Brick column design for 80 kN axial load

Data and assumptions

• Brick $200\times100\times100$ mm (height/width $=1.0$, so $k_p>1$ from Table 10). Load 80 kN axial, height 3.5 m. RCC beams and slab give lateral support in both directions, so the effective height is the actual height $h_{ef}=H=3.5$ m (IS 1905 cl. 4.3.2, Note 2).
• Column in cement mortar; self weight added (19 kN/m$^3$). Maximum slenderness ratio 12 (Table 7).

Step 1: Trial sections

For each size: $SR=H/t$, $k_a=0.7+1.5A$ (A < 0.2 m$^2$), $k_s$ from Table 9 ($e=0$), $P=80+$ self weight, $\sigma=P/A$, and the unit strength needed so that $\sigma\le f_b k_sk_ak_p$ ($k_p$ taken at the unit strength):

A 300 mm square column needs very strong bricks (17.5 N/mm$^2$ or more), so it is not practical for locally made bricks.

Step 2: Adopt 400 mm $\times$ 400 mm column

$$\begin{aligned} SR &= \frac{3500}{400} = 8.75\ (<12)\\ A &= 0.16\ \text{m}^2,\ k_a=0.7+1.5\times0.16=0.940\\ k_s &= 0.927\\ \text{Self weight} &= 0.16\times3.5\times19 = 10.64\ \text{kN}\\ P &= 80+10.64 = 90.64\ \text{kN}\\ \sigma &= \frac{90.64\times10^3}{160000} = 0.567\ \text{N/mm}^2 \end{aligned}$$

For 7.5 N/mm$^2$ bricks in 1:5 mortar (M1): $f_b=0.74$, $k_p=1.1$:

$$f_c = 0.74\times0.927\times0.940\times1.1 = 0.710\ \text{N/mm}^2 \ \ge\ 0.567$$

(5 N/mm$^2$ bricks give only 0.523 N/mm$^2$, not enough.)

Answer: provide a 400 mm $\times$ 400 mm brick column (two bricks by two bricks) built with bricks of crushing strength not less than 7.5 N/mm$^2$ in cement mortar 1:5 (M1), with all courses broken-bonded and the column tied to the RCC beam above. Allowable stress 0.710 N/mm$^2$ against actual 0.567 N/mm$^2$.