Intermediate Sal timber joist, 2.5 m span

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):

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.