Question: Design a R.C. Circular water tank for $350 m^3$ capacity. The tank is regarding on the ground having fixed base and it is free at the top.
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The height of tank is restricted to 4m including free board. Use concrete M25 & steel of grade Fe415. Also draw reinforcement details. Use IS code method.

Mumbai University > Civil Engineering > Sem 8 > Design and Drawing of Reinforced Concrete Structure

Marks : 14

Year : DEC 2014

mumbai university ddrcs • 1.9k views
 modified 3.1 years ago  • written 3.1 years ago by Aksh_31 • 630
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Data = $v = 350 m^3$ ; Ht.=4m

Assume free board = 0.3 m

$M_{25} , Fe_{415} \\ ∴H=4-0.3=3.7 m \\ v=\dfrac {πD^2}4×H \\ =350=\dfrac {πD^2}4×3.7 \\ ∴D=10.97≈11m$

Assume $t_1=150mm \\ t_2=30H+50 \\ =30×3.7+50=165mm$

∴Whichever greater should be used as thickness

$\dfrac {H^2}{Dt} =\dfrac {3.7^2}{11×165}=7.54$

Form IS code ∴3370:1967:-Page 35, table 9

T=Coefficient × wHR

T= Hoop tension.

Evaluate the coefficient for the particular value of $\dfrac {H^2}{Dt}$ Hoop tension (T) $=0.56×9810×3.7×\dfrac {11}2 \\ =111795 N/mm \\ ????????????????= \dfrac ????{????????????} = \dfrac {111795}{150}=745.3????????^2$

Since for hoop tension the reinforcement is alone inner as well as outer side of reinforcement.

$∴????????????????= \dfrac {745.3}2 = 372.62????????^2$

Now for calculation of ast min , evaluate the min % reinforcement with the help of t=164mm using the relation. $∴Ast min = \dfrac {0.28}{100} ×1000×165 \\ = 462 mm^2 \lt Astr (total) =745.3$

Assume 12mm ∅ base

∴ Spacing $=\dfrac {1000×113}{372.65} =303.23 mm \\ ≈300mm \\ Astp=\dfrac {1000×113}{300} =376.66 mm^2 \\ \text {Total Astp} = 376.66×2=753.33mm^2$

Check:-

$σst= \dfrac T{1000t+(m-1)Astp} \\ m= \dfrac {280}{3σ_cbc } = \dfrac {280}{3×8.5→(from table)} =10.98 \\ σst = \dfrac {111795}{1000×165+(10.98-1)753.33} \\ =0.64 N/mm^2 \lt 1.3N/mm^2(from table)$

∴safe

Vertical Steel:-

Form IS code : 3370:1967, page 36 table 10

B.M. = Coefficient $× wH^3$ Both are less than Astmin i.e. 462 mm2

∴ProvideAstp=$462 mm^2$

∴Assume 8 mm ∅ steel

∴Spacing $=\dfrac {1000×50}{462}=108 mm\\ ≈100 mm c/c$

For Shear:-

From IS code 3370 : 1967 page 37

V= coefficient $×WH^2$ (triangular) $∴V=0.179×9810×3.7^2 \\ =24039.SN ≈24.03 KN (V_{UP} ) \\ Ast= \dfrac {100^0 Asv}{Spacing} \\ =\dfrac {1000×50}{100} \\ =500 mm^2 \\ pt.=\dfrac {100Astp.}{1000×134} \\ (b×d) \\ pt.= \dfrac {1000×500}{134} \\ =0.37\%$

From page 84 of IS code 456:2000 by interpolation Zuc = ? $V_{us} = K\space Z_{uc}\space bd = (\text {for k value page 72} ) \\ =1.3×0.27×1000×134 \\ =47.03 KN \gt V_{UD}$

∴Safe

Reinforcement details:- For base reinforcement of water tank provide minimum reinforcement. In this case minimum reinforcement is similar to that of vertical steel.

∴Provide 8 mm ∅ 100 mm c/c at the base of water tank.

Important Notes:-

Permissible tensile stress in concrete:$(σ_ct)$ Permissible stress in in steel : - (σ_st) Minimum reinforcement criteria :-  written 3.1 years ago by Aksh_31 • 630