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8/9/2019 Lessen of Steel Design
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Eurocode 3-1-1: 2.2.2.2 (3)
The self-weight of the structure may, in most
cases, be calculated on the basis of the nominal
dimensions and mean unit masses.
Eurocode 3-1-1: 2.3.2.4 (6)
The self-weight of any unrelated structural or
non-structural elements made of different
construction materials should be treated as
different permanent actions.
Self-weights calculationSelf-weights calculation
Calculating the self-weight of the following floor constructionCalculating the self-weight of the following floor construction
Floor constructions of commercial building
consists of
50 mm sand/cement screed (tasoituslaasti)
150 mm normal weight reinforcement concrete
slab
12 mm plaster(gypsum mortar) ceiling
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Materials Density γγγγ
[kN/m3]
concrete (see ENV 206)
Lightweight (varies with density class) 9 - 20
normal weight *24
heavyweight >28
reinforced and prestressed concrete;
unhardened concrete
+1
mortar
cement mortar 19 - 23
gypsum mortar; lime mortar 12 - 18
lime-cement mortar 18 - 20
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Component Density
(kN/m3)
Thickness
(m)
Load intensity= density
×××× thickness (kN/m2)
Screed 20
(19 - 23)
0,05 20 ×××× 0,05 = 1,0
Slab 24 0,15 24 ×××× 0,15 = 3,6
Plaster 15(12 - 18)
0,012 15 ×××× 0,012 = 0,2
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A building in Helsinki with duo pitched roof,calculating the snow load distributions.
αααα 1 24
αααα 2 36
s = µ·sks = µ·sk
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αααα [0° 15°] [15° 30°] [30° 60°] ≥≥≥≥
60°
µµµµ1 0.8 0.8 0.8(60 - αααα)/30 0.0
µµµµ2 0.8 0.8 + 0.6(αααα-15)/30 1.1(60-αααα)/30 0.0
µµµµ 2.24 0.8 0.6
24 15−( )
30⋅+:= µµµµ 2.24 0.98
µµµµ 1.36 0.8 60 36−( )
30⋅:= µµµµ 1.36 0.64
µµµµ 1.24 0.8
µµµµ2.36
0.88µµµµ 2.36 1.1
60 36−
30⋅:=
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0.980.98 0.640.64
0.5x0.8=0.40.5x0.8=0.4
0.80.8
0.880.88
0.5x0.64=0.320.5x0.64=0.32
The characteristic snow load on the ground sk is
2.0 kN/m2. The possible snow loads on the roof
are:
0.98x2=1.960.98x2=1.96 0.64x2=1.280.64x2=1.28
0.4x2=0.80.4x2=0.8
0.8x2=1.6
0.8x2=1.6
0.88x2=1.760.88x2=1.76
0.32x2=0.640.32x2=0.64
In most cases it may be conservatively assumed
that maximum snow load intensity is applieduniformly across the full width
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Persistent and transient design situations for
verifications other than those relating to fatigue
(fundamental combination)
Σγ G.j Gk.j + γ Q.1Qk.1 + Σγ Q.j Ψ0.iQk.i (1)
Accidental design situation
Σγ G.A Gk.j + Ad + Ψ1.1 Qk.1 + ΣΨ2..iQk.i (2)
For building structures, as a simplification, (1)
can be replaced by whichever of the following
combination gives the larger value:
Σγ G.j Gk.j + γ Q.1Qk.1
Σγ G.j Gk.j + 0.9Σγ Q.j Qk.i
Combination of actions for ultimate limit designCombination of actions for ultimate limit design
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Three combinations of actions are defined:
Rare combination
ΣGk.j + Qk.1 + ΣΨ0.iQk.i (1)
Frequent combinationΣGk.j + Ψ1.1 Qk.1 + ΣΨ2..iQk.i (2)
Quasi-permanent combinations
ΣG
k.j+ΣΨ2..i
Qk.i
(3)
For building structures, as a simplification, (1)
can be replaced by whichever of the following
combination gives the larger value:
ΣGk.j + Qk.1
ΣGk.j + 0.9Σ Qk.i
Combination of actions for serviceability limit designCombination of actions for serviceability limit design
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Eurocode 3-1-1: 2.3.2.3 (5)
For continuous beams and frames, the same
design value of the self-weight of the structure
may be applied to all spans, except of cases
involving the static equilibrium of cantilevers
Load combinationsLoad combinations
Write the critical load combination for continuous beam with
permanent load and imposed load
Write the critical load combination for continuous beam withpermanent load and imposed load
Eurocode 3-1-1: 2.3.2.4 (3)
Variable actions should be applied where they
increase the destabilizing effects but omitted
where they would increase the stabilizing
effects
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Load combination 1: maximum bending in span 1
Load combination 1: maximum bending in span 1
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Load combination 2: maximum bending in span 2Load combination 2: maximum bending in span 2
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Span LSpan L
SpacingSpacing
SpacingSpacing
snow load ssnow load s
purlinspurlins
purlinpurlin
purlinpurlin
Sheeting gSheeting gSheeting gSheeting g
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Write load combinations for ultimate limit state
Basic load combination ruleBasic load combination rule
Point load and
distributed load
Point load and
distributed load
Two variable loadingsTwo variable loadings
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Combination
Case aCase a
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Final critical load combinationFinal critical load combination
Case bCase b
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Load combinationLoad combination
aa
bb
1.35 g⋅ 1.5 q F+( )⋅+
1.35 g⋅ 0.9 1.5⋅ q F+( )⋅+
F is assumed to be variable load in
this example and can be added to q
F is assumed to be variable load in
this example and can be added to q
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Final critical load combinationFinal critical load combination
Case aCase a
Case bCase b
M max max M max_a M max_b,( )
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aa
Determine the maximum
bending moment in Member 1
Determine the maximum
bending moment in Member 1
11
22
the moment of inertia of member
1 and 2 are the same
the moment of inertia of member
1 and 2 are the same
L = hL = h
1.35 g⋅ 1.5 s⋅+ snow load dominatessnow load dominates
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bb 1.35 g⋅ 1.5⋅+ wind load dominateswind load dominates
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cc 1.35 g⋅ 1.35 s W+( )⋅+
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The final critical load will be the one with the maximum
bending moment from the following figures
The final critical load will be the one with the maximum
bending moment from the following figures
aa
bb
cc
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General
• Applied to ultimate Limit State
Design
• Benefits of classifying cross-sections
– guide selection of globe analysis – determine the design criteria of
member
• Rules that guide the classification
– Width-to-thickness ratio – yield strength
– loading: bending, compression, bending+compression
• Based on normal stress. Shear buckling is considered separately indesign rules
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outstand flangeoutstand flangenormal stress
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loading cases: bending or compression
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(d)
(c)
(b)(a)
Simply supported onall four edges
t
L
b
Simply supported
edge
Free
edge
b
L
1
2
3
4
5
1 2 30 4 5
Plate aspect ratio L / b
Buckling coefficient k
b
L FreeExact
k = 0.425 + (b/L)2
0.425
( )2
2
2
112
−=
b
t k cr
σ
width to thickness ratio
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Class 1Class 2
Class 3
Euler Buckling Stress
0,5 0,6 0,9
1
1,0 λ p
N f p
u
y
= σ
==
k 4.28
/5.0
σ
λ t b
cr
( )2
2
2
112
−=
b
t k cr
σ
Np = σu / f y
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Outstand
Internal
Web
Flange
Web
Internal
Flange
(a) Rolled I-section (b) Hollow section
Flange
(c) Welded box section
InternalOutstand
InternalWeb
Internal: webs of open beams, flanges of boxes
Outstand: flanges of open section, legs of angles
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a. Webs: (internal elements perpendicular to axis of bending)
tw twd
tw d tw
tf
hdAxis of
Bending
d = h-3t (t = tf = tw)
ClassWeb subject to
bending
Web subject to
compression
Web subject to bending
and compression
Stressdistribution in
element(compression positive)
+ f y
f y -
d h
+ f y + f y
f y - f y -
d h hdαd
when α > 0,5:d/t w < 396ε/(13α − 1)when α < 0,5:
d/t w < 36ε/α
_
_
1 d/t w < 72ε _ d/t w < 33 ε _
d/tw < 83 ε _ d/t w < 38ε _ 2
when α > 0,5:d/t w < 456ε/(13α − 1) _ when α < 0,5:d/t w < 41,5ε/α _
Stress
distribution inelement
(compression positive)
+ f y
f y -
+ f y
+ f y
ψ f y -
d/2
d/2h
+
d h d h
3 d/t w < 124 ε _ d/t w < 42 ε _
when ψ > −1:d/t w < 42ε/(0,67 + 0,33ψ) _
when ψ < −1: _
d/t w < 62ε/(1 − ψ) (− )ψ
ε = f y/235f y
ε
235 275 355
1 0,92 0,81
_
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b. Internal flange elements: (internal elements parallel to axis of bending)
axis of
bending
b tf t f b
tf
b
b tf
Class TypeSection in bending
Section in compression
Stress distribution
in element andacross section
(compression
positive)
+-
f y
- +
+-
f y
- +
1
Rolled hollow section
Other
(b - 3t f )/ t f
b / t f
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ε= √235/355 = 0.8136
Welded profile, flange
9ε = 7.3 10ε = 8.1 14ε = 11.4
Web
33ε = 26.8 38ε = 30.9 42ε = 34.2
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(a) Class 4 cross-sections - axial force
Gross cross-section
Gross cross-section
Centroidal axis of
gross cross-section
Centroidal axis ofgross cross-section
Centroidal axis ofeffective cross-section
Non-effective zones
Non-effective zone
Non-effective zone
Centroidal axis
Centroidal axisCentroidal axis of
effective section
e N
eM
e M
Centroidal axis ofeffective section