Design of Flexure Member

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: 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 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). Total UDL $w = 0.000 + 0.902 = 0.902$ kN/m.

Reaction $R = 76.91$ kN (shear $V = 76.91$ kN); maximum bending moment $M = 107.77$ 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).

$$\delta = 5.21\ \text{mm} < \frac{L}{240} = 17.62\ \text{mm}$$

(UDL term $5wL^4/384EI$ plus the point-load term). 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.

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 $= 75$ 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). 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 $R = 2.78$ kN (shear $V = 2.78$ kN); maximum bending moment $M = 1.79$ 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, $I = bd^3/12 = 10\times10^6$ mm⁴. Limit $= L/240 = 10.73$ mm (members not carrying brittle finishes).

$$\delta = 9.53\ \text{mm} < \frac{L}{240} = 10.73\ \text{mm}$$

(UDL term $5wL^4/384EI$). Safe.

Answer

Provide a Sal beam 63 mm × 125 mm (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.

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.

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.

Simple timber beam

A simple beam is a single rectangular piece of sawn timber, simply supported on two supports, carrying the load by bending.

        load
   v  v  v  v  v  v
  +------------------+
  |  solid timber    |   b x d
  +------------------+
  ^                  ^
 support          support

Used for short spans up to 4 to 5 m. Section $Z = bd^2/6$, and the design checks bending, shear, bearing and deflection (IS 883:1994 cl. 7.5).

Flitched beam

A flitched beam is a built-up beam in which one or more steel (or timber) plates are sandwiched, "flitched", between two timber members and bolted through to act as one. The steel plate adds strength and stiffness without increasing the depth.

   timber | steel | timber
   +-----+-+-----+-+-----+       bolts through all
   |     | |     | |     |       parts at intervals
   |     | |     | |     |
   +-----+-+-----+-+-----+

• Timber and steel have the same strain at the same level, so with modular ratio $m = E_s/E_t$ the steel plate is replaced by a timber of width $m\,t_s$ (equivalent section). The load shared is in proportion to the stiffness $EI$.
• Bending stresses: $f_t = \dfrac{M y}{I_{eq}}$ for timber and $f_s = m\,\dfrac{M y}{I_{eq}}$ for steel, with $I_{eq} = I_t + m\,I_s$; each must not exceed its permissible stress.
• The bolts (or screws) at pitch keep the parts together and take the horizontal shear.
• Used where a deeper timber beam is not available, for long spans or heavy loads.