FCP Short Course. Ductile and Brittle Fracture. Stephen D. Downing. Mechanical Science and Engineering

Plastic Moment y s Elastic: M e ys c I ys bh 6 y s Fully Plastic: M dy y o ys 4 M M o e h 0 1.5 b ydy ys bh FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 4 of 4

Fully Plastic Moment F M o Beam can be loaded until M o is reached after which a plastic hinge is formed and the beam is free to rotate. FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 5 of 4

Collapse F right end support prevents collapse M = M o F M = M o F left end support prevents collapse collapse M = M o M = M o FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 7 of 4

Lower Bound Theorem If an equilibrium distribution of stress can be found which balances the applied load and is everywhere below or at yield, the structure will not collapse or just be at the point of collapse. Guaranteed load capacity where system will not have large deformations FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 8 of 4

Upper Bound Theorem The structure must collapse if there is any compatible pattern of plastic deformation for which the rate at which external work is equal to or greater than the rate of internal dissipation. Guaranteed load capacity where the system will have large deformations. FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 9 of 4

Application to a Cracked Plate Assume a stress distribution that satisfied equilibrium P LB y b z x a W y = ys x = 0 z = 0 P LB = ys ( W a ) b xy = 0 FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 10 of 4

Upper Bound Solution (continued) F s P UB du Slip plane area: A s W a b cos Force on shear plane: F s ys 3 W a b cos FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 1 of 4

Upper Bound Solution (continued) P UB Internal work: F s du duf du ys External work: s du 3 W a b cos du dw P du sin UB FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 13 of 4

Upper Bound Solution (continued) Upper Bound Theorem: du = dw P UB ys W a b du PUB dusin 3 cos ys W ab ys W ab 3 sin cos 3 sin Lowest upper bound for sin = 1 or = 45 P UB = 1.15 ys ( W a ) b P LB = ys ( W a ) b FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 14 of 4

Plane Stress Plane Strain thick plate plane strain z = 0 z = 0 thin plate plane stress z = 0 z = 0 FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 16 of 4

Transverse Strains Longitudinal Tensile Strain Transverse Compression Strain 0 0.00 0.004 0.006 0.008 0.010 0.01 Thickness 0.001 0.00 0.003 40 mm 0 mm 10 mm = 0.5 0.004 0.005 = 0.3 5 mm y z x 5 D 100 FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 18 of 4

Notch Stresses y z x t x z x z 7 0.01-0.005 63.5 0 15 0.01-0.003 70.6 14.1 30 0.01-0.00 73.0 1.8 50 0.01-0.001 75.1 9.3 FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 19 of 4

Flow Stress Ultimate strength Stress (MPa) Yield strength Strain Flow Stress: flow ys uts k flow 3 FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 3 of 4

Edge Notch Plane Strain From the punch problem: n = k P =.96 flow By analogy: n =.96 flow.96 flow 1.81 flow.38 flow FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 6 of 4

What does all this tell us? A B C D Lower Bound Theorem: P A = P B = P C = P D Upper Bound Theorem: P A = P B = P C < P D Slip Line Theory: P A = P B = P C < P D No effect of stress concentration on static strength! Some stress concentrations can increase the strength! FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 9 of 4

Failure Loads Edge Center Plane Stress P A net 1.15 flow P A net flow Plane Strain P A net.96 flow P A net 1.15 flow FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 30 of 4

Conclusion Net section area, state of stress and material strength control the failure load in a structure. FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 31 of 4

Stress or Strain Control? elastic material plastic zone Elastic material surrounding the plastic zone forces the displacements to be compatible, I.e. no gaps form in the structure. Boundary conditions acting on the plastic zone boundary are displacements. Strains are the first derivative of displacement FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 3 of 4

Define K and K S, e, Define: nominal stress, S nominal strain, e notch stress, notch strain, Stress concentration Strain concentration K K S e FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 33 of 4

Stress and Strain Concentration Stress/Strain Concentration First yielding K K T K T K 1 Nominal Stress FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 35 of 4

1018 Stress-Strain Curve 500 400 300 00 100 true fracture strain, f = 0.86 true fracture stress, f = 770 MPa 0 0 0.1 0. 0.3 0.4 0.5 Strain FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 38 of 4

7075-T6 Stress-Strain Curve 700 600 500 400 300 00 true fracture strain, f = 0.3 true fracture stress, f = 70 MPa 100 0 0 0.05 0.1 0.15 0. Strain FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 39 of 4

1018 Steel Test Data 100 80 load, kn 60 40 edge diamond slot hole 0 0 0 4 6 8 10 displacement, mm FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 41 of 4

Bridgeman Analysis (1943) ys z ys r z r a r a r Elastic stress distribution Plastic stress distribution z r cons tan t FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 44 of 4

FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 45 of 4 Stresses a r a a ln 1 o z a 0 z z dr r P a 1 ln a 1 a P flow max P max = A net flow CF CF constraint factor

Effect of Constraint Stress, z Increasing constraint Strain, z Higher strength and lower ductility FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 47 of 4

Failures from Stress Concentrations Net section stresses must be below the flow stress Notch strains must be below the fracture strain FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 48 of 4

Typical Stress Concentration K T ductile brittle K T brittle ductile Stress distribution Strain distribution A ductile material has the capacity for very large strains and it reaches the strength limit first A brittle material reaches the strain limit first FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 49 of 4

Strain Distribution Nominal strain will reach the yield strain when the strength limit is reached Strain distribution K T K T yield yield ductile transition K T f K yield T brittle yield f K f T yield K T yield FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 50 of 4

Conclusion Net section area, state of stress and material strength control the failure load in a structure only in ductile materials. In brittle materials, cracks will form before the maximum load capacity of the structure is reached. FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 54 of 4

Brittle Fracture Failure of structures by the rapid growth of cracks (or crack-like defects) until loss of structural integrity. FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 56 of 4

Stress Concentration applied Inglis Equation local a local 1 applied K t applied Crack dimensions a = 10-3 = 10-9 local Does this make sense? applied a Kt 000 FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 58 of 4

Griffith Approach l applied a Uncracked plate elastic energy 1 Ue w*l*t w*l*t E Introduce a crack which Reduces elastic energy by: a Ue a *t E Increases total surface energy by: w at s applied thickness, t For a crack to grow, the energy provided by new surfaces must equal that lost by elastic relief a U e FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 59 of 4

Griffith Approach applied Minimum criterion for stable crack growth: Strain energy goes into surface energy a U l a a a a a E t a s a 4 a t w a E s applied thickness, t s E a FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 60 of 4

Plastic Energy Term l a a w applied Previous derivation is for purely elastic material Most metals and polymers experience some plastic deformation Strain energy (U) goes into surface energy () & plastic energy (P) U P a a a Orowan introduced plastic deformation energy, γ p E ( s a p ) p s applied thickness, t Materials which exhibit plastic deformation absorb much more energy, removing it from the crack tip FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 61 of 4

Energy Release Rate, G l a a w applied applied thickness, t Irwin chose to define a term, G, that characterizes the energy per unit crack area required to extend the crack: G G 1 U t a a E Comparing to the previous expression, we see that: G s p Works well when plastic zone is small ( fraction of crack dimensions ) EG a Change in crack area FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 6 of 4

GExperimentally Measuring G a F F x x t Load ICa Ea U a + a Displacement atfracture Load cracked sample in elastic range Fix grips at given displacement Allow crack to grow length a Unload sample Compare energy under the curves G is the energy per unit crack area needed to extend a crack Experimentally measure combinations of stress and crack size at fracture to determine G IC FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 63 of 4

Stress Based Crack Analysis Westergaard, Irwin analyzed fracture of cracked components using elastic-based stress theory Three modes for crack loading Mode I opening Mode II in-plane shear Mode III out-of-plane shear From: Socie FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 64 of 4

Stress Intensity Factor, K af astress intensity factor flaw size operating stresses b Kspecimen geometry Critical parameter is now based on: Stress Flaw Size Specimen geometry included using correction factor crack shape specimen size and shape type of loading (i.e. tensile, bending, etc) FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 65 of 4

Stress Intensity Factor, K Stress Fields (near crack tip) x K 3 cos 1 sin sin r y K 3 cos 1 sin sin r xy K 3 cos sin cos r Note: as r 0, stress fields Plane stress - too thin to support stress through thickness z 0 z 0 Plane strain - so thick that constrains strain through thickness z 0 x y z FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 66 of 4

FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 67 of 4 Stress Intensity Factor, K Displacement Fields sin 1 1 3 cos r G K u cos 1 1 3 sin r G K v x u x x v y x v y u xy Displacement in x-direction Displacement in y-direction

Linear Elastic Fracture Mechanics How can we use an elastic concept (K) when there is plastic deformation? Small scale yielding Blunts the crack tip At elastic-plastic boundary, y r r* K 3 1 sin sin y cos r* Note: zone of yielding will vary with plastic zone r p Consider stress at = 0 K y r* FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 69 of 4

Size Requirements t,w a,a.5 K ys 50r ys K Ic t, mm 04- T3 345 44 40.7 7075 -T6 495 5 6.4 Ti-6Al-4V 910 105 33.3 Ti-6Al-4V 1035 55 7.1 4340 860 99 33.1 4340 1510 60 3.9 17-7 PH 1435 77 7. 5100 070 14 0.1 p FCP Fall 015 015 Stephen Downing, University of Illinois at Urbana-Champaign, All Rights Reserved 77 of 4