Sal timber beam, 2.5 m span, 20 kN/m

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):

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.