Drive Shaft Failure of Z-Drive Propulsion System Suzanne Higgins
Background Bunkering vessel MV Southern Valour Commissioned in 2008 in Singapore Operating in Cape Town harbour since August 2008
Z-Drive System Bunker barge powered by z-type propulsion system Power transmitted to bevel gears of above-water gear box Horizontal gears redirect power vertically to power transmission shaft Drives bevel gear of underwater gear box Power redirected to horizontal propeller shaft
Z-Drive System Underwater gear box attached to steering tube Worm wheel mounted on steering tube Worm shaft engages with worm wheel Steering motor attached to worm-geared steering drive Two propellers
Failure Catastrophic failure of drive shaft in splined area 2400 hours of service One Eighty appointed to determine cause of failure
Materials Testing Need three tests to fully classify steel Spectrographic analysis: Low alloy carbon steel AISI 4340 Microstructure: Quench and tempered; no anomalies. Hardness: HV 300 Material properties used to determine ultimate tensile strength for fatigue calculations
Fracture surfaces Original fracture surface Characterised by beach marks Indicative of fatigue failure Angle of fracture surface indicates failure due to bending stress Fractured further on cutting
Fracture surfaces Fracture surface characterised by beach marks Indicative of fatigue failure Angle of fracture surface indicates failure due to tangential stress Crack along root of spline Width of spline inconsistent
Fracture surfaces Fracture of remaining piece of shaft on cutting Fatigue failure due to torsional stress
Fracture surfaces Single area of initiation A: Bending stress Normal operation B: Tangential stress Under normal operating conditions, spline acts as a uniform shaft. In this case, due to inconsistent spline width, it acts as a gear i.e. The force of the mating part worked to open up the spline. Only occurs due to deformation C: Torsional stress Normal operation
Fatigue Fatigue is slow and insidious propagation of cracks under cyclic loading. Two requirements: stress and cycles. Low cycle fatigue typically 10 3/4 cycles. High cycle fatigue typically >10 6 cycles Shaft rotates and each rotation represents one stress cycle. Number of cycles of drive shaft calculated as: Cycles = Input rpm x Gear ratio x (Service hours)/60 = 1.88 x 10 7 cycles Category of high cycle fatigue
Fatigue Endurance limit is the stress limit below which fatigue will not occur for infinitely may cycles.
Stress limits For high cycle fatigue, endurance limit is 20-50% of UTS Material properties used to find UTS Endurance limit further modified to take into account various factors: Surface factor Size factor Stress concentration factor Temperature factor
Stress limits Operating Condition Ultimate Tensile Strength Endurance limit Endurance limit - modified Failure stress 1000 MPa 200 500 MPa 54 134 MPa
Bending moment calcs Need to determine bending moment at point of failure. Two known forces calculated from torque using power and speed data. Distances a,b,c,d measured on site. Three unknown reaction forces at bearings
Bending moment calcs Two equations Sum of forces: F1+F2=R1+R2+R3 Moment equilibrium: F2a+R2b+R3(b+c)-F1(b+c+d)=0 Statically indeterminate system of degree 1 Method of superposition The combined effect of the loads is the sum of the loads acting separately. Compatibility equation Sum of deflections is zero at reaction force Designate one of reactions as redundant force (R2)
= y R2 + y
Bending moment calcs Step 1: Choose R2 as redundant force and remove from system Two unknowns and two equations Determinate system Solve for deflection at R2 position due to applied loading (y)
Bending moment calcs Step 2: Remove applied forces F1 and F2 Solve for deflection at R2 position due to redundant force (y ) Two equations, three unknowns Deflection expression in terms of R2 R2
Bending moment calcs Step 3: Deflection at R2 is sum of the deflection due to applied loading and deflection due to redundant force. Deflection at R2 = 0 Use compatibility equation to solve for R2: y+y = 0
Bending moment calcs Step 4: Original bending moment diagram Only two unknowns R1, R3 Determinate system Resolve all forces Step 5: Calculate bending moment at point of failure
Deflection calculations M/I = σ/y Bending moment at point of failure known. Second moment of Inertia calculated from spline diameter: I = π(d 4 o d 4) i )/64 Know minimum stress required for fatigue failure (modified endurance limit) Can either substitute maximum deflection values to determine if stress levels are high enough for failure. Do not know maximum deflection value.
Deflection calculations Calculate deflection required for failure at each stress level and determine if it is reasonable. Stress type (Mpa) Deflection required (cm) Yield stress 13.63 Endurance limit 2.90 Modified endurance limit 0.45 Damaged spline leads to fatigue failure.
Visual inspection No signs excessive wear on bearings or other parts
Visual inspection Large dent of Kort nozzle (houses propeller)
Visual inspection Using typical material properties, calculated that force in order of 20kN is required to cause deformation. Given the vessel s tonnage and speed, feasible that this could be a result of a collision during operation.
Possible Sources of Fatigue Fatigue source Consequence Evidence found Cavitation Vibration on the drive assembly No pitting on propeller surface Misalignment Bending moment causing wear on bearing High stress event Localised damage on shaft creating initiation sites for fatigue under normal operating conditions No excessive wear on bearings Damage on Kort nozzle
Conclusions Shaft has multiple fracture surfaces characterised by beach marks indicating failure by fatigue. Single initiation site indicating high stress event, such as knock to the Kort nozzle. High stress event resulted in deformation of spline, causing it to act as a gear. Tangential force acts to open up spline. Likely that this occurred first. Opening of this crack increased stress concentrations.
Conclusions Fatigue cracks initiated under bending and torsional stress, present under normal operating conditions. As cracks propagated, load transferred to intact areas, increasing stress. Stress becomes greater than the UTS of the material. Remaining areas fail by fast fracture.
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