Wall with openings: pier loads and check

Principle

The load over a window opening is carried by the lintel/band and shared by the piers on either side, so the piers carry their own length of wall load plus half of each adjacent opening. Area factor $k_a$ is applied to the smaller piers.

Step 1: Slenderness and $k_s$

$H_{eff} = 0.75 \times 3000 = 2250$ mm, $SR = 2250/230 = 9.78 < 27$.

$$k_s = 0.89 + \frac{10 - 9.78}{2}\times 0.06 = 0.897$$

Step 2: Load on piers

Check: $2 \times 127.5 + 255.0 = 510 = 6.0 \times 85$ kN. OK.

Step 3: Self-weight (to base)

• End pier: $19 \times 0.23 \times 0.75 \times 3.0 = 9.83$ kN, total $P = 137.3$ kN.
• Middle pier: $19 \times 0.23 \times 1.5 \times 3.0 = 19.67$ kN, total $P = 274.7$ kN.

Step 4: Area factor and permissible stress

• End pier: $A = 0.23 \times 0.75 = 0.1725$ m² $< 0.2$, so $k_a = 0.7 + 1.5 \times 0.1725 = 0.959$.
• Middle pier: $A = 0.345$ m² $> 0.2$, $k_a = 1.0$.

$$f_{c,end} = 1.0 \times 0.897 \times 0.959 = 0.860\ \text{N/mm}^2, \qquad f_{c,mid} = 1.0 \times 0.897 = 0.897\ \text{N/mm}^2$$

Step 5: Actual stress

$$\begin{aligned} f_{end} &= \frac{137\,330}{0.1725\times10^{6}} = 0.796\ \text{N/mm}^2 < 0.860 \\ f_{mid} &= \frac{274\,670}{0.345\times10^{6}} = 0.796\ \text{N/mm}^2 < 0.897 \end{aligned}$$

The end pier has the smallest margin (93 percent utilisation); it should not be reduced in width further. Pier width must also satisfy seismic opening rules.

Answer: Both piers are safe: stress 0.80 N/mm² against permissible 0.86 N/mm² (end pier) and 0.90 N/mm² (middle pier).