4.2 - VSL

Figure 59: Plan showing dimensions
Figure 60: Tendon profile in longitudinal direction (internal span) Figure 61: Tendon profile in transverse direction (internal span)
on 6.60 m width: n p = 6.60 . 17.8 = 15.1
7.80
→ 16 monostrands 0 15 mm on 6.60 m width
on 2.40 m width: n p= 2.40 . 17.8 = 5.5
7.80
→ 6 monostrands ∅ 15 mm on 2.40 m width
b) Transverse direction:
g+q
=1.83;κ=
0.24 . 1000
= 0.158
g 7.802 . 25
hence from Fig. 58
→ u = 1.41;u=1.41 . 6 = 8.46kN/m2 g
hp=0.135 . 7.802 = 0.167 m (Fig. 61)
3.51 2
P= 8.46 . 7.80 2
= 385 kN/m
8 . 0.167
on 8.40 m width: P=8.40 . 385=3234 kN
Number of strands: np= 3234 =17.9
181
→18 monostrands 0 15 mm on 8.40 m width
on 7.20 m width: np= 7.20 . 17.9 =15.3
8,40
→16 monostrands 0 15 mm on 7.20 m width
- Determination of ordinary reinforcement:
a) Top reinforcement:
In the region of the punching cone:
ps =0.3% (Fig. 35)
Average of effective depth of reinforcement in both directions:
dsc = 240 - 15 - 15 = 210 mm (approx. value)
Width bcd (Fig. 30):
bcd = bc +3dsc = 450 + 3 . 210 = 1080 mm
→ ASS = 0.003 . 210 . 1080 = 680 mm2 chosen: 7 ∅12 mm (Ass = 791 mm2 )
In column strip:
ps= 0.15% (Fig. 35)
longitudinally:
bg = 0.4 . 7800 -1080 = 2040 mm
Asg =0.0015 . 210 . 2040 = 643 mm2 chosen: 6 ∅12 mm (Asg=678 mm 2)
21