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RC wall footing, 1000 kN total load

Short answer

For 1 m length of wall, required width 1.85 m including 10% footing weight. Factored pressure 202.7 kN/m2 gives Mu=60.87 kNm/m; provide depth 375 mm, main steel 10 mm @ 140 mm c/c and distribution 8 mm @ 110 mm c/c.

Design a RC concrete spread footing of an RC wall having 4 m length and 300 mm width. Total load on wall is equal to 1000 kN at service state. Take M20 and Fe415 grades of concrete and steel respectively, and safe bearing capacity of soil = 150 kN/m2^2.

Data and assumptions

  • Wall 300 mm thick and 4 m long carrying a total service load of 1000 kN, so the load per metre run is w=1000/4=250w = 1000/4 = 250 kN/m. M20 concrete, Fe415 steel, qa=150q_a = 150 kN/m². The footing is designed for a 1 m length of wall. Cover 50 mm.

Width of footing

B=1.1 wqa=1.1×250150=1.833 m⇒provide B=1.85 mB = \frac{1.1\,w}{q_a} = \frac{1.1 \times 250}{150} = 1.833\ \text{m} \Rightarrow \text{provide } B = 1.85\ \text{m}

(10% is added for the weight of the footing and backfill.)

Factored pressure and bending moment

wu=1.5×250=375w_u = 1.5 \times 250 = 375 kN/m, pu=wuB=3751.85=202.70p_u = \dfrac{w_u}{B} = \dfrac{375}{1.85} = 202.70 kN/m² Projection beyond the wall face =(1850−300)/2= (1850 - 300)/2 =775= 775 mm.

Mu=pu a22=202.70×0.77522=60.87 kN⋅m per metreM_u = \frac{p_u\, a^2}{2} = \frac{202.70 \times 0.775^2}{2} = 60.87\ \text{kN·m per metre}

Depth

The depth is governed by one-way shear at dd from the wall face (checked below); trial D=375D = 375 mm, d=375−50−6=319d = 375 - 50 - 6 = 319 mm (conservative). Flexure check: Mu,lim=0.138fckbd2=280.9M_{u,lim} = 0.138 f_{ck} b d^2 = 280.9 kN·m >60.87> 60.87 kN·m, so the depth is adequate and the section is under-reinforced.

Reinforcement

Ast=548A_{st} = 548 mm²/m from MuM_u (minimum 0.12% of bDbD = 450 mm²/m). Adopt 548 mm²/m → provide 10 mm @ 140 mm c/c across the width (561 mm²/m, pt=0.176%p_t = 0.176\%). Distribution steel along the wall (0.12% of bDbD) = 450 mm²/m → 8 mm @ 110 mm c/c (457 mm²/m).

Check for shear

Vu=pu(a−d)=202.70×(0.775−0.319)=92.43V_u = p_u (a - d) = 202.70 \times (0.775 - 0.319) = 92.43 kN/m; τv=924321000×319=0.290\tau_v = \dfrac{92432}{1000 \times 319} = 0.290 MPa < τc=0.30\tau_c = 0.30 MPa (M20, pt=0.176%p_t = 0.176\%, Table 19). Safe.

Development length

Ld=0.87×415×104×1.92=470L_d = \dfrac{0.87 \times 415 \times 10}{4 \times 1.92} = 470 mm; available length =775−50=725= 775 - 50 = 725 mm. Satisfied.

Detailing

SECTION ACROSS WALL
        |  wall 300  |
        |            |
   _____|            |_____
  |                       |  D = 375 mm
  |_______________________|
   o o o o o o o o o o o o   <- 10 mm @ 140 c/c
  <---------- 1.85 m ------->

Distribution bars of 8 mm @ 110 mm c/c are tied above the main bars, parallel to the wall.

Answer: Strip footing 1.85 m wide and 375 mm thick; main bars 10 mm @ 140 mm c/c across the wall, distribution bars 8 mm @ 110 mm c/c.

✓ Answer checked by Claude Opus 5.5 (high effort)

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