EUROCODE 2 WORKED EXAMPLES - Federbeton

EC2 – worked examples 7-3
Working with constant normal force (N = N 0) the ultimate limit state for the concrete tension
leads to
and then
Solving with respect to yn 2
y + 60y − 84795 = 0
n n
Table of Content
N( −y
)
=−0.6f
0 n
*
Syn
2
yn =− 30.25 + 30.25 + 84795 = 262.51mm
' (50 − 262.51)
σ s = 0.6⋅30⋅ ⋅ 15 =−218.57MPa
262.51
(550 − 262.51)
σ s = 0.6⋅30⋅ ⋅ 15 = 295.69MPa
262.51
and then
−yn0.6⋅30 =
400 2
− y
800 10
n + 15⋅1884[ 600 −2⋅y ⋅
n]
2
⎡
⎢
⎣
400 ⋅262.51
⎛262.51 ⎜
2 ⎝ 3
⎞
⎟
⎠
⎤
⎥
⎦
6
442.56⋅10 e=- =−553.2mm
3
800⋅10 Keeping constant the bending moment (M = M0), the limit state condition for the concrete
stress is
N( −y
n )
=−0.6f
*
ck
S
and then
M0 M 0(y n)
e = =
*
N 0.6f ⋅ S
−6
M = −0.6⋅30 ⋅ ⋅ − 300 + 1884 ⋅ (295.69 + 218.57) ⋅250 ⋅ 10 = 442.56kNm
and
yn
I M(y ) h
− + yn
=
S 0.6 f S 2
*
yn 0 n
*
yn ⋅ ck ⋅
*
yn
As
ck
ck
3
ck yn
6
M0 400 ⋅10
= = 22.22 ⋅10mm
0.6⋅f0.6⋅30 the previous numeric form becomes
400 3 2 2
6
yn + 15⋅1884 ⎡( 550 − yn) + ( 50 −yn) ⎤−22.22⋅10⋅yn
3 ⎣ ⎦
+ yn= 300
400 2
− yn + 15⋅1884[ 600 −2⋅yn] 2
and iteratively solving
6 3