--- title: "Spaced Sal column, 3.5 m, 120 kN" subject: "Design of Steel and Timber Structure" chapter: 11 key: "semester_6/design_of_steel_and_timber_structure:timber-spaced-column-3-5m" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-spaced-column-3-5m site: BCT Engineering --- # Spaced Sal column, 3.5 m, 120 kN Design of Steel and Timber Structure · Chapter 11: Design of Compression Members - Asked in: 2078 Chaitra, 10 marks - Video: [design spaced deodar column | steel & timber numerical | 2017 PU | TU.PU.KU.PoU](https://www.youtube.com/watch?v=r2o0LhbEpkA) — Sudip Dhakal - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-spaced-column-3-5m - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-11 ## Question Design a spaced column 3.5m long of Sal timber to carry 120kN load. ## Short answer Provide spaced column of two 75 mm × 200 mm Sal members with 75 mm clear spacing and spacer blocks at ends and mid-height. Restraint factor 2.5, permissible stress 4.79 N/mm², capacity 144 kN > 120 kN. ## Answer **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): | Property | Table 1 value (N/mm²) | Used (N/mm²) | |---|---|---| | Bending / tension along grain $f_b$ | 16.9 | 16.90 | | Compression parallel to grain $f_{cp}$ | 10.6 | 10.60 | | Compression perpendicular to grain $f_{cn}$ | 4.6 | 4.60 | | Horizontal shear $H$ | 0.94 | 0.94 | | Modulus of elasticity $E$ | 12670 | 12670 | 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. ```text | | | | 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What restraint factor is used for the spaced column here, and when is it larger? A. 2.5; it becomes 3 if end connectors are placed at S/10 to S/20 from the ends B. 1.0; it becomes 2.5 with spacers C. 3.0; it becomes 2.5 with end connectors D. 2.5 always; it never changes Answer: A. IS 883 cl. 7.6.3 uses 2.5, raised to 3 when the end fasteners are placed at S/10 to S/20 from the ends. (Key point: Restraint factor) 2. How is the slenderness of a spaced column defined, and what is the limit? A. S/d of the whole group, limit 50 B. S/d1 of one individual member, limit 80 C. S divided by the spacing, limit 80 D. S/d of one member, limit 50 Answer: B. Each member is checked by S/d1 using its own least dimension, and this must not exceed 80 (cl. 7.6.3.3). (Key point: Slenderness of a spaced column) 3. S/d1 = 35 exceeds K10 = 31.9. Which stress formula applies? A. fc = fcp B. fc = fcp[1 - (1/3)(S/d / K10)^4] C. fc = 0.329E x 2.5/(S/d1)^2 D. fc = 0.329E/(S/d1)^2 without the factor Answer: C. In the long range the elastic buckling stress is raised by the restraint factor: 0.329 x 12670 x 2.5/35^2 = 8.51 N/mm^2. (Key point: Long spaced column stress) 4. What area and capacity are used for the two 100 x 100 mm members? A. A = 10,000 mm^2, P = 85 kN B. A = 20,000 mm^2, P = 170 kN > 120 kN C. A = 20,000 mm^2, P = 170 kN < 120 kN D. A = 40,000 mm^2 including the gap, P = 340 kN Answer: B. Only the two timber members count: A = 20,000 mm^2, so P = 8.51 x 20,000 = 170 kN, above 120 kN. 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