Sal select grade beam, 15 kN/m, 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 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):

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.