ME111 Lecture ME111 Instructor: Peter Pinsky Class #16 November 1, 2000


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1 Toda s Topics ME111 Instructor: Peter Pinsk Class #16 November 1, 2000 Introduction to linear elastic fracture mechanics (LEFM). Problem 3 A centercracked plate of AII 1144 steel ( C = 115MPa m ) has dimensions b = 40 mm, t = 15 mm and h = 20 mm. For a factor of safet of three against crack growth, what is the maximum permissible load on the plate if the crack halflength a is: (a) 10 mm, and (b) 24 mm? thicknesst tate of stress near a crack tip. The stress intensit factor and fracture toughness 2a 2b Factor of safet for fracture h h Reading Assignment Juvinall Problem et #6 Due in class 11/8/00. Juvinall 6.2, 6.7 Problems continued on next pages Problem 4 A rectangular beam made of AB plastic ( C = 3MPa m ) has dimensions b = 20 mm deep and t = 10 mm thick. Loads on the beam cause a bending moment of 10 N.m. What is the largest through thickness edge crack that can be permitted if a factor of safet of 2.5 against fracture is required? 11/8/00 ME111 Lecture /8/00 ME111 Lecture 16 2 ME111 Lecture 16 1
2 Problem 5 A 50 mm diameter shaft has a circumferential surface crack as shown below, with crack depth a = 5 mm. The shaft is made of 18Ni maraging steel ( C = 123MPa m ). (a) If the shaft is loaded with a bending moment of 1.5 kn.m, what is the factor of safet against crack propagation? (b) If an axial tensile load of 120 kn is combined with the above bending moment, what is the factor of safet now? Note: The stress intensit factor for a case of combined loading is found b simpl summing the stress intensit factors found b considering each loading case separatel. This superposition works because we are combining linear elastic solutions. M P Problem 6 Two plates of A533B1 steel are placed together and then welded from one side, with the weld penetrating halfwa, as shown below. A uniform tension stress is applied during service. Determine the strength of this joint, as a percentage of its strength if the joint were solid, as limited b (i) fracture, taking into account plasticit at the crack tip, and (ii) full plastic ielding, for: (a) A service temperature of C when the properties are: = 550 MPa, C = 55MPa (b) A service temperature of C when the properties are: = 400 MPa, C = 200MPa (c) Comment on the suitabilit of this steel for use at these temperatures. Notes: 1. Fracture toughness generall increases with temperature while ield strength diminishes. 2. Assume the weld metal has the same properties as the plates. m m a = 5mm 50mm 5cm weld 10cm 11/8/00 ME111 Lecture /8/00 ME111 Lecture 16 4 ME111 Lecture 16 2
3 16.1 What is Fracture Mechanics? tructures will frequentl have sizeable existing cracks which might or might not grow, depending on the load level. To predict failure b crack growth we need a crack meter, which is provided b fracture mechanics When a material has an EXITING crack, flaw, inclusion or defect of unknown small radius, the stress concentration factor approaches infinit, making it practicall useless for predicting stress. Infinite stresses are predicted b Elasticit theor a b max = t nom Material will ield and stress remain finite a max = 1+ 2 b a t = 1+ 2 b t As b Then max / a 0 a max = 1+ 2 = 3 a = 3 Drill hole at crack tip reduces stress concentration and arrests crack growth (e.g. skin of airplane wing) t b/ a 11/8/00 ME111 Lecture /8/00 ME111 Lecture 16 6 ME111 Lecture 16 3
4 Linear elastic fracture mechanics (LEFM) analzes the gross elastic changes in a component that occur as a sharp crack grows and compares this to the energ required to produce new fracture surfaces tress tate Near the Crack Tip Using this approach, it is possible to calculate the average stress which will cause growth of an existing crack. r θ x Mode I inplane tension 16.2 Fracture Conditions There exist three possible fracture modes, as shown below: For Mode I fracture, the stress components at the crack tip are: Mode I inplane tension Mode II inplane shear Mode III outofplane shear τ θ θ 3θ = cos 1 sin sin 2π r x + θ θ 3θ = cos 1 + sin sin 2π r x + θ θ 3θ sin cos cos 2π r x = + O( r O( r O( r 1/ 2 ) 1/ 2 1/ 2 ) ) = β π a Mode I is most important (will not consider the others here) A crack generates its own stress field, which differs from an other crack tip stress field onl b the scaling factor, which we call the stress intensit factor. 11/8/00 ME111 Lecture /8/00 ME111 Lecture 16 8 ME111 Lecture 16 4
5 16.4 tate of tress Near a Crack Tip The units of the stress intensit factor are, for example, Consider a crack in a plate as shown: MPa m, or psi in, etc. 2a 2b Values of β = / 0 have been determined from the theor of elasticit for man cases of practical importance and some representative cases are plotted in the next few pages (taken from Mechanical Engineering Design, b higle and Mischke) Fracture Toughness h h The stress intensit factor describes the state of stress near a crack tip. If h b >> 1, >> 1 b a elastic analsis shows that the conditions for crack growth are controlled b the magnitude of the elastic stress stress intensit factor 0 0 = πa If the plate has finite dimensions relative to the crack length a, then the value of must be modified: 0 β = β = 0 πa where β depends on the geometr of the component and crack. 0 = πa is the base value of the stress intensit factor = β 0 is the true stress intensit factor It is found experimentall that existing cracks will propagate (I.e. grow) when the stress intensit factor reaches a critical value called the fracture toughness. The fracture toughness C is a material propert that The fracture toughness can be measured and tabulated is denoted C Plane strain conditions are the most conservative We have failure b fracture when grows to = < C Crack will not propagate C 16.6 Factor of afet in Fracture Crack will propagate C N = C 11/8/00 ME111 Lecture /8/00 ME111 Lecture ME111 Lecture 16 5
6 16.7 ummar of LEFM 16.8 Values of C for ome Metals Elastic analsis, based on the assumption of elastic behavior at the crack tip (I.e. no largescale plastic deformations occur), shows that the conditions for crack propagation are controlled b the magnitude of the elastic stress intensit factor Material C, MPa m, MPa where β β = β πa = 0 depends on the geometr of the component and crack. Aluminum It has been found experimentall that a preexisting crack will propagate when the stress intensit factor reaches a critical value called the fracture toughness C Titanium Ti6AL4V < C C Crack will not propagate Crack will propagate Ti6AL4V teel The factor of safet for a given stress intensit factor is: C N = /8/00 ME111 Lecture /8/00 ME111 Lecture ME111 Lecture 16 6
7 Example 16.1 A steel ship deck is 30 mm thick, 12 m wide and 20 m long and has a fracture toughness of C = 28.3 MPa.m 1/2. If a 65 mm long central transverse crack is discovered, calculate the nominal tensile stress that will cause catastrophic failure. Compare the stress found to the ield strength of = 240 MPa. First compute the stress intensit factor: a/b = = 0.005, h/ b = = From, Fig. 519, we find β = 1 (essentiall an infinite plate) 0 For failure we have: 3 = β πa = ( 1.0)( )( π ( )) 3 = C ( = 1.0)( )( π ( )) = = 88.6 MPa 3 π ( ) The uniaxial stress that will cause ielding is given b We note that = 240 = = 240MPa (fracture occurs well before ielding) 11/8/00 ME111 Lecture Example 16.2 A plate of width 1.4 m and length 2.8 m is required to support atensile load of 4 MN (in the long direction). Inspection procedures are capable of detecting throughthickness edge cracks larger than 2.7 mm. The two titanium allos in the table on page are being considered for this application. For a factor of safet of N = 1.3 against ielding and fracture, which one of the two allos will give the lightest weight solution? We start b determining thickness based on ielding: P / wt N = = t = PN w For the weaker allo (call it allo A) 6 PN ( 4 10 )(1.3) 910 t = = = 4.08mm = = = 700MPa w (1,400)(910) N 1.3 For the stronger allo (call it allo B) 6 PN ( 4 10 )(1.3) 1035 t = = = 3.59mm = = = 796MPa w (1,400)(1035) N 1.3 We now find the thickness to prevent crack growth (see Fig. 522): h = b 2.8/2 = 1, 1.4 a b = 2.7 = ,400 11/8/00 ME111 Lecture β = 3 = β0 = β πa = ( 1.1) π ( ) = We also have, N = C C = = = N N C so that, 0 = 1.1 ME111 Lecture 16 7
8 For the weaker allo (allo A) C 115 = = = MPa N (0.1013)(1.3) 6 P 4 10 t = = = 3. 27mm w (1,400)(873.3) For the stronger allo (allo B) ummar: C 55 = = = MPa N (0.1013)(1.3) 6 P 4 10 t = = = 6. 84mm w (1,400)(417.6) Weak allo (A): Yielding Fracture trong allo (B): Yielding Fracture t = 4.08 mm = 700 MPa t = 3.27mm = MPa t = 3.59mm = 796MPa t = 6.84 mm = 418 MPa Best design solution: use weak allo (A) with governed b ielding t = Fracture toughness limits design mm Example 16.3 A long rectangular plate has a width of 100 mm, thickness of 5 mm and an axial load of 50 kn. If the plate is made of aluminum 2014T651, ( C = 24MPa m ) what is the critical crack length for failure (N = 1)? β = / C 2a 50kN 2 b = 100 mm a /b At the critical crack length, C = = β πa But β depends on a / b, making a direct calculation of a impossible. 50, = β (100)(5) β Using Fig. 521: or a = meters! π a a = 0.04, a / b = 0.8, β = 1.85, β a = 0.37 a = 0.01, a / b = 0.2, β = 1.02, β a = M a = 0.015, a / b = 0.3, β = 1.06, β a = 0.13 Critical crack length a = 15 mm. 11/8/00 ME111 Lecture /8/00 ME111 Lecture ME111 Lecture 16 8
9 Example 16.4 For the proceeding problem, what is the factor of safet against fracture for a crack of length10 mm which is: (a) Centered in the plate (b) On the edge of the plate (a) Determine the stress intensit factor from Fig. 521: 16.9 Effects of mall cale Yielding at the Crack Tip tress are predicted to become infinite near the crack tip according to LEFM. For ductile materials, stresses are limited b local ielding in the vicinit of the crack tip. a = 5 mm, a / b = 0.1, β = ,000 = β πa = ( 1.02) 0.01π = 18. 1MPa (100)(5) m stress based on elasticit calculate the factor of safet: 24 = C N = = r ielded, redistribu ted stress (b) Determine the stress intensit factor from Fig. 522: r a = 10 mm, a / b = 0.1, β = 1.02 Which is the same as (a) N = 1.33 crack tip plasticzone x 11/8/00 ME111 Lecture /8/00 ME111 Lecture ME111 Lecture 16 9
10 The size of the plastic zone can be investigated as follows: It ma be shown that 1 r = 2π tress redistribution accompanies plastic ielding and causes the plastic zone to to extend approximatel 2r 1 = π 2 2 ahead of the real crack tip to satisf equilibrium conditions (see figure on previous page). To account for this, we think of the plastic zone as providing a virtual crack tip extension to a new effectivecrack size: a eff We observe that: = a + 2r 1 = a + π = β πa eff 2 the use of which now requires an iterative solution because β depends on a eff. Concluding Remarks: In man problems, the plastic zone is ver small and can be neglected in determining the stress intensit factor. However, when the stress intensit factor approaches the fracture toughness value, the plastic zone correction becomes significant and should be investigated. If the plastic zone size becomes large relative to the crack size the accurac of LEFM becomes questionable, and an elasticplastic fracture mechanics approach should be considered. 11/8/00 ME111 Lecture /8/00 ME111 Lecture ME111 Lecture 16 10
11 Example 16.5 A long rectangular plate has a width of 100 mm, thickness of 5 mm and an axial load of 50 kn. If the plate is made of titanium Ti6AL4V, ( = 910 MPa, C = 115 MPa m ) what is the factor of safet against crack growth for a crack of length a = 20 mm. 2a 175 kn We wish to compute the stress intensit factor so that we can evaluate: Now N C =, = β πa eff Example 16.6 The round bar of aluminum 2024T851 has a sharp notch around its circumference. Assume: Find: = 455 MPa, C = MPa (a) The size of the plastic zone at the crack tip. (b) The fracture load. P m 2 b = 100 mm i+ 1 = β a eff ( a eff 175,000 ) (100)(5) 1 = a + π 2 i π a eff 30 mm a = 2 mm 11/8/00 ME111 Lecture /8/00 ME111 Lecture ME111 Lecture 16 11
12 Example 16.7 For the steel beam shown, a crack of length a = 0.25 in will be detectable. Find the beam thickness to provide a factor of safet N = 2, (a) ignoring the plastic zone at the crack tip, (b) taking the plastic zone into account. Take: = 220 kpsi, C = 55 kpsi in 5 lb 5 lb 1.5 in 6 in 15 in 6 in 11/8/00 ME111 Lecture ME111 Lecture 16 12
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