theory of plates - imechanica of plates.pdf · 2014-06-04 · thin plate – out of plane shear...
TRANSCRIPT
R&DE (Engineers), DRDO
Theory of Plates
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Ramadas Chennamsetti
Theory of Plates
R&DE (Engineers), DRDO
Introduction
� “When a body is bounded by surfaces, flat in geometry, whose lateral dimensions are large compared to the separation between the surfaces is called a PLATE”
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Ramadas Chennamsetti2
� Plates are initially flat structural elements
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Introduction � Plates are subjected to transverse loads – loads
normal to its mid-surface � Transverse loads supported by combined bending
and shear action� Plates may be subjected to in-plane loading also
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Ramadas Chennamsetti3
� Plates may be subjected to in-plane loading also => uniform stress distribution => membrane
� Membrane action – in-plane loading or pronounced curvature & slope
� Plate bending – plate’s mid-surface doesn’t experience appreciable stretching or contraction
� In-plane loads cause stretching and/or contraction of mid-surface
R&DE (Engineers), DRDO
Introduction
� Plate stretching
NxNxx
zt
Uniform stretching of the plate => uo
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Ramadas Chennamsetti4
Uniform stretching of the plate => uoq
Axial deformation due to transverse load
Net deformation = Algebraic sum of uniform stretching and axial deformation due to bending load
R&DE (Engineers), DRDO
Introduction
� For plates -� Thin & thick plates –
� Thin plate => b = smallest side � Thick plate =>
bt 20<bt 20>
2000
1
10
1 ≥≥b
t
t≤
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Ramadas Chennamsetti5
� Small deflections –� Thin plate theory – Kirchoff’s Classical
Plate Theory (KCPT)� Thick plate theory – Reissner – Mindlin
Plate Theory (MPT)
5
tw ≤
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KCPT - AssumptionsAssumptions –� Thickness is much smaller than the other physical
dimensions � vertical deflection w(x, y, z) = w(x, y)
� Displacements u, v & w are small compared to plate thickness
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Ramadas Chennamsetti6
plate thickness � Governing equations are derived based on undeformed
geometry
� In plane strains are small compared to unity –consider only linear strains
� Normal stresses in transverse direction are small compared with other stresses – neglected
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KCPT - Assumptions
� Material – linear elastic – Hooke’s law holds good
� Middle surface remains unstrained during bending – neutral surface
� Normals to the middle surface before deformation remain normal to the same surface after
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remain normal to the same surface after deformation => doesn’t imply shear across section is zero – transverse shear strain makes a negligible contribution to deflections.� Transverse shear strains are negligible
� Rotary inertia is neglected
R&DE (Engineers), DRDO
Sign convention
� Following sign convention will be followed
z
Mz
z
ψψψψz
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Ramadas Chennamsetti8
y
x
Mx
My
Mz
x
ψψψψx
ψψψψy
ψψψψz
y
Positive moments Positive rotations
R&DE (Engineers), DRDO
Bending deformations
y
z
D
Bending takes place in both planes
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Ramadas Chennamsetti9
x
A
B
z
x
A
B
A’
B’
View ‘D’
Rotation => ψy
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Bending deformations
� Deformation in ‘x’ direction
� Deformation in ‘y’ direction
( )yxzzyxu y ,) , ,( ψ−=
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Ramadas Chennamsetti10
� Deformation in ‘y’ direction
� Vertical deformation
( )yxzzyxv x ,) , ,( ψ−=
) ,( yxww =
R&DE (Engineers), DRDO
Strains
� Assumption – out of plane shear strain -negligible
wwx
w
z
uxz
∂∂
=∂∂+
∂∂=γ 0
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Ramadas Chennamsetti11
y
w
y
w
y
w
z
vx
w
x
w
xxyz
yz
yyxz
∂∂==>=
∂∂+−==>
=∂∂+
∂∂=
∂∂==>=
∂∂+−==>
ψψγ
γ
ψψγ
0
0
0
R&DE (Engineers), DRDO
Strains
� Non-zero strains
wv
x
wz
xz
x
u yxx
∂∂∂∂∂−=
∂∂
−=∂∂=
2
2
2
ψ
ψε
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Ramadas Chennamsetti12
yx
wz
x
v
y
u
y
wz
yz
y
v
xy
xyy
∂∂∂−=
∂∂+
∂∂=
∂∂−=
∂∂−=
∂∂=
2
2
2
2γ
ψε
R&DE (Engineers), DRDO
Stresses
� Thin plate – out of plane shear strains vanish – out of plane shear stresses also vanish
0 ,0 == yzxz ττz
τ
z
τxz
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Ramadas Chennamsetti13
� Out of plane normal stress is also assumed to be zero – logical – thin structure – plane stress conditions
yτyz
τyz
x
τxz
τxz
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Stresses
� Non-zero stress components
y
z σxx
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Ramadas Chennamsetti14
x
σyy
σyy
σxx
τxy
τxy
All three stress components, σxx, σyy, τxy – in-plane
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Curvatures
� Curvature – reciprocal of radius of bending
� Rate of change of slope
z
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Ramadas Chennamsetti15
Rx
dx
x
ww*
dx
ψy
dxx
w
∂∂−
x
wy ∂
∂=== - rotation slope ψ xxy
x
w
xκ
ψ=
∂∂−=
∂∂
=2
2
curvatue
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Curvatures
� Similarly bending in yzplane introduces a curvature
2
2
y
wyy ∂
∂−=κ
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� Twisting of plate
y∂
yx
wxy ∂∂
∂−=2
2κ
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Constitutive law
� Linear elastic isotropic – Hookean material
� Three stress and strain components
συσε −=
EEyyxx
xx
Writing all three equations in matrix form
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Ramadas Chennamsetti17
( )υττ
γ
συσ
ε
υε
+==
−=
−=
12EG
EE
EE
xyxyxy
xxyyyy
xx
−−=
+−
−=
xy
yy
xx
xy
yy
xx
xy
yy
xx
xy
yy
xx
E
E
γεε
υυ
υ
υτσσ
τσσ
υυ
υ
γεε
2
100
01
01
1
)1(200
01
011
2
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Constitutive law
� Express strains in terms of curvatures
∂∂∂∂
+−−=
y
wx
w
Ezyy
xx
2
2
2
2
21
00
01
01
1 υυ
υ
υτσσ
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Ramadas Chennamsetti18
∂∂∂
∂
+−
yx
wy
xy 2
22
100
1 υυτ
Variation of stresses across thickness is linear
Basis – thin plates – plane section remain plane after bending – variation of axial deflection is linear across thickness – strains also vary linearly
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Equilibrium equations
� Equilibrium of an infinitesimal element
y
z
dx
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Ramadas Chennamsetti19
x
dx
dy
t
τ*yz
dx
dy
σ*yyσyy
σxx
σ*xx
τxy
τ*xy
τxy
τ*xy
τ*xz
τ*yz
Forces acting on an infinitesimal element dx dy dz
y
x
R&DE (Engineers), DRDO
Equilibrium equations
� Equilibrium in ‘x’ direction
∂
=−−−
++
dzdxdxdydydz
dzdxdxdydydz
yxzxxx
yxzxxx
σ
ττσττσ ***
0
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Ramadas Chennamsetti20
∂∂
+=
∂∂+=
∂∂+=
dyy
dzz
dxx
yxyxyx
zxzxzx
xxxxxx
τττ
τττ
σσσ
*
*
*
R&DE (Engineers), DRDO
Equilibrium equations
� Substitute – the following equation is obtained
(1) - 0=∂
∂+∂
∂+
∂∂
zyxxzxyxx τττ
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Ramadas Chennamsetti21
Similarly take equilibrium in ‘y’ and ‘z’ directions
(2) - 0=∂
∂+
∂∂
+∂
∂zyxyzyyxy τττ
(3) - 0=∂
∂+∂
∂+
∂∂
zyxzzyzxz τττ
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Shear stresses
� From equation (1) – shear stress τxz can be computed
Use stress deflection/curvature relations
wEzwwEz 222 ∂ ∂∂ ∂∂∂ τ
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Ramadas Chennamsetti22
( )
( )wx
Ez
z
yx
w
yx
w
x
wEz
z
zyx
wEz
yy
w
x
wEz
x
xz
xz
xz
22
2
3
2
3
3
3
2
2
2
2
2
2
2
1
11
011
∇∂∂
−=
∂∂
∂∂∂−+
∂∂∂+
∂∂
−=
∂∂
=∂
∂+
∂∂∂
+−
∂∂+
∂∂+
∂∂
−−
∂∂
υτ
υυυ
τ
τυ
υυ
Integrate this across thickness to get shear stress
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Shear stresses
� Integrate from mid plane to top surface of plate
( ) ( )∂∇∂
−=
∂∇∂
−= ∫∫∫
11
22
2
2 2
2
0
dzzx
wEdz
x
wEzd
hh
xz υυτ
z
z z = h/2
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Ramadas Chennamsetti23
( )
( )( )
( )( )
−
∂∇∂
−==>
−
∂∇∂
−=−=>
∂∇∂
−=−
∂−∂ − ∫∫∫
412
412
21
11
22
2
2
222
2
222
2
20
2
hz
x
wE
zh
x
wE
z
x
wE
xx
xz
xz
h
z
xz
zxz
υτ
υτ
υτ
υυτ
Parabolic variation
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Moments
� Moment wrt ‘y’ axis
∂∂+
∂∂
−−=
===
2
2
2
2
21 x
w
x
wEz
dzzzdFdM
dzdF
xx
xxxxy
xxxx
υυ
σ
σσ
Neutral planez
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Ramadas Chennamsetti24
( )
∂∂+
∂∂
−−=
∂∂+
∂∂
−−=
∂∂+
∂∂
−−==>
∂∂−
∫+
−
2
2
2
2
2
3
2
2
22
2
2
2
2
2
2
2
2
2
2
112
1
1
1
y
w
x
wEhM
dzzy
w
x
wEM
y
w
x
wEzdM
xx
y
h
h
y
y
υυ
υυ
υυ
υ
x
yσxx
z
dz
dy
dx
R&DE (Engineers), DRDO
Moments
� Moment wrt ‘x’ axis
∂∂+
∂∂
−−=
2
2
2
2
2
3
)1(12 y
w
x
wEhM x υ
υNeutral plane
z
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Ramadas Chennamsetti25
x
y
z
τxyz
dz
dy
dx
Twisting moment due to shear stress τxy
( )yx
wEhM xy ∂∂
∂−−
−=2
2
3
1)1(12
υυ
)1(12 2
3
υ−= Eh
DCompare with section modulus of beam
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Shear forces
� Vertical equilibrium of plate
z
q
dxx
QQQ
dxQdyQqdxdydxQdyQ
xxx
yxyx
∂∂
∂+=
=−−++
*
** 0
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Ramadas Chennamsetti26
x
y
q
dy
dx
Q*x
Q*y
Qy
Qx
dyy
QQQ y
yy ∂∂
+=*
Substitute these and simplify
qy
Q
x
Q yx −=∂
∂+
∂∂
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Shear forces
� Shear forces across thickness can be computed by integrating shear stress across thickness
dzQh
h
yzy = ∫+
−
2
2
τ per unit width
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Ramadas Chennamsetti27y
wD
y
wEhQ
dzh
zy
wEQ
hz
y
wE
y
h
h
y
yz
h
∂∇∂−=
∂∇∂
−−=
−
∂∇∂
−=
−
∂∇∂
−=
∫+
−
−
)()(
)1(12
4
)(
)1(2
4
)(
)1(2
22
2
3
2
2
22
2
2
22
2
2
2
υ
υ
υτ
R&DE (Engineers), DRDO
Governing equation
� Similar expression for Qx
x
wDQ x ∂
∇∂−= )( 2
Substitute Qx and Qy in the following expression – vertical equilibrium
qQQ yx −=
∂+∂
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Ramadas Chennamsetti28
( ) ( )
( ) ( )
( )D
qw
yx
D
qw
yw
x
qwy
Dy
wx
Dx
qy
Q
x
Q yx
=∇
∂∂+
∂∂=>
=∇∂∂+∇
∂∂=>
−=
∇
∂∂−
∂∂+
∇∂∂−
∂∂
−=∂
∂+
∂∂
22
2
2
2
22
22
2
2
22
R&DE (Engineers), DRDO
Governing equation
� Governing equation
( )D
qw
D
qw
yx
=∇∇=>
=∇
∂∂+
∂∂
22
22
2
2
2
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Ramadas Chennamsetti29
( )
D
qD
w
=∇=>
=∇∇=>
4
D
q
y
w
yx
w
x
w =∂∂+
∂∂∂+
∂∂
4
4
22
4
4
4
2
Bi-harmonic equationCompare with beam equation
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Boundary conditions
� Well posed problem – Governing equations and boundary conditions
� Three basic boundary conditions � Simply supported
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Ramadas Chennamsetti30
� Simply supported
� Clamped and
� Free edge
� Vertical deflection and their derivatives
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Simply supported
� Simply supported – for eg beam => vertical deflection = 0 and moment = 0
( ) 0, =yxw
Simply supportedx = constFor edge, x = const
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02
2
2
2
=
∂∂+
∂∂−=
y
w
x
wDM y υ
Simply supportedx
y
y = const
w = 0 implies second derivative in the direction tangent to this line is zero
02
2
=∂∂
x
w
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Simply supported
� Simply supported condition along edge y = const
∂+∂−=
=22
0
wwDM
w
υ
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Ramadas Chennamsetti32
∂∂+
∂∂−=
22 y
w
x
wDM x υ
w = 0 implies second derivative in the direction tangent to this line is zero
02
2
=∂∂
y
w
R&DE (Engineers), DRDO
Clamped
� Deflection and slope in normal directions vanish
0
0
=∂=
w
wx = constant
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Ramadas Chennamsetti33
0=∂x
0
0
=∂∂
=
y
w
wy = constant
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Free edges
� Free edge – Free from any external loads –Natural boundary conditions
� Bending moment and Shear force vanish
� Bending moment
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Ramadas Chennamsetti34
� Bending moment
const y at 0
const at x 0
2
2
2
2
2
2
2
2
==
∂∂+
∂∂−=
==
∂∂+
∂∂−=
y
w
x
wDM
y
w
x
wDM
x
y
υ
υ
R&DE (Engineers), DRDO
Free edges
� Moment Mxy and shear forces
Mxy
M
dy
The net force acting on the face
MM
MQ xy +∂
−−='
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Ramadas Chennamsetti35
dy
Mxy
dxMxy
Mxy
dy
dyy
MM xy
xy ∂∂
+
dyy
MM xy
xy ∂∂
+
y
MQ
My
MMQ
xyx
xyxy
xyx
∂∂
−==>
+∂
∂−−=
'
'
Total shear force,
Vx = Qx + Qx’
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Free edges
� Total shear force,
( ) ( )
∂∂
∂∂−−∇
∂∂−==>
+=
x
w
xDw
xDV
QQV
x
xxx
2
22
'
1 υ
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Ramadas Chennamsetti36
( ) ( )
( )
( )
∂∂∂−+
∂∂−=
∂∂∂−+
∂∂−==>
∂∂
−−∇∂
−==>
yx
w
y
wDV
yx
w
x
wDV
xxDw
xDV
y
x
x
2
3
3
3
2
3
3
3
2
2
2
1
υ
υ
υ
R&DE (Engineers), DRDO
Free edges
� The forces, Vx and Vy => reduced, or Kirchoff’s or effective shear forces
� In case of a free edge,
33 ∂∂
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( )
( ) const y at 02
const at x 02
2
3
3
3
2
3
3
3
==
∂∂∂−+
∂∂−=
==
∂∂∂−+
∂∂−=
yx
w
y
wDV
yx
w
x
wDV
y
x
υ
υ
R&DE (Engineers), DRDO
Bending of plates
� Governing equation of plate rectangular plate bending
( )D
qw =∇∇ 22
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Ramadas Chennamsetti38
� Solution to this equation – product of two functions – Assume
w = vertical deflection = w(x, y)
external loading, q = q(x, y)
( ) ( ) ( )yGxFyxww == ,
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Bending of plates
� Choice of functions – algebraic, trigonometric, hyperbolic etc or combination of these function
� Selection of a function – depends on
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Ramadas Chennamsetti39
� Selection of a function – depends on boundary conditions
� Simply supported edges – trigonometric function – Navier solution
� Deflection of a plate can be written as sum of infinite trigonometric functions
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Bending of plates
� Edges, x = 0 and x = a simply supported
Vertical deflection vanish
w(x=0, y)=0, w(x=a, y)=0
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Ramadas Chennamsetti40
x
y
x = 0 x = aSSSS
Possible form of solution
( ) ∑∞
=m
m a
xmFxF
sin
π
F1, F2,…….F∞ are coefficients
Symmetric loading wrt x = a/2 Maximum deflection at x = a/2
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Bending of plates
� Other edges simply supported
y = b
SSVertical deflection vanish
w(x, y=0)=0, w(x, y=b)=0
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Ramadas Chennamsetti41
x
y
x = 0, y = 0
x = aSSSS
SS
Possible form of solution
( ) ∑∞
=n
n b
ynGyG
sin
π
G1, G2,…….G∞ are coefficients
Selection of functions based on BCs
R&DE (Engineers), DRDO
Bending of plates
� Final solution,
( )
( )
b
yn
a
xmGFyxw
m nnm
sin
sin,
ππ
=
∑∑
∑∑∞ ∞
∞ ∞
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Ramadas Chennamsetti42
( )
b
n
a
m
yxwyxw
nm
nm n
mmn
,
sin sin,
πβπα
βα
==
==> ∑∑∞ ∞
Coefficients, Fm, Gn and wmn computed using Fourier Series
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Coefficients
� Any periodic function can be expanded into a sine or cosine function using Fourier expansionFunction of one variable
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Ramadas Chennamsetti43
( )
( )a
xm
a
xmf
a
xmxf
a
xmfxf
m
mm
sin
sin
sin
sin
'' πππ
π
==>
=∑∞
Integrate the above from limits 0 to a
R&DE (Engineers), DRDO
Coefficients
� Integration
( )
( ) ( )a
ma
m
a
m
a
dxmmπx
mmπxf
dxxmxmf
dxa
xm
a
xmfdx
a
xmxf
'''
0
'
0
'
coscos sin sin2
sin sin sin
+−−==>
=
∫∫
∫∫
ππ
πππ
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Ramadas Chennamsetti44
( ) ( )
( )( )
( )( )
a
m
mm
mmπ
mmπx
mmπ
mmπx
f
dxmmπx
mmπxf
dxa
xm
a
xmf
0
'
'
'
'
0
''
0
a
asin
a
asin
2
a
cosa
cos2
sin sin22
+
+−
−
−=>
+−−==> ∫∫ππ
Lower limit vanishes, evaluate upper limit
R&DE (Engineers), DRDO
Coefficients
� Upper limit
( )( )
( )mma
mma
xf
dxa
xmxf m
a
∫−
−=
'
'
0
' sin
2
sin π
ππ
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Ramadas Chennamsetti45
( )
( )
( ) dxa
xmxf
af
mm
mmaf
dxa
xmxf
mma
a
m
ma
ax
∫
∫
=
≠=
===>
−=
0
'
'
0
'
sin
2
for 0
for 2
sin
π
π
R&DE (Engineers), DRDO
Coefficients
� Function of variable ‘y’
( )b
yngyg
nn
sin
π∑=
∞
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Ramadas Chennamsetti46
( )
( ) dyb
ynyg
bg
b
yn
b
yng
b
ynyg
b
n
n
sin
2
sin
sin
sin
0
''
π
πππ
∫=
==>
R&DE (Engineers), DRDO
Coefficients
� Function of two variables
( )
( ) dxa
xm
a
xm
b
ynwdx
a
xmyxw
b
yn
a
xmwyxw
ax
xm nmn
ax
x
m nmn
∫∑∑∫
∑∑=
=
∞ ∞=
=
∞ ∞
=
=>
=
0
''
0
sin
sin
sin
sin,
sin
sin,
ππππ
ππ
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Ramadas Chennamsetti47
( )
( )
( ) dxdyb
yn
a
xmyxw
abw
dyb
yn
b
ynw
adxdy
b
yn
a
xmyxw
b
ynw
adx
a
xmyxw
aaba
by
y
ax
x
mn
by
ym nmn
by
y
ax
x
m nmn
ax
x
xm nx
∫ ∫
∫∑∑∫ ∫
∑∑∫
=
=
=
=
=
=
∞ ∞=
=
=
=
∞ ∞=
=
==
==>
=
=>
=
0 0
'
00 0
''
'
0
00
sin
sin,
4
sin
sin
2
sin
sin,
sin
2
sin,
ππ
ππππ
ππ
R&DE (Engineers), DRDO
Simply supported plate
� Assume loading over plate
( )b
yn
a
xmqyxqq
m nmn
sin
sin,
ππ∑∑
∞ ∞
==
Solution
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Ramadas Chennamsetti48
Solution
( )b
yn
a
xmwyxww
m nmn
sin
sin,
ππ∑∑
∞ ∞
==
Governing equation
D
q
y
w
yx
w
x
w =∂∂+
∂∂∂+
∂∂
4
4
22
4
4
4
2
R&DE (Engineers), DRDO
Simply supported plate
� Differentiating vertical displacement
=∂∂
∑∑∞ ∞
b
yn
a
xmw
a
m
x
w
m nmn
sin
sin
4
4
4 πππ
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Ramadas Chennamsetti49
=∂∂
∂
=∂∂
∑∑
∑∑∞ ∞
∞ ∞
b
yn
a
xmw
b
n
a
m
yx
w
b
yn
a
xmw
b
n
y
w
m nmn
m nmn
sin
sin22
sin
sin
22
22
4
4
4
4
ππππ
πππ
R&DE (Engineers), DRDO
Simply supported plate
� Plug in governing equation
=
+
+
q
D
qw
b
n
b
n
a
m
a
m mnmn2
4224 ππππ
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Ramadas Chennamsetti50
( )
+
==
+
==>
∑∑∞ ∞
b
yn
a
xm
b
n
a
m
q
Dyxww
b
n
a
mD
qw
m n
mn
mnmn
sin
sin
1, 2
2
2
2
24
2
2
2
2
24
πππ
π
R&DE (Engineers), DRDO
UDL
� Uniformly distributed load qo
Computation of coefficients
( )
= ∑ ∑
∞ ∞ ynxmqyxq mn
sin
sin,
ππ
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Ramadas Chennamsetti51
( ) dxdyb
yn
a
xmyxq
abq
a b
mn
sin
sin,
4
0 0
ππ∫ ∫=
( )
= ∑ ∑ baqyxq
m nmn sinsin,
q(x, y) = qo
R&DE (Engineers), DRDO
UDL
� Coefficients qmn
0 0
sin
sin
4 ππdxdy
b
ym
a
xmq
abq
a b
omn = ∫ ∫
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Ramadas Chennamsetti52
22
00
1644
sin
sin
4
ππ
ππ
mn
q
mn
ab
ab
dyb
ymdx
a
xm
ab
oomn
bao
mn
=
==>
= ∫∫
R&DE (Engineers), DRDO
UDL
� Vertical deflection
( ) ∑∑∞ ∞
+
==m n
mn
nm
b
yn
a
xmq
Dyxww 2224
sin
sin
1,
ππ
π
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Ramadas Chennamsetti53
( ) ∑∑∞ ∞
+
==>
+
m n
o
b
n
a
m
b
yn
a
xm
D
qyxw
b
n
a
m
2226
sin
sin
16,
ππ
π
R&DE (Engineers), DRDO
Patch load
� A patch load applied over an area u x v
( )
=∑∑∞ ∞
b
yn
a
xmqyxq mn
sin
sin,
ππvqo
Centroid at (xo, yo)
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Ramadas Chennamsetti54
==>
∑∑
∑∑∞ ∞
b
yn
a
xmqq
ba
m nmno
m nmn
sin
sin
ππ u
vqo
dxdyb
ym
a
xm
ab
ux
ux
vy
vy
omn
o
o
o
o
sin
sin
4 2
2
2
2
ππ∫ ∫+
−
+
−
=
Patch load
R&DE (Engineers), DRDO
Patch load
� Evaluating integrals
= ∫ ∫+
−
+
−
dxdyb
ym
a
xm
ab
ux
ux
vy
vy
omn
o
o
o
o
sin
sin
4 2 2 ππ
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Ramadas Chennamsetti55
==>
=∫+
−
− −
b
vm
b
yn
a
um
a
xmq
mnq
a
um
a
xm
m
adx
a
xm
ooomn
o
ux
ux
x y
o
o
o o
2
sin
sin
2
sin
sin
16
2
sin
sin
2 sin
2
2
2
2 2
πππππ
πππ
π
R&DE (Engineers), DRDO
Patch load
� Deflection
( )
+
= ∑∑∞ ∞
nmmn
b
yn
a
xmS
D
qyxw
m n
mno
sin
sin
16,
226
ππ
π
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Ramadas Chennamsetti56
=
+
b
vm
b
yn
a
um
a
xmS
b
n
a
mmn
D
oomn
m n
2
sin
sin
2
sin
sin
where,
ππππ
π
R&DE (Engineers), DRDO
Point load
� Point load
uv
Pqo = P
(xo, yo)
Assume the point load acts over an infinitesimal area u x v
Corresponding UDL Pq =
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Ramadas Chennamsetti57
From the earlier analysis
b
vn
a
um
b
yn
a
xm
uv
P
mnq
b
vn
a
um
b
yn
a
xmq
mnq
oomn
ooomn
2
sin
2
sin
sin
sin
162
sin
2
sin
sin
sin
16
2
2
πππππ
πππππ
==>
=
uvqo =
R&DE (Engineers), DRDO
Point load
� Simplifying
vmb
vn
uma
um
b
yn
a
xm
ab
Pq oo
mn 2
sin
2
sin
sin
sin4
π
π
π
πππ
==>
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Ramadas Chennamsetti58
b
yn
a
xm
ab
Pq
b
vm
a
umbaabq
oomn
mn
sin
sin
42
2
sinsin
ππ
ππ
==>
==>
R&DE (Engineers), DRDO
Point load
� Deflection due to point load
( ) b
ym
a
xmS
abD
Pyxw
mn
sin
sin4
, 2
'
4
ππ
π = ∑∑
∞ ∞
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Ramadas Chennamsetti59
( )
b
ym
a
xmS
b
n
a
mabDyxw
oomn
m m
sin
sin
,
'
2224
ππ
π
=
+
= ∑∑
R&DE (Engineers), DRDO
Bending & in-plane loading
� Plates are subjected to in-plane loading also – in addition to lateral / transverse loads
� In-plane loading – tensile or compressive
� Large in-plane compressive loads – Buckling takes place
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Ramadas Chennamsetti60
place
� Buckling – non-linear phenomenon –disproportionate increase of displacement with load
� Critical load – ability to resist axial load ceases –change in deformation shape
R&DE (Engineers), DRDO
Bending & in-plane loading
� Thin walled members – cross-sections like ‘I’, ‘L’, ‘H’, ‘C’ etc – undergo buckling –thin plates of small widths
� Combined loading of a rectangular plate –
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Ramadas Chennamsetti61
� Combined loading of a rectangular plate –loads � In-plane forces: Nx, Ny, Nxy and Nyx
� Transverse forces / moments: Mx, My, Mxy, Myx, Qx and Qy
� Small deformation and large deformation
R&DE (Engineers), DRDO
In-plane forces
� Infinitesimal element => dA = dx dy
Nx*
dx
Nxy*
θx*
Qx*
θx*
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Ramadas Chennamsetti62
Nx
dx
x
z θx
Qx
θx
In-plane loading Transverse/out-of-plane loading
For small angles => Sinθ ≈ θ and Cosθ ≈ 1
R&DE (Engineers), DRDO
In-plane forces
� Force equilibrium in x-direction
sincos
cos cos cos
**
*
∂+−
∂
+
+−
∂∂++−
dydxQ
QdxdyN
N
dxNdydxx
NNdyN
xxy
yxyxx
xxx
θθ
θθθ
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Ramadas Chennamsetti63
0
0sinsinsin
sincos
*
**
=∂
∂+
∂∂=>
=+
∂∂
+−+
∂∂+−
∂∂
+
y
N
x
N
dxQdxdyy
QQdyQ
dydxx
QQdxdy
y
NN
xyx
yyyy
yxx
xx
xyxy
xy
θθθ
θθ
If there are no in-plane forces – equation vanishes
R&DE (Engineers), DRDO
In-plane forces
� Force equilibrium in y-direction
� Angle is not equal to zero, but, small
0=∂
∂+
∂∂
y
N
x
N yxy
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Ramadas Chennamsetti64
� Angle is not equal to zero, but, small
y
w
x
wyx ∂
∂=∂∂=
≈≈
θθ
θθθ
,
1cos and sin
R&DE (Engineers), DRDO
In-plane forces
� Force equilibrium in x-direction
2
∂+∂
∂+−
∂
+
+−
∂∂++−
dxww
dydxQ
QdxdyN
N
dxNdydxx
NNdyN
xxy
xyx
xx
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Ramadas Chennamsetti65
0
sorder termhigher Neglect
02
2
2
=∂
∂+
∂∂=>
=∂∂+
∂∂+
∂∂
∂∂
+−∂∂+
∂∂+
∂∂
∂∂+−
∂∂
+
y
N
x
N
y
wdxQdy
y
w
y
wdxdy
y
x
wdyQ
dxx
w
x
wdydx
x
QQdxdy
y
NN
xyx
yy
yx
xx
xyxy
No change in in-plane equilibrium equations
R&DE (Engineers), DRDO
In-plane forces
� Similar expression for force equilibrium in y-direction
� Force equilibrium in z-direction� Contribution from in-plane normal forces, Nx
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Ramadas Chennamsetti66
� Contribution from in-plane normal forces, Nxand Ny and shear force, Nxy
� Contribution from shear force, Qx and Qy
� Contribution from externally applied load, q
R&DE (Engineers), DRDO
Z-direction
sinsin
sinsin
sinsin
∂∂+∂
∂
++∂−
∂∂
∂∂+
∂∂
∂∂
++∂∂−
∂∂
∂∂+
∂∂
∂∂++
∂∂−
dyww
dxdyN
Nw
dxN
dyy
w
yy
wdxdy
y
NN
y
wdxN
dxx
w
xx
wdydx
x
NN
x
wdyN
xy
yyy
xxx
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Ramadas Chennamsetti67
0coscos
coscos
sinsin
sinsin
=+
∂∂
∂∂+
∂∂
∂∂
++∂∂−
∂∂
∂∂+
∂∂
∂∂++
∂∂−
∂∂
∂∂+
∂∂
∂∂
++∂∂−
∂∂
∂∂+
∂∂
∂∂
++∂∂−
qdxdydyy
w
yy
wdy
y
y
wdxQ
dxx
w
xx
wdydx
x
x
wdyQ
dxy
w
xy
wdydx
x
NN
y
wdyN
dyx
w
yx
wdxdy
y
NN
x
wdxN
yyy
xxx
xyxyxy
xyxyxy
R&DE (Engineers), DRDO
Z-direction
� If no in-plane forces acting
qy
Q
x
Q yx −=∂
∂+
∂∂
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Ramadas Chennamsetti68
� Presence of in-plane forces
022
22
2
2
=+∂
∂+
∂∂+
∂∂+
∂∂∂+
∂∂
qy
Q
x
Q
y
wN
yx
wN
x
wN yx
yxyx
R&DE (Engineers), DRDO
Z-direction
� Substituting Qx and Qy
( ) ( )wy
DQwx
DQ yx22 , ∇
∂∂−=∇
∂∂−=
∂+∂+∂ wN
wN
wN
222
2
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Ramadas Chennamsetti69
( ) ( )
+
∂∂+
∂∂∂+
∂∂=∇=>
=+
∇
∂∂−
∂∂+
∇∂∂−
∂∂+
∂+
∂∂+
∂
qy
wN
yx
wN
x
wN
Dw
qwy
Dy
wx
Dx
yN
yxN
xN
yxyx
yxyx
2
22
2
24
22
22
21
0
2
R&DE (Engineers), DRDO
Plate buckling
� Buckling of thin plate
∂∂+
∂∂∂+
∂∂+=∇
2
22
2
24 2
1
y
wN
xy
wN
x
wNq
Dw yxyx
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Ramadas Chennamsetti70
Assume, q = 0, in-plane load, Nx = N1, rest zero
2
2
4
4
22
4
4
4
2x
w
D
N
y
w
yx
w
x
w x
∂∂=
∂∂+
∂∂∂+
∂∂
Assume all four edges are simply supported –Navier solution to plate bending
R&DE (Engineers), DRDO
Plate buckling
� Plate deflection
( )
nm
yxwyxwwm n
nmmn
πβπα
βα
==
== ∑∑∞ ∞
,
sin sin,
Substitute in
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Ramadas Chennamsetti71
b
n
a
mnm
πβπα == , Substitute in governing equation
( ) 21222
214224 02
mnm
mnnmm
D
ND
N
αβα
αββαα
=+=>
=−++ Characteristic equation
R&DE (Engineers), DRDO
Plate buckling
� From characteristic equation
( )2222
1
21222
nmmN
D
Nmnm
+
=
=>
=+
πππ
αβα
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Ramadas Chennamsetti72
ratioAspect - /
22
2
2
1
222
2
22
1
1
bac
nm
c
c
m
b
DN
b
n
a
m
m
aDN
baaD
=
+==>
+
==>
+
=
=>
π
π
R&DE (Engineers), DRDO
Plate buckling
� Critical load – smallest value
� Increase in N1 with n2 – Minimum value of n is equal to one – buckled shape in y-direction – single half sine wave
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Ramadas Chennamsetti73
direction – single half sine wave
2
2
2
1
+=m
c
c
m
b
DN
π
N1 is a function of variable ‘m’ – for minimum value of ‘m’, differentiate N1 wrt ‘m’
R&DE (Engineers), DRDO
Plate buckling
� Differentiate
22
21
1
012
c
m
c
cm
c
c
m
b
D
dm
dN π
=
−
+=
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Ramadas Chennamsetti742
2
1
2
4
01
b
DN
b
acm
m
c
c
cr
π=∴
===>
=
−=>
Whole number
R&DE (Engineers), DRDO
Plate buckling
� Number of half sine waves can’t be a whole number – it should be an integer
� Equation for critical loadfor n = 1
2π 2
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Ramadas Chennamsetti75
2
2
1 b
DKN
π=2
+=m
c
c
mK
Plotting ‘K’ vs aspect ratio = c = a/b for various integer values of ‘m’
R&DE (Engineers), DRDO
Plate buckling
� K vs aspect ratio
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Ramadas Chennamsetti76
1.414 2.449
R&DE (Engineers), DRDO
Plate buckling
� From figure, value of ‘K’ is same as intersection points of ‘m’ and ‘m+1’N1 critical load at m – when load is increased, buckled form changes from ‘m’ to ‘m+1’
Curves for m = 1 and
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Ramadas Chennamsetti77
At transition from ‘m’ to ‘m+1’
( )( )1
1
1
1
2
22
+=
+==>
+++=
+
mmc
mmc
m
c
c
m
m
c
c
m
Curves for m = 1 and m = 2meet at c = √2
Aspect ratio less than √2 => m = 1
Aspect ratio from √2to √6, m = 2
C>4 => K ≈ 4
R&DE (Engineers), DRDO
Plate buckling
� Estimation of buckling loadt = 1 cm, a = 2.3 m, b = 1 m, E = 200 GPa, υ = 0.30
Flexural modulus
( )101102002293
=×××==−Et
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Ramadas Chennamsetti78
( )( )
( )
0786.42
3.2
3.2
2
23.21
3.2
96.173.0112
10110200
112
22
2
29
2
3
=
+=
+==>
==>===
=−
×××=−
=−
m
c
c
mK
mb
ac
kNmEt
Dυ
R&DE (Engineers), DRDO
Plate buckling
� Minimum critical load
MNbNN
mMNb
D
b
DKN
cr
cr
72.73
/72.73086.4
1
2
2
2
2
1
=×=
=×== ππ
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Ramadas Chennamsetti79
MNbNN cr 72.731 =×=
In the above expression, for a given width and elastic properties of plate, critical load depends on ‘K’. In turn ‘K’ depends on ‘m’ for a given aspect ratio, c
2890.4 ,3
0786.4 ,2
4790.7 ,1 ,3.2
=======
Km
Km
KmcMinimum value of ‘K’ is considered for estimation of Ncr
R&DE (Engineers), DRDO
Plate and column buckling
� Uniaxial load for plate buckling
( )322
1
2
2
1
ππσ
π
===
=
EtK
DK
Nb
DKN
crp
crp
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Ramadas Chennamsetti80
( )
( ) ( )
( ) ( )212
2
11
2
2
21
222
11
112 ,
112
112
υπσ
πυ
σ
υππσ
−===>
−==>
−===
KC
tb
EC
tb
EK
Et
tbK
tb
DK
t
crp
crp
crpcrp
R&DE (Engineers), DRDO
Plate and column buckling
� Buckling of a column
( )2
2
22
2
2
2
2
EAC
EAkCP
l
EICP
cr
cr
ππ
π
===>
=
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Ramadas Chennamsetti81
( )
( )
( )22
1
2
2
2
2222
tb
EC
kl
EC
A
P
kl
Cl
CP
crp
crcrc
cr
πσ
πσ
=
===>
===>
Column
Plate
R&DE (Engineers), DRDO
Plate and column buckling
� Critical stress for plate depends on thinness ratio = t/b – not on the length � depends on width
� More thinner plate – lesser bucking load
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Ramadas Chennamsetti82
� More thinner plate – lesser bucking load
� Critical stress in column depends on slenderness ratio
� Longer columns – lower critical load
R&DE (Engineers), DRDO
Strain energy
� In thin plate theory, out-of-plane shear stresses vanish => τxz, τyz and τzz
� Stress components contributing to strain energy => σxx, σyy and τxy
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Ramadas Chennamsetti83
energy => σxx, σyy and τxy
� Strain energy,
( ) VdUV
xyxyyyyyxxxx 2
1∫ ++= γτεσεσ
Linear elastic material
R&DE (Engineers), DRDO
Strain energy
� Strain energy,
{ }{ }∫=V
Txyyyxxxyyyxx VdU γεετσσ
2
1
w ∂2
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Ramadas Chennamsetti84
{ } [ ]{ }εσ
υυ
υ
υγεε
υυ
υ
υτσσ
C
yx
wy
wx
w
EzEz
xy
yy
xx
xy
yy
xx
==>
∂∂∂∂∂∂∂
+−−=
+−−=
2
2
2
2
2
22
22
100
01
01
1
2
100
01
01
1
R&DE (Engineers), DRDO
Strain energy
� Strain energy,
{ } { } { } [ ]{ }
www
VdCVdUV
T
V
T
222
2
1
2
1∫∫
∂
∂∂
== εεεσ
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Ramadas Chennamsetti85
( )
( )
Vdz
yx
w
x
w
y
w
x
w
x
w
y
w
x
w
EU
V
2
22
2
2
2
2
2
2
2
2
2
2
2
2
2
12
12 ∫
∂∂∂−+
∂∂
∂∂+
∂∂+
∂∂
∂∂+
∂∂
−==>
υ
υ
υ
υ
R&DE (Engineers), DRDO
Strain energy
� Infinitesimal volume, dV = dxdydz
� Carry out integration over thickness => dz
322 =∫
+h
h
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Ramadas Chennamsetti86
( )2
3
2
2
2
112
12
υ−=
=∫−
EhD
hdzz
h
R&DE (Engineers), DRDO
Strain energy
� Simplify
dxdyx
w
x
w
DU ∫
∂∂+
∂∂
=
2
2
2
2
2
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Ramadas Chennamsetti87
( )dxdy
yx
w
x
w
x
wU
A∫
∂∂∂−
∂∂
∂∂−−
=22
2
2
2
2
122
υ
Strain energy – Finite Element Method – Total potential approach
R&DE (Engineers), DRDO
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Ramadas Chennamsetti88