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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.

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2 Modeling with SteelSmart System

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3 Statical System, Loads for Strength, and Forces

8 ft50 8 ft50

19.2019.20

0.400.40

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)

(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).

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From the previous analysis, we can conclude that there is no need for more iterations because

Ycg(final location of the neutral axis measured from the bottom fiber of the tension flange) will be

equal to Ycg0, and the effective section modulus at initiation of yielding (Sxe) can be calculated as

follows:

Ixe = Ix

= 1.474698 in4

Sxe = Ixe/ (0.5D)

= 1.474698/ 2 = 0.7373492 in3

4.1.2.2 Calculation of the flexural strength for Initiation of Yielding (Mx-Yielding)

Mnx-Yielding = SxeFy (Eq. C3.1.1-1)

= 0.7373492 * 50 = 36.8675 kips.in

Mx-Yielding = bMnx-Yielding

= 0.95 * 36.8675 = 35.0241 kips.in

4.1.3 Lateral-Torsional Buckling (Section C3.1.2.1 in the 2001

Specification)

4.1.3.1 Calculation of the lateral-tors ional buckling stress (Fc)

ey = ( )2yyy

2

rLK

E (Eq. C3.1.2.1-8)

=( )2

2

7824748.024

29500*= 309.4853 ksi

t =( )

+

2tt

W2

20 LK

ECGJ

Ar

1 (Eq. C3.1.2.1-9)

=( ) ( )

+

2

24

224

728378.1*29500*10*351894.8*11300

533302.2*593034.0

1

= 232.0336 ksi

Sf = Ix/ (0.5D)

= 1.474698/ (0.5*4) = 0.7373492 in3

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Fe =tey

f

ob

S

ArC (Eq. C3.1.2.1-5)

= 232.0336*309.48530.7373492

593034.0*533302.2*1

= 545.995 ksi > (2.78Fy= 139 ksi)

Fc = Fy (Eq. C3.1.2.1-2)

Since Fc= Fy, the section will not be subjected to lateral-torsional buckling.

4.1.4 Determination of the Allowable Flexural Strength (Mx-all) and Check

of Safety

Mx-all = Mx-Yielding= 35.0241 kips.in

Mx-max = 19.2 Kips.in < Mx-allsafe

Act/ All = 19.2/ 35.0241 = 0.5482

4.2 Design for shear (Vy)

The major shear strength (Vy-all) can be calculated in accordance with Section C3.2.1 in the

2001 specificationas discussed in the following sections.

4.2.1 Calculation of the Nominal Shear Stress (Fv)

h = flat portion of the solid web

= D 2(t + R) = 4 2(0.065 + 0.1875) = 3.495 in

h/ t = 3.495/ 0.065 = 53.76923

Web is unreinforcedKv= shear buckling coefficient = 5.34.

yv FEK = 5034.5*29500 = 56.1302058

(h/ t) allunsafe

Act/ All = 0.8135/ 0.5333 = 1.5254

5 Conclusion

It is unsafe to use a single (4CS2X065, 50Ksi) section as a typical stud for this curtain-wall

because the maximum actual lateral deflection will exceed its limit.

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6 Verification of SteelSmart System

6.1 Forces and Lateral Deflection Diagrams

6.1.1 Bending Moment Diagram (Kips.in)

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6.1.2 Shear Force Diagram (Kips)

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6.1.3 Lateral Deformation Diagram (in)

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6.2 Output

Cantilever Curtain Wall SteelSmart System 7 0 SP2