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ME111 Instructor: Peter Pinsky Class #16 Problem 3 November 1, 2000 A center-cracked plate of AISI 1144 steel ( K C = 115 MPa m ) has dimensions b = 40 mm, t = 15 mm and h = 20 mm. For a factor of safety of three against crack growth, what is the maximum Today’s Topics permissible load on the plate if the crack half-length a is: (a) 10 mm, and (b) 24 mm? • Introduction to linear elastic fracture mechanics (LEFM). thicknesst

• State of stress near a crack tip.

• The stress intensity factor and fracture toughness σ 2a 2b σ

• Factor of safety for fracture

h h

Reading Assignment

Juvinall 6.1 – 6.4 Problem 4 Problem Set #6 Due in class 11/8/00. A rectangular beam made of ABS plastic ( K C = 3 MPa 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 Juvinall 6.2, 6.7 largest through thickness edge crack that can be permitted if a factor of safety of 2.5 against fracture is required? Problems continued on next pages

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Problem 6 Problem 5 Two plates of A533B-1 steel are placed together and then welded A 50 mm diameter shaft has a circumferential surface crack as from one side, with the weld penetrating halfway, as shown below. shown below, with crack depth a = 5 mm. The shaft is made of A uniform tension stress is applied during service. Determine the 18-Ni maraging steel ( K = 123 MPa m ). C strength of this joint, as a percentage of its strength if the joint were solid, as limited by (i) fracture, taking into account plasticity at the (a) If the shaft is loaded with a bending moment of 1.5 kN.m, what crack tip, and (ii) fully plastic yielding, for: is the factor of safety against crack propagation? (b) If an axial tensile load of 120 kN is combined with the above (a) A service temperature of -750C when the properties are: bending moment, what is the factor of safety now? S = 550MPa, K = 55MPa m Note: The stress intensity factor for a case of “combined” loading y C is found by simply summing the stress intensity factors found (b) A service temperature of 2000C when the properties are: by considering each loading case separately. This S = 400MPa, K = 200MPa m “superposition” works because we are combining linear y C elastic solutions. (c) Comment on the suitability of this steel for use at these temperatures. Notes: M 1. Fracture toughness generally increases with temperature while P yield strength diminishes. 2. Assume the weld metal has the same properties as the plates.

a = 5mm

5cm 50mm

weld

10cm

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16.1 What is Fracture Mechanics?

• Structures will frequently have sizeable existing cracks which might or might not grow, depending on the load level.

• To predict failure by crack growth we need a “crack meter,” which is provided by fracture mechanics • When a material has an EXISTING crack, flaw, inclusion or defect of unknown small radius, the stress concentration factor approaches Infinite stresses are predicted by infinity, making it practically useless for predicting stress. Elasticity theory

a

b

σ max = Ktσ nom Material will yield and stress remain finite

æ a ö σ max = σ ç1+ 2 ÷ = 3σ K t è a ø

æ a ö As Kt = 3 σ max = σ ç1+ 2 ÷ è b ø b / a ® 0 a Then Kt = 1+ 2 Drill hole at crack tip – reduces stress concentration b σ max ® ¥ and arrests crack growth (e.g. skin of airplane wing)

b/ a

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16.3 Stress State Near the Crack Tip • Linear elastic fracture mechanics (LEFM) analyzes the gross elastic changes in a component that occur as a sharp crack grows and compares this to the energy required to produce new fracture surfaces. y

• Using this approach, it is possible to calculate the average stress r Mode I which will cause growth of an existing crack. θ x in-plane 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:

K θ é θ 3θ ù 1/ 2 σ x = cos 1- sin sin + O(r ) 2πr 2 ëê 2 2 ûú K θ é θ 3θ ù σ = + + 1/ 2 K = βσ π a x cos ê1 sin sin ú O(r ) 2πr 2 ë 2 2 û

K θ θ 3θ 1/ 2 Mode I Mode II Mode III τ xy = sin cos cos + O(r ) π in-plane tension in-plane shear out-of-plane shear 2 r 2 2 2

• A crack generates its own stress field, which differs from any other • Mode I is most important (will not consider the others here) crack tip stress field only by the scaling factor K , which we call the stress intensity factor.

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16.4 State of Stress Near a Crack Tip • The units of the stress intensity factor K are, for example,

• Consider a crack in a plate as shown: MPa m, or psi in, etc.

• Values of β = K / K 0 have been determined from the theory of elasticity for many cases of practical importance and some representative cases are plotted in the next few pages (taken σ 2a 2b σ from “Mechanical Engineering Design,” by Shigley and Mischke). 16.5 Fracture Toughness

• The stress intensity factor K describes the state of stress near a h h crack tip.

• It is found experimentally that existing cracks will propagate (I.e. grow) • If h b when the stress intensity factor K reaches a critical value called the >> 1, >> 1 fracture toughness. b a • The fracture toughness K C elastic analysis shows that the conditions for crack growth are is a material property that • The fracture toughness controlled by the magnitude of the elastic stress stress intensity can be measured and tabulated is denoted KC Plane strain conditions are factor K 0 the most conservative K 0 = σ πa

• We have failure by fracture when K grows toK = KC • If the plate has finite dimensions relative to the crack length a, K < K Crack will not propagate then the value of K 0 must be modified: C

K ³ K C Crack will propagate K = βK 0 = βσ πa where β depends on the geometry of the component and crack. 16.6 Factor of Safety in Fracture

K = σ πa is the base value of the stress intensity factor K 0 N = C is the true stress intensity factor K K = βK0

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16.7 Summary of LEFM 16.8 Values of K C for Some Metals

• Elastic analysis, based on the assumption of elastic behavior at the crack tip (I.e. no large-scale plastic deformations occur), shows that S , MPa the conditions for crack propagation are controlled by the magnitude Material K C , MPa m y of the elastic stress intensity factor K

Aluminum K = βK 0 = βσ πa 2024 26 455 where β depends on the geometry of the component and crack. 7075 24 495

7176 33 490

• It has been found experimentally that a pre-existing crack will Titanium propagate when the stress intensity factor reaches a critical value called the fracture toughness K C Ti-6AL-4V 115 910

K < K Crack will not propagate C Ti-6AL-4V 55 1035

K ³ K C Crack will propagate Steel

4340 99 860

• The factor of safety for a given stress intensity factor is: 4340 60 1515 K N = C 52100 14 2070 K

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Example 16.1 Example 16.2 A plate of width 1.4 m and length 2.8 m is required to support atensile A steel ship deck is 30 mm thick, 12 m wide and 20 m long and has load of 4 MN (in the long direction). Inspection procedures are capable a fracture toughness of K = 28.3 MPa.m1/2. If a 65 mm long central C of detecting through-thickness edge cracks larger than 2.7 mm. The two transverse crack is discovered, calculate the nominal tensile stress titanium alloys in the table on page are being considered for this that will cause catastrophic failure. Compare the stress found to the application. For a factor of safety of N = 1.3 against yielding and fracture, yield strength of S = 240 MPa. y which one of the two alloys will give the lightest weight solution?

• We start by determining thickness based on yielding: • First compute the stress intensity factor: Sy Sy PN 32.5 10 N = = Þ t = a/b = = 0.005, h/b = = 1.67 σ P / wt wS y 6000 6 For the weaker alloy (call it alloy A) 6 K PN (4´ 10 )(1.3) Sy 910 From, Fig. 5-19, we find β = » 1 (essentially an infinite plate) t = = = 4.08mm σ = = = 700MPa K0 wS y (1,400)(910) N 1.3

K = β σ πa = (1.0)(σ )( π (32.5´10- 3)) For the stronger alloy (call it alloy B) 6 PN (4´ 10 )(1.3) Sy 1035 For failure we have: t = = = 3.59mm σ = = = 796MPa wS y (1,400)(1035) N 1.3 - 3 K = KC Þ (1.0)(σ )( π (32.5´10 )) = 28.3 • We now find the thickness to prevent crack growth (see Fig. 5-22):

28.3 h 2.8/2 a 2.7 K σ = = 88.6 MPa = = 1, = = 0.00193 Þ β = = 1.1 - 3 π (32.5´10 ) b 1.4 b 1,400 K 0

- 3 K = βK0 = βσ πa = (1.1)σ π (2.7´ 10 ) = 0.1013σ The uniaxial stress that will cause yielding is given by S = 240MPa y K We also have, N = C so that, We note that K Sy 240 = = 2.71 (fracture occurs well before yielding) K K σ 88.6 K = 0.1013σ = C Þ σ = C N 0.1013 N

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For the weaker alloy (alloy A) Example 16.3 K 115 A long rectangular plate has a width of 100 mm, thickness of 5 mm σ = C = = 873 .3 MPa 0.1013N (0.1013)(1.3) and an axial load of 50 kN. If the plate is made of aluminum 2014-T651, ( K = 24 MPa m ) what is the critical crack length for failure (N = 1)? P 4´ 106 C t = = = 3.27mm wσ (1,400)(873.3) 50kN For the stronger alloy (alloy B) At the critical crack length,

K C 55 σ = = = 417.6 MPa K C = K = βσ πa 0.1013N (0.1013)(1.3) 2a But β depends on a / b , making a P 4´106 t = = = 6.84mm direct calculation of a impossible. wσ (1,400)(417.6) 2b = 100 mm æ 50,000 ö 24 = β ç ÷ π a è (100)(5) ø Summary: or Weak alloy (A): β a = 0.135 β = K / K C Yielding t = 4.08 mm σ = 700 MPa meters! Fracture t = 3.27mm σ = 873.3MPa Using Fig. 5-21: Fracture toughness Strong alloy (B): a = 0.04, a /b = 0.8, β = 1.85, β a = 0.37 limits design Yielding t = 3.59mm σ = 796MPa a = 0.01, a /b = 0.2, β = 1.02, β a = 0.102 Fracture t = 6.84 mm σ = 418 MPa M a = 0.015,a /b = 0.3, β = 1.06, β a = 0.13 Best design solution: use weak alloy (A) with t = 4.08 mm governed by yielding a /b Critical crack length a = 15 mm.

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16.9 Effects of Small Scale Yielding at the Crack Tip Example 16.4

For the proceeding problem, what is the factor of safety against fracture for a crack of length10 mm which is: • Stress are predicted to become infinite near the crack tip according to LEFM. (a) Centered in the plate (b) On the edge of the plate • For ductile materials, stresses are limited by local yielding in the (a) Determine the stress intensity factor from Fig. 5-21: vicinity of the crack tip.

a = 5 mm , a /b = 0.1, β = 1.02 æ 50,000 ö K = βσ πa = (1.02)ç ÷ 0.01π = 18.1MPa m σ stress based on elasticity è (100)(5)ø

calculate the factor of safety: 2ry K 24 N = C = = 1.33 K 18.1 yielded , redistribu ted stress

ry (b) Determine the stress intensity factor from Fig. 5-22:

a = 10 mm ,a /b = 0.1, β = 1.02

Which is the same as (a) crack tip x N = 1.33 plasticzone

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• To account for this, we think of the plastic zone as providing a virtual crack tip extension to a new effectivecrack size: • The size of the plastic zone can be investigated as follows: 2 1 æ K ö a = a + 2r = a + ç ÷ eff y ç ÷ π è S y ø • It may be shown that • We observe that:

2 æ ö 1 K = βσ π r = ç ÷ K aeff y π ç ÷ 2 è S y ø

the use of which now requires an iterative solution because β depends on a . • Stress redistribution accompanies plastic yielding and causes eff the plastic zone to to extend approximately

2 Concluding Remarks: 1 æ K ö 2r = ç ÷ y ç ÷ π è S y ø • In many problems, the plastic zone is very small and can be neglected in determining the stress intensity factor. ahead of the real crack tip to satisfy equilibrium conditions (see figure on previous page). • However, when the stress intensity 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 accuracy of LEFM becomes questionable, and an elastic-plastic fracture mechanics approach should be considered.

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Example 16.5 Example 16.6

A long rectangular plate has a width of 100 mm, thickness of 5 mm The round bar of aluminum 2024-T851 has a sharp notch around its and an axial load of 50 kN. If the plate is made of titanium Ti-6AL-4V, circumference. Assume: ( S y = 910 MPa , K C = 115 MPa m ) what is the factor of safety against crack growth for a crack of length a = 20 mm. S y = 455MPa, KC = 26.4 MPa m

Find: 175 kN We wish to compute the stress intensity factor so that we can evaluate: (a) The size of the plastic zone at the crack tip. K (b) The fracture load. N = C , K = βσ πa K eff 2a P Now

i+ 1 175,000 K = β (aeff ) π aeff 2b = 100 mm (100)(5) a = 2 mm 2 1 æ K i ö = + ç ÷ aeff a ç ÷ π è S y ø 30 mm

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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 safety N = 2, (a) ignoring the plastic zone at the crack tip, (b) taking the plastic zone into account. Take:

S y = 220 kpsi , KC = 55 kpsi in

5 lb 5 lb

1.5 in

6 in 15 in 6 in

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