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Solutions Manual for Theory and Analysis of Elastic Plates and Shells (2nd Edition) by J.N. Reddy — Complete Step-by-Step Solutions with Worked Examples, Derivations, and Explanations for All Major Topics in Plate and Shell Theory, Shear Deformation, and

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Escrito en
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The Solutions Manual for Theory and Analysis of Elastic Plates and Shells (2nd Edition) by J.N. Reddy is an authoritative guide that complements the textbook widely recognized as one of the most complete and rigorous treatments of plate and shell theory available today. This solutions resource is designed to support graduate-level coursework, self-study, and academic instruction in advanced solid mechanics and structural analysis. Reddy’s textbook introduces the theoretical foundations of plate and shell behavior, starting from elasticity principles and extending to practical analysis using classical and shear-deformation plate theories. The Solutions Manual follows the same chapter sequence, offering detailed, step-by-step derivations and numerical results for selected problems in each topic area. Topics covered include: Mathematical preliminaries of elasticity, stress-strain relations, and tensor notation; Classical thin-plate theory (Kirchhoff–Love formulation) for isotropic and orthotropic materials; First-order shear deformation theory (Mindlin/Reissner) for thick and laminated composite plates; Buckling, vibration, and stability analysis of rectangular, circular, and skew plates; Shell theory, including cylindrical and spherical geometries with various boundary conditions; Finite element formulations for plates and shells, variational energy methods, and Galerkin approximations. Every solution emphasizes clarity and precision, illustrating the assumptions behind simplified models, the transition between differential and matrix formulations, and the relationship between analytical and numerical methods. The manual bridges the gap between theoretical understanding and computational implementation, showing how to apply boundary conditions, evaluate deflections, compute stresses, and interpret results for engineering design. This companion is particularly useful for: Graduate students studying advanced structural mechanics, elasticity, or computational methods; Researchers developing finite element codes for plate and shell analysis; Instructors preparing lectures, assignments, and problem solutions for advanced mechanics courses. Using this manual, readers gain not only the correct numerical answers but also the logic and methodology required to derive them — a critical skill for anyone conducting design or simulation of thin-walled structures in aerospace, mechanical, or civil engineering applications. Key Benefits: Step-by-step analytical derivations. Verified finite element formulations for plates and shells. Illustrative MATLAB-style computational examples. Integration of classical and modern deformation theories. Perfect for academic coursework and professional reference. The Solutions Manual for Theory and Analysis of Elastic Plates and Shells (2nd Edition) provides a complete and practical foundation for mastering complex plate and shell behavior, reinforcing theoretical knowledge through worked examples and detailed mathematical exposition.

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Subido en
31 de octubre de 2025
Número de páginas
174
Escrito en
2025/2026
Tipo
Examen
Contiene
Preguntas y respuestas

Temas

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@SOLUTIONSSTUDY



@SOLUTIONSSTUDY

All 12 Chapters Covered




SOLUTIONS

, 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




@
@SSeeisismmicicisisoolalatitoionn

, @SOLUTIONSSTUDY




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) δij δij = 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

, 2 Theory and Analysis 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 Fij εijk = 0, then Fij = Fji . We have 0 =
Fij εijk
1
= (Fij εijk + Fij εijk)
2
1
= (Fijεijk − Fijεjik) (interchanged i and j)
2
1
= (Fij εijk − 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 êj (1)
Then the square of the magnitude of the position vector is
r2 = r · r = (x i êi ) · (x j êj ) = xixjδij
= xixi = xkxk (2)
Its derivative of r with respect to xi can be obtained from
∂r2 ∂
∂xi = (xkxk)
∂x
∂xk
i
∂xk
= x +x
∂xi k k
∂xi
∂xk
=2 xk = 2δikxk = 2xi
∂xi
Hence
∂r xi
= (3)
∂xi r
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