DIY tensile strength tester?

Things "rated" for 15,000 lbs may be out of range of a 20,000 lb measuring device, if they are "well-rated" (overbuilt, have a reasonable safety factor, etc).

As for turning push into pull, a lever would make sense. To remove side loading, you could make both the end of the lever where the ram connects and the mounting point for the base of the cylinder pivot; that mimics how things are mounted on my backhoe, so it's probably good design.

Don't get in the way of the shrapnel.

Glenn, Just a though or two: Modify a 20T H frame shop press (20T to be sure of a stiff frame as you only need 10T) that puts the bottle jack at the bottom area. The adjustable table will be above the jack ram and a lower shelf will be needed to catch the jack (at rest). A yolk assembly is made to hold the jack (bottom) that passes around the jack and table terminating in a pull eye or clamp above the table plus travel clearance required. When set up and operating, the jack ram presses against the table bottom and presses against the yolk pushing the yolk down, thus the DUT (device under test) will be stretched as the upper end is clamped (hung) on the H frame top bridge. The table has the support blocking and rods above and below. Next, add a pressure gauge to the jack cylinder calibrated based on the internal cyling bore. BEWARE!!! The external ram is **NOT** the cylinder diameter. Unscrew the top and pull out the ram/piston enough to measure the bore. Area times gauge = force. Be safe. Be careful. RichD, Atlanta (built my own 50T press after destroying a 20T marked as

Glenn Ashmore wrote:

I've used a 4 ton lever cable puller between two large oak trees to test tensile strength.

Next time I plan to use a 10-ton porta-power pullback cylinder mounted on a log-splitter beam.

"Glenn Ashmore" wrote

Can you use trigonometry to your advantage? Pushing _transversely_ on a string can put a lot more tension in it than the force your pushing device needs to put out. You do need a strong frame of course, and a way to know the _geometry_ at the moment the string breaks.

-- TP

That's a good idea!

Call the distance between rigid immovable supports X , and the perpendicular distance the rope is pulled away from a straight line as Y. Calculate an angle theta that is the arctangent of Y/X. With force F pulling sideways on the rope, tension in the rope is

F / (2 * sin(theta) )

Example: X = 10 feet (120 inches), Y = 6 inches For 100 lbf of force, the tension in the rope is 1001 lbf. The magnification doesn't drop as low as 2 (200 lbf tension for 100 lbf force) until Y =