Question: Design by approximate method a rectangular tank 6m x 4m in plan and 3.5 in height.
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Tank is resting on firm ground. Grade of concrete is M25 and steel is Fe415. Check the design for safe stresses.

Design the following

a) Side walls.

b) Base slab.

Draw neat sketches showing the reinforcement details.

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

ddrcs mumbai university • 265 views
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modified 11 weeks ago by gravatar for Yashbeer Yashbeer160 written 5 months ago by gravatar for tanya.tanyabarnwal tanya.tanyabarnwal0
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Step 1: Design constants

Rectangular tank: $6m \times 4m$

Height: 3.5m

Grade of concrete M25 Fe415

$6_{cbc} = 8.5 N/mm^2 \quad 6_{st} = 130 \ N/mm^2$

$ m = \frac{280}{36_{cbc}} = 11$

$ k = \frac{1}{ 1+ \frac{6_{st}}{m6_{cbc}}} = 0.48$

$j=1-\frac{k}{3}=0.86$

$Q=\frac{1}{2} 6_{cbc} k_j=1 \cdot 52$


Step 2: Bending Moment Calculation

$\frac{L}{B}=\frac{6}{4}=1 \cdot 5\lt2$

For $\frac{1}{B}\lt2 \quad h= \frac{H}{4} \text{ or lm}$ whichever is greater

$ = \frac{3.5}{4} \text{ or lm} \quad h = lm$

$p=w(4-b) \quad \text{ ... w = 9800 of water fixed}$

$p=9800(3.5-1)$

$p=24.5 \mathrm{kN} / \mathrm{m}$

Bending moment at support

a] Long wall $=\frac{p L^{2}}{12}=73.5 \mathrm{KNm}$

b] Short wall $=\frac{p B^{2}}{12}=\frac{24.5 \times y^{2}}{12}=32.67 \mathrm{kNm}$

Bending Moment at center

a] Long wall $=\frac{p L^{2}}{16}=55.125 \mathrm{KNm}$

b] Short wall $=\frac{p B^{2}}{16}=24.5 \mathrm{KNm}$

Tension Developed in

a] Long wall $T_{L}=\frac{P B}{2}=49 \mathrm{KN}$

b] Short wall $T_B=\frac{P L}{2}=73.5 \mathrm{kN}$


Step 3: Design of Long wall

a] Reinforcement at support

Moment at support = $73.5 kNm$

Tension $T_L = 49 kNm$

Calculation of depth

$M = Qhd^2$

$d=\sqrt{\frac{m}{Q b}}=\sqrt{\frac{73.5 \times 10^{6}}{1.52 \times 1000}}=49.9$

$\therefore D = 260 \ mm \quad Cover \ 35 \ mm$

$\therefore d = 225 \ mm$

$\overline{x}=\frac{260}{2}-35=95 \mathrm{mm}$

$Ast = \frac{M - T \overline{x}}{6st \ j \ d} = \frac{(73.5 \times 10^6) - (48 \times 10^3 \times 95)}{130 \times 0.86 \times 225}$

$Ast_1 = 2736.8 \ mm^2$

$Ast_2 = \frac{T}{6st} = \frac{49 \times 10^3}{130} = 376.93 \ mm^2$

$\text{Total Ast} = 3113.73 \ mm^2$

Provide 25mm $\phi$ ban

$\therefore Spacing = \frac{\pi \text{/} 4 \times 25^2}{3113.73} \times 100 = 157 mm$

$\therefore \text{# 25mm } \phi \text{ @ 155 mm c/c for inner face}$

b] Reinforcement at Mid Span

moment at mid span $= 55.125 KNm$

Tension $T_L = 49 KNm$

$Ast_1 = \frac{(55.125 \times 10^6) - (49 \times 10^3 \times 95)}{130 \times 0.86 \times 225} = 2006.36 mm^2$

$Ast_2 = \frac{T}{6st} = \frac{49 \times 10^3}{130} = 376.93 mm^2$

$Total Ast = 2383.3 mm^2$

$\text{Provide # 20mm @ 125mm c/c}$


Step 4: Design of shorter wall

a] Support:

Moment at supp of short = 32.67 KNm

Tension at short wall = 73.5 KN

$Ast_1 = \frac{(32.67 \times 10^6) - (73.5 \times 10^3 \times 95) }{B_0 \times 0.86 \times 225} = 1021.17 mm^2$

$Ast_2 = \frac{73.5 \times 10^3}{130} = 565.4 mm^2$

$Total Ast = 1586.57 mm^2$

$\text{Provide # 20mm @ 195mm c/c innerface}$

b] Midd span

Moment at mid $= 24.5 KNm$

$T = 73.5 KN$

$\text{Since difference is not much}$ $\therefore$ $\text{provide same as supp #20 mm @ 195mm c/c outerface}$


Step 5: Design for cantilever action

$\begin{aligned} BM &= \frac{1}{2} \times w H \times h \times \frac{h}{3} \\ &= \frac{1}{2} \times 9800 \times 3.5 \times 1 \times \frac{1}{3} \\ &= 5.717 KNm \end{aligned}$

$Ast=\frac{5 \cdot 77 \times 10^{6}}{130 \times 0.86 \times 185}=276 \mathrm{mm}^{2}$

$Ast_{min} = \frac{0.35}{100} \times 225 \times 1000 \times 787.5 mm^2$

$\text{Ast on each face = } 393.72$

$\therefore \text{provide #10mm @ 195mm c/c each face in verticle direction}$


Step 6: Provision of Haunches

It is customary to provide $150 \times 150mm$ haunches at junction at wall and base

Haunch ref of #10mm @ 195mm c/c is provided


Step 8: Design of base slab

Assume thck = 150mm

Provide #8mm @ 240 c/c in both direction

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modified 11 weeks ago  • written 11 weeks ago by gravatar for Yashbeer Yashbeer160
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