Ppt Lecture 6
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Transcript of Ppt Lecture 6
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ObjectivesStudents should understand:
Design and Materials Selection issues:Use of basic fracture mechanics equationYield before fractureLeak before break
Statistical nature of brittle fracture Wiebull modulus
Case study – Comet aircraft crashes
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Basic fracture mechanics equationK1c = σ Y √πa
combines three quantities, namely;a material property, K1c
the applied tensile stress or design stress, σan allowable flaw size, a
( Note: Y may also be a function of a).
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Yield before brittle fractureK1c = σ Y √πa
The stress required for a crack of size ac to propagate is σ = K1c / (Y √πac)
We want brittle failure stress < yield stress, σy , so in the limit set σ = σy
Rearranging π ac = (1/Y2) (K1c /σy)2
Safest vessel is when allowable flaw size is largest somaximise (K1c /σy)2
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Leak before breakK1c = σ Y √πa
For leak before break the critical flaw size needs to be ≥ t, the thickness of the vessel
t = (1/ π Y2) (K1c /σy)2
Thickness is also set so that vessel will not yield for spherical vessel, radius, r, σ = pr/2t
p = (2/ πr Y2) (K1c /σy)2
Maximise K1c2/σy for highest failure pressure
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M. F. Ashby, Materials Selection in Mechanical Design, Butterworth Hinemann, Oxford, 1999
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Statistics of brittle failureK1c = σ Y √πa
K1c is a material parameter ie constantBrittle failure stress thus depends on flaws sizeReal materials have a distribution of flaw sizesBrittle failure stress also has distribution Measure of breadth of distribution given by Wiebull modulus
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Wiebull Modulusfraction of samples of volume V0
that do not fracture under a tensile load σ
Ps(V0) = exp[-(σ/σ0)m]
σ0 is stress at which survival probability is 1/e
m is Wiebull Modulus
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Comet aircraft failuresRef: D. R. H. Jones, Engineering Materials 3 Materials Failure Analysis Case Studies and Design Implications, Pergamon Press Oxford, 1993 ch13
One of first pressurised cabin commercial aircraftBuilt by de Havillands, UK for BOAC in 19524x turbo-jet enginesCruising altitude of 35,000 ftAll-up weight of 49 tonnes
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Comet aircraft failures2nd May 1953 Plane crashed in tropical thunderstorm in Calcutta – structural failure of the aircraft10 January, 1954 Rome- London flight crashed in good weather – killed 29 passengers and 6 crew8 April, 1954 Rome-Cairo flight crashed again in reasonable weather
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Material dataAluminium alloy DTD546Obsolete – modern equivalent 2000 series alloys
3.5-4.8 wt% Cu< 1.0 wt% Fe<1.5 wt% Si<0.6 wt% Mg< 1.2 wt% Mn< 0.3 wt% TiBalance Al
min σy 325 MPaσuts 418 MPa, actual for failed material 450MPa
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dimensionsFuselage
3.7m diameter33m longThickness of Al sheet 0.91mmPressurised to 0.057MPa
What size crack would cause failure?
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Estimate K1c
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Critical flaw sizeFor pressure, p, in cylindrical pressure vesselHoop stress σ = pr/t
= 0.057*1.85/(0.91x10-3)= 116MPa
For plane strain, assume K1c = 30 MPam1/2
ac = 1/π (K1c/σy)2 = 21mm
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Plane strain?Require t ≥ 2.5 (K1c/σy)2
= 2.5 (30/350)2
= 18mm
Not plane strain, need plane stress analysis
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Plane stress Kc depends on thicknessassume same relationship as known Al alloy 7075-T6
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Critical flaw size – plane stressKc = 2.6 x plane strain value
= ~x80 MPam1/2
For plane stressac = 1/π (K1c/σy)2
= ~150mm
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Validity of using LEFM in plane stressNeed crack size , a ≥ 50 rp , plastic zone size
Plastic zone size in plane stressrp = 1/2p (K1c/σy)2 = 8.3 mm
So probably not valid But error low for crack in infinite sheet
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