Tutorial 1- Properties and Section Classification
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Transcript of Tutorial 1- Properties and Section Classification
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Tutorial 1
June 4 2013
Properties
Section Classification
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EXAMPLE 1Determine the shape factor for the I-section :
610 x 229 x 125 UB S 275
Section properties can be readily obtained from
tables for Universal beams (refer tables)
Wel,yy = 3220 cm3 Wpl,yy= 3680 cm3 Shape factor = Wpl,yy / Wel,yy= 3680/3220= 1.143 The value of shape factor for most I sections is
about 1.15
SECTIONS CLASSIFICATIONClassification of Steel Cross Sections is Based on :Eurocode 1993- 1-1 : Table 3.1 , Clause 5.5 and Table 5.2
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EXAMPLE 2 A rectangular section shown.
Assumefy=275 N/mm2. Find the following:
20 mm
150 mm
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Elastic section modulusWel= bd2/6= 20 x 1502/6 =75000mm3
Elastic Moment capacityMe = Welx = 75000 x 275 =20625000Nmm = 20.6kNm Plastic section modulus
Wpl = bd2/4=112500mm3 Plastic Moment capacity
Mp = Wplx = 112500 x 275 = 30937500 Nmm= 30.94kNm. Shape factor for section
Shape factor = Wpl / Wel= 112500/75000=1.5
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EXAMPLE 3Section 610 x 229 x 125 UB S 275 subject to pure bending
Section properties can be taken from published tables tf=19.6 mm tw = 11.9 mm B=229 mm r=12.7 mm
Table 3.1 : since tw=11.9 mm fyw=275 N/mm2
Since tf =19.6 mm fyf= 275 N/mm2
Therefore fyis taken as 275 N/mm2 ; the lower value
c = 229/211.9/212.7= 95.85
c/tf=95.85/19.6=4.89 d/tw= 46.0
Since the section is symmetrical about the major axis i.e the axis ofbending and consequently the neutral axis is atmid-height
92.0275
235
f
235 5.05.0
y
For flange;
For web; =0.92
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Outstand
compression
element
Actual
c/tf = 4.89 =0.92
Limits
9=9x0.92=8.28
10=10x0.92=9.214=14x0.92=12.88
Flange
outstand is
plastic orclass 1
Web
Since purebending;
neutral axis is
at mid depth
Actual
d/tw =46=0.92
Limits
72=72x0.92=66.2483=83x0.92=76.36
124=124x0.92=114.1
Since
d/t
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The cross section is plastic since all elements are
plastic
Note: if one element is having a different classification;then the section will take the lower one
Example: if flange is compact and web plastic ; thensection will be compact
Example: if flange is semi-compact and web plastic;then section is semi-compact
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EXAMPLE 4Section 610x229x125 UB S275 under pure compression load of
3200 kN
Section Properties: Take from tables
Design strength Py:
tw=11.9 mm therefore fyw=275 N/mm2using table 3.1
tf =19.6 mm Therefore fyf=275 N/mm2using table 3.1
Classifying flange
92.0275235
f235
5.05.0
yf
c/tf = 4.89
Since B/T
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Classifying the web
The limiting value of d/t is 33, 38 and 42 Actual d/t=46.0 Therefore the web is Slender
The section takes the lowest classification for flange andweb So the section is slender
92.0
275
235
f
235 5.05.0
yw
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EXAMPLE 5
Double angle strut composed of two 100 x 75 x 10 S 355 angles
connected to gusset plates by the long legs back to back and
subject to axial compression
10
0mm
b
75 mm
Connected leg
Outstand leg
d
t , thickness
See figure 5.2 sheet 3 of the codeDimensions and Properties of the section are availableB=75 mm; d = 100 mm; t=10 mmb/t=75/10=7.5; d/t=100/10=10; (b+d)/10=17.5
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Table 9: Since t=10 mm (< 40 mm) therefore py=355 N/mm2
Table 5.2 sheet 3of3): For double angles with components separated underaxial compression :Class 1 and 2 are not applicable; For class 3 threecriteria must be satisfied to comply with a semi-compact classificationnamely d/t
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Outstand b/t=7.5 Limit values15 =15x0.81=12.15 satisfied
Connected leg d/t=10 Limit values
15 =15x0.81=12.15
satisfied
Combined (b+d)/2t=8.75 Limit value
11.5=11.5x0.81=9.315 satisfied
Since all three criteria are satisfied; section is class 3
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EXAMPLE 6
Determine the classification for section 356 x 127 UB 33 Grade 275
subject bending
Wpl,yy = 539.8 cm3 ; Iyy= 8200 cm4; B = 125.4 mm ; D = 348.5 mm ;
t = 5.9 mm ; T = 8.5mm ; r = 10.2 mm (figure next slide)
Table 5.2 is to be used.
92.0275
235
f
235 5.05.0
y
CompressionFlange Actualb/T= 5.82 Limits : 9, 10 149x0.92=8.28Flangeis plastic
Web: subjectto bending d/t =52.7
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EXAMPLE 7
Determine the classification for the welded section of steelgrade 355 used as compression member, shown below
400
T=10mm
D=420
t=10
Flange classification:
81.0355
235
9.1810
62/102002/2/
T
stBw
Limits are 9, 10and 14for welded sections
(same as for rolled sections)
Since 14
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EXAMPLE 8
Classify a hot rolled circular section 273.0 x 6.3 CHS S275 subject to
compression due to bending.
CHS =circular hollow section
6.3 mm
273 mm
D/t = 273/6.3 = 43.33
Table 3.1: since t = 6.3 mm ; fy = 275 N/mm2
92.0
275
235
f
235 5.05.0
y
Therefore:
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EN 1993 1-1 Table 5.2: The limiting values of D/t
Limiting
value
Actual D/t
Class 1
Plastic
502 50x0.922= 42.32 43.33
Class 2
Compact
702 70x0.922= 59.25
Class 3
Semi
compact
902 140x0.922=118.496
Since 42.32< actual D/t < 59.25; the section is
classified as compact
EXAMPLE 9
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EXAMPLE 9
The figure below shows a hot formed rectangular hollow section.
500mm
300 mm
The properties of the section are:
t=10 mm,
A = 155 cm2,
d/t = 47.0,
b/t = 27.
Determine the following:
A - Classification for section under flexure
ANSWER: Assuming S275 steel
92.0
275
235
f
235 5.05.0
y
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FLANGE:
c/tf={300- 2x10}/10=28 Compression flange /internal element :
choose part subject to compression whose limit for plastic section is
33=33 x0.92=30.36.flange is plasticuse table 5.2
Also -d/t=80x 147 =33 ..since. b/T
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STEP 1
Assuming S275 steel
Thickness of flange = 10 mm < 40 mm.fyf= 275 N/mm2
Thickness of web = 10mm < 40mm.fyw=275 N/mm2
STEP 2
= (235/275)0.5= 0.92
STEP 3Flange aspect ratio , b/T = 27
Compression flange /internal element : choose part subject to
compression whose limit for plastic section is
33=33 x0.92=30.36.flange is plasticuse table 5.2
B- The axial compressive load below which the section is non-slender
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Web aspect ratio , d/t = 47
For web in compression, the limiting value
of d/t is 33, 38and 42
Actual d/t=47.0 Therefore the web is
Slender.
The section takes the lowest classification
for flange and web
So the section is slender.