Composite columns - Eurocodes

G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Eurocodes
Background and Applications
Dissemination of information for training
18-20 February 2008, Brussels
Eurocode 4
Composite Columns
Univ. - Prof. Dr.-Ing. Gerhard Hanswille
Institute for Steel and Composite Structures
University of Wuppertal
Germany
1

Contents
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Part 1: Introduction
Part 2: General method of design
Part 3: Plastic resistance of cross-sections and interaction curve
Part 4: Simplified design method
Part 5: Special aspects of columns with inner core profiles
Part 6: Load introduction and longitudinal shear
2

G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Part 1: Introduction
3

Composite columns
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
concrete encased
sections
concrete
filled hollow
sections
partially concrete
encased sections
4

Concrete encased sections
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
advantages:
• high bearing resistance
• high fire resistance
• economical solution with regard to
material costs
disadvantages:
• high costs for formwork
• difficult solutions for connections
with beams
• difficulties in case of later
strengthening of the column
• in special case edge protection is
necessary
5

Partially concrete encased sections
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
advantages:
• high bearing resistance, especially in case
of welded steel sections
• no formwork
• simple solution for joints and load
introduction
• easy solution for later strengthening and
additional later joints
• no edge protection
disadvantages:
• lower fire resistance in comparison with
concrete encased sections.
6

Concrete filled hollow sections
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
advantages:
• high resistance and slender columns
• advantages in case of biaxial bending
• no edge protection
disadvantages :
• high material costs for profiles
• difficult casting
• additional reinforcement is needed for fire
resistance
7

Concrete filled hollow sections with
additional inner profiles
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
advantages:
• extreme high bearing resistance in
combination with slender columns
• constant cross section for all stories is
possible in high rise buildings
• high fire resistance and no additional
reinforcement
• no edge protection
disadvantages:
• high material costs
• difficult casting
8

G. Hanswille
Composite columns with hollow sections Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
and additional inner core-profiles
University of Wuppertal-Germany
Commerzbank
Frankfurt
9

Design of composite columns
according to EN 1994-1-1
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Verifications for composite columns
Resistance of the member for
structural stability
General method
Simplified method
Resistance to local Buckling
Introduction of loads
Longitudinal shear outside the areas of load
introduction
10

Methods of verification in accordance
with EN 1994-1-1
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Methods of verification
general method:
• any type of cross-section and any
combination of materials
simplified method:
• double-symmetric cross-section
• uniform cross-section over the member length
• limited steel contribution factor δ
• related Slenderness smaller than 2,0
• limited reinforcement ratio
• limitation of b/t-values
11

Resistance to lokal buckling
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
concrete encased cross-sections
b
b c
c y
c y
c
⎧40
mm
z
Verification is not necessary where c z ≥ ⎨
⎩ b / 6
h h c
y
c
concrete filled hollow section
z
z
t
⎛ d⎞
2
t
max ⎜ ⎟ = 90 ε
d ⎛ d⎞
max ⎜ ⎟ = 52 ε
⎝ t ⎠
⎝ t ⎠
partially encased I sections
fyk,o
ε =
⎛ d⎞
max ⎜ ⎟ = 44 ε
fyk
⎝ t ⎠
f yk,o = 235 N/mm 2
b
d
t
12

G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Part 2:
General design method
13

General method
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Design for structural stability shall take account of
geometrical
imperfection
L
w o =
1000
residual
stresses due
to rolling or
welding
w o
σ E
+ -
-
L
+
• second-order effects including residual stresses,
• geometrical imperfections,
• local instability,
• cracking of concrete,
• creep and shrinkage of concrete
• yielding of structural steel and of reinforcement.
The design shall ensure that instability does not occur for the
most unfavourable combination of actions at the ultimate limit
state and that the resistance of individual cross-sections
subjected to bending, longitudinal force and shear is not
exceeded. Second-order effects shall be considered in any
direction in which failure might occur, if they affect the structural
stability significantly. Internal forces shall be determined by
elasto-plastic analysis. Plane sections may be assumed to
remain plane. Full composite action up to failure may be
assumed between the steel and concrete components of the
member. The tensile strength of concrete shall be neglected.
The influence of tension stiffening of concrete between cracks
on the flexural stiffness may be taken into account.
14

General method of design
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
F
e
f cm
σ c
σ s
w
0,4 f c
E cm
f ε c1
ε
ct
c1u
concrete
cracked concrete
plastic zones in structural steel
f y
f tm
-
f sm
reinforcement
stresses in structural steel section
+
-
E s
f s
ε c
ε s
ε a
f c
σ a
E v
f u
-
- - - -
stresses in concrete and reinforcement
+
-
f y
-
E a
structural
steel
ε v
15

Typical load-deformation behaviour of
composite columns in tests
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
F [kN]
1600
1200
800
A
e=100mm
e=160mm
B
w
e
F
F
A
B
concrete encased section and
bending about the strong axis:
Failure due to exceeding the
ultimate strain in concrete, buckling
of longitudinal reinforcement and
spalling of concrete.
concrete encased section and
bending about the weak axis :
Failure due to exceeding the
ultimate strain in concrete.
400
C e=130mm
C
0 20 40 60 80 100
Deflection w [mm]
concrete filled hollow section:
cross-section with high ductility
and rotation capacity. Fracture
of the steel profile in the tension
zone at high deformations and
local buckling in the
compression zone of the
structural steel section.
16

General Method – Safety concept based
on DIN 18800-5 (2004) and German
national Annex for EN 1994-1-1
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
N Ed
N
σ c
f cm
f ct
ε c1
ε c1u
0,4 f c
E cm
concrete
ε c
λ u
E d
σ s
f tm
f sm
reinforcement
E s
λ u
: amplification factor for ultimate
system capacity
Verification λ u
≥ γ R
E
f σ a
u E
f v
y
E a
ε
ε a
v
R pl,m γ R = R pl,d
+
E d
R pl,d
R pl,m
E d
w o =L/1000 w o
ε s
geometrical
Imperfection
e
structural
steel
Residual
stresses
-
-
+
M Ed
M
w u
w
+ -
17

Composite columns for the
central station in Berlin
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
χ
buckling curve a
t=25mm
S235
t=50mm
S355
800
550
700
1200
1,0
buckling curve b
buckling curve c
0,5
buckling curve d
-
Residual stresses
0,5 1,0 1,5 2,0
λ
18

G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Part IV-3:
Plastic resistance of cross-sections and
interaction curve
19

Resistance of cross-sections
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
f yd ν f cd
f sd
N N
pla,Rd
plc,Rd N pls,Rd
y
z
Design value of the plastic
resistance to compressive forces:
N
N
pl,Rd
pl,Rd =
= N
A
pla,Rd
a
f
yd
+ N
+
ν A
plc,Rd
c
f
cd
+ N
+ A
pls,Rd
s
f
sd
Characteristic value of the plastic
resistance to compressive forces:
N
pl,Rk
=
A
a
f
yk
+
A
s
f
sk
+ νA
c
f
ck
Design strength:
f
yd
=
f
yk
γ
a
f
sd
f
=
γ
sk
s
f
cd
=
f
γ
ck
c
Increase of concrete
strength due to better
curing conditions in case
of concrete filled hollow
sections:
ν =1,0
ν = 0,85
20

Confinement effects in case of concrete
filled tubes
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
t
σ aϕ
d
σ c,r
σ aϕ
σ
σ
η a
f yd
structural steel
2
a,
Rd
a,Rd
+ σ
=
2
a,
ϕ
η
a
f
− σ
yd
σ c,r
d-2t
a,Rd
σ
2
a,
ϕ
= f
2
yd
f
2.0
ck,c
f
1.5
1.25
1.0
0.5
0
0
ck
α 1
=1,00
α 2
= 5,0
concrete
α 1
=1,125
α 2
= 2,5
σ c,r
f ck, c =α1
fck
+ α2
σc,
r
f ck,c
0.05 0.10 0.15 0.20 0.25 0.30 0.35
σ
f
c,r
ck
For concrete stresses σ c >o,8 f ck the Poisson‘s ratio of concrete is higher than the
Poisson‘s ratio of structural steel. The confinement of the circular tube causes radial
compressive stresses σ c,r . This leads to an increased strength and higher ultimate
strains of the concrete. In addition the radial stresses cause friction in the interface
between the steel tube and the concrete and therefore to an increase of the
longitudinal shear resistance.
21

Confinement effect acc. to Eurocode 4-1-1
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Design value of the plastic resistance to compressive forces
taking into account the confinement effect:
⎛ t fyk
⎞
Npl,Rd
= ηa
fyd
Aa
+ Ac
fcd
⎜1
+ ηc
⎟
⎝ d fck
⎠
y
t
d
Basic values η for stocky columns
centrically loaded:
ηao = 0,25
ηco
=
4,9
M Ed
z
N Ed
e =
MEd
NEd
influence of
slenderness for
λ ≤ 0,5
η
η
a, λ
c, λ
=η
=
η
ao
co
+ 0,5 λ
K
− 18,5 λ
≤ 1,0
K
( 1−0,92
λ ) ≥0
K
influence of load
eccentricity :
η
a
=η
a, λ
e/d>0,1 : η a =1,0 and η c =0
+
e
10 (1 − η ao ) η c =η c,
d
λ
⎛ e
⎜1
− 10
⎝ d
⎟ ⎠
⎞
f c
f y
22

Plastic resistance to combined bending and
compression
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
y
N pl,Rd
N Ed
N
M pl,N,Rd = μ M pl,Rd
z
interaction curve
z pl
f yd
-
+
(1-ρ) f yd
M pl,Rd
0,85 f cd
f sd
M pl,N Rd
N Ed
-
M
The resistance of a cross-section to combined
compression and bending and the corresponding
interaction curve may be calculated assuming
rectangular stress blocks.
The tensile strength of the concrete should be
neglected.
The influence of transverse shear forces on the
resistance to bending and normal force should be
considered when determining the interaction curve, if
the shear force V a,Ed
on the steel section exceeds 50%
of the design shear resistance V pl,a,Rd
of the steel
section. The influence of the transverse shear on the
resistance in combined bending and compression
should be taken into account by a reduced design
steel strength (1 - ρ) f yd
in the shear area A v
.
V Ed
Va,Ed
≤0,5
Vpla,Rd
⇒ ρ= 0
2
⎡2 Va,Ed
⎤
Va,Ed
> 0,5 Vpla,Rd
⇒ ρ = ⎢ −1⎥
Vpla,Rd
⎢⎣
⎥⎦
23

Influence of vertical shear
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
V a,Ed
V c,Ed
z pl
-
f sd
f yd
-
0,85f cd
V Ed
M Rd
The shear force V a,Ed
should not exceed
the resistance to shear of the steel section.
The resistance to shear V c,Ed
of the
reinforced concrete part should be verified
in accordance with EN 1992-1-1, 6.2.
-
z pl
f yd
+
-
f yd
+
M a
N a
f sd
M Rd
= M a
+ M c+s N Ed =N a +N c+s
f sd -
f sd
N Ed
M c,+s
N c+s
Unless a more accurate analysis is
used, V Ed
may be distributed into
V a,Ed
acting on the structural steel
and V c,Ed
acting on the reinforced
concrete section by :
V
V
a,Ed
c,Ed
=
V
= V
Ed
Ed
M
M
− V
a
Rd
a,Ed
≈
M
M
pla,Rd
pl,Rd
Verification for vertical
shear:
V
a,Ed
≤
V
pla,Rd
V
c,Ed
≤
V
c, Rd
M pl,a,Rd
M pl,Rd
is the plastic resistance
moment of the steel section.
is the plastic resistance moment
of the composite section.
24

Determination of the resistance to normal
forces and bending (example)
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
t f
A s
y
h w
b
f yd N af
N aw,c
z pl
0,85f cd
- z
t aw,c
-
w
N s
N M pl,N,Rd
c
z
z s
c
V Ed
z aw,t
N
z Ed
N s
(1-ρ)f f sd
yd aw,t
+
N
N s
af
Position of the plastic neutral axis:
N + N −N
= N
c
w
aw,c
pl
aw,t
cd
w
Ed
pl
yd
w
∑ N i = NEd
( b − t )z 0,85 f + t z (1 − ρ)f
− t (h − z ) (1 − ρ)f
= N
Plastic resistance to bending M pl,N,Rd in case of the
simultaneously acting compression force N Ed and the
vertical shear V Ed :
M pl,N,Rd = Nc zc + Naw,c zaw,c + Naw,t zaw,t + Naf (hw + tf ) + 2Ns zs
w
pl
yd
Ed
z
pl
NEd
+ hw
t
=
(b − t )0,85 f
N
N
N
N
N
aw,c
aw,t
af
c
s
= b t
= 2A
f
s
w
= z
pl
= (h
w
f
= (b − t
f
t
yd
w
sd
w
− z
)z
pl
pl
)
cd
t
w
w
(1 − ρ)f
+
(1 − ρ)
0,85
f
2t
yd
f
cd
w
(1 − ρ)
yd
(1 − ρ)f
f
yd
yd
25

G. Hanswille
Simplified determination of the interaction curve
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
N pl,Rd
N
A
A
0,85f cd
f sd
f yd
- -
-
N pl,Rd
N pm, Rd
N pm,
Rd
2
B
C
Mpl,Rd
D
M max, Rd
As a simplification, the interaction
curve may be replaced by a polygonal
diagram given by the points A to D.
M
B
C
D
0,85f cd f sd
f yd
- - -
h n
+
0,85f cd
f sd
f yd
-
-
2hn
z pl
z pl
z pl
0,85f cd
-
f sd
f yd
-
M = M
+
f yd
+
f yd
+
B,Rd
M = M
C,Rd
N pm,Rd
M = M
f yd
D,Rd
pl,Rd
pl,Rd
max,Rd
0,5 N pm,Rd
26

Resistance at points A and D
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Point A
N pla,Rd
N plc,Rd
N pls,Rd
yd
Point D
h c
b c
f cd
- -
-
f sd
M pls,Rd
f yd
0,5 M plc,Rd
M pla,Rd
-
z si
- f yd
0,85 f cd +
f sd
+
z si
N
M
pl,Rd
A,Rd
= N
= 0
N
M
pla,Rd
D,Rd
D,Rd
+ N
= 0,5
=
t f
M
plc,Rd
N
plc,Rd
max,Rd
+ N
b
pls,Rd
t w
h
M = M + M + 1 2
max, Rd
W pl,a
W pl,s
W pl,c
pla,Rd
pls,Rd
M
plc,Rd
plastic section modulus of the
structural steel section
plastic section modulus of the crosssection
of reinforcement
plastic section modulus of the concrete
section
M
M
pls,Rd
M
pla,Rd
plc,Rd
= W
= W
= W
pl,a
pl,s
pl,c
f
f
yd
sd
=
0,85 f
[ ∑ A si z si ] f ys
cd
⎡b
= ⎢
⎢
⎣
c
h
4
2
⎡(h
− 2tf
)
= ⎢
⎢⎣
4
2
c
t
w
+ b t
− W
f
pl,a
(h − t
− W
f
)
pl,s
⎤
⎥
⎥⎦
f
⎤
⎥ 0,85 f
⎥
⎦
cd
27

Bending resistance at Point B (M pl,Rd )
- 0,85f cd 0,85f cd - 0,85
+ h f
n
cd
f sd
+
+ +
2 f sd -
f yd
+
h n
=
+
+
-
+ h n
+ 2f yd
N D,Rd
M pln,Rd
z pl h n
Point D
+ =
M D,Rd
N D,Rd
-
f yd
+
+
f sd
+
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
At point B is no resistance to
compression forces. Therefore
the resistance to compression
forces at point D results from the
additional cross-section zones in
compression. With N D,Rd the
depth h n and the position of the
plastic neutral axis at point B can
be determined. With the plastic
bending moment M n,Rd resulting
from the stress blocks within the
depth h n the plastic resistance
moment M pl,Rd at point B can be
calculated by:
M
pl,Rd
h n
z pl
=
M
D,Rd
M pl,Rd
−M
Point B
pln,Rd
28

Plastic resistance moment at Point C
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
- 0,85 f cd
-
- 2h n
+
+
-
2 f sd
+ 2h n =
f yd
+ +
f sd
0,85f cd 0,85f cd
2f yd
- - 2h -
+
n
f yd
+
fsd
+
+
+
The bending resistance at point C
is the same as the bending
resistance at point B.
M C,Rd = M pl,Rd
The normal force results from the
stress blocks in the zone 2h n.
N C,Rd =2N D,Rd =N cpl,Rd =N pm,Rd
Point B
Point C
h n
z pl
M pl,Rd
+ =
2h n
N C,Rd
2h n
N c,Rd
M c,Rd
29

G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Part 4:
Simplified design method
30

Simplified Method
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Methods of verification acc. to the simplified method
Axial
compression
Design based on the
European buckling curves
κ
λ K
Design based on second order
analysis with equivalent geometrical
bow imperfections
w o
Resistance
of member
in combined
compression
and bending
Design based on second order
analysis with equivalent geometrical
bow imperfections
w o
31

Scope of the simplified method
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
double symmetrical cross-section
uniform cross-sections over the member length with rolled,
cold-formed or welded steel sections
steel contribution ratio
A
0 ,2 ≤ δ ≤ 0,9 δ =
N
relative slenderness
Npl,Rk
λ = ≤ 2,0
N
cr
yd
pl,Rd
longitudinal reinforcement ratio
a
f
0 ,3% ≤ ρ ≤ 6,0 % ρ =
s
s
A
A
s
c
the ratio of the depth to the width of the composite crosssection
should be within the limits 0,2 and 5,0
32

Effects of creep of concrete
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
2000
1500
Load
F [kN]
F
F u = 2022 kN
short term test
F u = 1697 kN
long term test
e
w t
w o
L = 800 cm
The horizontal deflection and
the second order bending
moments increase under
permanent loads due to creep of
concrete. This leads to a
reduction of the ultimate load.
1000
30 cm
500
F v = 534 kN
permanent
load
30 cm
e=3 cm
The effects of creep of
concrete are taken into
account in design by a
reduced flexural stiffness of
the composite cross-section.
0
20 40 60 80 100
deflection w [mm]
33

Effects of creep on the flexural stiffness
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
The effects of creep of concrete are
taken into account by an effective
modulus of elasticity of concrete
E
c,eff
E
=
N
1+
N
G,Ed
Ed
cm
ϕ(t,t
notional size of the cross-section for
the determination of the creep
coefficient ϕ(t,t o )
h
o =
2 A
U
c
effective perimeter U of the crosssection
b
b
o
)
E cm
N Ed
N G,Ed
ϕ(t,t o )
Secant modulus of concrete
total design normal force
part of the total normal force that is
permanent
creep coefficient as a function of the time at
loading t o , the time t considered and the
notional size of the cross-section
In case of concrete filled hollow section the drying of the
concrete is significantly reduced by the steel section. A
good estimation of the creep coefficient can be
achieved, if 25% of that creep coefficient is used, which
results from a cross-section, where the notional size h o
is determined neglecting the steel hollow section.
h
h
ϕ t,eff
= 0,25 ϕ(t,t o
)
U = 2(b + h) U ≈2h
+ 0,5b
34

Verification for axial compression with the
European buckling curves
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
cross-section
buckling about
strong axis
ν = 0,85
buckling about
weak axis
ν = 0,85
ρ s ≤3% ν = 1,00
3%
ρ s ≤ 6%
ν = 1,00
buckling
curve
b
c
a
cr
0,2 0,6 1,0 1,4 1,8
< b
N
Verification:
Ed ≤ 1,0
NRd
ν = 1,00
ν
= 0,85
b
b
1,0
0,8
0,6
0,4
0,2
N
N
χ =
N
pl,Rd
Rd
pl,Rd
Design value of
resistance
= A
a
f
yd
b
+
a
N
A
c
= χ
Rd N pl,Rd
s
f
sd
λ
=
+ ν A
N
N
c
pl,Rk
cr
λ =
f
cd
≤
2,0
Npl,k
N
35

Relative slenderness
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
f
cd
f
=
γ
ck
c
ν = 0,85
relative slenderness:
λ
=
N
pl,Rk
N
cr
≤
2,0
characteristic value of the plastic
resistance to compressive forces
N = A f + A ν f + A
pl,Rk
a
yk
c
ck
s
f
sk
f
cd
f
=
γ
ck
c
ν = 1,00
elastic critical normal force
N
cr
π
=
2
(EJ)
( βL)
eff
2
β - buckling length factor
effective flexural stiffness
( EJ) eff = (EaJa
+ Ke
Ec,eff
Jc
+ EsJs
)
K e =0,6
36

Verification for combined
compression and bending
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
N pl,Rd
N Ed
N
w o
equivalent
geometrical bow
imperfection
Verification
max M
Ed
≤ M
Rd
= α
M
α M
= 0,9 for S235 and S355
μ
M
pl,Rd
w o
α M
= 0,8 for S420 and S460
α M
μ M pl,Rd
M pl,N,Rd
M
bending moments taking into
account second order effects:
f yd
(1-ρ) f yd
+
-
M Rd
0,85f cd
f sd
M pl,Rd
V Ed
M pl,N Rd
N Ed
The factor α M takes into account the
difference between the full plastic and the
elasto-plastic resistance of the cross-section
resulting from strain limitations for concrete.
-
max M
N
cr
(EI)
with
π
=
Ed
2
eff,II
K
= N
(E J)
β
2
= K
e,II
L
Ed
w
eff,II
2
o
=
(E
a
J
0,5
o
a
1
N
1−
N
+
K
K
e
o
Ed
Effective flexural stiffness
cr
E
c,eff
= 0,9
J
c
+
E
s
J
s
)
37

Equivalent initial bow imperfections
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Buckling curve
a b c
3%
< ρ s ≤ 6%
ρ s ≤3%
Member imperfection
w o = L/300 w o = L/200 w o = L/150
38

Imperfections for global analysis of frames
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
sway imperfection
Global initial sway imperfection acc. to EN 1993-1-1:
N Ed,1
N Ed,2
φ
=
φ
o
α
m
α
h
h
φ φ Φ o basic value with Φ o = 1/200
α h
reduction factor for the height h in [m]
α
h
=
2
h
but
2
3
≤ α
h
≤1,0
equivalent forces
α m
reduction factor for the number of columns in a row
Φ N Ed,1
N Ed,1
N Ed,2
Φ N Ed,2
α
m
=
0,5
⎢
⎡ 1 +
⎣
1
m
⎤
⎥
⎦
Φ N Ed,1
Φ N Ed,2
m is the number of columns in a row including only
those columns which carry a vertical load N Ed not
less than 50% of the average value of the column
in a vertical plane considered.
39

Frames sensitive against second order effects
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
N Ed,1
imperfections
λ > 0,5
N
N
pl,Rk
L Ed, 2
1
φ 1
φ 2
N Ed,2
L 2
Within a global analysis, member imperfections in
composite compression members may be
neglected where first-order analysis may be
used. Where second-order analysis should be
used, member imperfections may be neglected
within the global analysis if:
λ
≤ 0,5
N
N
pl,Rk
Ed,1
w 0
λ
≤
0,5
N
N
pl,Rk
Ed, i
N Ed,1
N Ed,2
w
4
o N
L2
N Ed,1 φ N Ed,2 φ 2
1
equivalent
w o
forces
q=8 N
2 Ed, 2
N L
Ed,1 φ 1 2
N Ed,2 φ 2
w
4
L
Ed,2
o NEd,2
2
N
λ =
cr
π
=
N
2
pl,Rk
N
cr
(EJ)
L
2
i
eff
( EJ) eff = (EaJa
+ 0,6E c,eff Jc
+ EsJs
)
40

Second order analysis
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
N
r M R
r M R
Bending moments including second order effects:
⎛ r sin ε (1− ξ)
+ sin ε ξ ⎞ ⎛ cos ε (0,5 − ξ)
⎞
M(
ξ)
= MR
⎜
⎟ + Mo
⎜
−1⎟
⎝ sin ε ⎠ ⎝ cos ( ε / 2) ⎠
EJ
V
z
M ε⎛
r cos ε (1− ξ)
+ cos ε ξ ⎞ ⎛ sin ε (0,5 − ξ)
⎞
( ξ)
=
R
⎜
⎟ + Mo
⎜
−1⎟
L ⎝ sin ε ⎠ ⎝ cos ( ε / 2) ⎠
q
L
ζ M
ζ
M max
M
o
=
(q L
2
+ 8Nw
o
1
)
2
ε
ε=L
Maximum bending moment at the point ξ M :
N
(E J)
Ed
eff,II
⎛
⎜
⎝
dM
=
dξ
⎞
0⎟
⎠
M
max
=
[ 0,5M(1 + r) + M ]
o
2
1+
c
cos(0,5 ε)
−
M
0
w o
M R
M R
c =
M (r −1)
M(1+
r) + 2M
o
1
tan(0,5ε
)
ξ
M
= 0,5
+
arctan
ε
c
41

Simplified calculation of second order effects
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
4,0
3,0
k
r=1,0
r=0,5
r=0
r M R
N
EJ
r M R
Exact solution:
M
max
= 0,5M
R
r −1
1
c =
1+
r tan(0,5ε
)
2
1+
c
(1+
r)
cos(0,5 ε)
2,0
r= - 0,5
L
ζ
ζ M
M max
ξ
M
= 0,5 +
arctan
ε
c
ε=L
N
(E J)
Ed
eff,II
M R
simplified solution:
1,0
exact method
simplified method
N
M R
k =
M
M
max
R
β
=
N
1−
N
Ed
cr
β = 0 ,66 +
β ≥
0,44
0,44r
0,25 0,50 0,75 1,00
N cr
42

Background of the member imperfections
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
N Rd
N
ε =L
N
(EJ)
Rd
Bending moment based on second
order analysis:
N Ed
=N Rd
eff,II
= 8 wo
(EJ) eff,II ⎡ 1 ⎤
M
⎢ −
2
⎥
L ⎣cos(
ε / 2)
1
⎦
wo Resistance to axial compression
μ M pl,Rd
based on the European buckling
curves:
N
= χ
Rd N pl,Rd
M Rd
M pl,Rd
M
Bending resistance:
M
Rd
= α
M
μ
M
pl,Rd
The initial bow imperfections were
recalculated from the resistance to
compression calculated with the
European buckling curves.
Determination of the equivalent bow
imperfection:
w
o
α
=
8(EJ)
M
eff,II
μ
d
M
pl,Rd
L
2
⎡ 1
⎢
⎣1−cos(
ε / 2)
⎤
−1⎥
⎦
43

Geometrical bow imperfections –
comparison with European buckling
curves for axial compression
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
j
500
400
1
N ( )
δ =
δ N (w )
Rd κ
Rd
o
1,2
1
1,1
2
3
1
2
3
C20/S235
C40/S355
C60/S355
300
200
j =
L
w o
1,0 2,0
2
3
λ
The initial bow imperfection is a
function of the related slenderness
and the resistance of cross-sections.
In Eurocode 4 constant values for w 0
are used.
1,0
0,9
0,8
w o = l/300
0,4 0,8 1,2 1,6 2,0
The use of constant values for w o leads to
maximum differences of 5% in
comparison with the calculation based on
the European buckling curves.
λ
44

Comparison of the simplified method with nonlinear
calculations for combined compression
and bending
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
1,0
0,8
N
N pl,Rd
Resistance as a function of the
related slenderness
Plastic cross-section
resistance
λ
=
N
pl,Rk
N
cr
λ k = 0,50
λ k = 1,00
λ k = 1,50
0,6
λ k = 2,00
0,4
general method
0,2
simplified method
M
0,2 0,4 0,6 0,8 1,0
M pl,Rd
45

Resistance to combined compression
and biaxial bending
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Interaction
M z , N
μ dz
μz,Ed
N
N pl,Rd
μ y, Ed
N
N
pl,Rd
μ dy
Ed
B
Interaction
M y , N
Interaction
M y , M z , N Ed
The resistance is given by a threedimensional
interaction relation. For
simplification a linear interaction
between the points A and B is used.
M
M
M
y,Rd
z,Rd
y,Ed
M
z,Ed
(N
(N
Ed
Ed
= μ
= μ
)
= μ
) = μ
y,Ed
z,Ed
M
dy
dz
M
M
M
y,Rd
y,Rd
pl,y,Rd
pl,y,Rd
A
Approximation:
μy,Ed
μz,Ed
+ ≤ 1,0
μ μ
dy
dz
approximation for the
interaction curve:
M
M
z
pl,z,Rd
Interaction
M y , M z
M
M
y
pl,y, Rd
μ
μ
μ
y,Ed
dy
dy
M
M
μ
+
μ
y,Ed
z,Ed
dz
pl,y,Rd
≤ 1,0
+
μ
dz
M
M
z,Ed
pl,z,Rd
≤ 1,0
46

Verification in case of compression an biaxial
bending
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
For both axis a separate verification
is necessary.
μ
μ
dy
dy
M
M
M
M
y,Ed
pl,y,Rd
y,Ed
pl,y,Rd
≤
+
α
μ
M
dz
M
M
z,Ed
μ
dz
pl,y,Rd
M
M
z,Ed
pl,y,Rd
Verification for the interaction of biaxial
bending.
≤ 1,0
Imperfections should be considered
only in the plane in which failure is
expected to occur. If it is not evident
which plane is the most critical,
checks should be made for both
planes.
≤
α
M
N
N pl,Rd
N
N
N
N pl,Rd
N
N
Ed
pl,Rd
Ed
pl,Rd
μ dy
μ dz
α M = 0,9 for S235 and S355
α M = 0,8 for S420 and S460
M
M
M
y,Rd
pl,y,Rd
M
z,Rd
pl,z,Rd
47

G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Part 5:
Special aspects of columns with inner core profiles
48

Composite columns – General Method
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Commerzbank Frankfurt
Millennium Tower Vienna
Highlight Center
Munich
New railway station in Berlin
(Lehrter Bahnhof)
49

Composite columns with concrete filled
tubes and steel cores – special effects
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Resistance based on stress blocks (plastic
resistance)
tube core concrete
f y f y f c
M
N
Non linear resistance with strain limitation for concrete
strains ε
f y f y
f c
M
α
M
=
M
μ M
Rd
pl,Rd
Cross-sections with
massive inner cores have
a very high plastic shape
factor and the cores can
have very high residual
stresses. Therefore these
columns can not be
design with the simplified
method according to EN
1944-1-1.
σ ED
N
r
50

Residual stresses and distribution of the
yield strength
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
300
σ ED [N/mm 2 ]
f yk
distribution of yield strenght
f y (r)
250
200
150
100
U
A
=
π
= π
d k
2
d
k
/ 4
d k
r k
f yk – characteristic value
of the yield strenght
fy(r)
⎛
= 0,95 + 0,1 ⎜
fyk
⎝
r
r
k
2
⎞
⎟
⎠
50
U/A [1/m]
10 20 30 40 50
d K
[mm]
residual stresses:
400 200 130
100
80
d K
d
σ ED
σ
E
(r) = σ
ED
⎡
2r
⎢1−
⎢
⎣
r
2
2
K
⎤
⎥
⎥
⎦
r, r k
51

General method – Finite Element Model
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
load introduction
stresses in the tube
initial bow imperfection
cross-section
stresses in concrete
52

G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Part 6:
Load introduction and
longitudinal shear
53

Load introduction over the steel section
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
N Ed
load introduction by headed studs within the
load introduction length L E
P D L E < 2,0 d
N c,Ed
N a,Ed
d
A a
N s,Ed
A s
A c
L E
⎧ 2d
≤ ⎨
⎩L / 3
d minimum transverse dimension of
the cross-section
L
member length of the column
sectional forces of the cross-section :
Npl,a
Npl,s
N a,Ed = NEd
Ns,Ed
= NEd
Nc,Ed
= N
N
N
pl,Rd
pl,Rd
required number of studs n resulting from the
sectional forces N Ed,c
+ N Ed,s
:
V
L,Ed
⎡ Npl,a
⎥ ⎥ ⎤
= Nc,Ed
+ Ns,Ed
= NEd
⎢1
−
⎢⎣
Npl,
Rd ⎦
P Rd – design resistance of studs
V = nP
L,Rd
Rd
Ed
N
N
pl,c
pl,Rd
54

G. Hanswille
Load introduction for combined comression and
Univ.-Prof. Dr.-Ing.
Institute for Steel and
bending
Composite Structures
University of Wuppertal-Germany
sectional forces due to N Ed
und M Ed
sectional forces based on plastic theory
M Ed
N Ed
M Rd = Ma,Rd
+ Mc+ s,Rd NRd
= Na,Rd
+ Nc+
s,Rd
N
z pl
-
M a;Rd
0,85f
f cd
sd
M
M Rd
-
c,+s,Rd
+ =
N pl,Rd
1,0
f yd
+
N a,Rd
f sd
N c+s,Rd
N Rd
N a,Ed
N c,Ed + N s,Ed
M c,Ed + M s,Ed
M a,Ed
N
N
N
N
Rd
pl,Rd
Ed
pl, Rd
E d
R d
E
R
d
d
N
=
N
M
M pl,Rd
a,Ed
a,Rd
R
=
d
M
M
=
a,Ed
a,Rd
⎛
⎜
M
⎝
M
=
Ed
pl,Rd
N
N
⎞
⎟
⎠
c+
s,Ed
c+
s,Rd
2
⎛ N
+ ⎜
⎝
N
Ed
=
pl,Rd
⎞
⎟
⎠
M
M
2
c+
s,Ed
c+
s,Rd
M
M
Ed
pl,Rd
M
M
Rd
pl,Rd
1,0
E
d
=
⎛
⎜
M
⎝
M
Rd
pl,Rd
⎞
⎟
⎠
2
⎛ N
+ ⎜
⎝
N
Rd
pl,Rd
⎞
⎟
⎠
2
55

Load introduction – Example
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
sectional forces based on stress blocks:
f yd
f f cd sd -N s,i
Nc
+ s,Rd = Nc
+ ∑Nsi
- M -N
a;Rd
Z
z pl - c s
M c,+s,Rd Mc
+ s,Rd = Nc
zc
+ ∑Nsi
zsi
z
N c
a,Rd
Z s
-N
+
c+s,Rd
f sd N s,i
shear forces of studs based on elastic theory shear forces of studs based on plastic theory
2
2
⎡N
c s,Ed M ⎤ ⎡
c s,Ed M ⎤
c s,Ed
Nc
+ s,Ed Mc
+ s,Ed
maxPEd
⎢
+ +
xi⎥
⎢
+
=
+
+ zi⎥
max PEd
= +
⎢ n
2
2
r ⎥ ⎢
n eh
0,5 n
i
r ⎥
⎣
∑ ⎦ ⎣ ∑
i ⎦
n – number of
P Ed (N)
z i
x studs within the
P i
ed,v r i
load introduction
e h
P length
ed,h P Ed (M)
N M
c+s,Ed M c+s,Ed
c+s,Ed
N c+s,Ed
56

Shear resistance of stud connectors welded
to the web of partially encased I-Sections
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
P Rd
D c
µP Rd
/ 2
µP Rd
/ 2
V = P + V
L,Rd
V
LR,Rd
Rd
= μ
P
LR, Rd
Rd
Where stud connectors are attached to the web of a fully or
partially concrete encased steel I-section or a similar
section, account may be taken of the frictional forces that
develop from the prevention of lateral expansion of the
concrete by the adjacent steel flanges. This resistance may
be added to the calculated resistance of the shear
connectors. The additional resistance may be assumed to
be on each flange and each horizontal row of studs, where μ
is the relevant coefficient of friction that may be assumed.
For steel sections without painting, μ may be taken as 0,5.
P Rd
is the resistance of a single stud.
57

Shear resistance of stud connectors welded
to the web of partially encased I-Sections
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
P Rd
P Rd
P Rd
P Rd
P Rd P Rd
V LR,Rd /2
< 300 < 400 < 600
V
L,Rd
=
nP
Rd
+ V
LR,Rd
V
LR,Rd
=
μ
P
Rd
P Rd = min
P
P
Rd,1
Rd,2
= 0,29 α
= 0,8 ⋅ f
u
d
2
⎛
⎜
π d
⎝ 4
2
f
ck
⎞
⎟
1
⎠ γv
E
cm
1
γ
v
In absence of better information from tests, the clear distance between
the flanges should not exceed the values given above.
58

Shear resistance of stud connectors welded
to the web of partially encased I-sections
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
F [kN]
F
F
F
F
3500
3000
2500
2000
test series S1
test series S2
1500
test S3/3
1000
test S1/3
500
w [mm]
0
0 2 4 6 8 10 12 14 16
59

Load introduction – longitudinal shear
forces in concrete
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
N Ed
b c
D c
D c
θ
Z s
Z s
Z θ
s
D c
N Ed
Longitudinal shear force in section I-I:
V
L,Ed
=
N
Ed
⎡ N
⎢1
−
⎢⎣
N
pl,a
pl,Rd
⎤
⎥
⎥⎦
A
A
c1
c
0,85 f
0,85 f
cd
cd
+ A
+ A
Longitudinal shear resistance of concrete struts:
V
L,Rd,max
= 4
c
y
ν 0,85
f
cd
cot θ + tan θ
L
E
θ =45 o
s1
s
f
f
sd
sd
L E
I
I
I
ν = 0,6
(1−
(f
ck
/ 250))
with
f
ck
in
N / mm
2
c y
c y
not directly connected concrete
area A s1
I
longitudinal shear resistance of the stirrups:
V
L,Rd,s
= 4
A
s
s
w
f
yd
s w - spacing of stirrups
cot θ
A s
- cross-section area of the stirrups
L
E
60

Load introduction – longitudinal shear
forces in concrete – test results
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
2000
F [kN]
F u
= 1608 kN
1500
1000
F
w
F
test I/1
500
0
0 2 4 6 8 10 12 14
w [mm]
61

Load introduction – Examples
(Airport Hannover)
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Load introduction with
gusset plates
62

Load introduction with partially loaded
end plates
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Load introduction with
partially loaded end plates
63

Load introduction with distance plates for
columns with inner steel cores
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
distance plates
Post Tower Bonn
64

Composite columns with hollow sections –
Load introduction
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
gusset plate
stiffeners and
end plates
distance plates
Stiffener
σ c
σ c
Distance
σ c
plate
65

Mechanical model
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
P
cR,m
= f
c
A
1
A
A
c
1
⎡
⎢1+ η
⎣
cL
t
d
f
f
y
c
⎤
⎥
⎦
σ c
σ a
A 1
Effect of partially
loaded area
Effect of
confinement by the
tube
A c
σ c,r
η c,L = 3,5 η c,L = 4,9
σ a,t
σ a,t
66

Typical load-deformation curves
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
P [MN]
2.5
series SXIII
P u
2.0
P u,stat
P
1.5
1.0
δ
0.5
δ [mm]
0 5 10 15 20 25 30 35
P [MN]
5.0
4.0
3.0
P
P u
P u,stat
series SV
2.0
1.0
δ
δ [mm]
0
5 10 15
20
67

Test evaluation according to EN 1990
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
r e [MN]
10,0
P c,Rm
8,0
6,0
41 tests
V r
= 0.14
4,0
P c,Rk = 0.78 P c,Rm
σ a,x
σ c
A 1
2,0
P c,Rd = 0.66 P c,Rm
r t [MN]
2,0 4,0 6,0 8,0 10,0
A c
P
cR,m
= f
c
A
1
⎡
⎢1+ η
⎢⎣
cL
t
d
f
f
y
c
⎤
⎥
⎥⎦
A
A
c
1
σ a,y
σ c,r
σ a,y
68

Load distribution by end plates
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
~1d
F [kN]
6000
F u
4000
F u = 6047 kN
F u,stat = 4750 kN
δ u = 7.5 mm
t s
t p
σ c
2000
b c
5,0 10,0 15,0
δ [mm]
b = t + 5 t
c
s
p
69

Design rules according to EN 1994-1-1
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
σ
c,Rd
= f
cd
⎡
⎢1+ η
⎣
cL
t
d
f
f
yk
ck
⎤
⎥
⎦
A
A
c
1
≤
Ac
f
A
1
cd
≤
f
yd
t s
Load distribution 1:2,5
t p
A 1
f ck
t
d
f yk
A 1
A c
η c,L
concrete cylinder strength A
wall thickness of the tube A1
diameter of the tube
yield strength of structural steel
loaded area
cross section area of the concrete
confinement factor
η c,L
= 4,9 (tube)
η c,L
= 3,5 (square hollow sections)
c ≤
20
σ c,Rd
b c
b = t + 5 t
c
s
p
b c
70

G. Hanswille
Univ.-Prof. Dr.-Ing.
Verification outside the areas of load introduction Institute for Steel and
Composite Structures
University of Wuppertal-Germany
F
δ
A
pure bond
(adhesion)
B
mechanical
interlock
σ r
C
friction
Outside the area of load introduction,
longitudinal shear at the interface
between concrete and steel should be
verified where it is caused by
transverse loads and / or end
moments. Shear connectors should
be provided, based on the distribution
of the design value of longitudinal
shear, where this exceeds the design
shear strength τ Rd
.
In absence of a more accurate
method, elastic analysis, considering
long term effects and cracking of
concrete may be used to determine
the longitudinal shear at the interface.
71

Design shear strength τ Rd
G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
b c
b
τ Rd,o = 0,30 N/mm 2
concrete
encased
sections
y
c y
c y
c z
h
c z
h c
τ
Rd
= τ
Rd,o
β
c
⎡ cz,min
⎤
β c = 1 + 0,02 cz
⎢1
− ≤2,5
c
⎥
⎣ z ⎦
c z - nominal concrete cover [mm]
z
c z,min =40mm (minimum value)
concrete filled tubes
τ Rd
= 0,55 N/mm 2
flanges of partially
encased I-sections
τ Rd
= 0,20 N/mm 2
concrete filled
rectangular hollow
sections
τ Rd
= 0,40 N/mm 2
webs of partially
encased I-sections
τ Rd
= 0,0 N/mm 2
72

G. Hanswille
Univ.-Prof. Dr.-Ing.
Institute for Steel and
Composite Structures
University of Wuppertal-Germany
Thank you very
much for your kind
attention
73