@SOLUTIONSSTUDY
@SOLUTIONSSTUDY
All 12 Chapters Covered
SOLUTIONS
, Contents
Preface.......................................................................................................................... iv
1. Vectors, Tensors, and Equations of Elasticity .................................................. 1
2. Energy Principles and Variational Methods ................................................... 19
3. Classical Theory of Plates ................................................................................ 51
4. Analysis of Plate Strips .................................................................................... 59
5. Analysis of Circular Plates ............................................................................... 75
6. Bending of Simply Supported Rectangular Plates ........................................ 91
7. Bending of Rectangular Plates with Various
Boundary Conditions .......................................................................................... 99
8. General Buckling of Rectangular Plates ....................................................... 115
9. Dynamic Analysis of Rectangular Plates ..................................................... 123
10. Shear Deformation Plate Theories ................................................................. 129
11. Theory and Analysis of Shells ....................................................................... 139
12. Finite Element Analysis of Plates ................................................................. 157
@
@SSeeisismmicicisisoolalatitoionn
, @SOLUTIONSSTUDY
1
Vectors, Tensors, and
Equations of Elasticity
1.1 Prove the following properties of δij and εijk (assume i, j = 1, 2, 3 when they are
dummy indices):
(a) Fijδjk = Fik
(b) δij δij = δii = 3
(c) εijkεijk = 6
(d) εijkFij = 0 whenever Fij = Fji (symmetric)
Solution:
1.1(a) Expanding the expression
Fij δjk = Fi1δ1k + Fi2δ2k + Fi3δ3k
Of the three terms on the right hand side, only one is nonzero. It is equal to Fi1 if
k = 1, Fi2 if k = 2, or Fi3 if k = 3. Thus, it is simply equal to Fik.
1.1(b) By actual expansion, we have
δij δij = δi1δi1 + δi2δi2 + δi3δi3
= (δ11δ11 + 0 + 0) + (0 + δ22δ22 + 0) + (0 + 0 + δ33δ33)
=3
and
δii = δ11 + δ22 + δ33 = 1 + 1 + 1 = 3
Alternatively, using Fij = δij in Problem 1.1a, we have δijδjk = δik, where i and k are
free indices that can any value. In particular, for i = k, we have the required result.
1.1(c) Using the ε-δ identity and the result of Problem 1.1(b), we obtain
εijkεijk = δiiδjj − δij δij = 9 − 3 = 6
, 2 Theory and Analysis of Elastic Plates and Shells
1.1(d) We have
Fijεijk = −Fijεjik (interchanged i and j)
= −Fjiεijk (renamed i as j and j as i)
Since Fji = Fij , we have
0 = (Fij + Fji) εijk
= 2Fij εijk
The converse also holds, i.e., if Fij εijk = 0, then Fij = Fji . We have 0 =
Fij εijk
1
= (Fij εijk + Fij εijk)
2
1
= (Fijεijk − Fijεjik) (interchanged i and j)
2
1
= (Fij εijk − Fjiεijk) (renamed i as j and j as i)
2
1
= (Fij − Fji) εijk
2
from which it follows that Fji = Fij.
♠ New Problem 1.1: Show that
∂r xi
=
∂xi r
Solution: Write the position vector in cartesian component form using the index notation
r = x j êj (1)
Then the square of the magnitude of the position vector is
r2 = r · r = (x i êi ) · (x j êj ) = xixjδij
= xixi = xkxk (2)
Its derivative of r with respect to xi can be obtained from
∂r2 ∂
∂xi = (xkxk)
∂x
∂xk
i
∂xk
= x +x
∂xi k k
∂xi
∂xk
=2 xk = 2δikxk = 2xi
∂xi
Hence
∂r xi
= (3)
∂xi r
@SOLUTIONSSTUDY
All 12 Chapters Covered
SOLUTIONS
, Contents
Preface.......................................................................................................................... iv
1. Vectors, Tensors, and Equations of Elasticity .................................................. 1
2. Energy Principles and Variational Methods ................................................... 19
3. Classical Theory of Plates ................................................................................ 51
4. Analysis of Plate Strips .................................................................................... 59
5. Analysis of Circular Plates ............................................................................... 75
6. Bending of Simply Supported Rectangular Plates ........................................ 91
7. Bending of Rectangular Plates with Various
Boundary Conditions .......................................................................................... 99
8. General Buckling of Rectangular Plates ....................................................... 115
9. Dynamic Analysis of Rectangular Plates ..................................................... 123
10. Shear Deformation Plate Theories ................................................................. 129
11. Theory and Analysis of Shells ....................................................................... 139
12. Finite Element Analysis of Plates ................................................................. 157
@
@SSeeisismmicicisisoolalatitoionn
, @SOLUTIONSSTUDY
1
Vectors, Tensors, and
Equations of Elasticity
1.1 Prove the following properties of δij and εijk (assume i, j = 1, 2, 3 when they are
dummy indices):
(a) Fijδjk = Fik
(b) δij δij = δii = 3
(c) εijkεijk = 6
(d) εijkFij = 0 whenever Fij = Fji (symmetric)
Solution:
1.1(a) Expanding the expression
Fij δjk = Fi1δ1k + Fi2δ2k + Fi3δ3k
Of the three terms on the right hand side, only one is nonzero. It is equal to Fi1 if
k = 1, Fi2 if k = 2, or Fi3 if k = 3. Thus, it is simply equal to Fik.
1.1(b) By actual expansion, we have
δij δij = δi1δi1 + δi2δi2 + δi3δi3
= (δ11δ11 + 0 + 0) + (0 + δ22δ22 + 0) + (0 + 0 + δ33δ33)
=3
and
δii = δ11 + δ22 + δ33 = 1 + 1 + 1 = 3
Alternatively, using Fij = δij in Problem 1.1a, we have δijδjk = δik, where i and k are
free indices that can any value. In particular, for i = k, we have the required result.
1.1(c) Using the ε-δ identity and the result of Problem 1.1(b), we obtain
εijkεijk = δiiδjj − δij δij = 9 − 3 = 6
, 2 Theory and Analysis of Elastic Plates and Shells
1.1(d) We have
Fijεijk = −Fijεjik (interchanged i and j)
= −Fjiεijk (renamed i as j and j as i)
Since Fji = Fij , we have
0 = (Fij + Fji) εijk
= 2Fij εijk
The converse also holds, i.e., if Fij εijk = 0, then Fij = Fji . We have 0 =
Fij εijk
1
= (Fij εijk + Fij εijk)
2
1
= (Fijεijk − Fijεjik) (interchanged i and j)
2
1
= (Fij εijk − Fjiεijk) (renamed i as j and j as i)
2
1
= (Fij − Fji) εijk
2
from which it follows that Fji = Fij.
♠ New Problem 1.1: Show that
∂r xi
=
∂xi r
Solution: Write the position vector in cartesian component form using the index notation
r = x j êj (1)
Then the square of the magnitude of the position vector is
r2 = r · r = (x i êi ) · (x j êj ) = xixjδij
= xixi = xkxk (2)
Its derivative of r with respect to xi can be obtained from
∂r2 ∂
∂xi = (xkxk)
∂x
∂xk
i
∂xk
= x +x
∂xi k k
∂xi
∂xk
=2 xk = 2δikxk = 2xi
∂xi
Hence
∂r xi
= (3)
∂xi r
