美国麻省理工大学：《结构分析与设计技术》教学资源（讲义）notes1

16.21 Techniques of Structural Analysis and sig 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 go,(see Fig. 1). To analyze the equilibrium of the beam we consider the free body diagram of an element of length A as shown in the figure and apply Newton's second law ∑F=0:V-△r=(V+△V)=0 ∑MB=0:-V△x-M+(M+△M)+(w△a) Dividing by△ r and taking the limit△x→0: dM

� 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 q0, (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 figure and apply Newton’s second law: Fy = 0 : V − q0Δx − (V + ΔV ) = 0 (1) � Δx MB = 0 : −V Δx − M + (M + ΔM) + (q0Δx) = 0 (2) 2 Dividing by Δx and taking the limit Δx → 0: dV = −q0 (3) dx dM dx = V (4) 1

A B q a dx Figure 1: Equilibrium of a simply supported beam Eliminating V, we obtain d 2 M +qo=0 Recall from Unified Engineering or 16.20(we'll cover this later in the course also) that the bending moment is related to the deflection of the beam w(a) by the equation d 2 M=El 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 dx2(ar2)+=0.0<x<L The boundary conditions of the beam are: The solution of equations 7 and 8 is given by w(a) 24EI (L-c)(L+Lc-a

� � x L q0 z x dx M M+dM V V+dV q0 A B A B Figure 1: Equilibrium of a simply supported beam Eliminating V , we obtain: d2M + q0 = 0 (5) dx2 Recall from Unified Engineering or 16.20 (we’ll cover this later in the course also) that the bending moment is related to the deflection of the beam w(x) by the equation: d2w M = EI (6) dx2 where E is the Young’s modulus and I is the moment of inertia of the beam. Combining 5 and 6, we obtain: d2 d2w dx2 EI dx2 + q0 = 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: q0 � 2 � w(x) = − x (L − x) L2 + 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 eI/dw dx2 +gow dz and requiring that the solution w(r) be the function minimizing it that also satisfies the displacement boundary conditions (0)=u(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 L which has a continuous second derivative and satisfies the boundary condi- tions 11. Substituting w(a)in 10 we obtain El (-2c1)2+gc1(Lx-x2) 2EL2+1 Note that our functional II now depends on CI only. w1(a) is an ap- proximate solution to our problem if CI minimizes II= II(c1). A necessary condition for this is: dIl de, eILC+ 90=0 or C1=-g0Fi, and the approximate solution becomes (L-x) 24EI In order to assess the accuracy of our approximate solution, lets compute the approximate deflection of the beam at the midpoint d1=w1(5) 90 2 3

� Corresponding variational formulation The same problem can be formulated in variational form by introducing the potential energy of the beam system: � L � EI �d2w�2 Π(w) = 0 2 dx2 + q0w dx (10) and requiring that the solution w(x) be the function minimizing it that also satisfies 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: w1(x) = c1x(L − x) (12) which has a continuous second derivative and satisfies the boundary condi￾tions 11. Substituting w1(x) in 10 we obtain: � L � � EI Π(c1) = 2 (−2c1) 2 + q0c1(Lx − x2 ) dx 0 (13) L3 = 2EILc2 1 + 6 q0c1 Note that our functional Π now depends on c1 only. w1(x) is an ap￾proximate solution to our problem if c1 minimizes Π = Π(c1). A necessary condition for this is: dΠ L3 = 4EILc1 + q0 = 0 dc1 6 or c1 = − q0L2 24EI , and the approximate solution becomes: w1(x) = − q0L2 x(L − x) 24EI In order to assess the accuracy of our approximate solution, let’s compute the approximate deflection of the beam at the midpoint δ1 = w1( L 2 ): δ = − q0L2 �L�2 = − q0L4 24EI 2 96EI 3

The exact value 8 =w()is obtained from eqn.9 as L 24EI2(-2 384EI We observe that 61 6 0.8 384 i. e the approximate solution underpredicts the maximum deflection by 20% However, if we consider the following approximation with 3 degrees of freedom (note it also satisfies the essential boundary conditions, eqn. 11) ar)=CiI(L-r)+C2. (L-r)+c3I(L-r) and require that II(c1, C2, c3) be a minimum aII 4cEⅠL+2c2EⅠL2+2c3EⅠL3 Lg0 0 6 2a1B2+42ED3+43E1+0=0 2clEIL+4c2EIL+<CEILs L5 q0 whose solution is El If you replace this values in eqn. 14 and evaluate the deflection at the midpoint of the beam you obtain the exact solution !!

� The exact value δ = w( L 2 ) is obtained from eqn.9 as: δ = − q0 L L − L� � L2 + LL − �L�2� = − 5 q0L4 24EI 2 2 2 2 384 EI We observe that: 1 δ1 96 4 = = = 0.8 δ 5 5 384 i.e. the approximate solution underpredicts the maximum deflection by 20%. However, if we consider the following approximation with 3 degrees of freedom (note it also satisfies the essential boundary conditions, eqn.11): w3(x) = c1x(L − x) + c2x2 (L − x) + c3x(L − x) 2 (14) and require that Π(c1, c2, c3) be a minimum: ∂Π ∂Π ∂Π = 0, = 0, = 0, ∂c1 ∂c2 ∂c3 i.e.: 4 c1 EI L + 2 c2 EI L2 + 2 c3 EI L3 + L3 q0 = 0 6 2 c1 EI L2 + 4 c2 EI L3 + 4 c3 EI L4 + L4 q0 = 0 12 2 c1 EI L3 + 4 c2 EI L4 + 24 c3 EI L5 + L5 q0 = 0 5 20 whose solution is: − (L2 q0) − (L q0) q0 c1 → , c2 → , c3 → 24 EI 24 EI 24 EI If you replace this values in eqn. 14 and evaluate the deflection at the midpoint of the beam you obtain the exact solution !!! 4