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    Verification Samples

    Cantilever Curtain Wall

    1 Objective

    To check the safety of using a single (4CS2X065, 50Ksi) section as a stud for a cantilever

    curtain wall, given the following data:

    The design is based on AISI-NASPEC 2001 with 2004 supplement, using the LRFD method.

    The wall height = 8 ft.

    The stud spacing = 24 in.

    The bridging members are located @ 24 o.c.

    The lateral bracing spacing is considered equal to the bridging members spacing.

    The torsional bracing spacing is considered equal to the bridging members spacing.

    The wind load intensity (IWL) = 20 psf.

    The wind load factor for strength = 1.25.

    The wind load factor for lateral deflection = 1.0.

    The web is unreinforced for shear strength calculation.

    The factor (Cb) = 1.0 for lateral-torsional buckling stress calculation.

    The web-crippling check is considered, and the bearing length (N) = 1.25 in.

    The effect of cold-work of forming is not considered.

    The effect of the standard punch-out (4" X 1.5") is not considered.

    The maximum allowed lateral deflection = L/ 360, where (L) is twice the wall height.

    The maximum absolute deflection is 0.6 in.

    Cantilever Curtain Wall SteelSmart System 7.0 SP2

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    2 Modeling with SteelSmart System

    Cantilever Curtain Wall SteelSmart System 7.0 SP2

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    3 Statical System, Loads for Strength, and Forces

    8 ft50 8 ft50

    19.2019.20

    0.400.40

    (a) Loading (lb/ft) (b) BMD (kips.in) (c) SFD (kips)

    Figure 1 Statical System, Loads for Strength, and Forces

    4 Safety Check

    4.1 Design for flexural moment (Mx)

    Figure 2 shows the cross-section elements with their corresponding numbers.

    2

    1

    1

    2'

    3

    Figure 2 Elements o f the Cross-Section and Their Numbering

    The major flexural strength of the section (Mx-all) is the least of the flexural strength at the initiation

    of the cross-section yielding and the lateral-torsional buckling strength.

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    4.1.1 Flat Widths Calculation

    Considering (D) the section depth, (B) the flange width, (C) the lip width, (t) the section

    thickness, and (R) the inside bend radius of the section corners; the flat widths of the elements

    subjected to compressive stresses shown in Figure 3 can be calculated as follows:-

    W1 = C (t + R)

    = 0.881 (0.065 + 0.1875) = 0.6285 in.

    W2 = B 2(t + R)

    = 2 2(0.065 + 0.1875) = 1.495 in.

    W3 = D 2(t + R)

    = 4 2(0.065 + 0.1875) = 3.495 in.

    W

    W

    2W

    1

    3

    Figure 3 Flat Widths of Elements Subjected to Compressive Stresses

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    4.1.2 Initiation of Yielding (Section C3.1.1 in the 2001 Specification)

    The effective properties are calculated by loading the section by a stress gradient

    (tension-compression) which has a value of (Fy) at the top fiber of the compression flange (Figure

    4).

    2 - 1f

    Fbottom top

    = F = 50 ksi

    = 50 ksitop

    1 - 1f and

    2 - 3f

    f2

    Mx

    N.A.

    F

    f1 - 3

    3

    2'

    1

    1'

    2

    Ycg0 = 2 in.

    Figure 4 Location of N.A., Stress Distribution, and Critical Stresses on the Fully EffectiveSection for Initiation of Yielding)

    Iteration (1)

    The section is assumed to be fully effective, i.e. the N.A. will be located at a distance from the

    bottom fiber of the tension flange equal to half the section depth (Figure 4).

    4.1.2.1 Effective properties calculation

    4.1.2.1.1 Element (1) and Element (2)

    Element (1) is classified as an unstiffened element under stress gradient, thus, the

    plate buckling coefficient (K) used for the calculation of the effective width of (W1) is determined in

    accordance with section B3.2 (a) in the 2004 supplementas follows:-

    f1-1 = compression stress at the upper end of W1, (Figure 4)

    =y

    cgo

    cgoF

    YD

    RtYD

    = 50*24

    1875.0065.024

    = 43.6875 ksi

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    f2-1 = compression stress at the lower end of W1, (Figure 4)

    = ycgo

    cgoF

    YD

    CYD

    = 50*24

    881.024

    = 27.975 ksi

    = f2-1/ f1-1= 0.6403434 (Eq. B3.2-1)1

    K = 0.578/ (+ 0.34) =0.5895894 (Eq. B3.2-2)1

    The effective width of (W1) is then calculated in accordance with Section B2.1 (a) in the 2001

    specificationas follows:

    Fcr =

    2

    1

    2

    2

    W

    t

    )1(12

    EK

    (Eq. B2.1-5)

    =2

    2

    2

    6285.0

    0.065

    )3.01(12

    29500**0.5895894

    = 168.1376 ksi

    =cr

    11

    F

    f (Eq. B2.1-4)

    = 0.509737 < 0.673

    d'1 = effective width of (W1)

    = W1 (Eq. B2.1-1)

    = 0.6285 in. (fully effective)

    Element (2) is classified as a uniformly compressed element with an edge stiffener, thus, the

    plate buckling coefficient (K) used for the calculation of the effective width of (W2) is determined in

    accordance with Section B4.2 (a) in the 2001 specificationas follows:

    f2 = compression stress at the centerline of W2, (Figure 4)

    =y

    cgo

    cgoF

    YD

    t5.0YD

    = 50*24

    065.0*5.024

    = 49.1875 ksi

    1Supplement 2004 to the NASPEC-AISI 2001 for the Design of Cold-Formed Steel Structural Members

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    S =2fE28.1 (Eq. B4-1)

    = 49.18752950028.1 = 31.34683

    W2/ t = 23 > (0.328S = 10.28176)

    Ia =3

    24 328.0S

    tWt399

    (Eq. B4.2-10)

    = ( )3

    4328.0

    31.34683

    23065.0399

    = 4.756928 * 10

    -4in

    4

    Ia-max =

    +5

    S

    tW115t 24

    = ( )

    +5

    31.34683

    23*115065.0

    4 = 1.595463 * 10-3

    in4> Ia

    Is =( )

    12

    SintW 23

    1 (Eq. B4-2)

    =( )

    12

    90Sin*065.0*6285.0 23

    = 1.34477 * 10-3

    in4

    RI =a

    s

    I

    I (Eq. B4.2-9)

    = 2.826971 >1.0 RI= 1.0

    n =

    S4

    tW582.0 2 (Eq. B4.2-11)

    =

    31.34683*4

    23582.0 = 0.3985684 > (1/3)

    2WC = 1.495881.0 = 0.5892977, > 0.25 and < 0.8

    Using Table B4-2;

    K = ( ) 43.0RWC5

    82.4 n

    I2

    +

    = ( ) 43.01495.1

    881.0*582.4

    0.3985684 +

    = 2.303512 < 4.0

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    The effective width of (W2) is then calculated in accordance with Section B2.1 (a) in the 2001

    specificationas follows:

    Fcr =2

    22

    2

    W

    t

    )1(12

    EK

    (Eq. B2.1-5)

    =2

    2

    2

    495.1

    0.065

    )3.01(12

    29500**2.303512

    = 116.1004 ksi

    =cr

    2

    F

    f (Eq. B2.1-4)

    =116.1004

    49.1875 = 0.6508942 < 0.673

    b2 = effective width of (W2)

    = W2 (Eq. B2.1-1)

    = 1.495 in. (fully effective)

    d1 = reduced effective width of (W1)

    = d'1RI (Eq. B4.2-7)

    = 0.6285 *1 = 0.6285 in.

    Rem1 = removed width of (W1) due to local buckling

    = W1 d

    1=0.6285 0.6285 = 0

    Rem2 = removed width of (W2) due to local buckling

    = W2 b2= 1.495 1.495 = 0

    4.1.2.1.2 Element (3)

    Element (3) is classified as a stiffened element under stress gradient (tension-

    compression), thus, the plate buckling coefficient (K) used for the calculation of the effective width

    of (W3) is determined in accordance with section B2.3 (a) in the 2001 specificationas follows:

    f1-3 = compression stress at the upper end of W3, (Figure 4)

    =y

    cgo

    cgoF

    YD

    RtYD

    = 50*24

    1875.0065.024

    = 43.6875 ksi

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    f2-3 = tensile stress at the lower end of W3, (Figure 4)

    =y

    cgo

    cgoF

    YD

    RtY

    = 50*24

    1875.0065.02

    = -43.6875 ksi

    = 3132 ff (Eq. B2.3-1)

    = 43.687543.6875- = 1.0

    K = 4 + 2(1 + )3+ 2(1 + ) (Eq. B2.3-2)

    = 4 + 2(1 + 1)3+ 2(1 + 1) = 24

    The effective width of (W3) is then calculated in accordance with Section B2.1 (a) in the 2001

    specificationas follows:-

    Fcr =2

    32

    2

    W

    t

    )1(12

    EK

    (Eq. B2.1-5)

    =2

    2

    2

    495.3

    0.065

    )3.01(12

    29500**42

    = 221.3313 ksi

    =cr

    31

    F

    f (Eq. B2.1-4)

    =221.331343.6875 = 0.4442805 < 0.673

    b3 = effective width of the compressed part of (W3)

    = W3-Comp (Eq. B2.1-1)

    = D Ycg0 t R = 1.7475 in. (fully effective)

    Rem3= 0

    Remt = the total removed length of the section due to local buckling

    = Rem1+ Rem2+ Rem3

    = 0 + 0 + 0 = 0

    Ae = effective area of the section

    = Ag (Remt* t) = 0.593034 in2(fully effective).

    Cantilever Curtain Wall SteelSmart System 7.0 SP2

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