16.21 Techniques of Structural Analysis and Design pring 2003 Unit#8-Principle of Virtual Displacements Principle of Virtual Displacements Consider a body in equilibrium. We know that the stress field must satisfy the differential equations of equilibrium. Multiply the differential equations of equilibrium by an "arbitrary"displacement field T (0+f1)1=0 Note that the field ii is noT the actual displacement field u; corresponding to the solution of the problem but a virtual displacement field. Therefore equation(1)can be interpreted as the local expression of virtual work done by the actual stresses and the body forces on the virtual displacement ii and that it must be zero. The total virtual work done on the body is obtained by integration over the volume 0jij+fiiidv=0
� � � � � 16.21 Techniques of Structural Analysis and Design Spring 2003 Unit #8 - Principle of Virtual Displacements Principle of Virtual Displacements Consider a body in equilibrium. We know that the stress field must satisfy the differential equations of equilibrium. Multiply the differential equations of equilibrium by an “arbitrary” displacement field u¯i: σji,j + fi u¯i = 0 (1) Note that the field u¯i is NOT the actual displacement field ui corresponding to the solution of the problem but a virtual displacement field. Therefore, equation (1) can be interpreted as the local expression of virtual work done by the actual stresses and the body forces on the virtual displacement u¯i and that it must be zero. The total virtual work done on the body is obtained by integration over the volume: σji,j + fi u¯idV = 0 (2) V 1
and it must also be zero since the integrand is zero every where in the domain oji,ju di fiuid=0 (a),-0nin,)d+/fa=0 o lin:ds dV+/fi dV=0 The integral over the surface can be decomposed into two: an integral over the portion of the boundary where the actual external surface loads(tractions Qre specified St and an integral over the portion of the boundary where the placements are specified(supports)Su. This assumes that these sets are disjoint and complementary, i.e S=SUS,S∩St= (6) Oije We will require that the virtual displacements i; vanish on Su, i.e., that the virtual displacement field satisfy the homogeneous essential boundary condi- a1(x)=0,ifx;∈S Then, the second integral vanishes. The resulting expression is a statement of the Principle of virtual Displacements(PVD) aijE;dv=ti ds+/fiii dl (9) It reads: The work done by the ecternal tractions and body forces on an ad missible(differentiable and satisfying the homogeneous boundary conditions but otherwise arbitrary) displacement field is equal to the work done by the equilibrated stresses(the actual solution of the problem)on the virtual strains ( the strains produced by the virtual field) Example: Consider the bar under a tensile load shown in the figure
¯ ¯ ¯ � � �� � � � � � � � � � and it must also be zero since the integrand is zero everywhere in the domain. σji,ju¯idV + fiu¯idV = 0 (3) � V V � σjiui ,j ¯ − σjiu¯i,j dV + fiu¯idV = 0 (4) V � V � � σjiu¯injdS − σij �ijdV + fiu¯idV = 0 (5) S V V The integral over the surface can be decomposed into two: an integral over the portion of the boundary where the actual external surface loads (tractions) are specified St and an integral over the portion of the boundary where the displacements are specified (supports) Su. This assumes that these sets are disjoint and complementary, i.e., S = Su ∪ St, Su ∩ St = ∅ (6) tiu¯idS + σjiu¯injdS − σij �ijdV + fiu¯idV = 0 (7) St Su V V We will require that the virtual displacements u¯i vanish on Su, i.e., that the virtual displacement field satisfy the homogeneous essential boundary conditions: u¯i(xj ) = 0, if xj ∈ Su (8) Then, the second integral vanishes. The resulting expression is a statement of the Principle of Virtual Displacements (PVD): σij �ijdV = tiu¯idS + fiu¯idV (9) V St V It reads: The work done by the external tractions and body forces on an admissible (differentiable and satisfying the homogeneous boundary conditions but otherwise arbitrary) displacement field is equal to the work done by the equilibrated stresses (the actual solution of the problem) on the virtual strains (the strains produced by the virtual field). Example: Consider the bar under a tensile load shown in the figure: 2
EA P L The PVD applied to this case is dv= pi a1=L P证 da ld dx EA,-元1 EA dindar,- pa The second term on the left hand side is zero because we have asked that i1=0 at the support. Note we have not asked for any condition on in at 1=L where the load is applied Ea li,dx The only way this expression can be satisfied for any admissible virtual dis- placement field in is if: P=EA EA 0 which represent the equilibrium conditions at the boundary and inside the bar, respectively P=AE E d 3
� � � L P E, A x1 The PVD applied to this case is: du¯1 � σ11 dV = Pu¯1� V dx1 x1=L � L � du1 du¯1 � A E dx1 = Pu¯1� 0 dx1 dx1 x1=L � L� d �du1 � d2u1 � � � EA ¯ ¯ u1� dx1 u1 − dx2 u1 dx1 = P ¯ 0 dx1 1 x1=L � du1 � � du1 � � L d2u1 � � EAdx1 u1 − EAdx1 u1 − EA dx2 ¯ ¯ u¯1dx1 = Pu¯1� x1=L x1=0 0 1 x1=L The second term on the left hand side is zero because we have asked that u¯1 = 0 at the support. Note we have not asked for any condition on u¯1 at x1 = L where the load is applied. � du1 � � � � � � L d2u1 EAdx1 � x1=L − P u1� = EA dx2 ¯ u¯1dx1 x1=L 0 1 The only way this expression can be satisfied for any admissible virtual displacement field u¯1 is if: du1 � P = EA � dx1 x1=L and d2u1 EA = 0 dx2 1 which represent the equilibrium conditions at the boundary and inside the bar, respectively: � du1 �� � � � P = A E � = Aσ11� dx1 x1=L x1=L and d � du1 � d EA = σ11 = 0 dx1 dx1 dx1 3
The solution of this problem is: the boundary conditions are P ea u EA ∈11 EA P Ee A Example: With the exact solution of the problem of the bar under a tensile load, verify the satisfaction of the PVD for the following virtual AE dc= Palo EA Pal= Pal g.e.d AE EA A EPAL Pal g.e.d
The solution of this problem is: u1(x1) = ax1 + b the boundary conditions are: u1(0) = 0 ⇒ b = 0 P = Ea A u1 = P EAx1 �11 = du1 dx1 = P EA σ11 = E�11 = P A Example: With the exact solution of the problem of the bar under a tensile load, verify the satisfaction of the PVD for the following virtual displacement fields: • u¯1 = ax1: � L 0 P adx1 = PaL(?) EA AE PaL = PaL q.e.d. ¯ 1 • u1 = ax : 2 � L 0 P 2ax1dx1 = PaL2 (?) EA AE L2 �A �EP �2a �2 = PaL q.e.d. �E �A Remarks: 4
Principle of Virtual Displacements enforces equilibrium (in weak form) enforces traction(natural) boundary conditions does NoT enforce displacement(essential) boundary conditions will be satisfied for all equilibrated solutions, compatible or in- compatible Unit dummy displacement method Another application of the PVD: provides a way to compute reactions(or dis- placements) in structures directly from PVD. Consider the concentrated reaction force at point 0 of a structure in equilibrium under a set of loads and supports. We can prescribe an arbitrary admissible displacement field ii and the PVD will hold. The unit dummy displacement method consists of choosing the virtual displacement field such that )=l in the direction of the reaction Ro we are interested in. Then the virtual work of the reaction is to Ro=Ro. The PVD then reads(in the absence of body forces Ro·0=/o;;dV (10) ij∈ij 11) where Ei are the virtual strains produced by the virtual displacement field uo Example: 5
¯ ¯ ¯ � � • Principle of Virtual Displacements: – enforces equilibrium (in weak form) – enforces traction (natural) boundary conditions – does NOT enforce displacement (essential) boundary conditions – will be satisfied for all equilibrated solutions, compatible or incompatible Unit dummy displacement method Another application of the PVD: provides a way to compute reactions (or displacements) in structures directly from PVD. Consider the concentrated reaction force at point ��0�� of a structure in equilibrium under a set of loads and supports. We can prescribe an arbitrary admissible displacement field u¯i and the PVD will hold. The unit dummy displacement method consists of choosing the virtual displacement field such that u¯i(x0) = 1 in the direction of the reaction R0 we are interested in. Then the virtual work of the reaction is u¯0 · R0 = |R0. The PVD then reads (in the absence of body forces): R0 · u¯0 = σij �ijdV (10) V R0 = V σij �ijdV (11) where �ij are the virtual strains produced by the virtual displacement field u¯0. Example: 5
E3 Different materials and areas of cross section: El, E2, E3, A1, A2, A3. For a truss element: o= Ee(uniaxial state A1L1011+A2L202+A3L30 Note: the indices in these expressions just identify the truss element number The goal is to provide expressions of the virtual strains Er in terms of the virtual displacement i so that they cancel out. From the figure, the strains ensued by the truss elements as a result of a tip displacement v are √(L2+)2+(L2tan) LI √2(1+tan2e)+2L2+n2-L1 L neglecting the higher order term u- and using 1+ tan8 cos 0+sin 0 we obtain 2L2-L1 were we have made use of the fact that: Lse=L1. We seek to extract the linear part of this strain, which should have a linear dependence on the displacement u. This can be done by doing a Taylor series expansion of 6
¯ ¯ ¯ ¯ � � � � � 2 θ L1 L2 L3 E1 A1 E2 A2 E3 A3 v P Different materials and areas of cross section: E1, E2, E3, A1, A2, A3. For a truss element: σ = E� (uniaxial state). Pv¯ = A1L1σ1�1 + A2L2σ2�2 + A3L3σ3�3 Note: the indices in these expressions just identify the truss element number. The goal is to provide expressions of the virtual strains �I in terms of the virtual displacement v¯ so that they cancel out. From the figure, the strains ensued by the truss elements as a result of a tip displacement v are: (L2 + v)2 + (L2 tan θ)2 − L1 L1 �1 = �3 = L2 2(1 + tan2 θ) + 2L2v + v2 − L1 = L1 sin2 θ neglecting the higher order term v2 and using 1 + tan2 θ = 1 + cos2 θ = cos2 θ+sin2 θ cos2 θ = 1 cos2 θ we obtain: L2 2 cos2 θ + 2L2v − L1 L1 1 + 2L2v L2 1 − L1 = 2L2v 1 + L2 1 − 1 L1 �1 = �3 = = L1 were we have made use of the fact that: L2 cos θ = L1. We seek to extract the linear part of this strain, which should have a linear dependence on the displacement v. This can be done by doing a Taylor series expansion of 6
the square root term v1+2.c=1+a+o[]2(Mathematica tip:Taylor series expansions can be obtained by using the Series function. In this case Series [Sqrt[1 +2x],x, 0, 3] 1=68=1+元2t-1 which is the sought expression. The expression for E2 can be obtained in a much more straightforward manner Applying the constitutive relation: o= Erer we can obtain the stresses in terms of the tip displacement v E1-=v E2 This expressions for the strains above also apply for the case of a virtual displacement field whose value at the tip is i. The resulting virtual strains L 2 L2 Replacing in PVD P=A1LE1720+A2L2E20 As expected the U's cancel out, as the principle must hold for all its admissible virtual values and we obtain an expression of the external load P and the resulting real displacement v. This expression can be simplified using: L2 8=L3 cos 0 A1ELi cos20 A3E3 L2 c0s28 A2e2 A1 El cos0 A2E2 A3E3 P e9> P=[41E1+43B3)coB+A2E2 PL2 (A1E1+Ag E3)cos 0+ A2 E2
¯ ¯ ¯ � √ the square root term 1 + 2x = 1 + x + O[x] 2 (Mathematica tip: Taylor series expansions can be obtained by using the Series function. In this case: Series[Sqrt[1 + 2x], x, 0, 3]. L2 L2 �1 = �3 = 1 + L2 v − 1 = L2 v 1 1 which is the sought expression. The expression for �2 can be obtained in a much more straightforward manner: v �2 = L2 Applying the constitutive relation: σI = EI �I we can obtain the stresses in terms of the tip displacement v: L2 L2 v σ1 = E1 L2 v, σ3 = E3 L2 v, σ2 = E2 1 1 L2 This expressions for the strains above also apply for the case of a virtual displacement field whose value at the tip is v¯. The resulting virtual strains are: L2 L2 v¯ �1 = L2 v¯, �3 = L2 v¯, �2 = 1 1 L2 Replacing in PVD: L2 L2 � �� � E2 � �� � v v¯ � �� � E3 L2 L2 Pv¯ = A� �� � 1L1 E1 L2 v L2 v¯ + A2L2 L2 L2 + A3L3 L2 v L2 v¯ 1 1 � ��1 � ���� ���� � ��1 � ���� As expected the v¯’s cancel out, as the principle must hold for all its admissible virtual values and we obtain an expression of the external load P and the resulting real displacement v. This expression can be simplified using: L2 = L1 cos θ = L3 cos θ: A1E1L2 1 cos2 θ v P = L3 v + A2E2 1 L2 �A1E1 cos2 θ A2E2 P = + + A3E3L2 3 cos2 θ + v L3 3 A3E3 cos2 θ v L1 L2 L3 � � v P = (A1E1 + A3E3) cos 3 θ + A2E2 L2 v = PL2 (A1E1 + A3E3) cos3 θ + A2E2 7
Example E1,A1 E2,A2 L2 P0=A1L1011+A2L202 L ∈2 E2E2, E2 Pio=A,LIE Lo to+A2L2E2 L2 L2 (-20)(-o) MIL Alei A2e2 P
¯ ¯ ¯ ¯ � Example: L1 L2 E1, A1 P/2 P/2 u0 E2, A2 PVD: Pu¯0 = A1L1σ1�1 + A2L2σ2�2 u¯0 L1 �1 = u0 , σ1 = E1�1, �1 = L1 u¯0 L2 u0 �2 = − , σ2 = E2�2, �2 = − L2 u¯0 (−u0) + A2L2E2 L1 L2 (−u¯0) L2 u0 Pu¯0 = A1L1E1 L1 �A1E1 L1 A2E2 L2 P = + u0 8
