MASSACHUSTTS INSTITUT OF TCHNOLOGY DPARTMNT OF MATRIALS SCINC AND NGINRING CAMBRIDG, MASSACHUSTTS 0239 322 MCHANICAL PROPRTIS OF MATRIALS PROBLM ST 4 SOLUTIONS Consider a 500 nm thick aluminum ilm on a 500 µm thick silicon waer, 200 mm in diameter Ater deposition at 60 C, the waer is cooled to room temperature What is the radius o curvature o the ilm? ( Si = 50 GPa, Si = 07, α Si = 3 0-6 C -, Al = 69 GPa, Al = 033, α Al = 23 0-6 C - ) The use o the Stoney ormula is appropriate or this system because the aluminum ilm is very thin compared to the relatively thick substrate The Stoney ormula is given by 2 h s s ρ =, 6 h where ρ is the radius o curvature, s is the biaxial modulus o the substrate, h s and h are the substrate and ilm thickness, and m is the stress at the interace The biaxial modulus o the substrate is calculated by s 50 GPa s = = = 807 GPa s -07 The stress at the interace in the aluminum ilm is given by m = m = m To ind the strain at the interace, we must calculate the mismatch strain due to the thermal expansion coeicient mismatch 6-6 - = ( α α ) = (3 0 C 23 0 C ) (20 C 60 C) = 00008 m s T Solving or the mismatch stress, 69GPa = m m = m 00008 8239 MPa = 033 = The radius o curvature o the ilm-on-substrate system can be calculated rom the Stoney ormula 2 6 2 h s s (807 GPa)(500 0 m) ρ = = = 828 m 9 6 h 6(008239 GPa)(500 0 m) m The positive curvature (concave) indicates that the ilm is in tension ater cooling to room temperature Because the thermal expansion coeicient or aluminum is larger than silicon, the aluminum attempts to contract more than the silicon during the cooling However, at the interace, the aluminum cannot contract ully due to the constraining silicon, and as a result the aluminum is in tension m
6 2 A bilyer is composed o a mm thick coating ( = 350 GPa, α = 9 0 C ) on a 5 mm substrate 6 ( = 70 GPa, α = 23 0 C ) and has a width o 5 mm The bilayer has ero curvature at the initial temperature o 20 C The bilayer is then heated to 00 C Calculate the resulting curvature and the maximum tensile and compressive stresses Where do the maximum internal stresses occur? To ind the curvature o the bilayer, we can use the equation or the most general bilayer derived in class ( α2 α)( T T0) κ = h 2( I + I 2 ) + + 2 h ab 2a2b where the coating properties are denoted by the subscript and the substrate properties by 2 To calculate the curvature, we need the moments o inertia or each layer 3 3 ba (0005 m)(000 m) 3 2 I = = = 4667 0 m 3 3 ba2 (0005 m)(0005 m) 4 I2 = = = 520833 0 m ntering the parameters o the bilayer system into the given equation, we ind that the curvature o the bilayer is κ = 025 m The maximum stresses or a bilayer were derived in class and exist at the interace o the two layers 2( I + I 2 2) a max, = κ + = 07 MPa hab 2 2( I κ + I ) a 2 2 2 2 max,2 = + = ha2b 2 564 MPa
3 The structure and mechanics o cork: The structure and properties o cork are approximately axisymmetric In the schematics on the next page, we see a macroscopic view o cork as it exists on a tree, a microscopic look at the arrangement o cork cells, and inally a view o an individual cork cell Ignoring the natural variations in cell structure, we can model this cellular material as a linear-elastic solid by = S ij ijkl kl Axisymmetry reduces the number o independent compliances to ive (note that the axis o symmetry is the x axis) = S + S222 + S233 22 = S2 + S222222 + S2233 33 33 = S2 + S223322 + S2222 33 23 = ( S2222 S2233) 23 3 = 2S3 = 2S Measurement o our o the ive compliances is straightorward Simple tensile or compressive tests are done on the cork along the axis o symmetry (x axis) and orthogonal to this direction (x 2 and x 3 axes) to measure the Young s modulus and Poisson s ratio in these directions = S 2 = 3 = S 2222 = 3 = 2 2 = 3 2 = S2 23 2 = 32 2 = S2233 The modulus G 23 in the x 2, x 3 -plane is obtained rom these measurements by G = 2( + ) = 2( S S ) 23 23 2 2222 2233
To determine the ith compliance value, S, which is related to the shear modulus, G, we rotate the cork through 45 about the x 3 -axis and cut a cube with one ace normal to the x 3 -axis, and the other two at 45 to the x 2 -axis and x -axis A simple compression test in the new x direction then gives a new Young s modulus, a What is the relationship between the shear modulus, G, and S? The shear modulus G relates the engineering shear strain and shear stress Using matrix notation, we have 6 = S666 = 6 G To convert between matrix and tensor notation, we have the relation that S66 = 4S Thus we ind that G = 4 S b By rotating the Sijkl through 45 about the x3-axis, obtain an expression or S in terms o the compliances in the original orientation The direction cosines or this change in orientation by rotation about the x 3 -axis is 0 cos 45 sin 45 0 2 2 sin 45 cos 45 0 ij = = 0 2 2 0 0 0 0 The tensor equation or the transormation o a ourth order tensor is S ijkl = mi nj ok plsmnop To ind the S component, we use the transormation rule S = S m n o p mnop Although the expanded orm o the tensor transormation has 8 terms per equation, we note that and 2 are the only nonero direction cosine terms that appear in the tensor transormation The resulting expanded orm is S = S + S + S + S + S + 2 2 2 2 2 S + S + S + 2 2 2 2 2 2 2 2 2 2S2 + 2 2 S2 + 2 2 S 22 + S + S + S + S + 2 2 2 222 2 2 2 22 2 2 2 22 2 2 2 22 S 2 2 2 2 2222 From the symmetry o the strain tensor and stress tensor that are related by the compliance tensor, we can simpliy the expression (Note that and S are ero) S 22 S = ( ) S + 4( ) ( ) S 4 3 2 + 2( ) ( ) S + 4( ) ( ) S + 4( ) ( ) S 2 2 2 2 3 2 2 2 2 222 + ( ) S 4 2 2222 In tensor notation, the expression or the compliance in the new direction is S = S + S + S + S 4 2 4 2 2222
c Use your answers rom parts (a) and (b) to ind an expression or the shear modulus G, in terms o experimental parameters, 2,,and From part (a) we have 4S G = From the answer to part (b) we can solve or S and substitute this expression into the equation above = 4 S S S S G 4 2 4 2 2222 Replacing the compliance values with the corresponding experimental values leads to the result (Note that the value o the compliance in the new direction is related to the new Young s modulus by S = ) = 4 G 4 2 4 2 4 = G 2 d For the isotropic case, what does the equation derived in part (c) become? For an isotropic material, which is more symmetric than the transversely isotropic symmetry assumed above, the moduli and Poisson s ratios do not vary with direction Hence, we can rewrite the answer in part (c) as Simple algebra leads to the result 4 = G 4 + = G 2+ = G G = 2( + ) Thus, the expression in part (c) correctly reduces to the relationship between the elastic modulus, shear modulus and Poisson s ratio or the isotropic case
4 An isotropic sample o material subjected to a compressive stress = p is conined so that it cannot deorm in either the x- or y-directions a Do the stresses occur in the material in the x- and y-directions? I so, how are they related to? I the sample were not conined in the x- and y-directions, we would expect the material to expand in the transverse direction upon compressive loading The act that the material is constrained in the transverse direction indicates that a stress must evolve to maintain ero strain in the transverse direction Looking at Hooke s law or an isotropic material in three dimensions, we ind x x = ( y + ) = 0 Due to the symmetry o the problem, we know x = y Thus, we can ind a relationship between the transverse stress and the applied compressive stress x = y = b Determine the stiness = in the direction o the applied stress in terms o the isotropic elastic constants and or the material Is equal to the elastic modulus rom uniaxial loading? Why or why not? Again using Hooke s law to ind the strain in the -direction, 2 = ( x + y) = Using this expression or the strain in the -direction, we can calculate the stiness in the - direction = = = 2 2 The stiness calculated here is not equal to the elastic modulus in uniaxial loading In uniaxial loading, the transverse stresses are ero and do not contribute to the uniaxial deormation However, in this case the transverse stresses do aect the strain in the direction o loading c What happens i the Poisson s ratio or the material approaches 05? As Poisson s ratio approaches 05, the strain in the -direction goes to ero or any applied compressive stress This can be understood without derivation We know that materials with a Poisson s ratio o 05 are incompressible ( + + = 0 ) Because the transverse directions are x y constrained (ero strain), the strain in the loading direction must be ero to satisy the incompressibility condition, regardless o the applied compressive stress
5 Consider a lat plate o isotropic material that lies in the x-y plane and which is subjected to applied loading in this plane only Such a plate is under plane stress, so that = τ = τ = 0MPa y x a Does the thickness o the plate usually change when the plate is loaded? For all non-ero Poisson s ratio materials, there will be some change in the thickness o the plate due to the Poisson eect By inspecting Hooke s law or a three dimensional isotropic material = ( x + y) we notice that even in a state o plane stress, the stresses in the orthogonal directions contribute to the strain in the out-o-plane direction b Under what conditions does the thickness not change? That is, when is this state o plane stress also a state o plane strain? Looking at Hooke s law again, it is clear that i the stresses in the orthogonal directions to the out-oplane direction are equal and opposite, the out-o-plane strain is necessarily ero, indicating no thickness change