Introduction to The Finite Element Method (FEM) - CI587 Unit 3 Topic: Euler-Bernoulli and Timoshenko Beams The Finite Element Method Euler-Bernoulli and Timoshenko Beams Read: Chapter 5 CONTENTS Euler-Bernoulli b e a m theory ➢ Governing Equations ➢ Finite element model ➢ Numerical examples Timoshenko b e a m theory ➢ Governing Equations ➢ Finite element model ➢ Shear locking ➢ Numerical example UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT Kinematics of the Linearized Euler-Bernoulli Beam Theory q(x ) Strains, displacements, and rotations are small z, w z f (x ) x x Undeformed Beam 𝑑𝑤 − 𝑑𝑥 Euler-Bernoulli Beam Theory (EBT) is based on the assumptions of (1) straightness, (2) inextensibility, and (3) normalitya u 90∘ Deformed Beam − 𝑑𝑤 𝑑𝑥 3 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT Kinematics of Deformation in the Euler-Bernoulli Beam Theory (EBT) 𝑑𝑤 − 𝑑𝑥 z Displacement field (constructed using the hypothesis) 𝑑𝑤 𝑢1 𝑥, 𝑧 = 𝑢 − 𝑧 , 𝑢2 = 0, 𝑢3 = 𝑤 𝑥 𝑑𝑥 u Linear strains w x z zz z y x −𝑧 zy zx yx xz xy xx 𝑑𝑤 𝑑𝑥 𝜕𝑢1 𝑑𝑢 𝑑2𝑤 𝜀11 = 𝜀𝑥𝑥 = = −𝑧 2 , 𝜕𝑥1 𝑑𝑥 𝑑𝑥 𝜕𝑢1 𝜕𝑢3 𝑑𝑤 𝑑𝑤 𝛾𝑥𝑧 = + =− + = 0. 𝜕𝑥3 𝜕𝑥1 𝑑𝑥 𝑑𝑥 Constitutive relations yy yz Notation for stress components 𝑑𝑢 𝑑2𝑤 𝜎𝑥𝑥 = 𝐸𝜀𝑥𝑥 = 𝐸 − 𝐸𝑧 2 𝑑𝑥 𝑑𝑥 𝜎𝑥𝑧 = 𝐺𝛾𝑥𝑧 = 0 4 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT Euler-Bernoulli Beam Theory z, w Beam cross section z F0 q(x) M0 • x y • 𝑞 𝑥 M cf L q(x ) V 𝜎𝑥𝑥 + ∆𝜎𝑥𝑥 𝜎𝑥𝑧 V V M cf w q(x ) 𝜎𝑥𝑧 + ∆𝜎𝑥𝑧 𝜎𝑥𝑥 +• N M 𝑓 𝑥 𝑐𝑓 𝑤 𝑉 + ∆𝑉 𝑁 + ∆𝑁 𝑀 + ∆𝑀 f (x ) x x cf w Definition of stress resultants 𝑁 = න 𝜎𝑥𝑥 𝑑𝐴, 𝐴 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS 𝑀 = න 𝜎𝑥𝑥 ∙ 𝑧𝑑𝐴, 𝐴 𝑉 = න 𝜎𝑥𝑧 𝑑𝐴. 𝐴 EBT AND TBT Euler-Bernoulli Beam Theory (Continued) Equilibrium equations 𝑑𝑁 + 𝑓 = 0, 𝑑𝑥 𝑑𝑀 − 𝑉 = 0, 𝑑𝑥 𝑑𝑉 + 𝑞 − 𝑐𝑓 𝑤 = 0 𝑑𝑥 න 1 ∙ 𝑑𝐴 = 𝐴, න 𝑧 ∙ 𝑑𝐴 = 0, න 𝑧 2 ∙ 𝑑𝐴 = 𝐼 𝐴 𝐴 𝐴 Stress resultants in terms of deflection 𝑑𝑢 𝑑2𝑤 𝑑𝑢 𝑁 = න 𝜎𝑥𝑥 𝑑𝐴 = න 𝐸 − 𝐸𝑧 2 𝑑𝐴 = 𝐸𝐴 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝐴 𝐴 𝑑𝑢 𝑑2𝑤 𝑑2𝑤 𝑀 = න 𝜎𝑥𝑥 × 𝑧 𝑑𝐴 = න 𝐸 − 𝐸𝑧 2 𝑧𝑑𝐴 = −𝐸𝐼 2 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝐴 𝐴 𝑑𝑀 𝑑 𝑑2𝑤 𝑉= = −𝐸𝐼 2 𝑑𝑥 𝑑𝑥 𝑑𝑥 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT Euler-Bernoulli Beam Theory (Continued) Governing equations in terms of the displacements 𝑑 𝑑𝑢 − 𝐸𝐴 − 𝑓 = 0, 𝑑𝑥 𝑑𝑥 Bars u 0≤𝑥≤𝐿 𝑑2 𝑑2𝑤 𝐸𝐼 2 + 𝑐𝑓 𝑤 − 𝑞 = 0, 𝑑𝑥 2 𝑑𝑥 0≤𝑥≤𝐿 Beams w z, w F0 q(x) M0 x cf L Axial deformation of a bar Bending of a beam Axial displacement is uncoupled from transverse displacement UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT Weak Form of the EB Beam Theory Governing equation 𝑑2 𝑑2 𝑤 𝐸𝐼 2 + 𝑐𝑓 𝑤 − 𝑞 = 0, 𝑑𝑥 2 𝑑𝑥 Weak form 𝑣𝑖 0≤𝑥≤𝐿 set of weight functions 𝑥𝑏 𝑑2 𝑑2𝑤 0 = න 𝑣𝑖 2 𝐸𝐼 𝑑𝑥 2 + 𝑐𝑓 𝑤 − 𝑞 𝑑𝑥 𝑑𝑥 𝑥𝑎 𝑥𝑏 =න 𝑥𝑎 𝑑𝑣𝑖 𝑑 𝑑 2 𝑤ℎ 𝑑 𝑑 2 𝑤ℎ − 𝐸𝐼 + 𝑐𝑓 𝑣𝑖 𝑤ℎ − 𝑣𝑖 𝑞 𝑑𝑥 + 𝑣𝑖 𝐸𝐼 𝑑𝑥 𝑑𝑥 𝑑𝑥 2 𝑑𝑥 𝑑𝑥 2 𝑥𝑏 𝑥𝑎 Implies that the primary variable is w Secondary variable (displacement) (shear force) 𝑥𝑏 =න 𝑥𝑎 𝑑𝑣 𝑑 𝑑2𝑤 − 𝐸𝐼 2 + 𝑐𝑓 𝑣𝑤 − 𝑣𝑞 𝑑𝑥 − 𝑣 𝑥𝑎 𝑄1 − 𝑣 𝑥𝑏 𝑄3 𝑑𝑥 𝑑𝑥 𝑑𝑥 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT Weak Form 𝑥𝑏 0=න 𝑥𝑎 𝑑 2 𝑣𝑖 𝑑 2 𝑤ℎ 𝐸𝐼 + 𝑐𝑓 𝑣𝑖 𝑤ℎ − 𝑣𝑖 𝑞 𝑑𝑥 − 𝑣𝑖 𝑥𝑎 𝑄1 − 𝑣𝑖 𝑥𝑏 𝑄3 𝑑𝑥 2 𝑑𝑥 2 𝑥𝑏 0=න 𝑥𝑎 Secondary variable (Bending Moment) 𝑥 Primary Variable, Slope/rotation (Continued) 𝑑𝑣𝑖 𝑑 2 𝑤ℎ 𝑏 + − ∙ 𝐸𝐼 𝑑𝑥 𝑑𝑥 2 𝑥 𝑎 𝑑 2 𝑣𝑖 𝑑 2 𝑤ℎ 𝐸𝐼 + 𝑐𝑓 𝑣𝑖 𝑤ℎ − 𝑣𝑖 𝑞 𝑑𝑥 − 𝑣𝑖 𝑥𝑎 𝑄1 − 𝑣𝑖 𝑥𝑏 𝑄3 𝑑𝑥 2 𝑑𝑥 2 − − 𝑑𝑣𝑖 𝑑𝑣𝑖 ∙ 𝑄2 − − ∙ 𝑄4 𝑑𝑥 𝑥 𝑑𝑥 𝑥 𝑎 𝑑 𝑑 2 𝑤ℎ 𝑄1 = 𝐸𝐼 𝑑𝑥 𝑑𝑥 2 = −𝑉ℎ 𝑥𝑎 , 𝑥𝑎 𝑑 2 𝑤ℎ 𝑄2 = 𝐸𝐼 = −𝑀ℎ 𝑥𝑎 , 𝑑𝑥 2 𝑥 𝑎 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS 𝑏 𝑑 𝑑 2 𝑤ℎ 𝑄3 = − 𝐸𝐼 𝑑𝑥 𝑑𝑥 2 = 𝑉ℎ 𝑥𝑏 𝑥𝑏 𝑑 2 𝑤ℎ 𝑄4 = −𝐸𝐼 = 𝑀ℎ 𝑥𝑎 𝑑𝑥 2 𝑥 𝑏 EBT AND TBT Beam Element Degrees of Freedom Generalized displacements ∆3 = 𝑤 𝑥𝑏 ∆1 = 𝑤 𝑥𝑎 2 =(xa) 1 2 he 4 =(xb) Generalized forces Q1 =−V(xa ) Q2 =−M(xa ) 1 he UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS Q3 =V(xb ) 2 Q4 =M(xb ) EBT AND TBT FINITE ELEMENT APPROXIMATION: Some Remarks ➢ Continuity requirement based on the weak form, which requires that the second derivative of w exists and square-integrable. ➢ Continuity based on the primary variables, which requires carrying w and its first derivative as the nodal variables, requires cubic approximation w. ➢ Post-computation of secondary variables (bending moment and shear force) requires the third derivative of w to exist. UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT FINITE ELEMENT APPROXIMATION Primary variables (serve as the nodal variables that must be 𝑑𝑤 continuous across elements) 𝑤, 𝜃 = − 𝑑𝑥 𝑤 𝑥 ≈ 𝑐0 + 𝑐1 𝑥 + 𝑐2 𝑥 2 + 𝑐3 𝑥 3 w w, 𝜃 ● ● ● × ● ● Hermite cubic polynomials ● 𝑤 𝑥𝑎 ≈ 𝑐0 + 𝑐1 𝑥𝑎 + 𝑐2 𝑥𝑎2 + 𝑐3 𝑥𝑎3 ≡ ∆1 𝑤 𝑥𝑏 ≈ 𝑐0 + 𝑐1 𝑥𝑏 + 𝑐2 𝑥𝑏2 + 𝑐3 𝑥𝑏3 ≡ ∆3 𝜃 𝑥𝑎 ≈ −𝑐1 − 2𝑐2 𝑥𝑎 − 3𝑐3 𝑥𝑎2 ≡ ∆2 𝜃 𝑥𝑏 ≈ −𝑐1 − 2𝑐2 𝑥𝑏 − 3𝑐3 𝑥𝑏2 ≡ ∆4 𝑥 − 𝑥𝑎 𝜑1𝑒 = 1 − 3 ℎ𝑒 2 𝑥 − 𝑥𝑎 +2 ℎ𝑒 𝑥 − 𝑥𝑎 𝜑2𝑒 = −(𝑥 − 𝑥𝑎 ) 1 − ℎ𝑒 𝑥 − 𝑥𝑎 𝑒 𝜑3 = 3 ℎ𝑒 𝜑4𝑒 = − 𝑥 − 𝑥𝑎 2 2 𝑥 − 𝑥𝑎 −2 ℎ𝑒 𝑥 − 𝑥𝑎 ℎ𝑒 2 3 3 𝑥 − 𝑥𝑎 − ℎ𝑒 4 𝑤 𝑥 ≈ 𝑐0 + 𝑐1 𝑥 + 𝑐2 𝑥 2 + 𝑐3 𝑥 3 = ∆𝑗 𝜑𝑗 𝑥 𝑗=1 12 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT HERMITE CUBIC INTERPOLATION FUNCTIONS 𝜙1 𝑥 1 𝜙2 𝑥 slope = 0 slope = 1 x he slope = 0 he x 𝜙3 𝑥 slope = 0 𝜙4 𝑥 1 he slope = 1 x slope = 0 he x 13 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT Finite Element Model 4 𝑤 𝑥 = ∆𝑗 𝜑𝑗 𝑥 𝑗=1 4 𝐾𝑖𝑗𝑒 ∆𝑗𝑒 − 𝐹𝑖𝑒 = 0 𝐾 𝑒 ∆𝑒 = 𝐹 𝑒 𝑗=1 𝑒 𝐾11 𝑒 𝐾21 𝑒 𝐾31 𝑒 𝐾41 𝑥𝑏 𝐾𝑖𝑗𝑒 = න 𝑥𝑎 𝑒 𝐾12 𝑒 𝐾22 𝑒 𝐾32 𝑒 𝐾42 𝑒 𝐾13 𝑒 𝐾23 𝑒 𝐾33 𝑒 𝐾43 𝑒 𝐾14 𝑒 𝐾24 𝑒 𝐾34 𝑒 𝐾44 ∆1𝑒 ∆𝑒2 ∆𝑒3 ∆𝑒4 𝑞1𝑒 𝑄1𝑒 𝑞2𝑒 𝑄2𝑒 = + 𝑞3𝑒 𝑄3𝑒 𝑞4𝑒 𝑄4𝑒 𝑑 2 𝜙𝑖𝑒 𝑑 2 𝜙𝑗𝑒 𝐸𝐼 + 𝑐𝑓 𝜙𝑖𝑒 𝜙𝑗𝑒 𝑑𝑥 2 2 𝑑𝑥 𝑑𝑥 w1e e1 1e e2 w2e e3 1 he Q3e , q3e Q1e , q1e 2 e 2 𝑥𝑏 𝑒 𝐹𝑖 = න 𝜙𝑖𝑒 𝑞𝑑𝑥 + 𝑄𝑖𝑒 𝑥𝑎 e 4 1e e2 Q2e , q2e 1 2 Q4e , q4e he 14 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT Finite Element Model For element-wise constant values of 𝐸𝑒 𝐼𝑒 and 𝑞𝑒 (and 𝑐𝑓 = 0) 6 2𝐸𝑒 𝐼𝑒 −3ℎ𝑒 𝐾𝑒 = −6 ℎ𝑒3 −3ℎ𝑒 −3ℎ𝑒 2ℎ𝑒2 3ℎ𝑒 ℎ𝑒2 −6 −3ℎ𝑒 3ℎ𝑒 ℎ𝑒2 3ℎ𝑒 6 3ℎ𝑒 2ℎ𝑒2 𝐹𝑒 = 𝑞𝑒 ℎ𝑒 12 (Continued) 𝑄1 6 −ℎ𝑒 𝑄2 + 6 𝑄3 ℎ𝑒 𝑄4 Postprocessing 4 2𝜑𝑒 𝑑 𝑑2𝑤 𝑗 𝑀 𝑥 = −𝐸𝐼 2 = −𝐸𝐼 ∆𝑗𝑒 𝑑𝑥 𝑑𝑥 2 𝑗=1 4 3𝜑𝑒 𝑑 𝑑𝑀 𝑑 𝑑2𝑤 𝑗 𝑉 𝑥 = =− 𝐸𝐼 2 = −𝐸𝐼 ∆𝑗𝑒 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 3 𝑗=1 4 2 𝑒 𝑑 𝜑𝑗 𝑥 𝑀 𝑥 𝑧 𝑑2𝑤 𝑒 𝜎𝑥 𝑥, 𝑧 = − = 𝐸𝑧 = 𝐸𝑧 ∆ 𝑗 𝐼 𝑑𝑥 2 𝑑𝑥 2 𝑗=1 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT ASSEMBLY OF TWO BEAM ELEMENTS connected end-to-end 6 −3ℎ 2𝐸𝐼 −6 ℎ3 −3ℎ 0 0 −3ℎ 2ℎ2 3ℎ ℎ2 0 0 −3ℎ −6 0 0 ℎ2 3ℎ 0 0 3ℎ − 3ℎ −6 −3ℎ 6+6 3ℎ − 3ℎ 2ℎ2 + 2ℎ2 3ℎ ℎ2 6 3ℎ −6 3ℎ 3ℎ 1 −3ℎ ℎ2 𝑄11 12 𝑄21 −𝐿 𝑞0 𝐿 24 𝑄31 + 𝑄12 + 48 0 𝑄41 + 𝑄22 12 𝑄32 𝐿 𝑄42 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS 𝑈1 𝑈2 𝑈3 𝑈4 𝑈5 𝑈6 ℎ = 𝐿 Τ2 𝑄41 + 𝑄22 = 0 𝑄31 + 𝑄12 = 0 𝑄31 , 𝑞31 𝑄11 , 𝑞11 𝑄21 , 𝑞21 1 • 1 𝑄41 , 𝑞41 1 𝑄12 , 𝑞12 𝑄23 , 𝑞23 𝑄22 , 𝑞22 • 2 2•1 2 2 •3 2 EBT AND TBT 𝑄42 , 𝑞42 A SIMPLE EXAMPLE - 1 P EA, EI Exact solution (according to the Euler-Bernoulli beam theory) L 𝑃𝐿3 𝑤 𝐿 = 3𝐸𝐼 Given problem One element discretization P EA, EI • • U 1, U 2 L U 3, U 4 Boundary conditions: 𝑈1 = 𝑈2 = 0, 𝑄3 = 𝑃, 𝑄4 = 0 𝐾 𝑒 ∆𝑒 = 𝑞 𝑒 + 𝑄𝑒 6 𝑞𝑒 ℎ𝑒 −ℎ𝑒 𝑒 𝐹 = + 6 12 ℎ𝑒 6 2𝐸𝑒 𝐼𝑒 −3ℎ𝑒 𝑒 𝐾 = −6 ℎ𝑒3 −3ℎ𝑒 0 −3ℎ𝑒 2ℎ𝑒2 3ℎ𝑒 ℎ𝑒2 ℎ𝑒 = 𝐿 𝑄1 𝑄2 𝑄3 𝑄4 −6 −3ℎ𝑒 3ℎ𝑒 ℎ𝑒2 3ℎ𝑒 6 3ℎ𝑒 2ℎ𝑒2 17 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT A SIMPLE EXAMPLE - 1 Solution using Cramer ’s rule (continued) 6𝐸𝐼 𝐿2 4𝑃𝐸𝐼 4𝐸𝐼 0 𝑃𝐿3 𝐿 𝐿 𝑃 = ⇒ 𝑈3 = = = 0 12 𝐸𝐼 2 3𝐸𝐼 12𝐸𝐼 6𝐸𝐼 𝐿4 𝐿3 𝐿2 6𝐸𝐼 4𝐸𝐼 𝐿 𝐿2 𝑃 12𝐸𝐼 𝐿3 6𝐸𝐼 𝐿2 6𝐸𝐼 𝐿2 𝑈3 4𝐸𝐼 𝑈4 𝐿 12𝐸𝐼 𝑃 𝐿3 6𝐸𝐼 0 6𝑃𝐸 𝐼 Τ𝐿2 𝑃𝐿3 𝐿2 𝑈4 = = =− 2 4 12 𝐸𝐼 Τ𝐿 2𝐸𝐼 12𝐸𝐼 6𝐸𝐼 𝐿3 𝐿2 6𝐸𝐼 4𝐸𝐼 EA, EI 𝐿 𝐿2 L 𝑈3 𝑃𝐿3 2𝐸𝐼 𝑃𝐿3 𝑤 𝐿 = 3𝐸𝐼 𝑑𝑤 𝑈4 = − ቤ 𝑑𝑥 𝑥=𝐿 18 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT EXAMPLE – 2: A determinate frame structure B A 1 A 1• a 1 b P 1 x 2 • 2 C 3 F F Given structure A • 2 1 a b C B •2 P Finite element discretization B Pb • P F P F Pb •B x 2 P •C F UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT EXAMPLE – 2 A • x 1 B a Pb Bar element, A B • (continued) P •B x 2 F 𝑄𝐴 −1 𝑢𝐴 = 𝑄𝐵 1 𝑢𝐵 𝑃𝑎 𝑢𝐴 = 0, 𝑄𝐵 = −𝑃 ⇒ 𝑢𝐵 = − 𝐸1 𝐴1 Beam element, A B P P •C F Bar element, B C 𝐹𝑏 𝑢𝑐 = 𝐸2 𝐴2 Beam element, B C −6 −3𝑎 3𝑎 𝑎2 6 3𝑎 3𝑎 2𝑎2 𝑄1𝐴 𝑤𝐴 𝑄2𝐴 𝜃𝐴 𝑤𝐵 = 𝑄𝐵 1 𝜃𝐵 𝑄2𝐵 𝑤𝐴 = 0, 𝜃𝐴 = 0, 𝑄1𝐵 = −𝐹, 𝑄1𝐵 = 𝑃𝑏 −3𝑎 2𝑎2 3𝑎 𝑎2 Pb Displacements at C relative to point B 𝐸1 𝐴1 1 𝑎3 −1 6 2𝐸1 𝐼1 −3𝑎 −6 𝑎3 −3𝑎 F 𝑄1𝐵 𝑤𝐵 𝑄2𝐵 𝜃𝐵 𝑤𝐶 = 𝑄𝐶 1 𝜃𝐶 𝑄2𝐶 𝑤𝐵 = 0, 𝜃𝐵 = 0, 𝑄1𝐶 = −𝑃, 𝑄2𝐶 = 0 6 2𝐸2 𝐼2 −3𝑏 −6 𝑏3 −3𝑏 −3𝑏 2𝑏2 3𝑏 𝑏2 −6 −3𝑏 3𝑏 𝑏2 6 3𝑏 3𝑏 2𝑏2 20 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT EXAMPLE – 3: Handling of a vertical spring z, w 𝑞0 𝓀𝑤(𝐿) 𝓀 1• 𝑄21 1 𝓀𝑤(𝐿) 𝓀 L 𝑄11 x 𝑈3 𝑄31 1 •2 𝑄4 = 0 𝑄31 = −𝑘𝑤 𝐿 = −𝑘𝑈3 𝑈4 ≠ 0 𝑈1 = 𝑈2 = 0 𝑘𝑤 𝐿 = 𝑘𝑈3 Alternatively, 1 −1 𝑘 −1 1 𝑢1𝑠 𝑄1𝑠 = , 𝑢2𝑠 𝑄2𝑠 𝑢1𝑠 = 0, 𝑢2𝑠 = 𝑈3 ⇒ 𝑄2𝑠 = 𝑘𝑈3 21 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT SOLUTION TO THE SPRINGSUPPORTED BEAM 6 2𝐸 𝐼 −3𝐿 𝐿3 −6 −3𝐿 −3𝐿 2𝐿2 3𝐿 𝐿2 −6 −3𝐿 3𝐿 𝐿2 6 3𝐿 3𝐿 2𝐿2 0 𝑈1 = 𝑤1 𝑄1 0 𝑞 𝐿 6 𝑈2 = 𝜃1 0 −𝐿 + 𝑄2 − 𝓀𝑈 = 3 𝑈3 = 𝑤2 𝑄3 12 6 𝐿 𝑄4 0 𝑈4 = 𝜃2 Boundary conditions 𝑤1 = 0, 𝜃1 = 0, 𝑄3 = −𝓀𝑈3 , 𝑄4 = 0 Condensed equations for the unknown generalized nodal displacements 12𝐸𝐼 +𝓀 𝐿3 6𝐸𝐼 𝐿2 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS 6𝐸𝐼 𝐿2 𝑈3 = 𝑞0 𝐿 6 4𝐸𝐼 𝑈4 12 −𝐿 𝐿 EBT AND TBT HANDLING OF A POINT SOURCES INSIDE AN ELEMENT ℎ s0 𝑞𝑖 = න 𝑞 𝑠 𝜑𝑖 𝑠 𝑑𝑠 1 0 𝑞 𝑠 = 𝐹0 𝛿(𝑠 − 𝑠0 ) 0 ℎ 𝑑𝜑𝑖 𝑞𝑖 = න 𝑞 𝑠 𝜑𝑖 𝑠 𝑑𝑠 = −𝑀0 ቤ , 𝑖 = 1,2,3,4 𝑑𝑠 0 𝑠=𝑠 s0 M 0 0 𝜑1 𝑠 = 1 − 3 𝑠 ℎ +2 𝑠 ℎ 𝑠 2 𝑠 3 𝜑3 𝑠 = 3 −2 , ℎ ℎ , 𝜑2 for F0 placed at s 0 = 0.5h UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS 𝜑2 𝑠 = −𝑠 1 − q = 1 q2 = - 𝑠 ℎ F0 2 F0 1 ● s 2 h 𝑞 𝑠 = 𝑀0 𝛿 ´ (𝑠 − 𝑠0 ) q3 = F0h 8 1 2 𝑠 2 𝑠 − ℎ ℎ 𝑠 = −𝑠 2 h 𝑞𝑖 = න 𝑞 𝑠 𝜑𝑖 𝑠 𝑑𝑠 = 𝐹0 𝜑𝑖 𝑠0 , 𝑖 = 1,2,3,4 3 ● s ℎ 2 F0 2 F0 2 q4 = F0h 8 EBT AND TBT EXAMPLE – 4: A simply-supported beam (a) Find the center deflection using one Euler-Bernoulli element in full beam 6 2𝐸 𝐼 −3𝐿 𝐿3 −6 −3𝐿 −3𝐿 2𝐿2 3𝐿 𝐿2 −6 −3𝐿 3𝐿 𝐿2 6 3𝐿 3𝐿 2𝐿2 0 𝑈1 = 𝑤1 𝑈2 = 𝜃1 𝑈3 = 𝑤2 𝑈4 = 𝜃2 𝑄1 𝑞1 𝑞 𝑄2 = 𝑞2 + 𝑄3 3 𝑞4 𝑄4 𝑈2 = 𝐿2 2𝐿2 𝐹0 𝐿2 16𝐸𝐼 , 𝑈2 𝑈4 𝐹0 𝐿 1 = 8 −1 𝑈4 = − 𝐹0 𝐿2 x L q2 = q3 = F0L F0L = q4 8 8 2 𝑤 𝑥 = 𝑈1 𝜑1 𝑥 + 𝑈2 𝜑2 𝑥 + 𝑈3 𝜑3 𝑥 + 𝑈4 𝜑4 (𝑥) = 𝑈2 𝜑2 𝑥 + 𝑈4 𝜑4 𝑥 16𝐸𝐼 𝐹0 𝐿2 16𝐸𝐼 𝑥 2 −𝑥 1 − +𝑥 𝐿 𝑥 2 𝑥 − 𝐿 𝐿 𝐹0 𝐿2 𝐿 𝐿 𝐹0 𝐿3 𝑤 0.5𝐿 = − − =− 16𝐸𝐼 8 8 64𝐸𝐼 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS F0 2 F0 2 q1 = 1 Condensed equations 2𝐸 𝐼 2𝐿2 𝐿3 𝐿2 F0 EBT AND TBT EXAMPLE – 4: A simply-supported beam (b) Find the center deflection using one Euler-Bernoulli element in half beam 0 6 16𝐸 𝐼 −1.5𝐿 −6 𝐿3 −3𝐿 −1.5𝐿 0.5𝐿2 1.5𝐿 0.25𝐿2 −6 −1.5𝐿 1.5𝐿 0.25𝐿2 6 1.5𝐿 1.5𝐿 0.5𝐿2 𝑈1 = 𝑤1 𝑈2 = 𝜃1 𝑈3 = 𝑤2 0 𝑈4 = 𝜃2 𝑄1 0 𝑄2 = − 0.5𝐹0 𝑄3 𝑄4 F0 Condensed equations 16𝐸 𝐼 0.5𝐿2 𝐿3 1.5𝐿 x L 1.5𝐿 𝑈2 = 0.5𝐹 0 0 𝑈3 −1 6 𝐹0 𝐿3 4 1.5𝐿 𝐹0 𝐿2 𝑈2 = = 32𝐸𝐼 3𝐿3 1 16𝐸𝐼 𝐹0 𝐿3 4 0.5𝐿 𝐹0 𝐿3 𝑈2 = = 32𝐸𝐼 3𝐿3 1 48𝐸𝐼 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS 𝐹0 2 1 2 𝑈1 = 0, 𝑄2 = 0, 𝑈4 = 0, 𝑄3 = −0.5𝐹0 EBT AND TBT EXERCISE PROBLEM Problem: Develop weak form and the finite element model of the following equation, where w and P are unknowns: 𝑑2 𝑑2 𝑤 𝑑2 𝑤 𝐸𝐼 2 + 𝑃 2 = 0, 𝑑𝑥 2 𝑑𝑥 𝑑𝑥 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS 0<𝑥<𝐿 EBT AND TBT EXERCISE PROBLEM d F0 Problem: Use the minimum number of E B T elements to find the compression in the spring, reactions at the fixed support, and spring force. q0 Rigid loading frame 2E I h EI h UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS Linear elastic spring, k EBT AND TBT TIMOSHENKO BEAM THEORY and its Finite Element Model ➢ Governing Equations ➢ Finite element model ➢ Shear locking ➢ Numerical example UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT Kinematics of Timoshenko Beam Theory 𝑞 𝑥 z, w z 𝑥, 𝑢 Undeformed Beam 𝑓 𝑥 x 90∘ 𝑑𝑤 − 𝑑𝑥 Deformed Beams Euler-Bernoulli 𝑑𝑤 Beam Theory (EBT) Straightness, − 𝑑𝑥 inextensibility, an d normality x − u UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS 𝑑𝑤 𝑑𝑥 Timoshenko Beam Theory (TBT) Straightness an d inextensibility EBT AND TBT Timoshenko Beam Theory Kinematic Relations z 𝑢1 𝑥, 𝑧 = 𝑢 𝑥 + 𝑧𝜙𝑥 (𝑥) 𝑢2 = 0, u 𝑢3 (𝑥, 𝑧) = 𝑤(𝑥) 𝜕𝑢1 𝑑𝑢 𝑑𝜙𝑥 𝜀𝑥𝑥 = = +𝑧 , 𝜕𝑥1 𝑑𝑥 𝑑𝑥 w z x 𝜕𝑢1 𝜕𝑢3 𝑑𝑤 𝛾𝑥𝑧 = + = 𝜙𝑥 + , 𝜕𝑥3 𝜕𝑥1 𝑑𝑥 Constitutive Equations 𝑑𝑢 𝑑𝜙𝑥 𝜎𝑥𝑥 = 𝐸𝜀𝑥𝑥 = 𝐸 +𝑧 𝑑𝑥 𝑑𝑥 𝑑𝑤 𝜎𝑥𝑧 = 𝐺𝛾𝑥𝑧 = 𝐺 𝜙𝑥 + 𝑑𝑥 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT Timoshenko Beam Theory 𝑑𝑁 𝑑𝑉 + 𝑓 = 0, − − 𝑞 + 𝑐𝑓 𝑤 = 0, 𝑑𝑥 𝑑𝑥 Equilibrium Equations (continued) − 𝑑𝑀 + 𝑉 = 0. 𝑑𝑥 Beam Constitutive Equations 𝑑𝑢 𝑑𝜙𝑥 𝑑𝑢 𝑁 = න 𝜎𝑥𝑥 𝑑𝐴 = න 𝐸 +𝑧 𝑑𝐴 = 𝐸𝐴 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝐴 𝐴 𝑑𝑢 𝑑𝜙 𝑑𝜙 𝑀 = න 𝜎𝑥𝑥 𝑧 𝑑𝐴 = න 𝐸 +𝑧 𝑧 𝑑𝐴 = 𝐸𝐼 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝐴 𝐴 𝑉 = 𝐾𝑠 න 𝜎𝑥𝑧 𝑑𝐴 = 𝐺𝐾𝑠 𝜙 + 𝐴 𝑑𝑤 𝑑𝑤 න 𝑑𝐴 = 𝐺𝐴𝐾𝑠 𝜙 + 𝑑𝑥 𝑑𝑥 𝐴 Governing Equations in terms of the displacements 𝑑 𝑑𝑤 − 𝐺𝐴𝐾𝑠 𝜙 + 𝑑𝑥 𝑑𝑥 𝑑 𝑑𝑤 − 𝐺𝐴𝐾𝑠 𝜙 + 𝑑𝑥 𝑑𝑥 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS + 𝑐𝑓 𝑤 = 𝑞 + 𝑐𝑓 𝑤 = 0 31 EBT AND TBT WEAK FORMS OF TBT Weak Form of Eq. (1) 𝑥𝑏 0=න 𝑥𝑎 𝑥𝑏 =න 𝑥𝑎 𝜈1 − 𝑑 𝑑𝑤 𝐺𝐴𝐾𝑠 𝜙 + 𝑑𝑥 𝑑𝑥 𝑑𝜈1 𝑑𝑤 𝐺𝐴𝐾𝑠 𝜙 + 𝑑𝑥 𝑑𝑥 + 𝑐𝑓 𝑤 − 𝑞 𝑑𝑥 + 𝑐𝑓 𝜈1 𝑤 − 𝜈1 𝑞 𝑑𝑥 − 𝜈1 ∙ 𝐺𝐴𝐾𝑠 𝑥𝑏 𝑑𝑤 𝜙+ 𝑑𝑥 𝑥𝑎 𝑥𝑏 𝑑𝜈1 𝑑𝑤 =න ቊ 𝐺𝐴𝐾𝑠 𝜙 + 𝑑𝑥 𝑑𝑥 𝑥𝑎 + 𝑐𝑓 𝜈1 𝑤 − 𝜈1 𝑞 ቋ 𝑑𝑥 − 𝜈1 𝑥𝑎 ∙ 𝐺𝐴𝐾𝑠 𝜙 + 𝑥𝑏 =න 𝑥𝑎 𝐺𝐴𝐾𝑠 𝑑𝑤 𝑑𝑥 − 𝜈1 (𝑥𝑏 ) ∙ 𝐺𝐴𝐾𝑠 𝜙 + 𝑥𝑎 𝑑𝑤 𝑑𝑥 𝑥𝑏 𝑑𝜈1 𝑑𝑤 𝜙+ + 𝑐𝑓 𝜈1 𝑤 − 𝜈1 𝑞 𝑑𝑥 − 𝜈1 𝑥𝑎 ∙ 𝑄1 − 𝜈1 𝑥𝑏 ∙ 𝑄3 𝑑𝑥 𝑑𝑥 32 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT WEAK FORMS OF TBT (continued) Weak Form of Eq. (2) 𝑥𝑏 0=න 𝑥𝑎 𝑥𝑏 =න 𝑥𝑎 𝑥𝑏 =න 𝑥𝑎 𝑥𝑏 0=න 𝑥𝑎 𝜈2 − 𝑑 𝑑𝜙 𝑑𝑤 𝐸𝐼 + 𝐺𝐴𝐾𝑠 𝜙 + 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑥 𝑑𝜈2 𝑑𝜙 𝑑𝑤 𝐸𝐼 + 𝐺𝐴𝐾𝑠 𝜈2 𝜙 + 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝜙 𝑏 𝑑𝑥 − 𝜈2 ∙ 𝐸𝐼 𝑑𝑥 𝑥 𝑑𝜈2 𝑑𝜙 𝑑𝑤 𝐸𝐼 + 𝐺𝐴𝐾𝑠 𝜈2 𝜙 + 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝜙 𝑑𝜙 𝑑𝑥 − 𝜈2 𝑥𝑎 ∙ −𝐸𝐼 − 𝜈2 𝑥𝑏 ∙ 𝐸𝐼 𝑑𝑥 𝑥 𝑑𝑥 𝑥 𝑑𝜈2 𝑑𝜙 𝑑𝑤 𝐸𝐼 + 𝐺𝐴𝐾𝑠 𝜈2 𝜙 + 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 − 𝜈2 𝑥𝑎 ∙ 𝑄2 − 𝜈2 𝑥𝑏 ∙ 𝑄4 𝑎 𝑎 𝑏 Total Potential Energy 𝑥𝑏 Π 𝑊, 𝜙𝑥 = න 𝑥𝑎 𝐸𝐼 𝑑𝜙 2 𝑑𝑥 2 𝐺𝐴𝐾𝑠 𝑑𝑤 + 𝜙+ 2 𝑑𝑥 2 + 𝑐𝑓 2 𝑤 𝑑𝑥 2 𝑥𝑏 − න 𝑤𝑞𝑑𝑥 + 𝑤 𝑥𝑎 𝑄1 + 𝑤 𝑥𝑏 𝑄3 + 𝜙 𝑥𝑎 𝑄2 + 𝜙 𝑥𝑏 𝑄4 𝑥𝑎 33 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT FINITE ELEMENT MODELS OF TIMOSHENKO BEAMS 𝑚 Finite Element Approximation w1 1 w1 1 s1 w2 2 he w2 2 he m=n=2 1 s1 w3 3 1 m=n=3 s2 he s2 𝑛 𝑤 ≈ 𝑤𝑗 𝜓𝑗 𝑥 , 𝑗=1 𝑗=1 2 s3 2 3 𝐾 11 𝐾 21 𝐾 12 𝐾 22 𝑤 𝑆 = 𝐹1 𝐹2 he 𝑥𝑏 𝑥𝑏 𝐾𝑖𝑗11 = න 𝑥𝑎 𝑑𝜓𝑖 𝑑𝜓𝑗 𝐺𝐴𝐾𝑠 + 𝑐𝑓 𝜓𝑖 𝜓𝑗 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝐾𝑖𝑗12 = න 𝑥𝑎 𝑥𝑏 22 𝐾𝑖𝑗 = න 𝑥𝑎 𝑑𝜑𝑖 𝑑𝜑𝑗 𝐸𝐼 + 𝐺𝐴𝐾𝑠 𝜑𝑖 𝜑𝑗 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑥𝑏 𝑥𝑏 1 𝐹𝑖 = න 𝑞𝜓𝑖 𝑑𝑥 + 𝜓𝑖 (𝑥 𝑎 ൯𝑄1 + 𝜓𝑖 (𝑥 𝑏 ൱ 𝑄3 , 𝑥𝑎 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS 𝜙 ≈ 𝑆𝑗 𝜑𝑗 𝑥 , 𝐾𝑖𝑗21 = න 𝑥𝑎 𝑑𝜓𝑖 𝐺𝐴𝐾𝑠 𝜑𝑗 𝑑𝑥 = 𝐾𝑖𝑗21 𝑑𝑥 𝐺𝐴𝐾𝑠 𝜑𝑖 𝑑𝜓𝑖 𝑑𝑥 𝑑𝑥 𝐹𝑖2 = 𝜑𝑖 (𝑥 𝑎 ൯𝑄2 + 𝜑𝑖 (𝑥 𝑏 ቁ 𝑄4 34 EBT AND TBT Shear Locking in Timoshenko Beams (1) Thick beam experiences shear deformation, 𝑑𝑤 𝜙𝑥 ≠ − 𝑑𝑥 (2) Shear deformation is negligible in thin beams, 𝜙𝑥 = − Linear interpolation of both: 𝑤, 𝜙𝑥 2 2 𝑤 ≈ 𝑤𝑗𝑒 𝜓𝑗𝑒 𝑥 , 𝜙 ≈ 𝑆𝑗𝑒 𝜑𝑗𝑒 𝑥 , 𝑗=1 𝑤 𝑥 = 𝑤1 𝜓1 𝑥 + 𝑤2 𝜓2 𝑥 , w2 w1 S2 2 1 he 𝑗=1 𝜙𝑥 𝑥 = 𝑆1 𝜓1 𝑥 + 𝑆2 𝜓2 𝑥 S1 2 1 𝑑𝑤 𝑑𝑥 he Thus, in the thin beam limit it is not possible for the element to realize the requirement 𝑑𝑤 𝜙𝑥 = − 𝑑𝑥 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT SHEAR LOCKING-REMEDY In the thin beam limit, should become constant so that it matches dw/dx. However, if is a constant then the bending energy becomes zero. If we can mimic the two states (constant and linear) in the formulation, we can overcome the problem. Numerical integration of the coefficients allows us to evaluate both and d /dx as constants. The terms highlighted should be evaluated using “reduced integration”. 𝑥𝑏 𝐾𝑖𝑗11 = න 𝑥𝑎 𝑥𝑏 1 1 𝑑𝜓𝑖 𝑑𝜓𝑗 1 1 𝐺𝐴𝐾𝑠 + 𝑐𝑓 𝜓𝑖 𝜓𝑗 𝑑𝑥 𝑑𝑥 1 𝑑𝜓 2 𝐾𝑖𝑗12 = න 𝐺𝐴𝐾𝑠 𝑖 𝜓𝑗 𝑑𝑥 = 𝐾𝑖𝑗21 𝑑𝑥 𝑥𝑎 1 𝑑𝜓 2 𝑥𝑏 𝑑𝜓 𝑗 1 2 𝐾𝑖𝑗22 = න 𝐸𝐼 𝑖 + 𝐺𝐴𝐾𝑠 𝜓𝑖 𝜓𝑗 𝑑𝑥 𝑑𝑥 𝑥𝑎 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS 𝑑𝑥 𝑑𝑥 EBT AND TBT STIFFNESS MATRICES OF TIMOSHENKO BEAM ELEMENT (for constant EI and GA) Reduced integration linear element (RIE) Linear approximation of 6 2𝐸𝑒 𝐼𝑒 −3ℎ𝑒 𝜇0𝑒 ℎ𝑒3 −6 −3ℎ𝑒 −3ℎ𝑒 ℎ𝑒2 𝜉𝑒 3ℎ𝑒 ℎ𝑒2 𝜁𝑒 𝜉𝑒 = 1.5 + 6Λ𝑒 , both w and f 𝑞1𝑒 𝑄1𝑒 𝑤1 𝑞2𝑒 𝑄2𝑒 𝜙1 𝑤2 = 𝑞 𝑒 + 𝑄𝑒 3 3 𝑒 𝜙2 𝑞4 𝑄4𝑒 𝐸𝑒 𝐼𝑒 𝑒 𝜁𝑒 = 1.5 − 6Λ𝑒 , Λ𝑒 = 2 , 𝜇0 = 12Λ 𝑒 𝐺𝑒 𝐴𝑒 𝐾𝑠 ℎ𝑒 −6 −3ℎ𝑒 3ℎ𝑒 ℎ𝑒2 𝜁𝑒 3ℎ𝑒 6 3ℎ𝑒 ℎ𝑒2 𝜉𝑒 Consistent interelement element (CIE) Hermite cubic approximation of w and dependent quadratic approximation of f 6 2𝐸𝑒 𝐼𝑒 −3ℎ𝑒 𝜇𝑒𝑒 ℎ𝑒3 −6 −3ℎ𝑒 −3ℎ𝑒 2ℎ𝑒2 Σ𝑒 3ℎ𝑒 ℎ𝑒2 Θ𝑒 −6 −3ℎ𝑒 3ℎ𝑒 ℎ𝑒2 Θ𝑒 3ℎ𝑒 6 3ℎ𝑒 2ℎ𝑒2 Σ𝑒 Σ𝑒 = 1.0 + 3Λ𝑒 , Θ𝑒 = 1.0 − 6Λ𝑒 , UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS 𝑞1𝑒 𝑄1𝑒 𝑤1 𝑞2𝑒 𝑄2𝑒 𝜙1 𝑤2 = 𝑞 𝑒 + 𝑄𝑒 3 3 𝑒 𝜙2 𝑞4 𝑄4𝑒 𝐸𝑒 𝐼𝑒 Λ𝑒 = , 𝜇 = 1 + 12Λ𝑒 𝐺𝑒 𝐴𝑒 𝐾𝑠 ℎ𝑒2 𝑒 EBT AND TBT AN EXAMPLE of TBT EA, EI • U1, U2 L F Exact solution (according to the E- B beam theory) • 𝐹𝐿3 𝑤 𝐿 = 3𝐸𝐼 U 3, U 4 One element discretization using the RIE element 6 2𝐸𝑒 𝐼𝑒 −3ℎ𝑒 𝜇0𝑒 ℎ𝑒3 −6 −3ℎ𝑒 𝜉𝑒 = 1.5 + 6Λ𝑒 , −3ℎ𝑒 ℎ𝑒2 𝜉𝑒 3ℎ𝑒 ℎ𝑒2 𝜁𝑒 −6 −3ℎ𝑒 3ℎ𝑒 ℎ𝑒2 𝜁𝑒 3ℎ𝑒 6 3ℎ𝑒 ℎ𝑒2 𝜉𝑒 𝑞1𝑒 𝑄1𝑒 𝑤1 𝑞2𝑒 𝑄2𝑒 𝜙1 𝑤2 = 𝑞 𝑒 + 𝑄𝑒 3 3 𝑒 𝜙2 𝑞4 𝑄4𝑒 𝜁𝑒 = 1.5 − 6Λ𝑒 , Λ𝑒 = 𝑈1 = 𝑈2 = 0, 𝑄3 = 𝐹, 𝑄4 = 0 𝐸𝑒 𝐼𝑒 𝑒 2 , 𝜇0 = 12Λ 𝑒 𝐺𝑒 𝐴𝑒 𝐾𝑠 ℎ𝑒 Boundary conditions: UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT AN EXAMPLE (TBT) 2𝐸𝐼 6 𝜇0 𝐿3 3𝐿 (continued) 𝜇0 𝐿3 𝐹𝐿2 𝜉 12Λ𝐹𝐿3 1.5 + 6Λ 3𝐿 𝑈3 𝐹 = ⇒ 𝑈3 = = 𝜉𝐿2 𝑈4 0 2𝐸𝐼 6𝐿2 𝜉 − 9𝐿2 6𝐸𝐼 12Λ When 𝐸𝐼 1.5𝐹𝐿3 𝐹𝐿3 Λ= = 0 ⇒ 𝑈3 = = 𝐺𝐴𝐾𝑠 𝐿2 6𝐸𝐼 4𝐸𝐼 When Λ ≠ 0 then 𝐹𝐿3 1.5 + 6Λ 𝐹𝐿3 𝑈3 = = 0.75 + 3Λ 6𝐸𝐼 3𝐸𝐼 𝐸𝐼 2 1 + 𝜈 𝐻2 1+𝜈 Λ= = = 𝐺𝐴𝐾𝑠 𝐿2 12𝐿2 𝐾𝑠 6𝐾𝑠 𝐻2 𝐿 2 1.3 𝐻 = 5 𝐿 2 𝐻 = 0.26 𝐿 2 39 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT AN EXAMPLE (TBT) (continued) One element discretization using the CIE element 6 2𝐸𝐼 −3𝐿 𝜇𝐿3 −6 −3𝐿 Σ = 1.0 + 3Λ , −3𝐿 2𝐿2 Σ𝑒 3𝐿 𝐿2 Θ𝑒 −6 −3𝐿 3𝐿 𝐿2 Θ𝑒 3𝐿 6 3𝐿 2𝐿2 Σ𝑒 Θ = 1.0 − 6Λ, Λ𝑒 = 𝑤1 𝑄1 𝜙1 𝑄2 = 𝑤2 𝑄3 𝜙2 𝑄4 𝐸𝐼 , 𝐺𝐴𝐾𝑠 𝐿2 𝜇 = 1 + 12Λ Condensed equations for the unknown displacements 2𝐸𝐼 6 𝜇𝐿3 3𝐿 𝜇𝐿3 2𝐹𝐿2 Σ 𝜇𝐹𝐿3 Σ 3𝐿 𝑈3 𝐹 = ⇒ 𝑈3 = = 2Σ𝐿2 𝑈4 0 2𝐸𝐼 12𝐿2 Σ − 9𝐿2 𝐸𝐼 12Σ − 9 40 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT AN EXAMPLE (TBT) When Λ = 0 ⇒ Σ = 1 and; 𝜇=1 (continued) then 𝜇𝐹𝐿3 Σ 𝐹𝐿3 𝑈3 = = 𝐸𝐼 12 σ −9 3𝐸𝐼 𝜇𝐹𝐿3 Σ 𝐹𝐿3 1 + 3Λ 1 + 12Λ 𝐹𝐿3 When Λ ≠ 0, 𝑈3 = = 𝑈3 = = 1 + 3Λ 𝐸𝐼 12 σ −9 1 + 12Λ 3𝐸𝐼 𝐸𝐼 2 1 + 𝜈 𝐻2 1+𝜈 Λ= = = 𝐺𝐴𝐾𝑠 𝐿2 12𝐿2 𝐾𝑠 6𝐾𝑠 𝐻2 𝐿 2 1.3 𝐻 = 5 𝐿 2 𝐻 = 0.26 𝐿 2 41 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT SUMMARY In this lecture we have covered the following topics: • Derived the governing equations of the Euler-Bernoulli beam theory • Derived the governing equations of the Timoshenko beam theory • Developed Weak forms of E B T and TBT • Developed Finite element models of E B T and TBT • Discussed shear locking in Timoshenko beam finite element • Discussed assembly of beam elements • Discussed examples 42 UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS EBT AND TBT
