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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
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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
Equilibrium equations
Cartesian coordinate system
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
Equilibrium equations
Cartesian coordinate system
In tensor notation is
4w(x,y) =q(x,y)
D
The above equilibrium equation is called Sophie Germain- Lagrange equation
9.21
Equilibrium equations
Sophie Germain (1776-1831)
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Joseph-Louis Lagrange (1736-1813)
Cartesian coordinate system
Sophie Germain- Lagrange equation
4w(x,y) =q(x,y)
D
9.22
Equilibrium equations
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
Principal values of the internal forces
Principal bending moments
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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Principal values of the internal forces
Principal bending moments
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
Loading and supports
Essential (displacement) BCs
w = 0
wx
= 0
9.28
Boundary conditions
Loading and supports
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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
Loading and supports
Mixed BCs
Mx = 0
w = 0
9.30
Boundary conditions
Loading and supports
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
L evys solution- single Fourier series
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
L evys solution- single Fourier series
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
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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
Example- FE analysis
Finite element mesh and supported nodes9.45
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Numerical methods
Example- FE analysis
Deflections9.46
Numerical methods
Example- FE analysis
Contour plot of the bending moments-Mx
9.47
Numerical methods
Example- FE analysis
Section plot of the bending moments-Mx
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Numerical methods
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Example- FE analysis
Contour plot of the bending moments-My9.49
Numerical methods
Example- FE analysis
Section plot of the bending moments-My9.50
Numerical methods
Example- FE analysis
Contour plot of the bending moments-Mxy 9.51
Numerical methods
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Example- FE analysis
Section plot of the bending moments-Mxy
9.52
Numerical methods
Example- FE analysis
Vector plot of the principal moments-M1
9.53
Numerical methods
Example- FE analysis
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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