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This is the Past Exam Paper of Mathematical Tripos which includes Solitons and Instantons, Smooth Function, Scalar Field Theory, Derrick Scaling Arguments, Bogomolny Equations, Topological Degree, Sigma Model Lumps etc. Key important points are: Nonlinear Continuum Mechanics, Isotropic Elastic Material, Constitutive Relation, Deformation Gradient, Simple Shear Deformation, Principal Stresses, Left Stretch Matrix, Undeformed Configuration, Shear Stress
Typology: Exams
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Wednesday 6 June 2007 1.30 to 4.
Attempt FOUR questions.
There are SIX questions in total.
The questions carry equal weight.
Cover sheet None Treasury Tag Script paper
1 Isotropic elastic material has constitutive relation σ = φ(F), where σ is Cauchy stress and F is the deformation gradient. Prove that σ is coaxial with FFT^ (or equivalently, with the left stretch matrix).
If the material is subjected to the simple shear deformation
1 γ 0 1
(the irrelevant 3-components being disregarded), show that
σ 11 − σ 22 = γσ 12 ,
regardless of the detailed form of φ.
[Express σij in terms of the principal stresses. There is no need to calculate the principal stretches; you may or may not find it convenient to do so, in calculating the principal axes of FFT^ .]
Paper 76
3 A cylinder composed of homogeneous isotropic elastic incompressible material occupies the domain a^20 < X 12 + X 22 < b^20 , 0 < X 3 < h 0 in its unstressed reference configuration. It has energy function per unit reference volume W (λ 1 , λ 2 , λ 3 ), where λi, i = 1, 2 , 3 denote the principal stretches. It is subjected to simultaneous twist and inflation, which can be viewed as first, the twist X → y:
y 1 = X 1 cos(αX 3 ) − X 2 sin(αX 3 ), y 2 = X 1 sin(αX 3 ) + X 2 cos(αX 3 ), y 3 = X 3 ,
followed by the inflation y → x: x 1 = f (ρ)y 1 /ρ, x 2 = f (ρ)y 2 /ρ, x 3 = y 3 ,
where ρ = (y^21 + y^22 )^1 /^2 ≡ (X 12 + X 22 )^1 /^2 and f (ρ) = (ρ^2 + a^2 − a^20 )^1 /^2. Thus, the inner surface is inflated to radius a, and α is the angle of twist per unit height. The outer curved surface is traction-free.
Calculate the principal stretches λi(ρ) at one representative location (such as y 1 = ρ, y 2 = 0), and deduce an expression for the total energy stored per unit height. The end couple has moment M about the 3-axis and the internal pressure is p(a). By considering the global balance of work-rate, show that
M = 2π
∂α
∫ (^) b 0
a 0
ρW (λ 1 , λ 2 , λ 3 ) dρ,
p(a) =
a
∂a
∫ (^) b 0
a 0
ρW (λ 1 , λ 2 , λ 3 ) dρ.
Find p(a) explicitly, for the case of neo-Hookean material for which W (λ 1 , λ 2 , λ 2 ) = 1 2 μ(λ
2 1 +^ λ
2 2 +^ λ
2 3 ).
Paper 76
4 (a) Write down the integral forms of the balance of energy and the entropy inequality, in the Lagrangian description, for a body with internal energy per unit mass U (F, η, ξ), where F is the deformation gradient, η is entropy per unit mass and ξ represents a collection of internal variables {ξr }. Deduce (under the usual assumptions) the constitutive relations
PIi = ρ 0
∂FiI
, θ =
∂η
where ρ 0 is the mass density in the undeformed configuration, P denotes the nominal stress tensor, and θ is the temperature. Deduce also that
ρ 0 θ η˙ = ρ 0 r − qI,I + fr ξ˙r ,
where r is heat supply per unit mass, q is the nominal (or Lagrangian) heat flux vector, and fr = −ρo∂U/∂ξr. Deduce also the inequality
fr ξ˙r −
qI θ,I θ
(b) The specific free energy ψ is defined so that ψ(F, θ, ξ) = U (F, η, ξ) − θη. Consider the particular case ψ = ψ(F∗, θ),
with F∗^ = FA−^1 : the internal variables ξr are now replaced by {AJI }, and fr are replaced by QIJ = −ρo∂ψ/∂AJI. Find P and Q in terms of P∗^ = ρ 0 ∂ψ/∂F∗.
For an isothermal process, given the dissipation potential Ω(Q, A), we have A˙JI = ∂Ω/∂QIJ. Find AJI as a function of time, in terms of the history of Q, in the case that
α n + 1
‖Q‖n+1^ −
τ
where α, τ and n are positive constants. Deduce a corresponding expression for the second Piola–Kirchhoff stress T = PF, as a functional of Q.
Paper 76 [TURN OVER
6 An incompressible non-hardening anisotropic plastic material, in plane strain deformation, has yield criterion
f (ξ, τ ) = 0; ξ =
σ 11 − σ 22 2
, τ = σ 12 ,
and it conforms to the associated flow law
Dij = λ∂f /∂σ˙ ij.
Expressing the yield criterion in the (ξ, τ ) plane in the form ξ = ξ(l), τ = τ (l), where l denotes arc length, let
dξ/dl = − cos(2φ), dτ /dl = − sin(2φ)
(so that the outward normal to the yield curve makes an angle − 2 φ to the τ -axis). Define also σ = (σ 11 + σ 22 )/2. Assuming yield, express the equations of equilibrium in terms of σ and l.
By considering dF (σ, l)/ds along a curve defined parametrically by (x 1 (s), x 2 (s)), show that F is constant along the curve, provided
(Fσ Fl)
x′ 1 cos(2φ) + x′ 2 sin(2φ) x′ 1 sin(2φ) − x′ 2 cos(2φ) x′ 1 x′ 2
Deduce that σ − l = constant on an α-line: dx 2 /dx 1 = tan φ, σ + l = constant on a β-line: dx 2 /dx 1 = − cot φ.
By locally choosing axes so that the x 1 -axis is tangent to the α-line, deduce from the flow law that du − vdφ = 0 along an α-line, dv + udφ = 0 along a β-line,
where (u, v) denote the components of velocity along the α- and β-lines.
Paper 76
