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

All12ChaptersCovered
t t t




SOLUTIONS

, Contents


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


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




2. Energy Principles and Variational Methods .............................................. 19
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3. Classical Theory of Plates.................................................................................51
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4. Analysis of Plate Strips .................................................................................... 59
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5. Analysis of Circular Plates .............................................................................. 75
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6. Bending of Simply Supported Rectangular Plates ................................. 91
t t t t t




7. Bending of Rectangular Plates with Various
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Boundary Conditions.......................................................................................... 99
t




8. General Buckling of Rectangular Plates.................................................... 115
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9. Dynamic Analysis of Rectangular Plates ................................................. 123
t t t t




10. Shear Deformation Plate Theories ............................................................. 129
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11. Theory and Analysis of Shells...................................................................... 139
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12. Finite Element Analysis of Plates ............................................................... 157
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, 1
Vectors, Tensors, and t t




t Equations of Elasticity t t




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




dummy indices):
t t




(a) Fijδjk = Fik t t




(b) δijδij =δii =3 t t t t t




(c) εijkεijk = 6 t t t




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




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




Fijδjk =Fi1δ1k +Fi2δ2k +Fi3δ3k
t t
t
t
t
t
t




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




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




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




δijδij = δi1δi1 +δi2δi2 +δi3δi3
t t
t
t
t
t
t




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




=3 t t




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




Alternatively,usingFij=δijinProblem1.1a,wehaveδijδjk=δik,whereiandk are free t t t t t t t t t t t t t t t t t t




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




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




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




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




1.1(d) We have t t




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




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




Since Fji = Fij, we have
t t t t t




0 = (Fij + Fji)εijk t t t t t




=2Fijεijk t
t




Theconverse alsoholds, i.e., ifFijεijk=0, then Fij=Fji. We have 0 = Fij
t t t t t t t t t t t t t t t
t t




εijk
t



1
= (Fijεijk +Fijεijk)
2
t t
t t t

t



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




1
t




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




1
t




= (Fij −Fji)εijk
2
t t t t


t




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




♠ New Problem 1.1: Showthat
t t t t t




∂r xi
= t



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




r= x j ê j (1) t t




Thenthe square of the magnitude of the position vector is
t t t t t t t t t t




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




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




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

t t t t




∂xi k k∂x
i t t t




∂xk
=2 xk = 2δikxk =2xi t t
t t t t


∂xi
Hence
∂r xi
= t (3)
∂xi r




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