Fracture Mechanics Solid Mechanics Solid Mechanics Presented by Calvin M. Stewart, PhD MECH 5390-6390 Fall 2020 Outline • Interatomic View of Fracture • Linear Elasticity • Equilibrium of Stress • Compatibility Equations of Strain • Airy Stress Functions • Stress Concentration Factors • Circular Hole • Elliptical Hole • Limitations of SCF Approach • Need for LEFM Interatomic View of Fracture Presented by Calvin M. Stewart, PhD MECH 5390-6390 Fall 2020 Interatomic View of Fracture • A material fractures when sufficient stress and work are applied on the atomic level to break the bonds that hold atoms together. The bond strength is supplied by the attractive forces between atoms. P(x) = Applied Force • Interatomic Forces ➢Attractive Forces, Fa ➢Repulsive Forces, Fr ➢Applied Force, P • Equilibrium Spacing, x0 Interatomic View of Fracture • Bond Energy, E b E= P x dx b ( ) xo • where x0 is the equilibrium spacing and P is the applied force. Idealize the applied force, P(x) function as x PP= c sin • For simplicity, small displacement, x PP= c Note: Distance, λ Interatomic View of Fracture • For small displacement, the Bond Stiffness, k is kP= c • Multiple Both sides by number of bonds per unit area, Ab and the gage length, x0 EAb Ab kx o= P c A b x o Note: k = x0 • Simplify to find Cohesive Stress, σc E E 2 = Ab E= P c A b x o c c =, x0 x0 Interatomic View of Fracture • Surface Energy 1 x == sin dx s c c 2 0 • Surface energy is equal to one-half of the fracture energy because two surfaces are created when a material fractures. • Replace λ to find Cohesive Stress, σc E s c = x0 Linear Elasticity Elasticity, Equilibrium Equations of Stress, Compatibility Equations of Strain, Airy Stress Functions Elasticity • Generalized Hooke’s Law −1 ij=CSSC ijkl kl,, ij = ijkl kl = nd 2 ij Stress Tensor 2 order 3 =9 terms nd 2 ij Strain Tensor 2 order 3 =9 terms th 4 Cijkl Stiffness Tensor 4 order 3 =81 terms th 4 Sijkl Compliance Tensor 4 order 3 =81 terms 11=CCC 1111 11 + 1112 12 +....... + 1133 33 12=CCC 1211 11 + 1212 12 +....... + 1233 33 ............................................................... 33=CCC 3311 11 + 3312 12 +....... + 33333 33 Elasticity • For isotropic, homogenous, elastic materials, the 3D form of Hooke’s laws can be written as E ij=+ ij kk ij 1+− 1 2 E Young’s Modulus 1+ =− Poisson’s Ratio ijEE ij kk ij • σij is the stress tensor • εij is the strain tensor Elasticity • Symmetry makes the tensors reduce to 6 terms which we will express with indices x, y, z 1 1+ x= x − ( y + z ) xy= xy E E 1 1+ y= y − ( x + z ) zx= zx E E 1 1+ z= z − ( x + y ) yz= yz E E • σx , σy , σz Normal stress components • τxy , τzx , τyz Shear stress components • εx , εy , εz Normal strain components • εxy , εzx , εyz Shear strain components Elasticity • Fracture mechanics mostly deals with 2-dimensional problems, in which case no quantity depends on the z coordinate. Two special cases are plane stress and plane strain conditions. Plane Stress Plane Strain Elasticity – Plane Stress Plane Stress – Stress is zero across a particular plane. In this case, the component is Thin in the z direction. 1 x=−( x y ) E z = 0 1 z= yz = zx = 0 − y=−( y x ) E z=+( x y ) E 1+ = xyE xy Elasticity – Plane Strain Plane Strain – Strain is zero across a particular plane. In this case, the component is Thick in the z direction. z= yz = zx = 0 xy Equilibrium Equations of Stress • When we speak of Equilibrium in Solid Mechanics, we are referring to the Newton Laws of Motion given as =Fam • Equilibrium only exists if the left hand side (LHS) and right hand side (RHS) of the equation are equal. In the case where a=0 and v≥0, =F0 • Which is called “static equilibrium” Equilibrium Equations of Stress • Consider a finite element of infinitesimal volume, dV subject to static equilibrium with stresses acting in the x direction. Equilibrium Equations of Stress • The equilibrium of forces in the x direction is • Repeating this process in the y and z direction and simplification furnishes the equilibrium equations of stress 2D case of plane stress and plane strain Compatibility Equations of Strain • There are six strain measurements εij that rise from three independent displacements, u, v, w. As such there exists six constraint equations, also called the Compatibility Conditions. • If Compatibility is not satisfied than gaps, overlaps, or discontinuities would exist in the strain field. • Let us assume the 2D case, where x xy du dv 1 du dv ij = = = =+ x y xy xy y dx dy 2 dy dx • Differentiate by x and then y d 2 ( xy ) dxdy Compatibility Equations of Strain • Start • Further Simplify to 1 du dv 2D Compatibility Equation xy =+ 2 dy dx 22 ddd 2 2 xy=+x y • Double Derivative dxdy dydy dxdx 2 33 d xy 1 d u d v =+ dxdy2 dxdydy dxdydx • Simplify d 2 d22 du d dv 2 xy =+ dxdy dydy dx dxdx dy Compatibility Equations of Strain 3D Compatibility Equations 222 2 ddd dyd d xy d yz d xyx y = + − zx 2 =+22 dxdy d y d x dzdx dy dz dx dy dd222 2 dd yz y d z d z d yzd zx xy 2 =+22 = + − dydz d z d y dxdy dz dx dy dz 222 2 ddzxd z x ddxd zx ddxy yz 2 =+22 = + − dzdx d x d z dydz dx dy dz dx George Biddell Airy • In 1862, Airy presented a new technique to determine the strain and stress field within a beam.[14] This technique, sometimes called the Airy stress function method, can be used to find solutions to many two- dimensional problems in solid mechanics. • For example, it was used by H. M. Westergaard to determine the stress and strain field around a crack tip and thereby this method contributed to the development of fracture mechanics. George Biddell Airy (1801-1892) Airy Stress Function • Any Stress field solution for an elastic problem must satisfy both equilibrium and compatibility, Airy (in 1863) introduced a function φ(x,y) to satisfy this requirement. d2 d 2 d 2 =,, = = − xxdy22 yy dx xy dxdy • Straightforward substation shows that this stress field • Always fulfils the equilibrium of stress • Only fulfils the compatibility equations of strain if the stress function is a solution of the so-called biharmonic equation. Airy Stress Function • This function automatically satisfies • Equilibrium: & Compatibility: d d xx +=xy 0 dx dy dd22d 2 2 xy=+x y dd 22 xy+= yy 0 dxdy d y d x dx dy • By Applying Hooke’s Law for a state of plane strain gives 222 2 2 dyyddxx xx d yy d xy (12−)2 + 2 − 2 + 2 = dx dy dx dy dxdy Airy Stress Function • Taking, • And Applying the Airy definition,d2 d 2 d 2 =,, = = − xxdy22 yy dx xy dxdy • We find, The Biharmonic Equation !!! Sweet!!! ☺ d4 d 4 d 4 +20 + = dx4 dx 2 dy 2 dy 4 Found in Continuum Mechanics of Linear Elastic Structures and d222 d 2 d 2 22 yyViscousddxx Incompressible xx Fluids yy xy = (12−)2 + 2 − 2 + 2 = dx dy dx dy dxdy =4 0 Note: φ is a real valued and has units of Force and can be written in alternative coordinate systems such as polar or cylindrical. Polar Coordinates • Sometimes it is better to express our answer in polar coordinates, particularly when examining the stress field in the vicinity of a crack. Strain-Displacement Stress-Strain Plane strain Plane stress Polar Coordinates • Similarly, equilibrium and compatibility can be expressed in polar coordinates as follows Equilibrium Compatibility Polar Coordinates • Even the Airy Stress function can be expressed in polar coordinates Airy Stress Function Stress Concentrations Circular Hole Stress Concentrations • Stress Concentrations are geometric discontinuities that lead to local increase in the stress field. Examples: holes, grooves, sharp corners, fillets, welds, surface defects • Stress Concentration Factor (SCF), KT characterizes the amplification of the stress SCF = local remote • Need: Stress field at the Notch!!!! Internal Force Lines Circular Hole • Elasticity problems can be solved by finding an Airy Stress function, φ(x,y) which satisfies the biharmonic equation and satisfies the boundary conditions of the posed problem. • Direction solutions of the governing equations are, for the most part, not available. • Consequently, an indirect approach called the semi-inverse method is often employed to solve a specific elasticity problem. Derivation See Tablet Derivation A more detailed derivation with explanations is given in “Principles of Fracture Mechanics” by R.J. Sanford – Chapter 2 Circular Hole • Using the semi-inverse method, we can find the stress concentration factor in an infinity wide plate with a circular hole. Assume uniaxial loading we can show that KT=3 remote local SCF = local remote Circular Hole • We can also find the solutions for the biaxial case, = −32 + = − =32 − = =34 + = = −34 − = − Elliptical Hole • The first quantitative evidence for the stress concentration effect of flaws was provided by Inglis, who analyzed elliptical holes in flat plates • The ratio σA/σ is defined as the stress concentration factor, KT. When a = b, the hole is circular and KT = 3.0, a well- known result. Derivation See Tablet Derivation A more detailed derivation with explanations is given in “Principles of Fracture Mechanics” by R.J. Sanford – Chapter 2 Elliptical Hole • As the major axis, a, increases relative to b, the elliptical hole begins to take on the appearance of a sharp crack. For this case, Inglis found it more convenient to express in terms of the radius of curvature, ρ: • When a>>b, Elliptical Hole • The previous equation predicts an infinite stress at the tip of an infinitely sharp crack, where ρ = 0.