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← Back to the Intze tank designer  ·  Design guide

Intze tank design formulas: where each one comes from

Every Theory link in the designer’s result sheet opens the matching entry on this page. Each entry gives the formula, what the symbols mean, how it is derived, and the code clause or theory it rests on. Symbols: R, D = radius and diameter of the cylindrical wall; r = radius of the bottom ring (girder); hw = water depth in the wall; h₂ = height of the cone; hb = rise of the bottom dome; Rc = radius of the column circle; n = number of columns; hs = staging height.

1. Capacity, proportions and loads

Water volume of an Intze container#

V = πR²hw + (πh₂/3)(R² + Rr + r²) − (πhb/6)(3r² + hb²)

The water occupies three simple solids: a cylinder of radius R and water depth hw (wall height less freeboard), a frustum of a cone of height h₂ between radii R and r, and the space above the bottom dome. The bottom dome arches upward into the water, so the volume of its spherical cap (rise hb, base radius r) is subtracted.

Frustum: V = (πh/3)(R² + Rr + r²). Spherical cap: V = (πh/6)(3a² + h²), with a the base radius. Both are standard solid-geometry results (integrate πx² over the height).

Source: Geometry; capacity requirement from the client / water-supply design.

Choosing the proportions#

Inputs: H₁/D, r/R, cone slope, freeboard → solve R so that V = required capacity

With the ratios fixed, every dimension is a multiple of R, so V is proportional to R³. The calculator finds R by iteration until the water volume equals the required capacity. Usual starting values: H₁/D ≈ 0.5–0.6, r/R ≈ 0.6–0.65, cone at 45°–50° to the horizontal, top-dome rise ≈ D/8, bottom-dome rise ≈ 0.22r–0.30r.

The cone and bottom-dome proportions are chosen so that the inward thrust of the cone and the outward thrust of the bottom dome on the girder nearly cancel (see the Intze balance).

Source: Practice (textbooks; RDSO standard designs).

Dead load#

DL = Σ(surface area × thickness × γc) + rings + finishes; γc = 25 kN/m³

Each shell weighs area × thickness × 25 kN/m³. Surface areas: spherical dome 2πRdh, cylinder 2πRH, cone π(R + r) × slant length. Ring beams are cross-section × 2π(radius) × 25. Staging adds the columns and bracings. A percentage is added for finishes and fittings.

Source: Unit weight of reinforced concrete: IS 875 (Part 1).

Weight of water#

WT = V × γw, γw = 9.81 kN/m³

1 m³ of water has a mass of 1 tonne, so its weight is 9.81 kN. The tank is checked both full (largest gravity load, largest seismic mass) and empty (least weight to resist overturning by wind or earthquake).

Source: Basic physics.

Roof imposed load#

LL = q × πR²

The live load on the top dome is taken on its plan (horizontal projection) area πR². Typical q: 1.5 kN/m² when the roof is accessible, 0.75 kN/m² when it is not.

Source: IS 875 (Part 2) — imposed loads on roofs.

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2. Wind (IS 875 Part 3:2015)

Design wind speed and pressure#

Vz = Vb k₁ k₂ k₃ k₄    pz = 0.6 Vz² (N/m²)

Vb is the basic wind speed of the site (map in the code). It is modified by k₁ (risk/design life), k₂ (terrain roughness and height — taken at the top of the container for the tank, at mid-height for the staging), k₃ (ground slope) and k₄ (importance for cyclonic regions).

The pressure comes from Bernoulli: p = ½ρV² with air density ρ ≈ 1.2 kg/m³, so p = 0.6V².

Source: IS 875 (Part 3):2015 — design wind speed and design wind pressure; k₂ from Table 2.

Force coefficient of the container#

A = D × (H₁ + ht)    Cf = f(h/D) for a circular section

The container is treated as a short circular cylinder. Its projected area is diameter × height of the wall plus the top dome rise. The force coefficient for a circular section rises with slenderness h/D (about 0.7 for squat shapes up to about 1.0 for very slender ones), because air can flow round the ends of a short cylinder.

Source: IS 875 (Part 3):2015 — force coefficients for circular sections.

Design wind pressure pd#

pd = Kd Ka Kc pz

Kd is the wind-directionality factor — 1.0 for a circular container (the wind can blow from any side and the shape is the same), 0.9 for the framed staging. Ka is the area-averaging factor (1.0 up to 10 m², 0.9 at 25 m², 0.8 at 100 m² or more): gusts do not hit a large area all at once. Kc is the combination factor, taken as 1.0.

Source: IS 875 (Part 3):2015 — design wind pressure; Ka from Table 4.

Wind force on the container#

F = Cf A pd

The force on each part is its force coefficient × projected area × design pressure, acting at the centroid of that area. The cone below the wall is added as a separate projected area. These heights are used to find the overturning moment at the foundation.

Source: IS 875 (Part 3):2015 — wind force from force coefficients.

Wind on the staging frame#

As = (n/2)·b·hs + nbr·2Rc·dbr    φ = As/(2Rchs)    F = Cf As pd (1 + η)

Seen in elevation, about half the columns face the wind, plus the bracings at each level. The solidity ratio φ = solid area / enclosed area decides the force coefficient of a frame. The leeward half of the staging is partly sheltered; its share is added through the shielding factor η, giving F = CfAspd(1 + η). For circular columns the round-section coefficient is used for the columns and the frame coefficient for the flat-sided bracings.

Source: IS 875 (Part 3):2015 — force coefficients for frames (Table 30) and shielding of multiple frames.

Total wind shear and moment#

Vwind = ΣF    Mwind = ΣF × z

The base shear is the sum of the forces; the overturning moment at the top of the foundation is each force times its height z above it. This moment is shared by the columns (column forces) and checked against soil pressure and uplift.

Source: Statics.

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3. Earthquake (IS 1893 Part 2:2014)

Equivalent circular tank#

h = V / (πD²/4)

The impulsive/convective mass formulas were derived for a flat-bottomed circular cylinder. For an Intze container the code allows an equivalent cylinder of the same diameter D at the water surface and the same volume V. Its depth h is the volume divided by the plan area.

Source: IS 1893 (Part 2):2014 — tanks of other shapes.

Impulsive and convective masses#

mi/m = tanh(0.866D/h) / (0.866D/h)    mc/m = 0.23 (D/h) tanh(3.68h/D)

When the tank shakes, the lower part of the water moves rigidly with the walls (impulsive mass mi) while the upper part sloshes slowly (convective mass mc). These expressions come from potential-flow theory for a rigid circular tank (Housner’s model). A tall, narrow tank is mostly impulsive; a wide, shallow tank is mostly convective. mi + mc is close to, but not exactly, m.

Source: IS 1893 (Part 2):2014 — spring-mass model for circular tanks.

Heights of the impulsive and convective masses#

hi/h = 0.375 (h/D ≤ 0.75), else 0.5 − 0.09375/(h/D)
hc/h = 1 − [cosh(3.68h/D) − 1] / [(3.68h/D) sinh(3.68h/D)]
hi*/h = 0.866(D/h) / [2 tanh(0.866D/h)] − 0.125 (h/D ≤ 1.33), else 0.45
hc*/h = 1 − [cosh(3.68h/D) − 2.01] / [(3.68h/D) sinh(3.68h/D)]

hi, hc place the masses so that the wall pressure moment is correct; they are used for the moment at the base of the wall. hi*, hc* (starred) also include the pressure on the base of the tank, and are used for the overturning moment of the whole tank on the staging.

Source: IS 1893 (Part 2):2014 — spring-mass model for circular tanks.

Structural mass#

ms = (Wcontainer + ⅓ Wstaging) / g

The empty container moves with the impulsive water, and about one-third of the staging mass participates in the first mode of a cantilever-like frame. This mass is added to mi for the impulsive period and base shear. It acts at the centre of gravity of the empty container (hcg).

Source: IS 1893 (Part 2):2014 — elevated tanks.

Lateral stiffness of the staging#

Ks = β · 12 n Ec I Np² / hs³,   Ec = 5000√fck

Between two bracing levels each column bends as a member fixed at both ends that sways by Δ; its stiffness is 12EI/hp³. With n columns in parallel and Np panels in series, each of height hp = hs/Np: K = 12nEI/(Nphp³) = 12nEI Np²/hs³. Bracings are not infinitely stiff, so the factor β (< 1) reduces it. Ec is the short-term modulus of concrete.

Source: Frame analysis; Ec = 5000√fck from IS 456:2000 Cl 6.2.3.1.

Impulsive time period#

Ti = 2π √[(mi + ms) / Ks]

The tank on its staging is a single-degree-of-freedom oscillator: mass on a spring. Its natural period is 2π√(m/k). The tank-empty period uses ms alone.

Source: IS 1893 (Part 2):2014 — time period of the impulsive mode.

Convective (sloshing) time period#

Tc = 2π √[D / (3.68 g tanh(3.68h/D))]

This is the period of the first sloshing mode of water in a circular tank, from linear wave theory (3.68 ≈ 2 × 1.841, the first root of J₁′). It depends only on the tank size and water depth, not on the staging. It is long (often 2–5 s), so the convective acceleration is small.

Source: IS 1893 (Part 2):2014 — time period of the convective mode.

Design horizontal acceleration coefficient#

Ah = (Z/2) (I/R) (Sa/g)

Z is the zone factor (0.10, 0.16, 0.24, 0.36); Z/2 converts the maximum considered earthquake to the design-basis earthquake. I is the importance factor and R the response-reduction factor of the staging. Sa/g is read from the response spectrum for the period and soil type: 1 + 15T below 0.1 s, a plateau of 2.5, then 1.00/T, 1.36/T or 1.67/T for hard, medium and soft soil.

The impulsive mode uses 5% damping. The convective mode uses 0.5% damping, so its Sa/g is multiplied by 1.75 (the factor can be changed in the inputs).

Source: IS 1893 (Part 2):2014 — design horizontal seismic coefficient; spectrum from IS 1893 (Part 1):2016.

Base shear#

Vi = Ah,i(mi + ms)g    Vc = Ah,c mc g    V = √(Vi² + Vc²)

Each mode gives a lateral force = acceleration coefficient × mass × g. The impulsive and convective peaks happen at different times because their periods are very different, so they are combined by the square root of the sum of squares (SRSS).

Source: IS 1893 (Part 2):2014 — base shear.

Overturning moment#

Mi = Ah,i[mi(hi* + hs) + mshcg]g    Mc = Ah,cmc(hc* + hs)g    M* = √(Mi² + Mc²)

Each force is multiplied by its height above the top of the foundation: the water masses act at hi* or hc* above the base of the container, which is itself hs above the foundation; the structural mass acts at hcg. Combined by SRSS as for the shear.

Source: IS 1893 (Part 2):2014 — base moment.

Sloshing wave height#

dmax = Ah,c R D/2

Here R is the response-reduction factor: Ah,c was divided by R, and the wave height is not reduced by the ductility of the staging, so R is multiplied back. The convective motion lifts the water surface at the wall. If the wave is higher than the freeboard it strikes the roof and the top ring, adding upward pressure that has not been designed for. So freeboard ≥ dmax.

Source: IS 1893 (Part 2):2014 — sloshing wave height.

Hydrodynamic pressure on the wall#

piw = 0.866 tanh(0.866D/h) · Ah,i γw h (at the base)    pcw ≈ 0.5625 Ah,c γw D (2/3) (at the surface)

During an earthquake the water pushes harder on the wall than its static pressure. The impulsive part is largest at the base; the convective part at the surface. They are combined by SRSS and compared with the hydrostatic pressure γwhw to show how much the hoop tension increases. The calculator reports this addition for the designer to check; the wall is otherwise designed for hydrostatic pressure.

Source: IS 1893 (Part 2):2014 — hydrodynamic pressure (circular tanks).

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4. Load combinations and column forces

Axial force in each column#

Pcol = P/n ± M Rc / Σy²,   Σy² = (n/2) Rc²

The columns act together like the fibres of one big section. Vertical load P is shared equally. The overturning moment M produces a linearly varying force, largest in the column furthest from the neutral axis, at distance Rc. For n columns equally spaced on a circle of radius Rc, Σy² = ΣRc²sin²θ = (n/2)Rc² whatever the wind direction.

Service combinations: DL + water with wind or earthquake, and tank empty with wind or earthquake (uplift). Ultimate: 1.5(DL + water), 1.2(DL + water ± EQ or wind), 1.5(DL ± EQ or wind) and 0.9DL ± 1.5(EQ or wind). The column bending moment Mu comes from portal action: column shear × half the panel height.

Source: Load factors: IS 456:2000 Table 18; IS 1893 (Part 2):2014 for the seismic combinations.

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5. Top dome and top ring beam B₁

Radius of a spherical dome#

Rd = (R² + h²) / 2h    sinθ₀ = R/Rd

From the right triangle formed by the dome radius, the base radius R and (Rd − h): Rd² = R² + (Rd − h)², which gives Rd = (R² + h²)/2h. θ₀ is the half-angle subtended at the centre by the springing.

Source: Geometry.

Load on the dome#

w = t × γc + live load (kN/m²)

Self-weight per square metre of shell surface plus the roof imposed load. Treating the live load as if it were spread over the curved surface (rather than on plan) is slightly conservative for the meridional thrust.

Source: IS 875 (Parts 1 and 2).

Membrane forces in a spherical dome#

Nφ = wRd / (1 + cosθ)    Nθ = wRd [cosθ − 1/(1 + cosθ)]    σ = N/t ≤ σcc

A thin dome carries load by in-plane forces only (membrane theory). Vertical equilibrium of the cap above angle θ: weight of cap 2πRd²(1 − cosθ)w = Nφ · 2πRdsinθ · sinθ, giving Nφ. Then the normal-equilibrium equation Nφ/R1 + Nθ/R2 = −w cosθ gives the hoop force Nθ.

Nφ is always compressive. Nθ is compressive while θ < 51.8° — which is why a shallow roof dome (rise about D/8) needs no tension design. The stress is compared with the permissible direct compression σcc.

Source: Membrane theory of shells (Timoshenko); σcc from IS 456:2000 Table 21.

Minimum reinforcement — IS 3370 (Part 2):2021#

ρmin = 0.28% (L ≤ 14 m) → 0.42% (L ≥ 28 m), linear between  (elevated tank, Fe 415/500)
surface zone = t/2, not more than 250 mm    As,min per face = ρmin × surface zone × 1000 (mm²/m)

Where a member is in compression (domes, cone meridian) or its tension steel is small, the steel is governed by the minimum needed to control cracking from shrinkage and temperature. When concrete shrinks or cools and is restrained, it cracks; the steel must be enough to spread this into many fine cracks rather than a few wide ones. That is why the minimum depends on the length of the member (longer members are more restrained) and on whether the tank is elevated (exposed to sun and air on all sides, so larger temperature changes) or on the ground.

TankSteelShort (L ≤ 14 m)Long
ElevatedFe 415 / Fe 5000.28%0.42% (L ≥ 28 m)
ElevatedFe 2500.44%0.66% (L ≥ 28 m)
Ground-supportedHYSD0.24%0.36% (L ≥ 22 m)

Length L is the horizontal length of the continuous member between movement joints, along the main steel. An Intze tank has no movement joints, so each ring-shaped member is continuous all the way round: the calculator takes L = circumference — 2πR for the top dome, B₁, the wall, the cone and B₂, and 2πr for the bottom dome and the girder. Most tanks of 100 kL and above therefore get 0.42%.

The percentage is of each surface zone (half the thickness, capped at 250 mm), in each direction, so the whole section gets twice the per-face value. Walls thinner than 160 mm and slabs thinner than 180 mm may have all of it in one layer (Cl 8.1.2). For ring beams the calculator applies the percentage to the whole cross-section.

Worked example: 100 mm top dome, D = 11.5 m → L = π × 11.5 = 36.1 m ≥ 28 m → 0.42% × 50 mm × 1000 = 210 mm²/m per face, 420 mm²/m in all in each direction; being under 180 mm thick it may be placed as a single layer.

Source: IS 3370 (Part 2):2021 Cl 8.1.1, Table 5 (Notes 1 and 2) and Cl 8.1.2.

Horizontal thrust at the springing#

H = Nφ cosθ₀ (kN/m)

At the edge the meridional thrust is inclined at θ₀ to the horizontal. Its vertical part goes down the wall; its horizontal part pushes outward on the ring beam B₁. Without a ring the dome would spread and crack at the edge.

Source: Statics.

Hoop tension in a ring beam#

P = H × R

A ring loaded by a uniform outward radial force H per metre carries a uniform hoop tension P = HR. Proof: cut the ring along a diameter. The resultant of the radial forces on a half ring is H × 2R (the projected length); it is resisted by two cut sections, each carrying P. So 2P = 2HR. The same result gives the hoop tension in a tank wall (with H = water pressure).

Source: Statics (thin-ring / “boiler formula”).

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6. Members in tension: rings, wall and cone hoops

Hoop tension in the wall#

F = γw hw R (kN per metre height)

At depth hw the water pressure is p = γwhw. A 1 m tall horizontal strip of the wall is a ring under uniform radial pressure, so F = pR (see ring tension). It is largest at the bottom of the wall. The wall is designed as a free-standing cylinder carrying all the pressure by hoop action (conservative near the base, which is partly restrained by B₂).

Source: Hydrostatics; IS 3370 (Part 4) gives coefficients for walls fixed or hinged at the base.

Hoop steel#

Ast = P / σst  (not less than the IS 3370 minimum)

In the uncracked design of water-retaining sections all the hoop tension is given to the steel at a low permissible stress σst (130 N/mm² for HYSD bars) so that cracks stay fine. The steel provided is the larger of this and the minimum reinforcement. Wall and cone steel is split equally between the two faces.

Source: Working-stress design, IS 3370 (Part 2):2009 — permissible steel stress for water-retaining parts.

Concrete tensile stress (no-crack check)#

σct = P / [Ac + (m − 1)Ast] ≤ permissible σct    m = 280/(3σcbc)

Before cracking, steel and concrete strain equally, so the steel behaves like m times its area of concrete. The transformed area is Ac + (m − 1)Ast (minus one because the bars displace their own area of concrete). The direct tensile stress on this area must stay below the permissible value (1.3, 1.5, 1.6, 1.7 N/mm² for M25–M40), so the section does not crack through and leak.

Rearranged, the smallest concrete area that passes is Ac,req = P/σct − (m − 1)Ast.

Source: Permissible direct tension: IS 3370 (Part 2):2009 Table 1; modular ratio: IS 456:2000 Annex B.

Thickness for hoop tension#

t ≥ F (1/σct − (m − 1)/σst)  (F in N/mm, t in mm)

Put Ast = F/σst into the crack check for a 1 m strip (Ac = 1000t): F/[1000t + (m − 1)F/σst] ≤ σct. Solving for t gives this formula. For the cone the result is floored at 200 mm for practical concreting on a slope.

Source: Derived from the crack check above.

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7. Conical slab and ring beam B₂

Cone angle#

α = 90° − (cone slope to the horizontal)   (angle of the generator with the vertical)

All the cone formulas use α, the angle between the sloping slab and the vertical. A slab at 45° has α = 45°; a steeper slab (smaller α) has less hoop force from water but more height.

Source: Geometry.

Meridional thrust in the cone#

Nφ = W / (2π ρ cosα)   (ρ = R at the top, r at the bottom)

The cone hangs the whole container down to the girder. At any horizontal cut the total vertical load W above it is carried by the meridional force Nφ around the circumference 2πρ; only its vertical component Nφcosα resists W. At the top W is the structure above (domes, wall, rings, finishes, roof live load). At the bottom add the cone’s own weight and the water standing directly on it, γw[V₁ + V₂ − πr²(hw + h₂)].

Source: Membrane theory of conical shells.

Hoop tension in the cone#

Nθ = (p/cosα + q tanα) × (d/2)

The hoop force in a shell of revolution is Nθ = pn R2, where R2 is the radius of curvature normal to the meridian. For a cone R2 = (d/2)/cosα. The water pressure p acts normal to the slab; the self-weight q (= tγc) has a normal component that adds the q tanα term. It is checked at the top (d = D, p = γwhw) and the bottom (d = 2r, p = γw(hw + h₂)), and the larger governs.

Source: Membrane theory of conical shells.

Ring beam B₂ at the wall–cone junction#

H = Nφ,top sinα + water on the beam face    P = H R

The cone is in compression and leans outward at the top, so its horizontal component Nφsinα pushes B₂ outward. Water pressure on the inner face of the beam (γwhw × beam depth) also pushes outward. The ring takes the total as hoop tension P = HR, designed with the hoop steel and crack check formulas.

Source: Statics.

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8. Bottom dome

Bottom dome geometry#

Rd = (r² + hb²) / 2hb    sinφ₀ = r/Rd

Same geometry as the top dome, with base radius r and rise hb. The dome rises upward into the water from the girder, so the water load presses it into compression — it works as an arch, not as a hanging shell.

Source: Geometry.

Load on the bottom dome#

W = γw[πr²(hw + h₂) − Vseg] + self weight

The water standing over the dome is the cylinder of radius r up to the water surface, less the volume of the dome bulge Vseg (water cannot be where the concrete is).

Source: Hydrostatics.

Membrane forces in the bottom dome#

Nφ = W / (2π r sinφ₀)    Nθ = p Rd − Nφ

At the springing, vertical equilibrium gives Nφ sinφ₀ × 2πr = W. For a sphere both principal radii equal Rd, so the normal equilibrium Nφ/Rd + Nθ/Rd = p gives Nθ = pRd − Nφ, with p the water pressure at dome level. Both are normally compressive and checked against σcc; steel is then the minimum.

Source: Membrane theory of shells.

Outward thrust on the girder#

H = Nφ cosφ₀

The horizontal component of the dome thrust at its springing pushes the girder outward. This is the force that balances the inward pull of the cone.

Source: Statics.

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9. Circular girder B₃

The Intze balance#

net = Hdome + water on face − Nφ,bot sinα    hoop force = net × r
exact balance when tanφ₀ = W / (2πr Hcone), i.e. hb = r tan(φ₀/2)

This is the idea behind the Intze tank (Otto Intze, 1880s). The cone, being compressed and sloping inward at the bottom, pulls the girder inward; the arched bottom dome pushes it outward. Choosing the rise of the bottom dome so the two nearly cancel leaves the girder with little hoop force, so it only has to carry vertical load to the columns.

The dome thrust is Nφcosφ₀ = W/(2πr tanφ₀); setting it equal to the cone thrust gives tanφ₀. For a spherical cap hb/r = tan(φ₀/2), which gives the exact rise.

Source: Statics; Intze principle.

Bending and torsion in a circular girder on n columns#

Mu,support = 1.5 C₁WRc   Mu,span = 1.5 C₂WRc   Tu = 1.5 C₃WRc
with θ = π/n: C₁ = (1 − θcotθ)/2π   C₂ = (θ/sinθ − 1)/2π   C₃ = (θsinψ/sinθ − ψ)/2π, cosψ = sinθ/θ

A ring beam on equally spaced supports with a uniform load W (total) is a classic curved-beam problem: because the beam is curved, a vertical load causes torsion as well as bending. By symmetry the shear and torsion are zero at mid-span and the rotation is zero at the supports; solving with the unit-load method gives these closed forms. ψ is measured from mid-span; the maximum torsion occurs at (θ − ψ) from the support.

nC₁ (support)C₂ (span)C₃ (torsion)Tmax from support
60.014820.007510.00150712.73°
80.008270.004170.0006289.53°
100.005270.002650.0003207.62°
120.003650.001830.0001846.35°

W = container dead load + water + roof live load; 1.5 is the limit-state load factor.

Source: Theory of curved beams (e.g. Timoshenko, Strength of Materials; Reynolds’s RC Designer’s Handbook); load factor IS 456:2000 Table 18.

Torsion design#

Mt = Tu(1 + D/b)/1.7    Me1 = Mu + Mt    Ve = Vu + 1.6Tu/b    τve = Ve/bd ≤ τc,max

IS 456 handles torsion by converting it into an equivalent bending moment (for longitudinal steel) and an equivalent shear (for stirrups). If τve exceeds the maximum shear stress τc,max the section must be enlarged. Closed stirrups are essential because torsion causes spiral cracking round all four faces.

Source: IS 456:2000 Cl 41.3.1 (equivalent shear), Cl 41.4.2 (equivalent moment), Table 20 (τc,max).

Flexural steel (limit state)#

Mu,lim = 0.138 fckbd² (Fe 415) / 0.133 fckbd² (Fe 500)    Mu = 0.87fyAstd(1 − Astfy/(bdfck))

The girder, bracings and foundation are not water-retaining surfaces (or are designed for strength), so they use limit-state design. If Mu exceeds Mu,lim the section would be over-reinforced — deepen it. Otherwise Ast is found by solving the quadratic. The girder gets top steel over the columns (hogging) and bottom steel at mid-span (sagging), each including the torsion equivalent moment.

Source: IS 456:2000 Annex G (G-1.1).

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10. Columns and bracings

Slenderness and additional moment#

le = 1.2 hp    slender if le/D > 12    Ma = PuD(le/D)²/2000

Between bracing levels a column is restrained against rotation at both ends but its ends can sway sideways, giving an effective length of 1.2 × panel height hp. A slender column bends further under its own axial load (P–Δ effect); IS 456 accounts for this with the additional moment Ma, added in the direction of slenderness.

Source: IS 456:2000 Table 28 (effective length), Cl 25.1.2 (slenderness), Cl 39.7.1 (additional moments).

Axial load–moment interaction#

For each neutral-axis depth: Pu = Cc + ΣfsiAsi    Mu = Cc(D/2 − x̄) + ΣfsiAsiyi    Puz = 0.45fckAc + 0.75fyAsc

A column under axial load and moment is checked against its interaction curve. The calculator builds the curve the way SP 16 charts are made: it sweeps the neutral axis through the section, takes the IS 456 parabolic-rectangular concrete stress block (0.446fck, strain 0.0035 at the edge) and the steel stress from its strain in each bar, and sums forces and moments. The steel percentage is raised until the point (Pu, Mu) lies inside the curve. Steel is kept between 0.8% and 4%.

Source: IS 456:2000 Cl 38.1 (assumptions), Cl 39.6 (Puz), Cl 26.5.3.1 (0.8–4%); SP 16.

Biaxial bending#

(Mux/Mux1)αn + (Muy/Muy1)αn ≤ 1,   αn = 1 to 2 as Pu/Puz goes from 0.2 to 0.8

Wind or earthquake can come from any direction, so the moment is resolved into components about the two axes of a rectangular column for every direction from 0° to 90°, and the worst case is checked with this interaction equation. Mux1, Muy1 are the uniaxial capacities at the same axial load.

Source: IS 456:2000 Cl 39.6.

Lateral ties#

tie diameter ≥ ¼ of the largest bar, ≥ 6 mm    pitch ≤ least lateral dimension, 16 × smallest bar, 300 mm

Ties hold the longitudinal bars in place so they do not buckle outward. In seismic zones the pitch is halved near the joints (over a length of at least the larger column dimension, 1/6 of the clear height and 450 mm) to confine the concrete where plastic hinges can form.

Source: IS 456:2000 Cl 26.5.3.2; IS 13920:2016 (special confining reinforcement).

Bracing moments and shear#

Vcol = V/n    joint moment = Vcol × hp    Vu = 2M/L

Under lateral load the staging works as a portal frame: each column takes an equal share of the shear and has a point of contraflexure at mid-height of each panel. The column moments above and below a joint (each Vcol × hp/2) are balanced by the bracings meeting at that joint. A bracing bent in double curvature, with M at each end over length L, carries a shear of 2M/L.

Source: Portal method of frame analysis.

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11. Foundation and quantities

Soil pressure under an annular raft#

p = P/A − γsoilDf ± M/Z    A = π(Ro² − Ri²)    Z = π(Ro⁴ − Ri⁴)/(4Ro)

The raft is a ring under the column circle. With the raft rigid, the pressure varies linearly: average P/A plus bending M/Z (Z = I/Ro of an annulus). The weight of soil removed at founding depth is subtracted to get the net pressure, compared with the net safe bearing capacity — increased by the permitted percentage for wind and earthquake cases. The minimum pressure must not be negative (no lifting off).

Source: Bending formula P/A ± M/Z; increase in allowable bearing pressure for seismic cases: IS 1893 (Part 1):2016.

Raft reinforcement#

Mu = 1.5 pmax L²/2 (cantilever projection L)

The part of the raft outside the column line acts as a cantilever pushed up by soil pressure. The moment per metre is pL²/2, factored by 1.5; steel then follows the limit-state flexure formula. Circumferential distribution steel of at least 0.12% is provided on both faces.

Source: IS 456:2000 Annex G; minimum 0.12% (HYSD) from Cl 26.5.2.1.

Isolated footings#

Areq = Pmax/SBCallowed    Mu = 1.5 p (L − a)²/8 (per metre)

Each column gets a square pad sized for the largest service load. The bending moment is taken at the face of the column: the projection (L − a)/2 acts as a cantilever under upward pressure p, giving p[(L − a)/2]²/2 = p(L − a)²/8. If any column has a net uplift (tank empty, full wind or EQ), a pad alone is unsuitable unless its weight plus soil exceeds the uplift with a factor of safety of 1.5.

Source: IS 456:2000 Cl 34 (footings).

Quantities#

Concrete = Σ volumes    steel ≈ Σ(area/m × length × 7850 kg/m³)    formwork = Σ contact areas

Approximate material quantities are summed from the same surface areas and thicknesses used for the dead load. They are for a preliminary estimate only; a bar bending schedule from the final drawings gives the actual steel.

Source: Unit weight of steel 7850 kg/m³.

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This reference explains the formulas used by the Nirmaanpro Intze tank designer for learning and preliminary design. Always refer to the current editions of IS 3370, IS 456, IS 875, IS 1893 and IS 13920, and have the final design checked by a licensed structural engineer.

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