Continuum mechanics Ian Hewitt, University of Oxford - PowerPoint PPT Presentation

Continuum mechanics Ian Hewitt, University of Oxford hewitt@maths.ox.ac.uk

Continuum mechanics A continuum approximation treats a material as having a continuous distribution of mass. It applies on scales much larger than inter-molecular distances. Each ‘point’ of the continuum can be ascribed properties, such as density, temperature, velocity, pressure, etc. ) p ( x , t ) ρ ( x , t ) ) T ( x , t ) ) u ( x , t ) Continuum mechanics provides a mathematical framework to describe how these properties vary in space and time. Continuum mechanics can be used to describe both ‘fluids’ and ‘solids’ - we focus on fluids.

Kinematics - Coordinate systems / derivatives - Strain rate Dynamics - Stress tensor - Constitutive laws Conservation laws - Conservation of mass - Conservation of momentum - Navier-Stokes equations - Conservation of energy Boundary conditions Depth-integrated models

Kinematics

Coordinate systems Eulerian description ( x , t ) Spatial coordinates, fixed in space x x = ( x, y, z ) = ( x 1 , x 2 , x 3 ) Lagrangian description ( X , t ) Spatial coordinates, fixed in material X We usually choose these as the coordinates of a reference configuration at t = 0 velocity u = ( u, v, w ) = ( u 1 , u 2 , u 3 ) ! D x Material paths are governed by x ( X , t ) x | t =0 = X Dt = u u x D (i.e. the derivative ‘following the fluid’) where is the time rate of change for fixed = X Dt

Coordinate systems A stake drilled into the ice tracks the ice motion in a Lagrangian system. A weather station on the ice surface measures atmospheric properties in a (roughly) Eulerian framework. Fluid models are usually written in an Eulerian coordinate system.

Material derivative Given some function of Eulerian coordinates (e.g. temperature) T = f ( x , t ) we can calculate the material derivative using the chain rule (recall ) x = x ( X , t ) DT Dt = ∂ T ∂ t + u · r T local term advective term rate of change with respect to time at fixed x ∂ T/ ∂ t rate of change with respect to x r T r rate of change of with respect to time at fixed = X u x The material derivative is also called the ‘convective’ derivative or ‘total’ derivative. DT Dt = ∂ T ∂ T = ∂ T ∂ t + u ∂ T ∂ x + v ∂ T ∂ y + w ∂ T In components, ∂ t + u i ∂ x i ∂ z 3 ∂ T ∂ T We use the summation convention (repeated indices imply a sum): X u i = u i ∂ x i ∂ x i i =1

Material derivative Example The rate of change of temperature as measured by a skier has components due to: - the temperature decreasing through the evening - the temperature increasing as they travel downhill DT Dt = ∂ T ∂ t + u · r T

6 Strain rate Strain is a measure of deformation. The strain rate is a measure of how fast strain is changing. One dimension Consider the rate of change of length of a small fluid element � � X � Dt ∂ x d x stretching Dt (d x ) = d u = ∂ u D 1 Dt (d x ) = ∂ u D ∂ x d x Time rate d x ∂ x Three dimensions The strain rate is now described by a rank-two tensor (a matrix) d x = ˆ s d s 1 Dt (d s ) = 1 D s T ( r u + r u T )ˆ d x = ˆ 2ˆ s = ˆ s i ˙ ε ij ˆ s d s s j d s ✓ ∂ u i x y z ε ij = 1 ◆ + ∂ u j where the strain rate tensor is ˙ x y 2 ∂ x j ∂ x i Time x y

Strain rate Examples q @ A � 0 1 x 0 1 1 0 0 u = 0 @ A ε ij = ˙ 0 0 0 @ A � z 0 0 � 1 0 @ A 0 1 z 1 0 1 0 0 2 u = 0 ε ij = ˙ 0 0 0 @ A @ A 0 1 0 0 2 0

Dynamics

Stress tensor Stress is force per unit area. The stress state is described by a rank-two tensor. At each point in the material, consider a small cube. = σ ij ij n j We define the Cauchy stress tensor as the force per σ = σ ij q unit area in the i direction on the face with normal in the j direction. 0 1 σ xx σ xy σ xz σ = σ ij = σ yx σ yy σ yz @ A σ zx σ zy σ zz Due to conservation of angular momentum, the stress tensor must be symmetric. s = σ · n p = � 1 ✓ ◆ We define the pressure by 3 σ ii or and the deviatoric stress tensor by + τ σ ij = � p δ ij + τ ij σ = � p δ + τ n The stress acting on a general surface with unit normal is n s or, in index notation, s = σ · n s i = σ ij n j

Constitutive law The constitutive law describes a relationship between stress and strain rates - it characterises the rheology of the material For a Newtonian fluid (e.g. water) η is the viscosity τ ij = 2 η ˙ ε ij For ice, it is common to use Glen’s flow law ≈ − × q ε ij = A ( T ) τ n − 1 τ ij A ≈ 2 . 4 × 10 � 24 Pa � 3 s � 1 at 0 � C 1 ˙ τ = n ≈ 3 2 τ ij τ ij This can be written in the form of a Newtonian fluid but with an effective (non-constant) viscosity 1 τ ij = 2 η ˙ ε ij η = 2 A τ n − 1 ε τ n = 1 ε τ n = 3 n = ∞ ˙ ε τ

Conservation laws

Conservation of mass A major concern at Karthaus … d M sources - sinks d t = Q Z Time

Conservation of mass d M Z Z d t = a d S + a b d S � Q c surface bed Z Z a d ace M � Q c a b bed

Conservation of mass Conservation of mass applies to each arbitrary (Eulerian) volume in the ice. Z Z V u · n d Z Z ‘sources and sinks’ here are due to ρ d V = � ρ u · n d S d t material flowing through the boundary V ∂ V V Z Z Z Z ∂ V Z ∂ρ Z Use divergence theorem ∂ t d V = � r · ( ρ u ) d V Z V V ∂ρ Since this is true for any V ∂ t + r · ( ρ u ) = 0 D ρ If the material is incompressible , , we obtain Dt = 0 r · u = 0

Conservation of mass An alternative derivation is to consider arbitrary material (Lagrangian) volumes V ( t ) u · n d Z since the boundary is now a material ρ d V = 0 Z d t surface, no mass crosses it. V ( t ) V ( t ) Z Z ∂ V ∂ρ Use Reynolds transport theorem ∂ t + r · ( ρ u ) d V = 0 Z V ( t ) ∂ρ Since this is true for any V ∂ t + r · ( ρ u ) = 0

Conservation of momentum We apply a similar argument to conserve momentum for each volume V Momentum conservation is equivalent to Newton’s second law : Rate of change of momentum is equal to the forces acting d Z Z Z Z u · n ρ u i d V = � ρ u i u j n j d S + σ ij n j d S + ρ g i d V d t ∂ V ∂ V V V flux of momentum surface forces body force V through boundary (gravity) ∂ V Z Z Z Z Z Z ( ρ u i u j ) + ∂σ ij ∂ � ∂ Apply divergence theorem Z Z ∂ t ( ρ u i ) d V = + ρ g i d V ∂ x j ∂ x j V V ∂ ∂ ( ρ u i u j ) = ∂σ ij Use that volume is arbitrary ✓ ◆ ∂ t ( ρ u i ) + + ρ g i ∂ x j ∂ x j ✓ ◆ ✓ ∂ u ◆ Use conservation of mass (in vector form) ∂ t + u · r u = r · σ + ρ g ρ

Navier-Stokes equations ✓ ◆ We have derived mass and momentum equations for an incompressible fluid ✓ ∂ u ◆ r · u = 0 ∂ t + u · r u = �r p + r · τ + ρ g ρ Combining with the Newtonian rheology gives the Navier-Stokes equations τ ij = 2 η ˙ ε ij ✓ ∂ u ◆ = �r p + η r 2 u + ρ g r · u = 0 ∂ t + u · r u ρ ✓ ◆ constant viscosity is used here this term is non linear!

Reynolds number ✓ ◆ ∂ t + u · r u = 1 ∂ u ρ r · σ + g Estimate the size of terms in the momentum equation for an ice sheet ⇠ ⇡ r ⇠ ⇠ u ⇠ U ⇡ 100 m y − 1 g ⇠ g ⇡ 9 . 8 m s − 2 x ⇠ L ⇡ 1000 m σ ⇠ ρ gz ⇠ ⇡ u · r u ⇠ 10 − 14 m s − 2 The inertial terms on the left are much much smaller than those on the right. Re = ρ UL More generally, the relative size of these terms is measured by the Reynolds number η - this is a measure of how ‘fast’ the flow is. For small Reynolds number (‘slow flow’) we have the Stokes equations ε ij = A ( T ) τ n − 1 τ ij r · u = 0 0 = �r p + r · τ + ρ g ˙

High Reynolds number flows ✓ ◆ ∂ t + u · r u = 1 ∂ u ρ r · σ + g For flows with high Reynolds number (e.g. most atmosphere and ocean processes) we can usually ignore the viscous terms. However, such flows are often turbulent , and there are Reynolds stresses (due to fluctuations in the velocity field) that have to be parameterised ∂ t + u · r u = � 1 ∂ u ⌦ u 0 u 0 ↵ ρ r p � r · + g Reynolds stresses ⌦ 0 0 ↵ When inertia is important we may also have to worry about the effects of Earth’s rotation D u D u becomes Dt + 2 Ω ∧ u + Ω ∧ ( Ω ∧ x ) Dt

Conservation of energy The same methods work to derive an energy equation. Rate of change of energy is equal to the work done by forces and net conductive heat transfer d Z Z Z Z Z 2 | u | 2 ) d V = � 2 | u | 2 ) u · n d S + ρ ( e + 1 ρ ( e + 1 k r T · n d S + u · σ · n d S + ρ u · g d V d t ∂ V ∂ V ∂ V V V u · n flux of energy conductive work done work done against through boundary transfer against gravity surface forces V ∂ V ✓ ∂ T ◆ De DT Applying the usual arguments … ∂ t + u · r T = r · ( k r T ) + τ ij ˙ ρ c p ε ij Dt = c p Dt ◆

Boundary conditions