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 --------------oOo-------------- | ||