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

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