04.05 Stick Systems

Introduction
The primary structural elements of a stick curtain walling system are vertical mullions, which run the full height of the cladding. They are normally supplied in one or two storey lengths, which are connected by sleeved joints. These joints are usually designed to permit vertical movements and allow transfer of shear loads but do not transmit axial loads or moments. Mullions are typically spaced at between 1.0 and 1.8m centres and support transoms. Transoms are connected to the mullions using angle cleats, sleeves, spigots or proprietary brackets.

The framework of mullions and transoms supports infill panels, which may be glazing units or insulated panels.
Mullions and transoms are usually made of extruded aluminium but may be steel. A number of manufacturers produce standard systems using aluminium profiles, which have been specially designed to support the infill panels with weathertight joints. Aluminium profiles are generally based on thin walled extruded box sections, which may require reinforcement to accommodate concentrated loads at fixings. Longitudinal ribs are often used to increase the wall thickness to accommodate fixing screws for mullion brackets and cross webs may be incorporated to prevent squashing of the section by bracket fixing bolts passing through the section. Enlarged fixing holes with bushes may be used to distribute concentrated loads from fixing bolts. A typical profile is shown here.

Structural systems
The structural behaviour of a curtain wall depends on the strength of the individual members and also on the way they are connected to each other and the supporting structure. Three ways of supporting the mullions may be used as described below. The calculation of bending moment, deflection and shear force for each support arrangement assumes that the only horizontal load is a uniformly distributed wind load with a total load of W on each storey height L.

The simplest method of construction is where the mullions are used as single span simply supported sections in which the ends of the members are free to rotate in response to the applied loads as shown here. Typically the top of the section of mullion would be supported by the floor allowing transfer of dead and live load and the bottom would be restrained by the section below via a sliding joint allowing transfer of horizontal live load only.

With this arrangement the maximum bending moment (M_max) in the mullion occurs at mid-span and is given by:

M_max = W L / 8

the maximum deflection (d_max) also occurs at mid-span and is given by:

d_max = 5 W L³ / 384 E I

and the maximum shear force (V_max) occurs at the ends of the mullion sections and is given by:

V_max = W / 2

This gives the least efficient structural behaviour. If the ends of the members are fixed to prevent rotation and hence transmit moments the member will generally be able to carry higher loads and give lower deflections.

This could be achieved by a rigid fixing arrangement however this would require more complicated connections and a more convenient method which achieves a similar effect is to make the mullion continuous over more than one span. This image shows a section of mullion continuous over two spans. The connections at the ends of the section of mullion transfer loads as in the previous case and the bracket at the intermediate floor transfers horizontal live load only.

With this arrangement the maximum bending moment (M_max) in the mullion occurs at the central support and is given by:

M_max = W L / 8

The maximum deflection (d_max) occurs at 0.5785L from the central support and is given by:

d_max = W L³ / 185 E I

and the maximum shear force (V_max) occurs at the central bracket and is given by;

V_max = 5 W / 8

Although the value of the maximum moment is the same as for the single span case, the maximum deflection is reduced to less than half. As the design of aluminium mullions is generally governed by deflection rather than strength this gives a more efficient design.

An alternative arrangement that gives the benefit of continuity over the supports with single storey height mullion sections is to arrange the joints between mullion sections to occur approximately a fifth of the storey height above the support as shown here.

With this arrangement the maximum bending moment (M_max) in the mullion occurs at the support and is given by:

M_max = W L / 12.5

The maximum deflection (d_max) occurs at 0.4628L from the support and is given by:

d_max = W L³ / 450 E I

and the maximum shear force (V_max) occurs at the support and is given by:

V_max = W / 2

This arrangement gives a further reduction in deflection but also gives a reduced maximum moment. The shear force is also less than for the two span mullion section.
It is clearly essential that the curtain wall is erected with the same structural arrangements as used for the design.

Fixings
Connections between transoms and mullions and between mullions and structural frame must also be designed. These connections normally consist of cleats or brackets screwed or bolted to the framing members and it is necessary to check the shear capacity of screws and bolts and the bearing capacity of the framing members. These connections often involve concentrated loads on thin webs of the framing members and it may be necessary to increase the bearing area by increasing the thickness of the webs or using larger diameter fixings.

Design examples

Example 1
Design a transom for a curtain wall with mullions spaced at 1.2m centre to centre (transom length of, say, 1.15m) and transoms spaced at 1.6m. The wall is glazed with 6/12/6 double glazed units.

Vertical loading
The transom must be designed to carry the dead load of the glass. The dimensioned assembly, together with the resulting bending moment and shear stress distribution in the transom due to dead loads, is illustrated here.

The dead load of the glass unit is:

W_{glass unit} = r g X Y t

where:

r is the density of glass, 2500 kg/m³,
g is the acceleration due to gravity, 9.81 m/s²,
t is the total thickness of glass, in m,
X & Y are the length and height dimensions of the glass, in m,

Substituting the known values into the above equation:

W_{glass unit} = 2500 x 9.81 x 1.2 x 1.6 x 0.012 = 565N

Note that the glass occupies less area than the grid dimensions used in the calculation above (1.2 by 1.6m) but this over-estimate is compensated by neglecting the self-weight of the transom.

Glass units are supported on setting blocks and thus the dead load is applied as two vertical loads (assumed as point loads, P) on the transom generating bending and shear stresses, here. Here, a load of 283N (565/2) is applied through each of the two setting blocks to the transom (and thence the mullion). The bending moment and deflection of the transom will depend on the distance between the setting blocks and the ends of the transom. The calculations below assume that the setting blocks are placed at the quarter points of the transom, which are the preferred points of support for the glazing unit. To reduce the moment and deflection of the transom, the setting blocks are normally placed closer to the ends of the transom. The minimum distance from the edge of the unit to the start of the setting block is normally 30mm which puts the centre of the setting block about 100mm from the end of the transom. It is clearly important to ensure that the actual positions of the setting blocks in the structure are not further from the ends of the transom than assumed in design.

For design in accordance with BS 8118 a safety factor (g_f) is applied to the load. In the case of strength calculations for dead load only g_f takes the value 1.2 giving a design load of 340N. For calculation of deflection g_f is taken as 1.0.

Stresses
The design bending moment in the transom due to the weight of the glass is given by:

M = g_f P L / 4 = (1.2 x 283 x 1.15) / 4 = 98Nm

The design shear force is:

V = g_f P = 340N

Bending resistance
Assuming elastic behaviour, the basic theory of bending gives:

M / I = s / y = E/ R

where:

M is the moment at the cross-section under consideration, Nmm,
I is the second moment of area, mm⁴,
s is the bending stress, N/mm^2, at a distance y mm from the neutral axis,
E is the Young’s Modulus of the material, N/mm²,
R is the radius of curvature of bending, mm.

Rearranging this equation the maximum moment the section can carry is given by:

M_max = s_max I / y_max

The maximum stress depends on the grade of aluminium and can be obtained from BS 8118 Table 4.2. The most common grade of aluminium for curtain wall sections is 6063 alloy in temper T6 for which the maximum stress in bending (p_o) given in BS 8118 is 160N/mm². For design purposes this value is reduced by dividing by a safety factor g_m which takes the value 1.2 from Table 3.3 of BS8118.

I and y_max depend on the transom dimensions and a suitable transom section can be selected to ensure that M_max is greater than the applied design moment.

For example a proprietary transom 50 mm wide and 28 mm deep with I_yy = 11 x 10⁴ mm⁴ and y_max = 25 mm would give

M_max = (160 / 1.2) x 11 x 10⁴ / 25 = 587 x 10³ Nmm

This greatly exceeds the applied moment and the section is satisfactory.

For the above equation to be valid the section must be able to develop the bending stresses. If the webs are too thin they may buckle before the maximum bending stress is reached and BS 8118 requires a check to ensure that buckling will not occur. This will be satisfied if;

b < b_o

where

b is the slenderness parameter for a part of the section (for example a flange or web) which is determined in accordance with clause 4.3.2 of BS 8118. b varies according to the stress distribution in the element.
b_o is a limiting value of b calculated in accordance with Table 4.3 of BS 8118.

In this case, b = 0.35 d / t for the web and b / t for the flange

where:

d is the depth of the web element of the section (i.e. the clear depth of web between flanges), 45.8mm,
b is the width of the flange (i.e. the clear width of the flange between webs), 18.5mm,
t is the element thickness, 2.1mm.

hence:

b_web = 0.35 x 45.8 / 2.1 = 7.6
b_flange = 18.5 / 2.1 = 8.8
b_o = 22 (250 / p_o)^1/2

where,

p_o is 160N/mm² as defined above and

b_o = 22 (250 / 160)^1/2 = 27.5

Both values of b are less than b_o and therefore the section can develop the full elastic moment.

BS 8118 allows for higher design moments to be obtained by taking into account plastic behaviour however this is unlikely to be justified for curtain walling as deflection limits will normally govern the design.

Shear resistance
It is first necessary to check whether the section is likely to be affected by buckling. A compact section is unaffected by buckling and a section is compact if:

d / t £ 49 (250 / p_o)^1/2
where the symbols are as defined above. By inspection the section considered above is compact.

Factored shear force resistance V_RS is given by:

V_RS = ( p_v A_v ) / g_m

where:

p_v is the limiting stress in shear, 95N/mm² for aluminium alloy 6063 of temper condition T6 (BS 8118: Part 1),
A_v is the effective shear area, mm²,
g_m is the material factor = 1.2.

A_v = 0.8 x N x D x t

where:

D is the overall depth of web measured to the outer surfaces of flanges, 50 mm in the example,
t is the web thickness, 2.1mm,
N is the number of webs, 2.

then:

A_v = 0.8 x 2 x 50 x 2.1 = 168mm²

V_RS = ( 95 x 168) / 1.2 = 13300N > 340N OK

The peak shear force and bending moment coexist at the same cross-section along the transom and the capacity of the section may, therefore, be reduced by the combination of stresses. However, since the applied shear force is not more than half of the factored shear resistance, the factored bending moment resistance of the transom section can be assumed to be unaffected (Clause 4.5.4, BS 8118: Part 1).

Check deflection
The maximum deflection, d_max, of a simply supported beam subject to two equal point loads is given by:

d_max = P L³[ 3 a / 4 L - ( a / L )³] / 6 E I

where:

P is the point load, N,
L is the span, mm,
E is Young’s Modulus of the material, 70 x 10³ N/mm² for grade 6063 aluminium (BS 8118: Part 1),
I is the second moment of area of the section about the axis of bending, 11 x 10⁴ mm⁴,
a is the distance from the support to the point load, mm.

BS 6262 states that for fixed windows, setting blocks should usually be positioned as near to the quarter points as possible. Here, this is 300mm (i.e. 1200/4), which is assumed to be the dimension ‘a’ in the above equation. The deflection of the transom is limited to 5mm to guard against the glazed transom touching the glass unit below. The maximum deflection is calculated as:

d_max = 2mm OK

Horizontal loading
The glass units deflect under positive and negative wind loading, compressing the inner/outer gaskets and causing bending about the major axis of the supporting transom and mullion members. A trapezoidal wind loading distribution across the panel may be assumed, where a portion of the wind load is resisted by the transom (and transferred to the mullions as point loads, image.

If a wind load of 1600Pa is assumed, the total lateral load on the transom is:

W_wind = Area (m²) x pressure (N/m²)

Loaded area below transom = (0.4 x 0.4) + (0.4 x 0.4) = 0.32 m²
Loaded area above transom = 0.6 x 0.6 = 0.36 m²

W_wind = Area (m²) x pressure (N/m²)
= 0.68 x 1600
= 1088N

This wind load can be assumed to be uniformly distributed over the length of the transom.

Stresses
The design bending moment is equal to:

M = g_f W L / 8 = (1.2 x 1088 x 1.15) / 8 = 188Nm

The design shear force is:

V = g_f W / 2 = 1.2 x 1088 / 2 = 653N

Bending resistance
For the transom selected above values for I and y_max for bending in the horizontal plane are 6.3 x 10⁴ and 22.7 respectively and the maximum moment is therefore given by:
M_max = (160 / 1.2) x 6.3 x 10⁴ / 22.7
= 370 Nm OK

Check for buckling

b_web =0.6 x d / t = 0.6 x 18.5 / 2.1 = 5.3
b_flange = b / t = 45.8 / 2.1 = 21.8
b_o = 22 (250 / 160)^1/2 = 27.5

Both values of b are less than b_o and the section is therefore satisfactory.

Shear resistance
Factored shear force resistance V_RS is given by:

V_RS = ( p_v A_v ) / g_m

where:

A_v = 0.8 x N x D x t

where:

D is the overall depth of web measured to the outer surfaces of flanges, 28mm,
t is the web thickness, 2.1mm,
N is the number of webs, 2,

then:

A_v = 0.8 x 2 x 28 x 2.1 = 94 mm²

V_RS =( 95 x 94 ) / 1.2 = 7441N > 653N OK

Since the shear force is not more than half of the factored shear resistance, the factored bending moment resistance of the mullion section can be assumed to be unaffected (Clause 4.5.4, BS 8118: Part 1).

Check lateral deflection
The maximum deflection of a simply supported, uniformly loaded beam is:

d_max = 5 W L³ / 384 E I

where:

W is the total load, in N,
other terms as previously defined

For framing members supporting double glazing units, the maximum frontal deflection under positive and negative peak wind load should not exceed the lesser of 1/175 of the length along the unit edge (here, 1200/175 = 6.9mm) or 15mm (CWCT, 1996). Therefore, the deflection limit is satisfied.

Combined loading
The above calculations have considered the effects of dead and wind loads separately. As these loads can act at the same time it is necessary to check the effect of combining the loads. The full procedure is given in BS 8118 clause 4.8 however in this case the following condition should be satisfied

M_xdesign/ M_xmax + M_ydesign/ M_ymax £ 1.0

substituting values determined above gives

188 / 370 + 98 / 587 = 0.68 OK

Example 2
Design a mullion which spans over two floors (say, 6.4m) to resist a wind load of 1600Pa (1.6kN/m2) and the dead loads transferred from the connecting transoms in Example 1. Assume a uniform distribution of wind loading on the facade.

Loading
The dimensioned assembly, together with the resulting bending moment and shear force distribution in the mullion due to the applied wind load, is illustrated here. This loading is assumed to be uniformly distributed along the length of the mullion. Assuming it is supported at the top, the mullion is also subject to tension due to dead loads.

For a mullion spacing of 1.2m the total wind load per span is:

W = Area (m²) x Pressure (N/m²)
= (1.2 x 3.2) x 1600
= 6144N

Stresses
The most highly stressed cross-section occurs at the intermediate support where the bending moment is given by:

M = g_f W L / 8 = (1.2 x 6144 x 3.2) / 8 = 2950Nm.

This is also the region of peak shear force, given by:

V = ( 5 / 8 ) g_f
W = (5/8) 1.2 x 6144 = 4608N.

Bending resistance
The mullion can be designed as a simply supported continuous beam, image, by following the procedure described above. For example a proprietary mullion 50mm wide and 105mm deep with I_xx = 165 x 10⁴ mm⁴ and y_max = 64 mm would give

M_max = (160 / 1.2) x 165 x 10⁴ / 64 = 3438 Nm

Check for buckling
b_web = 0.5 x d / t = 0.5 x 86 / 2.1 = 20.5
b_flange = b/t = 45.8 / 2.1 = 21.8
bo = 22 ( 250 / 160 )^1/2 = 27.5

Both values of b are less than bo and the section is therefore satisfactory.

Shear resistance
Section is compact. Factored shear force resistance V_RS is given by:

V_RS = ( p_v A_v ) / g_m

A_v = 0.8 x N x D x t
where:

D is the overall depth of web measured to the outer surfaces of flanges, 105mm;
t is the web thickness, 2.1mm
N is the number of webs, 2

then:

Av = 0.8 x 2 x 105 x 2.1 = 353 mm²

V_RS = ( 95 x 353 ) / 1.2 = 27900N > 4608N OK

Check deflection
The maximum deflection of a continuous beam with two equal spans subject to a uniformly distributed load occurs at 0.5785L, from the central support and is given by:

d_max = 5 W L³ / 185 E I

where:

W is the total load on one span, N,
L is the span between fixing points, mm,
Other terms as previously defined.

Substituting known values gives:

d_max = 9.4 mm

In this case because the lengths of the supported infill panels are much less than the overall span, Section 04.02, the deflection limit is span/200 or 20mm. Span/200 is 16mm and the deflection is therefore acceptable.

Check tensile stress
The transoms either side of the mullion transfer the dead load from the infill elements to the mullion as an axial load; there is no net moment induced where transoms of equal span and loading connect on either side of the mullion. With the grid dimensions of the example wall, four transoms connect through either side of each two-storey mullion. The total vertical load on each such mullion is therefore:

Half the weight of eight glass units

= ( 8 x 565 ) / 2 = 2300N

Half the weight of eight transoms (1.19kg/m)

= ( 1.19 x 9.81 x 1.35 x 8 ) / 2 = 63.0N

Self-weight of mullion (2.99kg/m)

= 2.99 x 6.4 = 19.1N

Total vertical load = 2382N

Allowing for a safety factor of 1.2 gives a design load of 2860N.This load, equal to approximately 1/5 of the wind load, causes tension in the mullion, assuming that the mullion is suspended. Note that this vertical load is transferred to the mullion at some eccentricity ‘e’ from its major axis, so inducing major axis bending in the member, however this moment is negligible compared with the maximum wind induced moment.

The factored tension resistance (P_RS) of the mullion is the lesser of the two values corresponding to:

General yielding along the member based on the general cross section;
Local failure at a critical section allowing for fixing holes.

General yielding along the member

P_RS = ( p_o A ) / g_m

where:

A is the gross section area (i.e. the area of aluminium ignoring and drilled holes), mm²

Substituting known values into the above equation gives:

P_RS = ( 160 x 954 ) / 1.2 = 127 x 10³ N

Local failure at critical section

P_RS= ( p_n A_n ) / g_m

where:

p_n is the limiting stress for local capacity of section under tension/compression, N/mm²,
A_n is the net section area, with deduction for holes, in mm².

The limiting tensile stress for an extrusion of alloy 6063, temper T4 is 175N/mm² (BS 8118: Part 1). If a 13mm diameter hole through both webs is assumed at the support, then the tension resistance is:

P_RS = 175 ( 954 - 2 ( 13 x 2.1)) / 1.2 = 131 x 10³ N

The lesser value of the factored tension resistance (P_RS) is 131kN which is far in excess of the tension caused by the transom loading (2.4kN). Therefore the mullion has adequate tension resistance in this case.

CWCT FACETS

04.05 Stick Systems