EUROCODE 2 WORKED EXAMPLES - Federbeton

EC2 – worked examples 6-38
EXAMPLE 6.14. 3500 kN concentrated load [EC2 clause 6.5]
3500 kN load on a 800x500 rectangular column by a 300x250 mm cushion
Materials:
concrete C30/37 f ck = 30 MPa,
steel B450C f yk = 450 MPa E s = 200000 MPa
0.85f 0.85⋅ 30
f 17 N/mm
1.5 1.5
ck
2
cd = = = ,
f 450
f 391.3 N / mm
1.15 1.15
loading area
Ac0 = 300·250 = 75000 mm 2
yk 2
yd = = = ,
dimensions of the load distribution area
d2 ≤ 3 d1 = 3·300 = 900 mm
b2 ≤ 3 b1 = 3·250 = 750 mm
maximal load distribution area
Ac1 = 900·750 = 675000 mm 2
load distribution height
Table of Content
( b − b ) = 750 − 250 = 500 mm
h ≥ 2 1
( d − d ) = 900 − 300 = 600 mm
h ≥ 2 1
Ultimate compressive stress
F = A f A / A = 75000⋅17 ⋅ 675000 / 75000 = 3825 kN ≤
Rdu c0 cd c1 c0
3.0 fcdA c0 3.0 17 75000
6
3.825 10 N
≤ = ⋅ ⋅ = ⋅
⇒ h = 600 mm
It is worth to observe that the FRdu upper limit corresponds to the the maximal value Ac1 = 3
Ac0 for the load distribution area, just as in this example; the 3500kN load results to be lower
than FRdu .
Reinforcement design
Point [6.7(4)-EC2] recommends the use of a suitable reinforcement capable to sustain the
transversal shrinkage stresses and point [6.7(1)P-EC2] sends the reader to paragraph [(6.5)-
EC2] to analyse this topic.
In this case there is a partial discontinuity, because the strut width (500 mm) is lower than the
distribution height (600 mm), then:
a = 250 mm