Lecture 8: Tissue Mechanics

Transcription

1 Computational Biology Group (CoBi), D-BSSE, ETHZ Lecture 8: Tissue Mechanics Prof Dagmar Iber, PhD DPhil MSc Computational Biology 2015/16

2 7. Mai / 57 Contents 1 Introduction to Elastic Materials Hooke s Law Uniform Strain Inside Elastic Materials The Tensor of Elasticity Isotropic Material 2 Tissue as a viscous Fluid 3 Models of Limb Bud Growth

3 Tissue as a visco-elastic material Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57

4 Tissue as a visco-elastic material The key differences between the behaviours of solids and fluids lies in how they respond to the application of a force: Elastic Solids Viscous Fluids The deformation is independent of the time over which the force is applied. The deformation disappears when the force is removed. Tissue as visco-elastic material A fluid continues to flow as long as the force is applied A fluid will not recover its original form when the force is removed. Tissue shows a mix of the two extreme properties. We will therefore now discuss the theory of visco-elastic fluids. Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57

5 Introduction to Elastic Materials Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57

6 7. Mai / 57 Definition of mechanical terms Stress. A coarse-grained description of the forces within a tissue. When a piece of tissue experiences a force from a neighboring tissue region (A), mechanical stress (σ) is defined as the ratio of the force (F ) to the area of contact (A) with that region (B, top equation). Tension and compression correspond to forces pointing respectively outwards from and inwards to the body. Deformation (also called strain ) is the relative change in size of an object subjected to a force. In one dimension, it is a dimensionless number, ɛ (B, bottom equation): the fraction of change in the object length (where L is the new length and L 0 the original length), which would be positive for elongation or negative for contraction.

7 7. Mai / 57 Hooke s Law According to Hooke s law, the force F needed to extend or compress a solid bar (spring) by some distance l is proportional to that distance. That is, F l. (1) The lengthening l of the bar will also depend on its length l. Thus, if the same forces act on each of two blocks that are put together, each will stretch by l. We thus see that we must have F l. (2) l

8 7. Mai / 57 Hooke s Law The force will also depend on the area of the block, and must be proportional to the cross-sectional area A (imagine to blocks side by side that are deformed by l). We then have F = YA l, (3) l where Y is the Young modulus. We can rewrite this as Hooke s Law F A }{{} stress = Y l l }{{} strain, (4)

9 7. Mai / 57 Hooke s Law When one stretches a block of material in one direction, it contracts at right angles to the stretch. We thus have for homogenous, isotropic materials w w = h l = σ, (5) h l where w and h refer to the width and height of the bar, and σ us called Poisson s ratio. Finally, since we are in a linear regime, we can use superposition, i.e. so forces and their effects are additive.

10 7. Mai / 57 Limits of Hooke s Law Hooke s law is only a first-order linear approximation to the real response of springs and other elastic bodies to applied forces. It must eventually fail once the forces exceed some limit, since no material can be compressed beyond a certain minimum size, or stretched beyond a maximum size, without some permanent deformation or change of state. Many materials will noticeably deviate from Hooke s law well before those elastic limits are reached.

11 7. Mai / 57 Uniform Strain Consider a rectangular block exposed to uniform hydrostatic pressure. If we push on the ends of the block with a pressure p, the compressional strain is p/y, and it is negative, l 1 = p }{{} l }{{} Y strain compressional strain (6) If we push on the two sides of the block with pressure p, the compressional strain is again p/y, but we now want the lengthwise strain, which we get according to Eq. 5, by multiplying with σ, l 2 l }{{} strain = l 3 = +σ p }{{} l }{{ Y} strain sideways strain (7)

12 7. Mai / 57 Uniform Strain Combined we have (recognizing that the argument is symmetric in all directions), l l = w w = h h = i l i l = p (1 2σ) } Y {{} strain (8) The change in volume, V = lwh, under hydrostatic pressure is then for small displacements V V = l l + w w + h h = 3 p (1 2σ) } Y {{} strain (9) V V is referred to as volume strain.

13 7. Mai / 57 Uniform Volume Strain The volume stress p is proportional to the volume strain Volume Strain p = K V V, K = Y 3(1 2σ). (10) The coefficient K is called the bulk modulus.

14 7. Mai / 57 Inside Elastic Materials We are now interested in what happens inside an elastic body, i.e. we want to determine the local stress and strain at every point in an elastic body. To this end, we define the displacement vector u = r r (11) of between the location of a point in the original and stretched material.

15 7. Mai / 57 Uniform Stretching We start by considering homogenous strain throughout the material as we would have in case of uniform stretching. Now in x-direction we would have u x x = l (12) l where u x refers to the x-entry of the displacement vector. We can then write u x = ε xx x, ε xx = l. (13) l

16 7. Mai / 57 Non-Uniform Stretch If the strain is not uniform, then the relation between u x and x will vary from place to place in the material and we have ε xx = u x x, ε yy = u y y, ε zz = u z z. (14) For shear-type strains we have ε xy = ε yx = 1 2 ( uy x + u ) x. (15) y For a pure rotation they are both zero, but for a pure shear we get that ε xy is equal to ε yx as required.

17 7. Mai / 57 Symmetric strain tensor We thus have the symmetric strain tensor ε ij = ε ji = 1 2 ( uj i + u ) i, (16) j and u x = ε xx x + ε xy y + ε xz z. (17) In principle, there could also be twisting (rotations), which we will ignore.

18 The Tensor of Elasticity According to Hooke s Law, each component of the stress tensor S ij is linearly related to each of the components of the strain. Here, the i, j component of the stress tensor S ij represents the ithe component of the force across a unit area perpendicular to the j-axis. Since S and ε each have nine components, there are 9 9 = 81 possible coefficients which describe the properties of elastic materials. They are constants if the material itself is homogeneous. If we write these coefficients as C ijkl we have 3 3 S i j = C i j k l ε k l. (18) k=1 l=1 i, j, k, l all take on the values 1,2, or 3. C ijkl is a tensor of the fourth rank, the tensor of elasticity. Since both S i j and ε k l are symmetric, each with only six different terms, there can be at most 36 different terms in C ijkl.there are, however, usually many fewer than this. Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57

19 7. Mai / 57 Units 3 3 S i j = C i j k l ε k l k=1 l=1 For continuous media, each element of the stress tensor S is a force divided by an area; it is therefore measured in units of pressure, namely pascals (Pa, or N/m 2, or kg/(ms 2 ). The elements of the strain tensor ɛ are dimensionless (displacements divided by distances). Therefore, the entries of C ijkl are also expressed in units of pressure.

20 7. Mai / 57 Isotropic Material For isotropic media (which have the same physical properties in any direction), C can be reduced to only two independent numbers: the bulk modulus K the shear modulus, that quantify the material s resistance to changes in volume and to shearing deformations.

21 7. Mai / 57 Isotropic Material Isotropic materials are characterized by properties which are independent of direction in space. Physical equations involving isotropic materials must therefore be independent of the coordinate system chosen to represent them. The strain tensor is a symmetric tensor. Since the trace of any tensor is independent of any coordinate system, the most complete coordinate-free decomposition of a symmetric tensor is to represent it as the sum of a constant tensor and a traceless symmetric tensor. In index notation: ε ij = ( 1 3 ε kkδ ij ) }{{} volumetric strain tensor where δ ij is the Kronecker delta. ) + (ε ij 1 3 ε kkδ ij }{{} shear tensor (19)

22 7. Mai / 57 Isotropic Material The most general form of Hooke s law for isotropic materials may now be written as a linear combination of these two tensors: ( ) ) S ij = 3K 1 3 ε kkδ ij + 2G (ε ij 1 3 ε kkδ ij (20) where K is the bulk modulus and G is the shear modulus. Thus, ( ( ) S 11 = 3K 1 3 (ε 22 + ε 33 )) + 2G ε (ε 22 + ε 33 ). (21)

23 Isotropic Material Differently said, for isotropic materials, the components of C must be the same for any choice of coordinate system. This is the case only if C xxxx = C xxyy + C xyxy (22) C xxyy = λ = Y ( ) σ (23) 1 + σ 1 2σ C xyxy = 2µ = Y 1 + σ C xxxx = 2µ + λ = Y 1 + σ ( 1 + σ ) 1 2σ (24) (25) (26) where Y is then Young modulus and σ the Poisson s ratio. We then have ( ) S i j = 2µε ij + λ ε kk δ ij. (27) k Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57

24 A Mechanical Feedback Restricts Sepal Growth and Shape in Arabidopsis A stereotypical growth pattern generates tensile stress at the sepal tip A supracellular microtubule alignment forms along maximal tension at the sepal tip The strength of the mechanical feedback can modulate sepal shape The microtubule response to tension acts as an Hervieux et al, Current Biology, 2016 organ shape-sensing mechanism Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil 7. Mai / 57 Lecture 8 MSc 2015/16

25 A Mechanical Feedback Restricts Sepal Growth and Shape in Arabidopsis Computational Biology Group (CoBi), D-BSSE, ETHZ 7. Mai / 57 Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16

26 Mechanical Measurements Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57

27 Tissue as a viscous Fluid Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57

28 7. Mai / 57 Behaviour of Fluids We are now interested in understanding how external forces affect the behaviour of fluids. Here, we will assume that the fluid behaves as a continuous substance rather than a set of discrete particles. The solution of the Navier-Stokes equations is a flow velocity. The flow velocity is a field, since it is defined at every point in a region of space and an interval of time.

29 7. Mai / 57 Behaviour of Fluids Once the velocity field is calculated, other quantities of interest, such as pressure or temperature, may be found. This is different from what one normally sees in classical mechanics, where solutions are typically trajectories of position of a particle or deflection of a continuum. Studying velocity instead of position makes more sense for a fluid; however for visualization purposes one can compute various trajectories.

30 7. Mai / 57 Navier-Stokes Equation The Navier-Stokes equations are derived from the basic principles of continuity of mass momentum energy. A continuity equation may be derived from conservation principles of mass, momentum, and energy. This is done via the Reynolds transport theorem.

31 7. Mai / 57 The Reynolds transport theorem The Reynolds transport theorem The Reynolds transport theorem is an integral solution relation stating that the sum of the changes of some intensive property (call it φ) defined over a control volume Ω must be equal to what is lost (or gained) through the boundaries of the volume plus what is created/consumed by sources and sinks inside the control volume.

32 Intensive and extensive properties Intensive and extensive properties Intensive property: a bulk property, i.e. it is a physical property of a system that does not depend on the system size or the amount of material in the system, e.g. temperature, T, density, ρ, and hardness of an object, η. Extensive property: is additive for subsystems, i.e. if the system could be divided into any number of subsystems, and the extensive property measured for each subsystem; the value of the property for the system would be the sum of the property for each subsystem (e.g. mass, m, volume, V ). The ratio of two extensive properties of the same object or system is an intensive property. For example, the ratio of an object s mass and volume, which are two extensive properties, is density, which is an intensive property. Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57

33 7. Mai / 57 The Reynolds transport theorem The Reynolds transport theorem for an intensive property φ is expressed by the following integral continuity equation: d φ dv = φu n da s dv (28) dt Ω Ω Ω where u is the flow velocity of the fluid and s represents the sources and sinks in the flow, taking the sinks as positive. Ω represents the control volume and Ω its bounding surface.

34 The Reynolds transport theorem The divergence theorem may be applied to the surface integral, changing it into a volume integral: d φ dv = (φu) dv s dv. (29) dt Ω Ω Applying Leibniz s rule to the integral on the left and then combining all of the integrals: φ Ω t Ω dv = (φu) dv s dv (30) Ω ( ) φ t + (φu) + s dv = 0. (31) Ω The integral must be zero for any control volume; this can only be true if the integrand itself is zero, so that: φ + (φu) + s = 0. (32) t Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57 Ω

35 7. Mai / 57 The Reynolds transport theorem φ + (φu) + s = 0. (33) t From this valuable relation (a very generic continuity equation), three important concepts may be concisely written: conservation of mass, conservation of momentum, and conservation of energy. Validity is retained if φ is a vector, in which case the vector-vector product in the second term will be a dyad.

36 7. Mai / 57 Conservation of Momentum A general momentum equation is obtained when the conservation relation is applied to momentum. If the intensive property φ considered is the mass flux (also momentum density), i.e. the product of mass density and flow velocity ρu, by substitution in the general continuum equation: (ρu) + (ρuu) = f (34) t which corresponds to ( ) ( ) ρ u u t + (ρu) + ρ t + u u = f (35)

37 7. Mai / 57 Conservation of Momentum & Mass ( ) ( ) ρ u u t + (ρu) + ρ t + u u = f (36) The leftmost expression relates to the conservation of mass ρ t + (ρu) = S (37) where ρ is the mass density (mass per unit volume), u is the flow velocity. S is a mass source. Typically, S = 0. However, in tissue models S 0 in case of growth. In case of incompressible fluids ρ t = 0 and ρ = 0.

38 7. Mai / 57 Conservation of Momentum & Mass ( ) ( ) ρ u u t + (ρu) + ρ t + u u = f (38) The second term corresponds to the fluid density times its acceleration. Thus, for a velocity v of the fluid, the acceleration at a fixed point in space is v t. However, we also need to take into account velocity changes because of translocations such that the acceleration is given as v t + v v x x + v v y y + v v z z = v t such that the fluid acceleration is given by + (v )v (39) ( ) v ρ t + (v )v. (40)

39 7. Mai / 57 Acceleration ( ) v ρ t + (v )v = m acceleration = F V V = f. (41) According to Newton s law, mass times acceleration is equal to the force on that particular volume element. So what forces f act on the fluid?

40 7. Mai / 57 Hydrostatics: Fluids at Rest We start with fluids at rest. When liquids are at rest, there are no shear forces (not even for viscous fluids). The law of hydrostatics, therefore, is that stresses are always normal to any surface inside the fluid. The normal force per uni area is called the pressure. Since there is no shear in a static fluid, it follows that the pressure stress is the same in all directions. The pressure in a resting fluid may, however, vary from place to place such that we have pressure p at x and pressure p + p dx at x + dx such that the resulting force is given by x F = ( p (p + p ) x dx) dydz = p dv (42) x and the force density per unit volume as f = F V = p. (43)

41 7. Mai / 57 Friction in Moving Fluids == Viscosity When a fluid is set in motion, different parts of the fluid move with different velocities. Just as there is friction when one surface of a solid slides over another, so there is friction when one layer of a fluid slides over another. This friction in fluids is called viscosity.. Imagine a layer of fluid of thickness s between two flat plates. The upper plate moves at velocity v. To maintain the moving plate at a constant speed, it is found experimentally that a force F is required which is directly proportional to the velocity v, inversely proportional to the separation s, and directly proportional to the area of the moving plate A, F = η Av s. (44) The proportionality constant η is called the coefficient of viscosity.

42 Hydrodynamics: Moving Fluids In a dynamic situation, shear stress can build up in a viscous fluid. If the coefficient of dynamic viscosity is constant we have f = p + µ u. (45) In the most general form (for compressible fluids) we have ( ) ( ) ρ u u t + (ρu) +ρ t + u u = p+µ( u+ 1 3 ( u))+f In case of incompressible flow ρ t = 0 ρ = 0 1 ( u) 3 = 0 [ 0 for spatially inhomogenuous mass source] Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57

43 7. Mai / 57 Units & Magnitudes Dynamic viscosity µ 1 Pa s for tissue. Density of water ρ 10 3 kg/m 3. Kinematic viscosity is then ν = µ m2 ρ 10 3 s. Reynolds number: Re = L U ν = with characteristic length scale L 10 3 m and characteristic speed U = s m/s = m/s.

44 Models of Limb Bud Growth Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57

45 A model of Limb Bud Growth Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57

46 Navier-Stokes Model of Limb Bud Growth Navier-Stokes equations: (Motion of a viscous fluid - Tissue as a viscous, incompressible fluid. ) ρ ( ) u t + u u = p + µ( u + 1 ( u)) + F 3 Continuity equation: (Conservation of mass - Volume increases due to proliferation) ρ u = S(c) Boundary Forces: (Forces induced by the Ectoderm) F = f (s, t)δ(x X(s, t)ds Reaction-diffusion-convection: (Signaling Network) Ω c + (uc) = D c + R(c) t Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57

47 Signalling-dependent Growth Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57

48 Signalling-dependent Growth Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57

49 3D Limb Bud Shapes from OPT Computational Biology Group (CoBi), D-BSSE, ETHZ 7. Mai / 57 Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16

50 Shape Changes as the result of proliferation? Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57

51 Change in Shape over 6h of development Computational Biology Group (CoBi), D-BSSE, ETHZ 7. Mai / 57 Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16

52 Proliferation Rates in the Limb Bud Computational Biology Group (CoBi), D-BSSE, ETHZ 7. Mai / 57 Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16

53 Proliferation insufficient predictor of shape change Computational Biology Group (CoBi), D-BSSE, ETHZ 7. Mai / 57 Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16

54 Numerical Optimization Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57

55 Optimized Rates include Shrinkage Computational Biology Group (CoBi), D-BSSE, ETHZ 7. Mai / 57 Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16

56 SUMMARY OF LIMB BUD GROWTH MODELS Computational Biology Group (CoBi), D-BSSE, ETHZ Prof Dagmar Iber, PhD DPhil Lecture 8 MSc 2015/16 7. Mai / 57

57 7. Mai / 57 Thanks!! Thanks for your attention! Slides for this talk will be available at: