1 Bending of beams Mindlin theory
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1 1 BENDNG OF BEAMS MNDLN THEORY 1 1 Bending of beams Mindlin theory Cross-section kinematics assumptions Distributed load acts in the xz plane, which is also a plane of symmetry of a body Ω v(x = 0 m Vertical displacement does not vary along the height of the beam (when compared to the value of the displacement w(x = w(x. The cross sections remain planar but not necessarily perpendicular to the deformed beam axis u(x = u(x, z = ϕ y (xz
![Timoshenko [6], Reissner [5] and Mindlin](/docs-images/78/78655769/images/2-1.jpg "[4]. J. Bernoulli J.-L. Lagrange C.-L. Navier R.")
2 2 STRAN-DSPLACEMENT EQUATONS 2 These hypotheses were independently proposed by Timoshenko [6], Reissner [5] and Mindlin [4]. J. Bernoulli J.-L. Lagrange C.-L. Navier R.-D. Mindlin B. F. de Veubeke 2 Strain-displacement equations Cross-section kinematics assumptions imply that only non-zero strain components are ε x (x = u(x x γ zx (x = w(x x = x (ϕ y(xz = + u(x z dϕ y(x z = κ y(xz = dw(x + z (ϕ y(xz = dw(x + ϕ y(x,
3 3 STRESS-STRAN RELATONS 3 when κ y denotes the pseudo-curvature of the deformed beam centerline. Bernoulli-Navier [7, kap..2] Mindlin Valid for h/l < 1/10 h/l < 1/3 Cross-section planar, perpendicular planar γ zx 0 0 (shear effects Unknowns w(x w(x, ϕ y (x ϕ y (x = dw(x independent 3 Stress-strain relations For simplicity, we will assume ε 0 = 0 σ x (x, z = E(xε x (x, z = E(xκ y (xz ( dw(x τ zx (x = G(xγ zx (x = G(x + ϕ y(x
4 3 STRESS-STRAN RELATONS 4 Non-zero internal forces: M y (x = σ x (x, zz dy dz = E(xκ y (x A(x A(x z 2 dy dz = E(x y (xκ y (x = E(x y (x dϕ y(x ( dw(x Q c z(x = τ zx (x dy dz = G(x A(x + ϕ y(x A(x ( dw(x = G(xA(x + ϕ y(x (1 dy dz Distribution of shear stresses τ zx for a rectangular cross-section Bernoulli-Navier Mindlin Constitutive eqs: τ = Gγ 0 constant Equilibrium eqs quadratic? [7, kap..2.5] Therefore, we modify the shear force relation in order to take into
5 3 STRESS-STRAN RELATONS 5 account equilibrium equations, at least in the sense of average work of shear components ( dw(x Q z (x = k(xq c z(x = k(xg(xa(x + ϕ y(x The multiplier k(x depends on a shape of a cross-section, for a rectangular cross-section, k = 5/6. (2 Homework 1. Derive the relation for the constant k for a general crosssection: k = y/(a 2 S 2 y (z A b 2 (z da.
6 4 EQULBRUM EQUATONS 6 4 Equilibrium equations (a Equilibrium equation of vertical forces (a (b dq z (x + f z (x = 0 (3 Equilibrium equation of moments (b dm y (x Q z (x = 0 (4 For a detailed derivation see Lecture 1, Homework 1.
7 5 GOVERNNG EQUATONS 7 5 Governing equations d d ( E(x y (x dϕ y(x ( ( dw(x k(xg(xa(x + ϕ y(x ( dw(x k(xg(xa(x + f z (x = 0 (5 + ϕ y(x = 0 (6 5.1 Kinematic boundary conditions: x u Pinned end: w = 0 Clamped end: w = 0, ϕ y = Static boundary conditions: x p Q z (x = Q z (x, M y (x = M y (x.
8 6 WEAK SOLUTON 8 6 Weak solution For notational simplicity, we will use relations (3 (4 instead of (5 (6. We will weight Eq. (3 by term δw(x, Eq. (4 by δϕ y (x and integrate them on. This leads to conditions ( dq z (x 0 = δw(x + f z (x, ( dm y (x 0 = δϕ y (x Q z (x, which are to be satisfied for all δw(x and δϕ y (x compatible with the kinematic boundary conditions.
9 6 WEAK SOLUTON 9 By parts integration 0 = [δw(xq z (x] b a 0 = [δϕ y (xm y (x] b a d(δw(x Q z (x + d(δϕ y (x M y (x δw(xf z (x δϕ y (xq z (x Enforcement of boundary conditions 0 = [ δw(xq z (x ] d(δw(x p Q z (x + δw(xf z (x 0 = [ δϕ y (xm y (x ] d(δϕ y (x p M y (x δϕ y (xq z (x
10 6 WEAK SOLUTON 10 The weak of equilibrium equations (we insert (1 for M y and (2 for Q z ( d(δw(x dw(x k(xg(xa(x [ δw(xqz (x ] p + + ϕ y(x = δw(xf z (x (7 d(δϕ y (x E(x y (x dϕ y(x + (8 ( dw(x δϕ y (xk(xg(xa(x + ϕ y(x = [ δϕ y (xm y (x ] p
11 7 FEM DSCRETZATON 11 7 FEM discretization We replace a continuous structure with n nodal points and (n 1 (finite elements. n every nodal point we introduce two independent quantities a deflection w i and a rotation ϕ yi of the i-th nodal point. On the level of whole structure, we collect the unknowns into vectors of deflections r w and rotations r ϕ. Discretization of unknown quantities and their derivatives w(x N w (xr w, ϕ y (x N ϕ (xr ϕ, dw(x dϕ y (x B w (xr w, B ϕ (xr ϕ.
12 7 FEM DSCRETZATON 12 Discretization of weight functions δw(x N w (xδr w δϕ y (x N ϕ (xδr ϕ d(δw(x B w (xδr w d(δϕ y (x B ϕ (xδr ϕ The linear system of discretized equilibrium equations K ww r w + K wϕ r ϕ = R w K ϕw r w + K ϕϕ r ϕ = R ϕ Compact notation K ww K ϕw K wϕ K ϕϕ r w r ϕ = R w R ϕ K (2n 2n r (2n 1 = R (2n 1
13 8 SHEAR LOCKNG 13 K ϕw = K wϕ T the stiffness matrix K is symmetric thanks to appearance of the terms (δw(x kga(xϕ y (x in (7 and δϕy (xkga(xw (x in (8. Homework 2. and vectors R w, R ϕ. Derive explicit relations for matrices K ww, K wϕ, K ϕw, K ϕϕ 8 Shear locking For h/l 0, the response of a Mindlin theory-based element should approach the classical slender beam (negligible shear effects. f the basis functions N w a N ϕ are chosen as piecewise linear, resulting response in too stiff excessive influence of shear terms, sc. shear locking.
14 8 SHEAR LOCKNG Statics-based analysis ( dw(x Shear force: Q z (x = k(xg(xa(x Bending moment: M y (x = E(x y (x dϕ y(x Severe violation of the Schwedler relation dm y (x Q z (x = Kinematics-based explanation + ϕ y(x constant linear
15 8 SHEAR LOCKNG 15 The approximate solution must be able to correctly reproduce the pure bending mode, see [3, Section 3.1]: κ y (x = dϕ y(x For the given discretization ( w(x w 1 1 x L ( ϕ y (x ϕ 1 1 x L = κ = const γ zx (x = dw(x + ϕ y(x = 0 + w 2 x L + ϕ 2 x L The requirement of zero shear strain leads to dw(x 1 L (w 2 w 1 dϕ y (x 1 L (ϕ 2 ϕ 1 γ zx (x 1 L (w 2 w 1 + ϕ 1 + x L (ϕ 2 ϕ 1 = 0. Therefore, the previous relation must be independent of the x coordinate ϕ 2 ϕ 1 = 0 κ y 1 L (ϕ 2 ϕ 1 = 0 κ
16 9 SELECTVE NTEGRATON 16 9 Selective integration The shear strain is assumed to be constant on a given interval, its value is derived from the value in the center of an interval γ zx (x 1 L (w 2 w 1 +ϕ (ϕ 2 ϕ 1 = 1 L (w 2 w (ϕ 1 + ϕ 2 Kinematics: the element behaves correctly, it enables to describe the pure bending mode. Statics: Q z (x = k(xg(xa(xγ xz (x constant, M y constant the Schwedler condition is not severely violated. 10 Bubble (hierarchical function t follows from analysis of the kinematics that the shear locking is caused by insufficient degree of polynomial approximation of the dis-
17 10 BUBBLE (HERARCHCAL FUNCTON 17 placement w(x. Therefore, we add a quadratic term to approximation of w(x: ( w(x w 1 1 x x + w 2 + αx(x L L L Pure bending mode requirement γ zx (x = dw(x + ϕ y(x 1 L (w 2 w 1 + α(2x L + ϕ 1 + x L (ϕ 2 ϕ 1 = 1 L (w 2 w 1 αl + ϕ 1 + x L (ϕ 2 ϕ 1 + 2αL = 0
18 10 BUBBLE (HERARCHCAL FUNCTON 18 Requirement of independence of coordinate x Final approximations ( w(x w 1 1 x L ( ϕ y (x ϕ 1 1 x L α = 1 2L (ϕ 1 ϕ 2 + w 2 x L + 1 2L (ϕ 1 ϕ 2 x(x L + ϕ 2 x L From the static point of view the element behaves similarly to previous formulation Q z is constant, M y is constant. Approximation of the w displacement not based not only on the values of deflections nodal, but also on the values of nodal rotations [2] sc. linked interpolation.
19 11 METHOD OF LAGRANGE MULTPLERS Method of Lagrange multipliers Recall the weak form of the bending moment equilibrium equations (8 for a beam with with M y = 0, constant values of E, G and a rectangular cross-section. 0 = E y = E bh d(δϕ y (x d(δϕ y (x E 2(1 + ν bh dϕ y (x d(δϕ y (x dϕ y (x ( dw(x δϕ y (x dϕ y (x ν + kga 1 h 2 + ϕ y(x ( dw(x δϕ y (x ( dw(x δϕ y (x / 12 Ebh 3 + ϕ y(x + ϕ y(x = 0 The condition of zero shear strain for h 0 is imposed via the sc. penalty term.
20 11 METHOD OF LAGRANGE MULTPLERS 20 For slender beams and linear-linear approximation this leads to the shear locking as {}}{ 1 h 2 arbitrary {}}{ δϕ y (x 0 for all x ({}} { dw(x + ϕ y(x = 0. f we introduce a new independent variable for imposing the condition γ xz = 0 for h 0, we suppress influence of the choice of approximation of unknowns w(x a ϕ y (x. Therefore, we have to add an additional condition to weak equilibrium equations (7 (8 ( δλ(x γ zx (x dw(x ϕ y(x = 0, (9 where γ(x is now a new variable independent of w and ϕ y and δλ(x is the corresponding weight function.
21 11 METHOD OF LAGRANGE MULTPLERS 21 Constitutive equations for the shear force Q z now simplify as Q z (x = k(xg(xa(xγ xz (x. Weak form of equilibrium of equations can now be rewritten as 0 = d(δw(x k(xg(xa(xγ zx (x [ δw(xq z (x ] p δw(xf z (x d(δϕ y (x 0 = E(x y (x dϕ y(x + δϕ y (xk(xg(xa(xγ zx (x [ δϕ y (xm y (x ] p ( 0 = δλ(x γ zx (x dw(x ϕ y(x Observe that is we choose the weight function in the specific form δλ(x = k(xg(xa(xδγ xz (x,
22 11 METHOD OF LAGRANGE MULTPLERS 22 we will finally obtain a symmetric stiffness matrix K. The last equation now can be modified as ( 0 = δγ xz (xk(xg(xa(x γ zx (x dw(x The additional variable γ xz needs to be discretized γ xz (x N γ (xr γ ϕ y(x. and inserted into the weak form of equilibrium equations. This yields, after standard manipulations, the following system of linear equations K ww K wϕ K wγ r w R w K ϕw K ϕϕ K ϕγ r ϕ = R ϕ K γw K γϕ K γγ r γ 0 The stiffness matrix, resulting from this discretization, is larger only virtually. t can be observed that parameters r γ only internal and can
23 11 METHOD OF LAGRANGE MULTPLERS 23 be eliminated (expressed via variables r w and r ϕ ; see, e.g. [1, pp ] for more details. This formulation works even for piecewise linear approximation of w and ϕ y ; it suffices to approximate γ as a piecewise constant on an element. Kinematics: shear locking avoided due to (9. Statics: the shear force Q z bending moment M y. is again (piecewise constant, so is the Homework 3. Derive the element stiffness matrix based on Lagrange multipliers. Assume the linear approximation of deflections w(x, linear approximation of rotations ϕ y (x and constant values of γ xz on a given elements. Show that this procedure yields results identical to the reduced integration and linked interpolation.
24 REFERENCES 24 A humble plea. Please feel free to any suggestions, errors and typos to zemanj@cml.fsv.cvut.cz. Version 000 References [1] Z. Bittnar and J. Šejnoha, Numerical method in structural mechanics, ASCE Press,???, [2] B. F. de Veubeke, Displacement and equilibrium models in the finite element method, nternational Journal for Numerical Methods in Engineering 52 (2001, , Classic Reprints Series, originally published in Stress Analysis (O. C. Zienkiewicz and G. S. Holister, editors, John Wiley & Sons, [3] A. brahimbegović and F. Frey, Finite element analysis of linear and non-linear planar deformations of elastic initially curved beams, n-
25 REFERENCES 25 ternational Journal for Numerical Methods in Engineering 36 (1993, [4] R. D. Mindlin, nfluence of rotatory inertia and shear in flexural motions of isotropic elastic plates, Journal of Applied Mechanics 18 (1951, [5] E. Reissner, The effect of transverse shear deformation on the bending of elastic plates, Journal of Applied Mechanics 12 (1945, [6] S. Timoshenko, On the correction for shear of the differential equation for transverse vibrations of prismatic bars, Philosophical Magazine 41 (1921, [7] J. Šejnoha and J. Bittnarová, Pružnost a pevnost 10, Vydavatelství ČVUT, Praha, opravit na anglickou verzi!!!, 1997.
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