Masonry pier under axial load and in-plane moment

Step 1: Section properties

$$\begin{aligned} A &= 230 \times 1500 = 3.45\times10^5\ \text{mm}^2 \\ Z &= \frac{230 \times 1500^2}{6} = 8.625\times10^{7}\ \text{mm}^3 \end{aligned}$$

Step 2: Stresses

$$\begin{aligned} \frac{N}{A} &= \frac{150\times10^3}{3.45\times10^5} = 0.435\ \text{N/mm}^2 \\ \frac{M}{Z} &= \frac{35\times10^6}{8.625\times10^7} = 0.406\ \text{N/mm}^2 \end{aligned}$$ $$\begin{aligned} \sigma_{max} &= 0.435 + 0.406 = 0.841\ \text{N/mm}^2 \ (\text{compression}) \\ \sigma_{min} &= 0.435 - 0.406 = +0.029\ \text{N/mm}^2 \ (\text{compression}) \end{aligned}$$

There is no tension (the resultant lies within the middle third).

Step 3: Permissible stress

$H_{eff} = 0.75 \times 3000 = 2250$ mm, $SR = 2250/230 = 9.78$ and $k_s = 0.89 + 0.22/2 \times 0.06 = 0.897$.

$A = 0.345$ m² $> 0.2$, so $k_a = 1.0$.

$$f_c = 1.2 \times 0.897 = 1.076\ \text{N/mm}^2, \qquad f_{c,seismic} = 1.33 \times 1.076 = 1.43\ \text{N/mm}^2$$

Step 4: Checks

• Maximum compression: $0.841 < 1.43$ N/mm². Safe.
• Gravity alone: $0.435 < 1.076$ N/mm². Safe.
• Tension: none.

The moment that just causes zero tension is $M_0 = \dfrac{N Z}{A} = N\dfrac{L}{6} = 150 \times 0.25 = 37.5$ kN·m, so the applied 35 kN·m is 93 percent of this limit. If the moment is larger, the pier will crack and tension must be taken by vertical bars (reinforced masonry), or the axial load must be increased or the pier lengthened.

Answer: $\sigma_{max} = 0.841$ N/mm² < 1.43 N/mm² and $\sigma_{min} = +0.029$ N/mm² (no tension); the pier is safe. Zero-tension limit moment = 37.5 kN·m.