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