take the derivative of i; with respect to ak, as required by equation(23) (en·e) and take the derivative of the reverse transformation of the components of the position vector x x== jej=kek (29) i(; ei)=ik(ekei) 0) (ek ei)=ski(ep e;)= Replacing equations(28) and(33) in(23) dui du; dck du ar ac ar a Replacing in equation(2 1「O =2a(e,66,)+m(e6) Exchange indices I and k in second term 1「out e:e)ee)+ek:e·e) 1/du duk 2axr+ ar (erei)(e,ek) Or. finally =(e)(e,·e) Compatibility of strains Given displacement field u, expression(21) allows to compute the strains components Eij. How does one answer the reverse question? Note analogy with potential-gradient field. Restrict the analysis to two dimensions dul a 11 5take the derivative of u˜i with respect to xk, as required by equation (23): ∂u˜i ∂ul = (el · ˜ei) (28) ∂xk ∂xk and take the derivative of the reverse transformation of the components of the position vector x: x = xjej = x˜k˜ek (29) xj (ej · ei) = x˜k(˜ek · ei) (30) xjδji = x˜k(˜ek · ei) (31) xi = x˜k(˜ek · ei) (32) ∂xi ∂x˜k = (˜ek · ei) = δkj (˜ek · ei) = (˜ej · ei) (33) ∂x˜j ∂x˜j Replacing equations (28) and (33) in (23): ∂u˜i ∂ui ∂xk ∂ul = = (el · ˜ei)(˜ej · ek) (34) ∂x˜j ∂xk ∂x˜j ∂xk Replacing in equation (22): 1� ∂ul ∂ul � �˜ij = ei)(˜ ej )(˜ 2 ∂xk (el · ˜ ej · ek) + ∂xk (el · ˜ ei · ek) (35) Exchange indices l and k in second term: 1� ∂ul ∂uk � �˜ij = ei)(˜ ej )(˜ 2 ∂xk (el · ˜ ej · ek) + ∂xl (ek · ˜ ei · el) 1� ∂ul ∂uk � (36) = 2 ∂xk + ∂xl (el · ˜ ej ei)(˜ · ek) Or, finally: �˜ij = �lk(el · ˜ei)(˜ej · ek) (37) Compatibility of strains Given displacement field u, expression (21) allows to compute the strains components �ij . How does one answer the reverse question? Note analogy with potential-gradient field. Restrict the analysis to two dimensions: ∂u1 ∂u2 ∂u1 ∂u2 �11 = , �22 = , 2�12 = + (38) ∂x1 ∂x2 ∂x2 ∂x1 5