--- title: "Intermediate Sal timber joist, 2.5 m span" subject: "Design of Steel and Timber Structure" chapter: 12 key: "semester_6/design_of_steel_and_timber_structure:timber-joist-sal-600mm" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-joist-sal-600mm site: BCT Engineering --- # Intermediate Sal timber joist, 2.5 m span Design of Steel and Timber Structure · Chapter 12: Design of Flexure Member - Asked in: 2079 Chaitra, 8 marks - Video: [How to design timber joists | Check for Bending, shear & deflection.](https://www.youtube.com/watch?v=WzBFqlIlOVQ) — Structural Engineer Calcs - 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-joist-sal-600mm - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-12 ## Question Design an intermediate Sal wood timber joist used in the roof of a hall from the following data: Spacing of joist = 600 mm Clear span of joist = 2.5 m Dead load of roof covering = 2 kN/m² Imposed load on roof = 1.5 kN/m² The timber is of standard grade and used in inside location. ## Short answer Use working stress design to IS 883. Provide 63 mm x 125 mm Sal joists at 600 mm centres. Bearing at supports 75 mm. Bending 10.92 N/mm², shear 0.53 N/mm², deflection 9.5 mm. ## Answer **Approach.** Working stress design of a timber joist (beam) to IS 883:1994 cl. 7.5. Checks: bending strength, horizontal shear, bearing stress at the supports and deflection. Self-weight is included (cl. 7.5.9.4). #### Data and permissible stresses Clear span $= 2.50$ m, bearing length $= 75$ 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 Joist spacing 0.6 m. Dead load of roof covering $2.0$ kN/m²; imposed load $1.5$ kN/m². Load per metre of joist $= (2.0 + 1.5)\times0.6 = 2.10$ kN/m (plus self-weight). The joists are inside, so inside stresses are used; a bearing of 75 mm on the wall is assumed (cl. 7.5.8.1). Effective span (cl. 7.5.2): clear span $+$ bearing length $= 2500 + 75 = 2575$ mm $= 2.575$ m. #### Trial section Width not less than 50 mm or $L/50 = 52$ mm (cl. 7.5.5), and depth not more than 3 times the width without lateral stiffening (cl. 7.5.6). Try **63 mm × 125 mm**. Self-weight $= 0.063\times0.125\times805\times9.81/1000 = 0.062$ kN/m (density 805 kg/m³). Total UDL $w = 2.100 + 0.062 = 2.162$ kN/m. Reaction $R = wL/2 = 2.78$ kN (shear $V = 2.78$ kN); maximum bending moment $M = wL^2/8 = 1.79$ kN·m. #### Bending $$ \begin{aligned} Z &= \frac{bd^2}{6} = \frac{63\times125^2}{6} = 0.164\times10^6\ \text{mm}^3 \\ f_{ab} &= \frac{M}{Z} = \frac{1.79\times10^6}{0.164\times10^6} = 10.92\ \text{N/mm}^2 \end{aligned} $$ Depth $\le 300$ mm, so no form factor is applied: $f_b = 16.90$ N/mm². $f_{ab} = 10.92 < 16.90$ N/mm². **Safe.** #### Horizontal shear (cl. 7.5.7) $$ H = \frac{3V}{2bd} = \frac{3\times2784}{2\times63\times125} = 0.530\ \text{N/mm}^2 \le 0.94\ \text{N/mm}^2 $$ (No reduction is taken for loads near the supports, which is on the safe side.) **Safe.** #### Bearing (cl. 7.5.8) $$ f_{cn}' = \frac{R}{b\times\text{bearing}} = \frac{2784}{63\times75} = 0.59\ \text{N/mm}^2 < f_{cn} = 4.60\ \text{N/mm}^2 $$ 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, $I = bd^3/12 = 10.25\times10^6$ mm⁴. Limit $= L/240 = 10.73$ mm (members not carrying brittle finishes). $$ \delta = \frac{5wL^4}{384EI} = 9.53\ \text{mm} < \frac{L}{240} = 10.73\ \text{mm} $$ **Safe.** #### Answer Provide Sal joists **63 mm × 125 mm** at 600 mm centres (Grade I (standard), inside), with a bearing of at least 75 mm at each support. $f_{ab} = 10.92$ N/mm², $H = 0.53$ N/mm², $\delta = 9.5$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the UDL on one joist found from the roof loads? A. (2.0 + 1.5) x span = 8.75 kN/m B. (2.0 + 1.5) x 0.6 = 2.10 kN/m plus self-weight C. 2.0 x 0.6 only, since imposed load is ignored D. (2.0 + 1.5) / 0.6 = 5.83 kN/m Answer: B. Area loads in kN/m^2 are multiplied by the joist spacing to get kN/m, then the joist self-weight is added. (Key point: Load per metre of joist) 2. Why is no form factor applied to the permissible bending stress of the 63 x 125 joist? A. Because the depth is not more than 300 mm B. Because Sal is a Group A species C. Because the joist is inside D. Because the load is a UDL Answer: A. The depth factor of cl. 7.5.4 applies only to depths over 300 mm, so f_b stays at 16.90 N/mm^2. (Key point: No form factor) 3. With M = 1.79 kN.m and Z = bd^2/6 = 0.164 x 10^6 mm^3, what is the result of the bending check? A. f_ab = 10.92 N/mm^2 < 16.90, safe B. f_ab = 10.92 N/mm^2 > 0.94, unsafe C. f_ab = 0.53 N/mm^2 < 16.90, safe D. f_ab = 21.8 N/mm^2 > 16.90, unsafe Answer: A. f_ab = M/Z = 1.79e6/0.164e6 = 10.92 N/mm^2, which is below the permissible 16.90 N/mm^2. (Key point: Bending check) 4. What deflection limit is used and does the joist satisfy it? A. L/360 = 7.2 mm, fails with 9.53 mm B. L/240 = 10.73 mm, passes with 9.53 mm C. L/240 = 10.73 mm, fails with 9.53 mm D. L/180 = 14.3 mm, passes with 9.53 mm Answer: B. For members without brittle finishes the limit is span/240 = 2575/240 = 10.73 mm, and the computed 5wL^4/384EI is 9.53 mm. 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