Nondestructive testing of defects in adhesive joints

It is an unsymmetrical two point tensor and is not completely related to material configuration
The force vector dh can be related to its initial configuration in a similar manner
dh = F.dH (2.15)
And rearranged as , dh = JF -1 F -T .dA = S.dA (2.16)
Where S is the second piola-kirchhoff stress tensor defined as S=JF -1 F -1 (2.17)
Hyperelasticity is type of elasticity where the stress at any point can be derived from the
deformation gradient and from an energy function. The behaviour of the material is said to be path
independent and P is work conjugate with the rate of deformation gradient ₣,a stored elastic potential Ψ
per unit undeformed volume can be established as the work done by the stresses form the initial to the
current position as, Ψ = P : ₣ =S : Ė= (∂Ψ/∂E) Ė (2.18)
Observing that ½ ċ=Ė is work conjugate to the second PK stress enables a totally lagrangian
constitutive equation to be constructed in the same manner as S = ∂Ψ/∂E = 2∂Ψ/∂c (2.19)
Based on the isotropic assumption, the strain energy density function W is expressed as a
function of the strain invariants .A better approach to modeling the response of rubbers comes from
assuming existence of strain energy which is a function of deformation tensor. Strain invariants can be
expressed in terms of the stretch ratios λ1, λ2, and λ3 as follows
Ψ = Ψ (I1, I2, I3)
I1 = λ1 2 + λ2 2 + λ3 2 = tr (c)
I2 = λ1 2 λ2 2 + λ2 2 λ3 2 + λ3 2 λ1 2 = ½ [(tr c) 2 -tr (c 2 )]
I3 = λ1 2 λ2 2 λ3 2 = det c (2.20)
For incompressible material I3=1.The strain energy function for Neo-Hookean material is given as
Ψ=C1 (I1-3) (2.21)
In many cases rubber is compressible to some extent and it is important to consider compressibility.
This work deals with the compressible form and introduces a strain energy contribution involving the
bulk modulus. For compressible form [4], the strain energy should combine both volumetric and
deviatoric term.
Ψ = Ψv+Ψd
Ψ = ½ λ (ln (detF)) 2 - µ ln (detF)+ ½ µ (tr (c) -3) (2.22)
PK stress and Cauchy stress can be easily be computed from (2.22) and given as
S = λ (ln(detF))c -1 + ½ µ (I –c -1 ) (2.25)
σ= λ/deft (ln (detF)) I+ (µ/detF) (b-I) (2.24)
The components of the fourth-order elasticity tensor are obtained as follows:
Cijkl=λcij -1 ckl -1 + (µ - λ ln (detF) (cik -1 cjl -1 - cil -1 ckj -1 ) i,j,k,l=1,2,3 (2.25)
III. FINITE ELEMENT FORMULATION
The finite element equations are non-linear in terms of displacement because of nonlinear
strain-displacement and stress-strain relationships. By the principle of virtual work, the weak form in
the current configuration is given as
∫ ∫ ∫ δe T σ dv = ∫ ∫ δu T q dA + ∫ ∫ ∫ δu T b dv (3.1)
V A V
For problems involving large displacement [6], consideration should be given to the
configuration. Since the current configuration is not known, we cannot compute directly from this equ
(3.1). Using the deformation gradient, the derivatives and integrals in the weak form are transformed to
the initial configuration (Total Langrangian formulation).
∫ ∫ ∫S δe dV o = ∫ ∫ δu T q o dA o + ∫ ∫ ∫ δu T b o dv o
V o A o V o
(3.2)