STUDY NOTES & REFERENCE GUIDE
All 12 Chapters Covered
SOLUTIONS
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, STUDY NOTES & REFERENCE GUIDE
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
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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 δ jk = Fi1δ 1k + F i2δ 2k + 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 + δ 33 = 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
εijkε ijk = δiiδ jj − δ ij δ ij = 9 − 3 = 6
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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 + F ji) εijk
= 2Fij εijk
The converse also holds, i.e., if Fijε ijk = 0, then Fij = Fji. We have 0
= Fij εijk
1
= (Fij εijk + F ij εijk)
2
1
= (Fijεijk − Fijεjik) (interchanged i and j)
2
1
= (Fijε ijk − Fjiε ijk) (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 ) = xix jδ ij
= xix i = 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
∂xi
∂xk
=2 xk = 2δ ikx k = 2xi
∂xi
Hence
∂r xi
= (3)
∂xi r
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Acadexas Stuvia | Page 4 of 184
All 12 Chapters Covered
SOLUTIONS
Acadexas Stuvia | Page 1 of 184
, STUDY NOTES & REFERENCE GUIDE
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
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@SSeeisismmicicisisoolalatitoionn
Acadexas Stuvia | Page 2 of 184
, STUDY NOTES & REFERENCE GUIDE
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 δ jk = Fi1δ 1k + F i2δ 2k + 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 + δ 33 = 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
εijkε ijk = δiiδ jj − δ ij δ ij = 9 − 3 = 6
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, STUDY NOTES & REFERENCE GUIDE
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 + F ji) εijk
= 2Fij εijk
The converse also holds, i.e., if Fijε ijk = 0, then Fij = Fji. We have 0
= Fij εijk
1
= (Fij εijk + F ij εijk)
2
1
= (Fijεijk − Fijεjik) (interchanged i and j)
2
1
= (Fijε ijk − Fjiε ijk) (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 ) = xix jδ ij
= xix i = 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
∂xi
∂xk
=2 xk = 2δ ikx k = 2xi
∂xi
Hence
∂r xi
= (3)
∂xi r
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Acadexas Stuvia | Page 4 of 184
