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Warehouse exterior wall, wind 860 N/m2, fixed base

Short answer

230 mm thick wall, bricks ≥3.5 N/mm² in M1 (1:5) mortar. Propped cantilever, base moment wH²/8 = 1.317 kNm/m. Maximum compressive stress 0.325 N/mm², no tension. Slenderness ratio 12.9 <27. Fix base with dowels, anchor at roof.

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 25 kN/m from the roof and wind pressure of 860 N/m2^2. The wall is fixed at base and tied with metal anchor at the roof level.

Data and assumptions

  • Exterior wall of a single-storey warehouse, 3.5 m high, fixed at base (continuous with the foundation) and tied with a metal anchor at roof level: a propped cantilever. P=25P=25 kN/m (axial); wind 860 N/m2^2.
  • Fixed at the base and laterally supported (not rotationally restrained) at the top: hef=0.85Hh_{ef}=0.85H (Table 4 Sl. 2). Maximum moment is at the fixed base, M=wH2/8M=wH^2/8 (hogging); sagging moment 9wH2/1289wH^2/128 is smaller.
  • 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=860 N/m2
  ------+-----------   load P=25 kN/m
        |   |
        | wall, H=3.5 m
        |   |  M = wH^2/8 at the fixed base
  ======+===========   foundation (fixed)

Step 1: Wind moment

w=0.860 kN/m (per m of wall)M=0.860×3.52/8=1.317 kNm/m\begin{aligned} w &= 0.860\ \text{kN/m (per m of wall)}\\ M &= 0.860\times3.5^2/8 = 1.317\ \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 base)=40.30 kN/mhef=0.85×3.5=2.975 mSR=2.9750.230=12.93 (<27)\begin{aligned} \text{Self weight of wall} &= 0.230\times3.50\times19 = 15.30\ \text{kN/m}\\ N\ (\text{at base}) &= 40.30\ \text{kN/m}\\ h_{ef} &= 0.85\times3.5=2.975\ \text{m} \\ SR &= \frac{2.975}{0.230} = 12.93\ (<27) \end{aligned}

Step 3: Stresses at the base section

σa=NA=40.30×103230×103=0.1752 N/mm2σb=6Mt2=6×1.317×1061000×2302=0.1494 N/mm2σmax=σa+σb=0.3246,σmin=σa−σb=0.0258 N/mm2\begin{aligned} \sigma_a &= \frac{N}{A}=\frac{40.30\times10^3}{230\times10^3} = 0.1752\ \text{N/mm}^2\\ \sigma_b &= \frac{6M}{t^2} = \frac{6\times1.317\times10^6}{1000\times230^2} = 0.1494\ \text{N/mm}^2\\ \sigma_{max} &= \sigma_a+\sigma_b = 0.3246,\qquad \sigma_{min} = \sigma_a-\sigma_b = 0.0258\ \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=0.0327e=M/N=0.0327 m, e/t=0.142<1/6e/t=0.142<1/6).

Step 4: Required basic compressive stress

Resultant eccentricity ratio e/t=M/(Nt)=0.142e/t = M/(Nt) = 0.142. From Table 9: ks=0.753k_s=0.753 (SR=12.93SR=12.93). 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=0.17520.753×1.25=0.186 N/mm2f_b \ge \frac{\sigma_a}{k_sk_ak_p\times1.25} = \frac{0.1752}{0.753\times1.25} = 0.186\ \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; bricks not less than 3.5 N/mm2^2 with cement mortar M1 (1:5) or richer (practically 5 N/mm2^2 bricks, 1:5 mortar). Base section: σmax=0.325\sigma_{max}=0.325, σmin=0.026\sigma_{min}=0.026 N/mm2^2 (all compression). Anchor the roof to the wall and tie the wall to the footing with dowels to ensure fixity.

✓ Answer checked by Claude Opus 5.5 (high effort)

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