Calculation of the deflection of a Bentley III yoke due to bending.
| Identifier | ExFiles\Box 100\1\ scan0156 | |
| Date | 27th May 1938 | |
| -3- STIFFNESS OF BENTLEY III YOKE. (1) Deflexion of A below O due to bending. Diagram showing a simply supported beam with forces and moments: - Point O: Moment M, Downward force 1400 lbs - Point A: Upward force 1,000 lbs at distance 'a' from O - Point B: Upward force 400 lbs at distance 10" from O - Moment indicated: 14500 lbs-ins Equations: Mx {John H Maddocks - Chief Proving Officer} = -14,500 + 1,000 (a - x) + 400 (10 - x) EI d²y/dx² = 14,500 - 1,000 (a - x) - 400 (10 - x) EI dy {F R Danby} /dx = 14,500x - 1,000 (ax - x²/2) - 400 (10x - x²/2) + α EI y = 14,500x²/2 - 1,000 (ax²/2 - x³/6) - 400 (10x²/2 - x³/6) + αx + β α = 0 because dy {F R Danby} /dx = 0 at O β = 0 " y = 0 at O EI yA = 7,250 a² - 333a³ - 2,000a² + 67a³ = 5,250 a² - 267a³ yA = 1/EI (5,250 a² - 267a³) | ||
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).