diameter " diameter m area m^2
0.5 0.0127 0.000126677

shear strength max force N max force lbs 2x max force (double shear) lbs
4.00E+08 5.07E+04 11375 22750


diameter " diameter m area m^2
0.625 0.015875 0.000197933

shear strength max force N max force lbs 2x max force (double shear) lbs
4.00E+08 7.92E+04 17774 35547

Shear stress = Force/Area

just took the shear strength of 1018 mild steel (400 MN/m^2) and multiplied by the area of the pin.

I had to convert the numbers to metric, then switch back to pounds. all I could find for shear strength of 1018 (mild) steel was in metric...don't feel like pulling the book out.

Also, note that I used 1018 steel. It's a pretty 'normal' used steel, so i'm probably not far off from what material they used, but am probably not right.

 

that is pretty close actualy and the math depends on the the alloy used but the pin will probably never break just because the force on the pin isnt strong enough its distributed through out the whole hitch and you should feel perfectly fine with its ablility to stand up under towing

 

That's my figures exactly.I'm busy putting my shoes back on,I ran out of fingers.

 

plus your shear force is for single point, there are 2 points of contact on the pin the load distributed between them.

 

thats why i multiplied by two at the end, and it says '2x max force'

unless you got a 10000lb trailer and a truck with either 1000 hp or really really big brakes, you're not pullin or stoppin 22000 lbs.

 

ah missed that part

 

it's in bold!

(post whoring)

 

Heh.. I started a math/engineering thread.

Like I said, I "know" that the pin has to be sufficient, but looking at it vs everything else in the system tickled the thought of "just how much?"

Thanks,

 

hey these are some big *****, i wonder what you tow with them- lol jk

http://www.stupidiotic.com/popup_ima...ID=129&image=2

http://www.stupidiotic.com/popup_ima...ID=129&image=1