Exam (elaborations) TEST BANK FOR Elasticity By J. R. Barber (Solution Manual)

Exam (elaborations) TEST BANK FOR Elasticity By J. R. Barber (Solution Manual) CHAPTER 1 1.1. Show that (i) ∂xi ∂xj = δij and (ii) R = √xixi , where R = |R| is the distance from the origin. Hence find ∂R/∂xj in index notation. Confirm your result by finding ∂R/∂x in x, y, z notation. For an orthogonal co¨ordinate system, ∂x ∂y = 0 (this is what is meant by orthogonality) and ∂x ∂x = 1 . In index notation, these results can be combined as ∂xi ∂xj = δij . The distance from the origin is R = q x2 1 + x2 2 + x2 3 = √xixi . Combining these results, we have ∂R ∂xj = ∂ ∂xj √xixi = 1 2√xixi ∂xi ∂xj xi + xi ∂xi ∂xj ! = 2xiδij 2√xixi = xj √xixi . In x, y, z notation, we would have R = √x2 + y2 + z2 and hence ∂R ∂x = (2x) 2√x2 + y2 + z2 = x R , which agrees. 1.2. Prove that the partial derivatives ∂2f/∂x2; ∂2f/∂x∂y; ∂2f/∂y2 of the scalar function f(x, y) transform into the rotated co¨ordinate system x′, y′ by rules similar to equations (1.15–1.17). We first note from equation (1.43) that ∂ ∂x′ = cos θ ∂ ∂x + sin θ ∂ ∂y and by a similar argument ∂ ∂y′ = ∇.j′ = i.j′ ∂ ∂x + j.j′ ∂ ∂y = −sin θ ∂ ∂x + cos θ ∂ ∂y . We then have ∂2f ∂x′2 = cos θ ∂ ∂x + sin θ ∂ ∂y ! cos θ ∂f ∂x + sin θ ∂f ∂y ! = cos2 θ ∂2f ∂x2 + sin2 θ ∂2f ∂y2 + 2 sin θ cos θ ∂2f ∂x∂y ∂2f ∂x′∂y′ = −sin θ ∂ ∂x + cos θ ∂ ∂y ! cos θ ∂f ∂x + sin θ ∂f ∂y ! = (cos2 θ − sin2 θ) ∂2f ∂x∂y + sin θ cos θ ∂2f ∂y2 − ∂2f ∂x2 ! ∂2f ∂y′2 = −sin θ ∂ ∂x + cos θ ∂ ∂y ! −sin θ ∂ ∂x + cos θ ∂ ∂y ! = cos2 θ ∂2f ∂y2 + sin2 θ ∂2f ∂x2 − 2 sin θ cos θ ∂2f ∂x∂y and these equations are clearly of the same form as (1.15–1.17). 1.3. Show that the direction cosines defined in (1.19) satisfy the identity lij lik = δjk . Hence or otherwise, show that the product σijσij is invariant under co¨ordinate transformation. For a given value of j, lij defines the components in x′ i co¨ordinates of a unit vector in the direction of the xj -axis. It follows that lij lik , is the dot product between two unit vectors defined in the x′ i-system. One of these vectors represents the xj -axis and the other the xk-axis. This dot product is unity if the axes are identical and zero if they are not, since the three axes are orthogonal. Hence lij lik = δjk . Now consider σ′ ij = lipljqσpq , from equation (1.22). We can write another version of the same quantity using different dummy indices as σ′ ij = lirljsσrs . We need to do this because otherwise when we take the product the same index would appear more than twice which leads to an ambiguity in terms of the summation convention. Taking the product of these quantities, including the implied summations, we then have σ′ ijσ′ ij = lipljqlirljsσpqσrs and using the identity we proved above, this gives σ′ ijσ′ ij = δprδqsσpqσrs = σpqσpq , showing that the product is invariant under co¨ordinate transformation. 1.4. By restricting the indices i, j etc. to the values 1,2 only, show that the two-dimensional stress transformation relations (1.15–1.17) can be obtained from (1.22) using the two-dimensional direction cosines (1.20). From (1.20), we have l11 = cos θ ; l12 = sin θ ; l21 = −sin θ ; l22 = cos θ and from (1.22), σ′ ij = lipljqσpq . Thus, σ′ 11 = l11l11σ11 + l12l11σ21 + l11l12σ12 + l12l12σ22 = σ11 cos2 θ + 2σ12 sin θ cos θ + σ22 sin2 θ σ′ 12 = l11l21σ11 + l12l21σ21 + l11l22σ12 + l12l22σ22 = −σ11 sin θ cos θ − σ21 sin2 θ + σ12 cos2 θ + σ22 sin θ cos θ σ′ 22 = l21l21σ11 + l22l21σ21 + l21l22σ12 + l22l22σ22 = σ11 sin2 θ − 2σ12 sin θ cos θ + σ22 cos2 θ and these are identical to (1.15–1.17). 1.5. Use the index notation to develop concise expressions for the three stress invariants I1, I2, I3 and the equivalent tensile stress σE. In index notation, I1 = σ11 + σ22 + σ33 = σii , see equation (1.3). I2 = σ11σ22 + σ22σ33 + σ33σ11 − σ2 12 − σ2 23 − σ2 31 . This is clearly a product of stresses. Two simple product terms in index notation are I2 1 = σiiσjj = σ2 11 + σ2 22 + σ2 33 + 2σ11σ22 + 2σ22σ33 + 2σ33σ11 and σijσij = σ2 11 + σ2 22 + σ2 33 + 2σ2 12 + 2σ2 23 + 2σ2 31 . Comparing these expressions, we see that I2 can be written I2 = σiiσjj − σijσij 2 . The third stress invariant I3 = σ11σ22σ33 − σ11σ2 23 − σ22σ2 31 − σ33σ2 12 + 2σ12σ23σ31 is the determinant of the stress matrix σij . It can be written I3 = 1 6 ǫijkǫpqrσipσjqσkr . From the above expressions, I2 1 − 3I2 = σiiσjj − 3(σiiσjj − σijσij) 2 = (3σijσij − σiiσjj) 2 . It follows that σE = s (3σijσij − σiiσjj) 2 , from equation (1.31). 1.6. Choosing a local co¨ordinate system x1, x2, x3 aligned with the three principal axes, determine the tractions on the octahedral plane defined by the unit vector n = ( 1 √3 , 1 √3 , 1 √3 )T which makes equal angles with all three principal axes, if the principal stresses are σ1, σ2, σ3. Hence show that the magnitude of the resultant shear stress on this plane is √2σE/3, where σE is given by equation (1.31). The traction on the octahedral plane is given by equation (1.21) as ti =   σ1 0 0 0 σ2 0 0 0 σ3     1/√3 1/√3 1/√3   =   σ1/√3 σ2/√3 σ3/√3   or ti = σi √3 . To find the resultant shear stress, we first find the normal compont of the vector t as t · n = tini = σ1 + σ2 + σ3 3 = ¯σ , from equation (1.75). The vector corresponding to this component is (t · n)n = ¯σni and the shear stress on the octahedral plane is a vector defined by ti − ¯σni = ( σ1 − ¯σ √3 , σ2 − ¯σ √3 , σ3 − ¯σ √3 )T . To find the magnitude τoct of this resultant, we take the square root of the sum the squares of the compontents, obtaining τoct = s (σ1 − ¯σ)2 + (σ2 − ¯σ)2 + (σ3 − ¯σ)2 3 = s (2σ1 − σ2 − σ3)2 + (2σ2 − σ3 − σ1)2 + (2σ3 − σ1 − σ2)2 27 . Expanding and simplifying, we obtain τoct = 1 3 q 2 (σ2 1 + σ2 2 + σ2 3 − σ1σ2 − σ2σ3 − σ3σ1) . Comparing this with equation (1.31), we see that τoct = √2σE 3 . 1.7. A rigid body is subjected to a small rotation ωz = ≪ 1 about the z-axis. If the displacement of the origin is zero, find expressions for the three displacement components ux, uy, uz as functions of x, y, z. Since the body is rigid, the strains are zero and hence ∂ux ∂x = ∂uy ∂y = ∂uz ∂z = 0 , (1) 1 2 ∂uy ∂x + ∂ux ∂y ! = 1 2 ∂uz ∂y + ∂uy ∂z ! = 1 2 ∂ux ∂z + ∂uz ∂x ! = 0 . (2) Also, from the given rotations, we have 1 2 ∂uy ∂x − ∂ux ∂y ! = ; 1 2 ∂uz ∂y − ∂uy ∂z ! = 1 2 ∂ux ∂z − ∂uz ∂x ! = 0 . (3) From equations (2,3), we conclude that ∂uy ∂x = ; ∂ux ∂y = − ; ∂uz ∂y = ∂uy ∂z = ∂ux ∂z = ∂uz ∂x = 0 (4) and hence uz is independent of x, y, z, ux is a function of y only and uy is a function of x only. Integrating the first two of equation (4) we then have uy = x + A ; ux = − y + B (5) where A,B are constants. Since the origin has zero displacement, the only permissible displacement field is therefore ux = − y ; uy = x ; uz = 0 . An alternative proof A shorter and more physical way of reaching these results is to note that the rigid body rotating through a small clockwise angle about the z-axis will experience the displacements ur = 0 ; u = r ; uz = 0 in polar co¨ordinates r, θ, z. We then use the transformation rules ux = ur cos θ − u sin θ ; uy = ur sin θ + u cos θ to obtain ux = −r sin θ = − y uy = r cos θ = x . 1.8. Use the index notation to develop a general expression for the derivative ∂ui ∂xj in terms of strains and rotations. It’s probably easier to start this question in x, y, z notation. If the indices are the same, we have (for example) ∂ux ∂x = exx If they are different and in cyclic order, we have ∂ux ∂y = 1 2 ∂uy ∂x + ∂ux ∂y ! − 1 2 ∂uy ∂x − ∂ux ∂y ! = exy − ωz . If they are in reverse cyclic order, we have ∂uy ∂x = 1 2 ∂uy ∂x + ∂ux ∂y ! + 1 2 ∂uy ∂x − ∂ux ∂y ! = exy + ωz . We can generalize these expressions using the index notation by writing ∂ui ∂xj = eij − ǫijkωk . Recall that if the indices are the same, ǫijk = 0 and hence the second term will be zero. If they are in cyclic order, ǫijk = 1, whereas for reverse cyclic order, ǫijk = −1. 1.9. Use the three-dimensional vector transformation rule (1.19) and the index notation to prove that the strain components (1.51) transform according to the equation e′ ij = lipljqepq . Hence show that the dilatation eii is invariant under co¨ordinate transformation. From (1.51) in x′ i co¨ordinates, we have e′ ij = 1 2 ∂u′ i ∂x′ j + ∂u′ j ∂x′ i ! Also, since u is a vector, u′ i = lipup ; u′ j = ljquq , from (1.19). The derivatives are components of the gradient operator which is also a vector, so ∂ ∂x′ i = lip ∂ ∂xp ; ∂ ∂x′ j = ljq ∂ ∂xq . Using these results, we then have e′ ij = 1 2 lipljq ∂up ∂xq + ljqlip ∂uq ∂xp ! = lipljq 2 ∂up ∂xq + ∂uq ∂xp ! = lipljqepq . Using this result, the dilatation is e′ ii = lipliqepq . Now for a given value of p, lip defines the components in x′ i co¨ordinates of a unit vector in the direction of the xp-axis. It follows that lipliq = δpq , since this defines the dot product between two unit vectors along two of a set of three orthogonal axes. It follows that e′ ii = δpqepq = eqq and hence that the dilatation is invariant under co¨ordinate transformation. 1.10 Find an index notation expression for the compliance tensor sijkl of equation (1.55) for the isotropic elastic material in terms of the elastic constants E, ν. We start in x, y, z co¨ordinates and write exx = σxx E − νσyy E − νσzz E = (1 + ν)σxx E − ν(σxx + σyy + σzz) E and exy = (1 + ν)σxy E . These results can be combined in index notation as eij = (1 + ν)σij E − νδijσmm E . Comparing with equation (1.55)2 and getting a hint from the transition from (1.71) to (1.72), we then see that sijkl = (1 + ν) 2E (δikδjl + δilδjk) − νδijδkl E . To verify this, we substitute it into (1.55), obtaining eij = (1 + ν) 2E δikδjlσkl + (1 + ν) 2E δilδjkσkl − νδijδkl E σkl = (1 + ν)σij 2E + (1 + ν)σji 2E − νδijσkk E = (1 + ν)σij E − νδijσkk E , since σij = σji. 1.11. Show that equations (1.58–1.60, 1.64) can be written in the concise form eij = (1 + ν)σij E − νδijσmm E .