Wall under axial and eccentric load

Data and assumptions

• Wall 230 mm thick; the joints are not raked, so the full thickness is effective. $SR=16$ (given, $<27$). Axial load $12$ kN/m and eccentric load $27$ kN/m at $7.33$ cm from the axis.
• Per metre length of wall; $k_a=1$ (area $0.23$ m$^2$/m $\ge 0.2$) and $k_p=1$ (modular bricks).
• Design by IS 1905:1987 (Tables 8, 9; cl. 5.4.1.4, 5.4.4).

Step 1: Resultant load and eccentricity

$$\begin{aligned} P &= 12+27 = 39\ \text{kN/m}\\ e &= \frac{12\times0+27\times7.33}{39} = 5.075\ \text{cm}\\ \frac{e}{t} &= \frac{5.075}{23} = 0.221 \end{aligned}$$

Since $e/t>1/6$ (0.167), tension would develop. Masonry cannot carry tension, so the area under tension is disregarded (cl. 5.4.1.4 b, 5.4.4). The stress over the compressed part is triangular, with a contact (compression) width of

$$3\left(\frac{t}{2}-e\right) = 3\times(11.5-5.075) = 19.3\ \text{cm}$$

Step 2: Maximum stress and stress reduction factor

For a triangular stress block, the maximum stress at the compressed face is twice the average over the contact width:

$$\sigma_{max} = \frac{2P}{3(t/2-e)} = \frac{2\times39}{0.1928} = 404.7\ \text{kN/m}^2 = 0.405\ \text{N/mm}^2$$

$k_s$ from Table 9 at $SR=16$, $e/t=0.221$ (between $1/6$, 0.65, and $1/4$, 0.61): $0.65-\frac{(0.221-0.1667)}{0.0833}\times(0.65-0.61)=0.624$.

Since $e/t>1/6$, a 25 percent increase in permissible stress is allowed, together with neglecting the tension area (cl. 5.4.1.4 b).

Step 3: Required basic compressive stress

$$f_b \ge \frac{\sigma_{max}}{k_s k_a k_p\times1.25} = \frac{0.405}{0.624\times1\times1\times1.25} = 0.519\ \text{N/mm}^2$$

IS 1905 Table 8: 5 N/mm$^2$ units give at most $f_b = 0.50$ (M1, H2, H1), which is not enough. Units of 7.5 N/mm$^2$ with M2 mortar give $f_b=0.59 \ge 0.519$. (Without the 25 percent increase, $f_b \ge 0.649$ N/mm$^2$, which needs 7.5 N/mm$^2$ units with M1 mortar, $f_b = 0.74$.)

Answer: use bricks of crushing strength not less than 7.5 N/mm$^2$ with cement mortar M2 (1:6), $f_b=0.59$ N/mm$^2$ (SR 16 < 27); use M1 (1:5) if the 25 percent increase is not taken. The eccentricity is large, so reduce it by centring the load, tie the wall to cross walls and provide an RCC band.