Consider the design of a suspended timber floor system in a domestic building in which the joists at 500 mm centres are simply supported by timber beams on load-bearing brickwork, as shown in Figure 7.16 (a). The support beams are notched at the location of the wall, as shown in Figure 7.16(b).

♦ Determine a suitable section size for the tongue and groove floor boards.

♦ Determine a suitable section size for the joists.

♦ Check the suitability of the main support beams.

Design data:

Centre of timber joists

500 mm

Distance between the centre-lines of the brickwork wall

4.5 m

Strength class of timber for joists and tongue and groove boarding and beams

Allowing for the continuity of the boards over the supports reduces the bending moment

M x , maximum

≈

2

wL

=

3.08

×

0.5 2

= 0.077 kNm

10

10

Minimum section modulus required:

Z min

≥

maximum bending moment

0.077

×

10 6

=

permissible stress

8.75

≈

8.8 × 10 3mm 3/metre width

Z

=

bh2

∴

h

≥

6Z

∴

h

≥

6

×

8.8

×

10 3

6

b

1000

h

≥

7.3 mm;

assume an additional 3 mm for wear

h

≥

7.3 + 3 =

10.3 mm

Try 16 mm thick

Clause 2.10.7

Deflection:

Clause 2.9

Since load-sharing exists use E meanto calculate deflection. Since the boards are continuous, assume the end span deflection (i.e. a propped cantilever) is approximately equal to 50% of a simply supported span:

δ max

≈

(0.5 × δ simply supported span )

=

1

2









5 W

total

384 EI

3

L









δ max

=

0.5 ×

5

×

3.08

×

0.5

×

10

3

×

500

3

≈

0.38 mm

δ permissible

≤

384

×

9700

0.003 × 500=

×









1000

×

12

16

1.5 mm

3









Adopt a minimum thickness of 16 mm for the tongue and groove boarding

δ max

<<

δ permissible

462

Design of Structural Elements

Contract : Solid Beams Job Ref. No. : Example 7.1

Calcs. by : W.McK. Checked by :

Part of Structure :

Suspended floor system

Calc. Sheet No. :

3

of

8

Date :

References

Calculations

Output

Joists at 500 centres:

Dead load due to self-weight of joist: assume

=

0.1 kN/m

Dead load due to t & g boarding = (0.08 × 0.5) =

0.04 kN/m

Imposed loading

=

1.5 kN/m

Total load

=

= (3.0 × 0.5) (0.1 + 0.04 + 1.5)

=

1.64 kN/m

Clause 2.10

Bending:

Permissible stress

σ m,adm,

=

σ m,g, × K 2 × K 3 × K 6 × K 7 × K 8

Clause 2.6.2 Clause 2.8 Table 17 (note a)

K 2− wet exposure K 3− load duration (long-term)

does not apply in this case for uniformly distributed imposed floor loads

K 3= 1.0

Clause 2.10.5

K 6− shape factor; K 7− depth of section;

does not apply in this case assume h ≤ 300 mm

Clause 2.10.6

K 7≈ 1.0

This assumption should be checked at a later stage and modified if necessary.