ECCS EC3-1!1!2013 LSS Compendiumv4 Kopia
ECCS EC3-1!1!2013 LSS Compendiumv4 Kopia
ECCS EC3-1!1!2013 LSS Compendiumv4 Kopia
M0
= 1.00;
M1
= 1.00 e
M2
= 1.25.
Introduction
Materials
Ductility properties
f
u
/ f
y
> 1.1;
Failure strain > 15%;
c
u
> 15 c
y
.
EN 10025
Steel Grade
S235 to S960
Steel Qualities
JR, J0, J2, K2
.
Introduction
Materials
II Properties of Materials
Table 2.1 of EN 1993-1-10
ensures adequate behaviour
against brittle fracture.
Introduction
Exercise 1
Yield Stress
Use:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
Check the variation of yield stress with thickness for S355
J2Waccording to EN 100025-5.
Specification t [mm]
>16 >40 >50 >63 >80 >100 >150 >200 >16
16 40 50 63 80 100 150 200 250
EN 10025-5 S355 J0W 355 345 335 335 325 315 295 - -
S355 J2W
S355 K2W
Maximum thickness - t (mm)
Introduction
Materials
II Properties of Materials
According to EC3-1-10, lamellar tearing may be neglected in a detail if Z
Ed
Z
Rd
,
Z
Ed
being the required design Z-value resulting from the magnitude of strains from
restrained metal shrinkage under the weld beads obtained as given in Table 3.2 of
EC3-1-10 and Z
Rd
is the available design Z-value for the material according to EN
10164 (Z15, Z25 and Z35 classes are established).
Through-thickness properties of materials should prevent lamellar tearing in
steel assemblies. The susceptibility of the material should be determined by
measuring the through-thickness ductility quality according to EN 10164, which is
expressed in terms of quality classes identified by Z-values.
Introduction
Materials
II Properties of Materials
1. General Observations Conceptual aspects
2. Codes of Practice and Normalization
3. Basis of Design
4. Materials
5. Geometric Characteristics and Tolerances
Introduction
III) Geometric Data
Dimensions, shape, ... - Characteristic or nominal values.
EN 1090 (and product standards) establishes two types of tolerances:
Fundamental tolerances required to ensure resistance and stability
of the structure;
Functional tolerances required to ensure aesthetical appearance of
the structure.
Cold-formed sections
Hot-rolled sections
Introduction
Geometric Characteristics and Tolerances
Introduction
Geometric Characteristics and Tolerances
Introduction
Geometric Characteristics and Tolerances
1. Structural Modeling
2. Global Analysis of Steel Structures
3. Classification of cross sections
Structural Analysis
The model should simulate real conditions (structural elements,
connections, loading, supports, ).
Beam elements
i) Type of element
- Modeling with linear, two-dimensional or
three-dimensional elements.
Structural Analysis
Structural Modeling
Bi-dimensional elements
Modeling with linear, two-dimensional or three-dimensional elements
Alternative ways of modeling slabs (stiffness in its own plan) in the
behaviour of the structure.
Structural Analysis
Structural Modeling
ii) Influence of member axis (normative expressions derived with
respect to the centroid of the section)
- If loads are not applied in the centroid of the sections it is
necessary to take into account additional forces due to the
eccentricity.
=
=
P N
e P M
C
C
'
'
=
=
P N
M
A
A
'
'
0
P
C A
C A
e
P
Structural Analysis
Structural Modeling
- Eccentricity in relation to shear centre.
y
z
C
c
C
g
e
p
L
M
z
V
z
M
y
V
y
M
t
pL
2
/8
pL/2
pLe/2
0
0
pL/2
pLe/2
Structural Analysis
Structural Modeling
iii) Influence of eccentricities and supports.
L
C
L
R
L
C
L
R
A A B B
h h
M
A
->M
A
AM = 9,75% for h/L
C
= 0.05; AM = 19% for h/L
C
= 0.1; AM = 36% for h/L
C
= 0.2.
Def. at span Ao
v
= 19% for h/L
C
= 0.05; Ao
v
= 34.4% for h/L
C
= 0.1; Ao
v
= 59% for h/L
C
= 0.2.
- Eccentricities in connections
Structural Analysis
Structural Modeling
- Influence of the dimension of the bearing
Rounding of bending moment taking into
account the distribution of shear force in
the bearing
- Section variation
Structural Analysis
Structural Modeling
iv) Elements with non-uniform sections or curved axis
h
o
c
p
L
Real structure
Mesh with 3 elements
Mesh with 10 elements
L = 10 m; p =10 kN/m; E = 210
GPa; b = 0.20 m; h
0
= 0.50 m
Structural Analysis
Structural Modeling
v) Influence of joints
M
z
M
y
V
z
V
y
N
M
t
M
y
|
y
M
z
|
z
M
t
N
c
Structural Analysis
Structural Modeling
|
j
|
Cd
M
j,Rd
M
j
S
j
i) Stiffness (S
j
)
ii) Bending resistance (M
j,Rd
)
iii) Rotation capacity (|
cd
)
L
j
EI=
S
j
j
j
j
S
M
= |
M
j
L
j
EI
j
j
j
j
j
L
I E
M
= |
M
j
=
The stiffness of a connection may be simulated by
a spring with rotational stiffness S
j
M
|
j,Ed
Ed
90
Structural Analysis
Structural Modeling
- Idealization of beam-to-column joint
S j,ini
M
Mj,Rd
S j,ini
M
Mj,Rd
S j,ini/ q
M
Mj,Rd
S j,ini
2/3 Mj,Rd
- Classification of beam-to-column joint
Structural Analysis
Structural Modeling
- Modeling of beam-to-
column joints
Structural Analysis
Structural Modeling
x x x
3 2 1
x x x
3 2 1
External node
Internal node
Consider the steel
frame (E = 210 GPa and S275)
subjected to the indicated loading,
already factored. Assume that the
column bases are fixed. Determine the
design internal forces and
displacements considering the
following situations.
a) Full-strength rigid joints between
the beams and the columns:
a.1) without considering the effect of
eccentricities;
a.2) considering the effect of
eccentricities;
b) Full-strength semi-rigid joints
between the beams and the column;
c) Partial-strength semi-rigid joints.
025 . 0 =
C
L h
10 m
5 m
5 m
HEA260
HEA260
HEA260
HEA260
IPE400
IPE400
Exc.
10 m
5 m
5 m
HEA260
HEA260
HEA260
HEA260
IPE400
IPE400
14.8 kN
204 kN
33.6 kN/m
45 kN/m
253.5 kN
204 kN
253.5 kN
18.2 kN
Structural Analysis
Structural Modeling
EXAMPLE 1
a.1) without eccentricities
10 m
5 m
5 m
1 2 3 3
4 4
5 6 6
1
o2
o5
o4
o1
o6
o3
M
z
(kNm) V
y
(kN) N
x
(kN) o (mm)
1 256.45 208.93 34.54 12.23
1 230.33 208.93 34.54 12.23
2 253.46 10.45 34.54 43.40
3 332.20 229.82 34.54 12.42
3 360.93 229.82 34.54 12.43
4 174.12 158.48 87.03 23.70
4 154.31 158.48 87.03 23.70
5 219.02 5.32 87.03 41.66
6 206.18 169.12 87.03 23.22
6 227.32 169.12 87.03 23.21
a.2) with eccentricities
M
z
(kNm) V
y
(kN) N
x
(kN) o (mm)
1 - - - -
1 235.85 208.80 35.24 11.96
2 247.31 10.58 35.24 41.87
3 338.99 229.95 35.24 12.15
3 - - - -
4 - - - -
4 158.04 158.51 88.47 22.97
5 215.45 5.29 88.47 40.77
6 209.59 169.09 88.47 22.48
6 - - - -
decrease of 2.4% and 1.6% at the midspan
moments of both beams.
decrease of 3.6% and 2.1% in the vertical
displacements.
10 m
5 m
5 m
1 2 3
4
5 6
o2
o5
o4
o1
o6
o3
M
y
V
z
M
y
V
z
Structural Analysis
Structural Modeling
Level 1
b.1) Full-strength joints
End-plate t =16 mm
M20 bolts, class 10.9
S
j,ini
= 416 125 kNm/rad and S
j
= 208 062
kNm/rad
Level 2
End-plate t =16 mm
M20 bolts, class 10.9
S
j,ini
= 389 646 kNm/rad and S
j
= 194 823
kNm/rad
Structural Analysis
Structural Modeling
M
z
(kNm) V
y
(kN) N
x
(kN)
o (mm)
1 246.22 209.16 33.13 12.75
1 220.08 209.16 33.13 12.75
2 264.85 10.22 33.13 46.33
3 319.68 229.59 33.13 12.93
3 348.38 229.59 33.13 12.94
4 166.86 158.43 84.25 25.09
4 147.05 158.43 84.25 25.09
5 226.02 5.37 84.25 43.46
6 199.43 169.17 84.25 24.62
6 220.57 169.17 84.25 24.62
The consideration of the flexibility of the joints leads to
- a decrease of 4.0%, 3.5%, 4.1% and 2.9% in the negative moments of cross sections 1, 3, 4
and 6, respectively, when compared to the rigid model without eccentricities
- an increase of 4.5% and 3.2% in the midspan moments of the beams (cross sections 2 and 5,
respectively) and of 6.8% and 4.3% of the corresponding vertical displacements.
10 m
5 m
5 m
1 2 3 3
4 4
5 6 6
1
o2
o5
o4
o1
o6
o3
M
y
V
z
Structural Analysis
Structural Modeling
b.1) Full-strength joints
Level 1
b.2) Partial strength, semi-rigid joints
End-plate: t =15 mm
M20 bolts, class 10.9
S
j,ini
= 39 699 kNm/rad and S
j
= 19 849
kNm/rad
Level 2
End-plate: t =15 mm
M20 bolts, class 10.9
S
j,ini
= 23 346 kNm/rad and S
j
= 11 673
kNm/rad
Structural Analysis
Structural Modeling
Initial stiffness
10 m
5 m
5 m
1 2 3 3
4 4
5 6 6
1
o2
o5
o4
o1
o6
o3
M
z
(kNm) V
y
(kN) N
x
(kN) o (mm)
1 207.89 209.80 23.84 14.90
1 181.66 209.80 23.84 14.90
2 306.39 9.58 23.84 57.02
3 275.02 228.95 23.84 15.03
3 303.64 228.95 23.84 15.03
4 129.63 158.53 71.02 31.73
4 109.81 158.53 71.02 31.73
5 263.78 5.27 71.02 53.18
6 161.15 169.07 71.02 31.34
6 182.28 169.07 71.02 31.34
Secant stiffness
M
z
(kNm) V
y
(kN) N
x
(kN) o (mm)
1 174.13 210.48 17.26 17.21
1 147.82 210.48 17.26 17.21
2 343.56 8.89 17.26 66.59
3 234.52 228.27 17.26 17.31
3 263.05 228.27 17.26 17.31
4 101.57 158.66 60.48 38.92
4 81.74 158.66 60.48 38.92
5 292.45 5.14 60.48 60.55
6 131.90 168.94 60.48 38.58
6 153.01 168.94 60.48 38.58
Comparative results
M
y
M
y
V
z
V
z
Structural Analysis
Structural Modeling
b.2) Partial strength,
semi-rigid joints
Consider the frame (E
= 210 Gpa and steel grade S275),
under the indicated loading. Consider
pinned column base joints, and full-
strength rigid joints at the external
nodes. At the internal node, the joint to
be used is shown in Figure 2.36. It has
an end plate 20 mm thick, and M24
class 10.9 bolts. Using an elastic
analysis and neglecting eccentricities
in the nodes, determine the forces and
displacements, considering the
following situations:
a) rigid internal node joint;
b) internal node joint modeled by two
rotational springs;
c) internal node joint modeled by three
rotational springs.
Structural Analysis
Structural Modeling
EXAMPLE 2
Structural Analysis
Structural Modeling
1. Structural Modeling
2. Global Analysis of Steel Structures
3. Classification of cross sections
Structural Analysis
Isostatic structures
Hiperstatic structures
Global elastic analysis
Global plastic analysis
- plastic,
- elastic perfectly plastic,
- elastic-plastic.
NOTES (EC3-1-1, Cl. 5.4):
- Although internal forces may be obtained from a global elastic analysis, the
design resistance may be quantified based on the plastic resistance of the section
(depending on the class of the section).
- Re-distribution of internal forces is allowed in global elastic analysis.
- Global plastic analysis entails the capacity for re-distribution of forces -
requirements: ductile material, compact sections, braced and symmetric.
Structural Analysis
Global Analysis of Steel Structures
Effects to consider in global analysis:
i) deformability and stiffness of the structure and supports;
ii) stability of the structure (global, members and local);
iii) behaviour of cross-sections (classification of sections);
iv) behaviour of joints (strength and stiffness);
v) imperfections (global and in members).
Structural Analysis
Global Analysis of Steel Structures
Global analysis of
continuous beam (S275).
a) global elastic analysis;
b) global plastic analysis.
Structural Analysis
Global Analysis of Steel Structures
42 kN/m 100 kN 100 kN
3 m
A
B
C E
F
G
3 m 3 m 3 m 6 m
D
EXAMPLE 3
D
89.7 kNm
A
C
F
120.6 kNm
68.4 kNm
B
E
G
120.6 kNm
89.7 kNm
Solution: IPE 300,
W
el,y
= 557.1 cm
3
.
M
pl
A
C
D
F
M
pl
<M
pl
B
E
G
M
pl
M
pl
3 100 6
3
= =
= =
A pl
esq
C
A pl
esq
B
V M M
V M M
. kNm 100
kN 3 . 33
=
=
pl
A
M
V
Solution: IPE 240,
W
pl,y
= 366.6 cm
3
.
Solution: IPE 270,
W
pl,y
= 484.0 cm
3
.
(Plastic resistance)
(Plastic resistance)
Plastic bending moment diagram
Elastic bending moment diagram
(Elastic resistance)
50 kN
280 kN
4 m
4 m 4 m
A
B
C
D
E
189.9 kNm
A
B
C
D
E
247.0 kNm
341.6 kNm
142.9 kNm
Elastic bending moment diagram
Solution: IPE 450,
W
el,y
= 1500 cm
3
.
50 kN
280 kN
M
pl
M
pl
M
pl
H
E
V
E
V
A
H
A
M
A
Rtulas Plsticas
A
B
C
D
E
Plastic bending moment diagram
pl
dir
B
pl
dir
C
pl
dir
D
M M
M M
M M
=
=
=
pl E E
pl E E
pl E
M V H
M V H
M H
= +
= +
=
4 280 8 4
4 4
4
kNm 280
kN 140
kN 70
=
=
=
pl
E
E
M
V
H
H
A
= 120 kN; V
A
= 140 kN and M
A
= 200 kNm (M
pl
= 280 kNm)
correct plastic mechanism.
Solution: IPE 360,
W
pl,y
= 1019 cm
3
.
(Elastic resistance)
Solution: IPE 400,
W
pl,y
= 1307.0 cm
3
.
(Plastic resistance)
Structural Analysis
Global Analysis of Steel Structures
Global analysis of a frame
(S275).
a) global elastic analysis;
b) global plastic analysis.
EXAMPLE 4
(Plastic resistance)
1
st
order analysis Internal forces and displacements are evaluated in relation to the
undeformed structure (EC3-1-1, cl. 5.2.1(1)).
2
nd
order analysis The deformation of the structure is considered in the evaluation of internal
forces and displacements (iterative procedure).
Structures sensitive to 2
nd
order effects structures with high compressed members and
structures with low stiffness (e.g.: structures with cables).
2
nd
order effects
P-o effects (local effects).
P-A effects (global effects).
o
P P
A
1
st
order analysis vs. 2
nd
order analysis
Structural Analysis
Global Analysis of Steel Structures
(elastic analysis 10 s =
Ed cr cr
F F o
15 s =
Ed cr cr
F F o
(plastic analysis)
F
Ed
: design loading for a given load combination;
F
cr
: elastic critical load.
Need to consider 2
a
order analysis - EC3-1-1 - cl. 5.2.1(3):
Structural Analysis
Global Analysis of Steel Structures
ii) NUMERICAL
CALCULATION:
Linear eigenvalue analysis
N
Ed
o
cr
a) b)
i) Analytical evaluation
ii) Numerical calculation
iii) Approximate methods (Horne, Wood, )
Structural Analysis
Global Analysis of Steel Structures
ELASTIC CRITICAL LOAD
Applicable for plane frames and one-storey frames with low inclination of the beams
( ), unbraced and with low axial force ( ):
|
|
.
|
\
|
=
Ed H
i
base Ed
top Ed
cr
h
V
H
, ) (
) (
o
o
o
cr
iii) APPROXIMATE METHODS (EC3, cl.5.2.1(4)B) (Horne, Wood,)
Ed
y
N
f A
3 , 0 s
26 s
o
H,Ed
V
Ed
H
Ed
h
i
HORNEs METHOD
Structural Analysis
Global Analysis of Steel Structures
WOODs METHOD
N
K11
K12
K21
K22
LE
q1
N
q2
Kc
K1
K2
N
K11
K12
K21
K22
q1
N
q2
Kc
K1
K2
No-sway
Sway
12 11 1
1
1
K K K K
K K
c
c
+ + +
+
= q
22 21 2
2
2
K K K K
K K
c
c
+ + +
+
= q
2
2
e
cr
L
EI
N
t
=
Ed
cr
cr
N
N
= o
Structural Analysis
Global Analysis of Steel Structures
NUMERICAL METHODS (EXACT)
i) Numerical methods (iterative procedures)
ii) Simplified methods
- Modeling
- Convergence
- Validation
carga, F
AF
2
AF
1
deslocamento, w
iteraes
i
t
e
r
a
e
s
Structural Analysis
Global Analysis of Steel Structures
2
nd
ORDER ANALYSIS
SIMPLIFIED METHODS (APPROX)
For regular structures, EC3-1-1 (clause 5.2.2), allows the inclusion of
secon-order effects associated with vertical loads in a simplified way.
Amplification of first-order effects associated with horizontal actions
(including imperfections), by:
( )
cr
o 1 1 1
If o
cr
>=3.0
- Amplified sway moment method (clause 5.2.2(4));
- Sway-mode buckling length method (clause 5.2.2(8)).
Amplified sway moment method
I
S
S cr
I
NS
II
ap
M M M
.
1
1
1
o
+ =
I
S
S cr
I
NS
II
ap
d d d
.
1
1
1
o
+ =
I
S
S cr
I
NS
II
ap
N N N
.
1
1
1
o
+ =
Structural Analysis
Global Analysis of Steel Structures
m h
o o | |
0
=
Equivalent horizontal forces
Global imperfections: lack of verticality
|
h
h
|
Structural Analysis
Global Analysis of Steel Structures
IMPERFECTIONS
Local imperfections: initial
curvature
e0
L
N
Ed
N
Ed
|
N
Ed
N
Ed
N
Ed
N
Ed
| N
Ed
| N
Ed
N
Ed
N
Ed
L
e N
Ed 0
4
L
e N
Ed 0
4
2
0
8
L
e N
Ed
e
0
/L
Equivalent geometrical imperfections
GLOBAL FRAME ANALYSIS
Choice between frame analyses regarding
the kind of member design:
Design by member buckling checks
Design by 2. order moments + cross-section checks
Methods depend on the account of
2. order effects imperfections and/or e
0
P
A
o
P
P
P
e
0
Structural Analysis
Global Analysis of Steel Structures
GOBAL ANALYSIS AND DESIGN WITH MEMBER BUCKLING
CHECKS
Global analysis
Check of components
and frame
Account for 2nd order
P-A effects
1st order analysis
Sway frame Non-sway frame
Sway Mode Buckling Length
Method
Amplified Sway Moment
Method
(o
cr
> 3)
2nd order analysis
Amplified sway
moments
No limitation
Cross-section resistances and local stability
Joint resistances
Out-of-plane stability of the members
In plane member stability
with non sway buckling length with sway buckling length
In plane member stability
1st order analysis
ocr < 10 resp. 15 ocr > 10 resp. 15
Account for
sway imperfection |
Account for
local bow imperfection
e
0,d
Yes Yes No No
No No
Yes, where the following conditions are met:
- at least one moment resistant joint at one
member end
-
Structural Analysis
Global Analysis of Steel Structures
GOBAL ANALYSIS AND DESIGN WITH MEMBER BUCKLING
CHECKS
Global analysis
Check of components
and frame
Account for 2nd order
P-A effects
1st order analysis
Sway frame Non-sway frame
Sway Mode Buckling Length
Method
Amplified Sway Moment
Method
(o
cr
> 3)
2nd order analysis
Amplified sway
moments
No limitation
Cross-section resistances and local stability
Joint resistances
Out-of-plane stability of the members
In plane member stability
with non sway buckling length with sway buckling length
In plane member stability
1st order analysis
ocr < 10 resp. 15 ocr > 10 resp. 15
Account for
sway imperfection |
Account for
local bow imperfection
e
0,d
Yes Yes No No
No No
Yes, where the following conditions are met:
- at least one moment resistant joint at one
member end
-
y Ed
0.5 A f N >
Structural Analysis
Global Analysis of Steel Structures
Global analysis
Check of components
and frame
Account for 2nd order
P A effects
-
2nd order
analysis
Sway frames
ocr < 10 resp. 15
Non-sway frames
ocr > 10 resp. 15
Alternative:
2nd order analysis
only in-plane
Amplified Sway Moment
Method
2nd order
analysis
Joint resistances
cross-section resistance
Account for
sway imperfection
Account for
local bow imperfection
e0,d
Yes
No
Yes Yes
Yes
Yes
Account for 2nd order
P o effects
Yes
Yes Yes Yes
Yes Yes Yes
Amplified sway
moments
Amplified local
moments
No
cross-section resistance
cross-section resistance
member buckling check
in plane
out of
plane
P
A
o
P
FRAME DESIGN WITH
FULL 2. ORDER
MOMENTS + CS-
CHECKS
Structural Analysis
Global Analysis of Steel Structures
C
Structural Analysis
Global Analysis of Steel Structures
Global elastic analysis of a 3-storey framed structure.
Structural Analysis
Global Analysis of Steel Structures
3D View of a module with 2 frames
Plan View of a module with 2 frames
Typical frame
Roof
Storey 2
Storey 1
EXAMPLE 5
Pinned supports Rigid joints
Beams: Levels 1 and 2 IPE 450
Level 3 IPE 360
Columns: External HEB 220
Internal HEB 260
Material: S 235
Consider a structure composed of a series of frames with 10 m spacing. Assume
that the structure is braced in the longitudinal direction.
2D analysis.
- Permanent Loading, including self-weight G: 20 kN/m (roof) and 30 kN/m (storeys).
- Imposed Loading I: 6 kN/m (roof) and 18 kN/m (storeys).
- Wind W: 9.5 kN (roof) and 19 kN (storeys).
G
W
I
I
1
I
2
I
3
Structural Analysis
Global Analysis of Steel Structures
Global Imperfections
m = 4 (number of columns)
h = 10.5 m:
Ex.
H
Ed, G (roof)
= 0.00264 x 20 x 19.5 =1.03 kN.
H
Ed, I1 (storey 1)
= 0.00264 x 18 x 19.5 =0.92 kN.
Load cases
Storey
h=10.5 m
|=0.00264 rad
Structural Analysis
Global Analysis of Steel Structures
Combinations of actions (ULS)
(Annex A EN 1990)
I ( )
W( ) 00 . 0 ; 20 . 0 ; 60 . 0
2 1 0
= = =
30 . 0 ; 50 . 0 ; 70 . 0
2 1 0
= = =
50 . 1
35 . 1
=
=
Q
G
Combination
ULS
Structural Analysis
Global Analysis of Steel Structures
Classification of the frame (o
cr
- EC3, cl.5.2.1(4)B)
Ex. Combination 1
H
Ed
=1.35x1.03+1.50x9.5=1.39+14.25=15.6 kN (roof)
H
Ed
=15.6+1.35x1.54+1.50x19=2=46.2 kN (storey 2)
H
Ed
=46.2+1.35x1.54+1.50x19=2.08+28.5=76.8 kN (storey 1)
V
Ed
=27x19.5=526.5 kN (roof)
V
Ed
=526.5+ 40.5x19.5=1316.3 kN (storey 2)
V
Ed
=1316.3+40.5x19.5=2106.0 kN (storey 1)
14.74 mm
18.52 mm
19.80 mm
2 . 81
28 . 1
3500
5 . 526
6 . 15
= |
.
|
\
|
=
cr
o 5 . 32
78 . 3
3500
3 . 1316
2 . 46
= |
.
|
\
|
=
cr
o 66 . 8
74 . 14
3500
0 . 2106
8 . 76
= |
.
|
\
|
=
cr
o
Combination 1
(Roof)
(Storey 2)
(Storey 1)
Structural Analysis
Global Analysis of Steel Structures
Classification of the frame
(o
cr
- EC3, cl.5.2.1(4)B)
Combination Storey
.
.
.
.
Structural Analysis
Global Analysis of Steel Structures
Amplification factors
Combination
Combination ULS
Combination
Structural Analysis
Global Analysis of Steel Structures
Combination 7 1 order bending moments
Combination 7 2 order bending moments
Combination 7 2 order bending moments (approx.)
Structural Analysis
Global Analysis of Steel Structures
1. Structural Modeling
2. Global Analysis of Steel Structures
3. Classification of cross sections
Structural Analysis
M
pl
M
el
M
|
s
Class 1
Class 2
Class 3
Class 4
Structural Analysis
Classification of Cross Sections
Class 1: plastic cross-sections
Class 2: compact cross-sections
Class 3: semi-compact cross-
sections
Class 4: slender cross-sections
M
Rd
Class 1 Class 3 Class 4 C
l
a
s
s
2
c/t
M
pl
M
el
Elastic global analysis
Plastic global analysis
M
pl
M
el
M
eff
discontinuity
Structural Analysis
Classification of Cross Sections
Classification of a cross-section
Classification of each of the plate elements (in full or partial
compression) composing the section
Local plate buckling
Dependence on the:
Slenderness c/t
Support conditions
Distribution of direct stresses
Remark:
Isolated plate conditions do not allow to account
for elastic restraints
for plastic redistribution
b
L
Free edge
Simple
support
t
between adjacent parts
Structural Analysis
Classification of Cross Sections
c/t-ratio for standard cross-section shapes
With higher steel grades sections tend to fall
into higher classes
Structural Analysis
Classification of Cross Sections
Classification is required for:
Selection of the global frame analysis:
Elastic frame analysis
Plastic frame analysis
Decision on the member buckling formulae with respect
to the degree of local plastic capacity
Plastic interaction : class 1, 2
Elastic interaction
Decision about the type of cross-section verification:
Elastic verification
Plastic verification
Effective cross-section properties
Structural Analysis
Classification of Cross Sections
Plate element
classification
acc. to
Table 5.2
235/
y
f c =
Structural Analysis
Classification of Cross Sections
Plate element
classification
acc. to Table 5.2
Structural Analysis
Classification of Cross Sections
Classification procedure of a cross-section
Principles: Classification must be made
for loading states including all internal forces/moments
N
Ed
+ M
y,Ed
+ M
z,Ed
for each load combination resulting from the global
analysis of the structure.
Cross-section class =
most unfavourable class of all
plate elements
M
y,Ed
N
Ed
M
z,Ed
Structural Analysis
Classification of Cross Sections
Classification for limit 3/4
basis is the elastic stress diagram for N
Ed
+ M
y,Ed
+ M
z,Ed
of the gross-section (A, I)
class 4 governs, if the limit value c/t (f
y
) of class 3 is
exceeded at any part of the cross-section
for class 4-sections the kind of classification (f
y
or o
com,Ed
)
is related to the kind of member check
M
y,Ed
-
-
-
N
Ed
M
z,Ed
+
o
com,Ed
Structural Analysis
Classification of Cross Sections
Particularities for Class 4 sections
Classification may be based on o
com,Ed
instead of f
y
potential upgrading to class 3
o
com,Ed
(instead of f
y
) may be used for A
eff
and I
eff
M
y,Ed
-
-
-
N
Ed
M
z,Ed
+
o
com,Ed
o = 0
y
M0
limit com,Ed
f
c c
t t
| |
|
|
\ .
s c
o
p
b/ t
28,4 k
o
=
c
o
=
com,Ed
p,red p
y M0
f
upgrade of
Structural Analysis
Classification of Cross Sections
Particularities for Class 4 sections
In Class 4 use of o
com,Ed
(instead of f
y
) may be made
for specific member design methods provided that
the 2. order effects on the member are negligible
the member check consists in verifying the cross-
sections resistance based on full 2. order analysis
the member buckling checks are made for pure N or
pure M
y
resp. M
z
, however not checks for N+M
y
+M
z
Structural Analysis
Classification of Cross Sections
Classification for limit 2/3 and 1/2:
basis is the plastic stress diagram for N
Ed
+ M
y,Ed
+ M
z,Ed
increase (N
Ed
+ M
y,Ed
+ M
z,Ed
) stepwise until the plastic limit is met
and determine the neutral axis stress blocks
class 3 governs, if the limit value c/t (f
y
) of class 2 is exceeded
class 2 governs, if the limit value c/t (f
y
) of class 1 is exceeded
M
pl,y
N
pl
M
pl,z
(plast.)
N
Ed
, M
y,Ed
, M
z,Ed
+
-
+
-
-
Structural Analysis
Classification of Cross Sections
Classify the following cross sections.
a) IPE 300 in steel grade S 235: i) subject to bending;
ii) subject to compression.
b) Square hollow section SHS 200x200x8 mm in steel
grade S 275, subject to compression.
Structural Analysis
Classification of Cross Sections
EXAMPLE 6
b) Web in compression (Table
2.21):
In a rectangular or square hollow
section, the net length of a web
can be approximated by c b 3 t
.
c/t (b 3t )/t = (200 3 8)/8 =
22.0 < 33 = 33 0.92 = 30.4 .
The cross section is class 1.
Structural Analysis
Exercise 2 Semi-Comp+
HEA 300 S355
Use:
- Semi-Comp Member Design (Windows OS)
Example 1: HEA 300 S355
Section in compression and biaxial bending N+M
y
+M
z
:
h
b
r
t
f
t
w
Example 1: HEA 300 S355
Section in compression and biaxial bending N+M
y
+M
z
:
,
, , , , ,
, , , ,
,
500 ( 0.125)
200 ( 0.407)
( 0.439)
100
= = =
= = =
= =
=
Ed Ed pl Rd
y Ed pl y y Ed pl y Rd
pl z z Ed pl z Rd
z Ed
N kN n N N
M kNm m M M
m M M
M kNm
h
b
r
t
f
t
w
HEA 300 S355
Fig. 1: Diagrams of elastic stress distribution (left) and plastic stress distribution (right) for the
determination of and o, Example 1
Result: Web Class 1
The cross-section is class 3.
Flange Class 3
PNA
o =1.0
o =146/208 =0.70
ENA
=0.57
= -0.44
2
0
8
m
m
1
4
6
m
m
-
+
+
+
-
-
-35.5 kN/cm
35.5 kN/cm
6.95 kN/cm
-15.8 kN/cm
-
-25.4 kN/cm
-44.1 kN/cm
-
3.5 kN/cm
35.2 kN/cm +
-
PNA
o =1.0
o =146/208 =0.70
ENA
=0.57
= -0.44
2
0
8
m
m
1
4
6
m
m
-
+
+
+
-
-
-35.5 kN/cm
35.5 kN/cm
6.95 kN/cm
-15.8 kN/cm
-
-25.4 kN/cm
-44.1 kN/cm
-
3.5 kN/cm
35.2 kN/cm +
-
Structural Analysis
Exercise 2 Semi-Comp+
Web in combined bending and compression:
2 2
290 214 227 208
24.5
8.5 8.5
42 42 0.81
24.5 64.8 ... the web is class 3 or better
0.67 0.33 0.67 0.33 0.44
61.5with new
f
w w
w
h t r
c
t t
c
t
c
= = = =
= < = =
+
=
( )
( )
limits
456 456 0.81
45.6 ... the web is class 2 or better
13 1 13 0.70 1
42.6 with new limits
396 396 0.81
39.6
13 1 13 0.70 1
c
o
c
o
< = =
=
< = =
( )
... the web is class 1
35.8with new limits =
Flange in combined bending and compression:
HEA 300 S 355
Structural Analysis
Exercise 2 Semi-Comp+
Web in combined bending and compression:
Flange in combined bending and compression:
( )
2 2
0.21 0.07 0.57 0.21 0.57 0.07 0.57 0.47
2 2 300 2 8.5 2 27
8.5
14
8.5 21 21 0.81 0.47 11.7 ... the flange is class 3 or better
10 8.1 ... the flange is class 3
w
f f
f
k
b t r c
t t
c
k
t
o
o
c
c
= + = + =
= = =
= < = =
> =
HEA 300 S 355
Fig. 1: Diagrams of elastic stress distribution (left) and plastic stress distribution (right) for the
determination of and o, Example 1
Result: Web Class 1
The cross-section is class 3.
Flange Class 3
Structural Analysis
Exercise 2 Semi-Comp+
Single span beam with forked ends and restraints
Cross-section and member checks for elasto-plastic
range (Class 1 3)
Includes I- and H- sections and tubular sections
Interface to LTBeam software developed
by CTICM (France)
Semi-Comp Member Design (Windows OS)
SCOPE OF THE SOFTWARE
Structural Analysis
Exercise 2 Semi-Comp+
Realised in form of an Excel sheet with code
behind
Advantage: Excel is widely used, the user interface
is self explanatory
Requirements: Excel 2007 or 2010, .NET
Framework 3.5 or higher, VSTO Runtime 4.0 (incl. in
the available package)
Semi-Comp Member Design (Windows OS)
THE SOFTWARE TOOL
Structural Analysis
Exercise 2 Semi-Comp+
5 sheets available:
0 Disclaimer
1.1 System data
1.2 Member check
1.2.1 Additional information
1.3 Cross-section check
Semi-Comp Member Design (Windows OS)
THE EXCEL FILE
Structural Analysis
Exercise 2 Semi-Comp+
Information about
cross-section,
boundary
conditions,
loading
LTBeam path
can be entered
Cross-section type
Finishing M0 = 1,00
Select fromlibrary M1 = 1,00
(optional)
Cross-section data Material My,Ed,max= 250,00 kNm
H = 500,0 [mm] Steel grade S235
B =
200,0 [mm]
fy = 235,0 N/mm
Tw = 10,2 [mm] E = 210000,0 N/mm
Tf = 16,0 [mm]
R = 21,0 [mm]
A [cm
2
] Iyy [cm
4
] Izz [cm
4
] Wel,y [cm
3
] Wel,z [cm
3
] Wpl,y [cm
3
] Wpl,z [cm
3
]
115,52 48198,53 2141,69 1927,94 214,17 2194,12 335,88
Vz,Ed,max= 50,00 kN
Boundary conditions It [cm
4
] Iw [cm
6
]
LBeam= 10,000 m 89,01 1254238,95
nfork= [-]
Loading in z-x-plane Loading in y-x-plane
NEd = -350,00 kN
qz,Ed
(*)
= 5,00 kN/m qy,Ed
(*)
= 4,50 kN/m Mz,Ed,max= 14,06 kNm
My,left,Ed = 250,00 kNm Mz,left,Ed = 0,00 kNm
My,right,Ed = 0,00 kNm Mz,right,Ed = 0,00 kNm
Pz,Ed
(*)
= 0,00 kN Py,Ed
(*)
= 0,00 kN
Distance of Loading to shear center
zSi = -250,00 mm
(*)
Mcr = 0,00 kNm Vy,Ed,max= 14,06 kN
Mcr,0 = 0,00 kNm
Specify path of LTBeam.exe file:
Definition of axes
oz,max= 22,15 mm
Worked examples
oy,max= 3,39 mm
oz [mm] (corresponding to design loads)
oy [mm] (corresponding to design loads)
SEMICOMP Member Design My,Ed [kNm]
Vz,Ed [kN]
Mz,Ed [kNm]
Vy,Ed [kN]
Partial factors M
C:\Users\Martin\LTBeam\LTBeam.exe
0,00
500,00
-100,00
0,00
-20,00
0,00
20,00
-20,00
0,00
20,00
0,00
50,00
0,00
5,00
Semi-Comp Member Design (Windows OS)
SHEET 1.1 SYSTEM DATA
Structural Analysis
Exercise 2 Semi-Comp+
Cross-section type
Finishing
M0
= 1,00
Select from library
M1
= 1,00
(optional)
Cross-section data Material
H = 500,0 [mm] Steel grade S235
B =
200,0 [mm]
f
y
= 235,0 N/mm
Tw = 10,2 [mm] E = 210000,0 N/mm
Tf = 16,0 [mm]
R = 21,0 [mm]
A [cm
2
] I
yy
[cm
4
] I
zz
[cm
4
] W
el,y
[cm
3
] W
el,z
[cm
3
] W
pl,y
[cm
3
] W
pl,z
[cm
3
]
115,52 48198,53 2141,69 1927,94 214,17 2194,12 335,88
SEMICOMP Member Design
Partial factors
M
Profile can be chosen from library or entered manually
List available CS-Types and Finishing types (Why does the finishing type have
to be considered)
Set partial factors. Set steel grade and E-modulus
CS- Values are calculated automatically
Semi-Comp Member Design (Windows OS)
SHEET 1.1 SYSTEM DATA
Structural Analysis
Exercise 2 Semi-Comp+
Definition of axes
Worked examples
Worked examples from
ECCS TC 8 Publication No. 119
SEMI-COMP final report included
Design of Members
Exercise 4 Semi-Comp+
Semi-Comp Member Design (Windows OS)
SHEET 1.1 SYSTEM DATA
Calculation using:
- Semi-Comp Member Design (Windows OS)
Structural Analysis
Exercise 2 Semi-Comp+
1. General Criteria
2. Tension
3. Cross Section Resistance
4. Torsion
5. Columns
6. Beams
7. Beam-columns
Design of Members
Main internal forces
Compression+Bending+Shear
Bending+Shear
Tension/Compression
Bending+Shear
Bending+Shear
Compression+Bending+Shear
Tension/Compression
Tension/Compression
Compression+Bending+Shear
Torsion Less common in steel structures
Design of Members
General Criteria
Design of Members
General Criteria
RESISTANCE OF CROSS SECTIONS
1 3
2
0 0
,
0
,
2
0
,
2
0
,
s
|
|
.
|
\
|
+
|
|
.
|
\
|
|
|
.
|
\
|
|
|
.
|
\
|
+
|
|
.
|
\
|
M y
Ed
M y
Ed z
M y
Ed x
M y
Ed z
M y
Ed x
f f f f f
t
o
1
,
,
,
,
s + +
Rd z
Ed z
Rd y
Ed y
Rd
Ed
M
M
M
M
N
N
Elastic resistance (6.2.1(5) do EC3-1-1):
Linearized interaction formulas (6.2.1(7) do EC3-1-1):
Nonlinear interaction formulas class 1/2 (6.2.1(6) do EC3-1-1):
BUCKLING RESISTANCE
Section properties gross and net section (6.2.2 do EC3-1-1):
G
G
e
N
G
Class 4 section in compression
G
G
Classe 4 section in bending
Design of Members
General Criteria
Classification and global analysis:
Limit class 1/2: class 1 plastic global analysis allowed
class 2 elastic global analysis
Limit class 3/4: class 3 elastic global analysis
class 4* elastic global analysis with
A,I (gross section) if > 0,5
A
eff
, I
eff
(effective section) if s 0,5
* Remark for class 4:
provided that real stresses are determined o
com,Ed
may be used for
c/t (o
com,Ed
) for calculation of the limit 3/4
and for calculation of A
eff
, I
eff p,red
Design of Members
General Criteria
Classification and cross-section design
Limit 3/4 class 4: effective A
eff
, W
eff
class 3: elastic W
el
Limit 2/3 class 2: full plastic W
pl
* Remark for class 4:
A
eff
, W
eff
may be based on different stress states:
pure N for A
eff
pure M for W
eff
or (N
Ed
+ M
y,Ed
+ M
z,Ed
) for A
eff
, W
eff
Iteration needed
Design of Members
General Criteria
Classification and member buckling design
To decide whether the elastic or plastic format of the beam-column
formulae should be used
N
Rd
, M
y,Rd
, k
yy
, k
zy
, _
LT
depend on the
cross-section classes: plastic behaviour in class 1,2
elastic behaviour in Class 3,4
+ s
_ _
y.Ed
Ed
yy
y pl .Rd LT y.Rd
M
N
k
N M
1
+ s
_ _
y.Ed
Ed
zy
z pl .Rd LT y.Rd
M
N
k
N M
1
(6.61)
(6.62)
Design of Members
General Criteria
Classification and member buckling design
Equivalent section class
Non-uniform loading states in the sections along the member
different classes
Need to define an equivalent cross-section
Proposal to consider the worst cross-section as the
decisive one ...
... and to define it as the one of maximum utilisation
Utilisation factor to be evaluated in 10
sections along the member length
Design of Members
General Criteria
Classification and member buckling design
Equivalent section class:
Example:
UF class
Use the class at maximum of
UF for the member check
UF utilisation factor
Design of Members
General Criteria
Classification and member buckling design
Utilisation factor UF:
Remark: e.g.
N
M
-
= = =
Ed Ed Ed
Rd N,Rd N,Rd
R M M
UF
R M M
Ed
M
Ed
N
-
N,Rd
M
N,Rd
M
Ed
N
UF
pl
N
pl
M
UF ... utilisation factor
R
Ed
R
Rd
2
Ed
Ed Ed
N,Rd pl,Rd
pl,Rd
N
M M
1 0,50 does not meanUF 0,50
M M
N
| |
| |
|
= = =
|
| |
|
\ .
\ .
Design of Members
General Criteria
1. General Criteria
2. Tension
3. Cross Section Resistance
4. Torsion
5. Columns
6. Beams
7. Beam-columns
Design of Members
Typical tension members
Typical cross sections in tension
Stress concentrations Eccentric joints
Reduction of resistance: Holes in bolted joints; 2nd order moments.
Welded tubular joints
Structural Analysis
Tension
0 . 1
.
s
Rd t
Ed
N
N
0 M y Rd . pl
f A N =
2 .
9 . 0
M u net Rd u
f A N =
Design plastic resistance of the gross cross section
Design ultimate resistance of the net cross section at holes for fasteners
(
M2
=1.25)
Design Resistance (6.2.3 do EC3-1-1):
(
M0
= 1.0)
0
2
. .
9 . 0
M
M
u
y
net
Rd pl Rd u
f
f
A
A
N N
0
d t A
|
|
.
|
\
|
+
i
i
p
s
t d t A
4
2
0
A
net
= min of:
(line 1)
t = plate thickness and d
o
= hole diameter.
Net area (A
net
)
(line 2)
s s
p
p
1
2
Structural Analysis
Tension
( )
2 M
u 0 2
Rd . u
f t d 5 . 0 e 0 . 2
N
=
2 M
u net 2
Rd . u
f A
N
|
=
2 M
u net 3
Rd . u
f A
N
|
=
d
0
e
2
e
1
e
1
e
1
p
1
p
1
p
1
(1 bolt)
(2 bolts)
(3 bolts)
Eccentricities in angles and U sections (3.10.1 do EC3-1-8).
Structural Analysis
Tension
a) Welded joints
Structural Analysis
Tension
Design chord AB for a design tensile axial force of N
Ed
= 220 kN
assuming two distinct possibilities for the connections:
a) welded connections;
b) bolted connections.
The cross section consists of two angles of equal legs, in steel grade S 235.
EXAMPLE 7
Structural Analysis
Tension
Design of Members
Exercise 3a
Design the lattice girder in steel grade S 275, supporting a
reinforced concrete floor. The loading, applied on the floor
and transmitted to the truss as concentrated loads applied in
the nodes, is defined by the following distributed loads.
Permanent action on the floor = 5.75 kN/m
2
(
G
= 1.35);
Variable action on the floor = 4.00 kN/m
2
(
Q
= 1.50).
Use:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
Design of Members
Exercise 3a
Use:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
The distance between lattice girders is 3.00 m; the nodes of
the truss are braced in the perpendicular direction to the
plane of the structure; the loading already includes the self
weight of the steel truss.
Design the members of the truss, assuming the following
alternatives:
a) Square hollow sections (SHS), and welded connections for
the members of the structure.
b) HEA profiles in the upper and lower chords (horizontal
members) and sections built up from 2 UPN channels in the
diagonals, bolted to gusset plates welded to the HEA profiles
in the upper and lower chords.
Design of Members
Exercise 3a
Design of Members
Exercise 3a
Design of Members
Exercise 3a
Calculation using:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
Structural Analysis
Exercise 3a
1. General Criteria
2. Tension
3. Cross Section Resistance
4. Torsion
5. Columns
6. Beams
7. Beam-columns
Design of Members
e.n.
M
c f
y
c f
y
f
y
f
y
f
y
f
y
M = M
el
M
el
< M < M
pl
M = M
pl
Shape factor: m = M
pl
/ M
el
Design of Members
Cross Section Resistance
ELASTIC AND PLASTIC RESISTANCE
I SECTION
y pl y y pl
f W f S d f
A
M = = = 2
2
2
d
c
e.n. elstico
M
f
y
f
y
f
y
f
y
M = M
el
M = M
pl
M
<f
y
f
y
f
y
f
y
M = M
el
M = M
pl
Seco I ou H
Seco em T
G
G
e.n. elstico
e.n. plstico
A
t
A
c
d
d
d
t
e.n. plstico
d
c
e.n. elstico
M
f
y
f
y
f
y
f
y
M = M
el
M = M
pl
M
<f
y
f
y
f
y
f
y
M = M
el
M = M
pl
Seco I ou H
Seco em T
G
G
e.n. elstico
e.n. plstico
A
t
A
c
d
d
d
t
e.n. plstico
( )
y pl y t c
t y t c y c pl
f W f S S
d f A d f A M
= + =
= + =
Design of Members
Cross Section Resistance
PLASTIC MOMENT RESISTANCE
T SECTION
0 . 1
.
s
Rd c
Ed
M
M
(clause 6.2.5 do EC3-1-1):
0 M y pl Rd . c
f W M =
0 min . . M y el Rd c
f W M =
0 min . . M y eff Rd c
f W M =
Class 1 or 2
Class 3
Class 4
Holes in tensile flange:
0 2 .
9 . 0
M y f M u net f
f A f A >
BENDING RESISTANCE
Design of Members
Cross Section Resistance
y
G
V
Ed
z
A
v
(clause 6.2.6 do EC3-1-1)
0 . 1
,
s
Rd c
Ed
V
V
PLASTIC RESISTANCE
V
pl.Rd
:
SHEAR
Design of Members
Cross Section Resistance
ELASTIC RESISTANCE
( )
0 . 1
3
0
s
M y
Ed
f
t
t I
S V
Ed
Ed
= t
e.n.
M
y
o
Tenses normais o
G
V
z
t
Tenses tangenciais t
t
( )
0 M y v Rd . pl
3 f A V =
BENDING AND SHEAR
Design of Members
Cross Section Resistance
Design of Members
Cross Section Resistance
I and H SECTIONS
PLASTIC INTERACTION BENDING AND SHEAR (approx.)
e.n.
M
y
o
Tenses normais o
G
V
z
t
Tenses tangenciais t
t
t
w
M
y
t
f
yr
f
yr
t (V
z
) o (M
y
)
V
z
h
m
z
y
Modelo 1
+
f
y
f
y
z pl
z
V
V
,
y pl
y
M
M
,
0
0
1.0
1.0
IPE ou HEA
IPE
HEA
(clause 6.2.8 EC3-1-1)
% 50 s
Ed
V
% 50 >
Ed
V
( )
y
f 1
( )
2
.
1 2 =
Rd pl Ed
V V
0
2
, . ,
4
M
y
w
w
y pl Rd V y
f
t
A
W M
|
|
.
|
\
|
= Rd c y Rd V y
M M
. , . ,
s
w w W
t h A =
NO REDUCTION
REDUCED MOMENT M
V.Rd
,
I and H SECTIONS
BENDING AND SHEAR INTERACTION
Design of Members
Cross Section Resistance
RECTANGULAR SECTION
f
y
M
y
f
y
N
f
y
a) 0 =
y
M , 0 = N ;
y y pl y
f S M M 2
,
= = b) 0 = N , 0 =
y
M ;
y pl
f A N N = =
M
y
z
en
N
f
y
f
y
2 f
y
+
f
y
f
y
c) 0 =
y
M , 0 = N
Design of Members
Cross Section Resistance
PLASTIC INTERACTION BENDING AND AXIAL FORCE
y pl
y
M
M
,
,
z pl
z
M
M
,
y pl
y
M
M
,
pl
N
N
pl
N
N
0
0
1.0
1.0 1.0
0
0
1.0
Seco
rectangular
HEA
Eixo de menor inrcia - z
Eixo de maior inrcia - y
Design of Members
Cross Section Resistance
PLASTIC INTERACTION BENDING AND AXIAL FORCE
I and H SECTIONS
Rd N Ed
M M
,
s
(
(
|
|
.
|
\
|
=
2
, ,
1
p
Rd pl Rd N
N
N
M M
-fy
N
M
M
pl
M
el
N
pl
+fy
-fy
-fy
+fy
-fy
M
N
-fy
Rectangular
section
BENDING AND AXIAL FORCE
Design of Members
Cross Section Resistance
Double-symmetric I or H sections
a
n
M M
Rd y pl Rd y N
=
5 . 0 1
1
, , , , Rd y pl Rd y N
M M
, , , ,
s
Rd z pl Rd z N
M M
, , , ,
=
a n s
(
(
|
.
|
\
|
=
2
, , , ,
1
1
a
a n
M M
Rd z pl Rd z N
a n >
( ) A t b A a
f
= 2
5 . 0 s a
but
if
if
but
Rd pl Ed
N N n
.
=
Rd pl Ed
N N
,
25 . 0 s
0
5 . 0
M y w w Ed
f t h N s
0 M y w w Ed
f t h N s
No reduction if
Circular hollow
sections
( )
7 . 1
, , , , ,
1 n M M M
Rd pl Rd z N Rd y N
= =
w
Rd y pl Rd y N
a
n
M M
=
5 . 0 1
1
, , , ,
f
Rd z pl Rd z N
a
n
M M
=
5 . 0 1
1
, , , ,
Rd y pl Rd y N
M M
, , , ,
s
Rd y pl Rd y N
M M
, , , ,
s
Rectangular hollow sections
but
z
en
y
-f
y
z
+f
y
e.n.
b/2
b/2
h/2
h/2
=
-f
y
+f
y
+
+2f
y
-2f
y
Design of Members
Cross Section Resistance
RECTANGULAR SECTION
PLASTIC INTERACTION BIAXIAL BENDING
0 . 1
. ,
,
. ,
,
s
(
(
+
(
(
| o
Rd z pl
Ed z
Rd y pl
Ed y
M
M
M
M
Class 3 or 4:
0
,
M
y
Ed x
f
o s
(6.2.9 do EC3-1-1)
Class 1 or 2:
BIAXIAL BENDING
Design of Members
Cross Section Resistance
BIAXIAL BENDING AND AXIAL FORCE
0 . 1
. ,
,
. ,
,
s
(
+
(
(
| o
Rd z N
Ed z
Rd y N
Ed y
M
M
M
M
n 5 ; 2 = = | o
2 = = | o
2
13 . 1 1
66 . 1
n
= = | o
6 s = | o
1 > |
but
but
I or H
Circular hollow sections
Rectangular hollow sections
Rd pl Ed
N N n
,
=
Remark: e.g.
N
M
-
= = =
Ed Ed Ed
Rd N,Rd N,Rd
R M M
UF
R M M
Ed
M
Ed
N
-
N,Rd
M
N,Rd
M
Ed
N
UF
pl
N
pl
M
UF ... utilisation factor
R
Ed
R
Rd
2
Ed
Ed Ed
N,Rd pl,Rd
pl,Rd
N
M M
1 0,50 does not meanUF 0,50
M M
N
| |
| |
|
= = =
|
| |
|
\ .
\ .
Design of Members
Cross Section Resistance
UTILISATION FACTOR : UF
Overall Member class
Equation for interaction equals 1.0 when load factor 1/q is
applied to the loading:
, ,
, , , ,
1.0
y Ed z Ed
pl y Rd pl z Rd
M M
M M
o |
q q
| | | |
+ =
| |
| |
\ . \ .
Design of Members
Cross Section Resistance
EVALUATION OF UTILISATION FACTOR
How should the plastic limit state be reached?
1. | N and | M
2. N = const. and | M
3. M = const. and | N
(| = increasing factor)
Design of Members
Cross Section Resistance
EVALUATION OF UTILISATION FACTOR
( )
2
2 2
, ,
2
, ,
4
1 1 1
4
2 2
Ed y pl w w y Ed
Ed
w
Ed w w y Ed y Ed w
N W c t M
N
c
N c c M M t
o
| |
| |
|
= + + +
|
| |
|
\ .
\ .
Increase N and M
y
:
( )
1
2 2
Ed
f
w w y
h N
t r
c t f
o
| |
= + +
|
|
\ .
Only increase M
y
:
+
M
y,Ed
N
Ed
Example: I-Profile with N + M
y
Design of Members
Cross Section Resistance
EVALUATION OF UTILISATION FACTOR
Classification: no problem to find ENA for class 3/4 border
Class 1/2 and 2/3 border: Plastic neutral axis (PNA)
Analytical solutions available, except for axial force + biaxial bending
Numerical solution for N+M
y
+M
z
My,Ed
-
-
-
N
Ed
Mz,Ed
+
o
com,Ed
Mpl,y
Npl
Mpl,z
(plast.)
NEd, My,Ed, Mz,Ed
+
-
+
-
-
Design of Members
Cross Section Resistance
EVALUATION OF UTILISATION FACTOR
Iterative determination of the PNA
Design of Members
Cross Section Resistance
EVALUATION OF UTILISATION FACTOR
Additional tool PNA-Iteration
z y
m m n
! !
= =
Boundary condition:
,( , )
1
Ed
pl p
N
n
N
o
|
= =
,
, ,( , )
1
y Ed
y
y pl p
M
m
M
o
|
= =
,
, ,( , )
1
z Ed
z
z pl p
M
m
M
o
|
= =
,( , ) Ed pl p
N N
o
| =
, , ,( , ) y Ed y pl p
M M
o
| =
, , ,( , ) z Ed z pl p
M M
o
| =
Design of Members
Cross Section Resistance
EVALUATION OF UTILISATION FACTOR
Comparison of INCA2.8 and SEMI-COMP results (PNA-Iteration)
IPE 500 S235
N
Ed
= -350kN
M
y,Ed
= 245,56kNm
M
z,Ed
= 7,25kNm
INCA 2.8 SEMI-COMP
o-web 0.744 0.734
Very good correspondence (1.34% difference)
Design of Members
Cross Section Resistance
EVALUATION OF UTILISATION FACTOR
Example: Equivalent cross-section class
IPE 500 S235
350
350
350
350
350
350
350
350
350
350
350
1
2
3
4
5
6
7
8
9
10
11
P
o
s
i
t
i
o
n
N
250
247,5
240
227,5
210
187,5
160
127,5
90
47,5
0
My
0,0
-6,2
-7,9
-5,1
2,3
14,0
2,3
-5,1
-7,9
-6,2
0,0
Mz
N
Ed
M
y
II
Ed
N
Ed
L
=
2
x
5
=
1
0
m
q
z,Ed
q
y,Ed
[kN] [kNm] [kNm]
Design of Members
Cross Section Resistance
EVALUATION OF UTILISATION FACTOR
Example contd.: Equivalent cross-section class
member buckling
check acc. class 3
(class at location
with highest
utilisation factor)
0.51
0.52
0.52
0.73
0.64
0.82
0.53
0.51
0.48
0.36
0.13
1
2
3
4
5
6
7
8
9
10
11
0.0 0.2 0.4 0.6 0.8 1.0
P
o
s
i
t
i
o
n
UF (EN 1993-1-1 with new c/t-limits for web)
UF
UF-EP
UF-EE
2
2
2
3
3
3
3
3
3
3
4
1
2
3
4
5
6
7
8
9
10
11
P
o
s
i
t
i
o
n
Class
Design of Members
Cross Section Resistance
EVALUATION OF UTILISATION FACTOR
Design of Members
Exercise 4
Evaluate the cross sectional resistance and the Utilization
Factor for the following case:
Use:
- Semi-Comp Software (Windows OS)
Example 1: HEA 300 S355
Section in compression and biaxial bending N+M
y
+M
z
:
,
, , , , ,
, , , ,
,
500 ( 0.125)
200 ( 0.407)
( 0.439)
100
= = =
= = =
= =
=
Ed Ed pl Rd
y Ed pl y y Ed pl y Rd
pl z z Ed pl z Rd
z Ed
N kN n N N
M kNm m M M
m M M
M kNm
h
b
r
t
f
t
w
HEA 300 S355
Example 1: HEA 300 S355
Section in compression and biaxial bending N+M
y
+M
z
:
h
b
r
t
f
t
w
Cross-section type
Finishing
M0
= 1,00
Select from library
M1
= 1,00
(optional)
Cross-section data Material
H = 500,0 [mm] Steel grade S235
B =
200,0 [mm]
f
y
= 235,0 N/mm
Tw = 10,2 [mm] E = 210000,0 N/mm
Tf = 16,0 [mm]
R = 21,0 [mm]
A [cm
2
] I
yy
[cm
4
] I
zz
[cm
4
] W
el,y
[cm
3
] W
el,z
[cm
3
] W
pl,y
[cm
3
] W
pl,z
[cm
3
]
115,52 48198,53 2141,69 1927,94 214,17 2194,12 335,88
SEMICOMP Member Design
Partial factors
M
Profile can be chosen from library or entered manually
List available CS-Types and Finishing types (Why does the finishing type have
to be considered)
Set partial factors. Set steel grade and E-modulus
CS- Values are calculated automatically
Design of Members
Exercise 4 Semi-Comp+
Semi-Comp Member Design (Windows OS)
SHEET 1.1 SYSTEM DATA
Section classification for member design check
Additional information
Point No.
x-coordinate
[m]
N
Ed
[kN]
M
y,Ed
[kNm]
M
z,Ed
[kNm]
Utilization
Factor
CS-Class
1 0,00 -350,00 250,00 0,00 0,506 2
2 0,17 -350,00 249,93 1,34 0,508 1
3 0,33 -350,00 249,72 2,56 0,511 1
4 0,50 -350,00 249,38 3,66 0,514 1
5 0,67 -350,00 248,89 4,63 0,517 1
6 0,83 -350,00 248,26 5,47 0,519 1
7 1,00 -350,00 247,50 6,19 0,521 1
8 1,17 -350,00 246,60 6,78 0,522 1
9 1,33 -350,00 245,56 7,25 0,523 1
10 1,50 -350,00 244,38 7,59 0,522 1
11 1,67 -350,00 243,06 7,81 0,521 1
12 1,83 -350,00 241,60 7,91 0,519 1
13 2,00 -350,00 240,00 7,88 0,516 1
14 2,17 -350,00 238,26 7,72 0,513 1
15 2,33 -350,00 236,39 7,44 0,508 1
Cross-section checks are performed at 61 points in the
beam
Design of Members
Exercise 4 Semi-Comp+
Semi-Comp Member Design (Windows OS)
SHEET 1.2.1 ADDITIONAL INFO
Choose method
NEd = -350,000 kN
My,Ed = 245,560 kNm Vz,Ed = 15,000 kN
Mz,Ed = 7,250 kNm Vy,Ed = 5,000 kN
Reference values for classification
c/tw = 41,765 o web = 0,734 web = -0,563 c = 1,000
c/tf = 4,619 o f lange = 1,000 f lange = 0,869
Boundaries Class 1 Class 2 Class 3
c/tw c/tw, max = 46,361 53,386 86,768
c/tf c/tf , max = 9,000 10,000 13,935
Cross section class = 1
Mpl,y,Rd= 515,62 kNm Mpl,z,Rd= 78,93 kNm Vpl,z,Rd= 812,35 kN
Mel,y,Rd= 453,07 kNm Mel,z,Rd= 50,33 kNm Vpl,z,Rd= 868,33 kN
NRd= 2714,76 kN
EN 1993-1-1, 6.2.3 / 6.2.4
Tension or compression
U(6.5)/(6.9) = 0,129 1,0 ok
EN 1993-1-1, 6.2.5
Bending moment
Ustrong A. = 0,476 1,0 ok
Uweak A. = 0,092 1,0 ok
EN 1993-1-1, 6.2.6
Shear
Uz-plastic= 0,018 1,0 ok Uy-plastic= 0,006 1,0 ok
U z-elastic= 0,025 1,0 ok U y-elastic= 0,015 1,0 ok
hw/tw = 47,020 72c /q = 60,000 ok
EN 1993-1-1, 6.2.9
Bending and axial force
U(6.41)/(6.42)= 0,319 1,0 ok UF = 0,523
Comment: Shear Interaction is not considered within this
software tool. If the utilization for shear is higher that 0.5,
an additional check for shear interaction is necessary.
SEMICOMP Cross-Section Check
Additional cross-section
check can be performed
Independent from member
check sheet
Design of Members
Exercise 4 Semi-Comp+
Semi-Comp Member Design (Windows OS)
SHEET 1.3 CROSS SECTION CHECK
Define cross-section loading and choose cross-
section resistance method
Press button to start calculation
Design of Members
Exercise 4 Semi-Comp+
Semi-Comp Member Design (Windows OS)
SHEET 1.3 CROSS SECTION CHECK
Choose method
NEd = -350,000 kN
My,Ed = 245,560 kNm Vz,Ed = 15,000 kN
Mz,Ed = 7,250 kNm Vy,Ed = 5,000 kN
Reference values for classification
c/tw = 41,765 o web = 0,734 web = -0,563 c = 1,000
c/tf = 4,619 o f lange = 1,000 f lange = 0,869
Boundaries Class 1 Class 2 Class 3
c/tw c/tw, max = 46,361 53,386 86,768
c/tf c/tf , max = 9,000 10,000 13,935
Cross section class = 1
Mpl,y,Rd= 515,62 kNm Mpl,z,Rd= 78,93 kNm Vpl,z,Rd= 812,35 kN
Mel,y,Rd= 453,07 kNm Mel,z,Rd= 50,33 kNm Vpl,z,Rd= 868,33 kN
NRd= 2714,76 kN
EN 1993-1-1, 6.2.3 / 6.2.4
Tension or compression
U(6.5)/(6.9) = 0,129 1,0 ok
EN 1993-1-1, 6.2.5
Bending moment
Ustrong A. = 0,476 1,0 ok
Uweak A. = 0,092 1,0 ok
EN 1993-1-1, 6.2.6
Shear
Uz-plastic= 0,018 1,0 ok Uy-plastic= 0,006 1,0 ok
U z-elastic= 0,025 1,0 ok U y-elastic= 0,015 1,0 ok
hw/tw = 47,020 72c /q = 60,000 ok
EN 1993-1-1, 6.2.9
Bending and axial force
U(6.41)/(6.42)= 0,319 1,0 ok UF = 0,523
Comment: Shear Interaction is not considered within this
software tool. If the utilization for shear is higher that 0.5,
an additional check for shear interaction is necessary.
SEMICOMP Cross-Section Check
Design of Members
Exercise 4 Semi-Comp+
Semi-Comp Member Design (Windows OS)
SHEET 1.3 CROSS SECTION CHECK
Classification of the cross-section
Note: If the SEMI-COMP method is chosen, modified
classification boundaries are used!
Reference values for classification
c/t
w
= 41,765 o
web
= 0,734
web
= -0,563 c = 1,000
c/t
f
= 4,619 o
flange
= 1,000
flange
= 0,869
Boundaries Class 1 Class 2 Class 3
c/t
w
c/t
w, max
= 46,361 53,386 86,768
c/t
f
c/t
f, max
= 9,000 10,000 13,935
Cross section class = 1
Design of Members
Exercise 4 Semi-Comp+
Semi-Comp Member Design (Windows OS)
SHEET 1.3 CROSS SECTION CHECK
Choose method
NEd = -350,000 kN
My,Ed = 245,560 kNm Vz,Ed = 15,000 kN
Mz,Ed = 7,250 kNm Vy,Ed = 5,000 kN
Reference values for classification
c/tw = 41,765 o web = 0,734 web = -0,563 c = 1,000
c/tf = 4,619 o f lange = 1,000 f lange = 0,869
Boundaries Class 1 Class 2 Class 3
c/tw c/tw, max = 46,361 53,386 86,768
c/tf c/tf , max = 9,000 10,000 13,935
Cross section class = 1
Mpl,y,Rd= 515,62 kNm Mpl,z,Rd= 78,93 kNm Vpl,z,Rd= 812,35 kN
Mel,y,Rd= 453,07 kNm Mel,z,Rd= 50,33 kNm Vpl,z,Rd= 868,33 kN
NRd= 2714,76 kN
EN 1993-1-1, 6.2.3 / 6.2.4
Tension or compression
U(6.5)/(6.9) = 0,129 1,0 ok
EN 1993-1-1, 6.2.5
Bending moment
Ustrong A. = 0,476 1,0 ok
Uweak A. = 0,092 1,0 ok
EN 1993-1-1, 6.2.6
Shear
Uz-plastic= 0,018 1,0 ok Uy-plastic= 0,006 1,0 ok
U z-elastic= 0,025 1,0 ok U y-elastic= 0,015 1,0 ok
hw/tw = 47,020 72c /q = 60,000 ok
EN 1993-1-1, 6.2.9
Bending and axial force
U(6.41)/(6.42)= 0,319 1,0 ok UF = 0,523
Comment: Shear Interaction is not considered within this
software tool. If the utilization for shear is higher that 0.5,
an additional check for shear interaction is necessary.
SEMICOMP Cross-Section Check
Design of Members
Exercise 4 Semi-Comp+
Semi-Comp Member Design (Windows OS)
SHEET 1.3 CROSS SECTION CHECK
Utilization factors
for all checks
Shear interaction
not considered
M
pl,y,Rd
=
515,62 kNm
M
pl,z,Rd
=
78,93 kNm
V
pl,z,Rd
=
812,35 kN
M
el,y,Rd
=
453,07 kNm
M
el,z,Rd
=
50,33 kNm
V
pl,z,Rd
=
868,33 kN
N
Rd
= 2714,76 kN
EN 1993-1-1, 6.2.3 / 6.2.4
Tension or compression
U
(6.5)/(6.9)
=
0,129 1,0 ok
EN 1993-1-1, 6.2.5
Bending moment
U
strong A.
=
0,476 1,0 ok
U
weak A.
=
0,092 1,0 ok
EN 1993-1-1, 6.2.6
Shear
U
z-plastic
=
0,018 1,0 ok
U
y-plastic
=
0,006 1,0 ok
U
z-elastic
=
0,025 1,0 ok
U
y-elastic
=
0,015 1,0 ok
h
w
/t
w
=
47,020
72c /q =
60,000 ok
EN 1993-1-1, 6.2.9
Bending and axial force
U
(6.41)/(6.42)
=
0,319 1,0 ok UF = 0,523
Comment: Shear Interaction is not considered within this
software tool. If the utilization for shear is higher that 0.5,
an additional check for shear interaction is necessary.
Design of Members
Exercise 4 Semi-Comp+
Semi-Comp Member Design (Windows OS)
SHEET 1.3 CROSS SECTION CHECK
Calculation using:
- Semi-Comp Member Design (Windows OS)
Structural Analysis
Exercise 4 Semi-Comp+
Calculation using:
- Semi-Comp Member Design (Windows OS)
Structural Analysis
Exercise 4 Semi-Comp+
1. General Criteria
2. Tension
3. Cross Section Resistance
4. Torsion
5. Columns
6. Beams
7. Beam-columns
Design of Members
St. Venant Torsion
Non-uniform Torsion:
T=T
t
x
z
T
T
t
C
y
z
C
T=T
t
+T
w
x
z
T
w
T
t
T
L
h
m
y
w t
T T T + =
t
T T =
Structural Analysis
Torsion
ST. VENANT TORSION
T
t
I G
T
dx
d
=
t
t
T
t
R
T
t
t
T
t
t
t
t
i
C
C
t
t
C
t
t,mx
t
i,mx
Structural Analysis
Torsion
v
sup
(x)
h
m
/2
h
m
/2
v
inf
(x)
(x)
z
y
T
V
sup
x
z
T
x
h
m
o
w
t
w
M
sup
M
inf
V
inf
y
( ) ( )
2
sup
m
h
x x v =
( )
fz
I E
dx
x v d
dx
dM
V
3
sup
3
sup
sup
= =
( )
2
2
2
sup
m
fz
h
I E
dx
x d
M
=
( )
( )
2
3
3
3
sup
3
sup
m
fz fz
h
I E
dx
x d
I E
dx
x v d
V
= =
( ) ( )
3
3
2
3
3
sup
2 dx
x d
I E
h
I E
dx
x d
h V T
W
m
fz m w
= = =
m
h M B
sup
=
(bi-moment)
m m w
h V h V T
inf sup
= =
Structural Analysis
Torsion
NON-UNIFORM TORSION
( ) 2
2
m fz W
h I I =
4
2
m z W
h I I ~
( )
2
1
m z f f W
h I I | | =
( )
ft fc fc f
I I I + = |
( )
( ) ( )
3
3
) (
dx
x d
I E
dx
x d
I G x T x T T
W T w t
= + =
0 . 1 s
Rd
Ed
T
T
DESIGN FOR TORSION (clause 6.2.7 EC3-1-1)
Ed w Ed t Ed
T T T
, ,
+ =
Structural Analysis
Torsion
NON-UNIFORM TORSION
1. General Criteria
2. Tension
3. Cross Section Resistance
4. Torsion
5. Columns
6. Beams
7. Beam-columns
Design of Members
y(x)
N
y(x)
L
N
y
x
N
0
N
cr
(z)
0
2
2
= + y N
dx
y d
I E
) ( cos ) ( sin
2 1
x k D x k D y + =
2
2
L
I E
N
cr
t
=
( ) I E N k =
2
0 0 ) 0 (
2
= = = D x y
t n L k D
L k D L x y
= v =
= = =
0
0 ) ( sin 0 ) (
1
1
I E
N
L
n
k n L k = = =
2
2 2
2
t
t
Design of Members
Columns
BUCKLING
Buckling length
- Isolated members
L
0.7 L
2 L
0.5 L
L
2
2
E
cr
L
I E
N
t
=
Buckling length trussed structures (Annex BB.1 EC3-1-1)
Buckling length frames
Design of Members
Columns
BUCKLING
C G
G
y
z
C
|
|
.
|
\
|
+ =
2
2
2
,
1
ET
W
T
C
T cr
L
I E
I G
i
N
t
( ) ( )
(
+ + =
T cr y cr T cr y cr T cr y cr TF cr
N N N N N N N
, ,
2
, , , , ,
4
2
1
|
|
( ) A I I y i
z y C C
+ + =
2 2
( )
2
1
C C
i y = |
Design of Members
Columns
BUCKLING
Torsional or Flexural-torsional Buckling Load
2
2
t
o
E
=
A
N
= o
i
L
E
=
c
E
o
t =
1
c
o o =
cr
y
N
Af
= =
1
2
2
2
2
t t
o
E
L A
I E
E
cr
=
=
i L
E
=
A I i =
1
=
Critical stress
Design of Members
Columns
BUCKLING
Imperfections
e
0
e
0
y(x)
L
N
y
x
N
(z)
e
0
e
N
e
N
cr
0.3o
c
0.2o
c
Compresso
Traco
c
o
o
1.0
1.0
Curva de Euler
0 . 1
,
s
Rd c
Ed
N
N
0 , M y Rd c
f A N =
0 , M y eff Rd c
f A N =
(Class 1, 2 or 3)
(Class 4)
1 . M y Rd b
f A N _ =
1 . M y eff Rd b
f A N _ =
Design of Members
Columns
BUCKLING RESISTANCE
(Class 1, 2 or 3)
(Class 4)
cr y T
N f A =
cr y eff T
N f A =
Torsional or Flexural-torsional Resistance
2
2
1
| |
_
+
=
0 . 1 s _ but
( )
(
+ + =
2
2 . 0 1 5 . 0 o |
1
1
= =
i
L
N f A
cr
cr y
1
A A
i
L
N f A
eff
cr
cr y eff
= =
(Class 1, 2 or 3)
(Class 4)
c t = = 9 . 93
1 y
f E
y
f 235 = c
2 . 0 s
04 . 0 s
cr Ed
N N Neglect BUCKLING if: ou
Design of Members
Columns
BUCKLING RESISTANCE
Design of Members
Exercise 3b
Design the lattice girder in steel grade S 275, supporting a
reinforced concrete floor. The loading, applied on the floor
and transmitted to the truss as concentrated loads applied in
the nodes, is defined by the following distributed loads.
Permanent action on the floor = 5.75 kN/m
2
(
G
= 1.35);
Variable action on the floor = 4.00 kN/m
2
(
Q
= 1.50).
Use:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
Design of Members
Exercise 3b
Use:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
The distance between lattice girders is 3.00 m; the nodes of
the truss are braced in the perpendicular direction to the
plane of the structure; the loading already includes the self
weight of the steel truss.
Design the members of the truss, assuming the following
alternatives:
a) Square hollow sections (SHS), and welded connections for
the members of the structure.
b) HEA profiles in the upper and lower chords (horizontal
members) and sections built up from 2 UPN channels in the
diagonals, bolted to gusset plates welded to the HEA profiles
in the upper and lower chords.
Design of Members
Exercise 3b
Design of Members
Exercise 3b
Design of Members
Exercise 3b
Design of Members
Exercise 3b
Design of Members
Exercise 3b
Design of Members
Exercise 3b
Design of Members
Exercise 2b
Design of Members
Exercise 2b
Design of Members
Exercise 3b
Calculation using:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
Calculation of HE 180 A Compression resistance
SUMMARY
Cross-section: HE 180 A
Loading: NEd = 742.60 kN
Steel Grade: S275 J0
M0: 1.00
M1: 1.00
Member length: L = 3.00 [m]
Design resistance: NRd = 851.73 kN
The column is safe!
RESULTS
Cross-section class
Internal compression parts
c/t
lim
: 66.56
c/t: 20.33
Class: 1
Outstand flanges
c/t
lim
: 66.56
c/t: 7.58
Class: 1
Resultant cross-section class: 1
Type of Verification: Plastic
Cross-section resistance
N
c,Rd
[kN]: 1244.38
Design of Members
Exercise 3b
Calculation using:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
Flexural Buckling Resistance
Out-of-Plane Buckling, z-z axis
k
z
: 1.00
L
cr,z
[m]: 3.00
N
cr,z
[kN]: 2129.27
z
: 0.76
Buckling curve: c
z
: 0.49
z
: 0.68
N
b,z,Rd
[kN]: 851.73
Flexural Buckling Resistance
In-Plane Buckling, y-y axis
k
y
: 1.00
L
cr,y
[m]: 3.00
N
cr,y
[kN]: 5780.30
y
: 0.46
Buckling curve: b
y
: 0.34
y
: 0.90
N
b,y,Rd
[kN]: 1119.73
Design Buckling resistance of the column N
b,Rd
[kN]: 851.73
Design resistance N
Rd
[kN]: 851.73
Utilization Ratio Cross-section (N) = N
Ed
/ N
c,Rd
: 0.60
Flexural Buckling (N
b
) = N
Ed
/ N
b,Rd
: 0.87
The column is safe Sent from my iPad
Design of Members
Exercise 3b
Design of Members
Exercise 3b
1. General Criteria
2. Tension
3. Cross Section Resistance
4. Torsion
5. Columns
6. Beams
7. Beam-columns
Design of Members
Design of Members
Beams
Design of Members
Beams
LATERAL-TORSIONAL BUCKLING
|
|
.
|
\
|
+ =
T
W
z T
E
cr
I G L
I E
I E I G
L
M
2
2
1
t t
M
cr
P
C
M
cr,2
<M
cr
P
C P
M
cr,1
>M
cr
C
Elastic critical moment
E
cr m cr
M M o =
Design of Members
Beams
LATERAL-TORSIONAL BUCKLING
y
z
C~G
y
z
G
C
( )
( )
( ) ( )
(
(
+ +
|
|
.
|
\
|
=
j g j g
z
T z
z
W
w
z
z
z
cr
z C z C z C z C
I E
I G L k
I
I
k
k
L k
I E
C M
3 2
5 . 0
2
3 2
2
2
2
2
2
1
t
t
( )
s a g
z z z =
( )
y
A
s j
I dA z z y z z
|
|
.
|
\
|
+ =
}
2 2
5 . 0
Design of Members
Beams
LATERAL-TORSIONAL BUCKLING
Elastic critical moment
Design of Members
Beams
LATERAL-TORSIONAL
BUCKLING
Elastic critical moment
Design of Members
Beams
LATERAL-TORSIONAL BUCKLING RESISTANCE
0 . 1
.
s
Rd b
Ed
M
M
1 . M y y LT Rd b
f W M _ =
W
y
= W
pl.y
Class 1 and 2;
W
y
= W
el.y
Class 3;
W
y
= W
eff.y
Class 4;
i) General method
( )
5 . 0
2
LT
2
LT LT
LT
1
| |
_
+
=
( ) | |
2
LT LT
LT LT
2 . 0 1 5 . 0 o | + + =
| |
5 . 0
cr y y
LT M f W =
0 . 1
LT
s _
ii) Special method
| |
5 . 0
cr y y
LT M f W =
( )
5 . 0
2 2
1
LT
LT LT
LT
| | |
_
+
=
2
1
0 . 1
LT
LT
LT
_
_
s
s
( ) | |
2
0 , 1 5 . 0 LT LT LT
LT LT
| o | + + =
4 . 0 0 , s LT
75 . 0 > |
Design of Members
Beams
LATERAL-TORSIONAL BUCKLING RESISTANCE
f
LT
LT
_
_ =
mod ,
0 . 1
mod ,
s
LT
_
( ) ( ) | |
2
8 . 0 0 . 2 1 1 5 . 0 1 = LT
c
k f
0 . 1 s f
0 , LT LT
s
2
0 , LT
cr Ed
M M s
Neglect LTB if:
Design of Members
Beams
LATERAL-TORSIONAL BUCKLING RESISTANCE
Design of Members
Exercise 5
Design the beam using a HEA profile in S 235 steel, supported
by web cleats and loaded by two concentrated loads, P = 70.0
kN (design loads). Consider free rotation at the supports (y-
axis and z-axis). Assume free warping at the supports but
consider that the web cleats do not allow rotation around the
axis of the beam (x-axis). Assume: a) unrestrained beam; b)
beam braced at the points of application of the concentrated
loads.
Use:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
- LTBeam software (Windows OS)
Design of Members
Exercise 5
Design of Members
Exercise 5
Design of Members
Exercise 5
Design of Members
Exercise 5
Design of Members
Exercise 5
Design of Members
Exercise 5
Design of Members
Exercise 5
Design of Members
Exercise 5
Design of Members
Exercise 5
Design of Members
Exercise 5
Calculation using:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
Calculation of HE 240 A Bending resistance
SUMMARY
Cross-section: HE 240 A
Loading: M
y,Ed
= 105.00 kNm
Loading: V
Ed
= 70.00 kN
Steel Grade: S235 J0
M0: 1.00
M1: 1.00
Member length: L = 6.00 [m]
CASE I) Load applied at centroid, C
1
= C
2
= 1.0
CASE II) Load applied at centroid, C
1
= 1.04, C
2
= 0.42
CASE III) Load applied at top flange, C
1
= 1.04, C
2
= 0.42
Design of Members
Exercise 5
Calculation using:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
RESULTS
Cross-section class
Internal compression parts
c/t
lim
: 72.00
c/t: 21.87
Class: 1
Outstand flanges
c/t
lim
: 72.00
c/t: 7.94
Class: 1
Resultant cross-section class: 1
Type of Verification: Plastic
Cross-section resistance
Bending moment
W
pl,y
[cm
3
]: 744.60
M
y,c,Rd
[kNm]: 174.98
Shear
V
c,Rd
[kN]: 341.64
(h
w
/t
w
)
lim
: 72.00
h
w
/t
w
: 27.47
Bending and Shear
p: 0.35
M
y,V,Rd
[kNm]: 174.98
Design of Members
Exercise 5
Calculation using:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
L
e,LT
[m]: 6.00
C1: 1.00
C2: 1.00
k
z
: 1.00
k
w
: 1.00
z
a
[mm]: 0.00
M
cr
[kNm]: 289.22
LT
: 0.78
Buckling curve: a
LT
: 0.21
LT
: 0.81
M
b,Rd
[kNm]: 141.41
CASE I) CASE II) CASE III)
Lateral-torsional buckling resistance
L
e,LT
[m]: 6.00
C1: 1.04
C2: 0.42
k
z
: 1.00
k
w
: 1.00
z
a
[mm]: 0.00
M
cr
[kNm]: 300.79
LT
: 0.76
Buckling curve: a
LT
: 0.21
LT
: 0.82
M
b,Rd
[kNm]: 142.84
L
e,LT
[m]: 6.00
C1: 1.04
C2: 0.42
k
z
: 1.00
k
w
: 1.00
z
a
[mm]: 115.00
M
cr
[kNm]: 231.19
LT
: 0.87
Buckling curve: a
LT
: 0.21
LT
: 0.75
M
b,Rd
[kNm]: 131.83
Design of Members
Exercise 5
Calculation (M
cr
only) using:
- LTBeam (Windows OS)
CASE II) Load applied at centroid
Critical Moment
Critical value of maximum moment Mcr = 298.61 kN.m
CASE III) Load applied at top flange
Critical Moment
Critical value of maximum moment Mcr = 228.41 kN.m
Design of Members
Exercise 5
Calculation using:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
Design Bending resistance of the beam
M
y,Rd
[kNm]: 131.83
Utilization Ratio
Cross-section - Bending Moment
(M
y
) = M
y,Ed
/ M
y,c,Rd
: 0.60
Cross-section - Shear
(V) = V
Ed
/ V
c,Rd
: 0.20
Cross-section - Bending and Shear
(M
y
+V) = M
y,Ed
/ M
y,V,Rd
: 0.60
Lateral-torsional buckling verification
(Mb) = M
y,Ed
/ M
b,Rd
: 0.80
Calculation of HE 240 A Bending resistance
1. General Criteria
2. Tension
3. Cross Section Resistance
4. Torsion
5. Columns
6. Beams
7. Beam-columns
Design of Members
Members with high slenderness subjected to bending and
compression, may fail by flexural buckling or lateral-torsional
buckling
Standard interaction expression
(M-N)
0 . 1 ) , , ( s
uz
z
uy
y
u
M
M
M
M
N
N
f
N, M
y
e M
z
design values of the forces,
N
u
, M
uy
e M
uz
design resistances.
Design of Members
Beam-columns
Flexural buckling and lateral-torsional buckling (doubly-
symmetric cross-section):
0 . 1
1 ,
, .
1 ,
, .
1
s
A +
+
A +
+
M Rk z
Ed z Ed z
yz
M Rk y LT
Ed y Ed y
yy
M Rk y
Ed
M
M M
k
M
M M
k
N
N
_ _
0 . 1
1 ,
, .
1 ,
, .
1
s
A +
+
A +
+
M Rk z
Ed z Ed z
zz
M Rk y LT
Ed y Ed y
zy
M Rk z
Ed
M
M M
k
M
M M
k
N
N
_ _
k
yy
, k
yz
, k
zy
e k
zz
- interaction factors that are dependent of instability
phenomena and plasticity Annex A of EC3-1-1 (Method 1) or Annex B
(Method 2).
e
N,y
N
Ed
Design of Members
Beam-columns
MEMBER STABILITY
Members not susceptible to torsional deformation checking of
flexural buckling against y-axis and z-axis, considering expressions
(6.61) and (6.62) with _
LT
= 1.0 and interaction factors k
yy
, k
yz
, k
zy
e k
zz
in members not susceptible to torsional deformation.
Members susceptible to torsional deformation checking of
lateral-torsional buckling, considering expressions (6.61) and (6.62)
with _
LT
according to 6.3.2 of EC3-1-1 and interaction factors k
yy
, k
yz
,
k
zy
e k
zz
in members susceptible to torsional deformation.
i) Members with closed hollow sections or open sections restrained
to torsion are not susceptible to torsional deformation.
ii) Members with open sections (I or H sections) are susceptible to
torsional deformation.
Design of Members
Beam-columns
MEMBER STABILITY
pode
Design of Members
Beam-columns
MEMBER STABILITY
Method 2
Interaction factors
for members not
susceptible to
torsional
deformations
Interaction factors
for members
susceptible to
torsional
deformations
Design of Members
Beam-columns
MEMBER STABILITY
Method 2
Design of Members
Beam-columns
MEMBER
STABILITY
Method 2
Design of Members
Exercise 6
Beam-column (IPE 360, S355)
Use:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
- LTBeam software (Windows OS)
- Semi-Comp Member Design Software (Windows OS)
Design of Members
Exercise 6
Design of Members
Exercise 6
Cross-section type
Finishing
M0
= 1,00
Select from library
M1
= 1,00
(optional)
Cross-section data Material
H = 500,0 [mm] Steel grade S235
B =
200,0 [mm]
f
y
= 235,0 N/mm
Tw = 10,2 [mm] E = 210000,0 N/mm
Tf = 16,0 [mm]
R = 21,0 [mm]
A [cm
2
] I
yy
[cm
4
] I
zz
[cm
4
] W
el,y
[cm
3
] W
el,z
[cm
3
] W
pl,y
[cm
3
] W
pl,z
[cm
3
]
115,52 48198,53 2141,69 1927,94 214,17 2194,12 335,88
SEMICOMP Member Design
Partial factors
M
Profile can be chosen from library or entered manually
List available CS-Types and Finishing types (Why does the finishing type have
to be considered)
Set partial factors. Set steel grade and E-modulus
CS- Values are calculated automatically
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.1 SYSTEM DATA
Cross-section type
Finishing M0 = 1,00
Select fromlibrary M1 = 1,00
(optional)
Cross-section data Material My,Ed,max= 250,00 kNm
H = 500,0 [mm] Steel grade S235
B =
200,0 [mm]
fy = 235,0 N/mm
Tw = 10,2 [mm] E = 210000,0 N/mm
Tf = 16,0 [mm]
R = 21,0 [mm]
A [cm
2
] Iyy [cm
4
] Izz [cm
4
] Wel,y [cm
3
] Wel,z [cm
3
] Wpl,y [cm
3
] Wpl,z [cm
3
]
115,52 48198,53 2141,69 1927,94 214,17 2194,12 335,88
Vz,Ed,max= 50,00 kN
Boundary conditions It [cm
4
] Iw [cm
6
]
LBeam= 10,000 m 89,01 1254238,95
nfork= [-]
Loading in z-x-plane Loading in y-x-plane
NEd = -350,00 kN
qz,Ed
(*)
= 5,00 kN/m qy,Ed
(*)
= 4,50 kN/m Mz,Ed,max= 14,06 kNm
My,left,Ed = 250,00 kNm Mz,left,Ed = 0,00 kNm
My,right,Ed = 0,00 kNm Mz,right,Ed = 0,00 kNm
Pz,Ed
(*)
= 0,00 kN Py,Ed
(*)
= 0,00 kN
Distance of Loading to shear center
zSi = -250,00 mm
(*)
Mcr = 0,00 kNm
Vy,Ed,max= 14,06 kN
Mcr,0 = 0,00 kNm
Specify path of LTBeam.exe file:
Definition of axes
oz,max= 22,15 mm
Worked examples
oy,max= 3,39 mm
o
z
[mm] (corresponding to design loads)
o
y
[mm] (corresponding to design loads)
SEMICOMP Member Design M
y,Ed
[kNm]
V
z,Ed
[kN]
M
z,Ed
[kNm]
V
y,Ed
[kN]
Partial factors M
C:\Users\Martin\LTBeam\LTBeam.exe
0,00
500,00
-100,00
0,00
-20,00
0,00
20,00
-20,00
0,00
20,00
0,00
50,00
0,00
5,00
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.1 SYSTEM DATA
Boundary conditions I
t
[cm
4
] I
w
[cm
6
]
L
Beam
= 10,000 m 89,01 1254238,95
n
fork
=
[-]
Loading in z-x-plane Loading in y-x-plane
N
Ed
= -350,00 kN
q
z,Ed
(*)
=
5,00 kN/m
q
y,Ed
(*)
=
4,50 kN/m
M
y,left,Ed
= 250,00 kNm M
z,left,Ed
= 0,00 kNm
M
y,right,Ed
= 0,00 kNm M
z,right,Ed
= 0,00 kNm
P
z,Ed
(*)
= 0,00 kN P
y,Ed
(*)
= 0,00 kN
Distance of Loading to shear center
z
Si
= -250,00 mm
(*)
M
cr
= 0,00 kNm
M
cr,0
= 0,00 kNm
Specify path of LTBeam.exe file:
C:\Users\Martin\LTBeam\LTBeam.exe
m
[-]
Choose lateral
restraints
Torsion restraint by
default for tubular
sections
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.1 SYSTEM DATA
Cross-section type
Finishing M0 = 1,00
Select fromlibrary M1 = 1,00
(optional)
Cross-section data Material My,Ed,max= 250,00 kNm
H = 500,0 [mm] Steel grade S235
B =
200,0 [mm]
fy = 235,0 N/mm
Tw = 10,2 [mm] E = 210000,0 N/mm
Tf = 16,0 [mm]
R = 21,0 [mm]
A [cm
2
] Iyy [cm
4
] Izz [cm
4
] Wel,y [cm
3
] Wel,z [cm
3
] Wpl,y [cm
3
] Wpl,z [cm
3
]
115,52 48198,53 2141,69 1927,94 214,17 2194,12 335,88
Vz,Ed,max= 50,00 kN
Boundary conditions It [cm
4
] Iw [cm
6
]
LBeam= 10,000 m 89,01 1254238,95
nfork= [-]
Loading in z-x-plane Loading in y-x-plane
NEd = -350,00 kN
qz,Ed
(*)
= 5,00 kN/m qy,Ed
(*)
= 4,50 kN/m Mz,Ed,max= 14,06 kNm
My,left,Ed = 250,00 kNm Mz,left,Ed = 0,00 kNm
My,right,Ed = 0,00 kNm Mz,right,Ed = 0,00 kNm
Pz,Ed
(*)
= 0,00 kN Py,Ed
(*)
= 0,00 kN
Distance of Loading to shear center
zSi = -250,00 mm
(*)
Mcr = 0,00 kNm
Vy,Ed,max= 14,06 kN
Mcr,0 = 0,00 kNm
Specify path of LTBeam.exe file:
Definition of axes
oz,max= 22,15 mm
Worked examples
oy,max= 3,39 mm
o
z
[mm] (corresponding to design loads)
o
y
[mm] (corresponding to design loads)
SEMICOMP Member Design M
y,Ed
[kNm]
V
z,Ed
[kN]
M
z,Ed
[kNm]
V
y,Ed
[kN]
Partial factors M
C:\Users\Martin\LTBeam\LTBeam.exe
0,00
500,00
-100,00
0,00
-20,00
0,00
20,00
-20,00
0,00
20,00
0,00
50,00
0,00
5,00
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.1 SYSTEM DATA
Cross-section type
Finishing M0 = 1,00
Select fromlibrary M1 = 1,00
(optional)
Cross-section data Material My,Ed,max= 250,00 kNm
H = 500,0 [mm] Steel grade S235
B =
200,0 [mm]
fy = 235,0 N/mm
Tw = 10,2 [mm] E = 210000,0 N/mm
Tf = 16,0 [mm]
R = 21,0 [mm]
A [cm
2
] Iyy [cm
4
] Izz [cm
4
] Wel,y [cm
3
] Wel,z [cm
3
] Wpl,y [cm
3
] Wpl,z [cm
3
]
115,52 48198,53 2141,69 1927,94 214,17 2194,12 335,88
Vz,Ed,max= 50,00 kN
Boundary conditions It [cm
4
] Iw [cm
6
]
LBeam= 10,000 m 89,01 1254238,95
nfork= [-]
Loading in z-x-plane Loading in y-x-plane
NEd = -350,00 kN
qz,Ed
(*)
= 5,00 kN/m qy,Ed
(*)
= 4,50 kN/m Mz,Ed,max= 14,06 kNm
My,left,Ed = 250,00 kNm Mz,left,Ed = 0,00 kNm
My,right,Ed = 0,00 kNm Mz,right,Ed = 0,00 kNm
Pz,Ed
(*)
= 0,00 kN Py,Ed
(*)
= 0,00 kN
Distance of Loading to shear center
zSi = -250,00 mm
(*)
Mcr = 0,00 kNm
Vy,Ed,max= 14,06 kN
Mcr,0 = 0,00 kNm
Specify path of LTBeam.exe file:
Definition of axes
oz,max= 22,15 mm
Worked examples
oy,max= 3,39 mm
o
z
[mm] (corresponding to design loads)
o
y
[mm] (corresponding to design loads)
SEMICOMP Member Design M
y,Ed
[kNm]
V
z,Ed
[kN]
M
z,Ed
[kNm]
V
y,Ed
[kN]
Partial factors M
C:\Users\Martin\LTBeam\LTBeam.exe
0,00
500,00
-100,00
0,00
-20,00
0,00
20,00
-20,00
0,00
20,00
0,00
50,00
0,00
5,00
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.1 SYSTEM DATA
M
y,Ed,max
= 250,00 kNm
V
z,Ed,max
= 50,00 kN
M
y,Ed
[kNm]
V
z,Ed
[kN]
0,00
500,00
-100,00
0,00
o
z,max
= 22,15 mm
o
z
[mm] (corresponding to design loads)
0,00
50,00
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.1 SYSTEM DATA
Design of Members
Exercise 6
Calculation using:
- Semi-Comp Member Design (Windows OS)
Design of Members
Exercise 6
Design of Members
Exercise 6
Calculation using:
- Semi-Comp Member Design (Windows OS)
Design of Members
Exercise 6
Calculation using:
- Semi-Comp Member Design (Windows OS)
Reference values
Correction factor k c table 6.6
kc = 0,969 [-]
(based on I. order cross-section forces)
Reference values for classification in the worst section along the member
c/tw = 41,765 o web = 0,734 web = -0,563 c = 1,000 y = 0,997 [-] max = 1,237 [-]
c/tf = 4,619 o f lange = 1,000 f lange = 0,869 z = 0,883 [-] 0 = 0,961 [-]
wy = 1,138 [-] 0,lim = 0,191 [-]
Boundaries Class 1 Class 2 Class 3 wz = 1,500 [-] LT = 0,856 [-]
c/tw c/tw, max = 46,361 53,386 86,768 npl = 0,129 [-] c y = 4,280 [-]
c/tf c/tf , max = 9,000 10,000 13,935 aLT = 0,998 [-] Cmy,0 = 0,996 [-]
bLT = 0,063 [-] Cmz,0 = 0,887 [-]
Member class = 1 cLT = 0,972 [-] Cmy = 0,999 [-]
dLT = 0,122 [-] Cmz = 0,887 [-]
Member Check eLT = 0,518 [-] CmLT = 1,000 [-]
Mpl,y,Rd = 515,618 kNm Mcr,0 = 558,807 [-] C1 = 1,065 [-]
NRd = 2714,757 kN Mpl,z,Rd = 78,932 kNm NEd = -350,000 kN
My,Rd = 515,618 kNm Mel,y,Rd = 453,066 kNm My,Ed,max = 250,000 kNm Cyy = 0,958 [-] Czy = 0,820 [-]
Mz,Rd = 78,932 kNm Mel,z,Rd = 50,330 kNm Mz,Ed,max = 14,063 kNm Cyz = 0,500 [-] Czz = 0,946 [-]
Lcr,y = 10,000 m Lcr,z = 5,000 m Mcr = 704,079 kNm
Ncr,y = 9989,710 kN Ncr,z = 1775,559 kN o LT = 0,49 [-] Cmy = 0,000 [-]
o y = 0,21 [-] o z = 0,34 [-] LT = 0,856 [-] Cmz = 0,000 [-]
y = 0,521 [-] z = 1,237 [-] _LT = 0,740 [-] CmLT = 0,000 [-]
_y = 0,918 [-] _z = 0,459 [-] fmod = 0,985 [-]
EN 1993-1-1, 6.3.3
Uniform member in bending and axial compression Global interaction factors
kyy = 1,077
Eq. (6.61): U = 1,113 1,0 not ok kyz = 1,517
Eq. (6.62): U = 0,845 1,0 ok kzy = 0,582
kzz = 1,031
Cross-section check at each end of the member
Left end: U = 0,506 1,0 ok
Right end: U = 0,129 1,0 ok
Additional member checks
EN 1993-1-1, 6.3.1
Strong axis flexural buckling check
Eq. (6.46): NEd/Nb,Rd = 0,141 1,0 ok
Weak axis flexural buckling check
Eq. (6.46): NEd/Nb,Rd = 0,281 1,0 ok
EN 1993-1-1, 6.3.2
Lateral torsional buckling
Eq. (6.54): MEd/Mb,Rd = 0,655 1,0 ok
SEMICOMP Member Check
Method 2 auxiliary terms (if applicable):
Choose method for
cross-section resistance
Choose method for
member check
Method 1 auxiliary terms (if applicable):
Lateral torsional buckling Strong axis buckling Weak axis buckling
Section classification for member design check
Member check is performed here
Semi-Comp Member Design (Windows OS)
SHEET 1.2 MEMBER CHECK
Design of Members
Exercise 6
Different available methods for the member check and
cross-section resistance
SEMICOMP Member Check
Choose method for
cross-section resistance
Choose method for
member check
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.2 MEMBER CHECK
Reference values
Correction factor k c table 6.6
kc = 0,969 [-]
(based on I. order cross-section forces)
Reference values for classification in the worst section along the member
c/tw = 41,765 o web = 0,734 web = -0,563 c = 1,000 y = 0,997 [-] max = 1,237 [-]
c/tf = 4,619 o f lange = 1,000 f lange = 0,869 z = 0,883 [-] 0 = 0,961 [-]
wy = 1,138 [-] 0,lim = 0,191 [-]
Boundaries Class 1 Class 2 Class 3 wz = 1,500 [-] LT = 0,856 [-]
c/tw c/tw, max = 46,361 53,386 86,768 npl = 0,129 [-] c y = 4,280 [-]
c/tf c/tf , max = 9,000 10,000 13,935 aLT = 0,998 [-] Cmy,0 = 0,996 [-]
bLT = 0,063 [-] Cmz,0 = 0,887 [-]
Member class = 1 cLT = 0,972 [-] Cmy = 0,999 [-]
dLT = 0,122 [-] Cmz = 0,887 [-]
Member Check eLT = 0,518 [-] CmLT = 1,000 [-]
Mpl,y,Rd = 515,618 kNm Mcr,0 = 558,807 [-] C1 = 1,065 [-]
NRd = 2714,757 kN Mpl,z,Rd = 78,932 kNm NEd = -350,000 kN
My,Rd = 515,618 kNm Mel,y,Rd = 453,066 kNm My,Ed,max = 250,000 kNm Cyy = 0,958 [-] Czy = 0,820 [-]
Mz,Rd = 78,932 kNm Mel,z,Rd = 50,330 kNm Mz,Ed,max = 14,063 kNm Cyz = 0,500 [-] Czz = 0,946 [-]
Lcr,y = 10,000 m Lcr,z = 5,000 m Mcr = 704,079 kNm
Ncr,y = 9989,710 kN Ncr,z = 1775,559 kN o LT = 0,49 [-] Cmy = 0,000 [-]
o y = 0,21 [-] o z = 0,34 [-] LT = 0,856 [-] Cmz = 0,000 [-]
y = 0,521 [-] z = 1,237 [-] _LT = 0,740 [-] CmLT = 0,000 [-]
_y = 0,918 [-] _z = 0,459 [-] fmod = 0,985 [-]
EN 1993-1-1, 6.3.3
Uniform member in bending and axial compression Global interaction factors
kyy = 1,077
Eq. (6.61): U = 1,113 1,0 not ok kyz = 1,517
Eq. (6.62): U = 0,845 1,0 ok kzy = 0,582
kzz = 1,031
Cross-section check at each end of the member
Left end: U = 0,506 1,0 ok
Right end: U = 0,129 1,0 ok
Additional member checks
EN 1993-1-1, 6.3.1
Strong axis flexural buckling check
Eq. (6.46): NEd/Nb,Rd = 0,141 1,0 ok
Weak axis flexural buckling check
Eq. (6.46): NEd/Nb,Rd = 0,281 1,0 ok
EN 1993-1-1, 6.3.2
Lateral torsional buckling
Eq. (6.54): MEd/Mb,Rd = 0,655 1,0 ok
SEMICOMP Member Check
Method 2 auxiliary terms (if applicable):
Choose method for
cross-section resistance
Choose method for
member check
Method 1 auxiliary terms (if applicable):
Lateral torsional buckling Strong axis buckling Weak axis buckling
Section classification for member design check
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.2 MEMBER CHECK
(based on I. order cross-section forces)
Reference values for classification in the worst section along the member
c/t
w
= 41,765 o
web
= 0,734
web
= -0,563 c = 1,000
c/t
f
= 4,619 o
flange
= 1,000
flange
= 0,869
Boundaries Class 1 Class 2 Class 3
c/t
w
c/t
w, max
= 46,361 53,386 86,768
c/t
f
c/t
f, max
= 9,000 10,000 13,935
Member class = 1
Section classification for member design check
Classification of the member, how to do?
Worst cross-section along the member has to be
found (highest utilisation factor)
Overall member
class
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.2 MEMBER CHECK
Section classification for member design check
Additional information
Point No.
x-coordinate
[m]
N
Ed
[kN]
M
y,Ed
[kNm]
M
z,Ed
[kNm]
Utilization
Factor
CS-Class
1 0,00 -350,00 250,00 0,00 0,506 2
2 0,17 -350,00 249,93 1,34 0,508 1
3 0,33 -350,00 249,72 2,56 0,511 1
4 0,50 -350,00 249,38 3,66 0,514 1
5 0,67 -350,00 248,89 4,63 0,517 1
6 0,83 -350,00 248,26 5,47 0,519 1
7 1,00 -350,00 247,50 6,19 0,521 1
8 1,17 -350,00 246,60 6,78 0,522 1
9 1,33 -350,00 245,56 7,25 0,523 1
10 1,50 -350,00 244,38 7,59 0,522 1
11 1,67 -350,00 243,06 7,81 0,521 1
12 1,83 -350,00 241,60 7,91 0,519 1
13 2,00 -350,00 240,00 7,88 0,516 1
14 2,17 -350,00 238,26 7,72 0,513 1
15 2,33 -350,00 236,39 7,44 0,508 1
Cross-section checks are performed at 61 points in the
beam
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.2.1 ADDITIONAL INFO
Reference values
Correction factor k c table 6.6
kc = 0,969 [-]
(based on I. order cross-section forces)
Reference values for classification in the worst section along the member
c/tw = 41,765 o web = 0,734 web = -0,563 c = 1,000 y = 0,997 [-] max = 1,237 [-]
c/tf = 4,619 o f lange = 1,000 f lange = 0,869 z = 0,883 [-] 0 = 0,961 [-]
wy = 1,138 [-] 0,lim = 0,191 [-]
Boundaries Class 1 Class 2 Class 3 wz = 1,500 [-] LT = 0,856 [-]
c/tw c/tw, max = 46,361 53,386 86,768 npl = 0,129 [-] c y = 4,280 [-]
c/tf c/tf , max = 9,000 10,000 13,935 aLT = 0,998 [-] Cmy,0 = 0,996 [-]
bLT = 0,063 [-] Cmz,0 = 0,887 [-]
Member class = 1 cLT = 0,972 [-] Cmy = 0,999 [-]
dLT = 0,122 [-] Cmz = 0,887 [-]
Member Check eLT = 0,518 [-] CmLT = 1,000 [-]
Mpl,y,Rd = 515,618 kNm Mcr,0 = 558,807 [-] C1 = 1,065 [-]
NRd = 2714,757 kN Mpl,z,Rd = 78,932 kNm NEd = -350,000 kN
My,Rd = 515,618 kNm Mel,y,Rd = 453,066 kNm My,Ed,max = 250,000 kNm Cyy = 0,958 [-] Czy = 0,820 [-]
Mz,Rd = 78,932 kNm Mel,z,Rd = 50,330 kNm Mz,Ed,max = 14,063 kNm Cyz = 0,500 [-] Czz = 0,946 [-]
Lcr,y = 10,000 m Lcr,z = 5,000 m Mcr = 704,079 kNm
Ncr,y = 9989,710 kN Ncr,z = 1775,559 kN o LT = 0,49 [-] Cmy = 0,000 [-]
o y = 0,21 [-] o z = 0,34 [-] LT = 0,856 [-] Cmz = 0,000 [-]
y = 0,521 [-] z = 1,237 [-] _LT = 0,740 [-] CmLT = 0,000 [-]
_y = 0,918 [-] _z = 0,459 [-] fmod = 0,985 [-]
EN 1993-1-1, 6.3.3
Uniform member in bending and axial compression Global interaction factors
kyy = 1,077
Eq. (6.61): U = 1,113 1,0 not ok kyz = 1,517
Eq. (6.62): U = 0,845 1,0 ok kzy = 0,582
kzz = 1,031
Cross-section check at each end of the member
Left end: U = 0,506 1,0 ok
Right end: U = 0,129 1,0 ok
Additional member checks
EN 1993-1-1, 6.3.1
Strong axis flexural buckling check
Eq. (6.46): NEd/Nb,Rd = 0,141 1,0 ok
Weak axis flexural buckling check
Eq. (6.46): NEd/Nb,Rd = 0,281 1,0 ok
EN 1993-1-1, 6.3.2
Lateral torsional buckling
Eq. (6.54): MEd/Mb,Rd = 0,655 1,0 ok
SEMICOMP Member Check
Method 2 auxiliary terms (if applicable):
Choose method for
cross-section resistance
Choose method for
member check
Method 1 auxiliary terms (if applicable):
Lateral torsional buckling Strong axis buckling Weak axis buckling
Section classification for member design check
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.2 MEMBER CHECK
All relevant values for member check
Member Check
M
pl,y,Rd
= 515,618 kNm
N
Rd
= 2714,757 kN M
pl,z,Rd
= 78,932 kNm N
Ed
= -350,000 kN
M
y,Rd
= 515,618 kNm M
el,y,Rd
= 453,066 kNm M
y,Ed,max
= 250,000 kNm
M
z,Rd
= 78,932 kNm M
el,z,Rd
= 50,330 kNm M
z,Ed,max
= 14,063 kNm
L
cr,y
= 10,000 m L
cr,z
= 5,000 m M
cr
= 704,079 kNm
N
cr,y
= 9989,710 kN N
cr,z
= 1775,559 kN o
LT
= 0,49 [-]
o
y
= 0,21 [-] o
z
= 0,34 [-]
LT
= 0,856 [-]
y
= 0,521 [-]
z
= 1,237 [-] _
LT
= 0,740 [-]
_
y
= 0,918 [-] _
z
= 0,459 [-] f
mod
= 0,985 [-]
Lateral torsional buckling Strong axis buckling Weak axis buckling
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.2 MEMBER CHECK
Reference values
Correction factor k c table 6.6
kc = 0,969 [-]
(based on I. order cross-section forces)
Reference values for classification in the worst section along the member
c/tw = 41,765 o web = 0,734 web = -0,563 c = 1,000 y = 0,997 [-] max = 1,237 [-]
c/tf = 4,619 o f lange = 1,000 f lange = 0,869 z = 0,883 [-] 0 = 0,961 [-]
wy = 1,138 [-] 0,lim = 0,191 [-]
Boundaries Class 1 Class 2 Class 3 wz = 1,500 [-] LT = 0,856 [-]
c/tw c/tw, max = 46,361 53,386 86,768 npl = 0,129 [-] c y = 4,280 [-]
c/tf c/tf , max = 9,000 10,000 13,935 aLT = 0,998 [-] Cmy,0 = 0,996 [-]
bLT = 0,063 [-] Cmz,0 = 0,887 [-]
Member class = 1 cLT = 0,972 [-] Cmy = 0,999 [-]
dLT = 0,122 [-] Cmz = 0,887 [-]
Member Check eLT = 0,518 [-] CmLT = 1,000 [-]
Mpl,y,Rd = 515,618 kNm Mcr,0 = 558,807 [-] C1 = 1,065 [-]
NRd = 2714,757 kN Mpl,z,Rd = 78,932 kNm NEd = -350,000 kN
My,Rd = 515,618 kNm Mel,y,Rd = 453,066 kNm My,Ed,max = 250,000 kNm Cyy = 0,958 [-] Czy = 0,820 [-]
Mz,Rd = 78,932 kNm Mel,z,Rd = 50,330 kNm Mz,Ed,max = 14,063 kNm Cyz = 0,500 [-] Czz = 0,946 [-]
Lcr,y = 10,000 m Lcr,z = 5,000 m Mcr = 704,079 kNm
Ncr,y = 9989,710 kN Ncr,z = 1775,559 kN o LT = 0,49 [-] Cmy = 0,000 [-]
o y = 0,21 [-] o z = 0,34 [-] LT = 0,856 [-] Cmz = 0,000 [-]
y = 0,521 [-] z = 1,237 [-] _LT = 0,740 [-] CmLT = 0,000 [-]
_y = 0,918 [-] _z = 0,459 [-] fmod = 0,985 [-]
EN 1993-1-1, 6.3.3
Uniform member in bending and axial compression Global interaction factors
kyy = 1,077
Eq. (6.61): U = 1,113 1,0 not ok kyz = 1,517
Eq. (6.62): U = 0,845 1,0 ok kzy = 0,582
kzz = 1,031
Cross-section check at each end of the member
Left end: U = 0,506 1,0 ok
Right end: U = 0,129 1,0 ok
Additional member checks
EN 1993-1-1, 6.3.1
Strong axis flexural buckling check
Eq. (6.46): NEd/Nb,Rd = 0,141 1,0 ok
Weak axis flexural buckling check
Eq. (6.46): NEd/Nb,Rd = 0,281 1,0 ok
EN 1993-1-1, 6.3.2
Lateral torsional buckling
Eq. (6.54): MEd/Mb,Rd = 0,655 1,0 ok
SEMICOMP Member Check
Method 2 auxiliary terms (if applicable):
Choose method for
cross-section resistance
Choose method for
member check
Method 1 auxiliary terms (if applicable):
Lateral torsional buckling Strong axis buckling Weak axis buckling
Section classification for member design check
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.2 MEMBER CHECK
Member check and cross-section check at each end of
the member
EN 1993-1-1, 6.3.3
Uniform member in bending and axial compression Global interaction factors
k
yy
= 1,077
Eq. (6.61):
U =
1,113 1,0 not ok k
yz
= 1,517
Eq. (6.62):
U =
0,845 1,0 ok k
zy
= 0,582
k
zz
= 1,031
Cross-section check at each end of the member
Left end: U = 0,506 1,0 ok
Right end: U = 0,129 1,0 ok
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.2 MEMBER CHECK
Reference values
Correction factor k c table 6.6
kc = 0,969 [-]
(based on I. order cross-section forces)
Reference values for classification in the worst section along the member
c/tw = 41,765 o web = 0,734 web = -0,563 c = 1,000 y = 0,997 [-] max = 1,237 [-]
c/tf = 4,619 o f lange = 1,000 f lange = 0,869 z = 0,883 [-] 0 = 0,961 [-]
wy = 1,138 [-] 0,lim = 0,191 [-]
Boundaries Class 1 Class 2 Class 3 wz = 1,500 [-] LT = 0,856 [-]
c/tw c/tw, max = 46,361 53,386 86,768 npl = 0,129 [-] c y = 4,280 [-]
c/tf c/tf , max = 9,000 10,000 13,935 aLT = 0,998 [-] Cmy,0 = 0,996 [-]
bLT = 0,063 [-] Cmz,0 = 0,887 [-]
Member class = 1 cLT = 0,972 [-] Cmy = 0,999 [-]
dLT = 0,122 [-] Cmz = 0,887 [-]
Member Check eLT = 0,518 [-] CmLT = 1,000 [-]
Mpl,y,Rd = 515,618 kNm Mcr,0 = 558,807 [-] C1 = 1,065 [-]
NRd = 2714,757 kN Mpl,z,Rd = 78,932 kNm NEd = -350,000 kN
My,Rd = 515,618 kNm Mel,y,Rd = 453,066 kNm My,Ed,max = 250,000 kNm Cyy = 0,958 [-] Czy = 0,820 [-]
Mz,Rd = 78,932 kNm Mel,z,Rd = 50,330 kNm Mz,Ed,max = 14,063 kNm Cyz = 0,500 [-] Czz = 0,946 [-]
Lcr,y = 10,000 m Lcr,z = 5,000 m Mcr = 704,079 kNm
Ncr,y = 9989,710 kN Ncr,z = 1775,559 kN o LT = 0,49 [-] Cmy = 0,000 [-]
o y = 0,21 [-] o z = 0,34 [-] LT = 0,856 [-] Cmz = 0,000 [-]
y = 0,521 [-] z = 1,237 [-] _LT = 0,740 [-] CmLT = 0,000 [-]
_y = 0,918 [-] _z = 0,459 [-] fmod = 0,985 [-]
EN 1993-1-1, 6.3.3
Uniform member in bending and axial compression Global interaction factors
kyy = 1,077
Eq. (6.61): U = 1,113 1,0 not ok kyz = 1,517
Eq. (6.62): U = 0,845 1,0 ok kzy = 0,582
kzz = 1,031
Cross-section check at each end of the member
Left end: U = 0,506 1,0 ok
Right end: U = 0,129 1,0 ok
Additional member checks
EN 1993-1-1, 6.3.1
Strong axis flexural buckling check
Eq. (6.46): NEd/Nb,Rd = 0,141 1,0 ok
Weak axis flexural buckling check
Eq. (6.46): NEd/Nb,Rd = 0,281 1,0 ok
EN 1993-1-1, 6.3.2
Lateral torsional buckling
Eq. (6.54): MEd/Mb,Rd = 0,655 1,0 ok
SEMICOMP Member Check
Method 2 auxiliary terms (if applicable):
Choose method for
cross-section resistance
Choose method for
member check
Method 1 auxiliary terms (if applicable):
Lateral torsional buckling Strong axis buckling Weak axis buckling
Section classification for member design check
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.2 MEMBER CHECK
Additional member checks
Additional member checks
EN 1993-1-1, 6.3.1
Strong axis flexural buckling check
Eq. (6.46): N
Ed
/N
b,Rd
= 0,141 1,0 ok
Weak axis flexural buckling check
Eq. (6.46): N
Ed
/N
b,Rd
= 0,281 1,0 ok
EN 1993-1-1, 6.3.2
Lateral torsional buckling
Eq. (6.54): M
Ed
/M
b,Rd
= 0,655 1,0 ok
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.2 MEMBER CHECK
Reference values
Correction factor k c table 6.6
kc = 0,969 [-]
(based on I. order cross-section forces)
Reference values for classification in the worst section along the member
c/tw = 41,765 o web = 0,734 web = -0,563 c = 1,000 y = 0,997 [-] max = 1,237 [-]
c/tf = 4,619 o f lange = 1,000 f lange = 0,869 z = 0,883 [-] 0 = 0,961 [-]
wy = 1,138 [-] 0,lim = 0,191 [-]
Boundaries Class 1 Class 2 Class 3 wz = 1,500 [-] LT = 0,856 [-]
c/tw c/tw, max = 46,361 53,386 86,768 npl = 0,129 [-] c y = 4,280 [-]
c/tf c/tf , max = 9,000 10,000 13,935 aLT = 0,998 [-] Cmy,0 = 0,996 [-]
bLT = 0,063 [-] Cmz,0 = 0,887 [-]
Member class = 1 cLT = 0,972 [-] Cmy = 0,999 [-]
dLT = 0,122 [-] Cmz = 0,887 [-]
Member Check eLT = 0,518 [-] CmLT = 1,000 [-]
Mpl,y,Rd = 515,618 kNm Mcr,0 = 558,807 [-] C1 = 1,065 [-]
NRd = 2714,757 kN Mpl,z,Rd = 78,932 kNm NEd = -350,000 kN
My,Rd = 515,618 kNm Mel,y,Rd = 453,066 kNm My,Ed,max = 250,000 kNm Cyy = 0,958 [-] Czy = 0,820 [-]
Mz,Rd = 78,932 kNm Mel,z,Rd = 50,330 kNm Mz,Ed,max = 14,063 kNm Cyz = 0,500 [-] Czz = 0,946 [-]
Lcr,y = 10,000 m Lcr,z = 5,000 m Mcr = 704,079 kNm
Ncr,y = 9989,710 kN Ncr,z = 1775,559 kN o LT = 0,49 [-] Cmy = 0,000 [-]
o y = 0,21 [-] o z = 0,34 [-] LT = 0,856 [-] Cmz = 0,000 [-]
y = 0,521 [-] z = 1,237 [-] _LT = 0,740 [-] CmLT = 0,000 [-]
_y = 0,918 [-] _z = 0,459 [-] fmod = 0,985 [-]
EN 1993-1-1, 6.3.3
Uniform member in bending and axial compression Global interaction factors
kyy = 1,077
Eq. (6.61): U = 1,113 1,0 not ok kyz = 1,517
Eq. (6.62): U = 0,845 1,0 ok kzy = 0,582
kzz = 1,031
Cross-section check at each end of the member
Left end: U = 0,506 1,0 ok
Right end: U = 0,129 1,0 ok
Additional member checks
EN 1993-1-1, 6.3.1
Strong axis flexural buckling check
Eq. (6.46): NEd/Nb,Rd = 0,141 1,0 ok
Weak axis flexural buckling check
Eq. (6.46): NEd/Nb,Rd = 0,281 1,0 ok
EN 1993-1-1, 6.3.2
Lateral torsional buckling
Eq. (6.54): MEd/Mb,Rd = 0,655 1,0 ok
SEMICOMP Member Check
Method 2 auxiliary terms (if applicable):
Choose method for
cross-section resistance
Choose method for
member check
Method 1 auxiliary terms (if applicable):
Lateral torsional buckling Strong axis buckling Weak axis buckling
Section classification for member design check
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.2 MEMBER CHECK
Auxiliary terms are displayed
Reference values
Correction factor k
c
table 6.6
k
c
= 0,969 [-]
y
= 0,997 [-]
max
= 1,237 [-]
z
= 0,883 [-]
0
= 0,961 [-]
w
y
= 1,138 [-]
0,lim
= 0,191 [-]
w
z
= 1,500 [-]
LT
= 0,856 [-]
n
pl
= 0,129 [-] c
y
= 4,280 [-]
a
LT
= 0,998 [-] C
my,0
= 0,996 [-]
b
LT
= 0,063 [-] C
mz,0
= 0,887 [-]
c
LT
= 0,972 [-] C
my
= 0,999 [-]
d
LT
= 0,122 [-] C
mz
= 0,887 [-]
e
LT
= 0,518 [-] C
mLT
= 1,000 [-]
M
cr,0
= 558,807 [-] C
1
= 1,065 [-]
C
yy
= 0,958 [-] C
zy
= 0,820 [-]
C
yz
= 0,500 [-] C
zz
= 0,946 [-]
Method 1 auxiliary terms (if applicable):
C
my
= 0,000 [-]
C
mz
= 0,000 [-]
C
mLT
= 0,000 [-]
Method 2 auxiliary terms (if applicable):
Design of Members
Exercise 6
Semi-Comp Member Design (Windows OS)
SHEET 1.2 MEMBER CHECK
Design of Members
Exercise 6
Design of Members
Exercise 6
Design of Members
Exercise 6
Design of Members
Exercise 6
Calculation (M
cr
only) using:
- LTBeam (Windows OS)
CASE I) Evaluation for top segment
Critical Moment
Critical value of maximum moment Mcr = ??? kN.m
CASE III) Evaluation for complete column type of bracing b)
Critical Moment
Critical value of maximum moment Mcr = ??? kN.m
CASE II) Evaluation for complete column type of bracing a)
Critical Moment
Critical value of maximum moment Mcr = ??? kN.m
Design of Members
Exercise 6
Calculation using:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
Design of Members
Exercise 6
Calculation using:
- Semi-Comp Member Design (Windows OS)
Design of Members
Exercise 6
CORRECTION!!!
C
my
= 0.6 + 0.4 x 0 = 0.6
Design of Members
Exercise 6
CORRECTION!!!
k
yy
= 0.623
CORRECTION!!!
eq. (6.&1) = 0.561
Design of Members
Exercise 6
Calculation using:
- Semi-Comp Member Design (Windows OS)
Design of Members
Exercise 7 SemiComp+
IPE 500 under bending and compression
Use:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
- LTBeam software (Windows OS)
- Semi-Comp Member Design Software (Windows OS)
1,
2,
700
250
75
Ed
y Ed
y Ed
N kN
M kNm
M kNm
=
=
=
Steel Grade: S 355
t
f
h
b
r
t
w
500
200
10.2
16
21
w
f
h mm
b mm
t mm
t mm
r mm
=
=
=
=
=
2 4
4 3 6
3
,
3
,
4
3
,
3
,
116 89.29
48200 1249 10
20.43
1928
4.31
2194
2142
214
336
t
y w
y
el y
z
pl y
z
el z
pl z
A cm I cm
I cm I cm
i cm
W cm
i cm
W cm
I cm
W cm
W cm
= =
= =
=
=
=
=
=
=
=
IPE 500
,
0
2
, , ,
0
2
, , ,
0
35.5
116 4118
1.0
35.5
2194 10 778.87
1.0
35.5
1928 10 684.44
1.0
y
pl Rd
M
y
pl y Rd pl y
M
y
el y Rd el y
M
f
N A kN
f
M W kNm
f
M W kNm
= = =
= = =
= = =
Design of Members
Exercise 7 SemiComp+
N
Ed
= 700kN
M
y1,Ed
= 250kNm
[ N
Ed
]
[ M
y,Ed
]
N
Ed
=
700kN
M
y,Ed,max
=
250kNm
Class 3 Class 4
M
y2,Ed
= 75kNm
M
y,Ed,min
= -75kNm
Decisive cross-section
Equivalent section class
for member design
is class 3
Design of Members
Exercise 7 SemiComp+
Maximum ratios
Outstand flanges under
compression
Internal compression parts under
bending and compression
Plastic limit (Class 2-3) 10 c t c s if 0.5 o > ;
188
6.53 1
c t
c
o
s
Elastic limit (Class 3-4) 14 c t c s if 1 > ;
38
0.653 0.347
c t
c
s
+
c
t
c
t
Maximum width-to-thickness ratios for compression
parts:
Design of Members
Exercise 7 SemiComp+
200 10.2
21
2 2 2 2
4.62 14 11.39 ... the flange is Class 3 or better
16
<10 8.14 ... the flange is Class 2 or better
w
f f
t b
r
c
t t
c
c
= = = < =
=
<9 7.32 ... the flange is Class 1
2 2
500 216 221 38
41.76 56.08
10.2 0.653 0.347
f
w w web
h t r
c
t t
c
c
=
= = = < =
+
... the web is Class 3 or better
>34 27.66 ... the web is Class 3 c =
-
+
-17.08 kN/cm
2
5.01 kN/cm
2
-
+
= 1.0
web
= -0.29
Elastic distribution Plastic distribution
Decisive cross-section at the left support:
235 235
0.8136
355
y
f
c = = =
Cross-Section is Class 3
Design of Members
Exercise 7 SemiComp+
2, ,
10 8.14
y f
| c = =
3, ,
14 11.39
y f
| c = =
2, ,
83 67.53
y w
| c = =
3, ,
124 100.89
y w
| c = =
Interpolation:
Step 1:
Cross-Section design check at the left support:
Step 2:
( )
( )
( )
( )
( )
( )
( )
( )
2, , 2, ,
,
3, , 2, , 3, , 2, ,
/ max ; ; 0
4.62 8.14 41.76 67.53
max ; ; 0 0
11.39 8.14 100.89 67.53
f y f w y w
ref y
y f y f y w y w
c t c t
c t
| |
| | | |
= =
`
)
= =
`
)
( ) ( )
3, , , , , , , , ,
778.87 778.87 684.44 0 778.87
y Rd pl y Rd pl y Rd el y Rd ref y
M M M M c t kNm = = =
,
700
0.17
4118
Ed
pl Rd
N
n
N
= = =
( ) ( )
,3, , 3, ,
1 778.87 1 0.17 646.46
N y Rd y Rd
M M n kNm = = =
M
pl
M
el
c/t
ref
M
M
3,y
0
1
Design of Members
Exercise 7 SemiComp+
The utilisation factors
3
and
el
can be calculated
by increasing the applied loading (N+M
y
) proportionally:
Level of loading
Current Eurocode
New Design Model
,
3
, , ,
,3
700 250
0.49 1
4118 778.87
2.04
y Ed
Ed
pl Rd pl y Rd
u
M
N
N M
q = + = + = <
A =
,
, , ,
,
700 250
0.54 1
4118 684.44
1.85
y Ed
Ed
el
pl Rd el y Rd
u el
M
N
N M
q = + = + = <
A =
N
M
Ed
M
Ed
N
-
N,Rd
M
pl
M
Ed
N
UF
pl
N
+104%
+156 %
N,Rd
M
New Design Model
With only moment in y-axis being present,
the design check simplifies as follows:
1,
,3, ,
250
0.39 1
646.46
y Ed
N y Rd
M
M
= = <
Design of Members
Exercise 7 SemiComp+
Design of Members
Exercise 7
Calculation using:
- Semi-Comp Member Design (Windows OS)
Design of Members
Exercise 7
Calculation using:
- Semi-Comp Member Design (Windows OS)
Member design check:
1
21000
76.41
35.5
y
E
f
t t = = =
,
, 1
1000
0.641
76.41 20.43
pl cr y curve
y
a
cr y y
N L
N i
= = = =
0.21
0.874
y
y
o
_
=
,
, 1
500
1.518
76.41 4.31
pl cr z curve
z
b
cr z z
N L
N i
= = = =
0.34
0.336
z
z
o
_
=
2 2
, 1
2 2
w t z
cr y
z z
I L G I E I
M C
L I E I
t
t
= +
`
)
2 3 2
2
,
2 2
21000 2142 1249 10 500 8100 89.29
1.44 10 804.8 ( .)
500 2142 21000 2142
cons
cr y
M kNm approx
t
t
= + =
`
)
, ,
811.1
cons
cr y LTBeam
M kNm = (Note: )
Left span (L=5m)
approximation
Design of Members
Exercise 7 SemiComp+
Design of Members
Exercise 7
Calculation using:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
( )
( )
( )
( ) ( )
2
2
,mod
1 0.3
1 1
0.35 0.823
2 1.33 0.33 1.33 0.33 0.35
1 0.5 1 1 2 0.8
0.714
1 0.5 1 0.823 1 2 0.879 0.8 0.913 1 0.782
0.913
c
LT
c
LT
LT
k
f k
f
f
_
_
+
= = = = =
(
=
(
(
= = < = = =
3,
,
0.49
778.87
0.879
0.714 1008.3
y y LT curve
LT
c
LT cr y
W f
M
o
_
=
= = =
, , ,
1008.3 ( )
cr y cr y LTBeam
M M kNm exact = =
Whole member
(L=2x5m)
exact
for k
c
Design of Members
Exercise 7 SemiComp+
( )
0.3
0.6 0.4 0.6 0.4 0.3 0.48
my
C
=
= + = + =
( )
1 0.3
0.35
2
0.6 0.4 0.6 0.4 0.35 0.74
mLT
C
+
= =
= + = + =
,
700
0.194
0.874 4118
Ed
y
y pl Rd
N
n
N _
= = =
,
700
0.506
0.336 4118
Ed
z
z pl Rd
N
n
N _
= = =
M
y,Ed,max
=
250kNm
M
y,Ed,min
= -75kNm
for C
mLT
for C
my
Design of Members
Exercise 7 SemiComp+
( )
( ) ( )
( ) ( )
,
,
1 0.2 0.48 1 0.641 0.2 0.194 0.521
0.6 0
0.1 0.1
1 1 0.506 0.896
0.25 0.74 0.25
0
Ed
y
yy my
y pl Rd
yz zz
Ed
zy
mLT z pl Rd
zz
N
k C
N
k k
N
k
C N
k
_
_
| |
= + = + =
|
|
\ .
= =
= = =
=
,
, 3, ,
250
0.194 0.521 0.408 1
0.782 778.87
y Ed
Ed
yy
y pl Rd LT y Rd
M
N
k
N M _ _
+ =
= + = <
,
, 3, ,
250
0.506 0.896 0.874 1
0.782 778.87
y Ed
Ed
zy
z pl Rd LT y Rd
M
N
k
N M _ _
+ =
= + = <
Strong axis:
Weak axis:
Equivalent section class
for member design
is class 3
0.430 < 1
0.930 < 1
Results acc. to
EN 1993-1-1:
Design of Members
Exercise 7 SemiComp+
Design of Members
Exercise 8
IPE 500 under biaxial bending and compression
Use:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
- LTBeam software (Windows OS)
- Semi-Comp Member Design Software (Windows OS)
Steel Grade: S 275
Design of Members
Exercise 8
Design of Members
Exercise 8
CORRECTION!!!
o = 0.783 instead of 0.718
Design of Members
Exercise 8
Calculation using:
- Semi-Comp Member Design (Windows OS)
CORRECTION!!!
Member class is 3 (see UF)
Design of Members
Exercise 8
Design of Members
Exercise 8
Design of Members
Exercise 8
Design of Members
Exercise 8
Design of Members
Exercise 8
Calculation (M
cr
only) using:
- LTBeam (Windows OS)
CASE I) Load applied at top flange
Critical Moment
Critical value of maximum moment Mcr = 495.26 kN.m
CASE II) Load applied at centroid
Critical Moment
Critical value of maximum moment Mcr = 1392 kN.m
Design of Members
Exercise 8
Design of Members
Exercise 8
Design of Members
Exercise 8
Design of Members
Exercise 8
Calculation using:
- Semi-Comp Member Design (Windows OS)
Design of Members
Exercise 8
Calculation using:
- Semi-Comp Member Design (Windows OS)
Design of Members
Exercise 8
Calculation using:
- ECCS EC3 Steel Member Calculator (iPhone/iPad)
Elastic Design of Steel Structures
Global elastic analysis of a 3-storey framed structure. Check of
column
3D View of a module with 2 frames
Plan View of a module with 2 frames
Typical frame
Roof
Storey 2
Storey 1
EXAMPLE 5b
Pinned supports Rigid joints
Beams: Levels 1 and 2 IPE 450
Level 3 IPE 360
Columns: External HEB 220
Internal HEB 260
Material: S 235
Elastic Design of Steel Structures
Combination 7 2 order axial force (approx.)
Combination 7 2 order bending moments (aprox.)
N
Ed
= 884.91 kN (compression);
M
y,Ed
= 115.13 kNm;
V
z,Ed
- negligible
Elastic Design of Steel Structures
3
1
63 . 42
350
14920
) ( cm
L
I
K K
c
c
c
= = = =
3
22 12
86 . 77
650
33740
50 . 1 5 . 1 cm
L
I
K K
b
b
= = = =
354 . 0
86 . 77 86 . 77 63 . 42 63 . 42
63 . 42 63 . 42
12 11 1
1
1
=
+ + +
+
=
+ + +
+
=
K K K K
K K
c
c
q
1
2
= q
Buckling lenghts
N
K
11
K
12
K
21
K
22
L
E
q
1
N
q
2
K
c
K
1
K
2
Columns
Double curvature beams (I
y,IPE 450
= 33740 cm
4
)
N
K11
K12
K21
K22
LE
q1
N
q2
Kc
K1
K2
Pinned base
m L L
y E
73 . 2 50 . 3 78 . 0 78 . 0
,
= = =
m L L
z E
50 . 3
,
= =
m L L
LT
50 . 3 = =
Braced structure
L
E,y
- no sway structure (Woods method)
Elastic Design of Steel Structures
Cross section resistance
Classification
Web:
Flange:
( ) 0 . 9 0 . 1 9 9 77 . 5 5 . 17 24 2 10 2 260 = = < = = c t c
Class 1
Class 1
33 0 . 1 33 33 7 . 17 10 177 = = < = = c t c
Cross section resistance
a
n
M M
Rd y pl Rd y N
5 . 0 1
1
, , , ,
=
318 . 0
4 . 2782
91 . 884
,
= = =
Rd pl
Ed
N
N
n
kN f A N
M y Rd pl
4 . 2782
0 ,
= =
( )
( )
) 50 . 0 ( 231 . 0
11840
5 . 17 260 2 11840
2
< =
=
=
A
t b A
a
f
kNm M kNm M
Ed y Rd y N
13 . 115 5 . 232
231 . 0 5 . 0 1
318 . 0 1
5 . 301
, , ,
= > =
=
kNm
f
W M
M
y
y pl Rd y pl
5 . 301
0
, , ,
= =
Section O.K.
Class 1
Elastic Design of Steel Structures
M
z,Ed
= 0:
0 . 1
1 ,
,
1
s +
M Rk y LT
Ed y
yy
M Rk y
Ed
M
M
k
N
N
_ _
0 . 1
1 ,
,
1
s +
M Rk y LT
Ed y
zy
M Rk z
Ed
M
M
k
N
N
_ _
Member stability (Method 2)
_
y
and _
z
Plan xz - L
E,y
= 2.73 m
259 . 0
00 . 1 9 . 93
1
10 22 . 11
73 . 2 1
2
1
,
=
= =
y
y E
y
i
L
o = 0.34 - Curva b
979 . 0 544 . 0 = =
y
_ |
Plan xy - L
E,z
= 3.50 m
566 . 0
00 . 1 9 . 93
1
10 58 . 6
5 . 3 1
2
1
,
=
= =
z
z E
z
i
L
o = 0.49 - Curva c
805 . 0 750 . 0 = =
z
_ |
kN
f A N
y Rk
4 . 2782
10 235 10 4 . 118
3 4
=
= =
kNm
f W M
y y pl Rk y
5 . 301
10 235 10 1283
3 6
, ,
=
= =
Elastic Design of Steel Structures
_
LT
general method, L = 3.50 m
kNm M M
E
cr cr
2488 77 . 1 = = 348 . 0 = =
cr
y y
LT
M
f W
+ =
(
(
+ =
M Rk y
Ed
y my yy
N
N
C k
_
61 . 0 =
yy
k
hence :
76 . 0 8 . 0 1 61 . 0
1
=
(
(
+ < =
M Rk y
Ed
my yy
N
N
C k
_
Elastic Design of Steel Structures
( ) ( )
94 . 0
4 . 2782 805 . 0
91 . 884
25 . 0 60 . 0
566 . 0 1 . 0
1
25 . 0
1 . 0
1
1
=
(
=
(
=
M Rk z
Ed
mLT
z
zy
N
N
C
k
_
Logo : 94 . 0 =
zy
k
0 . 1 566 . 0 241 . 0 325 . 0
0 . 1 5 . 301 966 . 0
13 . 115
61 . 0
0 . 1 4 . 2782 979 . 0
91 . 884
< = + =
( )
89 . 0
4 . 2782 805 . 0
91 . 884
25 . 0 60 . 0
1 . 0
1 94 . 0 =
(
> =
zy
k
Section O.K.
Elastic Design of Steel Structures
310
Thank you for your attention
luisss@dec.uc.pt