Isotropic linear elastic response

Transcription

1 Isotropic linear elastic response

2 Range of Young s modulus 2 E [GPa = 0 9 N/m 2 ] Metals Alloys Tungsten Molybdenum Steel, Ni Tantalum Platinum Cu alloys Zinc, Ti Silver, Gold Aluminum Magnesium, Tin Graphite Ceramics Semicond. Diamond Si carbide Al oxide Si nitride <> Si crystal <00> Glass-soda Concrete Graphite Polymers Polyester PET PS PC PP HDPE PTFE LDPE Composites Fibers Carbon fibers only CFRE( fibers)* Aramid fibers only AFRE( fibers)* Glass fibers only GFRE( fibers)* GFRE* CFRE* GFRE( fibers)* CFRE( fibers)* AFRE( fibers)* Epoxy only Wood( grain) What is the source of both universality and range in modulus? Based on data in Table B2, Callister 6Ed. Composite data based on reinforced epoxy with 60 vol% of aligned carbon (CFRE), aramid (AFRE), or glass (GFRE) fibers. From Callister, Intro to Eng. Matls., 6Ed

3 Universality of linear elastic response 3 Materials are made of atoms, held together by atomic interactions covalent and ionic bonding: ceramics, semiconductors (~200 N/m) metallic bonding: metals (~ 20 N/m) van der Waals interaction: polymers (~ 0.5 N/m) Materials are made of many atoms, governed by thermodynamics materials choose structures, phase variables (such as density) that minimize free energy: A = U TS A: Helmholtz free energy U: internal energy (bonding) T: (absolute) temperature S: entropy (disorder: kb log Ω)

4 Thermodynamic equation of state 4 E Bond energy Equilibrium bond length V

5 Universality of linear elastic response 5 Materials are made of atoms, held together by atomic interactions covalent and ionic bonding: ceramics, semiconductors (~200 N/m) metallic bonding: metals (~ 20 N/m) van der Waals interaction: polymers (~ 0.5 N/m) Materials are made of many atoms, governed by thermodynamics materials choose structures, phase variables (such as density) that minimize free energy: A = U TS A: Helmholtz free energy U: internal energy (bonding) T: (absolute) temperature S: entropy (disorder: kb log Ω) V P( V + V 0 ) + A 2 2 A V 3 0 V 0 0 = 0 + V V 0 = V K B@ V 2 2 V 0 CA +

6 Superposition principle 6 For small stresses the strains are linearly related to stresses: k = E k? = E k = G We can generalize these results by considering superposition.each stress component (σx σy σz τxy τxz τyz) is considered individually 2.All of the strains from each stress component computed 3.Sum of all strains = material response to stress x = E [ x ( y + z )] y = E [ y ( x + z )] z = E [ z ( x + y )] xy = G xy xz = G xz yz = G yz

7 Material property relationships 7 The superposition principle can relate our elastic moduli: E: Young s modulus (normal strain from uniaxial stress) ν: Poisson s ratio (perpendicular normal strain from uniaxial stress) G: shear modulus (shear strain from shear stress) K: bulk modulus (volume change from hydrostatic pressure)! 0 0 $ 0!! 0 /2 0 /2 0 $! /2 0 0 /2 = G 2 = E E ( ) = 2( + ) E G = E 2( + )

8 Material property relationships 8 The superposition principle can relate our elastic moduli: E: Young s modulus (normal strain from uniaxial stress) ν: Poisson s ratio (perpendicular normal strain from uniaxial stress) G: shear modulus (shear strain from shear stress) K: bulk modulus (volume change from hydrostatic pressure) = 0 B@ p p 0 CA 0 0 p = 0 B@ p E + 2 p E 0 0 p 0 E + 2 p E 0 p 0 0 E + 2 p CA E V = V( + x )( + y )( + z ) V V V = V( + ( x + y + z ) + ) V x + y + z = K = 3( 2 ) E E 3( 2 ) p = p K

9 Anisotropic linear elastic response 9

10 Isotropic stress/strain relations 0 x = E [ x ( y + z )] y = E [ y ( x + z )] z = E [ z ( x + y )] xy = G xy xz = G xz yz = G yz x = y = z = E ( + )( 2 ) E ( + )( 2 ) E ( + )( 2 ) h ( h ( h ( ) x + ( y + z ) i ) y + ( x + z ) i ) z + ( x + y ) i xy = G xy xz = G xz yz = G yz Is there a way to extend this to anisotropic response?

11 General state of stress Each point in a body has normal and shear stress components. We can section a cubic volume of material that represents the state of stress acting around the chosen point. As the cube is at equilibrium the total forces and moments are zero: Infinitesimal cube = equal and opposite forces on opposite sides of cube Note also: the values of all the components depend on how the cube is oriented in the material (we ll talk later about relating those values) The combination of the state of stress for every point in the domain is called the stress field.

12 Graphical stress tensor components 2 Stress area = force Fi = Σj=xyz σij Aj σxx=σx σxy=τxy σxz=τxz σyx=τxy σyy=σy σyz=τyz σzx=τxz σzy=τyz σzz=σz

13 Defining strain 3 We want to describe the dimension and shape change in a continuous cohesive body In a sufficiently small element, deformations of the element are all proportional to the size of the element length / length = unitless, % (0 2 ), mm/mm, µm/m (0 6 ), or in/in Requires that we capture both the orientation of original vector and change in that vector Original relative position is a vector: one index = 3 numbers to describe New relative position is a vector: one index = 3 numbers to describe Strain is a tensor: two indices (coordinates) = 3 3 numbers to describe

14 Normal strain and shear strain 4 Normal strain describes a length change in a vector Shear strain describes an orientation change in a vector Be aware: whether deformation changes length or changes orientation also depends on the original orientation s lim 0 s " = lim 0 AB,AC!0 s!0 2s

15 Normal strain and shear strain 5 We can describe all the strains on an element in a body: deformation " ij = 0 B@ x 2 xy 2 xz 2 xy y 2 yz 2 xz 2 yz z CA shear strains normal strains xy = (angle change in xy plane) = " xy + " yx xy = (rotation in xy plane) = " yx " xy

16 Graphical strain tensor components 6 Strain length = length change δli = Σj=xyz εij lj εxx=ϵx εxy=½γxy εxz=½γxz εyx=½γxy εyy=ϵy εyz=½γyz εzx=½γxz εzy=½γyz εzz=ϵz

17 Elastic constants: stiffnesses and compliances 7 Just as stress relates a vector (area) to another vector (force), and strain relates a vector (position) to another vector (change in position), our elastic constants relate stresses to strains: 4th rank tensors ij = X kl S ijkl kl ij = compliance [GPa ] X kl C ijkl kl stiffness [GPa] = 8 components! But first two and last two are symmetric: xyzz = yxzz and zzxy = zzyx And first pair and last pair can be swapped: xyzz = zzyx Stiffness is a second derivative of energy: Cijkl = d 2 U/dϵij dϵkl Results in 2 unique elastic constants. Better written with Voigt notation: 0 B@ CA 0 e 2 e 6 2 e 5 2 B@ e 6 e 2 2 e 5 2 e 4 e 3 2 e 4 CA e i = i = 6X j= 6X j= S ij j C ij e j 2 3 xx yy zz yz xz xz

18 Elastic constants: stiffnesses and compliances 8 The S and C matrices are inverses of each other The 2 stiffness and compliance matrix entries have factors of 2 and 4 to convert to tensor components: Cab = Cijkl for a=..6, b=..6 Sab = Sijkl for a=..3 and b=..3 Sab = 2Sijkl for a=..3 and b=..6 or a=4..6 and b=..3 or Sab = 4Sijkl for a=4..6 and b=4..6 Crystalline symmetry reduces the number of unique and nonzero entries

19 Stiffness / Compliance symmetry 0 xx xx xx yy!!!!! xx zz xx yz xx zy xx zx xx xz xx xy xx yx yy xx zz xx yz xx zy xx zx xx xz xx xy xx yx xx yy yy zz zz!!!! yy zz yy yz yy zy yy zx yy xz yy xy yy yx zz yy yz yy zy yy zx yy xz yy xy yy yx yy! zz yz zz zy! zz zx zz xz! zz xy zz yx yz zz zy zz zx zz xz zz xy zz yx zz!!! yz yz yz zy yz zx yz xz zx yz xz yz yz xy yz yx xy yz yx yz zy yz zy zy zy zx zy xz zx zy xz zy zy xy zy yx xy zy yx zy! zx zx zx xz! zx xy zx yx xy zx yx zx xz zx xz xz xz xy xz yx xy xz yx xz! xy xy xy yx yx xy yx yx CA

20 Symmetry operations 2-fold axis 3-fold axis 4-fold axis Rotating a cube around the body diagonal? mirror plane mirror plane 3-fold axis

21 Elastic constants: stiffnesses and compliances 2 The S and C matrices are inverses of each other The 2 stiffness and compliance matrix entries have factors of 2 and 4 to convert to tensor components: Cab = Cijkl for a=..6, b=..6 Sab = Sijkl for a=..3 and b=..3 Sab = 2Sijkl for a=..3 and b=..6 or a=4..6 and b=..3 or Sab = 4Sijkl for a=4..6 and b=4..6 Crystalline symmetry reduces the number of unique and nonzero entries Cubic symmetry is the most common for structural materials: C = C22 = C33 C2 = C3 = C23 C44 = C55 = C66 all others zero Isotropic materials are cubic and C C2 = 2C44 (or S S2 = S44/2) Hexagonal materials and aligned fiber composites have lower symmetry: C = C22 C33; C2 C3 = C23; C44 = C55 C66 Isotropic in basal plane: 2C66 = C C2 all others zero

22 Graphical compliance components εij = Σk=xyz Σl=xyz Sijkl σkl εxx = + E σxx σyy σzz - ν E - ν E Sxxxx = /E Sxxyy = -ν/e Sxxzz = -ν/e Sxxkl = 0 for all other kl

23 Voigt and Reuss averages Voigt average = isostrain Randomly oriented grains... E Voigt = C C 2 + 3C 44 C + 2C 2 2C + 3C 2 + C 44 C = 3 (C + C 22 + C 33 ) C 2 = 3 (C 2 + C 3 + C 23 ) C 44 = 3 (C 44 + C 55 + C 66 )

24 Voigt and Reuss averages Reuss average = isostress Randomly oriented grains E Reuss = S + 2S 2 + S 44 S = 3 (S + S 22 + S 33 ) S 2 = 3 (S 2 + S 3 + S 23 ) S 44 = 3 (S 44 + S 55 + S 66 ) EVoigt > Erandom polycrystal > EReuss

25 Grain structure and texture Ni alloy grain structure Each grain has a different orientation, and responds differently to applied stress Polycrystalline response is an average of individual grain responses. Texture is a preferential orientation of grains. M. Groeber et al., AIP Conf. Proc. 72, 72 (2004).

26 Ti / TiB metal-matrix composite SEM backscatter: polish C ij = TiB: orthorhombic crystal B GPa C A SEM secondary e - : deep etch EVoigt = 442GPa EReuss = 435GPa ETi = 0GPa ETi+20%vol TiB = 53GPa S. Gorsse et al., Mat. Sci. Eng. A340, (2003) D. R. Trinkle, Scripta Mater. 56, (2007)

27 Bovine femural bone: elastic constants longitudinal (3) C ij = transverse (,2) bone orthotropic stiffnesses: B GPa C A along length: E3 = 2.7GPa transverse: E =.6GPa Extracted from sound-speed measurements: C = r(v ) 2 C 22 = r(v 22 ) 2 C 44 = r(v 23 ) 2 W. C. Buskirk et al., J. Biomech. Eng. 03, (98)

28 Composite behavior 28

29 Composite = matrix + reinforcement 29 Matrix: continuous phase transfers load to reinforcement protects reinforcement from environment Types of matrix: MMC metal matrix composite: designed for plastic strain better yield stress, tensile strength, creep resistance CMC ceramic matrix composite: designed for fracture better toughness PMC polymer matrix composite: designed for elastic and plastic strain better modulus, yield stress, tensile strength, creep inexpensive, temperature range limited by polymer decomposition Reinforcement: stronger, discontinuous phase carries significant portion of load classified by geometry

30 Particle reinforcements 30 ferrite (bcc-fe) cementite (Fe3C) cemented carbide 0µm Spheroidite steel Co matrix (Vm=0-5%) WC particles From Callister, Intro to Eng. Matls., 6Ed 00µm

31 Particle reinforcements 3 rubber carbon particles Ti/TiB MMC 00nm Tire rubber alpha-ti (hcp) TiB needles From Callister, Intro to Eng. Matls., 6Ed S. Gorsse et al., Mat. Sci. Eng. A340, (2003).

32 Fiber reinforcements: continuous, aligned 32 Glass (E=76GPa) with SiC fibers (E=400GPa) fracture surface Ni3Al (gamma ) Mo + Ni3Al alpha-mo (bcc) From F.L. Matthews and R.L. Rawlings, Composite Materials; Engineering and Science, Reprint ed., CRC Press, Boca Raton, FL, (a) Fig. 4.22, p. 45 (photo by J. Davies); (b) Fig..20, p. 349 (micrograph by H.S. Kim, P.S. Rodgers, and R.D. Rawlings). W. Funk et al., Met. Trans A9, (988).

33 Fiber reinforcements: discontinuous, random 33 C fibers C matrix processed by laying down fibers in binder (pitch); high heat converts binder to C matrix carbon-carbon composite Randomly oriented fibers layered in 2D, not continuous with composite

34 Determining mechanical behavior 34 Stress / strain response depends on material properties (E, TS, σy) of matrix + reinforcement amount of matrix + reinforcement (Vm, Vr= Vm) orientation of reinforcement relative to load size and distribution of reinforcement geometry (length of fibers, cross-sectional shape, aspect ratio) Two limiting cases for analysis: isoload/isostress isostrain equal load in phases equal strain in phases

35 Isostrain 35 equal length/strain in phases lreinforcement = lmatrix = lcomposite εreinforcement = εmatrix = εcomposite shared load: F c = F m + F r F c A = F m A + F r A c = F m A m A m A + F r A r A r A c = V m m + V r r ROM for stresses Similar to Voigt average

36 Isoload / isostress 36 equal load/stress in phases Freinforcement = Fmatrix = Fcomposite σreinforcement = σmatrix = σcomposite shared length: `0c = `0m + `0r `c( + " c ) = `m( + " m ) + `r( + " r ) + " c = V m ( + " m ) + V r ( + " r ) " c = V m " m + V r " r ROM for strains Similar to Reuss average

37 Rule of mixtures: Cu particles + W matrix 37 Elastic moduli of the composite are constrained by two limits: isostrain and isoload E [GPa] isostrain isoload Empirical relations: E c = V m E m + K c V p E p (T.S.) c = V m (T.S.) m + K s V p (T.S.) p Kc Ks < 00 Cu 20% 40% 60% 80% W volume fraction (W)

38 Orientation effects on tensile strength 38 Tensile stress not parallel to fibers has complex stress state: 3 limiting cases: cos 2 = cos sin! cos sin sin 2.small misorientation: limited by fiber failure (σ = σ cos 2 θ) (T.S.) c =? k cos 2 2.large misorientation: limited by matrix tensile failure (σ = σ sin 2 θ) (T.S.) c =?? sin 2 3.medium misorientation: limited by matrix shear failure (τ = σ cos θ sin θ) (T.S.) c = m,y cos sin

39 Orientation effects on tensile strength 39 Tensile stress not parallel to fibers has complex stress state: 3 limiting cases: cos 2 = cos sin! cos sin sin 2 (T.S.) c =? k cos 2 Tensile strength [MPa] small misorientation (T.S.) c = intermediate misorientation large misorientation Tsai-Hill m,y cos sin θ [degrees] (T.S.) c =?? sin 2

40 Orientation effects on tensile strength 40 Tensile stress not parallel to fibers has complex stress state: Some limitations: cos 2 = cos sin! cos sin sin 2.Predicts that tensile strength increases for small misorientation. 2.Predicts cusps in strength vs. misorientation angle. 3.Doesn t account for multiaxial loading effects. Solution: Tsai-Hill failure criterion: k? k!2 k?? 2?! +???!2 + m,y!2 = (T.S.) c = 2 64 cos4? 2 k + sin4? 2? + cos 2 sin 2 2 m,y? 2 k!37 5 /2

41 Orientation effects on tensile strength 4 Tensile stress not parallel to fibers has complex stress state: cos 2 = cos sin! cos sin sin 2 Tensile strength [MPa] small misorientation intermediate misorientation (T.S.) c = large misorientation Tsai-Hill 2 64 cos4? 2 k + sin4? 2? θ [degrees] Tsai-Hill smooths out cusps Never exceeds aligned T.S. + cos 2 sin 2 2 m,y? 2 k!37 5 /2