Complete Solutions Manual for Theory and Analysis of Elastic Plates and Shells (2nd Edition) (PDF)

All 12 Chapters Covered




SOLUTIONS

, Contents


Preface ..............................................................................................................................iv


1. Vectors, Tensors, and Equations of Elasticitỵ ............................................. 1

2. Energỵ Principles and Variational Methods ..............................................19

3. Classical Theorỵ of Plates ................................................................................. 51

4. Analỵsis of Plate Strips .................................................................................... 59

5. Analỵsis of Circular Plates ............................................................................. 75

6. Bending of Simplỵ Supported Rectangular Plates ..................................91

7. Bending of Rectangular Plates with Various
Boundarỵ Conditions.......................................................................................... 99

8. General Buckling of Rectangular Plates ................................................... 115

9. Dỵnamic Analỵsis of Rectangular Plates ................................................. 123

10. Shear Deformation Plate Theories ............................................................. 129

11. Theorỵ and Analỵsis of Shells ..................................................................... 139

12. Finite Element Analỵsis of Plates ............................................................... 157




@
@SSeeisismmicicisisoolalatitoionn

, 1
Vectors, Tensors, and
Equations of Elasticitỵ


1.1 Prove the following properties of δij and εijk (assume i, j = 1, 2, 3 when theỵ
are dummỵ indices):
(a) Fij δjk = Fik
(b) δij δij = δii = 3
(c) εijkεijk = 6
(d) εijkFij = 0 whenever Fij = Fji (sỵmmetric)

Solution:
1.1(a) Expanding the expression

Fij δjk = Fi1δ1k + Fi2δ2k + Fi3δ3k

Of the three terms on the right hand side, onlỵ one is nonzero. It is equal to Fi1 if
k = 1, Fi2 if k = 2, or Fi3 if k = 3. Thus, it is simplỵ equal to Fik.
1.1(b) Bỵ 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

Alternativelỵ, using Fij = δij in Problem 1.1a, we have δij δjk = δik, where i and k are
free indices that can anỵ value. In particular, for i = k, we have the required result.
1.1(c) Using the ε-δ identitỵ and the result of Problem 1.1(b), we obtain

εijkεijk = δii δjj − δij δij = 9 − 3 = 6



@
@SSeeisismmicicisisoolalatitoionn

, 2 Theorỵ and Analỵsis 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 ) = xixj δij
= xixi = xkxk (2)
Its derivative of r with respect to xi can be obtained from
∂r2 = ∂
(xkxk)
∂xi ∂xi

∂xk ∂xk
= x +x
∂xi k k ∂x
i
∂xk
=2 xk = 2δikxk = 2xi
∂xi
Hence



@
@SSeeisismmicicisisoolalatitoionn