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Past exam of the solid mechanics course of the University of Oxford / 2023 / 3 / a / Solution by

Codex  0 2026-10-25

The first equation is the local balance of linear momentum. Use

M, L, T

for the dimensions of mass, length, and time, to distinguish dimensional time from the Cauchy stress tensor

T

$x (X, t)$ is the current position of the particle labelled by reference position $X$ ; both have dimension $L$ . The deformation map specifies the particle's trajectory.
$\ddot{x} = \partial_{t}^{2} x ∣_{X}$ is its material acceleration, with dimension $L T^{- 2}$ . The dots follow the particle and are not spatially fixed time derivatives.
$ρ$ is current mass density, with dimension $M L^{- 3}$ . For a homogeneous material with incompressibility, mass conservation makes it constant in space and time.
$b$ is body force per unit mass, such as gravitational acceleration, with dimension $L T^{- 2}$ . Hence $ρ b$ is force per unit volume.
$T$ is the Cauchy stress tensor, mapping a unit normal vector $n$ to surface traction $Tn$ . Each component has dimension $M L^{- 1} T^{- 2}$ , or force per unit area.
$div$ differentiates in the current configuration, with dimension $L^{- 1}$ . In Cartesian coordinates, $(div T)_{i} = \partial T_{ij} / \partial x_{j}$ . Thus $div T$ , $ρ b$ , and $ρ \ddot{x}$ all have dimension $M L^{- 2} T^{- 2}$ .
The superscript $T$ denotes the dimension-preserving matrix transpose. The relation $T = T^{T}$ expresses angular momentum conservation in a continuum without body couples or couple stresses.

Here

t

has dimension

T

. The hypothesis

\partial_{t} b = 0

means that the prescribed spatial body force field is the same for every member of the static family and the resulting motion.

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Past exam of the solid mechanics course of the University of Oxford / 2023 / 2 / c / Solution by

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First let

A ↓ 0

at fixed positive cavity radius

a

and fixed reference outer radius

B

. Then

λ_{a} = \frac{a}{A} ⟶ \infty, λ_{b} ⟶ (1 + \frac{a ^{3}}{B ^{3}})^{1/3},

(1)

and the pressure sustaining a finite cavity is

P (a) = - \int_{(1 + a^{3} / B^{3})^{1/3}}^{\infty} \frac{h ^{'} ( λ )}{λ ^{3} - 1} d λ .

(2)

The onset of cavitation is the additional limit $a ↓ 0$ . The outer stretch then tends to one, and the signed critical pressure is

P_{crit} = - \int_{1}^{\infty} \frac{h ^{'} ( λ )}{λ ^{3} - 1} d λ .

(3)

When this improper integral is positive and finite before the minus sign, the tensile loading magnitude is

- P_{crit} > 0

. Thus the cavity opens under tension, even though the load variable is named pressure. The critical value is independent of the original sphere radius

B

For

A = 0

and

a > 0

, the deformation map is smooth away from

R = 0

but singular at the original centre:

r = (R^{3} + a^{3})^{1/3}

approaches

a

, rather than zero, as

R ↓ 0

. This is the elastic cavitation mechanism; it cannot be represented by a smooth deformation map through the centre.

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The radial unit vector in spherical coordinates is

e_{R} = X / R

, so

X \otimes X / R^{2} = e_{R} \otimes e_{R}

. The identity matrix is the sum of the three basis outer products. Substitution in the given deformation gradient gives

F = f^{'} (R) e_{R} \otimes e_{R} + \frac{f ( R )}{R} (e_{Θ} \otimes e_{Θ} + e_{Φ} \otimes e_{Φ}) = diag (f^{'} (R), \frac{f ( R )}{R}, \frac{f ( R )}{R}) .

(1)

The reference and current radial directions coincide for this spherically symmetric deformation. The positive principal stretches are therefore

λ_{r} = f^{'} (R)

and

λ_{θ} = λ_{ϕ} = f (R) / R

. Incompressibility imposes

f^{'} (R) \frac{f ( R ) ^{2}}{R ^{2}} = 1, \frac{d}{d R} f (R)^{3} = 3 R^{2} .

(2)

Integrating and using

f (A) = a

gives

r = f (R) = (R^{3} + a^{3} - A^{3})^{1/3}, b^{3} = B^{3} + a^{3} - A^{3} .

(3)

In particular

b^{3} - a^{3} = B^{3} - A^{3}

, expressing conservation of shell volume. Writing

λ = r / R

also gives

λ_{r} = λ^{- 2}

and

λ_{θ} = λ_{ϕ} = λ

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Past exam of the solid mechanics course of the University of Oxford / 2023 / 1 / d / ii by

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Multiplying the deformation gradient by its matrix transpose gives the left Cauchy-Green deformation tensor and its trace:

B = 100 0 1 + τ^{2} r^{2} τ r 0 τ r 1, I_{1} = 3 + τ^{2} r^{2} .

(1)

The derivative of the strain energy density is

\overset{w}{^}^{'} (I_{1}) = μ_{1} + μ_{2} I_{1}^{n} .

(2)

For brevity put

g (r) = μ_{1} + μ_{2} (3 + τ^{2} r^{2})^{n}

. Part (a) gives the Cauchy stress tensor, in the current cylindrical coordinate system orthonormal basis, as

T = 2 g - p 00 0 2 g (1 + τ^{2} r^{2}) - p 2 gτ r 0 2 gτ r 2 g - p .

(3)

In particular, the end-face shear stress is

T_{θ z} = 2 gτ r

. The undetermined pressure

p

affects the normal stresses but does not affect this shear stress. No normal traction conditions have been specified to determine

p

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Use separate moving orthonormal bases in the reference and current cylindrical coordinate systems. For pure torsion of an incompressible cylinder, the deformation map is

x (R, Θ, Z) = R e_{r} (Θ + τ Z) + Z e_{z} .

(1)

The unit vectors depend on the angle:

\partial_{θ} e_{r} = e_{θ}

and

\partial_{θ} e_{θ} = - e_{r}

. Consequently

\partial_{R} x = e_{r}, \partial_{Θ} x = R e_{θ}, \partial_{Z} x = τ R e_{θ} + e_{z} .

(2)

In particular, the last derivative includes the rotation of

e_{r}

along the cylinder. The material gradient has the reference coordinate scale factor

1/ R

F = (\partial_{R} x) \otimes E_{R} + \frac{1}{R} (\partial_{Θ} x) \otimes E_{Θ} + (\partial_{Z} x) \otimes E_{Z} = e_{r} \otimes E_{R} + e_{θ} \otimes E_{Θ} + (τ r e_{θ} + e_{z}) \otimes E_{Z} .

(3)

The outer products specify the images of the three reference unit vectors. Thus, with current basis vectors indexing rows and reference basis vectors indexing columns,

F = 100010 0 τ r 1 .

(4)

Its determinant is one, checking incompressibility. The radius is unchanged, so the current outer radius in part (d) is

a = A

. The formula at the coordinate axis is understood by continuity in Cartesian coordinates.

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Write the first strain invariant in components of the deformation gradient:

I_{1} = tr (F F^{T}) = \sum_{i, I} F_{i I}^{2} .

(1)

Here lower-case and upper-case indices refer to the current configuration and reference configuration, respectively. Differentiating gives

\partial I_{1} / \partial F_{i I} = 2 F_{i I}

. The chain rule therefore gives

\frac{\partial W}{\partial F _{i I}} = 2 w^{'} (I_{1}) F_{i I}, \partial_{F} W = 2 w^{'} (I_{1}) F .

(2)

Substitution into the given hyperelastic material law, with its indicated matrix transpose, yields

T = F (\partial_{F} W)^{T} - p 1 = 2 w^{'} (I_{1}) F F^{T} - p 1 = 2 \frac{\partial w}{\partial I _{1}} B - p 1 .

(3)

The left Cauchy-Green deformation tensor is

B = F F^{T}

. The scalar

p

is the Lagrange multiplier enforcing incompressibility, rather than a function fixed by the strain energy density alone. The identity matrix

1

makes this an isotropic constraint reaction in hyperelastic stress.

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Past exam of the simulation and statistical programming course of the University of Oxford / 2023 / 3 / c / iv by

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Past exam of the simulation and statistical programming course of the University of Oxford / 2023 / 3 / b / Solution by

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An inverse matrix exists only if the lower triangular matrix has no zero diagonal entries; this necessary assumption is implicit in the request for

L^{- 1}

. Partition

L = (a b 0 C) .

(1)

Writing its inverse in the same block form and multiplying gives the recursive inverse of a lower triangular matrix:

L^{- 1} = (a^{- 1} - C^{- 1} b / a 0 C^{- 1}) .

(2)

Indeed the lower-left block of the product is

b / a - C (C^{- 1} b / a) = 0

, and the diagonal blocks are identities. This gives an implementation with no call to solve:

ltinv <- function(L) {
  n <- nrow(L)
  if (n == 1L) return(matrix(1 / L[1L, 1L], 1L, 1L))
  j <- 2L:n
  Cinv <- ltinv(L[j, j, drop = FALSE])
  B <- matrix(0, n, n)
  B[1L, 1L] <- 1 / L[1L, 1L]
  B[j, j] <- Cinv
  B[j, 1L] <- -as.vector(Cinv %*% L[j, 1L, drop = FALSE]) / L[1L, 1L]
  B
}

The preserved matrix dimensions also handle

n = 2

. A mathematical induction using the block identity proves that every returned matrix is the desired inverse. Applied to the Cholesky decomposition from (a), all diagonal entries are positive, so all divisions are valid.

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The following R code implements the given recursion for the Cholesky decomposition. The input is assumed to be a real symmetric matrix that is a positive-definite matrix.

mychol <- function(A) {
  n <- nrow(A)
  if (n == 1L) return(matrix(sqrt(A[1L, 1L]), 1L, 1L))
  j <- 2L:n
  L <- matrix(0, n, n)
  L[1L, 1L] <- sqrt(A[1L, 1L])
  b <- A[j, 1L, drop = FALSE] / L[1L, 1L]
  L[j, 1L] <- b
  S <- A[j, j, drop = FALSE] - tcrossprod(b)
  L[j, j] <- mychol(S)
  L
}

Here tcrossprod(b) is the outer product

b b^{T}

. The drop = FALSE arguments preserve matrix dimensions, including the final one-by-one subproblem, and the base case also returns a matrix.

To verify correctness, partition

A = (a r r^{T} B)

. The code sets

ℓ = a

b = r / ℓ

and recursively factors

S = B - b b^{T}

. This Schur complement is a positive-definite matrix: for

z \neq = 0

z^{T} S z = (- r^{T} z / a z)^{T} A (- r^{T} z / a z) > 0.

(1)

C C^{T} = S

is the recursive answer, then the returned lower triangular matrix satisfies

L = (ℓ b 0 C), L L^{T} = (a ℓ b ℓ b^{T} b b^{T} + C C^{T}) = (a r r^{T} B) = A .

(2)

The one-dimensional identity starts this mathematical induction proof. Each recursive call reduces the matrix size by one and therefore terminates.

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Video 1.

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. Source.

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This feature makes it possible for readers to find better explanations of any topic created by other writers. And it allows writers to create an explanation in a place that readers might actually find it.
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Video 2.
OurBigBook Web topics demo
. Source.
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Figure 2.
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