? ? ? ? ? 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 ?eld must satisfy the di?erential equations of equilibrium. Multiply the di?erential equations of equilibrium by an “arbitrary” displacement ?eld ˉu i : ˉσ ji,j + f i u i = 0 (1) Note that the ?eld ˉu i is NOT the actual displacement ?eld u i corresponding to the solution of the problem but a virtual displacement ?eld. 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 andˉ that it must be zero. The total virtual work done on the body is obtained by integration over the volume: ˉσ ji,j + f i u i dV = 0 (2) V 1 ˉ ˉ ˉ ? ? ?? ? ? ? ? ? ? ? ? ? and it must also be zero since the integrand is zero everywhere in the domain. ˉ ˉσ ji,j u i dV + f i u i dV = 0 (3) ? V V ? σ ji u i ,j ? σ ji ˉ ˉˉ u i,j dV + f i u i dV = 0 (4) V ? V ? ? ˉ ˉσ ji u i n j dS ? σ ij ? ij dV + f i u i dV = 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 speci?ed S t and an integral over the portion of the boundary where the displacements are speci?ed (supports) S u . This assumes that these sets are disjoint and complementary, i.e., S = S u ∪ S t , S u ∩ S t = ? (6) ˉ ˉ ˉt i u i dS + σ ji u i n j dS ? σ ij ? ij dV + f i u i dV = 0 (7) S t S u V V We will require that the virtual displacements ˉu i vanish on S u , i.e., that the virtual displacement ?eld satisfy the homogeneous essential boundary condi- tions: ˉu i (x j ) = 0, if x j ∈ S u (8) Then, the second integral vanishes. The resulting expression is a statement of the Principle of Virtual Displacements (PVD): ˉ ˉσ ij ? ij dV = t i u i dS + f i u i dV (9) V S t V It reads: The work done by the external tractions and body forces on an ad- missible (di?erentiable and satisfying the homogeneous boundary conditions but otherwise arbitrary) displacement ?eld 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 ?eld). Example: Consider the bar under a tensile load shown in the ?gure: 2 ? ? ? L PE, A x1 The PVD applied to this case is: dˉu 1 ? σ 11 dV = P ˉu 1 ? V dx 1 x 1 =L ? L ? du 1 dˉu 1 ? A E dx 1 = P ˉu 1 ? 0 dx 1 dx 1 x 1 =L ? L? d ? du 1 ? d 2 u 1 ? ? ? ˉ ˉ u 1 ?EA dx 1 u 1 ? dx 2 u 1 dx 1 = P ˉ 0 dx 1 1 x 1 =L ? du 1 ? ? du 1 ? ? L d 2 u 1 ? ? EA dx 1 u 1 ? EA dx 1 u 1 ? EA dx 2 u 1 dx 1 = P ˉˉ ˉ ˉ u 1 ? x 1 =L x 1 =0 0 1 x 1 =L The second term on the left hand side is zero because we have asked that ˉ ˉu 1 = 0 at the support. Note we have not asked for any condition on u 1 at x 1 = L where the load is applied. ? du 1 ? ? ? ? ? ? L d 2 u 1 EA dx 1 ? x 1 =L ? P u 1 ? = EA dx 2 u 1 dx 1 ˉ ˉ x 1 =L 0 1 The only way this expression can be satis?ed for any admissible virtual dis- placement ?eld ˉu 1 is if: du 1 ? P = EA ? dx 1 x 1 =L and d 2 u 1 EA = 0 dx 2 1 which represent the equilibrium conditions at the boundary and inside the bar, respectively: ? du 1 ? ? ? ? ? P = A E ? = Aσ 11 ? dx 1 x 1 =L x 1 =L and d ? du 1 ? d EA = σ 11 = 0 dx 1 dx 1 dx 1 3 The solution of this problem is: u 1 (x 1 ) = ax 1 + b the boundary conditions are: u 1 (0) = 0 ? b = 0 P = Ea A u 1 = P EA x 1 ? 11 = du 1 dx 1 = P EA σ 11 = 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 ?elds: ˉ? u 1 = ax 1 : ? L 0 P adx 1 = PaL(?) EA AE PaL = P aL q.e.d. ˉ 1 :? u 1 = ax 2 ? L 0 P 2ax 1 dx 1 = PaL 2 (?) EA AE L 2 ? A ? EP ? 2a ? 2 = P aL q.e.d. ? E ? A Remarks: 4 ˉ ˉ ˉ ? ? ? Principle of Virtual Displacements: – enforces equilibrium (in weak form) – enforces traction (natural) boundary conditions – does NOT enforce displacement (essential) boundary conditions – will be satis?ed 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 ?eld ˉu i and the PVD will hold. The unit dummy displacement method consists of choosing the virtual displacement ?eld such that ˉu i (x 0 ) = 1 in the direction of the reaction R 0 we are interested in. Then the virtual work of the reaction is ˉu 0 · R 0 = |R 0 . The PVD then reads (in the absence of body forces): R 0 · ˉu 0 = σ ij ? ij dV (10) V R 0 = V σ ij ? ij dV (11) where ? ij are the virtual strains produced by the virtual displacement ?eld ˉu 0 . Example: 5 ˉ ˉ ˉ ˉ ? ? ? ? ? q2 L1 L2 L3 E1A1 E2A2 E3A3 v P Di?erent materials and areas of cross section: E 1 ,E 2 ,E 3 ,A 1 ,A 2 ,A 3 . For a truss element: σ = E? (uniaxial state). Pvˉ = A 1 L 1 σ 1 ? 1 + A 2 L 2 σ 2 ? 2 + A 3 L 3 σ 3 ? 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 vˉ so that they cancel out. From the ?gure, the strains ensued by the truss elements as a result of a tip displacement v are: (L 2 + v) 2 + (L 2 tan θ) 2 ? L 1 L1 ? 1 = ? 3 = L 2 2 (1 + tan 2 θ) + 2L 2 v + v 2 ? L 1 = L 1 sin 2 θ neglecting the higher order term v 2 and using 1 + tan 2 θ = 1 + cos 2 θ = cos 2 θ+sin 2 θ cos 2 θ = 1 cos 2 θ we obtain: L 2 2 cos 2 θ + 2L 2 v ? L 1 L 1 1 + 2L 2 v L 2 1 ? L 1 = 2L 2 v 1 + L 2 1 ? 1 L 1 ? 1 = ? 3 = = L 1 were we have made use of the fact that: L 2 cos θ = L 1 . 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 ˉ ˉ ˉ ? √ 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]. L 2 L 2 ? 1 = ? 3 = 1 + L 2 v ? 1 = L 2 v 1 1 which is the sought expression. The expression for ? 2 can be obtained in a much more straightforward manner: v ? 2 = L 2 Applying the constitutive relation: σ I = E I ? I we can obtain the stresses in terms of the tip displacement v: L 2 L 2 v σ 1 = E 1 L 2 v, σ 3 = E 3 L 2 v, σ 2 = E 2 L 2 1 1 This expressions for the strains above also apply for the case of a virtual displacement ?eld whose value at the tip is vˉ. The resulting virtual strains are: L 2 L 2 vˉ ? 1 = L 2 vˉ, ? 3 = L 2 vˉ, ? 2 = L 2 1 1 Replacing in PVD: L 2 L 2 ? ?? ? E 2 ? ?? ? v vˉ ? ?? ? E 3 L 2 L 2 Pvˉ = A ? ?? ? E 11 L 1 L 2 v L 2 vˉ+ A 2 L 2 L 2 L 2 + A 3 L 3 L 2 v L 2 vˉ 1 1 ? ?? 1 ? ???? ???? ? ?? 1 ????? As expected the vˉ’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 simpli?ed using: L 2 = L 1 cos θ = L 3 cos θ: A 1 E 1 L 2 1 cos 2 θ v P = L 3 v + A 2 E 2 L 2 1 ? A 1 E 1 cos 2 θ A 2 E 2 P = + + A 3 E 3 L 2 3 cos 2 θ + v L 3 3 A 3 E 3 cos 2 θ v L 1 L 2 L 3 ? ? v P = (A 1 E 1 + A 3 E 3 )cos 3 θ + A 2 E 2 L 2 v = P L 2 (A 1 E 1 + A 3 E 3 ) cos 3 θ + A 2 E 2 7 ˉ ˉ ˉ ˉ ? Example: L1 L2 E1, A1 P/2 P/2 u0 E2, A2 PVD: P ˉu 0 = A 1 L 1 σ 1 ? 1 + A 2 L 2 σ 2 ? 2 ˉu 0 L 1 ? 1 = u 0 , σ 1 = E 1 ? 1 , ? 1 = L 1 ˉu 0 L 2 u 0 ? 2 = ? , σ 2 = E 2 ? 2 , ? 2 = ? L 2 ˉu 0 (?u 0 ) + A 2 L 2 E 2 L 1 L 2 (?ˉu 0 ) L 2 u 0 P ˉu 0 = A 1 L 1 E 1 L 1 ? A 1 E 1 L 1 A 2 E 2 L 2 u 0 P = + 8