CHS CORUS

/1 Last revised 30/08/2002 11:35

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Design of SHS Welded Joints: Supplement No. 1

CHS joints with high d0/t0 ratio

Large diameter CHS causes the d0/t0 ratio to be easily exceeded in the joint capacity limits (‘Design of SHS Welded Joints’, ENV 1993-1-1:1992/A1:1994 ‘Eurocode 3: Design Of Steel Structures, Part 1.1, Annex K’ and the CIDECT Design Guides). CT16 section 2.1.1 states ‘...if just one of these validity limits is slightly violated, and all of the joint’s other geometric parameters are well inside the limits, then we would suggest that the actual joint capacity should be reduced to about 0.85 times the capacity calculated using the design formulae.’ But when the d0/t0 limit is greatly exceeded a greater reduction may be required. Hence the following method is suggested.

Method If the only parameter exceeded is d0/t0 then a reduced chord yield strength should be used in the calculated joint capacity. This reduced chord yield strength should be based on BS5950-1:2000 clause 3.6.6 and Table 12 but should not be taken greater than 0.85 py. e.g. From table 12, limiting proportions for Class 3 (semi-compact) CHS in axial compression; D/t ≤ 80ε2 where ε = (275/py)0.5

Assuming limiting proportions for Class 3 sections are met; pyr = 80/(D/t) x 275 = 22000/(D/t)

Therefore, if joint d0/t0 limit (typically 40 or 50 depending on joint type) is exceeded; fy0 = pyr but ≤ 0.85 py where fy0 is the chord yield strength in the joint formula

Example In both cases, d0/t0 limit for relevant joint capacity is exceeded, hence the need to reduce fy0 to equal pyr ≤ 0.85 py. Example 1 Example 2

Chord Section CHS 1810 x 12.5 CHS 762 x 12.5

Chord Design Strength 355 N/mm2 275 N/mm2

D/t ≡ d0/t0 = 1810/12.5 = 144.8 762/12.5 = 60.96

pyr = 22000/(D/t) = 22000/144.8 = 152 N/mm2 22000/60.96 = 361 N/mm2

fy0 = pyr but ≤ 0.85 py fy0 = 152 but ≤ 302 N/mm2 fy0 = 361 but ≤ 234 N/mm2

Hence fy0 = 152 N/mm2 234 N/mm2

Corus Tubes, PO Box 101, Corby, Northamptonshire NN17 5UA Tel: +44 (0)1536 404120 Fax: +44 (0)1536 404049

Care has been taken to ensure that this information is accurate, but Corus Group plc, including its subsidiaries, does not accept responsibility or liability for errors or information which is found to be misleading.

/3 Last revised 06/09/2002 11:04

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KT-joints

KT-joint often occur in trusses and while some advice is provided by CIDECT and Eurocode 3 it only covers one brace load direction combination. As gap/overlap, section type and load direction combination affects the method of KT-joint capacity assessment further advice was required. The following suggested method is based on recommendations in CIDECT’s 'Design Of Rectangular Hollow Section (RHS) Joints Under Predominantly Static Loading' and 'Design Of Circular Hollow Section (CHS) Joints Under Predominantly Static Loading' which have been incorporated into Eurocode 3: ENV 1993-1-1:1992/A1:1994, Annex K.

1 13 3

2 2

1 13 32 2

(a) (b)

(e) (f)

1 13 32 2

1 13 32 2

(c) (d)

(g) (h)

Fig. 1- Eight types of KT joint

Fig. 1(a,b,e,f): Vertical brace opposite direction to both diagonal bracings Check brace 1and 3 as a normal K- and N-joint using the standard K- and N- joint formulae for gap or overlap. Repeat for brace 2 & 3. For types Fig. 1(e) and 1(f) check cross chord loading as described later.

Fig. 1(c,d,g,h): One diagonal brace opposite direction to other two bracings Follow the advice below depending on gap/overlap and section type. For types Fig. 1(g) and (h) check cross chord loading as described later.

KT-gap joints, CHS & RHS The resistance of gap joints can be related to K- and N-joints by replacing d1/d0 with (d1 + d2 + d3)/3 d0 for CHS, and for RHS with (b1 + b2 + b3 + h1 + h2 + h3)/6 b0 in the chord face deformation formula. The joint should be checked for all the relevant joint failure modes. The gap should be taken as the largest gap between two bracing members having significant forces acting in the opposite sense. If the vertical bracing member in a gap KT-joint shown in Fig. 1 has no force

/3 Last revised 06/09/2002 11:04

iij beov

beff bior

i

jj

beov

in it, the gap should be taken as the distance between the toes of members 1 and 2, and the joint treated as a standard K- and N-joint using d1/d0 for CHS and (b1 + b2 + h1 + h2)/4 b0 for RHS in the chord face deformation formula. The force components, normal to the chord, of the two members acting in the same sense are added together to represent the load. This should be less than, or equal to, the joint resistance component, normal to the chord, of the most highly loaded compressive member, normally N1. The single member acting in the opposite sense to the other two members should also be checked to ensure its force component is less than or equal to this joint resistance component. For example Fig. 1(c); N1,App sinθ1 + N3,App sinθ3 ≤ N1 sinθ1

N2,App sinθ2 ≤ N1 sinθ1

For example Fig. 1(d); N2,App sinθ2 + N3,App sinθ3 ≤ N1 sinθ1

N1,App sinθ1 ≤ N1 sinθ1

where; N1 is the calculated joint resistance.

Overlap KT-Joints, CHS Overlaps are more likely to occur in KT-joints. CHS overlap KT-joints are treated in the same way as CHS gap KT-joints but use the smallest overlap (in mm) between bracings. The procedure is easier than RHS KT-joints as they only need be checked for chord face deformation, calculated for the most highly loaded compressive bracing, usually N1.

Overlap KT-Joints, RHS RHS overlap KT-joint resistance can be determined by checking each overlapping bracing member and ensuring that Ni

≥ Ni,App where Ni is the calculated joint resistance.

The efficiency of the overlapped bracing member, subscript j, should be taken as equal to that of the overlapping member, i.e. Nj = Ni (Aj fyj) / (Ai fyi). For the overlapping bracing member effective width formulae, care should be taken to ensure that the member sequence of overlapping is properly accounted for. The overlapping bracing faces are designated as;

bi or beff is the face locating onto the chord beov is the face locating onto the overlapped bracing. The terms at the end of the overlapping bracing capacity formulae add these two faces together, i.e. ...+ beff + beov (or bi + beov for overlaps ≥ 80%), and assumes that only one face is overlapping. So, if the overlapping bracing is in the middle, and overlaps both diagonals, the formulae needs modifying accordingly, i.e.

Ni = fyi ti [(Ov / 50) (2 hi - 4 ti) + 2 beov] 25% ≤ Ov < 50%

Ni = fyi ti [2 hi - 4 ti + 2 beov] Ov ≥ 50%

This configuration uses the same formulae as a K-Joint: Ni = fyi ti [(Ov / 50) (2 hi - 4 ti) + beff + beov] 25% ≤ Ov < 50%

Ni = fyi ti [2 hi - 4 ti + beff + beov] 50% ≤ Ov < 80%

Ni = fyi ti [2 hi - 4 ti + bi + beov] Ov ≥ 80%

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Cross Chord Loading: CHS & RHS KT-joints with cross chord loading (i.e. X-joint loading), as in Figs. 1(e) to 1(h), should be treated as per the advice above depending on the gap/overlap, section type and brace load direction combination. However, an additional check should be made for the purlin or hanger joint. The joint should be checked as an X-joint using an equivalent bracing member size for the KT bracings (see Design of SHS Welded Joints: Supplement No.3 - Unidirectional K- and N-joints). Generally, in this case, the purlin or hanger side will be the critical part.



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Unidirectional K- and N-joints

If both bracings of a K- and N-joint act in the same sense, e.g. both in compression, the standard K- and N-joint formulae are not valid. In this situation it is suggested to check the K- and N-joint as a T-joint using one equivalent bracing on the chord. The equivalent bracing size is based on the footprint produced by the two bracings on the chord. The following suggests a method of calculating this equivalent bracing size and then calculating the joint capacity.

Equivalent bracing for CHS bracings with CHS or RHS chords Calculate footprint perimeter for each bracing, (see SHS Welding: Appendix 1: Table 2A).

Bracing footprint perimeter, Pi = di/2 [1 + Cosec θi + 3√(1 + Cosec2 θi)]

where: Cosec θi = 1/sin θi

Calculate a single equivalent bracing perimeter based on the overall footprint of both bracings;

Equivalent bracing perimeter, Peq = P1/2 + P2/2 + 2[d1/(2 sinθ1) + g + d2/(2 sinθ2)]

where; g = gap (+ve) or overlap (-ve)

For overlap ≥ 80%, Peq = Pi

(where subscript ‘i’ is the overlapped bracing only)

Effective diameter of equivalent single bracing,

Deff = Peq/π

Minimum diameter of equivalent single bracing,

Dmin = (d1 + d2)/2

Equivalent diameter of a single bracing in a T-joint,

Deq = (Deff + Dmin)/2 but ≤ d0 or b0

Equivalent length of a single bracing in a T-joint,

Heq = d1/(2 sinθ1) + g + d2/(2 sinθ2)

Equivalent bracing for RHS bracings with RHS chords Equivalent width and breadth of equivalent single bracing,

Beq = (b1 + b2)/2

Heq = h1/sinθ1 + g + h2/sinθ2

Fig. 1 - Unidirectional joint

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Peq = 2(Beq + Heq)

where; g = gap (+ve) or overlap (-ve)

Joint capacity of equivalent bracing T-joint with CHS chord The bracing is assumed to be at 90°, hence sinθi = 1.0

Chord face deformation capacity, Neq = fy0 t02(2.8 + 14.2β2) γ0.2 f(np)

Chord punching shear capacity (when Dmin ≤ d0 - 2t0),

Neq = fy0 t0 Peq/√3

where; β = Deq/d0 and γ = d0/(2t0)

Joint capacity of equivalent bracing T-joint with RHS chord The bracing is assumed to be at 90°, hence sinθi = 1.0

Chord face deformation capacity (for β ≤ 0.85),

Neq = fy0 t02/(1-β) (2η + 4(1-β)0.5) f(n)

Chord side wall buckling capacity (for β = 1.0),

Neq = f(fb) t0(2Heq + 10t0)

For 0.85 ≤ β ≤ 1.0 use linear interpolation between the chord face deformation and chord side wall buckling capacity above. Chord punching shear capacity (for 0.85 ≤ β ≤ (1-2t0/b0)),

Neq = fy0 t0 Peq/√3

where; β = Dmin/b0 or Beq/b0

η = Heq/b0

Note, it is not necessary to multiply the joint capacities for CHS bracings onto RHS chords by π/4 in the above formulas as the method is based upon the CHS bracing perimeters.

Proportioning equivalent bracing T-joint capacity into individual bracings The equivalent single bracing T-joint capacity needs to be divided between the two actual bracings in proportion to their individual applied forces, Ni,App, and bracing to chord angles. Joint capacity for individual bracings,

N1 = Neq N1,App/(N1,App sinθ1 + N2,App sinθ2)

N2 = Neq N2,App/(N1,App sinθ1 + N2,App sinθ2)

Additional checks required In addition to the above, each individual bracing must be checked as a T- or Y-joint using the standard formulae. In this check the CHS bracing on an RHS chord will need to be multiplied by π/4. In overlap joints, the overlapping bracing should also be checked as a T- or Y-joint with the overlapped bracing as the chord, using the standard T- or Y-joint formulae.



/3 Last revised 02/09/2002 11:02

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I-Section chord joints with chord stiffeners

I-Section chord T-, Y-, X-, K- and N-joints with SHS bracings often suffer low joint capacity due to effective width failure or chord web yielding. Both failure modes can be improved by the use of chord stiffener plates.

Fig. 1- Layout of chord stiffeners and symbols used

θ i

b i

b s

t f

t s c s

t w b ws

2

b w 2

a s

t w

b eff

h eff,s

(a) Bracing effective widths (b) Web effective widths

t i h i

The effect of stiffeners on effective width is well documented in the SCI’s Advisory Desk Note AD229, which can be referred to for further detail. Here the effects on effective width and chord web stability is summarised based on joint formulas in 'Design Of SHS Welded Joints'.

/3 Last revised 02/09/2002 11:02

Fig. 2- Alternative chord stiffeners

h ≥ (b -t )/2 s w s h ≥ (b -t )/2 s w s

When stiffeners of the type shown in Fig. 1 are used, then the formula for the resistance of the bracing member in a T-, Y-, X-, K- and N-gap or overlap joint (section 5.4) can be modified, as shown below, to increase the resistance to this type of failure. It is recommended that the stiffeners should be at least as thick as the I-section web. Suffix ‘s’ refers to the stiffener. Other symbols are as defined in ‘Design of welded joints’.

Effective width For effective width, as well as the two bracing out of plane faces (bi) carrying load, the stiffeners will act rather like an extra web, resulting in the faces along the length (hi) of the I-section also carrying some load, as shown by the shaded bracing areas in Fig. 1(a).

The standard effective width formula at the chord web is;

beff = tw + 2r + 7tf fy0 / fyi but ≤ (bi + hi - 2ti)

Calculate the effective width at the stiffener;

heff,s = ts + 2as + 7tf fy0 / fyi but ≤ (bi + hi - 2ti)

as = stiffener weld throat thickness, (‘2as’ becomes ‘as’ if single sided fillet welds are used)

T-, Y-, X-, K- and N-gap joints Check; peff = 2(beff + heff,s) but ≤ 2(bi + hi - 2ti)

Replace the standard effective width formula;

Ni = 2 fyi ti beff

with; Ni = fyi ti peff

Stiffener plates should be positioned on the bracing centre line.

K- and N-overlap joints (25% ≤ Ov < 50%) Check; leff = beff + 2heff,s but ≤ [2hi - 2ti - (Ov/50) (hi - 2ti)] + bi


Ni = fyi ti [(Ov/50) (hi - 2ti) + beff + beov]

with; Ni = fyi ti [(Ov/50) (hi - 2ti) + leff + beov]

Fig. 3 - K- & N-overlap joint (25% ≤ Ov < 50%) chord stiffeners position

x 1

x 1

x 2

x 2

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The overlapping bracing stiffener plates should be positioned as shown in Fig. 3. The overlapped bracing should be stiffened the same as the overlapping bracing, but the stiffener plates should be positioned on the overlapped bracing centre line.

K- and N-overlap joints (50% ≤ Ov < 80%) When Ov ≥ 50% the bracing sides, hi, are already fully effective so stiffener plates should be positioned as shown in Fig. 4 to make the bracing width fully effective.


Ni = fyi ti [hi - 2ti + beff + beov]

with; Ni = fyi ti [hi - 2ti + bi + beov]

Fig. 4 - K- & N-overlap joint (50% ≤ Ov < 80%) chord stiffeners position = =

The overlapping bracing stiffener plates should be positioned as shown in Fig. 4, inline with the bracing heel. The overlapped bracing should be stiffened the same as the overlapping bracing, but the stiffener plates should be positioned on the overlapped bracing centre line.

Chord web yielding Extra load will be carried by the stiffeners acting as an extra web, in the same way as for effective width. Replace the standard I-section chord web yielding formula;

Ni = fy0 tw bw / sinθi

with Ni = fy0 (tw bw + ts bws)/sinθi

where; bws = 2(ti + 5(tf + as)) but ≤ bs - tw - 2cs



/2 Last revised 06/05/2005 09:59

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Static design of stiffened and unstiffened CHS L-joints (knee joints)

Although design information exists in ‘Design of SHS Welded Joints’ for RHS, no information was available for CHS at the time. This supplement is based on a paper entitled ‘The static design of stiffened and unstiffened CHS L-joints’ presented at the ‘Tubular Structures IX’ conference by Professor R.Puthli (Faculty of Civil Engineering, University of Karlsruhe, Germany) which is from a PhD Thesis by Mr D.Karcher (KLIB Ingenieur-Partnerschaft, Ottersweier, Germany). Due to its profile, CHS L-joints suffer lower moment capacity than equivalent RHS L-joints. The following provides design information to enable CHS L-joints to be assessed.

To assess the L-joint capacity a reduction factor, κ must be calculated, this is dependant on d/t and steel grade. Angles less than 90° are not recommended especially as the weld integrity would need to be proved. Different diameter CHS should not be used for unstiffened L-joints and is not recommended for stiffened L-joints due to aesthetic reasons and fabrication problems. If different thickness CHS is to be used the thinner tube thickness should be used in the following formulae. Loads should be predominantly moment with the factored applied axial load no greater than 20% of the member tension capacity.

Stiffened Application limits: 90° ≤ θ <180°

235 N/mm2 ≤ fy ≤ 355 N/mm2

8 ≤ d/t ≤ 90ε2 (Class 1, 2 or 3)

tp ≥ 1.5 t but tp ≥ 10 mm

Fig. 1- CHS L-joints and symbols used

t

θ

(a) Stiffened (b) Unstiffened

d t

θ

d

t p

/2 Last revised 06/05/2005 09:59

As stiffened L-joints can carry the full moment capacity of the CHS the reduction factor, κ = 1.0 (plastic for class 1 & 2, elastic for class 3). Section Class Section loaded by bending or compression1 d/t < 50 ε2

2 d/t < 70 ε2 3 d/t < 90 ε2

fy 235 275 355 ε 1.00 0.92 0.81 ε = √(235/fy) ε2 1.00 0.85 0.66

Table 1: Section classes (EC3) and values for ε

Unstiffened Application limits: 90° ≤ θ <180°

235 N/mm2 ≤ fy ≤ 355 N/mm2

8 ≤ d/t ≤ 50ε2 (Class 1)

For unstiffened L-joints the following formula for the reduction of the plastic tube cross-section moment in plane capacity in the joint area is;

ε

+=κ

− 19.1

77.0t20

d

The joint capacity utilisation check is;

κ≤+ypl

app

y

app

fWM

fAN

Fig. 1: Graph of reduction factor, κ

Example CHS 273.0x16 S355, fy = 355 N/mm2, A = 129 cm2, θ = 135°, Wpl = 1058 cm3, Factored applied loads; Napp = 300 kN, Mapp = 70 kNm

d/t = 17 < 50 ε2 (Class 1 limit) = 50 x 0.66 = 33 therefore class 1 section

46.081.077.01620

27377.0t20

d 19.119.1

=

+

×=κε

+=κ

−−

κ≤+ypl

app

y

app

fWM

fAN

PASS 46.025.03551058

10007035510129

10003002 <=

××

+××

×



0.00.10.20.30.40.50.60.70.80.91.01.1

0 10 20 30 40 50CHS d/t ratio

Red

uctio

n fa

ctor

, k S235S275S355Stiffened

S235

S355S275

Stiffened

/1 Last revised 09/08/2005 10:01

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CHS/RHS chords and I-, H- bracing with flange in-plane with chord

CHS/RHS chords I-, H- bracings in the ‘Design of SHS Welded Joints’ publication is only intended for I-sections orientated with the web in-plane with the CHS/RHS chord. The formulae assumes that the load is carried by the flanges, and this supplement shows how the formulae can be modified to allow for the flanges being in-plane with the chord. Flange of I-section in plane with CHS chord

T-Joint Chord face Deformation

( ) ( ) ( )p0122

0y01 nf/dh0.251β204tfN ++= N1 for axial capacity formula does not change as yield lines are just a rectangular shape on the CHS regardless of either orientation of an RHS or I-section. M ( )01111,ip d/h25.01/Nh5.0 += M 111,op Nb= X-joint chord face deformation

( ) ( ) ( )p01

20y0

1 nf/dh0.251β0.811

tf5N +

−=

N1 for axial capacity formula does not change as yield lines are just a rectangular shape on the CHS regardless of either orientation of an RHS or I section. M ( )01111,ip d/h25.01/Nh5.0 += M 111,op Nb= Where N1 is chord face deformation joint capacity check from above.

Fig. 1 I-section, with flanges in-plane with CHS

b1

h1

t1

/2 Last revised 09/08/2005 10:01

T- and X- joint chord punching shear

In all cases the following check must be made to ensure that any factored applied loads and moments do not exceed the chord punching shear capacity ( ) 3/tf2tW/MA/N 00y11.elapp1app ≤+

Flange of I-section in plane with RHS chord

Plate Effective Width Check: Not Required

Chord Face Deformation Check: Use the formula for RHS chords and RHS bracings in 5.2.3 on page 27 ‘Design of SHS Welded Joints’ for

T-, Y- & X-joints when β ≤ 0.85 only.

Chord Side Wall Buckling: Use the formula for RHS chords and RHS bracings in 5.2.3 on page 27 ‘Design of SHS Welded Joints’ for

T-, Y- & X-joints when β = 1.0

When 0.85 < β < 1.0 interpolate between chord face deformation with β = 0.85 and chord side wall

buckling where β = 1.0.

Chord Punching Shear:

When 001 t2bb −<= only

( )1100y1 h2t2

3

tf2N+

=

In plane Moment:

M 111,ip Nh5.0= where N1 is lowest N1 joint capacity from applicable failure checks above Out if plane Moment:

M 111,op bN= where N1 is lowest N1 joint capacity from applicable failure checks above



Fig. 2 I-section, with flanges in-plane with RHS

b1

h1

t1

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