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

John Lynch, my hangar neighbour kindly did a SW FEA with animation on
the input shaft  extended point set screw coupling.

This would be used on the coupling to the
Ford sun gear connection and it is based on a 1.6 inch diameter sun gear
shaft.

It is very very conservative as we figured 1000 foot pounds out of
the engine. It does not come close to that but we are erring on the overkill
side as it adds little or no weight over the old one inch spline shaft.
It is much much stronger and no back lash or slop. Note the stress never
exceeds 5600 PSI in the shaft which is nothing for the Ford hardened alloy
steel sun gear.

John is also machining the billet silicon rubber coupling for the 650 HP time to
climb project.


Paul Lamar



That formula is R squared x Pi. It seems  you have an error of an order of magnitude there.

Robert LeClercq


Ah so You are right.

>>>>.156÷2×3.1416 <<<<

>>>>> = 0.2450448 square inches <<<<<<


(.156÷2)×(.156÷2)×3.1416 = 0.019113494


>>>>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

0.019113494×150000 = 2867.0241 Correct answer


Two 1/4 20 EPSS shear strength would be

5734.0482

we need 6000 pounds sheer force so two  1/4 28 EPSS  will still work but just barely.

We have 15 in this design so the factor of safety is about seven.

Thanks for catching that Robert.


Paul Lamar

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