1/du Green-Lagrange strain tensor: Ej2 ax dum When the absolute values of the derivatives of the displacement field are much smaller than 1, their products(nonlinear part of the strain) are even smaller and we' ll neglect them. We will make this assumption throughout this course(See accompanying Mathematica notebook evaluating the limits of this assumption). Mathematically aum au ax: a We will define the linear part of the green-Lagrange strain tensor as the small strain tensor Transformation of strain components Given: Eij, ei and a new basis ek, determine the components of strain in the new basis Ekl 1/0i We want to express the expressions with tilde on the right-hand side with their non-tilde counterparts. Start by applying the chain rule of differentia- (23) Transform the displacement components um(emei )=u(er ei (en·e) ui=uel ei� � Green-Lagrange strain tensor : (19) When the absolute values of the derivatives of the displacement field are much smaller than 1, their products (nonlinear part of the strain) are even smaller and we’ll neglect them. We will make this assumption throughout this course (See accompanying Mathematica notebook evaluating the limits of this assumption). Mathematically: � ∂ui � ∂um ∂um � � � 1 ⇒ ∼ 0 (20) ∂xj ∂xi ∂xj We will define the linear part of the Green-Lagrange strain tensor as the small strain tensor : �ij = 1 2 �∂ui ∂xj + ∂uj ∂xi � (21) Transformation of strain components Given: �ij , ei and a new basis ˜ek, determine the components of strain in the new basis �˜kl ˜ ˜ �˜ij = 1�∂ui + ∂uj � (22) 2 ∂x˜j ∂x˜i We want to express the expressions with tilde on the right-hand side with their non-tilde counterparts. Start by applying the chain rule of differentia￾tion: ∂u˜i ∂u˜i ∂xk = (23) ∂x˜j ∂xk ∂x˜j Transform the displacement components: u = u˜m˜em = ulel (24) um(˜ ei) = ul(el ˜ em · ˜ · ˜ei) (25) umδmi = ul(el ˜ · ˜ei) (26) ui = ul(el ˜ · ˜ei) (27) �ij = 1 2 �∂ui ∂xj + ∂uj ∂xi + ∂um ∂xi ∂um ∂xj � 4