 Solution

ID: past-exam-of-the-solid-mechanics-course-of-the-university-of-oxford/2023/3/a/solution

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

T

\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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