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

All12 Chapters Covered
d d d




SOLUTIONS

, Contents


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


1. Vectors, Tensors, and Equations of Elasticity............................................... 1
d d d d d




2. Energy Principles and Variational Methods ............................................. 19
d d d d




3. Classical Theory of Plates ................................................................................51
d d d




4. Analysis of Plate Strips ................................................................................... 59
d d d




5. Analysis of Circular Plates ............................................................................. 75
d d d




6. Bending of Simply Supported Rectangular Plates ................................ 91
d d d d d




7. Bending of Rectangular Plates with Various
d d d d d




Boundary Conditions ......................................................................................... 99
d




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




9. Dynamic Analysis of Rectangular Plates ................................................ 123
d d d d




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




11. Theory and Analysis of Shells ..................................................................... 139
d d d d




12. Finite Element Analysis of Plates .............................................................. 157
d d d d




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, 1
Vectors, Tensors, and d d




d Equations of Elasticity d d




1.1 Prove the following properties of δij and εijk (assume i,j = 1,2,3 when they are
d d d d d d d d d d d d d d d d d




dummy indices):
d d




(a) Fijδjk = Fik d d




(b) δijδij = δii = 3 d d d d d




(c) εijkεijk = 6 d d d




(d) εijkFij = 0 whenever Fij = Fji (symmetric) d d d d d d d




Solution:
1.1(a) Expanding the expression
d d d




Fijδjk =Fi1δ1k + Fi2δ2k +Fi3δ3k
d d
d
d
d
d
d




Of the three terms on the right hand side, only one is nonzero. It is equal to Fi1 if
d d d d d d d d d d d d d d d d d d




k = 1, Fi2 if k = 2, or Fi3 if k = 3. Thus, it is simply equal to Fik.
d d d d d d d d d d d d d d d d d d d d




1.1(b) By actual expansion, we have
d d d d d




δijδij = δi1δi1 + δi2δi2 + δi3δi3
d d
d
d
d
d
d




= (δ11δ11 + 0 + 0) + (0 + δ22δ22 + 0) + (0 + 0 + δ33δ33)
d d d d d d d d d d d d d d d




=3 d d




and
δii = δ11 + δ22 + δ33 = 1+ 1+ 1 = 3
d d d d d d d d d d d d d d d




Alternatively, using Fij = δij in Problem 1.1a, we have δijδjk = δik, where i and k are
d d d d d d d d d d d d d d d d d




free indices that can any value. In particular, for i = k, we have the required result.
d d d d d d d d d d d d d d d d d




1.1(c) Using the ε-δ identity and the result of Problem 1.1(b), we obtain
d d d d d d d d d d d d




εijkεijk = δiiδjj − δijδij = 9 − 3 = 6 d
d d d d d d d d d d d




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, 2 Theory and Analysis of Elastic Plates and Shells d d d d d d d




1.1(d) We have d d




Fijεijk = −Fijεjik (interchanged i and j)
d d d d d d




=−Fjiεijk (renamed i as j and j as i) d d d d d d d d d




Since Fji = Fij, we have
d d d d d




0 = (Fij + Fji)εijk
d d d d d




= 2Fijεijk d
d




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




Fijεijk d
d



1
= (Fijεijk +Fijεijk)
2
d d
d d d

d



1
= (Fijεijk − Fijεjik) (interchanged i and j)
2
d d d d d d d d




1
d




= (Fijεijk − Fjiεijk) (renamed i as j and j as i)
2
d d d d d d d d d d d




1
d




= (Fij − Fji)εijk
2
d d d d


d




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




♠ New Problem 1.1: Show that
d d d d d




∂r xi
= d



∂xi r
Solution: Write the position vector in cartesian component form using the index
d d d d d d d d d d d




notation
d




r = x j ê j (1) d d




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




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




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




∂r2 = ∂
(xkxk)
∂xi ∂x
∂xik ∂xk
= x +x d
d

d d d d




∂xi k k ∂x
i d d d




∂xk
=2 xk = 2δikxk = 2xi d d
d d d d


∂xi
Hence
∂r xi
= d (3)
∂xi r



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