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From the Rolls-Royce experimental archive: a quarter of a million communications from Rolls-Royce, 1906 to 1960's. Documents from the Sir Henry Royce Memorial Foundation (SHRMF).
Page of calculations for moments of torsion and stress on a 'Stabifix' component.

Identifier  ExFiles\Box 15\6\  Scan159
Date  1st January 1931 guessed
  
-2-

thus the moment of torsion

Mor = F . b

and at the point O1 will be the stress

F3 = F . b/a

with the moment of torsion

Mol = 0

all conditionally on the left part of the Stabifix shall be free in its movement.

Following this argument, we find at the point - C -

the stress

Fc = F

and the moment of torsion

Mc = F ( b + a/2 )

The moment distributes itself to the arms of the Stabifix as follows:-

Mcr = F2 . a/2 = F ( a/2 + b/2 )

Mcl = F3 . a/2 = F . b/2

These two moments of torsion give together

Mc = Mcl + Mcr = F ( b + a/2 )

These moments Mcr and Mcl as the reaction Fc give in the points Ur and Ul the stresses F'2 and F'3

F'2 = Fc + 1/a . Mcl = F ( 1 + b/a )

F'3 = Fc + 1/a . Mcr = F b/a

The stresses that come into play on the lower bar give in the lower fixing brackets the pressure F'r and F'l where

F'r = 1 / (a + 2c) { ( F2 - F3 ) . c + F2a} = ( a + b + c) / (a + 2c) F

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