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3 m wall under wind: free top, fixed bottom

Short answer

By calculation, a 230 mm thick wall is adequate (σmax=0.063 N/mm², stability factor 28). Use bricks of at least 3.5 N/mm² in cement mortar M1 (1:5). Without calculation, adopt 345 mm per IS 1905 Table 11 (H/t ≤ 10).

Design a wall 3.0 m high to resist a wind load. The wall is free at top and fixed at bottom supports. Assume uniformly distributed wind pressure of 12 N/m2^2.

Data and assumptions

  • Wall 3.0 m high, free at the top and fixed at the base: a cantilever (free-standing) wall. Uniform wind pressure taken as stated, 12 N/m2^2 (0.012 kN/m2^2) [reading: given value is very small, so stability is easily satisfied; if the wind pressure is larger, repeat with the same steps].
  • Design by IS 1905:1987 cl. 5.5.2.1 and Table 11, Table 4 Sl. 4. Per metre run; masonry 19 kN/m3^3; trial thickness 230 mm.
        ^ free top
        |<-- w = 12 N/m2
        |  wall H = 3.0 m
        |  M = wH^2/2 at base
  ======+======  fixed base

Step 1: Slenderness (cl. 4.3, Table 4 Sl. 4)

For a free top and a fixed base hef=1.5H=4.5h_{ef}=1.5H=4.5 m, so SR=4.5/0.23=19.6<27SR = 4.5/0.23 = 19.6<27 (cement mortar), but the slenderness check alone would give t≥4.5/27=0.167t\ge4.5/27=0.167 m.

Step 2: Bending and stresses

M=wH22=0.012×322=0.054 kNm/mSelf weight=0.23×3×19=13.11 kN/mσa=13.11×103230×103=0.0570 N/mm2σb=6Mt2=6×0.054×1061000×2302=0.0061 N/mm2σmax=0.0631,  σmin=0.0509 N/mm2\begin{aligned} M &= \frac{wH^2}{2} = \frac{0.012\times3^2}{2} = 0.054\ \text{kNm/m}\\ \text{Self weight} &= 0.23\times3\times19 = 13.11\ \text{kN/m}\\ \sigma_a &= \frac{13.11\times10^3}{230\times10^3} = 0.0570\ \text{N/mm}^2\\ \sigma_b &= \frac{6M}{t^2} = \frac{6\times0.054\times10^6}{1000\times230^2} = 0.0061\ \text{N/mm}^2\\ \sigma_{max} &= 0.0631,\ \ \sigma_{min} = 0.0509\ \text{N/mm}^2 \end{aligned}

No tension at the base (σmin>0\sigma_{min}>0), and the compressive stress is far below the allowable value for the lowest brick grade.

Step 3: Stability (cl. 4.2.2.4 a)

Free-standing wall: stability moment must be at least 1.5 times the overturning moment:

Mstab=Wt2=13.11×0.115=1.51 kNm/m ≫ 1.5×0.054=0.081 kNm/mM_{stab}=W\frac{t}{2} = 13.11\times0.115 = 1.51\ \text{kNm/m}\ \gg\ 1.5\times0.054 = 0.081\ \text{kNm/m}

Step 4: Table 11 check

IS 1905 Table 11 allows a free-standing wall without calculation (seismic zones I and II, mortar not leaner than M1) when H/t≤10H/t \le 10 for pressures up to 285 N/m2^2: t≥3.0/10=0.30t\ge3.0/10=0.30 m. Our wall satisfies stress and stability by calculation; if the simplified table is preferred, adopt 345 mm (one and a half brick).

Answer: the wall is safe with 230 mm thickness by calculation (σmax=0.063\sigma_{max}=0.063 N/mm2^2, stability factor 28); bricks of at least 3.5 N/mm2^2 in cement mortar M1 (1:5), practical 5 N/mm2^2. If the wall is built without calculation use t=345t=345 mm (Table 11, H/t≤10H/t\le10). Provide a damp-proof course and a coping to protect the top.

✓ Answer checked by Claude Opus 5.5 (high effort)

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