Mechanics of nanoparticle adhesion — A continuum approach

Mechanics of nanoparticle adhesion — A continuum approach 23AK=FN+ FH0 A2 1 pl pf⋅ + ⋅ −3 3pAK VdW(53)Next, the elastic–plastic contact area coefficient κ A is introduced. This dimensionlesscoefficient represents the ratio of plastic particle contact deformation areaA pl to total contact deformation area AK = Apl + A and includes a certain elasticeldisplacement:κ 2 1 plA= + ⋅ A 3 3 A(54)The solely elastic contact deformation A pl = 0, κ A = 2/3, has only minor relevancefor cohesive powders in loading (Fig. 2), but for the complete plastic contactdeformation (A pl = A K ) the coefficient κ A = 1 is obtained.From Eqs. (43), (53) and (54) the sum of contact normal forces is obtained as:K( )2N+H0= ⋅K⋅ f⋅A−pF F π r p κ κ (55)From Eq. (5) the transition radius of elastic-plastic model r K,f,el-pl (index el-pl)and from Eq. (6) the particle center approach of the two particles h K,f,el-pl are calculatedas:h( )3⋅π ⋅r⋅ p ⋅ κ −κ1,2 f A p=K,f,el−pl *r2⋅E( ) 22 29⋅π ⋅r⋅ p ⋅ κ −κ1,2 f A pK,f,el−=pl* 24⋅E(56)(57)Checking this model, Eq. (56), with pure elastic contact deformation, i.e., κ p →0 and κ A = 2/3, the elastic transition radius r K,f , Eq. (29), is also obtained. For example,nanodisperse titania particles (d 50,3 = 610 nm is the median diameter onmass basis (index 3), E = 50 kN/mm 2 modulus of elasticity, ν = 0.28 Poisson ratio,p f = 400 N/mm 2 micro-yield strength, κ A ≈ 5/6 contact area ratio, κ p = 0.44plastic repulsion coefficient) a contact radius of r K,f,el-pl = 2.1 nm and, from Eq.(57), a homeopathic center approach of only h K,f,el-pl = 0.03 nm are obtained. Thisis a very small indentation calculated, in principle, by means of a continuum approach.The contact deformation is equivalent to a microscopic force F N = 2.1 nNor to a small macroscopic pressure level of about σ = 1.4 kPa (porosity ε = 0.8) inpowder handling and processing.