5.6 Reissner–Mindlin plates

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A Reissner–Mindlin plate forms a kink if it is loaded with line loads, as illustrated

by analogy with the Timoshenko beam in Fig. 5.16 b. The shearing

strain γ = w/0.5 l activates the shear stress τ = Gγ, which keeps the balance

with the applied load, P/2 = τA = GAγ. If the load is evenly distributed, the

shearing strain varies linearly and the result is a well-rounded deflection curve.

Because kinks are a natural feature of a Reissner–Mindlin plate C0-elements

are sufficient for such plates; see Fig. 5.17.

The deformations of a Reissner–Mindlin plate are described by the deflection

and rotations of the planes x = constant and y = constant (see Fig.

5.18):

w, θx, θy . (5.43)

In a Kirchhoff plate the rotations θx and θy are not independent quantities,

because the planes maintain their position with respect to the midsurface

(which deflects) and the planes rotate by the same angle by which the deflection

surface w rotates: θx = −w,x and θy = −w,y. The expressions

γx = w,x +θx γy = w,y +θy (5.44)

432 5 Slabs

Fig. 5.15. Singularangle

of a Kirchhoff

moments, b) shear

forces

Fig. 5.16. Timoshenko beam a) subjected to a distributed load and b) a single

force

are the shearing strains. In a Kirchhoff plate the shearing strains are zero.

The system of differential equations for w, θx, θy in indicial notation is

ities at the obtuse

plate: a) bending

5.6 Reissner–Mindlin plates 433

5.17. A Reissner–

kinks

K(1 − ν){−(

1

2

(θα,β + θβ,α) + ν

1 − ν

θγ,γ δαβ),β

+ ￣λ2(θα + w,α )} = ν

1 − ν

1

￣λ

2

p,α α = 1, 2 (5.45)

− 1

2K(1 − ν)￣λ2(θα + w,α ),α = p

where

K = Eh3

12(1 − ν2) , ￣λ2 =

10

h2, h= slab thickness . (5.46)

The terms on the right-hand side are the gradient ∇p = [p,x , p,y ]T of the

vertical load and the vertical load p itself.

In the same way as the displacement components of an elastic solid form

a vector u the displacement components of a Reissner–Mindlin plate can be

assembled into a vector

u(x, y) = [w(x, y), θx(x, y), θy(x, y)]T . (5.47)

This is not a true displacement vector because x + u is not the position of

the point x after the deformation; the coordinates of the new position x_ are

instead

x

_ = θxz y

_ = θyz z

_ = w . (5.48)

Fig.

Fig. 5.18. The shearing strain γ

Mindlin plate can have

434 5 Slabs

5.19.

of a Mindlin type

beam element. a) Type 1:

w = ax, θ = 0; b) Type 2:

w = 0, θ = ax

mxx = K (θx,x + ν θy,y), myy = K (θy,y + ν θx,x) ,

mxy = (1 − ν)K (θx,y + θy,x) , (5.49)

while the shear forces also depend on the gradient of the deflection w:

qx = K

1 − ν

2

￣λ

2 (θx + w,x ), qy = K

1 − ν

2

￣λ

2 (θy + w,y ) . (5.50)

Unlike a Kirchhoff plate, a Reissner–Mindlin plate does not sustain the attack

of a single force. It shares this property with a wall beam or simply with 2-D

and 3-D elastic solids.

But otherwise the structural analysis of a Reissner–Mindlin plate is not

different from a Kirchhoff plate, because as long as the thickness to length

ratio is small, the shear deformations are small, and then it hardly matters

(apart perhaps from the results in a zone close to the boundary) which of the

two slab models is used; see Sect. 5.20, p. 480.

Because the differential system of equations (5.45) is of second order, the

symmetric strain energy

a(u,ˆu) =

_

Ω

[mxx ˆθx,x +mxy ˆθx,y +myx ˆθy,x +myy ˆθy,y

+qx (ˆθx + ˆw,x ) + qy (ˆθy + ˆw,y )] dΩ

contains only first-order derivatives.

Formally the solution of a Reissner–Mindlin plate is a vector-valued function

(see Eq. (5.47)), hence the FE solution becomes

uh = u1

⎡

⎣

ψ1

00

⎤

⎦ + u2

⎡

⎣

0

ψ2

0

⎤

⎦ + u3

⎡

⎣

00

ψ3

⎤

⎦

_ _ _

node 1

+u4

⎡

⎣

ψ4

00

⎤

⎦ + u5

⎡

⎣

0

ψ5

0

⎤

⎦ + u6

⎡

⎣

00

ψ6

⎤

⎦

_ _ _

node 2

. . . , (5.51)

Fig. Possible move-

The bending moment depends only on the rotations of the planes,

ments

5.6 Reissner–Mindlin plates 435

Fig. 5.20. Boundary conditions

where the three degrees of freedomui and the three associated shape functions

ψi(x, y) at each node are the deflection and the rotations in x-direction and

y-direction, respectively, so that the basic pattern which repeats at each node

is a sequence of three special unit deformations:

⎡

⎣

w00

⎤

⎦

⎡

⎣

0

θx

0

⎤

⎦

⎡

⎣

00

θy

⎤

⎦ . (5.52)

First the element inclines (see Fig. 5.19) but the planes remain vertical, because

θx = θy = 0 means that the shearing strains γx = w,x and γy = w,y

counterbalance the rotations of the plate mid-surface. In the second and third

deformation the element remains flat, w = 0, but the planes rotate; see Fig.

5.19.

436 5 Slabs

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