Deodar beam with two point loads, 4 m

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):

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.)