--- title: "Sal timber beam, 2.5 m span, 20 kN/m" subject: "Design of Steel and Timber Structure" chapter: 12 key: "semester_6/design_of_steel_and_timber_structure:timber-beam-sal-2-5m-20knm" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-beam-sal-2-5m-20knm site: BCT Engineering --- # Sal timber beam, 2.5 m span, 20 kN/m Design of Steel and Timber Structure · Chapter 12: Design of Flexure Member - Asked in: 2073 Bhadra, 10 marks - Video: [Steel and timber numerical 2081/sal timber numerical](https://www.youtube.com/watch?v=BGZ65AQOerk) — Rabin silwal - 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-beam-sal-2-5m-20knm - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-12 ## Question Design a timber beam of sal wood having clear span 2.5 m, support width 300 mm and subjected to imposed load of 20 KN/m. ## Short answer Adopt a 125 mm × 375 mm Sal wood beam (Grade I, inside) with 300 mm bearing. Bending stress 6.81 N/mm², shear 0.91 N/mm², deflection 2.3 mm; all within permissible limits. ## Answer **Approach.** Working stress design of a timber beam to IS 883:1994 cl. 7.5. Checks: bending strength, horizontal shear, bearing stress at supports and deflection. Self-weight is included (cl. 7.5.9.4). #### Data and permissible stresses Clear span $= 2.50$ m, bearing length $= 300$ mm. **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 | #### Loads and effective span Imposed UDL $20$ kN/m; self-weight is added to it. Support width 300 mm, so bearing length 300 mm. Effective span (cl. 7.5.2): clear span $+$ bearing length $= 2500 + 300 = 2800$ mm $= 2.800$ m. #### Trial section Width not less than 50 mm or $L/50 = 56$ mm (cl. 7.5.5) and depth not more than 3 times the width without lateral stiffening (cl. 7.5.6). Try **125 mm × 375 mm**. Self-weight $= 0.125\times0.375\times805\times9.81/1000 = 0.370$ kN/m (density 805 kg/m³). Total UDL $w = 20.000 + 0.370 = 20.370$ kN/m. Reaction $R = 28.52$ kN (shear $V = 28.52$ kN); maximum bending moment $M = 19.96$ kN·m. #### Bending $$ \begin{aligned} Z &= \frac{bd^2}{6} = \frac{125\times375^2}{6} = 2.930\times10^6\ \text{mm}^3 \\ f_{ab} &= \frac{M}{Z} = \frac{19.96\times10^6}{2.930\times10^6} = 6.81\ \text{N/mm}^2 \end{aligned} $$ The depth exceeds 300 mm, so the form factor (cl. 7.5.4) $K = 0.81\dfrac{d^2 + 89400}{d^2 + 55000} = 0.952$ applies: permissible $f_b = 16.90\times0.952 = 16.10$ N/mm². $f_{ab} = 6.81 < 16.10$ N/mm². **Safe.** #### Horizontal shear (cl. 7.5.7) $$ H = \frac{3V}{2bd} = \frac{3\times28518}{2\times125\times375} = 0.913\ \text{N/mm}^2 \le 0.94\ \text{N/mm}^2 $$ (no reduction is taken for loads near the supports, which is on the safe side). **Safe.** #### Bearing (cl. 7.5.8) $$ f_{cn}' = \frac{R}{b\times\text{bearing}} = \frac{28518}{125\times300} = 0.76\ \text{N/mm}^2 < f_{cn} = 4.60\ \text{N/mm}^2 $$ The permissible stress perpendicular to the grain applies for any bearing length at the ends of a member. **Safe.** #### Deflection (cl. 7.5.9) $E = 12670$ N/mm², $I = bd^3/12 = 549\times10^6$ mm⁴. Limit $= L/240 = 11.67$ mm (members not carrying brittle finishes). $$ \delta = 2.34\ \text{mm} < \frac{L}{240} = 11.67\ \text{mm} $$ (UDL term $5wL^4/384EI$). **Safe.** #### Answer Provide a Sal beam **125 mm × 375 mm** (Grade I (standard), inside), with a bearing of at least 300 mm at each support. $f_{ab} = 6.81$ N/mm², $H = 0.91$ N/mm², $\delta = 2.3$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The 20 kN/m is only the imposed load. How is the design UDL found? A. w = 20 kN/m only B. w = 20 x 1.5 factored load C. w = 20 + self-weight of trial section = 20.37 kN/m D. w = 20 - self-weight Answer: C. The given load excludes self-weight, so the beam's own weight (0.370 kN/m) is added. (Key point: Total load) 2. With clear span 2.5 m and support width 300 mm, what effective span is used? A. 2.5 m B. 2.8 m (clear span plus bearing) C. 3.1 m D. 2.2 m Answer: B. Effective span = clear span + bearing length = 2500 + 300 = 2800 mm. (Key point: Effective span) 3. What is the bending stress outcome for 125 x 375 mm? A. f_ab = 6.81 N/mm^2, within 16.90 N/mm^2 B. f_ab = 6.81 N/mm^2, above 0.94 N/mm^2, so unsafe C. f_ab = 16.10 N/mm^2, which is the actual stress D. f_ab = 19.96 N/mm^2, unsafe Answer: A. f_ab = M/Z = 19.96e6/2.930e6 = 6.81 N/mm^2, below the permissible 16.10 N/mm^2 after the form factor. (Key point: Bending check) 4. Which check is closest to its limit for this section? A. Bending B. Deflection C. Bearing D. Horizontal shear, 0.913 against 0.94 N/mm^2 Answer: D. Shear stress is 97% of the permissible value, while the other checks have large margins, so shear controls the section size. 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