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.