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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.

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 • 2.4k  views
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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 $\dfrac {H^2}{Dt}$

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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.

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$∴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 $× wH^3$

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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)

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

From page 84 of IS code 456:2000 by interpolation Zuc = ?

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$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:-

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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)$

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Permissible stress in in steel : - (σ_st)

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Minimum reinforcement criteria :-

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