Lecture 9_ Plate Analysis

8/4/2019 Lecture 9_ Plate Analysis

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Lecture 9

Plate analysisFloor and deck slabsPrint version Lecture on Theory of Elasticity and Plasticity of

Dr. D. Dinev, Department of Structural Mechanics, UACEG

9.1

Contents

1 Introduction 1

2 Assumptions 3

3 Field equations 4

4 Equilibrium equations 6

5 Principal values of the internal forces 8

6 Boundary conditions 9

7 Analytical solutions 11

7.1 Naviers solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

7.2 L evys solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

8 Numerical methods 14

9 Engineering methods 17 9.2

1 Introduction

Introduction

Civil engineering

The floor slabs in buildings

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Introduction

Bridge engineering

The deck slabs of bridges

9.4

Introduction

Marine engineering

The ship decks and hull

9.5

Introduction

Aircraft engineering

The floor panels and fuselage9.6

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Introduction

Automotive engineering

The car panels

9.7

Introduction

a

b

q

t

Definition

Plates are plane, 2-D structural components of which one dimension, called thickness t is

much smaller than the other dimensions

The plate loads are mainly transversal to the plane surface

They are carried by internal bending and twisting moments and shear forces The plate edges can be simply supported, fixed or elastically restrained

9.8

2 Assumptions

Assumptions

Classical plate theory

1. The plate is thin- t a,b and

ta, t

b

= 1

10

150

2. The in-plane strains are small compared to the unity- xx,yy,xy 13. The transverse normal strain is negligible- zz 0

4. The transverse shear stresses are negligible- xz,yz 0

Note

Applying the above assumptions we can reduce the 3-D problem to a 2-D plate bending

problem

This theory is known as a Kirchhoff-Love plate theory

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Assumptions

Gustav Kirchhoff (1824-1887)

Augustus Love (1863-1940)Classical plate theory

Kirchhoff-Love plate theory9.10

3 Field equations

Field equations

Displacements

The assumption # 3 implies that w(x,y,z) = w0(x,y) where w0 is the transverse displace-ment of the mid-plane (z = 0)

Applying the assumption of the Kirchhoff hypothesis (plane section, normal to the mid-

surface before deflection remains plane and normal to the deformed surface) gives

u(x,y,z) = zx

The assumption # 4 gives xz = yz = 0

Therefore x =wx and y =

wy

9.11

Introductionx, u

x, u

z, wy, v

z, w

q(x,y)

w0

z

z

x

w

0

P

P

Displacements

Plate kinematics9.12

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Field equations

Plate kinematics

Displacement field

u(x,y,z) = zw0x

v(x,y,z) = zw0

yw(x,y,z) = w0(x,y)

Strain-displacement field

xx =zw2

x2

yy = zw2

y2

yy = 2zw2

xy

9.13

Field equations

Plate kinematics

Curvature definition

xxyy

xy

=

w2

x2

w2

y2

2 w2

xy

Strain-displacement relation becomes

xxyyxy

= z xxyyxy

Or

= z

9.14

Field equations

Stresses

Constitutive equations- = E xxyy

xy

= E

12

1 0 1 0

0 0 12

z

xxyy

xy

Or

= zE

9.15

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Consider the equilibrium of a differential element Z= 0

q = Qxx

Qy

y

9.18

Equilibrium equations

Cartesian coordinate system

The moment equilibrium equations of a differential element lead to

Qx =Mxx

+Myx

y

Qy =My

y+

Mxy

x

Mxy = Myx

9.19



Using the above equilibrium relations we may obtain a single equation of the plate equi-

librium in terms of the internal forces

2Mxx2

+ 22Mxy

xy+

2My

y2= q

Replacement of the moments-displacements relations gives the equilibrium equation in

terms of the transversal displacement

4w

x4+ 2

4w

x2y2+

4w

y4=

q

D

9.20



In tensor notation is

4w(x,y) =q(x,y)

D

The above equilibrium equation is called Sophie Germain- Lagrange equation

9.21


Sophie Germain (1776-1831)

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Joseph-Louis Lagrange (1736-1813)


Sophie Germain- Lagrange equation

4w(x,y) =q(x,y)

D

9.22


Shear forces

The shear forces also can be expressed in terms of the displacements

Qx =

D

x2w

x2 +

2w

y2

Qy =D

y

2w

x2+

2w

y2

9.23

5 Principal values of the internal forces

Principal values of the internal forces

Principal bending moments

Consider the internal forces acting on plate with arbitrary section cut

Applying the equilibrium equation z = 0 we have

Qn = Qx cos+ Qy sin

9.24



The moment equations (Ms = 0 and Mn = 0) gives

Mnn = Mxx cos2 + 2Mxy sincos+Myy sin

2

Mns = (MyyMxx) sincos+Mxy(cos2 sin2 )

The infinitesimal parts ofQx, Qy and q are neglected9.25

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The extremum condition Mnn = 0 gives the principal direction

tan2=2Mxy

MxMy

The principal moments are

M1,2 =1

2(Mx +My)

1

2

(Mx +My)

2 + 4M2xy1/2

9.26

6 Boundary conditions

Boundary conditions

a

b

q

t

Loading and supports

An exact solution of the governing plate equations must simultaneously satisfy the differ-

ential equations and BCs of any given plate bending problem

Since the 8-th-order differential equation require two boundary conditions at each plate

edge

9.27

Boundary conditions


Essential (displacement) BCs

w = 0

wx

= 0

9.28

Boundary conditions


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Natural (force) BCs

Mx = 0

Vx = 0

Modified shear force (Kirchhoff equivalent force)

Vx = Qx +Mxyy

Vy = Qy +Mxy

x

9.29

Boundary conditions


Mixed BCs

Mx = 0

w = 0

9.30

Boundary conditions


Elastic restrains

w =Vx

kw

w

x=

Mx

k

9.31

Boundary conditions

Corner forces

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When we have a corner at the plate boundary the twisting moments jump from +Mxy toMxy

The jump of the twisting moment is called corner force Rc

Rc = 2Mxy

9.32

Boundary conditions

Corner forces

This effect appears at plates with corners with simply supported edges

Question

What happens when the edges are fixed or free?

9.33

7 Analytical solutions

7.1 Naviers solutionAnalytical solution

x

y

z

a

b

Naviers solution- double Fourier series

Naviers solution- solution by double trigonometric series

Rectangular plate

Boundary conditions

w = 0 at x = 0,x = a

w = 0 at y = 0,y = b

Mx = 0 at x = 0,x = a

My = 0 at y = 0,y = b

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Analytical solution

Naviers solution- double Fourier series

Shear forces

Qx =3D

a

m=1

n=1

m

a

2+

n

b

2+ (1)

n

b

2mwmn cos

m

ax

sin

n

by

Qy =3D

b

m=1

n=1

m

a

2+

n

b

2+ (1)

m

a

2nwmn sin

m

ax

cos

n

by

9.39

7.2 L evys solution

Analytical solution

L evys solution- single Fourier series

L evys solution- solution by single trigonometric series

Applicable to rectangular plates, simply supported at two opposite edges

The solution of the equilibrium equation is given by

w(x,y) = wh(x,y) + wp(x)

The particular solution is

wp(x) =

i=1

wm sinm

ax

The homogeneous solution is given by

wh(x,y) =

i=1

fm(y) sinm

ax

9.40

Analytical solution


The replacement of the above expressions into the equilibrium equation gives a differential

equation for ym

m44

a4fm(y)2

m22

a2fm(y) + f

m (y) = 0

The solution of the above equation is

fm(y) =Am coshm

ay +Bm

m

ay sinh

m

ay

+Cm sinhm

ay +Dm

m

ay cosh

m

ay

where Am, Bm, Cm = 0 and Dm = 0 are constants and can be determined from the BCs.9.41

Analytical solution


The final solution is

w(x,y) =

i=1

wm sinm

ax

+

i=1

Am cosh

m

ay +Bm

m

ay sinh

m

ay

sinm

ax

9.42

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8 Numerical methods

Numerical methods

Approximate solutions

The universal analytical solution of the governing plate bending equations for complex

domain geometry, BCs and loading is not possible to find

The engineering practice needs to use approximate solutions to solve the above mentioned

problems

The approximate solution are based on the energy and variational methods of structural

mechanics

Ritz method

Galerkin method

Kantorovich method

Numerical methods

Finite differences method

Gridwork method

Finite elements method

Finite strip method

9.43

Numerical methods

5 m

6

6

1 . 4

q = 5 q = 5 k N / m 2

q = 1 2

Example- FE analysis

Determine the displacements and the moments resultants for the given problem

E = 20000000kPa, = 1/4, t = 0.1m

9.44

Numerical methods


Finite element mesh and supported nodes9.45

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Numerical methods


Deflections9.46

Numerical methods


Contour plot of the bending moments-Mx

9.47

Numerical methods


Section plot of the bending moments-Mx

9.48

Numerical methods

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Contour plot of the bending moments-My9.49

Numerical methods


Section plot of the bending moments-My9.50

Numerical methods


Contour plot of the bending moments-Mxy 9.51

Numerical methods

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Section plot of the bending moments-Mxy

9.52

Numerical methods


Vector plot of the principal moments-M1

9.53

Numerical methods


Vector plot of the principal moments-M2

9.54

9 Engineering methods

Engineering methods

Elastic web analogy

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The engineering approach known as Marcus method (1924)

The plate is considered as an elastic web consisting of plate strips located at mid-spans of

the individual panels

The application of this method is limited to

Uniform load

The size difference of the neighboring panels less than 50%

The Poissons ratio is = 16

9.55

Engineering methods

Elastic web analogy

Mid-span deflections of the webs

(wx)x/2 =1

384

qx4

x

Dx

(wy)y/2 =2

384

qy4

y

Dy

Because ofDx = Dy = D and

(wx

)x/2

= (wy

)y/2

q = qx + qy

9.56

Engineering methods

Elastic web analogy

We may obtain the directional loads

qx =24y

4x + 24

y

q

qy = qqx

In general form

qx =Cy

4y

Cx4x +Cy4

y

q

where the factors Cx and Cy depend of the BCs9.57

Engineering methods

L

x

L

y

C a s e 1

Elastic web analogy

Case 1

Cx = 1

Cy = 19.58

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Engineering methods

L

x

L

y

C a s e 2

Elastic web analogy

Case 2

Cx = 2

Cy = 5

9.59

Engineering methods

L

x

L

y

C a s e 3

Elastic web analogy Case 3

Cx = 1

Cy = 5

9.60

Engineering methods

L

x

L

y

C a s e 4

Elastic web analogy

Case 4

Cx = 1

Cy = 19.61

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Engineering methods

L

x

L

y

C a s e 5

Elastic web analogy

Case 5

Cx = 1

Cy = 2

9.62

Engineering methods

L

x

L

y

C a s e 6

Elastic web analogy

Case 6

Cx = 1

Cy = 1

9.63

Engineering methods

Elastic web analogy

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The fundamental difference between the grillage and the plate is the presence of shear

forces between individual strips which produce a torsional resistance an reduce the deflec-

tions9.64

Engineering methods

Elastic web analogy

The approximated maximum span moments in plate are

Mx = Mx

1

5

6

2x2y

Mx

M0x

My = My

1

5

6

2y2x

My

M0y

where Mx and My are maximum span moments in strips and

M0x =q2x

8, M0y =

q2y8

9.65

Engineering methods

Elastic web analogy

The edge moments are calculated as a strip supported with the same type of supports as a

plate and loaded with directional load qx or qy

Msuppx =1

12qx

2x

Msuppy =1

8qy

2y

9.66

Engineering methods

Elastic web analogy

When the support moments of the neighboring panels does not match the bending moments

can be averaged or calculated as support moments in a continuous beam

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Engineering methods

5 m

6

6

1 . 4

q = 5 q = 5 k N / m 2

q = 1 2

Elastic web analogy-example

Calculate the bending moments of the slab

9.68

Engineering methods

The End

Any questions, opinions, discussions?

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Lecture 9_ Plate Analysis

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