****INSTANT DOWNLOAD****PDF***Solutions Manual for Theory and Analysis of Elastic Plates and Shells 2nd Edition

All12 Chapters Covered
b b b




SOLUTIONS

, Contents


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


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




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




3. Classical Theory of Plates ................................................................................51
b b b




4. Analysis of Plate Strips ................................................................................... 59
b b b




5. Analysis of Circular Plates ............................................................................. 75
b b b




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




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




Boundary Conditions ......................................................................................... 99
b




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




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




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




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




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




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, 1
Vectors, Tensors, and b b




b Equations of Elasticity b b




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




dummy indices):
b b




(a) Fijδjk = Fik b b




(b) δijδij = δii = 3 b b b b b




(c) εijkεijk = 6 b b b




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




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




Fijδjk =Fi1δ1k + Fi2δ2k +Fi3δ3k
b b
b
b
b
b
b




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




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




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




δijδij = δi1δi1 + δi2δi2 + δi3δi3
b b
b
b
b
b
b




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




=3 b b




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




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




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




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




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




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




1.1(d) We have b b




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




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




Since Fji = Fij, we have
b b b b b




0 = (Fij + Fji)εijk
b b b b b




= 2Fijεijk b
b




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




Fijεijk b
b



1
= (Fijεijk +Fijεijk)
2
b b
b b b

b



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




1
b




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




1
b




= (Fij − Fji)εijk
2
b b b b


b




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




♠ New Problem 1.1: Show that
b b b b b




∂r xi
= b



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




notation
b




r = x j ê j (1) b b




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




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




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




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

b b b b




∂xi k k ∂x
i b b b




∂xk
=2 xk = 2δikxk = 2xi b b
b b b b


∂xi
Hence
∂r xi
= b (3)
∂xi r



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