Chapter 5
The equations of linear elasticity
5.1 Summary of equations
• Strain-displacement relations: 1 e = (u + u ) (5.1) ij 2 i,j j,i • Equilibrium equations/equations of motion: ∂2u τ + F = ρ i (5.2) ij,j i ∂t2 • Constitutive equations: τij = λδij ekk + 2µeij (5.3)
5.2 Displacement formulation: The Navier-Lam´e equations
• Solve for the displacements: ∂2u (λ + µ)u + µu + F = ρ i (5.4) k,ki i,kk i ∂t2 or symbolically: ∂2u (λ + µ) grad div u + µ∇2u + F = ρ , (5.5) ∂t2 which is equivalent to: ∂2u (λ + 2µ) grad div u − µ curl curl u + F = ρ . (5.6) ∂t2
• This is a system of three coupled linear elliptic PDEs for the three displacements ui(xj ).
5.3 Stress formulation: The static Beltrami-Mitchell equations
• For static deformations, we have 1 − ν 1 − ν τ +F = 0 or symbolically ∇2θ + div F = 0. (5.7) 1 + ν ii,jj i,i 1 + ν θ,jj and the stresses fulfil the|{zBeltrami-Mitc} hell equations: 1 ν τ + τ + δ F +F + F = 0. (5.8) ij,kk 1 + ν kk,ij 1 − ν ij k,k j,i i,j ∇2 τij θ,ij div F | {z } | {z } |{z} • (5.8) represents a system of six coupled linear elliptic PDEs for the six stress components τij (xj ). When these have been determined, the strains can be recovered from (4.6) or (4.16). Then the displacements follow from (5.1). They are only determined up to arbitrary rigid body motions.
12 MT30271 Elasticity: The equations of linear elasticity 13
5.4 Simplifications for F = const.:
• For constant (or vanishing!) body force, the stress, strain and displacement components are bihar- monic functions, ui,jjkk = 0 τij,kkll = 0 eij,kkll = 0 (5.9) or symbolically: 4 4 4 ∇ u = 0 ∇ τij = 0 ∇ eij = 0. (5.10)
• The dilation and the trace of the stress tensor are harmonic functions:
uj,jkk = d,kk = 0 τjj,kk = θ,kk = 0 (5.11)
or symbolically: ∇2d = 0 ∇2θ = 0 (5.12)
• Note that in (5.4) – (5.8) F acts as an inhomogeneity in a system of linear equations. The system can be transformed into a homogeneous system for uh = u−up (with different boundary conditions) if a particular solution up (which does not have to fulfil the boundary conditions) can be found.
5.5 Boundary conditions:
(0) • Displacement (Dirichlet) boundary conditions: Prescribed displacement field ui .
(0) ui|∂D = ui (5.13)
(0) • Stress (Neumann) boundary conditions: Prescribed (applied) traction ti on boundary. Note that nj is the outer unit normal vector on the elastic body.
(0) ti|∂D = τij nj |∂D = ti (5.14)
• Mixed (Robin) boundary conditions – ‘elastic foundation’ represented by the stiffness tensor kij . Physically, this implies that the traction which the elastic foundation exerts on the body is propor- (0) tional to the boundary displacement. This can be combined with an applied traction ti as in the Neumann case. (0) (ti + kij uj ) |∂D = (τij nj + kij uj ) |∂D = ti (5.15) MT30271 Elasticity: The equations of linear elasticity 14
Governing Equations in Cylindrical Polar Coordinates
• x1 = x = r cos θ, x2 = y = r sin θ, x3 = z = z.
u = (ur, uθ, uz), e = (eij ), τ = (τij ), where i, j = r, θ, z.
• Vector calculus: ∂f 1 ∂f ∂f 1 ∂(ru ) 1 ∂u ∂u grad f = ˆr + θˆ + zˆ, div u = r + θ + z , ∂r r ∂θ ∂z r ∂r r ∂θ ∂z 1 ∂u ∂u ∂u ∂u 1 ∂(ru ) 1 ∂u curl u = z − θ ˆr + r − z θˆ + θ − r zˆ. r ∂θ ∂z ∂z ∂r r ∂r r ∂θ
• Stress-strain relations have the same form as in Cartesian coordinates:
τij = λδij div u + 2µeij , i, j = r, θ, z.
• Stress-displacement relations:
∂ur 1 ∂uθ ur ∂uz τrr = λ div u + 2µ , τθθ = λ div u + 2µ + , τzz = λ div u + 2µ , ∂r r ∂θ r ∂z
τ τ ∂u u 1 ∂u τ τ ∂u ∂u τ τ 1 ∂u ∂u rθ = θr = θ − θ + r , rz = zr = z + r , θz = zθ = z + θ . µ µ ∂r r r ∂θ µ µ ∂r ∂z µ µ r ∂θ ∂z
• Strain-displacement relations: ∂u 1 ∂u u ∂u e = r , e = θ + r , e = z , rr ∂r θθ r ∂θ r zz ∂z ∂u u 1 ∂u ∂u ∂u 1 ∂u ∂u 2e = 2e = θ − θ + r , 2e = 2e = r + z , 2e = 2e = z + θ . rθ θr ∂r r r ∂θ rz zr ∂z ∂r zθ θz r ∂θ ∂z • Equilibrium equations (statics): for the displacement formulation, use Navier’s equation,
(λ + 2µ) grad div u − µ curl curl u + F = 0,
whereas for the stress formulation, use ∂τ 1 ∂τ ∂τ τ − τ rr + rθ + rz + rr θθ + F = 0 ∂r r ∂θ ∂z r r ∂τ 1 ∂τ ∂τ 2 rθ + θθ + θz + τ + F = 0 ∂r r ∂θ ∂z r rθ θ ∂τ 1 ∂τ ∂τ 1 rz + θz + zz + τ + F = 0. ∂r r ∂θ ∂z r rz z
• Stress boundary conditions: these are when t is prescribed. We have, from ti = nˆj τij ,
tr = nˆrτrr + nˆθτrθ + nˆzτrz
tθ = nˆrτrθ + nˆθτθθ + nˆzτθz
tz = nˆrτrz + nˆθτθz + nˆzτzz MT30271 Elasticity: The equations of linear elasticity 15
Governing Equations in Spherical Polar Coordinates
• x1 = x = r sin θ cos φ, x2 = y = r sin θ sin φ, x3 = z = r cos θ.
u = (ur, uθ, uφ), e = (eij ), τ = (τij ), where i, j = r, θ, φ.
• Vector calculus: ∂f 1 ∂f 1 ∂f grad f = ˆr + θˆ + φˆ, ∂r r ∂θ r sin θ ∂φ
1 ∂ 2 ∂ ∂ div u = (r sin θ ur) + (r sin θ uθ) + (ruφ) , r2 sin θ ∂r ∂θ ∂φ ˆr rθˆ r sin θ φˆ 1 curl u = ∂ ∂ ∂ . r2 sin θ ∂r ∂θ ∂φ u ru r sin φ u r θ φ
• Stress-strain relations have the same form as in Cartesian coordinates:
τij = λδij div u + 2µeij, i, j = r, θ, φ.
• Stress-displacement relations:
∂ur 2µ ∂uθ τrr = λ div u + 2µ , τθθ = λ div u + + ur , ∂r r ∂θ
2µ 1 ∂uφ τrθ τθr ∂uθ uθ 1 ∂ur τφφ = λ div u + + ur + uθ cot θ , = = − + , r sin θ ∂φ µ µ ∂r r r ∂θ τ τ 1 ∂u ∂u u τ τ 1 ∂u 1 ∂u u cot θ rφ = φr = r + φ − φ , θφ = φθ = θ + φ − φ . µ µ r sin θ ∂φ ∂r r µ µ r sin θ ∂φ r ∂θ r • Strain-displacement relations: ∂u 1 ∂u u 1 ∂u u u cot θ e = r , e = θ + r , e = φ + r + θ , rr ∂r θθ r ∂θ r φφ r sin θ ∂φ r r ∂u u 1 ∂u 1 ∂u ∂u u 2e = 2e = θ − θ + r , 2e = 2e = r + φ − φ , rθ θr ∂r r r ∂θ rφ φr r sin θ ∂φ ∂r r 1 ∂u 1 ∂u u cot θ 2e = 2e = θ + φ − φ . φθ θφ r sin θ ∂φ r ∂θ r • Equilibrium equations (statics): for the displacement formulation, use Navier’s equation,
(λ + 2µ) grad div u − µ curl curl u + F = 0,
whereas for the stress formulation, use ∂τ 1 ∂τ 1 ∂τ 2τ − τ − τ + cot θ τ rr + rθ + rφ + rr θθ φφ rθ + F = 0 ∂r r ∂θ r sin θ ∂φ r r ∂τ 1 ∂τ 1 ∂τ 3τ + (τ − τ ) cot θ rθ + θθ + θφ + rθ θθ φφ + F = 0 ∂r r ∂θ r sin θ ∂φ r θ ∂τ 1 ∂τ 1 ∂τ 3τ + 2τ cot θ rφ + θφ + φφ + rφ θφ + F = 0. ∂r r ∂θ r sin θ ∂φ r φ
• Stress boundary conditions: these are when t is prescribed. We have, from ti = nˆj τij ,
tr = nˆrτrr + nˆθτrθ + nˆφτrφ
tθ = nˆrτrθ + nˆθτθθ + nˆφτθφ
tφ = nˆrτrφ + nˆθτθφ + nˆφτφφ