Design of the mezzanine floor system for seasoned hard wood

Structural Design
Design of timber structure
(ii) Design of the floor system for the mezzanine floor:-
a) Design for joists:-
Given:
Imposed Loading =4kN/m2
Headroom space mezzanine floor =3.2 m
Stiffness design based on (criterion - 2)
Use = E0.5
Mezzanine flooring = 19 mm seasoned hard wood
Maximum spanning = 600 mm
Design of deflection for joints:-
Check deflection:-
W =ψs Q
ψs =others ( Table 4.1 )=1.0
Q=Imposed Loading=4 kN /m2
W = ( spacing btwn joint ) × ψs Q

¿ 1.34 × (1 × 4 )¿ 5.36
Now calculating deflecting:-
δ=J2
5 w l4
384 EI
Let us assume the joist F17 seasoned messmate in size “190 x 45”
∴ E0.5=14000 , J 2=2 seasoned wood
δ= 2 ×5 ×5.36 × 23304
384 ×14000 × ( 45 ×1903
12 )δ = 1.57 ×1015
1.38 ×1014 =11.7> 9 mm∴ change∈¿ ¿
Now let us assume 240 x 45
δ= 2 ×5 ×5.36 × 23304
384 ×14000 × ( 45 ×2403
12 ) δ = 1.57 ×1015
2.78 ×1014 ¿ 5.33Spanlimit = 2330
360 =6.47
∴ As per area criteria2 , δ ≤ 9 mm δ=5.33 ≤ 9 mm
Condition proved. Size: (240 x 45) for joist
Design of strength limit state for joist:-
Design equation for bending strength
M d ≥ M¿
∴ M d =∅ K1 K 4 K 6 K9 K12 f b
' z
∅=capacity factor=Table 2.1 ,Category −1 , stress grades F17
∅=0.95
Duration of Load factor for strength:-
For,
Load case−1:−K1=0.57Load case−2:−K1=0.57Load case−3 :−K 1=0.80

Partial Seasoning factor K4:-
Indoor seasoning timber
∴ K 4=1
Ambient Temperature Factor K6 :−¿
Melbourne, Tropical Region
∴ K 6=1
Strength sharing factors for beams K g :−¿
K9 =g31+(g¿¿ 31−g31 ) [1−2 s
L ]¿
ncom=1∴ g31=1 g32=1.33 ¿ 1+ ( 1.33−1 ) [ 1−2 ( 1.34 )
2.33 ]K9 =0.95
Size effect K11:-
¿ 300 mm K11=1
Stability Factor K12
S1=discrete lateralrestraints ¿ 1.25 d
b ( Lay
d )
0.5
¿ 1.25 ( 240
45 )( 1340
240 )0.5
¿ 15.7
Pb=0.98 for seasoned F17 ∴10< PB <20 ( ¿ between ) K12=1.5−0.05 Pb S1 ¿ 1.5−0.05× 15.7
¿ 0.71
Section of modulus z:-
zx= b d2
6 = 45 ×2402
6 =432 000
Bending strength:-
f b
' =f 17=45
Dead Load:-

Load due ¿ flooring=950× 19× 10−3Thick 19 mm=18.05 kg
m2 ¿ 0.177 kN
m2
Load due ¿ joist=750 × 45
100 × 240
1000 =0.079 kN /m
Floor load by joist=0.177 ×0.6=0.1064Total Dead Load=0.1064 +0.079=0.185
Live load:-
Live load due ¿ floor =4 kN /m2
Load taken by joist=4 × 0.6=2.4 kN /m
Total live load =2.4 kN /m
Load case 1:-
md =0.95 ×0.57 ×1 ×1 ×0.95 × 1× 0.71× 45 × ( 432000× 10−6 )
¿ 7.10
P¿=1.35 G ¿ 1.35 ×0.89 ¿ 1.21
M = P¿ L
4 =0.70 ∴ M d 7.10> M 0.70
Load case 2:-
M d =7.10P=1.2 G+1.5 ψs Q¿ 1.2 ×0.89+ ( 1.5 ×1 ×2.4 )=4.668M = 4.668 ×2.33
4 =2.71
∴ M d 7.10> M 2.71
Load case 3:-
M d =0.95 ×0.80 ×1 ×1 ×0.93 ×1 × 45× 0.71 × ( 432000× 10−6 )
¿ 9.96
P¿= ( 1.2× 0.89 ) + ( 1.5 ×2.4 )
¿ 4.668
M =2.7
∴ M d 9.96> M 2.7