Base condition b) - Pinned

Base condition b) - Pinned

From PCA Table 9/30A - Tension in circular rings (hinged) : p Lv = 9.81*5.125 = 50.3 KN/m2 Max. tension at depth 0.7Lv

= Ft = coeff. * p * Lv * R = 0.707 * 50.3 * 6.125 = 218 KN

From PCA Table 9/30B - Moments in circular walls (hinged) :

pLv3 = 9.81 * 5.1253 = 1321 KNm/m

Max. moment at depth 0.8Lv

= M = coeff. * p * Lv3 * R = 0.0053 * 1321 = +7.0 KNm/m

For design of reinforced concrete for :

(i) Min. reinforcement (ii) Min. reinforcement (3 day) (iii) Ring tension

(iv) Bond (v) Bending (vi) Shear

refer condition a

Base condition c) - Fixed

P.C.A 9/296 - Moments in circular walls (fixed) :Max. +ve M at depth 0.7Lv = 0.0035 * 1321 = +4.7 KN-m

Max. -ve M at base = -0.0138 * 1321 = -18.2 KN-m

P.C.A. 9/35a - shear at base of concrete walls (fixed) :

v = 0.169 * 9.81 * 5.1252 = 44 KN

Ultimate load conditions

Moment (Mv)

Shear (Vo)

service load conditions -moment ms

Using P.C.A. Tables 9/30

From the figure

Lv = 5.125m

r = 6.125m

t = h = 0.250m

( EMBED Equation.2

Using P.C.A. Tables 9/35a

Shear at base of circular walls (hinged)

Max. shear,

v = coeff. * p * Lv2

= 0.093 * 9.81 * 5.1252

= 24 KN

Using P.C.A. Tables 9/29

EMBED Equation.2

P.C.A. 9/29A - Tension in circular rings (fixed)

Max. tension at depth 0.6Lv

= 0.585 * 50.3 * 6.125

= 180 KN

Ultimate Forces

Direct Tension (Ft) = 180 * 1.4 = 252 KN

Moment = -18.2 * 1.4 = -25.5 KN-m

Shear = 44 * 1.4 = 62 KN

h = 250 mm

d = 250 - 45 - 16/2 = 197 mm

Elastic theory equations

EMBED Equation.2

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_916940971.unknown

_916941109.unknown

_916747449.unknown

_916937977.unknown

_916744401.unknown