BahayPlanNSCP 2015 · PD 1096

NSCP 2015 · Section 413

Footing Solver

Enter the given loads, dimensions and soil data. The footing size, thickness, bar count, spacing and development lengths are solved step by step.

Therefore, use

  • 3 m wide × 0.5 m thick wall footing
  • 20 mm ∅ main bars @ 200 mm O.C. (transverse)
  • 15 pcs − 16 mm ∅ temperature bars, uniformly O.C.
  • 10 mm ∅ dowels, lap splice 400 mm
  • No hook needed at bar ends
Section — wall footing
NGLB = 3 m500Df 1.2 m123
  • 120 mm ∅ main bars @ 200 mm O.C.
  • 215 pcs − 16 mm ∅ temperature bars, uniformly O.C.
  • 3Wall dowels with 400 mm lap splice
Plan — 1 m strip of footing
B = 3 mwall 300 mm · main bars @ 200 mm

Step-by-step solution

Step 1 · Service and factored loads

  • Ps

    PDL + PLL = 290 + 220

    510 kN/m
  • Pu

    1.2PDL + 1.6PLL = 1.2(290) + 1.6(220)

    700 kN/m ← governs
  • Pu

    1.4PDL = 1.4(290)

    406 kN/m

Step 2 · Initial size and hc (assume hc = 0)

  • qe

    qa − γc·hc − γs(Df − hc) − qs = 193 − 23.5(0) − 15.74(1.2 − 0) − 0

    174.112 kPa
  • B

    Ps / qe = 510 / 174.112

    2.929 m → 3 m
  • qu

    Pu / B = 700 / 3

    233.333 kPa
  • One-way shear

    φ(1/6)λ√fc'·b·d = qu((B − a)/2 − d) · 0.75(1/6)(1)√20.7(1)(d)(1000) = 233.333(1.35 − d)

    d = 392.744 mm
  • hc

    d + cc + ½db = 392.744 + 75 + 10

    477.744 mm → 500 mm

Step 3 · Check using hc = 500 mm

  • qe

    qa − γc·hc − γs(Df − hc) − qs = 193 − 23.5(0.5) − 15.74(1.2 − 0.5) − 0

    170.232 kPa
  • B

    Ps / qe = 510 / 170.232

    2.996 m → 3 m
  • qu

    Pu / B

    233.333 kPa
  • d req

    one-way shear: φVc = qu(1.35 − d)

    392.744 mm
  • d provided

    hc − cc − ½db = 500 − 75 − 10

    415 mm > d req ∴ OK

∴ Use 3 m wide × 0.5 m thick wall footing

Step 4 · Mu and As (1 m strip)

  • Mu

    qu·x²/2, x = (B − a)/2 = 1.35 m

    212.625 kN·m
  • φMnt

    (459/1600)β1·fc'·b·d²(1 − 3β1/16), β1 = 0.85

    730.77 kN·m > Mu → tension-controlled, φ = 0.90
  • Rn

    Mu / (φbd²) = 212.625×10⁶ / (0.90 × 1,000 × 415²)

    1.372 MPa
  • ρ

    0.85fc'/fy [1 − √(1 − 2Rn/0.85fc')]

    0.00345
  • ρmin

    1.4 / fy = 1.4 / 415

    0.00337 → use ρ = 0.00345
  • As

    ρbd = 0.00345(1,000)(415)

    1,429.85 mm²

Step 5 · Spacing of main bars

  • s

    1000·Ab / As = 1000(π/4·20²) / 1,429.85

    219.715 mm → 200 mm
  • smax

    lesser of 3hc = 1,500 mm and 450 mm

    450 mm
  • Check

    s provided ≤ smax

    OK

∴ Use 20 mm ∅ main bars spaced 200 mm O.C.

Step 6 · Temperature reinforcement (along the wall)

  • As min

    0.002·B·hc = 0.002(3,000)(500)

    3,000 mm²
  • N

    As / Ab = 3,000 / (π/4·16²)

    14.921 → 15 pcs
  • s

    (B − 2cc − db) / (N − 1)

    202.429 mm
  • smax

    lesser of 5hc = 2,500 and 450 mm

    450 mm ∴ OK

∴ Use 15 pcs − 16 mm ∅ temperature bars spaced uniformly O.C.

Step 7 · Development length of main bars

  • ψt, ψe, ψs, Ktr

    bottom bars, uncoated

    1.0, 1.0, 0.8, 0
  • cb

    lesser of cc + db/2 = 85 and s/2 = 100

    85 mm
  • (cb + Ktr)/db

    85 / 20 = 4.25

    > 2.5 → use 2.5
  • Ld req

    (1/1.1)(fy/λ√fc')(ψtψeψs/((cb+Ktr)/db))·db ≥ 300

    530.701 mm
  • Ld provided

    from critical section to bar end, less cover

    1,275 mm > Ld ∴ no hook needed

Step 8 · Ldc of 10 mm wall dowels

  • Ldc (a)

    0.24fyψr/(λ√fc')·db = 0.24(415)/√20.7 (10)

    218.914 mm
  • Ldc (b)

    0.043fyψr·db = 0.043(415)(10)

    178.45 mm
  • Ldc (c)

    minimum

    200 mm
  • Ldc provided

    hc − cc − bar layers

    389 mm ≥ 218.914 ∴ OK

∴ The height of footing is adequate

Step 9 · Lap splice of wall vertical bars

  • Lsc

    greater of 0.071fy·db = 294.65 mm and 300 mm, × 4/3 since fc' < 21 MPa

    400 mm → use 400 mm

∴ Use lap splice = 400 mm

Design aid based on NSCP 2015 Vol. 1 (Sections 406, 413, 422, 425). Results should be reviewed and signed by a licensed civil engineer before construction.