shear on 2 bolts = 4)

(3/4)^2*pi/4 = .442 in^2

3kip / .442 in^2 = 6.8ksi

36ksi * .4 = 14.4ksi allowable based on A36 steel rod

6.8 / 14.4 = .472 < 1 therefore you'd be fine.

The only problem I see is with bearing... if the bearing area is too small you may see some localized yielding... which might bugger up the rods beyond use after a while...

3 kip / 14.4 ksi = .208 in^2

.208 in^2 / .75in = .277in

This means you need to support the bolt for a minimum of .277" each place it is in shear on both sides of the shear plane.

That's what I'd do..I'm sure someone could argue something different. But my 12 ton only has 3/4" bar, although that is higher strength material than A36 which is pretty much any run of the mill steel. So go with some 3/4" bar and you should be fine...or upgrade to some nice alloy steel and never think twice about it.

Wow.... first post ever and it's load calculations.......

Welcome to the board!!

[pirana]

I cut some rods from the 4 way tire iron I mentioned in my previous post which I don't remember how I aquired it, but the dia. of the rod is 3/4". What ever it's made from (chrome vanadium?) it's tough. My cutting torch wouldn't cut it. I'll try it anyway.