Safe axial load on a timber column

Formulas (IS 883)

Slenderness ratio $\lambda = l_e/d$, with $d$ the least dimension. Define

$$K_8 = 0.702\sqrt{\frac{E}{\sigma_{cc}}}$$

• Short column: $l_e/d \le 11$: $\sigma_c' = \sigma_{cc}$
• Intermediate: $11 < l_e/d < K_8$: $\sigma_c' = \sigma_{cc}\left[1 - \frac{1}{3}\left(\frac{l_e/d}{K_8}\right)^4\right]$
• Long: $l_e/d \ge K_8$: $\sigma_c' = \dfrac{0.329\,E}{(l_e/d)^2}$

Value of $K_8$

$$K_8 = 0.702\sqrt{\frac{9500}{7.5}} = 0.702 \times 35.59 = 24.98$$

Area $A = 100 \times 150 = 15000$ mm²; least dimension $d = 100$ mm.

(a) Both ends hinged

$l_e = L = 3000$ mm, so $l_e/d = 30 > K_8 = 24.98$: long column.

$$\begin{aligned} \sigma_c' &= \frac{0.329 \times 9500}{30^2} = 3.473\ \text{N/mm}^2 \\ P &= \sigma_c' A = 3.473 \times 15000 \approx 52\,100\ \text{N} \approx 52.1\ \text{kN} \end{aligned}$$

(b) One end fixed, other hinged

$l_e = 0.8 \times 3000 = 2400$ mm, so $l_e/d = 24 < 24.98$: intermediate column.

$$\begin{aligned} \sigma_c' &= 7.5\left[1 - \frac{1}{3}\left(\frac{24}{24.98}\right)^4\right] = 7.5\,[1 - 0.2838] = 5.37\ \text{N/mm}^2 \\ P &= 5.3715 \times 15000 = 80\,570\ \text{N} \approx 80.6\ \text{kN} \end{aligned}$$

Answer: (a) 52.1 kN (long column); (b) 80.6 kN (intermediate column).