Finite Element Method: Beam Element Beam Element ¨ Beam: A long thin structure that is subject to the vertical loads. A beam shows more evident bending deformation than the torsion and/or axial deformation. ¨ Bending strain is measured by the lateral deflection and the rotation -> Lateral deflection and rotation determine the number of DoF (Degree of Freedom). Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element ¨ Sign convention 1. Positive bending moment: Anti-clock wise rotation. 2. Positive load: y$ -direction. 3. Positive displacement: y$ -direction. Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element ¨ The governing equation dVdV -w-wdxdx$ +ˆ -dV dV= 0 =0or or w w= =- - dxdx$ ˆ dMdM VVdxdx$ ˆ+-dM dM= =0 0or or VV == dxdx$ˆ Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element ¨ Relation between beam curvature (k ) and bending moment 1 M d 2v$ M k = = or = r EI dx$ 2 EI d 2v$ Curvature for small slope (q = dv$ / dx$ ): k = ¨ dx$ 2 r : Radius of the deflection curve, v$: Lateral displacement function along the y$ -axis direction E : Stiffness coefficient, I : Moment of inertia along the z$-axis Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element Solving the equation with M , d 2 F d 2v$I EI = - wax$f dx$ 2 HG dx$ 2 KJ When EI is constant, and force and moment are only applied at nodes, d 4v$ EI = 0 dx$ 4 Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element Step 1: To select beam element type Step 2: To select displacement function Assumption of lateral displacement 3 2 vˆ() xˆ = a1 xˆ + a 2 xˆ + a 3 xˆ + a 4 - Complete 3-order displacement function is suitable, because it has four degree of freedom (one lateral displacement and one small rotation at each node) - The function is proper, because it satisfies the fundamental differential equation of a beam. - The function satisfies continuity of both displacement and slope at each node. $ $ $ ˆ Representing v$ with functions of d1y , d2 y , f1, f2 $ v$ (0) = d1y = a4 dv$a0f $ = f1 = a3 dv$ $ 3 2 v$aLf = d2 y = a1 L + a2 L + a3 L + a4 $ dvaLf $ 2 = f 2 = 3a1L + 2a2 L + a3 dx$ Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element $ $ $ ˆ Replacing a1~a4 with d1y , d2 y , f1, f2 2 1 v$ = L d$ - d$ + f$ +f$ O x$ 3 3 e 1y 2 y j 2 d 1 2 i NM L L QP 3 1 + L- d$ - d$ - 2f$ +f$ O x$ 2 +f$ x$ + d$ 2 e 1y 2 y j d 1 2 i 1 1y NM L L QP $ Representing it in matrix form, v$ = N odt $ Rd1y U | f$ | d$ = | 1 | o t S $ V d N = N1 N2 N3 N4 | 2 y | |f$ | T 2 W 1 1 N = 2x$ 3 - 3x$ 2 L + L3 N = x$ 3 L - 2x$ 2 L2 + x$ L3 1 L3 c h 2 L3 c h 1 1 N = -2x$ 3 + 3x$ 2 L N = x$ 3 L - x$ 2 L2 3 L3 c h 4 L3 c h where N1, N2, N3 , N4 : shape functions of the beam element Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element Step 3: To define strain-displacement relation and stress-strain relation Assume that the equation of strain-displacement relation is valid du$ dv$ d 2v$ $ $ $ $ $ $ $ e x bx, yg = , u = - y => e x bx, yg = - y 2 dx$ dx$ dx$ Basic assumption: Cross-section of the beam sustains its shape after deformation by bending, and generally rotates by degree of bdv$ / dx$g . Bending moment-lateral displacement relation and shear force-lateral displacement relation d 2v$ d 3v$ $ $ $ mbxg = EI 2 V = EI 3 dx$ dx$ Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element Step 4: To derive an element stiffness matrix and governing equations by direct stiffness method ¨ Element stiffness matrix and governing equations d 3v$ 0 EI f$ = V$ = EI b g = 12d$ + 6Lf$ -12d$ + 6Lf$ 1y dx$3 L3 e 1y 1 2 y 2 j d 2v$ 0 EI m$ = - m$ = - EI b g = 6Ld$ + 4L2f$ - 6d$ + 2L2f$ 1 dx$2 L3 e 1y 1 2 y 2 j d 3v$ L EI f$ = -V$ = - EI b g = -12d$ - 6Lf$ +12d$ - 6Lf$ 2 y dx$3 L3 e 1y 1 2 y 2 j d 2v$ L EI m$ = m$ = EI b g = 6Ld$ + 2L2f$ - 6Ld$ + 4L2f$ 2 dx$2 L3 e 1y 1 2 y 2 j - Matrix Form $ 12 6L -12 6L $ R f1y U L ORd1y U L 12 6L -12 6L O | | 2 2 | | 2 2 m$ EI M 6L 4L -6L 2L P f$ EI M 6L 4L -6L 2L P | 1 | = M P| 1 | $ M P S V 3 S V k = 3 f$ L -12 -6L 12 -6L d$ L -12 -6L 12 -6L | 2 y | M P| 2 y | M P |m$ | M 6L 2L2 -6L 4L2 P|f$ | M 6L 2L2 -6L 4L2 P T 2 W N QT 2 W N Q Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element Step 5: To constitute a global stiffness matrix using boundary conditions Assemble example Assume EI of the beam element is constant. 1000 lb load and 1000 lb - ft moment are applied at the center of the beam. Assume load and moment were only applied at nodes. Left end of the beam is fixed and right end is pin-connected. The beam is divided into two elements (nodes 1, 2, and 3 as shown above figure). Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element d1y f1 d2 y f 2 d2 y f 2 d3 y f 3 L 12 6L -12 6L O L 12 6L -12 6L O EI M 6L 4L2 -6L 2L2 P EI M 6L 4L2 -6L 2L2 P k (1) = M P k (2) = M P L3 M-12 -6L 12 -6LP L3 M-12 -6L 12 -6LP M 2 2 P M 2 2 P N 6L 2L -6L 4L Q N 6L 2L -6L 4L Q R F1y U L 12 6L -12 6L 0 0 O Rd1y U | M | M 6L 4L2 -6L 2L2 0 0 P | f | | 1 | M P | 1 | |F2 y | EI M-12 -6L 12 +12 -6L + 6L -12 6L P |d2 y | = ´ SM V L3 M 6L 2L2 -6L + 6L 4L2 + 4L2 -6L 2L2 P Sf V | 2 | M P | 2 | |F3y | M 0 0 -12 -6L 12 -6LP |d3y | | M | M 0 0 6L 2L2 -6L 4L2 P | f | T 3 W N Q T 3 W Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element Boundary conditions and constrains at the node 1(fixed) and node 3(pin-connected) are f1 = 0 d1y = 0 d3y = 0 R-1000U L24 0 6L ORd2 y U EI | 1000 | = M 0 8L2 2L2 P|f | S V L3 M PS 2 V (5.2.5) | 0 | 6L 2L2 4L2 | f | T W NM QPT 3 W Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element Example: Beam analysis using direct stiffness method Cantilever beam supported by roller at the center Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element Load P is applied at node 1. Length: 2L ,Stiffness: EI Constrains: (1) Roller at node 2, (2) Fixed at node 3. - Global stiffness matrix d1y f1 d2 y f 2 d3 y f 3 L12 6L -12 6L 0 0 O M 2 2 P M 4L -6L 2L 0 0 P EI M 12 +12 -6L + 6L -12 6L P K = L3 M 4L2 + 4L2 -6L 2L2 P M P M 12 -6LP M 2 P N Symmetry 4L Q Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element RF1y U L 12 6L -12 6L 0 0 ORd1y U |M | M 6L 4L2 -6L 2L2 0 0 P| f | | 1 | M P| 1 | |F2 y | EI M-12 -6L 24 0 -12 6L P|d2 y | = SM V L3 M 6L 2L2 0 8L2 -6L 2L2 PSf V | 2 | M P| 2 | |F3y | M 0 0 -12 -6L 12 -6LP|d3y | |M | M 2 2 P|f | T 3 W N 0 0 6L 2L -6L 4L QT 3 W Boundary conditions d2 y = 0 d3y = 0 f 3 = 0 R-PU L12 6L 6L ORd1y U EI | 0 | = M6L 4L2 2L2 P|f | S V L3 M PS 1 V | 0 | 6L 2L2 4L2 |f | T W NM QPT 2 W 7PL3 3PL2 PL2 d1y = - f1 = f 2 = 12EI 4EI 4EI Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element ¨ Substituting the obtained values to the final equation, R 7PL3 U F 12 6L -12 6L 0 0 - R 1y U L O| 12EI | |M | M 6L 4L2 -6L 2L2 0 0 P| 3PL2 | | 1 | M P| | |F2 y | EI M-12 -6L 24 0 -12 6L P| 4EI | = 2 2 2 0 S V 3 M PS 2 V M2 L 6L 2L 0 8L -6L 2L PL | | M P| | |F3y | M 0 0 -12 -6L 12 -6LP| 4EI | |M | M 0 0 6L 2L2 -6L 4L2 P| 0 | T 3 W N Q| | T 0 W 5 F = - P M = 0 F = P 1y 1 2 y 2 3 1 M = 0 F = - P M = PL 2 3y 2 3 2 F1y = - P : Force at node 1 F2 y , F3y , M3 : Reacting forces and moment at nodes M1 , M2 : Zero Mechanics and Design SNU School of Mechanical and Aerospace Engineering Finite Element Method: Beam Element ¨ Calculating local nodal loads Force at the element 1.