MECH3300 : Fracture Mechanics - lecture 6

Dr Rowan Truss(Bldg 45-rm 215, R.Truss@uq.edu.au)

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

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).

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

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

M. F. Ashby, Materials Selection in Mechanical Design, Butterworth Hinemann, Oxford, 1999

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

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

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

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

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

dimensionsFuselage

3.7m diameter33m longThickness of Al sheet 0.91mmPressurised to 0.057MPa

What size crack would cause failure?

Estimate K1c

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

Plane strain?Require t ≥ 2.5 (K1c/σy)2

= 2.5 (30/350)2

= 18mm

Not plane strain, need plane stress analysis

Plane stress Kc depends on thicknessassume same relationship as known Al alloy 7075-T6

Critical flaw size – plane stressKc = 2.6 x plane strain value

= ~x80 MPam1/2

For plane stressac = 1/π (K1c/σy)2

= ~150mm

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