1 BENDING OF BEAMS MINDLIN 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 symmetryof a body v(x) = 0 m Vertical displacement does not vary along the height of the beam (whencompared to the value of the displacement) w(x) = w(x). The cross sections remain planar but not necessarily perpendicular tothe deformed beam axis u(x) = u(x, z) = y(x)z

2 STRAIN-DISPLACEMENT EQUATIONS 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 com-ponents are

x(x) =u(x)x

=

x(y(x)z)) =

dy(x)dx

z = y(x)z

zx(x) =w(x)x

+u(x)z

=dw(x)dx

+

z(y(x)z) =

dw(x)dx

+ y(x) ,

3 STRESS-STRAIN RELATIONS 3

when y denotes the pseudo-curvature of the deformed beam centerline.

Bernoulli-Navier [7, kap. II.2] Mindlin

Valid for h/L < 1/10 h/L < 1/3

Cross-section planar, perpendicular planar

zx 0 6= 0 (shear effects)Unknowns w(x) w(x), y(x)

y(x) = dw(x)dx independent

3 Stress-strain relations

For simplicity, we will assume 0 = 0x(x, z) = E(x)x(x, z) = E(x)y(x)z

zx(x) = G(x)zx(x) = G(x)(dw(x)dx

+ y(x))

3 STRESS-STRAIN RELATIONS 4

Non-zero internal forces:My(x) =

A(x)

x(x, z)z dy dz = E(x)y(x)A(x)

z2 dy dz

= E(x)Iy(x)y(x) = E(x)Iy(x)dy(x)dx

(1)

Qcz(x) =A(x)

zx(x) dy dz = G(x)(dw(x)dx

+ y(x))

A(x)

dy dz

= G(x)A(x)(dw(x)dx

+ y(x))

Distribution of shear stresses zx for a rectangular cross-sectionBernoulli-Navier Mindlin

Constitutive eqs: = G 0 constant

Equilibrium eqs quadratic ?

[7, kap. II.2.5]

Therefore, we modify the shear force relation in order to take into

3 STRESS-STRAIN RELATIONS 5

account equilibrium equations, at least in the sense of average work ofshear components

Qz(x) = k(x)Qcz(x) = k(x)G(x)A(x)(dw(x)dx

+ y(x))

(2)

The multiplier k(x) depends on a shape of a cross-section, for a rect-angular cross-section, k = 5/6.

Homework 1. Derive the relation for the constant k for a general cross-

section: k = I2y/(AA

S2y(z)

b2(z) dA).

4 EQUILIBRIUM EQUATIONS 6

4 Equilibrium equations

(a) (b)

Equilibrium equation of vertical forces (a)dQz(x)dx

+ fz(x) = 0 (3)

Equilibrium equation of moments (b)dMy(x)

dxQz(x) = 0 (4)

For a detailed derivation see Lecture 1, Homework 1.

5 GOVERNING EQUATIONS 7

5 Governing equations

ddx

(k(x)G(x)A(x)

(dw(x)dx

+ y(x)))

+ fz(x) = 0 (5)

ddx

(E(x)Iy(x)

dy(x)dx

) k(x)G(x)A(x)

(dw(x)dx

+ y(x))

= 0 (6)

5.1 Kinematic boundary conditions: x Iu

Pinned end: w = 0

Clamped end: w = 0, y = 0

5.2 Static boundary conditions: x Ip

Qz(x) = Qz(x) , My(x) =My(x) .

6 WEAK SOLUTION 8

6 Weak solution

For notational simplicity, we will use relations (3)(4) instead of (5)(6).

We will weight Eq. (3) by term w(x), Eq. (4) by y(x) and inte-grate them on I. This leads to conditions

0 =I

w(x)

(dQz(x)dx

+ fz(x)

)dx,

0 =I

y(x)

(dMy(x)

dxQz(x)

)dx,

which are to be satisfied for all w(x) and y(x) compatible with thekinematic boundary conditions.

6 WEAK SOLUTION 9

By parts integration

0 = [w(x)Qz(x)]ba

I

d(w(x))dx

Qz(x) dx+I

w(x)fz(x) dx

0 = [y(x)My(x)]ba

I

d(y(x))dx

My(x) dxI

y(x)Qz(x) dx

Enforcement of boundary conditions

0 =[w(x)Qz(x)

]IpI

d(w(x))dx

Qz(x) dx+I

w(x)fz(x) dx

0 =[y(x)My(x)

]IpI

d(y(x))dx

My(x) dxI

y(x)Qz(x) dx

6 WEAK SOLUTION 10

The weak of equilibrium equations (we insert (1) for My and (2) forQz)

I

d(w(x))dx

k(x)G(x)A(x)(dw(x)dx

+ y(x))dx =

[w(x)Qz(x)

]Ip+I

w(x)fz(x) dx (7)

I

d(y(x))dx

E(x)Iy(x)dy(x)dx

dx + (8)

I

y(x)k(x)G(x)A(x)(dw(x)dx

+ y(x))dx =

[y(x)My(x)

]Ip

7 FEM DISCRETIZATION 11

7 FEM discretization

We replace a continuous structure with n nodal points and (n 1)(finite) elements.

In every nodal point we introduce two independent quantities a de-flection wi and a rotation yi of the i-th nodal point.

On the level of whole structure, we collect the unknowns into vectorsof deflections rw and rotations r.

Discretization of unknown quantities and their derivatives

w(x) Nw(x)rw , dw(x)dx Bw(x)rw ,

y(x) N(x)r , dy(x)dx B(x)r .

7 FEM DISCRETIZATION 12

Discretization of weight functions

w(x) Nw(x)rw d(w(x))dx Bw(x)rw

y(x) N(x)r d(y(x))dx B(x)r The linear system of discretized equilibrium equations

Kww rw +Kw r = Rw

Kw rw +K r = R

Compact notation Kww KwKw K

rwr =

RwR

K(2n2n)r(2n1) = R(2n1)

8 SHEAR LOCKING 13

Kw = KwT the stiffness matrix K is symmetric thanks to ap-pearance of the terms

I(w(x))kGA(x)y(x) dx in (7) and

y(x)kGA(x)w(x) dx in (8).

Homework 2. Derive explicit relations for matricesKww,Kw,Kw,Kand vectors Rw, R.

8 Shear locking

For h/L 0, the response of a Mindlin theory-based element shouldapproach the classical slender beam (negligible shear effects).

If the basis functions Nw a N are chosen as piecewise linear, resultingresponse in too stiff excessive influence of shear terms, sc. shearlocking.

8 SHEAR LOCKING 14

8.1 Statics-based analysis

Shear force: Qz(x) = k(x)G(x)A(x)(dw(x)dx

+ y(x)) linear

Bending moment: My(x) = E(x)Iy(x) dy(x)dx constant Severe violation of the Schwedler relation

dMy(x)dx

Qz(x) = 0

8.2 Kinematics-based explanation

8 SHEAR LOCKING 15

The approximate solution must be able to correctly reproduce the purebending mode, see [3, Section 3.1]):

y(x) =dy(x)dx

= = const zx(x) =dw(x)dx

+ y(x) = 0

For the given discretization

w(x) w1(1 x

L

)+ w2

x

L

dw(x)dx

1L(w2 w1)

y(x) 1(1 x

L

)+ 2

x

L

dy(x)dx

1L(2 1)

The requirement of zero shear strain leads to

zx(x) 1L(w2 w1) + 1 + x

L(2 1) = 0.

Therefore, the previous relation must be independent of the x coordi-nate

2 1 = 0 y 1L(2 1) = 0 6=

9 SELECTIVE INTEGRATION 16

9 Selective integration

The shear strain is assumed to be constant on a given interval, its valueis derived from the value in the center of an interval

zx(x) 1L(w2w1)+1+12 (2 1) =

1L(w2 w1) + 12 (1 + 2)

Kinematics: the element behaves correctly, it enables to describe thepure bending mode.

Statics: Qz(x) = k(x)G(x)A(x)xz(x) constant, My constant the Schwedler condition is not severely violated.

10 Bubble (hierarchical) function

It follows from analysis of the kinematics that the shear locking iscaused by insufficient degree of polynomial approximation of the dis-

10 BUBBLE (HIERARCHICAL) FUNCTION 17

placement w(x).

Therefore, we add a quadratic term to approximation of w(x):

w(x) w1(1 x

L

)+ w2

x

L+ x(x L)

Pure bending mode requirement

zx(x) =dw(x)dx

+ y(x)

1L(w2 w1) + (2x L) + 1 + x

L(2 1)

=1L(w2 w1) L+ 1 + x

L(2 1 + 2L) = 0

10 BUBBLE (HIERARCHICAL) FUNCTION 18

Requirement of independence of coordinate x

=12L

(1 2)

Final approximations

w(x) w1(1 x

L

)+ w2

x

L+

12L

(1 2)x(x L)

y(x) 1(1 x

L

)+ 2

x

L

From the static point of view the element behaves similarly to pre-vious formulation Qz is constant, My is constant.

Approximation of the w displacement not based not only on the valuesof deflections nodal, but also on the values of nodal rotations [2] sc.linked interpolation.

11 METHOD OF LAGRANGE MULTIPLIERS 19

11 Method of Lagrange multipliers

Recall the weak form of the bending moment equilibrium equations (8)for a beam with withMy = 0, constant values of E, G and a rectangularcross-section.

0 = EIyI

d(y(x))dx

dy(x)dx

dx+ kGAI

y(x)(dw(x)dx

+ y(x))dx

= Ebh3

12

I

d(y(x))dx

dy(x)dx

dx

+56

E

2(1 + )bh

I

y(x)(dw(x)dx

+ y(x))dx/

12Ebh3

I

d(y(x))dx

dy(x)dx

dx+5

1 + 1h2

I

y(x)(dw(x)dx

+ y(x))dx = 0

The condition of zero shear strain for h 0 is imposed via thesc. penalty term.

11 METHOD OF LAGRANGE MULTIPLIERS 20

For slender beams and linear-linear approximation this leads to theshear locking as

1h2

I

arbitrary y(x)

0 for all xI (dw(x)dx

+ y(x))dx = 0.

If we introduce a new independent variable for imposing the conditionxz = 0 for h 0, we suppress influence of the choice of approximationof unknowns w(x) a y(x).

Therefore, we have to add an additional condition to weak equilibriumequations (7)(8)

I

(x)

(zx(x) dw(x)dx y(x)

)dx = 0 , (9)

where (x) is now a new variable independent of w and y and (x)is the corresponding weight function.

11 METHOD OF LAGRANGE MULTIPLIERS 21

Constitutive equations for the shear force Qz now simplify asQz(x) = k(x)G(x)A(x)xz(x) .

Weak form of equilibrium of equations can now be rewritten as

0 =I

d(w(x))dx

k(x)G(x)A(x)zx(x) dx[w(x)Qz(x)

]Ip

I

w(x)fz(x) dx

0 =I

d(y(x))dx

E(x)Iy(x)dy(x)dx

dx

+I

y(x)k(x)G(x)A(x)zx(x) dx[y(x)My(x)

]Ip

0 =I

(x)(zx(x) dw(x)dx y(x)

)dx

Observe that is we choose the weight function in the specific form(x) = k(x)G(x)A(x)xz(x) ,

11 METHOD OF LAGRANGE MULTIPLIERS 22

we will finally obtain a symmetric stiffness matrix K.

The last equation now can be modified as

0 =I

xz(x)k(x)G(x)A(x)(zx(x) dw(x)dx y(x)

)dx.

The additional variable xz needs to be discretizedxz(x) N(x)r

and inserted into the weak form of equilibrium equations. This yields,after standard manipulations, the following system of linear equations

Kww Kw Kw

Kw K K

Kw K K

rw

r

r

=

Rw

R

0

The stiffness matrix, resulting from this discretization, is larger onlyvirtually. It can be observed that parameters r only internal and can

11 METHOD OF LAGRANGE MULTIPLIERS 23

be eliminated (expressed via variables rw and r); see, e.g. [1, pp. 234235] for more details.

This formulation works even for piecewise linear approximation of wand y; it suffices to approximate as a piecewise constant on anelement.

Kinematics: shear locking avoided due to (9). Statics: the shear force Qz is again (piecewise) constant, so is thebending moment My.

Homework 3. Derive the element stiffness matrix based on Lagrangemultipliers. Assume the linear approximation of deflections w(x), linearapproximation of rotations y(x) and constant values of xz on a givenelements. Show that this procedure yields results identical to the reducedintegration and linked interpolation.

2

REFERENCES 24

A humble plea. Please feel free to e-mail any suggestions, errors andtypos to zemanj@cml.fsv.cvut.cz.

Version 000

References

[1] Z. Bittnar and J. Sejnoha, Numerical method in structural mechanics,ASCE Press, ???, 1996.

[2] B. F. de Veubeke, Displacement and equilibrium models in the finiteelement method, International Journal for Numerical Methods in Engi-neering 52 (2001), 287342, Classic Reprints Series, originally publishedin Stress Analysis (O. C. Zienkiewicz and G. S. Holister, editors), JohnWiley & Sons, 1965.

[3] A. Ibrahimbegovic and F. Frey, Finite element analysis of linear andnon-linear planar deformations of elastic initially curved beams, In-

REFERENCES 25

ternational Journal for Numerical Methods in Engineering 36 (1993),32393258.

[4] R. D. Mindlin, Influence of rotatory inertia and shear in flexural mo-tions of isotropic elastic plates, Journal of Applied Mechanics 18 (1951),3138.

[5] E. Reissner, The effect of transverse shear deformation on the bendingof elastic plates, Journal of Applied Mechanics 12 (1945), 6976.

[6] S. Timoshenko, On the correction for shear of the differential equationfor transverse vibrations of prismatic bars, Philosophical Magazine 41(1921), 744746.

[7] J. Sejnoha and J. Bittnarova, Pruznost a pevnost 10, VydavatelstvCVUT, Praha, opravit na anglickou verzi!!!, 1997.

Bending of beams -- Mindlin theoryStrain-displacement equationsStress-strain relationsEquilibrium equationsGoverning equationsKinematic boundary conditions: xIuStatic boundary conditions: x Ip

Weak solutionFEM discretizationShear lockingStatics-based analysisKinematics-based explanation

Selective integrationBubble (hierarchical) functionMethod of Lagrange multipliers