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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).
Part one of a treatise on the mathematical analysis and frequency calculation of engine crankshaft vibrations.

Identifier  ExFiles\Box 132\1\  scan0111
Date  18th March 1939
  
RM{William Robotham - Chief Engineer}
373
Part One
Owing to space limitations it is necessary to print Part Two of this treatise in next week's issue.
Engine Crankshafts
called deflection curves, or normal elastic curves, which terms will be defined later. When a shaft of a one-mass system as represented in Fig. 3 is deflected and released, it will vibrate through gradually decreasing angles. This is the simplest form of harmonic motion, and the maximum displacement of point D along line xx is called the amplitude of vibration and is usually designed by the symbol ±h.{Arthur M. Hanbury - Head Complaints} The instantaneous angular displacement of point D is
δ = ωt
where ω is the phase velocity of vibration (velocity of point D) in radians per second and t the time in seconds occupied in moving from A to D.{John DeLooze - Company Secretary}
The period T in seconds of one cycle (360 deg. or 2 π) is
T = 2π / ω
The number of complete cycles per second, also called the frequency of vibration, is
f = 1/T = ω / 2π
From Fig. 3 it can be seen that the displacement of point D during the time interval t is
p = r sin ωt.
The velocity, which is represented by the vertical projection of vector r ω, 90 deg. ahead of vector r, is
v = r ω cos ωt.
The acceleration is represented by a vertical projection of vector r ω², 180 deg. ahead of vector r, and is
q = r ω² sin ωt.
The natural frequency of vibration of a single-mass system is
ω / 2π = (1 / 2π) * √(M / I) cycles per second (1)
or f = (60 / 2π) * √(M / I) = 9.55 * √(M / I) cycles per minute
The natural frequency of a two-mass system with one node is
(1 / 2π) * √(M(I₁ + I₂) / (I₁ I₂)) cycles per second (2)
or f = (60 / 2π) * √(M(I₁ + I₂) / (I₁ I₂)) = 9.55 * √(M(I₁ + I₂) / (I₁ I₂)) cycles per minute
where:
M = Gπd⁴ / 32L = GJ / L lb-in. per radian (3)
M is the torsional rigidity or stiffness of the shaft, or the torque required to deflect the shaft one radian torsionally.
J = (π d⁴)/32, the polar moment of inertia of a circular shaft, in in.⁴
L, the length of the shaft in in.
G, the modulus of rigidity (11.8 × 10⁶ to 12.3 × 10⁶ lb. per sq. in. for steel). In our calculations we will take G to be equal to 12 × 10⁶ lb. per sq. in.
I = (W R²)/g, the moment of inertia of the disc, ............(4)
where W is the weight of the disc in lb.; R, the radius of gyration in in., and g the constant of gravity in in.-sec.⁻² units (386).
The phase velocity:
ω = 2πf radians per sec. (5)
ω = 2πf / 60 radians per min.
ω² = M(I₁ + I₂) / (I₁ I₂) rad²-sec.⁻²
See Fig. 1 for the two-mass system.
Six-Cylinder Engine Frequency Calculation One-Node Vibration
TABLE I
Figs. 10-11-12.
f₁ = 18440 cycles per min.
ω₁² = 3,700,000 rad²-sec.⁻²
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Cyl. Mass No. | d | l | I (Eq. 14) | Iω² | ε (Eq. 9) | Iω²ε | ΣIω²ε | M = GJ/l (Eq. 3) | ΣIω²ε / M | I ε² | S (Eq. 13) |
| 1 | 3 | 7 | 0.22 | 0.814 × 10⁶ | 1.0000 | 0.814 × 10⁶ | 0.814 × 10⁶ | 13.7 × 10⁶ | 0.0595 | 0.2200 | ±0.2680 |
| 2 | 3 | 7 | 0.22 | 0.814 × 10⁶ | 0.9405 | 0.765 × 10⁶ | 1.579 × 10⁶ | 13.7 × 10⁶ | 0.1150 | 0.1944 | 0.5210 |
| 3 | 3 | 7.5 | 0.22 | 0.814 × 10⁶ | 0.8255 | 0.672 × 10⁶ | 2.250 × 10⁶ | 12.8 × 10⁶ | 0.1758 | 0.1500 | 0.7400 |
| 4 | 3 | 7 | 0.22 | 0.814 × 10⁶ | 0.6497 | 0.526 × 10⁶ | 2.777 × 10⁶ | 13.7 × 10⁶ | 0.2020 | 0.0928 | 0.9100 |
| 5 | 3 | 7 | 0.22 | 0.814 × 10⁶ | 0.4477 | 0.364 × 10⁶ | 3.140 × 10⁶ | 13.7 × 10⁶ | 0.2290 | 0.0440 | 1.0310 |
| 6 | 3 | 9 | 0.22 | 0.814 × 10⁶ | 0.2180 | 0.177 × 10⁶ | 3.317 × 10⁶ | 10.7 × 10⁶ | 0.3085 | 0.0104 | 1.0850 |
| FLW | | | 10.0 | 37 × 10⁶ | –0.0905 | –3.317 × 10⁶ | 0 | ... | ... | 0.0810 | ... |
ΣI ε² = 0.7926
Automotive Industries
March 18, 1939
  
  


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