Unit and mortar for 200 x 400 column

Data and assumptions

• Column 200 mm $\times$ 400 mm, clear height 3 m, fully restrained at both ends, so $h_{ef}=0.75H=2.25$ m (IS 1905 Table 4 Sl. 1). Load 120 kN with an eccentricity of 50 mm, taken along the 400 mm side ($e/D=0.125$).
• Units with height/width $\le0.75$ are assumed ($k_p=1$). Self weight is added, 19 kN/m$^3$.

Step 1: Slenderness, factors and stress

$$\begin{aligned} SR &= \frac{2.25}{0.20} = 11.25\ (<12)\\ A &= 0.2\times0.4 = 0.08\ \text{m}^2\ \Rightarrow\ k_a = 0.7+1.5\times0.08=0.82\\ \frac{e}{D} &= \frac{50}{400}=0.125\ (1/24<e/D<1/6)\ \Rightarrow\ k_s = 0.823,\ \text{increase }1.25\\ \text{Self weight} &= 0.08\times3\times19 = 4.56\ \text{kN}\\ P &= 120+4.56 = 124.56\ \text{kN}\\ \sigma &= \frac{124.56\times10^3}{80000} = 1.557\ \text{N/mm}^2 \end{aligned}$$

($k_s$ from Table 9: at $e/D=0.125$ it is 0.86 for $SR=10$ and 0.80 for $SR=12$; interpolating for $SR=11.25$ gives 0.823.)

Step 2: Required basic compressive stress

$$f_b \ge \frac{\sigma}{k_s k_a k_p \times 1.25} = \frac{1.557}{0.823\times0.82\times1.0\times1.25} = 1.846\ \text{N/mm}^2$$

Step 3: Select unit and mortar (IS 1905 Table 8)

Answer: required $f_b = 1.846$ N/mm$^2$; the column needs masonry units of at least 25 N/mm$^2$ crushing strength with H1 mortar (1:3), or 30 N/mm$^2$ units with H2 mortar (1:4). Such strong units are uncommon in brickwork, so in practice enlarge the section (e.g. 300 mm $\times$ 400 mm, or 400 mm square) or use an RCC column.