Theory of Plates and Shells

Lecture notes on Theory of Plates and Shells (Civil and Structural Engineering)

1. Theory of Plates and Shells
For
Post Graduate Students of Structural Engineering


Prof. D
R. Atteshamuddin S. Sayyad
B.E. (Civil), M.E. (Structures), Ph.D ( Structural Engineering)
Associate Professor
Department of Civil Engineering
College of Engineering
Kopargaon, Ahmednagar -423601
Maharashtra, India

Rs: 300/-

2. Dedicated to
My Parents, Teachers and Students

3. Preface
I take an opportunity to write this book titled as “Theory of Plates and Shells” for the P ost
Graduate Students of Structural Engineering. While writing this book I have constantly kept in
mind the requirement of all the post graduates students. S o I have written this book by covering
the syllabus of structural engineering course of University of Pune & Dr. Babasaheb Ambedkar
Marathwada University, and also questions which will ask in the exami nation is given in the
exercise. The main object of the book is to present the subject matter in the most concise,
compact, to the point and lucid manner. Although every care has been taken to check mistakes,
yet it is difficult to claim perfection. Any error, omission and suggestions for the improvement of
this volume, brought to my notice, will be thankfully acknowledged and incorporated in next
edition.


D
R. A. S. Sayyad

4. CONTENTS

1 Basic Equations Page No
1.1 State of Stress at a Point 1
1.2 State of Strain at a Point 1
1.3 Strain Displacement Relationship 2
1.4 Equilibrium Equations 2
1.5 Strain Compatibility Equations 3
1.6 Stress Strain Relationship 4
1.7 Plane Stress Problems 6
1.7.1 State of Stress at a Point 6
1.7.2 State of Strain at a Point 6
1.7.3 Strain Displacement Relationship 7
1.7.4 Equilibrium Equations 7
1.7.5 Strain Compatibility Equations 7
1.7.6 Stress Strain Relationship 7
1.8 Plane Strain Problems 7
1.8.1 State of Stress at a Point 8
1.8.2 State of Strain at a Point 8
1.8.3 Strain Displacement Relationship 8
1.8.4 Equilibrium Equations 8
1.8.5 Strain Compatibility Equations 9
1.8.6 Stress Strain Relationship 9
2 Bending of Rectangular Thin Plate
2.1 Introduction 10
2.2 Small Deflection Theory / Kirchhoff’s Thin Plate Bending Theory /
Classical Plate Theory /
11
2.2.1 Assumptions in small deflection theory 11
2.2.2 Associate boundary conditions 19
2.3 Navier’s Solution for Lateral Deflection of Rectangular Plate 21
2.3.1 Simply supported Plate carrying a uniformly distributed load of
intensity q
0
24
2.3.2 Simply supported Plate carrying a sinusoidal load 30
2.3.3 Simply supported Plate carrying a patch load 34
2.3.4 Simply supported Plate carrying a concentrated load 37
2.3.5 Simply supported Plate carrying a linearly varying load 39
2.4 Levy’s solution for the lateral deflection of rectangular plate 41
2.4.1 Levy’s solution for rectangular plate with at least two opposite
edges simply supported carrying a uniformly distributed load.
41
2.4.2 Levy’s solution for rectangular plate with at least two opposite
edges simply supported carrying a linearly varying load.
49

5. 2.4.3 Levy’s solution for rectangular plate with at least two opposite
edges simply supported carrying a moments along the edges.
53
2.4.4 Levy’s solution for rectangular plate with at least two opposite
edges clamped carrying a uniformly distributed load.
56
3 Pure Bending of Plate
3.1 Slopes and curvatures of bent plate. 61
3.2 Moment curvature Relation. 67
3.3 Particular cases in Pure Bending 69
4 Bending of Circular Plates.
4.1 Symmetrical Bending of Circular Plate. 76
4.2 Equation of Deflection for Uniformly Loaded Circular Plate. 81
4.2.1 Simply supported circular plate subjected to uniformly distributed
load
82
4.2.2 Fixed / Clamped circular plate subjected to uniformly distributed
load
88
4.3 Equation of Deflection for Circular Plate subjected to centre Concentrated
Load.
93
4.3.1 Simply supported circular plate subjected to Centre Concentrated
load
94
4.3.2 Fixed / Clamped circular plate subjected to Centre Concentrated
load
100
4.4 Problem 104
4.5 Problem 108
4.6 Circular Plate with Circular Hole at the Center. 109
4.6.1 Bending of a plate by moments M
1 and M2 uniformly distributed
along inner and outer boundaries
109
4.6.2 Bending of a plate by moment M
1 uniformly distributed along inner
boundary
114
4.6.3 Bending of a plate by shearing forces along inner boundaries 115
5 General Theory of Cylindrical Shell
5.1 Definition. 122
5.2 Some important terms used in shells. 122
5.3 Advantage of Shell structures. 123
5.4 Disadvantage of Shell structures. 123
5.5 Classification of Shell structures. 123
5.6 Assumptions made in theory of thin elastic shell. 125
5.7 Determination of Stress Resultants 125
5.8 General theory of cylindrical shell. 132
5.8 Membrane theory of cylindrical shell. 137
5.8.1 Horizontal cylinder with closed ends filled with liquid and
supported at ends with simple supports
139

6. 5.8.2 Cylindrical Roof Shell 143
6 Bending Theory of Cylindrical Shell
6.1 The need for bending theory. 147
6.2 Strains in cylindrical shell. 147
6.3 The Finsterwalder Theory 148
6.3.1 Assumptions made in Finsterwalder theory. 148
6.3.2 Equations of Equilibrium 148
6.3.3 Derivation of Finsterwalder eighth (8
t
h
) order deferential equation. 151
6.4 The D-K-J Theory (Donnell – Karman – Jenkins Theory) 158
6.4.1 Equations of Equilibrium 158
6.4.2 Flugge’s simultaneous equations 161
6.4.3 The D-K-J Equations 164
6.5 The Schorer Theory. 167
6.6 VLASOV Bending Theory. 170
6.6.1 Assumptions. 170
6.6.2 Equations of Equilibrium. 170
6.6.3 Derivation of VLASOV Equation 173
6.7 Beam Theory for Cylindrical Shell. 175
6.7.1 Advantages. 175
6.7.2 Assumptions. 175
6.7.3 Beam Analysis. 176
6.7.4 Arch analysis. 178
7 Shells of Revolution.
7.1 Surface of Revolution with Axisymmetric Loading 179
7.2 Surface of Revolution with Unisymmetric Loading 185
7.3 Particular Cases of Shells in the form of Surface of Revolution 187
7.3.1 Example: Spherical dome of constant thickness under its own
weight
187
7.3.2 Example: Spherical Tank filled with Liquid 191
7.3.3 Example: Conical Sh ell filled with liquid 195
7.3.4 Example: Shell in the form of an Ellipsoid of Revolution 199
7.3.5 Example: Shell in the form of Torus 201
8 Shells of Double Curvature
8.1 Membrane Theory for Shells of Double Curvature other than Surface of
Revolution
204
8.1.1 Pseudo Stress Resultants 204
8.1.2 Equations of Equilibrium 205
8.2 Membrane Theory for Rectangular hyperbolic Paraboloid with Straight line
Generators and Boundaries
208
8.3 The Umbrella Roof 210
 

7. NOTATIONS

x, y, z Rectangular Coordinates
,r
θ Polar Coordinates
,
xy
rr Radii of curvature of the middle surface of a plate in xz and yz planes respectively
a, b Edge lengths of plate in x and y directions, respectively
h Thickness of a plate or a shell
D Flexural rigidity of plate or shell.

µ Poisson’s ratio
λ Lami’s Constant.
E Young’s modulus.
G Shear modulus.
q Intensity of a continuously distributed load.
p Pressure
P Point load
X, Y, Z Body forces.
u, v,w Displacements in x, y, z direction.
,
xy
σσ,
z
σ Normal stresses in x, y, z directions, respectively.
nσ Normal stress in n direction
rσ Radial stress in polar coordinates
,
tθ
σσ Tangential Stresses in polar coordinates
xyτ, ,
zx yz
ττ Shear stresses in xy, zx, yz plane.
,
xyεε,
z
ε Normal strains in x, y, z directions, respectively.
nε Normal strain in n direction
rε Radial strain in polar coordinates
,
tθ
εε Tangential strains in polar coordinates
xyγ, ,
zx yz
γγ Shearing strains in xy, zx, yz plane.
rθ
γ Shearing strains in polar coordinates.
,
xyMM Bending moments on a plane normal to the and axes respectivelyxy
xy
M Twisting moments on a plane normal to the axes in the direction x y
n
M,
t
M Bending moments on a plane normal to the , axes respectivelynt
ntM Twisting moments on a plane normal to the axes in the direction nt
,
xy
QQ Transverse shear forces on a plane normal to the and axes respectivelyxy

8. n
Q Transverse shear forces on a plane normal to the and axes respectivelyxy
,
xy
NN Normal forces to the and axes respectivelyxy
xy
N Normal forces perpandicular to the axis in direction.xy
12
,rr Radii of curvature of a shell in the form of surface of revolution in meridianal plane
and in the normal plane perpendicular to meridian respectively. φ
χ Changes of curvature of a shell in meridianal plane.
,,NNN
φθθφ
Membrane forces of a shell
M
θ
, M
φ
Bending moments in a shell of meridianal section and a section perpendicular to
meridian respectively.

,
xx
NN
φ
Membrane forces per unit length of axial section and a section perpendicular to the
axis of cylindrical shell.
,
xx
MM
φ
Bending moments per unit length of axial section and a section perpendicular to the
axis of cylindrical shell.
x
M
φ
Twisting moment per unit length of an axial section of cylindrical shell.
,
x
QQ
φ
Shearing forces parallel to z axis per unit length of an axial section and a section
perpendicular to the axis of a cylindrical shell respectively.

9. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     1 


Chapter-1
Basic Equations

1.1 State of stress at a point: At a point there are three mutually perpendicular planes i.e.
orthogonal planes. Three stresses components, one normal and two shears, act on each plane.
Therefore, nine stress components must be known at each point to define completely state of
stress at a point. Therefore six stress components in the Cartesian (
,,
xyz) co-ordinate system.
[]
xxyxz
yx y yz
zx zy z
σττ
στστ
ττσ
⎛⎞
⎜⎟
=
⎜⎟
⎜⎟
⎝⎠

But it is proved that shear stresses are complementary i.e. and
xy yx yz zy xz zx
;
ττττ ττ= ==
Therefore there are only six components of stress at a point, three normal stresses and three shear
stresses. Therefore stress at a point is specified as
[]
xxyxz
xy y yz
xz yz z
σττ
στστ
ττσ
⎛⎞
⎜⎟
=
⎜⎟
⎜⎟
⎝⎠
(1.1)
Similarly, six stress components in the cylindrical ( ,,rz
θ) co-ordinate system.
[]
rr rz
rz
rz z zθ
θθθ
θ
σττ
στστ
ττσ
⎛⎞
⎜⎟
=
⎜⎟
⎜⎟
⎝⎠
(1.2)
1.2
State of strain at a point: Similar to stress considerations, the nine components of strain
are reduced to six independent components, three linear strains corresponding to the change in

10. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     2 

line elements and three shear strains corresponding to the change in right angles formed by line
elements. Thus the state of strain at a point of a body in the Cartesian (
,,
xyz) co-ordinate
system can be expressed in the matrix form as


[]
xxyxz
xy y yz
xz yz z
εγγ
εγεγ
γγε
⎛⎞
⎜⎟
=
⎜⎟
⎜⎟
⎝⎠
(1.3)
Similarly, six strain components in the cylindrical (
,,rz
θ) co-ordinate system
[]
rr rz
rz
rz z zθ
θθθ
θ
εεε
εεεε
εεε
⎛⎞
⎜⎟
=
⎜⎟
⎜⎟
⎝⎠
(1.4)
1.3
Strain Displacement relationship: The six strain components, three linear strain and
three shear strains, at a point of the body are related to the three displacements u, v, and w by the
following expressions in the Cartesian ( , ,
xyz) co-ordinate system
Normal strain:
Shear strain:
xyz
xy yz xz
uvw
,,
xyz
uv vw uw
,,
yx zy zx
εεε
γγγ
∂∂∂ ⎤
===
⎥
∂∂∂
⎥
∂∂ ∂∂ ∂∂ ⎥
=+ =+ =+
⎥
∂∂ ∂∂ ∂∂ ⎦
(1.5)
Strain displacement relationship for cylindrical ( ,,rzθ) co-ordinate system
1
Normal strain:
11
Shear strain:
rz
rzrz
uvuw
,,
rrr z
uvu v w uw
,,
rrr zr zr
θ
θθ
εε ε
θ
γγγ
θθ
∂∂ ∂ ⎤
==+=
⎥
∂∂ ∂
⎥
∂∂ ∂ ∂ ∂∂
⎥
=+− =+ =+
⎥∂∂ ∂ ∂ ∂∂ ⎦

(1.6)
1.4
Equilibrium Equations: Consideration of the variation of the state of stress from point
to point leads to the equilibrium equations in the Cartesian (
,,
xyz) co-ordinate system are given
by

11. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     3 

00and0
xy xy y yz yzxxz xz z
X; Y Z
xyz xy z xyz
ττστ τστ τ σ∂∂∂∂ ∂ ⎤∂∂ ∂ ∂
+++= ++ += +++=
⎥
∂∂∂ ∂∂ ∂ ∂∂∂ ⎦


(1.7)
Where , and
XYZ are the components of body force such as gravitational, centrifugal, or other
inertia forces. The equilibrium equations for a body referred in cylindrical co-ordinates ( , ,rzθ)
system
211
00
1
and 0
rr r zrrrz
r
zrz z rz
z
P; P
rr z r rr z r
P
rr z r
θθθθθθ
θ
θ
τσσ τσττστ
θθ
ττστ
θ∂−∂∂∂ ⎤∂∂ ⎛⎞
+++ += ++++=
⎜⎟
⎥
∂∂∂ ∂∂∂ ⎝⎠
⎥
⎥∂∂∂
++++=
⎥
∂∂∂ ⎦
(1.8)
Where , and
rz
PP P
θ
are the components of body force such as gravitational, centrifugal, or
other inertia forces.
1.5 Strain compatibility equations:
It is clear from the strain displacement relationship that if the three displacement
components are given, then the strain components can be uniquely determined. If, on the other
hand, the six strain components are arbitrarily specified at a point, then the displacement
components cannot be uniquely determined. This is because the six strain components are related
to only three displacement components , andvizu v w. Hence if displacement components are to
be single valued and continuous, then there must exist certain interrelationship among the strain
components. These relations are called the strain compatibility equations. For three dimensional
bodies there exist six strain compatibility equations.
In the Cartesian ( , ,xyz) co-ordinate system.

22 2 22 2222
22 22 22
and
yxy y yzx xxzzz
;
yxxyzyyz zxxz
εγ ε γ
ε εγεε ⎤∂∂ ∂ ∂∂∂∂ ∂∂
+= += += ⎥
∂∂∂∂∂∂∂∂ ∂∂∂∂ ⎥⎦
(1.9)

12. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     4 

And

22
2
22
2
xy yz y xy yzxxz xz
yz xyxzz
;
yz x z y x xz y z y x
xy z y x z
γγεγγεγ γ
γγγε ⎤∂∂∂∂∂⎡⎤⎡⎤∂∂ ∂∂∂
=+− =−+ ⎥⎢⎥⎢⎥
∂∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂⎣⎦⎣⎦ ⎥
⎥
∂∂⎡⎤∂∂ ∂
⎥=+−
⎢⎥
∂∂ ∂ ∂ ∂ ∂ ⎥⎣⎦ ⎦
(1.10)
Similarly strain compatibility equations, for the case of small displacements, in terms of
cylindrical coordinates ( , ,rz
θ) can be obtained as

() ()
22222 2
22 2 2
22 2
2
22
rrzrz rr
zzzrz
rr
;r r
zrrz r r r
rr rr
zr z z
θθ
θθ
εγεεγ εε
θθ
εγ εεγ
θθ ⎤∂∂∂∂∂ ∂∂
+= −++ = ⎥
∂∂∂∂ ∂∂ ∂ ∂∂
⎥
⎥∂∂ ∂∂ ∂
++−= ⎥
∂∂∂∂∂∂ ⎦
(1.11)
And

() () ()
()
() ()
222
2
1
2
11
2
rz
rz r
rrrz
z
rz rz z
rrrr
zrzr
rrr
r
rr r rz z
r
zz rr r rr
θθ θ
θ
θ
θθ
γ
γγ εε
θθ θ
γεγ
γ
θθ
γγ γε
θθ ⎤∂∂∂ ∂ ∂ ∂∂ ⎡⎤⎡⎤ ⎡ ⎤
+−= − ⎥⎢⎥⎢⎥ ⎢ ⎥
∂∂ ∂ ∂ ∂ ∂ ∂⎣⎦ ⎣ ⎦ ⎣⎦
⎥
⎥∂∂⎡∂⎤∂∂⎛⎞
⎥−− =
⎜⎟⎢⎥
∂∂ ∂ ∂∂ ∂∂⎝⎠ ⎥⎣⎦
⎥
⎡∂ ⎤ ∂∂∂∂ ∂ ⎛⎞ ⎛⎞
⎥−−=−
⎜⎟⎜⎟⎢⎥
⎥∂∂ ∂ ∂ ∂ ∂ ⎝⎠⎝⎠⎣⎦ ⎦
(1.12)
1.6 Stress strain relationship:
The stresses and strains cannot be independent when we consider physical problem of the
theory of elasticity which is concerned with the determination of stress components and
deformation due to external loads acting on an elastic body. Hence the stresses need to be
related to strain through a physical law. For isotropic material, generalized Hook’s law
gives the following stress strain relations.

() ()
()
11
;;
1
and , ,
xxyzyyxz
xy yz xz
z z y x xy yz xz
EE
EGGG
εσυσσεσυσσ
ττ
τ
εσυσσ γ γ γ
⎤
⎡⎤ ⎡⎤=−+ =−+
⎣⎦ ⎥⎣⎦
⎥
⎥
⎡⎤=−+ = = =
⎣⎦ ⎥
⎦
(1.13)

Where ,and
E Gυ are the elastic properties of the material.

13. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     5 

Similarly in terms of cylindrical coordinates ( , ,rzθ) can be obtained as

() ()
()
11
;
1
and , ,
rr z rz
rz rz
zz r r z rz
EE
EGGG
θθθ
θθ
θθθ
εσυσσεσυσσ
ττ
τ
εσυσσ γ γ γ
⎤
=−+ =−+⎡⎤⎡⎤
⎣⎦⎣⎦ ⎥
⎥
⎥
=−+ = = =⎡⎤
⎣⎦ ⎥⎦
(1.14)
Alternately stress-strain relation for isotropic material can be written as,
()( )
()
()
()
()
()
()
1000
1000
1000
12
000 0 0
2112
12
000 0 0
2
12
000 0 0
2
x x
y y
z z
xy xy
zy yz
zx xz
E
υυ υ
υυυ
σ ε
υυ υσ ε
υσ ε
τ γυυ
υ
τ γ
τ γ
υ
−⎡⎤
⎢⎥
−⎢⎥⎧⎫ ⎧⎫
⎢⎥⎪⎪ ⎪⎪
−
⎢⎥⎪⎪ ⎪⎪
⎢⎥⎪⎪ ⎪⎪ −
⎪⎪ ⎪⎪
⎢⎥=⎨⎬ ⎨⎬
⎢⎥+−
⎪⎪ ⎪⎪
⎢⎥ −⎪⎪ ⎪⎪
⎢⎥
⎪⎪ ⎪⎪
⎢⎥
⎪⎪ ⎪⎪
⎩⎭ ⎩⎭
⎢⎥ −
⎢⎥
⎣⎦
(1.15)
[
] []{}Dσ ε∴ =
OR
()
()
() 2
2
2
xxyz x
yxyz y
zxyz z
G
G
G
σλε ε ε ε
σλε ε ε ε
σλε ε ε ε
=+++
=+++
=+++
(1.16)
Where
()( )
()
Lame's constant = and
112 21
EE
G
υ
λ
υυυ
==
− −+

Similarly in terms of cylindrical coordinates ( ,,rzθ) can be obtained as
()
()
() 2
2
2
rr z r
rz
zr z z
G
G
G
θ
θ θθ
θ
σλε ε ε ε
σλε ε ε ε
σλε ε ε ε
=+++
=+++
=+++
(1.17)

14. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     6 

1.7 Plane Stress Problems: The two dimensional elasticity problems under the following
conditions, is considered as plane stress problem.
1.
The thickness of body i.e. the dimension in one direction, say z direction, is small in
comparison with other dimensions in x and y directions.
2.
The load acting on the body are in the plane perpendicular to the thickness of the body
i.e. z axis. The loads are distributed uniformly over the thickness.
3.
The stresses in the direction of thickness of the body must be zero on free boundary.
Therefore it is assumed that the stress components in the direction of thickness are zero
i.e.
0
zxzyz
σττ===
The plane stress problem is characterized by the following basic concepts and relations of theory
of elasticity.
1.7.1 State of a stress at a point: The three stress components in z direction are zero
() 0
zxzyz
στ τ=== . Therefore only three independent stress components
xy
,
σσand
xy
τ exist
at a point. Note that
xy yx
ττ=. Such state of stress in known as plane stress.
[]
x
y
xy
σ
σσ
τ
⎧⎫
⎪⎪
=⎨⎬
⎪⎪
⎩⎭
(1.18)
1.7.2 State of a strain at a point: Strain at a point is defined by three independent strain
components, two linear strain, and
xy
εε, and one shear strain
xy
γ.
[]
x
y
xy
ε
εε
γ
⎧⎫
⎪⎪
=⎨⎬
⎪⎪
⎩⎭
(1.19)
It may be noted that 0
z
ε≠, but can be obtained by the condition 0
z
σ=,
()xy
z
E
υσσ
ε+∴ =−

15. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     7 

But and
xz yz
γγare zero.
1.7.3 Strain Displacement Relationship: Three strain components are related to two
displacement components by the following expressions:

and
xyxy
uv uv
;
xy yx
εεγ
⎤∂∂ ∂∂
===+
⎥
∂∂ ∂∂ ⎦
(1.20)
1.7.4 Equations of equilibrium: Conditions of equilibrium 0, 0
xy
FF∑=∑= , for the
components of body force and internal forces due to stresses give the following equations.

0and 0
xy xy yx
XY
xy xy
ττσσ∂∂∂ ⎤∂
++= ++=
⎥
∂∂ ∂∂ ⎦
(1.21)
1.7.5 Strain Compatibility Equations: The strain components , and
xy xy
εεγ are related by
eliminating displacements from the strain displacement relations as seen previously. Therefore
for plane stress problem, the strain compatibility equation is expressed as
222
22
yxyx
yxxy
εγε∂∂∂
+=
∂∂∂∂
(1.22)
1.7.6 Stress Strain Relationship: Stresses can be expressed in terms of strains as per Hook’s
law
2
10
10
1
1
00
2
xx
yy
xy xy
E
σ ευ
σ υε
υ
υτ γ
⎛⎞
⎧⎫ ⎡⎤ ⎜⎟
⎪⎪ ⎢⎥ ⎜⎟
=⎨⎬ ⎢⎥⎜⎟
−
⎪⎪ ⎢⎥⎜⎟
−
⎩⎭ ⎣⎦ ⎜⎟
⎝⎠
(1.23)
1.8 Plane Strain Problems: The two dimensional elasticity problems under the following
conditions, is considered as plane strain problem.
1.
The thickness of body i.e. the dimension in one direction, say z direction, is very large in
comparison with other dimensions in x and y directions.
2.
External forces are perpendicular to the z axis.

16. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     8 

3.
The strains in the z direction of the length of the body are zero 0
zxzyz
εεε===
The plane strain problem is characterized by the following basic concepts and relations of theory
of elasticity.
1.8.1 State of a stress at a point: Stress at a point is defined by three independent stress
components, two normal stress, and
xy
σσ, and one shear stress
xy
τ.
[]
x
y
xy
σ
σσ
τ
⎧⎫
⎪⎪
=⎨⎬
⎪⎪
⎩⎭
But 0
z
σ≠ (1.24)
It may be noted that 0
z
σ≠, but can be obtained by the condition 0
z
ε=, ()zxy
συσ σ∴ =+
But and
xz yz
ττare zero.
1.8.2 State of a strain at a point: The three strain components in z direction are
zero
() 0
zxzyz
εγ γ=== . Therefore only three i ndependent strain components
and
xy xy
,
εεγ exist at a point. Note that
xy yx
γγ=. Such state of strain in known as plane strain.
[]
x
y
xy
ε
εε
γ
⎧⎫
⎪⎪
=⎨⎬
⎪⎪
⎩⎭
(1.25)
1.8.3 Strain Displacement Relationship:

The displacement in z direction is zero. Therefore the two displacement components ,uvin
x and y directions, exists at a point. The relation between strain components and displacement
components are same as before,
i.e.

and
xyxy
uv uv
;
x yyx
εεγ
⎤∂∂ ∂∂
===+
⎥
∂∂ ∂∂ ⎦
(1.26)
1.8.4 Equations of equilibrium: Corresponding to three stress components in xy plane, the
equilibrium equations are also same and restated as

17. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     9 

0and 0
xy xy yx
XY
xy xy
ττσσ∂∂∂ ⎤∂
++= ++=
⎥
∂∂ ∂∂ ⎦
(1.27)
1.8.5 Strain Compatibility Equations: eliminating displacements components, the three strain
components are related by one equation as proved already.

222
22
yxyx
yxxy
εγε∂∂∂
+=
∂ ∂∂∂
(1.28)
1.8.6 Stress Strain Relationship: These relations are different than that of plane stress problem
and can be derived from the strain condition 0
z
ε=. Stresses can be expressed in terms of strains
as per Hook’s law
()( )
()
()
()
10
10
112
12
00
2
x x
y y
xy xy
E
σ ευυ
σ υυ ε
υυ
υτ γ
⎛⎞
⎜⎟⎧⎫ ⎧⎫ −
⎜⎟⎪⎪ ⎪⎪
=− ⎜⎟⎨⎬ ⎨⎬
+−
⎜⎟⎪⎪ ⎪⎪
−
⎜⎟⎩⎭ ⎩⎭
⎜⎟
⎝⎠
(1.29)

18. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     10 


Chapter 2
Bending of Rectangular Thin Plates

2.1 Introduction:
Flat plates are extensively used in many engineering application like roof and floor of
buildings, deck slab of bridges, foundation of footing, water tanks, bulk heads, turbine disks etc.
plates used in such applications are normally subjected to lateral loads, causing bending of the
plate. The geometry of the plate normally defined by the middle plane which is plane
equidistance from the top and bottom faces of the plate. The thickness of the plate (
h) is
measured in a direction normal to the middle plane of the plate. The flexural properties of the
plate largely depend on its thickness rather than its two dimensions (length and width). Plate is
subjected to moments and forces as shown in figure, transverse shear forces acts perpendicular to
the plane of the plate whereas central shear forces acts in the plane of the plate. In general, plate
problems can be classified into three major categories
viz. thin plate, moderately thick plate and
thick plate, depending upon the thickness of the plate.

19. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     11 

If the thickness of the plate is very small as compared to the other two dimensions, then such a
plate can be called as thin plate. A plate can be considered as thin if the ratio of thickness to the
lesser of other two dimensions is less than 0.05. The simplest and most widely used plate theory
is classical small deflection theory of thin isotropic and anisotropic plates.
2.2 Small Deflection Theory / Kirchhoff’s Thin Plate Bending Theory / Classical Plate
Theory:
The classical thin plate theory is based on Love-Kirchhoff’s hypothesis which makes
assumptions similar to those made by the Bernoulli-Navier hypothesis used in the theory of thin
or shallow beams. If the deflection is not small, then the bending of the plate is accompanied by
in plane stresses. The nature and magnitude of these in plane stresses depends not only
magnitude of the deflection but also type of support provided at the edges of the plate.
2.2.1 Assumptions in small deflection theory:
The following fundamental assumptions are made in the classical small deflection theory of
thin homogenous elastic plates.
1.
Straight line initially normal to the middle surface to the plate remains straight and normal to
the deformed middle surface of the plate and unchanged in length.
Note: This assumption corresponds to the Bernoulli’s melier hypothesis for the deflection of
beam.
2. Displacement w is assumed to be very small. This means the slope of the deflected surface is
small and hence square of the slope would be negligible in comparison with unity
3. The normal stresses
x
σand
y
σ, inplane shear stress
xy
τ are assumed to be zero at middle
surface of the plate
Note: This assumption is valid if deflection w is very small as compared to the thickness of
plate
()wh〈〈
4. Stress
z
σ i.e. transverse normal stress is small as compared to other stress components and
may be neglected in stress strain relationship.

20. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     12 

( )zxyxy
,,σσστ〈〈
5. The midplane remains unstrained after bending.
The above assumptions, known as Kirchhoff’s hypothesis, reduce the three dimensional plate
problems to two dimensions. Hence in Cartesian coordinate system, only normal stresses
and
xy
σσ and shear stress
xy
τ would exists in the plate, and these stresses are function of two
coordinates
viz x and y


Under the second assumption, points on the middle surface of the plate can be assumed to be
displaced only in the
z direction. For other points of plate, the corresponding u and v
sin
sin
sin
u
z
wu
xz
www
uz
x xx
θ=−
∂
=−
∂
∂∂∂ ⎛⎞
∴=− ≈
⎜⎟
∂∂∂ ⎝⎠

21. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     13 

displacements in the x and y direction are proportional to their distances from the middle surface.
Thus point is displaced by the amount
w
uz
x
∂
=−
∂
in the x direction and by the amount
w
vz
y
∂
=−
∂
in the y direction. Therefore displacements in the x, y and z directions are given as.
)
)
() )
w
uz a
x
w
vz b
y
wwx,y c
∂ ⎫
=−
⎪
∂
⎪
∂ ⎪
=− ⎬
∂
⎪
⎪
=
⎪
⎭
(2.1)
To obtain components of strain put equation (2.1) in the (1.20)
)
)
)
2
2
2
2
2
2
x
y
xy
w
za
x
w
zb
y
w
zc
xy
ε
ε
γ
⎫∂
=−
⎪
∂
⎪
⎪∂ ⎪
∴ =− ⎬
∂
⎪
⎪
∂
=− ⎪
∂∂ ⎪⎭
(2.2)
To find out corresponding stresses put (2.2) in the equation (1.23)
)
)
)
22
22 2
22
22 2
22
1
1
2
1
x
y
xy
zE w w
a
xy
zE w w
b
yx
wEw
Gc
xy xy
συ
υ
συ
υ
τ
υ
⎫⎡⎤−∂ ∂
∴ =+ ⎪⎢⎥
−∂ ∂⎣⎦ ⎪
⎪
⎡⎤−∂ ∂ ⎪
∴ =+ ⎬⎢⎥
−∂ ∂⎣⎦ ⎪
⎪
∂∂
⎪∴ =− =−
∂∂ + ∂∂ ⎪
⎭
(2.3)
Let us consider an element of a rectangular plate which is subjected to a system of
internal forces and external load
q per unit area. For thin plates, the dead load of the plate can be
combined with lateral load
q. since the plate element considered is very small we will not

22. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     14 

consider the variation of forces on each edge, but instead assume that the mean value of forces
would be acting at the centre of each face or edge. The bending and twisting moments and shear
forces acting on the plate area of unit width and height
h are shown in following figure.
Where
, Bending moments on a plane normal to the and axes respectively
xy
MM x y=
Twisting moments on a plane normal to the axes in the direction
xy
M xy=
, Transverse shear forces on a plane normal to the and axes respectively
xy
QQ x y=
It is assumed that positive moments produce tension in the fibres located at the bottom part of
plate. Now, since moments are resultant of the stresses developed in the plate, these are called
stress resultant and are forces per unit length of the plate. Which are given by
22
22
h/ h/
xx x
h/ h/
M .b.z.dz.dy .z.dzσσ
++
−−
==∫∫
(2.4)
Substitute the value of
x
σ from equation (2.3 a) in the equation (2.4) to obtain
x
M

23. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     15 

2 22
2
222
2
1
h/
x
h/
Eww
M zdz.
xy υ
υ
+
−
⎛⎞∂∂
∴ =− + ⎜⎟
−∂∂ ⎝⎠ ∫

2
223
22 2
2
13
h/
x
h/
Ew wz
M
xy
υ
υ
+
−
⎛⎞⎡⎤∂∂
∴ =− + ⎜⎟ ⎢⎥
−∂ ∂⎝⎠⎣⎦

()
32 2
222
12 1
x
Eh w w
M
xy
υ
υ
⎛⎞∂∂
∴ =− + ⎜⎟
∂∂− ⎝⎠
22
22x
ww
MD
x y
υ
⎛⎞∂∂
∴ =− +⎜⎟
∂∂⎝⎠ (2.5)
Where
()
3
2
Flexural Regidity of Plate
12 1
Eh
D
υ
==
−
and w denotes small deflection of the plate
in z direction. Now in the same manner we can calculate moment in y direction,
22
22
h/ h/
yy y
h/ h/
M .b.z .dz .dy .z .dzσσ
++
−−
==∫∫
(2.6)
Substitute the value of
y
σ from equation (2.3 b) into the equation (2.6) to obtain
y
M, hence
22
22y
ww
MD
yx
υ
⎛⎞∂∂
∴ =− +⎜⎟
∂∂⎝⎠ (2.7)
Now, the shearing stresses
xy
τgive the twisting moment. Therefore twisting moment will be
obtain by making following integration.
2
2
h/
xy xy
h/
M .z.dzτ
+
−
=∫
(2.8)
Substitute the value of
xy
τ from equation (2.3 c) into the equation (2.8) to obtain
xy
M, hence we
get
2 2
2
2
1
h/
xy
h/
Ew
M zdz.
xy
υ
+
−
∂
∴ =−
+∂∂∫

24. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     16 

2
23
2
13
h/
xy
h/
Ewz
M
xy
υ
−
⎡⎤∂
∴ =−
⎢⎥
+∂∂ ⎣⎦

()
()
()
()
()
32
3 2
2
1
12 1 1
1
12 1
xy
xy
Eh w
M
xy
Eh w
M
xy υ
υυ
υ
υ−∂
∴ =−
+−∂∂
− ∂
∴ =−
∂∂−

()
2
1
xy
w
MD
xy
υ
∂
∴ =− −
∂∂
(2.9)
Now taking moments of all forces on the element with respect to x axis and equating to zero, we
obtained the equation of equilibrium
() 0
x
M∑=
()
0
222
yxy
y y xy xy
y x
yxx x,y
MM
M dy dx M dx M dx dy M dy
yx
Q Qdy dy dy
Q dy dx dy Q dy Q dx dy q dx dy
yx
∂∂⎛⎞⎛ ⎞
+−++ −⎜⎟⎜ ⎟
∂∂⎝⎠⎝ ⎠
∂⎛⎞ ∂⎛⎞
−+ + −+ − =⎜⎟ ⎜⎟
∂∂ ⎝⎠⎝⎠

0
yxy
y
MM
dx dy dx dy Q dx dy
yx
∂∂∴ +−=
∂∂
0
yxy
y
MM
Q
yx
∂ ∂
∴ +−=
∂∂ (2.10)
While obtaining above equilibrium equation, the moment due to lateral load q and the moment
due to the rate of change of shear forces viz
y
Q
y
∂
∂
have been neglected since these quantities are
multiplied by terms which are very small.
22
22
yxy
y
MM ww
QD
yx yxy
∂∂ ⎛⎞∂∂ ∂
∴ =+=− + ⎜⎟
∂∂ ∂∂∂ ⎝⎠ (2.11)

25. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     17 

In the same manner, by taking moments with respect to the y axis (0
y
M
∑ =), we get
0
xyx
x
MM
Q
xy
∂∂∴ +−=
∂∂ (2.12)
22
22
xyx
x
MM ww
QD
x yxxy
∂ ⎛⎞∂ ∂∂ ∂
∴ =+=− + ⎜⎟
∂∂ ∂∂∂ ⎝⎠ (2.13)
Considering all the forces acting on the element in the z direction we obtain the following
equation of equilibrium. If
(),xy
q = intensity of distributed external load on the plate surface, then
considering equilibrium of forces in the z direction,
0
z
F
∴ ∑=
()
0
yx
xxy y x,y
QQ
Q dx dy Q dy Q dy dx Q dx q dxdy
xy
∂⎛⎞∂⎛⎞
+−++−+= ⎜⎟⎜⎟
∂∂⎝⎠ ⎝⎠

()
0
yx
x,y
QQ
dx dy dxdy q dxdy
xy
∂∂
++=
∂∂

(),
0
yx
xy
QQ
q
xy
∂∂∴ ++=
∂∂ (2.14)
Since there are no forces in the x and y directions and no moments with respect to the z axis, the
three equations (2.10), (2.12) and (2.14) completely define the equilibrium of the element. Let us
eliminate the shearing forces and
xy
QQ from these equations by determining them from
equations (2.10) and (2.12) and substitute into the equation (2.14). Hence

()
0
xy y xyx
x,y
MMMM
q
xx y y y x
∂∂∂⎡⎤⎡⎤∂∂∂∴ ++ ++=
⎢⎥⎢⎥
∂∂ ∂ ∂ ∂ ∂⎣⎦⎣⎦
()
2222
22
0
xy y xyx
x,y
MMMM
q
xxyyxy
∂∂∂∂
∴ ++++=
∂∂∂∂∂∂

()
222
22
20
xy yx
x,y
MMM
q
xxyy
∂∂∂
∴ +++=
∂∂∂∂
(2.15)

26. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     18 

To represent this equation in terms of the deflections w of the plate, use equations (2.5), (2.7) and
(2.9), we obtain
()
()
222 2 2
222
222
222
21
0
x,y
ww w
DD
xxyxy xy
ww
Dq
yyx
⎡⎤⎛⎞⎡ ⎤∂∂∂ ∂ ∂
∴ −+ +−− +⎢⎥⎜⎟ ⎢⎥
∂∂∂∂∂ ∂∂⎝⎠⎣ ⎦⎣⎦
⎡⎤⎛⎞∂∂∂
− ++=⎢⎥⎜⎟
∂∂∂ ⎝⎠⎣⎦
υυ
υ

()
()
44 44 4
422 224 22
21 0
x,y
ww ww w
Dq
xxy xyy xy
⎡⎤∂∂ ∂∂ ∂
∴ −+ +− ++ +=
⎢⎥
∂∂∂ ∂∂∂ ∂∂⎣⎦ υυ υ

()
444
4224
2
x,y
qwwwx xy y D
∂∂∂
∴ ++=
∂∂∂∂
(2.16)
This latter equation can also be written in the symbolic form
()
2
22
22
x,y
q
wx yD
⎛⎞∂∂
∴ +=⎜⎟
∂∂⎝⎠

()
()
2
2 x,yq
w
D
∴ ∇=
Where
2
22
2
22
Laplacian operator
⎛⎞∂∂
∇= + =⎜⎟
∂∂⎝⎠xy


()
,4 xy
q
w
D
∴ ∇= (2.17)
Which is the Lagrange equilibrium equation for the bending of thin plate with constant
thickness. Therefore the solution of the problem of bending of plates by a lateral load reduces to
the integration of equation (2.17). If, for a particular case, a solution of this equation is found that
satisfies the conditions at the boundaries of the plate, then bending and twisting moments can be
calculated from the equations (2.5), (2.7) and (2.9). The corresponding normal and shearing
stresses are obtained from equation (2.3).

27. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     19 

()
23
6
2/12
xx x
x
M MM h
Z
I hh
σ
⎛⎞
== =
⎜⎟
⎝⎠

()
()
max
2max
6
x
x
M
h
σ
∴ = (2.18)
Similarly
()
()
22
max max
66
and
yxy
yxy
M M
hh
στ== (2.19)
Equations (2.11) and (2.13) are used to determine the shearing forces and
xy
QQ . The shearing
stresses and
xz yz
ττcan now be determined by assuming that they are distributed across the
thickness of the plate according to the parabolic law. Then,
()
()
max max
33
and
22 yx
xz yz
VV
hh
ττ== (2.20)
2.2.2 Associate boundary conditions:
1) Simply supported edge conditions:

Plate boundaries that is prevented from deflecting but free to rotate about a line along the
boundary edges such as hinge is defined as simply supported edge. Along the simply supported
edge the bending moment and deflection would be zero.
The condition on a simply supported edge parallel to y axis at x = a
22
22
0
0
xa
xxa
xa
w
ww
MD
xy
υ
=
=
=
=
⎡⎤∂∂
=− + =
⎢⎥
∂∂⎣⎦
(2.21)
Since the changes of w w.r.t. y co-ordinate vanishes along the edge
2
2
0
xxa
xa
w
MD
x
=
=
⎡⎤∂
=− =
⎢⎥
∂⎣⎦
(2.22)

The condition on a simply supported edge parallel to x axis at y = b

28. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     20 

22
22
0
0
yb
y
yb
yb
w
ww
MD
yx
υ
=
=
=
=
⎡⎤∂∂
=− + =
⎢⎥
∂∂⎣⎦
(2.23)
Since the changes of w w.r.t. x co-ordinate vanishes along the edge
2
2
0
y
ya
yb
w
MD
y
=
=
⎡⎤∂
=− =
⎢⎥
∂⎣⎦
(2.24)
2) Clamped edge conditions:
If a plate is clamped the deflection and slope of the middle surface must vanish at the boundary.
on a clamped edge parallel to y axis at x = a the boundary conditions are
00
xa
xa
w
w
x
=
=
∂
==
∂
(2.25)
on a clamped edge parallel to x axis at y = b the boundary conditions are 00
yb
yb
w
w
y
=
=
∂
==
∂
(2.26)
3) Free edge condition: In the most general case a twisting moment, bending moment and
transverse shear force acts on an edge of the plate. An edge on which all these three stresses
vanishes is defined as free edge.

0
0
xxyx
xa
yyxy
yb
MM Q
MM Q
=
=
== =
== =
(2.27)
Kirchhoff Paradox: Later on Kirchhoff provided that three boundary conditions are too
many and so that two conditions are sufficient for the complete determination of w
satisfy the
equation
4
∇=
q
w
D . Kirchhoff pointed out that the two conditions prescribing and
xy x
M Q can
be replaced by a single one. The reason is that the twisting moment acting on an element of the

29. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     21 

edge of the plate may be replaced by two statically equivalent vertical forces, which can then
combined with the vertical shearing forces.
Note: owing to such replacement the stress distribution in the intermediate neighborhood of the
edge will naturally be affected but the stress distribution in the rest of the plate essentially the
same.


R = concentrated force at the corners or it is also called as corner reaction.
Therefore concentrated force R at corners of rectangular plate supported around the edges in
some manner and under the action of transverse load to prevent middle surface deflection at the
corners.
2
xy
RM∴ = (2.28)
The corner of rectangular plate under the action of uniformly distributed load tends to
rise. This action is prevented by the concentrated reactions at the corners.

2.3 Navier’s Solution for Lateral Deflection of Simply Supported Rectangular Plate:
The solution of preceding article can be used in calculating deflections produced in a
simply supported rectangular plate y any kind of loading given by the equation
(),xy
qq
= .

30. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     22 

Consider a rectangular plate of sides ‘
a’ and ‘b’ which is simply supported on four sides and
subjected to distributed load
q , which is function of two variables x and y can be expressed in
the domain 0 , 0
xayb≤≤ ≤≤ by a double trigonometric series given by
11
sin sin
mn
mn
mx ny
ww
abππ
∞∞
==
⎛⎞⎛⎞
=
⎜⎟⎜⎟
⎝⎠⎝⎠
∑∑ (2.29)
()
11
sin sin where 1 3 5
mnx,y
mn
mx ny
q q m,n , , .........
abππ
∞∞
==
⎛⎞⎛⎞
==
⎜⎟⎜⎟
⎝⎠⎝⎠
∑∑ (2.30)

To calculate any particular coefficient
mn
q of this series, we multiply both sides of equation
(2.30) by
()()sin sinjx/a ky/bππ and integrate with respective x and y from 0 to a and 0 to
b. observing that
()
00 00
sin sin sin sin sin sin
xayb xayb
mnx,y
xy xy
jx ky mx ny jx ky
q dxdy q dxdy
ab ababππ ππππ
== ==
== ==
∴ =∫∫ ∫∫
(2.31)
0
0
Performing integration with respect to , we get sin sin 0 at
Performing integration with respect to , we get sin sin 0 at
xa
x
yb
y
mx jx
x dx m j
aa
ny ky
ydynk
bb
ππ
ππ
=
=
=
=
=≠
=≠∫
∫

at
mj=, performing integration with respect to x, we get
2
00
sin sin sin
xa xa
xx
mx jx mx
dx dx
aa aππ π
==
==
⎛⎞⎛⎞ ⎛⎞
=
⎜⎟⎜⎟ ⎜⎟
⎝⎠⎝⎠ ⎝⎠
∫∫

31. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     23 

0
0
1
1cos2
2
1
sin 2
22
xa
x
xa
x
mx
dx
a
mx a
x
amπ
π
π
=
=
=
=
⎡⎤ ⎛⎞
=+
⎜⎟⎢⎥
⎝⎠⎣⎦
⎡ ⎤⎛⎞
=+
⎜⎟⎢ ⎥
⎝⎠⎣ ⎦
∫

0
sin sin
2
xa
x
mx jx a
dx
aaππ
=
=
⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟
⎝⎠⎝⎠∫
(2.32)
Similarly,atnk=, performing integration with respect to y, we get
0
sin sin
2
ya
y
ny ky b
dy
bbππ
=
=
⎛⎞⎛⎞
=
⎜⎟⎜⎟
⎝⎠⎝⎠
∫
(2.33)
Substitute equation (2.32) and (2.33) into the equation (2.31)
()
00
sin sin
4
xayb
mnx,y
xy
mx ny ab
qdxdyq
abππ
==
==
⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟
⎝⎠⎝⎠∫∫


(),
00
4
sin sin
xayb
mn xy
xy
mx ny
q q dx dy
ab a bππ
==
==
⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟
⎝⎠⎝⎠∫∫
(2.34)
Performing the integration indicated in equation (2.34) for a given load distribution,
i.e., for a
given
()x,y
q, we find the coefficient of series (2.30) and represent in this way the given load as a
sum of partial sinusoidal loadings. Now, Substitute the solution of
w (2.29) and
()x,y
q (2.30) in
the fourth order governing differential equation (2.16).
44 224 44
4224
2
mn
mn
qm mn n mx mx mx mx
w sin sin sin sin
aabb a aDa aπππ ππ ππ⎡⎤ ⎛⎞⎛⎞ ⎛⎞⎛⎞
∴ ++ =
⎜⎟⎜⎟ ⎜⎟⎜⎟⎢⎥
⎝⎠⎝⎠ ⎝⎠⎝⎠⎣⎦
44 224 44
4224
2
mn
mn
q
w
mmnn
D
aabb
π ππ
∴ =
⎡⎤
++
⎢⎥
⎣⎦
(2.35)
Substitute equation (2.35) in the equation (2.29)

32. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     24 

44 224 44
11
4224
sin sin
2
mn
mn
q mx ny
w
abmmnn
D
aabb
π π
πππ
∞∞
==
⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟
⎡⎤ ⎝⎠⎝⎠
++
⎢⎥
⎣⎦∑∑ (2.36)

2.3.1 Simply supported Plate carrying a uniformly distributed load of intensity
0
q
Integrate equation (2.34) w.r.t. x and y to find the value of
mn
q
∴ Performing Integration with respect to x we get
0
2
sin
=
=
⎛⎞
=
⎜⎟
⎝⎠
∫
xa
x
mx a
dx
amπ
π
(2.37)
In the same manner Performing Integration with respect to y we get
0
2
sin
=
=
⎛⎞
=
⎜⎟
⎝⎠
∫
yb
y
ny b
dy
bnπ
π
(2.38)
Substitute equation (2.37) and (2.38) in the equation (2.34) we obtain
42 2
mn
ab
q
ab m nππ
⎡⎤⎡ ⎤
∴ =
⎢⎥⎢ ⎥
⎣⎦⎣ ⎦

0
2
16
for 1 3 5
0for246
mn
q
m, n , , ..........
q mn
m, n , , ..........
π
⎧
= ∞⎪
∴ =⎨
⎪ = ∞⎩
(2.39)
Substitute equation (2.39) in the equation (2.35) to get
0
2 44 224 44
4224
16 1
2
mn
q
w
mn mmnn
D
aabb
π
π ππ
∴ =
⎡ ⎤
++
⎢ ⎥
⎣ ⎦
(2.40)
Substitute magnitude of
mn
w from equation (2.40) into the equation (2.36) to obtain the
deflection of the plate, therefore the expression for the deflected shape of the plate can now be
written as

33. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     25 

0
26
22
11
22
16 1
sin sin
mn
q mx ny
w
mn a b
mn
D
ab
π π
π
∞∞
==
⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟
⎝⎠⎝⎠⎡⎤
+
⎢⎥
⎣⎦∑∑ (2.41)
Equation of Deflection for rectangular plate subjected to uniformly distributed load
In case of uniform load we have a deflection surface symmetrical with respect to axes
/2, /2;xa yb== and quit naturally all terms with even numbers for m or n series (2.41)
vanish, since they are unsymmetrical with respect to the above mentioned axes. The maximum
deflection of plate is at its center and is found by substituting
/2, /2;xa yb
= = in the equation
(2.41).
()
1
0
2
26
22
11
2216 1
1
mn
mn
q
w
mn
mn
D
ab
π
∞∞ +
−
==
∴ =−
⎡⎤
+
⎢⎥
⎣⎦∑∑ (2.42)
This is a rapidly converging series, and a satisfactory approximation is obtained by taking only
the first term of the series, which, for example in the case of square plate gives
()
44
00
6max
4 0.0416qa qa
w
DD
π
∴ == (2.43)
The expressions for bending moments and twisting moments can be obtained from equation
(2.5), (2.7) and (2.9). Since the moments are expressed by second derivatives of series,
differentiate equation (2.41) w.r.t. x and y.

2 22
0
226 2
22
11
22
sin sin
16
mn
mx ny
qwm ab
xmn a
mn
D
ab
∞∞
==
⎛⎞⎛⎞
⎜⎟⎜⎟
⎛⎞∂− ⎝⎠⎝⎠
∴ = ⎜⎟
∂ ⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑
ππ
π
π
(2.44)

3 33
0
236 3
22
11
22
cos sin
16
mn
mx ny
qwm ab
xmn a
mn
D
ab
∞∞
==
⎛⎞⎛⎞
⎜⎟⎜⎟
⎛⎞∂− ⎝⎠⎝⎠
∴ = ⎜⎟
∂ ⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑
ππ
π
π
(2.45)

34. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     26 


2 22
0
226 2
22
11
22
sin sin
16
mn
mx ny
qwn ab
ymn b
mn
D
ab
∞∞
==
⎛⎞⎛⎞
⎜⎟⎜⎟
⎛⎞∂− ⎝⎠⎝⎠
∴ = ⎜⎟
∂ ⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑
ππ
π
π
(2.46)

2 2
0
26
22
11
22
cos cos
16
mn
mx ny
qwmn ab
xymn ab
mn
D
ab
∞∞
==
⎛⎞⎛⎞
⎜⎟⎜⎟
⎛⎞∂ ⎝⎠⎝⎠
∴ = ⎜⎟
∂∂ ⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑
ππ
π
π
(2.47)

3 23
0
226 2
22
11
22
cos sin
16
mn
mx ny
qwmn ab
xy mn ab
mn
D
ab
∞∞
==
⎛⎞⎛⎞
⎜⎟⎜⎟
⎛⎞∂ ⎝⎠⎝⎠
∴ =− ⎜⎟
∂∂ ⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑
ππ
π
π
(2.48)
Substitute derivatives from equations (2.44) and (2.46) into the equation (2.5) to obtain bending
moment in x direction i.e.
x
M
22 22
0
2622
22
11
22
16 1
sin sin
x
mn
q mn mxny
MD
mn a b a b
mn
D
ab ππ π π
υ
π
∞∞
==
⎛⎞ ⎛⎞⎛⎞
∴ =− − − ⎜⎟ ⎜⎟⎜⎟
⎝⎠⎝⎠⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑

22
0
2422
22
11
22
16 1
sin sin
x
mn
q mn mxny
M
mn a b a b
mn
ab ππ
υ
π
∞∞
==
⎛⎞ ⎛⎞⎛⎞
∴ =+ ⎜⎟ ⎜⎟⎜⎟
⎝⎠⎝⎠⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑ (2.49)
Similarly Substitute derivatives from equations (2.44) and (2.46) into the equation (2.7) to obtain
bending moment in y direction i.e.
y
M
22
0
2422
22
11
22
16 1
sin sin
y
mn
q nm mxny
M
mn b a a b
mn
ab
π π
υ
π
∞∞
==
⎛⎞ ⎛⎞⎛⎞
∴ =+ ⎜⎟ ⎜⎟⎜⎟
⎝⎠⎝⎠⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑ (2.50)
It is seen that maximum bending moments accurse at the centre of plate. Substituting
/2
xa=

35. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     27 

and / 2yb= in the equations (2.49) and (2.50) we get

()
22
0
2422max
22
11
22
16 1
x
mn
q mn
M
mn a b
mn
ab
υ
π
∞∞
==
⎛⎞
∴ =+ ⎜⎟
⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑ (2.51)

()
22
0
2422
max 22
11
22
16 1
y
mn
q nm
M
mn b a
mn
ab
υ
π
∞∞
==
⎛⎞
∴ =+ ⎜⎟
⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑ (2.52)
Twisting moment
from equation (2.9)

()
2
0
26
22
11
22
cos cos
16
1
xy
mn
mx ny
q mnab
MD
mn ab
mn
D
ab
∞∞
==
⎛⎞⎛⎞
⎜⎟⎜⎟
⎛⎞⎝⎠⎝⎠
∴ =− − ⎜⎟
⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦ ∑∑
ππ
π
υ
π
()
0
24
22
11
22
16 1
1coscos
xy
mn
q mx ny
M
ab a b
mn
ab ππ
υ
π
∞∞
==
⎛⎞⎛⎞
∴ =− −
⎜⎟⎜⎟
⎝⎠⎝⎠⎡⎤
+
⎢⎥
⎣⎦ ∑∑ (2.53)
The expression for vertical shear forces can be obtained from equation (2.11) and (2.13)

xy
xx
M
QV
y
∂
∴ =+
∂
And
xy
yy
M
QV
x
∂
=+
∂
(2.54)
In which and
xy
VV are the reactive forces at the supported edges of the plate. Therefore the
expressions for resultant shears or reactive forces at the supported edges of the plate can be
obtained from following equations. For the edges x = a we find

()
33
32
2
xy
xx
xa
M ww
VQ D
yxxy
υ
=
∂⎛⎞
⎡ ⎤∂∂
=− =− +−⎜⎟ ⎢ ⎥
∂∂∂∂ ⎣ ⎦⎝⎠
(2.55)
In the same manner reactive force for the edge y = b,

()
33
32
2
xy
yy
yb
M ww
VQ D
x yxy
υ
=
∂⎛⎞ ⎡ ⎤∂∂
=− =− +−⎜⎟ ⎢ ⎥
∂∂∂∂ ⎣ ⎦⎝⎠
(2.56)

36. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     28 

Therefore reactive force in x direction is given by substituting equation (2.45) and (2.48) into the
equation (2.55), we obtain
()
33
0
263
22
11
22
23
0
262
22
11
22
16 1
cos sin
16 1
2cossin
mn
x
mn
q mx ny m
mn a b a
mn
D
ab
VD
q mx ny mn
mn a b ab
mn
D
ab πππ
π
πππ
υ
π
∞∞
==
∞∞
==
⎧⎫ ⎛⎞⎛⎞⎛⎞
−⎪⎪ ⎜⎟⎜⎟⎜⎟
⎝⎠⎝⎠⎡⎤ ⎝⎠⎪⎪
+
⎢⎥⎪⎪
⎪⎣⎦ ⎪
∴ =−⎨⎬
⎛⎞⎛⎞⎛⎞⎪⎪
−− ⎜⎟⎜⎟⎜⎟⎪⎪
⎝⎠⎝⎠⎡⎤ ⎝⎠
⎪⎪ +
⎢⎥
⎪⎪
⎣⎦⎩⎭
∑∑
∑∑

()
32
0
2332
22
11
22
16 1
2cossin
x
mn
q mmnmxny
V
mn a ab a b
mn
ab
π π
υ
π
∞∞
==
⎛⎞ ⎛⎞⎛⎞
∴ =− + − ⎜⎟ ⎜⎟⎜⎟
⎝⎠⎝⎠⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑ (2.57)
()
()
32
0
2332
22
11
22
16 1
2sin
xxa
mn
q mmnny
V
mn a ab b
mn
ab π
υ
π
∞∞
=
==
⎛⎞ ⎛⎞
∴ =− + − ⎜⎟ ⎜⎟
⎝⎠⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑ (2.58)
Similarly, reactive force in y direction is given by using equation (2.56)

()
32
0
2332
22
11
22
16 1
2sincos
y
mn
q nmnmxny
V
mn b a b a b
mn
ab
π π
υ
π
∞∞
==
⎛⎞ ⎛⎞⎛⎞
∴ =− + − ⎜⎟ ⎜⎟⎜⎟
⎝⎠⎝⎠⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑ (2.59)

()
()
32
0
2332
22
11
22
16 1
2sin
y
yb
mn
q nmnmx
V
mn b a b a
mn
ab
π
υ
π
∞∞
=
==
⎛⎞ ⎛⎞
∴ =− + − ⎜⎟ ⎜⎟
⎝⎠⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑ (2.60)
The minus sign indicates that the reactions on the plate act upward. From symmetry it may be
concluded that equations (2.58) and (2.60) also represents pressure distribution along the
sides 0 and 0xy== , respectively.
From equation (2.28) the concentrated reaction at each corner can be determined as

37. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     29 

()
0
24
22
11
22
32 1
1coscos
mn
q mx ny
R
ab a b
mn
ab
π π
υ
π
∞∞
==
⎛⎞⎛⎞
∴ =−
⎜⎟⎜⎟
⎝⎠⎝⎠⎡⎤
+
⎢⎥
⎣⎦ ∑∑ (2.61)
∴ Total downward force = 4R
()
0
24
22
11
22
128 1
41 cos cos
mn
q mx ny
R
ab a b
mn
ab
∞∞
==
⎛⎞⎛⎞
∴ =−
⎜⎟⎜⎟
⎝⎠⎝⎠⎡⎤
+
⎢⎥
⎣⎦ ∑∑
π π
υ
π
(2.62)
Now bending stresses in x and y directions are determined using equation (2.18), therefore
bending stress in x direction is given by
22
0
224 2 2
22
11
22
96 1
sin sin
x
mn
q mn mxny
hmn a b a b
mn
ab ππ
συ
π
∞∞
==
⎛⎞ ⎛⎞⎛⎞
∴ =+ ⎜⎟ ⎜⎟⎜⎟
⎝⎠⎝⎠⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑ (2.63)
The maximum bending stress is at the center of the plate i.e./2and /2
xayb= = , hence the
maximum bending stress in x direction is
()
22
0
224 2 2max
22
11
22
96 1
x
mn
q mn
hmn a b
mn
ab
συ
π
∞∞
==
⎛⎞
∴ =+ ⎜⎟
⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑ (2.64)
Similarly, Bending stress in y direction

22
0
224 2 2
22
11
22
96 1
sin sin
y
mn
q nm mxny
hmn b a a b
mn
ab ππ
συ
π
∞∞
==
⎛⎞ ⎛⎞⎛⎞
∴ =+ ⎜⎟ ⎜⎟⎜⎟
⎝⎠⎝⎠⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑ (2.65)
The maximum bending stress is at the center of the plate i.e./2and /2
xayb= = , hence the
maximum bending stress in x direction is
()
22
0
224 2 2
max 22
11
22
96 1
y
mn
q nm
hmn b a
mn
ab
συ
π
∞∞
==
⎛⎞
∴ =+ ⎜⎟
⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑ (2.66)

38. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     30 

The maximum shearing stress will be at the middle of the sides of the plate. Observing that the
total transverse force and
xy
VV is distributed along the thickness of plate according to parabolic
law and using equation (2.58) and (2.60), we obtain

()
()
32
0
2332max
22
11
22
24 1
2
xz
mn
q mmn
mn a ab
mn
ab
τυ
π
∞∞
==
⎛⎞
∴ =+− ⎜⎟
⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑ (2.67)

()
()
32
0
2332
max 22
11
22
24 1
2
yz
mn
q nmn
mn b a b
mn
ab
τυ
π
∞∞
==
⎛⎞
∴ =+− ⎜⎟
⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦∑∑ (2.68)
2.3.2 Simply supported Plate carrying Sinusoidal load:

Taking coordinate axes as shown in figure, we assume that the load distributed over the surface
of the plate is given by the expression

()
0
sin sin
x y
qx,y q
ab
π π⎛⎞⎛⎞
=
⎜⎟⎜⎟
⎝⎠⎝⎠
(2.69)

In which
0
q represents the intensity of load at the centre of plate as shown in figure. Substituting
equation (2.69) in the equation (2.34) and after integration we get Fourier coefficient

0
for 1 1
0for11
mn
qmn
q
mn
==⎧
=⎨
〉〉⎩
(2.70)

39. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     31 

The boundary conditions for simply supported edges are

00for0and
00for0and
x
y
wM xxa
wM yyb
====
====
(2.71)
It may seen from the boundary conditions are satisfied if we take for deflection the expression,

1
sin sin
x y
ww
ab
π π⎛⎞⎛⎞
=
⎜⎟⎜⎟
⎝⎠⎝⎠
(2.72)
Substitute
0mn
qq= in the equation (2.35) we get

0
1 2
4
22
11
q
w
D
ab
π
∴ =
⎡ ⎤
+
⎢ ⎥
⎣ ⎦
(2.73)
Substitute equation (2.73) in the equation (2.36) and we conclude that the deflection surface
satisfying governing differential equation and boundary conditions.

0
2
4
22
sin sin
11
q
x y
w
ab
D
abππ
π⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟
⎝⎠⎝⎠⎡⎤
+
⎢⎥
⎣⎦ (2.74)
The expressions for bending moments and twisting moments can be obtained from equation
(2.5), (2.7) and (2.9). Differentiate equation (2.74) w.r.t.
x and y we get
0
2 22
2
22 11
sin sin
11
x
q
x y
M
ab a b
ab
π π
υ
π⎛⎞⎛⎞⎛⎞
∴ =+
⎜⎟⎜⎟⎜⎟
⎝⎠⎝⎠⎝⎠⎡⎤
+
⎢⎥
⎣⎦ (2.75)
0
2 22
2
22 11
sin sin
11
y
q
x y
M
ba a b
abππ
υ
π⎛⎞⎛⎞⎛⎞
∴ =+
⎜⎟⎜⎟⎜⎟
⎝⎠⎝⎠⎝⎠⎡⎤
+
⎢⎥
⎣⎦ (2.76)
()
0
2
2
22
1coscos
11
xy
q
x y
M
ab
abππ
υ
π⎛⎞ ⎛⎞
∴ =− −
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠⎡⎤
+
⎢⎥
⎣⎦ (2.77)

40. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     32 

It is seen that the maximum deflection and the maximum bending moments are at the centre of
plate. Substituting / 2 and / 2xayb== in the equations (2.75), (2.76) and (2.77), we obtain

()
0
2max
4
22
11
q
w
D
ab
π
∴ =
⎡ ⎤
+
⎢ ⎥
⎣ ⎦
(2.78)
()
0
2 22max
2
22 1
11
x
q
M
ab
ab
υ
π
⎛⎞
∴ =+
⎜⎟
⎝⎠⎡⎤
+
⎢⎥
⎣⎦ (2.79)
()
0
2 22
max
2
22 1
11
y
q
M
ba
ab
υ
π
⎛⎞
∴ =+
⎜⎟
⎝⎠⎡⎤
+
⎢⎥
⎣⎦ (2.80)
In the particular cases of a square plate,
a = b, and the forgiving formulae becomes
()
4
0
4max
4
qa
w
D
π
∴ = () ()
()
2
0
2max max
1
4
xy
qa
MMυ
π+
==
(2.81)
We use equations (2.11) and (2.13) to calculate the shearing forces and obtained
0
2
22
cos sin
11
x
q
x y
Q
ab
a
abππ
π⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟
⎝⎠⎝⎠⎡⎤
+
⎢⎥
⎣⎦ (2.82)
0
2
22
cos sin
11
y
q
x y
Q
ab
b
abππ
π⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟
⎝⎠⎝⎠⎡⎤
+
⎢⎥
⎣⎦ (2.83)
Therefore the expressions for resultant shears or reactive forces at the supported edges of the
plate can be obtained from the equations (2.11) and (2.13). ()
0
2 32
22 11
2cossin
11xy
xx
M q
x y
VQ
yaabab
ab
π π
υ
π∂⎛⎞ − ⎛⎞⎛⎞⎛⎞
∴ =− = +−⎜⎟ ⎜⎟⎜⎟⎜⎟
∂ ⎝⎠⎝⎠⎝⎠⎡⎤⎝⎠
+
⎢⎥
⎣⎦ (2.84)
For the edges
x = a we find

41. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     33 


()
()
0
2 32
22
21
sin
11
xxa
q y
V
aab b
ab υ π
π
=
−⎛⎞ ⎛⎞
∴ =− + ⎜⎟ ⎜⎟
⎝⎠⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦ (2.85)
Similarly, ()
0
2 32
22 11
2sincos
11xy
yy
M q
x y
VQ
x babab
ab
ππ
υ
π∂⎛⎞ ⎛⎞⎛⎞⎛⎞
∴ =− =− +−⎜⎟ ⎜⎟⎜⎟⎜⎟
∂ ⎝⎠⎝⎠⎝⎠⎡⎤⎝⎠
+
⎢⎥
⎣⎦ (2.86)
For the edges
y = b we find
()
()
0
2 32
22
21
sin
11
y
yb
q
x
V
bab a
ab
υ
π
π
=
−⎛⎞ ⎛⎞
∴ =− + ⎜⎟ ⎜⎟
⎝⎠⎡⎤ ⎝⎠
+
⎢⎥
⎣⎦ (2.87)
Hence the pressure distribution follows the sinusoidal law. The minus sign indicates that the
reactions on the plate act upward. From symmetry it may be concluded that equations (2.85) and
(2.87) also represents pressure distribution along the sides 0 and 0
xy
= =, respectively.
Now bending stresses in x and y directions are determined
0
2222
2
22611
sin sin
11
x
q
x y
habab
ab
π π
συ
π⎛⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞
∴ =+
⎜⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎡⎤
+
⎢⎥
⎣⎦ (2.88)
0
2222
2
22611
sin sin
11
y
q
x y
hbaab
ab
π π
συ
π⎛⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞
∴ =+
⎜⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎡⎤
+
⎢⎥
⎣⎦ (2.89)
The maximum bending stress is at the centre of the plate. Hence substitute / 2 and / 2
xayb==
To fine the maximum bending stress.
()
0
2222max
2
2261
11
x
q
hab
ab
υ
σ
π
⎛⎞ ⎛ ⎞
∴ =+
⎜⎟ ⎜ ⎟
⎝⎠ ⎝ ⎠⎡⎤
+
⎢⎥
⎣⎦ (2.90)

42. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     34 

0
2222
2
2261
11
y
q
hba
ab
υ
σ
π
⎛⎞ ⎛ ⎞
∴ =+
⎜⎟ ⎜ ⎟
⎝⎠ ⎝ ⎠⎡⎤
+
⎢⎥
⎣⎦ (2.91)

2.3.3 Simply supported Plate carrying a patch load:
Consider rectangular plate of size
×ab is simply supported as shown in figure. The plate
is loaded with patch load of size
uv× with intensity
2
0
q/m.

The centre of the patch load will be at a distance
0
x from the origin in the x direction and
0
y
from the origin in the
y direction. Patch load is acting over the area
()uv, therefore the intensity
of the load is given by.
()
x,y
P
q
uv
=
Where P = Total load
The Fourier coefficient
mn
q for this case can be determined from equation (2.34). Therefore
substitute value of
(),xy
qin the equation (2.34) and integrate from
( )
0
/2xu− to()
0
/2xu+ in x
direction and
()
0
/2yv− to()
0
/2yv+ in y direction.

43. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     35 

00
00
22
22
4
sin sin
xu/ yv/
mn
xu/ yv/
Pmxny
qdxdy
abuv a b ππ
++
−−
⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟
⎝⎠⎝⎠∫∫
(2.92)
Performing integrate with respect to x over the limit
( )
0
/2xu− to( )
0
/2xu+ , hence we obtain
00
0 0
22
2 2
sin cos
xu/xu/
xu/ xu/
mx a mx
dx
amaππ
π
++
− −
⎡⎤⎛⎞ ⎛⎞
=−
⎜⎟ ⎜⎟ ⎢⎥
⎝⎠ ⎝⎠ ⎣⎦
∫

00
0
cos cos
22
2
2
mx mxamumu
maaaa
mxamu
sin sin
maa
ππ ππ
π
π π
π
⎡ ⎤⎛⎞⎛⎞
=− + − −
⎜⎟⎜⎟⎢ ⎥
⎝⎠⎝⎠⎣ ⎦
⎡⎤⎛⎞ ⎛⎞
=− −
⎜⎟⎜⎟⎢⎥
⎝⎠⎝⎠⎣⎦

0
0
2
0
2
2
sin
2
xu/
xu/
mxmx a mu
dx sin sin
am a aπππ
π
+
−
⎡ ⎤⎛⎞⎛⎞ ⎛⎞
∴ =
⎜⎟ ⎜⎟ ⎜⎟⎢ ⎥
⎝⎠ ⎝⎠ ⎝⎠⎣ ⎦
∫
(2.93)
Similarly performing integrate with respect to
y over the limit
( )
0
/2yv−to()
0
/2yv+, we obtain
0
0
2
0
2
2
sin
2
yv/
yv/
nyny b nv
dy sin sin
bnb bπππ
π
+
−
⎡ ⎤⎛⎞⎛⎞ ⎛⎞
∴ =
⎜⎟ ⎜⎟ ⎜⎟⎢ ⎥
⎝⎠ ⎝⎠ ⎝⎠⎣ ⎦
∫
(2.94)
Therefore substitute equations (2.93) and (2.94) into the equation (2.34) to determined value of
coefficient
mn
q
0042 2
22
mn
mx nyPa mu b nv
q sin sin sin sin
abuv m a a n b b
⎡⎤⎡⎤⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞
∴ =
⎜⎟ ⎜⎟⎜⎟ ⎜⎟⎢⎥⎢⎥
⎝⎠ ⎝⎠⎝⎠ ⎝⎠⎣⎦⎣⎦
ππ ππ
ππ

Hence the Fourier coefficient
mn
q for a patch load can be obtained as
00
216
22
mn
mx nyPmunv
q sin sin sin sin
mnu v a b a bππ ππ
π⎡⎤⎛⎞⎛⎞ ⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟⎜⎟⎜⎟⎢⎥
⎝⎠⎝⎠⎝⎠⎝⎠⎣⎦ (2.95)
If patch load is acting at the centre of the plate,
i.e.
00
2and 2
xa/ y b/= = , put into the equation
(2.95), we obtain

44. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     36 

2
16
sin sin
22
mn
Pmunv
q
mnu v a b
⎡⎤⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟⎢⎥
⎝⎠⎝⎠⎣⎦
ππ
π
(2.96)
If and
ua vb== , equation (2.96) becomes
2
16
mn
P
q
mn ab
π
= which is similar to Fourier
coefficient obtained for uniformly distributed load. Substitute the value of
mn
q in the above
equation of
mn
w (2.35) we get
2 44 224 44
11
4224
16 1
sin sin
22
2
mn
mn
Pmunv
w
mnu v a bmmnn
D
aabb
π π
π πππ
∞∞
==
⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟
⎡⎤ ⎝⎠⎝⎠
++
⎢⎥
⎣⎦∑∑ (2.97)
Substitute magnitude of
mn
w from equation (2.97) into the equation (2.36) to determined
deflection
w of the plate
2 44 224 44
11
4224
sin sin
16 22
sin sin
2
mn
mu nv
Pmxny ab
w
mnu v a bmmnn
D
aabbππ
ππ
π πππ
∞∞
==
⎛⎞⎛⎞
⎜⎟⎜⎟
⎛⎞⎛⎞⎝⎠⎝⎠
∴=
⎜⎟⎜⎟
⎡⎤ ⎝⎠⎝⎠
++
⎢⎥
⎣⎦∑∑ (2.98)
The series converge rapidly, and we can obtain the deflection at any point on the plate with
sufficient accuracy by taking only first few terms of the series. Let us, for example, calculate the
maximum deflection,
i.e. when patch load at the center of the plate i.e. 2and 2
xa/ y b/= =
()
6 4224
11
4224
16 1
sin sin
22
2
max
mn
Pmunv
w
mnu v a bmmnn
D
aabb
π π
π
∞∞
==
⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟
⎡⎤ ⎝⎠⎝⎠
++
⎢⎥
⎣⎦∑∑ (2.99)

45. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     37 

2.3.4 Simply supported plate carrying a concentrated load:

Fourier coefficient for the plate carrying concentrated load can be determined from equation
(2.95)
by making some mathematical adjustment. Hence using following mathematical
adjustment in the equation (2.95), we obtain
00
216 22
22
22
mn
mu nv
sin sin
mx nyPmunv ab
qsinsin
mu nvmnu v a b a b
ab
⎡⎤ ⎛⎞ ⎛⎞
⎜⎟ ⎜⎟⎢⎥
⎛⎞⎛⎞ ⎛⎞ ⎛⎞⎝⎠ ⎝⎠
⎢⎥
∴ =
⎜⎟ ⎜⎟⎜⎟⎜⎟
⎛⎞ ⎛⎞⎝⎠ ⎝⎠⎝⎠⎝⎠⎢⎥
⎜⎟ ⎜⎟
⎢⎥
⎝⎠ ⎝⎠⎣⎦
ππ
ππ ππ
πππ


00
4 22
22
mn
mu nv
sin sin
mx nyP ab
qsinsin
mu nvab a b
ab
⎡⎤ ⎛⎞⎛⎞
⎜⎟⎜⎟⎢⎥
⎛⎞⎛⎞ ⎝⎠⎝⎠
⎢⎥
∴ =
⎜⎟⎜⎟
⎛⎞⎛⎞⎝⎠⎝⎠⎢⎥
⎜⎟⎜⎟
⎢⎥
⎝⎠⎝⎠⎣⎦
ππ
ππ
ππ

Hence after simplification the Fourier coefficient
mn
q for a concentrated load can be obtained as

004
sin sin
mn
mx nyP
q
ab a bππ⎡⎤⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟⎢⎥
⎝⎠⎝⎠⎣⎦ …………
0
1
sin
lim
θ
θ
θ
→
⎛⎞
=
⎜⎟
⎝⎠
(2.100)
Substitute the value of Fourier coefficient
mn
q in the above equation (2.35) to determine value of
constant
mn
w. Hence we get

46. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     38 

00
44 224 44
11
4224
41
2
mn
mn
mx nyP
wsinsin
ab a bmmnn
D
aabbππ
πππ
∞∞
==
⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟
⎡⎤ ⎝⎠⎝⎠
++
⎢⎥
⎣⎦∑∑ (2.101)
Substitute magnitude of
mn
w from equation (2.101) into the equation (2.31) to determined
deflection
w of the plate.
00
44 224 44
11
4224
sin sin
4
sin sin
2
mn
mx ny
Pmxny ab
w
ab a bmmnn
D
aabbππ
ππ
πππ
∞∞
==
⎛⎞⎛⎞
⎜⎟⎜⎟
⎛⎞⎛⎞⎝⎠⎝⎠
∴ =
⎜⎟⎜⎟
⎡⎤ ⎝⎠⎝⎠
++
⎢⎥
⎣⎦∑∑ (2.102)
If concentrated load is acting at the centre of the plate, put
00
2and 2
xa/ y b/= = in the above
equation, we get
44 224 44
11
4224
41
sin sin
2
mn
Pmxny
w
ab a bmmnn
D
aabb ππ
πππ
∞∞
==
⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟
⎡⎤ ⎝⎠⎝⎠
++
⎢⎥
⎣⎦∑∑ (2.103)
Deflection is maximum at the centre of plate i.e. at
2and 2
xa/ y b/= = , hence equation
(2.103) becomes

max 4 4224
11
4224
41
2
mn
P
w
ab mmnn
D
aabb
π
∞∞
==
∴ =
⎡ ⎤
++
⎢ ⎥
⎣ ⎦
∑∑ (2.104)
This is a rapidly converging series, and a satisfactory approximation is obtained by taking only
the first term of the series, which, for example in the case of square plate gives

22
max 4
0.0112Pa Pa
w
DD
π
∴ == (2.105)

47. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     39 

2.3.5 Simply supported plate carrying linearly varying load (hydrostatic pressure):
In this case let assumed simply supported rectangular plate subjected to linearly varying
load as shown in figure. Intensity of loading is zero at
0x
= and maximum at xa=i.e.
0
q.

But for linearly varying load the value of
(),xy
qat any point at a distance of x is given by the
expression
()
0
x,y
qx
q
a
=

The Fourier coefficient
mn
q for this case can be determined by substituting value of
(),xy
q into the
equation (2.34)
0
2
00
4
sin sin
xa yb
mn
xy
q mx ny
q x dx dy
ab a b ππ
==
==
⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟
⎝⎠⎝⎠∫∫
(2.106)
After performing integration with respect to x and
y in the equation (2.106), we get
2
0
sin
xa
x
mx a
xdx
amπ
π
=
=
⎛⎞
=
⎜⎟
⎝⎠
∫
And
0
2
sin
yb
y
ny b
dy
bnπ
π
=
=
⎛⎞
=
⎜⎟
⎝⎠
∫
(2.107)
Substitute values of integration from equation (2.107) into the equation (2.106) to obtain the
Fourier coefficient
mn
q.
2
0
2
4 2
mn
qab
q
ab m n
ππ
⎛⎞⎛⎞
∴ = ⎜⎟⎜⎟
⎝⎠⎝⎠

48. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     40 

0
2
8
mn
q
q
mn
π
∴ = (2.108)
Substitute magnitude of Fourier coefficient
mn
q from equation (2.108) in the equation (2.35) to
obtain constant of deflection
mn
w.
0
2 44 224 44
11
4224
8 1
2
mn
mn
q
w
mn mmnn
D
aabb
π πππ
∞∞
==
∴ =
⎡ ⎤
++
⎢ ⎥
⎣ ⎦
∑∑ (2.109)
Put magnitude of
mn
w into the equation (2.36) to obtained expression for deflection of the plate.
0
2 44 224 44
11
4224
8 1
sin sin
2
mn
q mx ny
w
mn a bmmnn
D
aabb ππ
π πππ
∞∞
==
⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟
⎡⎤ ⎝⎠⎝⎠
++
⎢⎥
⎣⎦∑∑
0
26
22
11
22
8 1
sin sin
mn
q mx ny
w
Dmn a b
mn
ab ππ
π
∞∞
==
⎛⎞⎛⎞
∴ =
⎜⎟⎜⎟
⎝⎠⎝⎠⎡⎤
+
⎢⎥
⎣⎦∑∑ (2.110)
Deflection is maximum at the center of the plate, therefore substitute 2 and 2
xa/ y b/= = in the
above equation, hence we get
()
0
26max
22
11
22
8 1
mn
q
w
Dmn
mn
ab
π
∞∞
==
∴ =
⎡ ⎤
+
⎢ ⎥
⎣ ⎦
∑∑ (2.111)
This is a rapidly converging series, and a satisfactory approximation is obtained by taking only
the first term of the series, which, for example in the case of square plate gives

()
4
0
6max
2qa
w
D
π
∴ = (2.112)

49. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     41 

2.4 Levy’s Solution for the Lateral Deflection of Rectangular Plates:
The Navier’s solution is very straight forward; it applies only to the limited category of
simply supported rectangular plate. A more general technique which yields the lateral deflection
of plate with boundary conditions other than simply supported was developed by Levy.
2.4.1 Levy’s solution for rectangular plate with at least two opposite edges simply
supported carrying a uniformly distributed load.
Assumptions:
1. M Levy assumed that two opposite edges are simply supported and other two edges with
arbitrary supports or any type of supports.
2.
It is assumed that the sides x = 0 and x = a are simply supported.
3.
It is assumed that the whole load q is shared along x direction producing deflection
()
1
wx
4.
hence load along y direction is zero and deflection in y direction is given by ()
2
wx,y
5.
This method used single trigonometric series

In figure a plate
ab× bounded by x-y co-ordinates. The origin is o is taken at the midpoint of the
side
b. the boundary condition at 0 and
x xa= = are simply supported whereas those at
/2
yb=± May arbitrary. A typical plate strip of unit width AB spanning along x is assumed to
behave as a simply supported beam of span ‘a’ with the difference that the flexural rigidity of

50. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     42 

this strip is D not EI. The applied load (),qqxy= is considered as prismatic i.e. along the x
direction, the variation of
q may be of any y direction is constant.
Thus, the solution
(),wwxy=of the plate can be taken as sum of complimentary
solution
()
22
,wwxy= and particular integral
()
11
wwx= .
()()
12
,wwxwxy∴ =+ (2.113)
Solution for
()
1
wx
For isotropic plate the particular solution is function of x only. Since it is solution of any
arbitrary plate strip AB as a beam satisfying the differential equation (2.16)
Since
()
1
wx is a function of x only derivative with respective y vanishes from the equation.
()
()
4
1
1
4
x
qwxx D
∂
∴ =
∂ (2.114)
Where D is the flexural rigidity of the plate
It is not essential that particular solution has to satisfy all the boundary conditions at the four
edges. Here it satisfy all the boundary conditions at the two edges
viz at x = a and x = a both of
which simply supported, i.e.,
1
2
1
2
0at0and
0at0andwxxa
w
x xa
x
===
∂
===
∂
(2.115)
The expression for
1
w is obtained from equation (2.114). Let the solution of
11
andwq can be
expressed interms of sine series
1
1
sin
m
m
mx
ww
a
π
∞
=
⎛⎞
=
⎜⎟
⎝⎠
∑ (2.116)
And
1
1
sin
m
m
mx
qq
a
π
∞
=
⎛⎞
=
⎜⎟
⎝⎠
∑ (2.117)

51. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     43 

Substitute equation (2.116) and (2.117) in the equation (2.114) to obtained the constant
m
w
44
4
11
1
sin sin
mm
mm
mx m mx
wq
aa D a
πππ
∞∞
==
⎛⎞ ⎛⎞
∴ =
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠∑∑
4
44
1
m
m
m
qa
w
mD
π
∞
=
∴ =∑ (2.118)
To determine the value of
m
q, multiply equation (2.117) by ( )sin /jxaπ both sides we get
1
00
sin sin sin
xa xa
m
xx
jx jx mx
qdxq dx
aaaπππ
==
==
⎛⎞ ⎛⎞⎛ ⎞
=
⎜⎟ ⎜⎟⎜ ⎟
⎝⎠ ⎝⎠⎝ ⎠
∫∫
(2.119)
Performing integration at
jm≠, we get
0
sin sin 0
xa
x
jx mx
dx
aaππ
=
=
⎛⎞⎛ ⎞
=
⎜⎟⎜ ⎟
⎝⎠⎝ ⎠
∫

Therefore performing integration at
jm
=
2
00 0 1
sin sin sin sin 2
222
xaxa xa
xx x
jx mx mx mx a a
dx dx x
aa a amππ π π
π
===
== =
⎡⎤⎛⎞⎛⎞ ⎛⎞ ⎛ ⎞
==+ =
⎜⎟⎜⎟ ⎜⎟ ⎜ ⎟ ⎢⎥
⎝⎠⎝⎠ ⎝⎠ ⎝ ⎠ ⎣⎦
∫∫
(2.120)
Put magnitude of above integration from equation (2.120) into the equation (2.119) to obtain the
value of Fourier coefficient
m
q. Hence
1
0
sin
2
xa
m
x
qamx
qdx
aπ
=
=
⎛⎞
∴ =
⎜⎟
⎝⎠∫

1
0
2
sin
xa
m
x
mx
qq dx
aaπ
=
=
⎛⎞
∴ =
⎜⎟
⎝⎠∫
(2.121)
For the constant intensity of loading
()
()10 , xy
qx q q= =

0
4
m
q
q
mπ
∴ = (2.122)
Substitute (2.122) into the equation (2.118) we get

52. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     44 

4
0
55
1
4
m
m
qa
w
mD
π
∞
=
∴ =∑ (2.123)
Put equation (2.123) into the equation (2.116) to obtained
1
()wx
4
0
1 55
1
4
sin
m
qa mx
w
mD a
π
π
∞
=
⎛⎞
∴ =
⎜⎟
⎝⎠∑
4
0
1 55
1
4 1
sin
m
qa mx
w
Dm a
π
π
∞
=
⎛⎞
∴ =
⎜⎟
⎝⎠∑ (2.124)
This should satisfy boundary conditions and governing equation of plate.
Solution for ()
2
wx,
y
Let assume the solution of
()
2
wx,y which is product of two functions, one is function of y i.e.
m
Yand second is function of x i.e.
( )sin /mxaπ
()
2
1
,sin
m
m
mx
wxy Y
a
π
∞
=
⎛⎞
=
⎜⎟
⎝⎠
∑ (2.125)
This should satisfy boundary condition at / 2yb
=± and plate equation.
Differentiate equation (2.125) with respective
x and y we get
4 44
2
44
1
422
II2
22 2
1
4
IV2
4
1
sin
sin
sin
m
m
m
m
m
m
wm mx
Y
xa a
w mmx
Y
xy a a
w mx
Y
ya ππ
ππ
π
∞
=
∞
=
∞
=
⎤∂ ⎛⎞
=
⎥⎜⎟
∂ ⎝⎠
⎥
⎥
∂ ⎛⎞
⎥=
⎜⎟
∂∂ ⎝⎠ ⎥
⎥
∂ ⎛⎞
⎥=
⎜⎟
⎥∂ ⎝⎠ ⎦
∑
∑
∑
(2.126)
Substitute in the governing differential equation (2.16) we obtained
44 22
II IV 0
42
111
sin 2 sin sin
mmm
mmm
qm mx m mx mx
YYY
aaaa aDππππ π
∞∞∞
===
⎛⎞ ⎛⎞ ⎛⎞
−+=
⎜⎟ ⎜⎟ ⎜⎟
⎝⎠ ⎝⎠ ⎝⎠
∑∑∑

53. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     45 

22 44
IV II 0
24
1
2sin
mmm
m
qmm mx
YYY
aa aDππ π
∞
=
⎛⎞ ⎛⎞
∴ −+ =⎜⎟ ⎜⎟
⎝⎠⎝⎠∑ (2.127)
The general solution for the above fourth order differential equation is taken in the form as
..
c
yy PI
=+
4
cosh sinh sinh
cosh
mm m
m
m
my my my my
AB C
aaa aqa
Y
D my my
D
aaπππ π
ππ⎡⎤ ⎛⎞ ⎛⎞ ⎛⎞
++
⎜⎟ ⎜⎟ ⎜⎟⎢⎥
⎝⎠ ⎝⎠ ⎝⎠
⎢⎥
∴ =
⎢⎥ ⎛⎞
+
⎢⎥ ⎜⎟
⎝⎠⎣⎦ (2.128)
Since the loading is considered prismatic along y, identical boundary conditions at / 2 yb
=±
renders the problem symmetric about x axis. In that case the constant
mm
C and D will vanish
().. 0
mm
ie C D== in the equation (2.83) and for antisymmetric problems in which loading on
the plate maintains antisymmetry about
x- axis even if the boundary conditions at /2yb=± are
identical therefore
mm
A and B Will vanishes
( ).. 0
mm
ie A B==.
Therefore
For the symmetric problems:
4
cosh sinh
mm m
qa my my my
YA B
Daaa πππ⎡⎤ ⎛⎞ ⎛⎞
=+
⎜⎟ ⎜⎟⎢⎥
⎝⎠ ⎝⎠⎣⎦
(2.129)
And for the antisymmetric problems:
4
sinh cosh
mm m
qa my my my
YC D
Daaa πππ⎡⎤ ⎛⎞ ⎛⎞
=+
⎜⎟ ⎜⎟⎢⎥
⎝⎠ ⎝⎠⎣⎦
(2.130)
Therefore solution of symmetric problems is given by substituting equation (2.129) in the
equation (2.125) we get
()
4
2
1
,coshsinhsin
mm
m
qa my my my mx
wxy A B
Daaaa ππππ
∞
=
⎡⎤ ⎛⎞ ⎛⎞⎛⎞
∴ =+
⎜⎟ ⎜⎟⎜⎟⎢⎥
⎝⎠ ⎝⎠⎝⎠⎣⎦∑ (2.131)

54. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     46 

Therefore to find expression for deflection of surface substitute equation (2.124) and (2.131) in
the equation (2.113) we get
4 4
0
55
11
4
sin cosh sinh sin
mm
mm
qa mx qa my my my mx
wAB
mD a D a a a aπππππ
π
∞∞
==
⎡⎤⎛⎞ ⎛⎞ ⎛⎞⎛⎞
=+ +
⎜⎟ ⎜⎟ ⎜⎟⎜⎟ ⎢⎥
⎝⎠ ⎝⎠ ⎝⎠⎝⎠ ⎣⎦
∑∑
4
55
1
4
cosh sinh sin
mm
m
qa my my my mx
wAB
Dm a a a a πππ π
π
∞
=
⎡⎤ ⎛⎞ ⎛⎞⎛⎞
∴ =++
⎜⎟ ⎜⎟⎜⎟⎢⎥
⎝⎠ ⎝⎠⎝⎠⎣⎦∑ (2.132)
Above equation satisfy the boundary condition and equation of equilibrium of the plate.
Now to determine the constants in the equation (2.132) and
mm
AB use boundary conditions
given below
()0for/2wyb==± (2.133)
()
2
2
0for /2
w
yb
y
∂
==±
∂ (2.134)
Therefore using boundary condition from equation (2.133), put
/2yb
=±into the equation
(2.132) and equate with zero, we get

4
55
1
4
cosh sinh sin 0
mm
m
qa my my my mx
AB
Dm a a a a ππππ
π
∞
=
⎡⎤ ⎛⎞ ⎛⎞⎛⎞
∴ ++ =
⎜⎟ ⎜⎟⎜⎟⎢⎥
⎝⎠ ⎝⎠⎝⎠⎣⎦∑
55
4
cosh sinh 0
222
mm
mb mb mb
AB
maaaπππ
π⎛⎞ ⎛⎞
∴ ++ =
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠
Let 2
m
mb
a
π
α=
55
4
cosh sinh 0
mmmmm
AB
mααα
π
∴ ++ = (2.135)
Differentiate equation (2.132) w.r.t.
y upto second order we get
4
1
sinh cosh sinh sin
mm m
m
w qa mym my my m mym mx
AB B
yD aa a a a aa aππ π ππ ππ π
∞
=
∂⎡ ⎤ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞
=++
⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎢⎥
∂ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠⎣⎦
∑

55. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     47 

22 22
2224
2 22 22
1
22
cosh sinh
sin
cosh cosh
mm
m
mm
my m my my m
AB
aa a aawqa mx
yD a my m my m
BB
aa aaππ π ππ
π
ππ ππ
∞
=
⎡⎤ ⎛⎞ ⎛⎞
+
⎜⎟ ⎜⎟⎢⎥
∂ ⎝⎠ ⎝⎠ ⎛⎞
⎢⎥=
⎜⎟
⎢⎥∂ ⎝⎠⎛⎞ ⎛⎞
+⎢⎥ ⎜⎟ ⎜⎟
⎝⎠ ⎝⎠⎣⎦
∑ (2.136)
Substitute / 2
yb=± into the equation (2.136) and use boundary condition from equation (2.134)
and equate with zero, hence
cosh sinh 2 cosh 0
mmmmmmm
AB B
α αα α∴ ++= (2.137)
Subtract equation (2.137) from (2.135) we get
55
4
cosh sinh cosh sinh 2 cosh 0
mmmmmmmmmmmm
AB AB B
mααα ααα α
π
∴ ++ −− − =

55
4
2cosh
mm
B
mα
π
∴ =
55
2
cosh
m
m
B
m
πα
∴ = (2.138)
Substitute value of
m
B from equation (2.138) into the equation (2.135) to obtained constant
m
A
55 55
42
cosh sinh 0
cosh
mm mm
m
A
mmααα
ππα
∴ ++ =
55 55
2tanh 4
cosh
mm
mm
A
mm
αα
α
π π
∴ =− −
55
2tanh 4
cosh
mm
mm
A
m
αα
α
π +
⎡ ⎤
∴ =−
⎢ ⎥
⎣ ⎦

55
2tanh 4
cosh
mm
m
m
A
m
αα
πα⎡⎤ +
∴ =−
⎢⎥
⎣⎦ (2.139)
Substitute constants and
mm
AB from equations (2.138) and (2.139) into the equation (2.132) to
calculate the deflection of the plate.

56. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     48 

55 554
1
55
2tanh 44
cosh
cosh
sin
2
sinh
cosh
mm
m
m
m my
mm aqa m x
w
Da my my
maaαα π
ππα π
ππ
πα
∞
=
⎡⎤ ⎡⎤ + ⎛⎞
−⎢⎥ ⎢⎥ ⎜⎟
⎝⎠ ⎛⎞⎣⎦⎢⎥
∴ =
⎜⎟
⎢⎥
⎝⎠⎛⎞
⎢⎥+
⎜⎟
⎢⎥ ⎝⎠⎣⎦∑
554
55
1
55
tanh 2
1cosh
2cosh4
sin
2
sinh
cosh
mm
m
m
m my
maqa m x
w
mD a my my
maaαα π
πα π
π ππ
πα
∞
=
⎡⎤⎡⎤ + ⎛⎞
−⎢⎥⎢⎥ ⎜⎟
⎝⎠ ⎛⎞⎣⎦⎢⎥
∴ =
⎜⎟
⎢⎥
⎝⎠⎛⎞
⎢⎥+
⎜⎟
⎢⎥ ⎝⎠⎣⎦∑
5544
55 55
1
55
tanh 2
cosh
2cosh44
sin
2
sinh
cosh
mm
m
m
m my
maqa qa m x
w
mD mD a my my
maaαα π
πα π
ππ ππ
πα
∞
=
⎡⎤ ⎧⎫⎡⎤ + ⎛⎞
⎢⎥ ⎪⎪⎢⎥ ⎜⎟
⎝⎠⎪⎪ ⎛⎞⎣⎦⎢⎥
∴ =− ⎨⎬ ⎜⎟⎢⎥
⎝⎠⎛⎞⎪⎪
⎢⎥ +
⎜⎟⎪⎪
⎢⎥ ⎝⎠⎩⎭⎣⎦∑
5544
55
1
55
tanh 2
cosh
2cosh54
sin
384 2
sinh
cosh
mm
m
m
m my
maqa qa m x
w
DmD a my my
maaαα π
πα π
π ππ
πα
∞
=
⎡⎤ ⎧⎫⎡⎤ + ⎛⎞
⎢⎥ ⎪⎪⎢⎥ ⎜⎟
⎝⎠⎪⎪ ⎛⎞⎣⎦⎢⎥
∴=− ⎨⎬ ⎜⎟⎢⎥
⎝⎠⎛⎞⎪⎪
⎢⎥ +
⎜⎟⎪⎪
⎢⎥ ⎝⎠⎩⎭⎣⎦ ∑ (2.140)
This series in this expression rapidly converge and sufficient accuracy is obtained by taking only
the first term. Differentiate above equation with respective
x and y to determine bending
moments and bending stresses.
Differentiate equation (2.140) with respect to
x upto second order, we get
552224 4
22 55
1
55
tanh 2
cosh
2cosh54
sin
384 2
sinh
cosh
mm
m
m
m my
maw m qa qa m x
xa DmD a my my
maaαα π
παππ
π ππ
πα
∞
=
⎡⎤ ⎧⎫⎡⎤ + ⎛⎞
⎢⎥ ⎪⎪⎢⎥ ⎜⎟
∂ ⎝⎠⎪⎪ ⎛⎞⎣⎦⎢⎥
=− − ⎨⎬ ⎜⎟⎢⎥
∂ ⎝⎠⎛⎞⎪⎪
⎢⎥ +
⎜⎟⎪⎪
⎢⎥ ⎝⎠⎩⎭⎣⎦
∑ (2.141)
Similarly differentiate equation (2.140) with respect to
y upto second order, we get

57. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     49 

55
44
55 55
1
55
tanh 2
sinh
2cosh
54 2
cosh sin
384 cosh
2
sinh
cosh
mm
m
m m
m mym
maa
wqa qa my mym mx
yDmDm a aa a
mmy
maaαα ππ
πα
ππππ
ππα
ππ
πα
∞
=
⎡⎤ ⎧⎫⎡⎤ + ⎛⎞
⎢⎥ ⎪⎪⎢⎥ ⎜⎟
⎝⎠⎣⎦⎢⎥ ⎪⎪
⎢⎥ ⎪⎪
∂ ⎪⎪ ⎛⎞ ⎛⎞
⎢⎥=− + ⎨⎬ ⎜⎟ ⎜⎟
∂⎢⎥ ⎝⎠ ⎝⎠⎪⎪
⎢⎥
⎪⎪
⎛⎞
⎢⎥
+⎪⎪ ⎜⎟
⎢⎥ ⎝⎠⎪⎪⎩⎭⎣⎦
∑ (2.142)
22
55 2
24 4 22
25555 2
1
22
55 2
tanh 2
cosh
2cosh
54 2
sinh sin
384 cosh
4
cosh
cosh
mm
m
m m
m mym
maa
wqa qa my mym mx
yDmDm a aa a
mym
maaαα ππ
πα
πππ π
ππα
ππ
πα
∞
=
⎡⎤ ⎧⎫⎡⎤ + ⎛⎞
⎢⎥ ⎪⎪⎢⎥ ⎜⎟
⎝⎠⎣⎦⎢⎥ ⎪⎪
⎢⎥ ⎪⎪
∂ ⎪⎪ ⎛⎞ ⎛⎞
⎢⎥=− + ⎨⎬ ⎜⎟ ⎜⎟
∂ ⎢⎥ ⎝⎠ ⎝⎠⎪⎪
⎢⎥
⎪⎪
⎛⎞⎢⎥
⎪⎪+
⎜⎟
⎢⎥
⎝⎠⎪⎪
⎩⎭⎣⎦
∑ (2.143)
Bending and twisting moments can be calculated using equations (2.5), (2.7) and (2.9) and
corresponding normal stresses are found from equation (2.18) and (2.19)

2.4.2 Levy’s solution for rectangular plate with at least two opposite edges simply
supported carrying linearly varying load:
In this case also the solution
(),wwxy=of the plate can be taken as sum of complimentary
solution
()
22
,wwxy= and particular integral
()
11
wwx= given by the equation (2.113)
Solution for ()
1
wx
Let assume the solution scheme for w and q from equation (2.116) and (2.117). In case of
uniformly varying load, value of load
1
q at any point at a distance of x is given by
0
1
qx
q
a
=
Substitute value of
1
qinto the equation (2.121) to find out the value of Fourier coefficient
m
q.

58. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     50 


1
0
2
sin
xa
m
x
mx
qq dx
aaπ
=
=
⎛⎞
∴ =
⎜⎟
⎝⎠∫


0
0
2
sin
xa
m
x
qx mx
qdx
aa aπ
=
=
⎛⎞
∴ =
⎜⎟
⎝⎠∫

0
2
0
2
sin
xa
m
x
q mx
qxdx
aa π
=
=
⎛⎞
∴ =
⎜⎟
⎝⎠∫

0
2
00
2
cos sin
aa
m
q mx a mx a
qx
aamam ππ
ππ
⎡ ⎤
⎛⎞ ⎛⎞
∴ =+ ⎢ ⎥⎜⎟ ⎜⎟
⎝⎠ ⎝⎠⎢ ⎥⎣ ⎦

0
2
m
q
q
mπ
∴ = (2.144)
Therefore substitute value of Fourier coefficient in the equation (2.118) to obtain constant
m
w
4
0
55
1
2
m
m
qa
w
mD
π
∞
=
∴ =∑ (2.145)
Put equation (2.145) into the equation (2.116) to obtained deflection of the plate
1
()wx
()
4
0
1 55
1
2
sin
m
qa mx
wx
mD a
π
π
∞
=
⎛⎞
∴ =
⎜⎟
⎝⎠∑ (2.146)
This should satisfy the plate equation and boundary condition at the edges / 2yb=±

59. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     51 

Solution for ()
2
,wxy
Solution of
()
2
,wxy for the symmetric problems is given by the equation (2.131)
()
4
2
1
,coshsinhsin
mm
m
qa my my my mx
wxy A B
Daaaa ππππ
∞
=
⎡⎤ ⎛⎞ ⎛⎞⎛⎞
=+
⎜⎟ ⎜⎟⎜⎟⎢⎥
⎝⎠ ⎝⎠⎝⎠⎣⎦
∑
Therefore the total deflection of surface is given by the equation (2.72)
4 4
0
55
11
2
sin cosh sinh sin
mm
mm
qa mx qa my my my mx
wAB
mD a D a a a aπππππ
π
∞∞
==
⎡⎤⎛⎞ ⎛⎞ ⎛⎞⎛⎞
∴=+ +
⎜⎟ ⎜⎟ ⎜⎟⎜⎟ ⎢⎥
⎝⎠ ⎝⎠ ⎝⎠⎝⎠ ⎣⎦∑∑
4
55
1
2
cosh sinh sin
mm
m
qa my my my mx
wAB
Dm a a a a ππππ
π
∞
=
⎡⎤ ⎛⎞ ⎛⎞⎛⎞
∴ =++
⎜⎟ ⎜⎟⎜⎟⎢⎥
⎝⎠ ⎝⎠⎝⎠⎣⎦∑ (2.147)
Above equation (2.147) will satisfy the boundary condition and equation of equilibrium.
To determine the constant and
mm
AB use boundary conditions from equation (2.133) and
(2.134).
Therefore using boundary condition from equation (2.133) substitutes
/2yb
=±into the equation
(2.147) and equate with zero, hence we get
55
2
cosh sinh 0
222
mm
mb mb mb
AB
maaaπππ
π⎛⎞ ⎛⎞
∴ ++ =
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠

Let
2
m
mb
a
π
α=
55
2
cosh sinh 0
mmmmm
AB
mααα
π
∴ ++ = (2.148)
Differentiate equation (2.147) with respect to y upto second order we get
4
1
sinh cosh sinh sin
mm m
m
w qa mym my mym mym mx
AB B
yD a a a a a a a aππ π ππ ππ π
∞
=
∂⎡ ⎤ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞
=++
⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎢⎥
∂ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠⎣⎦
∑

(2.149)

60. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     52 

22 22
2224
2 22 22
1
22
cosh sinh
sin
cosh cosh
mm
m
mm
my m my my m
AB
aa a aawqa mx
yD a my m my m
BB
aa aaππ π ππ
π
ππ ππ
∞
=
⎡⎤ ⎛⎞ ⎛⎞
+
⎜⎟ ⎜⎟⎢⎥
∂ ⎝⎠ ⎝⎠ ⎛⎞
⎢⎥=
⎜⎟
⎢⎥∂ ⎝⎠⎛⎞ ⎛⎞
+⎢⎥ ⎜⎟ ⎜⎟
⎝⎠ ⎝⎠⎣⎦
∑ (2.150)
Using boundary condition from equation (2.134) substitutes / 2yb
=±into the equation (2.150)
and equate with zero, hence we get
cosh sinh 2 cosh 0
mmmmmmm
AB B
α αα α∴ ++= (2.151)
Subtract equation (2.151) from (2.148)
55
2
cosh sinh cosh sinh 2 cosh 0
mmmmmmmmmmmm
AB AB B
mααα ααα α
π
∴ ++ −− − =
55
2
2cosh
mm
B
mα
π
∴ =
55
1
cosh
m
m
B
m
πα
∴ = (2.152)
Substitute value of constant
m
B from equation (2.152) into the equation (2.148) to obtain
m
A
55 55
21
cosh sinh 0
cosh
mm mm
m
A
mmααα
ππα
∴ ++ =
55 55
12
cosh sinh
cosh
mm mm
m
A
mmααα
παπ
∴ =− −
55 55
tanh 2
cosh
mm
mm
A
mm
αα
α
π π
∴ =− −
55
tanh 2
cosh
mm
mm
A
m
αα
α
π +
⎡ ⎤
∴ =−
⎢ ⎥
⎣ ⎦

55
tanh 2
cosh
mm
m
m
A
m
αα
πα⎡⎤ +
∴ =−
⎢⎥
⎣⎦ (2.153)

61. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     53 

Substitute values of constants and
mm
AB from equations (2.153) and (2.152) into the equation
(2.147) to obtained deflection of the plate
4
55 55 55
1
sinh
tanh 22
cosh sin
cosh cosh
mm
m mm
my
qa m y m y m x a
w
Dm m am a aπ
αα
π ππ
ππα πα
∞
=
⎡⎤ ⎛⎞
⎜⎟⎢⎥
⎡⎤ + ⎛⎞ ⎛⎞⎛⎞ ⎝⎠
⎢⎥
∴ =− +
⎢⎥ ⎜⎟ ⎜⎟⎜⎟
⎝⎠ ⎝⎠⎝⎠⎢⎥ ⎣⎦
⎢⎥
⎣⎦∑
(2.154)

Differentiate above equation (2.154) with respect to x and y to determine bending and twisting
moments from equations (2.5), (2.7) and (2.9), and corresponding stresses from equations (2.18),
(2.19) and (2.20)
2.4.3 Levy’s solution for rectangular plate carrying moments along edges:
Consider a rectangular plate simply supported all along the edges is acted on by moment
M distributed along the edges at
/2yb
=±. The solution can be efficiently carried out using
levy’s method. Since the acting load is the moment and not distributed loading the particular
integral part
()
1
wxis absent. Let assume the solution of
()
2
wx,y from equation (2.125)

()
2
1
,sin
m
m
mx
wwxy Y
a
π
∞
=
== ∑
If the moment acting at the edges / 2yb
=± are M 1 and M2 resp. and are an unequal in
magnitude.

62. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     54 

Note: since acting load is moment all the four constants (,,and
mmm m
ABC D ) will be
operative in the equation of deflection.
The evaluation of constants can be done satisfying the edge condition viz
2
12
/2 0;
y
w
at y b w M D M
y
∂
===−=
∂
(2.155)
2
22
/2 0;
y
w
at y b w M D M
y
∂
=− = =− =
∂
(2.156)
If
12
==MMM then solution is given by equation (2.84) (symmetrical problem) and
If
12
±= =∓MMM then solution is given by equation (2.85) (unsymmetrical problem)
Let consider solution for symmetric problems (i.e.
12
= =MMM ) is given by equation
(2.129)
1
cosh sinh sin
mm
m
my my my mx
wA B
aaaaππππ
∞
=
⎡⎤ ⎛⎞ ⎛⎞⎛⎞
∴ =+
⎜⎟ ⎜⎟⎜⎟⎢⎥
⎝⎠ ⎝⎠⎝⎠⎣⎦∑ (2.157)
Let assume
m
m
a
π
α=
() ()()
1
cosh sinh sin
mmmm m m
m
wAyByyx αααα
∞
=
∴ =+ ⎡⎤
⎣⎦∑ (2.158)
To evaluate constants and
mm
AB use boundary conditions
0for/2wyb==± (2.159)

2
2
for / 2
w
DM yb
y
∂
−= =±
∂ (2.160)
Using boundary condition (2.159), Substitute / 2yb
=±in the equation (2.158) and equate with
zero, we obtain

cosh sinh 0
222
mmm
mm
bbb
ABααα⎛⎞ ⎛⎞
∴ +=
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠

63. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     55 

Let assume
2
m
m
bα
β⎛⎞
=
⎜⎟
⎝⎠
, therefore equation becomes
cosh sinh 0
mmmmm
AB
β ββ∴ += (2.161)
Now differentiate equation (2.158) with respective y upto second order
() ()() () ()
1
sinh cosh sinh sin
mmmmm mmmmmm
m
w
AyByyBy x
y
αα α αα αα α
∞
=
∂
=++⎡⎤
⎣⎦
∂
∑
() ()() () ()
2
222
2
1
cosh sinh 2 cosh sin
mmmmm mmmmmm
m
w
AyByyBy x
y
αα α αα αα α
∞
=
∂
⎡⎤=+ +
⎣⎦
∂
∑ (2.162)
Using boundary condition (2.160), Substitute / 2yb
=±in the equation (2.162) and equate
withM, hence we get
()
222
cosh sinh 2 cosh sin
222 2
mmm m
mmm mmmm
bbb b
M
AB B x
D
ααα α
αααα⎡⎤ ⎛⎞ ⎛⎞⎛⎞ ⎛⎞
∴ ++ =−
⎜⎟ ⎜⎟⎜⎟ ⎜⎟⎢⎥
⎝⎠ ⎝⎠⎝⎠ ⎝⎠⎣⎦

Where M is uniformly distributed over the edges and therefore its value is given by
0
4M
M
mπ
= ,
Substitute in the above equation we get
() ()
22
0
2
cosh sinh
222 4
sin sin
2cosh
2
mmm
mmm m
mm
m
mm
bbb
AB
M
x x
Dmb
B
ααα
αα
αα
πα
α⎡⎤ ⎛⎞ ⎛⎞⎛⎞
+
⎜⎟ ⎜⎟⎜⎟⎢⎥
⎝⎠ ⎝⎠⎝⎠
⎢⎥
∴ =−
⎢⎥ ⎛⎞
+⎢⎥ ⎜⎟
⎝⎠⎣⎦

[]
0
2
4
cosh sinh 2 cosh
mmmmmmm
m
M
AB B
mD
βββ β
πα
∴ ++=− (2.163)
Subtracting equation (2.163) from (2.160), we get

0
2
4
cosh sinh cosh sinh 2 cosh
mmmmmmmmmmmm
m
M
AB AB B
mD
βββ βββ β
πα+−−−=
0
2
4
2cosh
mm
m
M
B
mD
β
πα
∴ =−

64. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     56 

0
2
2
cosh
m
mm
M
B
mD
παβ
∴ =− (2.164)
Substitute magnitude of constant
m
B from equation (2.164) into the equation (2.161) to find
value of constant
m
A, hence
0
2
2
cosh sinh 0
cosh
mm mm
mm
M
A
mD
βββ
πα β
∴ −=
0
2
2tanh
cosh
mm
m
mm
M
A
mD
ββ
παβ
∴ = (2.165)
Therefore substitute values of constants and
mm
AB from equations (2.165) and (2.164) into the
equation (2.158) to find out deflection of the plate.
() ()()
00
22
1
2tanh 2
cosh sinh sin
cosh cosh
mm
mmmm
m mm m m
MM
wyyyx
mD mDββ
αααα
πα β πα β
∞
=
⎡⎤
∴ =−
⎢⎥
⎣⎦∑
() ()()
0
2
1
2
tanh cosh sinh sin
cosh
mm mm m m
m mm
M
wyyyx
mD
ββ αα α α
πα β
∞
=
∴ =− ⎡⎤
⎣⎦∑ (2.166)

2.4.4 Levy’s solution for rectangular plate with two opposite edges Clamped subjected to
uniformly distributed load:
When two opposite edges of the plate are clamped and subjected to uniformly distributed load,
the deflection of the plate is given by the equation (2.132)
4
55
1
4
cosh sinh sin
mm
m
qa my my my mx
wAB
Dm a a a a πππ π
π
∞
=
⎡⎤ ⎛⎞ ⎛⎞⎛⎞
∴ =++
⎜⎟ ⎜⎟⎜⎟⎢⎥
⎝⎠ ⎝⎠⎝⎠⎣⎦∑
Applying the boundary conditions at clamed edges to find out unknown constants and
mm
AB
0for/2wyb== (2.167)
0for/2
w
yb
y
∂
==
∂
(2.168)

65. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     57 

Therefore substitute / 2yb= in the above equation of deflection and equate with zero using
boundary condition (2.167)
55
4
cosh sinh 0
222
mm
mb mb mb
AB
maaaπππ
π⎛⎞ ⎛⎞
∴ ++ =
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠

Let
2
m
mb
a
π
α=
55
4
cosh sinh 0
mmmmm
AB
mααα
π
∴ ++ = (2.169)
Now differentiate equation (2.132) with respect to y, substitute / 2yb= in the
w
y
∂
∂
and equate
with zero using boundary condition (2.168)
4
1
sinh cosh
222
sin 0
sinh
2
mm
m
m
mb m mb mb m
AB
aa a aaqa m x
Da mb m
B
aaππ π ππ
π
ππ
∞
=
⎡⎤ ⎛⎞ ⎛⎞
+
⎜⎟ ⎜⎟⎢⎥
⎝⎠ ⎝⎠ ⎛⎞
⎢⎥
=
⎜⎟
⎢⎥ ⎝⎠⎛⎞
+
⎢⎥ ⎜⎟
⎝⎠⎣⎦∑
sinh cosh sinh 0
222 2
mm m
mb m mb mb m mb m
AB B
aa a aa aaππ π ππ ππ⎛⎞ ⎛⎞ ⎛⎞
∴ ++=
⎜⎟ ⎜⎟ ⎜⎟
⎝⎠ ⎝⎠ ⎝⎠

() () ()sinh cosh sinh 0
mmmm mmm
AB Bααα α∴ ++= (2.170)
Solving equation (2.169) and (2.170) for the constants and
mm
AB and substitute in the equation
(2.132) to obtain the expression for the deflection of the plate.

Exercise
Que. State the assumptions in the small deflection theory of thin plate. [P.U., Ans. Article 2.2.1]
Que. Distinguish between thin and thick plate bending.
[P.U., Ans. Article 2.2]
Que. Sketch the free body diagram of a plate element representing lateral loads, moments and
shears.
[P.U., Ans. Article 2.2, Figure, Page No. 14]

66. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     58 

Que. Write the equations defining equilibrium of the element.
[P.U., Ans. Article 2.2, Eq. 2.10,
2.12 and 2.14]
Que. Starting from the first principles derive the governing differential equation in Cartesian co-
ordinates for thin plate under bending.
4
∇=
q
w
D

The symbols carry the usual meaning.
[P.U., Ans. Derivation of Article 2.2]
Derive the boundary conditions to be used for
1)
simply supported edge 2) a clamped edge 3) a free edge [P.U., Ans. Article 2.2.2]
Que. Derive the governing differential equation of thin rectangular plate subjected to transverse
load
(),qxy per unit area, according to Kirchhoff’s theory. Hence explain how Kirchhoff’s
reduces three boundary conditions to two per edge. Discuss boundary condition of
cantilever plate.
[Dr. B.A.M.U., Ans. Derivation of Article 2.2]
Que. Explain the stepwise procedure in Navier’s method for bending analysis of a thin
rectangular plate simply supported at all four edges.
[P.U., Ans. Derivation of Article 2.3]
Que. Discuss Navier’s solution of simply supported rectangular plate subjected to UDL
(),qxy.
Hence find the expressions for transverse deflection(),wxy, bending and twisting
moments, bending and shear stresses, shear force and reactive forces along the boundaries.

[Dr. B.A.M.U., Ans. Derivation of Article 2.3.1]

Que. Discuss Navier’s solution of simply supported rectangular plate subjected to sinusoidal
loading. Hence find the expressions for transverse deflection
(),wxy, bending and twisting
moments, bending and shear stresses, shear force and reactive forces along the boundaries.

[Dr. B.A.M.U., Ans. Derivation of Article 2.3.2]

Que. Using Navier’s method, find for a square plate of side 3 m, thickness 12 cm under uniform
load of 3 KN/m
2
values for
a)
Maximum deflection in mm

67. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     59 

b) Maximum B.M. in KNm. E = 210 GPa & u = 0.1
The plate is simply supported on all edges [P.U., Ans. Problem based on Derivation of
Article 2.3.1]

Que:
A thin rectangular plate of size 2aa
× and simply supported on all edges carries a
uniformly distributed load of intensity
0
q per unit area. Using Navier’s method computes the
deflection and the principal moments at the centre of plate. [
P.U., Ans. Problem based on
Derivation of Article 2.3.1]

Que.
Using Navier’s method, obtain expression for lateral displacement w of the plate ab
×
subjected to hydrostatic pressure which varies in x direction. Take origin at the centre of left
edge find the central deflection if q
0 is the peak intensity of loading on the edge x = a.
[
P.U., Ans. Problem based on Derivation of Article 2.3.5]
Que. A thin square plate of size aa× and simply supported on all edges carries a patch load.
Using Navier’s method computes the deflection and the principal moments at the centre of
plate.
[Ans., Derivation of Article 2.3.3]
Que. A thin square plate of size aa× and simply supported on all edges carries a point load.
Using Navier’s method computes the deflection and the principal moments at the centre of
plate.
[Ans., Derivation of Article 2.3.4]
Que. Describe the stepwise procedure in the levy’s method for thin plate bending analysis.
Illustrate the example where two opposite edges of a rectangular plate are simply supported
[
P.U., Ans. Problem based on Derivation of Article 2.4.1]
Que. A square plate
×aa with all four edges simply supported, carries a uniformly distributed
load of intensity q
0. Using levy’s method, compute the maximum deflection in the plate.
[
P.U., Ans. Problem based on Derivation of Article 2.4.1]
Que. Discuss Levy’s solution of simply supported rectangular plate subjected to UDL
(),qxy.
Hence find the expressions for transverse deflection(),wxy, bending and twisting

68. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     60 

moments, bending and shear stresses, shear force and reactive forces along the boundaries.

[Dr. B.A.M.U., Ans. Derivation of Article 2.4.1]

Que.
Using Levy’s method, obtain expression for lateral displacement w of the plate ab
×
subjected to hydrostatic pressure which varies in x direction. Take origin at the centre of
left edge find the central deflection if q
0 is the peak intensity of loading on the edge x = a.
[
P.U., Ans. Problem based on Derivation of Article 2.4.2]
Que. Using Levy’s method, obtain expression for lateral displacement w of the plate ab
×
subjected to moments at the edges. Take origin at the centre of left edge find the central
deflection.
[Ans., Derivation of Article 2.4.3]
Que. Discuss Levy’s solution of simply supported rectangular plate subjected to UDL
(),qxy.
Hence find the expressions for transverse deflection (),wxy[Ans., Derivation of Article
2.4.4]

69. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     61 


Chapter 3
Pure Bending of Plate


3.1 Slope and Curvatures of Bent Plate:
Consider the middle plane of the plate before bending occurs, as the xy plane. During
bending, the particles that were in the xy plane undergo small displacements w perpendicular to
xy plane and form the middle surface of the plate. These displacements of the middle planes are
called as deflections of the plate. Taking normal section of the plate parallel to xz plane as shown
in figure (a) we find that the slope of middle surface in the x direction is /
x
idwdx= and in y
direction is /
y
idwdy= .

Taking now any direction an in the xy plane as shown in figure (b) making an angle α with the x
axis. We find that the difference in the deflection of the two adjacent points a and a
1 in the an
direction is

70. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     62 



ww
dw dx dy
x y
∂∂
=+
∂∂

And slope in the same direction is

wwxwy
nxnyn
∂∂∂∂∂
=+
∂∂∂ ∂∂
(3.1)
cos sin
ww w
nx y
α α
∂∂∂
∴ =+
∂∂ ∂
(3.2)
cos sin
nx y
α α
∂∂ ∂
∴ =+
∂∂ ∂
(3.3)
Similarly

wwxwy
txtyt
∂∂∂∂∂
=+
∂∂∂∂∂
(3.4)


() ()cos 90 sin 90
ww w
tx yαα
∂∂∂
∴ =+++
∂∂ ∂

sin cos
ww w
tx y
α α
∂ ∂∂
∴ =− +
∂∂ ∂
(3.5)
sin cos
tx y
α α
∂∂ ∂
∴ =− +
∂∂ ∂
(3.6)
To find out the direction in which the slope of the surface is maximum, we differentiate equation
(3.2) w.r.t. α and equate it with zero. Hence the direction of slope can be determine as
,,
usingchain rule
wxynt
wwxwy
nxnyn
→→
∂∂∂ ∂∂
=+
∂∂∂ ∂∂

71. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     63 

sin cos 0
ww
xyαα
∂∂∴ −+ =
∂∂


1
/
tan
/
wy
wx
α
∂∂⎛⎞
∴ =
⎜⎟
∂∂⎝⎠
(3.7)
We can also determine the direction in which the slope of surface would be minimum by
equating equation (3.2) to zero. This direction can be obtained as
cos sin 0
ww
xyαα
∂ ∂
∴ +=
∂∂


2
/
tan
/
wx
wy
α
⎛⎞∂∂
∴ =−
⎜⎟
∂∂⎝⎠
(3.8)
Therefore from equation (3.7) and (3.8) we note that

()()
12
tan tan 1αα
=− (3.9)
Which indicates that the directions of maximum and minimum slopes are orthogonal.
Now to obtain the relations of curvatures,
In considering the curvature of the middle surface in any direction an we obtain

2
2
1∂∂∂ ⎛⎞
∴ ==−
⎜⎟
∂∂∂ ⎝⎠
n
ww
rn nn

Therefore from equation (3.2) and (3.3) we get

2
2
cos sin cos sin
www
nx y x y
α ααα
⎡⎤⎛ ⎞∂∂ ∂∂ ∂
∴ =− + +
⎜⎟⎢⎥
∂∂ ∂ ∂ ∂⎣⎦⎝ ⎠


22 2 2 2
2 2
22 2
cos sin cos sin cos sin
ww w w w
nx yx yx y
α αα αα α
⎛⎞∂∂ ∂ ∂ ∂
∴ =− + + +⎜⎟
∂ ∂ ∂∂ ∂∂ ∂⎝⎠

22 2 2
22
22 2
cos 2 sin cos sin
ww w w
nx yx y
α αα α
⎛⎞∂∂ ∂ ∂
∴ =− + +⎜⎟
∂∂ ∂∂ ∂⎝⎠

2
22
2
11 1
cos 2sin cos sin
xxy y
w
nr r r
α αα α
⎛⎞∂
∴ =− +⎜⎟
⎜⎟∂
⎝⎠

72. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     64 


2211 1 1
cos sin 2 sin
⎛⎞
∴ =−+⎜⎟
⎜⎟
⎝⎠
nx xy y
rr r r
α αα (3.10)
Where

2
2
1
Bending curvature of thesurfacein planeparallelto plane
x
w
xz
rx
∂
=− =
∂


2
2
1
Bending curvature of thesurfacein planeparallelto plane
y
w
yz
ry
∂
=− =
∂


2
1
Twisting curvature or twist of thesurfacew.r.t. & axis.
xy
w
xy
ryx
∂
==
∂∂


2
2
1
curvature of themiddle surfacein any direction
n
w
an
rn
∂
==
∂

Note: curvatures are the second order derivatives of slope. The curvature is considered positive
if it is convex downward, the minus sign is taken in the equation, since for the deflection
convex downward, as shown in figure, the second derivative
22
/
∂∂wx is negative.
Similarly, instead of the direction ‘an’ if we take the direction perpendicular to an, the curvature
in this new direction will be obtained from equation (3.10) by substituting
2
π
α
⎛⎞
+
⎜⎟
⎝⎠
forα. Thus
we get

2211 1 1
cos sin 2 sin
222
tx xy y
rr r r
πππ
α αα
⎛⎞
⎛⎞ ⎛⎞ ⎛⎞
∴ =+−+++⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟
⎜⎟
⎝⎠ ⎝⎠ ⎝⎠
⎝⎠


2211 1 1
cos sin 2 sin
ty xy x
rr r r
α αα
⎛⎞
∴ =++⎜⎟
⎜⎟
⎝⎠
(3.11)
Where

2
2
1
curvature of themiddle surfacein any direction
∂
==
∂
t
w
at
rt

73. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     65 

OR

2
2
ww
ttt
∂ ∂∂⎛⎞
∴ =−
⎜⎟
∂∂∂ ⎝⎠

2
2
sin cos sin cos
⎡⎤⎛ ⎞∂∂∂ ∂∂
∴ =− − + − +
⎜⎟⎢⎥
∂∂∂ ∂ ∂⎣⎦⎝ ⎠
www
txy xy
α ααα

22 2 2 2
2 2
22 2
cos sin cos sin cos sin
⎛⎞∂∂ ∂ ∂ ∂
∴ =− − − +⎜⎟
∂∂ ∂∂ ∂∂ ∂⎝⎠
ww w w w
t y yx yx x
α αα αα α

22 2 2
22
22 2
cos 2 sin cos sin
⎛⎞∂∂ ∂ ∂
∴ =− − +⎜⎟
∂∂ ∂∂ ∂⎝⎠
ww w w
ty yx x
α αα α

2
22
2
11 1
cos 2sin cos sin
⎛⎞∂
∴ =+ +⎜⎟
⎜⎟∂
⎝⎠
yxy x
w
tr r r
α αα α

2211 1 1
cos sin 2 sin
⎛⎞
∴ =++⎜⎟
⎜⎟
⎝⎠
ty xy x
rr r r
α αα
Adding equation (3.10) and (3.11) we obtain
222211 1 1 1 1 1 1
cos sin 2 sin cos sin 2 sin
⎛⎞⎛⎞
∴ += − + + + +⎜⎟⎜⎟
⎜⎟⎜⎟
⎝⎠⎝⎠
nt x xy y y xy x
rr r r r r r r
α αα α αα
() ()
22 2211 1 1
sin cos sin cos
⎡⎤
∴ += + + +⎢⎥
⎢⎥⎣⎦
nt x y
rr r r
α ααα

11 11
∴ += +
nt xy
rrrr
(3.12)
Which shows that, at any point of the middle surface the sum of the curvature in two
perpendicular directions is independent of the angle α and this is usually called as the
average curvature on this surface at a point.
The twist of surface at a with respect to the an and at direction is

74. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     66 


1
nt
w
rtn∂∂⎛⎞
=
⎜⎟
∂∂⎝⎠


1
sin cos cos sin
nt
ww
rx y x y
α ααα
⎡⎤⎛⎞∂∂∂ ∂
∴ =− + +
⎜⎟⎢⎥
∂∂∂ ∂⎣⎦⎝⎠

()
22 2
22
22
1
sin cos cos sin cos sin
nt
ww w
rx xy y
αααααα
⎛⎞∂∂ ∂
∴ =− + − +⎜⎟
∂∂∂ ∂⎝⎠

222
22
1sin2 sin2
cos 2
22
nt
www
rx xy y
α α
α⎛⎞∂∂∂∴ =− + +⎜⎟
∂∂∂∂⎝⎠

1 1 sin 2 1 1 sin 2
cos2
22
nt x xy y
rr r r
α α
α⎛⎞∴ =+−⎜⎟
⎜⎟
⎝⎠


111sin21
cos2
2
nt x y xy
rrr r
α
α
⎛⎞⎡⎤
∴ =− +⎜⎟⎢⎥
⎜⎟
⎢⎥⎣⎦⎝⎠
(3.13)
In our further discussion we shall be interested in finding, in terms of α the direction in which
the curvature of the surface is a maximum or a minimum and in finding the corresponding values
of the curvature. We obtain the necessary equation for determining α by equating the derivative
of equation (3.10)
w.r.t.
α equating with zero which gives


111
2sin cos 2cos2 2sin cos 0
xxyy
rrr
αα α αα
∴ −−+ =

11 1
sin 2 2cos2 sin 2 0
xxy y
rr r
ααα
∴ +−=

11 1
sin 2 2cos2
xy xy
rr r
α α
⎛⎞
∴ −=−⎜⎟
⎜⎟
⎝⎠

75. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     67 


sin 2 2 1
cos2 11
xy
xy
r
rr
α
α
∴ =−
⎛⎞
−⎜⎟
⎜⎟
⎝⎠


2
tan 2
11
xy
xy
r
rr
α
−
∴ =
⎛⎞
−⎜⎟
⎜⎟
⎝⎠
(3.14)
From this equation we find two values of α differing by/2π. Substituting these in equation
(3.10) we find two values of1/
n
r, One representing maximum and one representing minimum
curvature at a point ‘a’ of the surface. These two curvatures are called as principal curvatures
from the surface and corresponding planes defined by
α and /2πα+ are called principal
planes of curvatures.
Note: when the co ordinate planes xz and yz are taken parallel to the principle planes of
curvature at a point
a then twisting curvature or twist
1
0
xy
r
= i.e.
1
0
nt
r
=
3.2 Moment Curvature Relations:
As mentioned earlier, the state of stress in a plate will be two dimensional according to
thin plate theory. In Cartesian co ordinates, the state of stress may be identified as normal
stresses and shear stresses. These stresses are distributed over the thickness of the plate and cause
bending and twisting moments as well as vertical shear forces. Since those moments are
resultants of the stresses developed in the plate, these are called as stress resultants.


/2 /2
/2 /2
.
hh
xx xx
hh
Mdy dz dy M dzσσ
−−
=⇒=∫∫


/2 /2
/2 /2
.
hh
yy yy
hh
Mdx dz dx M dzσσ
−−
=⇒=∫∫

76. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     68 


/2 /2
/2 /2
.
hh
xy xy xy xy
hh
Mdy dz dy M dzττ
−−
=⇒=∫∫


/2 /2
/2 /2
.
hh
yx yx yx yx
hh
Mdx dz dx M dzττ
−−
=⇒=∫∫

From moment equilibrium equations we have
xy yx
ττ= hence it is obvious from that
xy yx
M M=
Using strain displacement relationship and stress-strain relationship we obtain following moment
curvature relationship for isotropic plate.

2
2
2
2
2
10
10
1
00
2 2
x
y
xy
w
x
M
w
MD
y
M
w
xy
υ
υ
υ
⎧ ⎫∂
⎛⎞ −⎪ ⎪
∂⎜⎟ ⎪ ⎪⎧⎫
⎜⎟ ⎪ ⎪⎪⎪ ∂⎪ ⎪
⎜⎟=−⎨⎬ ⎨ ⎬
∂⎜⎟⎪⎪ ⎪ ⎪
−⎛⎞
⎜⎟⎩⎭ ⎪ ⎪
∂⎜⎟⎜⎟
−⎪ ⎪⎝⎠⎝⎠
∂∂⎪ ⎪⎩⎭
(3.15)
i.e.

22
22
11
xx
xy
ww
MD MD
x yrr
υυ
⎛⎞⎛⎞∂∂
=− + ⇒ = + ⎜⎟⎜⎟
⎜⎟∂∂⎝⎠ ⎝⎠


22
22
11
yy
yx
ww
MD MD
yx rr
υυ
⎛⎞⎛⎞∂∂
=− + ⇒ = + ⎜⎟⎜⎟
⎜⎟∂∂⎝⎠ ⎝⎠


()
()
2
1
11
xy xy
xy
w
MD MD
xyr
υυ
∂
=− − ⇒ =− −
∂∂

Where D is the flexural rigidity of plate.
Similarly, it can be shown that the following will be the moment curvature relationship for an
isotropic plate in polar co-ordinates

77. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     69 


()
2
2
2
22
2
2
10
11
10
00 1
11
r
r
w
r
M
ww
MD
rr r
M
ww
rr r
θ
θ
υ
υ
θ
υ
θθ
⎧ ⎫
∂
⎪ ⎪−
∂⎪ ⎪
⎛⎞⎧⎫
⎪ ⎪
⎛⎞∂∂⎜⎟⎪⎪ ⎪ ⎪
=−+⎨⎬ ⎨ ⎬ ⎜⎟⎜⎟
∂∂⎝⎠⎪⎪ ⎪ ⎪ ⎜⎟
−
⎩⎭ ⎝⎠ ⎪ ⎪
⎛⎞∂∂
⎪ ⎪−−⎜⎟
∂∂ ∂⎪ ⎪⎝⎠⎩⎭
(3.16)
3.3 Particular cases in Pure Bending:

Case I) If and 0
xy xy
MMM M== =
From equation (3.15)
11
x
xy
MD
rr υ
⎛⎞
=+⎜⎟
⎜⎟
⎝⎠
And
11
y
yx
MD
rr υ
⎛⎞
=+⎜⎟
⎜⎟
⎝⎠

fand0
xy xy
iM M M M== =
11 11
xy yx
DD
rr rrυυ
⎛⎞⎛⎞∴ +=+⎜⎟⎜⎟
⎜⎟⎜⎟
⎝⎠⎝⎠

11 11
xx yy
rr rr
υυ
⎛⎞⎛⎞
∴ −=− ⎜⎟⎜⎟
⎜⎟
⎝⎠ ⎝⎠

() ()
11
11
xy
rr
υ υ∴ −= −

78. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     70 

11
xy
rr
∴ = (3.17)
Equation of moments becomes
11
xy
xx
MMMD
rr υ
⎛⎞
∴ === + ⎜⎟
⎝⎠
()
1
1
x
MD
r
υ∴ =+
()
1
1
x
M
rD
υ
∴ =
+ (3.18)
Similarly
()
1
1
y
M
rD
υ
∴ =
+ (3.19)
i.e., the plate in this case is bent to a spherical surface the curvature which is given by
equation (3.19).
From equation (3.18)
()
2
2
1
wM
xDυ
⎡⎤∂
=−⎢⎥
∂+
⎣⎦

()
2
2
1
wM
dx dx
xDυ
∂
∴ =−
∂+∫∫

()
1
1
wM
xC
xD
υ
⎡⎤∂ ∴ =− +⎢⎥
∂+
⎣⎦

Integrate w.r.t. x
()
2
12
12
Mx
wCxC
Dυ
⎡⎤
∴ =− + +⎢⎥
+
⎣⎦

79. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     71 

()
2
1
12
Mx
wCx
Dυ
⎡⎤
∴ =− +⎢⎥
+
⎣⎦
(3.20)
Similarly from equation (3.19)
()
2
2
1
wM
yDυ
⎡⎤∂
=−⎢⎥
∂+
⎣⎦

()
2
3
12
My
wCy
Dυ
⎡⎤
∴ =− +⎢⎥
+
⎣⎦
(3.21)
Therefore adding equation (3.20) and (3.21) we obtain
()
()
22
13
12 12
Mx My
wCxCy
DDυυ
⎡⎤⎡⎤
∴ =− + − +⎢⎥⎢⎥
++
⎣⎦⎣⎦

()()
()
22
31
12 12
Mx My
wCyCx
DDυυ
⎡⎤⎡⎤
∴ =− − + +⎢⎥⎢⎥
++
⎣⎦⎣⎦
(3.22)
Where
13
andCC are constants of integration and define the plane from which deflection are
measured. If this plane taken tangent to the middle surface of plate then
13
0CC==
Therefore equation (3.22) becomes

()
()
22
12 12
M xMy
w
DD
υυ
⎡⎤⎡⎤
∴ =− −⎢⎥⎢⎥
++
⎣⎦⎣⎦


()
()
22
21
M
wxy
D
υ
∴ =− +
+ (3.23 a)
()
22
wCx y
∴ =+

2
wCr
∴ = (3.23 b)
This is the equation of paraboloid of revolution.
This gives synclastic surface.
Case II) If and 0
xy xy
MM M≠=
From equation (3.15)

80. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     72 

11
x
xy
MD
rr υ
⎛⎞
=+⎜⎟
⎜⎟
⎝⎠
And
11
y
yx
MD
rr υ
⎛⎞
=+⎜⎟
⎜⎟
⎝⎠

12
If Let and
xy x y
M MMM MM≠= =
1
11 11
x
xy xy
MD MD
rr rr υυ
⎛⎞ ⎛⎞
∴ =+ ⇒=+⎜⎟ ⎜⎟
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠
(3.24)
And
2
11 11
y
yx yx
MD MD
rr rr υυ
⎛⎞ ⎛⎞
∴ =+ ⇒=+⎜⎟ ⎜⎟
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠
(3.25)
To find value of
1
y
r
⎛⎞
⎜⎟
⎜⎟
⎝⎠
multiply equation (3.24) by υ and subtract from (3.25)
21
11 11
yx xy
MMD D
rr rrυυυυ
⎛⎞⎛⎞∴ −= + − +⎜⎟⎜⎟
⎜⎟⎜⎟
⎝⎠⎝⎠

2
2111 1 1
yx x y
MMD D D D
rrr rυυυυ
∴ −= + − −
()
2
211
1
y
MMD
r
υ υ∴ −= −
()
21
2
1
1
y
M M
rD
υ
υ
−
∴ =
−
(3.26)
Similarly
()
12
2
1
1
x
M M
rD
υ
υ
−
∴ =
−
(3.27)
Since
1
0; 0
xy
xy
M
r== (3.28)
From equation (3.27)

81. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     73 


()
2
12
2 2
1
MMw
x D
υ
υ
⎡⎤
−∂
⎢⎥=−
∂ −⎢⎥
⎣⎦

Integrate both sides of the above equation w.r.t.
x, we get
()
12
12
1
MMw xC
x D
υ
υ
⎡⎤
−∂
⎢⎥
∴ =− +
∂ −⎢⎥
⎣⎦
In which
1
C is the constant of integration, now making integrate again w.r.t. x, we obtained
()
2
12
122
21
MMx
wCxC
Dυ
υ
⎡⎤
−
⎢⎥
∴ =− + +
−⎢⎥
⎣⎦
Where constant of integration
2
C is to be determine using boundary condition 0at 0wx== we
get
2
0C=, substitute in the above equation, hence
()
2
12
12
21
MM x
wCx
Dυ
υ
⎡⎤
−
⎢⎥
∴ =− +
−⎢⎥
⎣⎦ (3.29)
Similarly from equation (3.26)
()
2
21
342
21
MM y
wCyC
Dυ
υ
⎡⎤
−
⎢⎥
∴ =− + +
−⎢⎥
⎣⎦
Using similar boundary condition, 0 at 0
wy
= = we get
4
0C=, substitute in the above equation,
hence
()
2
21
32
21
MMy
wCy
Dυ
υ
⎡⎤
−
⎢⎥
∴ =− +
−⎢⎥
⎣⎦ (3.30)
From equation (3.28) of twisting curvature
1
0
xy
r
= we obtain
5
wC= (3.31)
Adding equations (3.29), (3.30) and (3.31) we get

82. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     74 

() ()
22
12 21
13522
2211
MM MM xy
wCxCyC
DDυυ
υυ
⎡⎤⎡⎤
−−
⎢⎥⎢⎥
∴ =− + − + +
−−⎢⎥⎢⎥
⎣⎦⎣⎦
()
()
()
22
12 21
13 522
2211
MM MM xy
wCxCyC
DDυυ
υυ
⎡⎤⎡⎤
−−
⎢⎥⎢⎥
∴ =− − + + +
−−⎢⎥⎢⎥
⎣⎦⎣⎦ (3.32)
Where
13 5
,andCC C are constants of integration and define the plane from which deflection are
measured. If this plane taken tangent to the middle surface of plate then
13 5
0CC C===
Therefore equation (3.32) becomes
()
()
22
12 21
22
2211
MM MM
x y
w
DDυυ
υυ
⎧⎫
−−⎪⎪
∴ =− +⎨⎬
−−
⎪⎪⎩⎭ (3.33)
21
fiM M=−
Put in the equation (3.33) we obtain
()
()
22
11 11
22
2211
MM MM
x y
w
DDυυ
υυ
⎧⎫
++⎪⎪
∴ =− −⎨⎬
−−
⎪⎪⎩⎭
()
()
()
()
22
1 22
11
2211
⎧⎫
++⎪⎪
∴ =− ⎨⎬
−−
⎪⎪⎩⎭
yx
wM
DDυυ
υυ

()
()
221
21
∴ =−
−
M
wyx
D
υ
(3.34)
Which is called as Anticlastic surface.

Exercise
Que. Derive the relations between moments, curvatures and deflections in pure bending of plates.
[P.U. Ans., Article 3.2]

83. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     75 

Que. Show that curvature 1/
n
r in general direction
θ can be obtained with the help
of1/ ,1/ and 1/
xy xy
rr r . [P.U. Ans., Article 3.1., Derivation upto Equation (3.14)]
Que. Prove that the directions of maximum and minimum slopes are orthogonal in pure bent
plate.
[Dr. B.A.M.U. & P.U. Ans., Article 3.1., Derivation upto Equation (3.9)]
Que. Shows that, at any point of the middle surface of the bent plate, the sum of the curvature in
two perpendicular directions is independent of the angle
α. [Dr. B.A.M.U. & P.U. Ans.,
Article 3.1., Derivation upto Equation (3.12)]
Que. Using first principles determine central deflection of square plate subjected to pure couple
‘M’ distributed along all edges. Corner points are simply supported. Couple on edges
produces synclastic bending surface.
[P.U. Ans., Article 3.3, Case I]
Que. A square plate of 400 mm side and 10 mm thick is simply supported at its four corners. If is
subjected to uniformly distributed pure couple M
1 acting on edges parallel to x axis
creating hogging and uniformly distributed couple M
2 on edges parallel to y axis creating
sagging. If M
1 = M2 = 500 Nm, E = 200 GPa and 0.3
υ= find lateral displacement of
midpoints of edges.
[P.U. Ans., Article 3.3, Case II]

84. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     76 


Chapter 4
Bending of Circular Plates


4.1 Differential Equation for Symmetrical bending of circular plates:
Consider Symmetrical bending of circular plates shown in figure. Let us take origin of
coordinates ‘
O’ at the centre of deflected plate as shown in figure. Let r denoted the radial
distances of points in the middle plane of the plate and
w be the deflection of the plate in z
direction at any point A.


Figure: Bending of circular plate

85. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     77 

Then for the small value of
w, the maximum slope of deflected surface at A is given by
w
r
φ
∂
=−
∂
and the curvature of plate in diametral section
rz is
2
2
1
n
w
rrr
φ∂ ∂
=− =
∂ ∂

Where,
φ = Small angle between the normal to the deflection surface at ‘ A’ and the axis of
symmetry
‘OB’.
From the symmetry we conclude that 1/
n
r is one of the principle curvatures of the deflected
surface at A and the second principle curvature will be in the section through normal AB and
perpendicular to rz plane. Therefore we conclude that AB is the radius of second principle
curvature which is denoted by
t
r
11
t
w
rr rr
φ ∂
∴ ==−
∂
(4.1)
Bending moments per unit length along
mn is
2
2
r
dw dw d
MD D
dr r dr dr rυφυ
φ
⎡⎤ ⎡ ⎤
=− + = +
⎢⎥ ⎢ ⎥
⎣ ⎦⎣⎦
(4.2)
And
2
2
1dw dw d
MD D
dr r dr dr r
θ
φφ
υυ⎡⎤
⎡ ⎤
=− + = +
⎢⎥ ⎢ ⎥
⎣ ⎦⎣⎦
(4.3)
Where and
r
M M
θ
denotes bending moments per unit length. The moment
r
Macts along
circumferential section of the plate, such as the section made by conical surface with the apex at
B, and
M
θ
acts along the diametral section rz of the plate.
Equations (4.2) and (4.3) contains only one variable, or
w
φ , which can be determine by
considering equilibrium of an element of the plate such as element ‘
abcd’ as shown in figure cut
out from the plate by two cylindrical sections
ab and cd and by two diametral sections ad and bc.
The couple acting on the side
cd of the element is
r
Mrdθ
The corresponding couple on the side
ab is

86. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     78 

()
r
r
dMM dr r dr d
dr
θ
⎛⎞
++
⎜⎟
⎝⎠

The couples on the sides’
ad and bc of the element are each
Mdr
θ
and they give a resultant
couple in the plane
roz
Mdr d
θ
θ


From the symmetry it can be concluded that the shearing forces that may act on the element must
vanish on diametral sections of the plate but that they are usually present on cylindrical sections
such as sides
cd and ab of the element. Denoting by Q the shearing force per unit length of the
cylindrical section of radius
r, the total shearing force acting on the sides cd of the element is
Qrd
θ, and the corresponding force on the side ab is ()
dQ
Qdrrdrd
dr
θ
⎡⎤⎛⎞
++
⎜⎟⎢⎥
⎝⎠⎣⎦
. Neglecting
the small difference between the shearing forces on the two opposite sides of the element, we can
state that these forces give a couple in the
rz plane equal to Qrd dr
θ

87. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     79 

Summing up the moments with proper sign and neglecting the moment due to external
load on the element as a small quantity of higher order, we obtain the following equation of
equilibrium of the element
abcd.
() 0
r
rr
dM
M dr r dr d M rd M drd Qrd dr
dr
θ
θθ θθ
⎛⎞
∴ ++−−+=
⎜⎟
⎝⎠

()() 0
r
rr
dM
M r dr d dr r dr d M rd M drd Qrd dr
dr
θ
θθθθθ
∴ ++ +− − + =
0
rr
rr r
dM dM
M r d M dr d r dr d dr dr d M r d M dr d Q r d dr
dr dr
θ
θθ θ θθθθ
∴ ++ + −−+=
Since and
dr d
θ are very small its higher power is neglected.
0
r
r
dM
M drd r drd M drd Qrd dr
dr
θ
θθθθ
∴ +−+=
0
r
r
dM
Mr MQr
dr
θ
∴ +−+= (a)
Put value of and
r
M M
θ
from equation (4.2) and (4.3) in (a)
0
ddd d
DrDDQr
dr r dr dr r dr rφυ φυ φφ
φφυ⎡⎤ ⎡⎤⎡ ⎤
∴ ++ +− ++=
⎢⎥ ⎢⎥⎢ ⎥
⎣⎦ ⎣⎦⎣ ⎦

2
22
1ddd dQr
rrr
dr r dr r dr r dr r Dφυ φ υφ φφ
φυφυ ⎛⎞ ⎡⎤
∴ ++ + + − − −=−
⎜⎟ ⎢⎥
⎣⎦⎝⎠

2
2
ddddQr
r
dr r dr dr r dr r Dφυ φ φυφ φφ
φυυ
∴ ++ + −− −=−
2
22
1dd Q
rdr dr r Dφφφ
∴ +−=− (4.4)
Put
dw
dr
φ=−
2
22
11d dw d dw dw Q
rdr dr dr dr r dr D⎡⎤ ⎡⎤ ⎡⎤
∴ −+ − −−=−
⎢⎥ ⎢⎥ ⎢⎥
⎣⎦ ⎣⎦ ⎣⎦

88. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     80 

32
322
11dw dw dw Q
dr r dr r dr D
∴ −− + =−
32
322
11dw dw dw Q
dr r dr r dr D
∴ +−= (4.5)
In any particular case of symmetrically loaded circular plate the shearing force
Q can easily
being calculated by dividing the load distributed the circle of radius
r by2r
π; then equation
(4.4) or (4.5) can be used to determine the slope φ and the deflection w of the plate. The
integration of these equations is simplified if we observe that they can be put in the following
form.
1dddwQ
r
dr r dr dr D⎡⎤⎛⎞
=
⎜⎟⎢⎥
⎝⎠⎣⎦
(4.6)
If
Q is represented by a function r, this equation can be integrated without any difficulty in each
particular case. Sometimes it is advantageous to represent the right hand side of equation (4.6) as
a function of intensity
q of the load distributed over the plate.
For this purpose we multiply both sides of the equation by2
r
π. Then, observing that
0
22
r
rQ rqdrππ∴ =∫

0
r
Qqdr∴ =∫

Put in the equation (4.6), and multiply both sides of the equation by
r, we obtained
0
11
r
dddw
rr qrdr
dr r dr dr D⎡⎤⎛⎞
∴ =
⎜⎟⎢⎥
⎝⎠⎣⎦∫

Differentiate the term inside the bracket with respect to
r, we get
2
2
0
11
r
ddwdw
rr qrdr
dr r dr dr D⎡⎤⎛⎞
∴ +=⎢⎥⎜⎟
⎝⎠⎣⎦ ∫

89. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     81 

Again differentiate left hand side of the above equation with respect to
r, we obtained
32 2 2
32 22 2
0
11
r
dw dw dw dw dw q
rr r rdr
rdrdr dr rdrdr D⎡⎤⎛⎞⎧⎫ ⎛⎞
∴ ++− +=⎢⎥⎜⎟⎨⎬ ⎜⎟
⎢⎥⎩⎭ ⎝⎠⎝⎠⎣⎦ ∫

322
322
0
1
2
r
dw dw dw dw q
rrdr
dr dr dr r dr D
∴ +−−= ∫

Differentiate both sides of the equation w.r.t. r to eliminate integral term
43 3 2
43 3 2 2
11dw dw dw dw dw qr
r
dr dr dr r dr dr r D⎛⎞⎛ ⎞ ⎡⎤
∴ ++− +−=⎜⎟⎜ ⎟ ⎢⎥
⎣⎦⎝⎠⎝ ⎠
432
4323
21 1dw dw dw dw q
dr r dr r dr r dr D
∴ +−+= (4.7)
This can also be written as
11ddddw q
rr
r dr dr r dr dr D⎡⎤⎧⎫⎛⎞
=⎨⎬⎢⎥ ⎜⎟
⎝⎠⎩⎭⎣⎦ (4.8)
This equation can be easily integrated if the intensity of the load q is given as a function of r
4.2 Equation of Deflection for Uniformly Loaded Circular Plate: If a circular plate of radius
a carries a load of intensity q uniformly distributed over the entire surface of the plate,
Now multiply equation (4.8) by
r
1ddddw qr
rr
dr dr r dr dr D⎡⎤⎧⎫⎛⎞
=⎨⎬⎢⎥ ⎜⎟
⎝⎠⎩⎭⎣⎦

Integrate w.r.t
r
2
1
1
2dddw qr
rr C
dr r dr dr D⎡⎤⎧⎫⎛⎞
= +⎨⎬⎢⎥ ⎜⎟
⎝⎠⎩⎭⎣⎦
where
1
C is a constant of integration to be found later from the conditions at the center and at the
edge of the plate. Dividing both sides of above equation by
r and making second integration, we
find

90. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     82 

2
12
1
log
4ddwqr
rCrC
rdr dr D⎛⎞
= ++
⎜⎟
⎝⎠
Multiply by
r to the both sides of the above equation and performing integration with respect to
r, we obtained
42 2 2
123
log
16 2 4 2
dw q r r r r
rCrCC
dr D ⎡⎤
=+ −++
⎢⎥
⎣⎦

Divide by
r to the both sides of the equation, we get
3
3
12
log
16 2 4 2
Cdw q r r r r
Cr C
dr D r⎡⎤
=+ −++
⎢⎥
⎣⎦

Integrate w.r.t.
r
42 2 2
1234
log log
64 4 4 4
qr r r r
wCrCCrC
D ⎡⎤
=+ −++ +
⎢⎥
⎣⎦
(4.9)
Rearranging the term we get
4
22
12 34
log log .
64
qr
wC r C rr Cr C
D
=+ +++ (4.10)
The above equation gives the deflection of plate subjected to uniformly distributed load.


Let us now calculate the constants of integration for various particular cases.

4.2.1 Simply supported circular plate subjected to uniformly distributed load:
From generalized expression (4.10) for deflection of deflected surface of circular plate
subjected to uniformly distributed load.
4
22
12 34
log log .
64
qr
wC r C rr Cr C
D
=+ +++

91. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     83 


Note: since the deflection, moment and transverse shear are to be finite at the centre of
plate
()0=r . Therefore the constants C1 and C2 have to remain zero.
Therefore above equation becomes

4
2
34
64
qr
wCrC
D
∴ =++ (4.11)
The constants of integration are now to be determined from the conditions at the edges of the
plate.
0forwra== (4.12a)
0for
r
M ra== (4.12b)
Using boundary condition (4.12a), substitute
r = a in the equation (4.11) and equate with zero,
we obtain
4
2
34
0
64
qa
Ca C
D
∴ ++ = (4.13)
Differentiate equation (4.11) with respect to
r upto second order to find the moment of the plate
33
33
4
22
64 16dw q r q r
Cr Cr
dr D D
∴ =+=+ (4.14)

92. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     84 

22
32
3
2
16dw q r
C
dr D
∴ =+ (4.15)
Put equation (4.14) and (4.15) into the equation (4.2)
23
33
3
22
16 16
r
qr qr
MDC Cr
Dr Dυ⎡⎤ ⎛⎞
=− + + +⎢⎥ ⎜⎟
⎝⎠⎣⎦
(4.15 a)
Since this bending moment will vanish at the edges
i.e. at r = a, Use equation (4.12b) to obtained
constant
3
C by equating equation (4.15 a) with zero
23
33
3
220
16 16qa qa
DC Ca
Da Dυ⎡⎤ ⎛⎞
∴ −++ +=⎢⎥ ⎜⎟
⎝⎠⎣⎦

22
33
3
22 0
16 16qa qa
CC
DD
υ
∴ ++ +=
()()
2
3
21 3 0
16
qa
C
D
υυ
∴ ++ +=
()
()
2
3
3
32 1qa
C
D
υ
υ
+
∴ =−
+
(4.16)
Put constant
3
C from equation (4.16) into the equation (4.13) to obtain constant
4
C
()
()
24
2
4
3
0
32 1 64qa qa
aC
DD υ
υ+
∴ −++=
+

()
()
44
4
3
32 1 64qa qa
C
DD υ
υ+
∴ =−
+

()
()
4
4
23
1
64 1qa
C
D υ
υ⎛⎞+
∴ =− ⎜⎟
⎜⎟
+
⎝⎠

()
4
4
62 1
64 1qa
C
D
υυ
υ⎛⎞+−−
∴ = ⎜⎟
⎜⎟
+
⎝⎠

93. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     85 

()
()
4
4
5
64 1qa
C
D
υ
υ
+
∴ =
+
(4.17)
Substitute constants
34
andCC from equations (4.16) and (4.17) into the equation (4.11) to
obtain deflection of the plate, hence
()
()
()
()
244
2
35
32 1 64 1 64qa qa qr
wr
DDDυυ
υυ++
∴ =− + +
++

()
()
()
()
22 4 4
23 5
64 1 1q
warar
D υυ
υυ⎛⎞ ++
∴ =− + +⎜⎟
⎜⎟
++
⎝⎠

()
()
()
()
424
24
23 5
64 1 1qa r r
w
Da a υυ
υυ⎛⎞ ++
∴ =− + +⎜⎟
⎜⎟
++
⎝⎠

()
()
()
()
42
4
23 5
64 1 1qa r r
w
Da a
υ υ
υ υ
⎡ ⎤++⎛⎞ ⎛⎞
∴ =− +⎢ ⎥⎜⎟ ⎜⎟
++⎝⎠ ⎝⎠⎢ ⎥⎣ ⎦
(4.18)
Deflection is maximum at the center of the plate
i.e. at 0r
=
()
()
4
5
64 1qa
w
D
υ
υ
+
∴ =
+
(4.19)
Substitute C3 in equation (4.15 a) to find out bending moment
r
M, we get
()
()
()
()
2223
33 3
16 1 16 16 1 16
r
qa q r qa qr
MD r
DDrD Dυυ υ
υυ⎡⎤ ⎛⎞++
∴ =− − + + − +⎢⎥ ⎜⎟
⎜⎟
++⎢⎥ ⎝⎠⎣⎦

()
()
()
()
422
2424
343412 4
64 1 1
r
qa r r
M
aaaa υυυ υ
υυ⎡⎤ ++
∴ =− + − +⎢⎥
++
⎣⎦

()
()()
()
42
42
41 3
12 4
64 1
r
qa r
M
aa υυ
υ
υ⎡⎤ ++
∴ =− + −⎢⎥
+
⎣⎦

() ()
()
()
4
22
4
3
3441
64 1
r
qa
Mra
a υ
υυ
υ
⎡ ⎤+
∴ =− + − +⎢ ⎥
+
⎣ ⎦

94. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     86 

() ()
22
3443
64
r
q
Mra
υ υ⎡⎤∴ =− + − +
⎣⎦

() ()
22
3
16
r
q
Mra
υ⎡⎤
∴ =− + −
⎣⎦

() ()
22
3
16
r
q
Mar
υ⎡⎤
∴ =+−
⎣⎦
(4.20)
Use equation (4.3) to find bending momentM
θ
, therefore differentiate equation (4.18) w.r.t. r
upto second order
()
()
43
42
434
64 1dw qa r r
dr D a a υ
υ⎡⎤ +
∴ =− ⎢⎥
+
⎣⎦

()
()
242
242
4312
64 1dw qa r
dr D a a υ
υ
⎡ ⎤+
∴ =− ⎢ ⎥
+
⎣ ⎦

Substitute in the equation (4.3) ofM
θ
, we obtain
()
()
()
()
42 3
42 4 2
43 4312 1 4
64 1 1qa r r r
M
aa ra a
θ
υυ
υ
υυ⎡⎤⎛⎞⎛ ⎞ ++
∴ =− − + −⎢⎥⎜⎟⎜ ⎟
⎜⎟⎜ ⎟
++⎢⎥⎝⎠⎝ ⎠⎣⎦

()
()
()
()
42 2
42 4 2
43 4 3412
64 1 1qa r r
M
aa a a
θ
υ υυυ
υ υ
⎡⎤ ++
∴ =− − + −⎢⎥
++
⎣⎦

()
()
()
()
224
22
4
334
3
64 1 1aaqa
Mr r
a
θ
υ υυ
υ
υυ
⎡ ⎤++
∴ =− − + −⎢ ⎥
++
⎣ ⎦

()()
()
()
2
2
3
13 1
16 1aq
Mr
θ
υ
υυ
υ⎡⎤ +
∴ =− + − +⎢⎥
+
⎣⎦

() ()
22
13 3
16
q
Mra
θ
υ υ⎡⎤∴ =− + − +
⎣⎦
(4.21)
Bending stresses of the plate are to be found out from the following equations

95. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     87 

()
33
............. 1
12 12
r
r
M bh h
zI b
I
σ==== (4.22)
Substitute equation (4.20) into the above equation (4.22), we get
() ()
22
3
16
r
qz
ar I
συ ⎡ ⎤∴ =+−
⎣ ⎦

() ()
22
3 12
3
16
r
q
ar z
h
συ ⎡⎤
∴ =+−
⎣⎦
() ()
22
33
3
4
r
z
qar
h
συ
⎡ ⎤∴ =+−
⎣ ⎦
(4.23)
Similarly
()
33
............. 1
12 12
M bh h
zI b
I
θ
θ
σ==== (4.24)
Substitute equation (4.21) into the above equation (4.24), we obtain
() ()
22
13 3
16
qz
ra I
θ
συυ ⎡ ⎤∴ =− + − +
⎣ ⎦

()( )
22
3 12
313
16q
arz
h
θ
συυ⎡⎤
∴ =+−+
⎣⎦

()( )
22
33
313
4z
qa r
h
θ
συυ
⎡ ⎤∴ =+−+
⎣ ⎦
(4.25)
Maximum stress will occur at centre of plate
i.e. r = 0, therefore put r = 0 in equation (4.23)
()
()
2
3max 03
3
4
r
r
z
qa
h
συ
=
∴ =+
()()
2
3max3
3
42
r
h
qa
h
συ
∴ =+
()()
2
max
3
3
8
r
a
q
h
σ υ
⎛⎞
∴ =+
⎜⎟
⎝⎠

96. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     88 

() ()
2
max3
3 Aspect Ratio
8
r
a
q
h
σηυ η
⎛⎞
∴ =+ ==
⎜⎟
⎝⎠ (4.26)
Similarly, Put
r = 0 in equation (4.25) bending stress
M
θ

() ()
2
3max 03
3
4
r
z
qa
h
θ
συ
=
∴ =+
()()
2
3max3
3
42h
qa
h
θ
συ
∴ =+
()()
2
max
3
3
8a
q
h
θ
σ υ
⎛⎞
∴ =+
⎜⎟
⎝⎠
()
()
2
max3
3AspectRatio
8 a
q
h
θ
σηυ η
⎛⎞
∴ =+ ==
⎜⎟
⎝⎠ (4.27)
Therefore maximum stresses in andr
θ are same.
() ()
max max rθ
σσ∴ = (4.28)
4.2.2 Fixed / Clamped circular plate subjected to uniformly distributed load:

From generalized expression (4.10) for deflection of deflected surface of circular plate subjected
to uniformly distributed load.

97. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     89 

4
22
12 34
log log .
64
qr
wC r C rr Cr C
D
=+ +++
Note: since the deflection, moment and transverse shear are to be finite at the centre of
plate
()0=r . Therefore the constants C1 and C2 have to remain zero.
Therefore above equation becomes

4
2
34
64
qr
wCrC
D
∴ =++ (4.29)
Integration constants
34
andCC are to be finding out using boundary conditions at the edges of
the plate. In this case the slope of the deflected surface in the radial direction must be zero for

0andrra== and deflection must be zero at the edges of plate i.e. at ra=
0forwra== (4.30a)
0for
dw
ra
dr
== (4.30b)
Using boundary condition (4.30a), substitute
r = a in the equation (4.29) and equate with zero,
we obtain

4
2
34
0
64
qa
Ca C
D
∴ ++ = (4.31)
Now using boundary condition (4.30b), differentiate equation (4.29) with respect to
r and equate
with zero, we obtain
33
33
4
22
64 16dw q r q r
Cr Cr
dr D D
∴ =+=+ (4.32)
3
3
2
16
ra
dw q a
Ca
dr D
=
∴ =+
3
3
20
16
qa
Ca
D
∴ +=

98. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     90 

2
3
32
qa
C
D
∴ =− (4.33)
Substitute value of C
3 in the equation (4.31)
24
2
4
0
32 64
qa qa
aC
DD
∴ −++=
24
2
4
32 64
qa qa
Ca
DD
∴ =−
4
4
64
qa
C
D
∴ = (4.34)
Substitute C
3 and C4 in the equation (4.29)
244
2
32 64 64
qa qa qr
wr
DDD
∴ =− + +
()
4224
2
64
q
waarr
D
∴ =−+
()
2
22
64
q
war
D
∴ =− (4.35)
Deflection is maximum at the center of the plate i.e. 0r=
()
4
0
max
64
r
qa
w
D
=
∴ = (4.36)
This deflection is equal to three-eighths of the deflection of a uniformly loaded strip with built in
ends having flexural rigidity equal to
D, a width of unity, and a length of the diameter of the
plate.
Now differentiate equation (4.35) with respect to
r upto second order and substitute in the
equations (4.2) and (4.3) to obtain the bending moments and
r
M M
θ

() ()
22
22
64
dw q
ar r
dr D
∴ =−−

99. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     91 

()
22
16
dw q
ra r
dr D
∴ =− − (4.37)
()
2
222
2
2
16
dw q
rar
dr D
⎡ ⎤∴ =− − + −
⎣ ⎦

2
22
2
3
16
dw q
ra
dr D
⎡ ⎤∴ =−
⎣ ⎦
(4.38)
Substitute equation (4.37) and (4.38) into the equation (4.2) to obtain the expression for
r
M
() ()
22 22
3
16 16
r
qqM Dra rar
DrD
υ⎡⎤ ⎛⎞∴ =− − − −
⎜⎟⎢⎥
⎝⎠⎣⎦

()()
22 22
3
16
r
qM ra arυ⎡ ⎤∴ =− − − −
⎣ ⎦

()()
22
31
16
r
q
Mr a
υ υ⎡ ⎤∴ =− + − +
⎣ ⎦

() ( )
22
13
16
r
q
Ma r
υυ
⎡ ⎤∴ =+−+
⎣ ⎦
(4.39)
In the same manner substitute equation (4.37) and (4.38) into the equation (4.3) to obtain M
θ

()
22 22 1
3
16 16qq
MD ra rar
DrD
θ
υ
⎡⎤⎛⎞⎛ ⎞
⎡⎤
∴ =− − + − −
⎜⎟⎜ ⎟⎢⎥ ⎣⎦
⎝⎠⎝ ⎠⎣⎦

2222
3
16
qM raar
θ
υυ⎡ ⎤∴ =− − − +
⎣ ⎦

()()
22
13 1
16
qM ra
θ
υυ⎡ ⎤∴ =− + − +
⎣ ⎦
(4.40)
Substituting
ra= in the equation (4.39) and (4.40), we find for the bending moments at the
boundaries of the plate
() ( )
22
13
16
rra
q
Maa
υ υ
=
⎡ ⎤∴ =+−+
⎣ ⎦

100. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     92 

22 22
3
16
rra
q
M aa aaυυ
=
⎡ ⎤∴ =+−−
⎣ ⎦

()
2
max
8
rra
qa
M
=
∴ =− (4.41)
And
()()
22
13 1
16
ra
qM aa
θ
υυ
=
⎡ ⎤∴ =− + − +
⎣ ⎦

2222
3
16
ra
qM aaaa
θ
υυ
=
⎡ ⎤∴ =− + − −
⎣ ⎦

()
2
max
8
ra
qa
M
θ
υ
=
∴ =− (4.42)
Bending moments are maximum at the centre of the plate i.e. 0
r
=
()
2
0
1
16
rr
q
Ma
υ
=
∴ =+
()
2
0
1
16
r
q
M a
θ
υ
=
∴ =+
00 rrr
MM
θ==
∴ = (4.43)
Corresponding Bending Stresses are given by the following equation

()
33
& ............. 1
12 12 2
r
r
M bh h h
zIzb
I
σ===== (4.44)
Put equation (4.41) in the equation (4.44) to find maximum bending stress
()
2
3max
12
82
r
qa h
h
σ
∴ =−
()
2
max
3
4
r
qa
h
σ
⎛⎞
∴ =−
⎜⎟
⎝⎠
()
2
max3
4
r
q
σ η∴ =− (4.45)

101. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     93 

Similarly
()
33
& ............. 1
12 12 2
M bh h h
zIzb
I
θ
θ
σ===== (4.46)
Put equation (4.45) in the equation (4.46)
()
2
3max
12
82qa h
h
θ
σ
∴ =−
()
2
max
3
4qa
h
θ
σ
⎛⎞
∴ =−
⎜⎟
⎝⎠
()
2
max3
4q
θ
σ η∴ =− (4.47)
From equations (4.45) and (4.47) it is observed that
()
()
max maxr θ
σσ∴ = (4.48)
4.3 Equation of Deflection for Circular Plate subjected to Centre Concentrated Load

Consider a circular plate of radius a carrying a concentrated load q at the centre.
2
2
q
rQ q Q
r
π
π
=∴=
Substitute above value of shearing force
Q in the equation (4.6), we obtained
1
2dddw q
r
dr r dr dr r D
π
⎡⎤⎛⎞
∴ =
⎜⎟⎢⎥
⎝⎠⎣⎦

102. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     94 

Integrate both sides of above equation with respect to r
, we get
1
1
log
2ddw q
rrC
rdr dr D
π
⎡⎤⎛⎞
∴ =+
⎜⎟⎢⎥
⎝⎠⎣⎦

where
1
C is a constant of integration to be found from the condition at the center and at the edge
of the plate. Now multiply both sides of above equation by
r
1
log
2
ddw q
rrrCr
dr dr D
π
⎛⎞
∴ =+
⎜⎟
⎝⎠
Integrate both sides of above equation with respect to
r, we get
222
12
log
22 4 2
dw q r r r
rrCC
dr D
π
⎡⎤⎛⎞
∴ =−++
⎜⎟ ⎢⎥
⎝⎠ ⎣⎦
2
1
log
22 4 2
Cdw q r r r
rC
dr D r
π
⎡⎤
∴ =−++
⎢⎥
⎣⎦
Integrate w.r.t.
r
2222
12 3
1
log log
222 48 4qr rrr
wrCCrC
D
π
⎡⎤⎛⎞
∴ =−−+++⎢⎥⎜⎟
⎝⎠⎣⎦

()
22
12 3
log 1 log
84
qr r
wrCCrC
D
π
∴ =−+++ (4.49)
4.3.1 Simply supported circular plate subjected to centre concentrated / point load:
To find the expressions for deflection, moments and corresponding stresses for the
simply supported plate subjected to concentrated load, consider generalized expression (4.49) for
deflection of deflected surface of circular plate subjected to concentrated load.
()
22
12 3
log 1 log
84
qr r
wrCCrC
D
π
∴ =−+++
Note: since the deflection, moment and transverse shear are to be finite at the centre of
plate
()0=r . Therefore the constants C2 have to remain zero.

103. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     95 



()
22
13
log 1
84
qr r
wrCC
D
π
∴ =−++ (4.50)
Integration constants
34
andCC are to be finding out using boundary conditions at the edges of
the plate. In this case deflection and moment must be zero at the edges of plate i.e. at
ra
=
0forwra== (4.51a)
0for
r
M ra== (4.51b)
Therefore using boundary condition (4.51a), substitute
r = a into the equation (4.50) and equate
with zero
()
22
13
log 1
84
ra
qa a
waCC
D
π
=
∴ =−++
()
22
13
log 1 0
84
qa a
aCC
D
π
∴ −+ + = (4.52)
Now, differentiate equation (4.50) with respect to r upto second order to find out moment
r
M
()
2
11
log 1 2
82dw q r
rrrC
dr D r
π
⎡⎤
∴ =+−+
⎢⎥
⎣⎦

104. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     96 

()
1
2log 1
82
dw qr r
rC
dr D
π
∴ =−+ (4.53)
()
2
1
2
2
2log 1
82 Cdw q
rr
dr D r
π
⎡⎤
∴ =+−+
⎢⎥
⎣⎦

()
2
1
2
2log 1
82
Cdw q
r
dr D
π
∴ =++ (4.54)
Put equations (4.53) and (4.54) into the equation (4.2), we get
() ()
1
1
2log 1 2log 1
828 2
r
Cqqrr
MD r r C
DrDυ
ππ⎧⎫⎡⎤⎡ ⎤⎪⎪
∴ =− + + + − +⎨⎬⎢⎥⎢ ⎥
⎪⎪⎣⎦⎣ ⎦⎩⎭

()() ()
1
2log 1 1 1
82
r
Cq
MD r
D
υ υυ
π
⎧⎫ ∴ =− + + − + +⎡⎤⎨⎬ ⎣⎦
⎩⎭
Therefore using boundary condition (4.51b), substitute
r = a in the above equation and equate
with zero, we obtain
()() ()
1
2log 1 1 1
82
rra
Cq
MD a
D
υ υυ
π
=
⎧ ⎫
∴ =− + + − + +⎡⎤⎨ ⎬⎣⎦
⎩⎭
()() ()
1
2log 1 1 1 0
82
Cq
a
D
υυ υ
π
∴ ++− + +=⎡⎤
⎣⎦

() ()()
1
12log11
28
Cq
a
D
υ υυ
π
∴ +=− ++− ⎡ ⎤
⎣ ⎦

()
()
1
1
2log
41q
Ca
D
υ
π υ
⎡ ⎤−
∴ =− + ⎢ ⎥
+
⎣ ⎦
(4.55)
Substitute value of constant
1
Cfrom equation (4.55) into the equation (4.52) to obtain constant
3
C, we obtain
()
()
()
22
3
1
log 1 2log 0
8414qa q a
aaC
DD υ
ππυ⎡⎤ −
∴ −− + += ⎢⎥
+
⎣⎦

105. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     97 

()
()
()
22
3
1
2log log 1
16 1 8qa qa
Ca a
DD υ
πυπ⎡⎤ −
∴ =+−−⎢⎥
+
⎣⎦

()
()
()
2
3
11
2log log 1
82 1qa
Caa
D υ
πυ
⎧ ⎫⎡⎤ −⎪ ⎪
∴ =+−−⎨ ⎬⎢⎥
+
⎪ ⎪⎣⎦⎩⎭

()
()
2
3
11
1
821qa
C
D
υ
π υ
⎡ ⎤−
∴ =+ ⎢ ⎥
+
⎣ ⎦
(4.56)

Substitute in the equation (4.50) of deflection of the plate, we get
()
()
()
()
()
222
11 1
log 1 2log 1
8414821qr q r qa
wr a
DD D
υ υ
π πυπυ
⎡ ⎤⎡ ⎤−−
∴ =−− +++ ⎢ ⎥⎢ ⎥
++
⎣ ⎦⎣ ⎦

()
()
()
()
()
22 2
11 1
2log 1 2log 2 1
16 1 2 1q
wrrra a
D υυ
πυυ⎧⎫
⎡ ⎤⎡ ⎤−−⎪⎪
∴ =−−+++⎨⎬ ⎢ ⎥⎢ ⎥
++
⎪⎪ ⎣ ⎦⎣ ⎦⎩⎭

()
()
()
()
()
22222
11
2log log 2 2
16 1 1q
wrrarara
D
υ υ
π υυ
⎡⎤ −−
∴ =−−+−+⎢⎥
++
⎣⎦

()
222 22 1
2log
16 1qr
wr ar
Da υυ
πυ⎧++−⎫⎛⎞ ⎡ ⎤
∴ =+− ⎨⎬ ⎜⎟⎢⎥
+⎝⎠ ⎣ ⎦⎩⎭

()
222 3
2log
16 1qr
wr ar
Da υ
πυ⎧+⎫⎛⎞ ⎛ ⎞
∴ =+− ⎨⎬ ⎜⎟ ⎜ ⎟
+⎝⎠ ⎝ ⎠⎩⎭
(4.57)
Deflection of the plate is maximum at the center of the plate
i.e. r = 0 ()
2
0
max3
16 1
r
q
wa
D υ
πυ
=
+⎛⎞
∴ =
⎜⎟
+⎝⎠ (4.58)
Differentiate equation (4.57) w.r.t.
r upto second order
213
2log4 2
16 1dw q a r
rrr
dr D r a a υ
πυ
⎧ ⎫⎡⎤+ ⎛⎞ ⎛ ⎞
∴ =+− ⎨ ⎬⎜⎟ ⎜ ⎟⎢⎥
+⎝⎠ ⎝ ⎠⎣⎦⎩⎭

106. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     98 

3
2log 4 2
16 1dw q r
rrr
dr D a υ
πυ
⎧ +⎫⎛⎞ ⎛ ⎞
∴ =+− ⎨ ⎬⎜⎟ ⎜ ⎟
+⎝⎠ ⎝ ⎠⎩⎭

3
log 4 2 1
16 1dw q r
rr
dr D a υ
πυ
⎧ ⎫⎡ +⎤⎛⎞ ⎛ ⎞
∴ =+− ⎨ ⎬⎜⎟ ⎜ ⎟ ⎢ ⎥
+⎝⎠ ⎝ ⎠ ⎣ ⎦⎩⎭

log
41
dw q r r
r
dr D a
π υ
⎧ ⎫⎛⎞
∴ =− ⎨ ⎬⎜⎟
+⎝⎠⎩⎭
(4.59)
2
2
11
log
41dw q a r
r
dr D r a a
π υ
⎧ ⎫⎛⎞
∴ =+− ⎨ ⎬⎜⎟
+⎝⎠⎩⎭

2
2
11
log
41dw q a r
r
dr D r a a
π υ
⎧ ⎫⎛⎞
∴ =+− ⎨ ⎬⎜⎟
+⎝⎠⎩⎭

2
2
log
41
dw q r
dr D a
υ
π υ
⎧ ⎫⎛⎞
∴ =+ ⎨ ⎬⎜⎟
+⎝⎠⎩⎭
(4.60)
Now, to find the bending moments of the plate, Put equation (4.59) and (4.60) into the equation
(4.2) to obtain
r
M
log log
414 1
r
qr q rr
MD r
Da rD a υυ
π υπ υ
⎡⎤⎧⎫⎧ ⎫⎛⎞ ⎛⎞
∴ =− + + −⎨⎬⎨ ⎬⎢⎥ ⎜⎟ ⎜⎟
++⎝⎠ ⎝⎠⎩⎭⎩ ⎭⎣⎦
log log
41 1
r
qr r
MD
Da a
υ υ
υ
π υυ
⎡ ⎤⎛⎞ ⎛⎞
∴ =− + + −
⎜⎟ ⎜⎟⎢ ⎥
+ +⎝⎠ ⎝⎠⎣ ⎦

log log
4
r
qr r
M
aa
υ
π
⎡⎤⎛⎞ ⎛⎞
∴ =− +
⎜⎟ ⎜⎟⎢⎥
⎝⎠ ⎝⎠⎣⎦

()log 1
4
r
qr
M
a
υ
π
⎡⎤⎛⎞
∴ =− +
⎜⎟⎢⎥
⎝⎠⎣⎦
(4.61)
Similarly put equation (4.59) and (4.60) in the equation (4.3) to obtain M
θ

1
log log
414 1qr q rr
MD r
Da rD a
θ
υ
υ
π υπ υ
⎡⎤ ⎧⎫⎧ ⎫⎛⎞ ⎛⎞
∴ =− + + − ⎨⎬⎨ ⎬⎢⎥ ⎜⎟ ⎜⎟
++⎝⎠ ⎝⎠⎩⎭⎩ ⎭⎣⎦

107. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     99 

1
log log
41 1qr rr
MD r
Da ra
θ
υ
υ
π υυ
⎡⎤⎧⎫⎧ ⎫⎛⎞ ⎛⎞
∴ =− + + − ⎨⎬⎨ ⎬⎢⎥ ⎜⎟ ⎜⎟
++⎝⎠ ⎝⎠⎩⎭⎩ ⎭⎣⎦
2
1
log log
41 1qr r
M
aa
θ
υ
υ
π υυ
⎡⎤⎛⎞ ⎛⎞
∴ =− + + −
⎜⎟ ⎜⎟⎢⎥
+ +⎝⎠ ⎝⎠⎣⎦

()
2
1
1log
411qr
M
a
θ
υ
υ
π υυ
⎡⎤ ⎛⎞⎛⎞
∴ =− + + −⎢⎥ ⎜⎟⎜⎟
++⎝⎠⎝⎠⎣⎦

()
2
1
1log
41qr
M
a
θ
υ
υ
πυ⎡⎤ ⎛⎞−⎛⎞
∴ =− + +⎢⎥ ⎜⎟⎜⎟
+⎝⎠⎝⎠⎣⎦

()
()()11
1log
41qr
M
a
θ
υυ
υ
πυ⎡⎤ +−⎛⎞⎛⎞
∴ =− + +⎢⎥ ⎜⎟⎜⎟
+⎝⎠⎢⎥ ⎝⎠⎣⎦

() ()1log 1
4
qr
M
a
θ
υ υ
π
⎡⎤ ⎛⎞∴ =+ +−
⎜⎟⎢⎥
⎝⎠⎣⎦
(4.62)
Corresponding bending stresses are to be find out using following equations
()
()
3
2
12
log 1
42
3
log 1
2
r
r
r
M qr h
z
I ah
qa
hr
συ
π
συ
π
⎡⎤⎛⎞
=⇒− +
⎜⎟⎢⎥
⎝⎠⎣⎦
⎡⎤⎛⎞
⇒+
⎜⎟⎢⎥
⎝⎠⎣⎦
(4.63)
Similarly
() ()
() ()
3
2
12
1log 1
42
3
1log 1
2
M qr h
z
I ah
qr
ha
θ
θ
θ
συυ
π
συυ
π
⎡⎤ ⎛⎞
=⇒ + +−
⎜⎟⎢⎥
⎝⎠⎣⎦
⎡ ⎤⎛⎞
⇒++−
⎜⎟⎢ ⎥
⎝⎠⎣ ⎦
(4.64)

108. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     100 

4.3.2 Fixed / clamped circular plate subjected to centre concentrated / point load:

Consider generalized expression (4.49) for deflection of deflected surface of circular plate
subjected to concentrated load.
()
22
12 3
log 1 log
84
qr r
wrCCrC
D
π
∴ =−+++
Note: since the deflection, moment and transverse shear are to be finite at the centre of
plate
()0=r . Therefore the constants C2 have to remain zero.
()
22
13
log 1
84
qr r
wrCC
D
π
∴ =−++ (4.65)
Integration constants
13
andCC are to be finding out using boundary conditions at the edges of
the plate. In this case deflection and slope must be zero at the edges of plate i.e. at
ra=
0forwra== (4.66a)
0for
dw
ra
dr
== (4.66b)
Therefore using boundary condition (4.66a), substitute
r = a into the equation (4.65) and equate
with zero

109. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     101 


()
22
13
log 1
84
ra
qa a
waCC
D
π
=
∴ =−++
()
22
13
log 1 0
84
qa a
aCC
D
π
∴ −+ + = (4.67)
Now, differentiate equation (4.65) with respect to
r ()
2
11
log 1 2
82
dw q r
rrrC
dr D r
π
⎡⎤
∴ =+−+
⎢⎥
⎣⎦
()
1
2log 1
82
dw qr r
rC
dr D
π
∴ =−+
Now using boundary condition (4.66b), substitute
r = a in the above equation and equate with
zero ()
1
2log 1
82
ra
dw qa a
aC
dr D
π
=
∴ =−+
()
1
2log 1 0
82
qa a
aC
D
π
∴ −+ =
()
1
2log 1
28
aqa
Ca
D
π
∴ =− −
()
1
2log 1
4
q
Ca
D
π
∴ =− − (4.68)
Substitute value of constant C
1 from equation (4.68) into the equation (4.67) to obtain constant
3
C we get () ()
22
3
log 1 2log 1 0
844
qa q a
aaC
DD
ππ
∴ −− − + =
() ()
22
3
2log 1 log 1
448
qaqa
Ca a
DD
ππ
∴ =−−−

110. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     102 

()()
2
3
2log 1 2 log 1
16
qa
Caa
D
π
∴ =−−−⎡ ⎤
⎣ ⎦

2
3
16
qa
C
D
π
∴ = (4.69)
Therefore substitute values of integration constants
13
andCC into the equation (4.65) to find the
expression for deflection of the plate.
() ()
222
1
log 1 2log 1
84 416
qr q r qa
wr a
DD D
ππ π
∴ =−− −+
()( )
22
21
2log 1 2log 1
16
qr a
wra
Dr
π
⎡⎤
∴ =−−−+
⎢⎥
⎣⎦
22
2
2log 1
16
qr r a
w
Dar
π
⎡ ⎤⎛⎞
∴ =−+
⎜⎟⎢ ⎥
⎝⎠⎣ ⎦
(4.70)
Deflection is maximum at the center of the plate
i.e. at r = 0, therefore put r = 0 into the
equation (4.70)
()
2
max
16
qa
w
D
π
∴ = (4.71)
To find the bending moments differentiate equation (4.70) with respect to
r upto second order
2 1
4log 2 2
16dw q r a
rr r
dr D a r a
π
⎡ ⎤⎛⎞ ⎛⎞⎛⎞
∴ =+−
⎜⎟ ⎜⎟⎜⎟⎢ ⎥
⎝⎠ ⎝⎠⎝⎠⎣ ⎦

4log
16
dw q r
r
dr D a
π
⎡ ⎤⎛⎞
∴ =
⎜⎟⎢ ⎥
⎝⎠⎣ ⎦

log
4
dw qr r
dr D a
π
⎛⎞∴ =
⎜⎟
⎝⎠ (4.72)
2
2
1
log
4dw q a r
r
dr D r a a
π
⎡ ⎤⎛⎞⎛⎞ ⎛⎞
∴ =+
⎜⎟⎜⎟ ⎜⎟⎢ ⎥
⎝⎠⎝⎠ ⎝⎠⎣ ⎦

111. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     103 

2
2
1log
4
dw q r
dr D a
π
⎡ ⎤⎛⎞
∴ =+
⎜⎟⎢ ⎥
⎝⎠⎣ ⎦
(4.73)
Therefore substitute equations (4.72) and (4.73) into the equation (4.2) we obtain
1log log
44
r
qrqrr
MD
DarDa υ
ππ
⎡ ⎤⎡⎤ ⎛⎞ ⎛⎞
∴ =− + +⎢ ⎥⎜⎟ ⎜⎟⎢⎥
⎝⎠ ⎝⎠⎣⎦⎣ ⎦

()11 log
4
r
qr
M
a
υ
π
⎡ ⎤⎛⎞
∴ =− + +
⎜⎟⎢ ⎥
⎝⎠⎣ ⎦
(4.74)
Similarly substitute equations (4.72) and (4.73) into the equation (4.3) we obtain
1
1 log 4 log
416
qrqr
MD r
DarDa
θ
υ
ππ
⎡⎤
⎡ ⎤⎡⎤⎛⎞ ⎛⎞
∴ =− + +⎢⎥ ⎜⎟ ⎜⎟⎢ ⎥⎢⎥
⎝⎠ ⎝⎠⎣ ⎦⎣⎦⎣⎦

log log
4
qrr
MD
Daa
θ
υυ
π
⎡ ⎤⎛⎞ ⎛⎞
∴ =− + +
⎜⎟ ⎜⎟⎢ ⎥
⎝⎠ ⎝⎠⎣ ⎦

()1log
4
qr
M
a
θ
υυ
π
⎡ ⎤⎛⎞
∴ =− + +
⎜⎟⎢ ⎥
⎝⎠⎣ ⎦
(4.75)
Corresponding bending stresses are given as follows
()
3
12
11 log
42
r
rr
M qrh
z
I ah
σσυ
π
⎡⎤ ⎛⎞
=⇒=−++
⎜⎟⎢⎥
⎝⎠⎣⎦

()
2
3
11 log
2
r
qr
ha
συ
π
⎡ ⎤⎛⎞
∴ =− + +
⎜⎟⎢ ⎥
⎝⎠⎣ ⎦
(4.76)
Similarly
()
3
12
1log
42M qrh
z
I ah
θ
θθ
σσυυ
π
⎡⎤ ⎛⎞
=⇒=−++
⎜⎟⎢⎥
⎝⎠⎣⎦

()
2
3
11 log
2qr
ha
θ
συ
π
⎡ ⎤⎛⎞
∴ =− + +
⎜⎟⎢ ⎥
⎝⎠⎣ ⎦
(4.77)

112. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     104 

Equation (4.76) and (4.77) shows that stresses and
r
θ
σσ become infinite at the centre of plate
()0r= where concentrated load is applied. Therefore the theory which has been presented is
valid near the point of application of the concentrated load.
At
()ra=
()
()
2max max
3
42
rr
qq
M
h
σ
ππ=− ∴ = − (4.78)
()
()
2max max
3
42qq
M
h
θθ
υ υ
σ
ππ
=− ∴ = − (4.79)

4.4 Problem
A circular plate of radius R is clamped along boundary and carries a load whose
intensity / unit area
q is varies according to the relation
0
1
⎛⎞
=−
⎜⎟
⎝⎠
r
qq R
obtain an expression for
deflection of plate at the radius
R and hence calculate maximum deflection.
Solution:
Total load supported by plate at radial distance r
W = Total volume over radius r
= Volume of cylinder + Volume of cone

113. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     105 

()
22
01
3
Wrq rqqππ
∴ =+ −

2 0 2
33q
Wr q
π
⎡ ⎤
∴ =+
⎢ ⎥
⎣ ⎦


2 0
0 2
1
33q r
Wr q
R
π
⎡ ⎤⎛⎞
∴ =+−
⎜⎟⎢ ⎥
⎝⎠⎣ ⎦


2 0
0 2
1
33q r
Wr q
R
π
⎡ ⎤⎛⎞
∴ =+−
⎜⎟⎢ ⎥
⎝⎠⎣ ⎦


20
121
3
q r
Wr R
π
⎡ ⎤⎛⎞
∴ =+−
⎜⎟⎢ ⎥
⎝⎠⎣ ⎦


20 2
3
3q r
Wr
R
π
⎛⎞
∴ =−
⎜⎟
⎝⎠

()
20
32
3
q
Wr Rr
R
π
∴ =−
Therefore total load on the plate

2rQ W
π=
()
20
232
3
q
rQ r R r
R
ππ
∴ =−
()
20
32
6
q
QRrr
R
∴ =− (4.80)
Substitute in the governing equation (4.6) of the plate, we get

1dddwQ
r
dr r dr dr D⎡⎤⎛⎞
=
⎜⎟⎢⎥
⎝⎠⎣⎦

()
201
32
6 qdddw
rRrr
dr r dr dr RD⎡⎤⎛⎞
∴ =−
⎜⎟⎢⎥
⎝⎠⎣⎦

Integrate w.r.t.
r

114. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     106 


23
0
1
132
623
qddw Rrr
rC
rdr dr RD
⎛⎞⎡⎤⎛⎞
∴ =−+⎜⎟⎜⎟⎢⎥
⎝⎠⎣⎦ ⎝⎠

34
0
1
32
623
qddw Rrr
rCr
dr dr RD
⎛⎞⎛⎞
∴ =−+⎜⎟⎜⎟
⎝⎠ ⎝⎠
Integrate w.r.t.
r

45 2
0
12
32
6815 2
qdw R r r r
rCC
dr RD
⎛⎞⎛⎞
∴ =−++⎜⎟⎜⎟
⎝⎠ ⎝⎠

34
0 2
1
32
68152
q Cdw R r r r
C
dr RD r
⎛⎞
∴ =−++⎜⎟
⎝⎠

45 2
0
12 3
32
log
63275 4
q Rr r r
wCCrC
RD
⎛⎞
∴ =−+++⎜⎟
⎝⎠ (4.81)
Since deflection is finite at the centre
( )0r=, C2 = 0

45 2
0
13
32
63275 4
q Rr r r
wCC
RD
⎛⎞
∴ =−++⎜⎟
⎝⎠ (4.82)
Boundary conditions

0and 0
rR
rR
dw
w
dr
=
=
= = (4.83)

55 2
0
13
32
63275 4
rR
q RR R
wCC
RD
=
⎛⎞
∴ =−++⎜⎟
⎝⎠

4 2
0
13
161
6 2400 4
rR
qR R
wCC
D
=
⎛⎞
∴ =++
⎜⎟
⎝⎠

4 2
0
13
161
0
6 2400 4
qR R
CC
D
⎛⎞
∴ ++=
⎜⎟
⎝⎠ (4.84)
Differentiate equation (4.82) w.r.t.
r

115. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     107 


34
0
1
32
68152
qdw R r r r
C
dr RD
⎛⎞
∴ =−+⎜⎟
⎝⎠

44
0
1
32
6815 2
rR
qdw R R R
C
dr RD
=
⎛⎞
∴ =−+⎜⎟
⎝⎠

3
0
1
29
6 120 2
rR
qRdw R
C
dr D
=
⎛⎞
∴ =+
⎜⎟
⎝⎠
From boundary condition (4.83) we obtain

3
0
1
29
0
6 120 2
qR R
C
D
⎛⎞
∴ +=
⎜⎟
⎝⎠

3
0
1
29
2 6 120
qRR
C
D
⎛⎞
∴ =−
⎜⎟
⎝⎠

22
00
11
29
0.0806
360
qR qR
CC
DD
⎛⎞
∴ =− ⇒ =−
⎜⎟
⎝⎠ (4.85)
Substitute in the equation (4.84) we get

42 2
00
3
161 29
0
6 2400 360 4
qR qR R
C
DD
⎛⎞ ⎛⎞
∴ −+=
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠

24 2
00
3
29 161
360 4 6 2400
qR qR R
C
DD
⎛⎞ ⎛ ⎞
∴ =−
⎜⎟ ⎜ ⎟
⎝⎠ ⎝ ⎠

4
0
3
29 161
1440 14400qR
C
D
⎡ ⎤⎛⎞⎛ ⎞
∴ =−
⎜⎟⎜ ⎟⎢ ⎥
⎝⎠⎝ ⎠⎣ ⎦


4
0
3
0.00896
qR
C
D
∴ = (4.86)
Substitute C
1 and C2 in the equation (4.82) of deflection

2445 2
000
32
0.0806 0.00896
63275 4
qqRqRRr r r
w RDDD
⎛⎞
∴ =−− +⎜⎟
⎝⎠

116. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     108 


2445 2
000
32
0.0806 0.00896
63275 4
qqRqRRr r r
w RDDD
⎛⎞
∴ =−− +⎜⎟
⎝⎠
Deflection is maximum at
r = 0
()
4
0
max 0
0.00896
r
qR
w
D
=
∴ = (4.87)
4.5 Problem
A circular plate of radius R is simply supported boundary and carries a load
whose intensity / unit area
q is varies according to the relation
0
1
⎛⎞
=−
⎜⎟
⎝⎠
r
qq R
obtain an
expression for deflection of plate at the radius
R and hence calculate maximum deflection.


Solution: Refer previous problem. Solution upto equation (4.82) is same i.e.

45 2
0
13
32
63275 4
q Rr r r
wCC
RD
⎛⎞
∴ =−++⎜⎟
⎝⎠
To find constants C
1 and C3 use boundary conditions of simply supported plate.
i.e.
0and 0
rrR rR
wM
==
==

117. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     109 

4.6 Circular Plate with Circular Hole at the Centre.

4.6.1 Bending of a plate by moments M
1 and M2 uniformly distributed along inner and
outer boundaries:

Consider bending of a circular plate by moments M 1 and M2 uniformly distributed along inner
and outer boundaries. Since plate is subjected to pure bending moments the shearing force
vanishes i.e. Q = 0 in the differential equation (4.6) of deflection.

1dddwQ
r
dr r dr dr D⎡⎤⎛⎞
=
⎜⎟⎢⎥
⎝⎠⎣⎦


1
0dddw
r
dr r dr dr⎡⎤⎛⎞
∴ =
⎜⎟⎢⎥
⎝⎠⎣⎦
(4.88)
Integrate both sides of above equation with respect to
r, we get

1
1ddw
rC
rdr dr⎡⎤⎛⎞
∴ =
⎜⎟⎢⎥
⎝⎠⎣⎦

Multiply by
r to the both sides of above equation and making integration with respect to r again

2
12
2
dw r
rCC
dr
∴ =+
Divide by
r to the both sides and integrate with respect to r

2
12 3
log
4
r
wC C rC
∴ =+ + (4.89)
This is also written as

118. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     110 


2
12 3
log
4
rr
wC C C
a
⎛⎞
∴ =+ +
⎜⎟
⎝⎠ (4.90)
The constants of integration are now to be determined from the conditions at the edges. Since
plate is simply supported along the outer edge, we have


2
1
0at
at
at
r
r
wra
M Mra
M Mrb
==
==
==
(4.91)
Since deflection is zero at the supporting edges, using above boundary condition put
r = a in the
equation of deflection (4.90) and equate with zero, we get

2
12 3
log
4
ra
aa
wC C C
a
=
⎛⎞
∴ =+ +
⎜⎟
⎝⎠

2
13
0
4
a
CC
∴ += (4.91 a)
Differentiate equation (4.90) twice with respect to r to determine bending moments, hence

2
1
2
Cdw r
C
dr r
∴ =+ (4.92)

2
12
22
2
CCdw
dr r
∴ =− (4.93)
Therefore put equation (4.92) and (4.93) in equation (4.2) to find bending moment
r
M

12 2
12
22
r
CC C r
MD C
rr rυ
⎡ ⎤⎛⎞
∴ =− − + +
⎜⎟⎢ ⎥
⎝⎠⎣ ⎦


()
()
12
2
11
2
r
CC
MD
r
υ υ
⎡ ⎤
∴ =− + − −
⎢ ⎥
⎣ ⎦
(4.94)
This moment must be equal to
1
M for r = b and equal to
2
Mfor r = a, hence equation (4.94)
becomes

119. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     111 

() ()
12
2
11
2
rrb
CC
MD
b
υ υ
=
⎡ ⎤
∴ =− + − −
⎢ ⎥
⎣ ⎦


()
()
12
1 2
11
2
CC
MD
b
υ υ
⎡ ⎤
∴ =− + − −
⎢ ⎥
⎣ ⎦
(4.95)
() ()
12
2
11
2
rra
CC
MD
a
υ υ
=
⎡ ⎤
∴ =− + − −
⎢ ⎥
⎣ ⎦


()
()
12
2 2
11
2
CC
MD
a
υ υ
⎡ ⎤
∴ =− + − −
⎢ ⎥
⎣ ⎦
(4.96)

Subtract equation (4.96) from equation (4.95) we get

()
() () ()
12 12
12 22
1111
22
CC CC
MM D D
ba
υ υυ υ
⎡ ⎤⎡ ⎤
∴ −=−+−−++−−
⎢ ⎥⎢ ⎥
⎣ ⎦⎣ ⎦


()
12 2 22
11
1
MMDC
ba υ
⎡ ⎤
∴ −= − −
⎢ ⎥
⎣ ⎦


()
()
12
2
22
11
1
MM
C
D
ba
υ
−
∴ =
⎡⎤
−−
⎢⎥
⎣⎦


( )
()
()
22
12
2
22
1
ab M M
C
Da b
υ
−
∴ =
⎡ ⎤−−
⎣ ⎦
(4.97)
Put in the equation (4.95) to find out C
1

()
()
()
()
()
22
121
1 2 22
1
11
2 1ab M MC
MD
b Da b
υ υ
υ
⎡⎤
−
⎢⎥
∴ =− + − −
⎢⎥ ⎡⎤−−
⎣⎦⎣⎦


()
()
()
2
121
1
22
1
2
aM MC
MD
Da b
υ
⎡ ⎤
−
⎢ ⎥∴ =− + −
⎢ ⎥⎡ ⎤−
⎣ ⎦⎣ ⎦


( )
()
()
2
12 1
1
22
1
2
aM M C
MD
ab
υ
−
∴ −=−+
⎡⎤−
⎣⎦

120. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     112 


()
( ) ( )
()
22 2
112
1 22
2
1
Ma b aM M
C
D ab
υ
⎡ ⎤−− −
⎢ ⎥∴ −=
+ −⎢ ⎥
⎣ ⎦


()
()
22
21
1 22
2
1
aM bM
C
D ab
υ
⎡ ⎤
−
⎢ ⎥∴ −=
+ −⎢ ⎥
⎣ ⎦


()
()
22
12
1 22
2
1
bM a M
C
D ab
υ
⎡ ⎤
−
⎢ ⎥∴ =
+ −⎢ ⎥
⎣ ⎦
(4.98)
To determine the constant
3
C in equation (4.90), the deflection at the edges of plate must be
considered. Therefore put integration constant
1
C from equation (4.98) in the equation (4.91 a).

2
31
4
a
CC
∴ =−

()
()
222
21
3 22
21
aM bMa
C
D abυ
⎡ ⎤
−
⎢ ⎥∴ =
+ −⎢ ⎥
⎣ ⎦
(4.99)
Substitute equations (4.97), (4.98) and (4.99) in the equation (4.90) to obtain the expression for
deflection of the plate.

()
()
()
()
()
() ()
22222
1212
22 22
222
21
22
2
log
14 1
21
ab M MbM a Mrr
Da ab Da b
w
aM bMa
D ab
υ υ
υ
⎧⎫ ⎡⎤
−− ⎛⎞
⎪⎪ ⎢⎥ +
⎜⎟
+ ⎡⎤−⎪⎪ ⎝⎠−−⎢⎥
⎣⎦⎪⎪ ⎣⎦
∴ =⎨⎬
⎡⎤⎪⎪ −
⎢⎥+⎪⎪
+ −⎢⎥⎪⎪ ⎣⎦⎩⎭

()
()
()
( )
() ()
2222 22
12 21
2222
log
211
arab M M aM bMr
w
aD abDa b
υυ
⎧⎫
⎡ ⎤−− −⎪⎪ ⎛⎞
⎢ ⎥∴ =+ +⎨⎬ ⎜⎟
+⎡⎤ −⎝⎠−− ⎢ ⎥⎪⎪ ⎣ ⎦⎣⎦⎩⎭
(4.100)
Substitute integration constant
12
andCC from equation (4.97) and (4.98) in the equation (4.94)
for bending moments of the plate.

121. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     113 

()
() ()
()
()
()
()
2222
1212
222 22
112
1
21 1
r
ab M MbM a M
MD
Dr ab Da b υ
υ
υ υ
⎡⎤
⎧ ⎫⎧⎫ ⎡⎤
−−−⎪⎪⎪⎪
⎢⎥ ⎢⎥∴ =− + −⎨⎬⎨⎬
⎢⎥ + ⎡⎤− −−⎢⎥⎪⎪⎪⎪ ⎣⎦⎩⎭⎣⎦ ⎩⎭⎣⎦

()
()
()
22 22
12 12
2222 2
r
ab M M bM a M
M
abra b
⎧⎫⎡⎤
⎡ ⎤
− −⎪⎪
⎢⎥ ⎢ ⎥∴ =−⎨⎬
⎢⎥⎡⎤ −− ⎢ ⎥⎪⎪ ⎣ ⎦⎣⎦⎣⎦⎩⎭
(4.101)
Now simplify equations (4.92) and (4.93) by substituting constants of integration
12 3
,andCC C
and then put into the equation (4.3) to obtain bending moment
M
θ


12 2
12
1
22CC C r
MD C
rr r
θ
υ
⎡⎤⎛⎞⎛ ⎞
=− − + +
⎜⎟⎜ ⎟⎢⎥
⎝⎠⎝ ⎠⎣⎦


()
()
12
2
11
2
CC
MD
r
θ
υ υ
⎡ ⎤
∴ =− + + −
⎢ ⎥
⎣ ⎦
(4.102)

Put equation (4.97) and (4.98) in the equation (4.102) we get

()
() ()
() ()
()
()
2222
1212
222 22
112
1
21 1 ab M MbM a M
MD
Dr ab Da b
θ
υ
υ
υ υ
⎡⎤ ⎡⎤
−−−
⎢⎥ ⎢⎥
∴ =− + +
+⎢⎥ ⎡⎤− −−⎢⎥
⎣⎦ ⎣⎦⎣⎦


()
()
()
2222
1212
22 22 2
ab M MbM a M
M
ab ra b
θ
⎧⎫
⎡ ⎤⎡⎤
−−⎪⎪
⎢ ⎥⎢⎥∴ =− +⎨⎬
⎢ ⎥⎡⎤− −⎢⎥⎪⎪⎣⎦ ⎣⎦⎣ ⎦⎩⎭
(4.103)
Corresponding bending stresses are given as follows
and
r
r
MM
zzI I
θ
θ
σσ==
And maximum values of these stresses are given by
()
()()
()
max max
max max
and
r
r
MM
zz
II
θ
θ
σσ==

122. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     114 

4.6.2 Bending of a plate by moment M1 uniformly distributed along inner boundary:

Note: In the previous derivation substitute ( )
2
0=M in the equations (4.97), (4.98), (4.99),
(4.100), (4.101) and (4.103) we obtain following equations.

()
()
2
1
1 22
2
1
⎡ ⎤
⎢ ⎥∴ =
+ −⎢ ⎥
⎣ ⎦
bM
C
D ab
υ
(4.104)

() ()
22
1
2 22
1
∴ =
−−
abM
C
Da b
υ
(4.105)

()
()
22
1
3 22
21
⎡ ⎤
−
⎢ ⎥∴ =
+ −⎢ ⎥
⎣ ⎦
bMa
C
D ab
υ
(4.106)
()
()
()
() ()
222
2211
22 22
1
log
21 1
⎧⎫ ⎡⎤
⎪⎪ ⎛⎞
⎢⎥
∴ =−+⎨⎬ ⎜⎟
+ ⎡⎤− ⎝⎠−−⎢⎥⎪⎪ ⎣⎦ ⎣⎦⎩⎭
bM a bM r
wra
Da ab Da b
υ υ
(4.107)

()
()
22 2
11
22 2 2 2⎡⎤⎡⎤
⎢⎥⎢⎥
∴ =−
−−⎢⎥⎢⎥
⎣⎦⎣⎦
r
abM bM
M
ra b a b
(4.108)

()
()
222
11
22 22 2⎧⎫
⎡ ⎤⎡⎤
⎪⎪
⎢ ⎥⎢⎥∴ =− +⎨⎬
⎢ ⎥⎡⎤− −⎢⎥⎪⎪⎣⎦ ⎣⎦⎣ ⎦⎩⎭
bM a bM
M
ab ra b
θ
(4.109)
Corresponding bending stresses are given as follows
and
r
r
MM
zzI I
θ
θ
σσ==

123. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     115 

And maximum values of these stresses are given by
()
()
()
()
max max
max max
and
r
r
MM
zz
II
θ
θ
σσ==
4.6.3 Bending of a plate by shearing forces along inner boundaries:

Now consider the case of bending of a plate by shearing forces
Q0 uniformly distributed along
the inner edges as shown in figure. The shearing force per unit length of a circumference of
radius
r is

0
2
Qb P
Q
rrπ
==
where
0
2PbQ
π= denotes the total load applied to the inner boundaries of the plate. Therefore
substitute value of shearing force in the equation (4.6) we get

1
2dddw P
r
dr r dr dr rD
π
⎡⎤⎛⎞
∴ =
⎜⎟⎢⎥
⎝⎠⎣⎦

Integrate both sides of above equation w.r.t.
r

1
1
log
2ddw P
rrC
rdr dr D
π
⎡⎤⎛⎞
∴ =+
⎜⎟⎢⎥
⎝⎠⎣⎦

Multiply both sides of above equation by
r, and making integration with respect to r we get

222
12
log
22 4 2
dw P r r r
rrCC
dr D
π
⎡⎤
∴ =−++
⎢⎥
⎣⎦

124. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     116 


2
1
log
22 4 2
Cdw P r r r
rC
dr D r
π
⎡⎤
∴ =−++
⎢⎥
⎣⎦

Integrate w.r.t.
r

2222
12 3
1
log log
222 48 4Pr rr r
wrCCrC
D
π
⎡⎤⎛⎞
∴ =−++++⎢⎥⎜⎟
⎝⎠⎣⎦

()
2
2
12 3
P
log 1 log
84 r
wrrCCrC
D
π
∴ =−+++ (4.110)
This is also written as

2
2
12 3
log 1 log
84
Pr r r
wr CC C
Da a
π
⎡⎤⎛⎞ ⎛⎞
∴ =−+++
⎜⎟ ⎜⎟⎢⎥
⎝⎠ ⎝⎠⎣⎦
(4.111)
The constants of integration will now be calculated from the boundary conditions. Since plate is
simply supported along the outer edge, we have

0at
0at
0at
r
r
wra
M ra
M rb
==
==
==
(4.112)
Therefore from equation (4.111)

22
13
84
ra
Pa a
wCC
D
π
=
−
∴ =++

22
13
0
84
Pa a
CC
D
π
−
∴ ++= (4.113)
Now differentiate equation (4.111) w.r.t. r upto second order

2
1
log
22 4 2
Cdw P r r r r
C
dr D a r
π
⎡⎤⎛⎞
∴ =−++
⎜⎟⎢⎥
⎝⎠⎣⎦
(4.114)

2
2
122
111
log 1
22 4 2 Cdw P r
C
dr D a r
π
⎡⎤⎛⎞⎛⎞
∴ =+−+−
⎢⎥ ⎜⎟⎜⎟
⎝⎠⎝⎠⎣⎦ (4.115)
Put equation (4.114) and (4.115) in the equation (4.2)

125. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     117 


2
1 2
2
1
111
log 1
22 4 2
log
22 4 2
r
CPr
C
Da r
MD
CPr r r r
C
rD a r
π
υ
π
⎧⎫⎡⎤⎡⎤⎛⎞⎛⎞
+− + −⎪⎪⎢⎥⎢⎥ ⎜⎟⎜⎟
⎝⎠⎝⎠⎪⎣ ⎦ ⎪⎣⎦
∴ =−⎨⎬
⎡⎤⎡⎤⎪⎪ ⎛⎞
+−++
⎢⎥ ⎜⎟⎢⎥⎪⎪
⎝⎠⎣⎦⎣⎦⎩⎭

() ()
12
2
11 1
log log 1 1
222 42 42
r
CCPr r
MD
Da a rυυ
υ υ
π
⎧⎫⎡⎤ ⎛⎞ ⎛⎞ ∴ =− +−−−++−−⎨⎬ ⎜⎟ ⎜⎟⎢⎥
⎝⎠ ⎝⎠⎣⎦⎩⎭
() () () ()
12
2
11
1log1 1 1
24 2 2
r
CCPr
MD
Da r
υ υυυ
π
⎧⎫⎡⎤ ⎛⎞∴ =− − + + + + − −⎨⎬ ⎜⎟⎢⎥
⎝⎠⎣⎦⎩⎭ (4.116)

() () ()
12
2
111
82
rra
CCP
MD
Da
υ υυ
π
=
⎧⎫∴ =− − + + − −⎨⎬
⎩⎭

() () ()
12
2
1110
82
CCP
Da
υυυ
π
⎧⎫
∴ −+ +− − =⎨⎬
⎩⎭
(4.117)
Similarly
() () () ()
12
2
11
1log1 1 1
24 2 2
rrb
CCPb
MD
Da b
υ υυυ
π
=
⎧⎫⎡⎤ ⎛⎞∴ =− − + + + + − −⎨⎬ ⎜⎟⎢⎥
⎝⎠⎣⎦⎩⎭

() () () ()
12
2
11
1log1 1 10
24 2 2 CCPb
Da b
υυυυ
π
⎧⎫⎡⎤ ⎛⎞
∴ −+ + + +− − =⎨⎬ ⎜⎟⎢⎥
⎝⎠⎣⎦⎩⎭ (4.118)
Subtract equation (4.118) from (4.117)

() () ()
() () () ()
12
2
12
2
111
82
11
1log1 1 10
24 2 2
CCP
Da
CCPb
Da b
υυυ
π
υυυυ
π
⎧⎫
∴ −+ +− −⎨⎬
⎩⎭
⎧⎫⎡⎤ ⎛⎞
−−++−++−=⎨⎬ ⎜⎟⎢⎥
⎝⎠⎣⎦⎩⎭


() () ()
22
22
log 1 1 1 0
4
CCPb
Da a b
υυυ
π
⎧⎫ ⎛⎞
∴ −+−−+−=⎨⎬ ⎜⎟
⎝⎠⎩⎭

() () ()
22
22
log 1 1 1
4
CCPb
Da a b
υ υυ
π
⎛⎞ ∴ − +=−−−
⎜⎟
⎝⎠

126. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     118 

() ()
2 22
11
log 1 1
4Pb
C
Da ab
υυ
π
⎛⎞ ⎡ ⎤
∴ −+=−−
⎜⎟ ⎢ ⎥
⎝⎠ ⎣ ⎦


()
()
22
222
1
log
41Pb ba
C
Da ba υ
πυ+⎛⎞
∴ −=
⎜⎟
−−⎝⎠


()
()()
22
2 22
1
log
41Pb ba
C
Da ab υ
πυ+⎛⎞
∴ =
⎜⎟
− −⎝⎠
(4.119)
Put C2 in equation (4.117) to obtained C1
() ()
() ()
()()
22
1
2 22
11
11 log 0
82 41
CPPbba
DaDa abυυ
υυ
ππυ
⎧⎫ ⎡⎤−+⎪⎪ ⎛⎞
⎢⎥
∴ −+ +− =⎨⎬ ⎜⎟
− −⎝⎠⎢⎥⎪⎪ ⎣⎦⎩⎭

() () ()
()
2
1
22
11 log1 0
824
CPPbb
DDa abυυ υ
ππ
⎧⎫ ⎡⎤
⎪⎪ ⎛⎞
⎢⎥
∴ −+ +− + =⎨⎬ ⎜⎟
−⎝⎠⎢⎥⎪⎪ ⎣⎦⎩⎭


()
() ()
()
2
1 22
11
log
14 2Pbb
C
Da ab υυ
πυ
⎧ ⎫
−+⎪ ⎪⎛⎞
∴ −= − ⎨ ⎬⎜⎟
+ −⎝⎠⎪ ⎪⎩⎭


()
()
()
2
1 22
12
log
41Pbb
C
Da ab υ
πυ
⎧⎫
−⎪⎪⎛⎞
∴ =− ⎨⎬⎜⎟
+−⎝⎠⎪⎪⎩⎭ (4.120)
Now from equation (4.113)

22
13
0
84
Pa a
CC
D
π
−
∴ ++=

()
()
()
222
322
12
log 0
84 1 4Pa P b b a
C
DDa ab υ
ππ υ
⎧⎫
−− ⎪⎪⎛⎞
∴ +−+=⎨⎬⎜⎟
+−⎝⎠⎪⎪⎩⎭
()
()
()
22 2
3 22
12
log
816 1Pa Pa b b
C
DD a ab υ
ππ υ
⎧ ⎫
−⎪ ⎪⎛⎞
∴ =− − ⎨ ⎬⎜⎟
+−⎝⎠⎪ ⎪⎩⎭

127. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     119 


()
()
()
22 2
3 22
12
log
816 1Pa Pa b b
C
DD a ab υ
ππ υ
⎧ ⎫
−⎪ ⎪⎛⎞
∴ =− − ⎨ ⎬⎜⎟
+−⎝⎠⎪ ⎪⎩⎭


()
()
()
22
3 22
11
1log
821Pa b b
C
Da ab υ
πυ
⎧⎫
−⎪⎪ ⎛⎞
∴ =− +⎨⎬ ⎜⎟
+−⎝⎠⎪⎪⎩⎭ (4.121)
Therefore from the equation (4.111) of deflection

()
()
()
()
()() ()
()
()
22
2
22
22 2 2
22 22
12
log 1 log
816 1
11 1
log log 1 log
41 8 21
Pr Prbb
wr
Da Da ab
rP b ba Pa b b
aDa D a ab abυ
ππ υ
υυ
πυπ υ
⎧⎫
−⎡⎤ ⎪⎪⎛⎞ ⎛⎞
∴ =−+ − ⎨⎬⎜⎟ ⎜⎟⎢⎥
+−⎝⎠ ⎝⎠⎣⎦ ⎪⎪⎩⎭
⎧ ⎫
+− ⎪ ⎪⎛⎞ ⎛⎞ ⎛⎞
++−+ ⎨ ⎬⎜⎟ ⎜⎟ ⎜⎟
−+ −−⎝⎠ ⎝⎠ ⎝⎠ ⎪ ⎪⎩⎭
(4.122)
But when b is infinitely small then ()log / 0ba→

()
()
()
()
22
2
11 1
log 1 1
8161821
Pr Pr Pa
wr
Da D Dυυ
ππυπυ
⎧ ⎫⎧ ⎫−−⎡⎤ ⎪ ⎪⎪ ⎪⎛⎞
∴ =−− ++ ⎨ ⎬⎨ ⎬⎜⎟⎢⎥
++⎝⎠⎣⎦⎪⎪⎪ ⎪ ⎩⎭⎩ ⎭


()
()
()
()
22
22 2
11
log
82121
Prr a
wr r a
Da
υ υ
π υυ
⎧⎫ −−⎪⎪⎛⎞
∴ =−−++⎨⎬⎜⎟
++⎝⎠⎪⎪⎩⎭

()
()
()
22222
11
log
821
Pr
wr arar
Daυ
πυ⎧⎫ −⎪⎪⎛⎞
∴ =+−+−⎨⎬⎜⎟
+⎝⎠⎪⎪⎩⎭

()
()
()
222
11
log 1
821
Pr
wr ar
Daυ
πυ⎧⎫
⎡ ⎤−⎪⎪⎛⎞
∴ =+−+⎨⎬ ⎢ ⎥⎜⎟
+⎝⎠⎪⎪ ⎣ ⎦⎩⎭
(4.123)
Differentiate above equation twice w.r.t. r and substitute in the equations (4.2) and (4.3) to obtain
bending moments
r
M andM
θ
.
Corresponding bending stresses are given as follows
and
r
r
MM
zzI I
θ
θ
σσ==

128. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     120 

And maximum values of these stresses are given by
()
()
()
()
max max
max max
and
r
r
MM
zz
II
θ
θ
σσ==
Exercise:
Que. Develop from first principles, governing differential equation for circular plate under
axisymmetric loading.
[P.U. Article 4.2]
Que. Obtain expressions for radial moments at the centre and at the edge of clamped circular
plate under uniformly distributed loading
q[P.U. & Dr. B.A.M.U. Article 4.2.2]
Que. Find the transverse deflectionwfor the simply supported circular plate of radius a subjected
to uniformly distributed loadq. Hence find expressions for , , and
rr
MM
θ θ
σσ and their
maximum values.
[P.U. & Dr. B.A.M.U. Article 4.2.1]
Que. Find the transverse deflection w for the simply supported circular plate of radius a
subjected to centre point load P. Hence find expressions for , , and
rr
MM
θ θ
σσ and their
maximum values.
[P.U. & Dr. B.A.M.U. Article 4.3.1]
Que. A circular plate of radius a is clamped at edges. The plate carries a load of intensity
quniformly distributed over the entire surface of the plate. The thickness of the plate is h
Analyze the plate from the basic principles and obtain the expressions for
[P.U. & Dr.
B.A.M.U. Article 4.3.2]
i) the maximum deflection
ii)
bending moments at the boundary of the plate
iii)
bending moments at the centre of the plate
iv)
Variation of stresses at the inner face of the plate along the radius of the plate.
Que. A solid circular slab of concrete with radius R = 3.5 m and uniform thickness of 120mm
carries distributed load where intensity qvaries according to relation
61
r
q
R
⎛⎞
=−
⎜⎟
⎝⎠
where
q is in KN/m at a radial distance r meters from the centre. Assuming the edge of the slab

129. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     121 

as clamped, compute the maximum deflection in the slab. Assume
5
0.17 10E=× MPa and
Poission’s ratio = 0.17 for the slab material.
[P.U. Article 4.4]
Que. A solid circular slab of concrete with radius R = 3.0 m and uniform thickness of 120mm
carries distributed load where intensity
qvaries according to relation
4000 1
r
q
R
⎛⎞
=−
⎜⎟
⎝⎠

where
q is in KN/m at a radial distance r meters from the centre. Assuming the edge of the
slab as simply supported, compute the maximum deflection in the slab. Assume
5
0.15 10E=× MPa and Poission’s ratio = 0.17 for the slab material. [P.U. Article 4.5]
Que. A cylindrical R.C.C. water tank with radius 3m stores water to a depth 2m. The bottom slab
is flat with uniform thickness of 180 mm. the slab may be considered to be simply
supported along its edges. Assuming
5
0.25 10E=× MPa and Poission’s ratio = 0.15, find
the maximum deflection and maximum bending stress.
[P.U. Article 4.2.1]
Que. A solid circular slab of concrete with radius 3m and uniform thickness 120mm carries a
uniformly distributed load of 3000 N/m
2
assuming the edges of slab as simply supported,
compute the maximum bending moment and deflection in the slab. Starting from
fundamentals, derive the relations you use. For concrete assume
5
0.15 10E=× MPa and
Poisson’s ratio = 0.17.
[P.U. Article 4.2.1]
Que. Find the transverse deflectionwfor the simply supported circular plate with hole of radius a
subjected to moments
12
and
M Mdistributed uniformly along inner and outer edges. Hence
find expressions for , , and
rr
MM
θ θ
σσ and their maximum values. [Article 4.6.1]
Que. Find the transverse deflectionwfor the simply supported circular plate with hole of radius a
subjected to shearing forces along the inner boundaries. Hence find expressions for
,,and
rr
MM
θ θ
σσ and their maximum values. [Article 4.6.3]

130. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     122 


Chapter 5
General Theory of Cylindrical shell

5.1 Definition
“A shell can be defined as a curved structure of which one dimension, the thickness, is small
in comparison with the other two dimensions,”
•
Shell bears the same relation to plates, as curved beams to straight beam
•
In general shells are termed as curved plates
•
There are two different classes of shell i.e. Thick shell and Thin shell
•
A shell will be called thin if the maximum value of ratio h/R can be neglected in
comparison with unity
Where h = thickness of shell
R = Radius of curvature of middle surface
•
Correspondingly shells will be called thick shell whenever such terms can not be
neglected.
•
OR h/R > 1/10 = Thick shell
h/R = 1/10 to 1/50 = Thin shell
h/R < 1/50 = Shell is too thin to be used as load carrying member.
5.2 Some important Terms used in shells:
Ruled surface:
A ruled surface may be defined as a surface formed by the motion of a straight
line which is known as the generator or ruling.
Singly Ruled surface: A surface is said to be singly ruled if at every point only a single straight
line can be ruled. e.g. Conical shells, conoids and cylinders.

131. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     123 

Doubly Ruled surface: A surface is said to be doubly ruled if at every point two straight lines
can be ruled. Hyperbolic paraboloid and hyperboloid of Revolution of one sheet.
Principal curvatures: The curvatures of a point along the direction of maximum and minimum
curvatures are called principal curvatures.
Middle surface: The surface that bisects the thickness of shell or a locus of a point bisecting
thickness of a shell is called middle surface of the shell.
Membrane Action: A shell which carries load entirely by direct stresses lying on its plane is
called as membrane. For membrane action to be possible, shell has to be thin.
Membrane State of Stress: A state of stress in which the stresses in the shell are constant over
its thickness may be defined as ‘Membrane state’. A more mathematical approach would be to
regard the membrane theory as a particular case of the more exact bending theory. Thus
membrane theory results if certain effects in the bending theory are ignored.
5.3 Advantages of shell structures:
1. Major load is carried through membrane action and not through bending.
2.
Due to this small thickness can be used, requires less material and is economical.
3.
Shapes are architecturally beautiful and streamlined.
4.
Large floor area uninterrupted by supports is obtained with shells.
5.4 Disadvantage of shell structures:
1. Difficult to analyze.
2.
Difficult to construct due to complex geometry.
3.
Cannot be used as a floor.
5.5 Classification of shell: Shell surfaces may be broadly classified as singly curved and doubly
curved. Singly curved surfaces are developable. Thus a cylinder can be developed into a plane
rectangle without stretching, shrinking or tearing. Similarly a cone may be developed into a
sector of circle. Doubly curved surfaces are nondevelopable. Hence they will not tend to flatten
out under loads. Further classification of shell surfaces can be attempted on the basis of Gauss

132. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     124 

curvature. Subclassification is based upon whether a shell is a translational surface. Ruled
surface or a surface of revolution.





Thin Shell
Singly Curved
Gauss Curvature Zero
Membrane equation parabolic
Doubly Curved
Anticlastic
Gauss Curvature Negative Membrane equation Hyperbolic
Synclastic
Gauss Curvature Positive Membrane equation elliptic
Shells of
Revolution
e.g. Conical Shell
Shells of
Translation e.g.
Cylindrical Shell Ruled Surface, e.g. Conical & Cylindrical Shell
Shells of Revolution e.g. Circular Domes, paraboloid, ellipsoid of
Revolution
Shells of
Translation
e.g. elliptic
paraboloid.
Circular
Paraboloi
d
Shells of Revolution e.g. Conical
Shell
Shells of
Translation
e.g.
Cylindrical
Shell
Ruled Surface e.g. Conical & Cylindrical Shell

133. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     125 

5.6 Assumptions made in theory of thin elastic shell:
1. Stresses in z direction is neglected in comparison with and
xy
σσ
2.
Straight line normal to undeformed middle surface remains straight and normal to
deformed middle surface.
3.
Displacements are small enough so that the changes in geometry of shell are negligible
for equilibrium.
4.
The material is linearly elastic, homogenous.
Due to assumption (1) (called Love’s Hypothesis) the 3-D problems are converted into 2-D
problems. In assumption (2) effect of shear deformation is neglected. Assumption (3) makes
equations simple and (4) avoid orthotropiy, non-linearity and discontinuities.
5.7 Determination of Stress Resultants:

Consider an infinitesimal small element of shell cut out by sections parallel to x and y axes and
normal to mid plane. Let x and y be the directions of principal curvatures and ,
xy
rr be the radii
of principal curvatures for the element. Let the stresses , , , and
x y xy yx xz yz
σστ τ γ γ= be acting at a
point in the material at distance z from the mid surface as shown in figure.

134. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     126 

Using similarities of triangle
1
xx
x
xx x
Lrz
L
rrz r
−
=⇒=
−

1
x
x
z
L
r
∴ =−
Similarly 1
y
y
z
L
r
=−
Hence along x direction length of arc will be
1
x
z
r
⎛⎞
−⎜⎟
⎝⎠
and along y direction
1
y
z
r
⎛⎞
−⎜⎟
⎜⎟
⎝⎠

Therefore stress resultants per unit length of edge of shell become
/2
/2
/2
/2
/2
/2
/2
/2
1
1
1
1
h
xx
yh
h
yy
xh
h
xy xy
yh
h
yx xy
xh
z
Ndz
r
z
Ndz
r
z
Ndz
r
z
Ndz
r
σ
σ
τ
τ
+
−
+
−
+
−
+
− ⎫⎛⎞
=− ⎪⎜⎟
⎜⎟
⎪⎝⎠
⎪
⎛⎞
⎪
=− ⎜⎟
⎪
⎝⎠⎪
⎬
⎛⎞ ⎪
=− ⎜⎟
⎪⎜⎟
⎝⎠ ⎪
⎪
⎛⎞
⎪=− ⎜⎟
⎪
⎝⎠ ⎭
∫
∫
∫
∫


/2 /2
/2 /2
/2 /2
/2 /2
/2 /2
/2 /2
11
11
11
hh
xxz yyz
yyhh
hh
xx yy
yxhh
hh
xy xy yx xy
yxhh
zz
QdzQdz
rr
zz
M zdz M zdz
rr
zz
M zdz M zdz
rr
ττ
σσ
ττ
++
−−
++
−−
++
−− ⎫⎛⎞ ⎛⎞
=− =− ⎪⎜⎟ ⎜⎟
⎜⎟ ⎜⎟
⎪⎝⎠ ⎝⎠
⎪
⎛⎞ ⎛⎞ ⎪
=− =− ⎜⎟ ⎬⎜⎟
⎜⎟
⎝⎠ ⎪⎝⎠
⎪
⎛⎞ ⎛⎞
⎪
=− =− ⎜⎟ ⎜⎟
⎪⎜⎟
⎝⎠⎝⎠ ⎭
∫∫
∫∫
∫∫
(5.1)

135. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     127 

(Note: Though
xyy x
ττ= but
xyy x
NN≠ because and
x y
rr can be different.
xyy x
NN= Only
for spherical shell because in case of spherical shell
x
y
rr=)
Rule used in determining the directions of the moments is same as in the case of plates. Since
thickness h of shell is very small in comparison with ,, and
xy
xy
zz
rr
rr
may be omitted from
above expressions then
xy yx
NN= similar to in case of plate.
Strains in shell due to bending and stretching:
In considering the bending of shell, we assume that linear elements, such as AD and BC,
which are normal to the middle surface of the shell, remain straight and become normal to the
deformed middle surface of the shell. During bending, the lateral faces of the element ABCD
rotate only with respect to their lines of intersection with the middle surface.
Due to bending let the original radius of curvature
x
r reduced to
'
x
r and
y
r reduced to
'
y
r.
Therefore original length of curvature
1
x
z
r
⎛⎞
−⎜⎟
⎝⎠
reduced to
'
1
x
z
r
⎛⎞
−⎜⎟
⎝⎠
.
Change in length of fibre at a distance z from mid surface =
''
11
11
xxxx
zz
z
rrrr
⎛⎞⎛⎞⎛ ⎞
−−−=− −⎜⎟⎜⎟⎜ ⎟
⎝⎠⎝⎠⎝ ⎠

Original length of fibre =
1
x
z
r
⎛⎞
−⎜⎟
⎝⎠

Strain in x direction =
'
11
change in length
orignal length
1
xx
x
x
z
rr
z
r
ε
⎛⎞
−−⎜⎟
⎝⎠
==
⎛⎞
−⎜⎟
⎝⎠

''
11 11
and
1 1
xy
xx yy
x y
zz
rr rrz z
r r
εε
⎛⎞⎛⎞−−
∴ =− =− ⎜⎟⎜⎟
⎜⎟
⎛⎞ ⎛⎞
⎝⎠ ⎝⎠
− −⎜⎟ ⎜⎟
⎜⎟
⎝⎠ ⎝⎠
(5.2)

136. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     128 

In considering stretching of the shell then, let the strain in x and y direction due to stretching is
12
and
εε respectively. For a midplane length ‘dx’ in x direction,
12
andll is the length of fibre
at a distance z from midplane before and after bending.
1
1
x
z
ldx
r
⎡⎤
∴ =−⎢⎥
⎣⎦ And ()
21 '
11
x
z
ldx
r
ε
⎡⎤
∴ =+− ⎢⎥
⎣⎦
()
1 '
21
1
11 1
1
xx
x
x
zz
dx dx
rrlll
ll z
dx
r
ε
δ
ε
⎡ ⎤⎡⎤
+−−−
⎢ ⎥⎢⎥
− ⎣ ⎦⎣⎦
∴ == =
⎡⎤
−
⎢⎥
⎣⎦
()()
11 11 '''
11 1
11
xx xxx
x
xx
zz zzz
rr rrr
zz
rr
εε εε
ε+−+ −+ −− +
∴ ==
−−

1''
1
111
11
xxx
x
xx
z
rrr
zz
rrε
ε
ε
⎡⎤
−+
⎢⎥
⎣⎦
∴ =−
−−
Similarly
2''
2
111
11
yyy
y
yy
z
rrr
zz
rrε
ε
ε
⎡ ⎤
−+⎢ ⎥
⎢ ⎥⎣ ⎦
=−
−−
(5.3)
The thickness of the shell will be always assumed small in comparison with the radii of
curvature. In such a case the quantities like / and /
xy
zr zr can be neglected in comparison with
unity. We shall neglect also the effect of elongations
12
and
εε on the curvature. Then the
expressions are simplified as,
11 '
11
xx
xx
zz
rr
εεεχ
⎡⎤
∴ =− −=−
⎢⎥
⎣⎦ (5.4)
22 '
11
yy
yy
zz
rrεε εχ
⎡⎤
∴ =− − =−⎢⎥
⎢⎥⎣⎦
(5.5)

137. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     129 

In which
12
and
εε are membrane strains and ()()and
xy
χ χ−− are bending strains. Where
and
xy
χχ denotes changes of curvatures. Using these expressions for the components of strain
of a lamina and assuming that there are no normal stresses between laminae
()0
z
σ
=, the
following expressions for the components of stress are obtained
Stresses interms of strains:
()
2
1
xxyz
E
σ ευεε
υ
⎡ ⎤∴ =++
⎣ ⎦
−
(5.6)
But since ()
2
1
zzyx
E
σ ευεε
υ⎡⎤=++
⎣⎦
−
is assumed as zero we obtain,
()
()
2
0
1
zyx
zyx
E
ευεε
υ
ευε ε
⎡⎤∴ ++=
⎣⎦
−
∴ =− +
(5.7)
Therefore substitute value of
z
εfrom equation (5.7) into the equation (5.6) we get
()( )
2
1
xxyyx
E
σευευεε
υ
⎡ ⎤∴ =+−+
⎣ ⎦−

Substitute value of and
xy
εε from equations (5.4) and (5.5) respectively into the above
equation. We get
() ()122
1
xxy
E
zz
σεχυεχ
υ
⎡ ⎤∴ =−+−
⎣ ⎦
−

Note: () 0 since it is ver
y small
xy
υε ε+=
()122
1
xxy
E
z
σευεχυχ
υ
⎡ ⎤∴ =+−+
⎣ ⎦
−
(5.8)
Similarly ()212
1
yyx
E
zσ ευε χυχ
υ
⎡ ⎤=+−+
⎣ ⎦
−
(5.9)

138. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     130 

Substituting these expressions in equation (5.1) and neglecting small quantities /and/
xy
zr zr in
comparison with unity, we obtained stress resultants quantities
()
/2 /2
122
/2 /2
1
hh
xx xy
hh
E
Ndz z dz
σευεχυχ
υ
++
−−
⎡ ⎤∴ == +−+
⎣ ⎦
−∫∫

()
/2
2
122
/2
12
h
xxy
h
Ez
N
ευε χυχ
υ
+
−
⎡ ⎤
∴ =+−+
⎢ ⎥
−⎣ ⎦

[]
122
1
x
E
Nhh
ευε
υ
∴ =+
−

[]
122
1
x
Eh
N ευε
υ
∴ =+
−
(5.10)
Similarly []
212
1
y
Eh
N ευε
υ=+
−
(5.11)
In the same manner bending moments and
xy
M Mcan be determined using equations (5.8) and
(5.9).
()
/2 /2
2
122
/2 /2
1
hh
xx xy
hh
E
M zdz z z z dzσευεχυχ
υ
++
−−
⎡ ⎤∴ == +−+
⎣ ⎦
−∫∫

()
/2
223
122
/2
1223
h
xxy
h
Ez zz
M
ευε χυχ
υ
+
−
⎡ ⎤
∴ =+−+
⎢ ⎥
−⎣ ⎦

()
3
2
1
xxy
Eh
M
χυχ
υ
−∴ =+
−

( )xxy
MD χυχ∴ =− + (5.12)
Similarly ( )yyx
MD χυχ=− + (5.13)
A more general case of deformation of the element is obtained if we assume that, in addition to
normal stresses, shearing stresses also are acting on the lateral sides of the element. Denoting by

139. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     131 

γ the shearing strain in the middle surface of the shell and by
xy
dxχ the rotation of the edge BC
relative to Oz about the x axis, we obtained
( )2
xy xy
zGτγ χ∴ =− (5.14)
Substituting this equation (5.14) into the equation (5.1), we obtained
()
/2 /2
/2 /2
2
hh
xy xy xy
hh
Ndz zGdzτγχ
++
−−
∴ ==−∫∫

/2
2
/2
2
2
h
xy xy
h
z
NGz
γχ
+
−
⎡ ⎤
∴ =−
⎢ ⎥
⎣ ⎦

xy
NGh
γ∴ = ( )xy yx
NN= (5.15)
In the same manner twisting moment
xy
Malso be find out from the expression (5.14), we obtain
()
/2 /2
/2 /2
2
hh
xy xy xy
hh
M zdz z Gzdzτγχ
+
−−
∴ ==−∫∫

/2
23
/2
2
23
h
xy xy
h
zz
MG
γχ
+
−
⎛⎞
∴ =−⎜⎟
⎝⎠

3
6
xy
xy
Gh
M
χ
∴ =
()
()
3
2
1
12 1
xy
xy yx xy
Eh
MM Dχ
υχ
υ
∴ =− = = −
− (5.16)
Thus we can express the resultant forces per unit length , and
xy xy
NN N and the moments
,and
xy xy
MMM interms of six quantities of strain. The three components of strain
12
,&εεγ
of middle surface of the shell and three quantities , &
xy xy
χχχ representing the changes of
curvature and the twist of the middle surface. In many practical problems bending may be

140. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     132 

neglected, as bending if any may be localized in some small portion of shell remaining major
part being subjected to membrane action only.

5.7 General theory of cylindrical shell:
A Circular Cylindrical Shell with Axisymmetric Loadings.

In practical applications we frequently encounter problems in which a circular cylindrical
shell is submitted to the action of forces distributed symmetrically with respect to the axis of the
cylinder. The stress distribution in cylindrical boilers submitted to the action of steam pressure,
stresses in cylindrical containers having a vertical axis and submitted to internal liquid pressure,
and stresses in circular pipes under uniform internal pressure are examples of such problems.
To establish the equations required for the solution of these problems, consider the
various actions on a small element of a shell are as shown in figure. Size of element isdx a d
φ× .
Here we have by symmetry, N
φ
constant, 0
xx
NN
φφ
= = and 0Q
φ
= also M
φ
constant,
and 0
xx
MM
φφ
== . Thus out of six equations of equilibrium 0
y
F
∑=, 0
x
M∑= and
0
z
M∑= are identically satisfied and are of no use.

141. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     133 

Therefore remaining equations of equilibrium are 0
x
F
∑=, 0
z
F∑= and 0
y
M∑= .
Therefore now summing up the forces along x direction we get,
0
x
xx
N
Nad N dxad
x
φφ
∂⎛⎞
∴ −++ =
⎜⎟
∂⎝⎠

0
x
N
dx a d
x
φ
∂
∴ =
∂
(5.17)
Above expression represents that the forces
x
Nare constant, and we take them equal to zero. If
they are different from zero, the deformation and stress corresponding to such constant force can
be easily calculated and superposed on stresses and deformations produced by lateral loads.
Note: Body force X along x direction is not considered
Similarly now summing up the forces along z direction we get,

Therefore component in z direction
sin sin
22
dd
Ndx Ndx
φφ
φ φ
=+
2sin
2
d
Ndx
φ
φ
=
2
2
d
Ndx
φ
φ
=

142. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     134 

Ndxd
φ
φ= (5.18)
Therefore 0
z
F∑=
0
x
xx
Q
Q ad Q dx ad N dxd Zdxad
x
φ
φφφφ
∂⎛⎞
∴ −++ + + =
⎜⎟
∂⎝⎠

0
x
Q
dx ad N dxd Zdxad
x
φ
φφφ
∂
∴ ++=
∂

0
x
NQ
Z
xa
φ∂
∴ ++=
∂

x
NQ
Z
xa
φ∂
∴ +=−
∂
(5.19)
Now taking moment @y i.e. at a distance of dx

0
x
xx x
M
Mad M dxad Qaddx
x
φφφ
∂⎛⎞
∴ −+ + =
⎜⎟
∂⎝⎠

0
x
x
M
dx a d Q a d dx
x
φφ
∂
∴ −+=
∂

0
x
x
M
dx a d Q a d dx
x
φφ
∂
∴ −=
∂

143. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     135 

0
x
x
M
Q
x
∂∴ −=
∂
(5.20)
These are two equations and three unknowns quantities: , and
xx
NQ M
φ
, thus problem in
indeterminate. Therefore two solve the problem consider displacement of a point in the mid
surface of shell. From symmetry we conclude that the component of v of the displacement in
circumferential direction vanishes. We thus have to consider only the components u and w in the
x and z directions. The expressions fir the strain components then become x
uw
and
x a
φ
εε
∂
==−
∂

Therefore using hook’s law and integrating over the thickness we get
()
/2 /2
2
/2 /2
1
hh
xx x
hh
E
Ndz dz
φ
σευε
υ
++
−−
∴ == +
−∫∫

/2
2
/2
1
h
x
h
Euw
Ndz
xa
υ
υ
+
−
∂⎛⎞
∴ =−
⎜⎟
−∂ ⎝⎠ ∫

[]
/2
2 /2
1
h
x hEuw
Nz
xa
υ
υ
+
−∂⎛⎞
∴ =−
⎜⎟
−∂⎝⎠

2
1
x
Eh u w
N
x a
υ
υ
∂⎛⎞∴ =−
⎜⎟
−∂⎝⎠
(5.21)
Similarly
()
/2 /2
2
/2 /2
1
hh
x
hh
E
Ndz dz
φφ φ
σευε
υ
++
−−
∴ == +
−∫∫

2
1
Eh u w
N
xa
φ
υ
υ
∂⎛⎞∴ =−
⎜⎟
−∂⎝⎠
(5.22)
Since 0
x
N= (Assumed)
2
0
1
Eh u w
xaυ
υ
∂⎛⎞
∴ −=
⎜⎟
−∂⎝⎠

144. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     136 

uw
x a
υ
∂
∴ =
∂

Substitute in the equation (5.22)
2
2
1
Eh w w
N
aa
φ
υ
υ
⎛⎞
∴ =−
⎜⎟
−⎝⎠

Ehw
N
a
φ
∴ =− (5.23)
Considering the bending moments, we conclude from symmetry that there is no bending
curvature in circumferential direction, the curvature along x direction is
2
2
dw
dx
⎛⎞
−⎜⎟
⎝⎠
. Therefore from
equation of plate, bending moment in x direction is given by
2
2
x
w
MD
x
∂
∴ =−
∂
(5.24)
Where
()
3
2
12 1
Eh
D
υ
=
−
= flexural rigidity of the shell
Substitute value of
x
Q from equation (5.20) into the equation (5.19) we get
2
x
NM
Z
xa
φ∂
∴ +=−
∂
(5.25)
Put value of and
x
M N
φ
from equation (5.24) and (5.23) into the equation (5.25) to obtained
22
22
wEhw
DZ
xxa
⎡⎤∂∂
∴ −−=−
⎢⎥
∂∂⎣⎦ (5.26)
All problems of symmetrical deformation of circular cylindrical shells thus reduced to the
integration of equation (5.26). The simplest application of this equation is obtained when the
thickness of shell is constant. Under such conditions equation (5.26) becomes
4
42
wEhw
DZ
xa
∂
∴ +=−
∂

145. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     137 

4
44
4 2
44
wZ Eh
w
x DDa
ββ
⎛⎞∂
∴ += = ⎜⎟
∂ ⎝⎠ (5.27)
General solution of this differential equation (5.27) is
()
( )()
12 34
cos sin cos sin
xx
we c xc x e c xc x fx
ββ
ββ ββ
−
=++ ++ (5.28)
Where f(x) is particular solution of above equation
5.8 Membrane theory of cylindrical shell:

Let consider, an element is cut from the shell by two adjacent generators and two
cross sections perpendicular to the x axis, and its position is defined by the coordinates x and the
angle
φ. In addition a load will be distributed over the surface of the element, the components of
the intensity of this load being denoted, by X, Y and Z.
Let x axis be taken along the length, y tangent to the cross-section and z is along normal
to the surface. The stress resultants acting on edges of a small element having size
.dx rd
φ are as
shown in figure. Considering the equilibrium of the element and summing up the forces in x
direction we obtained

146. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     138 

0
xx
xx xx
NN
N r d N dx r d N dx N d dx X r d dx
x
φ
φφ
φφ φφ
φ
∂⎡⎤∂⎡⎤
∴ −++ −++ + =
⎢⎥⎢⎥
∂∂⎣⎦ ⎣⎦
0
xx
NN
dx r d d dx X r d dx
x
φ
φφ φ
φ
∂∂
∴ ++=
∂∂
1
0 xx
NN
X
xr
φ
φ
∂∂
∴ ++=
∂∂ (5.29)
Similarly, the force in the direction of the tangent to the normal cross section, i.e. in the y
direction gives the corresponding equation of equilibrium, i.e. 0
y
F
∑=
0
x
xx
NN
N dx N d dx N rd N dx rd Yrd dx
x
φφ
φφ φ φ
φφ φφ
φ
∂∂⎡⎤ ⎡ ⎤
∴ −++ − ++ + =
⎢⎥ ⎢ ⎥
∂∂⎣⎦ ⎣ ⎦
0
x
NN
ddx rddx Yrddx
x
φφ
φφφ
φ
∂ ∂
∴ ++=
∂∂
1
0 x
NN
Y
rx
φφ
φ
∂∂
∴ ++=
∂∂ (5.30)
The forces acting in the direction of the normal to the shell, i.e. in the z direction

147. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     139 

Total downward force in z direction = sin sin
22
dd
Ndx Ndx
φφ
φ φ⎛⎞ ⎛⎞
+
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠

= 2
2
d
Ndx
φ
φ⎛⎞
⎜⎟
⎝⎠
= Ndxd
φ
φ
0Ndxd Zrd dx
φ
φ φ∴ +=
0NZr
φ
∴ +=
NZr
φ
∴ =− (5.31)
Note: We find N
φ
from equation (5.31) and
x
N
φ
,
x
Nby integration of equation (5.30) and
(5.29)
5.8.1 Example: Horizontal cylinder with closed ends filled with liquid and supported at
ends with simple supports.

Solution:

If
0
P be the pressure at the axis of the tube, the pressure at any point is given by,

148. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     140 

0
cosPzPa γφ=− = − (5.32)
Where, γ is the density of liquid. Thus ( )
0
0; 0 and cosXY ZPa γφ== =−−
Therefore put value of pressure at any point from equation (5.32) into the equation (5.31) to
findN
φ
.
NZr NZa
φφ
∴ =− ⇒ =−
( )
0
cosNPaa
φ
γφ∴ =− −
( )
0
cosNPaa
φ
γφ∴ =−− +
( )
2
0
cosNPaa
φ
γφ∴ =− (5.33)
Now substitute value of N
φ
from equation (5.33) into the equation (5.30) to find out
x
N
φ
.
()
2
01
cos 0 x
N
Pa a Y
ax
φ
γφ
φ
∂∂
⎡⎤∴ −++=
⎣⎦
∂∂
()
()sin 0 0
x
N
aY
x
φ
γφ
∂
∴ += =
∂

()sin
x
N
a
x
φ
γφ
∂
∴ =−
∂

Therefore integrating above equation w.r.t. x we get
()sin
x
N
adx
x
φ
γφ
∂
∴ =−
∂∫∫

( )sin
x
Nadx
φ
γφ∴ =−∫

()
1
sin
x
NaxC
φ
γφ φ∴ =− + (5.34)
Now put value of
x
N
φ
from equation (5.34) into the equation (5.29) to find out
x
N
()
1
1
sin 0
x
N
axC X
xr
γφ φ
φ
∂
∂
∴ +− ++=⎡⎤
⎣⎦
∂∂

149. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     141 

()
1
11
cos 0
x
N
ax C
xa a
γφ φ
φ
∂
∂
∴ −+=
∂∂

()
1
1
cos
x
N
xC
xa
γφφ
φ
∂
∂
∴ =−
∂∂

Integrate w.r.t. x to find out
x
N ()
1
1
cos
x
N
xC
xa
γφφ
φ
∂⎡ ⎤ ∂∴ =−
⎢ ⎥
∂∂ ⎣ ⎦
∫∫

() ()
2
12
cos
2
x
xx
NCC
a
γφφφ
φ
∂∴ =− +
∂ (5.35)
Here integration constant must be determined by using known values of and
xx
NN
φ
i.e.
boundary conditions. For example consider ends horizontally free i.e. 0
x
N
= at x = 0 and x = l
Therefore from equation (5.35)
At x = 0
()
2
0Cφ
∴ = (5.36)
At x = l
()
2
1
cos 0
2
ll
C
a
γφ φ
φ
∂
∴ −=
∂
()
2
1
cos
2
ll
C
a
γφφ
φ
∂∴ =
∂
()
1
cos
2
al
C
γ
φ φ
φ
∂ ∴ =
∂

Integrate w.r.t. φ
()
1
cos
2
al
C
γ
φ φ
φ
∂ ∴ =
∂∫∫

()
1
sin
2
al
CC
γ
φφ∴ =+ (5.37)

150. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     142 

Substitute in the equation (5.35) we get
x
N
φ

sin sin
2
x
al
Nax C
φ
γ
γφ φ∴ =− + + (5.38)
It is seen from equation (5.34) that the constant C represents forces
x
N
φ
uniformly distributed
around the edges of the tube, as is the case when tube is subjected to torsion. If there is no torque
applied, we must take C = 0. Then the above equation becomes
sin sin
2
x
al
Nax
φ
γ
γφφ∴ =− + (5.39)
0
sin
2
x
x
al
N
φ
γ
φ
=
∴ = And sin
2
x
xl
al
N
φ
γ
φ
=
∴ = (5.40)
By substituting values of
()
()
21
andCCφ φ in the equation (5.35) we get
2
cos sin
22
x
xx al
N
a γ
γφφ
φ
∂∴ =−
∂
2
cos cos
22
x
xxal
N
a γ
γφφ∴ =−
()cos
2
x
x
Nlx
γ
φ∴ =− − (5.41)
And from equation (5.33)
( )
2
0
cosNPaa
φ
γφ∴ =− (5.42)
Here and
xx
NN
φ
are the shear stresses and bending stresses for the simply supported beam of
hallow circular cross section for the weight of liquid i.e.
2
a
πγ per unit length. These are
independent of P
0

151. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     143 

5.8.2 Example: Cylindrical Roof Shell

Consider shell ABCD with semicircular edges AD & BC simply supported on gable walls &
edges AB & CD free. The uniformly distributed load of self weight P acts uniformly over the
middle surface. Area of shell at any point (),xφ we have load components
X = 0,
sinYP
φ= and cosZPφ=
Therefore from equation (5.31)
cosNzrPa
φ
φ∴ =− =− (5.43)
At or
22
π π
φφ
==− , 0N
φ
= which is as required as there are no external forces at straight
edges, AB and CD. Therefore from equation (5.30)
1x
NN
Y
xr
φφ
φ
∂∂
⎡ ⎤
∴ =− +
⎢ ⎥
∂∂ ⎣ ⎦

Substitute value of andNY
φ
we get, ()
1
cos sinx
N
Pa P
xa
φ
φ φ
φ
∂
⎡ ⎤∂
∴ =− − +
⎢ ⎥
∂∂ ⎣ ⎦

()sin sin
x
N
PP
x
φ
φ φ
∂
∴ =− +⎡ ⎤
⎣ ⎦
∂

152. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     144 

2sin
x
N
P
x
φ
φ
∂
∴ =−
∂

Integrating both sides with respective x
2sin
x
N
dx P dx
x
φ
φ
∂
∴ =−
∂∫∫

()
1
2sin
x
NPxC
φ
φ φ∴ =− + (5.44)
Since
x
N
φ
must be symmetric at the two supported ends and hence constant term ()
1
Cφ must be
zero i.e. at x = 0, 0
x
N
φ
= therefore
()
1
0Cφ=, Hence put in the equation (5.44), we get
2sin
x
NPx
φ
φ∴ =− (5.45)
It is seen that this solution does not vanish along the edges AB and CD as it should for free
edges. In structural applications, however, the edges are usually reinforced by longitudinal
members string enough to resist the tension produced by shearing force (5.45). Substituting
equation (5.45) in the equation (5.29), we obtained
1 xx
NN
X
xr
φ
φ
∂⎡ ⎤∂
∴ =− +
⎢ ⎥
∂∂ ⎣ ⎦

Substitute value of
x
N
φ
from equation (5.45) and X to obtain the value of
x
N, Hence
()
1
2sin
x
N
Px
xa
φ
φ
∂⎡ ⎤ ∂
∴ =− −
⎢ ⎥
∂∂ ⎣ ⎦

2
cos
x
N P
x
xa
φ
∂ ⎡ ⎤
∴ =
⎢ ⎥
∂ ⎣ ⎦

Integrate with respective x we get
2
cos
x
N Px dx
xaφ
∂ ∴ =
∂∫∫

()
2
2
2
cos
2
x
Px
NC
a
φ φ∴ =+ (5.46)

153. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     145 

If the ends of the shell are supported in such a manner that the reactions act in the planes of the
end cross sections, the force
x
N must vanish at the ends, i.e.
/2 /2
0
xxll
NN
−
= =, Hence we get
()
2
2
2
0cos
4
Pl
C
a
φ φ∴ =+
()
2
2
cos
4
Pl
C
a
φ φ∴ =− (5.47)
Put equation (5.47) in the equation (5.46) to obtained
22
2
cos cos
24
x
PxPl
N
aa
φ φ∴ =−
()
22cos
4
4
x
P
Nxl
a
φ
∴ =− (5.48)
These expressions
(),and
xx
NN N
φφ
are satisfactory except for the fact that they do not satisfy
the condition that
x
N
φ
must be equal to zero at the free edges. We get some value for
x
N
φ
and to
resist this, a beam or thick strip must be provided at edges AB and CD.

Exercise:
Que. State the advantages and disadvantages of shell structures compared to plates. [P.U. Article
5.3 & 5.4]
Que. Classify thin shell into various types based on shell geometry and curvature. [P.U. & Dr.
B.A.M.U. Article 5.5]
Que. Write the assumptions made in the general theory of thin elastic shells, stating implication
of each.
[P.U. Article 5.6]
Que. Derive expressions for the strains and
xy
εε at a point due to the bending and membrane
(stretching) action in a shell. Hence obtain expressions for the stress resultants interms of
strain.
[P.U. Article 5.7, Derivation of Equation 5.2 & 5.3]

154. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     146 

Que.
In a thin shell of thickness h, if ,,,and
x y xy xz yz
σστ τ τ are the stresses at a point z from
midplane, write expressions for stress resultants at the section, interms of these.
[P.U.
Article 5.6, Derivation of Equation 5.6 to 5.16]
Que. sketch cylindrical shell. Show the stress resultants for a small element on the shell surface.
Derive equations of equilibrium for this element.
[P.U. Article 5.7]
Que. Derive the governing differential equations for general theory of cylindrical shell,
considering actions on an infinitesimal element.
[P.U. Article 5.7]
Que. Write equilibrium equations for membrane analysis of thin cylindrical shells. Using these,
analyse a semicircular cylindrical shell roof of uniform thickness h under a self weight ‘P’
per unit area, to obtain membrane stress resultants
( ),and
xx
NN N
φ φ
in the shell. Assumed
shell has curved edges simply supported and straight edges free.
[P.U. Article 5.7.2]
Que. A horizontal cylindrical shell with closed ends is filled with liquid of density
γ and is
simply supported at ends. Derive expressions for stress resultants, , and
xx
NN N
φ φ
for a
meridianal angle φ in the shell. [P.U. Article 5.7.1]

155. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     147 

Chapter 6
Bending theory of Cylindrical Shells

6.1 The need for bending theory:
Most cylindrical concrete shells used in practice are
not behaved as a membrane. Along the edges of
shell stresses and displacements are different from
those given by membrane theory usually exists. It
depends on support conditions or physical boundary
conditions.
Let consider above figure of shell element ABCD
which is not supported along edges AB and CD
(free edges) but membrane theory gives that stresses and
x
NN
φ φ
are present at this edges. The
actual boundary conditions are realized by applying corrective line loads. But application of such
corrective line load would bend shell and depart from its membrane state. The shell now seek a
new equilibrium and in that process brings into play bending moments, twisting moments and
radial shears. A bending theory is essential to account for these effects.
6.2 Strains in cylindrical shell:
Displacements in x, y and z directions are ,anduv w
respectively.
Therefore strain in x direction is given by
x
u
x
ε
∂
=
∂
(6.1)
Similarly strain in y direction is sum of strain
corresponding to plane state of stress and circumferential

156. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     148 

strain.

1vw
aa
φ
ε
φ
∂
= −
∂
(6.2)
Where
1
strain corresponding to plane state of stress
v
a
φ
∂
=
∂

And
circumferential strain
w
a
=
And shear strain
x
uv
ax
φ
γ
φ
∂ ∂
=+
∂∂
(6.3)
6.3 The Finsterwalder Theory:
Starting from 1932 several rigorous and approximate bending theories have been put
forward for the analysis of reinforced concrete cylindrical shell. The earliest of these was due to
Finsterwalder. By making a few simplifications, finsterwalder was able to develop, for the first
time, a theory that the engineer could use for the analysis of shell roofs. The assumptions
underlying all shell theories and the additional simplifying assumption made by the finsterwalder
theory listed below:
6.3.1 Assumptions made in Finsterwalder theory:
1. Material is homogenous, isotropic and obeys hooks law.
2.
Stresses normal to shell surfaces are neglected.
3.
All dimensions are very small.
4.
A rectilinear element normal to middle surface remains rectilinear after deformation
5.
0
xxx
MM Q
φ
===
Note: First four assumptions are common to all bending theories of cylindrical shells. Last
Assumption was introduced by finsterwalder to simplify the problem.
6.3.2 Equations of equilibrium:
It is possible to derive four equations of equilibrium for an element of the unloaded shell acted
upon by the stress resultant shown in figure below.

157. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     149 

It is noted that ,and
xx x
MMQ
φ
are not to be considered.

Equating all forces in the x direction to zero we get
0
x
F∑=
0
xx
xx xx
NN
N ad N dx ad N dx N ad dx
xa
φ
φφ
φφ φ
φ
∂⎛⎞∂⎛⎞
∴ −++ −++ = ⎜⎟⎜⎟
∂∂⎝⎠ ⎝⎠
0
xx
NN
dxad ad dx
xa
φ
φφ
φ
∂∂
∴ +=
∂∂

0
xx
NN
a
x
φ
φ
∂∂
∴ +=
∂∂ (6.4)
Now summing up the forces in
φ direction, i.e., the direction of the tangent to the shell element
at its midpoint pointing in the direction of increasing φ and equating them to zero, we get
0F
φ
∑=
sin sin 0
22
x
xx
NN
N dx N a d dx N a d N dx a d
ax
Qdd
Qdx Q ad dx
a
φφ
φφ φ φ
φ
φφ
φφ φ
φ
φφ
φ
φ
⎡∂⎤⎡ ∂⎤⎛⎞ ⎛⎞
∴ −++ +− ++⎢⎥⎢ ⎥ ⎜⎟ ⎜⎟
∂∂
⎝⎠ ⎝⎠⎣⎦⎣ ⎦
⎡∂⎤ ⎛⎞
+−−+ =⎢⎥ ⎜⎟
∂⎝⎠⎣⎦

158. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     150 

Note: First two terms of this equation are the same as those appearing in the membrane
theory. The additional term is the resolved components of shear forces
Q
φ
in tangential
()φdirection, because Q
φ
is vertical.
Neglecting higher powers of anddx dφ, on simplification we obtained
0
x
NN
a d dx dx a d Q dx a d
ax
φφ
φ
φφφ
φ
∂∂
∴ +−=
∂∂

0
x
NN
aQ
x
φφ
φ
φ
∂∂
∴ +−=
∂∂ (6.5)
Now equating to zero the sum of all forces in the direction of the inward normal drawn at the
midpoint of the shell element.

sin sin 0
22
N Qdd
N dx N ad dx Q dx Q ad dx
aa
φ φ
φφ φφφφ
φφ
φφ ∂ ∂⎛⎞ ⎛⎞
++ − ++ =⎜⎟ ⎜⎟
∂∂ ⎝⎠⎝⎠

Neglecting higher powers of
anddx d
φ we obtained
2sin 0
2
Qd
Ndx d dx
φ
φφ
φ
φ
∂
∴ +=
∂
20
2
Qd
Ndx d dx
φ
φφ
φ
φ
∂
∴ +=
∂

159. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     151 

On simplifying, we get
0
Q
N
φ
φ
φ
∂
∴ +=
∂ (6.6)
Another equation of equilibrium results from equating the sum of moments of all forces about
the generatrix AD.

0
AD
M
∴ ∑=
0
MQ
Mdx M ad dx Q ad dxad
aa
φφ
φφ φ
φφφ
φφ
∂∂⎛⎞⎛⎞
∴ −++ −+ =⎜⎟⎜⎟
∂∂⎝⎠⎝⎠
Neglecting higher powers of anddx d
φ we obtained
0
M
aQ
φ
φ
φ
∂
∴ −=
∂
1
0M
Q
a
φ
φ
φ
∂
∴ −=
∂ (6.7)
6.3.3 Derivation of Finsterwalder eighth (8
th
) order differential equation:
Let us introduce a function ()f
φ which is such that
()cos
nx
Mf
L
φ
π
φ=−

160. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     152 

Or
()
cos
n
n
x na
Mf
aL
φ
λ
π
φλ
⎛⎞
=− =
⎜⎟
⎝⎠
(6.8)
From equation (6.8) into the equation of equilibrium (6.7) to obtain value of Q
φ
.
1M
Q
a
φ
φ
φ
∂
∴ =
∂

()
1
cos
n
x
Qf
aa
φ
λ
φ
φ∂
⎡ ⎤
∴ =−
⎢ ⎥
∂⎣ ⎦

()1
cos
n
f
x
Q
aa
φ
φ
λ
φ
∂
∴ =−
∂ (6.9)
Put equation (6.9) into the equation (6.6) to obtain value ofN
φ

Q
N
φ
φ
φ
∂
∴ =−
∂

()1
cos
n
f
x
N
aa
φ
φ λ
φφ∂
⎡ ⎤∂
∴ =− −⎢ ⎥
∂∂
⎣ ⎦

()
2
2
1
cos
n
f
x
N
aa
φ
φ
λ
φ
∂
∴ =
∂
(6.10)
Substitute equation (6.10) into the equation (6.5) to obtain
x
N
φ

() ()
2
2
11 1
cos cosx nn
N ff
x x
xaa a a aφ φφλλ
φφφ
⎡ ⎤∂ ⎛⎞∂∂ ∂
∴ =− −⎢ ⎥⎜⎟
∂∂∂∂⎢ ⎥⎝⎠⎣ ⎦

() ()
3
23
1
cosx n
N ff
x
x aa
φ φφ λ
φφ
∂ ⎡⎤∂∂
∴ =− +⎢⎥
∂∂∂
⎣⎦

Integrate both sides with respective x we get
() ()
3
23
1
cosx n
N ff
x
dx
xa a
φ φφ λ
φφ∂ ⎡⎤∂∂
∴ =− +⎢⎥
∂∂∂
⎣⎦∫∫

161. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     153 

() ()
3
23
1
sin
n
x
n
ff xa
N
aa
φ
φφ λ
φ φλ
⎡⎤∂∂
∴ =− +⎢⎥
∂∂
⎣⎦

() ()
3
3
1
sin
n
x
n
ff
x
N
aa
φ
φφ
λ
λφ φ
⎡⎤∂∂
∴ =− +⎢⎥
∂∂
⎣⎦
(6.11)
Substitute value of
x
N
φ
from equation (6.11) into the equation (6.4) to find out value of
x
N,
hence
1 xx
NN
xa
φ
φ
∂⎡ ⎤∂
∴ =−
⎢ ⎥
∂∂
⎣ ⎦

() ()
3
3
11
sin
xn
n
ffNx
x aa a
φφ λ
φλ φ φ⎛⎞ ⎡⎤∂∂∂ ∂
∴ =− − +⎜⎟ ⎢⎥
⎜⎟∂∂ ∂∂
⎣⎦⎝⎠

() ()
24
22 4
1
sin
xn
n
ffNx
x aa
φφ λ
λφ φ
⎡⎤∂∂∂
∴ =+ ⎢⎥
∂∂∂
⎣⎦

Integrating both sides with respective x we get
() ()
24
22 4
1
sin
xn
n
ffNx
dx
xa aφφ λ
λφ φ⎡⎤∂∂∂
∴ =+ ⎢⎥
∂∂∂
⎣⎦∫∫

() ()
24
22 4
1
cos
n
x
n
ff
x
N
aa
φφ
λ
λφ φ
⎡⎤∂∂
∴ =− +⎢⎥
∂∂
⎣⎦
(6.12)
Expressions for displacements u, v and w are derived from stress-strain relationship
()
()
()
1
1
21
xx
x
xx
x
NN
Ed
NN
Ed
NN
Gd Ed
φ
φφ
φ φ
φ
ευ
ευ
υ
γ=−
=−
+
==
(6.13)
Put values of strains from equations (6.1), (6.2) and (6.3) into the equation (6.13), we get

162. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     154 

()
()
()
1
11
21
x
x
xx
u
NN
xEd
vw
NN
aaEd
NNuv
axGdEd
φ
φ
φ φ
υ
υ
φ
υ
φ
∂
=−
∂
⎛⎞∂
−= −
⎜⎟
∂⎝⎠
+∂∂
+= =
∂∂
(6.14)
Expressions for
,anduv w may now be derived by making use of the stress strain relationship
(6.14). Before we do so, we may set v, the Poisson’s ratio to zero
( )0υ=. We get
x
Nu
xEd
∂
=
∂
(6.15)
1 Nvw
aaEd
φ
φ
∂
−=
∂
(6.16)
2
x
Nuv
axEd
φ
φ
∂∂
+=
∂∂
(6.17)
Now consider equation (6.15) and substitute value of
x
N from equation (6.12) to obtain value of
displacement u
() ()
24
22 4
11
cos
n
n
ff
xu
x Ed a a
φφ λ
λφ φ
⎡⎤∂∂∂
∴ =− + ⎢⎥
∂∂∂
⎣⎦

Integrating above equation with respect to x we get
() ()
24
22 4
11
cos
n
n
ff
xu
x Ed a a
φφ λ
λφ φ
⎡⎤∂∂∂
∴ =− + ⎢⎥
∂∂∂
⎣⎦∫∫

() ()
24
32 4
1
sin
n
n
ff
x
u
Ed a
φφ
λ
λφ φ
⎡⎤∂∂
∴ =− + ⎢⎥
∂∂
⎣⎦
(6.18)
From equation (6.17) after simplification we get

163. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     155 

2
x
Nvu
xEd a
φ
φ
∂∂
=−
∂∂

Substitute values of and
x
Nu
φ
from equation (6.11) and (6.18) into the above equation to find
value if displacement v, we get
() () () ()
. ... .. ::
321 1
sin sin
nn
nn
x xv
ff ff
xEd a a a Ed a
λλ
φφ φφ
λφλ⎧⎫⎧ ⎫∂∂
⎡⎤ ⎡⎤
∴=− + −− +⎨⎬⎨ ⎬⎣⎦ ⎣⎦
∂∂
⎩⎭⎩ ⎭
Where
.
fstands for ()fφ
φ
∂
∂

() () () ()
. ... ... ::.
321
sin
n
nn
xv
ff f f
x Ed a Ed a a
λ
φφ φφ
λλ
⎧⎫∂
⎡⎤⎡ ⎤
∴ =− + + +⎨⎬ ⎣⎦⎣ ⎦
∂
⎩⎭
Integrate with respective x to obtained v
() () () ()
. ... ... ::.
321
sin
n
nn
xv
f fffdx
xEda Eda a
λ
φφ φφ
λλ⎧⎫∂
⎡⎤⎡ ⎤
∴ =− + + +⎨⎬ ⎣⎦⎣ ⎦
∂
⎩⎭∫∫

() () () ()
.... ...::.
2421
cos
n
nn
x
vff ff
Ed Ed a
λ
φφ φφ
λλ
⎧⎫
⎡⎤⎡ ⎤
∴=+−+⎨⎬ ⎣⎦⎣ ⎦
⎩⎭ (6.19)
Now from equation (6.16)
aNv
w
Ed
φ
φ
∂
=−
∂
(6.20)
Substitute values of andNv
φ
from equations (6.10) and (6.19) into the equation (6.20) to obtain
the displacement w, we get
() () () () ()
. ... ... ::. ..
2421 1
cos cos
nn
nn
x xa
wff ff f
Ed Ed a Ed a aλλ
φφ φφ φ
φλ λ⎧⎫∂ ⎧⎫
⎡⎤⎡ ⎤=+−+−⎨⎬⎨⎬⎣⎦⎣ ⎦
∂ ⎩⎭⎩⎭

() () () () ()
.. :: :: ::: ..
24211
cos
n
nn
x
wff ff f
Ed Ed Ed a
λ
φφ φφ φ
λλ
⎧⎫
⎡⎤⎡ ⎤=+−+−⎨⎬ ⎣⎦⎣ ⎦
⎩⎭
(6.21)
Knowing v and w we may write,

164. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     156 

1 w
v
a
ϑ
φ
⎡ ⎤∂
=+
⎢ ⎥
∂⎣ ⎦
(6.22)
Therefore put equations (6.19) and (6.21) into the equation (6.22) we get
() () () ()
() () () () ()
. ... ... ::.
24
.. :: :: ::: ..
2421
1
cos
211
nn
n
nn
ff f f
Ed Ed
x
a a
ff ff f
Ed Ed Ed
φφ φφ
λλ
λ
ϑ
φφ φφ φ
φλ λ
⎡⎤
⎡⎤⎡ ⎤+− +
⎣⎦⎣ ⎦⎢⎥
⎢⎥
=
⎢⎥⎧⎫∂
⎡⎤⎡ ⎤++−+−⎢⎥⎨⎬ ⎣⎦⎣ ⎦
∂⎢⎥⎩⎭⎣⎦

() () () ()
() () () () ()
. ... ... ::.
24
... ::. ::. :::. ...
2421
1
cos
211
nn
n
nn
ff f f
Ed Ed
x
a a
ff ff f
Ed Ed Ed
φφ φφ
λλ
λ
ϑ
φφ φφ φ
φλ λ
⎡⎤
⎡⎤⎡ ⎤+− +
⎣⎦⎣ ⎦⎢⎥
⎢⎥
=
⎢⎥⎧⎫∂
⎡⎤⎡⎤++−+−⎢⎥⎨⎬ ⎣⎦⎣⎦
∂⎢⎥⎩⎭⎣⎦

() () () ()
() () ()
()
()
. ... ... ::.
24
7:::.
... ::. ::. ...
24721
1
cos
21
nn
n
nn
ff f f
x
faEd a
ff f f
φφ φφ
λλ
λ
ϑ
φ
φφ φ φ
λλφ
⎡⎤
⎡⎤⎡ ⎤+− +
⎣⎦⎣ ⎦⎢⎥
⎢⎥
=
⎢⎥⎧⎫ ⎡⎤ ∂⎪⎪
⎡⎤⎢⎥++−+−⎨⎬ ⎢⎥⎣⎦
∂⎢⎥⎪⎪ ⎣⎦⎩⎭⎣⎦
(6.23)
We may next find
φ
χ (change in curvature)
2
22
1 w
w
a
φ
χ
φ
⎡ ⎤∂
=+
⎢ ⎥
∂⎣ ⎦
(6.24)
Put equation (6.21) in the equation (6.24)
() () () () ()
() () () () ()
.. :: :: ::: ..
24
2 2
.. :: :: ::: ..
22 4211
1
cos
211
nn
n
nn
ff ff f
Ed Ed Ed
x
a a
ff ff f
Ed Ed Ed
φ
φφ φφ φ
λλ
λ
χ
φφ φφ φ
φλ λ
⎡⎤
⎡⎤⎡ ⎤+− + −
⎣⎦⎣ ⎦⎢⎥
⎢⎥
=
⎢⎥⎧⎫∂
⎡⎤⎡ ⎤++−+−⎢⎥⎨⎬ ⎣⎦⎣ ⎦
∂⎢⎥⎩⎭⎣⎦

165. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     157 

() () () () ()
() () () () ()
.. :: :: ::: ..
24
2
:: ::: ::: :::: ::
24211
1
cos
21
nn
n
nn
ff ff f
Ed Ed Ed
x
Eda a
ff ff f
φ
φφ φφ φ
λλ
λ
χ
φφ φφ φ
λλ
⎡⎤
⎡⎤⎡ ⎤+− + −
⎣⎦⎣ ⎦⎢⎥
⎢⎥
=
⎢⎥⎧⎫
⎡⎤⎡ ⎤++−+−⎢⎥⎨⎬⎣⎦⎣ ⎦
⎢⎥⎩⎭⎣⎦
(6.25)
Put in the moment curvature equation
M D
φφ
χ=− where
()
33
2
1212 1
Ed Ed
Dυ
==
−

() () () () ()
() () () () ()
.. :: :: ::: ..
24
2
:: ::: ::: :::: ::
24211
cos
21
nn
n
nn
ff ff f
Ed Ed Ed
xD
M
Eda a
ff f f f
φ
φφ φφ φ
λλ
λ
φφ φφφ
λλ
⎡⎤
⎡⎤⎡ ⎤+− + −
⎣⎦⎣ ⎦⎢⎥
−
⎢⎥
=
⎢⎥⎧⎫
⎡⎤⎡ ⎤++−+−⎢⎥⎨⎬⎣⎦⎣ ⎦
⎢⎥⎩⎭⎣⎦
(6.26)
Put equation (7.26) into above equation (7.8)
()
() () () () ()
() () () () ()
.. :: :: ::: ..
24
2
2
:: ::: ::: :::: ::
24211
12 21
nn
nn
ff ff f
Ed Ed Ed
d
f
a
ff f f f
φ φφφφ
λλ
φ
φφ φφ φ
λλ
⎡⎤
⎡⎤⎡ ⎤+− + −
⎣⎦⎣ ⎦⎢⎥
⎢⎥
=
⎢⎥⎧⎫
⎡⎤⎡ ⎤++−+−⎢⎥⎨⎬⎣⎦⎣ ⎦
⎢⎥⎩⎭⎣⎦

Rearranging all the terms we get
() () () () ()
()
()
() () () ()
.. :: :: ::: ..
24
6 2
:: ::: :::: ::
264 2211
0
21
12
nn
nn
ff ff f
Ed Ed Ed
f a
fffff
dφφ φφ φ
λλ
φ
φφφφφ
λφλ
⎡⎤
⎡⎤⎡ ⎤+− + −
⎣⎦⎣ ⎦⎢⎥
⎢⎥
=
⎢⎥⎧⎫⎡⎤ ∂⎪⎪
⎡⎤⎢⎥++−+−+⎨⎬⎢⎥ ⎣⎦
∂⎢⎥⎪⎪⎣⎦⎩⎭⎣⎦
(6.27)
This is called as finsterwalder 8th order differential equation.
Which is also written as
() () () () ()
() () () () () ()
.. :: :: ::: ..
24
2
:: ::: ::: :::: ::
24 2211
0
21
12
nn
nn
ff ff f
Ed Ed Ed
a
ff f f f f
dφφ φφ φ
λλ
φφ φφ φ φ
λλ
⎡⎤
⎡⎤⎡ ⎤+− + −
⎣⎦⎣ ⎦⎢⎥
⎢⎥
=
⎢⎥⎧⎫
⎡⎤⎡ ⎤++−+−+⎢⎥⎨⎬⎣⎦⎣ ⎦
⎢⎥⎩⎭⎣⎦
(6.28)

166. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     158 


6.4 The D-K-J Theory (Donnell – Karman – Jenkins Theory)
The simplest among the so-called exact theories which takes into account ,
xx
MM
φ

and
x
Q ignored in the finsterwalder theory is the D-K-J theory in which the three displacements
.anduv w appear in uncoupled form. The theory appears to be due to Donnell, who first used it
in connection with his studies on the stability of thin-walled circular cylinders in 1933-1934.
Karman and Tsien also employed the same theory in 1941 in their investigation on the buckling
of cylindrical shells. Its presentation in a form suitable for the analysis of cylindrical shell roofs
appeared in a book by Jenkins published in 1947. The theory is appropriately called as Donnell –
Karman – Jenkins theory.
6.4.1 Equations of Equilibrium:
The equilibrium equations already derived in finsterwalder theory will now be modified to take
into account ,
xx
MM
φ
and
x
Q. It will be assumed that
xx
M M
φφ
= and
xx
NN
φφ
=

Referring to above figure, the following equations of equilibrium may be set up. Equating all
forces in the x direction to zero we get
0
x
F∑=

167. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     159 

0
xx
xx xx
NN
N a d N dx a d N dx N a d dx
xa
φ
φφ
φφ φ
φ
∂⎛⎞∂⎛⎞
∴ −++ −++ = ⎜⎟⎜⎟
∂∂⎝⎠ ⎝⎠
0
xx
NN
dxad ad dx
xa
φ
φφ
φ
∂∂
∴ +=
∂∂

1
0 xx
NN
xa
φ
φ
∂∂
∴ +=
∂∂ (6.29)
Equating the sum of all the forces acting in
φ direction to zero, we get
0F
φ
∑=
0
x
xx
NN
N dx N a d dx N a d N dx a d
ax
φφ
φφ φ φ
φφ φ
φ
∂∂⎛⎞⎛⎞
∴ −++ − ++ =⎜⎟⎜⎟
∂∂
⎝⎠⎝⎠
0
x
NN
a d dx dx a d
ax
φφ
φφ
φ
∂∂
∴ +=
∂∂

0
x
NN
a
x
φφ
φ
∂∂
∴ +=
∂∂ (6.30)
The third equation of equilibrium is derived by equating all forces in the direction of the inward
normal drawn at the midpoint of the shell element to zero.

168. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     160 

sin sin
22
0
x
xx
N Qdd
N dx N ad dx Q dx Q ad dx
aa
Q
Qad Q dxad
x
φ φ
φφ φφφφ
φφ
φφ
φφ ∂ ∂⎛⎞ ⎛⎞
++ − ++⎜⎟ ⎜⎟
∂∂ ⎝⎠⎝⎠
∂⎛⎞
−++ =
⎜⎟
∂⎝⎠

Neglecting higher powers of anddx d
φ we obtained
2sin 0
2
x
Q Qd
N dx d dx dx a d
x
φ
φφ
φφ
φ
∂ ∂
∴ ++=
∂∂
0
x
Q Q
Ndxd d dx dxad
x
φ
φ
φφ φ
φ
∂ ∂
∴ ++=
∂∂
On simplifying, we get
0
x
Q Q
Na
x
φ
φ
φ
∂ ∂
∴ ++ =
∂∂ (6.31)
Another equation of equilibrium results from equating the sum of moments of all forces about
the generatrix. Moments of all forces about the generatrix AD.

0
AD
M
∴ ∑=

169. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     161 

0
x
xx
MM
M dx M a d dx M a d M dx a d
ax
Q
Qaddxad
a
φφ
φφ φ φ
φ
φ
φφ φ
φ
φφ
φ
∂∂⎛⎞⎛⎞
∴−++ − ++⎜⎟⎜⎟
∂∂⎝⎠⎝⎠
∂⎛⎞
− +=⎜⎟
∂⎝⎠

Neglecting higher powers of anddx d
φ we obtained
0
x
MM
dx d a dx d aQ dx d
x
φφ
φ
φφφ
φ
∂∂
∴ +−=
∂∂
0
x
MM
aaQ
x
φφ
φ
φ
∂∂
∴ +−=
∂∂ (6.32)
Moments of all forces about the generatrix AB.
0
AB
M
∴ ∑=
0
x
xx x
x x
xx
M
Mad M dxad M dx
x
M Q
M a d dx Q dx a d dx
ax
φ
φ
φ
φφ
φφ
φ
∂⎛⎞
∴ −++ −
⎜⎟
∂⎝⎠
∂⎛⎞ ∂⎛⎞
++ −+ =⎜⎟ ⎜⎟
∂∂ ⎝⎠⎝⎠

0
xx
x
MM
a dxd dxd aQ dxd
x
φ
φφφ
φ
∂∂
∴ +−=
∂∂

0
xx
x
MM
aaQ
x
φ
φ
∂∂
∴ +−=
∂∂ (6.33)
7.4.2 Flugge’s simultaneous equations:
Flugge was the first to derive three simultaneous differential equations in u, v and w for most
concrete reinforced shells.
From equation (6.15)
x
u
NEd
x
∂
=
∂

From equation (6.16)
Ed v
Nw
a
φ
φ
⎛⎞∂
= −
⎜⎟
∂⎝⎠

170. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     162 

And from equation (6.17)
1
2
x
Ed u v
N
ax
φ
φ
⎡ ⎤∂∂
=+
⎢ ⎥
∂∂⎣ ⎦

Substitute and
xx
NN
φ
in the equation (6.29) we get
11
0
2uEduv
Ed
xxa ax
φφ
⎛⎞⎡⎤∂∂ ∂ ∂∂⎛⎞
∴ ++= ⎜⎟⎜⎟ ⎢⎥
∂∂ ∂ ∂∂⎝⎠ ⎣⎦⎝⎠
222
22
1
0
2uEd u v
aEd
xax
φφ
⎡⎤∂∂∂
∴ ++=
⎢⎥
∂∂∂∂ ⎣⎦
222
22
11
0
2uuv
a
xax
φφ
⎡⎤∂∂∂
∴ ++=
⎢⎥
∂∂∂∂⎣⎦ (6.34)
Similarly substitute and
x
NN
φφ
in the equation (6.30) we get
1
0
2Ed v Ed u v
wa
axax
φφ φ
⎡⎤⎧ ⎫⎛⎞ ⎡ ⎤∂∂ ∂ ∂∂
∴ −+ + = ⎨⎬⎢⎥⎜⎟ ⎢⎥
∂∂ ∂ ∂∂⎝⎠ ⎣ ⎦⎣⎦⎩ ⎭
222
22
1
0
2Ed v w Ed u v
a
aaxx
φφ φ
⎛⎞⎡ ⎤∂∂ ∂ ∂
∴ −+ + =⎜⎟ ⎢⎥
∂∂ ∂∂∂⎝⎠⎣ ⎦
222
2
22
1
0
2vw u v
aa
xx
φφ φ
⎛⎞⎡ ⎤∂∂ ∂ ∂
∴ −+ + =⎜⎟ ⎢⎥
∂∂ ∂∂ ∂⎝⎠⎣ ⎦ (6.35)
Now from equation (6.32) and (6.33) we obtained
1 x
MM
Q
ax
φφ
φ
φ
∂∂⎡⎤
=+
⎢⎥
∂∂⎣⎦
(6.36)
And
1 xx
x
MM
Q
xa
φ
φ
∂⎡⎤∂
=+
⎢⎥
∂∂⎣⎦
(6.37)
Now using moment curvature relation
22 2
222
,and
x x
wDw Dw
MD M M
x aax
φφ
φ φ
∂∂ ∂
=− =− =−
∂∂ ∂∂
(6.38)
Substitute equation (6.38) in the equation (6.36) we get

171. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     163 

22
22
1 Dw Dw
Q
aa xax
φ
φ φφ
⎛⎞⎛ ⎞∂∂∂ ∂
∴ =− +−⎜⎟⎜ ⎟
∂∂∂∂∂⎝⎠⎝ ⎠
33
33 2
DwD w
Q
aax
φ
φ φ
∂∂
∴ =− −
∂∂∂
33
23 2
1Dww
Q
aa x
φ
φ φ
⎡⎤∂∂
∴ =− +
⎢⎥
∂ ∂∂⎣⎦
(6.39)
Similarly from equation (6.37) and (6.38) we get
22
2
1
x
wDw
QD
xxaax
φ φ
⎡⎤⎛⎞⎛ ⎞∂∂ ∂ ∂
∴ =− + −⎢⎥⎜⎟⎜ ⎟
∂∂ ∂ ∂∂⎝⎠⎝ ⎠⎣⎦
33
32 2
x
wD w
QD
xax
φ
∂∂
∴ =− −
∂∂∂
33
32
1
x
Dw w
Qa
axax
φ
⎡⎤∂∂
∴ =− +
⎢⎥
∂∂∂⎣⎦ (6.40)
Put equation (6.39) and (6.40) into the equation (6.31)
x
Q Q
Na
x
φ
φ
φ
∂⎡⎤ ∂
∴ =− +
⎢⎥
∂∂⎣⎦
33 3 3
23 2 3 2
11Dww Dw w
Naa
aa x x a x ax
φ
φφφ φ
⎡⎤⎧⎫⎧ ⎫⎡⎤⎡ ⎤∂∂∂∂∂∂⎪⎪⎪ ⎪
∴ =− − + + − +⎢⎥⎨⎬⎨ ⎬⎢⎥⎢ ⎥
∂∂∂∂∂∂∂∂⎪⎪⎪ ⎪⎢⎥ ⎣⎦⎣ ⎦⎩⎭⎩ ⎭⎣⎦

44 4 4
24 24 4 22
11Dww Dw w
Naa
aa x a x ax
φ
φφ φ
⎡⎤⎧⎫⎧ ⎫⎡⎤⎡ ⎤∂∂ ∂ ∂⎪⎪⎪ ⎪
∴ =− − + + − +⎢⎥⎨⎬⎨ ⎬⎢⎥⎢ ⎥
∂∂∂ ∂ ∂∂⎪⎪⎪ ⎪⎢⎥⎣⎦⎣ ⎦⎩⎭⎩ ⎭⎣⎦

44 4
2
24 24 4
1
2Dw w w
Na
aa x x
φ
φφ
⎡ ⎤∂∂ ∂
∴ =++
⎢ ⎥
∂∂∂ ∂⎣ ⎦

But from equation (6.16)
Ed v
Nw
a
φ
φ
⎛⎞∂
=−
⎜⎟
∂⎝⎠

172. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     164 

44 4
2
24 24 4
1
2Ed v D w w w
wa
aaaxx
φφφ
⎡ ⎤⎛⎞∂∂∂∂
∴ −= + +
⎜⎟ ⎢ ⎥
∂∂∂∂∂⎝⎠ ⎣ ⎦

34 4 4
2
24 24 4
1
2
12vEdwww
wa
Ed a x x
φφφ
⎡ ⎤⎛⎞∂∂∂∂
∴ −= + +
⎜⎟ ⎢ ⎥
∂∂∂∂∂⎝⎠ ⎣ ⎦

24 4 4
24
24 24 4
2
12
vdwww
waa
axx
φφφ
⎡ ⎤⎛⎞∂∂∂∂
∴ −= + +
⎜⎟ ⎢ ⎥
∂∂∂∂∂⎝⎠ ⎣ ⎦

Or
2
22 2
2
22 2
0
12
vdww
wa
ax
φφ
⎡⎤⎛⎞∂∂∂
−− + =
⎜⎟ ⎢⎥
∂∂∂⎝⎠ ⎣⎦
(6.41)
Equations (6.34), (6.35) and (6.41) are the simplified version of simultaneous differential
equations due to Flugge’s in the three displacements u, v and w.

6.4.3 The D-K-J Equations:
To derive the differential equations interms of w by eliminating u and v from equations (6.34),
(6.35) and (6.41).
Therefore differentiate equation (6.35) w.r.t. operator
2
2
x
⎛⎞∂
⎜⎟
∂⎝⎠
43 4 4
2
22 2 3 4
1
0
2vw u v
aa
xx x x
φφ φ
⎛⎞⎡ ⎤∂∂ ∂ ∂
∴ −+ +=⎜⎟ ⎢⎥
∂∂ ∂∂ ∂∂ ∂⎝⎠⎣ ⎦ (6.42)
Differentiate equation (6.34) w.r.t. operator
2
a
xφ
⎛⎞∂
⎜⎟
∂∂⎝⎠

444
32
3322
1
0
2uuv
aaa
xxx
φφ φ
⎡⎤∂∂∂
∴ ++=
⎢⎥
∂∂ ∂∂ ∂∂⎣⎦
44 4
23
322 3
1
2 uv u
aa a
xx x
φ φφ
⎡⎤∂∂ ∂
∴ +=−
⎢⎥
∂∂ ∂∂ ∂∂⎣⎦ (6.43)

173. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     165 

Now differentiate equation (6.35) w.r.t. operator
2
2
φ
⎛⎞∂
⎜⎟
∂⎝⎠
43 4 4
2
43 3 22
1
0
2vw u v
aa
xx
φφ φ φ
⎛⎞⎡ ⎤∂∂ ∂ ∂
∴ −+ + =⎜⎟ ⎢⎥
∂∂ ∂∂ ∂∂⎝⎠⎣ ⎦
44 43
2
32243
1
2 uv vw
aa
xx
φ φφφ
⎡⎤⎛⎞∂∂ ∂∂
∴ +=−− ⎜⎟⎢⎥
∂∂ ∂∂ ∂ ∂⎣⎦⎝⎠ (6.44)
Therefore from equation (6.43) and (6.44) we get
443
3
343
uvw
a
x
φφφ
∂∂∂
∴ =−
∂∂ ∂ ∂
(6.45)
Put equation (6.45) into equation (6.42) we obtain
43 43 4
2
22 2 2 4 3 4
11
0
2
vw vw v
a
xx a x
φφ φφ
⎡⎤⎛⎞⎛⎞∂∂ ∂∂ ∂
∴ −+ −+= ⎢⎥⎜⎟⎜⎟
∂∂ ∂∂ ∂ ∂ ∂⎝⎠⎝⎠ ⎣⎦

424 4 3 3
24 4 22 2 23
11
22
va v v w w
axxxa
φ φφ φ
∂∂∂∂ ∂
∴ ++=+
∂∂∂∂∂∂∂

44433
24
24 22 4 2 23
11
2
2 vvvww
aa
axxxa
φ φφφ
⎡⎤∂∂∂∂∂
∴ ++=+
⎢⎥
∂∂∂∂∂∂∂⎣⎦
2
22 3 3
2
222 223
11
2
ww
av
ax xa
φ φφ
⎡⎤∂∂ ∂ ∂
∴ +=+
⎢⎥
∂∂ ∂∂ ∂⎣⎦
(6.46)
Differentiate above equation w.r.t. φ we get
2
22 4 4
22
22 2224
1
2
vww
aa
xxa
φφφ φ
⎡⎤⎡ ⎤∂∂∂ ∂ ∂
∴ += +
⎢⎥⎢ ⎥
∂∂∂ ∂∂ ∂⎣⎦⎣ ⎦

2
22 4 4 4 4
2244
22 22 4 4 4
22
vwwww
aaaa
x xxxφφ φ φ
⎡⎤⎡ ⎤∂∂∂ ∂ ∂ ∂ ∂
∴ += ++−
⎢⎥⎢ ⎥
∂∂∂ ∂∂ ∂ ∂ ∂⎣⎦⎣ ⎦

174. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     166 

22
22 22 4
224
22 22 4
vw
aawa
x xxφφ φ
⎡⎤⎡⎤∂∂∂ ∂∂ ∂
∴ +=+−
⎢⎥⎢⎥
∂∂∂ ∂∂ ∂⎣⎦⎣⎦

2
22 4
24
22 4
vw
awa
x xφφ
⎡⎤ ⎛⎞∂∂ ∂ ∂
∴ +−=−
⎜⎟⎢⎥
∂∂ ∂ ∂ ⎝⎠⎣⎦
(6.47)
Now differentiate equation (6.41) w.r.t. operator
2
22
2
22
a
x
φ
⎡ ⎤∂∂
+
⎢ ⎥
∂∂⎣ ⎦

i.e. w.r.t. operator
2
22 4 4 4
242
22 4 224
i.e. 2aaa
xxx
φ φφ
⎡⎤⎡ ⎤∂∂ ∂ ∂ ∂
+++
⎢⎥⎢ ⎥
∂∂ ∂ ∂∂∂⎣⎦⎣ ⎦
we obtained
2
22
2
22
2
22 2 4 4 4
224
22 2 22 4 4
20
12
v
aw
x
dwww
aaa
xaxx
φφ
φφφ
⎡⎤ ⎛⎞∂∂ ∂
∴ +−
⎜⎟⎢⎥
∂∂ ∂ ⎝⎠⎣⎦
⎡⎤⎡ ⎤∂∂ ∂ ∂∂
−+ ++=
⎢⎥⎢ ⎥
∂∂ ∂∂ ∂∂⎣⎦⎣ ⎦
24
22 2 22
22
22 2 22
0
12
vd
awaw
xax
φφ φ
⎡⎤ ⎡⎤ ⎛⎞∂∂∂ ∂∂
∴ +−− +=
⎜⎟⎢⎥ ⎢⎥
∂∂ ∂ ∂∂⎝⎠⎣⎦ ⎣⎦
(6.48)
From equation (6.47) and (6.48) we get
4
42 22
42
42 22
0
12
wd
aaw
xax
φ
⎡⎤∂∂∂
∴ −− + =
⎢⎥
∂∂∂ ⎣⎦

4
22 64
2
22 24
12
aw
aw
x dxφ
⎡⎤∂∂ ∂
∴ +=−
⎢⎥
∂∂ ∂⎣⎦

4
22 64
2
22 24
12 0
aw
aw
xdx
φ
⎡⎤∂∂ ∂
∴ ++ =
⎢⎥
∂∂ ∂⎣⎦
(6.49)
Or
4
22 44
2
22 4
0
aw
aw
xkx
φ
⎡⎤∂∂ ∂
++=
⎢⎥
∂∂ ∂⎣⎦
(6.50)
This is Donnell – Karman – Jenkins equation in terms of w

175. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     167 


6.5 The Schorer Theory:
The schorer theory published in 1936 has merit of extreme simplicity. Schorer also
assumed 0
xxx
MM Q
φ
=== like finsterwalder but another assumption implement in this theory
is that tangential strain
1v
w
a
φ
ε
φ
⎛⎞∂
=−
⎜⎟
∂⎝⎠
and shear strain
1
x
uv
ax
φ
γ
φ
⎛⎞
∂∂
=+
⎜⎟
∂∂⎝⎠
are both very
small as compared to longitudinal strain
x
u
x
ε
∂
=
∂

Therefore this assumption leads to following relations
1
0
vv
ww
a
φφ
⎛⎞∂∂
∴ −= ⇒ =
⎜⎟
∂∂⎝⎠
(6.51)
11
0
uv v u
ax xaφ φ
⎛⎞∂∂ ∂ ∂
∴ += ⇒ =−
⎜⎟
∂∂ ∂ ∂⎝⎠
(6.52)
Note: Schorer theory is applicable to only long shell.
From equation (6.32) we get
1M
Q
a
φ
φ
φ
∂
∴ =
∂

22
22 22
1 Dw Dw
QM
aa a
φφ
φ φφ
⎛⎞ ⎛ ⎞∂∂ ∂
∴ =− =−⎜⎟ ⎜ ⎟
∂∂ ∂⎝⎠ ⎝ ⎠
3
33
Dw
Q
a
φ
φ
∂
∴ =−
∂
(6.53)
From equation (6.31) we get
3
33
Q Dw
NN
a
φ
φφ
φ φφ
∂ ⎛⎞∂∂
∴ =− ⇒ =− − ⎜⎟
∂∂∂ ⎝⎠
4
44
Dw
N
a
φ
φ
∂
∴ =
∂
(6.54)

176. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     168 

From equation (6.30) we obtained
1x
NN
xa
φ φ
φ
∂∂
∴ =−
∂∂
5
45
x
N Dw
xa
φ
φ
∂ ∂
∴ =−
∂∂ (6.55)
From equation (6.29) we obtained
1 xx
NN
xa
φ
φ
∂∂
∴ =−
∂∂
22
2
1 xx
NN
xax
φ
φ
⎛⎞∂∂
∴ =−⎜⎟
⎜⎟
∂∂∂
⎝⎠

2
2
1 xx
NN x ax
φ
φ
∂⎛⎞∂ ∂
∴ =− ⎜⎟
∂∂∂ ⎝⎠
Therefore using equation (6.55) we get
2 6
256
x
N Dw
xa
φ
∂ ∂
∴ =
∂∂ (6.56)
Now from equation (6.15)
x
u
NEd
x
∂
∴ =
∂

Differentiate with respective x twice we get
2 3
23
x
N u
Ed
x x
∂ ∂
∴ =
∂∂
(6.57)
Therefore equation (6.56) and (6.57) we obtained
36
356
uDw
xEda
φ
∂∂
∴ =
∂∂
(6.58)
From relation (6.52)

177. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     169 

1vu
xa φ
∂∂
∴ =−
∂∂

Differentiate with respective x we get
44
43
1vux axφ
∂∂
∴ =−
∂∂∂
43
43
1vux axφ
⎛⎞∂∂∂
∴ =− ⎜⎟
∂∂∂ ⎝⎠ (6.59)
From equation (6.58) we get
47
467
vDw
xEda
φ
∂∂
∴ =−
∂∂
(6.60)
Again from equation (6.51) we obtained
v
w
φ
∂
∴ =
∂

Differentiate with respective x we get
45
44
wvx xφ
∂∂
∴ =
∂∂∂

44
44
wvx xφ
⎛⎞∂∂∂
∴ =⎜⎟
∂∂∂ ⎝⎠
Therefore from equation (6.60) we get
48
468
wDw
xEda
φ
∂∂
∴ =−
∂∂

84
68 4
0
Dw w
Eda x
φ
∂∂
∴ +=
∂∂

()
38 4
68 4
00
12
Ed w w
Eda x υ
φ
∂∂
∴ += =
∂∂
(6.61)

178. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     170 

844 2
84 2
0
12
waw d
k
kx a
φ
⎛⎞∂∂
∴ += = ⎜⎟
∂∂ ⎝⎠ (6.62)
This is schorer form of the differential equation for cylindrical shell.

6.6 VLASOV Bending Theory (Applicable to shallow shell)
6.6.1 Assumptions:
1. The squares and products of surface derivatives / and /
pdz dx q dz dy= = are
negligible in comparison with unity.
2.
The gauss curvature of the undeformed middle surface of the shell is very small and can
be assumed to be equal to zero.
3.
The derivative of r, s and t may be neglected. This in effect amount assuming that the
principal curvature of the shell remains constant.
6.6.2 Equations of Equilibrium:
Let the equation of doubly curved shell surface be z = z(x, y); we usually denotes
222
22
;;
zzz
rst
xxyy
∂∂∂
===
∂∂∂∂
(6.63)
Equating all forces in the x direction to zero we get
0
x
F∑=
0
yxx
x x yx yx
NN
N dy N dx dy N dx N dy dx X dx dy
xy
∂⎛⎞∂⎛⎞ ∴ −++ −++ + = ⎜⎟⎜⎟
∂∂⎝⎠ ⎝⎠
0
yxx
NN
dx dy dy dx X dxdy
xy
∂∂
∴ ++=
∂∂

179. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     171 


0
yxx
NN
X
xy
∂∂
∴ ++=
∂∂ (6.64)
0
y
F∑=
0
yxy
yy xyxy
NN
N dx N dy dx N dy N dx dy Y dx dy
yx
∂∂⎛⎞⎛ ⎞∴ −++ −++ + =⎜⎟⎜ ⎟
∂∂
⎝⎠⎝ ⎠
0
yxy
NN
dy dx dxdy Y dx dy
yx
∂∂
∴ ++=
∂∂
0
yxy
NN
Y
yx
∂∂
∴ ++=
∂∂ (6.65)

180. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     172 

For the sake of convenience denote
xy yx
HM M
= =− . Now summing up the moments about
edge DC and equate with zero we get
0
DC
M∑=
0
yxy
y y xy xy
y
y
MM
Mdx M dy dx M dy M dx dy
yx
Q
Qdydxdy
y
∂∂⎛⎞⎛ ⎞
∴ −++ −++⎜⎟⎜ ⎟
∂∂⎝⎠⎝ ⎠
∂⎛⎞
−+ =⎜⎟
∂⎝⎠

0
yxy
y
MM
dy dx dxdy Q dxdy
yx
∂∂∴ +−=
∂∂
0
yxy
y
MM
Q
yx
∂∂∴ +−=
∂∂
0
y
y
M H
Q
yx
∂ ∂∴ −−=
∂∂ (6.66)
Similarly summing up the moments about edge AB and equate with zero we get
0
x
x
M H
Q
xy
∂ ∂∴ −−=
∂∂ (6.67)
Now resolving forces in z direction we get
222
22
20
yx
xxy y
QQ zzz
NN Nz
xy x xy y
∂∂ ∂∂∂
∴ ++ + + +=
∂∂ ∂ ∂∂ ∂

From equation (6.66) and (6.67) substitute value of and
xy
QQ we obtained
222
22
20
xy
xxy y
MM HHzzz
NN Nz
xx y yy x x xy y
∂∂⎡⎤⎡⎤∂∂∂∂∂∂∂
∴ −+ −+ + + +=
⎢⎥⎢⎥
∂∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂
⎣⎦⎣⎦
22 2222
222 2
220
xy
xxy y
MM Hzzz
NN Nz
xxyy x xyy
∂∂ ∂∂∂∂
∴ −+++ ++=
∂∂∂∂ ∂ ∂∂∂
(6.68)
Strain displacement relations:

181. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     173 

2
xyxy
uuuv
rw tw sw
xyyx
εεγ
∂∂∂∂
=− =− =+−
∂∂∂∂
(6.69)
Moment curvature relations: 22 2
22
111
xyxy
ww w
rx ry r xy
∂∂ ∂
===−
∂∂ ∂∂
(6.70)
Stress strain relations:
()
()
2
2
1
1
1
xxy
yyx
xy yx xy xy
Ed
N
Ed
N
Ed
NNG
ευε
υ
ευε
υ
γ γ
υ
⎫
=+
⎪
−
⎪
⎪
=+ ⎬
−
⎪
⎪
== =
⎪
+ ⎭
(6.71)
()
22
22
22
22
2
1
x
y
xy
ww
MD
xy
ww
MD
yx
w
MHD
xy
υ
υ
υ
⎫
⎡ ⎤∂∂
=− + ⎪⎢ ⎥
∂∂
⎣ ⎦⎪
⎪
⎡ ⎤∂∂ ⎪
=− + ⎬⎢ ⎥
∂∂⎣ ⎦⎪
⎪
∂
⎪== −
∂∂⎪
⎭
(6.72)
6.6.3 Derivation of VLASOV Equation:
From equation (6.69)
2 32
222
x
uw
r
yxy yε∂ ∂∂
=−
∂∂∂ ∂
;
2 32
222
y vw
t
x yx x
ε∂ ∂ ∂
=−
∂ ∂∂ ∂

22 33 22
22 2 2 2 2
yx uv ww
rt
yxxyyx y xεε∂∂
∂ ∂∂∂
∴ +=+−−
∂∂∂∂∂∂ ∂ ∂
(6.73)
2233 22
2222 22
yxuv ww
rt
xyyx y x y x
εε∂∂∂∂ ∂∂
∴ +=+++
∂∂ ∂∂ ∂ ∂ ∂ ∂
(6.74)
Again from equation (6.69)

182. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     174 

2 33 2
22
2
xy uv w
s
xyyx xy xy
γ∂ ∂∂ ∂
∴ =+ −
∂∂∂∂ ∂∂ ∂∂
(6.75)
Therefore put equation (6.74) into the equation (6.75) we get
22 2 22 2
22 2 2
2
xy y x ww w
rt s
xyy x y x xy
γε ε∂∂ ∂ ∂∂ ∂
∴ =++ + −
∂∂∂ ∂ ∂ ∂ ∂∂
(6.76)
Let us consider influence of vertical force only i.e. X = Y = 0
Introduce a stress function φ such that
22 2
22
;and
yx xy
NN N
xy xy
φφ φ
⎡ ⎤∂∂ ∂
==−=
⎢ ⎥
∂∂ ∂∂⎣ ⎦
(6.77)
Rearranging the term in the equation (6.76) we get
222 22 2
22 2 2
20
yxyx ww w
rt s
yxxy y x xyεγε⎡⎤ ∂∂ ⎡⎤∂ ∂∂ ∂
∴ +− + + − =⎢⎥ ⎢⎥
∂∂∂∂ ∂ ∂ ∂∂⎢⎥ ⎣⎦⎣⎦
(6.78)
From equation (6.71) if
()0υ= we get
2
yxyx
xyxy
NNN
Ed Ed Ed
εεγ
∴ == = (6.79)
22 222
2222
112 yy xyxx
xy
NNN
yEdy xEdx Edxyεε
γ∂∂ ∂∂∂
∴ ===
∂∂∂∂ ∂∂
(6.80)
Put equation (6.77) into the equation (6.80) we get
22 44 4
2424 22
112 yx
xy
yEdy xEdx Edxy
εε
φ φφ
γ∂∂ ∂∂ ∂∴ ===−
∂∂∂∂ ∂∂
(6.81)
Put in the equation (6.78) we obtained
44 4 2 2 2
44 22 2 2
1
220
rt sw
Ed y x x y y x x y
φφ φ⎡⎤⎡ ⎤∂∂ ∂ ∂ ∂ ∂
∴ ++ + + − =
⎢⎥⎢ ⎥
∂∂ ∂∂ ∂ ∂ ∂∂⎣⎦⎣ ⎦
421
0
k
w
Edφ
∴ ∇+∇ =
42
0
k
Ed wφ∴ ∇+ ∇ = (6.82)

183. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     175 

In the equation () substitute ,and
xy xy
MMM interms of w and ,and
xy xy
NN N interms of φ we
get
42
k
Dw z φ
∴ ∇−∇ = (6.83)
6.7 Beam Theory for Cylindrical Shell (Lundgren’s Beam Theory):
Lundgren’s beam theory provides a simple approach of analysis for cylindrical shell of long
span, supported at ends. In this approach, shells are treated simply as a beam, and bending and
shear stresses for this beam is calculated.
6.7.1 Advantages:
1. It brings shell analysis within the reach of those who are not familiar with advance
mathematics.
2.
It is also applicable to shell having nonuniform thickness (along the directrix).
3.
Line load acting on the shell also treated by this.
4.
Structural action of shell is easily visualized.
5.
It can handle shell strengthened by ribs in longitudinal and transverse direction
(stiffening Beams).
6.
Application to shell with non-circular directrices is possible.
6.7.2 Assumptions:
1. Transverse deflection of shell in its plane is neglected. This assumption is replaced by
Bernoulli’s assumption (plane section before deformation remains plane after
deformation) if shell is subjected to pure bending.
2.
0
xxyx
MM Q=== (true only for a long shell)
3.
Shear strain
x
φ
γ caused by
x
N
φ
is neglected
All these assumptions will be linearly valid only for long beams. It is observed that beam theory
gives fairly acceptable answers for cylindrical roof shells without edge beams if L/a > 5, and
shells with edge beams if L/a > 3. These limits may be used as guidelines for a designer. The
shell must be uniformly loaded.

184. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     176 

This method is carried out in two distinct steps
a)
Beam analysis
b)
Arch Analysis
6.7.3 Beam Analysis:
Shell regarded as a beam supported at curved edges. Bending stresses and shearing stresses are
found using the beam formulae.

d = thickness of shell at any c/s

At any c/s value of stress resultants Nx is given by
..
yy
x
yy
M
NZd
I
= (6.84)
yy
M = Bending Moment at any c/s
yy
I = Moment of inertia @ y-y axis
2
0
sin
2cos
c
c
yy
c
Idada a
φφ
φ
φ
φφ
φ
=
=
⎛⎞
∴ =− ⎜⎟
⎝⎠∫

2
32
2
0
sin sin
2 cos 2cos
c
cc
yy
cc
I ad d
φφ
φ
φφ
φ φφ
φφ
=
=
⎡ ⎤
∴ =−+ ⎢ ⎥
⎣ ⎦
∫

2
3
2
0
cos sin sin1cos2
22
2
c
cc
yy
cc
I ad d
φφ
φ
φφ φφ
φ
φφ
=
=
⎡ ⎤+
∴ =−+ ⎢ ⎥
⎣ ⎦
∫

185. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     177 

2
3
2
0
cos sin sin1cos2
22
2
c
cc
yy
cc
I ad d
φφ
φ
φφ φφ
φ
φφ
=
=
⎡ ⎤+
∴ =−+ ⎢ ⎥
⎣ ⎦
∫

2
3
2
0
sin sin sinsin 2
22
24
c
cc
yy
cc
Iad
φ
φφ φφφ
φ
φφ⎡⎤
∴ =+− +
⎢⎥
⎣⎦
22
3
sin 2 sin sin
22
24
cc c c
yy
cc
Iad
φ φφφ
φφ⎡⎤
∴ =+−+⎢⎥
⎣⎦
2
3
sin 2 2sin
2
cc
yy c
c
Iad
φ φ
φ
φ⎡⎤
∴ =+−
⎢⎥
⎣⎦
3 2sin
sin cos
c
yy c c c
c
Iad
φ
φφφ
φ⎡⎤ ⎛⎞
∴ =+ −⎢⎥ ⎜⎟
⎢⎥ ⎝⎠⎣⎦ (6.85)
Equation (6.84) and (6.85) are applicable to only shells of symmetric cross/section.
The beam analysis also enables
x
N
φ
to be found out by the use of well known
.VQ
Ib
formula for
the shell subjected to only vertical loading symmetrically distributed over the cross/section.
.. .. .
22
x
yy yy yy
VQd VQd VQ
N
IbId I
φ
∴ === (6.86)
Where, V = vertical shearing force and Q = moment of Area
This is given by
sinsin
2
c
c
Qa ad
φφ
φ
φφ
⎡⎤
=−
⎢⎥
⎣⎦

2 sin
2sin
c
c
Qad
φ
φφ
φ
⎡⎤∴ =− ⎢⎥
⎣⎦ (6.87)

186. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     178 

6.7.4 Arch Analysis: in this second part of analysis, a slice unit width is considered and is
analysed as an arch, to find the stress resultants
,andMQN
φφφ

The object of Arch analysis is to find ,MQ
φφ
andN
φ

in the shell.
The equilibrium of the arch is maintained by two set of
forces, namely the load acting on the element and
x
N
x
φ
∂
∂
(Known as specific shear).
This specific shear is resolved into horizontal and vertical components. Sum of vertical
component balance load on shell and sum of horizontal components which are symmetrically
disposed on crown balance themselves.
0
2 sin Total load on the arch of unit width
x
N
dP
x
φ
φ
φ
φφ
∂
∴ =
∂∫
(6.88)
Exercise:
Que.
Derive the 8
th
order governing differential equations for cylindrical shell subjected to
bending, according to the Finsterwalder theory.
[Dr. B.A.M.U. Article 6.3]
Que. Write short note on D-K-J theory for cylindrical shell. [Dr. B.A.M.U. Article 6.4]
Que.
Derive equations of VLASOV’s bending theory of shallow shells of double curvature from
first principal.
[Dr. B.A.M.U. Article 6.6]
Que. Derive Schorer form of differential equations for cylindrical shell.
[Dr. B.A.M.U. Article
6.5]

Que. Derive Flugge’s simultaneous differential equations.
[Dr. B.A.M.U. Article 6.4.2]
Que. Describe in brief, the Lundgren’s beam theory for thin shells. What are its limitations?
[P.U. & Dr. B.A.M.U. Article 6.7]
Que. Explain the beam method of analysis of cylindrical shell with advantages and limitations of
this method.
[Dr. B.A.M.U. Article 6.7]

187. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     179 

Chapter 7
Shells of Surface of Revolution

7.1 Membrane Theory for Surface of Revolution with Axisymmetric Loading:
Surface of revolution is the surface obtained by revolving a plane curve about an axis in
its own plane. The curve is called as ‘meridian’ and plane of the curve is called as ‘meridianal
plane’ Intersections of the shell with planes normal to axis of revolution are called ‘parallel
circles’. Position of a parallel circle is defined by angle
φ made by the normal to the surface and
axis of revolution. Meridianal plane and the plane perpendicular to the meridian at a point are the
planes of principal curvature at the point. The corresponding radii of curvatures are
1
r and
2
r
respectively. Radius of parallel circle is
0
r. Let the angle subtended by the element at the centers
of curvature be d
φ anddθ.

Fig. (7.1)

188. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     180 

The dimensions of the element are
1
rd
φ in y direction and
0
rdθ in x direction.
Therefore the area of element is =
10
rrd d
θφ (7.1)

But from figure (7.1)
02
sinrd r d
θ φθ∴ =
The surface area of the element is then

12
sinrr d d
φθφ (7.2)

In writing the equations of equilibrium of the element, let us begin with the forces in the
direction of the tangent to the meridian. On the upper edge of the element the force N
φ
is acting
over the length
0
rd
θ therefore total force on the upper edge
0
Nrd
φ
θ =
2
sinNr d
φ
φφ (7.3)

189. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     181 

Similarly corresponding force on the lower edge of the element, is
N
Nd
φ
φ
φ
φ
∂
+
∂ acting over the
length
0
0
r
rd
φ
φ
⎛∂ ⎞
+
⎜⎟
∂⎝⎠
. Therefore total force on the lower edge is
0
0
N r
Ndrd
φ
φ
φ φ
φφ
∂⎡⎤ ⎛∂ ⎞
++
⎜⎟⎢⎥
∂∂ ⎝⎠⎣⎦
(7.4)
From equation (7.3) and (7.4), by neglecting small quantities of second order we find the
resultant in the y direction to be equal to
0
00
N r
Nrd N d r d d
φ
φφ
θ φφθ
φφ
∂⎛⎞ ⎛∂ ⎞
=− + + + ⎜⎟ ⎜⎟
∂∂ ⎝⎠⎝⎠

00
00 0
NN rr
Nrd Nrd d rd N d d d d d
φφ
φφ φ
θ θφθ φθφφθ
φφφφ
∂ ∂∂∂
=− + + + +
∂∂∂∂

0
0
N r
drd N dd
φ
φ
φθφθ
φφ
∂ ∂
=+
∂∂

() 0
Nr d d
φ
φθ
φ
∂
=
∂
(7.5)

190. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     182 

The forces acting on the lateral edges of the element are equal to ‘
1
Nrd
θ
φ’ and have a resultant
in the direction of the radius of parallel circle equals to ‘
1
Nrdd
θ
φθ’. Component of this force
in y direction is ‘
1
cosNr dd
θ
φφθ− ’. (7.6)
The component of external force/body force in the same direction is
10
Yrr d d
φθ
Therefore now sum up the all forces in y direction, from equation (7.5), (7.6) and considering
body force, we get the equation of equilibrium in the direction of the meridian becomes,
0
y
F∑=
() 01 10
cos 0Nr dd Nr dd Yrrdd
φθ
φθ φφθ φθ
φ
∂ ∴ −+=
∂

() 01 10
cos 0Nr Nr Yrr
φθ
φ
φ
∂ ∴ −+=
∂
(7.7)
The second equation of equilibrium is obtained by summing up the projection of the forces in the
z direction. The forces acting on the upper and lower edges of the element have a resultant in z
direction equal to
0
Nrd d
φ
θφ (7.8)
The forces acting on the lateral sides of the element having the resultant
1
Nrdd
θ
φθ in the
radial direction of the parallel circle gives a component in the z direction of the magnitude

1
sinNr dd
θ
φφθ (7.9)
The external load acting on the element has in the same direction a component
10
zrr d d
θφ (7.10)
Therefore sum up the forces in z direction; we obtained the second equation of equilibrium
Therefore from equations (7.8), (7.9) and (7.10) we get
0
z
F∑=

191. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     183 

01 10
sin 0Nrd d Nr d d zrrd d
φθ
θφφφθθφ∴ ++=
01 10
sin 0Nr Nr zrr
φθ
φ∴ ++= (7.11)
Divide by
10
rr we get
0 110
10 10 10 sin
0
Nr Nr zrr
rr rr rr
φ θ φ
∴ ++=
10
sin
0
N N
z
rr
φ θ φ
∴ ++=
Since
02
sinrr
φ=
12
sin
0
sin
N N
z
rr
φ θ φ
φ
∴ ++=
12
0
N N
z
rr
φ θ
∴ ++= (7.12)
Due to Axisymmetric, equilibrium in x direction is obvious. From above two equilibrium
equations andNN
θ φ
can be found if
01
,,,,rrYZφ are known.
From equation (7.7)
()10 1 0
cosYr r N r N r
θφ
φ
φ
∂
∴ =−
∂
(7.13)
Solving equation (7.12) for N
θ

21
NN
z
rr
φθ
⎡ ⎤
∴ =− +
⎢ ⎥
⎣ ⎦

2
1
N
Nr z
r
φ
θ
⎡ ⎤
∴ =− +
⎢ ⎥
⎣ ⎦
(7.14)
Substitute equation (7.14) in the equation (7.13) we obtained

192. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     184 

()10 2 1 0
1
cos
N
Yr r r z r N r
r
φ
φ
φ
φ
⎡⎤ ∂
∴ =− + −
⎢⎥
∂
⎣⎦
()10 21 2 0
cos cosYr r zr r r N N r
φφ
φφ
φ
∂
∴ =− − −
∂

()10 21 2 0
cos cosYr r zr r r N N r
φφ
φφ
φ
∂
∴ +=−−
∂

Multiply by sinφ this equation may be written as
()201021
cos sin sin sin cos sinrN Nr Yrr zrr
φφ
φφφφφφ
φ
∂
∴ −− =+
∂

()
2
201221
cos sin sin sin cos sinrN Nr Yrr zrr
φφ
φφφφφφ
φ
∂
∴ −− =+
∂

() ()
2012
cos sin sin sin sin cosrN Nr rr Y z
φφ
φφφφφφ
φ
∂
∴ −− = +
∂
(7.15)
Which is also written as, () ()
2
212
sin sin cos sinNr rr Y Z
φ
φ φφ
φ
∂
=− +
∂
(7.16)








Therefore integrating equation (7.16) w.r.t.
φ to find out value ofN
φ
, we get
() ()
2
212
sin sin cos sinNr d rr Y Z d
φ
φφφφφ
φ
∂ ∴ =− +
∂∫∫

()
22
22 2
sin 2 sin cos sinNr Nr r N
φ φφ
φφφφ
φφ
∂∂
=+
∂∂

And
() ()2022
sin cos sin sin cos sin sinNr Nr Nr Nr
φφφφ
φφφφφφφ
φφ
∂∂
+= +
∂∂

222
sin cos cos sin sinNr Nr r N
φφ φ
φφφφφ
φφ
⎛⎞∂∂
=++
⎜⎟
∂∂⎝⎠

2
22
2 sin cos sinNr r N
φ φ
φφ φ
φ
∂
=+
∂

Therefore we can write equation (7.15) as equation (7.16)

193. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     185 

( )
2
212
sin sin cos sinNr rr Y Z d C
φ
φφφφ
∴ =− + +∫

()
122
2
1
sin cos sin
sin
NrrYZdC
r
φ
φφφ
φ
∴ =− + +∫

()
122
2
1
2 sin cos sin
2sin
NrrYZdC
r
φ
πφ φφ
πφ
∴ =− + + ∫
(7.17)
The term
12
2sinrr d
π φφ stands for surface area of an elemental strip of the dome.
Alternatively we may consider equilibrium of the whole cap with total load R (downwards)
Therefore equation of equilibrium along axial direction
0
2sin 0rN R
φ
π φ∴ +=
0
2sin
R
N
r
φ
πφ
∴ =− (7.18)


This equation can be used instead of differential equation (7.7), from which it can be
obtained by integration (7.17). It can be seen that when N
φ
is obtained from equation (7.18), the
force N
θ
can be calculated from equation (7.12). Hence the problem of membrane stresses can
be readily solved in each particular case.
7.2 Membrane Theory for Surface of Revolution with Unsymmetrical Loading:

194. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     186 

Considering again an element cut from a shell by two adjacent meridians and two parallel
circles as shown in figure, in general case not only normal forces andNN
φ θ
but also shearing
forces NN
φθθφ
= will act on the sides of the element. Area of the element is given in equation
(7.2).
Taking the sum of the projections in the y direction of all forces acting on the element, we must
add to the forces considered in the equations (7.5), (7.6) and External force. The shearing force
1
N
drd
θφ
θφ
θ
∂
∂
(7.19)
Representing the difference in the shearing forces acting on the lateral sides of the element,
hence by adding equation (7.5), (7.6), External force and equation (7.19), we obtain the equation
()10 1 10
cos 0
N
drd Nr dd Nr dd Yrrdd
θφ
φθ
θφ φθ φφθ φθ
θφ
∂ ∂
∴ +−+=
∂∂
()10110
cos 0
N
rNrNr Yrr
θφ
φθ
φ
θφ
∂ ∂
∴ +−+=
∂∂ (7.20)

The second equation of equilibrium is obtained by summing up the projection of the forces in the
x direction; we must include the difference of the shearing forces acting on the top and bottom of
the element as given by the expression

195. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     187 

()
0
00
Nr
N dd rdd rN dd
φθ
φθ φθ
φθφθ φθ
φφ φ
∂∂ ∂
+=
∂∂ ∂
(7.21)
The force
1
N
rd d
θ
θφ
θ
∂
∂
(7.22)
Due to variation of the force N
θ
and the force
1
cosNr dd
θφ
φθφ (7.23)
Due to the small angle cosdφθbetween the shearing forces N
θφ
acting on the lateral sides of
the element. The component in x direction of the external load acting on the element is
01
Xrrd dθφ (7.24)
Summing up all these forces, we obtain the equation
()01101
cos 0
N
rN d d rd d N r d d Xrrd d
θ
φθ θφ
φθ θ φ φθ φ θφ
φθ
∂∂
++ +=
∂∂

()01101
cos 0
N
rN r N r Xrr
θ
φθ θφ
φ
φθ
∂∂
++ +=
∂∂
(7.25)
The third equation of equilibrium is obtained by projecting the forces on the z axis. Since the
projection of shearing forces on this axis vanishes, the third equation confirms with equation
(7.11), which was derived for symmetrical bending.
The problem of determining membrane stresses under unsymmetrical loading reduces to
the solution of equations (7.20), (7.25) and (7.11) for given values of components X, Y and Z of
the intensity of the external load.

7.3 Particular cases of Shells in the form of Surface of Revolution:
7.3.1 Example: Spherical dome of constant thickness under its own weight.

196. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     188 


Solution: - Consider a spherical dome of constant thickness under its own weight as shown in
figure. Let
q be the constant weight of dome per unit area and ‘a’ be the radius of the sphere.
Let total load R on that part of the spherical dome subtended by angle
φ.
Consider a section at an angle φ from axis


Therefore
0
sin
r
aφ
=
0
sinra φ∴ =


The total load R on that part of the spherical dome subtended by angle φ is given by

0
0
2
R rad q
φ
π φ∴ =∫

Substitute value of
0
rand simplify, we get
2
0
2sin
R adq
φ
π φφ∴ =∫

197. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     189 

2
0
2sin
R aq d
φ
π φφ∴ = ∫

[ ]
2
0
2cosRaq
φ
π φ∴ =−
[ ]
2
2coscos0Raqπφ∴ =−+
[ ]
2
21cosRaqπ φ∴ =− (7.26)
Put this value in the equation (7.18) to determine the value of N
φ

[ ]
2
0
21cos
2sin
aq
N
r
φ
π φ
πφ−
∴ =−
[ ]
()
2
02
21cos
sin
2sin
aq
Nra
a
φ
πφ
φ
πφ
−
∴ =− =
[ ]
2
1cos
sinaq
N
φ
φ
φ
−
∴ =−
( )
()
2
1cos
1cos
aq
N
φ
φ
φ
−
∴ =−
−

( )
()()
() ()(){}
22
1cos
1 cos 1 cos
aq
Nababab
φ
φ
φφ−
∴ =− − = − +
−+

()1cos
aq
N
φ φ
∴ =−
+ (7.27)
Now from equation (7.12) we obtained value of N
θ

0
N N
z
aa
φ θ
∴ ++=
12
for spherical dome
rra
⎛⎞
⎜⎟
==
⎝⎠

198. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     190 

Where coszq φ=
cos 0
N N
q
aa
φ θ
φ
∴ ++ =
Substitute value of N
φ
from equation (8.20)
()
cos 0
1cos
Naq
q
aa
θ
φ
φ
∴ −++=
+
()
cos
1cos
q
Na q
θ
φ
φ
⎡ ⎤
∴ =−⎢ ⎥
+
⎣ ⎦

()
1
cos
1cos
Naq
θ
φ
φ
⎡ ⎤−
∴ =− +⎢ ⎥
+
⎣ ⎦
(7.28)
Now
At
0
φ=
2
aq
N
φ
=− (compressive) and
2
aq
N
θ
=− (compressive)
(Negative sign indicates compressive force and Positive sign indicates tensile force)
Similarly at
2
π
φ= Naq
φ
=− (compressive) and Naq
θ
= (Tensile)
Which indicates that N
φ
is compressive throughout from 0 90toφ φ= = but N
θ
is
compressive at top i.e. 0φ= and tensile at 90φ=.
Therefore to find out magnitude of φ where N
θ
changes its sign from compressive to tensile
equate equation (7.28) of N
θ
with zero we get,
() ()
11
cos 0 cos 0
1cos 1cos
aq
φφ
φφ
⎡⎤−−
∴ −+=⇒∴ +=⎢⎥
++
⎣⎦

2
cos cos 1 0φφ∴ +−=
Solving quadratic equation for cosφ we get

199. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     191 

114
cos
2
φ
−±−
∴ =
0'
51.8 51 50φφ∴ =⇒= (7.29)

In above expression it is assumed that shell gets reaction tangential at the support. If reaction is
not tangential, bending must occur near the support. Generally for shell angles other
than
0
90φ=, it is difficult to provide tangential reaction. Hence for such angles usually a ring
beam is used to take horizontal components of the force N
φ
and only vertical components are
transferred to the supports. In such cases hoop strain in the beam is differed from strain along
angle θ in the shell at the beam. Hence there is bending induced locally in a small portion near
the ring beam.
7.3.2 Example: Spherical Tank filled with Liquid.

200. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     192 

Solution: Consider a spherical tank with radius a supported along a parallel circle A-A as shown
in figure and filled with liquid of specific weightγ. Consider hydraulic pressure of the liquid of
specific weight γ at an angleφ.

Therefore the pressure at an angle φ is given by
()cosRz aaγ φ∴ =− = − (7.30)
Over the
0
Area 2 2 sinrad a ad
πφπ φφ== ( )
0
sinra φ= (7.31)
Therefore resultant R of the pressure for the portion of shell defined by an angle φ is
()
0
cos cos 2 sinR aa a ad
φ
γ φφπφφ∴ =− −∫

()
3
0
21coscossin
R ad
φ
πγφφφφ∴ =− − ∫

Put
cos
sin
t
ddt
φ
φφ
=
∴ −=

()
3
0
21
R attdt
φ
πγ∴ =− ∫

23
3
0
2
23
tt
Ra
φ
πγ
⎡⎤
∴ =−
⎢⎥
⎣⎦

201. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     193 

23
3
0
cos cos
2
23
Ra
φ
φ φ
πγ⎡⎤ ∴ =−
⎢⎥
⎣⎦

23
3
cos cos 1 1
2
2323
Ra
φφ
πγ⎡⎤
∴ =−−+
⎢⎥
⎣⎦
23
3
cos cos 1
2
236
Ra
φφ
πγ⎡⎤
∴ =−−
⎢⎥
⎣⎦
2
3
cos 2cos 1
21
236
Ra
φφ
πγ⎡⎤ ⎛⎞
∴ =−−
⎜⎟⎢⎥
⎝⎠⎣⎦
2
3
1 cos 2cos
21
62 3
Ra
φ φ
πγ
⎡ ⎤⎛⎞
∴ =− − −
⎜⎟⎢ ⎥
⎝⎠⎣ ⎦
(7.32)
Substitute equation (7.32) in the equation (7.18) to find the value of N
φ

2
3
2
0
1 cos 2cos
21
62 3
2sin 2sin
a
R
N
ra
φ
φ φ
πγ
πφ π φ
⎡ ⎤⎛⎞
−− ⎜⎟⎢ ⎥
− ⎝⎠⎣ ⎦
∴ ==
22
2
1 cos 2cos
1
sin 6 2 3a
N
φ
γ φφ
φ
⎡ ⎤⎛⎞
∴ =−−
⎜⎟⎢ ⎥
⎝⎠⎣ ⎦

2
2
2
132cos
13cos
sin 6 3
a
N
φ
γ φ
φ
φ
⎡ −⎤⎛⎞
∴ =−
⎜⎟⎢ ⎥
⎝⎠⎣ ⎦

()
2
2
2
1cos 32cos
6sin
a
N
φ
γ
φ φ
φ
⎡ ⎤∴ =−−
⎣ ⎦

23
2
2
13cos 2cos
6sina
N
φ
φ φγ
φ⎡⎤−+
⎣⎦
∴ =
2223
22
1 cos 2cos 2cos
6sin sina
N
φ
γ φφφ
φφ⎡⎤−−
∴ =−
⎢⎥
⎣⎦

202. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     194 

()
()
22
2
2cos 1 cos
1
6 1cosa
N
φ
φ φγ
φ
⎡⎤
−
⎢⎥
∴ =−
−⎢⎥
⎣⎦
( )
()()
22
2cos 1 cos
1
61cos1cos
a
N
φ
φφγ
φ φ
⎡⎤ −
∴ =− ⎢⎥
−+
⎣⎦

()
22
2cos
1
61cos
a
N
φ
γ φ
φ
⎡⎤
∴ =− ⎢⎥
+
⎣⎦
(7.33)
Now put value of N
φ
from equation (7.33) into the equation (7.12) to determine the value of N
θ

0
N N
z
aa
φ θ
∴ ++=
N
Na z
a
φ
θ⎛⎞
∴ =− +⎜⎟
⎝⎠
Put value of N
φ
and z from equation (7.33) and (7.30) respectively we get
()
()
22
2cos
1cos
61cos
a
Na aa
a
θ
γφ
γ φ
φ
⎛⎞⎡⎤∴ =− − − −⎜⎟⎢⎥
⎜⎟ +
⎣⎦⎝⎠

()
()
22
2
2cos
11cos
61cos
a
Na
θ
γφ
γ φ
φ
⎛⎞ ⎡⎤∴ =− − + −⎜⎟ ⎢⎥
⎜⎟ +
⎣⎦⎝⎠

()
22
2cos
166cos
61cos
a
N
θ
γφ φ
φ
⎛⎞
∴ =−+ +−⎜⎟
⎜⎟
+
⎝⎠

()
22
2cos
56cos
61cos
a
N
θ
γφ φ
φ
⎛⎞
∴ =+ −⎜⎟
⎜⎟
+
⎝⎠
(7.34)
These expressions for andNN
θφ
are valid upto
0
φφ≤
For calculating the resultant R for
0
φφ〉, we must take into account the sum of the vertical
reactions along the ring A-A also in addition to the internal pressure.

203. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     195 

Weight of sphere = γ× volume of sphere
34
Weight of sphere
3
rγπ=
34
Weight of sphere
3
aγπ=
2
33
4 1 cos 2cos
21
3623
Raa
φ φ
γπ πγ
⎡ ⎤⎛⎞
∴ =− −−
⎜⎟⎢ ⎥
⎝⎠⎣ ⎦
(7.35)
Solving again for andNN
θφ
we get
()
22
2cos
5
61cos
a
N
φ
γ φ
φ
⎡⎤
∴ =+ ⎢⎥
+
⎣⎦
(7.36)
And
()
22
2cos
16cos
61cos
a
N
θ
γφ φ
φ
⎛⎞
∴ =− −⎜⎟
⎜⎟
+
⎝⎠
(7.37)
7.3.3 Example: Conical Shell filled with liquid.

Solution: Let consider a conical shell subjected to load P in the direction of the axis of cone as
shown in figure. Let the semi apex angle of the cone beα. Therefore at point where radius of
parallel circle is
0
r

204. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     196 

180
2
πφ α
⎛⎞
=−+
⎜⎟
⎝⎠

2
π
φ α∴ =− (7.38)
If the load P is applied in the direction of the axis of cone, then the stress distribution is
symmetrical.
We obtain value of N
φ
from equation (8.18)
0
0
2sin
2sin
2
RP
N
r
r
φ
ππφ
π α
− −
∴ ==
⎛⎞
−
⎜⎟
⎝⎠

0
2cos
P
N
r
φ
πα
−
∴ = (7.39)
Since the curvature of the meridian in the case of a cone is zero,
1
r
=∞ from equation (7.12)
2
Nzr
θ
∴ =−
0
sin
zr
N
θ
φ
−
∴ = ()
02
sinrrφ= (7.40)
For the above case of loading z = 0 0 N
θ
∴ = (7.41)

205. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     197 

Now, Let us consider a conical tank to be filled with a liquid of specific weight γ as shown in
figure. Measuring the distances y from the bottom of the tank and denoting by the total depth of
liquid in the tank, the pressure at any parallel circle mn is
Pz
=−
( )Pdyγ∴ =−
( )zdyγ∴ =− − (7.42)
Also for such tank
2
π
φ α=+ (7.43)
And
0
tanry
α= (7.44)
Where, α is the semi-cone angle of the tank. Substituting value of φ from equation (7.43) in the
equation (7.40) we obtained
0
sin
zr
N
θ
φ
−
∴ =
0
sin
2
zr
N
θ
π
α
−
∴ =
⎛⎞
+
⎜⎟
⎝⎠

0
cos
zr
N
θ
α
−
∴ =
Put equation (7.44)
( )tan
cos
dyy
N
θ
γ α
α−
∴ = (7.45)
at, 0 and 0yydN
θ
== =
And at / 2 is maximumyd N
θ
=
()
max /2
tan
22
cos
yd
dd
d
N
θ
γ α
α
=
⎛⎞
−−
⎜⎟
⎝⎠
∴ =

206. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     198 

()
2
max /2
tan
4cos
yd
d
N
θ
γα
α
=
∴ = (7.46)
For calculating N
φ
we need total load R
Therefore Total load R = Weight of liquid in the conical part mno
+ Weight of liquid in the cylindrical part mnst
()
()Volumeof cylidrical portion × density
Volume of conical portion × density
R
⎧⎫
⎪⎪
∴ =⎨⎬
+⎪⎪⎩⎭

()
22
00 1
r × r y×
3
Rdy
π γπγ
⎧⎫ ⎛⎞
∴ =− − +⎨⎬ ⎜⎟
⎝⎠⎩⎭

()
22 22 1
tan × tan y×
3
Ry dy yπ αγπαγ
⎧⎫ ⎛⎞
∴ =− − +⎨⎬ ⎜⎟
⎝⎠⎩⎭
()
0
tanryα=
()
22
tan
3
y
Rydyγπα
⎧⎫
∴ =− − +⎨⎬
⎩⎭
(8.40)
Substitute above equation in the equation to determine N
φ

()
22
01
tan
2cos 3
y
Nydy
r
φ
γπα
πα
⎡ ⎤⎧⎫
∴ =−+ ⎨⎬⎢ ⎥
⎩⎭⎣ ⎦

222
tan
3
2 tan cos
yd y
N
y
φ
γπα
παα
⎛⎞
−
⎜⎟
⎝⎠
∴ =
2
tan
3
2cos
yd y
N
φ
γ α
α
⎛⎞
−
⎜⎟
⎝⎠
∴ = (7.47)
0 at 0Ny
φ
==
But 0 atNyd
φ
≠ =
3
is maximum at
4
Nyd
φ
=

207. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     199 


()
max3/4
323
tan
434
2cos
yd
dd d
N
φ
γ α
α
=
⎛⎞
−×
⎜⎟
⎝⎠
∴ =
()
2
max3/4
3tan
16 cos
yd
d
N
φ
γα
α
=
∴ = (7.48)
7.3.4 Example: Shell in the form of an Ellipsoid of Revolution

Solution: such shells are used as tops/bottom of vertical tanks, or boiler shells. Let a and b be the
semi axes of ellipsoid any angle φ the principle radius if curvature are given by
()
22
1 3/2
22 2 2
sin cos
ab
r
abφφ
=
+
And
()
2
2 3/2
22 2 2
sin cos
a
r
abφφ
=
+
(7.49)
Or in terms of x and y coordinates
()
3/2
42 42
1 44
ay bx
r
ab
+
= And
( )
1/ 2
42 42
2 2
ay bx
r
b
+
= (7.50)

208. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     200 

If P is the uniform pressure acting on the shell, then the resultant force R corresponding to an
angle φ is (Uniform pressure P acting over the area
2
0
r
π)
Pz∴ =−
2
0
R zrPπ∴ =− =− (7.51)
From equation (7.18) we get
2
0
0
2sin
rP
N
r
φ
π
πφ
∴ =
()
22
2
022
2
sin
sin
2sin
rP
Nrr
r
φ
πφ
φ
πφ
∴ ==
2
2
rP
N
φ
∴ = (7.52)
And from equation (7.12)

2
2
1
2
rP
Nr z
r
θ
⎛⎞
∴ =− +⎜⎟
⎝⎠
2
2
2
1
2
rP
NrP
r
θ
∴ =− + ( )zp=−
2
2
1
1
2
r
NrP
r
θ
⎡ ⎤
∴ =−
⎢ ⎥
⎣ ⎦
(7.53)
At 0φ= from equation (A)
2
12
a
rr
b
==
2
2
Pa
NN
b
φθ
∴ == (7.54)
Similarly
At / 2φπ=
2
12
and
b
rra
a
= =

209. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     201 

2
Pa
N
φ
∴ = And
2
2
1
2
a
NPa
b
θ
⎡ ⎤
=−
⎢ ⎥
⎣ ⎦
(7.55)
N
φ
Is always positive and N
θ
may be positive or negative depending upon a and b
Therefore N
θ
negative in some part near the equator if
22
2ab〉. Sphere is a spherical case of
ellipsoid and has a = b from which we have at all points
2
Pa
NN
φθ
== for steam pressure P
7.3.5 Example: Shell in the form of Torus.

Solution: Rotating a circle @ axis of symmetry we get ‘torus’ or ‘toroid’. Let ‘b’ be the mean
radius of torus and ‘a’ be the radius of circular section. Consider equilibrium of a ring shaped
portion formed by rotating portion AB of the circle @ axis of symmetry. Forces N
φ
at edge B are
in horizontal plane and being same throughout, these being canceled mutually
(i.e.cos & cosNN
φφ
φ φ− ). To find forces N
φ
at edge A consider the external load.

Let the shell be subjected to internal pressure P throughout. zP∴ =−

210. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     202 

Therefore vertical component of external force over the Surface area ( )
22
0
rbπ− is given by

( )
22
0
zrbPπ
∴ =− − (7.56)
Vertical component of force N
φ
is
0
2sinrN
φ
π φ (7.57)
Therefore for equilibrium from equation (7.56) and (7.57) we get
( )
22
00
2sin 0rN r b P
φ
πφπ
∴ −−=
( )
22
00
2sinrN r b P
φ
πφπ
∴ =−
( )
()
22
0
0
0
2
rbP
N
rb
r
a
φ
−
∴ =
−
()
0
sin
rb
a
φ
−⎡⎤
=⎢⎥
⎣⎦

( )( )
()
00
00
2
ar b r bP
N
rr b
φ
−+
∴ =
−

( )
0
0
2
ar bP
N
r
φ
+
∴ = (7.58)
Now from equation (7.12)
2
1
N
Nr z
r
φ
θ
⎡ ⎤
∴ =− +
⎢ ⎥
⎣ ⎦

( )
0
2
01
2
ar bP
Nr P
rr
θ
+
⎡ ⎤
∴ =− −⎢ ⎥
⎣ ⎦

()
2
00
0
2
2
rP
Nrbr
r
θ
∴ =− + −⎡ ⎤
⎣ ⎦

[]
2
0
2
2sin
Pr
Nrb
r
θ
φ
∴ =− ()
02
sinrrφ=

211. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     203 

[ ]
0
2sin
rbP
N
θ
φ
−
∴ = (7.59)
Note: we can analyze toroids with elliptical section in a similar manner

Exercise:
Que. In terms of coordinates , andr
θφ, derive equations of equilibrium for a small element of
axisymmetric thin shell.
[P.U. Article 7.1]
Que. For a small element of an axisymmetric shell subjected to axisymmetric loading; derive
equations of equilibrium interms of stress resultants andNN
φ θ
at a point defined by the
coordinates
and
φθ using membrane theory. [P.U. Article 7.1]
Que. For a thin spherical dome subjected to distributed gravity load ‘q’ per unit area, prove that
there will be no tension in the shell id the meridianal angle φ is limited to 51 degrees.
[P.U. Article 7.3.1]
Que. A conical shell with apex at top and axis vertical carries a downward load ‘P’ at the apex.
The semi apex angle isα. Obtain expressions for meridianal and hoop stress resultants at
any point in the shell.
[P.U. Article 7.3.3]
Que. A thin toroidal shell of circular section of radius ‘a’ and mean radius of the torus ‘b’. The
shell is subjected to internal pressure of intensity ‘p’ per unit area. Write equilibrium
equations for this shell and hence obtain expressions for meridianal and hoop stress
resultants.
[P.U. Article 7.3.5]
Que. An ellipsoidal shell is used as a boiler shell. Its cross-section has semi major axis ‘a’ and
semi-minor axis ‘b’. For a steam pressure ‘p’ per unit area, derive expressions for stress
resultants in the shell. Comment on the nature of stresses.
[P.U. Article 7.3.4]
Que. For a conical shell storing liquid of density
γ for a depth‘d’, obtain expressions for
maximum values of andNN
φ θ
in the shell. [P.U. Article 7.3.3]

212. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     204 

Chapter 8
Shells of Double Curvature

8.1 Membrane Theory for Shells of Double curvature other than Surface of Revolution:

,,
xyxy
NNN are real stress resultants and , ,
xyz
FFF are real loads.
8.1.1 Pseudo Stress Resultants:

213. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     205 

,,
xyxy
NNN are pseudo resultants and X, Y, Z are pseudo loads.





It is expedient to introduce “pseudo’ stress resultants
,and
xy xy
NN N in place of the real stress
resultants , and
xy xy
NN N . Similarly also introduce pseudo loads X, Y and Z in the directions x, y
and z in place of the real loads,and
xy z
FF F . The pseudo stress resultant
x
N is such that it exerts
same force in the x direction as the real stress resultant does.
2
2
2
2
1cos
cos 1
1
similarly
1
xx xx
yy
xy xy
p
Ndy N dy N N
q
q
NN
p
NNφ
ψ +
∴ =⇒=
+
+
⇒=
+
⇒=
(8.1)
Also, the fictitious loads X, Y and Z are so defined that
()( )( )
( )
''' '
Real loads area fictitious loads areaABCD A B C D×= ×
Apply this relationship in all the three directions x, y and z thus we get
22
1
x
F p q dx dy X dx dy++ =
Therefore,
22 22 22
1Similarly 1and1
xyz
XFpq YFpq ZFpq=++ =++ =++ (8.2)
8.1.2 Equations of Equilibrium:
Referring to the element
''' '
ABCD, the equations of equilibrium in x and y direction may be
formulated as follows,
Equating all forces in the x direction to zero we get
()
n
22 2
22
, Eq of surface of shell
;&;;
zzxy
zz z z z
pq r s t
xy x xy y
=
∂∂ ∂ ∂ ∂
∴ == = = =
∂∂ ∂ ∂∂ ∂

214. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     206 

0
x
F∑=
0
yxx
xx yxyx
NN
N dy N dx dy N dx N dy dx X
xy
⎛⎞ ∂⎛⎞∂ ∴ −++ −++ += ⎜⎟⎜⎟ ⎜⎟
∂∂⎝⎠ ⎝⎠

0
yxx
NN
dx dy dy dx X
xy
∂∂
∴ ++=
∂∂

0
yxx
NN
X
xy
∂∂
∴ ++=
∂∂
(8.3)
Similarly equating all forces in the y direction to zero we get
0
xyy
NN
Y
yx
∂∂
∴ ++=
∂∂
(8.4)
For formulating the equations of equilibrium in the z direction, we have consider element
ABCD
Vertical component of the normal force acting on AD =
2
2
1
1
x
pz
Ndy
xq
+∂
∂+

= ()x
z
Ndy
x
∂
↑
∂

Vertical component of the normal force acting on BC = ()xx
zz
N dy N dx dy
xxx
∂∂∂ ⎛⎞
+ ↓
⎜⎟
∂∂∂ ⎝⎠

Therefore resultant of the vertical forces on the pair of sides AD and BC
xx x
zz z
NdyNdy Ndxdy
xxxx
∂∂∂∂ ⎛⎞
=− + +
⎜⎟
∂∂∂∂ ⎝⎠

x
z
Ndxdy
xx
∂∂⎛⎞
=
⎜⎟
∂∂⎝⎠
(8.5)
Similarly, Vertical component of the shear force acting on AD = ()xy
z
Ndy
y
∂
↑
∂

Vertical component of the shear force acting on BC = ()xy xy
zz
N dy N dx dy
yxy
⎛⎞∂∂∂
+ ↓
⎜⎟
∂∂∂ ⎝⎠

215. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     207 

Therefore resultant of the shear forces on the pair of sides AD and BC
xy xy xy
zz z
NdyNdy N dxdy
yyxy
⎛⎞∂∂∂∂
=− + +
⎜⎟
∂∂∂∂ ⎝⎠

xy
z
Ndxdy
xy
⎛⎞∂∂
=
⎜⎟
∂∂⎝⎠
(8.6)
Therefore summing up all the forces acting on AD and BC we get
xxy
zz
N dx dy N dx dy
xx x y
⎛⎞∂∂ ∂ ∂⎛⎞
=+
⎜⎟⎜⎟
∂∂ ∂ ∂⎝⎠ ⎝⎠

xxy
zz
NN
x xx y
⎛⎞∂∂ ∂ ∂⎛⎞
=+
⎜⎟⎜⎟
∂∂∂ ∂⎝⎠ ⎝⎠
(8.7)
Similarly summing up all the forces acting on AB and CD we get,
yyx
zz
NN
yyy x
⎛⎞∂∂∂ ∂ ⎛⎞
=+
⎜⎟ ⎜⎟
∂∂∂ ∂ ⎝⎠⎝⎠
(8.8)
The load z contributes a downward force of Z dx dy. summing up all the forces acting on the
element in the z direction we get,
From equation (8.7) and (8.8)
0
xxy y yx
zz zz
NN NNz
xxx y yyy x
⎛⎞⎛⎞∂∂ ∂ ∂ ∂∂ ∂ ∂⎛⎞ ⎛ ⎞
∴ ++++=
⎜⎟⎜⎟⎜⎟ ⎜ ⎟
∂∂∂ ∂ ∂∂∂ ∂⎝⎠ ⎝ ⎠ ⎝⎠⎝⎠

This may be expanded as,
22 22
22
0
yx xy yx
xxy y y
NN NNzz zz
NN NN p qz
xxyyxyxy yx
⎛⎞⎛⎞∂∂ ∂∂∂∂ ∂∂
∴ ++++++++= ⎜⎟⎜⎟
⎜⎟⎜⎟
∂∂∂∂∂∂∂∂ ∂∂
⎝⎠⎝⎠

222
22
20
yx xy yx
xxy y
NN NNzzz
NN N p qz
xxyyxyyx
⎛⎞⎛⎞∂∂ ∂∂∂∂∂
∴ +++++++= ⎜⎟⎜⎟
⎜⎟⎜⎟
∂∂∂∂∂∂∂∂
⎝⎠⎝⎠

Making use of equation () and () we can write

216. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     208 

222
22
20
xxy y
zzz
NN NXpYqz
xxyy
∂∂∂
∴ ++−−+=
∂∂∂∂

Therefore the equation of equilibrium in the z direction takes the following form
2
xxyy
rN s N tN X p Yq z∴ ++=+− (8.9)
8.2 Membrane Theory for rectangular hyperbolic paraboloid with straight line generators
and boundaries:

Let, consider equation for a rectangular hyperbolic paraboloid, we may find p, q, r, s and t of
the surface by differentiation. Thus

222
22
let
;
1
&0;;0
xy
z
c
zy zx
pq
xc yc
zzz
rs t
xxycy
=
∂∂
∴ == ==
∂∂
∂∂∂
=== = ==
∂∂∂∂

Put this value in the equation (), we get
2
xy
NXpYqz
C
∴ =+−
Denoting right hand side of the equation by L,

217. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     209 

()
2
xy
CL
NLXpYqz∴ ==+− (8.10)
Inserting this value in the equation (8.3) and integrate with respective x, we get
()
1
yxx
NN
Xdx f y
xy
⎡⎤∂∂
∴ =− + +⎢⎥
∂∂
⎣⎦∫∫
(8.11)
()
1
2
x
CdL
NXdxfy
dy
⎡⎤
∴ =− + +
⎢⎥
⎣⎦∫
(8.12)
Similarly making use of equation (8.4) to obtain
y
N
()
2
2
y
CdL
NYdyfx
dx
⎡⎤
∴ =− + +
⎢⎥
⎣⎦∫
(8.13)
Consider a rectangular hyperboloid under the action of dead weight ‘g’ of the shell. The
value of L corresponding to this value is
22 22 2
1
g
LZ g pq Cxy
C
∴ =− =− + + = − + +
Put in the equation (8.10) we obtained
22 2
2
xy
Cg
NCxy
C
∴ =− + +
22 2
2
xy
g
NCxy∴ =− + +
Therefore from equation (8.12)
()
22 2
1
2
x
N g
Cxy dxfy
xy
⎡⎤∂ ∂⎛⎞∴ =− − + + +
⎜⎟⎢⎥
∂∂ ⎝⎠⎣⎦∫∫

()
1
22 2
2
x
gy
Ndxfy
Cxy
∴ =+
++∫
(8.14)
()
2
22 2
2
y
gx
Ndyfx
Cxy
∴ =+
++∫
(8.15)

218. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     210 

Therefore from equation (8.14) and (8.15) we get
( )()
22 2
1
log
2
x
gy
NxCxyfy
∴ =++++ (8.16)
And ( ) ()
22 2
2
log
2
y
gx
NyCxyfx
∴ =++++ (8.17)
() ()
12
andfyfx are to be evaluated from the boundary conditions which depend upon the
manner in which the shell is supported.
8.3 The Umbrella Roof:
Consider the arrangement of the umbrella roof formed by four abutting hyperbolic
paraboloids resting on four trusses along their edges. Taking any one of the hyperbolic
paraboloids, say OABCD, it abuts against the adjacent hyperbolic paraboloids along the edges
OA and OB. Along the remaining two edges AC and BC, it is supported on trusses which are stiff
in their own planes but are incapable of resisting any loads applied normal to their planes.

Choosing the origin at O, we may formulate the boundary conditions as follows:
0
x
Natxa== (8.18)
0
y
Natyb== (8.19)
These two enables the arbitrary functions () ()
12
andfyfx appearing in the equation (8.16) and
(8.17) to be evaluated.

219. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     211 

Therefore using equation (8.18) in the equation (8.16)
( )()
222
1
0log
2
gy
aCay fy
∴ =++++
()( )
222
1
log
2
gy fyaCay∴ =− + + + (8.20)
Put in the equation (8.16) we get,
( ) ( )
22 2 22 2
log log
22
x
gy gy
NxCxyaCay
∴ =+++−+++
22 2
222
log
2
x
xCxygy
N
aCay
⎛⎞
+++
⎜⎟∴ =
⎜⎟
+++
⎝⎠ (8.21)
Similarly using equation (8.19) in the equation (8.17)
()
( )
222
2
log
2
gy fxbCxb∴ =− + + + (8.22)
Therefore,
22 2
222
log
2
y
yCxygx
N
bCxb
⎛⎞
+++
⎜⎟
∴ =
⎜⎟
+++
⎝⎠ (8.23)
Equation (), () and (), define the state of stress in the shell. We may note, in passing, that no
boundary conditions were applied along the edges OA and OB which are open boundaries. Along
these edges, one should expect both a normal stress as well as a shear.
Let us consider shallow shell,
For shallow shell
22
&
pq are neglected in comparison with unity
zg∴ −≈−
Hence from equation (8.10)
2
xy
gc
N∴ =−
Similarly from equation (8.21) and (8.23) 0& 0
xy
NN= =
Which we arrived at important conclusion:

220. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     212 

“A shallow hyperbolic paraboloid submitted to the action of dead weight develops a state of
pure shear unaccompanished by normal stresses”
Let us consider, Hyperbolic paraboloid may visualize as being made up of suspension
cables and arches placed at right angles to each other. At any point on the edge, the normal
components of the tension and thrust of the cable and arch, being equal, cancel each other; the
tangential components add up and give rise to shear. This interpretation also helps us in deciding
on the direction in which the shell is to be reinforced.

Obviously tension steel is to be arranged on the direction of cables, because cables
carried a load in pure tension and arches carries a load in pure compression
.

Exercise:
Que.
Explain how the state of pure shear is developed in s shallow hyperbolic paraboloid under
the action of its dead weight. How is direction of placement of tension steel in such shell
decided?
[Dr. B.A.M.U.] [Ans. Article 8.2]
Que. Derive expression for membrane shear for rectangular hyper shell with straight line
generators as boundaries subjected to dead weight ‘g’ according to membrane theory.
[Dr. B.A.M.U.] [Ans. Article 8.3]

221. Theory of Plates and Shells 

Prof. Atteshamuddin S. Sayyad     213 

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Timoshenko S. P. and Woinowsky-Krieger S., 1959, “Theory of Plates and Shells”,
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Chandrashekhara K., 2001, “Theory of Plates”, Universities Press (India) Limited,
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Bairagi N. K., 1986, “Plate Analysis”, Khanna Publishers., Delhi
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Ugural, A. C., 1981, “Stresses in Plates and Shells”, McGraw-Hill Book Company, Inc.,
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Bairagi N. K., 1986, “Shell Analysis”, Khanna Publication., Delhi
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Ramaswamy G. S., “Design and Construction of Concrete Shell Roofs”, CBS
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