Solutions Manual for Theory and Analysis of Elastic Plates and Shells 2nd Edition by J. N. Reddy PDF | Complete Step-by-Step Solutions to All Chapters | Covers Classical Plate Theory, Bending, Shear Deformation, Finite Element Formulations, and Shell Stru

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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) 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, F i2 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 = δi1 δi1 + δi2 δi2 + δi3 δi3
= (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
Fij εijk = −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 Fijεijk = 0, then Fij = Fji. We have 0
= Fij εijk
1
= (Fij εijk + Fij εijk)
2
1
= (Fijεijk − Fijεjik) (interchanged i and j)
2
1
= (Fijεijk − Fji εijk) (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 ) · (xj êj ) = xixjδij
= xixi = xkxk (2)
Its derivative of r with respect to xi can be obtained from
∂r2 ∂
= (xkxk)
∂xi ∂x i
∂x ∂xk
= kx +x
∂xi k k ∂xi
∂xk
=2 xk = 2δikxk = 2xi
∂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
(b) ∇2(rn ) = n(n + 1)rn−2
(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
∂ ∂r = nrn−2 x ê
∇(rn ) = ê (rn ) = nr n − 1 ê = nrn−2r
i ∂xi i i i
∂xi
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 ) = (∇ · ∇) (r n) = ∇ · [∇(r n)]
à !
∂ ³ ´ ³ ´
∂
= ê · nrn−2x iê i = n(ê j · êi) rn−2x i
j ∂xj ∂xj
∙ ¸
xj
= nδij (n − 2)rn−3 x i + rn−2δij
h
r 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= ê j · (x iê i) = (ê · ê ) =δ δ =3
∂xj j i ij ij
∂xj

1.2(d) We obtain

culr(f r) = ∇ × [f r]
à !
∂ ∂
= ê × (f (r)x ê
i )i = (ê j × ê i) [f(r)x i]
j ∂x"j # ∂xj
∂r
=ε ê f 0(r) x + f (r)δ
jik k i ij
xj
∙ ¸
f 0(r)
= εjik êk xjxi + f (r)δij
r
The two terms in the square brackets, xixj and δij are symmetric, hence, by Problem
1.1(d) the expression in the last line is zero.

, 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
n n
X X
cji = ajkbki = bkiajk
k=1 k=1
Xn
= (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]T = [C]


♠ 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 =A i êi · Bj Ck εjkm êm = AiBjCkεjkmδim
=Ai Bj Ck εjki = Ai Bj Ckεijk = A × B · C

Since i, j, and k can be permuted in a cyclic order, it also follows that

Ai Bj Ckεijk = 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 = ê i , B = êj , C = ê m , A = ê n

We obtain

(êi × êj ) · (êm × ên ) = (êi · êm )(êj · ên ) − (êi · en)(êj · êm )

εijk êk · ε mnp êp = δimδjn − δinδjm

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
ê1 ê2 ê 3
B × C = ¯ B1 B2 B
¯ (3)
3
¯ ¯
C1 C2 C3

and the “scalar triple product” of
vectors
A1 A2 A3
¯ ¯
A · (B × C) = B2 B3 (4)
B1 ¯
¯
C1 C2 C3

Now
let A = c1i êi ≡ C1, B = c2j êj ≡ C2, C = c3k êk ≡ C3

in Eq. (3). We obtain

C1 · (C2 × C3) = c 1 i êi · (c 2j êj × c3k êk )
c11 c12 c13
¯ ¯
= ¯c21 c22 c23 ¯ ≡|C|
c31 c32 c33

,6 Theory and Analysis of Elastic Plates and Shells


or
|C| = c 1i êi · (c2j êj × c3k êk )
= c1ic2jc3kεijk
which is the same as Eq. (1). Next consider the product Cr · (Cs × Ct):
Cr · (Cs × Ct) = cricsj ctkεijk
Multiplying both sides with εrst and expanding, we arrive at
cricsj ctkεrstεijk = 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)
⎡ ⎤⎧ ⎫ ⎧ ⎫
2b ⎣ 12 0 3h ⎦ ⎨ x1 ⎬ f h ⎨ 12 ⎬
0 4h2 h2 x2 = 0 0
h3 ⎩ ⎭ 12 ⎩ ⎭
3h h2 2h2
x3 h
where b, f0, and h are constants
Solution:
1.6(a) The matrix form of the equations is
⎡ ⎤⎧ ⎫ ⎧ ⎫
2 −1 0 ⎨ x1 ⎬ ⎨ 1 ⎬
⎣ −1 2 −1 ⎦ x2 = 2
⎩ ⎭
0 −1 2 ⎩ x3 ⎭ 2
Using Cramer’s rule we obtain
1 ¯ 1 −1 0¯ 1 9
x1 = 2 2 −1 = [(4 − 1) + (4 + 2) − 0] =
|A| ¯ |A| |A|
2 −1 2¯
1 ¯ 2 1 0¯ 1 14
x2 = −1 2 −1 = [2(4 + 2) − (−2 − 0) − 0] =
|A| ¯ |A| |A|
0 2 2¯
1 ¯ 2 −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 ⎬ 4 ⎨ 12 ⎬
⎣ 0 4h2 h 2 ⎦ x2 = f24b0h 0
3h h2 2h2 ⎩ x3 ⎭ ⎩ ⎭
h
The determinant of the coefficient matrix is |A| = 12(8h−
4 h4)+ 3h( 12h3) = 48h4.
−
Using Cramer’s rule we obtain

α ¯ 12 0 3h ¯ α 3 72α
x1 = 0 4h2 h2 = [12(8h 4 — h4 ) + h(−12h )] =
|A| ¯ h h2
¯
2h2 ¯ |A |A|
¯ ¯ |
α ¯ 12 12 3h2 ¯ α 2 24α
x2 = ¯ 0 0 h ¯= [12(−h 3) + 3h(12h )] =
|A| 3h h 2h 2 |A| |A|
¯ ¯
1 ¯ 12 0 12 ¯ α 2 96α
x3 = ¯ 0 4h
2 0¯ = [12(4h3 )+ 3h(−48h )] = −
|A| ¯ 3h h 2 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 − I1 λ2 + I2 λ − I3

wher 1
e I1 = cii, I2 = (cii cjj − cij cij), I3 = |C|
2
Solution: Using the result of Problem 1.5(b) and the ε-δ identity, we obtain
1
|C − λI| = εijkεrst(cir − λδir)(cjs − λδjs )(ckt − λδkt)
6 h
1 3 2
= εijkεrst −λ δirδjsδ kt + λ (cirδjsδkt + cktδirδjs + cjsδirδkt)
6
— λ (circjsδkt + cirδjsckt + δircjsckt)+ circjsckt ]
2
= −λ3 + λ (cirεijkεrjk + cktεijkεijt + εijkεiskcjs)
6
λ
+ (εijkεrskcircjs + εijkεrjtcirckt + εijkεistcjsckt)
6
1
+ εijkεrstcircjsckt
6
λ
= −λ3 + c iiλ2 + (ciic jj − cij cji )+ |C|
2

, 8 Theory and Analysis of Elastic Plates and Shells


1.8 If we identify a second-order tensor A associated with the direction cosines
aij = ê
¯i · ê j [see Eq. (1.2.57)]
A = aij ê¯ i êj
show that (a) A · A = A, (b) A · AT = I, and (c) A : A = 3.
Solution:
1.8(a) We have
A · A = aij akp akj ê¯i êp = a i p ēˆ i êp = A
where we have used the identity in Eq. (1.2.61).
1.8(b) We have
A · AT = aij akp δjp ˆēiˆēk = aij akj ê
¯i ēˆk = δ i p ê¯ i ê¯ k = I
where we have used the identity in Eq. (1.2.61).
1.8(c) We have
µ ¶µ ¶
A : A = aij amn ê¯i · ên ê¯ j · êm

= aij amnainamj = δjnδnj = δjj = 3
where we have used the identity in Eq. (1.2.61) repeatedly.

1.9 Use the definition ∇2 = ∇ · ∇ to show that the Laplacian operator in the
cylindrical coordinate system is given by
2 1 h ∂ ³r ∂ ´ 1 ∂2 ∂2 i
∇ = + 2
+r 2
r ∂r ∂r r ∂θ ∂z
Solution: Using the definition (1.2.30) of ∇ and the derivatives of the basis
vectors (1.2.29)
∂êr ∂êθ
∇ = ê ∂ + êθ ∂ + ê ∂ , = ê , = −ê
r ∂r r ∂θ z ∂z ∂θ θ ∂θ r

we obtain
µ ¶ µ ¶
∂ êθ ∂ ∂ ∂ ê θ ∂ ∂
∇ · ∇ = êr ∂r + · êr ∂r +
r ∂θ + ê z ∂z r ∂θ + ê z ∂z
µ ¶
∂ ∂ ∂
= êr · êr ∂r + êθ ∂ + ê z ∂z
∂r r ∂θ
µ ¶
+ 1eθˆ · ∂ êr ∂ + ê θ ∂ + êz ∂
r ∂θµ ∂r r ∂θ ∂z
∂ ∂ ∂ ¶
ê êθ ∂
+ êz · r ∂r + r ∂θ + ê z ∂z
∂z à !
∂2 1 ∂êr ∂ + êθ ∂2 ∂2
= 2 + eˆ θ · + 2
∂r r ∂θ ∂r r ∂θ2 ∂z
∂2 1 ∂ 1 ∂2 ∂2
= 2+ + 2 2 + 2
∂r r ∂r r ∂θ ∂z
µ ¶
= 1 ∂ r
∂
+
1 ∂2 ∂2
+ 2
2
r ∂θ 2 ∂z
r ∂r ∂r


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