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 crankshaft vibration for four-cylinder and six-cylinder diesel engines, showing stresses at different revolution speeds.
| Identifier | ExFiles\Box 132\1\ scan0120 | |
| Date | 25th March 1939 | |
| 404 DIESEL CRANKSHAFT VIBRATION continuous service, the diameter of the shaft must be increased, or else the moment of inertia of the rotating masses (reciprocating weights) diminished. Col. 22 of Table III represents the amplitude Amax. = A × γ in degrees, at the free end of the crankshaft—furthest from the flywheel. Within the speed range of the engine, up to 2600 r.p.m., the maximum amplitude Amax. = 0.94 deg. for the 9 order critical and Amax. = 0.85 deg. for the 7½ order critical, and both of these correspond to crankshaft stresses within the safe limit. Values of Amax. are plotted in Fig. 16. Four-Cylinder Engine The equivalent length of the crank unit Fig. 8 will be the same as in the six-cylinder engine. The equivalent length of the whole shaft is L = 2 × 7 + (1.5 or 2)d = 19 in. The moment of inertia per cylinder is also the same, namely, 0.22 lb.-in.-sec.² A flywheel of larger size must be used than in the six-cylinder engine, and while the outside and inside diameters of the rim remain the same, 15 and 10 in., respectively, the weight will be increased to 108 lb. The square of the radius of gyration remains the same, 40.6. The moment of inertia of the flywheel is I₅ = (W × m²) / 386 = (108 × 40.6) / 386 = 11.5 lb-in-sec.² The torsional rigidities of the shaft sections are as follows (Equation 3, Fig. 18) : M₁ = M₃ = GJ / l₁-₃ = (12 × 10⁶ × 8) / 7 = 13.7 × 10⁶ lb-in. per radian. M₂ = (12 × 10⁶ × 8) / 7.5 = 12.8 × 10⁶ lb-in. per radian. M₄ = (12 × 10⁶ × 8) / 8 = 12 × 10⁶ lb-in per radian. M₅ = (12 × 10⁶ × 8) / 19 = 5.05 × 10⁶ lb-in. per radian. (18) The natural frequency of an equivalent two-mass system with one node (Fig. 17, Equation 2) is f = 9.55 √(M₅(I + I₅) / II₅) or f = 9.55 √( (5.05 × 10⁶ (0.88 + 11.5)) / (0.88 × 11.5) ) = 23600 cycles per minute --- GRAPH (Figure 16): Y-Axis: VIBRATION STRESS - LB. PER SQ. IN. X-Axis: REVOLUTIONS PER MIN. Annotations: - ω² = 3.7 × 10⁶ RAD.²-SEC.⁻² - f = 18,440 CYCLES PER MIN. - 6 CYL. ENGINE - I.M.E.P = 100 LB. PER SQ. IN. - MAX. TENSILE STRESS OF STEEL 120,000 LB. PER SQ. IN. - SPEED RANGE FOR ENGINE - ENDURANCE LIMIT - REVERSE BENDING & TORSION - SAFE STRESSES - FIRING ORDER 1-4-2-6-3-5 - Harmonic Orders on Peaks: 12, 11½, 10½, 10, 9½, 9, 8½, 8, 7½, 7, 6½, 6, 5½, 5, 4½, 4, 3½ - Amplitude Labels: 0.5°, 1.0°, 1.5°, 2.0° --- The phase velocity is given by Equation (5): ω² = M₅(I + I₅) / II₅ = (5.05 × 10⁶(0.88 + 11.5)) / (0.88 × 11.5) = 6,252,000 rad²-sec.⁻² ω² = 6,252,000 rad²-sec.⁻² This is the approximate value for first-node vibration, which we will use in calculating the frequency of such vibration in the four-cylinder engine. After first trying ω² = 6,000,000 rad²-sec.⁻² and then ω² = 7,840,000 rad²-sec.⁻², we find by interpolation that ω² = 7,500,000 rad²-sec.⁻² will make the last torque equal to zero. (Table IV, col. 8.) deg. of deflection at cylinder No. 1, as calculated by means of Equation (13), are given in col. 12 of Table IV and are graphically represented in Fig. 19. The natural frequency of two-node vibration, f₁ = 73,000 cycles per min., and the lowest order harmonic that can come into resonance is 73,000/3000 = 24. This order is of no practical importance. However, if there is any possibility of resonance with orders of less than the twelfth, the stresses at resonance for two-node vibration should be investigated. The amplitude of vibration of the crank unit farthest from the flywheel is given by Equation (11) as follows: A = (5230 U Σe) / (ΣIe² f²) or A = (5230 × 11 × 2.25 × PΣe) / (0.5419 × 683717904) = 0.00035 PΣe deg. Amax. = Aγ (19) Values for A and Amax. for the different harmonic orders are given in Table VI, cols. 18 and 22. Inserting the values of Σe from Table VII in Table VI in the same way as was done in the case of Table III for the six-cylinder engine, we obtain the torsional vibration stresses at resonance speed with hysteresis damping for the firing order 1-2-4-3. The stresses, in lb. per sq. in., are plotted in Fig. 20. Col. 22 of Table VI represents the amplitude Amax. = Aγ in deg. at the free end of the crankshaft farthest from the flywheel. Within the speed range of the engine, up to 2600 r.p.m., the amplitude Amax. = 0.46 --- March 25, 1939 Automotive Industries | ||