appendix a -...

Appendix AFourier Series

A.1 General Equations

A periodic function (Wylie 1972) can be represented by a series that is expressed as

f x = 0 5Ao +A1 cosx+A2 cos2x + +Am cosmx+B1 sinx

+B2 sin2x+ +Bm sinmx

or

f x = 0 5Ao +∞

m= 1

Am cosmx+∞

m= 1

Bm sinmx (A.1)

The series given in Eq. (A.1) is known as a Fourier series and is used to express periodicfunctions such as those shown in Figure A.1. The coefficients A and B in Eq. (A.1) are evalu-ated over a 2π period starting at a given point d. The value of Ao can be obtained by integratingEq. (A.1) from x = d to x = d + 2π.Thus,

d + 2π

df x dx= 0 5Ao

d + 2π

ddx+A1

d + 2π

dcosx dx+

+Am

d + 2π

dcosmx dx+B1

d + 2π

dsinx dx+

+Bm

d + 2π

dsinmx dx

Stress in ASME Pressure Vessels, Boilers, and Nuclear Components, First Edition. Maan H. Jawad.© 2018, The American Society of Mechanical Engineers (ASME), 2 Park Avenue,New York, NY, 10016, USA (www.asme.org). Published 2018 by John Wiley & Sons, Inc.

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The first term in the right-hand side of the equation gives πAo. All other terms on the right-handside are zero because of the relationships

d + 2π

dcosmx dx= 0 m 0

d + 2π

dsinmx dx= 0

Hence,

Ao =1π

d + 2π

df x dx (A.2)

The Am term in Eq. (A.1) can be obtained by multiplying both sides of the equation by cos mx.

d + 2π

df x cosmx dx=

12Ao

d + 2π

dcosmx dx+A1

d + 2π

dcosx cosmx dx+

+Am

d + 2π

dcosmx cosmx dx+B1

d + 2π

dsinx cosmx dx+

+Bm

d + 2π

dsinmx cosmx dx

(A.3)

Since

d + 2π

dcosmxcosnx dx= 0 m n

d + 2π

dcos2mx dx= π m 0

d + 2π

dcosmxsinnx dx= 0

x

x

Figure A.1 Periodic functions

310 Appendix A

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Equation (A.3) becomes

d + 2π

df x cosmx dx=Amπ

or

Am =1π

d + 2π

df x cosmx dx (A.4)

Similarly the values of Bm can be found by multiplying both sides of Eq. (A.1) by sin mx.Using the expressions

d + 2π

dsinmxsinnx dx= 0 m n

and

d + 2π

dsin2mx dx= π

the equation becomes

Bm =1π

d + 2π

df x sin mx dx (A.5)

Accordingly, we can state that for a given periodic function f(x), a Fourier expansion can bewritten as shown in Eq. (A.1) with the various constants obtained from Eqs. (A.2), (A.4),and (A.5).

Example A.1Express the function shown in Figure A.2 in a Fourier series.

Solution

f x = 0 −π < x < 0

f x = po 0 < x < π

d = −π

From Eq. (A.2),

Ao =1π

0

−π0 dx+

1π

π

0podx

311Appendix A

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or

Ao = po

From Eq. (A.4),

Am =1π

π

0po cosmx dx

Am = 0

From Eq. (A.5),

Bm =1π

π

0po sinmx dx

=pomπ

−cosmx π0

=−pomπ

cosmπ−1

Bm =2pomπ

when m is odd

= 0 when m is even

–2π –π π 2π 3πx

x

x

x

1.0 m= 1

m= 1,3

m= 1,3,5

po

popo

0.5

1.0

0.5

1.0

0.5

–π π2

π

–π π2– π

2π

–π π2– π

2π

π2–

Figure A.2 Periodic rectangular function

312 Appendix A

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Therefore, the expansion of the function shown in Figure A.2 is expressed as

f x = 0 5po +2poπ

∞

m = 1 3,…

1msinmx

A plot of this equation with m = 1, 3, 5 is shown in Figure A.2.

A.2 Interval Change

In applying the Fourier series to plate and shell problems, it is more convenient to specifyintervals other than 2π. Defining the new interval as 2p, Eqs. (A.2), (A.4), and (A.5) can bewritten as

Ao =1p

d + 2p

df x dx (A.6)

Am =1p

d + 2p

df x cos

mπx

pdx (A.7)

Bm =1p

d + 2p

df x sin

mπx

pdx (A.8)

where 2p = period of function, and the series can be written as

f x =12Ao +

∞

m = 1

Am cosmπx

p+

∞

m = 1

Bm sinmπx

p(A.9)

Example A.2Find the Fourier expansion of the function f(x) = cos x as shown in Figure A.3.

SolutionThe period 2p is equal to π. Thus, p = π/2 and d = −π/2.

–π π2–

xπ2

π

Figure A.3 Periodic cosine function

313Appendix A

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Ao =1

π 2

π 2

−π 2cosx dx=

4π

Am =2π

π 2

−π 2cosx cos

mπx

π 2dx=

4π

∞

m = 1

−1 m + 1

4m2−1

Bm = 0

and from Eq. (A.9),

f x =2π+4π

∞

m = 1

−1 m+ 1

4m2−1cos2mx

A.3 Half-Range Expansions

If a function is symmetric with respect to the axis of reference as shown in Figure A.4, then thecoefficient integral can be simplified by integrating over one-half the period. This integrationcan be performed as an even or odd function.Hence, for an even periodic function,

Ao =2p

p

0f x dx

Am =2p

p

0f x cos

mπx

pdx

Bm = 0

(A.10)

(a) (b)

Odd

x

x

x

Odd

Even

(c)

Figure A.4 Periodic symmetric function. (a) Odd step function, (b) odd continuous function, and(c) even step function

314 Appendix A

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For an odd periodic function,

Ao =Am = 0

Bm =2p

p

0f x sin

mπx

pdx

(A.11)

It should be noted that the even and odd functions defined by Eqs. (A.10) and (A.11) andFigure A.4 do not refer necessarily to the shape of the function but rather to the reference linefrom which they are defined. This can best be illustrated by Example A.3.

Example A.3Figure A.5 shows a plot of the function y = x − x2. Obtain and plot the Fourier series expansionof this function (a) from y = −1 to y = 1; (b) as an even series from y = 0 to y = 1; and (c) as anodd series from y = 0 to y = 1.

2.0

1.0 1.0

y

2.0

–1.0 1.0Continuous function

Even function

2.0

–1.0 1.0 2.0

2.01.0Odd function

–1.0

2.0x

Figure A.5 Function y = x − x2

315Appendix A

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Solutiona. d = −1, 2p = 2 or p = 1

Ao =1

−1x−x2 dx= −

23

Am =1

−1x−x2 cos

mπx

1dx= −

4cosmπm2π2

Bm =1

−1x−x2 sin

mπx

1dx= −

2cosmπmπ

f x = −13−

∞

m= 1

4cosmπm2π2

cosmπx−∞

m = 1

2cosmπmπ

sinmπx

= −13−4π2

∞

m= 1

−1 m

m2cosmπx−

2π

∞

m= 1

−1 m

msinmπx

b.Ao = 2

1

0x−x2 dx=

13

Am = 21

0x−x2 cos

mπx

1dx= −

2 1 + cosmπm2π2

Bm = 0

f x =16−

∞

m= 1

2 1 + cosmπm2π2

cosmπx

c. Ao =Am = 0

Bm = 21

0x−x2 sin

mπx

1dx=

4 1−cosmπm3π3

f x =∞

m= 1

4 1−cosmπm3π3

sinmπx

A.4 Double Fourier Series

In solving rectangular plate problems of length a and width b, it is customary to express theapplied loads in terms of a single or double series. The double Fourier series is normallyexpressed as an odd periodic function with a half-range period given between 0 and a forone side of the plate and 0 and b for the other side. Thus,

316 Appendix A

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f x,y =∞

m= 1

∞

n= 1

Bmnsinmπx

asin

nπy

b(A.12)

where

Bmn =4ab

b

0

a

0f x,y sin

mπx

asin

nπy

bdxdy (A.13)

Example A.4The rectangular plate shown in Figure A.6 is subjected to a uniform pressure po. Determine theFourier expansion for the pressure.

SolutionFrom Eq. (A.13),

Bmn =4poab

b

0

a

0sin

mπx

asin

nπy

bdxdy

=16poπ2mn

m,n are odd functions

f x,y =16poπ2

∞

m= 1 3,…

∞

n = 1 3,…

1mn

sinmπx

asin

nπy

b

y

b

xa

Figure A.6 Rectangular plate

317Appendix A

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Appendix BBessel Functions

B.1 General Equations

In many plate and shell applications involving circular symmetry, the resulting differentialequations are solved by means of a power series known as Bessel functions. Some of thesefunctions are discussed in this appendix.The differential equation

d2y

dx2+1x

dy

dx+ y = 0 (B.1)

is referred to as Bessel’s equation of zero order. Its solution (Bowman 1977) is given by thefollowing power series:

y=C1Jo x +C2Yo x (B.2)

where C1 and C2 = constants obtained from boundary conditions and Jo(x) = Bessel function ofthe first kind of zero order.

Jo x = 1−x2

22+

x4

22 42−

x6

22 42 62+

=∞

m= 0

−1 m

m 2

x

2

2m

Yo(x) = Bessel function of the second kind of zero order.


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Yo x = Jo xdx

x J2o x

= Jo x lnx+x2

22−

x4

22 421 +

12

+x6

22 42 621 +

12+13

−

Equation (B.1) is usually encountered in a more general form as

x2d2y

dx2+ x

dy

dx+ x2−k2 y = 0 (B.3)

The solution of this equation (Hildebrand 1964) is

y=C1Jk x +C2J−k x

when k is not zero or a positive integer or

y=C1Jk x +C2Yk x

when k is zero or a positive integer and where

Jk x =Bessel function of the first kind of order k

=∞

m = 0

−1 m

m m+ kx

2

2m+ k

J−k x =Bessel function of the first kind of order k

=∞

m = 0

−1 m

m m−k

x

2

2m−k

Yk x =Bessel function of the second kind of order k

=2π

lnx

2+ γ Jk x −

12

k−1

m = 0

k−m−1m

x

2

2m−k

+12

∞

m= 0

−1 m+ 1 h m + h m + kx 2 2m+ k

m m + k

h m =m

r = 1

1r

m > 1

γ = 0 5772

A plot of Jo(x), J1(x), Yo(x), and Y1(x) is shown in Figure B.1. Also, Table B.1 gives some valuesof J(x) and Y(x).

320 Appendix B

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–1.0

1.0

1.0

–1.0

2 4 6 8 10

Yo(x)Y1(x)

xY(x)

2 4

(a)

(b)

J1(x)

Jo(x)

6 8 10xJ(x)

Figure B.1 A plot of various Bessel functions: (a) Jo and J1 functions and (b) Yo and Y1 functions

Table B.1 Values of Jo, J1, Yo, and Y1

(x) Jo(x) J1(x) Yo(x) Y1(x)

0.0 1.0000 0.0000 −∞ −∞0.5 0.9385 0.2423 −0.4445 −1.47151.0 0.7652 0.4401 0.0883 −0.78121.5 0.5118 0.5579 0.3825 −0.41232.0 0.2239 0.5767 0.5104 −0.10702.5 −0.0484 0.4971 0.4981 0.14593.0 −0.2601 0.3391 0.3769 0.32473.5 −0.3801 0.1374 0.1890 0.41024.0 −0.3972 −0.0660 −0.0169 0.39794.5 −0.3205 −0.2311 −0.1947 0.30105.0 −0.1776 −0.3276 −0.3085 0.14795.5 −0.0068 −0.3414 −0.3395 −0.02386.0 0.1507 −0.2767 −0.2882 −0.17506.5 0.2601 −0.1538 −0.1732 −0.27417.0 0.3001 −0.0047 −0.0260 −0.30277.5 0.2663 0.1353 0.1173 −0.25918.0 0.1717 0.2346 0.2235 −0.15818.5 0.0419 0.2731 0.2702 −0.02629.0 −0.0903 0.2453 0.2499 0.10439.5 −0.1939 0.1613 0.1712 0.203210.0 −0.2459 0.0435 0.0557 0.2490

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A different form of Eq. (B.3) that is encountered often in plate and shell theory is

x2d2y

dx2+ x

dy

dx− x2 + k2 y = 0 (B.4)

The solution of this equation (Dwight 1972) is

y=C1Ik x +C2I−k x

when k is not zero or a positive integer or

y=C1Ik x +C2Kk x

when k is zero or a positive integer and where

Ik x =modified Bessel function of the first kind of order k

=∞

m= 1

x 2 2m + k

m m + k

Kk x =modified Bessel function of the second kind of order k

= −1 k + 1 lnx

2+ γ Ik x

+12

k−1

m = 0

−1 m k−m−1m

x

2

2m−k

+12

∞

m = 0

−1 k x 2 2m + k

m m+ k1 +

12+ +

1m

+ 1 +12+ +

1m+ k

A plot of Io(x), I1(x), Ko(x), and K1(x) is shown in Figure B.2.

6

y

y

x x

5

4

3

2

2

3

1

1

0

J2(x)

J1(x)K1(x)

K2(x)

1 2 0 1 23 4

Figure B.2 A plot of modified Bessel functions (Wylie 1972)

322 Appendix B

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Another equation that is often encountered in plate and shell theory is given by

x2d2y

dx2+ x

dy

dx− ix2 + k2 y = 0 (B.5)

The solution of this equation (Hetenyi 1964) for the important case of k = 0 is given by

y=C1Z1 x +C2Z2 x +C3Z3 x +C4Z4 x (B.6)

where

Z1 x = ber x =∞

m= 0

−1 m x 2 4m

2m 2

Z2 x = −bei x = −∞

m = 0

−1 m x 2 4m + 2

2m+ 1 2

Z3 x = −2πkei x =

Z1 x

2−2π

R1 + lnγx

2Z2 x

Z4 x = −2πker x =

Z2 x

2−2π

R2 + lnγx

2Z1 x

R1 =x

2

2−h 3

3 2

x

2

6+h 5

5 2

x

2

10−

R2 =h 2

2 2

x

2

4−h 4

4 2

x

2

8+h 6

6 2

x

2

12−

h n = 1 +12+13+ +

1n

γ = 0 5772

A plot of Z1(x), Z2(x), Z3(x), Z4(x), and their derivatives is shown in Figure B.3.

B.2 Some Bessel Identities

The derivatives and integrals of Bessel functions follow a certain pattern. The identities givenhere are needed to solve some of the problems given in this text.

d

dxxJ1 x = xJo x

d

dxJo x = −J1 x

d

dxxnJn x = xnJn−1 x

d

dx

Jn x

xn=−Jn+ 1 x

xn

323Appendix B

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–10.0

–5.0

0

Z2(x)

+5.0

+10.0

65310

x

x

dx

dZ2(x)

dx

dZ1(x)

dx

dZ4(x)

dx

dZ3(x)

Z1(x)

Z3(x)

Z4(x)

2 4

65310

0.1

0

–0.1

–0.2

–0.3

–0.4

–0.5

+0.2

+0.3

+0.4

+0.5

2 4

Figure B.3 Plot of the Z functions (Hetenyi 1964)

324 Appendix B

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d2Z1 x

dx2= Z2 x −

1x

dZ1 x

dx

d2Z2 x

dx2= −Z1 x −

1x

dZ2 x

dx

d2Z3 x

dx2= Z4 x −

1x

dZ3 x

dx

d2Z4 x

dx2= −Z3 x −

1x

dZ4 x

dx

The last four equations are needed in the solution of circular plates on elastic foundation. Inthese equations the value of (kx) is needed rather than (x) in the Z functions. In this case, theseequations take on the form

k2Z1 kx = k2Z2 x −k

xZ1 kx

k2Z2 kx = −k2Z1 x −k

xZ2 kx

k2Z3 kx = k2Z4 x −k

xZ3 kx

k2Z4 kx = −k2Z3 x −k

xZ4 kx

B.3 Simplified Bessel Functions

As x approaches zero, the various Bessel functions can be expressed as

Jk x =xk

2k k

Yk x =−2k k−1

πx−k k 0

Yo x =2πlnx

Ik x =xk

2kk

Kk x = 2k−1 k−1 x−k k 0

Ko x = − lnx

Z1 x = 1 0 Z2 x = −x2

4

Z3 x =12

Z4 x =2πlnγx

2

325Appendix B

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dZ1 x

dx= −

x3

16dZ3 x

dx=x

πlnγx

2

dZ2 x

dx= −

x

2dZ4 x

dx=

2πx

As x approaches infinity, the various Bessel functions can be expressed as

Jk x =2πxcos x−ξk ξk = 2k + 1

π

4

Yk x =2πxsin x−ξk

Ik x =ex

2πx

Kk x =e−x

2x π

Z1 x = ηcosσ Z2 x = −ηsinσ

Z3 x = β sinτ Z4 x = −βcosτ

η=1

2πxex 2 β =

2πxe−x 2

σ =x

2−π

8τ =

x

2+π

8

dZ1 x

dx=

η

2cosσ−sinσ

dZ2 x

dx=−η

2cosσ + sinσ

dZ3 x

dx=

β

2cosτ−sinτ

dZ4 x

dx=−β

2cosτ + sinτ

326 Appendix B

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Appendix CConversion Factors

Pressure units

1 psi 1 N/mm2 1 bar 1 kPa 1 kgf/cm2

psi 1.0000 145.0 14.50 0.1450 14.22N/mm2 0.006895 1.000 0.1000 0.0010 0.09807bars 0.06895 10.000 1.000 0.0100 0.9807kPa 6.895 1000.0 100.00 1.000 98.07kgf/cm2 0.0703 10.20 1.020 0.0102 1.000

1 N/mm2 = 1 MPa.

Stress units

1 ksi 1 kN/mm2 1MPa 1 kgf/mm2

ksi 1.000 145.0 0.1450 1.422kN/mm2 0.006895 1.000 0.001 0.009807MPa 6.895 1000.00 1.000 9.807kgf/mm2 0.7033 102.0 0.1020 1.000

Force units

1 lb 1 kgf 1 N

lb 1.000 2.205 0.2248kgf 0.454 1.000 0.1020N 4.448 9.807 1.0000


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0003135128.3D 332 24/8/2017 4:28:05 PM


Answers to Selected Problems

1.2 Nϕ = −poR/2Nθ = −poR(cos

2 ϕ − 1/2)1.3

Nϕ =−γR2

63HR

+ 1−2cos2ϕ1 + cosϕ

Nθ =−γR2

63HR

−1−6−4cos2ϕ1 + cosϕ

1.6 t = 0.90 inch1.8

maxNs =−847γL2 sinα

432at s=

L

12

maxNθ =γL2

4sinα at s=

L

2

2.4 A = 5.76 inch2

2.5t1 = 0 21 inch, t2 = 0 82 inch

t3 = 0 98 inch, t4 = 0 69 inch

A= 1 22inch2

3.3 max Mx = 0.322 Qo/β at x= 0 61 rt3.4 At section a–a, Mo = 0 and Ho = 0.0195Dβ3


0003135129.3D 333 24/8/2017 4:29:17 PM


3.5 Ma = 14.95/β and Mb = 44.97/β4.1 t = 5/16 inch6.2 t = 0.73 inch6.3 p = 148.9 psi7.2 σx = 92.2 MPa, σy = 124.9 MPa8.3 = 8670 psi, max w = 0.23 inch8.4 = 12,330 psi8.6

Mr =p

163 + μ a2−r2 −

pb2

4K1 +K2−K3−K4−1

where

K1 = 1 + μ3 + μ

2 1 + μ−

b2

a2−b2lnb

a−12

K2 =1−μr

1 + μ1−μ

a2b2

a2−b2lnb

a

K3 = 1 + μ lnr

a

K4 =3 + μ4

a2−r2

r2

Mt =p

16a2 3 + μ −r2 1 + 3μ −

pb2

4K1 +K2−K3 +K5−μ

K5 =3 + μ4

a2 + r2

r2

9.1 Mp = pL2/89.3 Mp = pL2/1449.8 Mp = 157.3 p

334 Answers to Selected Problems

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Index

AAmerican Society of Mechanical

Engineers (ASME)boiler code, 69nuclear code, 70pressure vessel code, 71

Analysiscylindrical shells, 51multicomponent structures 27

Answers to selected problems, 333Approximate analysis of plates, 239Axial compression

ASME code, 127empirical equations, 130

BBending of

circular plates, 215conical shells, 165cylindrical shells, 71rectangular plates, 187shells of revolution, 151spherical shells, 156

Bessel functions, 319Buckling of

circular plates, 259conical shells, 181cylindrical shells, 103

axial compression, 117lateral and end pressure, 114lateral pressure, 108

ellipsoidal shells, 181finite difference equation, 275rectangular plates, 263, 279shallow heads, 16spherical shells, 175stiffened spherical shells, 180thermal buckling, 277

CCircular platesASME code, 225bending of, 215buckling of, 259design of, 253on elastic foundation, 227uniform loads in the θ-direction, 215variable boundary conditions, 231yield line theory, 239

Conical shells, 18bending, 165membrane, 18

Conversion factors, 327Creep analysisthick wall cylinders, 65thick wall spherical sections, 150


0003201645.3D 335 24/8/2017 4:30:03 PM


Cylindrical shells, 20ASME

boiler code, 69nuclear code, 70pressure vessel code, 71

axial compression, 127bending of, 71buckling of, 103elliptical cross section 22external pressure, 120long, 76, 82short, 88thermal stress

axial direction, 89radial direction, 91

DDeflection, 42–47

shells of revolution due to axisymmetricloads, 42

Design ofcircular plates, 225cylindrical shells under external pressure, 179rectangular plates, 212shells of revolution, 23

Discontinuity stresses, 98

EEdge loads on shells

conical, 165cylindrical, 76spherical, 156

Elastic analysisthick-wall cylinders, 51thick-wall spherical sections, 141

Elastic foundation, 227–231, 237, 325bending of circular plates, 227

Ellipsoidal shells 6, 145, 146internal pressure, 15

Empirical external pressure equation, 124Equivalent stress limits, 57External pressure chart, 123

FFinite difference method

buckling equation, 275differential equation, 275

Finite elementanalysis, 283axisymmetric triangular linear, 302higher order, 305

linear triangular elements, 295nonlinear, 307one-dimensional, 287

Fitting reinforcement, 43Fourier series, 309

GGeometric chart for external pressure, 121

HHigher order finite elements, 305

IIsochronous stress-strain curves, 68

JJoint efficiency, 70, 144, 148, 225

KKnuckle of formed heads, 16, 145

LLinear triangular finite elements, 295Long cylindrical shells, 76

MMembrane theory of shells of revolution, 1Modulus of elasticity, table of various

materials, 195

NNozzle reinforcement, 38

OOne-dimensional finite elements, 287

PPeak stress, 57Plastic analysisthick wall cyliners, 63thick wall spherical sections, 150

Plates on elastic foundation, 325circular, 325

Poisson’s ratio, table of variousmaterials, 195

Pressure-area method, 35Primary stress, 57

RRectangular platesASME equations, 212

336 Index

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bending of, 187buckling of, 263design of, 212thermal stress, 277yield line theory, 246

Reinforcementconduit, 44fittings, 43nozzles, 38

SSecondary stress, 57Shallow heads, buckling of, 145Shells of revolution

bending, 151membrane, 1

Short cylindrical shells, 88Spherical shells, 7–14, 37–39, 57, 72, 141–145,

150, 156–166, 168, 175–178, 180–181ASME code, 14, 142, 179axisymmetric loading, 6bending, 151, 156buckling, 175creep analysis, 150internal pressure 7

plastic analysis, 150stiffened, 180various loading conditions, 8

Stress categories, 57

TThermal buckling of plates, 277Thermal stresscylindrical shells, 89rectangular plates, 277

Thick wall cylinderscreep analysis, 65elastic analysis, 51off-center bore, 56plastic analysis 63

Triangular plates, 248Tubesheets, 227

VVariable boundary conditions, 231

YYield line theory, 239–252circular plates, 239, 247

337Index

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