--- title: "Sal select grade beam, 15 kN/m, 4 m" subject: "Design of Steel and Timber Structure" chapter: 12 key: "semester_6/design_of_steel_and_timber_structure:timber-beam-sal-15knm-4m" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-beam-sal-15knm-4m site: BCT Engineering --- # Sal select grade beam, 15 kN/m, 4 m Design of Steel and Timber Structure · Chapter 12: Design of Flexure Member - Asked in: 2074 Bhadra, 12 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-beam-sal-15knm-4m - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-12 ## Question A timber beam of Sal of select grade carries an udl of 15 KN/m inclusive of its self-weight. The clear span of beam is 4m. Design the timber beam. Take bearing length of support = 230 mm. ## Short answer Use working stress design to IS 883, select grade (1.16 factor). Provide 125 mm x 350 mm Sal beam. Bending 13.15 N/mm², shear 1.09 N/mm², deflection 11.1 mm. ## 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 $= 4.00$ m, bearing length $= 230$ mm. **Species:** Sal (Shorea robusta), Group A. **Location:** inside. **Grade:** Select grade (Grade I values multiplied by 1.16, IS 883:1994 cl. 6.3). 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 | 19.60 | | Compression parallel to grain $f_{cp}$ | 10.6 | 12.30 | | Compression perpendicular to grain $f_{cn}$ | 4.6 | 5.34 | | Horizontal shear $H$ | 0.94 | 1.09 | | Modulus of elasticity $E$ | 12670 | 12670 | #### Loads and effective span UDL $w = 15$ kN/m inclusive of self-weight (so self-weight is not added again). Bearing length 230 mm. Effective span (cl. 7.5.2): clear span $+$ bearing length $= 4000 + 230 = 4230$ mm $= 4.230$ m. #### Trial section Width not less than 50 mm or $L/50 = 85$ mm (cl. 7.5.5) and depth not more than 3 times the width without lateral stiffening (cl. 7.5.6). Try **125 mm × 350 mm**. The given UDL already includes the self-weight, so $w = 15.000$ kN/m (the self-weight of the trial section, 0.345 kN/m, is within this). Reaction $R = 31.73$ kN (shear $V = 31.73$ kN); maximum bending moment $M = 33.55$ kN·m. #### Bending $$ \begin{aligned} Z &= \frac{bd^2}{6} = \frac{125\times350^2}{6} = 2.552\times10^6\ \text{mm}^3 \\ f_{ab} &= \frac{M}{Z} = \frac{33.55\times10^6}{2.552\times10^6} = 13.15\ \text{N/mm}^2 \end{aligned} $$ The depth exceeds 300 mm, so the form factor (cl. 7.5.4) $K = 0.81\dfrac{d^2 + 89400}{d^2 + 55000} = 0.967$ applies: permissible $f_b = 19.60\times0.967 = 18.96$ N/mm². $f_{ab} = 13.15 < 18.96$ N/mm². **Safe.** #### Horizontal shear (cl. 7.5.7) $$ H = \frac{3V}{2bd} = \frac{3\times31725}{2\times125\times350} = 1.088\ \text{N/mm}^2 \le 1.09\ \text{N/mm}^2 $$ (no reduction is taken for loads near the supports, which is on the safe side). **Safe.** #### Bearing (cl. 7.5.8) $$ f_{cn}' = \frac{R}{b\times\text{bearing}} = \frac{31725}{125\times230} = 1.10\ \text{N/mm}^2 < f_{cn} = 5.34\ \text{N/mm}^2 $$ 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, $I = bd^3/12 = 447\times10^6$ mm⁴. Limit $= L/240 = 17.62$ mm (members not carrying brittle finishes). $$ \delta = 11.05\ \text{mm} < \frac{L}{240} = 17.62\ \text{mm} $$ (UDL term $5wL^4/384EI$). **Safe.** #### Answer Provide a Sal beam **125 mm × 350 mm** (Select grade, inside), with a bearing of at least 230 mm at each support. $f_{ab} = 13.15$ N/mm², $H = 1.09$ N/mm², $\delta = 11.1$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The 15 kN/m UDL is stated to include self-weight. What does this mean for the design? A. Self-weight is added again to 15 kN/m B. Self-weight is subtracted from 15 kN/m C. w = 15 kN/m is used directly without adding self-weight D. Self-weight is ignored in deflection but added in bending Answer: C. Since the load already contains self-weight, adding it again would double count it. (Key point: Self-weight handling) 2. How are the permissible stresses for select grade obtained? A. Grade I Table 1 values divided by 1.16 B. Grade I values multiplied by 1.16 (e.g., 16.9 to 19.6 N/mm^2) C. Grade I values multiplied by 0.80 D. Table 1 values for outside location Answer: B. IS 883 cl. 6.3 raises Grade I values by a factor of 1.16 for select grade; E is unchanged. (Key point: Select grade stresses) 3. For the 125 x 350 section, why is the permissible bending stress 18.96 rather than 19.60 N/mm^2? A. Because of the moment at midspan B. Because the width is below 50 mm C. Because select grade is reduced for bearing D. Because depth exceeds 300 mm, so K = 0.967 multiplies 19.60 Answer: D. The depth factor K = 0.81(d^2+89400)/(d^2+55000) = 0.967 applies above 300 mm depth, giving 19.60 x 0.967 = 18.96. (Key point: Bending with form factor) 4. Which check is nearly critical for this trial section? A. Bearing: 1.10 against 5.34 N/mm^2 B. Bending: 13.15 against 18.96 N/mm^2 C. Deflection: 11.05 against 17.62 mm D. Horizontal shear: 1.088 against 1.09 N/mm^2 Answer: D. H = 3V/(2bd) = 1.088 N/mm^2 is almost equal to the permissible 1.09 N/mm^2, so shear is the governing check. (Key point: Shear governs) ## Related in Design of Steel and Timber Structure - [Sal timber beam, 2.5 m span, 20 kN/m](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atimber-beam-sal-2-5m-20knm) — Design of Steel and Timber Structure, chapter 12 - [Intermediate Sal timber joist, 2.5 m span](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atimber-joist-sal-600mm) — Design of Steel and Timber Structure, chapter 12 - [Beam 6 m span, 46 kN/m, Fe 410](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-udl-46knm-6m) — Design of Steel and Timber Structure, chapter 7 - [Select Sal column, 200 kN + 30 kNm](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atimber-column-sal-select-bending-3-2m) — Design of Steel and Timber Structure, chapter 11 - [Beam 6 m span with 100 kN secondary beam load](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-20knm-100kn-midspan) — Design of Steel and Timber Structure, chapter 7 - [Sal timber column, 50 kN, 4 m](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atimber-column-sal-50kn-4m) — Design of Steel and Timber Structure, chapter 11 ## Related in other subjects - [Design of a simply supported timber beam](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Adesign-timber-floor-beam) — Design of Timber and Masonry Structures, chapter 3 - [Beam with torsion, 4.5 m clear span](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Abeam-with-torsion-4-5-m) — Design of RCC Structure, chapter 5 - [T-beam decking quantities (6 m clear span)](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3At-beam-decking-all-quantities) — Estimating and Costing, chapter 6 - [Brick and mortar for 230 mm wall (100 kN/m)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awall-eccentric-230mm-100kn-5m) — Concrete Technology and Masonry Structure, chapter 8, asked 7×