design philosophy.pdf

Upload: rufus-cheng

Post on 04-Apr-2018

225 views

Category:

Documents


0 download

TRANSCRIPT

  • 7/30/2019 Design philosophy.pdf

    1/12

    CSE 362 Design of Steel StructuresDesign philosophy, allowable stress design, design examples,limit stress design, comparison

    Professor K.F. ChungProfessor K.F. Chung

    Department of Civil and Structural EngineeringThe Hong Kong Polytechnic University

    Hong Kong SAR, China

    2

    Overall stability / equilibrium considerations / Linear elastic design

    ACTION RESISTANCE

    Section capacityAxial forces

    Tension

    Compression

    where F is the applied forceA is the cross-sectional areaDesign is the design strength

    Shear force

    Bending moment

    DesignA

    F =

    Design

    A

    F =

    DesigntI

    yAS =

    DesignZ

    M =

    Allowable stress design

  • 7/30/2019 Design philosophy.pdf

    2/12

    3

    F = 100 kN

    DataDesign strength of mild steely,ASD = 165 N/mm

    2

    Consider

    DesignAF =

    A

    FDesign

    165

    10x100

    FA

    3

    Design

    required =

    Use a 25 x 25 mm square rod with the following cross-sectional area:

    A = 625 mm2 Arequired = 606 mm2

    Design of a tie

    = 606 mm2

    4

    DataP = 10 kNDesign strength of mild steelDesign = 165 N/mm

    2 or 0.6 x 275 N/mm2

    Density of steel = 7850 kg/m3

    Unit cost= $20 /kg

    Assume no instability.

    Truss 1

    AB

    C

    P1.0 m

    1.0 m

    -P

    P2+

    LABLAC

    Truss 2

    A

    B

    C

    P

    1.0 m

    1.0 m

    +P

    P2LABLAC

    2.0 m

    Truss 3

    A

    C

    P

    LABLAC

    B

    2

    P+

    2P

    1.0 m

    Design of a two-member truss

  • 7/30/2019 Design philosophy.pdf

    3/12

    5

    Member Length (m) Total length (m) Force (kN)

    Tensionmembers

    AB 2.8

    7.6

    141

    BC 2.0 200

    CE 2.8 283

    Compressionmembers

    AD 4.0

    10.8

    100

    BD 2.8 141CD 2.0 200

    ED 2.0 200

    2.0 m 4.0 m

    P = 100 kN

    P

    2.0 m

    -2P -P

    -2P

    +2P

    P2+P2

    P22+

    A

    BC

    DE

    2.0 m

    Design of a cantilever truss

    6

    (N/mm2)

    Design = 275 N/mm2

    t = 183 N/mm2 with load factor 1.5 tension

    c = 100 N/mm2 compression due to buckling

    Spare strength / reserved strength

    PlasticElastic

    Simplified stress-strain curve

    Design of a cantilever truss

  • 7/30/2019 Design philosophy.pdf

    4/12

    7

    For S275 steel, the yield strength is 275 N/mm2. With a factor of safetyequal to 1.5, the design strength of the members under tension is

    For those members under compression, buckling may occur, and hence,the design strength is further reduced to

    c = 100 N/mm2 for I section

    No reduction in the design strength for square hollow section incompression.

    tensionmm/N1835.1

    275 2t ==

    SchemeS275 steel

    Design strength (N/mm2)

    Tension Compression

    I Solid bar 183 100

    II Joist 183 100

    III Square hollow section 183 183

    Design of a cantilever truss

    8

    Design for member under tension (Scheme I)

    F = 283 kN (largest tension force in member CE)

    t = 183 N/mm2

    The cross sectional area, A, of the member required is

    Arequired = F /t = 283 x 103 / 183

    = 1546 mm2

    Use 40 x 40 mm solid bar

    Aprovided = 40 x 40

    = 1600 mm2 > Areq = 1546 mm2

    or the stress,= F / Aprovided= 283 x 103 / 1600

    = 176.9 N/mm2

  • 7/30/2019 Design philosophy.pdf

    5/12

    9

    F = 200 kN (largest compression force in members CD & ED)

    c = 100 N/mm2

    The cross sectional area, A, of the member required is

    Arequired = F /c = 200 x 103 / 100= 2000 mm2

    Use 45 x 45 mm solid bar

    Aprovided = 45 x 45

    = 2025 mm2 > Areq = 2000 mm2

    or the stress,= F / Aprovided= 200 x 103 / 2025

    = 98.8 N/mm2 Areq = 1546 mm2

    or the stress,= F / Aprovided= 283 x 103 / 1620

    = 174.7 N/mm2

  • 7/30/2019 Design philosophy.pdf

    6/12

    11

    F = 200 kN (largest compression force in members CD & ED)

    c = 100 N/mm2

    The cross sectional area, A, of the member required is

    Arequired = F /c = 200 x 103 / 100= 2000 mm2

    Use 89 x 89 x 19.4 kg/m Joist

    Aprovided = 24.9 cm2

    = 2490 mm2 > Areq = 2000 mm2

    or the stress,= F / Aprovided= 200 x 103 / 2490

    = 80.3 N/mm2 Areq = 1546 mm

    2

    or the stress,= F / Aprovided= 283 x 103 / 1690

    = 167.5 N/mm2

  • 7/30/2019 Design philosophy.pdf

    7/12

    13

    F = 200 kN (largest tension force in member CE)

    c = 183 N/mm2 (=t =y / 1.5)

    No reduction for buckling is necessary.

    The cross sectional area, A, of the member required isArequired = F /c = 200 x 10

    3 / 183

    = 1093 mm2

    Use 90 x 90 x 3.6 SHS, 9.3 kg/m

    Aprovided = 1240 mm2 > Areq = 1093 mm

    2

    or the stress,= F / Aprovided= 200 x 103 / 1240

    = 161.3 N/mm2

  • 7/30/2019 Design philosophy.pdf

    8/12

    15

    Scheme

    Section designationMemberlength

    (m)

    *Weight

    (kg)

    Totalweight

    (kg)

    **Unitcost

    ($/kg)

    Totalcost

    ($)DimensionsUnit

    weight(kg/m)

    IT

    C

    40 x 40 Solid bar

    45 x 45 Solid bar

    12.6

    15.9

    7.6

    10.8

    96

    172268 30 8040

    IIT

    C

    76 x 76 x 12.7 Joist

    89 x 89 x 19.4 Joist

    12.7

    19.4

    7.6

    10.8

    97

    210307 15 4605

    IIIT

    C

    90 x 90 x 5.0 SHS

    90 x 90 x 3.6 SHS

    13.3

    9.3

    7.6

    10.8

    102

    101203 18 3654

    * Minimum amount of materials (weight/tonnage) does NOT always meansmost economical design.

    ** The unit cost also includes cost of fabrication / welding / erection.

    Design of a cantilever trussTwo section sizes

    16

    Scheme One sectionsize

    Two sectionsize

    Ratio

    One : Two

    I 8790 (2.0) 8040 (2.2) 1.09 : 1.0

    II 5355 (1.2) 4605 (1.3) 1.16 : 1.0

    III 4410 (1.0) 3654 (1.0) 1.21 : 1.0

    Scheme III is always preferred based on the above assumption and

    data according to design strength unit cost (material / fabrication / welding / erection) availability of materials

    Although there is saving of about 20% in using two section sizes,only one section size may be used in practice due to small quantity.The normal unit of purchase is one ton.

    Design of a cantilever truss

  • 7/30/2019 Design philosophy.pdf

    9/12

    17

    As demonstrated in the above example Design of a truss, thefollowing data are very important:

    section properties of members, such as cross-sectional area

    design strength of materials under different types of internal forces- t tension- c compression (buckling ?)- s shear- b bending- tor torsion

    partial safety factor for materials under different types of internalforces

    unit cost of material / fabrication / welding / erection / time

    Structural design

    18

    Linear elastic design, i.e. all members of a structure operate withinthe linear elastic range of the stress-strain curve of the steelmaterial. Moreover, the structural response of all the membersoperate within the linear elastic range of the load-deformation curvesof the structure.

    DataLoading, F obtained from loading allowance

    Strength of material, fy obtained from tests or codesSafety factor, = 1.5 ~ 2.0

    Design strength ft =

    Section properties of the members, A

    fy

    Linearelasticrange

    LOAD RESISTANCE

    Applied stress Design strength

    F/A ft = fy/

    Allowable stress designWorking stress design, or permissible stress design

  • 7/30/2019 Design philosophy.pdf

    10/12

    19

    Partial safety factors to allow for extreme situations

    -f on loading G for permanent actions

    Q for variable actions

    Loads U.K. H.K. EuropeDead load (D)Imposed load (L)Wind load (W)

    1.41.61.4

    1.41.61.4

    1.31.51.3

    D + L + W 1.2 1.2 1.1

    Dead load restraining upliftand overturning

    1.0 1.0 1.0

    -M on materials/resistance

    Steel M = 1.0 or 1.05 (tension)

    = 1.1 (local buckling)

    = 1.2 (connection)

    Concrete M = 1.5 (compression)

    Partial safety factors for extreme situations

    Limit state designUltimate limit state

    20

    DataLoad = FPartial safety factor,f = 1.0 ~ 1.6Factored design load =f x F (increased)Strength of material = fy from tests or codes (y ormax)Partial safety factor,M = 1.0 ~ 1.5Section properties, A

    Fullrange

    LOAD RESISTANCE

    Internal forces due toapplied load

    Section capacityMember resistance

    ftA = fyA /

    ULS - strength (yielding, buckling, rupture)- stability against overturning & sway- fracture due to fatigue- brittle fracture

    Ff M

    Limit state designUltimate limit state

  • 7/30/2019 Design philosophy.pdf

    11/12

    21

    Long term behaviour (not extreme situation) tends to happenduring the life time of the structure.

    deflection / deformation (most common)

    vibration, fatigue, corrosion & durability

    From deflection, the design criterion is

    Expected deflection < Limits of deflection

    Compared to load/resistance calculation, the deflection is always veryapproximate. It is the order of magnitude other than the exact value moremeaningful.

    Typical deflection limits

    deflection due to unfactored imposed load

    beams carrying plaster or other brittle finish

    all other beams

    180

    Span

    360

    Span

    200

    Span

    Limit state designServiceability limit state

    22

    Allowable stress design Limit state design

    Stress Cross-section capacity

    Axial force F P = pyA

    Shear force Fv P = pvAv

    Bending moment M Mc = py S or 1.2 py Z

    Interaction

    yc pA

    Fp =

    [ ]vp

    tI

    yAS =

    yb pzMp =

    ypz

    M

    A

    P+ 1

    M

    M

    P

    F

    c

    +

    Allowable stress design vs Limit state design

  • 7/30/2019 Design philosophy.pdf

    12/12

    23

    ASDDesign on the allowable

    LSDDesign on the limit

    One overall factor of safety on each loading type and load combinationon each material, failure mode and otherconstruction effects such as connection.

    All the members of a structureoperate within the linearelastic range.

    Design for failureSome materials may be yielded in service.

    Deflection check Deflection check in serviceability limit state withdifferent set of partial safety factor (usually equal to1.0)

    M

    f

    Allowable stress design vs Limit state design