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Slab design

Guided step by step as per IS 456:2000: one-way slabs and two-way slabs with the Annex D moment coefficients, including steel, spacing, shear, deflection and corner torsion steel.

Guided · step 1 of 9

One-way action and the data

A slab supported on two opposite sides (or with long side more than twice the short side) bends in one direction only, like a wide beam. We design a 1 m wide strip.

Clear span 3.5 m, simply supported, supports 230 mm wide

Trial thickness D = 160 mm; M20, Fe415; live load 3 kN/m², finishes 1 kN/m²

Slab

Strip along the span

wᵤ = 12 kN/mL = 3.635 m10 mm bars at the bottom · not to scale
Mu
19.82kN·m/m
Main steel
10 @ 180mm
436 mm²/m
Shear
OK
τv 0.162 ≤ 0.506
Deflection
OK
L/d 26.9 ≤ 28.9

1 m section through the slab

D = 160d1000 mm strip (width to scale, depth ×1.6)
Full solution, all steps at once
  1. One-way action and the data

    Clear span 3.5 m, simply supported, supports 230 mm wide

    Trial thickness D = 160 mm; M20, Fe415; live load 3 kN/m², finishes 1 kN/m²

  2. Effective depth and effective spand = D − cover − φ2L = min(Lc + d, Lc + bw)

    d = 160 − 20 − 102 = 135 mm

    L = min(3.5 + 0.135, 3.5 + 0.23) = 3.635 m

  3. Loads and the factored loadw = 25D + finishes + livewu = 1.5·w

    Self weight = 25 × 0.16 = 4 kN/m²

    w = 4 + 1 + 3 = 8 kN/m²

    wu = 1.5 × 8 = 12 kN/m per metre width

  4. Design bending momentMu = wuL²8

    Mu = 12 × 3.635² × 18 = 19.82 kN·m per m

  5. Depth needed for bendingdreq = √Mu0.138·fck·b

    dreq = √198198380.138 × 20 × 1000 = 84.8 mm

    Provided d = 135 mm ✓

  6. Main steelAst = 0.5fckfy·[1 − √1 − 4.6Mufckbd²]·bd

    4.6Mufckbd² = 4.6 × 1981983820 × 1000 × 135² = 0.2501

    Ast = 436.1 mm²/m; minimum = 0.12 % × 1000 × 160 = 192 mm²/m

  7. Bar spacing and distribution steels = 1000·aφAst

    a = π × 10²4 = 78.5 mm²; s = 1000 × 78.5436.1 = 180.1 mm

    Provide 10 mm bars @ 180 mm c/c (Ast = 436 mm²/m)

    Distribution: 8 mm @ 260 mm c/c (192 mm²/m needed)

  8. Shear checkVu = wuL2τv = Vubd

    Vu = 0.5 × 12 × 3.635 = 21.81 kN

    τv = 218101000 × 135 = 0.162 N/mm²

    pt = 100 × 4361000 × 135 = 0.323 % → τc = 0.395; k = 1.28; kτc = 0.506 ✓

  9. Deflection checkfs = 0.58fy·A_{st,req}A_{st,prov}(L/d)allowed = basic × MF

    fs = 0.58 × 415 × 436436 = 240.5 N/mm²; pt = 0.323 % → MF ≈ 1.44

    Allowed = 20 × 1.44 = 28.9; actual L/d = 3635135 = 26.9 ✓