3 m wall under wind: fixed top and bottom

Data and assumptions

• The wall is fixed at top and bottom, so $h_{ef}=0.75H=2.25$ m (full restraint, Table 4 Sl. 1). No vertical load other than self weight is given. The pressure is taken as stated, 16 N/m$^2$ (0.016 kN/m$^2$) [reading: a very small wind pressure, so stresses are tiny and slenderness controls the thickness].
• End moment of a wall fixed at both ends under uniform load: $M=wH^2/12$ at each end (sagging $wH^2/24$ at mid-height). Section checked at the base, where self weight is the largest.
• 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.

  ======+===========  fixed (top)
        |<-- wind w
        |  wall H=3 m
        |  M = wH^2/12 at ends
  ======+===========  fixed (bottom)

Step 1: Wind moment

$$\begin{aligned} w &= 0.016\ \text{kN/m (per m of wall)}\\ M &= 0.016\times3.0^2/12 = 0.012\ \text{kNm/m} \end{aligned}$$

Step 2: Trial thickness 115 mm, load and slenderness

$$\begin{aligned} \text{Self weight of wall} &= 0.115\times3.00\times19 = 6.56\ \text{kN/m}\\ N\ (\text{at base}) &= 6.56\ \text{kN/m}\\ h_{ef} &= 0.75\times3.0=2.25\ \text{m} \\ SR &= \frac{2.250}{0.115} = 19.57\ (<27) \end{aligned}$$

Step 3: Stresses at the base section

$$\begin{aligned} \sigma_a &= \frac{N}{A}=\frac{6.56\times10^3}{115\times10^3} = 0.0570\ \text{N/mm}^2\\ \sigma_b &= \frac{6M}{t^2} = \frac{6\times0.012\times10^6}{1000\times115^2} = 0.0054\ \text{N/mm}^2\\ \sigma_{max} &= \sigma_a+\sigma_b = 0.0624,\qquad \sigma_{min} = \sigma_a-\sigma_b = 0.0516\ \text{N/mm}^2 \end{aligned}$$

$\sigma_{min}>0$: the whole section is in compression, no tension (the resultant eccentricity $e=M/N=0.0018$ m, $e/t=0.016<1/6$).

Step 4: Required basic compressive stress

Resultant eccentricity ratio $e/t = M/(Nt) = 0.016$. From Table 9: $k_s=0.619$ ($SR=19.57$). $k_a=1$, $k_p=1$.

$$f_b \ge \frac{\sigma_a}{k_sk_ak_p\times1.00} = \frac{0.0570}{0.619\times1.00} = 0.092\ \text{N/mm}^2$$

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

Answer: a half-brick wall of 115 mm is sufficient for stress (SR = 19.6 < 27), but for stability and practical robustness adopt a 230 mm (one-brick) wall; bricks of 3.5 N/mm$^2$ with cement mortar M1 (1:5). Stresses are very small: $\sigma_{max}=0.0624$, $\sigma_{min}=0.0516$ N/mm$^2$, far below the permissible values.