Question: Design a circular water tank of 13m internal diameter &amp; 5m height. The tank has flexible base &amp; it rests on the ground.
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Use M30 concrete & Fe415 steel. Use Working Stress

approach. The permissible stress in steel under direct tension is 130 MPa. The permissible stress in concrete under direct tension is 1.5 MPa. The following table can be referred to fix the thickness of the tank wall or, alternate approaches can be adopted. Draw reinforcement details. Use Working Stress approach.

Thickness of Members in Direct Tension (Uncracked Condition).

Civil Eng > Sem 8 > Design and Drawing of Reinforced Concrete structure

ddrcs mumbai university • 149 views
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 modified 7 weeks ago by Ankit Pandey ★ 60 written 4 months ago by
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Ans:

Given,

$D=13 \mathrm{m}$

$H=5 m$

$6_{at}=1.5 N/mm^2$

$6_{st}=130 N/mm^2$

1) Volume of tank = $\pi R^2 H$

D=13m

$\therefore R = 6.5m$

$V= \pi \times6.5^2\times 5$

$V=663.66 m^3$

2) Hoop tension

$T=\frac{Lrw \times H \times D}{2}$

$T=\frac{10 \times 5 \times 13}{2}$

$T=325 \ KN$

3) Horizontal reinforcement

$A_{st}= \frac{T}{6_{st}}$

$A_{st}= \frac{325 \times 10^3}{130}$

$A_{st}= 2500 \ mm^2$

use 24 mm $\phi$ bar

$Spacing = \frac{\frac{\pi}{4} \times 24^2}{2500} \times 1000$

= 180.95mm

$\approx 175mm$

provide 24mm $\phi$ @ 175mm c/c

$A_{stp} = \frac{\frac{\pi}{4} \times 24^2}{175} \times 1000$

= $2585.08 mm^2$

4) Calculation of thickness

$6_{at}=\frac{T}{1000x + t(m-1) A_{stp}}$

$m=\frac{280}{3 \ 6{cbc}}$

but $6_{cbc}=\frac{30}{3} = 10$

$\therefore \frac{280}{3 \times 10}$

=$9.33$

$1 \cdot 5=\frac{325 \times 10^3}{1000 x+t(9 \cdot 33-1) \times 2585\cdot 08}$

$= 195 \cdot 13 \mathrm{mm}$

$t\approx 200 \mathrm{mm}$

5) Vertical reinforcement

$100 mm=0.3 %$

$200mm= ?$

$450mm = 0.2%$

$\therefore Ast_{min}= 0.271 % bt$

=$\frac{0.271}{100} \times 1000 \times 200$

$=542 mm^2$

use 12 mm $\phi$ bar

$Spacing = \frac{\frac{\pi}{4} \times 12^2}{542} \times 1000$

= 208.67 mm

$\approx 200mm$

provide 12mm $\phi$ bar @ 200mm c/c

6) Design of base slab

Assume thickens of base slab = 300 mm

300 mm - 0.242 %

$A_{st} = \frac{0.242}{100} \times 1000 \times 300$

$= 726 mm^2$

use 16mm $\phi$ bar

$Spacing = \frac{\frac{\pi}{4} \times 16^2}{726} \times 1000$

= 276.95 mm

$\approx 250mm$

provide 16 mm $\phi$ bar @ 250mm c/c

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 modified 7 weeks ago  • written 7 weeks ago by Ankit Pandey ★ 60
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