16.21 Techniques of Structural Analysis and sig Spring 2003 Unit #3 -Kinematics of deformation e Figure 1: Kinematics of deformable bodies Deformation described by deformation mapping x'=p(x (1)
� 16.21 Techniques of Structural Analysis and Design Spring 2003 Unit #3 - Kinematics of deformation Figure 1: Kinematics of deformable bodies Deformation described by deformation mapping: x = ϕ(x) (1) 1
We seek to characterize the local state of deformation of the material in a neighborhood of a point P. Consider two points P and Q in the undeformed P: x=Tle1+c2e2+3e3=Tie: and deformed x=91(x)e1+y2(x)e2+93(x)e3=9(x)e Q:x+dx'=(9(x)+d4)e configurations. In this expression pie Expressing the differentials do i in terms of the partial derivatives of the functions pi(a;ei) and similarly for dp2, dy 3, in index notation Replacing in equation (5) Q: x+d dx a dr:e We now try to compute the change in length of the segment PQ which deformed into segment PQ. Undeformed length(to the square Idxl=dx dx= dc:d ci
� � � � � We seek to characterize the local state of deformation of the material in a neighborhood of a point P. Consider two points P and Q in the undeformed: P : x = x1e1 + x2e2 + x3e3 = xiei (2) Q : x + dx = (xi + dxi)ei (3) and deformed P� : x = ϕ1(x)e1 + ϕ2(x)e2 + ϕ3(x)e3 = ϕi(x)ei (4) Q� : x� + dx� = ϕi(x) + dϕi ei (5) configurations. In this expression, dx� = dϕiei (6) Expressing the differentials dϕi in terms of the partial derivatives of the functions ϕi(xjej ): ∂ϕ1 ∂ϕ1 ∂ϕ1 dϕ1 = dx1 + dx2 + dx3, (7) ∂x1 ∂x2 ∂x3 and similarly for dϕ2, dϕ3, in index notation: ∂ϕi dϕi = dxj (8) ∂xj Replacing in equation (5): Q� : x� + dx� = ϕi + ∂ϕi dxj ei (9) ∂xj dx� i = ∂ϕi dxjei (10) ∂xj −→ We now try to compute the change in length of the segment PQ which −−→ deformed into segment P� Q� . Undeformed length (to the square): ds2 = �dx�2 = dx · dx = dxidxi (11) 2
Deformed length(to the square) (ds)2=|dx2=dx.dy_09,0 (12) The change in length of segment PQ is then given by the difference between quations(12)and(11) (13) We want to extract as common factor the differentials. To this end we observe The dak -d r i d c,d Sik; dck (15) dr, ack 2Eik: Green-Lagrange strain tensor Assume that the deformation mapping p(x) has the form x+u where u is the displacement field. Then api ari a au and the green- Lagrange strain tensor becomes dui aui aum aum 3
� �� � Deformed length (to the square): (ds� ) 2 = �dx� �2 = dx� · dx� = ∂ϕi dxj ∂ϕi dxk (12) ∂xj ∂xk −→ The change in length of segment PQ is then given by the difference between equations (12) and (11): (ds� ) 2 − ds2 = ∂ϕi dxj ∂ϕi dxk − dxidxi (13) ∂xj ∂xk We want to extract as common factor the differentials. To this end we observe that: dxidxi = dxjdxkδjk (14) Then: (ds� ) 2 − ds2 = ∂ϕi dxj ∂ϕi dxk − dxjdxkδjk ∂xj ∂xk �∂ϕi ∂ϕi � = − δjk dxjdxk (15) ∂xj ∂xk 2�jk: Green-Lagrange strain tensor Assume that the deformation mapping ϕ(x) has the form: ϕ(x) = x + u (16) where u is the displacement field. Then, ∂ϕi ∂xi ∂ui ∂ui = + = δij + (17) ∂xj ∂xj ∂xj ∂xj and the Green-Lagrange strain tensor becomes: � ∂um �� ∂um � 2�ij = δmi + ∂xi δmj + ∂xj − δij (18) ∂ui ∂uj =�δij + ∂xj + ∂xi + ∂um ∂um − �δij ∂xi ∂xj 3
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 differentiation: ∂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
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 5
take 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
Differentiate the strain components as follows Oc,a Oc1a x2 ac1ax2 4 0r20r20x2 (41) and conclude that 02 02∈1102 Oxidr ax2 axi 6
Differentiate the strain components as follows: ∂2 � � ∂3u1 ∂3u2 ∂x1∂x2 2�12 = ∂x1∂x2 + ∂x2 (39) 2 1∂x2 ∂2�11 ∂3u1 = ∂x2 ∂x1∂x2 (40) 2 2 ∂2�22 ∂3u2 = ∂x2 ∂x2∂x2 (41) 1 1 and conclude that: ∂2�12 ∂2�11 ∂2�22 2 = + ∂x1∂x2 ∂x2 ∂x2 (42) 2 1 6
