Design of Reinforced Tank(Hilal)

5 8-l2m

from top

22.8

28.5

cm2

11.4

l4.~

cm2

7 ¢16

m:n/m

3.5 ¢16

I I

+

3 .5¢! 13

i

1 x 12 x 8 x 0.4­ 7.5

:he fixing moment at the base is given by

= 5.1

m.t.

:.:a.:(. thickr.ess of wall required to resist this moment can be determi­ ~ei fro~

equation (9) as folloW3 :

t~ax = V';df

/ 3

=

V5100 /

~

-

41 em

22

i.e.

the maximum chosen thickness of 40

c~

is convenicnt.

Vertical tension stcel on the inside surface of the wall can be determined from equation (10), :::

= 0.26.x: 41

0.26 t

thus chosen

=

Details of Reinforcements of a Deep Circular Tank

-,.....

~

.

.

o c8m

6 q> 16 ( 12 cm2)

e

~ ~

f

rt)

:;

...

0 0 C'oI

E

8 N

e

o

o

~

j

%

I')

6.-e"ln/ m

.... ~

6 ~ 8 "'nIm.

I')

+ !Q

....

E

~

~ .

~

€ lD

..;.

...

e

8


l

][

­

... ~~~

~

_... 10040'

~

_~--

\J/

at its base

-

..

• ~

5e

•

·c 0

;

~

a::

---.~

~_. 6,,10 m""I'l ' \[7

Fig. IIl-6

23 Other vertical reinforcements necessary ·to bind the rings and to re­ sist shr-Lnkage stresses and eventual field bending moments may be chosen:> 20 % of max. ring reinforcements and not less than 5¢l 8 mmlm. u~

In our case wall

t~ickness

~

=

Uf/5

5100/5

at position of max. field moment

= _..;:1~02::.:0=--_ 1300 x: .30

=

=33

1020 kg m

ems

= = 14 x 0.2 = 2.8 cm2

20 % of max. rings ( 7

Choose 6

The details

of reinforcements in the lower part of the wall are shown in fib\U"e IIL6. :~

IIL3. THEORY OF THICK CYLINDERS

If the thickness of the wall t of a circular tank is not small in prci.'ortion to the radius H, the stress crt due to T will not be uni­

formly distributed over the cross-section. Lamb has solved this blem in 1852 as follows

pro­

Fig. IlL7 f, .. df,

ft dR

~ftdRd\jl Fig. III-7 The stress distribution

at

" r!It

on the cross-section of a thick wall

of a circular tank subject to hydrostatic pressure p can be determined if we consider the equilibrium of an element of' a tauk: enclosing angle d'll as shown.

Ass\IDe

H

= radius of any fiber inside the wall.

Ri

= radius of inner face of wall.

Ro

=

radius of outer face of wall.

"

an

24

Or

& E:

Vt

&

~

r

E

t

= 1m

= radial stress and strain = tangential stres::; and strain

1 1 = Poisson's ratio = 5" .;. b for concrete

Due to symmetry in shape and loading we have in tangential direction

the shearing force

in radial direction

dOt

=0

Q

= 0

and

Due to equilibrium in radial direction, we get the following relation : ( Or + dO~ ) (R + dR) Ur R dljl

+

dO'r

d'i' - Or R d~ -

+ Or dR dljl

R dljl

- at

dR dljl

Reduc ang by dljl end neglecting

cond degree,

=

at

=

dB dill

+ dar dR dIP

0

- err Ii dljl­

0

dO'r dR being a small value of the se­

we get : dar R +

"r dR

at

=

= crt

dR

dear R)

(15)

dE

According to theory of elasticity, one can express the relation bet­ ween stresses and strains in the following manner

e:r = -1

( iJ r

~at

1

( O't -

~Or )

E

E:t = -E

?

)

l

'(16)

These i;wo equations give Or

E = ~(Er HEt 1 -~

°t

E = ----..:' (E t 1 -.r

)

+~~ )

) )

~

)

(17)

Due to 1ihe deformation of the wall of the tank caused by the hylU'osta­ tie pressure, the radius Ii will be increased by the radial displacement

y

i.e. it will be ( R

c.irect~on

Therefore the strain in

+ J)'

t~e

rUdial

E r is given by

(18) ':Ie

have further ;

= = =

Circumference of tank before deformation Circumference of tank after deformation Increase in length of circumference

2 rt R

21t(R+y)

2" y

Strain in tangential direction

E:t =

y/

2 It

(19)

= y/R

Et

Substituting the values of 19

or

2 It R

in equations 17,

Cit

as given by equations 18

E: t

(....s;r

E

1

-.,2

:r ) ) ) )

R

1- + ~~ )

E ( 1 _~2

=

+ .)

dR

em

R

we get

~s!z =--!!(R9J:+-,y)

+

1.+

.)

R

&

=

+ R

dR

dR R2

9::£.

d2~ dR

+

or

dR

dR

.dR

R

(20)

~

Introducing these values in equation 15, 1..

&

we get

=

Vr

and

Er

2 9 dR

R.9ii: dR

+ .}

9:I.

i.e •

dR

_

s

= 0

(21)

The solution of this differential equation is (22) in which

01 and C 2

are the intesration constants. Differentiating equation 22

with respect to R,

we get :

26

Substituting these values in the elements of equation 21,

we get:

Which means i?hat the solution is correct.

Deter~tion

i)

For

R.

ii) For

R

of

= R.:L = Re

01

& 02

Or

=-

p

Or

=

0 (no

Fig. IIL8

outSide pressure)

therefore

Or =

'.

=

Fig. III-8

E

1 _~2 E

1

(

~+

z

~

dR

R

2 [ °1

( 1 +

=

)

e)

-.)

-

c

E

1

°2 :::-2" R

( 1 - ~

we have

o

(1 - ~ )

- P

=

i

7"

-~

Because of condition ii,

Because of condition

2

2 ( °1 -

+ e

°1 + ~

c

2)

I?

)J

and

we have

E 1

-i

These two equations give °1 (1+,)) ( 1 -

R~

-::t)=p R

and

°2

=p =P

1 E

R2 i

e

1 + :? E

70

R2 0

R~-

R~:L

Rf

R~:L

IT

E

o

°1

~2

1 ­

or

0

R~:l.

(23)

(24)

27

Therefcre O't

=

= =

:::

( il. + R

1 -,~ 2

r

E

1

LC1

-::l '"'

_~c.

=

~

( 1 + ~

P

R20

R2i

[

·E

-l

1

°2

)

+~

R

R2

E

or O't

d.R

[ P 1 -

E

1

&.) =

~

o

,,2

tt

( 1 -

~

C 22

(1+"') +

i

R

)

]

yJ .,

R2

i -

C2

+ ~ (C1 1 + -:::2 R

C

Ri

J.....ti.. H~ ~ ;.> R

E

H- - R2i 0

(l-~)J

2

2

( 1 +

- Ri

R -;T ) R

(25) Hyperbola

.lhich gives the formula of Lame for stress distribution on the cross­ section of thick cylinders subject to internal pressure p.

There fure

the stress at the inner surface of the tank : 2 R +

o

R~~

(25a)

=P~2'

Ro - R.~

and at the outer surface:

=P

R

Z 2 R i R2 _ R~

o

Assuming

Ro

we get

crt o

)..2 _ 1

A

=1 = 1.2

;\=2 ~hc

.. -."

....

=P

2

-...."...-? 1 :\-

-

"

"

:1

"

=

.

cr'ti and - at o

1

,2

= "

+ 1

2

ab:bc

1.22 difference not big

0",

stresses are Graphically represented

in fiB;ure III. 9

~

(25b)

~

= A Ri

2 A + 1

For:\

= Ro

?i;. 1II-9

28

For

R

R~

~

= co

=

1/2

CT\;O

This means that the stresses are bigger at the inside surface of the tank. In reinforced concrete circular walls with one mesh , the

ring

reinforcement is to be placed nearer to the inner surface of the wall and in walls with double meshes the inner rings may be bigger than the outer one s . In reinforced concrete circular tanka Ro

< 1.2

Ri ( 1. e.

= Tit)

which means that the error by use of cylinder formula ( crt less than.

x < 1.2) is

10'%.

For very thin cylinders

R

i

::: R o

:!

R

the formula of Lam€l

gives

the same result as the cylinder formula.

R~

) =p

~

2 R

t

i

Ri

=P

t

=

:/hereas in relatively thick plain concrete circular walls,

! t th~

avera­

ge tensile stresses calculated according to the cylinder formula give serious not allowed errors. III.4.

TANK WALLS FIXED TO FLOOR

111.4.1 Theory of Reissner and Lewe

Fig. III.10

D = 2R

I

! ...

\X I

H

----

'iill~l-t liq:.l.iJ. prcs suz-e ,

~='''''E j--r:,-..:.x:-'.-'• .:.;;....-­.~

Fig. Ill-IO

t""-V~

29 The liquid

press~

Px at any depth x will be counteracted by the

COlllbined resistance of rings and cantilevers,

thus (26)

Pr

=

in which

the part of the_liquid pressure resisted in the horizontal di­ rection by ring action.

Pc

=

the part of the liquid pressure resisted in the vertical direc­ tion by cantilever action.

The value of Pr can be determined as follows :

t =

i

Strain in tangential direction

E

Stress in tangential directicn

at

=E t

E

Ring ten.::; ion

T

= at

tx

Pressure Z'eaisted by ring action

p

=

r

(refer to equation 19)

=

:I.E

(27)

R

= y. t R

(28)

E

x

1:.

R

This equation means that i f we know too equation of the elastic line of the wall. we can find P r and respec:tively Pc • The of the elastic line can be determined as follows :

equation·

It is known from the general theory of elasticity that the rela­ tion between the external force q, the shearing force Q. , the bending moment M and the deflection y are

6~ __ -

M/ E I ,

dx

in which

dM dx

q

=Q

= wxR d'4l

- at t x d'4l

q

= wx:::\ ill

-

I

=

d'~

12

to'x

~=-q&4=

dx

q / E I

(0)

d.x

= force acting on wall in radial ( Fig. IlL 11) •

q

R

4­

t

bu'';

at

1.E t dlj) = ( wxR- 1.. t x R x R

= 1.E R E ) dlj)

direction

then but

Fig. III-ll

Substituting these values in the differential equation of elastic line - 30 R dljl t 3

x

E

we Bet :

Z t E) dljl R x

( wx R d4-

w R2

dx

E t

x

A.ssuming a tank with constant wall thickness .n

=4F

+

y -

12 E4­

:

or

~=-x-y

12

K

the

~

.-4

H4-

-;;;:r

dx

4­

R2

~

and

w . t

x

E

= =

t

and putt ing

~

we

(31)

get:

H

(2)

0

This equation gi. ves the final form of the differential

equation

of the elastic line. Its solution gives the value of :: and respecti­ vely Pr

and

Pc

or the ring tension T and the cantilever

moment M.

The solution of the differential equation 32 is given by

y

=

w R2

3: t

+ ";'2

) cos nf + ) sin

in which A

) (3)

~

& B are 4- integration constants to be deter­ l 2 from the conditions at the supports. The derivatives o~ y are

l

~ined

n~

)

given by

I

' A

2

,B

. -.... ,..,~

- It.2

-

\ I

(B 1 e ~g + B2 e-n f ) C03

d2

H2

~ =-

H3

d3.}'

~.

~.

~=

(A en~ - A e-ne) sin l 2

(A

l

en~ +

+

w a2 H E t 11

n~

+ (n~ Dl e

n~

- B2 e -n~) . ~ coz

A.:2 e-n~) sin nf - (A eng - A e-n~) l 2

C03

n~

n'f

R2 = - (y - w ---- x)

E t

which means tL:..t

~~Je

s o Lutri.o n Ls correc t.

'.rhe intesration co necancs

f~!'

an open t ank t'Lv::ed at t;;e

baa-s c ar, be

deter:uined fro:u tl-r.> following co::.ditions : a)

at base x = Ei

b)

at top

Y

1.: = ~ =

x = 0

From condition b

= Co • S! dx

;'.: ec

2

d "

:ix

2 o d-, = I d)t

or

2= 0

= 0

~~3

= 0

~A

13

1

-

=0

32

and

r.'3'

Q oc .:::.....,.f = 0

or

dJ~

~':, ::::r:

t::e~e

:'8 La t

i.ons

',':e

-Q :::""'

+4

~l

= E2

(11.

1

-

and

l,.2) + (3 1 + B2) .= 0

A1 = ~

-2

";)

-'1

• thus

nr +:a 1 ( e" R

y =

) ..

2

+ \'/ - -

sin n

+ e-nE

or

X

E t

A (enf + e-n~ ) cos n~ +B [ (en~+ e-n t ) 1 1 nt . cos n~ sin nf - 2

e-

J

\7

R2

+--x E t

and

n~J +

sin

a. t

tr:e bas e x

= H,

~

=1,

li

> 10

Y

" Ass1.Uning li

"" 1

R

Y

:_:1:

n

= 0 = A1

e

n

t

cos n + B1 e

. £:.Y. = 0 = A1 dx

= 0,

~

= O.

...

n

=

and

e -n

dx

1/3::X: 1 x 100

= 0.018

e - 4-.2 n

2 w R H E t n

sin n +

= 4-.2

, neglected I

w R 2 H

and

t

E

2

en (cos n - sin n) + B1 en (sin · n + cos n) + w R H Etn

The solution of these two equations gives

11.

w R2 H· (cos n + sin n- ~) = 1 E t en L 2

) ) )

(34- )

~

B = + w R H (cos n - sin n -~) )) 1 E t en n

I

~ubsti tuting

these constants in the equation of y, we can find

rino tension and respectively the load distribution.

the

33 The second derivative of y gives the bending 1D.0ment because

~ =dx

.

M ,

E.I

The moment at the toot can be

For x

=H

M

2 n = fiT" 2 n2

=7

E I ant

[­

E I

-[ 2 n

2

(A1 sin n - B1 cos n)

w R2 H (cos n E t 2

wR

H

=-

lI

=- 2

(cos n - sin n -

~)

=- E I

J

cos n

J.

n

cos

n - sin n cos n _

t3

W

12

t H

R2·

n)

n

. but

I

3

t =-12'

·or

or then

!L=.-.1

or

'n

2 n (n - 1)

Calculating the max. shearing force at the base Q

sin n

w R2 (cos n sin n + sin2 Sin2 n n - """'""=--= t H n 2 cos 2

I.

w R2 t 2 12 H

=

" ::11'

n2

sin n)

sin n -

n

2 1 ) 2 W R ( 2 n ' I t r 1 -n

'"'f' f

+

E t

+

.

n~

we get: 2

=-

from the

relation cos

f

determined

~

from the

rel~tion

we get : :...' .

o

. ~2 t2

_.7 -,

"'maX -

12 H2

•2

n 2 (2 n - 1) .

In the same way other cases can be calculated ed 6e conditionse.s.

(6) acco~iIl8

to

their

;4 i

)

ii)

the edge conditions for an open ta.'lk hinGed at the base axe at base

x :: H

y = 0

at top

x

=0

M = 0

and

.

&

Q

=0

M

=

or

d2

~ =

or

0

d 2 'V '

'

.~=

0

dx

0

3

an~B = 0

the edge conditions for a wall fixed at top & bottom are at base

x

=H

y

=0

and

at top

x

=0

y = 0

and

2;i dx ~ dx

=0 =0

The shape of the elastic lL?1e, load distribution curve aDd the ben­

ding moment diagram for the case of an open tank fixed at the base are as shown in

~ig.

111.12

H

\

Elcstic- line

Bending moments

Load distribution Fif,. III-12

H the wall vtes:« fixed at its upper ed,se, the main ordinates 0:1: the

diagra~s

will be practically not affected, and a small fixing mo­

ment - shown dotted - will be created at the top. Nben exa.miuing the load distribution curve. we n,otice that

the

pressure at the bottom of the vIall is res:Lsted by the cantillever ac­ tion &ld as the deflection at the base by full fixation :Ls equal

to

z.:ro, therefore the load resisted horizontally by rint5 action must be equal to zero. In the upper portion of the wall Pc assumes negative values. If no such load of

op~osite siG~

were acting near the top

the wall. the cantilevers should the"'", r',: c ea v.e

thei~

of

r;reatest 'deflec­

35 tion, but obviously this cannot be the case owing to the

restraillt

afforded by the rings, and ill consequence of the load being zero the water level the deflection

~or

at

the upper most ring must be nil.

Dimensionillg : The maximum bending moment at the foot of the wall causes ten­

sile'stresses on the water side., In order no e to have cracks causing rusting of the steel, the working tensile bending stress in concrete must not

exeed its tensile strength i.e•. for a rectangular section with breadth b = Lm , crt b = 6 Mf / t 2 Substitutillg for Mf the value given in equation 35 0't b

n

2 R

= ilK 4

or

- . n (n - 1 ) H

= 0.5 +

n

Knowing that

=w

we get:

v

a 'H

.

~ + 0.25

RG

w

(37)

.

x:

where .

=y;

or

R n

t

"

then

. (38)

= 1.73

Reissner &: Lewe found that all tanks With equivalent

values

of

K ( or n ) have similar load distribution curves. Accordingly they .

.

gave a series of curves far load distribution and respectively

the

ri:c.g tension as well as the cantilever moments in walls of open cir­ cular tanks fixed to floor and having rectangular or triangular sec­ tion as a factor of K or n. in. the following manner ( Figure Ring tension. Bend Lng m.oment

T

= Co

wHR

= Co

P R

2 2 / 12 H M = 15 w R t

where

Co

=y

~ A-

III.13 ) t

.t

.. ~

3

where

(3

2t 4 = d~ • K-;:·7x

36 and

in which

In order to use these curves proceed as follows :

Deteriine the value of n from equation 37, namely

1)

~

= 0.5

+

l

otb H

~

R w

2 18 kg/cm for normal cases

O'tb

c:

2)

Calculate

and

+

w

0.25

= 0.001

in which

kg/err?

for water •

the thickness at the foot of the wall from equation 38

which gives t

= 1.73

then choose a convenient profile for the wall. 3) To determine the ring tension in the'wall draw for the calculated

.value of ( n ) a vertical line to meet the projection

line

for

interpolation in a point - a -; through - a - draw a·horizontal line to meet the maxima curve in - b -. This last point - b ­ gives the apex of the distribution curve i f the wall were rectan­ gular. For trapezoidal walls, linear interpolation is sufficient.

Having drawn the load distribution curve, the ring tension in eve­ ry section can be determined from the relation : T

c:

Co P R

4) To draw the bending moment diagram, it is recommended to determi­

" "'f=-

from equation 35, namely:

~

ne the maximum fixing moment

2

wR

t

2

12R

• 2n(n ­

1)

,

In the same way·as given under (3) determine the apex of

positive part of the bending moment diagram for a rectangular

the wall.

I f the wall we~~ trapezoidal multiply the maximum positive value

( t x I t)3

where t x is the thickness of the wall at pos~tion of

by

OPE)I CIRClIl.I1R wI/as

FUEO

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l:::-::::: J:=.:::'

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o~p

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w. W£I(fltT P'- Llt?/J

MOMEMTS

EEIJD/IJQ So

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ol1

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"1' "

tJ

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t

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n­ :Pig. 111-13 l

3

"

S

37 maximum positive moment. fhe curves show that the cantilever

action

increases for diminishing values of K ( or n ) i.e. the scaller

the

height, the big.;er the radius and the stiffer the wall. By R =

,

cz>

K = 0 ( the limiting case of a straight cantilever) the water pres­

sure is resisted totally by cantilever action.

Example It is required to design the wall of a cylind.erical reinforced concrete open water tank, 5.0

IDS

deep and 20

IDS

diameter.

Assume the

wall to be fixed to the floor. Solution

H

= 5.00

n

= 0.5

R

IDS

= 10.00 me

_! O'tb Y'-R""'2<-=--- + 0.25 II

+

=0.5

w

t =

l~ x 59 0

+

1000

-

ClllS

T

max

~

3-.53 Co

'II

H R

t

= 3.53 = 35

ems

and at bottom 35 ems.

From the curves of Reissner ( Fig. III.13 ),

= =

+ 0.25

or

1.73

Choose thickness at top 20

for n

x .001

= 0.5 and = 0.4-25 X 1

=

Co X

5

X

find

at

Tma.x '

= 21.25

tons/m

0.4-25 10

~

thus

acting at 2.5 ms from top surface. The max. ring rei..n:foreement is given by

max. As

=

chosen 12

Tmax as ~

=

21.25 1.4

13 mm/m ( As

=

15.2

em2;m

=

15.9

em2 ) on both sides Le. 6

on each side. Thickness of wall at position of max:. ring tension given by : t

x

= 20 ~ 35 = 2

27.5 ems

is

;8 The max. tensile stress i.n concrete taking shrinkage i.oto con.sidera­ tion is given by °t E s h

= 0.25 =

mm/m

E

s

= =

Tmax + E:s h Es As . A + n As

c

2100 t/cm

2

i.n which

n

Be

21250 + 0.00025 x 2100000 x 15.9 27.5 x 100 + 10 x 15.9

=

= 10

then

21250 + 8350 2909

= 10.2

kg/cm 2

The m.x. fix:ing moment is given by

=

w R2 t 2 . 2 n ( n - l )

2 2 1 x 10 x .35 12 x 5

=­

12 H

( 3.53 - 1) ~~en

x

2 x

= - 3.67 m t

calculating the necessary vertical reinforcement

in

wall of a tank, the effect of the compression due to the own of the wall may be

neglec~ed

the

weight

without a£fecting the sa:rety of the wall

as it causes a reduction of the thickness of the concrete

in sections

sUbject to ten.si1e stresses on the water side and reduces the area of the tension reinforcement which can be determined at the foot

of the

wall as follows "t

=

35 cms

32

=:

kl

Assuming

-­ 11r3670

2, \(g/cm

°c = 31

2 kg/cm

~ccording

=~ =

to equation 10,

n

= 15

ness

and

3670 1284 x .}2

a =0

and and

~

we get

2 = 1284 kg/cm so that

= 9.0

cm'2:

= 9.1

2 cm -the same result !

chosen

7

~

l} mm/m

we get

AS = 0.26 t = 0.26 x 35 ~'or

= 35-3 = 32 cms

k l = 0.53

i.e.

as = 1400

As

d

therefore

max. positive bending moment, we get for a wall of constant thick­

and lies a-t;· ~ =

P = 5.5

e'.6

= 5~5

where

=29

tx

C~,

2 2 x 1 x 10 X.O.35 -- ·x 12 x 5

t~erefore

~

(

35

= 0.64 m.t.

26

= kl

~

As

=

640 2 = 1.9 cm 1300 x .26

k

= 0.975

l

ac

~

low,

chosen

= 1300 5

¢l

8

)

kg/cm-

=/m

It is however possible to construct the wall with constant thick­ ness of 25 ems and provide it with a haunch 10 x lj.O ems at base. this case the effect of the haunch on the ring tive bending mom.ents may be neglected

te~sicn

and the posi­

determined in tt.e

~~~

II:.

follo'Ni~g

manner : For t

= 25

ems 12 x 54­

=1200

and

n

4-: =Vr:;;;: K/4- =V 1200 / lj. ~

= 4..16

.,

For max. ring tension, the curves of Reissner sive = . 0.48

&

~

_= 0.• 545

= 0.48 X 5 x 10

= 17.2

=

cm2 /m

2 chosen 7 ¢ 13 mm/m on each side ( 18.5 cm

= 24-000+

(J

t

= 24000 Fo~

0.00025 x 2100000 x 18.5 100 x 25 + 10 x 18.5

+ 9750 2680

= 12.6

max positive bending

kg/cm2

moment

= 4-.16

and

tons

-

f1 ~. I

e

E_ I)~ ~ ~ , ei .

­

""

.....

l"j..

fJ ~:

.g.,.go.

lD!

G:

i

'

e~_ ~ .e­ -

....

_.

lieI~ ! I

I

we get : For n

24-

I

qj-9­

:

-,..,

ll~

II n !'I "\

-l-. T eol~ 0

~
III-l4­ 0.10

~ ~~ Jr~:zt!"'I L, \!t13 ~~;;~=.::~

4-0 1 x 10 2 X 0.25 2 12 x 5

= ~he

740

1300 x 0.22

=

=

mt

0.74

chosen 6

2.6

9

mm/m

8

details of reinforcement are shown in Fig. 111.14

111.4.2 Simplified Methods for Determining the }<'ixing iJoment,

th~

Shearing Force and the Thickness of the Wall

base.

at the 2

R = L 73 .~

I t has been proved in equation 38 that. t

from

R n

which one can determine the value of n , n

=

thus :

1.315 H

~

The max. i'i:d.ng moment is given according to equation 35 by the relation : 2n(n-1)

SUbstituting for t the value given in equation 38,

mf

Assuming we get

2

2

=-

wR

=

wH (2 R) t

°1

=

?J.

=

f

2

we get

(1.73~) 2

t

R ne:. .

12 H

n (n -

1)

n - 1 n

6.95 6.95

1.

1 _

n wB D t

(40)

°1 The shearing force at the base of the wall is given accor _

ding to equation 36 by the relation :

~=

w R2 t 2

12 H2

But acccr-dang to e quatron

38

• 2 t2

2 n

= ~ R

(2 n - 1 )

a:

n

41: Substituting this value of t 2, in the equation of

Assuming

2 n - 1 n 2

and substituting,

= C2

~ ,

we

ge-c

we get

:;::~

w H2 ~ = C2 2

(41)

It is further known that the bending tensile stress at the foot of the wall is given by

Assuming crt b = 18 kg/cm 2 the value given in equation 40, 180

=

AssUl'ling further

6

VI

H D

~

Cl

t

=

6 ;'If

=

: O"tb

t2

180 t/m2

= Vie

get

12 w H R

Cl t

,

_1_ '15 Cl

=

and sUbstituting for ill f

C: 3

or

j;h€:n

, t =03

In order to get t in ems for w in t/m3 (equals. 1

"

wHR

for

water) ,

H and R in meters, we should have : (42)

in which C

3

,

= 100 C3

=

_;;;;.10~0",-_

15 Cl

The values of r.lf ' ~ and t at the base of a wall fiXed to the floor as given by equations 40, 41 and 42 depend on the value of n which in turn depends on the magnitude of the thickness t. In this manner a method of trial and error is to be applied i.e. we have first to assume a convenient value for t, then determine the corresponding value of n according to equation 39. Applying equation 42, it is pos­ sible to check the assumed value of t. It is however possible to avoid these trials in special e.g. water tanks in the following

~er.

Equation 3?'gives :

cases

'42 /'

Assuassuming

-!Otb H ~ R w

n

= 0.5

+

°tb

= 180

t/m2

n

= 0.5

+ f180

w

and

:2

is a factor of. ~

n

0.25

+

=1

0.25

+

tim}

,

which

then means

that

no; H

or

7

R

The following table and curves, fig. 111.1.5 give the values of 01 ' 02 & 03 cs factors of n in general or as factors of H/R2 for the special cases in which w

=1

LW-t f

t/m3 ( e.g. water tanks ).

I

I Ii IIII I I

I

I

'I I II

i .I' III II~ 'Jlllll i Iii I II I I Tit-Tf.Nr-l.

lj

'rH-~ I 1

1

I \ Ho'-l.

1'1

1 I

4 I

I,

I

II

I

2

i

Fig. ill- 15 1).1

(JJ

'2 /R

i I

1,4

1.2

1,0

0.8 10.6

,

16.4 15.2 13.9

0.2

0.1

0.08 0.06 0.04 0.02 0.01

I

I

n

0.4

/.11

12.5110.9 9.00 6.5014.77 4.32 3.82 3.2} 2.46 1.9}

\

.

°1 7.40 7.44­ 7.49

7.55[7. 6517.82 8.22 8.80 9.04 9.4

I

10.1 11.7 14.4

I

°2 .118 .127 .• 138 .154 .175 .211 .284 .375 .409 .455 .525 .646 .769 °3 .900 .895

.890 .882 .870 .852 .812 .756 .736 .708 .663 .570 .462

Circular tanks may be divided into 3 categories :

a)

Deep tanks with big depths, small diameters and thin walls, where

43 n is bigger than 8. The convenient cross section for the wall is tra­ pezoidal with a min. thickness at the top of 20 - 25 cms

and a max.

thickness at the bottom

H R.

t max = 0.8

to

0.85

= 0.8

to

0.90 w H R at 0.8 to 0.90 H f;rom top

~

The max. ring tension

t max

The max. fixing moment at base : , ,1'

''''''f

=­

w H D t max 7.5 to 8

The max. field moment is about :

b)

Shallow tanks with small depths, big diameters and thick

n is smaller than ca 2.5.

Walls,

A small part of the water pressure is re­

sisted by ring tension and the wall behaves mainly as a simple ca.'1.ti-,

lever.

c)

Medium tanks which lie between the previous two categories. 'rhe

convenient section for the wall is rectangular and provided with

a

haunch at the base. The max. ring tension T = 0.5 to 0.75 w H Rat 0.5 to 0.75 H from top max Cms 0.8 T tons The corresponding thickness of wall : t

=

The max. thickness at base The max. fixing monenf at base The max. field mom.ent

t max ~ff

: M

max

= 0.4 = 0.8

to 0.6 w H R to 0.7 w H R

=

w H D t max 8 to 10

~-

Uf / 5

The max. ring tension and its position in rectangular walls

of

circular tanks fixed at the base and subject to liquid _pressure have

been Biven by 'N.S. Gray in his b?ok Bunkers ~ Silos and Gantries· vRwe s

O~ MI1X

1 .IrS

ft

Reinf'orced Concrete ',7ater Towers,

in the following curves (Fig. 111.16):

ft

RIIJ(f ­ TEIJP0/,l

POStr,OIJ

Fig. III-16

l.tH' IT

u

IJ If

!Z- I -

L-- ~

~

\

·7S

\ \

\1\ V X

J<.>

.SD

V

/'

./

/ ' <,

V\ """'-

f-' Ill- 10-

~ l-

--

.........

I--"

I

r_ •. • t"'I'.R

,

.,-t!

\

IX..

)(

- -- ......

1<

'"S

Il ~}

1-- .......

I--

_i.-l

r

(l

~

/.0

()

~ ""..

...

~/

'r---.&-w.l4J.

t--

o

~~t-, 1<1 i

L> ~

--

l---

./

......... <, I>< /I 1/ .,/'" ........... ..... <, ......r-.. 1/ / I /

I

•'IS

V

./' ./

V V V

I--"

,

II/.D e»

....

.3.0

,4.0

Applications We show in the following

ex~~ples,

the direct application

the given simplified method for determining the internal forces . open circular water tanks with walls fixed at base. 1). Depth E 10m Diameter D 10 m H I R2 = 10 I 25 = 0.4

=

=

the corresponding n According to table 01

=9

w

=1

of in,

tim;

i.e.' case of deep tank.

= 7.8 ~

Wall is chosen trapezoidal:

20 cos thick at its top end, the thick­

ness at the base is given by tI:lB.X

.,

"'f

= °3 wHR

=

= 0.855

wHD tIlla%

°1

=

= 42.75

x 1 x 10 x 5

1 x 10 x 10 x 42 7.8 ;'1 -+

max

= i.lf

/

5

cms

= 5.40 = 1.08

chosen m.t, m.t.

42

em.

45

=

X1

0.211

x 10

2

=

2

According to Gray

Fig. III.16

Average thickness of wall

g

=10-

t

O.}l

10.5 t

!

=}2

=

·D

= }1 cms

t

10 -= 1,

,

Co = 0.8

10

Tmax = Co w H R = 0.8 x 1 x 10 x 5

= 40t

x'

IDS

2)

= 0.225 x 10 =. 2.25

=CH

For section at base ~

=~ 10

According to table C1

=

= C3

max

wH R

c

1

t/~

= 0.5

medium tank:

0.685 x 1 x 5 x 10

= 0.685 = 34.25

.

t

trom base

w=

Depth H = 5 m , diameter D = 20 m

=-

R'::

=-

9.7

-

C = .225

.=1...:x::.....5'-=x:...=.20"-"x~.'""345

9.7

cms chosen 35

cms

=- }.6 m.t.

= -~

= 0.72 m.t. =..l.:.§. 5 5 Tba wall ,Will be chosen of constant thicknes:3 25 ems and provided with

Mmtx

....

a haunch at the base of 10 x 4-0 cms , The max. ring tension position can be determined according to curves ot Gray,

nt = 2.:..92.­ = 20 0.25 Tmax The

=Co w H R =0.51

g, = i D

= 0.25

...

=25.5

tons

20

x Ix 5 x 10

shear at the base for a \vall 25

ems

H = 1.315 y' R t

The corresponding value of 2 Q w H "'maX

=

°2

2

100x 25

C2

is given by 2

= 0.4-25

x

thus :

thick can be calculated

= 1.315 x 5

= 4.16 0.4-25

and

2: = 5.} tons 2

its

at 2.35 ms trom base

follows :

n

and

as

The use of thiS method for detElrmin.i.ng the internal

forces in

walls of circular tanks is very convenient, because when one is used to it, he can easily estimate

t~e

coefficients Cl required fo~ deter~ required for c.eterminiug the max •. thick­

::lining the fixing moment, C 3 ness of the w~ll, Co required for determining the max. ring· tension

without appreciable errors which may affect the design as 'carr be seen from the following example

5)

It is required to design the wall of an open circular

tank 7

IllS

deep and 15 ms diameter. Make a quick estims.te and check the results:

a)

Estimate: This tank is a medium one, its diameter ~ tWice its depth then

the max. ring tension may be estimated by : T~ax

= 0.65

= 0.55

w ~.R

x 1 x 7 x 7.5

=

34 ton

=

27 cms

'Wall thickness ( equation 7 ) t

= 0.8

T ..

= 0.8

max

x 34

Max. ring reinforcements =

...2:!:­ = 1.4

24.5

ems

6

7.5

=

4>

16 mm/mon each side

t.la.x. thickness at base

t max

=

0.8 wE R

=

0.8 x 1 x

7

x

42 cms

The wall is chosen 27 cms and provided with a haunch

15 x 50 cms at

the base. The fixing mome nt :

=

wED t max 8

=

1 x 7 x 15 x .42 8

=-

Thickness requu-ed to resist this moment safely

t rnax

=i' ~f

13 =

Y55

00

1;

= 42.7

ems

i.e. chosen thickness of 42 cms is convenient Vartical t~nsion steel required at base:

5.50

m.t.

41 A.s

u

:I

5500 1300 x .385

z

~d

= 11

2

6

C't:l.

cP

161m

Max. field moment

M+max

:I

5

Uf /

=

= 1100 kgm

5$00 / 5

Vertical reinforcement required

=

M+max ~

=

d

1100

=,; .50

1300 x .24

2 em

7

¢

8 mm/m

~:

~R ~ ~-.: 0.125

Section aD base

7.5

t max

=03

wHR

= 0.77 7 x 7.5 = 40.5

03

A.ecording "to "table

= 0.77

x 1 x

cm

Chosen thickness eu: 42 cms is convenient. The corresponding value of n is given b1' :

=

n

1.315

VR

H

"tmax

=

1.315 x 7

= 5.20

17.5 x 42

giving

C1

=8.65

So "tha"t

=-

wH D t

=_ 2

8.65

x 15 x .42 = 5.10

m"t

8.65

The "thickness required to resis"t this momen"t is given by : "t

max

=

1

5100

3

=

A.ccording "to Gray t

g, t

-L

=

.27

<

41 cms

( t

= 27

= 26. g, D

42

cms·

cms )

=-L 15

c0

= 0.465

Tmax

=

Co w H R

=

0.65 x 1 x 7 x 7.5

=

34 t

Xl

=

C

H

=

0.34 x 7

=

2.4

= 0.65

c

= .,;4 at

JJlS

from base

The internal forces are shown in figure IIL17 The details of reinforcements are similar to those shown in IIL14

figure

48

If the wall of the tank were free at top and hinged at bottom, the maximUJ:l ring tension will be increased by ca 10% in deep

tanks

and 20 to 25 % in medium tanks and its position moves a SI!lB.ll distan­ ce towards the base of the wall. In this case

•

moment

l.l:r

-r-"

I c

o. ~

.: ::t:

Bending Mom.

I

I

Diag.

I.

Nf.

Ring Tension

-----=tE'

Diag.

'2

-u~

!It t<--·------""""'--­

Fig. III-17

will be equal to zero and the field moment will be increased ca io.

~

to

I 3 • The distribution of the ring tension and cantilever moments

will have the form shown in fig. 111.19. 111.4.; -

Tables of the Portland

Ce~nt

Association

The American Portland Cement A.ssociation has published the following series of tables E giVing the ring tension and cantilever mo­ ~nts

in rectangular walls of cylindertcal tankS free at top and fi­

xed or hinged at bottom for different cases of loading. The tables mel ude f'lI!'ther a lot of data very useful. in the design of circular tanks. Illustrative Example 'rhe use of the " Portland Cement Association" tables

will be

explained in the follOWing example : Figure III.1B shows an open cylinderical reinforced concrete wa­ ter tank 5 ms deep

and 20 ms diameter. It is requi=ed to determine

the internal forces ror the E

f~llowing

cases :

"Circular Concrete Tanks without Prestressing ". Concrete Informa­ tion of the Portland Cement Association. ST - 57 - 1.

Table.

Table /I

~ ~ ii

Tenalon In clrcutar ring. Triangular load

rind b ••• , free top

r ...

: .,

.':..: "I

tlle")( " . " "

CMfnc.,nla

,

.0.1'10 .0.21 ~ .0.2$4 .0.268 .0.21J

·0.012 ,0160 ..0.209 .O.2!1O .0.21$

.0.OAa .0.110 .0.110 ..0.210 .0.214

.0,04' .0.091 .0.142 .O.'U .0.231

.0.029 .0.0Il .. 0.099 .0.13"

.tI.172

.0.134 .0.20J .0.261 .0.322 .0.357 .0.017 .0.11.... 0.2~1 .0.)39 .04U) .0.02'" .rl.137 .O:H:' .0•.).41 .0.421 .0.0'1 .0.lIt .0.2J. . . 0.3 ..... 0..... 1 -0.011 .0.104 .0.211 .. O.JJ~ .0.4"3

.0.3&2 .0.429 .0."11 .0.!lO" .0.!I34

.0.))0 .0."0' .0.469 .0 ~1" .0.!!1!>

.0.262 .0.))4 .0.39'1 .0.4047

10.0 ·0.011 .0.098 .0.201 12.0 -O.OO~ .0.091 .0.202 1".0 -0.002 .0.098 .0.200 1e.0 0.000 .0.099 .0.1"

.0.101 .0.190 .0.21" .0.U6 .O.2U

.0.323 .0.312 .0.S06 .0.304

..0."31 .0."29 .0."20 .0.412

.0.~2

.0.601 .0.~43 .0.121 .0.539 .0.13' .O.~)I .0.&.41

Table III

-­- .

D.IH

0.9H

.0.004 .0.010 .0.0,. .0.011

.0.589 .0 0 .0.1331.0 94 .0.666 .O.~.. I .0.681 .O.~"

.0.'79 .0.2'1 .0..... ' ,021!1

0.011

0.111

0.211

.

0."11

O.~II

.1.160 .1.0U .I,OJ1 6.0 .1.010 1,0 .. 0.989

.1.112 .1.013 .1.0014 .. 1.024 .1.005

.1.081 .1.051 .1.041 .1.0)lI .1.022

.0.991 .0.'12 .0.796 .0.&.48 .0.4!1, ,O.2~8 .O.OSI .1.029 .0.971 .0.U1 .0.748 .0.!lS3 .0.322 .0.IO!l .1.0"2 .1.01~ .0.'4' .O.12~ .0.629 ..0.H9 .0.111 .1.G4~ .1,034 .0.98l1 .0.879 .0.194 .0.4)0 .0.149 .1.038 .1.0014 .1.025 ,O.'~J .0.78S .. 0.~1' .0.189

10.0 .0.919 12.01.0.994 14.0 .0.991 16.0 .1.000

.0.991 .0.997 .O.t'l .0.99'

.1.010 .1.003 .1.000 .O.t,t

.. 1.023 .1.039 .1.014 .1,031 .1.001 .1.022 .1.00) .1.01$

Tension In clrcul~r rlnga Shear / M V, applied fit top rlxed bas_, ,,.ee toP. T - eeet. )( VJt/1I

Po.ill.... 110" I,.dlclt.. t.",io""

.0.S!lt 10.~91 .0.226 .0.911 .0.6~2 .0.262 .0.94' .0.10~ ..0.294 .0,916 .0.1~ .0.J21

-"­ ID .

,0

.

.0.~21

Table IV Ton51on In circular ,.Inga Rectangular load Hinged ba ..e , 'reo top

Vi

"0"" i',...tiC"1I1I

1.~7

0." 0.1 1.2 1.6 2.0

.. -

J.O 4.0 S.O 6.0 8.0

- 8.J2 - 1.J4 - 8.22 - '.0'2 -10.42

10.0 12.0 U.O 16.0

J.O' 3.n 4.51 !I.12

I

0.111 .. -

I.J2 3.11 3.~4 J.lIJ 2.$~

0.211 1.01 2.0 .. 2..... l.GO 2.&lI

I O.JH I 0.411 I O.~II i 0.811 I 0.7/f I O.s"'-; l~~~i' _ _ -

0.88 -

O.I~

- 0.41

1.ST

.. 1.1~ - 0.10

1.19 1.10 1.14

.. 1.2S - 0.11 • t,11 - 0.61 _ 1.02 - 0.52

_ -

".J1 - '2.10 - 1.4J _ 0.~8 - 0.02 - 4.13 - 2.liO - LID , 0.1' 0.28 - 4." - 2...... - 0.1' • 0.11 • 0.41 • - !I.11 - '1.27 - O.~O 0.)4 • 0.!l9 .. - S.36 - I.I~ - 0.02 0.63 0.66.

-11.61 .:. 6.4) - 1.4J .o.J6 .. 0.11 -12.16 - ~.41 - 1.0J .. 0.63 0.13 ~IJ.17 - ~.)4 - 0.81 • 0.80 • 0.11 -14.14 - ~.22 - 0.33 .. 0.'6 0.16

0.82 0.$2 0."2 0.J2

CIMKlCi."..

.1.J"O .1.I9~ .1.0~2 .1.J02 ,,,.Iel .LO!.I .1.'l)~ .1.181 .1.062 .1.20J .1.1"1 .1.OG9 .1.160 .1.121 .I,01J

"

0

','

.­

, ..

.­

'

at peNnt

>.0 .1,014 '.0 .1.017 .0.992 ',0 .0.t89 '.0 .0.'8' '.0

.1.01' • 1.031 .1,014 .. 1.00) .0.'96

.1.08'

.0.99S .0.991 .0.991 .1.000

.1.000 .0."7 .0.997 .0.998

10.0 12.0 14.0 18.0

.0.992 .0.99l1 .1.000 .t.00'1

.0.311

.1.01~ .1.11'" .1.006 .0.919 '0.119 .I.O~) .1.061 .1.0'9 .I.QoI~ .0.91' .O.I~J .I.on .1.0)6 .1.06t .1.062 .1.0U .0.9OG .1.02J .1.043 • LOlJJ .'.066 .1.039 .0.9H .1.001 .. 1.024 .1.043 .1.01S.. .. 1.06' .0.997

.o.!ln .0.6'" .0.103 .0.741 .0.821

..1.011 .1.002 .0.'99 .0.199

.1.021 .1.011 .1.001 .1.003

.1.0$2 .1.041 .1.0Jl .1.021

Table VI Ten.lon In circular ,.Ing_ Momont / m ., M. appllod at b... H Ingod ba5e, 'ree top

n." )(

T loiN/I" Po.,I, .... 1'01" l""dlC:II•• I."".lon

J.O 4.0 !I.O 6.0 S.O

1.78 1.11 1.!l4 1.04 0-,24

0.00 0.00 0.01 0.02

·When 11'1'11 t.ble It uud ror ah .., apphod It Ihe b ....... l'Iite the loC) II n...d. O.OH i. the bonum o/lhe Will a.." 1.0H 1.11'1. lop. Shur ...:lItlO In"'ltd I. pCIII""", ol.ll ...lrd i' ""·O'l......

.I.OJO .1.0!lO ,1.061 • t .Of4

.O.JIO .O.J~• .O.lt4 .0.427 .0.... 6

.0.171 .0.&J3 .0.120 .0.~11 .o.tn .0.61J 1.0.911 .0.lIJG

'm .,

O

,,

'

E

:

..

,

.

..

!l-O-.O-II---·--~-===:;:..:::~--,-----;-.. ---l 0.711 O.lIlI 0.'" 0.$11 I 0.811 0.11/ 0.211 0.311 0.4"

0,19 • O.IJ 0.04 0.J8. 0.3J • 0.19 0.06 0.20 • 0.0(1 0.50 0.37 017 0.01 O.!lJ O.H 0.48 0.24 • 0.0' 0.01 0.02 0.0'2 0.03 ~ 0.05 -

.. 1.06t ..1.064 .'.0$9 ... O~

...

2.~ 2.JO .. 2.12 1.91 2.10 2.02 .. 2.06 • 2.10 .. 2.14 .. 2.10 1.42 1.79. 2.0J .. 2.48 1.01 2.04. 2.72 0.12 .. 0.7t' • 1.4J 0.22 1.10 2.0'2. 2.!l0 O.SI

O.I~

.O.It.4 .0.J4~ .0.11,., .O.JAG .U.19lj .. 0.43J .0.2.... .0.4S0 ·0.2~1

.0,'01 .0.764 .0.015 .0.4U .0.'1.30 .0.791 ..0.149 .0.~2 .0.'51 .0.143 .0.109 .0.558 .0.ge~ .0.8l1~ .0.1:.4 .0.614 .1.011 .O.1l4 .0.819 .0.(;09

0." 0.1 1.2 1.6 '.0 ..

O.~I

0.12. 0.04 0.00 0.04 -

O

,

- 0.11 - 0.01 .. 0.02 - 0.21 - 0.1l - 0.03 0.48 - 0.25 - 0.10 - 0.02 O.JIS _ 0.15 - 0,0:' - 0.01 0.21 "O.O!l 0.01 0.01

0.)1

• 0.33 • 0.21 .O.IJ • O.O!l .

.0.327 .0.Jlb

m o

e-

1.....'0"

0.4 .1.41" 01 .1.42J 1.2 .1.HO 1.6 .1.'171 2.0 .1,20S

'

CooK,cl.tll. AI poinl'

0.011

.0.210 .0.2!!'

.0.""

.O.UO .0.11" .0.171 .0.IU6

.

•

111

O.IH

.0.111

.0,14~ .0.2"9 .00UI

.0.309 "O.ll2 .0.JI4 .0.2JJ .0.41' .O.la' .0.210

', "

c:u.I. )( I,1t

H"

:.

~.

.o.,n .0.N2 .o.ne

.0.211 .0,J1~ '0.4'" .O.~ .O,~I. .0.479 .0."') .0.)61 .0... 19 .0.!>4~ .0.~79 .0.~!lJ .O."!! .O.HI .0.419 .0.:'62 .0.117 .0.601 .0.2U .0.J4J .0.dJ .0.~6I .0.639 .0.1") .0.617

1611 '0.00" ·0.100 .0.191 .0.299 -O."OJ

o

.:

.0.791 .0.J"J \.0.3.8' .0.4)4

0,1" 10.1"

0.111

.O,J.:I61.0.284 ..0.21 $

Iii ·0.011 10.111 10.211 I 0.311 ,1_0_.'_I_I,.0_'_"':"'.J.1_0_'"_I-J,.I_0._'_"_1;.-0_.• ",_, ~ I 0.911 I-- - - . - -..- " , "

J.O 4.0

Table V

.0.119 .0.131 .0.114 .O.IOJ

.t po&~1

I O.4H I G.SH

O.:-H

ii :iEi ii:E! I~im Iii·i;i i;i~ ii'Hr iE~ :H* ~m~ : : :; r -

0.611 I 0.111

..0."8 .. 1.022 .I.OU .1.040

6.0

.0.Ci7" .0.011 -0.001 ·0.011

,....0 ""....

.0.271 .0.206 .O.I"~ ..0.092 .0.~2 .0.41!! .0.269 .0.119 .0.66~ .0.~19 .0.Jl1 .0.241 '0.742 .O.GOO .0.H9 .0.294 .0.lI06 .0.661 .O.!l14 .O,34~

.1.0010 .1.0013 .1.040 .1.032

~.O

:.

.0.~11 .O.!IO~ ..0.4JI .1.0~2 .0.921 .. 0.195 .1.211 .1,011 .0.941 .. 1.2~1 .1.1 .. 1 • 1.00' .1,2~J .1.144 .1,0011

s.e

I a.IH I O.2H

"I

.

0." 0.8 1.2 1.6 2.0

.:J.HJ .O.GGO .0.101 .0.6111 .0.'29

CMm,•• ,,11

G.GH

.O.l~ .0.330 1.2 .O.HO .O.J~~ .0.Jl51 .0.JlI2 .o.nl .0.271 .0..10) .0.).. 1 .0.369 .0.3U .0.20~ .0.260 .c.aa ••• 0.37J .0.411

:,':

,

O.JII

I

'

• C)

elMt )( willi

"7:":;;;; .0.4401.0 JU 1.0.]$2 O • • 9.423 _0.401 .0.311 J.O ".0

• I-.---------~-:-----~---

~I

m

'O.O~J

1..':"

T - coet'. )( pR

H'

.0.tH .0.0~2 .0.2'0 .0.073 .0.2~' .0.092 .O.JO' .0,112 .O.~JO .0.311 .O.UI

O

POttll"'. lion Indic,I •• l.n,lon

r -

Po ••l' .... 'Ill" In.lin"" I"",ton

2.0

m

Ten.lon In olrcular rlngl Rectangular load rind ltase. free top

Ten.lon In circular rln.,

Trlangutar load

Hinged b••e, fre. top

•.

at poInt

O.•H I O.'H O.>H l.~,~ ..L..·," IUH 10.'H

11,4 ~ .0.'4' .0.1J4 U l!. .0.283 ..0.239 . !,} .0.11l ..0.211 I 6 .O.2U .0.2611 2.U .0.234 ~O.2SI

J.O 4.0 !I.O 1.0 1.0

~~

-,

Po,itj\ll "0" j"dlul.. 1111110"

£i ,. O.OH

QI

-

_ -

\0.0 • 0.21 12.0 • 0.)1 _ 14.0 .. 0.28 1".0 0.22.

0.11 0.4J 1.00 a 0.01 I.OJ - 0."2 0.16 - O.!I' 0.~1 - 0.73 0.23 O.O~ 0.001 0.01

_ -

0.14 0.46 0.21 0,01

....

1.04 .. 0.4S

-

O.O~

- 0.87 .. _

I,p • 0.10 .. 0.44 1.6t 1.4' I.7S .. l.lI .. 0.10 .. 2.02 1.9S • 2.U • 2.10 • 2.8(1 • 7.22 .. I.n J.!lt

.. J.2I 4.S4 • 3.61 4.JO

J."

'."

: ~:~~ t ::~~

lI,SS • 4.13 S.S6 6.!l1 .. 1.61 4.31 • I.l" • 8." .. 2.41 "Ill • J.1l .. 6,60 • '.41 .11.0.1 • 8.02 6.~4 .10.21 ..13.01 .11,41 U< 1.21 2.0!l • 5.87 .11.J2 .16.~2 .. ll1.06 - 0.02 2.'~

0.14 - 0.13 0.82. 4.11 .. II.IJ 0.111 1.15 - 0.18 J.52 .11.27 1.2t - 0.S1 .2.2'9 .IO.!I$ 0.18 0.64 - 1.28 - 1.)0 1.'2. t.67

• I." .1t.41 ..21.10 .. 23.~ .24.$3

.20.87 .07S.13 .)0.J4 .34.i~

.Whlll thl. I.bl. i. uud tor mom.n' IC)plt.d It Ilielop.... hlle Ihe lop I. hl"OIMt. 0.011 III ond 1.011 II the lOp. Mo",.tli .",hed I' ." ed9- II ta.llI.... ou, rd rotatlo" At that e"o ••

i. Il'Ie bollo'" of Ih

.. he" 11 C.~'"

'.

,UU,tf Will

W

Mom.ntl In cyllndrlc:a. w.1I Te!.. ngul.r load Fixed base, tre. to~

.. ', _

0

',

Mom.. - cMf. X vII'

POlili~. Ii'll' I"dicd•• t• .,.lon In th. o",t.ld.

x

.-.l1t

•

•

"----­

... OOO:!! •. 0014 •. 0011 -e, ·.0012 •. 0042 •. 0011 •. 00 .. 1 •. 0010 ..OO]:!!

1.2

I." z.o

...

3.o

•. 0021 •. 0001 -.1)042 006] +.0010 •. 0010 •. 0071 •. 0103 •. 0112 ·.001) •. 0107 •• 0121 ·.0061 •. 0099 •. 0120

•. 0006 ..0024 •. 0047 ·.OUOJ ..ool~ •. 0028 ·.0002 ..0008 •. 00IG '.0001 •. ()())J •. 0008 9.0002

, .o

••.o

•. Qf}90 •. 0060 9.00~ •. 00046 •. 001' •. 00.12 +.OOOB •. 0016

.0000

L'i

·.04)6

•. 00211 -.0012 .. 0112 •. 0021i ".OUO' \ ...... •. 0(1) -.0001 •. 0090 •. 0019 •. Q/'Cll •. 0019

dIU6..· 1.01l3 •. OO~. • .MII •. 00·44 », 0091 •. ooJ3 •. 007]

... 3. .......

,..•

-.0~05

I 0.311 ! 0.411

•. 0018 ..0040 •.(08) •. 00!)2 •. 0121 -.01~1 • OUOI •. nOO7 •. 0016 9.ooJJ •. 00"7 ..00lll •. 0109 .0000 •. 00''' •.unoo .Q(II'i .00)4 e , 00~7 •. onlo .0000 .OOOU ..ootl2 '.IlVl:'1 •. 0019 •. OOJ!) ••0062 .e000 .O('W ·.0002 •. 0010 •. OU)8

"

••

.oocc

1.0

10,1) 110 14.0 1$.0

I·.·· .<:

11£ •.

.:.:

POlil,w. algn lndu;.' •• 101'.101' in tho "",I,i,lo

1 .1

0 r .

,

.

.O
.0000 .0000

.onoo

.00"0

.1.k)uO

.

'Z ~=

I

-'"

-.0001

.OOW ·.0001

-.0001

.0n~)2

-.:

Pll~II'w.

"II"

1"11'~IIU

••• •• •••

•. 01.1 •. OU7 e . Oln

••OUtt ..0145

•• 0Ui] ..0111 •. 0118 •. 0012 •. OQCJ4 +.0071 •. 0011 •. 0061 •. 00!»7 •. OU5"

•• 0

••

•.nun

.000.

•

.1OE

3:<"­ Q

:

ten,'o" In oul."I.

Q._-_.

x

.

Mo .... _ tll'·f. X "11

.'

1.011

0.111

0.'"

•. 0011 ..002$ •. 004] •• 00"5 •. 00lJ) •. 0011 +.0032 •. 0012 •. 0076 _.00]] •• 0i)(J" ••0008 •• 0012 •. 0071

Moments In cylindric... wall ShoAr / m V, applied at tDp rbed b;tSO, froe top

- :.

0.711

•. OH1 •. 0.\12 •. 0]'" •. 0]19 •. 03'2'9 •. 0192 -..01"0 •. 02'f6 •. O~] •. 0211 •. 0:l~5 •.02]'2 •. 0199 •. 0219 ·.0205

Table X

::~:.

Mom. - (.oof. X pili

f-­

._0101 •. 0201 '.OUI e , 0111 •. 0131 •. Ou~ •. O"~ +.0114 ·.OI:!!1

Table IX Momenta In cyllndrlc:a. wall Rectangular load n:lled base, froe top

0.611

O.~II

"oo~;T7,;; '"0~O;3~

•• 002'0 •. 0019 ...1)(\18 • <1")1' ·.0001

I.,

•. 0602

:::~: I:::;~.

•. 0001 •• 0004 •. 0007 •• 0019 •.no'1'
.0000

Id.U 12.0 1".0 16.0

-.01?~

0.2/1

0.111

~.120~

•. 0097 •. 0077 •. 0012 ·.0119 -.Oll] •• 1.1077 -.0ll1l0 ·.01€111 •. 00~9 -.\ll.I~U -.02~" +.1I04G _.()(I)I ".ool'J -.004' ~.0'a7 •. OtI76 •. OO:JU •. 0029 ·.0022 ·.014ti

•. 0071 •. 0047

OOOOI'~'

I

-.O'SO -.0302 ·.0:\29 .01"6 •. 0(2) -.0061 -.0224 -,1)46~ •. 00')0 •. 0022 -.0101 •• 0) II •. 0111 "oo~. -.OO!'!I -.U7]} +.0'15 ·.001~ -.0021 -.Oll!'!

•

Co.Rici.n' ••, poll"

..

0." 0.1

::."_X 01 w"::'

+

M.m - co-'. )( ( .. 1/ 1 pili' POliU .... ign Indic.'.. '.1'.101' III th. o",l.'d.

_:

-'-/1']- - - - -------------_._-.~.. -_. /I' ~ .~~~1~2/~. L~~~J..?~/!,..J..2~-~~~J0.1~J~L~.:~~.!:~ '" I 0.• oon •. ... Coo .., •••,••,

ItI

Mom.nta In ~yllndrlo.1 •• 11 Trapezoidal lo ..d Hinged base. free top

'.

.

.

,

--------·------------"---f 1--------....,------------=--1

I':;;;1'

~ o.;-;r.·211 -1.'1/ fD.W ci·~':;:·;".:i;;T~:,;, I-;;;'-I~::;/I

~~;

0.8 1.2 I.G '.0

J.O .•.0007 •. 0026 00,0~ •. 00Q4 •. 001~ ).0 ~ •. 0002 e , ()(.O8 ... 0004 •• 0001

':l= 17.0 I ... d 16.0

.0000 .0000 .0000

I ,;,;;'

:.;).j;' .~;;;;. ~;()'~3; ~:-oi~~ ~-~;; ··~;~·:.7."·~:7~~· ~~~:J -:;;·I'~

.0000 -.1)IJt.1G -.00'2!» -.0083 '.000' ,.0020 •. 00]7 •. OO:!') •. 0011 9.00J6 •• 0062 •. 0077 •. 0010 •. OO)G •• OOG' _.0018 •• 00!»1 •. 00JJ +.oo,!) ·.0011 •. 000l

-.011:; -.OJ62 •. n',')4 •. l)')l7 _.1 J1!) -.000'> -.OOU9 -.UV7 -.04GI ·.llltl~ •. OQr,8 9.0011 -.0093 -.0"~1 -.0!'!2') •. 00111 "OU~~ -.OOH~ ~.01'7 _.OJII')

.-.1 K.1!1 -.1178 ·.081G -.011']

-.un)

_.1)48.\ - 016!,

... (t()!)1 •. 008]

••0014

•• OO~2 •. 0066

-.OO1~

•. 0035 e . GO:!!I ·.0061 •• 002'2 9.0036 •. 00 ..9 ... oooll ••0018 •• OOJI

•. (1042 ·.0051 •. OU~3 -.lKU] .'(IQ~1. ,.0007

•.00." •. 00lll

-.014~

•• OUI7

-.0101 -.0073

•. 002"

~.OO40

,

.0.171 .1)2"Q ,O]OU .0.)~,4 .0... n2 .0.44' .0.4!)~ .O.~:t!\ .O ..I"~ .O."'~ .0.2011 .o.no .0.224 .0.'21l .n.219 .0.214 11 .0.082 .0.13~ .0.1:17 .0.164 .0.1~!) 90.14) .0.117 .0.1116 90.0114 1.1; .0.079 d).I22 .0.139 ..0.138 .O.12~ 90.10~ .0.OB1 90.U~ .0.0.\0 2.U .0.071 .(u.~ ..0.116 .0.11') .0.10) .0.080 .O.OM .0.OJ1 .0.COIi

Table XI

.O.~/N

0'"!'0,L)8~

.U:OUIl .0.(1'.:' .0004 - ••1.01'1

~·~f~:~: :~:~: :~:~~ :~:~~ I:~::~ :~.~~ :~:~~~ :~.~:: ~~:~:~ ~.~~~ 5.0/,0.I)G-4

li Q .0.OG2

- 01 .... -.018·1

r--'---------.--------------,

'~"'-

01~·.IU),)]

-. O~ "').1

+.0021 •• 0030 •• 0021j -.001? -.Ot47 •• 0014 •. 001 .. •. 00:'2 ..(,0.2 -.OI:!J •. 0010 •. 0018 •. 00'21 .. h007 .. 010~ •. OOOG •. 001'2 ·.0020 -.oon!! ~.oo')'

... 0001 .0000 1'.000'2 •• O(t()!) .0000 -,OUOI .0000 •• 000" .0000 .0000 .0000 •. 0002 .0000 .0000 ~.oool •. 0001

C""/It,.",,,. _t p(1I"I· I~'; ~ o'.~-'-;-f o.;~/·I~.i~~·I;;;;-rO.~1I 10 .6 11 10.111 0.9I1...1.-!.01l .,.. .. .. _. __ ... _. __ . . . . _. -.. ..­ .......,_.,jWll .. - -". ,---­ tI' ':

,u.07lJ .0.067 .O.1l41 .0.018 .0.01l +0.00) .e.oea _n.OO7 0.011 ,O.Q70 .0.O!loG ..I).OJ.' .0.018 .0.006 0.000 -0.00.1 _O.Ou~ ·0.00" ,0.0'4'1 .0.00110000 -0.002 -0.00] -u.oca 0.001

"01'0'O~7 .o.ose .0.011 10.0 .O.oc.] 11.0 .0.0..9

.0.041 .O.tJ-t2

.0,0r.1 .0.012

.0.012 ..0002 -0.002 -0002 -0.001 ~O.OOI .0.007 0000 -0.002 ·0001 -0.001 O.OOU

O.()I'IO 0.0l1O

:::~ I:~:~:~ :~:~~~ :~:~g :~:~ :~.:~ ~~:~ :::: -~.= ~:= ::~~ ·Wl\.n thi. 1.111. I, I"l"~ /Uf .h.a, .ppli.d .llhe b........ hilo 'h. 1.. ,.1. n ...... 0.01111 Itl" hull"ul\'U' til ..... illl .ltd 1.01/ i. thu toop. Sh.ar acllng In ....rd I. PO,IUW•• Quh ...,d II l1 ..

u""ive.

Moments In cylindrical wall

Momont/ M • M, applied at l;Jaao

Hlngod baae, fre~ top .

Mom. - eeet, X '" Po.,Iiv•••"n 1",liCo1lI.lI 10,.,iol\ In ouhldo

Table XII

--_._---~------ - - - ' - " - ' //'/

ui 'O~II I 0.211-1

CoalfiClGnh at poin'·

I r.~II. i.-0.811·1.~wG·O!i: O;I·.~O~;·I-:O"O~·~--:;-;-~; :;.~~-6 :O~~~ .O~~;~ ~.~:;"1· .O.G!):! .U.tl41 .!.OOU OlJ .OUO'! 1.2 •• 0006 1r.;.0001 10'.0 lI'12

\lIf

'.l~ )0 (,0 10

100 11.0\ 140 lG.U

.0.041> .0011 .0011 _OIW/'1

0007 002:! 1I00Ill ~00}(j 1)007 -0011 000" 4UOIII 0001 ·000')

0.311

I ~.-411 LO:~/~ .. 0.611 I

.0.090 .0.1(,. . . 0.'2!.3 .0.l1) .tJ.~] .O.c.~l) 10.8::'4 .0.003 .0.1'2'~ .0:206 .O.JIG .0."~4 ..0.6Ilo .0.802 .0.0]'\ .0.078 .0.1)2 .U.1~.1 .0.39] .0.~10 .I).n~ .0.012 +O.OJ" .0.096 .0.1~.J .0.340 .0.~19 .0.7..8 ·0.030 ·0.CI-I4

.n.olo .0.Oft7 0.014 .0.l)~IJ ~O'.I 0.01) 0.0"0 .(\.1"': ·u.lIG~, ·U.OJ7 ·0.022 -O.(JoI" ·O.Utoti ·O.OlI4' ('''_"9

·O.O~I

O.().I~ -OOlOj

0000 000' ·0.00') -0.028 -0.0:>3 OlJ("O ·0.000 -O.OlIJ ·0.016 -0.0..01 0.000 0.000 0.000 -0.001l -0 O;'!) 0.000 0.000 .0.002 ·0.00) -0.0"/'

'a,

-O.OG. ".064

.1.000 .1.000 .1.000 .1.001)

.0,2;'7 .n..... !i .r!;')" .100tJ .O.I~O .O.J"I" 'U'-4~I'I.OOI' .0.09r· ..0.1'"16 ,O.fil.l+i ,1.rlOO .UO~1 .O~:.1 .11.">7/ •• (Illl' .0.04:01 .0.l/d .O.~I,) .1.I)Il(.l

-OOJ' 0.1>151 ·0.0~9 _0.060 ·O.O~l -0.066

.1l.121 .U.4G1 .1.QoJO .0.Oltl ,0."7" .1.000 .0.0"1l .. 0.)87 .. 1.000 .0.01~ ·.O.J~~ .1.000

.WtUIr'l U....",III. II u..oll mum.l"t ;lP(lhOll alll.o lup, ....h'l .. '"elup I~ h.n')..,1, 0011 i, Ihe t,-,lIom 0' the ....olff afl." 1.0ll it. 1110 Mom.'" a"plleu al ;Ill lulQIt it "u.. t .....

."1).

....hon 11'.""0' 0"'1....01,.1foloillon e' Il\ill (IlJ·]O.

Moments In circular slab wIthout center .upp.ort Uniform 10ilLl

~lo·;.d:~~;'

X ,.HI

Po,iliwe "0" in.lie.l). «Pm"' .... lUn in ...,I.u;.

~~

le""'.d!~

.n.~_~,L~~_O/~ I ~;!-o~/.I.~~~~~.~·.~~!~ .L~~~. t.~.~.': L~~~).". !.~)~ ~~~/J:J~~' .1 :.:.~.!!... R.rt, ...1 ,lIl1molll,. "',.

,.•" I ..• 73 1•.067[.:.0"1 •.••, I ..0"1 ••• 7>

I ..•74

I ..071

I ..... I

0 •• "

•.0031 '.'" 1 ....

1 •.• '>0 I •.•39

I

,.-r.:o" T

1..•

.087

I "'"

.008 1 -.0»

Table XIII Morn."u I" clroular sl.h with e."t4r tupport

Uniform load F'hrnd edge
Mo ... _ -

)(' rRI

Po.,I1 ....ID" I"dlul,. 00".,.,..,1," I" .~".OI I".d.. d

O.JOR

0.06 0.10

-O.I41J

--_ . . 0.05 0.10 O.H!

~O.O72'

-0.2100

0.16 0.20 0.7&

_­ I -0.0417

I

0.'0 O.:!~

-0.0700 -0.0787

I

0"01~_1

....R

I

-0.0275 -0.01;'4

-a.lou

I

-D.OM1 ·0.04'1

-0.01.'"

Tabl" XIV MCU11enh h, olroulAr ._.b with oont.. r eupport

Uniform load

lUnged ada.

Mu,.... - CO"" X pRI

Po.ltI.... Illt;l" 1"111,..,,,. oump,."lnn I., ....

,1....., 1""....1

••• 0.10

O.I~

- I----f'--

0.70 O.2~

- 0,1040

00' 0.10

-0.0768 -0.OJ74

O.I~

0.01'1511

.lJ.n~r.9

U.nti1l4

·0.0'1)') -0.0470 -0.Ol67 .0.0734

-0.(1)16

-0.021J

0.'0 U.1~

r-~·~·1nl':!!..I~--"·--I·· -O.Onl -0.0·1'"

-0.0171

.O.~11\1

-O.ot!»

-0.0.133 -0.0263

-0.01114 _O.OnI4

,0,00'0 -0.001" -O.OCM'

80.0J7~

_0.017~

Table XV Mnments In etre ...I;1r .lAbwlth center sUf),lort Moment /,..., I M, applied at eoduo Hln~ed edge Mu",. - co.'.

><

M

Po.iIIv. -.iO" Indlc.ll)1 com.,r ...lon In lop lul'.c.

-O.~l

. _-- -.---.-------" -_.0.",.." I

.O.OI7~

.(1.1)1,)4

.O.02!'1

.O.uIJ2 .O.IHOJ

.O.ftI~8

.n.otJ.4!,o.0I971 .0.0163

.0.(JO'n

.0.006~

.0.ooJ8

,0,01111 .0.01118

.0.1.11.'\2

/

.O.U??II

.. 0 Illl""

.. .n.OI71 .0.0141 .O.O":!

,O.IU4!J .n.012.1 .0.fIl0' .0.,,,.1'.

Table XVIII

Table XVI

StlHl'leu of c)'lIndrlcal wall Near edge hJng~dl far edge fn.

Shul" ilt ba'. of cylindrical •• 11 ,- - c_I.X

r"

,.11

(I,ia"v",I..,. (t.ctlilnout..

• _ eoel. X 1:",11

t'

iJi

1'01,11... ,i",,, iftd.cat•• • h• ., aelll1g 1,,_lIcj

----

,,,

I

T ,i."v"".' 0, ',cl.""",I., 10••1.

12

f,l .. nyul",luad, "ud b....

0.' 0.'

.o.ns .o.n. .0.JJ9 • 0.117 .0.299

.0.407

.0104

·1.00 -2.2 •

'.0

.0,l10

.0.189

·2.~1

'.0 '.0 '.0 '.0 '.0

.0.262

.O.JIO

.O.l~'

.O.2J6 .0.'l13 .0.197 .0.174

.0.211

.0.:143 .0.22'1 .O,lt3

.0,lll

.0,121

10.0 12.0

..0.16. 'O.IH

140 •• 0

.O.lltt

...

I.'

R'cllf1Ql"'" lOla, 'Iud boll.

.O.~!l'2

.0.460

.. 0.17'1 IO.U' .0.147 .0.111

.0.127

0.13Y 0.'2'70 O.J..u 0.399

1.~

.1.dU·

, .e

.0.1'"

·1.~1

.0.ll4 .0.120

·1.7~

'.0 '0

hU'I,.d lI...

.o.ns

I

~ tn

I ,

Coerrlt:i.nt

0.' 0.' MtUlnll.1

H'

-

I

n»

(""-'". a' ba..,

"'III

C•• mel.nt 0.11.) 0.111 0.,0) 1.010 1.101 1.1" 1.:11

••

.0

0.44~

12

O.~'"

14

'.

0.6JS

-J.II

-J.n ·4.10

.0.110

-4."9

.0.098

-).1'

.0.011

·Ul ·6 ) •

• O.Or., .O,01J

-6.n

·7.}6

.0.06'

Table XIX Table XVII StlHneu of cirClul8r pliltea With C4l'lter aupport

LOolld on center aupport for c:lrc:ul .. r .Iab Lu.u' -

till',

X { ~~:"

I---

I

(/I)

O.O~

HI.,,,..IJ

.11 1.1

1.317

o,n,

I

."lIa

·_-o·-,,--r--':;O-j--o-.,-,­

I

0.10

1.320

F ... iJ

.. - co.l, X K"/H

:~::::~::~ ::;:.~I

O.tl~

6.16

1.66

I

0.10 •...:!.!:.._I__o._o'_!_-:-:"-:--f_=-:--t~~:--;-~~:-_I

0.1~

0,1.)

O:ZO

1-483 1.007 9.:9

1.!lo42

1.62~

1,1t.l1

1.100

10.1'

...

C".l.

I

I

0.'2'0

0.3u,

0.33'2

I

O.HI

(

0.317

Without center 'upport COfO'. _ 0.104

.;

Tabl. XX. Suppl.m~nlary Co~ff1d.nh fo, Values of W'/DI G,oal., than 16 (Ext.n,ion 01 TaLf., 1'0 XI, XVI anJ XVIII)·

II'

I

TABl£ "

TABLE I .

Oi Coerr!cle.,lt "t point _. _~ __~~.IIJ~~~/I ..

'20

.0116 .U.746 ,0,781 .0.800 .U.191 .0.76..]

..."

J2

.0

I .0.S20

.0.M.4

I.o.m

.0'0' .0.lGa

.0.643 .(1.131 .0.121 .0.1l!l~ .0lJl!l. •0.I-Z4

.O.IO~

.Q.J'2S .Q.3n .. 0.4~ .O.~

.0."3 .0.636

CootrICl4nl1 a, polnl .9~"

.7~1I

.1011

..O.ln .0.U7 .0.112 ,0.211

.0.112 .0.111 .0.114

.0.117 .0.8.3' .0.861 .0.860

.O.2~

•0.791 .0.111

.0.21J~

.o.m

......

.O.a;lj

TABLE \I

3~

r.. .

41

.2!L~

-10.11 -10.67

be

-1'.~

-IO.U

,,,

!E.

.0011

20 24 40

-1'.04

-~.'"

-2J.14

.1011

- •.aa

• U(I

-10.14

-4.~

• -

-Io.n

3.70 1.11 1.00 I.JI

.I~"

- 1.~'

• 1.00 - 0...

.o.n

.. 1.'21 • 1.10

CooI'fI.cI.nl.a .t

.2011 .. • • •

0.2'2: 0.61 1.21 1.6a

.. '.68

• 1.12

.nll

.10/1

.8~II

.IS.JO .13.20 .. 1.10 • 3.11 - 0.70 - 3.40

• '11.' .2S.' • 23.'l 'I 1•. 1 .14.'

• 16.9 • 40.1 .. 4~.9 .. 41.~ .4S.1 .. 41.2

TADLE IX III Lei

'­

'0

24

.. J2

.0

be

Cc.oetrIe'.nl. at point

·1D"l.a~II ·.0015 •. 0012 •. 0008 •. OOO~ •. 0004 •. 00U2

0.0013 .,00" •• lXlO!t

,iOII

1.. 0002 •. 0004 e , 0006

•.OV07 •. 0007 ••0Cl0G ••0006 ••0004

.9~II

'.0011

-.00'24 -,0013 -.0011 -.roel ·.0010 -.0041 -.OOOS -.0037 ·.OOOJ ·.0031

I •.OOO~ -.000'

·fQl l,lQlnt. not.hown in Ih_ ....

I .90'~II .0.603 .0.647 .O.nl

.O.l'" .0.3n .0.4]8 ·0.413

.0.'20 .0.8!'1'2

.0.~11

.o.n.

.O.~IJ

I

j

I .1~J .il.,4, .0.911 .'.0'26 ,1,D-l0 .1.0·4) .1-().I0

In

-.00~5

n"

I

.O~II

· '.'

.0.1'2~

.O.IU'J .O.'it~

,09')6 .1.0n .l.O3~

.I~"

I .toll I

.!lSII

.0.]1, .0.00

.0.12' .0.149 .0.189

.1.061 .1.0M .1.064

.0.26'2 .0,2'i4

.1.041 .1.021

.0.~19 .O.~l .0.6~'2

.0. loS

~1i1t

.'JOII

.I.O~~

.1.011 .1.039 .1.061 ...1.061

"'.0G4

"1.0~

• • • .. .. •

43,3 61.8 lIS.4 17.9 17.2 84.0

• H.3 .. 4S.3 • 13.8 • '3.1 .. 10).0

. •.oon .10/1

•. 001'2 •. 0007 0.0002

.0000 .0000

.. I~I.O

.111/

•. 0014 ...0011 •• OOO?

••ooo~

.,0001 .0000

I

.9011

,UI/

,2011

:J~II

.0.033 .0.011 .0.020

.0.0'l1 .0.011 .0.011 .0.0Q0I .. O.OOJ .0.002

.0.014 .0.009 .0.004 .0.001

0.000 0.000

I

.9~"

•• CJOiJ~ -.0018 •. 0..>01 -.0011 _.0001 ...0007 •. 0006 _.C,JOS •.0006 _.rJOO) •. 0004 •. 0001

.&011

.8~"

.'011

-
.0.095 .O.OH .0.0Q2 -0.0)1

.0.'296

-0.037 -0.061 -0.017

-
~.~,

'ao~'1

-0 ....

.I~"

I

.9011

I

"J~/I

·-I···r~

.0.'4) .0.991 .1.L1.3O • i.eso .1061

.0J'47 .0.8:PI .0.111 .0.92'0 .. 0.9!Q

l

.0.477 ..0.411 .. O.W .0.617 " 0.113

COIlnc;I."I, ., poInt

CLlGlf!ci."11 .at POlnl

svn

Coel'ficlgn!l al pollli

.0.01~

I .IDII I

TABLE VIII

eoemc:l.nla "t p
.0.011 .0.001

.o.ne

.7>11

TAOLE vu

TABLE XI

.1011

Co.ftki.nta at point

.06'2'9 .0.'904 .0.111 ..0.1:" .0.911 .0.'4')

TABLE X

.0.012 .O.OJl:l • 0.031 .0.035 .0.0211 .0.0'29 .0.0'211 .0.02S .0.014 .0.011 .0.013 .0.011

TABLE IY

TABLE 111 CG6I'fiC10.,(. at paJnl

TA(jl£ \II

Co-If,cl."u .at point .0~1I

.I~" .0.1~

.0793 .0.147 .0.1160 ,0.900 .0.911

I

1.0011

.tvu

-.OOGJ

.,0006 •. 000$ .0000 .0000

...0014

.0000

e ,OO(U

.-.OO~

-.0040 -.OOJ'2 -.00'10 ~.0013

.8011 ••noIO •. 000S •. 0003

.0000

l

,'I'"

t.OOli

.0.601

.1.000 .1.000 .1.000

.O.2~

.o.sn

.0.111 .0.12) .0.04l .O.G-II

.0.S1~

.0.457 .0.4'24 .0.387

.1.000 .1.000

.,.000

..0.1'4 .0.101 10.089 .0.080

.o.un

.0.061

......

••OOOJ

...00'24

..0020

•. 002Il

•• 0017 ...001] ...0011 ...0010 0.0001

•. 0014 •. 0011 •• 0001 ...DOU7

TABLE XVI

I

T'L1t fl••

.0000

.I$"j~~ •. 0020 .,001$ ••0000 0.0006

I

flud

.0.1'12 to.l11

.0.090 .0.011 ..O.Oli .0.07.

,.~-

TABLE XVIII

Hlng_d at Edv_ Stinntl ..

.0.012 .O.OM .0.04. .0.04) .0.039 .0.038

-&.20 - 8.9-4

1.4310 I ....

~1t.62

' '101 2.026

-12.16 -1171

2.220 2.400

-io.JO

..

49 1)

~ottom

2)

"

3)

II

" "

-

t

edge sliding

o • zR

hinged

fixed

.20

1ft

\ Fig. III-IS 1)

Top Edge Free and Bottom Edge Slid±ng H

5.00

::

R :: 10 ms

and

IDS

The maximum wall thickness can be determined according to

equation 7

from the relation t

thus

max

t max

= 0.8 ~

wH R

0.8 x 1 x 5 x 10 :: 40 cms :" ..

As the water pressure increases with the depth,

a~d

.i.e!.fully

resisted by ring action, it i.e! convenient to choose a trapezoidal wall with. a minimum thiek:ness t :: 20 ems at top and a manmum thickness of 40 ems at bottom. Table 1

Ring Tension in Cylinderieal Tank Hall with· Free Top

& 8li­

T::xR

ding Base

..

..

Distance from [top edge x

0

0.2H

0.4R

0.6H

0.8H

1.OH

!Ring Tension ~ tim

0

10.000

20.000·

30.000

40.000

5O •.0CO

2)

Tou Edge Free Bottom EdGe Hinged As the max. ring tension and bendi.ng !!lo:,:ent take place at the mid­

dle part of the wall, it is recomL1ended to choose a wall of constant

thic mess with (26)

In our ease

1; .;".

{).5x .5%10 "=25

2. H .D t

52 =---"--­ =5 20 x 0.25

.. i

Therefore ,..

ems

Ring tension per meter height will be computed by mUltip1ying"p~

'coefficie~'l;s for 10 = 50 tim

by the

5 x

B2 I Dt' = 5.0 taken from table ir

Bending moments in the vertical wall strips 1

m

t

PmaxE

wide are computed

=

by

multiplying the coefficient given in table VIII by :.:

I t has to be noted. that :

*

The coefficients are given and :point

3£

su~h

that point O.OB denotes the top

l.OH "the. base of the wa1.1.

The positive ring forces are tensile while the negative are com­ pressive •..

*

Positive bending moments cause tensile stresses in ,the outer sur­ face and compressive stresses in the inner surface of the wall.

.

'.

Table 2 : RiDg Tension & Bending MO!:!eI:ts iIi. Cy1ind.erica1 .. Tank Wall with Free Top and Hinged Base.

~':i::ef. :.OH :.lH

~enII .008 .114 sian T tim _ + . 0.4 5.70 Coet.

VIII

p mt/m

+ .00 .00 +' ~OO

.00

0.2H +

.235

0.3li !0.4H II

+

+

0.5H. 0.6H +

+

0.7H 1+"

.617 I .606

.'356

O.SH +

.. +

r , ::;03

11.7517.8

+

1+

23.45 28.1 30.85 30w5 25.1

+

+

+

+

+

+.

+ +

+ +.

+ '. +

0.9B .2Cl4 +

14.7 +

.0001 .0006 .0016 .0034 :0057 .008' .0094· .007c +

.O~

+

+

'+

+

+

.075

.20

1.425

.712

1.00

+ + 1.:).72 .975

51 aeaction at base ( coefficient table XvI )

0.121

::

w g2

....

0.121 xl x25 :: 3.0 tim'

::

The max. ring tension in the wall Tmai '== 30.85 tonS' and

acta' at

o .6H

:: 3.(;0 as frol!l the top edge: The required "ring reinforcement o.t

x

::

As

=

0.6H is given by T

30850 1400

=

O's

=

22 ems

16 q:> 13 Dlln/I:1 ( 8eP 13 ID.I!l/m on each side) 21.2 cI!I 2

The tensile stress in concrete, including shrinkage, is

=

T + £ sh E5 As Ac + Do As

::

30850 + 0.00025 x 2100.')00 x 21.2 100 x 25 x 10 x 21.2

=

;0850 + 11100 2500 + 212

::

419'30 2712

::

1172 kgm and acts at x

(J

t

u The max • moment "max + from top

d

::

22

::

For

(J

t - 3 k

1

V1172 tit

2

,et= 0 ,

O'c

k

= 25

1

safe.

::

O.SH

4

InS.

or

:: 0.G43 2 &.

-. ,­

ltz = 1,00 choose

1300 x .22

=

= k1"V'M

kg/cm

1172

=­

d

d

i.e

kg/cm2

by

:: 15.5 kg/cm2

25 - 3 = 22 ems

::

= 1400

s

=k :»

~ven

.;)

and

q; 10

mm/~

To";) Edge Free. Bottom Edge Fixed : In this case, the water pressure will be resisted by ring ten­

sion in the

torizc~tal

cal dir~ction. The

direction and cantilever action in the verti­

rna.x.:'

ring tensic!J. and max. positive moment '.'lill btl

smaller thall b. case 2. wall hine;ed. at base I While relatively

big

fiXing mouent s Ylill be induced at the fued bottom ed[;e of the wall. For

~lis

reasen, it is

wall of constant th i.ckne s s t 10 x 40

cr.~)

at its foot.

reco~~elldcd

= 25

to choose for this case a

cms sti.ffened \'lith a haunch ( ca

Due to the existance of the haunch, it is recommended to deter­ 1Iiil:l~

the fixing mo:nen-c and respectively the thickness of tha wall at

its foot either by method of Reissner i;~

given before while mined from the

p.e .A.

Or

by the simplified

method

,ring tension aDd. positive moments can be deter­ tables neglecting the effect of the haunch.

ThUS, For section at base

Refer to examp'le2 in the simplliied

me­

thods 3600 1284- x .32

= 8.8 cm2

. 7 e.p 13 mm.

The ring tension per meter height will be computed by multiply­ ing P ~ 5 %,10 50 tim by the coefficie~ts for H2 I D t taken max from table 1. The positive bending moments in the vertical';'wall! strips

=

are 'computed

1 m wide >,

VII by

=

by multiplying the coefficient given in table

Pmax H2 = 5 x 52 :: 125 mt/m. The ring tension and the cantilever moments at. the 'different

depths of the wall

are

given in table 3 in which all the values are determined from the P.C.A. tables for H2 I D t = 5 with the exception

of the fixing moment at the foot of the wall which equals - 3.6 mt as given in example 2 page 45 • Ta.ble 3.

Ring Tension & Bending

~10ments

in Cylinderical Tank Wall

with Free Top & Fixed Base.

Point O.OE r

Ring Ten.

Coef + I .025 T tim

fiLM.

+ 1.25

Coef. +. II .00 M

1!lt/m .00

O.lH 0.2H 0.3H

0.4H

0.5H

0.6H

0.7H

0.8H

0.9H

+ .428

+ .477

+ .398

+ .259

+

.346

.+ + + 6.8512.25 J.7.3

+ 121.4

+ + + 23.85 23.45 19.9

+ 12.9

+ 4.6

+ + + + + + + .0002 .OOOE .0016 .0029 .0046 .0059 .006

+

+ .006

+ + .025 j".10

+ .3.5

+ + .137 .245

+

+ .20

+

+

.36

•58

+ .469

+ .74

+

.74- .

.092

,

.003

1­

1.72

~he

shear at base of wall equals p • H x coef. of 0.21; for HID t from table XVI.

ta~en

i.e.

~

= 0.21;

x 5 x 5

= 5.;;

=5

tim.

-,._....,-~-:;;;;;;;;;;-~--

I

!

e: OJ

.' 0, "

:

eo

. 50 tim

Cantilever moments Fig. III-19 Ring tension rhe ring ten.'lion and cantilever moments !'or the 3 cases

are

shown

in figure III.19. In cases of tank walls monolithically cast with the floor sup­ ported on clayey soils liable to unknown rotations, it is

reco~~ended

to design the section at the foot of the wall for the max. possible negative bending moment M = - 3600 kgm - case of total fixation, the max. positive bending moment Mt = 1172 kgm. and the T = 31 t max

~ax.

ring tension

are to be taken from the case of a inged base i.e . .we ma­

ke the design for the hatched values shown in figure. The tension the floor due to cantilever action 111.5.

~ax

= 5.33

in

tim'

Tank 3alls Continuous with Roof or Floor ifhen the bottom of the wall and the edge of the floor slab

are made continuous as shown in figure III.20, the elastic deformation of the floor slab tends to rotate the lower point of the wall and in­ troduces a moment M in the corner. is so much like

mo~ent

p.e.A. , a:re

The data in tables nuzaber-s XVIII

stii'fnesS~s

which denote

red to cause a unit rotation at the edge of the yrail slab.

M,

distribution applied to continuous frames, on­

ly there is no ca:rry over m9m&ntS. and XIX of the

'The procedure for obtaining

moc:~nts

requi­

and tile floor

Only relative values of stiffness are required in this appli­

cution. The

eXR~ple

sho\vn

~

fig. 111.20 gives a good explanation for

tl~

problem. The nain steps for the solution are

1)

Determine the fixing moment

P-.t the bottom of. th~ wall and at ..

the

edge of the floor, aasumi.ng

.2!l

D-2R

2)

The difference between

0

~I

%:

\

distribution

method

-. "Ct '"

10

the

tributed between wall and floor rno~ent

.

2.5 m

I

1

I

by

~

ci

0

t/

~

:

rwo fixing moments will be dis ­

.2

E

joint between wall and floor to­ . taly fixed. The two fi.xi.ng mome­ f

nts are generally not equal.

_to.o m

-;->0.,

e

.-?:"

IeI

/

-

rf

I.~m

I .

rtf

Fig. 111-20 ~ I-- .

,

according to the relative stiffness and distribution factors of both wall and floor • .3)

iind out the effects of the distributed mo:o.ents on .both and

add

it to the solution of fixed· joiJit. l~)

Determine the concrete dimensions and steel reinforcements such

"tha.t there is adequate strength and suf'ficient safety against crac -

Exanple A reinforced concrete water tank 10 me diameter and 5.0 rns deep is supported oIl( a cy1inderical wall at its outside edge & on a central column at the centre as shown in fig. III. 20. The wall is free at it.'3 top edge and continuous with the floor slab, at its bottom edge. column capital is 1.5 m diameter, and the drop panel, 50

CIllS

The

thick l

is 2.5ms diameter.

It is reqUired "to clarify- the effect of continui"ty between vaL and floor on both of "tliem.

55 3tep 1:

Fixed End

~o~ents

of Wall & Floor

31~

The fixed end aonenb of the 'Mall is c01!1!-,uted by multiplyino the" coefi. given in table VII for the corresponding ~ID t at P:nax h.2

~

=5 x

=

Dt M

10

=-

52

= 125

52 .25

%

0.0122

1.0 H by

%

= 10

coeff. ( table VII ) at 1.0 H

=- 1.53

125

=- 0.0122

mt.

The radial moments of the floor slab are efiected by the drop

panel,

but since the ratio of the panel area to the total slab area is about 1/20, its effect on the loads and moments may be neglected coefficients given in table XIII, XIV &. 'l:.V

and

the

can be used.

load/m2 floor slab = weight of water + own weight p = 5.0 + 0.35 x 2.5 = 5.875 t/m

2

The radial fixed end moment equals the coefficient of 0.049 from ta­ ble XIII for clD = 1.5/10 = 0.15 at point 1.0 R multiplied by p R2 i.e.

M • - 0.049 x 5.875 x 52 = - 7.2 mt.

Step 2:

The Connecting I.lornent

The fixed end moments of the wall ( - 1.53 mt ) and

floor

( - 7.2 mt ) being not equal, the difference will be diatributed bet­ ween wall and floer slab according to their relative stiffness, thus: The relative stiffness of the wall equals t 3 / H = 25 3/500 = 31.25 multiplied by coefficient taken fro:ll table XVIII for i.e.

multi~lied

by coefficient taken

fro~

= 0.15

i.e.

t = 10.

31.25 x 1.01 = 31.56

The relative stiffness of the floor slab equals 85.75

2 H / D

85.75 x 0.332

= 28.47

The distribution factors are

therefor~

t3 /

R = 35

3 ,I 500

=

table XIX for clD = 1.5/10

56 for wall

:

for floor

31.56 31.56 + 25~47

=

~

28.47 31.56 + 28.47

=

23.47. 60.03

60.03

The. final connecting moment will be

= 0.523

= 0.477 deter~ined

by moment

dis­

tribution as follows: ( fig. 111.21 ).

Distribution Factor

-Vall

Floor

0.523

0.477 ­

/"

Fixed end

~.loment

Distributed I.1ownt

Tbe

,­

quic~

+1.53

~7.20

+2.97

+2.70

+4.50

-4.50

Fixer,. end moments ~L53 mt

l2:2)

Connecting moments

,.--.4. 5

mt

~)mt

FiC;.1E.21

damping of the bending moments makes the simple

ment distribution shown su.f1'icient

1'01'

mt

mo­

deteJ-'Jlinins the connecting mo­

ment in the corner. .3tep 3

Ring TenSion & Cantilever

Ring tension per meter beight

:~oment

~

of ;'iall

cantilever moment for ln strip trian~~ar

will be calculated 1'or a wall wi+-h fixed base, subject to

water pressure plus t~e effect of an induced moment equal to - 2.97 a)

Ring tension due to triangular water pressure wili be

by IDultiplyin~ coefficients f~om table 1 for ( ~ / Dt )

computed

=

~ R / H2

=_

2.97 x 5

52

=

= 10

by

0.594

Can·tilever moaent due tc triangular water pressure will be cca­

puted by t1ul t iplying coefi ic ieot from table VII for E2 / Dt 2 ~ PtIaX H 5 x 5~ 125

=

d)

by

Ring tension due to an induced moment of - 2.97 mt will be com­

puted by multiplying coefficient from table VI for ~ / Dt

c)

10

= 5 x 5 = 25.

Pmax. R b)

m~

= 10

by

=

Ca.:'ltil<:~er ::lo~ent due to an induced t:lo::ent of - 2.97 mt

Will be

57

=10

computed by multiplying coefficient from table XI for H2 I Dt by

M

=-

2,. 97 mt.

The final value will be as given in. the following

table and figure

Ill.22 :

'-"',-.,---------------,----------------, Ring

Tension

i .

s:l

o

*'

p., .~.

oH

I

F'

0.265 0.124

0.098

0.23

+ 2.46

+ 0.1}6

2.596

+ 0 •208

0.64

+ 5.20

+ 0.38

5.58

0.011

.1H

1

- I

+ 0.56

.3H

+ 0.323

0.94

'~4H

+ 0.437

+ 0.73 Po1.00

0.43

+

+ + 0.82 13.55

+ 0.608

+

+ 8.10

+

+ 8.66

0.179

20.87

4.47

o

o

a

+

o.ooelo.oool

.oco +

.000

'

0.002'

+

b.ooo

+ + .0001 0.009 p.Ol}

+

"

I

+, +1! 0.027: 0.0'+0,

+ + 0.15 e 0.24'7

+ + .0019 0.067 0.238

0.200 0.438

+

+

+

+

0.55

+ + + .0028 0.123 0.350

7.93

.0012 0.467 0.150

o

.0122 1.0CO 1.530

+

+

+

+

0.369 0.019

+

+

12.4-0

+

H

I

0.006 0.006 1

+ + .0007 0.053 0.088

+ I + 19.48 11.00 11.55 +

I

+ 0.49 '13.06

+ 7.91

.9H

0.000

11.43

+

+ 0.440

lo.ooq

+ ! + 0.084 0.134

.7H 0.589 11.63 14.73 ' 6.82 .8H

0.000 ~.OOO

+ .0004 0.028b.050

12.36

+

!I

.000

+

+

+

5.20

0.38~

2.84

+

4.79

I

I

+ 0.21

+

.6H

I

I

Coeff. Coeff. , T T T poefi. Coeff. M I M ill Table Table tim tim total able Table mt/mmt/m total VI due due in VII XI due due, in to .eo tim i to to I nt/t t:. load moment load IDOlOOnt I load moment A load mome~:t

'f"I

.2H

I

Cantilever :';[oment

o

2.97C 4.500

1

58 SheltI' Q

at

0.158

= =

or

base PH

-+-

0.158 x5 x 5

5.81 -+-

U / H

5.81 x 2.97 I 5

=

,.95 + ,.45 = 7.4 tim - --, 6cpa"hvrnI

I

"'"'1m

-l

:Ti lIll

6¢>IO

I

e

..

u.ce

0

456

E

....

II

"500

-'

E

..... -e­

CD OIl

E .....

.... OIl OIl

2l\

Cantilever

Rino tension

moment Fig. 111-22

Step 4

Radial and Tangential

~OI1ents

in

Floor Slab :

The radial moments are computed by selecting coefficients for clD = 0.15

5.875 x 25

from tables nIl

= 147 mt/m

&

xv ,

and multiplying them by

( for fixed edge ) and by M

= 2.7

mt/m

moment at edge : Radial moments in the last

line are for a segment havi.:lg an arc 1 m long at the edge ( point

1.OR ). They are obtained b J mul­ tiplying the original moI:ient per meter by the fraction indicating its distance from the centre,for illustration 5.3, x 0.6 ~he

= 3.2

mt

moments in the 2 last lines

of the table are plotted in

Fi~.

p R2

Fig. II1-23

= for

59 III.22 • The radial I:1o::ent around the high, and toe actual value is oniy

= 14

mt.

c~ntral

= 0.7

th~

colUI:1Il is relatively

theoretical 0.7(16

+

4-.})

It is however recolltilended, in this case to enlar,;e the co­

lumn capital as shown by the dotted line fig. III.2} 0.1 H

at thickness

Point

75 cm

0.15 H

0.2 H

66.66 cm

58.33 cm

so that :

0.25 H 50 CI:1

.15R .20B .25B .30R AOR .50R .60R .70P'

- -

-

- -

-

.80~

.9GR 1.OR

-

-

Coef.f • T .XIII + .+ + + + + Fixed, 10-2x 10.9 5.21 2.00 0.02 2.20 2.93 2.69 1.69 0.06 2.16 4-.9

-

Coef! • Table + + + + + + + I M at edge 1.59 .930 .545 .280 .0.78 , .323 .510 .6'53 .7se .90(1 1.0

,xv,

!::adial

3ixed

- - -

-

-

Lloment + + + + + + Edge 16.0 7.67 2.94­ 0.03 }.23 4.}1 3.95 2.48 0.09 3.18 7.2

-

-

-

-

+ Radial ~oment + + + + + + - I ­ , I 2.14­ 0.21 1.38 2.43 2.7 0.76 0.87 1.47 1.79 edge 14.302.51 1:.1 at

I',eotal Radial :.:oment I:J:,;/m Total Radia:l. !,~oZl/Segm.ent

I ­ -

Itt-.O 8.50 3.50 0.73

-

-

+

+

+

-

5.18 5.33 4.27 2.23 0.75 4.5

) .I~

-

-

+'

+

+ +

+

+

..

'

+

+

+

-

3.05 2.Q4. 1.10 0.22 1.38 2.59 }.20 3.00 1.78 0.68 tt-.5

The tansential moments are computed by selecting coefficients 2 for clD 0.15 from tables XIII & Y::'{ and multiplying them by w R =

=

147 kgm/m ( for fixed edge ), and by M

= 2.7

ge ).

mt/m

(tor mO!ll.ent at ed­

'.In.-.llm

The total tangential ben­ ding moments are plotted

.~f

in

figure 111.24. The effective depth within the drop

panel

is 46 centimeters instead of

32 ems. in tlle rest of

the

floor slab, and if' the moze ntrs

I

J.

Fig. 1II-24

I

...

\

r

60 ill that region arc reduced in the ratio 32/46 it is seen critical

~oment, ~.822

occurs at the edge of the drop panel.

.­ I Coeff. tTl XIII) 2.18 2.84 2.43 Fued, 10-2 x

..

'rang. at '"

-

­

-

Goeff. Table XV, rt. at edge Fixed

the

1.01~1

.15R .20P. .25R .30R .40R .50R .60R .70R .80R .90R

Point

~ang.

that

-

...

-

+

+

+

+

-

-

+

+

1.77 0.51 0.31 0.80 0.86 0.57 0.06 0.98

- -

-

-

-

-

-

.+

+

+

.319 .472 .463 .404 .251 .100 .035 .157 .263 .363 .451

-

-

-

-

+ lIoment I ­ + + + edge .21 4.18 3.57 2.60 0.75 0.47 1.18 1.27 0.84 0.09 Moment !I + + +

I;,

­

Total Tang. ::oment mt/m

a) 3a11;

-

-

-

-

-.

-

.I.

-

+

+

I +

+

-

iO. 86 1.27 1.25 1.09 0.68 0.27 0.10 0.42 0.71 0.98 1.22

edge

Step 5;

-

1.4LJ

-

+

+

+

4.07 5.45 4.82 3.69 1.43 0.20 1.28 1.69 1.55 0.8S 0.22

Design of the Different E1ement'!i f~.

rillg tension

Tmax ; 13.06 t at 0.5 H t =25 cm

max. ring reinforcements 13060 1400

s :: 1400

(0'

= 9 .3

6 ¢ 10

kg/cri)

1m on each side ( 9.4 cm2 )

max. tensile stress ill concrete ( including shrinkage ) !ssuming Es h = 0.00025 Es = 2100 000 kg/cm 2

=

13060 + 0.00025 x 2100000 x 9.4 100 x 25 + 10 x 9.4

=

13060 + 5740 2594

'fhe max. positive mo!:.ent in wall

=

=

456 kgm

7.27 at 0.7 H from top

Vertical reinforcement on outside surface: A = 456 1.6 cm2 chosen 5 ~ 8 mID (min). s 1300 x .22

=

Section at foot of wall : max. negative coment M - = 4500 max

kgm.

61

t

=

V ~5yO

= 38.7 c::s

3

37 = le1

As =

-.I L:-500

kl

4-50\:: 1280 x .;7

= 0.553 =

b)

1550 1;00 x .22

=

C:lS

=1400

~s

co.2

9.5

chosen

kg/co. 2

7

~

13 tmn/m

5.44- co.2 chosen 7

~

13

M = 1550

at top end of haunch As =

for

chosen 40

kgm

rmn/m

O.K.

Floor Slab The column load is determined by multiplying coefficients taken

from table XVII for clD 2.7 mt/m, for induced

=0.15

by P R2 , for edge fixed, and. by

l.~

morae nt

Col. load when edge is fixed Col. load due to induce mooent at edge

9.29

=

147

t

=

34

t

'1'0tal

181

t

x 2.7

Diam. of critical section for shear around theoretical capital 1.50 + 0.50

= 2.00 wit~in

is

ms

x 2.00 _ 6.28 ms

~

Length of this section is given by Load on area

=

x n x 12

5.675

the section Q

= 18.5·t

162500 25600

=

.87 b d

Diam. of critical section for shear around drop panel::: 250 + 35=285 co.

Length of this section

=rt

=

x 2.85

Load on area within the s:ection . 9.875 Shear stress "t =

Q

.87 b d

=

9.

ms

X.Tt'

.x: 1.425

18100G - 27500 •87 x 900 x 32

=

2

= ;7.5 t

143500 25000 .

?

R- - column load

Shear at edge of wall

= P

Length: of this section

= 5.375 x 3.14 x 5 - 181 = 460- 181=269 t = 3.14.x: 10 = 31.4 mn

It"

.

2

62

Shear stresses 1 The

=

.87 b d

horiz. reaction of

the

Q=

wall

Jq;lcm 2

= 3.10

269000 .87 x 3140 x 32

=

Q

7.4 tim will be resisted by ring

reinforcement at the foot of the wall.

=

10

The max. positive radial moment

~

5330 1240 x .32

10

llax. - va B .M. at wall surface

,

32= kl

cm

mm/m

13

,.

M - = 3500

y'3500

= 17

kgm

"

kg/cm2

.. k1 = 0.54

0'

c

= _ _3",,5.:.;:0:.;:0:...-_ = 8.6 cm2

AS

= 35

=, 5.33 mt.t

e

19 mm ,

"

6 x 350000

100 x 35 2

O't =

~

,=

30

=

.~

~

chosen - 7

1280 x .32'

1280

13 _ =/m

Top radial reinforcements at middle "max - = 14mt

t = 65.66

x 14 x 10 5 100 x 66.662

=6

o 'C

= 19

14000 1250 x 63 If.

-= 8.5 mt

t

6 x 2.5 x 10 5 '

at = 100 x 58.33 2 As

=

:.: -= 3.5 mt

at

= 17 cms2

t

r

= 75

ems

kg/cm2

2/m = 18 cm

total

As

= 58.33 ems = 15 kg/cm2

at

r

= 1.00

m

total.A s

= 12.3

x

= 12.; em2 1m

8500 1250 x .55

at

CI:lS

= 35

cas

at

= 18

z ,Tt X 1.5

Tt

= 85cm2

x 2 = ?8cm2

r = 1.25 ms

total As

= 8.6

x n oX 25 =67.5cri

63 :'he above calcul:l.tL~s S~O"l t~•._t the total aao:..l I.t 0; the to:, dial reinfo::.'cet:e.:lt does ::.ot

VaI'J'

..

,;.,­

much in the ce~tral part of tee 51.'ll:>

uced,

around the c o Lumn head so t:-.at the same nunbez- o£ bars :na"v be

at snaller distWl.ces ii:I. the midJ.le and. at bL;;.,;er di5tanc~s as we move outwards. Therefore :

Use 32 fi' 19

mIll

arranged as shown in fig. III.25 in which we .use 4 t~;­

pes of bars of· the following form:

~4·fi'19 ~4~19 ~~

4

l<:'

19

4

~

19

32

a

19

/

. P'1g~ III.25

i-,:ax. ring .reinfo::'cementG in lower

fiber at x .. ,

= 0.75

r

lZ:tang 1690 = 4.4. cm2 = k d 1300 x 0.3 2 Ring reinforcements in upper fiber at A.

s

4822 fi' 16 m.m 1250 x G.} Ring reinforcements in· upper fiber at 2750 1250 x O.}

c)

fi' 10 mm

:::

Central Colu."1ll

¢ I} m.m

@

= 0.25

x:

@

.r

15 cms

x .= 0.;5 @

cm

15

r

15 cms

P = 181 t, c:P = 60 cms,

A.

s

= 12 fi' 16 mm

P = O"c ( A.c + n.A. s ) or 181000 = O'c ( 2850 + 15 x 24 ) = 24 CI:lS or ~c = 56 kg/c~2 O.K.

The details of reinforcements are shown in fig. 111.26 It has to be noted. that in the given details, tent up bars have been avoided because the shear stresses are low (~. value

=

2 kg/cm ) , mo re ove r , such bars in walls are liable to move from

6.35

their

position during concreting operations and the laying of straight bars .~ circular ~lements is easier.

64

R- D'

1.$

/1.

I

•

..;•

~.. I~..f

aPLlNo; SHO/Y/,i/tf BCTTOH R£/1oI1': Or Pi 00,(;

4~A$

LRP3 /N AU. ItO.'!

• ~~,I.,. II00K$

AND 1'D IE

$rRG.,r~~D.

~/.et:UI.l1J! WnTER

rR,oRIt"/TY

.rRN~

J7S,,'/

FiQ. IIr -26

65

TV . R 0 0 F S

.A. N D

FLOOR S

0 F

C IRe U L .A. R

TAN K S

IV.l. INTRODUCTION Roofs and floors of circular tanks iIJaY be of the beam and slab type as shown in figures IV - 1 a and b

-Ltl I IJ~

.

.I

' .

.

"

Fig. IV-la

Fig. IV-lb or of the flat slab type as shown in figure IV- 2. .

:Fig. IV-2

66 The determination of the internal forces in such sys tems inclu­ des no special difficulty as they follow the general rules of the theo­ ry and design of reinforced concrete. They may also be composed of one circular slab with or

Without

central support. The internal forces in these two types are given in CROSS

seCT/Oil R- R

SECTIOAI

S£CT/ON

$- B

SECT/O'" C-C

L'.~g.

IV-3a

67 . the table of the P.C.A.

'l?he method of

ap~·licaticn·is

shown i=:.

the

numerical exazapLe just eX;Jlained. Circular slabs ma;y however be used in circular tanks und.er different conditiOIlS of supports as shown. in figures

IV-3 a &. b.

Figure a gives a

~'later

'to'lier in which circula:tl

slabs with and Without cantilevers and central holes are used for the roof and the floor of the tank, the intermediate floors of the tower and the foundation. Figure b shows a circular tank in which the roof circular slabs are supported on circular beams and co Luzns along radial axes.

·O
The determination of the internal forces in such

i

j

[I

J

!

'--.

.­

.~Y;J."":f'r'

I

Ii

-

I .­

arranged

,.... .....,

,-J!

L-,

r- '-, "

Fig. IV-3b cases are given in article IV-2. In :nany cases,

the choice of a rei::..forced

of revolution - for exazpl.e a dome or a .tiJ:Ja te

saving

in

cost

even

wnen

the

concr3te

cone - results ,p:-eater

surface in

expenses

an

ul­

of the'

, 68·'

shuttering are taken in consideration. Figure IV-4 tank cubed

500q cubic meters capacity designed by the at

El Nasr City,

Cairo.

Figure

shows

BJ+

author

Inze

and

IV-5 gives another

exe­ examp~e

"'f

.8 o o

...: ,·v.

Reinforced concrete

Inze water tower with conical roof Fig. IV-4 of a circular reinf.'orced concrete reservoir covered by an B

spherical dome.

&.­ 9

s D

Reinforced concrete dome covering a circular tank

Fig. IV-5

CI!lS

thick

69 IV -2.

INTERlTAL FORCES L'i CI ?CUL.l.R FLAT PU. TES I t Will be assuned that :

p

= Uniform

P

= Total

load, concentrated load or live load per unit

t

= Plate

thickness.

load per unit area.

E = Modulus of elasticity ( generally = 210 000

len~~h.

kg/ca2 for concre­

te.) V

= l/m =

Poisson's ratio.

and 1/5. 1/6." D =

For concrete, it varies between 1/10

It is generally assumed in the follOWing equal

Some neglected its effect

12 (1 -

~~d

to

assume it equal to zero.

Flexural rigidity of a plate.

v2) =

6

=

Deflection of the middle surface of the plate.

If!

=

Slope angle"

A

=

Roaction at the support.

"

"

"

"

"

"

~ = Radial shearinG force. r = Radial bending moment.

M

tit

= Tangential bending moment.

a

= Radius of plate from its middle ax:is to the support. = Span of a circular strip of a conti:Q.uouS

'p La be ,

b

=P a =

radius of a specified circle on the plate.

r

=pa

radius of any circle on the plate.

=:

All logarithms appearing in the following equationsJE are natu­ ral (i.e.

to the base

e) •

Reference Beton-Kalender 1952 • Volune II. "ililhelm. Ernst & Sonn. Berlin.

Pa@& 271.

~:Unchen.

Published by

Dusseldorf.

70 Circul~r

1) -Solid

?late Subjected to a Unifornly Distributed Load. (Fig.IV.6)

Load ~

~

p

Total load

const~t

P =­ --.r> 2

~':b:ed

s.)

l.:r -

A

P- -[ 16 P Q2 64 It D

P .; =__

....r

16

a

2

2

a

TC

Edr.:e

1 +v-

C3

+ v )P

11:

o ~

It

=-L

a

It

P == P

•

2

-

:--<.'!!

2J

(1 + 3 v ) p2J

U - _P_- [1 + v too: _-16 -rr

2

P a

(l-p ) (.2....::!:.l':' p ) 1 + 'I

16 n D '. p "';t=-

(:2:;,.I +

16

Tt

It

• p ( ~ - P 2) 1 +v

+ v

~_

(1 +

.3 v)

p

.,

-

0.007

p -­

_Oef2 3

.

or

p

Fi~.

IV....5

P~

71 2) Solid Circular Plate Subjected to a

Conce~trated

Load P

a~

Center

(Fig. Dr. 7)

~ =-

-p-.

1:.

for

P

>

G.r

p,

2 It a p a) Fixed. Edge 2 P a 2 2 6 = (1 - P. + 2 p In p) 16

It.

:':t

=-

!1r

6

..

~r

=

J)

-~. v)

~

It 1::>

=- -'...-

In p ]

r

~'r

16 It D

[ ~(1_p2) + 1 + I{

= -P- (1 + v ) 1n 4 r;,

_

-

:'.f

~'t

A

=--L

p • 1n p

for for

2jlnP] ) )

~ p

It

= -P- [1 - (1+ v) In 4

p

=C

Edge

'.ot = - P- "[1 -v- (1 + v) 1n ~

)

(1 + v) In 13

Su~norted

P a 2

4

= 0,

,2 It a

41t

b) Simnly

p

., = - -Fa.

+ (1 + v ) In p

- p - [ v + (1

= Mt

for

40ltD

= _.-L [1 4-Tr- 4

0

D

M

r

=

It

p

J)

If'

= -Pa- , • p I

-'-I- It D

for

P::>

for

p =0

1

( - - 1n p) 1+ v

'13

)

)

pJ

"

~)

= P

Total load P

~=-

2

= P n: a

?

2

=-

~ ~"ixed

'Tt b

a

rt

if

2

:Uo3.d at t:Je

~:idd.le

IV.B)

(:Fi.:;.

a)

Unifor~

.30lid Circular l-'late S·.lb,jected to Partial

2

p

A = P /

2 r:: a

for

P

P

1

2 Tta

p

for

<- P:<

p

1

Ed.)e

i) I:mer loaded Part

'__'

6

P a2

[

64 rr D

Pa • = 16'yr D

~

1\

r

=--l:.-

1611

2

4-3P+4~

P ( f3

2

- 4 In

[ (1 + v)

(~2

2

lnp-2

j3 -

2

~)

p

- 4 In 0) ...

T

11 - -P ­[ (1 + v) (p2_ A,ln 13) - 1 -t - '16 It ii) Outer unloaded 'Dart

=

5

P a

2

[(2

tr D

32

Pa

-p2;

'. p [a, 2

? )-

=

l.~

P =- [- 4 + (1 + v)

16

D

It

16

:i~Dly

(1 -

It

,y-

2

4 In P

,...IS + P

6

-

J

~

-<

1

f'

pJ

J

J

(1+ v) (p 2- 4 In. p)

~2, 2 (1 - v ) --:-2 + (1+ 'tI) (13 _ 4 In P)J P

Supnorted Edge

0

ID:ler loaded part

i)

3vp2

P

(1 _p2) + 2 ({J2+ 2/) In

'P

'I ' t = P - [ -4 : 16 It

b)

p

'2

p2 ]

'"

p

~

J3

P a 2 , - 1- { 4 (3 + v) - (7 + 3 v) f3 2+ 4 (1+ v ) P --m ~ p

- 64

It

D • 1 +v

- 2 [ 4 -(1 - v)P

2

- 4 (1+ v,) In

P]

-;z­

p2 + 1 +v

73

=

Pa 16 It' D

.-L., 1 +V

[4- -

(1·_11) 13

2

~(1

:; .

" PA 2 I"'

+v) In p - l +

J

P

16

Tt

ii) Outer unloaded part

P a2

= 32' IT D

. 1

~

p

'\I { [ 2

2

(3 +v) - ( 1 -,,) 13 ]

"

(l-lh +

= 16

Pa It

D

-L {[ 4 1 + ....

-(1-'1)

P~

p

<

1

2 (1+")!3

4-

2

lnp

(l+v) p2 In p }

2

P -(1 +v)

Lp

- 4 (1+ \/)

P in P }

P

16

IT

P

16 Tt "

.....'

r=a

a

a

0.0 3 p

p

~. (a) (b)

Solid Circular Plate Subjected to Partial Uni.fOr!ll Load

at the Middle

Fig! IV-B

74 301id CircuJ.ar Plate Loaded b·,r a :.in:;

4,)

Pic

;ci...:.r~ load

'<.r=

0 .(. p

fo!:'

0

A

c

13

Q-r

L::l~d

= P P PI3 =-­ -?

p 4

for

Fi....ced Edi>;e

c.)

i) Inner

-

I; a'5 p [ 1 8 D

C =

:':r =

0

p~t

L:t

= - ~a

p

f+

~

v)

(1 +

p21n 13 + (1 +

(1 -

i

<

p

~

p

p~ 2 In p) p2

J

+ 2 In 13)

ii) Outer part

e '.\. 'P

~.

3

LL P [(1

=

,pti

-- p p 4 D

p" )

'.

[2~ (1 -

(1 -

r?\

+ 2 ( ~2+ p2) In p ]

1 ] - 2 ) - 2 1n P p

fJ [2 -(1 -

"'r = - P 4- a

0)

+

8 D

~-

v)

(1 +

v)

.p

2­

2 In p )

J

.Sim.Dly Supported Edge i) I:n.r.er part

{j

(13

=

? e. GD

3

= ~ ~2

0

.-L l'+v

.

{C3 + v) .

(1 _

~2)

c

p


+ 2 (1 + Y) p2

2 '

In 13

. -.[(1 - v)(l - jl ) - 2 (1 + V) In j3

• : : v • [(1 - v)(l - j!2) - 2 (l + v)

In:

]

p]

p

2

}

p

.c ...,

75

3 6 = P a

8 D

? a2 'P = - -

.L.L

./i

r

[ 2 -

1 +v

4 D

P a = - ~ [(1 -v) 4

2

~p-

(1 - v) j3 -- (1 + v)

1

~ (~- 1) -

p

~ !

"J

)

M:=:"

2 ( 1 + v)

P= o. 50

I

2 (1 + v)

~1p ]

pJ

In

'P

Pi

~----l.i

....l...-_ I

+

r=ap I"""'" I

a

6

I

!

. 0.188

M

r

IP ,

lIllllJITJ;l]lJ11'rmmrrmIlJl:l"'1,

I

I

I

I Mt

1~-"'l I lIl• • •[lIlIlIl"'-1

I

[ I i

0.031

0.156

i rrmmmrrm'J'l"l'!1T!'mTmTiiilT1l'11T!1Ti'7 P a I

!,"""'WIIIII:IIII'

II 0.50 )

I

I

! ..

I~l.oo: . I ~,

0~50:!,' ,~I I

(

'

I.

.

1',';; ,11:

I ( b )

Solid Circular Plate Loaded' by a Ring Load Fig. IV-9

;

P

76 5) Solid Ci:'Cul3.1' Plate 2u'tl;1t>ct",d to an Inter::.ediate :':J.a.ial R:'n;- ::c::e:rt

(Fig. IV-10) .~

= C

A = 0

r'ixed :!X.r'e

a)

i) !nnel' part

.. 2 B =- ~ 4 D

.• • :'":r = '~t

=­1.:2

r l

2

2 ~ • Ln !J + (1 -

2 p)

P

lfl

2

.

2 .

lfl

4 D

r.

= -

(1 -

2 D

P
6 = - - - • P (1 - P + 2 J.n p)

Ll

=-t:a

~ P-)

P

2 (1 + v ) (1 - ~ )

ii) Cuter pa:::'t ~l a 2

2]

~2 p2 [1

l! =.2

~

a D

1

• p2,(=P

- p)

2

+ v + (1 - v) 1') ] P'­

.,.·t=

-

1: ~ [1 + -;-

- 0.042

( 8.»

( b )

Fig. IV-10

v -

7J

( 1- " ) 1 ¥

??

[) = lj.

D

"

­

.-J:......

"P =:::......::::.

2 D

t.~

1 + v

~

= Mt =

[1 + v + (1 _ 'oJ)

~2

J

ii) Oute!' part

B =

1:: a 2 4- D

'/I

~! a 2 D

6)

L 1 + v

L 1 +V

2 [(1 - v) (1 -p ) - 2 (1 + y) In

[(1 - v)

p + (1 + 'J) 1 P

pJ

1

Sirm1;y Supported Solid Circular Plate Subjected to

Radial Ring t:Or,)9l!ts at its Edges Fig. IV-ll

M

,- kg:;

Radial mcraerrt

.:.0...

I!l

B

=2


=

2 " a

(1

D (1 + v)

.... a D (1 +v)

2

-p)

'I.!'

.'''r, = .~

=

v

-t

0

P

= .... ~

A = 0

Fig.

rv-ri

78 7) Circ'ilar Plate with a Circular Hole at the Center to a Uniforoly Dis~ributed Load p/o2

-.I,:

a) Fixed Edfje k

assuming .

o = -'::l

0.4-

[

64 D

__ n a}

Ijl

[

.0;,.....=...

16 D

-

1

~··t

[

(1 - v){32 + (1 + v) (1 + 1 - v + (1 + v)

(1 - k 1) P :-

-

2

2

:,

(J+

+v) (1- k

2

1

k 1 -p + 4- j3

1)

+4~~ C3

2

P l.n P ]

+ ... )

v) k 1•

?

1)

+ 4-" ~-- (1 +

Jr

+ 4- (1 + v) 13

(1 + v) (1 - k

assunung

=

6:D

2

+ v + 4- (1 + v)

1) a.3 =16 D

'P

16

2

'? 0.

.

.- = ­ ~

~

[0

(1 -

p~

[_1 1 +v·.

.. =:L£ [ "r

. 2

1 +v -

p2

3

o

+ v) (1 _ 2 ~2) + k ] 2 4-

1 -v + v .

2

p

v )

In p ]

~ 1

a4- {

2? ] - 4- k 1 In p -813 p""ln p

4-

b) 3im,'ly 3uP'::lorted Edge

"0

then

f3

In

2 'In p ] (1 - v) k 1• 1 p2 + 4- (1 + v ) p.

+ (1 -

5

~i

P

2 ~ ) (1 - p ) + p

1 + 2 (1 - k 1 -

[0

" :: d16

=p~

13 > 1

:for

~ (r - £ ) 2 p

=-

Q...

(Fig. IV-12)

<1

~

for

~,1 SUbject~d

~

f)

(1 -

4~

2

.3

k

2 + k 2) p- P + - 1 - v 2

(1 -

Li p

(1 +

~)

1

-+4-

P

+ 4- (1 + v )

(1 - v) (1 _ 213 2 ) + (1 +.3 ,,) (1

+

then

• k 2• In p- 8 P2 p 2 In f' }

(3 + v ) (1 - p2) .+ k

.1.0

P]

In

-fJ

+ 4- (1 + v)

~2

?

-s ) ?

P-

ln P ]

P2

pln p

In p

J

J

79

Circular Plate with a Circular Hole at 1ihe Center Subjected to a Uniformly Distributed Load p / m2

-_.-._--

p

p r=ap, a

p=aB

~

...

..-\:>.00 i

.~

1 ~pa "

... 0.144­

+

1.225

1.r""I""""""--+-­__~ pa

2

'1"

N

I

l-mi

0 .1]1,.;..8

,""""""

I

II!.:"'",!' t

:

i~.50p I

Fig. IV-12

,

pa 2

I ~pa

80 8)

CireulaI' P1a.te with a CirelliaI' Hole at' the Center and Subjected to a Ring Load a1onp, the

I~~er

Edge e

=p

(Fig .: IV-1, )

Load P / m' A

=P

:>

13

:f.'or

j3

1

:f.'or

=-P 13

A.

13

> 1

a) Fixed Edge

B =

p ~,

8 D

l' a 2

= p2

k

ASSUI:ling .

1 + (1 + v) In P 1 - v + (1 + v) ~2

,

'.,

l~j)

13 [(1 + 2

p[

1 k, (p - -

2

(1 -p,) + 4 k

-

,

then

1n 'p + 2 'p2ln P ]

]

'4>

=

!J~

=

~ 2

P[ - 1

·r.!t

=

~ 2

1

J3 [ .- v + (1 + v) k, - (1 - v) k, p2 - (1 + v) 1n

2 D

)

P

+ (1 + -,v)

p 1n P

k,

+ (1 .,

II )

1 -3 "'p­="'

k

-

(1 + V). In p

J

p]

b) Si:::-:,l;y Sup'Dorted Edse ASSU!ni.~g

5

=

a'

P

8 D


!':r

~.

• =

·,'t =

P a2 2 D P a

2

P a 2

k4

13

[

=( 1 + v)

then

3 +"v - 2 k 4 . (1 -fl) + -4 k 4 In p+ 2 P 21n p ]

1 + 1 _ k

p [

1n~

L 1-rf

1 +

p [ k4

v

1 - v

tc - - -4 . -1

4 p

1 - v

II

(~2"-

p

-

P In P ]

1) - (1 + II) 1n P

i3 [ 1 - v - k 4

(~+

1) -

J

(1 + v) lnp

]

81 .

CircuJ.ar Plate with a Circular Hole at the Center

SUb;)ec'ted to a. lUng Load. along the Inner Ring

=r

Pa3

B

"""':'""~~"TI!Il!1I'~u

,0.229

! !

0.034

Ioora:rib----i--__""-' P -to

.1 1"----9 ,,14 0, " I

1

--"""'1-

a

i

i

0.038, 0.718 Inmn=--+--.mrmnmr' P a

T

" rr'l

~

I

~i 'll!i:I~';;rf:

p

I

( a )

( b )

p I---+==:I==~~

t? P=. o

o=a

"

t

Pt P ~ x. ==F==~! . -­ 'I' •b=a @

I

6

"II

P '( b )

( a ) F1g~IV-l~

·

.,

82

9)

Circul~r

Plate

Radial Ring

a Circular Hole at

~it~

Uo~nts

on the

I~~er

Center ana Subjected to

t~e

Edge.

p

= ~

(Fib' IV-l4)

~

Assuning

k

2

5

t: a = ----. 2 D

5

=

=0

A

= 0

2

'3 1 - \/ + (1 +v) ~ 2

2

k5 (- 1 +P-2lnp)

r = - :.I k 5 [ 1 + V +. (1 - v )

M

then

" a

~

-

~~

1~

=-

• k5

c;1 -

11 k 5 [ 1+ V' -

p)

(1- v)

6

M M

M

M

( a )

" Fig. IV-14

( b )

?J.

83 b)

Simply Sup:'()l'ted Ed-se Assuming

.;

2 !.l' a 2 _Sk (l-p B =-2 D

= t!

a D

( p

'then,<

2 1 + v lnp)

1 - v

1 + v

1 +.... 1 +_.1 - v

1 +v

,

)

P

iO) Si..m.nly Sm)'()orted Circular Plate with a Circular Hole at the Center, . and Subjected to Radial Ring Uorrents alorg the Sup'Jorted Edse • t 1 then (Fig.IV-15) Assume rad.ial moment l:l an kg/m and k ~ 7 .1. :-~ 2 I ~( 1 _ p2 _ 2 ~ ?2 In p)

(5 =M-a,1 - y .

2 D 1 +v -, i

=

tp

= l:!

a

D

~ 1 +v

(

p

+.L.±.:!... ~ 2 1.)'

1 - y9

6

<'

Fig. IV-15

II. RADIAL 1m:':E:1TS

(1'~r)' TA.\GE:·~TI.AL MO;m~TS

(M t ) lUll) REACTIOnS OF CIRCU­

LAR FLll.T. PLA. TES CmlTI NUOUS OVER RING SUPPORTS AT EQU..u. DISTA1\CES (a) ,

a) Three equal I

P=1.5·

Case l£?

0.5

III'

"1'

I!

snan~

l' t/m2

0 I"

:1"

.

II!,

'1

b) Four egual snanEl 1.0 p=2.0 0 l' t/m2 l 1 £LUlll ! I I X' ! I l' I I ! '1. pase 2 j

I

Case 2 ~

I I I I "

a

-1

'1'

1

2

1

?

I I

I

I' , 'i? ' , , , " l a I a l

pase 2

F

j' ," '1" , , 'i' a l a I ~

P::o,

v =1/6

j

'i~ I a I

I "

a

Mr =M-~ t­

v =1/6

..

BElIDIHG '.

p

Case

Case

1

o·

Case

2

r

"

2.0

1.3 +.0578 +.0271 h0132 +.0049

1.6

1.1 rt.0889 +.0225 f+.0330 +.0119

1.5

1.0 fr.092;;

1.!~

+~0151

Mt

M r

1:...:

.LOi.

1+ .0188 "".C835 -.013S

1.5

Case

1

2

P

~,it

I·!r.

BE:IDI!IG ::miEHTS

Llm.~aTS

+.0455 +.0109

0'

M.

jlr

"

+.0129 -.0811 -.0135

.0847 +.0179 +.0546 +.0093 ~.084B

+.0127

+.04"~4

+.0085

.0748 +.0053 +.04-49 +.0057

.­

0.9 It·OB75 +.0042 +.0506 +.0068

1.0 ....1069 -.0314 -.0769 -.0172

0.7 1+.0411 '-.0267 rt· 0300 -:.0097

0.6 ....0;:,08 +.0080 +.0179 +.0109

0.5

-.0591

~.0552

-.0291

0.5 ft-.0137 +.OllO +.0284 +.0099

0.; h0535 -.04-+1

~.0235

-.0141

C.4 l!.O225 +.•0104 +.0322- +.0044

.0077 -.0066

0.1 \-.•0035 -.0310 - ..0244­ -.0676 . 2 2 2 2 L:ult p a ,1' a l' a 1'a

0.1

~.0851

.0376 1-.03G5

0.0 t-.03:57 1-.0357 I.~ult

p a2

~.O057

2 l' 0.

p a2

-.0057 l' a

2

REACTIOHS tiup.

Case 1

REACTIOHS Sup. 2

1+0.3655

+0.5612

+1.20C1

+O.·~131

+ 0.3518

+ 0.4535

1

1

+ 1.1642

+ 0.5898

0

pa

Case 2

Case .2

2

::ult

Case 1

1'a

+0.-+330 2 ;.:ul-:; 1'a l' a

+0.670<;: 1'a

P a<::

85 R.:illIAL MQ:.4E:i'1S (M) TALiGENTIAL MO~NTS (M ) AND REACTIONS OF CI"=L<.rJIJ..i-. r ' t FLAT ?LATES OVEH. RING . SUP.20RTS AT EQUAL DISTAiWES ( a )

cozrrsuous

I

c) Five equal spans

p =2.5 1.5 .5 pasel £' , , 'E' , , 'X" , 3 2 1

Case2

~, ,! '1:' 'I a I a

i ,

'Ok'

I

b

, '.x'

p tim

I

'!' 'X'

I'

a

a

I

2.5 2.1

I

~

, pase21' , L;&' , 'X' , 'X' , '&' "12 ' , ,

a

I

l

=1/6

Case

1

i.1 r

r"'lf' '~1' , ~1' , '}" 'ii' "10 ' '"

2

P

-r

Case .,

~

I

f a

v· =1/6

00 '

, r:rm.1EHTS

Case

1

"'r

LT.1;

0

+.C081

3.0

+.0098 -.0787 -.0131

·1 a I a

f a

, Mr::M"t;= ­ BE~IDING·

r.: t

~J

~'~t

0

Ia

fa

P =0,

Zl!OL:EITTS

BElIDIUG·

Case

, '1

'£' , , !

V

p

.

,, 'I"

t

d) Six equal spans p =3 2 l O p tim

2

.,

~,!

"'t

~'r

:".C7~9

-.0153

2.6 It .0814 +.0153 + .0354­ +.0076·

.0818 +.0159 +.03+8 +.0076

·2.5 1t.078';1 +.0115 +.0426 +.e073 .

2.0

.0799 +.0118 +.0425 +.0075

2.4 ft-.0659 +.0066 +.0401 +.0052

1.9

.0:),/6 +.0058 +.0403 +.0051

2.0 1-.1106 -.0232 -.0852 -.0149 '

1.5

~.1l30

1.6 +.0164 +.0062 .+.0521 +. 008~J1.

-.0254 -.0875 -.0155

1.1

.0132 +.0082 +.0302 +.0115

LO

.0289 +.0094 +.0432 +.0107

1.5 4-.0296 +.0069 +.0422 +.0084­ 1.4­ .• 0340 +.0058 ..

+.0057

+.0·~33

·LO 1-.0660 -.0121 -.0753 -.0164­

+.0069

0.6 fi-.0247 +.0120 +.0189 +eOllO

0.5

.O'j46 h0186 -.053'? -.0277

0.5 1-.0338 +.0095 +.0292 +.00;:8

0.1

.0029 fr.0039 -.0062 -.0052

0.4 fr.0357 +.0022 +.0327 +.004-0

0.9

0.0 Uult

\r.0375 fj-.OO'77

+.OL~83

ft-.C04-8 +.0048 -.0043 p a2

,2

P a

·2

P a

-.00L~3

2 p&

0.1 '-.0320 -.0809 -.0255 -.C6S:5 . 2 2 l('ult P a 2 pa pa pc.2 REAC·rrmrS Case' 2 Oase 1

REACTIO~~S

:3up.

Case 1

3

Case 2

3up.

+ 0.3675

+ 0.4594

3

Lf-O.3725

+0.4675

2

+ 1.2§5

2 + 1.0255

+1.1500

+0 ~9733

1

O.9~22

+1.0420

1

+ 0.7931

+ 0.8765

,'.iult

pa

"­

pa

0

+0.7373

tpa

pa

2

pa

+0.683!1­ J P a­

86

IV-~. :.IZMBRAliE FORCES D~ SlJRFA.CES OF REVOLUTION

It is assumed here that the

thicl~ess

3[

of the shell is so small

that it can beconsiderdd as a membrane which can resist meridian and ring forces, in the plane of the surface,

only

i.e. the bending mo­

:rents due to fixa"tion at supports, uns;ymmetrical loading and

similar

effects are neglected •

IV-.5 .1. Notations. It will be assumed that

( !ig. IV-16 )

...

.

'.

\.

..is

of

rOt8tic.n

Fig. IV-16

:e

= radius normal to axis of revolution of any circular ring at any plane z - z

Rl = radius of cu-'"Vatu:re of meridian

R2 = cross radius curvatrure al.ong the normal - to axis of rotation.

U.p (eve , T1)

=resultant

meridian force per unit length of circumfe

.z-eace ,

Ne (evt. T2)

=resultant

ring force per unit le~~h of meridian

H

= horizontal thrust of shell per unit length of circUl!lference

'.7 011

=Sum

if

of vertical forces above z - z' (expressed through the ansle19)

Refer to M~ Hilal ~dition.

"Design of Reinforced Concrete HallS " First

Published by J.

~arco

& Co. Cairo

87 .: /

IV-3.2.The t.ieridia.'1 Force ~n order to have equilibrium at an:y horizontal section z z ,

the vertical component of the meridian rorces N~

~t be equal to the

vertical load above z - z pee meter run circumferio'.nce •. Hence we

get:

r.

( Fig. IV-17 )

N~J\N'

11 · r

,N~

/

~
Nipsin~

\

I I

But

R

= R2

sin

~

N.pCos~

11=

z---!!=-!e Fig! IV-17

Z

z

-----

w~

/

N~

=

, so that

2 Tt R

=

\Vii' / 2

N'P

sin 'P

R

sin ~

It

or

N Ill· can also be given in the form.

(44)

The horizontal thrust H per unit length of circumference is

=

H

. IV":",,,.,,

Will / 2 tt R tan

\j'

=N

cos

ljl

(45)

The Ring Force Assuming that the radial component of the external loads

square meter suri'ace is p .

r

per

and considering the equilibrium of the ex­

ternal and internal forces normal to the surface, one can prove that: (46)

For a spheriCal surface

Rl

=

R2

N'9

+

Me

=a = Pra

and

"88

For a conical surface "'

= = Pr

Rl

and

Cl;l

Ne

~2

(48)

IV-3.4 ADplication to Popular Reinforced Concrete Surfaces of Revo­

lution. a) Spherical Shells The relation between a, R and y is given by (Fig. IV-18 )

:I

2 a =. R + 2 Y The surface area of a spherical

(49)

shell

z

is A.

= 2 na y

(50)

i. e. it is equal to the surface

area

H

of a cylinder baving the same radius a and height

s.

R

The internal forces due to a vertical dead load glm2 surface is shown

Fig. IV-18

in figure IV- 19 W If'

H

= =

2 nay g a cos Ij? = g a --.!..­ l+cos III a + z

g.

:>ad = g/m2 surface

~ ;----r,

1 ga

(51) . (52)

~

~ ,

~ H

p

I

. ' ga

r-i}'

~

'

-I

Comp.- • tension+ Fig. IV-19

~

Comp.-., tension+ ~ig.

IV-20

89 a

Ne

=

(53) 1

g a ( cos Ijl -

z -

(54)

a

+ Z

The internal forces due to a vertical load shown in fiSJre

p/m 2

horizontal is

IV- 20 (55)

H

C OS I.O _ =p a

=,p z/2

=p =p

= constant = -L (2 z2

(56)

2

NIjl Ne

a/2 a cos24l

_ a2 )

(57)

2 a

2

The internal forces due to a liquid pressure as that shown in Fig. IV-21

is given by :

w

Assuming P

r

= weight

=w ( h

l

3 w: [3

. a [ H..p = -'I.

hl+a

=+

h 1 +a

6"

Ne

I

1 m3 liqUid, 'then

+ a - a cos It)

.I

. _ \1h 1

(58)

:-IT I a

l~os p (1+2 cos 41)J(59 ~ 1+c06 ..p

L-c oe p

1+Oos

I

~ --_.

(5+4 cos'!')] (60)

--'--'­

<;>

Fig. IV-21 b)

Conical Shells The surface area of a conical shell (fig. IV-22 ) is given by :

A

=21t3.s/2

~.~

(61)

The meridian force is Ns =

V\p 1

2 rr Y cos ..p

(62)

The ring force is Ne = p . y cos ..p 1 Sin2..p r Interna~·forces

h

due to a vertical dead load 2 rY'/ m suriace are : O

~~1

Fig. IV-22

4>

I

"90 (64)

N

e

= s R2

p/m2 horizontal are :

Internal forces due to a vertical load

e =

N

(66)

(65) .

I y

p

y 'cos 3
The internal forces in the conical floor of an Inze tank (fig. IV-23 ) are

t

A.ssuming w

=weight

I m'} liquid t then Pr

=W

(68)

( h 2 - s sin IP )

p

"

-P/sin'll

N

e

N = .. s

£2li 6 s

[2 (l'}

where p is 'the

ma:r

h

(2- - s ) cos sin
s'}) sin.p - '}

weig~t

1mal loads that

= w. s

~

(1

2

- s2)


J

pi

(70)

s sin.p

of 'the cylindrical wall of the tank and the even­

be applied to it.

Since 'the cone can absorb forces only in the direction of its ge­ nerators, a ring must be provided at the foot of the

\~ll.

It receives

a radial load P cot