Cantilever Curtain Wall
Transcript of Cantilever Curtain Wall

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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 AISINASPEC 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 lateraltorsional buckling stress calculation.
The webcrippling check is considered, and the bearing length (N) = 1.25 in.
The effect of coldwork of forming is not considered.
The effect of the standard punchout (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
(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 crosssection elements with their corresponding numbers.
2
1
1
2'
3
Figure 2 Elements o f the CrossSection and Their Numbering
The major flexural strength of the section (Mxall) is the least of the flexural strength at the initiation
of the crosssection yielding and the lateraltorsional 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
(tensioncompression) 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:
f11 = 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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f21 = compression stress at the lower end of W1, (Figure 4)
= ycgo
cgoF
YD
CYD
= 50*24
881.024
= 27.975 ksi
= f21/ f11= 0.6403434 (Eq. B3.21)1
K = 0.578/ (+ 0.34) =0.5895894 (Eq. B3.22)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. B2.15)
=2
2
2
6285.0
0.065
)3.01(12
29500**0.5895894
= 168.1376 ksi
=cr
11
F
f (Eq. B2.14)
= 0.509737 < 0.673
d'1 = effective width of (W1)
= W1 (Eq. B2.11)
= 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 NASPECAISI 2001 for the Design of ColdFormed Steel Structural Members
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S =2fE28.1 (Eq. B41)
= 49.18752950028.1 = 31.34683
W2/ t = 23 > (0.328S = 10.28176)
Ia =3
24 328.0S
tWt399
(Eq. B4.210)
= ( )3
4328.0
31.34683
23065.0399
= 4.756928 * 10
4in
4
Iamax =
+5
S
tW115t 24
= ( )
+5
31.34683
23*115065.0
4 = 1.595463 * 103
in4> Ia
Is =( )
12
SintW 23
1 (Eq. B42)
=( )
12
90Sin*065.0*6285.0 23
= 1.34477 * 103
in4
RI =a
s
I
I (Eq. B4.29)
= 2.826971 >1.0 RI= 1.0
n =
S4
tW582.0 2 (Eq. B4.211)
=
31.34683*4
23582.0 = 0.3985684 > (1/3)
2WC = 1.495881.0 = 0.5892977, > 0.25 and < 0.8
Using Table B42;
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. B2.15)
=2
2
2
495.1
0.065
)3.01(12
29500**2.303512
= 116.1004 ksi
=cr
2
F
f (Eq. B2.14)
=116.1004
49.1875 = 0.6508942 < 0.673
b2 = effective width of (W2)
= W2 (Eq. B2.11)
= 1.495 in. (fully effective)
d1 = reduced effective width of (W1)
= d'1RI (Eq. B4.27)
= 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:
f13 = 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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f23 = 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. B2.31)
= 43.687543.6875 = 1.0
K = 4 + 2(1 + )3+ 2(1 + ) (Eq. B2.32)
= 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. B2.15)
=2
2
2
495.3
0.065
)3.01(12
29500**42
= 221.3313 ksi
=cr
31
F
f (Eq. B2.14)
=221.331343.6875 = 0.4442805 < 0.673
b3 = effective width of the compressed part of (W3)
= W3Comp (Eq. B2.11)
= 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 (MxYielding)
MnxYielding = SxeFy (Eq. C3.1.11)
= 0.7373492 * 50 = 36.8675 kips.in
MxYielding = bMnxYielding
= 0.95 * 36.8675 = 35.0241 kips.in
4.1.3 LateralTorsional Buckling (Section C3.1.2.1 in the 2001
Specification)
4.1.3.1 Calculation of the lateraltors ional buckling stress (Fc)
ey = ( )2yyy
2
rLK
E (Eq. C3.1.2.18)
=( )2
2
7824748.024
29500*= 309.4853 ksi
t =( )
+
2tt
W2
20 LK
ECGJ
Ar
1 (Eq. C3.1.2.19)
=( ) ( )
+
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. C3.1.2.15)
= 232.0336*309.48530.7373492
593034.0*533302.2*1
= 545.995 ksi > (2.78Fy= 139 ksi)
Fc = Fy (Eq. C3.1.2.12)
Since Fc= Fy, the section will not be subjected to lateraltorsional buckling.
4.1.4 Determination of the Allowable Flexural Strength (Mxall) and Check
of Safety
Mxall = MxYielding= 35.0241 kips.in
Mxmax = 19.2 Kips.in < Mxallsafe
Act/ All = 19.2/ 35.0241 = 0.5482
4.2 Design for shear (Vy)
The major shear strength (Vyall) 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 curtainwall
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
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