Prop torque 1000 foot pounds

12000 inch pounds

shear force 2 inch diameter. 6000 pounds

1/4 28 set screw

Diameter of extended point is .156

Shear area of of extended tip is .156/2 X Pi

.156÷2×3.1416

= 0.2450448 square inches

Yield stress of EPSS alloy steel 150,000 psi

Sheer strength of 1/4 20 EPSS alloy steel

Alloy steel is 150,000 psi

.245×150000

= 36750

Two 1/4 20 EPSS shear strength would be

36750×2

= 73,500 pounds sheer strength

So in theory all we need are two 1/4 20 EPSS

Let's calculate the torsional strength of a 2 inch prop shaft
with a .125 wall heat treated and ground to needle bearing
size.


1×1×3.1416−(.875×.875×3.1416)

= 0.7363125 square inches

Alloy steel is 150000 psi

so the strength is

0.7363125×150000

= 110446.875 inch pounds

110446.875÷12

= 9203.90625 foot pounds

Factor of safety with out the set screw holes is 9
shaft alone.

Lets try 4 set screws. Gives us a factor
of safety based on set screw shear only of 2

four set screw area is

four .156 diameter holes .08 deep

= 0.04992 square inches

0.7363125−0.04992


= 0.6863925 new area

alloy steel 150,000 psi

0.6863925×150000

= 102958.875 inch pounds

102958.875÷12

= 8579.90625

Factor of safety drops to 8.5

This surprised even me.

If we up it to eight set screws in two rows of 4
the factor of safety based on set screw shear
alone doubles to 4 while the shaft gets no weaker
in torsion. Still with a factor of safety of  8.5 shaft alone.

Check my arithmetic.

Paul Lamar

The Rotary Engine News Letter. Powered by Linux.
ACRE NL web site. http://www.rotaryeng.net
You Tube http://tinyurl.com/beqqxas
Store http://www.rotarpower.com/#!
Copyright 1998-2015
All world wide rights reserved.