Brick and mortar for 230 mm wall (100 kN/m)

Data and assumptions

• Wall 230 mm thick, modular bricks, joints not raked, so the full thickness 230 mm is effective (IS 1905 cl. 5.5.1.1).
• Load 100 kN/m at top, resultant $e/t = 1/12$. Wall 5 m long between cross walls, clear height 3.5 m between RCC slabs (slab thickness not given, so $H = 3.5$ m taken).
• RCC slabs give lateral and rotational restraint at top and bottom.
• Design code: IS 1905:1987 (Tables 4 to 10); NBC 109 follows the same permissible-stress method.
• Brick masonry unit weight 19 kN/m$^3$ (IS 875 Part 1); modular brick, height/width $\le 0.75$, so $k_p=1$.

Step 1: Effective height, length and slenderness ratio (IS 1905 cl. 4.3, 4.4, 4.6)

Height between slab centres $H = 3.50$ m. Effective height $h_{ef} = 0.75 H = 0.75\times3.50 = 2.625$ m. Effective length (wall supported by a cross wall at each end, Table 5 Sl. 3) $L_{ef} = 1.0L = 5.00$ m. The lesser value governs:

$$SR = \frac{h_{ef}}{t} = \frac{2.625}{0.230} = 11.41\ (<27,\ \text{Table 7, cement mortar})$$

Step 2: Design load per metre run at the base

$$\begin{aligned} \text{Self weight} &= t\,H_c\,\gamma = 0.230\times3.50\times19 = 15.30\ \text{kN/m}\\ P &= 100.00 + 15.30 = 115.30\ \text{kN/m} \end{aligned}$$

Stress on the wall:

$$\sigma = \frac{P}{A} = \frac{115.30\times10^3}{230\times10^3\ \text{mm}^2/\text{m}} = 0.501\ \text{N/mm}^2$$

Step 3: Reduction factors

• Stress reduction factor $k_s$ (Table 9, $SR=11.41$, $e/t=0.083$, by linear interpolation) $= 0.828$
• Area reduction factor $k_a$ (cl. 5.4.1.2): area per metre 0.23 m$^2$ is not less than 0.2 m$^2$, so $k_a = 1.0$
• Shape factor $k_p = 1.0$ (modular bricks, height/width $\le 0.75$, Table 10)
• Since $e/t > 1/24$, a 25 percent increase in permissible stress is allowed (cl. 5.4.1.4 a), factor 1.25

Step 4: Required basic compressive stress

$$\sigma \le f_b\,k_s\,k_a\,k_p\times1.25 \Rightarrow f_b \ge \frac{0.501}{0.828\times1.0\times1.0\times1.25} = 0.485\ \text{N/mm}^2$$

(Without the 25 percent increase the requirement would be $f_b \ge 0.606$ N/mm$^2$.)

Step 5: Strength of brick and grade of mortar (IS 1905 Table 8)

Smallest standard unit strength giving $f_b \ge 0.485$ N/mm$^2$:

Answer: use bricks of crushing strength not less than 7.5 N/mm$^2$ with mortar M2 (cement:sand 1:6) ($f_b = 0.59$ N/mm$^2$ $\ge$ 0.485). Other combinations in the table are also acceptable; a stronger mortar allows a weaker brick.