4.2 - VSL

Figure 62: Influence zone column 1 Figure 63: Tendon profile in critical shear periphery
transversely:
bg = 0.4 ⋅ 8400 -1080 = 2280 mm
Asg=0.0015 ⋅ 210 ⋅ 2280=718 mm2 chosen: 4+4 ∅ 12 mm (Asg = 904 mm2) b) Bottom reinforcement:
• Internal spans: none
• Edge spans: p s ≥ 0.15 - 0.50 ⋅ p p (Formula 4.2.)
longitudinally:
p np ⋅ Ap 18 ⋅ 146
p= =
dp ⋅ b 200 ⋅ 7800
= 0.17%
→ p s ≥ 0.15-0.50 ⋅ 0.17 = 0.065%
→ A s ≥ 0.065 ⋅ 220 ⋅ 10 = 143 mm 2 /m
chosen:∅ 6 mm, spacing 175 mm
transversely:
pp ≥ 18 ⋅ 146
= = 0.16%
200 ⋅ 8400
→ps ≥ 0.07%
→As ≥ 0.07⋅ 220 ⋅ 10 = 154 mm2 /m
chosen:∅ 6 mm, spacing 175 mm
Check for punching:
Determining column 1 (Fig. 62):
g+q = 11 kN/m2
V g+q = 11 . 7.60 . 8.60 = 719kN
Prestress:
50% within the critical shear periphery, i.e. 9 monostrands in each
direction
Point of inflection:
According to Fig. 30 the point of inflection ideally lies at a distance d s/ 2
from the column edge. In Figs. 60 and 61 it is assumed that the
dimensions of the column are not yet known and the point of inflection
is adopted at a distance 0.051 from the column axis (value from
experience). In Fig. 62 the dimensions of the column have been
established. Thus the real position of the point of inflection is known.
The values given in Figs. 60 and 61 change accordingly (Fig. 63).
longitudinally :
tga 2 ⋅ 13
= = 0.078 = sina (Fig. 63a)
332
Vp =2 . 0.078 . 181 . 9 . 09=229kN
(Factor 0.9=10% long-term losses)
transversely :
tga 2 ⋅ 12
= = 0.072 = sina (Fig. 63b)
332
Vp =2 . 0.072 . 181 . 9 . 0.9 = 211 kN
ΣVp =440 kN
22
Figure 64: Tendon profile in longitudinal direction (edge span)