Applied Mechanics and Structures II: Vibrations and Machine Design

The second chapter of NM Section 2 takes its other two subjects. Vibrations: free and forced vibration of damped and undamped systems, with one and with several degrees of freedom — the natural frequency, the damping ratio and logarithmic decrement, resonance and the magnification factor, transmissibility and isolation, two-degree-of-freedom modes and the vibration absorber, and the whirling of shafts. Machine Design: design for static and for dynamic (fatigue) loading; the design of shafts, gears, and rolling and sliding contact bearings; and the joining techniques of bolting, riveting and welding. They are one chapter because they share a question: what a component does under a load that changes, whether the change is a vibration it must not resonate with or a cycle it must survive a hundred million times. The same ideas return in the engine-dynamics part of Section 5, where the shaft is a propeller shaft.

1. Free vibration, undamped and damped

A mass m on a spring of stiffness k, displaced and released, oscillates at the natural frequency ωₙ = √(k/m) rad/s, fₙ = ωₙ/(2π) Hz. Equivalently ωₙ = √(g/δ_st), where δ_st is the static deflection under the weight. Springs in parallel add, k = k₁ + k₂; in series their flexibilities add, 1/k = 1/k₁ + 1/k₂. The energy (Rayleigh) method equates the maximum kinetic and potential energies and gives ωₙ directly for systems that are awkward to draw.

With a viscous damper c, the damping ratio ζ = c/c_c = c/(2√(km)). If ζ < 1 the system is underdamped and oscillates at the damped frequency ω_d = ωₙ√(1 − ζ²) with an exponentially decaying amplitude; ζ = 1 is critically damped, the fastest return without oscillation; ζ > 1 is overdamped. The logarithmic decrement δ = ln(xₙ/xₙ₊₁) = 2πζ/√(1 − ζ²) measures damping from two successive peaks, so ζ = δ/√(4π² + δ²), close to δ/(2π) for light damping. Amplitudes 10 mm and 8 mm give δ = ln 1.25 = 0.223 and ζ = 0.0355.

2. Forced vibration, resonance, transmissibility and several degrees of freedom

Under a harmonic force F₀ sin ωt, the steady-state amplitude is X = (F₀/k)·MF with the magnification factor MF = 1/√[(1 − r²)² + (2ζr)²], r = ω/ωₙ. At resonance (r = 1) MF = 1/(2ζ), unbounded without damping, and the response lags the force by 90°. A rotating unbalance m₀e gives a force m₀eω², so its amplitude grows with speed and tends to m₀e/m well above resonance.

The transmissibility — force transmitted to the foundation over force applied — is TR = √[1 + (2ζr)²]/√[(1 − r²)² + (2ζr)²]. It equals 1 at r = √2 whatever the damping, so isolation (TR < 1) requires r > √2: the mount must be soft enough that the natural frequency is well below the forcing frequency. Above √2, damping increases the transmitted force, so isolators are lightly damped, with just enough damping to pass through resonance on start-up. Undamped at r = 2, TR = 1/(r² − 1) = 1/3.

A system with n degrees of freedom has n natural frequencies, found from det(K − ω²M) = 0, and n mode shapes, the fixed amplitude ratios in which it can vibrate at each. Two masses joined by springs have an in-phase lower mode and an out-of-phase higher mode. The dynamic vibration absorber adds a small mass–spring tuned to the forcing frequency, √(k₂/m₂) = ω, which brings the main mass to rest at that frequency at the cost of two new resonances either side. A rotating shaft whirls at its critical speed ω_c = √(k/m) = √(g/δ), the natural frequency of its lateral vibration; a shaft whose static deflection at the disc is 1 mm has ω_c = √(9.81/0.001) = 99 rad/s, about 946 rpm.

⚠️ Damping helps at resonance and hurts in the isolation range
Adding damping always lowers the resonant peak of MF and TR, but beyond r = √2 it raises TR. A question asking how to reduce the force reaching a ship’s structure from an engine running well above the mount’s natural frequency wants "softer mounts", not "more damping".

3. Design for static and dynamic loading

Under static loading, a ductile part is designed against yield with a factor of safety N = S_y/σ (using a failure theory for combined stresses) and a brittle part against the ultimate strength; stress concentrations matter little for a ductile part under static load, because local yielding redistributes the stress, but they matter for brittle parts and for fatigue. Under dynamic (fluctuating) loading, a part fails by fatigue at stresses well below yield. The S–N curve of a steel flattens at the endurance limit Sₑ, below which life is effectively infinite; for many steels the specimen endurance limit is roughly half the ultimate strength, and it is reduced for the real part by surface finish, size, reliability, temperature and load-type (Marin) factors, and by the fatigue stress-concentration factor K_f = 1 + q(K_t − 1), where q is the notch sensitivity.

A fluctuating stress has a mean σₘ and an alternating amplitude σₐ. The Goodman line, σₐ/Sₑ + σₘ/S_ut = 1/N, joins the endurance limit to the ultimate strength; the more conservative Soderberg line, σₐ/Sₑ + σₘ/S_y = 1/N, joins it to the yield strength; the Gerber parabola lies outside both. With σₐ = σₘ = 100 MPa, Sₑ = 300 MPa, S_ut = 600 MPa and S_y = 400 MPa, Goodman gives 1/N = 0.333 + 0.167 = 0.5, N = 2, and Soderberg gives 1/N = 0.333 + 0.25, N = 1.71. Miner’s rule Σnᵢ/Nᵢ = 1 sums the damage of blocks at different stress levels.

4. Shafts, gears and bearings

A shaft under combined bending M and torque T is sized by the equivalent torque Tₑ = √(M² + T²) (maximum shear stress theory, τ = 16Tₑ/(πd³)) and the equivalent bending moment Mₑ = ½[M + √(M² + T²)] (maximum principal stress theory, σ = 32Mₑ/(πd³)); with M = 3 kN·m and T = 4 kN·m, Tₑ = 5 kN·m and Mₑ = 4 kN·m. Shock and fatigue factors multiply M and T in the design codes, and the shaft must also be checked for deflection and critical speed. Power and torque are related by P = 2πNT/60.

Gears transmit a constant velocity ratio when they satisfy the law of gearing — the common normal at the point of contact passes through the fixed pitch point — which the involute profile does for any centre distance. Pitch diameter d = mz (module m, teeth z), centre distance (d₁ + d₂)/2, velocity ratio N₁/N₂ = z₂/z₁; the Lewis equation F_t = σ_b b m Y sizes the tooth as a cantilever in bending, and the Buckingham and wear checks cover dynamic load and surface fatigue. Spur, helical (smoother, with an axial thrust), double-helical, bevel and worm gears serve parallel, intersecting and crossed shafts; a marine reduction gearbox between a medium-speed engine and the propeller is double-helical.

Rolling-contact bearings are rated by the basic dynamic load rating C, the load giving a life of one million revolutions with 90% reliability; the rating life is L₁₀ = (C/P)ᵏ million revolutions, with k = 3 for ball and 10/3 for roller bearings, P being the equivalent load. Sliding-contact (journal) bearings run on a hydrodynamic oil film built by the shaft’s rotation; Petroff’s equation gives the friction of a lightly loaded bearing, and the Sommerfeld number S = (r/c)²(μN/p) collects the design variables. The propeller shaft’s stern-tube and line-shaft bearings are sliding bearings; the thrust block carries the propeller thrust through tilting pads.

5. Joining techniques: bolting, riveting and welding

A bolted joint is tightened to a preload so that the members stay in contact under the working load; the bolt then carries only a fraction of the external load, set by the relative stiffness of bolt and members, and fatigue life improves. Stress is taken on the tensile stress area of the thread, and a bolt in shear is checked on its shank area. A riveted joint (lap or butt, single or multiple rows) can fail by shearing of the rivets, tearing of the plate between rivets, and crushing (bearing) of the plate or rivet; its efficiency is the least of these strengths divided by the strength of the solid plate — for plate tearing alone, (p − d)/p, so a 50 mm pitch with 20 mm holes gives 60%.

Welding is the joining technique of modern shipbuilding. A butt weld with full penetration is as strong as the plate; a fillet weld of leg size s fails in shear across its throat t = s cos 45° = 0.707s, so a parallel fillet of length L carries P = 0.707 s L τ. A 10 mm fillet 200 mm long at 100 MPa permissible shear carries 141.4 kN. Welds bring residual stresses, distortion, heat-affected zones and the risk of cracking, controlled by procedure qualification, preheating and inspection.

⚠️ The throat, not the leg
Fillet-weld strength uses the throat 0.707s, not the leg s. Using the leg over-predicts the capacity by 41%.

Key takeaways

  • ωₙ = √(k/m) = √(g/δ); ζ = c/(2√(km)); ω_d = ωₙ√(1 − ζ²); ζ = δ/√(4π² + δ²) from the log decrement.
  • MF = 1/(2ζ) at resonance; isolation needs r > √2, where damping raises TR; undamped TR = 1/(r² − 1).
  • Goodman σₐ/Sₑ + σₘ/S_ut = 1/N; Soderberg uses S_y and is more conservative; Miner Σn/N = 1.
  • Shafts: Tₑ = √(M² + T²), Mₑ = ½(M + Tₑ); gears d = mz; bearings L₁₀ = (C/P)³ million revolutions for ball bearings.
  • Fillet welds fail across the throat 0.707s; riveted-joint efficiency is the weakest mode over the solid plate.

Practice questions (14)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A 10 kg mass is supported on a spring of stiffness 4000 N/m. Its natural frequency (in Hz), to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 3.18

    ωₙ = √(k/m) = √400 = 20 rad/s and fₙ = 20/(2π) = 3.18 Hz. Reporting 20 answers in rad/s rather than Hz; √(m/k) inverted gives 0.05.
  2. Two springs of stiffness 2000 N/m and 3000 N/m are connected in series. The equivalent stiffness (in N/m) is

    1. 1200
    2. 5000
    3. 2500
    4. 6000
    Show answer

    Answer: A — 1200

    In series the flexibilities add: 1/k = 1/2000 + 1/3000 = 5/6000, so k = 1200 N/m, less than either spring. 5000 is the parallel sum; 2500 is the arithmetic mean.
  3. In a free-vibration test, two successive peak amplitudes are 10 mm and 8 mm. The damping ratio, to four decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.0355

    δ = ln(10/8) = 0.2231; ζ = δ/√(4π² + δ²) = 0.2231/6.287 = 0.03549 ≈ 0.0355. The light-damping shortcut δ/(2π) = 0.03551 rounds to the same 0.0355. Using the ratio 8/10 gives a negative decrement.
  4. An undamped machine on its mounts runs at twice the natural frequency of the mounting. The transmissibility, to three decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.333

    Undamped TR = 1/|1 − r²| = 1/(4 − 1) = 0.333; since r = 2 > √2 the mount isolates, passing a third of the force. Writing 1/(1 + r²) = 0.2 uses the wrong sign; the magnification factor has the same undamped value but describes displacement, not force.
  5. Which statements about forced vibration are correct?

    1. transmissibility is 1 at a frequency ratio of √2 for every damping ratio
    2. at resonance the displacement lags the force by 90°
    3. above a frequency ratio of √2, adding damping reduces the transmitted force
    4. a tuned absorber brings the main mass to rest at the forcing frequency
    Show answer

    Answer: A — transmissibility is 1 at a frequency ratio of √2 for every damping ratio; B — at resonance the displacement lags the force by 90°; D — a tuned absorber brings the main mass to rest at the forcing frequency

    All TR curves cross at r = √2, TR = 1; at r = 1 the phase is 90° whatever ζ; an absorber tuned to ω makes the main mass stationary at that ω. Above √2 damping INCREASES TR, which is why isolators are lightly damped.
  6. A shaft carrying a disc has a static deflection of 1 mm at the disc. Taking g = 9.81 m/s², its critical (whirling) speed in rpm, to the nearest integer, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 946

    ω_c = √(g/δ) = √(9.81/0.001) = 99.05 rad/s; N = 60ω/(2π) = 945.8 ≈ 946 rpm. Stopping at 99 gives rad/s; using δ in mm as 1 gives 3.13 rad/s.
  7. A steel part has endurance limit 300 MPa (corrected) and ultimate strength 600 MPa, and carries a mean stress of 100 MPa with an alternating stress of 100 MPa. By the Goodman criterion the factor of safety is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 2

    1/N = σₐ/Sₑ + σₘ/S_ut = 100/300 + 100/600 = 0.333 + 0.167 = 0.5, so N = 2. Soderberg with a yield strength of 400 MPa would give 1/N = 0.583 and N = 1.71; ignoring the mean stress gives N = 3.
  8. For the same fluctuating stress, which criterion gives the smallest factor of safety?

    1. Soderberg
    2. Goodman
    3. Gerber
    4. all three give the same value
    Show answer

    Answer: A — Soderberg

    Soderberg joins the endurance limit to the yield strength, which is below the ultimate strength that Goodman uses, so its safe region is smallest and its factor of safety lowest; the Gerber parabola lies outside the Goodman line and is the least conservative.
  9. A shaft section carries a bending moment of 3 kN·m and a torque of 4 kN·m. The equivalent torque by the maximum shear stress theory (in kN·m) is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 5

    Tₑ = √(M² + T²) = √(9 + 16) = 5 kN·m. The equivalent bending moment ½(M + Tₑ) = 4 kN·m belongs to the maximum principal stress theory; adding M and T gives 7.
  10. A ball bearing has a basic dynamic load rating of 30 kN and carries an equivalent load of 5 kN. Its rating life (in millions of revolutions) is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 216

    L₁₀ = (C/P)³ = 6³ = 216 million revolutions. The roller-bearing exponent 10/3 would give 392; forgetting the cube gives 6.
  11. A pinion of 20 teeth meshes with a gear of 60 teeth, both of module 4 mm. The centre distance (in mm) is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 160

    Pitch diameters are 4 × 20 = 80 mm and 4 × 60 = 240 mm, so the centre distance is (80 + 240)/2 = 160 mm. Adding the diameters without halving gives 320 mm.
  12. A parallel fillet weld of leg size 10 mm and length 200 mm has a permissible shear stress of 100 MPa. The load it can carry (in kN), to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 141.4

    Throat t = 0.707 × 10 = 7.07 mm, area 7.07 × 200 = 1414 mm², load 1414 × 100 = 141 400 N = 141.4 kN. Using the leg instead of the throat gives 200 kN.
  13. A riveted joint has a pitch of 50 mm and rivet holes of 20 mm diameter. Its efficiency against tearing of the plate between rivets (in per cent) is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 60

    Tearing efficiency = (p − d)/p = (50 − 20)/50 = 0.6 = 60%. d/p = 40% is the fraction of plate removed; the joint’s overall efficiency would be the least of the tearing, shearing and crushing values.
  14. Which statements about bolted and welded joints are correct?

    1. a preloaded bolt carries only part of an external tensile load while the joint stays closed
    2. a fillet weld is designed on its throat area
    3. a full-penetration butt weld is weaker than the plate by the factor 0.707
    4. welding leaves residual stresses in the joint
    Show answer

    Answer: A — a preloaded bolt carries only part of an external tensile load while the joint stays closed; B — a fillet weld is designed on its throat area; D — welding leaves residual stresses in the joint

    Preload makes the external load share between bolt and members; fillet welds fail across the throat; welding shrinkage leaves residual stresses. The 0.707 factor belongs to the fillet throat — a sound full-penetration butt weld develops the full plate strength.