Solutions Manual for Theory and Analysis of Elastic Plates and Shells (2nd Edition) by J.N. Reddy

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




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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) Fij δ jk = Fik
(b) δij δij = δii = 3
(c) εijkε ijk = 6
(d) εijkFij = 0 whenever Fij = Fji (symmetric)

Solution:
1.1(a) Expanding the expression

Fijδjk = 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

δijδij = 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 δij δjk = 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
εijkε ijk = δiiδjj − δij δij = 9 − 3 = 6




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,2 Theory and Analysis of Elastic Plates and Shells


1.1(d) We have

Fijεijk = −Fijεjik (interchanged i and j)
= −Fji εijk (renamed i as j and j as i)
Since Fji = Fij, we have

0 = (Fij + Fji) εijk
= 2Fijεijk

The converse also holds, i.e., if Fijεijk = 0, then Fij = Fji . We have
0 = Fij εijk 1
= (Fij εijk + Fij ε ijk)
21
= (Fij εijk − Fij εjik) (interchanged i and j)
21
= (Fijεijk − Fjiε ijk) (renamed i as j and j as i)
21
= (Fij − F ji) 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 = xjêj (1)
Then the square of the magnitude of the position vector is
r2 = r · r = (xiêi) · (xjêj ) = xixj δij
= xixi = xkxk (2)
Its derivative of r with respect to xi can be obtained from
∂r2 = ∂
∂xi ∂xi (x kxk )
∂x ∂xk
= ∂xk x + x ∂x
k k i
i
∂xk xk = 2δikxk = 2xi
=2
∂xi
Hence
∂r xi (3)
∂xi = r




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,1. Vectors, Tensors, and Equations of Elasticity 3

1.2 Let r denote a position vector. Show that:
(a) grad (rn ) = nrn−2r
2 n ) = n(n + 1)r n−2
(b) (r
∇
(c) div (r) = 3
(d) curl (rf(r)) = 0, where f(r) is an arbitrary continuous function of r with
continuous first derivatives

Solution:

1.2(a) We have

∇(rn ) =êi ∂ (rn ) = nrn−1 êi ∂r = nrn−2 xiêi = nr n−2 r
∂x ∂x
i i

where the result from Eq. (3) of Problem 1.1 is used in arriving at the last step.
1.2(b) From the result of the above exercise, we have
∇2 (rn ) = (∇
Ã
· ∇) (rn! ) = ∇ · [∇(rn )]
∂ ³ ´ ∂ ³ n −2 ´
n— 2
= êj j · nr x i êi = n(êj · êi ) r xi
∂x xj ∂x j
∙ ¸
= nδij (n − 2)rn−3 xi + rn−2δij
r
h i
= n (n − 2)rn−2 + 3rn−2 = n(n + 1)rn−2


1.2(c) Using Eq. (3) of Problem 1.1(b), we obtain

à !
∂xi
∇· r = êj ∂ = δij δij = 3
∂xj · (xiêi) = (êj · êi ) j
∂x

1.2(d) We obtain

culr(f r) = ∇ × [fr]
à !
∂ ∂ [f(r)xi]
= êj j × (f(r)xi êi ) = (êj × êi)
∂x ∂x j
" #
0
= εjik êk f (r) ∂r xi + f(r)δij
xj ¸
∙ 0
f (r)
= εjik êk xjxi + f(r)δij
r
The twoterms in the square brackets, xixj and δij are symmetric, hence, byProblem
1.1(d) the expression in the last line is zero.




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,4 Theory and Analysis of Elastic Plates and Shells


♠ New Problem 1.2: Let [A] and [B] be m × n and n × p matrices, respectively. Show
that
([A][B])T = [B]T[A]T (1)

Solution: By definition of the product of two matrices, we have [A][B] = [C] with
n
X
cij = aikbkj
k=1

Then the transpose of [C] has the coefficients
Xn X n
cji = a jkb ki = bki ajk
k=1 k=1
n
X
= (bik)T (akj)T
k=1

which implies the result in Eq. (1).

1.3 If [B] is a symmetric n × n matrix and [C] is any n × n matrix, show that
[C]T[B][C] is symmetric.
Solution: Let [A] = [B][C]. Using Eq. (1) of New Problem 1.2, we have
³ ´T
[C]T[A] = [A] T[C] = [C]T[B][C]
where we have also used the identity
³ ´T = [C]
[C] T


♠ New Problem 1.3: Show that the dot and cross can be interchanged without
changing the value in the scalar triple product

A·B×C = A×B·C (1)


Solution: We have

A · B × C =Aiêi · Bj Ck εjkmêm = Ai Bj Ck εjkm δim
=AiBj Ckε jki = AiBj Ckεijk = A × B · C
Since i, j, and k can be permuted in a cyclic order, it also follows that

AiBj Ckε ijk = C · A × B = B · C × A
and A · B × C = A × B · C = C · A × B = C × A · B = B · C × A = B × C · A.




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,1. Vectors, Tensors, and Equations of Elasticity 5

1.4 Establish the ε-δ identity of Eq. (1.2.15).
Solution: The ε-δ identity follows directly from the vector identity

(A × B) · (C × D) = (A · C)(B · D) − (A · D)(B · C) (1)
by letting
A = êi, B = êj, C = êm , A = ên
We obtain

(êi × êj ) · (êm × ên ) = (êi · êm)(êj · ên ) − (êi · en )(êj · êm )
εijk êk · εmnp êp = δim δjn − δin δj m


or εijkεmnk = δimδjn − δinδjm
which was to be proved. Note that εijk = εkij = εjki.

1.5 Prove that the determinant of a 3 × 3 matrix [C] can be expressed in the form
|C| = εijk c1i c2j c3k (a)

and, thus, prove
1
|C| = εijk εrst cir cjs ckt (b)
6
where cij is the element occupying the ith row and the jth column of [C].

Solution: First we note the definition of the cross product of two vectors
¯ ê ê3 ¯
¯ 1
ê2
¯ ¯¯
B × C = B1¯ B2 B3 ¯ (3)
¯ ¯
C1 C2 C3
and the “scalar triple product” of vectors
¯A A A ¯
¯ ¯ 1 2 3¯¯
A · (B × C) = B1¯ B2 B3 ¯ (4)
¯ ¯
C1 C2 C3
Now let
A = c1iêi ≡ C1 , B = c2jêj ≡ C2 , C = c3k êk ≡ C3
in Eq. (3). We obtain

C1 · (C2 × C3 ) = c1iêi · (c2jêj × c3k êk )
¯c c12 c13 ¯
11
¯ ¯¯ c21 c22 c23 ¯¯¯≡ |C|
= ¯ ¯
c31 c32 c33




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,6 Theory and Analysis of Elastic Plates and Shells


or
|C| = c1iêi · (c2jêj × c3k êk )
= c 1ic 2jc 3kεijk
which is the same as Eq. (1). Next consider the product Cr · (Cs × Ct):
Cr · (Cs × Ct) = cricsj c tkεijk Multiplying
both sides with εrst and expanding, we arrive at
cric sj c tkεrstεijk = εrst[Cr · (Cs × Ct)]
= ε1st[C1 · (Cs × Ct)] + ε2st[C2 · (Cs × Ct)]
+ ε3st[C3 · (Cs × Ct)]
= C1 · (C2 × C3) − C1 · (C3 × C2)
+ C2 · (C3 × C1) − C2 · (C1 × C3)
+ C3 · (C1 × C2) − C3 · (C2 × C1)
= 6[C1 · (C2 × C3)] = 6|C|
where we have used the identity in Eq. (1) of New Problem 1.3.

1.6 Using Cramer’s rule determine the solution to the following equations:
(a)
2x1 − x2 = 1
−x1 + 2x2 − x3 = 2
−x2 + 2x3 = 2

(b) ⎡ ⎤⎧ ⎫
12 0 3h x ⎧ ⎫
2b 2 ⎨ f h ⎨ 12 ⎬
1⎬
⎣ 0
4h h2 ⎦ = 0
x2 0
h33h h2 2h2 ⎩ ⎭ 12 ⎩ ⎭
x3 h
where b, f0, and h are constants
Solution:
1.6(a) The matrix form of ⎡the equations is⎤ ⎧ ⎫
2 −1 0 x ⎧ ⎫
⎣ −1 ⎨ 1⎬ 1
2 −1⎦ x ⎨ ⎬
2
0 −1 2 ⎩ ⎭ = ⎩2 ⎭
x3 2
Using Cramer’s rule we ¯ obtain ¯
¯ 1 −1 0¯ ¯
1 ¯ 1 9
x1 = 2¯ 2 −1 =¯ [(4 − 1) + (4 + 2) − 0] =
|A|¯ 2 1 2 |A| |A|
¯¯ ¯¯
−
1 ¯ 2 1 0¯ ¯ 1 14
x2 = ¯−1 2 −1 =¯ [2(4 + 2) − (−2 − 0) − 0] =
¯ ¯
|A| ¯ 0 2 2¯¯ |A| |A|
¯
¯¯ 2 −1 1¯ ¯
1 1 11
x3 = ¯ −1 2 2 ¯= [2(4 + 2) + (−2 + 1) + 0] =
|A| ¯ 0 1 2 |A| |A|
¯ − ¯




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,1. Vectors, Tensors, and Equations of Elasticity 7

where the determinant |A| of the coefficient matrix is
¯ ¯
¯¯ 2 −1 0¯ ¯
|A| = −1
¯ 2 −1 =¯ 2(4 − 1) + (−2 − 0) − 0 = 4
¯ ¯
¯ 0 −1 2¯
Hence, x1 = 9/4, x2 = 14/4, and x3 = 11/4.
1.6(b) We have ⎡ ⎤⎧ ⎫
12 0 3h ⎨ x1 ⎬ ⎧ ⎫
4h2 h2 ⎦ x2 = f h4 ⎨ 12 ⎬
⎣ 0 0

3h h2 2h2 ⎩ ⎭ 24b ⎩ 0 ⎭
x3 h
The determinant of the coefficient matrix is |A| = 12(8h4 −h 4 )+ 3h(−12h3) = 48h4. Using
Cramer’s rule we obtain
¯ ¯
12 0 3h ¯
¯¯¯ ¯ 72α
x1 = α 0 4h2 h2 ¯¯= α [12(8h4 − h4 ) + h(−12h3)] =
|A| ¯h h2 2h |A| |A|
¯¯ ¯2 ¯
12 12 3h¯ ¯¯
x2 = α ¯ ¯ 0 0
3
h2 = α [12(−h ) + 3h(12h )] =
2
24α
|A| 3h h 2h ¯ |A|
¯ |A|
¯¯ 2¯
¯
x3 = 1 ¯ 0 ¯ 12 0 12 ¯ 3 2
4h2 0¯ = α [12(4h ) + 3h(−48h )] = − 96α
|A| 3h
¯ h2 h ¯¯ |A| |A|
¯
Hence, x1 = 3α/2, x2 = α/(2h), and x3 = −2α/h, where α = (f0h4/24b).
1.7 Let [C] be a 3× 3 matrix, [I] be a 3× 3 identity matrix, and λ be a scalar. Show that
det[C − λI] = λ3 − I1λ2 + I2λ − I3

where 1
I1 = cii, I2 = (ciicjj − cij cij ), I3 = |C|
2
Solution: Using the result of Problem 1.5(b) and the ε-δ identity, we obtain 1
|C − λI| = ε1ijkεrst(cirh− λδir)(cjs − λδjs)(c kt − λδkt)
=6 εijkεrst 3 2
ir js kt ir js kt kt ir js js ir kt
−λ δ δ δ + λ (c δ δ + c δ δ +c δ δ )
6
— λ (circjsδkt + cir δjsc kt + δir cjsc kt ) + circjsc kt ]
λ2
3
= −λ + 6 (cirε ijkεrjk + cktεijkε ijt + εijkεisk c js)
λ
+ (εijkεrskcircjs + ε ijkε rjtc irckt + εijkε istc jsckt)
6
1
+ εijkεrstcirc jsc kt
6 λ
3 2
= −λ + ciiλ + (ciicjj − cij cji ) + |C|
2




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, 1. Vectors, Tensors, and Equations of Elasticity 7




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