16.21 Techniques of Structural Analysis and Design Spring 2005 Unit #8 Principle of Virtual Displacements

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1 6. Techniques of Structural Analysis and Design Spring 005 Unit #8 rinciple of irtual Displacements Raúl Radovitzky March 3, 005 rinciple of irtual 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 : ( ) σ ji,j + f i u i = 0 () Note that the field u i is NOT the actual displacement field u i corresponding to the solution of the problem but virtual a displacement field. Therefore, equation () can be interpreted as the local expression virtual of 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: ( ) σ ji,j + f i u i d = 0 ()

2 and it must also be zero since the integrand is zero everywhere in the domain. σ ji,j u i d + f i u i d = 0 (3) [( ) ) σ ji u i,j σ ji u i,j d + f i u i d = 0 (4) σ ji u i n j ds σ ij ɛ ij d + f i u i d = 0 (5) S 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 S t and an integral over the portion of the boundary where the displacements are specified (supports) S u. This assumes that these sets are disjoint and complementary, i.e., S = S u S t, S u S t = (6) t i ū i ds + σ ji ū i n j ds σ ij ɛ ij d + S t S u σ ij ɛ ij d = t i u i ds + f i u i d (9) f i ū i d = 0 (7) We will require that the virtual displacements u i vanish on S u, i.e., that the virtual displacement field satisfy the homogeneous essential boundary conditions : ū i (x j ) = 0, if x j S u (8) Then, the second integral vanishes. The resulting expression is a statement of the rinciple of irtual Displacements (D): S t It reads: The work done by the external tractions and body forces on ad amissible (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:

3 x E, A L The D applied to this case is: dū σ d = ū dx x =L L du dū A E dx = ū 0 dx dx x =L L [ d ( du ) d ] u EA ū u dx = u dx 0 dx dx x =L [ ] [ ] du L d du u EA ū EA ū EA u dx = u =L dx x =0 dx 0 dx x x =L The second term on the left hand side is zero because we have asked that u = 0 at the support. Note we have not asked for any condition u aton x = L where the load is applied. [ ] du L d u EA ū = EA ū dx dx x =L dx x =L 0 The only way this expression can be satisfied for admissible any virtual displacement field u is if: du = EA dx x =L and d u EA = 0 dx and d ( ) EA du d = σ = 0 dx dx dx 3 which represent the equilibrium conditions at the boundary and inside the bar, respectively: ( ) = A E du = Aσ dx x =L x =L

4 The solution of this problem is: the boundary conditions are: u (x ) = ax + b u (0) = 0 b = 0 = Ea A u = x EA du ɛ = = dx EA σ = Eɛ = A Example: With the exact solution of the problem of the bar under a tensile load, verify the satisfaction of the D for the following virtual displacement fields: u = ax : u = ax : L AE adx = al(?) 0 EA al = al q.e.d. L AE ax dx = al (?) 0 EA A E a L = al q.e.d. E A Remarks: 4

5 rinciple of irtual 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 D: provides a way to compute reactions (or displacements) in structures directly from D. 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 ū i and the D will hold. The unit dummy displacement method consists of choosing the virtual displacement field such that u i (x 0 ) = in the direction of the reaction R 0 we are interested in. Then the virtual work of the reaction is ū 0 R 0 = R 0. The D then reads (in the absence of body forces): R 0 u 0 = σ ij ɛ ij d (0) R 0 = σ ij ɛ ij d () where ɛ ij are the virtual strains produced by the virtual displacement field ū 0. Example: 5

6 E E E3 A A A3 L L L3 θ v Different materials and areas of cross section: E, E, E 3, A, A, A 3, but require symmetry to simplify the problem: E 3 = E, A 3 = A. For a truss element: σ = Eɛ (uniaxial state). v = A L σ ɛ + A L σ ɛ + A 3 L 3 σ 3 ɛ 3 Note: the indices in these expressions just identify the truss element number. The goal is to provide expressions of the virtual strains ɛ I 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: (L + v) + (L tan θ) L ɛ = ɛ 3 = L L ( + tan θ) + L v + v L = L neglecting the higher order term v and using + tan θ = + sin θ = cos θ cos θ+sin θ = we obtain: cos θ cos θ L + L v L L + L v L cos θ L L v ɛ = ɛ 3 = = = + L L L L were we have made use of the fact that: = L. We seek to extract cos θ the linear part of this strain, which should have a linear dependence on the 6

7 displacement v. This can be done by doing a Taylor series expansion of the square root term + x = + x + O[x] (Mathematica tip: Taylor series expansions can be obtained by using Series the function. In this case: Series[Sqrt[ + x], x, 0,. 3] L L ɛ = ɛ 3 = + v = v L L which is the sought expression. The expression ɛfor can much more straightforward manner: v ɛ = L be obtained in a Applying the constitutive relation: σ I = E I ɛ I we can obtain the stresses in terms of the tip displacement v: L L v σ = E v, σ 3 = E 3 v, σ = E L L L This expressions for the strains above also apply for the case of a virtual displacement field whose value at the tip v. is The resulting virtual strains are: L L v ɛ = v, ɛ 3 = v, ɛ = L L L Replacing in D: L L v v L L v = A L E v v + A L E + A 3 L 3 E }{{} 3 L L v v L }{{} L L }{{} L }{{ }}{{}}{{}}{{}}{{ }}{{} 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 and the resulting real displacement v. This expression can be simplified using: L = L cos θ = L 3 cos θ: A E L cos θ v A 3 E 3 L 3 cos θ = v + A E + v L 3 L L 3 3 ( A E cos θ A E A 3 E 3 cos θ ) = + + v L L L 3 [ ] v = (A E + A 3 E 3 ) cos 3 θ + A E L L v = (A E + A 3 E 3 ) cos 3 θ + A E 7

8 Example: / E, A u0 E, A L / L D: u 0 = A L σ ɛ + A L σ ɛ u 0 u 0 ɛ =, σ = E ɛ, ɛ = L L u 0 u ɛ =, σ = E ɛ, ɛ = L 0 L u 0 u 0 ( u 0 ) ( u 0 ) u 0 = A L E + A L E L L L L ( A ) E A E = + u 0 L L 8