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 shaft bending, load distribution, and forces on bearings.
| Identifier | ExFiles\Box 115\2\ scan0381 | |
| Date | 10th January 1939 guessed | |
| - 3 - Assuming that the shaft bends so that it is supported by the edge of the main bearings the c.g. will be 0.685 from bearing pt. on No.7. bearing, 3.738 from bearing pt. on No.6. bearing. * (3.738 / 4.425) of load will be carried by No.7. bearing. * No.7. bearing carries 0.845 of balance weight load. = 1490 lbs. at 2600. 1770 lbs. at 2850. For the load on the bearing this is vectorially added to half the gas and inertia load of the rest of the system. ------ Bending due to balance weight alone. [Diagram of a simply supported beam. Left support 'R', right support 'S'. Total length 'l'. Downward force 'P' is applied at a distance 'a' from R and 'b' from S.] R = Pb/l - EI d²y/dx² = Pbx/l - P[x-a] - EI dy{F R Danby}/dx = Pbx²/2l - P/2 [x-a]² + A.{Mr Adams} - EI y = Pbx³/6l - P/6 [x-a]³ + Ax + B. but y = 0 when x = 0 or x = l. ∴ B=0 and Pbl²/6 - Pb³/6 + Al = 0. ∴ EI dy{F R Danby}/dx|x=0 = Pbl/6 - Pb³/6l = P[3.75*4.435/6 - 3.75³/(6*4.435)] = P [2.773 - 1.982] = 0.791 P This can be superposed on the bending due to gas and inertia forces in (A) above. The vector sum of the radial and tangential bending can then be applied to the condition that fouling does not occur. dy{F R Danby}/dx|x=0 = (0.791 P) / EI . EI = 7.56 * 10^6. dy{F R Danby}/dx|x=0 In line of web = 1.05 * 10^-7 P ⊥ " " " zero since balance weight force acts along line of web. | ||