--- title: "Circular Sal column, open shed, 200 kN" subject: "Design of Steel and Timber Structure" chapter: 11 key: "semester_6/design_of_steel_and_timber_structure:timber-circular-sal-3-5m-200kn" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-circular-sal-3-5m-200kn site: BCT Engineering --- # Circular Sal column, open shed, 200 kN Design of Steel and Timber Structure · Chapter 11: Design of Compression Members - Asked in: 2070 Bhadra, 8 marks - 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-circular-sal-3-5m-200kn - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-11 ## Question Design a circular salwood column to be used in an open shed, to carry an axial load of 200 kN. The column is 3.5 m high. ## Short answer Provide a circular Sal column of diameter 180 mm. Equivalent square side = 159.5 mm, S/a = 21.94 (long), fc = 8.66 N/mm², capacity = 220 kN > 200 kN. ## Answer **Approach.** IS 883:1994 cl. 7.6.1.6: the permissible load on a circular column must not exceed that of a **square column of equal cross-sectional area**. So the circular column is designed through its equivalent square side $a$, with $a^2 = \pi D^2/4$, i.e. $a = 0.886D$. #### Data and permissible stresses Height (unsupported length) $S = 3.5$ m, axial load $N = 200$ kN, pin ends. **Species:** Sal (Shorea robusta), Group A. **Location:** outside. **Grade:** Grade I (standard) (Table 1 values are for Grade I). Permissible stresses (IS 883:1994 Table 1, outside location): | Property | Table 1 value (N/mm²) | Used (N/mm²) | |---|---|---| | Bending / tension along grain $f_b$ | 14.0 | 14.00 | | Compression parallel to grain $f_{cp}$ | 9.4 | 9.40 | | Compression perpendicular to grain $f_{cn}$ | 3.5 | 3.50 | | Horizontal shear $H$ | 0.94 | 0.94 | | Modulus of elasticity $E$ | 12670 | 12670 | An open shed is an **outside** location; the grade is not stated, so Grade I is used. #### Trial section Try diameter $D = 180$ mm. Area $A = \pi D^2/4 = 25,447$ mm². Equivalent square side: $$ a = 0.886D = 159.5\ \text{mm} $$ #### Slenderness and permissible stress $$ \begin{aligned} S/a &= \frac{3500}{159.5} = 21.94 \\ K_8 &= 0.584\sqrt{\frac{E}{f_{cp}}} = 0.584\sqrt{\frac{12670}{9.40}} = 21.44 \end{aligned} $$ $S/a = 21.94 > K_8 = 21.44$, so it is a **long column** ($S/a < 50$, OK): $$ f_c = \frac{0.329E}{(S/a)^2} = \frac{0.329\times12670}{21.94^2} = 8.66\ \text{N/mm}^2 $$ #### Check $$ P = f_cA = 8.66\times25,447/10^3 = 220\ \text{kN} > 200\ \text{kN} $$ Safe. **Answer: provide a circular Sal column of diameter 180 mm.** (Its permissible load equals that of a 160 mm square column of the same area.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is a circular column designed under IS 883 cl. 7.6.1.6? A. Its permissible load must not exceed that of a square of equal area; a = 0.886D B. As a square of side D C. Using radius of gyration D/4 D. Circular columns are not allowed Answer: A. Equal-area square: a^2 = pi D^2/4, so a = 0.886D. (Key point: Circular column rule) 2. For D = 180 mm, what is a? A. 159.5 mm B. 180 mm C. 127 mm D. 141 mm Answer: A. 0.886 x 180 = 159.5 mm. (Key point: Equivalent side) 3. S/a = 21.94 and K8 = 21.44. What class and fc? A. Long; fc = 0.329E/(S/a)^2 = 8.66 N/mm^2 B. Intermediate; fc = 9.4 C. Short; fc = fcp D. Long; fc = fcp Answer: A. S/a exceeds K8, so the elastic buckling formula applies. (Key point: Class) 4. What is used for the open shed and what is the result? A. Outside values; P = 220 kN > 200 kN B. Inside values; P = 220 kN C. Outside values; P = 180 kN < 200 kN D. Inside values; P = 150 kN Answer: A. An open shed is outside; P = 8.66 x 25,447 = 220 kN. 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