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Warehouse exterior wall, wind 750 N/m2

Short answer

230 mm thick wall, bricks of minimum crushing strength 3.5 N/mm² in cement mortar M1 (1:5) or richer. Wind moment 1.148 kNm/m, maximum compressive stress 0.294 N/mm², no tension. Slenderness ratio 15.2 <27. Tie wall to roof and floor with metal anchors.

Design an exterior wall of a single storey warehouse of 3.5 m height. The loading on the wall consists of a vertical load of 30 kN/m from the roof and wind pressure of 750 N/m2^2. The wall is tied with metal anchor at the floor and roof levels.

Data and assumptions

  • Exterior wall of a single-storey warehouse, height 3.5 m, tied by metal anchors at floor and roof levels, so it spans vertically as simply supported. Vertical roof load P=30P=30 kN/m (assumed axial); wind pressure 750 N/m2^2.
  • Anchors give lateral restraint only (no rotational restraint), so the effective height is hef=1.0Hh_{ef}=1.0H (IS 1905 Table 4 Sl. 3).
  • Design by IS 1905:1987 cl. 5.4.1, 5.4.2, 5.5.2, 5.5.3 (NBC 109 follows the same method). Unit weight of masonry 19 kN/m3^3. Per metre run of wall.
     roof anchor (support)
        |<-- wind w
  ------+-----------   load P (roof)
        |   |
        | wall, H=3.5 m
        |   |  M = wH^2/8 at mid-height
  ------+-----------
        floor anchor (support)

Step 1: Wind moment

w=0.750 kN/m (per m of wall)M=0.750×3.52/8=1.148 kNm/m\begin{aligned} w &= 0.750\ \text{kN/m (per m of wall)}\\ M &= 0.750\times3.5^2/8 = 1.148\ \text{kNm/m} \end{aligned}

Step 2: Trial thickness 230 mm, load and slenderness

Self weight of wall=0.230×3.50×19=15.30 kN/mN (at mid-height)=37.65 kN/mhef=1.0×3.5=3.5 mSR=3.5000.230=15.22 (<27)\begin{aligned} \text{Self weight of wall} &= 0.230\times3.50\times19 = 15.30\ \text{kN/m}\\ N\ (\text{at mid-height}) &= 37.65\ \text{kN/m}\\ h_{ef} &= 1.0\times3.5=3.5\ \text{m} \\ SR &= \frac{3.500}{0.230} = 15.22\ (<27) \end{aligned}

Step 3: Stresses at the mid-height section

σa=NA=37.65×103230×103=0.1637 N/mm2σb=6Mt2=6×1.148×1061000×2302=0.1303 N/mm2σmax=σa+σb=0.2939,σmin=σa−σb=0.0334 N/mm2\begin{aligned} \sigma_a &= \frac{N}{A}=\frac{37.65\times10^3}{230\times10^3} = 0.1637\ \text{N/mm}^2\\ \sigma_b &= \frac{6M}{t^2} = \frac{6\times1.148\times10^6}{1000\times230^2} = 0.1303\ \text{N/mm}^2\\ \sigma_{max} &= \sigma_a+\sigma_b = 0.2939,\qquad \sigma_{min} = \sigma_a-\sigma_b = 0.0334\ \text{N/mm}^2 \end{aligned}

σmin>0\sigma_{min}>0: the whole section is in compression, no tension (the resultant eccentricity e=M/N=0.0305e=M/N=0.0305 m, e/t=0.133<1/6e/t=0.133<1/6).

Step 4: Required basic compressive stress

Resultant eccentricity ratio e/t=M/(Nt)=0.133e/t = M/(Nt) = 0.133. From Table 9: ks=0.676k_s=0.676 (SR=15.22SR=15.22). ka=1k_a=1, kp=1k_p=1. Since 1/24<e/t≤1/61/24<e/t\le1/6, a 25 percent increase is allowed (cl. 5.4.1.4 a).

fb≥σakskakp×1.25=0.16370.676×1.25=0.194 N/mm2f_b \ge \frac{\sigma_a}{k_sk_ak_p\times1.25} = \frac{0.1637}{0.676\times1.25} = 0.194\ \text{N/mm}^2

Step 5: Brick and mortar (IS 1905 Table 8)

Mortar (IS 1905 Table 1)Min. unit strength (N/mm²)Table 8 fbf_b (N/mm²)
M2 (cement:sand 1:6)3.50.35
M1 (cement:sand 1:5)3.50.35
H2 (cement:sand 1:4)3.50.35
H1 (cement:sand 1:3)3.50.35

Answer: 230 mm thick wall (one brick). Bricks of at least 3.5 N/mm2^2 crushing strength in cement mortar M1 (1:5) or richer; practically use 5 N/mm2^2 bricks with 1:5 mortar. Maximum stress 0.294 N/mm2^2 (compression) and minimum 0.033 N/mm2^2; slenderness ratio 15.2 < 27. Provide metal anchors at roof and floor at close spacing to pass the wind load into the roof diaphragm and floor.

✓ Answer checked by Claude Opus 5.5 (high effort)

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