Problem:
Design a rectangular water tank with fixed base for a given capacity in litres.
- Draw cross section of tank showing reinforcement details of vertical wall and base slabs.
- Draw plan of tank showing reinforcement details.
Design steps:
- Capacity of circular water tank ( V ) in liters.
- Length and breadth of tank ( L x B ) in m.
- Free board ( f ) in mm.
- Strength of concrete (ex : M20, M35)
- Strength of steel in (ex : Fe415, Fe-500)
IS: 3370 (Part II) Table 1, Clause 3.3.1 and IS: 456-2000, Clause B-2.1.1 and Table 21.
- Unit weight of water , γ = 9.8 kN/m3.
- Permissible stress in concrete , σcbc.
- Permissible tensile stress in steel , σst.
h = Depth of water tank. This can be found out using following equation.
\[h = \frac{{{W_{capacity}}}}{{L \times B}}\]
Choose required depth of water tank.
Total height tank = depth of the tank + Free board
H = h + f
Check for L/B ratio : If L / B ratio is the lesser than two. The cantilever action will be from the bottom of the base slab to the one metre above. The rest is affected by horizontal frame action. Hence, it is considered for analysis.
Calculate modular ratio.
\[m = \frac{{280}}{{3 \times {\sigma _{cbc}}}}\]
Neutral axis depth factor.
\[n = \frac{{m \times {\sigma _{cbc}}}}{{m \times {\sigma _{cbc}} + \times {\sigma _{st}}}}\]
Lever arm.
\[j = 1 - \frac{n}{3}\]
Moment of resistance.
\[k = \frac{1}{2} \times {\sigma _{cbc}} \times j \times n\]
The critical section hc is H/4 or 1m. Take higher value.
The water pressure, Ph.
\[{P_h} = \gamma \times \left( {H - {h_c}} \right)\]
Fixed end moment of long wall.
\[{F_l} = \frac{{{P_h} \times {L^2}}}{{12}}\]
Fixed end moment of short wall.
\[{F_s} = \frac{{{P_h} \times {B^2}}}{{12}}\]
Moment distribution taable :
| Members. | Stiffness. | Total joint stiffnes. | Distribution factor. |
|---|---|---|---|
| Short wall | a = 4.E.I / B | a + b | Ds = a / ( a + b ) |
| Short wall | b = 4.E.I / L | Dl = b / ( a + b ) |
Moment distribution taable with formula :
| Short wall. | Ds | Dl | Long wall. |
|---|---|---|---|
| Fs | Fl | ||
| Fs x Ds | Fl x Dl | ||
| Total |
Calculate effective depth.
\[d = \sqrt {\frac{M}{{k \times b}}}\]
Provide sufficient effective depth
\[d = D - {C_{effective}}\]
Direct pull on long wall.
\[{T_L} = {P_h} \times \frac{B}{2}\]
Direct pull on short wall.
\[{T_B} = {P_h} \times \frac{L}{2}\]
Eccentricity of reinforcement from centre of wall.
\[x = \frac{D}{2} - {C_{effective}}\]
Design moment at corner.
\[{M_{end}} = M - {T_L}.x\]
Area of steel due to design bending moment.
\[{A_{st1}} = \frac{{{M_{end}} \times {{10}^6}}}{{{\sigma _{st}} \times j \times d}}\]
Area of steel due to direct tension.
\[{A_{st2}} = \frac{{{T_L}}}{{{\sigma _{st}}}} \times 1000\]
Total area of steel.
\[{A_{st}} = {A_{st1}} + {A_{st2}}\]
Consider the desired bar diameter.
\[{S_v} = \frac{{As{t_{provided}}}}{{Ast}} \times 1000\]
Provide diameter of bar at distance C/C.
Moment at centre.
\[M = \frac{{{P_h} \times {L^2}}}{8} - {M_{corner}}\]
Design moment at centre.
\[{M_{middle}} = M - {T_L}.x\]
Area of steel due to design bending moment.
\[{A_{st1}} = \frac{{{M_{middle}} \times {{10}^6}}}{{{\sigma _{st}} \times j \times d}}\]
Area of steel due to direct tension.
\[{A_{st2}} = \frac{{{T_L}}}{{{\sigma _{st}}}} \times 1000\]
Total area of steel.
\[{A_{st}} = {A_{st1}} + {A_{st2}}\]
Consider the desired bar diameter.
\[{S_v} = \frac{{As{t_{provided}}}}{{Ast}} \times 1000\]
Provide diameter of bar at distance C/C.
Design moment at corner.
\[{M_{end}} = M - {T_B}.x\]
Area of steel due to design bending moment.
\[{A_{st1}} = \frac{{{M_{end}} \times {{10}^6}}}{{{\sigma _{st}} \times j \times d}}\]
Area of steel due to direct tension.
\[{A_{st2}} = \frac{{{T_B}}}{{{\sigma _{st}}}} \times 1000\]
Total area of steel.
\[{A_{st}} = {A_{st1}} + {A_{st2}}\]
Consider the desired bar diameter.
\[{S_v} = \frac{{As{t_{provided}}}}{{Ast}} \times 1000\]
Provide diameter of bar at distance C/C.
Moment at centre.
\[M = \frac{{{P_h} \times {L^2}}}{8} - {M_{corner}}\]
Design moment at centre.
\[{M_{middle}} = M - {T_B}.x\]
Area of steel due to design bending moment.
\[{A_{st1}} = \frac{{{M_{middle}} \times {{10}^6}}}{{{\sigma _{st}} \times j \times d}}\]
Area of steel due to direct tension.
\[{A_{st2}} = \frac{{{T_B}}}{{{\sigma _{st}}}} \times 1000\]
Total area of steel.
\[{A_{st}} = {A_{st1}} + {A_{st2}}\]
Consider the desired bar diameter.
\[{S_v} = \frac{{As{t_{provided}}}}{{Ast}} \times 1000\]
Provide diameter of bar at distance C/C.
Cantilever bending moment.
\[{M_{cantilever}} = \frac{{\gamma \times H \times {h^2}}}{6}\]
Area of steel due to cantilever bending moment.
\[{A_{st}} = \frac{{{M}}}{{{\sigma _{st}} \times j \times d}}\]
Only minimum reinforcement is required.
\[{A_{st}} = 0.3\% \times {t_{wall}} \times 1000\]
Consider the bar diameter.
\[{S_v} = \frac{{As{t_{provided}}}}{{As{t_{required}}}} \times 1000\]
Provide diameter of bar at distance C/C.
The base slab is laid on 75mm thick lean conrete mix. Provide thickness of 150 mm or more to avoid leakage of water.
Provide minimum nominal reinforcement in both direction.
Reinforcement bar should be continuous and bent in junction of rectangular water tank.
Sectional elevation and plan of rectangular water tank.
Indian standard code IS 3370-Part ( 2 ).
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