Check of a solid brick wall under gravity load

Step 1: Slenderness

Lateral and rotational restraint at both ends gives $H_{eff} = 0.75H$:

$$\begin{aligned} H_{eff} &= 0.75 \times 3200 = 2400\ \text{mm} \\ SR &= \frac{2400}{230} = 10.43 \end{aligned}$$

$SR < 27$, so OK.

Step 2: Stress reduction factor

Interpolate between $SR = 10$ ($k_s = 0.89$) and $SR = 12$ ($k_s = 0.84$):

$$k_s = 0.89 - \frac{0.43}{2}\times 0.05 = 0.879$$

Step 3: Permissible stress

Per metre length, area $A = 0.23 \times 1.0 = 0.23$ m² $> 0.2$ m², so $k_a = 1.0$; for bricks $k_p = 1.0$.

$$f_c = 1.0 \times 0.879 \times 1.0 \times 1.0 = 0.879\ \text{N/mm}^2$$

Step 4: Actual stress at base

$$\begin{aligned} W_{wall} &= 19 \times 0.23 \times 3.2 = 13.98\ \text{kN/m} \\ P &= 120 + 13.98 = 133.98\ \text{kN/m} \\ f &= \frac{133\,980}{230 \times 1000} = 0.583\ \text{N/mm}^2 \end{aligned}$$

$0.583 < 0.879$, so the wall is safe (utilisation 66 percent).

Step 5: Maximum load

$$P_{max} = f_c\,A = 0.879 \times 230 \times 1000 = 202\,170\ \text{N/m} \approx 202\ \text{kN/m}$$

Allowable additional slab load $= 202.2 - 14.0 = 188\ \text{kN/m}$.

Answer: Actual stress 0.58 N/mm² < 0.879 N/mm², safe. Maximum total load = 202 kN/m (slab load about 188 kN/m).