Warehouse exterior wall, wind 860 N/m2, anchored

Data and assumptions

• Exterior wall of a single-storey warehouse, 3.5 m high, tied by metal anchors at floor and roof levels (simply supported vertically). $P=25$ kN/m (axial); wind pressure 860 N/m$^2$.
• Anchors give lateral restraint only: $h_{ef}=1.0H$ (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/m$^3$. Per metre run of wall.

     roof anchor (support)
        |<-- wind w
  ------+-----------   load P=25
        |   |
        | wall, H=3.5 m
        |   |  M = wH^2/8 at mid-height
  ------+-----------
        floor anchor (support)

Step 1: Wind moment

$$\begin{aligned} w &= 0.860\ \text{kN/m (per m of wall)}\\ M &= 0.860\times3.5^2/8 = 1.317\ \text{kNm/m} \end{aligned}$$

Step 2: Trial thickness 230 mm, load and slenderness

$$\begin{aligned} \text{Self weight of wall} &= 0.230\times3.50\times19 = 15.30\ \text{kN/m}\\ N\ (\text{at mid-height}) &= 25 + 15.30/2 = 32.65\ \text{kN/m}\\ h_{ef} &= 1.0\times3.5=3.5\ \text{m} \\ SR &= \frac{3.500}{0.230} = 15.22\ (<27) \end{aligned}$$

Step 3: Stresses at the mid-height section

$$\begin{aligned} \sigma_a &= \frac{N}{A}=\frac{32.65\times10^3}{230\times10^3} = 0.1419\ \text{N/mm}^2\\ \sigma_b &= \frac{6M}{t^2} = \frac{6\times1.317\times10^6}{1000\times230^2} = 0.1494\ \text{N/mm}^2\\ \sigma_{max} &= \sigma_a+\sigma_b = 0.2913,\qquad \sigma_{min} = \sigma_a-\sigma_b = -0.0075\ \text{N/mm}^2 \end{aligned}$$

$\sigma_{min}$ is a small tension of about 0.0075 N/mm$^2$ (resultant $e/t=0.175>1/6$). This is below the permissible flexural tension of 0.07 N/mm$^2$ for vertical bending with mortar M1 or better (IS 1905 cl. 5.4.2), so the wall is safe against cracking provided the mortar is M1 or richer.

Step 4: Required basic compressive stress

Resultant eccentricity ratio $e/t = M/(Nt) = 0.175$. From Table 9: $k_s=0.653$ ($SR=15.22$, $e/t=0.175$; the factor is lower than for a smaller eccentricity at the same slenderness). $k_a=1$, $k_p=1$. Conservatively, the 25 percent increase allowed for $e/t>1/6$ (with the tension area disregarded) is not taken.

$$f_b \ge \frac{\sigma_a}{k_sk_ak_p\times1.00} = \frac{0.1419}{0.653\times1.00} = 0.217\ \text{N/mm}^2$$

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

For comparison, IS 1905 Table 11 (free-standing walls; 860 N/m$^2$ gives $H/t=5$) applies to walls without anchors: it would need $t=3.5/5=0.70$ m, which shows the importance of the anchors.

Answer: 230 mm thick wall; bricks not less than 3.5 N/mm$^2$ in cement mortar of grade M1 (1:5) or richer (needed for the 0.07 N/mm$^2$ flexural tension allowance); practical choice 5 N/mm$^2$ bricks with 1:5 mortar. Stresses at mid-height: $\sigma_{max}=0.291$ and $\sigma_{min}=-0.007$ N/mm$^2$; slenderness ratio 15.2.