-
Institute for Steel Structures and Shell Structures Lessingstr.
25/3 A-8010 Graz tel: +43-316/873-6200 (or 6206) fax:
+43-316/873-6707 [email protected] [email protected] home:
www.stahlbau.TUGraz.at
Valorisation Project: SEMI-COMP+ n RFS2-CT-2010-00023
"Valorisation Action of Plastic Member Capacity of Semi-Compact
Steel Sections a more Economic Design"
BACKGROUND INFORMATION - EXTRACT
9th October 2011
WORKED EXAMPLES
TABULATED M3,Rd-VALUES
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SEMI-COMP+ Table of Contents
2
TABLE OF CONTENTS
1. ANNEX A Worked Examples
..........................................................................................................
3
1.1. Example 1: HEAA 300 under biaxial bending and compression
.............................................. 3
1.2. Example 2: IPE 500 under bending and compression
........................................................... 12
1.3. Example 3: Welded section under pure bending
..................................................................
21
1.4. Example 4: RHS 250x150x6 under biaxial bending and
compression ................................... 27
1.5. Example 5: SHS 200x6.3 under bending and compression
.................................................... 36
1.6. Example 6: SHS 200x6.3 under pure bending
........................................................................
43
1.7. Example 7: IPE 500 under N+My+Mz including class for member
design .............................. 48
2. ANNEX B - Tabulated Values FOR W3,y and W3,z
............................................................................
52
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SEMI-COMP+ Worked Example 1
3
1. ANNEX A WORKED EXAMPLES
1.1. Example 1: HEAA 300 under biaxial bending and
compression
1.1.1. Loading
The first worked example detailed herein deals with an HEAA 300
in S355 steel. It is assumed as simply supported beam-column (L = 5
m), and is subjected to the following combined axial force and
biaxial bending moments:
1.1.2. 1.2. Material
According to EN 1993-1-1, the material properties of the
mentioned S355 steel are:
2
2
2 2
0
210000
81000
355 35.5
1.0
y
M
NEmmNG
mmN kNf
mm cm
=
=
= =
=
1.1.3. Geometry and section properties
The geometrical dimensions as well as relevant properties of the
studied cross-section HEAA 300 are:
,
,
946.887
34
Ed
y Ed
z Ed
N kNM kNmM kNm
=
=
=
2833007.510.5
27
w
f
h mmb mmt mmt mmr mm
=
=
=
=
=
h
b
r
tf
tw
2 4
4 3 6
3,
3,
4
3,
3,
88.91 49.3513800 877.2 10
12.469767.31065
4734
315.6
482.3
t
y w
yel y
zpl y
z
el z
pl z
A cm I cmI cm I cm
i cmW cmi cmW cm
I cmW cm
W cm
= =
= =
==
==
=
=
=
-
SEMI-COMP+
The plastic and elastic resistances of
,0
, , ,0
, , ,0
, , ,0
, , ,0
35.5 88.91 3151.0
35.5 1065 101.0
35.5 482.3 11.0
35.5 976 101.0
3
ypl Rd
M
ypl y Rd pl y
M
ypl z Rd pl z
M
yel y Rd el y
M
yel z Rd el z
M
fN A
fM W
fM W
fM W
fM W
= = =
= =
= =
= =
= =
35.515.6 101.0
1.1.4. Internal forces
Both major axis and minor axis ben
with y = z = 0) and applied at theinvestigated system and the
loading
87 kNm
34 kNm
My
Mz
W
a HEAA 300 are obtained as follows:
2
2
2
56.3
0 378.1
10 171.2
346.5
kN
kNm
kNm
kNm
=
=
=
20 112.0kNm =
ding moment distributions are supposed to be tr
e same end of the member. The following figurecase.
5 m
Worked Example 1
4
riangular (linear
e illustrates the
-
SEMI-COMP+ Worked Example 1
5
1.1.5. Stress distribution
The following figure shows the elastic and the plastic stress
distribution in the most critical section at the left support (the
location of the maximal utilisation).
1.1.6. Cross-section classification
The check whether the cross-section is Class 3 or better is done
with the actual elastic stress distribution. The following table
shows the maximum width-to-thickness ratios for compression parts
to clarify the new limits that determine the class of
cross-section.
2
235 235 0.8136355
0.57 0.21 0.72 0.07 0.72 0.455yf
k
= = =
= + =
300 7.5 272 2 2 2 11.36 21 11.53 ... the flange is Class 3 or
better
10.5
10 8.14 ... the flange is Class 3
w
f f
tb rc kt t
= = = < =
> =
Maximum ratios Outstand flanges under bending and compression
Internal compression parts under
bending and compression
Plastic limit (Class 2-3) 10c t
if 0.5 > ;
1886.53 1
c t
if 0.5 ; 41.5c t
Elastic limit (Class 3-4)
21c t k In this case:
20.57 0.21 0.07k = +
if 1 > ; 380.653 0.347
c t
+
if 1 ; ( ) ( )62 1c t
-
+
PNA
web = 1.0
flange = 1.0
-0.25 f y-0.86 f y
+0.26 f y-0.35 f y
-0.12 f y
-0.48 f y
-0.62 f yflange = 0.72
web = 0.24
-
SEMI-COMP+ Worked Example 1
6
2 2 283 210.5 227 3827.73 41.96 7.5 0.653 0.347
... the web is class 3 or better
f
w w web
h t rct t
= = = < =+
188 34 27.666.53 1
... the web is Clas
> = =
s 3
So the cross-section at the left support is found to be in Class
3.
1.1.7. Cross-section design check
The cross-section capacity is checked with the design model for
H-shaped sections from the Semi-Comp Design Guidelines.
The decisive cross-section check at the left support reads:
Interpolation:
2, , 10 8.14y f = = 3, , 14 11.39y f = =
2, , 83 67.53y w = = 3, , 124 100.89y w = =
2, , 10 8.14z f = = 3, , 16 13.02z f = =( )( )
( )( )
( )( )
( )( )
( )( )
( )
2, , 2, ,,
3, , 2, , 3, , 2, ,
2, ,,
3, , 2, ,
11.36 8.14 27.73 67.53/ max ; ;0 max ; ;0 0.99
11.39 8.14 100.89 67.53
11.36 8.14/ max ;0
13.02
f y f w y wref y
y f y f y w y w
f z fref z
z f z f
c t c tc t
c tc t
= = =
= =
( )0.66
8.14=
Step 1:
( ) ( )( ) ( )
3, , , , , , , , ,
3, , , , , , , , ,
378.1 378.1 346.5 0.99 346.8
171.2 171.2 112.0 0.66 132.2
y Rd pl y Rd pl y Rd el y Rd ref y
z Rd pl z Rd pl z Rd el z Rd ref z
M M M M c t kNm
M M M M c t kNm
= = == = =
Step 2:
,
946,8 0.33156
Ed
pl Rd
NnN
= = =
( ) ( )( ) ( )
,3, , 3, ,
2 2,3, , 3, ,
1 346.8 1 0.3 242.8
1 132.2 1 0.3 120.3N y Rd y Rd
N z Rd z Rd
M M n kNm
M M n kNm
= = == = =
Step 3:
, ,
,3, , ,3, ,
1y Ed z EdN y Rd N z Rd
M MM M
+
where 2 = and 5 1n =
-
SEMI-COMP+ Worked Example 1
7
The cross-section check for the given class-3 section then
is:
2 1.587 34 0.279 1242.8 120.3
+ =
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SEMI-COMP+
Utilisation factor along the member
1.1.8. Member design check
The member capacity is checked wfactors are those for compact
sectionmembers will be based on the classcorresponds to
cross-section at the l
, , 500cr y cr zL L cm= =
121000 76.435.5y
Ef
= = =
,
, 1
50076.41 12.4
pl cr yy
cr y y
N LN i
= = =
,
, 1
50076.41 7.3
pl cr zz
cr z z
N LN i
= = = =
W
length according to EN 1993-1-1:
with the existing Eurocode method 2 formulae. Tns and
M3,(y,z),Rd are used instead of Mpl,(y,z),Rd. The d of the
cross-section with maximum utilisation (i.eft support.
41
0.5256
curveb=
0.34
0.873y
y
==
0.490.602
z
z
==
0.896 curvec=
Worked Example 1
8
The interaction design check for .e. Class 3); this
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SEMI-COMP+ Worked Example 1
9
So as to calculate the normalized slenderness for
lateral-torsional buckling, it is necessary to calculate the
elastic critical moment for LT-buckling considering bending about
the major axis.
( )( ) ( ) ( )
2 22 2
, 1 2 3 2 32 2z twz z
cr y g j g jw z zz
k L G IIE I kM C C z C z C z C zk I E Ik L
= + +
For this case of study, the previous formula can be simplified
as follows:
22
, 1 2 2w tz
cr yz z
I L G IE IM CL I E I
= +
where C1=1.77 according to moment distribution.
2 3 22
, 2 2
21000 4734 877.2 10 500 8100 49.351.77 10 1177.2500 4734 21000
4734cr y
M kNm
= + =
Then, the normalized slenderness can be determined as:
3,
,
0.34346.8 0.5430.9421177.2
y y LTcurveLT b
LTcr y
W fM
=
= = = =
In order to check the accuracy of the elastic critical moment,
an additional calculation of Mcr,y using the software LTBeam was
carried out.
, , 1199.8cr y LTBeamM kNm=
The discrepancies between the analytical value and the LTBeam
value are acceptable for the case of study.
( ) ( ) ( ) ( )2 2,mod ,mod
1 10 0.7521.33 0.33 1.33 0.33 0
1 0.5 1 1 2 0.8 1 0.5 1 0.752 1 2 0.543 0.8 0.892 1
0.942 1.00.892
c
LTc
LTLT LT
k
f k
f
= = = =
= = =
-
SEMI-COMP+ Worked Example 1
10
Next step is to determine the interaction factors for members
susceptible to torsional deformations:
plastic interaction factors for Class 3 according to the
SEMI-COMP approach
,
946.8 0.3440.873 3156
Edy
y pl Rd
NnN
= = =
,
946.8 0.4980.602 3156
Edz
z pl Rd
NnN
= = =
( ) ( )( )
( ) ( )( )
,
,
,
1 0.2 0.6 1 0.525 0.2 0.344 0.667
0.6 0.6 0.956 0.574
0.1 0.1 0.8961 1 0.498 0.8730.25 0.6 0.25
1 2 0.6 0.
Edyyy my
y pl Rd
yz zz
z Edzy
mLT z pl Rd
Edzzz mz
z pl Rd
Nk CN
k k
NkC N
Nk CN
= + = + =
= = =
= = =
= + =
( )( )6 1 2 0.896 0.6 0.498 0.956 + =
For comparison, the elastic interaction factors according to EN
1993-1-1 lead to
0.665
0.761
0.936
0.761
=
=
=
=
yy
yz
zy
zz
kkkk
Finally, the member check for strong and weak axis results in
the following expressions:
Strong axis:
, ,
, 3, , 3, ,
1y Ed z EdEd yy yzy pl Rd LT y Rd z Rd
M MN k kN M M
+ +
87 340.344 0.667 0.574 0.659 11 346.8 132.2
+ + =
