Curtain Wall Glazing: Stick System: CALCULATION EXAMPLE PDF

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c Rui de S. Camposinhos 2019 GLAZING STICK FACADE SYSTEM UNDER WIND ACTION Calculation Example Abstract An example is given for the calculation of glazing curtain wall with insulated glass units under wind action. Limit states for breakage and deflection is performed for the double glazing panels ac...

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c Rui de S. Camposinhos 2019

GLAZING STICK FACADE SYSTEM UNDER WIND ACTION Calculation Example Abstract An example is given for the calculation of glazing curtain wall with insulated glass units under wind action. Limit states for breakage and deflection is performed for the double glazing panels according to the Standard prEN 16612. Limit states design is presented for the aluminum supporting stick system according to the Eurocode 1999.

Figure 1 depicts the elevation of a glazed facade stick system to realize a high 5.4 m building entrance. The glazed facade forms a square mesh with Individual components – mullions and transoms – installed in the field to receive as double glazing insulating panels (IGU). Mullions span the full height of the lobby entrance receiving the transoms at 1/3 of the span so that glazing panels have 1.8×1.8 m2 . Each of the double glazing panes is monolithic and annealed (AN) 10 mm thick. The gas space between the plates is 12 mm. Mullions and transoms have Class 1 cross-sections according EN 1999-1 definition [3].

Figure 1: Facade Elevation – Lobby Entrance The wind pressure acting on the external and internal surfaces is obtained from data in Table1 where the peak velocity pressure, the external and internal pressure coefficients are depicted. These values were obtained according to EC 1991-1-4 [1]. 1

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Table 1: Wind Action: Peak velocity pressure and pressure coefficients qp 2.0 kPa

cpe.10 -1.2

cpe.1 -1.4

cpi 0.2

The double insulating panels are produced at the same region of the site and the isochore pressure due to the effect of change in altitude is not relevant. Isochore pressure inside the IGU results solely from the temperature differences. Minimum and maximum values to consider are respectively Tmin = −10◦ and Tmax = +50◦ . i. Checking the breakage safety of the glazing panels according to prEN16612 [2] for a consequence class CC2. The surface area of the panes is Apan = 1.82 = 3.24 m2 and the external pressure coefficients is determined accordingly: cpe(3.24) = −1.4 − (−1.4 − (−1.2)) × log 3.24 ' −1.298 External pressure coefficients give the effect of the wind on the external surfaces of buildings and internal pressure coefficients give the effect of the wind on the internal surfaces of buildings. The corresponding characteristic valuesare: wSkext = qp · cpe(3.24) = 2.0 × (−1.298) ' 2.596 kPa wSkint = qp · cpi = 2.0 × (−1.2) = −0.4 kPa. For consequente class CC2 the partial coefficient for the wind action is γf = 1.5 so the design values for the wind action are: wSdext = −2.596 × 1.5 ' 3.894 kPa wSdint = −0.4 × 1.5 = −0.6 kPa. The design value of strength for annealed glass, fgd , depends on the characteristic value of the bending strength of for glass material and partial factors depending on the glass type and load duration (prEN 16612) [2]: fgd =

k mod · ksp · fgk 45 = = 25.0 MPa γM a 1.8

fgk – is the characteristic value of the bending strength equals to 45 MPa; γM a – is the material partial factor for annealed glass equals to 1,8; ksp – is the factor for the glass surface profile taken as equal to 1. k mod – is the factor for the load duration. The isochore pressure generated by difference of temperature the gas space is: 2

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pC,0 = cT · (TC − TP ) − (pa − pP ) cT = 0, 34 kPa/◦ K pC,0 = 0.34 × (±60) − (0) = 20.4 kPa In case of double glazing the distribution of external uniformly distributed wind loads is essentially determined by the relative stiffness of the panes as follows, δ, with t1 = t2 = 10 mm the pane’s thickness: δ1 =

t31

t31 = 0.5 + t32

δ2 = 1 − δ1 = 0.5 Additionally, the distribution (partition) of external loads as well as the effect of internal loads is determined by the insulating unit factor, ϕ, which depends on the characteristic length of the unit, a ¨, depending on the thickness of the glass panes and the gas space, s, and the shape of the unit. s s · t31 · t32
a ¨ = 28, 9 × 4 k5 · t31 + t32 s 12 · 103 · 103 (1) a ¨ = 28, 9 × 4 ' 0.68153 0.0194 · (103 + 103 ) 1 1 ϕ= ' 0.02014 4 = 1 + (a/¨ a) 1 + (1.8/0.68153)4 Coefficient value for k5 allows for determining large deflection volume changes as a function of the λ = a/b the ration between the short pane dimension (a) and the larger (b) according to Expression 21 . h i z1 (−6,8×λ1,33 ) ' 0.194 × 0, 4198 + 0, 22 × 10 (2) k5 = 16 × λ2 Partition of load carried by pane 1 (exterior) pd;1 in [kPa] is given by: pd;1 = (δ1 + ϕ · δ2 ) · wSdext − (1 − ϕ) · δ1 · wSdint − ϕ · p0 · γf

(3)

pd;1 = (0.5 + 0.02014 × 0.5) × 3.894 − (1 − 0.02014) × 0.5 × (−0.6) − 0.02014× → → (±20.4) × 1.5 ' 1.664 ou 2.896 Partition of load carried by pane 1 (exterior) pd;2 in [kPa] is given by: pd;2 = (1 − ϕ) · δ2 · wSdext − (δ2 + ϕ · δ1 ) · wSdint + ϕ · p0 · γf 1

k5 = 0.0194 may be read or interpolated in Table B.3 of the Standard prEN 16612 [2].

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pd;2 = (1 − 0.02014) × 0.5 × 3.894 − (0.5 + 0.02014 × 0.5) × (−0.6) + 0.02014 → → ×(±20.4) × 1.5 ' 2.652 ou 1.497 The greatest value is observed in pane 1 (exterior) – pd;1 ' 2.896 kPa and the corresponding maximum tensile bending stress is calculated: σmax = k1 ·

a2 · pSd t2

(4)

Coefficient k1 value depends on the ration λ = a/b, the non-dimensional load, p¨, and on the pane’s thickness t. The non-dimensional load depends on the panel Apan , the Young Modulus, E, and on the pane’s thickness t:  p¨ =

A 4 × t2

2 ·

pSd E

(5)

Substituting the respective values for the most unfavorable case, one has:  p¨ =

1.82 4 × 0.012

2 ×

2.896 ' 2.715kPa 70 000 000

(6)

O valor de p¨ = 2.715 permite determinar –pela Tabela B.1 da prEN 16612 – o valor de k1 a aplicar na express˜ao 4, obtendo-se k1 ' 0.27315. The maximum tensile bending stress value at the pane centre is: 1.82 × 2.896 ' 25.6 MPa 0.012 This value is practically equal to the maximum value of the design bending stress fgd = 25 MPa so breakage safety is checked2 . σmax = 0.27315 ×

ii. Checking deflection assuming as maximum value equals to 1/150 of the pane’s short size. The characteristic values of the action shall be considered. According to Expression B.2 of prEN 16612 [2] the maximum deflection value is: wmax = k4 ·

a4 pSd · t3 E

(7)

For pane 1 (external) pd;1 is [kPa]: pd;1 = (δ1 + ϕ · δ2 ) · wSdext − (1 − ϕ) · δ1 · wSdint − ϕ · p0 · γf 2

(8)

Note: For the other (inner) pane the maximum tensile bending stress value is approximately 21.7 MPa when a temperature increase is considered with p¨ = 2.65 and k1 = 0.2367.

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pd;1 = (0.5 + 0.02014 × 0.5) × 2.596 − (1 − 0.02014) × 0.5 × (−0.4) − 0.02014× → → (±20.4) × 1.0 ' 1.664 ou 1.931 For pane 2 (inner), pd;2 , is [kPa]: pd;2 = (1 − ϕ) · δ2 · wSdext − (δ2 + ϕ · δ1 ) · wSdint + ϕ · p0 · γf pd;2 = (1 − 0.02014) × 0.5 × 2.596 − (0.5 + 0.02014 × 0.5) × (−0.4) + 0.02014 → → ×(±20.4) × 1.5 ' 2.652 ou 1.065 Replacing the corresponding values for the most unfavorable case:  p¨ =

1.82 4 × 0.012

2 ×

1.931 ' 1.810kPa 70 000 000

(9)

with p¨ = 1.81 and k4 = 0.03654 determined according to Table B.2 of prEN 16612. Expression 7 gives: wmax = 0.03654 ×

1.84 1.931 × ' 0.01058 ≈ 10.6 mm 3 0.01 70 000 000

This values is smaller than fmax =

1800 = 12 mm3 150

Mullions and transoms are made of 6082.B aluminum in the T5 temper, rectangular effective tubular section with shorter section external size fixed to b = 70 mm and 6 mm wall thickness (Figure 2-a) of Class 1 cross-sections – which can form a plastic hinge with the rotation capacity required for plastic analysis without reduction of the resistance [3]. Transoms’s cross-section are equal to mullions [70 × h] (Figure2) with shims placed at 0,4 m from the ends. Transoms are simply supported in the mullions. iii. Calculation of cross-section dimension (h) according to EN 1999 [3]. The framing is attached to the building structure the floor, but the IGU self-weight is transferred trough the mullions in the upper fixed joints. For the sake of simplicity, it is assumed an approximate (trapezoidal) configuration for wind loads in the mullions. First concern shall be the mullions deformation once it has a greater span (5.4 m] than those of the mullions [1.8]. It’s assumed that the wind transferred load from the IGU has a trapezoidal shape.
2 pSk · 5 × l2 − 4 × a2 fmax = (10) 1920 × E · I with: 3

The maximum deflection value in the inner pane is 10.4 mm when a temperature increase is considered with p¨ = 1.768 e k4 = 0.0368.

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Figure 2: Section a-a (See elevation(Figure 1) – cpe = cpe10 − 1.2, exposed area is greater 10 m2 [1]; – pSk =

1.8 × 2 paineis × 2.0 kPa × (−1.2 − 0.2) = −5.04 kN/m; 2

– l = 5.4 m; – E = 70 GPa; – I – Second moment of area for the horizontal axis parallel to the facade’s plane. The maximum deflection value is [l/250; 15 mm] = 15 mm [3]. Replacing values and solving Expression10: 5.04 × 5 × 5.42 − 4 × 0.92 0.015 ≥ 1920 × 70000000 × I


2

resultando I ≥ 5.0807585 × 10−5 m4 , and from the following Expression h can be calculated: I=

b · h3 − b0 · h30 70 · h3 − (70 − 2 × 6) · (h − 2 × 6)3 = = 50807585 mm4 12 12

A value for h ' 322 mm is obtained and the following “standardized” are assumed: b = 70 × h = 325 ∧ t = 6 [mm]. Ultimate Limit States 6

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Bending moment The design value of the bending moment MSdmax at the relevant cross section shall satisfy: MSdmax ≤ MRd .
pSd · 3 × l2 − 4 × a2 24 = γf · pSk = 1.5 × 5.04 = 7.56 kN/m
7.56 = · 3 × 5.42 − 4 × 0.92 ' 26.54 kN m 24

MSdmax = pSd MSdmax

In the case of a Class 1 section and an alloy 6082 and T-5 temper, the design resistance for bending about on principal axis, MRd , is MRd = f0 · Wpl /γM 1 b · h2 − b0 · h20 0.07 × 0.3252 − 0.058 · 0.3132 = ' 4.2789 × 10− 04 m3 4 4 = 230000 × 4.2789 × 10−4 /1.1 ' 89.47 kN m

Wpl = MRd

MRd > MSdmax A mullion at a mid-height cross-section must also withstand the two lower IGU: NSd = γg × (0.01 + 0.01) × 1.8 × 1.8 × 25 kN/m3 × 2 = 1.35 × 3.24 ' 4.374 kN

Bending and axial force. For doubly symmetric cross-sections the following criteria shall be satisfied NSd MSd + Ag Wpl

≤

26.54 4.374 + = 62967.1 kPa ≤ 0.004596 4.2789 × 10− 4

f0 γM 1 230000 kPa 1.1

Shear. The design value of the shear force VSdmax at each cross-section shall satisfy: VRd ≥ VSdmax with pSd · (l − a) 7.56 × (5.4 − 0.9) VSdmax = = = 17.01 kN 2 2 f0 1 230000 VRd = Aν · √ · = (2 × 0.06 × 0.325) × √ ' 470.8 kN 3 γM 1 3 × 1.1

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Transoms limit state

for deformation and biaxial bending:

The transom maximum deflection, fmax depends on the self-weight, NSk of the glass action on the shims at a distance, d = 0.4 m from supports: NSk = (0.01 + 0.01) × 1.82 × 25/2 = 0.81 kN
NSk · d 3 · l2 − 4 · d2 24 × E · It b3 · h − b30 · h0 = = 4.200041 × 10−6 12
0.81 × 0.4 = × 3 × 1.82 − 4 × 0.42 −6 24 × 70 000 000 × 4.200 041 × 10 1800 = 3.6 mm ' 0.41 mm < 500

fmax = It fmax fmax

Transom biaxial bending: M⊥,Ed Mk,Ed + ≤1 M⊥,Rd Mk,Rd

(11)

Perpendicular direction to the facade: 1.83 l3 = 7.56 × ' 3.6742 kN.m 12 12 230000 M⊥,Rd = Wpl · 1.1 Wpl = 4.2789 × 10−4 ≡ mainel 230000 = 4.2789 × 10−4 × ' 89.47 kN.m 1.1

M⊥,Ed = psd ·

M⊥,Rd

Parallel direction to the facade:

Wplk

Mk,Ed = psd · d = (1.35 × 0.81) × 0.4 ' 0.44 kN.m 230000 Mk,Rd = Wplk · 1.1 b2 · h − b20 · h0 0.072 × 0.325 − 0.0582 · 0.313 = = ' 1.3489 × 10−4 m3 4 4 230000 Mk,Rd = 1.3489 × 10−4 × ' 28.20 kN.m 1.1

Substituting the calculated values in Expression11: 3.6742 0.44 + = 0.0566  1 89.47 28.2 8

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iv. In the mullions section receiving two transoms 10 mm holes in each mullion wall are defined in Figure 2-b. The ultimate limit state in these section must be check. The most unfavorable drilled section is located in the upper zone under weight of 3 IUG4 . The maximum tensile stress in the mullion drilled cross-section is given by Expression 12: NSd MSd fu + ≤ 0.9 × Anet α · Wnet γM 2

(12)

The IUG’s self-weight is: NSdp = γg × (0.01 + 0.01) × 1.8 × 1.8 × 25 kN/m3 × 3 = 1.35 × 4.86 ' 6.561 kN and that of the transoms is: NSdt = γg ·Ag ×1.8×27.4 kN/m3 ×3 = 1.35×0.004596×1.8×27.4×3 ' 0.681 kN Resulting in a total axial force of: NSd = 6.561 + 0.681 ' 7.48 kN The bending acting moment due to the wind action at the drilled crosssection located at 1.8 m from the fixed (upper) is determined5 : MSd−1.8 MSd−1.8

l 1.82 × 1.8 − pSd × 2 2 2 1.8 5.4 × 1.8 − 7.56 × ' 8.165 kNm = 7.56 × 2 2 = pSd ×

with: Anet = Ag − 4 × 0.01 × 0.006

Wnet

= 0.004596 − 0.00024 ' 0.004356 m2   0.325 − 0.1 2 = Wpl − 4 × (0.06 × 0.01) × 2 = 4.2789 × 10−4 − 1.35 × 10−4 ' 2.929 × 10−4

4 5

The self-weight of transoms is negligible. Assuming a uniformly distributed load

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and, using Expression 12 8.165 7.48 + ' 1908 + 27 877 0.9 × 0.004356 2.929 × 10−4 = 29 785 kPa ≤

10

270000 = 216 000 kPa 1.25

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References [1] EN 1991; Eurocode 1: Actions on structures – Part 1-4: General actions – Wind actions; CEN – Standardization European Committee, 2004; Brussels; [2] prEN16612; Glass in building – Determination of the load resistance of glass panes by calculation and testing; CEN – Standardization European Committee, 2013; Brussels; [3] EN 1999-1-1; Eurocode 9: Design of aluminium structures – Part 1-1: General structural rules; CEN – Standardization European Committee, 2007; Brussels;

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