EUROCODE 2 WORKED EXAMPLES - Federbeton

EC2 – worked examples 6-22
Assumed this statement, the expression for Fwd is:
Fwd = Fw1 a + Fw2 z
when the two conditions F wd (a = ) = 0 and Fwd (a = 2z) = FEd are imposed, some trivial
2
algebra leads to:
2 FEd
FEd
Fw1
= and Fw2
=− ;
3 z
3
in conclusion, the expression for Fwd as a function of a is the following:
2 FEd FEd
2a /z−1 Fwd = a− = FEd
.
3 z 3 3
Materials: concrete C35/45 fck = 35 MPa, steel B450C fyk = 450 MPa
0.85f 0,85⋅ 35
f 19.83 N / mm
1.5 1.5
ck
2
cd = = = ,
fyk 450
2
fyd = = = 391.3 N / mm
1.15 1.15
Nodes compression resistance (same values of the previous example):
Compressed nodes
σ = 20.12 N/mm
1Rd,max
Table of Content
2
tied-compressed nodes with tension rods in one direction
σ = 17.05 N/mm
2Rd,max
2
tied-compressed nodes with tension rods in different directions
σ = 15N/mm 3Rd,max
2
Actions:
FEd = 500 kN
Load eccentricity with respect to the column outer side: e = 200 mm
The column vertical strut width is evaluated setting the compressive stress equal to σ 1Rd,max:
F 500000
Ed x1= = ≅62
mm
σ1Rd,maxb 20.12⋅ 400
node 1 is located x1/2 = 31 mm from the outer side of the column;
the upper reinforcement is stated to be 40 mm from the cantilever outer side; the distance y1 of the node 1 from the lower border is calculated setting the internal drive arm z to be
0,8⋅d (z = 0,8⋅260 = 208 mm):
y1 = 0.2d = 0.2·260 = 52 mm