R&DE (Engineers), DRDO
rd_mech@yahoo.co.in
Theory of Plates
Ramadas Chennamsetti
Introduction
R&DE (Engineers), DRDO
Plates are initially flat structural elements
Ramadas Chennamsetti
2
rd_mech@yahoo.co.in
“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”
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
=> 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
Ramadas Chennamsetti
3
rd_mech@yahoo.co.in
R&DE (Engineers), DRDO
Introduction
R&DE (Engineers), DRDO
Plate stretching
z
Nx
t
Nx
x
Axial deformation due to transverse load
Net deformation = Algebraic sum of uniform stretching
and axial deformation
due to
bending load
Ramadas
Chennamsetti
4
rd_mech@yahoo.co.in
Uniform stretching of the plate => uo
q
Introduction
R&DE (Engineers), DRDO
1
t
1
≥ ≥
10 b 2000
For plates Thin & thick plates –
t < 20b b = smallest side
t > 20b
t
w ≤
5
Small deflections –
Thin plate theory – Kirchoff’s Classical
Plate Theory (KCPT)
Thick plate theory – Reissner – Mindlin
Plate Theory (MPT)
Ramadas Chennamsetti
5
rd_mech@yahoo.co.in
Thin plate =>
Thick plate =>
KCPT - Assumptions
R&DE (Engineers), DRDO
Assumptions –
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
In plane strains are small compared to unity –
consider only linear strains
Normal stresses in transverse direction are small
compared with other stresses – neglected
Ramadas Chennamsetti
6
rd_mech@yahoo.co.in
Governing equations are derived based on undeformed
geometry
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
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
Ramadas Chennamsetti
7
rd_mech@yahoo.co.in
R&DE (Engineers), DRDO
Sign convention
R&DE (Engineers), DRDO
Following sign convention will be followed
z
z
ψz
Mz
ψy
y
y
Mx
x
ψx
x
Positive moments
Positive rotations
Ramadas Chennamsetti
8
rd_mech@yahoo.co.in
My
Bending deformations
R&DE (Engineers), DRDO
z
Bending takes
place in both planes
y
D
x
z
’
BB
A
x
B
View ‘D’
Ramadas Chennamsetti
9
rd_mech@yahoo.co.in
A’
Rotation => ψy
A
Bending deformations
R&DE (Engineers), DRDO
Deformation in ‘x’ direction
u ( x , y , z ) = − zψ y (x , y )
v ( x, y , z ) = − zψ
x
(x , y )
Vertical deformation
w = w (x, y)
Ramadas Chennamsetti
10
rd_mech@yahoo.co.in
Deformation in ‘y’ direction
Strains
R&DE (Engineers), DRDO
=> γ xz
=> γ yz
∂z ∂x
∂w
∂w
= −ψ y +
= 0 => ψ y =
∂x
∂x
∂v ∂w
γ yz =
+
=0
∂z ∂y
∂w
∂w
= −ψ x +
= 0 => ψ x =
∂y
∂y
Ramadas Chennamsetti
11
rd_mech@yahoo.co.in
Assumption – out of plane shear strain negligible
∂u ∂w
γ xz =
+
=0
Strains
R&DE (Engineers), DRDO
Non-zero strains
ε yy
∂u
∂ w
y
=
= −z
= −z
2
∂x
∂x
∂x
2
∂v
∂ψ x
∂ w
=
= −z
= −z
∂y
∂y
∂y 2
γ xy
2
∂u ∂v
∂ w
=
+
= −2 z
∂y ∂x
∂x∂y
2
Ramadas Chennamsetti
12
rd_mech@yahoo.co.in
ε xx
∂ψ
Stresses
R&DE (Engineers), DRDO
Thin plate – out of plane shear strains vanish – out
of plane shear stresses also vanish
τ yz = 0
τxz
τyz
τyz
z
y
x
τxz
Out of plane normal stress is also assumed to be
zero – logical – thin structure – plane stress
conditions
Ramadas Chennamsetti
13
rd_mech@yahoo.co.in
z
τ xz = 0,
Stresses
R&DE (Engineers), DRDO
Non-zero stress components
z
σxx
y
σyy
σyy
x
τxy
σxx
All three stress components, σxx, σyy, τxy – in-plane
Ramadas Chennamsetti
14
rd_mech@yahoo.co.in
τxy
Curvatures
R&DE (Engineers), DRDO
Curvature – reciprocal of radius of bending
Rate of change of slope
z
x
dx w*
∂w
dx
∂x
x
ψy
dx
slope = rotation = ψ y = -
∂w
∂x
∂ψ
2
w
∂
y
curvatue =
=−
= κ xx
2
∂x
∂x
Ramadas Chennamsetti
15
rd_mech@yahoo.co.in
w
R
−
Curvatures
R&DE (Engineers), DRDO
Similarly bending in yz plane introduces a
curvature
κ
∂ w
= −
2
∂y
2
yy
κ xy
∂ w
= −2
∂x∂y
2
Ramadas Chennamsetti
16
rd_mech@yahoo.co.in
Twisting of plate
Constitutive law
R&DE (Engineers), DRDO
Linear elastic isotropic – Hookean material
Three stress and strain components
ε yy =
γ xy =
τ xy
G
E
σ yy
E
=
−υ
−υ
τ xy
E
σ yy
E
σ xx
E
2 (1 + υ )
εxx
0 σxx
 1 −υ
  1
σ 
ε
υ
1
0
=
−
 yy
 yy

E
γ 
τ 


υ
0
0
2
(
1
)
+

 xy 
 xy

 
σxx
υ
εxx
1
0


  E 
 

σ
υ
=
1
0
 yy
ε yy
2
τ  1−υ 0 0 1−υ γ 

 xy17
 xy 
2


Ramadas Chennamsetti
rd_mech@yahoo.co.in
ε xx =
σ xx
Writing all three equations in
matrix form
Constitutive law
R&DE (Engineers), DRDO
Express strains in terms of curvatures

Ez

 = −
1−υ


2

1
υ

0

υ
1
0









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
Ramadas Chennamsetti
18
rd_mech@yahoo.co.in
 σ xx

 σ yy
τ
 xy
 ∂ 2w
 ∂x 2
0 
∂ 2w

0 
2
1 + υ  ∂y
 ∂ 2w
2  2
 ∂x∂y

Equilibrium equations
R&DE (Engineers), DRDO
Equilibrium of an infinitesimal element
z
y
dx
σxx
t
τxy
x
Forces acting on an infinitesimal
element dx dy dz
y
τxy
x
dy
ττ**yz
yz
σyy
σ*yy
dx
τ*xz
τ*xy
τ*xy
Ramadas Chennamsetti
σ*xx
19
rd_mech@yahoo.co.in
dy
Equilibrium equations
R&DE (Engineers), DRDO
Equilibrium in ‘x’ direction
−σ
xx
dydz
dydz
σ
*
xx
τ
*
zx
τ
*
yx
+τ
−τ
zx
*
zx
dxdy
dxdy
+τ
−τ
yx
*
yx
dzdx
dzdx
∂ σ xx

dx 
xx +
∂x

∂ τ zx


=  τ zx +
dz 
∂z


∂ τ yx


=  τ yx +
dy 
∂y


= 0

= σ

Ramadas Chennamsetti
20
rd_mech@yahoo.co.in
σ
*
xx
Equilibrium equations
R&DE (Engineers), DRDO
Substitute – the following equation is
obtained
∂τ xx ∂τ xy ∂τ xz
∂x
+
∂y
+
∂z
= 0 - (1)
∂τ xy
∂x
+
∂τ yy
∂y
+
∂τ yz
∂z
= 0 - (2)
∂τ xz ∂τ yz ∂τ zz
+
+
= 0 - (3)
∂x
∂y
∂z
Ramadas Chennamsetti
21
rd_mech@yahoo.co.in
Similarly take equilibrium in ‘y’ and ‘z’ directions
Shear stresses
R&DE (Engineers), DRDO
From equation (1) – shear stress τxz can be
computed
Use stress deflection/curvature relations
 ∂2w
∂ 2 w  ∂
 2 + υ 2   +
∂ y   ∂y
 ∂x
∂τ xz  Ez
=
2
∂z
1−υ
 Ez ∂ 2 w  ∂τ xz
=0
−
+
 1 + υ ∂x∂y  ∂z
3
∂ 3w
∂ 3w
 ∂ w
+ (1 − υ )
 3 + υ
2
∂x∂y
∂x∂y 2
 ∂ x
∂τ xz  Ez  ∂
2
=
∇
w

2
∂z
 1 − υ  ∂x
(
)
Integrate this across thickness to get shear stress
Ramadas Chennamsetti



22
rd_mech@yahoo.co.in
∂ 
Ez
−
∂x  1 − υ 2
Shear stresses
R&DE (Engineers), DRDO
Integrate from mid plane to top surface of
z
plate
z = h/2
z
∫τ dτ
xz
xz
 Ez
= ∫
2
1
−
υ

0
− τ xz
=> −τ xz
(
)
(
2
E ∂ ∇2w
∂ ∇ w
dz =

1 − υ 2 ∂x
 ∂x
(
)
h 2
∫ z dz
z
h 2
E ∂ ∇ w  z2 
 
=
2
1−υ
∂x  2  z
2
(
)  h
(
)
∂ ∇2w
E
=
2 1−υ 2
∂x
=> τ xz =
)
(
E
2 1−υ 2
(

 4 − z 


∂ ∇2w  2 h2 
 z −
 Parabolic variation
∂x 
4 
23
)
)
2
Ramadas Chennamsetti
2
rd_mech@yahoo.co.in
h 2
0
Moments
R&DE (Engineers), DRDO
Moment wrt ‘y’ axis
dF xx = σ
dM
z
σ xx
dz
=> dM
z
σxx
dx
dy
y
M
y
x
M
y
= zdF xx = z σ
xx
dz
 ∂ 2w
∂ 2w 


+υ
2
2 
∂x 
 ∂x
∂ 2w
Ez 2  ∂ 2 w

=−
+υ
2 
2
∂y 2
1 − υ  ∂x
Ez
=−
1−υ 2
y
E
=−
1−υ 2
y
dz



+h 2
 ∂ 2w
∂ 2w 
2


υ
+
z
dz
2
2  ∫
∂y  −h 2
 ∂x
Eh 3
=−
12 1 − υ 2
(
Ramadas Chennamsetti
)
 ∂ 2w
∂ 2w 


+υ
2
2 
∂y 
 ∂x
24
rd_mech@yahoo.co.in
Neutral plane
xx
Moments
R&DE (Engineers), DRDO
Moment wrt ‘x’ axis
Neutral plane
 ∂2w ∂2w 
Eh 3
υ

Mx =−
+
2 
2
2 
12 (1 − υ )  ∂ x
∂y 
z
dz
τxy
z
M xy
dx
y
dy
x
Compare with section
modulus of beam
Eh 3
∂2w
(1 − υ )
=−
2
12(1 − υ )
∂x∂y
Eh 3
D =
12 ( 1 − υ
Ramadas Chennamsetti
2
)
25
rd_mech@yahoo.co.in
Twisting moment due to shear
stress τxy
Shear forces
R&DE (Engineers), DRDO
Vertical equilibrium of
plate
*
Q x dy + Q y* dx + qdxdy − Q x dy − Q y dx = 0
∂Q x
Q = Qx +
dx
∂x
∂Q y
*
Qy = Qy +
dy
∂y
q
Q*
Q *y
x
Qx
Qy
y
Substitute these and simplify
dx
dy
x
∂Q y
∂Q x
+
= −q
∂x
∂y
Ramadas Chennamsetti
26
rd_mech@yahoo.co.in
*
x
z
Shear forces
R&DE (Engineers), DRDO
Shear forces across thickness can be
computed by integrating shear stress across
+h 2
thickness
Q
y
=
∫τ
yz
dz per unit width
−h 2
Q
Q
y
y
yz
∂ (∇ 2 w )
E
=
2 (1 − υ 2 )
∂y
 2
h2 
 z −

4 

+h 2
 2
h2 
 z −
 dz
∫
4 
−h 2 
Eh 3
∂ (∇ 2 w )
∂ (∇ 2 w )
= −
= −D
2
12 (1 − υ )
∂y
∂y
Ramadas Chennamsetti
27
rd_mech@yahoo.co.in
τ
∂ (∇ 2 w )
E
=
2 (1 − υ 2 )
∂y
Governing equation
R&DE (Engineers), DRDO
Similar expression for Qx
∂ (∇ 2 w )
Qx = −D
∂x
Substitute Qx and Qy in the following expression – vertical
equilibrium
∂Q x ∂Q y
+
= −q
∂x
∂y
∂
∂

 ∂ 
2
2

D
w
D
w
−
∇
+
−
∇



∂x
∂y

 ∂y 
2
∂2
∂
q
2
2
=>
∇
w
+
∇
w
=
∂y 2
D
∂x 2
(
)
(
(
)
 ∂2
∂2
=>  2 +
∂y 2
 ∂x
(
(
) = − q

)
)
 2
q
 ∇ w =
D

Ramadas Chennamsetti
28
rd_mech@yahoo.co.in
∂
∂x
Governing equation
R&DE (Engineers), DRDO
Governing equation
 ∂2
∂2

+
2
2
∂
x
∂
y

 2
q
 ∇ w =
D

q
2
2
=> ∇ ∇ w =
D
q
4
=> ∇ =
D
)
∂ w
∂ w
∂ w
q
+2
+
=
4
2
2
4
∂x
∂x ∂y
∂y
D
4
4
4
Bi-harmonic equation
Compare with beam equation
Ramadas Chennamsetti
29
rd_mech@yahoo.co.in
(
Boundary conditions
R&DE (Engineers), DRDO
Well posed problem – Governing equations
and boundary conditions
Three basic boundary conditions
Vertical deflection and their derivatives
Ramadas Chennamsetti
30
rd_mech@yahoo.co.in
Simply supported
Clamped and
Free edge
Simply supported
R&DE (Engineers), DRDO
Simply supported – for eg beam => vertical
deflection = 0 and moment = 0
w( x, y ) = 0
 ∂2w
∂2w 
M y = − D 2 + υ 2  = 0
∂y 
 ∂x
y
w = 0 implies second derivative in the
direction tangent to this line is zero
∂ 2w
= 0
2
∂x
Ramadas Chennamsetti
x
Simply supported
y = const
31
rd_mech@yahoo.co.in
Simply supported
x = const
For edge, x = const
Simply supported
R&DE (Engineers), DRDO
Simply supported condition along edge y =
const
w=0
w = 0 implies second derivative in the
direction tangent to this line is zero
∂ w
=0
2
∂yChennamsetti
Ramadas
2
32
rd_mech@yahoo.co.in
 ∂2w ∂2w 
M x = − Dυ 2 + 2 
∂y 
 ∂x
Clamped
R&DE (Engineers), DRDO
Deflection and slope in normal directions
vanish
x = constant
w=0
∂w
=0
∂y
y = constant
Ramadas Chennamsetti
33
rd_mech@yahoo.co.in
w=0
∂w
=0
∂x
Free edges
R&DE (Engineers), DRDO
 ∂2w
∂2w 
M y = − D 2 + υ 2  = 0 at x = const
∂y 
 ∂x
 ∂2w ∂2w 
M x = − Dυ 2 + 2  = 0 at y = const
∂y 
 ∂x
Ramadas Chennamsetti
34
rd_mech@yahoo.co.in
Free edge – Free from any external loads –
Natural boundary conditions
Bending moment and Shear force vanish
Bending moment
Free edges
R&DE (Engineers), DRDO
Moment Mxy and shear forces
Mxy
The net force acting on
the face
Q = − M xy −
Mxy
'
x
dy
M xy +
Mxy
Mxy
dy
M xy +
∂M xy
∂y
∂M xy
∂y
dy
∂M xy
=> Qx' = −
+ M xy
∂y
∂M xy
∂y
Total shear force,
dy dx
Ramadas Chennamsetti
Vx = Qx + Qx’
35
rd_mech@yahoo.co.in
dy
Free edges
R&DE (Engineers), DRDO
Total shear force,
Vx = Qx + Qx'
∂ 2
∂ ∂ w
=> Vx = − D
∇ w − D(1 − υ )  2 
∂x
∂x  ∂x 
)
 ∂3w
∂3w 
=> Vx = − D  3 + (2 − υ )
2
∂x∂y 
 ∂x
3
3
∂ w
∂w 
V y = − D  3 + (2 − υ ) 2 
∂x ∂y 
 ∂y
Ramadas Chennamsetti
36
rd_mech@yahoo.co.in
(
2
Free edges
R&DE (Engineers), DRDO
The forces, Vx and Vy => reduced, or
Kirchoff’s or effective shear forces
In case of a free edge,
 ∂3w
∂3w 
V y = − D  3 + (2 − υ ) 2  = 0 at y = const
∂x ∂y 
 ∂y
Ramadas Chennamsetti
37
rd_mech@yahoo.co.in
 ∂3w
∂3w 
Vx = − D  3 + (2 − υ )
= 0 at x = const
2
∂x∂y 
 ∂x
Bending of plates
R&DE (Engineers), DRDO
Governing equation of plate rectangular
plate bending
(
)
q
∇ ∇ w =
D
2
2
w = vertical deflection = w(x, y)
Solution to this equation – product of two
functions – Assume
w = w( x , y ) = F ( x )G ( y )
Ramadas Chennamsetti
38
rd_mech@yahoo.co.in
external loading, q = q(x, y)
Bending of plates
Choice of functions – algebraic,
trigonometric, hyperbolic etc or
combination of these function
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
Ramadas Chennamsetti
39
rd_mech@yahoo.co.in
R&DE (Engineers), DRDO
Bending of plates
R&DE (Engineers), DRDO
Edges, x = 0 and x = a simply supported
Vertical deflection vanish
w(x=0, y)=0, w(x=a, y)=0
Possible form of solution
∞
SS
x=0
x
SS
x=a
F1, F2,…….F∞ are coefficients
Symmetric loading wrt x = a/2
Maximum deflection at x = a/2
Ramadas Chennamsetti
40
rd_mech@yahoo.co.in
 mπ x 
F ( x ) = ∑ Fm sin 

 a 
m
y
Bending of plates
R&DE (Engineers), DRDO
Other edges simply supported
Vertical deflection vanish
SS
y=b
w(x, y=0)=0, w(x, y=b)=0
Possible form of solution
∞
G1, G2,…….G∞ are coefficients
SS
x = 0,
y=0
x
SS
x=a
SS
Selection of functions based on BCs
Ramadas Chennamsetti
41
rd_mech@yahoo.co.in
 nπ y 
G ( y ) = ∑ Gn sin 

 b 
n
y
Bending of plates
R&DE (Engineers), DRDO
Final solution,
 mπ x   nπ y 
w( x, y ) = ∑∑ FmGn sin 
 sin 

 a   b 
m n
∞
∞
∞
∞
m
n
mπ
nπ
, βn =
αm =
a
b
Coefficients, Fm, Gn and wmn computed using
Fourier Series
Ramadas Chennamsetti
42
rd_mech@yahoo.co.in
=> w( x, y ) = ∑∑ wmn sin α m x sin β n y
Coefficients
R&DE (Engineers), DRDO
Any periodic function can be expanded into
a sine or cosine function using Fourier
expansion
Function of one variable
f (x ) = ∑
m
mπ x
f m sin
a
m 'π x
mπ x
m 'π x
sin
=> f (x )sin
= f m sin
a
a
a
Integrate the above from limits 0 to a
Ramadas Chennamsetti
43
rd_mech@yahoo.co.in
∞
Coefficients
R&DE (Engineers), DRDO
Integration
m'πx
mπx
m'πx
∫0 f (x)sin a dx = fm ∫0 sin a sin a dx
a
fm
f m  πx
mπx
m πx
πx
'
=> ∫ 2 sin
sin
dx = ∫ cos m − m − cos m + m'
2 0
a
a
2 0
a
a
a
(
a
'
(
)
)
(
πx
 πx
'
'
m
m
sin
m
m
sin
−
+
fm 
a
a
=> 
−
π
2  π m − m'
m + m'
a
 a
(
)
(
)
(
)
a




0
Lower limit vanishes, evaluate upper limit
Ramadas Chennamsetti
)

 dx

44
rd_mech@yahoo.co.in
a
Coefficients
R&DE (Engineers), DRDO
a
∫
0
m 'π x
fm
f (x )sin
dx =
a
2
a
=> ∫
0
sin
πx
π
(
m−m )
a
fma
m 'π x
f (x )sin
dx =
a
2
=0
(
m−m )
a
'
'
x=a
for m = m '
for m ≠ m
2
mπ x
f m = ∫ f (x )sin
dx
a
a
0
Ramadas Chennamsetti
'
a
45
rd_mech@yahoo.co.in
Upper limit
Coefficients
R&DE (Engineers), DRDO
Function of variable ‘y’
nπ y
g ( y ) = ∑ g n sin
b
n
∞
nπ y
nπ y
nπ y
=> g ( y )sin
= g n sin
sin
b
b
b
b
2
nπ y
g n = ∫ g ( y )sin
dy
b0
b
'
Ramadas Chennamsetti
46
rd_mech@yahoo.co.in
'
Coefficients
R&DE (Engineers), DRDO
Function of two variables
 mπ
(
)
w x, y = ∑ ∑ wmn sin 
∞
∞
m
n

x   nπ
 sin 
a   b
y


x=a
x=a
'
∞ ∞
 m 'π x 
 nπ y 
 mπ x   m π x 
 dx = ∑ ∑ wmn sin 
dx
=> ∫ w( x, y )sin 
 ∫ sin 
 sin 
 b  x =0  a   a 
m n
 a 
x =0
x=a
 m 'π x 
a ∞ ∞
 nπ y 


(
)
w
x
,
y
sin
dx
=
w
sin


∑
∑
mn
∫x =0
 a 
2 m n
 b 


y =b x = a
y =b
=> wmn
4
=
ab
y =b x = a
 m π x   nπ y 
(
)
w
x
y
,
sin

 sin 
dxdy
∫y =0 x∫=0
 a   b 
Ramadas Chennamsetti
47
rd_mech@yahoo.co.in
'
 m 'π x   n 'π y 
a ∞ ∞
 nπ y   n π y 
 sin 
dxdy = ∑ ∑ wmn ∫ sin 
 dy
=> ∫ ∫ w( x, y )sin 
 sin 
2 m n
 b   b 
 a   b 
y =0 x =0
y =0
Simply supported plate
R&DE (Engineers), DRDO
Assume loading over plate
mπ x
nπ y
q = q ( x, y ) = ∑ ∑ qmn sin
sin
a
b
m n
∞
∞
Solution
mπ x
nπ y
w = w( x, y ) = ∑∑ wmn sin
sin
a
b
m n
∞
Governing equation
∂ w
∂ w
∂ w
q
+2
+
=
4
2
2
4
∂x
∂x ∂y
∂y
D
4
4
Ramadas Chennamsetti
4
48
rd_mech@yahoo.co.in
∞
Simply supported plate
R&DE (Engineers), DRDO
Differentiating vertical displacement
∂ w
 mπ 
 mπ
= ∑∑ 
 wmn sin 
4
∂x
 a
m n  a 
4
∞
∞
4
x   nπ y 
 sin 

  b 
∂ w
 nπ 
 mπ x   nπ y 
= ∑∑ 
 wmn sin 
 sin

4
∂y
 a   b 
m n  b 
∞
∞
4
∂ w
 mπ   nπ 
 mπ
2 2 2 = 2∑∑ 
 
 wmn sin
∂x ∂y
 a
m n  a   b 
4
∞
∞
2
2
Ramadas Chennamsetti
x   nπ y 
 sin 

  b 
49
rd_mech@yahoo.co.in
4
Simply supported plate
R&DE (Engineers), DRDO
 mπ  4  mπ  2  nπ  2  nπ  4 
qmn
 + 2
 
 +
  wmn =

D
 a   b   b  
 a 
qmn
=> wmn =
2
2 2
n 
4 m
Dπ  2 + 2 
b 
a
1
w = w( x, y ) =
4
Dπ
 mπ x   nπ y 
sin 
 sin 

∑∑
2
2
2
 a   b 
m n m
n 
 2 + 2 
a
b 
50

Ramadas Chennamsetti
∞
∞
qmn
rd_mech@yahoo.co.in
Plug in governing equation
UDL
R&DE (Engineers), DRDO
Uniformly distributed load
qo
Computation of coefficients
∞
∑∑q
m
mn
n
 mπ x 
 nπ y 
sin 
 sin 

 a 
 b 
q(x, y) = qo
4
mπ x
nπ y
=
q( x, y )sin
sin
dxdy
∫
∫
ab 0 0
a
b
a b
qmn
Ramadas Chennamsetti
51
rd_mech@yahoo.co.in
q (x , y ) =
∞
UDL
R&DE (Engineers), DRDO
Coefficients qmn
q mn
mπ x
mπ y
∫0 ∫0 q o sin a sin b dxdy
4 qo
mπ x
mπ y
=
sin
dx ∫ sin
dy
∫
a
b
ab 0
0
a
q mn
=> q mn
b
4qo
=
ab
 4 ab

2
 mn π
Ramadas Chennamsetti
 16 q o
=
2
 mn π
52
rd_mech@yahoo.co.in
4
=
ab
a b
UDL
R&DE (Engineers), DRDO
 m π x   nπ y 
q mn sin 
 sin 

∞ ∞
1
a   b 

w = w( x , y ) = 4 ∑ ∑
2
2
2
π D m n
 m   n  
  +   
 a   b  
 m π x   nπ y 
sin 
 sin 

∞ ∞
16 qo
a   b 

=> w(x, y ) = 6
∑
∑
2
2
2
π D m n
 m   n  
  +   
 a   b  
53
Ramadas Chennamsetti
rd_mech@yahoo.co.in
Vertical deflection
Patch load
R&DE (Engineers), DRDO
A patch load applied over an area u x v
Centroid at (xo, yo)
 mπ x   nπ y 
q ( x , y ) = ∑ ∑ q mn sin 
 sin 

 a   b 
m
n
∞ ∞
 m π x   nπ y 
=> q o = ∑ ∑ q mn sin 
 sin 

 a   b 
m
n
q mn
∞
4qo
=
ab
u
v
xo + yo +
2
2
∫
xo −
qo
v
u
Patch load
mπ x
mπ y
∫ v sin a sin b dxdy
u
yo −
2 Ramadas
2
Chennamsetti
54
rd_mech@yahoo.co.in
∞
Patch load
R&DE (Engineers), DRDO
Evaluating integrals
xo +
u
2
o
2
∫
xo −
u
v
yo +
2
2
mπ x
mπ y
∫ v sin a sin b dxdy
u
yo −
2
2
mπ x
2a
 mπ xo   mπ u 
∫ u sin a dx = mπ sin a  sin 2a 
x −
=> qmn
rd_mech@yahoo.co.in
qmn
4 qo
=
ab
xo +
16 qo
 mπ xo   mπ u   nπ yo   mπ v 
=
sin 
 sin 
 sin 
 sin 

2
mn π
 a   2a   b   2b 
Ramadas Chennamsetti
55
Patch load
R&DE (Engineers), DRDO
Deflection
 m π x   nπ y 
S mn sin 
 sin 

∞ ∞
16 qo
a   b 

w( x , y ) = 6 ∑ ∑
π D m n
 m  2  n  2 
mn   +   
 a   b  
S mn
 m π xo
= sin 
 a
  m π u   nπ y o
 sin 
 sin 
  2a   b
Ramadas Chennamsetti
  mπ v 
 sin 

  2b 
56
rd_mech@yahoo.co.in
where,
Point load
R&DE (Engineers), DRDO
Point load
Assume the point load acts over
an infinitesimal area u x v
Corresponding UDL
P
qo =
uv
qo =
P
uv
P
(xo, yo)
mπ xo
nπ yo
16 qo
mπ u
nπ v
sin
sin
qmn =
sin
sin
2
mn π
a
b
2a
2b
mπ xo
nπ y o
16 P
mπ u
nπ v
sin
sin
sin
sin
=> qmn =
2
mn uvπ
a
b
2a
2b57
Ramadas Chennamsetti
rd_mech@yahoo.co.in
From the earlier analysis
Point load
R&DE (Engineers), DRDO
Simplifying
Ramadas Chennamsetti
58
rd_mech@yahoo.co.in
=> qmn
mπ u 
nπ v 

sin
sin




mπ xo
nπ yo
4P
2a 
2b 

=
sin
sin
ab
a
b  mπ u  mπ v 



 2a  2b 
4P
mπ xo
nπ yo
=> qmn =
sin
sin
ab
a
b
Point load
R&DE (Engineers), DRDO
Deflection due to point load
mπ x
mπ y
sin
sin
∞ ∞ S
4P
a
b
w( x, y ) = 4
∑∑
2
2
2
π abD m m
 m   n  
  +   
 a   b  
mπ xo
mπ yo
'
S mn = sin
sin
a
b
Ramadas Chennamsetti
59
rd_mech@yahoo.co.in
'
mn
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
Buckling – non-linear phenomenon –
disproportionate increase of displacement with
load
Critical load – ability to resist axial load ceases –
change in deformation shape
Ramadas Chennamsetti
60
rd_mech@yahoo.co.in
R&DE (Engineers), DRDO
Bending & in-plane loading
R&DE (Engineers), DRDO
In-plane forces: Nx, Ny, Nxy and Nyx
Transverse forces / moments: Mx, My, Mxy, Myx,
Qx and Qy
Small deformation and large deformation
Ramadas Chennamsetti
61
rd_mech@yahoo.co.in
Thin walled members – cross-sections like
‘I’, ‘L’, ‘H’, ‘C’ etc – undergo buckling –
thin plates of small widths
Combined loading of a rectangular plate –
loads
In-plane forces
R&DE (Engineers), DRDO
Infinitesimal element => dA = dx dy
Nxy
*
Qx*
Nx*
θx*
θx*
dx
z θx
Nx
θx
x
In-plane loading
Transverse/out-of-plane loading
For small angles => Sinθ ≈ θ and Cosθ ≈ 1
Ramadas Chennamsetti
62
rd_mech@yahoo.co.in
Qx
In-plane forces
R&DE (Engineers), DRDO
∂N x 

− N x dy cos θ x +  N x +
dx dy cos θ x* − N xy dx cos θ y +
∂x


∂N xy 

∂Qx 

*
 N xy +
dy dx cos θ y −  Qx +
dx dy sin θ x*
∂y
∂x




∂Q y 

+ Qx dy sin θ x −  Q y +
dy dx sin θ y* + Q y dx sin θ y = 0
∂y


∂N x ∂N xy
=>
+
=0
∂x
∂y
If there are no in-plane forces – equation vanishes
Ramadas Chennamsetti
63
rd_mech@yahoo.co.in
Force equilibrium in x-direction
In-plane forces
R&DE (Engineers), DRDO
Force equilibrium in y-direction
∂N xy ∂N y
+
=0
∂x
∂y
Angle is not equal to zero, but, small
∂w
θx =
,
∂x
∂w
θy =
∂y
Ramadas Chennamsetti
64
rd_mech@yahoo.co.in
sin θ ≈ θ and cos θ ≈ 1
In-plane forces
R&DE (Engineers), DRDO
∂N x 

dx  dy − N xy dx +
− N x dy +  N x +
∂x


∂N xy 

∂Q x   ∂ w ∂ 2 w 

 N xy +
dy  dx −  Qx +
dx  dy 
+ 2 dx 
∂y
∂x

  ∂x ∂ x



∂ Q y   ∂w ∂ 2 w 
∂w
∂w 
dy  dx 
=0
+ Qx dy
−  Q y +
+ 2 dy  + Q y dx
∂y
∂x 
∂y
  ∂y ∂ y

Neglect higher order term s
No change in in∂N x ∂N xy
=>
+
= 0 plane equilibrium
∂x
∂y
Ramadas Chennamsetti
equations
65
rd_mech@yahoo.co.in
Force equilibrium in x-direction
In-plane forces
R&DE (Engineers), DRDO
Contribution from in-plane normal forces, Nx
and Ny and shear force, Nxy
Contribution from shear force, Qx and Qy
Contribution from externally applied load, q
Ramadas Chennamsetti
66
rd_mech@yahoo.co.in
Similar expression for force equilibrium in
y-direction
Force equilibrium in z-direction
Z-direction
R&DE (Engineers), DRDO
 ∂w ∂  ∂w  
∂w 
∂N x

− N x dy sin
+ Nx +
dx  dy sin 
+

 dx 
∂x 
∂x

 ∂x ∂x  ∂x  
∂ N xy
 ∂w ∂  ∂w  

∂w 

 dx 
− N xy dy sin
+  N xy +
+
dx  dy sin 
∂y 
∂x

 ∂y ∂x  ∂y  
 ∂w ∂  ∂w  
∂w 
∂Q x

− Q x dy cos
+  Qx +
+
dx  dy cos 

 dx 
∂x 
∂x

 ∂x ∂x  ∂x  
∂Q y
 ∂w ∂  ∂w  

∂w 

 dy  + qdxdy = 0
− Q y dx cos
+  Q y +
dy  cos 
+
67
∂y 
∂Ramadas
y
 Chennamsetti
 ∂y ∂y  ∂y  
rd_mech@yahoo.co.in
∂N y
 ∂w ∂  ∂w  

∂w 



 dy 
dy  dx sin 
− N y dx sin
+Ny +
+
∂y 
∂y

 ∂y ∂y  ∂y  
∂ N xy

 ∂w ∂  ∂w  
∂w 
− N xy dx sin
+  N xy +
dy  dx sin 
+

 dy 
∂x 
∂y
 ∂x ∂y  ∂x  

Z-direction
R&DE (Engineers), DRDO
If no in-plane forces acting
∂Q x ∂Q y
+
= −q
∂x
∂y
∂ 2w
∂ 2w
∂ 2 w ∂Q x ∂Q y
Nx
+ 2 N xy
+ Ny
+
+
+q=0
2
2
∂x
∂x∂y
∂y
∂x
∂y
Ramadas Chennamsetti
68
rd_mech@yahoo.co.in
Presence of in-plane forces
Z-direction
R&DE (Engineers), DRDO
Substituting Qx and Qy
∂ 2
∂ 2
Qx = − D
∇ w , Qy = − D
∇ w
∂x
∂y
(
)
(
)
∂ 
∂
+ − D
∇2w
∂x 
∂x
(
∂
 ∂ 
2

+
−
D
∇
w


∂y
 ∂y 
)
(
)

 + q = 0

2
2
2


1
∂
∂
w
∂
w
w
4
=> ∇ w =  N x 2 + 2 N xy
+ N y 2 + q 
D
∂x
∂ x∂ y
∂y

Ramadas Chennamsetti
69
rd_mech@yahoo.co.in
∂2w
∂2w
∂2w
N x 2 + 2 N xy
+ Ny 2
∂x
∂ x∂ y
∂y
Plate buckling
R&DE (Engineers), DRDO
Buckling of thin plate
2
2
2

1
∂ w
∂ w
∂ w
4
∇ w = q + N x 2 + 2 N xy
+ Ny 2 
D
∂x
∂y∂x
∂y 
Assume, q = 0, in-plane load, Nx = N1, rest zero
∂ w
∂ w
∂ w Nx ∂ w
+2 2 2 +
=
4
4
2
∂x
∂x ∂y
∂y
D ∂x
4
4
2
Assume all four edges are simply supported –
Navier solution to plate bending
Ramadas Chennamsetti
70
rd_mech@yahoo.co.in
4
Plate buckling
R&DE (Engineers), DRDO
Plate deflection
∞
∞
m
n
w = w( x, y ) = ∑∑ wmn sin α m x sin β n y
N1 2
α + 2α β + β − α m = 0
D
N1 2
2
2 2
=> α m + β n =
αm
D
Ramadas Chennamsetti
4
m
2
m
(
2
n
4
n
Substitute in
governing equation
Characteristic
equation
)
71
rd_mech@yahoo.co.in
mπ
nπ
αm =
, βn =
a
b
Plate buckling
R&DE (Engineers), DRDO
From characteristic equation
2
m
+ β n2
)
2
=
N1 2
αm
D
 m π 
N1  mπ 
 nπ  
=>

 = 
 +
 
D  a 
 b  
  a 
2
2
Dπ a
=> N 1 =
m2
2
2
2
 m 
n 
  +   
 b  
  a 
2
Dπ 2  m
c 2
n 
=> N 1 =
+
2

b c m

c = a / b - Aspect ratio
Ramadas Chennamsetti
2
2
2
2
72
rd_mech@yahoo.co.in
(α
Plate buckling
R&DE (Engineers), DRDO
Critical load – smallest value
Increase in N1 with n2 – Minimum value of
n is equal to one – buckled shape in ydirection – single half sine wave
2
N1 is a function of variable ‘m’ – for minimum value of
‘m’, differentiate N1 wrt ‘m’
Ramadas Chennamsetti
73
rd_mech@yahoo.co.in
Dπ  m c 
N1 = 2  + 
b  c m
2
Plate buckling
R&DE (Engineers), DRDO
Differentiate
dN1 2π D  m c  1 c 
=
 +  − 2  = 0
2
dm
b  c m  c m 
1 c 
=>  − 2  = 0
c m 
a
=> m = c =
b
2
4 Dπ
∴ N1cr =
2
b
Ramadas Chennamsetti
Whole number
74
rd_mech@yahoo.co.in
2
Plate buckling
R&DE (Engineers), DRDO
Number of half sine waves can’t be a
whole number – it should be an integer
Equation for critical load for n = 1
m c 
K = + 
 c m
2
Plotting ‘K’ vs aspect ratio = c = a/b
for various integer values of ‘m’
Ramadas Chennamsetti
75
rd_mech@yahoo.co.in
Dπ
N1 = K 2
b
2
Plate buckling
R&DE (Engineers), DRDO
1.414
2.449
Ramadas Chennamsetti
76
rd_mech@yahoo.co.in
K vs aspect ratio
Plate buckling
R&DE (Engineers), DRDO
From figure, value of ‘K’ is same as
intersection points of ‘m’ and ‘m+1’
c 
 m c   m +1
+
 +  =

m +1
 c m  c
2
=> c = m(m + 1)
2
2
c = m(m + 1)
Ramadas Chennamsetti
Aspect ratio less than
√2 => m = 1
Aspect ratio from √2
to √6, m = 2
C>4 => K ≈ 4
77
rd_mech@yahoo.co.in
N1 critical load at m – when load is increased, buckled
form changes from ‘m’ to ‘m+1’
Curves for m = 1 and
At transition from ‘m’ to ‘m+1’
m = 2 meet at c = √2
Plate buckling
R&DE (Engineers), DRDO
Estimation of buckling load
t = 1 cm, a = 2.3 m, b = 1 m, E = 200 GPa, υ = 0.30
(
)
−2 2
Et
200 ×10 × 1×10
D=
=
= 17.96 kNm
2
2
12 1 − υ
12 1 − 0.3
a 2.3
c= =
= 2.3 => m = 2
b
1
3
(
9
)
(
2
)
2
 m c   2 2.3 
=> K =  +  = 
+
= 4.0786

2.3 2 
c m


Ramadas Chennamsetti
78
rd_mech@yahoo.co.in
Flexural modulus
Plate buckling
R&DE (Engineers), DRDO
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
c = 2.3, m = 1, K = 7.4790
m = 2, K = 4.0786
Minimum value of
‘K’ is considered for
estimation of Ncr
Chennamsetti
m = 3, K Ramadas
= 4.2890
79
rd_mech@yahoo.co.in
Minimum critical load
2
2
Dπ
Dπ
N1cr = K 2 = 4.086 × 2 = 73.72 MN / m
b
b
N = N1cr × b = 73.72 MN
Plate and column buckling
R&DE (Engineers), DRDO
Uniaxial load for plate buckling
t
(
K
π 2E
=> σ 1crp =
12 1 − υ 2 b 2
t
K
π 2E
=> σ 1crp = C1
, C1 =
2
12 1 − υ 2
b
t
(
)
)( )
( )
Ramadas Chennamsetti
(
)
80
rd_mech@yahoo.co.in
σ 1crp =
N1crp
Dπ 2
N1crp = K 2
b
Dπ 2
π 2 Et 3
=K 2 =K 2
bt
b t 12 1 − υ 2
Plate and column buckling
R&DE (Engineers), DRDO
Buckling of a column
=> Pcr = C 2
l2
π 2 EAk 2
l
2
= C2
π 2 EA
(l k )
2
Pcr
π 2E
=> σ crc =
= C2
Column
2
A
l
k
π 2E
Plate
σ crp = C1
2
b
t Chennamsetti
Ramadas
( )
( )
81
rd_mech@yahoo.co.in
Pcr = C 2
π 2 EI
Plate and column buckling
R&DE (Engineers), DRDO
Critical stress for plate depends on thinness
ratio = t/b – not on the length
More thinner plate – lesser bucking load
Critical stress in column depends on
slenderness ratio
Longer columns – lower critical load
Ramadas Chennamsetti
82
rd_mech@yahoo.co.in
depends on width
Strain energy
R&DE (Engineers), DRDO
Linear elastic material
1
U = ∫ (σ xxε xx + σ yyε yy + τ xyγ xy ) dV
2V
Ramadas Chennamsetti
83
rd_mech@yahoo.co.in
In thin plate theory, out-of-plane shear
stresses vanish => τxz, τyz and τzz
Stress components contributing to strain
energy => σxx, σyy and τxy
Strain energy,
Strain energy
R&DE (Engineers), DRDO
Strain energy,
ε yy γ xy } dV
T
 ∂2w 

2 




∂x 
σ xx 


1
0
1
0
υ
ε
υ

xx
 
 ∂ 2 w 
Ez 
Ez 
 
υ 1
υ 1
0  ε yy  = −
0 
σ yy  = −
2
2
2 
−
∂
y
1
1
−
υ
υ


1 + υ  
1 + υ 
τ 

γ
0
0
0
0
2

  xy 


 xy 
2 
2  2 ∂ w 


 ∂x∂y 


=> {σ } = [C ]{ε }
Ramadas Chennamsetti
84
rd_mech@yahoo.co.in
1
U = ∫ {σ xx σ yy τ xy }{ε xx
2V
Strain energy
R&DE (Engineers), DRDO
Strain energy,
 ∂ 2 w
∂2w  ∂2w 
 2 + υ 2  2 
∂y  ∂x 
 ∂x

2
2
2 
 ∂ w ∂ w  ∂ w 2
E

=> U =
+ υ 2 + 2  2 z dV
2 ∫ 
2 1 − υ V  ∂x
∂y  ∂x 


2
2


∂ w

+ 2(1 − υ )



 ∂x∂y 
(
)
Ramadas Chennamsetti
85
rd_mech@yahoo.co.in
1
1
T
T
U = ∫ {σ } {ε }dV = ∫ {ε } [C ]{ε }dV
2V
2V
Strain energy
R&DE (Engineers), DRDO
Infinitesimal volume, dV = dxdydz
Carry out integration over thickness => dz
h
+
2
3
h
∫h z dz = 12
−
2
3
Eh
D=
2
12 1 − υ
(
)
Ramadas Chennamsetti
86
rd_mech@yahoo.co.in
2
Strain energy
R&DE (Engineers), DRDO
Simplify



dxdy
2 
 
  
  
Strain energy – Finite Element Method – Total
potential approach
Ramadas Chennamsetti
87
rd_mech@yahoo.co.in
 ∂ 2 w ∂ 2 w  2

 
+
2
2 
∂x 
D   ∂ x
U =

∫
 ∂ 2w ∂ 2w  ∂ 2w
2 A
 − 2 (1 − υ ) ∂ x 2 ∂ x 2 −  ∂ x ∂ y



Ramadas Chennamsetti
88
rd_mech@yahoo.co.in
R&DE (Engineers), DRDO