Brick and mortar: 20 cm wall, 3.4 m

Data and assumptions

• Wall 20 cm thick, modular bricks, joints not raked. Load 100 kN/m at top, $e/t=1/12$.
• Wall 5 m long between cross walls, clear height 3.4 m between RCC slabs (slab thickness not given, $H=3.4$ m).
• RCC slabs give lateral and rotational restraint (Table 4 Sl. 1).
• 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 $H = 3.40$ m. Effective height $h_{ef} = 0.75 H = 0.75\times3.40 = 2.550$ 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:

$$SR = \frac{h_{ef}}{t} = \frac{2.550}{0.200} = 12.75\ (<27,\ \text{Table 7, cement mortar})$$

Step 2: Design load per metre run at the base

$$\begin{aligned} \text{Self weight} &= t\,H_c\,\gamma = 0.200\times3.40\times19 = 12.92\ \text{kN/m}\\ P &= 100.00 + 12.92 = 112.92\ \text{kN/m} \end{aligned}$$

Stress on the wall:

$$\sigma = \frac{P}{A} = \frac{112.92\times10^3}{200\times10^3\ \text{mm}^2/\text{m}} = 0.565\ \text{N/mm}^2$$

Step 3: Reduction factors

• Stress reduction factor $k_s$ (Table 9, $SR=12.75$, $e/t=0.083$, by linear interpolation) $= 0.784$
• Area reduction factor $k_a$ (cl. 5.4.1.2): area per metre 0.20 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

$$\sigma \le f_b\,k_s\,k_a\,k_p\times1.25 \Rightarrow f_b \ge \frac{0.565}{0.784\times1.0\times1.0\times1.25} = 0.576\ \text{N/mm}^2$$

(Without the 25 percent increase the requirement would be $f_b \ge 0.720$ N/mm$^2$.)

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

Smallest standard unit strength giving $f_b \ge 0.576$ 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$^2$ $\ge$ 0.576). Other combinations in the table are also acceptable; a stronger mortar allows a weaker brick.