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
, 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) δijδij = 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 Fijεijk = 0, then Fij = Fji. We have
0 = Fij εijk
1
= (Fij εijk + Fij εijk)
2
1
= (Fijεijk − Fijεjik) (interchanged i and j)
2
1
= (Fijεijk − 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 ê j (1)
Then the square of the magnitude of the position vector is
r2 = r · r = (x i ê i ) · (xj ê j ) = xixjδij
= xixi = xkxk (2)
Its derivative of r with respect to xi can be obtained from
∂r2 = ∂
(xkxk)
∂xi ∂x
∂xik ∂xk
= x +x
∂xi k k ∂x
i
∂xk
=2 xk = 2δikxk = 2xi
∂xi
Hence
∂r xi
= (3)
∂xi r
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
, 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) δijδij = 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 Fijεijk = 0, then Fij = Fji. We have
0 = Fij εijk
1
= (Fij εijk + Fij εijk)
2
1
= (Fijεijk − Fijεjik) (interchanged i and j)
2
1
= (Fijεijk − 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 ê j (1)
Then the square of the magnitude of the position vector is
r2 = r · r = (x i ê i ) · (xj ê j ) = xixjδij
= xixi = xkxk (2)
Its derivative of r with respect to xi can be obtained from
∂r2 = ∂
(xkxk)
∂xi ∂x
∂xik ∂xk
= x +x
∂xi k k ∂x
i
∂xk
=2 xk = 2δikxk = 2xi
∂xi
Hence
∂r xi
= (3)
∂xi r
