mroc3st05

1

3. MEMBRANE ANALYSIS OF ROTATIONAL SHELLS

A rotational shell or shell of revolution is formed by rotating a curve (meridian) about an axis of rotation. In this Section, the governing membrane equations of rotational shells will be derived and thenapplied for the analysis of a number of shell problems.

3.1 Geometrical Description

At any point on the middle surface of a rotational shell, one can define two principal radii of curvature. Fig. 3.1 shows two principal sections containing the normal to the shell at point P. These sections create two plane curves with two local principal radii of curvature and as shown in Fig. 3.1. One of these sections is called the meridion curve while the projection of the other section on the plane perpendicular to the axis of revolution creates theparallel circle (with radius of curvature denoted by ) on the shell surface. 0r

1r 2r

2

3.1 Geometrical Description (1)

Fig. 3.1 Two principal sections containing the normal to the shell at point P

(radius of curvature of meridian curve)

3


Owing to rotational symmetry, the centre of curvature of always lies on the axis of revolution. However, the centre of curvature of does not have to lie on this axis. r1 is the radius of curvature of the meridian curve at that point.

The angle between the normal to the surface at P with the axis of revolution will be denoted by . The horizontal angular position of P from some arbitrary origin will be denoted by the angle θ . The direction of the axis of revolution is assumed to coincide with the zaxis.

2r

φ

1r

4


For rotational shells, and (and hence ) are independent of , i.e. they are functions of only. Referring to Fig. 3.1, the radius of the parallel circle at point P can be written as

(3.1a)

Also the following relations exist among the shell geometrical parameters

1r 2r 0r θφ

0r

φsin20 rr =

5


Derivation of Curvature Expression

Curvature

Now

Also

Upon differentiation, we obtain

Note that

* *

1d d dzds dz dsθ θκ = =

222 1)()( ⎟

⎠⎞

⎜⎝⎛+=⇒+=

dzdr

dzdsdrdzds o

o

dzdro=*tanθ

2

2**2sec

dzrd

dzd o=θθ

2*2*2 1tan1sec ⎟

⎠⎞

⎜⎝⎛+=+=

dzdroθθ

s

r2

r0Θ*

Θ*

Z

θ

6


Therefore

Now

2 2

* * 2 2

1 3/ 22 * 2 20 0

1sec

1 1

o od r d rd d dz dz dzds dz ds dr dr

dz dz

θ θκθ

= = = =⎡ ⎤⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

20

0

20

00*

02 1

1

1cos⎟⎠⎞

⎜⎝⎛+=

⎟⎠⎞

⎜⎝⎛+

===dzdrr

dzdr

r

dsdzrrr

θ

7


The curvature in the meridian plane is given by

The curvature in the second principal plane is given by

2/320

20

2

11

1

1

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+

−==

dzdrdz

rd

rκ

(3.1h)

2/120

0

22

1

11

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+

==

dzdrr

rκ

(3.1i)

1κ

2κ

8

Example

Using Eqs. (3.1h) and (3.1i), show that for a spherical shell of radius R, the radiiof curvature and are equal to R.1r 2r

z

x

z R

r0

Rrzrr

Rzrrzr

dzdrr

dzrd

dzdr

r

rdzdr

dzrdTherefore

dzrdr

dzdr

zdzdrr

zRr

=⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+=

=+=⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+

−=

⎟⎠⎞

⎜⎝⎛−−

=

−=+⎟⎠⎞

⎜⎝⎛

−=

−=

2/12

002

220

2/12

00

2/120

0

20

2

2/320

1

0

20

20

2

20

2

0

20

00

2220

1

1

11

1

222

22

9

Consider various classes of rotational shells (1)

To better understand the meaning of , and , consider the various classes of rotational shells:(a) Singly curved shells of zero Gaussian curvature

▪ , developable shells▪ cylindrical and conical shells

∞=1r

0r 1r 2r

10


(b) Doubly curved shells having positive Gaussian curvature (synclastic shells)▪ are on the same side▪ domes, ellipsoid of revolution, etc.

21 randr

11


(c) Doubly curved shells having negative Gaussian curvature (anticlastic shells)▪ are on the opposite side▪ hyperboloid of revolutions etc.

21 randr

12

3.2 Shell Element

Consider a shell element taken from a rotational shell whose boundaries are the two adjacent meridian planes (first principalsections) and the two perpendicular planes (second principal sections) as shown in Fig. 3.2

A, B, C and D are on a horizontal planeB, C and E are on a meridian plane

Fig 3.2 Infinitesimal Element of a Rotational Shell

z

C

E

DA

r1dΦ

Φ + dΦ

13

3.3 Stresses and Stress-Resultants (1)

Positive directions of plane stresses acting on a shell element are shown in Fig. 3.3. Stresses are positive when they are in the positive directions on the positive surfaces.

Fig. 3.3 (a) Membrane Stresses and (b) Unit Cross-Section in Meridian Plane

dzrzdA ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

1

1

(a) (b)

h/2

h/2

dz

14

3.3 Stresses and Stress-Resultants (2)

Force per unit length is expressed as

For thin shells, and implying that and . Thus, Eq. (3.2) becomes

Similarly

Unit: force per unit length

dzNh

hφφ σ∫

−

=2/

2/

dzrzdAN

h

hA⎟⎟⎠

⎞⎜⎜⎝

⎛−== ∫∫

− 1

2/

2/

1θθθ σσ

θN

(3.2)

1rh << 2rh << 1/ 1 <<rz1/ 2 <<rz

dzNh

hθθ σ∫

−

=2/

2/

(3.3)

dzNNh

hφθθφφθ τ∫

−

==2/

2/

(3.4)

(3.5)

15

3.4 Equilibrium Equations (Membrane Forces) (1)

There are 3 equilibrium equations involving the 3 unknowns . The problem is statically determinate. To derive the equilibrium equations, consider the free body diagram of a rotational shell as shown in Fig. 3.5.

( )0,0,0 === ∑∑∑ zFFF φθ

( )φθφθ NNN ,,

Fig. 3.5 Free Body Diagram of a Rotational Shell Element

φθ drdrdA 10 .=φθθ drNN 1.=

φθφθφ drNN 1.=

θφθφθ drNN 0.=

θφφ drNN 0.=

0r is in the horizontal plane

16


θN

Neglecting

2sin θ

θdNd

Plan View

Elevation: Meridian Plane

zNNNN θφθθθθ ++=

Note thatθN has components in three directions z,,φθ

Contribution from each force

17


φN

zNNN φφφφ += Note that φN has components in two

directions z,φ

18


φθN

θφθφθ NN =

Elevation

19


φθφθθφθφ NNN +=

Plane Perpendicular to z AxisElevation:

Meridian Plane

θφN

20

Fig. 3.5 Free Body Diagram of a Rotational Shell Element

Contributions to : , ,F N N Nθ θ φθ θφ∑: , ,F N N Nφ φ θ θφ∑: ,zF N Nφ θ∑

21


______0=∑ θF

( ) ( )( ) 0

2sin

2sin

2cos

2cos

=++++

−++−+

dApNNdN

NNdNdNdNdN

θθφφθφθ

φθφθφθθθθ

αα

θθ

Since θd is small, we have 12

cos =θd

and.22

sin αα=

Thus, Eq. (3.6) becomes 02

2 =+++ dApNNdNd θθφφθθα

( ) ( ) ( )( ) 0cos 10101 =++

∂

∂+

∂∂ φθθφφφ

φθ

θθ

φθθφ

φθθ drdrpddrNddrN

ddrN

( )0cos 101

01 =++

∂

∂+

∂∂ rrprN

NrNr θθφφθθ φ

φθ(3.7)

Note that 01 =∂∂θr

, φd and θd are independent.

(3.6)

22


______0=∑ φF

( )( ) 0

2cos

2cos

2sincos2

2cos

2cos

=+−++

−−+

dApNNdN

dNdNdNdN

φθφθφθφ

θφφφ

αα

θφφφ

(3.8)

Since φd and θd are small, we have ,12

cos =φd

22sin θθ dd

=

and 12

cos =α

. Thus, Eq. (3.8) becomes0cos =++− dApNddNNd φθφθφ θφ

( ) ( ) ( )( ) 0cos 101

10 =+

∂

∂+−

∂

∂φθθ

θφ

θφφφφ

θφ

θφθ

φ drdrpddrN

ddrNddrN

( )0cos 1011

0 =+∂

∂+−

∂

∂rrp

NrrN

Nrφ

θφθ

φ

θφ

φ(3.9)

23


______

Eq. (3.11) indicates that shell element can resist the normal loadthrough its membrane actions provided that the curvatures exist without resorting to transverse shear forces and bending moments.

0=∑ zF

02

sinsin22

sin2 =++ dApdNdN nθφφ

θφ (3.10)

0sin =++ dApdNdN nθφφ θφ

( )( ) 0sin 1010 =++ φθθφφφθ θφ drdrpddrNddrN n

021

=++ nprN

rN θφ (3.11)

where φsin20 rr =

.

npθφ NandN

21 & kk

24


In summary, the three equilibrium equations for membrane forces in axisymmetric shells subject to general load are:

( )0cos 101

01 =++

∂

∂+

∂∂

=∑ rrprNNrNrF θθφ

φθθθ φ

φθ

( )0cos 101

01 =+−

∂

∂+

∂

∂=∑ rrprN

NrNrF φθ

φθφφ φ

φθ

021

=++=∑ nz prN

rN

F θφ

(E1)

(E2)

(E3)

25

3.5 Rotational Shells with Axisymmetric Loading (1)

In a number of important loading cases, such as dead weight and internal fluid pressure loading, geometrically complete shells of revolution have axisymmetric behaviour. Axisymmetric behaviour is independent of the variable . The loading, internal forces and deformations can vary with respect to .

Thus in this special case,→ All variables are independent of

→ as they would have produced unsymmetrical deformation with respect to the axis of rotation if they were nonzero.

Note that Eq. (3.7) is identically satisfied.

θφ

,0=θp ( ),φφφ pp = ( )φnn pp =

( )0.. =

∂∂θθei

θ

;0== φθθφ NN

26


The substitution of Eq. (3.11) into Eq. (3.9) yields

Multiplying by sin Φ and noting that r0 = r2 sin Φ leads to

( )0cos 10

121

0 =+⎟⎟⎠

⎞⎜⎜⎝

⎛++ rrpp

rN

rrd

Nrdn φ

φφ φφ

( ) ( ) φφφφ

φφ φφ

φ cossinsinsin 101000 rrprrp

ddNr

dNrd

n−−=+

(3.12)

( ) ( )φφφ

φφ

φ cossinsin

100

npprrd

Nrd+−= (3.13)

27


Integrating from to yields

Multiplying Eq. (3.14) by 2π leads to

1φ φ

( ) φφφφ φ

φ

φ

φ

φφ dpprrNr n cossinsin 100

11

+−= ∫

( )1

1

sincossinsin 0100 φφφ

φ

φφ φφφφφ NrdpprrNr n ++−= ∫ (3.14)

( ) ( )1

1

sin22cossinsin2 0100 φφφ

φ

φφ φπφπφφφπ NrdrrppNr n ++−= ∫

(3.15)

28

Denoting the total vertical component of all loads above the horizontal section as , Eq. (3.15) may be simply written as


Γ−=φπ φ sin2 0Nr (3.16)

φ Γ

1sin2 0 φφφ φπ

=− Nr

φNΦNΦ

r0

NΦ

Φ1

Φ

29


Thus the expression for at is given by

and from Eq. (E3),

Eqs. (3.17) and (3.18) are the equilibrium equations for rotationalshells with axisymmetric loading.

φN φ

φπφπφ 220 sin2sin2 rr

N Γ−=

Γ−= (3.17)

φπφ

θ 21

21

2 sin2 rpr

rN

prN nnΓ

+−=⎟⎟⎠

⎞⎜⎜⎝

⎛+−= (3.18)

30

Sign Convention (1) :

is the total vertical component of all loads above the horizontal section defined by and is positive when it acts in the same direction as the vertical component of .

is always taken to be positive

is taken to be negative if the two centres of curvatures O1, O2 are on the opposite sides of the shell (i.e. if the local region of the shell has anticlastic curvature) and positive if they are lying on the same side of the shell

Γφ

φN

2r

1r

31

Sign Convention (2) :

is positive if acting towards the axis of rotation and is always perpendicular to the middle surface.

is positive if acting away from the axis of rotation.φp

np

32

Consider a spherical shell of radius r, thickness h and subjected to a uniform pressure p.

Truncated spherical shell under uniform pressure

3.5 Example showing a shell resisting loads predominantly by membrane forces

33

3.5 Example - Solution (1)

By balancing the forces in the vertical direction, we have

(a)

where N is the in-plane force per unit of circumference. Note: N does not vary with respect to Φ.

In view of Eq. (a), the compressive direct stress σ

(b)

2sin202

0sin02 prprNrpNr ==⇒=

φπφπ

hpr

hN

2−=−=σ

34


The stress normal to the mid-surface of the shell is negligible and thus the direct strain, based on Hooke’s law, is

(c)

Under this strain, the reduced circumference is 2πr’ = 2π(r + rε) or r’ = r(1 + ε). The variation in the curvature κ is

(d)

( ) ( )Ehpr

E 211 ννσσε −−=−=

( )...11

1

11

1111

2 −+−−=⎟⎠⎞

⎜⎝⎛+

−=

⎟⎠⎞

⎜⎝⎛ −+

=−′

=

εεεε

εε

κ

rr

rrr

Now for the bending stress

35


Neglecting higher-order terms as they are small, and in view of Eq. (c), the curvature may be expressed as

(e)

The spherical shell bending moment M is related to the curvature by

(f)

Thus, the bending stress σb is given by

(g)

( )Eh

pr 2

1 νεκ −=−=

( ) ( )242

1122 ph

EhpDDM −=

−−=+−=

νκν

46

2

phM

b −==σ

36


The ratio of the bending stress to the direct stress is

(h)

It is clearly seen from Eq. (h) that the bending stress is very much smaller than the direct stress as h/(2r) << 1.

Therefore, the applied load is resisted predominantly by the in-plane stress of the shell.

rhb

2=

σσ

37

3.5.1. Spherical Domes (1.1)

Example 1: Determine the membrane forces in a spherical shell of constant thickness under its own weight of q kN/(unit area of its middle surface).

H

38

3.5.1. Spherical Domes (2)

In view of Eq. (3.15)

For the considered problem, , . Thus

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡++−=

=∫ 1

1

sincossinsin1

0100

φφφφ

φ

φφ φφφφ

φNrdpprr

rN n

( ) ⎥⎦

⎤⎢⎣

⎡−+−= ∫ 0sincossin

sin1 222

02 φφφφφ

φ

φ dqqaa

N

[ ]1coscos1

0sinsin 2

02 +−

−−=⎥

⎦

⎤⎢⎣

⎡−−= ∫ φ

φφφ

φ

φ qadqa

φcos1+−=

qa

(3.19)

0;cos;sin 1 === φφφφ qpqp nφsin; 021 ararr ===

(3.20)

39


Alternatively,

The vertical resultant Γ can also be determined through integration

( ) ( )22 2 1 cosq aH qaπ π φΓ = = −

( )φφπ

φπφπφ cos1sin2

cos12sin2 2

2

2

22 +

−=−

−=Γ

−=qa

rqa

rN

( )φπφφπφπφφ

cos12sin22 2

0

20

0

−===Γ ∫∫ qadqaadrq

40


Referring to Eq. (3.20), it can be observed that is always negative throughout the shell. Hence the meridional force in a dome under its own weight is always compressive. The hoop force , however, is compressive at the top but changes sign somewhere along the meridian and becomes tensile in the lower part of the shell. becomes zero when

( )

⎥⎦

⎤⎢⎣

⎡−

+=

⎥⎦

⎤⎢⎣

⎡+

−−=

Γ+−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+−=

φφ

φφ

φπφ

θ

coscos11

cos1cos

sin2 21

21

2

qa

aqaqa

rpr

rN

prN nn

(3.21)

o520coscos11

≈⇒=−+

φφφ (3.22)

φN

θN

θN

41


Observe that at , we have .

For a hemispherical dome, the absolute maximum valueof both membrane forces is qa.

It is interesting to note that ancient engineers were well aware of this structural behaviour of domes. When building domes with masonry materials, such as brick and stone, which are relatively weak intension, but strong in compression, they would confine their dome sector to the compression zone or for higher domes, would reinforce them in the tensile region. The hoop reinforcement would consist of wooden ties placed along parallel circles; when tied together they would form a closed strengthening ring capable of absorbing tensile forces.

0=φ2qaNN −== θφ

42


Stress Resultants for Constant Thickness, Spherical Domes under Selfweight

Problem: Determine the membrane forces of a submerged dome. The dome radius R = βH, depth of water is D = αH where the maximum height of the dome is H. Plot the variations of the membrane forces with respect to Φ for the case α = 2 and β = 1.

43

Example 2: (1)

Consider a spherical dome with its top sector been removed for natural lighting and ventilation. In such domes, a stiffening ring beam is provided at the top to reduce the internal forces in the shell body. The weight of this ring beam is applied to the shell as a uniformly distributed line loading as shown in the figure. Establish the membrane forces in this cut-out spherical dome of constant thickness under its own weight of and the ring load of.

2/ mkNqmkNP /

44

Example 2: (2)

φφπφπ daadrdA sin22 20 ==

( ) 02

0 sin2sin2sin20

φπφφπφπφ

φ

PadqaPaqdA +=+=Γ ∫∫( ) 00

2 sin2coscos2 φπφφπ Paqa +−=

( )φ

φφφφπφ

200

22

sinsincoscos

sin2Pqa

rN

+−−=

Γ−=

( )φ

φφφφ

φπθ

200

21

2

sinsincoscoscos

sin2Pqaqa

rprN n

+−+−=

Γ+−=

(3.23)

(3.24)

(3.25)

(3.26)

45

3.5.2. Conical Shells (1)

For a conical shell with an apex angle of ,

A more common space variable used is either or . They are related through

Thus,

α

∞==−= 1,2

rconstantαπφ

(3.28)

(3.27)

yy′ αcosyy ′=

ααπφαφπφ cos

2sinsin,tan,

sin2 00

=⎟⎠⎞

⎜⎝⎛ −==

Γ−= yr

rN

ααπαπφ cossin2sin2 yyN

′Γ

−=Γ

−=

46

3.5.2. Conical Shells (2)

Thus,

Equations (3.28) and (3.29) simplify the expressions for the membrane forces of conical shells. The simplification is due to the constancy of .

αα

φφπθ costan

sin,

sin20

2221

2yrrpr

rprN nn ==−=

Γ+−=

npyNαα

θ costan

−=

φ

(3.29)

47

Example 3: (1)

Example 3: Evaluate the membrane forces in an inverted conical tank filled with liquid of unit weight up to the depth d.γ

48

Example 3: (2)

( )αηρηαπηηπρ tan:,tan 222 === NotedddV

( )απαπηηαπ tan:,33

tantan 0

20

232

0

2 yrNoteryydVy

==== ∫

,32

31

20

20

20

⎟⎠⎞

⎜⎝⎛ −−=

⎟⎠⎞

⎜⎝⎛ +−=Γ

ydr

yrrp

γπ

γππ

(3.30)

hydrostatic pressure due to head of water above A-A

( )ydpNote −= γ:

49

Example 3: (3)

α

αγ

α

γ

απ

γπ

φπφ

cos2

tan32

cos232

cos232

sin2

0

0

20

0

⎟⎠⎞

⎜⎝⎛ −

=

⎟⎠⎞

⎜⎝⎛ −

=⎟⎠⎞

⎜⎝⎛ −

=Γ

−=

ydy

ydr

r

ydr

rN

Note that the maximum value of occurs when

Which gives ,

maxφN

223 0

d C yd y

dy

⎡ ⎤⎛ ⎞−⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ =

dy43

=

(3.31)

50

Example 3: (4)

Note that the maximum value of occurs when which gives

ααγ

α

αγφ cos16

tan3cos2

tan21

43

2

max

dddd

N =⎟⎠⎞

⎜⎝⎛ −

= (3.32)

(3.33)( ) ( )( )ydpNoteydy

pyN

n

n

−−=−=

−=

γααγ

αα

θ

:,costan

costan

maxθN 0=dydNθ

( )[ ] dyyddy

yydCd2102

2

=⇒=−=−′

51

Example 3: (5)

In view of this value of dy21

=

ααγ

α

αγθ cos4

tancos

tan21

21

2

max

dddd

N =⎟⎠⎞

⎜⎝⎛ −

= (3.34)

Distribution of membrane forces in inverted tankfilled with liquid up to depth d

52

3.5.2 Ogival Domes

An ogival dome is generated by rotating an arc of a circle about an axis of rotation which is parallel but away from the vertical radius of the arc by a distance, say as shown in the figure.

From geometrical considerations

r′

53

Example 4: (1)

Determine the membrane forces in an ogival dome under its own weight.

( )

( )

( ) ( )[ ]0002

02

0

sincoscos2

sinsin2

2

0

0

φφφφφπ

φφφπ

φπ

φ

φ

φ

φ

−−−−=

−=

=Γ

∫

∫

qa

dqa

adqr

( )[ ]( )( )[ ]

( ) φφφφφφφφ

φφφπφφφφφπ

φπφ

sinsinsinsincoscos

sinsinsin2sincoscos2

sin2

0

000

0

0002

0

−−+−

=

−−+−

=−=

qa

aqa

rRN

(3.35)

(3.36)

54

Example 4: (2)

[ ] ( )[ ]( ) ⎥

⎦

⎤⎢⎣

⎡−

−+−+

−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+−=

φφφφφφφφφ

φφφ

φ φφθ

sinsinsinsincoscoscos

sinsinsin

cos

0

0000

12

12

qa

rN

qrrN

prN n

(3.37)

55

3.5.3 Toroidal Shells

It is obvious that a normal to the surface defined by a meridian angle pierces through the surface at two points. There is not a one-to-

one correspondence between and a unique point on the surface.The shell should be treated separately in two parts, BAD and BCD. The subscripts e and i denote the exterior portion BAD and the interior surface BCD, respectively.

Toroidal Shell Geometry

φφ

r1e

r1i

r2e

r2i

56

Example 5: (1)

Establish the membrane forces in the toroidal shell under an internal pressure p.

Considering the exterior surface, one obtains from geometrical considerations

bar e += φsin0

ee

rrφsin

02 =

ar e =1

57

Example 5: (2)

Now

Thus

( )22 brp oe −−=Γ π

( )0

0

0 2sin2 rbrpa

rN

e

ee

+=

Γ−=

φπφ

( )

( )

2

sin2

2sin

00

0

0

00

12

pa

brr

prar

brpaprrN

prN

e

eenee

=

−=

⎥⎦

⎤⎢⎣

⎡ ++−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+−=

φ

φφ

θ

(3.38)

(3.39)

(3.40)

58

Example 5: (3)

Considering the interior surface, one obtains from geometrical considerations

( )iabr φπ −−= sin0

( )ii

rrφπ −

=sin

02

ar i −=1

59

Example 5: (4)

Now

Thus

( )20

2 rbpi −−=Γ π

( )0

0

0 2sin2 rbrpa

rN

i

ii

+=

Γ−=

φπφ

( )( )

( )

2

sin2

2sin

00

0

0

00

12

pa

rbr

prarbrpapr

rN

prN

i

iinii

=

−=

⎥⎦

⎤⎢⎣

⎡−+

+−=⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

φ

φφ

θ

(3.41)

(3.42)

(3.43)

60

Assignment No. 3.1 (1)

1. Determine the radii of principal curvatures r1 and r2 of an ellipsoid shell of revolution shown in Fig. 1 below.

Fig. 1

61


2. An elevated reinforced concrete conical water tank is filled with water up to level A as shown in Fig. 2. Establish the expressions for the membrane forces in the conical shell. The unit weight ofwater is 9.81 kN/m3.

Fig. 2

62

Assignment No. 3.1 (3.1)

3. Determine the membrane forces in a spherical tank of radius a, completely filled with liquid of unit weight and supported along its parallel circle at as shown in Fig. 3.

Fig. 3

γo1200 =φ

63


4. An Intze tank shown in Fig. 4 is to contain water up to the indicated level. The unit weight of the water is 9.81 kN/m3. Determine the membrane forces in the conical shell and the domed base as well as the axial force in the ring beam at the juncture A of the two shells. Neglect the dead load of the structure.

Fig. 4

64


5. An open ogival shell is subjected to its own weight of ofmiddle surface) and a ring load of of circumferential length) as shown in Fig. 5. Determine (a) the membrane forces inthe shell, (b) the reaction at the support and (c) the axial forces in the ring beams.

Fig. 5

2/(mkNqmkNqa /(

65

3.6 Shells of Revolution with Non-symmetric Load (1)

Shell structures can be subjected to loadings which are not axisymmetric. Examples of non-symmetric loadings are wind forces, earthquake forces, soil pressure on buried pipes and temperaturegradients in composite and metallic shells. To perform a membrane analysis of rotationally symmetric shells under non-symmetric loading, the 3 equilibrium equations must be used, i.e.

( )0cos 101

01 =++

∂

∂+

∂∂

=∑ rrprNNrNrF θθφ

φθθθ φ

φθ( )

0cos 1010

1 =+−∂

∂+

∂

∂=∑ rrprN

NrNrF φθ

φθφφ φ

φθ

021

=++=∑ nz prN

rN

F θφ

(3.44a)

(3.44b)

(3.44c)

66


If is eliminated from these equations by the substitution of Eq. (3.44c) into Eqs. (3.44a) and (3.44b), one obtains

If either or is eliminated from Eqs. (3.45), a second-order differential equation is obtained. However, it is usually simpler to solve Eqs. (3.45) directly.

θN

⎟⎟⎠

⎞⎜⎜⎝

⎛−

∂∂

=∂

∂−⎟⎟

⎠

⎞⎜⎜⎝

⎛++

∂

∂θ

φφθ

φθ

θφθφφ

φφppr

NN

rr

ddr

rN n

sin1

sin1cot1

12

10

0

( )φφθφ

φ φθ

φφφ

pprN

rrN

ddr

rN

n +−=∂

∂+⎟⎟

⎠

⎞⎜⎜⎝

⎛++

∂

∂cotcot1

10

10

0

(3.45a)

(3.45b)

φN φθN

67


For axisymmetric structures, the general loading can always be separated into two major components, viz. symmetric and antisymmetric with respect to the meridian plane θ = 0. The distributed loading can be expanded in terms of Fourier series as follows:

where , and are the symmetric and antisymmetric components (with respect to θ = 0) respectively.

nppp ,, θφ

( ) ( ) θφθφ θθθ nPnPp nn

nn

cossin01∑∑∞

=

∞

=

+=

( ) ( ) θφθφ φφφ nPnPp nn

nn

sincos10∑∑∞

=

∞

=

+=

( ) ( ) θφθφ nPnPp nnn

nnn

n sincos00∑∑∞

=

∞

=

+=

(3.46a)

(3.46b)

(3.46c)

nnnn PPP ,, θφ nnnn PPP ,, θφ

68


Corresponding to each set of loading components, there exists a set of stress-resultants as the solution to the governing equations (3.44). If a set of loading is chosen, say

( ) θφθθ nPp n sin=

( ) θφφφ nPp n cos=

( ) θφ nPp nnn cos=

(3.47a)

(3.47b)

(3.47c)

69


)0

(

sin)(1

=

=

θ

θφθθ

about

componentlsymmetrica

PPPlan

0=θ πθ =

θ0r

70


)0

(

cos)(1

=

=

θ

θφ

φ

φφ

about

pofcomponentlsymmetrica

PP

)0

(

cos)(1

=

=

θ

θφ

about

pofcomponentlsymmetrica

PP

n

nn

Elevation Elevation

φ φ

)(1φφp )(

1φnp

Plan Plan

)(1φnp

)(1φnp−

θθ0=θ 0=θπθ = πθ =

0r 0r

)(1φφp

)(1φφp−

71


The solution to Eq. (3.44) can be shown to be in the form

Hence the solution to the symmetric load can be expressed as

( ) θφθθ nNN n cos=

( ) θφφφ nNN n cos=

( ) θφθφθφ nNN n sin=

(3.48a)

(3.48b)

(3.48c)

( ) θφθθ nNN nn

cos0∑∞

=

=

( ) θφφφ nNN nn

cos0∑∞

=

=

( ) θφθφθφ nNN nn

sin1∑∞

=

=

(3.49a)

(3.49b)

(3.49c)

72

Simplified Wind Load on Spherical Dome (1)

Simplified Wind LoadActual Wind Load

ElevationElevation

Plan

P

A aφ

P

PPA

0=θ πθ =

2πθ =

23πθ =

θ

θcosp

φsinp

73


As the wind pressure is perpendicular to the shell surface,

The intensity of the simplified wind pressure on the spherical dome at point A can be expressed as

The loading is symmetrical about θ =0.Existing solution will take the form

(3.50)

(3.51)

(3.52a)

(3.52b)(3.53c)

0== φθ pp

φθ sincosppn =

( ) ( ) θφθφ θθ cos, NN =

( ) ( ) θφθφ φφ cos, NN =( ) ( ) θφθφ θφθφ sin, NN =

74


For spherical shell,

Eqn. (3.45) becomes

φsin, 021 ararr ===

( ) ( ) ( ) paNNd

dN−=++ φ

φφφ

φφ

φφθφθ

sin1cot2

( ) ( ) ( ) φφφ

φφφφ

φθφφ cos

sin1cot2 paNN

ddN

−=++

(3.53)

(3.54a)

(3.54b)

75

Derivation of Eq. 3.54 from Eq. 3.45

⎟⎠

⎞⎜⎝

⎛ −∂∂

=∂∂

−⎟⎟⎠

⎞⎜⎜⎝

⎛++

∂∂

θφ

φθφθ

θφθφφ

φφppr

NN

rr

ddr

rN n

sin1

sin1cot1

12

10

0

1 2 3 4

,21 arr == ,sin0 φar = ,sincos φθppn = 0== φθ pp

,cos)( θφφφ NN = θφφθφθ sin)(NN =

θφφ

θφφ φθφθ sin

)(sin

)(1

ddNN

=∂

∂⇒

θφφφφ φθ sin)(cotcos

sin12 N

aaa

a ⎥⎦

⎤⎢⎣

⎡+⇒ θφφ φθ sin)(cot2 N=

76

Contd.

θφφ

θφφ φφ sin)(

sin1)sin)((

sin13 NN =−−⇒

( ) θφθφ

sin0sinsinsin

14 papa −=⎥⎦

⎤⎢⎣

⎡−−⇒

paNNd

dN−=++∴ )(

sin1cot)(2

)(φ

φφφ

φφ

φφθφθ

⎟⎠

⎞⎜⎝

⎛ −∂∂

=∂∂

−⎟⎟⎠

⎞⎜⎜⎝

⎛++

∂∂

θφ

φθφθ

θφθφφ

φφppr

NN

rr

ddr

rN n

sin1

sin1cot1

12

10

0

1 2 3 4

,21 arr == ,sin0 φar = ,sincos φθppn = 0== φθ pp

,cos)( θφφφ NN = θφφθφθ sin)(NN =

77


To uncouple Eq. (3.54), we introduce

In view of Eq. (3.55), the addition and subtraction of Eqs. (3.54a) and (3.54b) lead respectively to

( ) ( )φφ φθφ NNU −=2

( )φφ

φφ

cos1sin

1cot2 11 +−=⎟⎟

⎠

⎞⎜⎜⎝

⎛++ paU

ddU

( )φφ

φφ

cos1sin

1cot2 22 −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−+ paU

ddU

(3.55a)

(3.56a)

(3.56b)

( ) ( )φφ φθφ NNU +=1

(3.55b)

( ) ( ) ( ) paNNd

dN−=++ φ

φφφ

φφ

φφθφθ

sin1cot2

( ) ( ) ( ) φφφ

φφφφ

φθφφ cos

sin1cot2 paNN

ddN

−=++

(3.54a)

(3.54b)

78

Note:

)()(: φφφ

QUpddUODE =+ … (a)

CdQeUeSolution pddp+∫=∫ ∫ φφφ: … (b)

Eq. (3.56a)

)cos1(sin

1cot2 11 φ

φφ

φ+−=⎟

⎠

⎞⎜⎝

⎛ ++ paUddU

φφφ

sin1cot2)( +=p

)cos1()( φφ +−= paQ

79

Note:

( )⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

⎟⎠

⎞⎜⎝

⎛ −+=∫

φφφ

φφφφ

sincos1sinln

sincos1ln)ln(sin2)(

2

p

Mathematical Handbook by Spiegel and Liu (pp92-93)

φφφφφ

cos1sin)cos1(sin

3)(

+=−=∫ pe

( ) ( )3 2

3

sin 1 cos cos

1cos cos3

pdQe d pa d pa d

pa

φφ φ φ φ φ

φ φ

∫ = − = − − −⎡ ⎤⎣ ⎦

⎡ ⎤= − − +⎢ ⎥⎣ ⎦

∫ ∫ ∫

80


The solutions to Eq. (3.56) are

Adding and subtracting Eqs. (3.55a) and (3.55b) give

( )φφNUU 221 =+

( )φφθNUU 221 =−

(3.57a)

(3.57b)

(3.58a)

(3.58b)

⎥⎦

⎤⎢⎣

⎡+⎟⎠⎞

⎜⎝⎛ −−

−=

⎥⎦

⎤⎢⎣

⎡+⎟⎠⎞

⎜⎝⎛ −

+=

23

32

13

31

cos31cos

sincos1

cos31cos

sincos1

CpaU

CpaU

φφφφ

φφφφ

( ) ( )φφ φθφ NNU −=2

(3.55a)( ) ( )φφ φθφ NNU +=1

(3.55b)

81



where

⎥⎦⎤

⎢⎣⎡

⎟⎠⎞

⎜⎝⎛ −++= φφφ

φφ

φθ3

343 cos31coscos

sinsin paCCN (3.59b)

2221

421

3CCCandCCC −

=+

= (3.60)

⎥⎦⎤

⎢⎣⎡

⎟⎠⎞

⎜⎝⎛ −++= φφφ

φφ

φ42

433 cos31coscos

sincos paCCN (3.59a)

82

The two constants of integration can be determined from the force boundary conditions. For the present case, can be evaluated from the condition that the forces at the crown, i.e. at ,

retain finite values. Since the denominator in Eq. (3.59) has a zero value of the third order, the non-singular condition requires that the values of the terms in the brackets and their derivatives up to at least the second order vanish. Setting to zero the second derivatives of the terms in the brackets of Eqs. (3.59a) and (3.59b) at yields respectively:


0and32

34 =−= CpaC

43 CandC43 CandC

0=φ φ3sin

0=φ

(3.61)

83

Note:

Consider the term in the bracket of Eq. (3.59a)

⎭⎬⎫

⎩⎨⎧ −−−+−=

∂∂ )sin(cos

34)sin(cos2)sin(][ 3

4 φφφφφφ

paC

φ2sin−

( )( ) ( )⎭⎬⎫

⎩⎨⎧ −−−−+−=

∂∂ φφφφφφφ

coscos34sincos3

342cos2)cos(][ 322

42

2paC

⎥⎦⎤

⎢⎣⎡ +−−+−= φφφφ 42

4 cos342sin2cos2cos paC

[ ] 0atvanishmust2

2

=∂∂ φφ

84

Note:

To ensure a finite value of at

0

424 cos

342sin2cos2

cos =⎭⎬⎫

⎩⎨⎧ +−−=

φφφφ

φpaC

papaC32

3424 −=⎥⎦⎤

⎢⎣⎡ +−=

φθN ,0=φ

[ ] ( ) 0toleadingvanishmust3.59bEq.of 32

2

=∂∂ Cφ

85



( )φφφφθ

φ3

3 coscos32sin

coscos3

+−−=paN

( )φφφθ

φθ3

3 coscos32sinsin

3+−−=

paN

(3.62a)

(3.62b)

⎥⎦⎤

⎢⎣⎡

⎟⎠⎞

⎜⎝⎛ −++= φφφ

φφ

φθ3

343 cos31coscos

sinsin paCCN (3.59b)

⎥⎦⎤

⎢⎣⎡

⎟⎠⎞

⎜⎝⎛ −++= φφφ

φφ

φ42

433 cos31coscos

sincos paCCN (3.59a)

0and32

34 =−= CpaC (3.61)

86


In view of Eqs. (3.44c), (3.51), (3.53) and (3.62a),

( )φφφφθ

φθ

423 cos2cos2sin3

sincos

3+−−=

+−=

paapNN n

(3.62c)

021

=++=∑ nz prN

rN

F θφ(3.44c)

(3.51)φθ sincosppn =

φsin, 021 ararr === (3.53)

(3.62a)( )φφφφθ

φ3

3 coscos32sin

coscos3

+−−=paN

87


Plots of the variations of the internal forces are given in the figure below. Note that are always in compression and is always in the direction against the wind load.

θφ NandN φθN

Elevation

φNφNa

0=θ πθ =

0=φ

2πφ =

φN φθN θN

pa667.0−

pa−

Variations of internal membrane forces in a hemispherical dome subjected to lateral wind loading

0=θ πθ =

2πθ =

23πθ =

φθN

Plan

88

3.7 Hemispherical Dome Supported on Columns (1)

A hemispherical shell supported on four columns is subjected to its own weight unit area of middle surface. The solutions are achieved by superimposing the solutions to the problems of Figs.(B) and (C).

/kNq

Plan

ElevationRing force (F/rad)

(A) (C)

(B)(A) = (B) + (C)

qq

column β

β

89


Consider case (C): Hemispherical shell subjected to boundary forces

The self equilibrated boundary force, can be expanded inFourier’s series as

φNa ′

2qa

βπ4

2qa− ββ2

θ0 2π

π 23π

π2

( )radFaN /'φ

90


whereθθφ nBnANa n

nn

nsincos

10∑∑∞

=

∞

=

+=′

01 2

00 =′= ∫ θπ φ

πdNaA

0sin1 2

0=′= ∫ θθ

π φ

πdnNaBn

{

⎭⎬⎫⎥⎦⎤+

⎢⎣⎡−=

′=

∫∫

∫∫

∫

−

+

−θθθθ

θθβ

πθθπ

θθπ

π

βπ

βπ

βπ

βπ

φ

π

dndn

dnqadnqa

dnNaAn

coscos

cos4

cos2

cos1

2/

2/

0

222

0

2

0

(3.64a)

(3.63)

(3.64b)

91


⎪⎪

⎩

⎪⎪

⎨

⎧

=−

==

,...12,8,4sin2

,....11,10,9,7,6,5,3,2,10

2

nfornnqa

nforAn

ββ

(3.64c)

Eq. (3.63) thus becomes

θββφ n

nnqaNa

ncossin2

..12,8,4

2

∑∞

=

−=′(3.65)

92

Note:

( ) ∑∑∞

=

∞

=+=

10sincos'

nn

nn nBnAaN θθφ

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡++−= ∫ ∫∫∫

−

+

−

β π

βπ

βπ

βπ

π

θθθθθθβ

πθθπ 0

2

2

2

0

2 coscoscos4

cos2 dndndnqadnqaAn

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

++−=−

+

−

π

βπ

βπ

βπ

βπ θθθβπθ

π nn

nn

nn

nnqa sinsinsin

4sin2 2

200

2

( )⎭⎬⎫

⎩⎨⎧ −−⎟

⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛ ++−= βπβπβπβ

βnnnn

nqa sin

2sin

2sinsin

42 2

93

[ ] 0sincoscossin2

;12

1 =−−+−== βββββ

qaAn

[ ] 02sin2sin2sin2sin4

;22

2 =−+−−== βββββ

qaAn

[ ] 03sin3cos3cos3sin6

;32

3 =−+−−== βββββ

qaAn

[ ]βββββ

4sin4sin4sin4sin)4(4

2;42

4 +++−==qaAn

ββ

nn

qaAn sin2 2−=⇒

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧++−== ∫∫∫∫

−

+

−

π

βπ

βπ

βπ

βπ

θθθβ

πθπ

dddaqdqaAn2

20

2

0

20 4

2;0

94

⎥⎦

⎤⎢⎣

⎡⎭⎬⎫

⎩⎨⎧ +−++−++−−−= βππβπβπβ

βππ

π 220

402 2qa

{ } 04

2 2=⎥⎦

⎤⎢⎣

⎡ +++−= βββββππ

πqa

⎪⎩

⎪⎨

⎧

⎢⎢⎢

⎣

⎡++−= ∫ ∫∫∫

+

−

+

−

β βπ

βπ

βπ

βπ

π

θθθθθθβ

πθθπ 0

2

2

22

0

2 sinsinsin4

sin1 dndndnqadnqaBn

⎪⎭

⎪⎬

⎫

⎥⎥⎥

⎦

⎤++ ∫∫

−

+

−

π

βπ

βπ

βπθθθθ

2

2

23

23

sinsin dndn

95

⎥⎥⎦

⎤

⎢⎢⎣

⎡+++++⎥

⎦

⎤⎢⎣

⎡−= −

+

−

+−

+

−

πβπ

βπ

βπβπβπ

βπ

βπβ

π

θθθθθβ

θπ

22

23

23

2

20

22

0

2coscoscoscoscos

4cos nnnnn

nqa

nnqa

( ) ( )⎢⎣⎡

⎟⎠⎞

⎜⎝⎛ ++−−++⎟

⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛ ++−= βπβπβπβπβπβ

β 23coscoscos

2cos

2cos1cos

4

2

nnnnnnn

qa

⎥⎦⎤−+⎟

⎠⎞

⎜⎝⎛ −− ββπ nn cos1

23cos

[ ] 0)sin(sinsinsin4

;12

1 =−−+−−== βββββn

qaBn

[ ] 02cos2cos2cos2cos4

;22

2 =+−+−== βββββn

qaBn

[ ] 03sin3sin3sin3sin4

;32

3 =−−+== βββββn

qaBn

[ ] 04cos4cos4cos4cos4

;42

4 =−+−== βββββn

qaBn 0=∴ nB

96


Solutions to Eq. (3.66) may be expressed as

become(3.44c)to(3.44a)Eqs.,0Since === nppp θφ

0sin

1cot1

2

10

0

=∂

∂−⎟⎟

⎠

⎞⎜⎜⎝

⎛++

∂

∂

θφφ

φφφ

φθφθ N

Nrr

ddr

rN

0cot1

0

10

0

=∂

∂+⎟⎟

⎠

⎞⎜⎜⎝

⎛++

∂

∂

θφ

φφφθ

φφ N

rrN

ddr

rN

01

2 =+ φθ NrrN

( ) θφφφ nNN nn

cos,...12,8,4

∑∞

=

=

( ) θφθθ nNN nn

cos,...12,8,4

∑∞

=

=

(3.66a)

(3.66b)

(3.66c)

(3.67a)

(3.67b)

( ) θφφθφθ nNN nn

sin,...12,8,4

∑∞

=

= (3.67c)

97


The substitution of Eq. (3.67) into Eqs. (3.66a) and (3.66b) furnishes

For spherical shells,

(3.68b)

(3.68a)

(3.69)

0sin

cot1

2

10

0

=−⎟⎟⎠

⎞⎜⎜⎝

⎛++ nn

n NnNrr

ddr

rddN

φφθφθ

φφ

φφ

0sin

cot1

2

10

0

=+⎟⎟⎠

⎞⎜⎜⎝

⎛++ nn

n Nr

nrNddr

rddN

φθφφ

φφ

φφ

φsin, 021 ararr ===

98


Eq. (3.68) becomes

Introducing the variables

0sin

cot2 =++ nnn NnN

ddN

φφθφθ

φφ

φ

0sin

cot2 =++ nnn NnN

ddN

φθφφ

φφ

φ

nnn NNU φθφ +=1

nnn NNU φθφ −=2

(3.70b)

(3.71a)

(3.71b)

(3.70a)

99


By adding and subtracting Eqs. (3.70a) and (3.70b), one obtains

The solutions to Eq. (3.72) are

0sin

cot2 11 =⎟⎟

⎠

⎞⎜⎜⎝

⎛++ n

n Und

dUφ

φφ

0sin

cot2 22 =⎟⎟

⎠

⎞⎜⎜⎝

⎛−+ n

n Und

dUφ

φφ

φ

φ

211 sin2

cotn

nn CU =

(3.72a)

(3.72b)

(3.73b)φ

φ

222 sin2

tan n

nn CU =

(3.73a)

100

)()(: φφφ

QUpddUODE =+

CdQeUeSolution dpdp+∫=∫ ∫ φφφ:

( ) 0sin

cot2:a72.3Eq. 11 =⎟⎟

⎠

⎞⎜⎜⎝

⎛++ n

n Und

dUφ

φφ

φφ

φφ dndp ∫∫ ⎟⎠

⎞⎜⎝

⎛ +=sin

cot2

( ) ⎟⎠

⎞⎜⎝

⎛ −+=

φφφ

sincos1lnsinln2 n

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

+=

2cos

2sin2

2sin2

lnsinln22

φφ

φ

φ n

2tanφ

⎥⎦⎤

⎢⎣⎡=

2tansinln 2 φφ n

2tansin2 φφφ ndpe =∫

φ

φ

211 sin2

cot ⎟⎠⎞

⎜⎝⎛

=⇒

n

nn CU

101

Adding and subtracting Eqs. (3.71a) and (3.71b) give


⎥⎦⎤

⎢⎣⎡ +=

2tan

2cot

sin12 212

φφφφ

nn

nnn CCN

⎥⎦⎤

⎢⎣⎡ −=

2tan

2cot

sin12 212

φφφφθ

nn

nnn CCN

(3.74a)

(3.74b)

0sin

cot2 =++ nnn NnN

ddN

φφθφθ

φφ

φ

0sin

cot2 =++ nnn NnN

ddN

φθφφ

φφ

φ(3.70b)

(3.70a)

φ

φ

211 sin2

cotn

nn CU = (3.73b)φ

φ

222 sin2

tan n

nn CU =(3.73a)

102

If there is no opening at the crown of the shell, the stress resultants must be finite at . This requires that . Thus Eq. (3.67) becomes


0=φ 01 =nC

θφφφ nCN n

nn

cos2

tansin2

12

,...12,8,42 ∑

∞

=

= (3.75a)

( ) θφφφ nNN nn

cos,...12,8,4

∑∞

=

= (3.67a)

( ) θφφθφθ nNN nn

sin,...12,8,4

∑∞

=

= (3.67c)

θφφφθ nCN n

nn

sin2

tansin2

12

,...12,8,42 ∑

∞

=

−= (3.75b)

103


For the hemispherical shell

Comparing Eq. (3.76) and Eq. (3.65)

2/πφφφ ==′ NN

θφ nCN nn

cos21

2,...12,8,4

∑∞

=

=′ (3.76)

nnqaC nβ

βsin4

2 −= (3.77)

θββφ n

nnqaNa

n

cossin2..12,8,4

2

∑∞

=

−=′ (3.65)

104


Eq. (3.75) becomes

Now

Eq. (3.78b) indicates that the values of at the boundary generally exist. This has to be taken into account in the design of the support either by providing a ring girder or otherwise, the shearing stress resultants be removed. The latter requires a general solution taking bending into consideration.

θφβφβφ n

nnqaN n

n

cos2

tansinsin2

,...12,8,42 ∑

∞

=

−=

θφβφβφθ n

nnqaN n

n

sin2

tansinsin2

,...12,8,42 ∑

∞

=

=

φθ

φθφθ

NN

NNNrrN

−=⇒

=+=+ 01

2

(3.78a)

(3.78b)

(3.78c)

φθN

105


Derive the expressions for the membrane stress resultants in a spherical shell supported on k equally spaced tangential columns and subjected to a live load p per unit area of horizontal projection as shown in Fig. 1. The boundary of the shell is defined by Φ = α and the angle, measured on the horizontal plane, subtended by the column width is 2β.

106


107

Assignment No. 3 (1.3)ρ

)(aCase

β

β

αsina

αφ =

)(bCase

)(cCase

columnsspacedequallyk

)()()( cCasebCaseaCase += --- (1)

108

Volume of Solids of Revolution (1)

(a) Axis of Revolution: y-axis

(b) Axis of Revolution: x-axis

Show that the volume of a hemisphere of radius R is .Hint: x =

3

32 Rπ

22 yR −

109

Surface Area of Solids of Revolution (2)

(a) Axis of Revolution: y-axis

b) Axis of Revolution: x-axis

Show that the surface area of a sphere of radius R is .Hint: x =

24 Rπ22 yR −

mroc3st05

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