13 Variational Formulation of Plane Beam Element IFEM Ch 13 Slide 1
Beams Resist Primarily Transverse Loads IFEM Ch 13 Slide 2
Transverse Loads are Transported to Supports by Flexural Action Neutral surface Compressive stress Tensile stress IFEM Ch 13 Slide 3
Beam Configuration Spatial (General Beams) Plane (This Chapter) Beam Models Bernoulli-Euler Timoshenko (advanced topic not covered in class) IFEM Ch 13 Slide 4
Plane Beam Terminology y,v q(x) y,v Beam cross section x, u z Neutral axis Neutral surface L Symmetry plane IFEM Ch 13 Slide 5
Common Support Conditions ;; ;; ;; Simply Supported Cantilever IFEM Ch 13 Slide 6
Basic Relations for Bernoulli-Euler Model of Plane Beam [ ] u(x, y ) v(x, y ) = [ y v(x) x v(x) ] [ y ] v = = v(x) [ ] y θ v(x) e = u y y y x = 2 v x = d2 v 2 dx = κ 2 σ = Ee = Ey d2 v y dx = E κ 2 M = E I κ Plus equilibrium equation M'' = q (not used specifically in FEM) IFEM Ch 13 Slide 7
Tonti Diagram for Bernoulli-Euler Model of Plane Beam (Strong Form) Prescribed end displacements Displacement BCs Transverse displacements v(x) Distributed transverse load q(x) Kinematic κ = v'' M''=q Equilibrium Curvature κ(x) M = EI κ Constitutive Bending moment M(x) Force BCs Prescribed end loads IFEM Ch 13 Slide 8
Total Potential Energy of Beam Member = U W L L ( 2 ) 2 v d x U = 1 2 V σ xx e xx dv = 1 2 0 Mκ dx = 1 2 0 EI x 2 = 1 2 L 0 EIκ 2 dx W = L 0 qv dx. IFEM Ch 13 Slide 9
Degrees of Freedom of Beam Element θ j θ i v i i v j j u (e) = v i θ i v j θ j IFEM Ch 13 Slide 10
Bernoulli-Euler Kinematics of Plane Beam Element y,v P (x + u, y + v) θ j v j E = E (e), I = I (e) θ i v i i j x, u x (e) l = L P(x, y) IFEM Ch 13 Slide 11
Plane Beam Element Shape Functions ξ = 1 ξ = 1 v (e) i = 1 θ (e) i = 1 N (e) vi (ξ) = 1 4 (1 ξ)2 (2 + ξ) N (e) θi (ξ) = 1 8 l(1 ξ)2 (1 + ξ) v (e) j = 1 θ (e) j = 1 N (e) vj (ξ) = 1 4 (1 + ξ)2 (2 ξ) N (e) θ j (ξ) = 1 8 l(1 + ξ)2 (1 ξ) IFEM Ch 13 Slide 12
Shape Functions in Terms of Natural Coordinate ξ v (e) = [ N (e) v i N (e) θ i N (e) v j N (e) θ j ] v (e) i θ (e) i v (e) j θ (e) j = Nu(e) ξ = 2x l 1 N v (e) i = 1 4 (1 ξ)2 (2 + ξ), N (e) θ i N v (e) j = 1 4 (1 + ξ)2 (2 ξ), N (e) θ j = 1 8 l(1 ξ)2 (1 + ξ), = 1 8 l(1 + ξ)2 (1 ξ). IFEM Ch 13 Slide 13
Element Stiffness and Consistent Node Forces B = 1 [ ξ 6 l l 3ξ 1 6 ξ l 3ξ + 1 ] (e) = 1 2 u(e)t K (e) u (e) u (e)t f (e) K (e) = l 0 EI B T B dx = 1 1 EI B T B 1 2 l dξ f (e) = l 0 N T qdx= 1 1 N T q 1 2 l dξ IFEM Ch 13 Slide 14
Analytical Computation of Prismatic Beam Element Stiffness K (e) = EI 2l 3 1 1 36ξ 2 6ξ(3ξ 1)l 36ξ 2 6ξ(3ξ + 1)l (3ξ 1) 2 l 2 6ξ(3ξ 1)l (9ξ 2 1)l 2 36ξ 2 6ξ(3ξ + 1)l dξ symm (3ξ + 1) 2 l 2 = EI l 3 12 6l 12 6l 4l 2 6l 2l 2 12 6l symm 4l 2 IFEM Ch 13 Slide 15
Mathematica Script for Symbolic Computation of Prismatic Plane Beam Element Stiffness ClearAll[EI,l,Ξ]; B={{6*Ξ,(3*Ξ-1)*l,-6*Ξ,(3*Ξ+1)*l}}/l^2; Ke=(EI*l/2)*Integrate[Transpose[B].B,{Ξ,-1,1}]; Ke=Simplify[Ke]; Print["Ke for prismatic beam:"]; Print[Ke//MatrixForm]; Ke for prismatic beam: 12 EI l 3 6EI l 2 12 EI l 3 6EI l 2 6EI l 2 4EI l 6EI l 2 2EI l 12 EI l 3 6EI l 2 12 EI l 3 6EI l 2 6EI l 2 2EI l 6EI l 2 4EI l Corroborates the result from hand integration. IFEM Ch 13 Slide 16
Analytical Computation of Consistent Node Force Vector for Uniform Load q f (e) = 1 2 ql 1 1 N dξ = 1 2 ql 1 1 1 2 1 12 = ql l 1 2 1 12 l 1 4 (1 ξ)2 (2 + ξ) 1 8 l(1 ξ)2 (1 + ξ) 1 4 (1 + ξ)2 (2 ξ) 1 8 l(1 + ξ)2 (1 ξ) "fixed end moments" dξ IFEM Ch 13 Slide 17
Mathematica Script for Computation of Consistent Node Force Vector for Uniform q ClearAll[q,l,Ξ] Ne={{2*(1-Ξ)^2*(2+Ξ), (1-Ξ)^2*(1+Ξ)*l, 2*(1+Ξ)^2*(2-Ξ),-(1+Ξ)^2*(1-Ξ)*l}}/8; fe=(q*l/2)*integrate[ne,{ξ,-1,1}]; fe=simplify[fe]; Print["fe^T for uniform load q:"]; Print[fe//MatrixForm]; fe^t for uniform load q: lq 2 l 2 q 12 lq 2 l2 q 12 Force vector printed as row vector to save space. IFEM Ch 13 Slide 18