Rolls-Royce Archives
         « Prev  Box Series  Next »        

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).
Analysis of one-node frequency and maximum vibration stresses at resonance speed for a six-cylinder engine crankshaft.

Identifier  ExFiles\Box 132\1\  scan0116
Date  18th March 1939
  
378
DIESEL CRANKSHAFT VIBRATION

Six-Cylinder Engine
TABLE III
ONE-NODE FREQUENCY
MAX. VIBRATION STRESSES AT RESONANCE SPEED WITH HYSTERESIS DAMPING

f₁ = 18440 cycles per min.
ω₁² = 3,700,000 rad²-sec.⁻²
Firing Order : 1-4-2-6-3-5
I.M.E.P. = 100 lb. per sq. in.

| 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 |
|---|---|---|---|---|---|---|---|---|---|
| Harmon Order K | Critical Speed R.P.M. f : K | P Fig. 9ab | Σe Table V | PΣe | A Eq. 12 | Q 10850 × A | γ Eq. 8 | S 1-4-2-6-3-5 | A max. Eq. 12 |
| 3 | 6200 | 25.00 | 4.0810 | 102.000 | ±0.051000 | ±555.00 | 90.0 | ±50000 | ±4.60 |
| 3½ | 5300 | 20.00 | 0.5000 | 10.000 | 0.005000 | 54.20 | 286.0 | 15600 | 1.43 |
| 4 | 4600 | 15.00 | 0.2500 | 3.750 | 0.001900 | 20.60 | 466.0 | 9900 | 0.89 |
| 4½ | 4100 | 10.10 | 1.4500 | 14.500 | 0.007300 | 80.00 | 236.0 | 19000 | 1.72 |
| 5 | 3720 | 8.00 | 0.2500 | 2.000 | 0.001000 | 10.85 | 656.0 | 7250 | 0.66 |
| 5½ | 3350 | 6.00 | 0.5000 | 3.000 | 0.001500 | 16.33 | 525.0 | 8600 | 0.79 |
| 6 | 3000 | 4.65 | 4.0814 | 19.000 | 0.009500 | 103.00 | 210.0 | 21600 | 2.00 |
| 6½ | 2840 | 3.50 | 0.2500 | 1.750 | 0.000900 | 9.80 | 700.0 | 6860 | 0.63 |
| 7 | 2640 | 2.80 | 0.5000 | 0.700 | 0.000350 | 3.80 | 1100.0 | 4200 | 0.38 |
| 7½ | 2460 | 2.50 | 1.4500 | 3.630 | 0.001800 | 19.50 | 470.0 | 9200 | 0.85 |
| 8 | 2300 | 1.60 | 0.2500 | 0.400 | 0.000200 | 2.17 | 1470.0 | 3200 | 0.30 |
| 8½ | 2170 | 1.50 | 0.5000 | 0.750 | 0.000380 | 4.15 | 1030.0 | 4300 | 0.39 |
| 9 | 2050 | 0.90 | 4.0814 | 4.080 | 0.002100 | 22.78 | 446.0 | 10200 | 0.94 |
| 9½ | 1940 | 1.10 | 0.5000 | 0.450 | 0.000220 | 2.40 | 1600.0 | 3880 | 0.35 |
| 10 | 1840 | 0.90 | 0.2500 | 0.170 | 0.000085 | 0.92 | 2200.0 | 2000 | 0.19 |
| 10½ | 1760 | 0.70 | 1.4500 | 0.870 | 0.000440 | 4.80 | 955.0 | 4600 | 0.42 |
| 11 | 1680 | 0.60 | 0.2500 | 0.125 | 0.000060 | 0.65 | 2620.0 | 1700 | 0.16 |
| 11½ | 1600 | 0.50 | 0.5000 | 0.200 | 0.000100 | 1.08 | 2000.0 | 2160 | 0.20 |
| 12 | 1530 | 0.33 | 4.0814 | 1.300 | 0.000650 | 6.50 | 840.0 | 5500 | 0.50 |

The outside diameter of the flywheel rim is 15 in., the inside diameter 10 in., the weight 95 lb., and the square of the radius of gyration, m² = (D² + d²)/8 = 40.6. Therefore, the moment of inertia of the flywheel is
I₇ = W × m² / 386 = 95 × 40.6 / 386 = 10 lb.-in.-sec.²

The torsional stiffness of the crankshaft is calculated by means of Equation (3) (see also Fig. 11)
M₁ = M₂ = M₄ = M₅ = GJ / l₁ = (12 × 10⁶ × 8) / 7 = 13.7 × 10⁶
M₃ = GJ / l₃ = (12 × 10⁶ × 8) / 7.5 = 12.8 × 10⁶
M₆ = GJ / l₆ = (12 × 10⁶ × 8) / 9 = 10.7 × 10⁶
M₇ = GJ / l₇ = (12 × 10⁶ × 8) / 26.75 = 3.6 × 10⁶ (15)

We now have a six-cylinder crankshaft reduced to the system illustrated in Fig. 11, with calculated inertias of 0.22 lb.-in.-sec.² for the moving mass of each cylinder, with a flywheel inertia of 10 lb.-in.-sec.², and torsional rigidities of the shaft units as calculated in Equation (15). For the purpose of calculating the natural frequencies this system is replaced by:
(a) A two-mass system (Fig. 10),

one mass representing the six crank units, with a moment of inertia I = 1.32 lb.-in.-sec.², the other representing the flywheel, with a moment of inertia of 10 lb.-in.-sec.² This is for the first mode of vibration with one node.
(b) A three-mass system (Figs. 13, 14, and 15), for second-mode vibration with two nodes.
The third, fourth, and higher modes of vibration are of no particular interest in our case.
The natural frequency of the two-mass system is according to Equation (5)
ω₁² = M₇(I + I₇) / II₇ = (3.6 × 10⁶(1.32 + 10)) / (1.32 × 10) = 1800 rad/sec.
ω₁² = 3,300,000 rad²-sec.⁻²

mass system with one node according to Equation (2) is
f₁ = 9.55 √(M₇(I + I₇) / II₇) or
f₁ = 9.55 √((3.6 × 10⁶(1.32 + 10)) / (1.32 × 10)) = 17200 cycles per minute

The phase velocity of the two-

Part Two
of this treatise by O.{Mr Oldham} Malychevitch will be concluded in AUTOMOTIVE INDUSTRIES in the issue of March 25.

March 18, 1939
Automotive Industries
  
  


Copyright Sustain 2025, All Rights Reserved.    whatever is rightly done, however humble, is noble
An unhandled error has occurred. Reload 🗙