? 16.21 Techniques of Structural Analysis and Design Spring 2003 Unit #1 In this course we are going to focus on energy and variational methods for structural analysis. To understand the overall approach we start by con- trasting it with the alternative vector mechanics approach: Example of Vector mechanics formulation: Consider a simply supported beam subjected to a uniformly-distributed load q 0 , (see Fig.1). To analyze the equilibrium of the beam we consider the free body diagram of an element of length Δx as shown in the ?gure and apply Newton’s second law: F y = 0 : V ? q 0 Δx ? (V + ΔV ) = 0 (1) ? Δx M B = 0 : ?V Δx ? M + (M + ΔM ) + (q 0 Δx) = 0 (2) 2 Dividing by Δx and taking the limit Δx → 0: dV = ?q 0 (3) dx dM dx = V (4) 1 ? ? x L q0 z x dx M+dMM V V+dV q0 A B A B Figure 1: Equilibrium of a simply supported beam Eliminating V , we obtain: d 2 M + q 0 = 0 (5) dx 2 Recall from Uni?ed Engineering or 16.20 (we’ll cover this later in the course also) that the bending moment is related to the de?ection of the beam w(x) by the equation: d 2 w M = EI (6) dx 2 where E is the Young’s modulus and I is the moment of inertia of the beam. Combining 5 and 6, we obtain: d 2 d 2 w dx 2 EI dx 2 + q 0 = 0, 0 < x < L (7) The boundary conditions of the beam are: w(0) = w(L) = 0, M (0) = M (L) = 0 (8) The solution of equations 7 and 8 is given by: q 0 ? 2 ? w(x) = ? x (L ? x) L 2 + Lx ? x (9) 24EI 2 ? Corresponding variational formulation The same problem can be formulated in variational form by introducing the potential energy of the beam system: ? L ? EI ? d 2 w ? 2 Π(w) = 0 2 dx 2 + q 0 w dx (10) and requiring that the solution w(x) be the function minimizing it that also satis?es the displacement boundary conditions: w(0) = w(L) = 0 (11) A particularly attractive use of the variational formulation lies in the determination of approximate solutions. Let’s seek an approximate solution to the previous beam example of the form: w 1 (x) = c 1 x(L ? x) (12) which has a continuous second derivative and satis?es the boundary condi- tions 11. Substituting w 1 (x) in 10 we obtain: ? L ? ? EI Π(c 1 ) = 2 (?2c 1 ) 2 + q 0 c 1 (Lx ? x 2 ) dx 0 (13) L 3 = 2EILc 2 1 + 6 q 0 c 1 Note that our functional Π now depends on c 1 only. w 1 (x) is an ap- proximate solution to our problem if c 1 minimizes Π = Π(c 1 ). A necessary condition for this is: dΠ L 3 = 4EILc 1 + q 0 = 0 dc 1 6 or c 1 = ? q 0 L 2 24EI , and the approximate solution becomes: w 1 (x) = ? q 0 L 2 x(L ? x) 24EI In order to assess the accuracy of our approximate solution, let’s compute the approximate de?ection of the beam at the midpoint δ 1 = w 1 ( L 2 ): δ = ? q 0 L 2 ? L ? 2 = ? q 0 L 4 24EI 2 96EI 3 ? The exact value δ = w( L 2 ) is obtained from eqn.9 as: δ = ? q 0 L L ? L ? ? L 2 + L L ? ? L ? 2 ? = ? 5 q 0 L 4 24EI 2 2 2 2 384 EI We observe that: 1 δ 1 96 4 = = = 0.8 δ 5 5 384 i.e. the approximate solution underpredicts the maximum de?ection by 20%. However, if we consider the following approximation with 3 degrees of freedom (note it also satis?es the essential boundary conditions, eqn.11): w 3 (x) = c 1 x(L ? x) + c 2 x 2 (L ? x) + c 3 x(L ? x) 2 (14) and require that Π(c 1 ,c 2 ,c 3 ) be a minimum: ?Π ?Π ?Π = 0, = 0, = 0, ?c 1 ?c 2 ?c 3 i.e.: 4 c1 EI L + 2 c2 EI L 2 + 2 c3 EI L 3 + L 3 q0 = 0 6 2 c1 EI L 2 + 4 c2 EI L 3 + 4 c3 EI L 4 + L 4 q0 = 0 12 2 c1 EI L 3 + 4 c2 EI L 4 + 24 c3 EI L 5 + L 5 q0 = 0 5 20 whose solution is: ? (L 2 q0) ? (Lq0) q0 c1 → ,c2 → ,c3 → 24 EI 24 EI 24 EI If you replace this values in eqn. 14 and evaluate the de?ection at the midpoint of the beam you obtain the exact solution !!! 4