Spaced Sal column, 3.5 m, 120 kN

Approach. A spaced column is two or more equal timber members kept apart by spacer blocks at the ends and at intermediate points and joined by bolts or nails. IS 883:1994 cl. 7.6.3 applies the solid-column formulas with a restraint factor of 2.5 (3 if the end connectors are placed at $S/20$ to $S/10$ from the ends) for buckling across the spacers, i.e. on the thickness $d_1$ of one member; $S/d_1$ must not exceed 80 (cl. 7.6.3.3). In the other direction (parallel to the spacers) the spacers give no help, so each member is also checked as a solid column on its width $d_2$, with $S/d_2 \le 50$.

Data and permissible stresses

Length $S = 3.5$ m, axial load $N = 120$ kN, ends pin-jointed.

Species: Sal (Shorea robusta), Group A. Location: inside. Grade: Grade I (standard) (Table 1 values are for Grade I).

Permissible stresses (IS 883:1994 Table 1, inside location):

Location and grade are not stated, so inside and Grade I are assumed.

Trial section

Square members such as 100 mm × 100 mm do not work: parallel to the spacers $S/d = 35 > K_8 = 20.19$, so $f_c = 0.329\times12670/35^2 = 3.40$ N/mm² and $P = 3.40\times20,000/10^3 = 68$ kN $< 120$ kN. Wider members are therefore used.

Two members, each 75 mm thick ($d_1$) × 200 mm wide ($d_2$). Area $A = 2\times75\times200 = 30,000$ mm².

Slenderness and permissible stress

(a) Buckling across the spacers (cl. 7.6.3), restraint factor 2.5:

$$\begin{aligned} S/d_1 &= \frac{3500}{75} = 46.7 \ (\le 80,\ \text{OK}) \\ K_{10} &= 0.584\sqrt{\frac{2.5E}{f_{cp}}} = 0.584\sqrt{\frac{2.5\times12670}{10.60}} = 31.9 \end{aligned}$$

$46.7 > K_{10} = 31.9$, so this is a long spaced column (cl. 7.6.3.2):

$$f_{c1} = \frac{0.329E\times2.5}{(S/d_1)^2} = \frac{0.329\times12670\times2.5}{46.7^2} = 4.79\ \text{N/mm}^2$$

(b) Buckling parallel to the spacers (each member as a solid column):

$$\begin{aligned} S/d_2 &= \frac{3500}{200} = 17.5 \ (\le 50,\ \text{OK}) \\ K_8 &= 0.584\sqrt{\frac{E}{f_{cp}}} = 0.584\sqrt{\frac{12670}{10.60}} = 20.19 \end{aligned}$$

$11 < 17.5 < K_8$, so it is intermediate:

$$f_{c2} = 10.60\left[1 - \frac{1}{3}\left(\frac{17.5}{20.19}\right)^4\right] = 8.61\ \text{N/mm}^2$$

The governing permissible stress is $f_c = 4.79$ N/mm².

Check

$$P = f_cA = 4.79\times30,000/10^3 = 143.6\ \text{kN} > 120\ \text{kN}$$

Safe.

Details of the spaced column

• Clear spacing between the two members is taken as $75$ mm (equal to the member thickness; this spacing is a detailing choice). It is kept by Sal spacer blocks 75 mm thick and 200 mm wide at both ends and at mid-height; the end blocks are placed within $S/20 = 175$ mm of the ends (so the factor 2.5 is used).
• Each spacer block is connected to both members with two bolts (12 mm) or staggered nails.

   |  |   |  |      two members 75 x 200 mm
   |  |===|  |      end block
   |  |   |  |
   |  |===|  |      spacer block (mid-height)
   |  |   |  |
   |  |===|  |      end block

Answer

Provide a spaced column of two 75 mm × 200 mm Sal members with 75 mm clear spacing and spacer blocks as shown. Capacity 144 kN $>$ 120 kN.