--- title: "Deodar beam with two point loads, 4 m" subject: "Design of Steel and Timber Structure" chapter: 12 key: "semester_6/design_of_steel_and_timber_structure:timber-beam-deodar-two-point-loads" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-beam-deodar-two-point-loads site: BCT Engineering --- # Deodar beam with two point loads, 4 m Design of Steel and Timber Structure · Chapter 12: Design of Flexure Member - Asked in: 2080 Chaitra, 6 marks - Video: [DESIGN OF TIMBER BEAM | TIMBER NUMERICAL | MOST ASKED PROBLEM | TU\PU\KU\PoU](https://www.youtube.com/watch?v=z1nVMWAzamY) — 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-beam-deodar-two-point-loads - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-12 ## Question Design a deodar wood beam to carry two-point loads 75 kN and 75 kN placed at 1/3rd and 2/3rd of the span. The beam has clear span of 4 meter with support width of 230 mm. Assume all necessary data. ## Short answer Use working stress design to IS 883. Bearing at supports 230 mm. Provide 300 mm x 550 mm deodar beam. Bending stress 7.13 N/mm², shear 0.70 N/mm², deflection 5.2 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 the 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:** Deodar (Cedrus deodara), Group C. **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 $f_b$ | 10.2 | 10.20 | | Compression parallel to grain $f_{cp}$ | 7.8 | 7.80 | | Compression perpendicular to grain $f_{cn}$ | 2.7 | 2.70 | | Horizontal shear $H$ | 0.7 | 0.70 | | Modulus of elasticity $E$ | 9480 | 9480 | #### Loads and effective span Two concentrated loads of 75 kN each at the one-third points ($a = L/3$ from each support), taken as service loads. Self-weight is added. 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 **300 mm × 550 mm**. Self-weight $= 0.3\times0.55\times557\times9.81/1000 = 0.902$ kN/m (density 557 kg/m³). This is the only UDL, so $w = 0.902$ kN/m. Reaction $R = 75 + wL/2 = 76.91$ kN (shear $V = 76.91$ kN); maximum bending moment $M = 75\times L/3 + wL^2/8 = 107.77$ kN·m. #### Bending $$ \begin{aligned} Z &= \frac{bd^2}{6} = \frac{300\times550^2}{6} = 15.125\times10^6\ \text{mm}^3 \\ f_{ab} &= \frac{M}{Z} = \frac{107.77\times10^6}{15.125\times10^6} = 7.13\ \text{N/mm}^2 \end{aligned} $$ The depth exceeds 300 mm, so the form factor (cl. 7.5.4) $K = 0.81\dfrac{d^2 + 89400}{d^2 + 55000} = 0.888$ applies: permissible $f_b = 10.20\times0.888 = 9.06$ N/mm². $f_{ab} = 7.13 < 9.06$ N/mm². **Safe.** #### Horizontal shear (cl. 7.5.7) $$ H = \frac{3V}{2bd} = \frac{3\times76907}{2\times300\times550} = 0.699\ \text{N/mm}^2 \le 0.70\ \text{N/mm}^2 $$ (No reduction is taken for loads near the supports, which is on the safe side.) **Safe.** #### Bearing (cl. 7.5.8) $$ f_{cn}' = \frac{R}{b\times\text{bearing}} = \frac{76907}{300\times230} = 1.11\ \text{N/mm}^2 < f_{cn} = 2.70\ \text{N/mm}^2 $$ The permissible stress perpendicular to the grain applies for any bearing length at the ends of a member. **Safe.** #### Deflection (cl. 7.5.9) $E = 9480$ N/mm², $I = bd^3/12 = 4159\times10^6$ mm⁴. Limit $= L/240 = 17.62$ mm (members not carrying brittle finishes). $$ \delta = \frac{23PL^3}{648EI} + \frac{5wL^4}{384EI} = 5.11 + 0.10 = 5.21\ \text{mm} < \frac{L}{240} = 17.62\ \text{mm} $$ **Safe.** #### Answer Provide a Deodar beam **300 mm × 550 mm** (Grade I (standard), inside), with a bearing of at least 230 mm at each support. $f_{ab} = 7.13$ N/mm², $H = 0.70$ N/mm², $\delta = 5.2$ mm. (Shear governs the size.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the effective span of the beam obtained here? A. Clear span only, 4000 mm B. Clear span plus bearing length, 4000 + 230 = 4230 mm C. Clear span plus twice the bearing length D. Clear span minus bearing length Answer: B. IS 883 cl. 7.5.2 takes effective span as clear span plus bearing length, here 4.23 m. (Key point: Effective span) 2. Why is the permissible bending stress reduced to 9.06 N/mm^2 for the 300 x 550 section? A. Because deodar is a weak species B. Because the beam is outside C. Because the depth exceeds 300 mm, so the form factor K = 0.888 applies D. Because the beam has two point loads Answer: C. For depths over 300 mm, cl. 7.5.4 gives a form factor K below 1; 10.2 x 0.888 = 9.06 N/mm^2. (Key point: Form factor for deep beams) 3. Which formula checks horizontal shear for the rectangular beam, and does it pass? A. H = 3V/(2bd) = 0.699 <= 0.70, passes (just) B. H = V/(bd) = 0.47, passes easily C. H = 3V/(2bd) = 0.699, fails because it is near 0.70 D. H = M/Z = 0.699, passes Answer: A. Maximum shear stress in a rectangle is 1.5 V/(bd) = 0.699 N/mm^2, which is within the permissible 0.70 N/mm^2. (Key point: Horizontal shear check) 4. Bearing at supports is checked as R/(b x bearing length) = 1.11 N/mm^2. Against which permissible stress is it compared? A. Compression parallel to grain, 7.8 N/mm^2 B. Horizontal shear, 0.70 N/mm^2 C. Bending stress, 10.2 N/mm^2 D. Compression perpendicular to grain, 2.7 N/mm^2 Answer: D. The support pushes across the grain, so the permissible compression perpendicular to grain (2.7 N/mm^2) governs. 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