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Welded joints examplesCelsius®355 NH

Welded Joints Examples Contents

2

CONTENTS

RHS T-Joint Moment in Plane 03CHS Gap K-Joint 09CHS Overlap K-Joint 12RHS Gap K-Joint 15RHS Overlap K-Joint 21RHS Chord CHS Bracings Gap K-Joint 23RHS Chord CHS Bracings Overlap K-Joint 82.7 29RHS Chord CHS Bracings Overlap K-Joint 102.7 33RHS Chord RHS Bracings Overlap KT-Joint 36CHS Chord CHS Bracings Gap KT-Joint 41CHS Chord CHS Bracings Overlap KT-Joint 46UB Chord RHS Bracings Gap K-Joint 52

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101112

Welded Joints Examples RHS T-Joint Moment In Plane

1 RHS T-Joint Moment In Plane

3

References are generally to Tata Steel publication ‘Design of Welded Joints’ (unless otherwise stated).

SHS 150 x 150 x 10All material Celsius 355 to EN 10210

RHS 150 x 150 x 8

90º

N1,Ed = 19.2 kN Mip,1,Ed = 54 kNm

Np,Ed = 350 kNMip,0,Ed = 16.6 kNm

N0,Ed = 136 kN Mip,0,Ed = 35.8 kNm

Dimensions b0 = 150 mm b1 = 150 mm h0 = 150 mm h1 = 150 mm t0 = 10.0 mm t1 = 8.0 mm Validity limits check Chord: (b0-3 t0)/t0 ; (h0-3 t0)/t0 ≤ 38ε (Class 1 or 2 compression) 38ε = 38 √(235/fy0) = 38√(235/355) = 30.92 (b0-3 t0)/t0 = (150-3 x 10)/10 = 12 ∴PASS (h0-3 t0)/t0 = (150-3 x 10)/10 = 12 ∴PASS b0/t0 ≤ 35 b0/t0 = 150/10 = 15 ∴PASS h0/t0 ≤ 35 h0/t0 = 150/10 = 15 ∴PASS Compression brace: bi/ti ; hi/ti ≤ 35 b1/t1 = 150/8 =18.75 ∴PASS h1/t1 = 150/8 =18.75 ∴PASS Compression brace: (b1-3 t1)/t1 ; (h1-3 t1)/t1 ≤ 38ε (Class 1 or 2 compression) 38ε = 38 √(235/fy0) = 38√(235/355) = 30.92 (b1-3 t1)/t1 = (150-3 x 8)/8 = 15.75 ∴PASS (h1-3 t1)/t1 = (150-3 x 8)/8 = 15.75 ∴PASS 0.25 ≤ bi/b0 ≤ 1.0 b1/b0 = 150/150 = 1.0 ∴PASS 0.5 ≤ h0/b0 ≤ 2.0 h0/b0 = 150/150 = 1.0 ∴PASS 0.5 ≤ hi/bi ≤ 2.0 h1/b1 = 150/150 = 1.0 ∴PASS 30° ≤ θi ≤ 90° θ1 = 90° ∴PASS

Reference

5.3.1 Figure 37

(EN 1993-1-1) Table 5.2

(EN 1993-1-1) Table 5.2


4

Axial: Chord face failure (deformation) (valid when β ≤ 0.85)

required not check0.1150150

bb

0

1 ∴===β

Although this check is not required in this particular case, the chord end stress factor calculation for is shown below for information as it includes moments;

RHS chord end stress factor, kn RHS chord most compressive stress, σ0,Ed;

0,op,el

Ed,0,op

0,ip,el

Ed,0,ip

0

Ed,0Ed,0 W

M

W

M

AN

++=σ

Note: Moment is additive to compressive stress which is positive for moments. For RHS chords use most compressive chord stress. A0 = 54.9 cm2 = 5490 mm2

32Ed,010236

100010008.35109.54

1000136×

××+

×

×=σ

2

Ed,0 mm/N47.176=σ Chord factored stress ratio, n;

⎟⎠

⎞⎜⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛ σ=

35547.176

fn

0y

Ed,0

497.0n =

Using formulae: Using graph: For n > 0 (compression):

β−=

n4.03.1kn but kn ≤ 1.0

0.10.1

497.04.03.1kn ≤×

−=

0.1k0.1but101.1k nn =∴≤=

From graph, for β = 1.0; kn = 1.0

(However, not required in this case as chord deformation not critical as β>0.85)

Reference 5.3.3

6.3

5.3.2.1

5.3.2.1

5.3.2.1Fig. 38

5.3.2.1


5

Axial: Chord shear (valid for X-joints with Cos θ1 > h1/h0) As this is a T-joint with θ1 = 90° this check is not required. Axial: Chord side wall buckling (valid when β = 1.0) β = 1.0 ∴check required

5M01

1

1

0bnRd,1 /t10

sinh2

sintfk

N γ⎟⎟⎠

⎞⎜⎜⎝

⎛+

θθ=

Chord side wall buckling strength, fb For compression brace, T-joint:

0y

i0

0

fE

sin12

th

46.3π

θ⎟⎟⎠

⎞⎜⎜⎝

⎛−

=λ where: E = 210000 N/mm2

589.0

355210000

90sin12

10150

46.3 =

π

°⎟⎠

⎞⎜⎝

⎛ −

=λ

Using formulae: Using graph: From EN 1993-1-1:2005, 6.3.1.2, flexural buckling reduction factor, χ; From EN 1993-1-1:2005, table 6.1, 21.0=α (curve ‘a’)

( )( )2 2.015.0 λ+−λα+=φ

( )( ) 714.0589.02.0589.021.015.0 2 =+−+=φ

1.0but122

≤χλ−φ+φ

=χ

1.0but589.0714.0714.0

122

≤χ−+

=χ

895.0=χ

From graph, for

589.0=λ ;

895.0=χ

0yb ff χ= (for T- and Y-joints with compression in bracing) 2

b N/mm318355895.0f =×=

0.1/101090sin1502

90sin103180.1N Rd,1 ⎟⎟

⎠

⎞⎜⎜⎝

⎛×+

°×

°××

=

Reference 5.3.3

5.3.3

5.3.3

5.3.2.3

6.1 (For ‘E’)

5.3.2.3

(EN 1993-1-1) Table 6.1 (EN 1993-1-1) 6.3.1.2

5.3.2.3 Fig. 40


6

Axial: Chord punching shear

(valid when 0.85 ≤ β ≤ 1 – 1/γ)

5.7102

150t2

b

0

0 =×

==γ

867.05.7

110.185.0 =−≤≤

∴Check not required Axial: Bracing failure (effective width) (valid when β ≥ 0.85) β = 1.0 ∴check required ( ) 5Mi,eff1i11yRd,1 /b2t4h2tfN γ+−= where:

ii,effiiyi

00y

0

0i,eff bbbutb

tftf

bt10b ≤×=

mm 150bbutmm 1251508355

10355150

1010b i,effi,eff ≤=×××

××

=

beff,i = 125 mm ( ) 0.1/12528415028355N Rd,1 ×+×−××= PASS kN 2.19kN1471N Rd,1 ∴>= Moment in plane: Chord face failure (deformation) (valid when β ≤ 0.85)

required not check0.1150150

bb

0

1 ∴===β

Reference 5.3.3

6.3

5.3.3

6.3

5.3.3

5.3.2.2

5.3.5.1

6.3


7

Moment in plane: Chord side wall crushing (valid when 0.85 < β ≤ 1.0)

required check0.1150150

bb

0

1 ∴===β

( ) 5M

2010ykRd,1,ip /t5htf5.0M γ+=

where: fyk = fy0 (for T-joints)

∴ fyk = 355 N/mm2 ( ) 0.1/105150103555.0M 2

Rd,1,ip ×+××= PASS kNm 54kNm71M Rd,1,ip ∴>= Moment in plane: Bracing failure (effective width) (valid when 0.85 < β ≤ 1.0)

required check0.1150150

bb

0

1 ∴===β

( ) 5M11111

1,eff1,ip,pl1yRd,1,ip /tthb

bb

1WfM γ⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=

where: Wpl,ip,1 = 237 cm3 = 237000 mm3 beff,1 = 125 mm

( ) 0.1/881501501501251237000355M Rd,1,ip ⎥

⎦

⎤⎢⎣

⎡−⎟

⎠

⎞⎜⎝

⎛ −−=

PASS kNm 54kNm1.74M Rd,1,ip ∴>= Summary Axial joint resistance for brace 1 limited by chord side wall buckling and moment in plane resistance by chord side wall crushing resistance. Axial joint resistance, PASS kN 2.19kN1272N Rd,1 ∴>= Moment in plane joint resistance, PASS kNm 54kNm71M Rd,1,ip ∴>=

Reference 5.3.5.1

6.3

5.3.5.1

5.3.2.4

5.3.5.1

6.3

5.3.5.1


8

Interaction check When more than one type of force exists, e.g. axial and moments in plane, an interaction check is required;

0.1M

M

M

M

N

N

Rd,i,op

Ed,i,op

Rd,i,ip

Ed,i,ip

Rd,i

Ed,i≤++ (for RHS chords)

As there are no moments out of plane;

776.0M

07154

12722.19

Rd,i,op=++

PASS 000.1776.0 ∴≤

Reference 2.3

Welded Joints Examples CHS Gap K-Joint

2 CHS Gap K-Joint

9


N1,Ed = 500 kN N2,Ed = - 400 kN

CHS 219.1x12.5All material CELSIUS 355 to EN 10210

CHS 139.7x5.0 CHS 114.3x3.6

45º45º

40

Np,Ed = 1000 kN N0,Ed = 1636 kN

Note: Brace 1 usually designated compression and brace 2 tension. Dimensions d0 = 219.1 mm d1 = 139.7 mm d2 = 114.3 mm t0 = 12.5 mm t1 = 5.0 mm t2 = 3.6 mm

Validity limits check 10 ≤ d0/t0 ≤ 50 d0/t0 = 219.1/12.5 = 17.53 ∴PASS d0/t0 ≤ 70ε2 (Class 1 or 2 for compression chord) 70ε2 = 70 [√(235/fy0)]

2 = 70(235/355) = 46.34 17.53 < 46.34 ∴PASS di/ti ≤ 70ε2 (Class 1 or 2 for compression brace) 70ε2 = 70 [√(235/fy0)]

2 = 70(235/355) = 46.34 d1/t1 = 139.7/5.0 = 27.94 < 46.34 ∴PASS di/ti ≤ 50 d1/t1 = 139.7/5.0 = 27.94 ∴PASS d2/t2 = 114.3/3.6 = 31.75 ∴PASS 0.2 ≤ di/d0 ≤ 1.0 d1/d0 = 139.7/219.1 = 0.64 ∴PASS d2/d0 = 114.3/219.1 = 0.52 ∴PASS -0.55 d0 ≤ e ≤ +0.25 d0 -0.55 x 219.1 ≤ e ≤ +0.25 x 219.1 -120.5 ≤ e ≤ +54.8 e = 0 mm ∴PASS g ≥ t1 + t2 t1 + t2 = 5 + 3.6 = 8.6 mm g = 40 mm ∴PASS 30° ≤ θ i ≤ 90° θ1 = 45° ∴PASS θ2 = 45° ∴PASS

Reference

5.1.1 Figure 27

(EN 1993-1-1) Table 5.2

(EN 1993-1-1)

Table 5.2


10

Chord face failure (deformation)

Compression brace (1):

5M0

1

1

200ypg

Rd,1 /dd2.108.1

sintfkk

Nfailure, face Chord γ⎟⎟⎠

⎞⎜⎜⎝

⎛+

θ=

Gap/lap function, kg

764.85.122

1.219t2

d

0

0 =×

==γ

Using formulae: Using graph:

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

−+γ

+γ=33.1t/g5.0exp1

024.01k0

2.12.0

g

Note: g is positive for a gap and negative for an overlap

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

−×+×

+=33.15.12/405.0exp1

764.8024.01764.8k2.1

2.0g

761.1kg =

from graph; g/t0 = 40/12.5 = 3.2 kg = 1.761

CHS chord end stress factor, kp CHS chord least compressive applied factored stress, σp,Ed;

0,op,el

Ed,0,op

0,ip,el

Ed,0,ip

0

Ed,pEd,p W

M

W

M

AN

++=σ

Note: Moment is additive to compressive stress which is positive for moments. For CHS chords use least compressive chord stress.

22Ed,p N/mm30.123

101.8110001000

=×

×=σ

Chord factored stress ratio, np; 347.0355

30.123f

n0y

Ed,pp =⎟

⎠

⎞⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛ σ=

Using formulae: Using graph: For np > 0 (compression): kp = 1 - 0.3 np (1 + np) but ≤ 1.0

( ) 0.1but347.01347.03.01kp ≤+×−=

860.0kp =

from graph; kp = 0.860

Reference 5.1.3

6.3

5.1.2.2

5.1.2.1

5.1.2.1

5.1.2.1

5.1.2.2 Fig.29

5.1.2.1 Fig.28


11

0.1/1.2197.1392.108.1

45sin5.12355860.0761.1N

2

Rd,1 ⎟⎠

⎞⎜⎝

⎛ +°

×××=

kN986N Rd,1 = Tension brace (2):

Rd,12

1Rd,2 N

sinsin

Nθθ

=

98645sin45sinN Rd,2 ×

°°

=

kN986N Rd,2 = Chord punching shear (valid when di ≤ d0 – 2 t0) di ≤ d0 – 2 t0

di ≤ 219.1 – 2 x 12.5 = 194.1 mm d1 = 139.7 mm < 194.1 mm ∴check chord punching shear d2 = 114.3 mm < 194.1 mm ∴check chord punching shear

5M

i2

ii0

0yRd,i /

sin2

sin1dt

3

fN γ

θ

θ+π=

Brace (1):

0.1/

45sin245sin17.1395.12

3355N

2Rd,1°

°+××π××=

kN1919N Rd,1 = Brace (2):

0.1/

45sin245sin13.1145.12

3355N

2Rd,2°

°+××π××=

kN1570N Rd,2 = Joint strength dictated by chord face failure for both bracings; PASS kN 500kN986N,resistance joint 1 Brace Rd,1 ∴>= PASS kN 400kN986N,resistance joint 2 Brace Rd,2 ∴>=

Reference

5.1.3

5.1.3

Welded Joints Examples CHS Overlap K-Joint

3 CHS Overlap K-Joint

12


N1,Ed = 500 kN

Np,Ed = 1000 kN CHS 219.1x12.5All material CELSIUS 355 to EN 10210

CHS 139.7x5.0

N2,Ed = - 400 kN

N0,Ed = 1636 kN

CHS 114.3x3.6

45º45º

- 45

e = - 42.2

Note: Brace 1 usually designated compression and brace 2 tension Dimensions

d0 = 219.1 mm d1 = 139.7 mm d2 = 114.3 mm t0 = 12.5 mm t1 = 5.0 mm t2 = 3.6 mm Validity limits check 10 ≤ d0/t0 ≤ 50 d0/t0 = 219.1/12.5 = 17.53 ∴PASS d0/t0 ≤ 70ε2 (Class 1 or 2 for compression chord) 70ε2 = 70 [√(235/fy0)]

2 = 70(235/355) = 46.34 17.53 < 46.34 ∴PASS di/ti ≤ 70ε2 (Class 1 or 2 for compression brace) 70ε2 = 70 [√(235/fy0)]

2 = 70(235/355) = 46.34 d1/t1 = 139.7/5.0 = 27.94 < 46.34 ∴PASS di/ti ≤ 50 d1/t1 = 139.7/5.0 = 27.94 ∴PASS d2/t2 = 114.3/3.6 = 31.75 ∴PASS 0.2 ≤ di/d0 ≤ 1.0 d1/d0 = 139.7/219.1 = 0.64 ∴PASS d2/d0 = 114.3/219.1 = 0.52 ∴PASS -0.55 d0 ≤ e ≤ +0.25 d0 -0.55 x 219.1 ≤ e ≤ +0.25 x 219.1 -120.5 ≤ e ≤ +54.8 e = -42.2 mm ∴PASS 25% ≤ λov ≤ 100% λov = |g| sin θ i /di x 100% ∴PASS λov = 45 sin 45°/114.3 x 100% λov = 27.8% ∴PASS 30° ≤ θ i ≤ 90° θ1 = 45° ∴PASS θ2 = 45° ∴PASS

Reference

5.1.1 Figure 27

(EN 1993-1-1)

Table 5.2 (EN 1993-1-1)

Table 5.2


13

Chord face failure (deformation)

Compression brace (1):

5M0

1

1

200ypg

Rd,1 /dd2.108.1

sintfkk

Nfailure, face Chord γ⎟⎟⎠

⎞⎜⎜⎝

⎛+

θ=

Gap/lap function, kg

764.85.122

1.219t2

d

0

0 =×

==γ

Using formulae: Using graph:

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

−+γ

+γ=33.1t/g5.0exp1

024.01k0

2.12.0

g

Note: g is positive for a gap and negative for an overlap

( )( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−×+×

+=33.15.12/455.0exp1

764.8024.01764.8k2.1

2.0g

024.2kg =

from graph; g/t0 = -45/12.5 = -3.6 kg = 2.024


0,op,el

Ed,0,op

0,ip,el

Ed,0,ip

0

Ed,pEd,p W

M

W

M

AN

++=σ


22Ed,p N/mm30.123

101.8110001000

=×

×=σ


30.123f

n0y

Ed,pp =⎟

⎠

⎞⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛ σ=


( ) 0.1but347.01347.03.01kp ≤+×−=

860.0kp =


Reference 5.1.3

6.3

5.1.2.2 5.1.2.1

5.1.2.1

5.1.2.2Fig. 29

5.1.2.1Fig. 28

5.1.2.1


14

0.1/1.2197.1392.108.1

45sin5.12355860.0024.2N

2

Rd,1 ⎟⎠

⎞⎜⎝

⎛ +°

×××=

kN1134N Rd,1 = Tension brace (2):

Rd,12

1Rd,2 N

sinsin

Nθθ

=

113445sin45sinN Rd,2 ×

°°

=

kN1134N Rd,2 = Chord punching shear check not required for overlapping bracings Localised shear check for overlapping bracings not required as overlap not greater than 60%. Therefore, joint strength dictated by chord face failure for both bracings; PASS kN 500kN1134N,resistance joint 1 Brace Rd,1 ∴>= PASS kN 400kN1134N,resistance joint 2 Brace Rd,2 ∴>=

Reference 5.1.3

Welded Joints Examples RHS Gap K-Joint

4 RHS Gap K-Joint

15


N1,Ed = 650 kN

SHS 200 x 200 x 10All material Celsius 355 to EN 10210

SHS 120 x 120 x 5

45º

5

45º

N2,Ed = - 650 kN

40

Np,Ed = 1000 kN N0,Ed = 1920 kN

Dimensions b0 = 200 mm b1 = 120 mm b2 = 120 mm h0 = 200 mm h1 = 120 mm h2 = 120 mm t0 = 10.0 mm t1 = 5.0 mm t2 = 5.0 mm

Validity limits check Chord: (b0-3 t0)/t0 ; (h0-3 t0)/t0 ≤ 38ε (Class 1 or 2 compression) 38ε = 38 √(235/fy0) = 38√(235/355) = 30.92 (b0-3 t0)/t0 = (200-3 x 10)/10 = 17 ∴PASS (h0-3 t0)/t0 = (200-3 x 10)/10 = 17 ∴PASS b0/t0 ≤ 35 b0/t0 = 200/10 = 20 ∴PASS h0/t0 ≤ 35 h0/t0 = 200/10 = 20 ∴PASS Compression brace: bi/ti ; hi/ti ≤ 35 b1/t1 = 120/5 =24 ∴PASS h1/t1 = 120/5 =24 ∴PASS Compression brace: (b1-3 t1)/t1 ; (h1-3 t1)/t1 ≤ 38ε (Class 1 or 2 compression) 38ε = 38 √(235/fy0) = 38√(235/355) = 30.92 (b1-3 t1)/t1 = (120-3 x 5)/5 = 21 ∴PASS (h1-3 t1)/t1 = (120-3 x 5)/5 = 21 ∴PASS Tension brace: bi/ti ; hi/ti ≤ 35 b2/t2 = 120/5 =24 ∴PASS h2/t2 = 120/5 =24 ∴PASS bi/b0 ≥ 0.35 b1/b0 = 120/200 = 0.6 ∴PASS b2/b0 = 120/200 = 0.6 ∴PASS 0.1+0.01 b0/t0 ≤ bi/b0 ≤ 1.0 0.1+0.01 b0/t0 = 0.3 b1/b0 = 120/200 = 0.6 ∴PASS b2/b0 = 120/200 = 0.6 ∴PASS

Reference

5.3.1 Figure 37 (EN 1993-1-1)

Table 5.2

(EN 1993-1-1)

Table 5.2


16

0.5 ≤ h0/b0 ≤ 2.0 h0/b0 = 200/200 = 1.0 ∴PASS 0.5 ≤ hi/bi ≤ 2.0 h1/b1 = 120/120 = 1.0 ∴PASS h2/b2 = 120/120 = 1.0 ∴PASS -0.55 h0 ≤ e ≤ +0.25 h0 -0.55 x 200 ≤ e ≤ +0.25 x 200 -110.0 ≤ e ≤ +50.0 e = +5.0 mm ∴PASS 30° ≤ θ i ≤ 90° θ 1 = 45° ∴PASS θ 2 = 45° ∴PASS g ≥ t1 + t2 40 ≥ 5 + 5 = 10 mm ∴PASS 0.5 b0(1-β) ≤ g ≤ 1.5 b0(1-β) g = 40 mm β = (b1+b2+h1+h2)/(4 b0) β = (120+120+120+120)/(4x200) = 0.6 0.5 x 200(1-0.6) ≤ g ≤ 1.5 x 200(1-0.6) 40 mm ≤ g ≤ 120 mm ∴PASS -0.55 h0 ≤ e ≤ +0.25 h0 e = 5 mm -0.55 x 200 ≤ e ≤ +0.25 x 200 -110 ≤ e ≤ +50 ∴PASS Chord face failure (deformation) – brace 1 Note: Brace 1 usually designated compression and brace 2 tension. Compression brace (1):

5M0

2121

i

200yn

Rd,i /b4

hhbbsin

tfk9.8N γ⎟⎟

⎠

⎞⎜⎜⎝

⎛ +++θ

γ=

Reference

5.3.3 5.3.3


17

RHS chord end stress factor, kn RHS chord most compressive applied factored stress, σ0,Ed;

0,op,el

Ed,0,op

0,ip,el

Ed,0,ip

0

Ed,0Ed,0 W

M

W

M

AN

++=σ

Note: Moment is additive to compressive stress which is positive for moments. For RHS chords use most compressive chord stress.

22Ed,0 N/mm34.256

109.7410001920

=×

×=σ

Chord factored stress ratio, n;

⎟⎠

⎞⎜⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛ σ=

35534.256

fn

0y

Ed,0

722.0n =


β−=

n4.03.1kn but kn ≤ 1.0

0.1but6.0

722.04.03.1kn ≤×

−=

819.0kn =


10102

200t2

b

0

0 =×

==γ

0.1/2004

12012012012045sin

1010355819.09.8N2

Rd,1 ⎥⎦

⎤⎢⎣

⎡×

+++×

°××××

=

kN694N Rd,1 =

Reference 5.3.2.1

5.3.2.1

6.3

5.3.2.1Fig. 38

5.3.2.1


18

Chord shear between bracings check – brace 1

5Mi

0,v0yRd,i /

sin3

AfN γ

θ=

where: for RHS bracings:

212.0

1034041

1

t3

g41

1

2

2

20

2=

×

×+

=

+

=α

Av,0 = (2 h0 + α b0) t0 ( ) 2

0,v mm 442410200212.02002A =×+×=

0.1/45sin3

4424355N Rd,i°

×=

kN1282N Rd,1 = Bracing Effective width – brace 1 ( ) 5MeffiiiiyiRd,i /bbt4h2tfN γ++−= where:

ii,effiiyi

00y

0

0i,eff bbbutb

tftf

bt10b ≤×=

mm 120bbutmm 1201205355

10355200


××

=

beff,i = 120 mm ( ) 0.1/1201205412025355N Rd,i ++×−××= kN817N Rd,1 =

Reference 5.3.3 5.3.2.5 5.3.2.5 5.3.3

5.3.2.2


19

Chord punching shear – brace 1 (valid when β ≤ 1 – 1/γ) β ≤ 1 – 1/γ

0.6 ≤ 1 – 1/10 = 0.9 mm ∴check chord punching shear

5Mi,p,eii

i

i

00yRd,i /bb

sinh2

sin3

tfN γ⎟⎟

⎠

⎞⎜⎜⎝

⎛++

θθ=

where:

ii,p,ei0

0i,p,e bbbutb

bt10b ≤=

mm 120bbutmm 60120200

1010b i,p,ei,p,e ≤=××

=

be,p,i = 60 mm

0.1/6012045sin1202

45sin310355N Rd,i ⎟⎟

⎠

⎞⎜⎜⎝

⎛++

°×

°

×=

kN1506N Rd,1 = Summary - brace 1 Joint strength for brace 1 dictated by chord face deformation. ∴ Brace 1 joint resistance, PASS kN 650kN694N Rd,1 ∴>= Brace 2 Repeat brace resistance formulae for brace 2. Note: As both braces are of same geometry, brace 2 resistance will be the same: Chord face deformation, kN694N Rd,2 = Chord shear check, kN1282N Rd,2 = Bracing effective width, kN817N Rd,2 = Punching shear, kN1506N Rd,2 = ∴ Brace 2 joint resistance, PASS kN 650kN694N Rd,2 ∴>=

Reference 5.3.3 5.3.3

5.3.2.2


20

Chord axial load resistance in gap Check: N0,gap,Rd ≥ N0,gap,Ed

( ) ( ) 5M2

Rd,0,plEd,00y0,v0y0,v0Rd,gap,0 /V/V1fAfAAN γ⎥⎦⎤

⎢⎣⎡ −+−=

where: A0 = 74.9 cm2 = 7490 mm2 Av,0 = 4424 mm2 V0,Ed = max( |N1,Ed| sin θ1 , |N2,Ed| sin θ2) V0,Ed = max( |650| sin 45° , |-650| sin 45°) = max (459.6, 459.6) V0,Ed = 459.6 kN

0M

0y0,vRd,0,pl

)3/f(AV

γ=

kN 7.9060.1

)3/355(4424V Rd,0,pl =×

=

650 - 650

Np,Ed = 1000 1000 460

460N0,Ed = 1920

1000 460

N0,gap,Ed = 1460

460 Horz 460 Horz

Change of chord axial force in K-joint N0,gap,Ed = max (Np,Ed + N1,Ed cos θ1, N0,Ed + N2,Ed cos θ2) N0,gap,Ed = max (1000 + 650 cos 45°, 1920 + (-650) cos 45°) N0,gap,Ed = max (1460, 1460) = 1460 kN

( ) ( ) 0.1/7.906/6.4591355442435544247490N 2Rd,gap,0 ⎥⎦

⎤⎢⎣⎡ −×+−=

PASS kN 1460kN2442N Rd,gap,0 ∴>=

Reference 5.3.3 5.3.3 5.3.2.5 5.3.3 5.3.3 (EN 1993-1-1)

6.2.6 (2)

5.3.3

Welded Joints Examples RHS Overlap K-Joint

5 RHS Overlap K-Joint

21


N1,Ed = 600 kN

Np,Ed = 1000 kN N0,Ed = 1849 kNSHS 200 x 200 x 10All material Celsius 355 to EN 10210

SHS 120 x 120 x 5 N2,Ed = - 600 kN

45º45º 70 i j

-50.1

Dimensions b0 = 200 mm b1 = 120 mm b2 = 120 mm h0 = 200 mm h1 = 120 mm h2 = 120 mm t0 = 10.0 mm t1 = 5.0 mm t2 = 5.0 mm Validity limits check Chord: (b0-3 t0)/t0 ; (h0-3 t0)/t0 ≤ 38ε (Class 1 or 2 compression) 38ε = 38 √(235/fy0) = 38√(235/355) = 30.92 (b0-3 t0)/t0 = (200-3 x 10)/10 = 17 ∴PASS (h0-3 t0)/t0 = (200-3 x 10)/10 = 17 ∴PASS Compression brace: (b1-3 t1)/t1 ; (h1-3 t1)/t1 ≤ 33ε (Class 1 compression) 33ε = 33 √(235/fy0) = 33√(235/355) = 26.85 (b1-3 t1)/t1 = (120-3 x 5)/5 = 21 ∴PASS (h1-3 t1)/t1 = (120-3 x 5)/5 = 21 ∴PASS Tension brace: bi/ti ; hi/ti ≤ 35 b2/t2 = 120/5 =24 ∴PASS h2/t2 = 120/5 =24 ∴PASS 0.25 ≤ bi/b0 ≤ 1.0 b1/b0 = 120/200 = 0.6 ∴PASS b2/b0 = 120/200 = 0.6 ∴PASS 0.5 ≤ h0/b0 ≤ 2.0 h0/b0 = 200/200 = 1.0 ∴PASS 0.5 ≤ hi/bi ≤ 2.0 h1/b1 = 120/120 = 1.0 ∴PASS h2/b2 = 120/120 = 1.0 ∴PASS -0.55 h0 ≤ e ≤ +0.25 h0 -0.55 x 200 ≤ e ≤ +0.25 x 200 -110.0 ≤ e ≤ +50.0 e = -50.1 mm ∴PASS 25% ≤ λ ov λ ov = |g| sin θ i /h i x 100% λ ov = 70 sin 45°/120 x 100% λ ov = 41.2% ∴PASS bi/bj ≥ 0.75 bi/bj = 120/120 = 1.0 ∴PASS

Reference

5.3.1 Figure 37 (EN 1993-1-1)

Table 5.2

(EN 1993-1-1) Table 5.2

Welded Joints Examples RHS Overlap K-Joint

22

Bracing Effective width – Overlapping brace, i (2)

Note: Only the overlapping brace (i) need be checked. The resistance of the overlapped brace (j) is based on an efficiency ratio to that of the overlapping brace. Overlapping brace, i (2): For 25% ≤ λov < 50%

5Miov

iov,ei,effiyiRd,i /t450

h2bbtfN γ⎟⎠

⎞⎜⎝

⎛ −λ

++=

where:

ii,effiiyi

00y

0

0i,eff bbbutb

tftf

bt10b ≤×=

mm 120bbutmm 1201205355

10355200


××

=

beff,i = 120 mm

iov,eiiyi

jyj

j

jov,e bbbutb

tftf

bt10

b ≤×=

120bbut12053555355

120510b ov,eov,e ≤×

××

××

=

mm 50b ov,e =

0.1/5450

2.411202501205355N Rd,i ⎟⎠

⎞⎜⎝

⎛ ×−×++×=

PASS kN 600kN617NN Rd,2Rd,i ∴>== Bracing Effective width – Overlapped brace, j (1)

( )( )yii

yjjRd,iRd,j fA

fANN =

where: Aj = A1 = 22.7 cm2 = 2270 mm2 Ai = A2 = 22.7 cm2 = 2270 mm2

( )( )3552270

3552270617N Rd,j ××

×=

PASS kN 600kN617NN Rd,1Rd,i ∴>==

Reference 5.3.3

5.3.3

5.3.2.2

5.3.2.2

5.3.3

Welded Joints Examples RHS Chord CHS Bracings Gap K-Joint

6 RHS Chord CHS Bracings Gap K-Joint

23


Dimensions b0 = 200 mm d1 = 114.3 mm d2 = 114.3 mm h0 = 200 mm t1 = 6.3 mm t2 = 6.3 mm t0 = 10.0 mm

Validity limits check Chord: (b0-3 t0)/t0 ; (h0-3 t0)/t0 ≤ 38ε (Class 1 or 2 compression) 38ε = 38 √(235/fy0) = 38√(235/355) = 30.92 (b0-3 t0)/t0 = (200-3 x 10)/10 = 17 ∴PASS (h0-3 t0)/t0 = (200-3 x 10)/10 = 17 ∴PASS b0/t0 ≤ 35 b0/t0 = 200/10 = 20 ∴PASS h0/t0 ≤ 35 h0/t0 = 200/10 = 20 ∴PASS

Compression brace: (d1/t1 ≤ 50ε (Class 1 or 2 compression) 50ε = 50 √(235/fy0) = 50√(235/355) = 40.7 d1/t1 = 114.3/6.3 = 18.1 ≤ 40.7 ∴PASS Tension brace: di/ti ≤ 50 d2/t2 = 114.3/6.3 = 18.1 ∴PASS 0.4 ≤ di/b0 ≤ 0.8 d1/b0 = 114.3/200 = 0.57 ∴PASS d2/b0 = 114.3/200 = 0.57 ∴PASS

Reference

5.3.1 Figure 37 (EN 1993-1-1)

Table 5.2

(EN 1993-1-1)

Table 5.2

g = 85

45° 45°

Np,Ed = 1212 kN

N2,Ed = - 500 kN

N1,Ed = 500 kN

N0,Ed = 1920 kN

CHS 114.3 x 6.3

SHS 200 x 200 x 10

ALL MATERIAL; CELSIUS 355 to EN 10210

e = 23.3


24

0.5 ≤ h0/b0 ≤ 2.0 h0/b0 = 200/200 = 1.0 ∴PASS

-0.55 h0 ≤ e ≤ +0.25 h0 -0.55 x 200 ≤ e ≤ +0.25 x 200 -110.0 ≤ e ≤ +50.0 e = +23.0 mm ∴PASS 30° ≤ θ i ≤ 90° θ 1 = 45° ∴PASS θ 2 = 45° ∴PASS g ≥ t1 + t2 85 ≥ 6.3 + 6.3 = 12.6 mm ∴PASS

Chord face failure (deformation) – brace 1 Note: Brace 1 usually designated compression and brace 2 tension. Compression brace (1):

5M0

2121

i

200yn

Rd,i /b4

ddddsin

tfk9.8N γ⎟⎟

⎠

⎞⎜⎜⎝

⎛ +++θ

γ=

RHS chord end stress factor, kn RHS chord most compressive applied factored stress, σ0,Ed;

0,op,el

Ed,0,op

0,ip,el

Ed,0,ip

0

Ed,0Ed,0 W

M

W

M

AN

++=σ

Note: Moment is additive to compressive stress which is positive for moments. For RHS chords use most compressive chord stress.

22Ed,0 N/mm34.256

109.7410001920

=×

×=σ

Chord factored stress ratio, n;

⎟⎠

⎞⎜⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛ σ=

35534.256

fn

0y

Ed,0

722.0n =

Reference

5.3.3

5.3.2.1

5.3.2.1


25


β−=

n4.03.1kn but kn ≤ 1.0

0.1but6.0

722.04.03.1kn ≤×

−=

819.0kn =


10102

200t2

b

0

0 =×

==γ

0.1/2004

3.1143.1143.1143.11445sin

1010355819.09.8N2

Rd,1 ⎥⎦

⎤⎢⎣

⎡×

+++×

°××××

=

kN5194

661......kN661N Rd,1 =π

×=

Chord shear between bracings check – brace 1

5Mi

0,v0yRd,i /

sin3

AfN γ

θ=

where: for CHS bracings: 0=α Av,0 = (2 h0 + α b0) t0 ( ) 2

0,v mm 40001020002002A =×+×=

0.1/45sin3

4000355N Rd,i°

×=

kN9104

......kN1159N Rd,1 =π

×=

Reference

6.3

5.3.3

5.3.2.5

5.3.2.5

5.3.2.1Fig. 38

5.3.2.1


26

Bracing Effective width – brace 1 ( ) 5MeffiiiiyiRd,i /bdt4d2tfN γ++−= where:

ii,effiiyi

00y

0

0i,eff dbbutd

tftf

bt10b ≤×=

mm 120bbutmm 2.951203.6355

10355200


××

=

beff,i = 95.2 mm ( ) 0.1/2.953.1143.643.11423.6355N Rd,i ++×−××=

kN7254

.....kN923N Rd,1 =π

×=

Chord punching shear – brace 1 (valid when β ≤ 1 – 1/γ) β ≤ 1 – 1/γ

0.6 ≤ 1 – 1/10 = 0.9 mm ∴check chord punching shear

5Mi,p,eii

i

i

00yRd,i /bd

sind2

sin3

tfN γ⎟⎟

⎠

⎞⎜⎜⎝

⎛++

θθ=

where:

ii,p,ei0

0i,p,e dbbutd

bt10b ≤=

mm 120bbutmm 2.573.114200

1010b i,p,ei,p,e ≤=××

=

be,p,i = 57.2 mm

0.1/2.573.11445sin

3.114245sin310355N Rd,i ⎟⎟

⎠

⎞⎜⎜⎝

⎛++

°×

°

×=

kN11264

........kN1434N Rd,1 =π

×=

Summary - brace 1 Joint strength for brace 1 dictated by chord face deformation. ∴ Brace 1 joint resistance, PASS kN 500kN519N Rd,1 ∴>=

Reference 5.3.3

5.3.2.2

5.3.3

5.3.3 5.3.2.2


27

Brace 2 Repeat brace resistance formulae for brace 2. Note: As both braces are of same geometry, brace 2 resistance will be the same: Chord face deformation, kN519N Rd,2 = Chord shear check, kN910N Rd,2 = Bracing effective width, kN725N Rd,2 = Punching shear, kN1126N Rd,2 = ∴ Brace 2 joint resistance, PASS kN 500kN519N Rd,2 ∴>= Chord axial load resistance in gap Check: N0,gap,Rd ≥ N0,gap,Ed

( ) ( ) 5M2

Rd,0,plEd,00y0,v0y0,v0Rd,gap,0 /V/V1fAfAAN γ⎥⎦⎤

⎢⎣⎡ −+−=

where: A0 = 74.9 cm2 = 7490 mm2 Av,0 = 4000 mm2 V0,Ed = max( |N1,Ed| sin θ1 , |N2,Ed| sin θ2) V0,Ed = max( |500| sin 45° , |-500| sin 45°) = max (354,354 ) V0,Ed = 354 kN

0M

0y0,vRd,0,pl

)3/f(AV

γ=

kN 8.8190.1

)3/355(4000V Rd,0,pl =×

=

Reference 5.3.3 5.3.2.5 5.3.3

5.3.2.2 5.3.3 (EN 1993-1-1)

6.2.6 (2)


28

500 - 500

Np,Ed = 1212 1212 354

354N0,Ed = 1920

1212 354

N0,gap,Ed = 1566

354 Horz 354 Horz

Change of chord axial force in K-joint N0,gap,Ed = max (Np,Ed + N1,Ed cos θ1, N0,Ed + N2,Ed cos θ2) N0,gap,Ed = max (1212 + 500 cos 45°, 1920 + (-500) cos 45°) N0,gap,Ed = max (1566, 1566) = 1566 kN

( ) ( ) 0.1/8.819/3541355400035540007490N 2Rd,gap,0 ⎥⎦

⎤⎢⎣⎡ −×+−=

PASS kN 1566kN2520N Rd,gap,0 ∴>=

Reference 5.3.3

Welded Joints Examples RHS Chord CHS Bracings Overlap K-Joint 82.7

7 RHS Chord CHS Bracings Overlap K-Joint 82.7

29


NOTE: BRACE ‘j’ USUALLY DESIGNATED COMPRESSION AND BRACE ‘i’ TENSION.

Dimensions

h0 = 100.0 mm b0 = 200.0 mm di = 114.3 mm dj = 114.3 mm t0 = 10.0 mm ti = 6.3 mm tj = 6.3 mm fu,i = 470 N/mm²

fu,j = 470 N/mm² (fu,i & fu,j from EN10210-1:2006 Table A.3) For Eccentricity;

( ) 2h

sinsinsin

gsin2d

sin2de 0

ji

ji

j

j

i

i −⎥⎥⎦

⎤

⎢⎢⎣

⎡

θ+θ

θ×θ×⎥⎥⎦

⎤

⎢⎢⎣

⎡+

θ×+

θ×=

( )mm1.109g

60sin1003.1147.82g

sin100dgwhere

i

iov

−=°×

×=

θ××λ

=

Reference

6.5 Figure 57

30° 60°e = - 19.5

CHS 114.3 x 6.3

CHS 114.3 x 6.3

Ni,Ed = - 400kN

Nj,Ed = 500kN

Np,Ed = 1000kN N0,Ed = 1636kN

RHS 200 x 100 x 10


-109.1


30

( ) ( ) ( ) ( ) ( )( )

mm5.19e

2100

3060sin30sin60sin1.109

30sin23.114

60sin23.114e

−=

−⎥⎦

⎤⎢⎣

⎡°+°°×°

×⎥⎦

⎤⎢⎣

⎡−+

°×+

°×=

Validity limits check

b0/t0 ≤ 38ε2 (Class 1 or 2 for compression chord) 38ε2 = 38 [√(235/fy0)]

2 = 38(235/355) = 25.15 b0/t0 = 200.0/10.0 = 20.0 < 25.15 ∴PASS

dj/tj ≤ 50ε2 (Class 1 for compression brace) 50ε2 = 50 [√(235/fy0)]

2 = 50(235/355) = 33.10 dj/tj = 114.3/6.3 = 18.14 < 33.10 ∴PASS di/ti ≤ 50 (Tension Brace) di/ti = 114.3/6.3 = 18.14 ≤ 50 ∴PASS 0.4 ≤ di/b0 ≤ 0.8 di/d0 = 114.3/200.0 = 0.57 ∴PASS dj/d0 = 114.3/200.0 = 0.57 ∴PASS 0.5 ≤ h0/ b0 ≤ 2.0 h0/ b0 = 100.0/200.0 = 0.50 ∴PASS -0.55 h0 ≤ e ≤ +0.25 h0 -0.55 x 100.0 ≤ e ≤ +0.25 x 100.0 -55 ≤ e ≤ +25 e = -19.5 mm ∴PASS λov ≥ 25% λov = |g| sin θ I /di x 100% ∴PASS λov = 109.1 sin 60°/114.3 x 100% λov = 82.7% ∴PASS 30° ≤ θI ≤ 90° θj = 30° ∴PASS θI = 60° ∴PASS di/dj ≥ 0.75 di/dj = 114.3/114.3 = 1.0 ∴PASS

Brace Effective Width;

Overlapping brace (i):

( ) 5Miiov,eiiyiRd,i /t4d2ddtfN γ−++=

where:

iov,eiiyi

jyj

jjov,e ddbutd

tftf

t/d10d ≤×

×

××=

Reference

(EN1993-1-1) (Table 5.2)

(EN1993-1-1) (Table 5.2)

5.3 Figure 37

5.3.3

5.3.2.2


31

mm63d3.114but63d

3.114dbut3.1143.63553.6355

3.6/3.11410d

ov,eov,e

ov,eov,e

=∴≤=

≤×××

×=

∴ Ni,Rd = 355 x 6.3 x (114.3 + 63 + 228.6 – 25.2) / 1.0

=Rd,iN 851 kN ----- 851 × π⁄4 = 668 kN > 400 kN ∴ PASS


Overlapped Brace (j):

( )( )yii

yjjRd,iRd,j fA

fANN

×

××=

where:

22

i

22j

mm2140cm4.21A

mm2140cm4.21A

==

==

( )( )3552140

3552140668N Rd,j ××

×=

PASSkN500kN668N Rd,j ∴>=

Overlapping Bracings Shear Check For CHS Bracings;

.weldingneedstoehidden%20componentverticalbracebetweendifferenceAs

%8.271004.3462504.346.diffcomponent.vert%

kN250)30(sin500NkN4.346)60(sin400N

)(sinNN&)(sinNN

;N&NofcomponentVertical

Ed,v,j

Ed,v,i

jEd,jEd,v,jiEd,iEd,v,i

Ed,jEd,i

>∴

=×−

=∴

=°×=∴

=°×=∴

θ×=θ×=∴

Reference

5.3.3


32

Shear Check valid when 80% < λov < 100% and overlapped brace hidden toe welded;

( )5Mj

jj,effsjj,u

i

ii,effiov

i,u

jEd,jiEd,i

1sin

tdcd2

3

f

sin

tdd2100

100

3

f4

......

...........cosNcosN

γ×

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

θ

××+×+

θ

×⎥⎥⎦

⎤

⎢⎢⎣

⎡⎜⎜⎝

⎛+×⎟

⎠

⎞λ−

××π

≤θ+θ

mm4.91d3.1144.91d

3.114but3.1143.6355

10355200

1010d

dbutdtftf

bt10d:where

i,effi,eff

i,eff

iiii,y

00,y

0

0i,eff

=∴≤=

≤×××

××

=

≤××

××=

mm4.91d3.1144.91d

3.114but3.1143.6355

10355200

1010d

dbutdtftf

bt10d:where

j,effj,eff

j,eff

jjjj,y

00,y

0

0j,eff

=∴≤=

≤×××

××

=

≤××

××=

)weldedtoehidden(2c:where s=

( )

( )( )( )

( ) 11

30sin3.64.9123.1142

3470

60sin

3.64.913.1142100

7.82100

3470

4......

...........cosNcosN jjii

×

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

°××+×

×+°

×⎥⎥⎦

⎤

⎢⎢⎣

⎡⎜⎜⎝

⎛+××⎟⎟

⎠

⎞−

××π

≤θ+θ∴

= 1307 kN

∴ ( ) ( ) PASSkN1307kN63330cos50060cos400 ∴≤=°×+°×

Reference

5.3.3

5.3.2.2

5.3.2.2

5.1.3


8 RHS Chord CHS Bracings Overlap K-Joint 102.7

33


NOTE: BRACE ‘j’ USUALLY DESIGNATED COMPRESSION AND BRACE ‘i’ TENSION.

Dimensions

h0 = 100.0 mm b0 = 200.0 mm di = 114.3 mm dj = 114.3 mm t0 = 10.0 mm ti = 6.3 mm tj = 6.3 mm fu,i = 470 N/mm²

fu,j = 470 N/mm² (fu,i & fu,j from EN10210-1:2006 Table A.3) For Eccentricity;

( ) 2h

sinsinsin

gsin2d

sin2de 0

ji

ji

j

j

i

i −⎥⎥⎦

⎤

⎢⎢⎣

⎡

θ+θ

θ×θ×

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

θ×+

θ×=

( )mm5.135g

60sin1003.1147.102g

sin100dgwhere

i

iov

−=°×

×=

θ××λ

=

Reference

6.5 Figure 57

30° 60°e = - 30.6

CHS 114.3 x 6.3

CHS 114.3 x 6.3

Ni,Ed = - 400kN

Nj,Ed = 500kN

Np,Ed = 1000kN N0,Ed = 1636kN

RHS 200 x 100 x 10


g = -135.5


34

( ) ( ) ( ) ( ) ( )( )

mm6.30e

2100


30sin23.114

60sin23.114e

−=

−⎥⎦

⎤⎢⎣

⎡°+°°×°

×⎥⎦

⎤⎢⎣

⎡−+

°×+

°×=

Validity limits check

b0/t0 ≤ 38ε2 (Class 1 or 2 for compression chord) 38ε2 = 38 [√(235/fy0)]

2 = 38(235/355) = 25.15 b0/t0 = 200.0/10.0 = 20.0 < 25.15 ∴PASS dj/tj ≤ 50ε2 (Class 1 for compression brace) 50ε2 = 50 [√(235/fy0)]

2 = 50(235/355) = 33.10 dj/tj = 114.3/6.3 = 18.14 < 33.10 ∴PASS di/ti ≤ 50 (Tension Brace) di/ti = 114.3/6.3 = 18.14 ≤ 50 ∴PASS 0.4 ≤ di/b0 ≤ 0.8 di/d0 = 114.3/200.0 = 0.57 ∴PASS dj/d0 = 114.3/200.0 = 0.57 ∴PASS 0.5 ≤ h0/ b0 ≤ 2.0 h0/ b0 = 100.0/200.0 = 0.50 ∴PASS -0.55 h0 ≤ e ≤ +0.25 h0 -0.55 x 100.0 ≤ e ≤ +0.25 x 100.0 -55 ≤ e ≤ +25 e = -30.6 mm ∴PASS λov ≥ 25% λov = |g| sin θ i /di x 100% ∴PASS λov = 135.5 sin 60°/114.3 x 100% λov = 102.7% ∴PASS 30° ≤ θi ≤ 90° θj = 30° ∴PASS θi = 60° ∴PASS di/dj ≥ 0.75 di/dj = 114.3/114.3 = 1.0 ∴PASS


Overlapping brace (i):

( ) 5Miiov,eiiyiRd,i /t4d2ddtfN γ−++= where λov ≥ 80%;

where:

mm63d3.114but63d

3.114dbut3.1143.63553.6355

3.6/3.11410d

ddbutdtftf

t/d10d

ov,eov,e

ov,eov,e

iov,eiiyi

jyj

jjov,e

=∴≤=

≤×××

×=

≤××

××=

Reference

(EN1993-1-1) Table 5.2

(EN1993-1-1) Table 5.2

5.3 Figure 37

5.3.3

5.3.2.2


35

Ni,Rd = 355 x 6.3 x (114.3 + 63 + 228.6 – 25.2) / 1.0

=Rd,iN 851 kN -------> 851 × π⁄4 = 668 kN > 400 kN ∴ PASS


Overlapped Brace (j):

( )( )yii

yjjRd,iRd,j fA

fANN

×

××=

where:

22

i

22j

mm2140cm4.21A

mm2140cm4.21A

==

==

( )( )3552140

3552140668N Rd,j ××

×=

PASSkN500kN668N Rd,j ∴>=

Overlapping Bracings Shear Check for CHS Bracings; where λov > 100%;

where:

∴

Reference

5.3.3

5.3.3

5.3.2.2

mm4.91d3.1144.91d

3.114but3.1143.6355

10355200

1010d

dbutdtftf

bt10d

j,effj,eff

j,eff

jjjj,y

00,y

0

0j,eff

=∴≤=

≤×××

××

=

≤××

××=

( )5M

jEd,jiEd,i

1)30sin(

3.64.913.11433

4704

......

...........cosNcosN

γ×

×+×××

π

≤θ+θ

( )5Mj

jj,effjj,u

jEd,jiEd,i

1sin

tdd3

3

f4

......

...........cosNcosN

γ×

θ

×+×××

π

≤θ+θ

kN673cosNcosN jjii ≤θ+θ

( ) ( ) PASSkN673kN63330cos50060cos400 ∴≤=°×+°×

Welded Joints Examples RHS Chord RHS Bracings Overlap KT-Joint

9 RHS Chord RHS Bracings Overlap KT-Joint

36


NOTE: Brace ‘j’ designated overlapped, Brace ‘i’ overlapping Brace ‘1’ SHS is 90 x 90 x 5.0, Brace ‘2’ and ‘3’ is SHS 100 x 100 x 6.3 Dimensions

h0 = 200.0 mm h1 = 90.0 mm h2 = 100.0 mm h3 = 100.0 mm b0 = 200.0 mm b1 = 90.0 mm b2 = 100.0 mm b3 = 100.0 mm t0 = 10.0 mm t1 = 5.0 mm t2 = 6.3 mm t3 = 6.3 mm γM5 = 1.0 A1 = 23.2cm2 A2 = 16.7cm2 A3 = 23.2cm2

This example is to show how to calculate a KT-Joint with SHS chord and braces. The method of checking a KT-Joint depends on the direction of the brace forces, gap or overlap and chord section profile. This example is for;

• One diagonal brace opposite direction to the other two bracings • Both diagonal bracings overlapping • Rectangular chord and bracings

With this configuration, the joint can be checked like two separate K-joints. Please be aware that different brace angle and overlap values may change how the joint is calculated and you should refer to the Tata Design of Welded Joints literature before tackling any joint to make sure you are using the correct method and formulae.The force combination of the braces can be seen at 5.6.2 Figure 51(c) (as a (from left to right) compression-compression-tension) of our Design Of Welded Joints literature.

Reference

45° 50°

e = - 19.5

N2,Ed = - 650kN N1,Ed = 492kN

RHS 200 x 100 x 10

ALL MATERIAL; CELSIUS 355 to EN 10210 N3,Ed = 150kN

SHS 200 x 200 x 10.0

Brace 1

Brace 3

Brace 2

e2= -40.4

e1 = -50.0


37

GEOMETRIC CALCULATIONS For Eccentricity; Brace 1 in relation to Brace 3;

( ) 2h

sinsinsing

sin2b

sin2be 0

31

31

3

3

1

11 −⎥

⎦

⎤⎢⎣

⎡

θ+θθ×θ

×⎥⎦

⎤⎢⎣

⎡+

θ×+

θ×=

Brace 1 to 3 overlap, λov = 50%;

where;

i

iovsin100

bgθ×

×λ=

( ) 145sin100

0.900.50g1 −×°×

×=

mm6.63g1 −= (gap is negative as it’s an overlap)

( ) ( ) ( ) ( ) ( )( )

mm0.50

20.200


90sin20.100

45sin20.90e1

−=

−⎥⎦

⎤⎢⎣

⎡°+°°×°

×⎥⎦

⎤⎢⎣

⎡−+

°×+

°×=

For Eccentricity, Brace 2 in relation to Brace 3;

( ) 2h

sinsinsing

sin2b

sin2be 0

32

32

3

3

2

22 −⎥

⎦

⎤⎢⎣

⎡

θ+θθ×θ

×⎥⎦

⎤⎢⎣

⎡+

θ×+

θ×=

Brace 2 to 3 overlap, λov = 50%; where;

2

2ov2 sin100

bgθ×

×λ=

( )°××

=50sin100

1000.50g2


( ) ( ) ( ) ( ) ( )( )

mm4.40

20.200


90sin20.100

50sin20.100e2

−=

−⎥⎦

⎤⎢⎣

⎡°+°°×°

×⎥⎦

⎤⎢⎣

⎡−+

°×+

°×=

Validity limits check for both K-Joints

(b0-3 t0)/t0 ; (h0-3 t0)/t0 ≤ 38ε (Class 1 or 2 for chord) 38ε = 38 [√(235/fy0)] = 38√ (235/355) = 30.92 (b0-3 t0)/t0 = (200-3 x 10)/10 = 17 ∴PASS (h0-3 t0)/t0 = (200-3 x 10)/10 = 17 ∴PASS

(b1-3 t1)/t1 ; (h1-3 t1)/t1 ≤ 33ε (Class 1 for compression braces)

Reference

6.5 Figure 57

6.5 Figure 57

5.3 Figure 37

(EN1993-1-1) Table 5.2

(EN1993-1-1) Table 5.2


38

33ε = 33 [√(235/fy0)] = 33√ (235/355) = 26.85 (b1-3 t1)/t1 = (90-3 x 5)/5 = 15 ∴PASS

(h1-3 t1)/t1 = (90-3 x 5)/5 = 15 ∴PASS (b3-3 t3)/t3 = (100-3 x 6.3)/6.3 = 12.9 ∴PASS (h3-3 t3)/t3 = (100-3 x 6.3)/6.3 = 12.9 ∴PASS bi/ti ; hi/ti ≤ 35 (Tension Brace) b2/t2 = 100.0/6.3 = 15.9 ∴PASS h2/t2 = 100.0/6.3 = 15.9 ∴PASS 0.25 ≤ bi/b0 ≤ 1.0 b1/b0 = 90.0/200.0 = 0.45 ∴PASS b2/b0 = 100.0/200.0 = 0.5 ∴PASS b3/b0 = 100.0/200.0 = 0.5 ∴PASS

0.5 ≤ h0/ b0 ≤ 2.0 h0/ b0 = 200.0/200.0 = 1.0 ∴PASS 0.5 ≤ hi/ bi ≤ 2.0 h1/b1 = 90.0/90.0 = 1.0 ∴PASS h2/b2 = 100.0/100.0 = 1.0 ∴PASS h3/b3 = 100.0/100.0 = 1.0 ∴PASS

-0.55 h0 ≤ e ≤ +0.25 h0 -0.55 x 200.0 ≤ e ≤ +0.25 x 200.0 -110 ≤ e ≤ +50 e1 = -50 mm ∴PASS e2 = -40.4 mm ∴PASS

λov ≥ 25% λov,1 = |g| sin θ 1 /h1 x 100%

λov,1 = 63.6 x sin 45°/90 x 100% λov,1 = 50% ∴PASS λov,2 = 65.3 x sin 50°/100 x 100% λov,2 = 50% ∴PASS

30° ≤ θi ≤ 90° θ1 = 45° ∴PASS

θ2 = 50° ∴PASS θ3 = 90° ∴PASS bi/bj ≥ 0.75 b1/b3 = 90.0/100.0 = 0.9 ∴PASS

b2/b3 = 100.0/100.0 = 1.0 ∴PASS Brace 1 & 3 as overlap K-Joint Brace Effective Width Failure for 50% ≤λov<80%;

( ) 5Miiov,ei,effiyiRd,i /t4h2bbtfN γ−++=

where;

11,ov,e111y

33y

3

31,ov,e bbbutb

tftf

bt10b ≤×

×

××=

Reference

6.5 Figure 57

5.3.3 (K- & N-

Overlap Sect)

5.3.2.2


39

mm4.71b0.90but4.71b

0.90bbut0.900.53553.6355

0.1003.610b

1,ov,e1,ov,e

1,ov,e1,ov,e

=∴≤=

≤×××

××

=

where;

11,eff111y

00y

0

01,eff bbbutb

tftf

bt10

b ≤××

××

×=

90bbut900.5355

10355200

1010b 1,eff1,eff ≤×

×

××

×=

mm90b90bbutmm90b 1,eff1,eff1,eff =∴≤=

N1,Rd = 355 x 5.0 x (90.0 + 71.4 + 2 x 90 – 4 x 5.0)/1.0

=Rd,1N 570 kN > 492 kN ∴PASS

For overlapped Brace (3):

where; ( )( )yii

yjjRd,iRd,j fA

fANN

×

××=

22

1

223

mm1670cm7.16A

mm2320cm2.23A

==

==

( )( )3551670

3552320570N Rd,3 ××

×=

=Rd,3N 792kN > 150kN ∴PASS

Brace 2 & 3 as overlap K-Joint

Brace Effective Width Failure for 50% ≤λov<80%;

22,ov,e222y

33y

3

32,ov,e bbbutb

tftf

bt10

b ≤××

××=

Reference

5.3.2.2

5.3.3 (K- & N-

Overlap Sect)

(Tata Technical

Guide TST02)

5.3.2.2


40

mm63b0.100but63b

0.100bbut0.1003.63553.6355

0.1003.610b

2,ov,e2,ov,e

2,ov,e2,ov,e

=∴≤=

≤×××

××

=

where;

22,eff222y

00y

0

02,eff bbbutb

tftf

bt10

b ≤××

××

×=

100bbut100

3.635510355

2001010

b 2,eff2,eff ≤××

××

×=

mm4.79b100bbutmm4.79b 2,eff2,eff2,eff =∴≤=

N2,Rd = 355 x 6.3 x (79.4 + 63 + 2 x 100 – 4 x 6.3) / 1.0

=Rd,2N 709 kN > 650 kN ∴PASS

For Overlapped Brace (3):

where: ( )( )yii

yjjRd,iRd,j fA

fANN

×

××=

22

2

223

mm2320cm2.23A

mm2320cm2.23A

==

==

( )( )3552320

3552320709N Rd,3 ××

×=

N3,Rd = 709kN > 150kN ∴PASS

SUMMARY

Here, the validity limits are combined to include all three braces and then for the calculations, due to the nature of the joint, it can be treated as two overlap K-Joints. This means Brace 3 (the vertical overlapped brace) is checked twice in relation to Brace 1 and Brace 2 and must pass on both of these checks. It is important to check the joint configuration and refer to our Design of Welded Joints literature to make sure the correct method and formulae is used.

Reference

5.3.2.2

5.3.3 (K- & N- Overlap Sect)

(Tata Technical Guide TST02)

Welded Joints Examples CHS Chord CHS Bracings Gap KT-Joint

10 CHS Chord CHS Bracings Gap KT-Joint

41

512kN 1087kN


NOTE: Brace ‘1’ CHS is 88.9 x 5.0, Brace ‘2’ and ‘3’ is CHS 101.6 x 6.3 Dimensions

d0 = 193.7 mm d1 = 88.9 mm d2 = 101.6 mm d3 = 101.6 mm t0 = 10.0 mm t1 = 5.0 mm t2 = 6.3 mm t3 = 6.3 mm γM5 = 1.0 A1 = 13.2cm2 A2 = 18.9cm2 A3 = 18.9cm2

This example is to show how to calculate a gap KT-Joint with CHS chord and braces. The recommendations for an all-CHS gap KT-Joint are in EN1993-1-8 Table 7.6. The method of checking a KT-Joint depends on the direction of the brace forces, gap or overlap and chord section profile. This example is for;

• One diagonal brace opposite direction to the other two bracings • Both diagonal bracings have a gap to the central brace • Circular chord and bracings

With this configuration, the joint capacity is based on the most highly loaded compression brace for chord face failure and punching shear failure for each brace. Please be aware that different brace angle and gap values may change how the joint is calculated and you should refer to the Tata Design of Welded Joints literature before tackling any joint to make sure you are using the correct method and formulae.The force combination of the braces can be seen at 5.6.2 Figure 51(c) (as a (from left to right) compression-compression-tension) of our Design Of Welded Joints literature.

Reference

g2,3 = 52mm g1,3 = 30mm

N3,Ed = 113kN

45° 30°

e = - 19.5

N2,Ed = - 488kN

N1,Ed = 369kN

RHS 200 x 100 x 10


CHS 193.7 x 10.0

Brace 1

Brace 3

Brace 2

e2 = 21.2mm e1 = 46.8mm


42

Geometric Calculations For Eccentricity; Brace 1 in relation to Brace 3;

( ) 2d

sinsinsing

sin2d

sin2de 0

31

31

3

3

1

11 −⎥

⎦

⎤⎢⎣

⎡

θ+θθ×θ

×⎥⎦

⎤⎢⎣

⎡+

θ×+

θ×=

Brace 1 to 3 gap, g1 = 30mm;

( ) ( )( ) ( )( )

mm8.46

27.193


90sin26.101

45sin29.88e1

=

−⎥⎦

⎤⎢⎣

⎡°+°°×°

×⎥⎦

⎤⎢⎣

⎡+

°×+

°×=


( ) 2d

sinsinsing

sin2d

sin2de 0

32

32

3

3

2

22 −⎥

⎦

⎤⎢⎣

⎡

θ+θθ×θ

×⎥⎦

⎤⎢⎣

⎡+

θ×+

θ×=

Brace 2 to 3 gap, g2 = 52mm;

( ) ( )( ) ( )( )

mm2.21

27.193


90sin26.101

30sin26.101e2

=

−⎥⎦

⎤⎢⎣

⎡°+°°×°

×⎥⎦

⎤⎢⎣

⎡+

°×+

°×=


10 ≤ d0/t0 ≤ 50 d0/t0 = 193.7/10.0 = 19.37 ∴PASS d0/t0 ≤ 70ε2 (Class 1 or 2 for chord) 70ε2 = 70 [√(235/fy0)]2 = 70√ (235/355)2 = 46.34 d0/t0 = 19.37 ≤ 46.34 ∴PASS

di/ti ≤ 50 d1/t1 = 88.9/5.0 = 17.78 ∴PASS d2/t2 = 101.6/6.3 = 16.13 ∴PASS d3/t3 = 101.6/6.3 = 16.13 ∴PASS di/ti ≤ 70ε2 (Class 1 or 2 for compression bracings) 70ε2 = 70 [√(235/fy0)]2 = 70√ (235/355)2 = 46.34 d1/t1= 17.78 ∴PASS d3/t3 = 16.13 ∴PASS 0.2 ≤ di/d0 ≤ 1.0 d1/d0 = 88.9/193.7 = 0.46 ∴PASS d2/d0 = 101.6/193.7 = 0.52 ∴PASS d3/d0 = 101.6/193.7 = 0.52 ∴PASS

-0.55 d0 ≤ e ≤ +0.25 d0 -0.55 x 193.7 ≤ e ≤ +0.25 x 193.7 -106.5 ≤ e ≤ +48.4 e1 = 46.8 mm ∴PASS e2 = 21.2 mm ∴PASS

Reference

5.1.1 Figure 27

(EN1993-1-1)

Table 5.2

(EN1993-1-1) Table 5.2


43

g ≥ t1 + t2 but ≤ 12t0 for g1,3 ≥ t1 + t3 = 5.0 + 6.3 = 11.3 mm but ≤ 12t0 g2,3 ≥ t2 + t3 = 6.3 + 6.3 = 12.6 mm but ≤ 12t0

12t0 = 12 x 10.0 = 120 mm g1,3 = 30 mm ∴PASS g2,3 = 52 mm ∴PASS 30° ≤ θi ≤ 90° θ1 = 45° ∴PASS

θ2 = 30° ∴PASS θ3 = 90° ∴PASS

Chord Face Failure (deformation); Based on compression brace with most compressive load; Compression brace (1);

5M0

321

1

200ypg

Rd,1 /d3

ddd2.108.1

sintfkk

N γ⎟⎟⎠

⎞⎜⎜⎝

⎛×

++×+×

θ

×××=

Gap/lap factor, kg;

Using Formulae:

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

−+γ

+γ=33.1t/g5.0exp1

024.01k0

2.12.0

g

where;

69.90.102

7.193t2

d

0

0 =×

==γ

Note: (g) is positive for gap and negative for overlap. Use the largest gap between two bracings having significant forces acting in opposite direction.

Using Formulae: Using Graph:

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

−×+×

+=33.10.10/525.0exp1

69.9024.0169.9k2.1

2.0g

70.1kg =

From graph;

g/t0 = 52/10.0 = 5.2

kg = 1.70

Reference

5.1.3 (K- & N- Joints)

5.1.2.2 (kg)

6.3 (γ)

For graph 5.1.2.2 Fig. 29


44

PASSkN369kN538N Rd,1 ∴>=


0,op,el

Ed,0,op

0,ip,el

Ed,0,ip

0

Ed,pEd,p W

M

W

M

AN

++=σ


22Ed,p N/mm7.88

107.571000512

=×

×=σ


7.88f

n0y

Ed,pp =⎟

⎠

⎞⎜⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛ σ=


( ) 0.1but25.0125.03.01kp ≤+×−=

91.0kp =


0.1/7.1933

6.1016.1019.882.108.145sin

0.1035591.070.1N 2

Rd,1 ⎟⎠

⎞⎜⎝

⎛×

++×+×

°×××

=

For this type of configuration (see 5.6.2 Figure 51(c));

1Rd,13Ed,31Ed,1 sinNsinNsinN θ×≤θ×+θ×

°×≤°×+°× 45sin53890sin11345sin369

PASSkN380kN374 ∴≤

Reference

5.1.2.1

5.1.2.1

5.1.2.1 (kp)


5.6.2


45

1Rd,12Ed,2 sinNsinN θ×≤θ×

°×≤°× 45sin53830sin448


Check for Chord Punching Shear failure;

Valid if: di ≤ d0 – 2t0 where d0 – 2t0 = 193.7 – (2 x 10.0) = 173.7 mm d1 = 88.9 mm d2 = 101.6 mm d3 = 101.6 mm Therefore check for Chord Punching Shear;

5Mi

2i

100y

Rd,i /sin2

sin1dt

3

fN γ

θ×

θ+××π××=

For Brace 1;

0.1/45sin2

45sin19.880.103

355N 2Rd,1

°×

°+××π××=


For Brace 2;

0.1/30sin2

30sin16.1010.103

355N 2Rd,2

°×

°+××π××=


For Brace 3;

0.1/

90sin290sin16.1010.10

3355N

2Rd,3°×

°+××π××=


Summary Here, the validity limits are combined to include all three braces. The eccentricity values use both brace 1 and 2 (diagonals) in relation to brace 3 (the vertical brace). It is important to check the joint configuration and refer to our Design of Welded Joints literature to make sure the correct method and formulae is used. To re-iterate this is a theoretical method which is not supported by research or tests.

Reference

5.1.3

5.1.3

Welded Joints Examples CHS Chord CHS Bracings Overlap KT-Joint

11 CHS Chord CHS Bracings Overlap KT-Joint

46

512kN 1087kN


NOTE: Brace ‘j’ designated overlapped, Brace ‘i’ overlapping Brace ‘1’ CHS is 88.9 x 5.0, Brace ‘2’ and ‘3’ is CHS 101.6 x 6.3 Dimensions

d0 = 193.7 mm d1 = 88.9 mm d2 = 101.6 mm d3 = 101.6 mm t0 = 10.0 mm t1 = 5.0 mm t2 = 6.3 mm t3 = 6.3 mm γM5 = 1.0 A1 = 13.2 cm2 A2 = 18.9 cm2 A3 = 18.9 cm2

fu1 = 470 N/mm2 fu2 = 470 N/mm2 fu3 = 470 N/mm2

This example is to show how to calculate an overlap KT-Joint with CHS chord and braces. Please be aware that there are no recommendations for an overlap KT- Joint in EN1993-1-8 or elsewhere as far as Tata Steel is aware. This is a Tata Steel method and is included in the Tata Steel Design of Welded Joints literature. It is a theoretical method which is not supported by research or tests. The method of checking a KT-Joint depends on the direction of the brace forces, gap or overlap and chord section profile. This example is for;

• One diagonal brace opposite direction to the other two bracings • Both diagonal bracings overlapping • Circular chord and bracings

With this configuration, the joint capacity is based on the most highly loaded compression brace for chord face failure. Please be aware that different overlap values may change how the joint is calculated and you should refer to the Tata Design of Welded Joints literature before tackling any joint to make sure you are using the correct method and formulae.The force combination of the braces can be seen at 5.6.2 Figure 51(c) (as a (from left to right) compression-compression-tension) of our Design Of Welded Joints literature. Due to one of the braces falling within the limits (overlapping percentage) for the localised brace-to-chord shear check, this calculation has also been completed.

Reference

‘fu‘ values taken from Product Standard EN10210-1 Table A.3 as required by UK National Annex

45° 50°

e = - 19.5

N2,Ed = - 488kN N1,Ed = 369kN

RHS 200 x 100 x 10

ALL MATERIAL; CELSIUS 355 to EN 10210 N3,Ed = 113kN

CHS 193.7 x 10.0

Brace 1

Brace 3

Brace 2

e2 = -60.4mm e1 = -73.2mm


47

Geometric Calculations For Eccentricity; Brace 1 in relation to Brace 3;

( ) 2d

sinsinsing

sin2d

sin2de 0

31

31

3

3

1

11 −⎥

⎦

⎤⎢⎣

⎡

θ+θθ×θ

×⎥⎦

⎤⎢⎣

⎡+

θ×+

θ×=

Brace 1 to 3 overlap, λov = 71.6%;

where;

1

1ovsin100

dgθ×

×λ=

( ) 145sin100

9.886.71g1 −×°×

×=

mm90g1 −= (gap is negative as it’s an overlap)

( ) ( ) ( ) ( ) ( )( )

mm2.73

27.193


90sin26.101

45sin29.88e1

−=

−⎥⎦

⎤⎢⎣

⎡°+°°×°

×⎥⎦

⎤⎢⎣

⎡−+

°×+

°×=


( ) 2d

sinsinsing

sin2d

sin2de 0

32

32

3

3

2

22 −⎥

⎦

⎤⎢⎣

⎡

θ+θθ×θ

×⎥⎦

⎤⎢⎣

⎡+

θ×+

θ×=

Brace 2 to 3 overlap, λov = 65.2%; where;

2

2ov2 sin100

dgθ×

×λ=

( )°××

=50sin100

6.1012.65g2


( ) ( ) ( ) ( ) ( )( )

mm4.60

27.193


90sin26.101

50sin26.101e2

−=

−⎥⎦

⎤⎢⎣

⎡°+°°×°

×⎥⎦

⎤⎢⎣

⎡−+

°×+

°×=


10 ≤ d0/t0 ≤ 50 d0/t0 = 193.7/10.0 = 19.37 ∴PASS d0/t0 ≤ 70ε2 (Class 1 or 2 for chord) 70ε2 = 70 [√(235/fy0)]2 = 70√ (235/355)2 = 46.34 d0/t0 = 19.37 ≤ 46.34 ∴PASS

Reference

6.5 Figure 57

6.5 Figure 57

5.1.1 Figure 27

(EN1993-1-1)

Table 5.2


48

di/ti ≤ 50 d1/t1 = 88.9/5.0 = 17.78 ∴PASS d2/t2 = 101.6/6.3 = 16.13 ∴PASS d3/t3 = 101.6/6.3 = 16.13 ∴PASS di/ti ≤ 70ε2 (Class 1 or 2 for compression bracings) 70ε2 = 70 [√(235/fy0)]2 = 70√ (235/355)2 = 46.34 d1/t1= 17.78 ∴PASS d3/t3 = 16.13 ∴PASS 0.2 ≤ di/d0 ≤ 1.0 d1/b0 = 88.9/193.7 = 0.46 ∴PASS d2/b0 = 101.6/193.7 = 0.52 ∴PASS d3/b0 = 101.6/193.7 = 0.52 ∴PASS

-0.55 d0 ≤ e ≤ +0.25 d0 -0.55 x 193.7 ≤ e ≤ +0.25 x 193.7 -106.5 ≤ e ≤ +48.4 e1 = -73.2 mm ∴PASS e2 = -60.4 mm ∴PASS

λov ≥ 25% λov,i = |g| sin θ i /di x 100%

λov,1 = 90.0 x sin 45°/88.9 x 100% λov,1 = 71.6% ∴PASS λov,2 = 86.5 x sin 50°/101.6 x 100% λov,2 = 65.2% ∴PASS

30° ≤ θi ≤ 90° θ1 = 45° ∴PASS

θ2 = 50° ∴PASS θ3 = 90° ∴PASS

Chord Face Failure (deformation); Compression brace (1);

5M0

1

1

200ypg

Rd,1 /dd

2.108.1sin

tfkkN γ⎟⎟

⎠

⎞⎜⎜⎝

⎛×+×

θ

×××=

Gap/lap factor, kg;

Reference

(EN1993-1-1)

Table 5.2

5.1.1 Figure 27 6.5 Figure 57

5.1.3 (K- & N- Joints)

5.1.2.2 (kg)

6.3 (γ)

Using Formulae:

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

−+γ

+γ=33.1t/g5.0exp1

024.01k0

2.12.0

g

where;

69.90.102

7.193t2

d

0

0 =×

==γ

Note: (g) is positive for gap and negative for overlap

Use the smallest overlap for (g)


49




0,op,el

Ed,0,op

0,ip,el

Ed,0,ip

0

Ed,pEd,p W

M

W

M

AN

++=σ


22Ed,p N/mm7.88

107.571000512

=×

×=σ


7.88f

n0y

Ed,pp =⎟

⎠

⎞⎜⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛ σ=


( ) 0.1but25.0125.03.01kp ≤+×−=

91.0kp =


0.1/7.193

9.882.108.145sin

0.1035591.015.2N 2

Rd,1 ⎟⎠

⎞⎜⎝

⎛ ×+×°

×××=

Tension Brace (2);

Rd,1

2

1Rd,2 N

sinsin

N ×θθ

=

637

50sin45sinN Rd,2 ×

°°

=

Chord Punching shear not required for overlapping bracings

Reference


5.1.2.1

5.1.2.1

5.1.2.1 (kp)


5.1.3

Using Formulae: Using Graph:

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−×+×

+=33.10.10/)5.86(5.0exp1

69.9024.0169.9k2.1

2.0g

15.2kg =

From graph;

g/t0 = -86.5/10.0 = -8.7

kg = 2.15


50

Localised Brace-to-Chord Shear Check for overlapped Joints;

Although this failure mode check was included as a corrigendum in section 7.1.2 (6) of EN 1993-1-8, no equations have been included in tables 7.2, 7.10, 7.21 or 7.24. In their absence, the following criteria for adequacy of the shear connection may be used. They are based on research by CIDECT but expressed using EN1993-1-8 symbols.

Modified equation only applicable when brace 3 is the overlapped brace (j) and diagonals 1 and 2 overlap (i).

For KT-Joint overlap joint check 60% < λov,1 or λov,2 < 100%;

( ) .......cosNcosNcosN 3Ed,32Ed,21Ed,1 ≤θ+θ−θ

........sin

tdd2100

100

3

fsin

tdd2100

100

3

f4 2

22,eff22,ov

2u

1

11,eff11,ov

1uθ

×⎥⎥⎦

⎤

⎢⎢⎣

⎡+×⎟⎟

⎠

⎞⎜⎜⎝

⎛ λ−

×+θ

×⎥⎥⎦

⎤

⎢⎢⎣

⎡+×⎟⎟

⎠

⎞⎜⎜⎝

⎛ λ−

×⎜⎜⎝

⎛×

π

( )5M3

33,eff2s3,eff1s3u 1sin

tdcdc

3

fγ

×⎟⎟⎠

⎞

θ

××+××+

where Brace Effective Width (overlapping braces);

ii,effiiyi

00y

0

0i,eff ddbutd

tftf

dt12

d ≤××

××=

9.88dbut9.88

0.53550.10355

7.1930.1012d 1,eff1,eff ≤×

××

××

=

9.88dbut1.110d 1,eff1,eff ≤=

mm9.88d 1,eff =

6.101dbut6.101

3.63550.10355

7.1930.1012d 2,eff2,eff ≤×

××

××

=

6.101dbut100d 2,eff2,eff ≤=

mm100d 2,eff =

Reference

5.1.3 (Modified for three braces)

5.1.2.3


51

where Brace Effective Width (overlapped brace);

jj,effj

jyj

00y

0

0j,eff ddbutd

tftf

dt12

d ≤××

××=

6.101dbut6.101

3.63550.10355

7.1930.1012d 3,eff3,eff ≤×

××

××

=

6.101dbut100d 3,eff3,eff ≤=

mm100d 3,eff =

Assume the hidden toes of braces 1 & 2 are NOT welded therefore;

( )2cweldeds/toehiddenIf1c

1c

s2s

1s==

=

Therefore;

( )( ) .......90cos11350cos48845cos369 ≤°+°−−°

........50sin

3.61006.1012100

2.65100

3470

45sin

0.59.889.882100

6.71100

3470

4 °

×⎥⎦

⎤⎢⎣

⎡+××⎟

⎠

⎞⎜⎝

⎛ −

×+°

×⎥⎦

⎤⎢⎣

⎡+××⎟

⎠

⎞⎜⎝

⎛ −

×⎜⎜⎝

⎛×

π

( )0.10.1

90sin3.610011001

3470

×⎟⎠

⎞°

××+××+

( ) 0.1341907380971267467

4kN575 ×++×

π≤

( )989745

4kN575 ×

π≤

Summary Here, the validity limits are combined to include all three braces. The eccentricity values use both brace 1 and 2 (diagonal) in relation to brace 3 (the vertical overlapped brace). The localised brace-to-chord shear check has been completed due to the high overlap. Please note that even if one brace is within limits for the brace-to-chord shear check, the check will still have to be calculated. In this example both diagonal braces overlap within the limits for the shear check. The brace-to-chord shear check equation sees the addition of the horizontal loads of the 3 bracings, the 2 angled braces having loads that result in the same horizontal direction. This means the brace 2 load (negative tension load) had to be subtracted so that the double-negative created a positive result so that the equation calculated correctly. It is important to check the joint configuration and refer to our Design of Welded Joints literature to make sure the correct method and formulae is used. To re-iterate this is a theoretical method which is not supported by research or tests.

Reference

5.1.2.3 5.1.3


Welded Joints Examples UB Chord RHS Bracings Gap K-Joint

12 UB Chord RHS Bracings Gap K-Joint

52

References are generally to Tata Steel publication ‘Design of Welded Joints’ (unless otherwise stated). Reference

Dimensions b0 = 125.3 mm h1 = 100.0 mm h2 = 100.0 mm h0 = 311.0 mm b1 = 100.0 mm b2 = 100.0 mm tw = 9.0 mm t1 = 6.3 mm t2 = 6.3 mm tf = 14.0 mm dw = h0 – 2 x (tf – r0) = 265.2 mm r0 = 8.9 mm Validity limits check 5.7.1 Figure 55 (EN1993-1-1 Chord: Web; Table 5.2) dw/tw ≤ 38ε (Class 1 or 2 compression) 38ε = 38 √(235/fy0) = 38 √(235/355) = 30.92 dw/tw = 265.2/9 = 29.5 ∴PASS dw ≤ 400 dw = 265.2 ∴PASS Chord: Flange; cf/tf ≤ 10ε (Class 1 or 2 compression) (EN1993-1-1 10ε = 10 √(235/fy0) = 10 √(235/355) = 8.1 Table 5.2) cf = (b0 – 2r0 - tw)/2 = (125.3 – 2 × 8.9 – 9)/2 = 49.25 cf/tf = 49.25/14.0 = 3.52 ∴PASS

Compression and Tension brace: hi/ti ≤ 35 hi/ti = 100/6.3 = 15.87 ∴PASS bi/ti ≤ 35 bi/ti = 100/6.3 = 15.87 ∴PASS hi/bi = 1.0 hi/bi = 100/100 = 1.0 ∴PASS Compression brace: (b1 – 3t1)/t1 ; (h1 – 3t1)/t1 ≤ 38ε (Class 1 or 2 compression) 38ε = 38 √(235/355) = 30.92 (b1 – 3t1)/ t1 = (100 – 3 × 6.3)/6.3 = 81.1/6.3 = 12.87 ∴PASS

g = 165.8mm

45° 45°

Np,Ed = 250 kN

N2,Ed = -150 kN N1,Ed = 150 kN

N0,Ed = 462 kN

SHS 100 x 100 x 6.3

UB 305 x 127 x 48 (S355)

MATERIAL; BRACES: CELSIUS S355 to EN 10210 UB CHORD: ADVANCE S355 to EN 10025


53

g ≥ t1 + t2 = 6.3 + 6.3 = 12.6 Reference g = 165.8 (zero eccentricity) ∴PASS 30° ≤ Ө1 ≤ 90° Ө1 = 45° ∴PASS 30° ≤ Ө2 ≤ 90° Ө2 = 45° ∴PASS Chord Web Yielding;

5M

i

i,ww0yRd,1 /

sinbtf

N γθ

××=

5.7.4

where… ( ) ( )0fi0fi

ii,w rt10t2butrt5

sinh

b ++≤++θ 5.7.2

( ) ( )9.80.14103.62but9.80.14545sin

100b 1,w ++×≤++°

=

bw,1 = 255.92 but ≤ 241.6 bw,1 = 241.6 mm and bw,2 = bw,1 as same brace and angle dimensions. Therefore;

kN10920.1/

45sin6.2410.9355N Rd,1 =

°××

=

and N2,Rd = N1,Rd as same brace and angle dimensions 1092 > 150 kN ∴PASS Chord Shear;

5Mi

0v0yRd,i /

sin3

AfN γ

θ×

×=

5.7.4

where… Av,0 = A0 – (2 - α) x b0 x tf + (tw + 2r) x tf 5.7.3

where…

⎟⎟⎠

⎞⎜⎜⎝

⎛

×

×+

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

×

×+

=α

2

2

2f

2

0.1438.16541

1

t3

g41

1

5.7.3

α = 0.073

∴ Av,0 = 6120 – (2 - 0.073) x 125.3 x 14.0 + (9.0 + 2 x 8.9) x 14.0

Av,0 = 3114.9 mm2


54

Reference

∴ 0.1/45sin3

9.3114355N Rd,1°×

×=

kN903N Rd,1 =

and N2,Rd = N1,Rd as same brace and angle dimensions 903kN > 150 kN ∴PASS Bracing Effective Width; 5Mi,effiyiRd,i /ptf2N γ×××= 5.7.4

Check if Bracing Effective Width required (Page 60 Welded Joints Literature); g / tf ≤ 20 - 28β 165.8 / 14.0 ≤ 20 – 28 (100+100+100+100/4 x 125.3) 11.8 ≤ -2.4 ∴ Check Effective Width; where… peff,i = tw + 2r + 7tf x (fy0 / fyi) 5.7.2 but ≤ bi + hi – 2ti for K or N gap joints peff,1 = 9.0 + 2 x 8.9 + 7 x 14.0 x (355/355)

but ≤ 100 + 100 – 2 x 6.3 peff,1 = 124.8 but ≤ 187.4 and peff,2 = peff,1 as same brace and angle dimensions 0.18.1243.63552N Rd,1 ÷×××= 558kN > 150kN ∴PASS

and N2,Rd = N1,Rd as same brace and angle dimensions Chord Axial Force Resistance in gap;

( ) 5MRd,0,pl

Ed,00y0,v0y0,v0Rd,gap,0 /

VV

1fAfAAN γ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−××+×−=

5.7.4

where… ( )2Ed,21Ed,1Ed,0 sinN,sinNmaxV θ×θ×=

5.7.4

( )°×−°×= 45sin150,45sin150maxV Ed,0


55

Vo,Ed = 106.1kN Reference

where… ( ) ( )0.1

3/3558.31413/fAV

0M

0y0,vRd,0,pl

×=

γ=

5.7.4

Vpl,0,Rd = 643.9kN

( ) 0.1/9.6431.10613558.31413558.31416120N Rd,gap,0

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−××+×−=

N0,gap,Rd = 2077kN

check; N0,gap,Rd ≥ N0,gap,Ed

250kN 250.0 +106.1 356.1

Therefore; N0,gap,Ed = 356.1kN 2077kN ≥ 356.1kN ∴PASS

150kN 150kN

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