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).
Calculations for the deflection of a freely supported beam with two different sections, loaded in the middle.
| Identifier | ExFiles\Box 112\1\ scan0113 | |
| Date | 10th January 1939 guessed | |
| DEFLECTION OF A BEAM OF TWO DIFFERENT SECTIONS, FREELY SUPPORTED AT ITS ENDS & LOADED IN THE MIDDLE. OMITTING THE ZERO TERMS :- 2 ((l₁ MBL / I₁) + (l₂ MBR / I₁)) + (a₂ W₂ / l₂ I₂) (l₂² - a₂²) = (6E/l₁) (δB - δA) + (6E/l₂) (δB - δC) LET MBL = MBR = M : AND (δB - δA) = - (δB - δC) = δ (2 l₁ M / I₁) + (2 l₂ M / I₂) = (2 a W / l₂ I₂) (l₂² - a²) = δ (1/l₁ - 1/l₂) 6 E.{Mr Elliott - Chief Engineer} NOW IF l₁ = 4'-0", l₂ = 6'-0", a = 5'-0", & M = +4 W. (32 W / I₁) + (48 W / I₂) + (110W / 6 I₁) = δ (¼ - ⅙) 6E = ½ δ E OR δ = (64 W / E I₁) + (132.6 W / E I₂) THEN IF I₁ = I₂, δ = (W / E I₂) x 196.6 BUT IF I₁ IS DOUBLED δ = (W / E I₂) x 164.6 | ||