Solutions Manual for Theory and Analysis of Elastic Plates and Shells (2nd Edition) by Reddy – Complete Worked Solutions

Solutions Manual for Theory
and Analysis of Elastic Plates
and Shells 2nd Edition by
Reddy




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, 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 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


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, 2 Theory and Analysis of Elastic Plates and Shells


1.1(d) We have n k




Fijεijk = −Fijεjik (interchanged i and j)
nk nk n k nk nk nk




= −Fjiεijk (renamed i as j and j as i) nk n k nk nk nk nk nk nk nk




Since Fji = Fij, we have
nk nk nk nk nk




0 = (Fij + Fji) εijk
nk nk nk nk nk




= 2Fij εijk nk
nk




The converse also holds, i.e., if Fijεijk = 0, then Fij = Fji. We have
nk nk nk nk nk nk nk nk nk nk nk nk nk nk




0 = Fij εijk nk
nk nk
nk


1
= (F ε + Fij εijk)
2 ij ijk
nk nk
nk nk nk
nk

1
= (Fijεijk − Fijεjik) (interchanged i and j)
2
nk nk nk n k nk nk nk nk



1
nk




= (Fijεijk − Fjiεijk) (renamed i as j and j as i)
2
nk nk nk n k nk nk nk nk nk nk nk



1
nk




= (Fij − Fji) εijk
2
nk nk nk n
k

nk


from which it follows that Fji = Fij.
nk nk nk nk nk nk nk




♠ New Problem 1.1: Show that
nk nk nk nk nk




∂r xi
= nk

∂xi r
Solution: Write the position vector in cartesian component form using the index
n k nk nk nk nk nk nk nk nk nk nk



notation
nk



r = x j ê j (1) nk nk




Then the square of the magnitude of the position vector is
nk nk nk nk nk nk nk nk nk nk




r2 = r · r = (x i ê i ) · (xj ê j ) = xixjδij
nk nk nk nk nk nk nk nk nk nk




= xixi = xkxk nk nk nk (2)
Its derivative of r with respect to xi can be obtained from
nk nk nk nk nk nk nk nk nk nk nk




∂r2 = ∂
(xkxk)
∂xi ∂xi
∂x k ∂xk
= x +x n k
nk
nk nk nk n k



∂xi k k ∂xi nk nk nk


∂xk
=2 xk = 2δikxk = 2xi
nk nk
nk nk nk nk

∂xi
Hence
∂r =
xi (3)
∂xi nk

r



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