TST56 Welded Joints Examples 12.13 Edit

8/17/2019 TST56 Welded Joints Examples 12.13 Edit

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Welded joints examples

Celsius®355 NH


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Welded Joints Examples Contents

2

CONTENTS

RHS T-Joint Moment in Plane 03

CHS Gap K-Joint 09

CHS Overlap K-Joint 12

RHS Gap K-Joint 15

RHS Overlap K-Joint 21

RHS Chord CHS Bracings Gap K-Joint 23

RHS Chord CHS Bracings Overlap K-Joint 82.7 29

RHS Chord CHS Bracings Overlap K-Joint 102.7 33

RHS Chord RHS Bracings Overlap KT-Joint 36CHS Chord CHS Bracings Gap KT-Joint 41

CHS Chord CHS Bracings Overlap KT-Joint 46

UB Chord RHS Bracings Gap K-Joint 52

1

2

3

4

5

6

7

8

910

11

12


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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 10

All 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 kN

Mip,0,Ed = 16.6 kNm

N0,Ed = 136 kN

Mip,0,Ed = 35.8 kNm

Dimensions

b0 = 150 mm b1 = 150 mmh0 = 150 mm h1 = 150 mmt0 = 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/f y0) = 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 ∴PASSh0/t0 ≤ 35 h0/t0 = 150/10 = 15 ∴PASS

Compression brace:

bi/ti ; hi/ti ≤ 35 b1/t1 = 150/8 =18.75 ∴PASSh1/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/f y0) = 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


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4

Axial : Chord face failure (deformat ion)

(valid when β ≤ 0.85)

required not check0.1150

150

b

b

0

1 ∴===β

Although this check is not required in this particular case, the chord end stressfactor 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

A

N++=σ

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

A0 = 54.9 cm2 = 5490 mm

2

32Ed,0 10236

100010008.35

109.54

1000136

×

××+

×

×=σ

2Ed,0 mm/N47.176=σ

Chord factored stress ratio, n;

⎟ ⎠

⎞⎜⎝

⎛ =

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ σ=

355

47.176

f n

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 asβ>0.85)

Reference5.3.3

6.3

5.3.2.1

5.3.2.1

5.3.2.1Fig. 38

5.3.2.1


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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 notrequired.

Axial : Chord s ide wal l buckl ing

(valid when β = 1.0) β = 1.0∴check required

5M01

1

1

0bnRd,1 /t10

sin

h2

sin

tf kN γ⎟⎟

⎠

⎞⎜⎜⎝

⎛ +

θθ=

Chord side wall buckling strength, f b

For compression brace, T-joint:

0y

i0

0

f

E

sin

12

t

h

46.3

π

θ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −

=λ where: E = 210000 N/mm2

589.0

355

210000

90sin

12

10

150

46.3 =

π

°⎟ ⎠

⎞⎜⎝

⎛ −

=λ


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’)

( ) 22.015.0 λ+−λα+=φ

( )( 714.0589.02.0589.021.015.0 2 =+−+=φ

1.0but

1

22 ≤χλ−φ+φ=χ

1.0but

589.0714.0714.0

1

22≤χ

−+=χ

895.0=χ

From graph, for

589.0=λ ;

895.0=χ

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

2b N/mm318355895.0f =×=

0.1/1010

90sin

1502

90sin

103180.1N Rd,1 ⎟⎟

⎠

⎞⎜⎜

⎝

⎛ ×+

°

×

°

××=

Reference5.3.3

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

5.3.2.3Fig. 40


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6

Ax ial : Chord punch ing shear

(valid when 0.85 ≤ β ≤ 1 – 1/γ)

5.7102

150

t2

b

0

0 =×

==γ

867.05.7

110.185.0 =−≤≤

∴Check not required

Axial : Brac ing failure (effect ive width)

(valid when β ≥ 0.85)

β = 1.0

∴check required

( ) 5Mi,eff 1i11yRd,1 /b2t4h2tf N γ+−=

where:

ii,eff iiyi

00y

0

0i,eff bbbutbtf

tf

bt10b ≤×=

mm150bbutmm1251508355

10355

150

1010b i,eff i,eff ≤=××

××

×=

beff,i = 125 mm

( ) 0.1/12528415028355N Rd,1 ×+×−××=

PASSkN2.19kN1471N Rd,1 ∴>=

Moment in plane: Chord face failure (deformation)

(valid when β ≤ 0.85)

required not check0.1150

150

b

b

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


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7

Moment in plane: Chord side wall crushing

(valid when 0.85 < β ≤ 1.0)

required check0.1150

150

b

b

0

1 ∴===β

( ) 5M2

010ykRd,1,ip /t5htf 5.0M γ+=

where:

f yk = f y0 (for T-joints)

∴ f yk = 355 N/mm2

( ) 0.1/105150103555.0M 2Rd,1,ip ×+××=

PASSkNm54kNm71M Rd,1,ip ∴>=

Moment in plane: Bracing failure (effective width)

(valid when 0.85 < β ≤ 1.0)

required check0.1150

150

b

b

0

1 ∴===β

( ) 5M11111

1,eff 1,ip,pl1yRd,1,ip /tthb

b

b1Wf M γ

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−=

where:

Wpl,ip,1 = 237 cm3 = 237000 mm

3

beff,1 = 125 mm

( ) 0.1/88150150150

1251237000355M

Rd,1,ip ⎥⎦

⎤

⎢⎣

⎡−⎟

⎠

⎞⎜⎝

⎛ −−=

PASSkNm54kNm1.74M Rd,1,ip ∴>=

Summary Axial joint resistance for brace 1 limited by chord side wall buckling andmoment in plane resistance by chord side wall crushing resistance.

Axial joint resistance, PASSkN2.19kN1272N Rd,1 ∴>=

Moment in plane joint resistance, PASSkNm54kNm71M 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


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8

Interaction checkWhen more than one type of force exists, e.g. axial and moments in plane, aninteraction 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=++

PASS000.1776.0 ∴≤

Reference2.3


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Welded Joints Examples CHS Gap K-Joint

2 CHS Gap K-Joint

9


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

CHS 219.1x12.5 All 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 mmt0 = 12.5 mm t1 = 5.0 mm t2 = 3.6 mm


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/f y0)]2 = 70(235/355) = 46.34

17.53 < 46.34 ∴PASS

di/ti ≤ 70ε2 (Class 1 or 2 for compression brace)

70ε2 = 70 [√(235/f y0)]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 ∴PASSd2/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 ≤ +54.8e = 0 mm ∴PASS

g ≥ t1 + t2 t1 + t2 = 5 + 3.6 = 8.6 mmg = 40 mm ∴PASS

30° ≤ θ i ≤ 90° θ1 = 45° ∴PASSθ2 = 45° ∴PASS

Reference

5.1.1 Figure 27

(EN 1993-1-1)Table 5.2

(EN 1993-1-1)Table 5.2


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10

Chord face failure (deformation)

Compression brace (1):

5M0

1

1

200ypg

Rd,1 /d

d2.108.1

sin

tf kkNfailure, face Chord γ⎟⎟

⎠

⎞⎜⎜⎝

⎛ +

θ=

Gap/lap function, kg

764.85.122

1.219

t2

d

0

0

=×==γ


( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

−+γ

+γ=33.1t/g5.0exp1

024.01k

0

2.12.0

g

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

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

−×+×

+=33.15.12/405.0exp1

764.8024.01764.8k

2.12.0

g

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

A

N++=σ

Note: Moment is additive to compressive stress which is positive for moments. For

CHS chords use least compressive chord stress.

2

2Ed,pN/mm30.123

101.81

10001000=

×

×=σ

Chord factored stress ratio, np; 347.0355

30.123

f n

0y

Ed,pp =⎟

⎠

⎞⎜⎝

⎛ =

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ σ=


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

5.1.2.2Fig.29

5.1.2.1Fig.28


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11

0.1/1.219

7.1392.108.1

45sin

5.12355860.0761.1N

2

Rd,1 ⎟ ⎠

⎞⎜⎝

⎛ +

°×××

=

kN986N Rd,1 =

Tension brace (2):

Rd,12

1Rd,2 N

sin

sinN

θ

θ=

98645sin

45sinN 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

f N γ

θ

θ+π=

Brace (1):

0.1/45sin2

45sin17.1395.12

3

355N

2Rd,1 °

°+××π××=

kN1919N Rd,1 =

Brace (2):

0.1/45sin2

45sin13.1145.12

3

355N

2Rd,2 °

°+××π××=

kN1570N Rd,2 =

Joint strength dictated by chord face failure for both bracings;

PASSkN500kN986N,resistance joint 1 Brace Rd,1 ∴>=


Reference

5.1.3

5.1.3


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Welded Joints Examples CHS Overlap K-Joint

3 CHS Overlap K-Joint

12


N1,Ed = 500 kN

Np,Ed = 1000 kN CHS 219.1x12.5 All 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 mmt0 = 12.5 mm t1 = 5.0 mm t2 = 3.6 mm


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/f y0)]

2 = 70(235/355) = 46.34

17.53 < 46.34 ∴PASS

di/ti ≤ 70ε2 (Class 1 or 2 for compression brace)

70ε2 = 70 [√(235/f y0)]

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 ∴PASSd2/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 ≤ +54.8e = -42.2 mm ∴PASS

25% ≤ λov ≤ 100% λov = |g| sin θi /di x 100% ∴PASSλov = 45 sin 45°/114.3 x 100%λov = 27.8% ∴PASS

30° ≤ θ i ≤ 90° θ1 = 45° ∴PASSθ2 = 45° ∴PASS

Reference

5.1.1 Figure 27

(EN 1993-1-1)Table 5.2

(EN 1993-1-1)Table 5.2


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13

Chord face failure (deformation)


5M0

1

1

200ypg

Rd,1 /d

d2.108.1

sin

tf kkNfailure, face Chord γ⎟⎟

⎠

⎞⎜⎜⎝

⎛ +

θ=

Gap/lap function, kg

764.85.122

1.219t2

d

0

0 =×

==γ


( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

−+γ

+γ=33.1t/g5.0exp1

024.01k

0

2.12.0

g

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

( )( )⎥

⎥

⎦

⎤

⎢

⎢

⎣

⎡

−−×+

×+=

33.15.12/455.0exp1

764.8024.01764.8k

2.12.0

g

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

A

N++=σ

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

2

2Ed,pN/mm30.123

101.81

10001000=

×

×=σ


30.123

f n

0y

Ed,pp =⎟

⎠

⎞⎜⎝

⎛ =

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ σ=



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

5.1.2.2Fig. 29

5.1.2.1Fig. 28

5.1.2.1


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14

0.1/1.219

7.1392.108.1

45sin

5.12355860.0024.2N

2

Rd,1 ⎟ ⎠

⎞⎜⎝

⎛ +

°×××

=

kN1134N Rd,1 =

Tension brace (2):

Rd,121Rd,2 Nsin

sinN θ

θ=

113445sin

45sinN 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 greaterthan 60%.

Therefore, joint strength dictated by chord face failure for both bracings;



Reference

5.1.3


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Welded Joints Examples RHS Gap K-Joint

4 RHS Gap K-Joint

15


N1,Ed = 650 kN

SHS 200 x 200 x 10 All 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 mmh0 = 200 mm h1 = 120 mm h2 = 120 mmt0 = 10.0 mm t1 = 5.0 mm t2 = 5.0 mm


Chord:



Compression brace:

bi/ti ; hi/ti ≤ 35 b1/t1 = 120/5 =24 ∴PASSh1/t1 = 120/5 =24 ∴PASS

Compression brace:


Tension brace:

bi/ti ; hi/ti ≤ 35 b2/t2 = 120/5 =24 ∴PASSh2/t2 = 120/5 =24 ∴PASS

bi/b0 ≥ 0.35 b1/b0 = 120/200 = 0.6 ∴PASSb2/b0 = 120/200 = 0.6 ∴PASS

0.1+0.01 b0/t0 ≤ bi/b0 ≤ 1.0 0.1+0.01 b0/t0 = 0.3b1/b0 = 120/200 = 0.6 ∴PASSb2/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


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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 ∴PASSh2/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 ≤ +50.0e = +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 mmβ = (b1+b2+h1+h2)/(4 b0)β = (120+120+120+120)/(4x200) = 0.60.5 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



5M0

2121

i

200yn

Rd,i /b4

hhbb

sin

tf k9.8N γ⎟⎟

⎠

⎞⎜⎜⎝

⎛ +++

θ

γ=

Reference

5.3.3

5.3.3


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17


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

A

N++=σ


2

2Ed,0N/mm34.256

109.74

10001920=

×

×=σ


⎟ ⎠

⎞⎜⎝

⎛ =

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ σ=

355

34.256

f n

0y

Ed,0

722.0n =



β−=

n4.03.1kn but kn ≤ 1.0

0.1but6.0

722.04.03.1kn ≤

×−=

819.0kn =


kn = 0.819

10

102

200

t2

b

0

0 =

×

==γ

0.1/2004

120120120120

45sin

1010355819.09.8N

2

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


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18

Chord shear between bracings check – brace 1

5M

i

0,v0yRd,i /

sin3

Af N γ

θ=

where:for RHS bracings:

212.0

103

4041

1

t3

g41

1

2

2

2

0

2 =

×

×+

=

+

=α

Av,0 = (2 h0 + α b0) t0

( ) 20,v mm442410200212.02002 A =×+×=

0.1/45sin3

4424355N Rd,i

°

×=

kN1282N Rd,1 =

Bracing Effective width – brace 1

( ) 5Meff iiiiyiRd,i /bbt4h2tf N γ++−=

where:

ii,eff iiyi

00y

0

0i,eff bbbutb

tf

tf

b

t10b ≤×=

mm120bbutmm1201205355

10355

200


××

×=

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


19/56


19

Chord punch ing 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

sin

h2

sin3

tf N γ⎟⎟

⎠

⎞⎜⎜⎝

⎛ ++

θθ=

where:

ii,p,ei0

0i,p,e bbbutb

b

t10b ≤=

mm120bbutmm60120200

1010b i,p,ei,p,e ≤=×

×=

be,p,i = 60 mm

0.1/6012045sin

1202

45sin3

10355N Rd,i ⎟⎟

⎠

⎞⎜⎜⎝

⎛ ++

°×

°

×=

kN1506N Rd,1 =

Summary - brace 1Joint strength for brace 1 dictated by chord face deformation.

∴ Brace 1 joint resistance, PASSkN650kN694N Rd,1 ∴>=

Brace 2Repeat 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 =


Reference 5.3.3

5.3.3

5.3.2.2


20/56


20

Chord axial load resistance in gap

Check: N0,gap,Rd ≥ N0,gap,Ed

( ) ( ) 5M2Rd,0,plEd,00y0,v0y0,v0Rd,gap,0 /V/V1f Af A AN γ⎥⎦⎤

⎢⎣

⎡−+−=

where:

A0 = 74.9 cm2 = 7490 mm

2

Av,0 = 4424 mm

2

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=

kN7.906

0.1

)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 ⎥⎦

⎤⎢⎣

⎡−×+−=

PASSkN1460kN2442N Rd,gap,0 ∴>=

Reference5.3.3

5.3.3

5.3.2.5

5.3.3

5.3.3(EN 1993-1-1)

6.2.6 (2)

5.3.3


21/56

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 10

All material Celsius 355 to EN 10210

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

45º45º

70 i

-50.1

Dimensions

b0 = 200 mm b1 = 120 mm b2 = 120 mmh0 = 200 mm h1 = 120 mm h2 = 120 mmt0 = 10.0 mm t1 = 5.0 mm t2 = 5.0 mm


Chord:

(b0-3 t0)/t0 ; (h0-3 t0)/t0 ≤ 38ε (Class 1 or 2 compression)38ε = 38 √(235/f y0) = 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 ∴PASSCompression brace:

(b1-3 t1)/t1 ; (h1-3 t1)/t1 ≤ 33ε (Class 1 compression)33ε = 33 √(235/f y0) = 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 ∴PASSh2/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 ≤ +50.0e = -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/b j ≥ 0.75 bi/b j = 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


22/56

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 theoverlapped brace (j) is based on an efficiency ratio to that of the overlappingbrace.

Overlapping brace, i (2):

For 25% ≤ λov < 50%

5Miov

iov,ei,eff iyiRd,i /t450

h2bbtf N γ⎟

⎠

⎞⎜

⎝

⎛ −

λ++=

where:

ii,eff iiyi

00y

0

0i,eff bbbutb

tf

tf

b

t10b ≤×=

mm120bbutmm1201205355

10355

200


××

×=

beff,i = 120 mm

iov,eiiyi

jyj

j

j

ov,e bbbutbtf

tf

b

t10

b ≤×=

120bbut1205355

5355

120

510b ov,eov,e ≤××

××

×=

mm50b ov,e =

0.1/5450

2.411202501205355N Rd,i ⎟

⎠

⎞⎜⎝

⎛ ×−×++×=

PASSkN600kN617NN Rd,2Rd,i ∴>==

Bracing Effective width – Overlapped brace, j (1)

)( )yii

yj jRd,iRd, j

f A

f ANN =

where:

A j = A1 = 22.7 cm2 = 2270 mm

2

Ai = A2 = 22.7 cm2 = 2270 mm

2

( )

( )3552270

3552270617N Rd, j

×

××=

PASSkN600kN617NN Rd,1Rd,i ∴>==

Reference

5.3.3

5.3.3

5.3.2.2

5.3.2.2

5.3.3


23/56

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 mmh0 = 200 mm t1 = 6.3 mm t2 = 6.3 mmt0 = 10.0 mm


Chord:



Compression brace:

(d1/t1 ≤ 50ε (Class 1 or 2 compression)50ε = 50 √(235/f y0) = 50√(235/355) = 40.7d1/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 ∴PASSd2/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


24/56


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 ≤ +50.0e = +23.0 mm ∴PASS

30° ≤ θ i ≤ 90° θ 1 = 45° ∴PASSθ 2 = 45° ∴PASS

g ≥ t1 + t2 85 ≥ 6.3 + 6.3 = 12.6 mm ∴PASS

Chord face failure (deformation) – brace 1



5M0

2121

i

200yn

Rd,i /b4

dddd

sin

tf k9.8N γ⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ +++

θ

γ=


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

A

N++=σ


2

2Ed,0N/mm34.256

109.74

10001920=

×

×=σ


⎟ ⎠

⎞⎜⎝

⎛ =⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ σ=

355

34.256

f n

0y

Ed,0

722.0n =

Reference

5.3.3

5.3.2.1

5.3.2.1


25/56


25



β−=

n4.03.1kn but kn ≤ 1.0

0.1but6.0

722.04.03.1kn ≤×

−=

819.0kn =


kn = 0.819

10102

200

t2

b

0

0 =×==γ

0.1/2004

3.1143.1143.1143.114

45sin

1010355819.09.8N

2

Rd,1 ⎥⎦

⎤⎢⎣

⎡×+++

×°××××

=

kN5194

661......kN661N Rd,1 =π×=

Chord shear between bracings check – brace 1

5M

i

0,v0yRd,i /

sin3

Af N γ

θ=

where:for CHS bracings:

0=α

Av,0 = (2 h0 + α b0) t0

( )2

0,v mm40001020002002 A =×+×=

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


26/56


26

Bracing Effective width – brace 1

( ) 5Meff iiiiyiRd,i /bdt4d2tf N γ++−=

where:

ii,eff iiyi

00y

0

0i,eff dbbutd

tf

tf

b

t10b ≤×=

mm120bbutmm2.95120

3.6355

10355

200

1010b i,eff i,eff ≤=×

×

×××=

beff,i = 95.2 mm

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

kN7254

.....kN923N Rd,1 =π×=

Chord punch ing 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

sin

d2

sin3

tf N γ⎟⎟

⎠

⎞⎜⎜⎝

⎛ ++

θθ=

where:

ii,p,ei0

0i,p,e dbbutd

b

t10b ≤=

mm120bbutmm2.573.114200

1010b i,p,ei,p,e ≤=××=

be,p,i = 57.2 mm

0.1/2.573.11445sin

3.1142

45sin3

10355N Rd,i ⎟⎟

⎠

⎞⎜⎜⎝

⎛ ++

°×

°

×=

kN11264

........kN1434N Rd,1 =π×=

Summary - brace 1

Joint strength for brace 1 dictated by chord face deformation.


Reference

5.3.3

5.3.2.2

5.3.3

5.3.3

5.3.2.2


27/56


27

Brace 2Repeat 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 =


Chord axial load resistance in gap

Check: N0,gap,Rd ≥ N0,gap,Ed

( ) ( ) 5M2Rd,0,plEd,00y0,v0y0,v0Rd,gap,0 /V/V1f Af A AN γ⎥⎦

⎤⎢

⎣

⎡ −+−=

where:

A0 = 74.9 cm2 = 7490 mm

2

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 ( AVγ

=

kN8.8190.1

)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)


28/56


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 ⎥⎦⎤

⎢⎣

⎡ −×+−=

PASSkN1566kN2520N Rd,gap,0 ∴>=

Reference

5.3.3


29/56

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 d j = 114.3 mmt0 = 10.0 mm ti = 6.3 mm t j = 6.3 mm f u,i = 470 N/mm²

f u,j = 470 N/mm²

(f u,i & f u,j from EN10210-1:2006 Table A.3)

For Eccentricity;

( ) 2h

sin

sinsing

sin2

d

sin2

de 0

ji

ji

j

j

i

i −⎥⎥⎦

⎤

⎢⎢⎣

⎡

θ+θ

θ×θ×

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

θ×+

θ×=

( )mm1.109g

60sin100

3.1147.82g

sin100

dgwhere

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

N j,Ed = 500kN

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

RHS 200 x 100 x 10


-109.1


30/56


30

( ) ( ) ( ) ( ) ( )

( )

mm5.19e

2

100

3060sin

30sin60sin1.109

30sin2

3.114

60sin2

3.114e

−=

−⎥⎦

⎤⎢⎣

⎡

°+°°×°

×⎥⎦

⎤⎢⎣

⎡−+

°×+

°×=


b0/t0 ≤ 38ε2 (Class 1 or 2 for compression chord)

38ε2

= 38 [√(235/f y0)]2

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

d j/t j ≤ 50ε2 (Class 1 for compression brace)

50ε2 = 50 [√(235/f y0)]2 = 50(235/355) = 33.10

d j/t j = 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 ∴PASSd j/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 ≤ +25e = -19.5 mm ∴PASS

λov ≥ 25% λov = |g| sin θI /di x 100% ∴PASSλov = 109.1 sin 60°/114.3 x 100%λov = 82.7% ∴PASS

30° ≤ θI ≤ 90° θ j = 30° ∴PASSθI = 60° ∴PASS

di/d j ≥ 0.75 di/d j = 114.3/114.3 = 1.0 ∴PASS

Brace Effective Width;

Overlapping brace (i):

( ) 5Miiov,eiiyiRd,i /t4d2ddtf N γ−++=

where:

iov,eiiyi

jyj

j jov,e ddbutdtf

tf

t/d

10

d ≤××

×

×=

Reference

(EN1993-1-1)

(Table 5.2)

(EN1993-1-1)(Table 5.2)

5.3 Figure 37

5.3.3

5.3.2.2


31/56


31

mm63d3.114but63d

3.114dbut3.1143.6355

3.6355

3.6/3.114

10d

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

yj jRd,iRd, j

f A

f ANN

×

××=

where:

22i

22 j

mm2140cm4.21 A

mm2140cm4.21 A

==

==

( )( )3552140

3552140668N Rd, j ×

××=

PASSkN500kN668N Rd, j ∴>=

Overlapping Bracings Shear Check For CHS Bracings;

.weldingneedstoe

hidden%20componentverticalbracebetweendifference As

%8.271004.346

2504.346.diff component.vert%

kN250)30(sin500N

kN4.346)60(sin400N

)(sinNN&)(sinNN

;N&Nof componentVertical

Ed,v, j

Ed,v,i

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

Ed, jEd,i

>∴

=×−

=∴

=°×=∴

=°×=∴

θ×=θ×=∴

Reference

5.3.3


32/56


32

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

( )5M j

j j,eff s j j,u

i

ii,eff iov

i,u

jEd, jiEd,i

1

sin

tdcd2

3

f

sin

tdd2100

100

3

f

4......

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

γ×

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

θ

××+×+

θ

×⎥⎥⎦

⎤

⎢⎢⎣

⎡⎜⎜⎝

⎛ +×⎟

⎠

⎞λ−

××π

≤θ+θ

mm4.91d3.1144.91d

3.114but3.1143.6355

10355

200

1010d

dbutdtf

tf

b

t10d:where

i,eff i,eff

i,eff

iiii,y

00,y

0

0i,eff

=∴≤=

≤×××

××

=

≤××

××=

mm4.91d3.1144.91d

3.114but3.1143.6355

10355

200

1010d

dbutdtf

tf

b

t10d:where

j,eff j,eff

j,eff

j j j j,y

00,y

0

0 j,eff

=∴≤=

≤××

××

×=

≤××

××=

)weldedtoehidden(2c:where s=

( )

( )( )( )

( ) 11

30sin

3.64.9123.1142

3

470

60sin

3.64.913.1142100

7.82100

3

470

4......

...........cosNcosN j jii

×

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

°××+×

×+°

×⎥⎥⎦

⎤

⎢⎢⎣

⎡⎜⎜⎝

⎛ +××⎟⎟

⎠

⎞−

××π

≤θ+θ∴

= 1307 kN

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

Reference

5.3.3

5.3.2.2

5.3.2.2

5.1.3


33/56


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 d j = 114.3 mmt0 = 10.0 mm ti = 6.3 mm t j = 6.3 mm f u,i = 470 N/mm²

f u,j = 470 N/mm²

(f u,i & f u,j from EN10210-1:2006 Table A.3)

For Eccentricity;

( ) 2h

sin

sinsing

sin2

d

sin2

de 0

ji

ji

j

j

i

i −⎥⎥⎦

⎤

⎢⎢⎣

⎡

θ+θ

θ×θ×

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

θ×+

θ×=

( )mm5.135g

60sin100

3.1147.102g

sin100

dgwhere

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

N j,Ed = 500kN

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

RHS 200 x 100 x 10


g = -135.5


34/56


34

( ) ( ) ( ) ( ) ( )

( )

mm6.30e

2

100

3060sin

30sin60sin5.135

30sin2

3.114

60sin2

3.114e

−=

−⎥⎦

⎤⎢⎣

⎡

°+°°×°

×⎥⎦

⎤⎢⎣

⎡−+

°×+

°×=


b0/t0 ≤ 38ε2 (Class 1 or 2 for compression chord)

38ε2 = 38 [√(235/f y0)] 2

= 38(235/355) = 25.15

b0/t0 = 200.0/10.0 = 20.0 < 25.15 ∴PASS

d j/t j ≤ 50ε2 (Class 1 for compression brace)

50ε2 = 50 [√(235/f y0)]2 = 50(235/355) = 33.10

d j/t j = 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 ∴PASSd j/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 ≤ +25e = -30.6 mm ∴PASS

λov ≥ 25% λov = |g| sin θ i /di x 100% ∴PASSλov = 135.5 sin 60°/114.3 x 100%λov = 102.7% ∴PASS

30° ≤ θi ≤ 90° θ j = 30° ∴PASSθi = 60° ∴PASS

di/d j ≥ 0.75 di/d j = 114.3/114.3 = 1.0 ∴PASS


Overlapping brace (i):

( ) 5Miiov,eiiyiRd,i /t4d2ddtf N γ−++= where λov ≥ 80%;

where:

mm63d3.114but63d

3.114dbut3.114

3.6355

3.6355

3.6/3.114

10d

ddbutdtf

tf

t/d

10d

ov,eov,e

ov,eov,e

iov,eiiyi

jyj

j jov,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


35/56


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):

( )yiiyj j

Rd,iRd, jf A

f ANN

×

××=

where:

22i

22 j

mm2140cm4.21 A

mm2140cm4.21 A

==

==

( )( )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

10355

200

1010d

dbutdtf

tf

b

t10d

j,eff j,eff

j,eff

j j j j,y

00,y

0

0 j,eff

=∴≤=

≤×××

××

=

≤××

××=

( )

5M

jEd, jiEd,i

1

)30sin(

3.64.913.1143

3

470

4......

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

γ×

×+×××

π

≤θ+θ

( )5M j

j j,eff j j,u

jEd, jiEd,i

1

sin

tdd3

3

f

4......

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

γ×

θ

×+×××

π

≤θ+θ

kN673cosNcosN j jii ≤θ+θ

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


36/56

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’ overlappingBrace ‘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 mmb0 = 200.0 mm b1 = 90.0 mm b2 = 100.0 mm b3 = 100.0 mmt0 = 10.0 mm t1 = 5.0 mm t2 = 6.3 mm t3 = 6.3 mm

γM5 = 1.0 A1 = 23.2cm2 A2 = 16.7cm

2 A3 = 23.2cm

2

This example is to show how to calculate a KT-Joint with SHS chord and braces. The method ofchecking a KT-Joint depends on the direction of the brace forces, gap or overlap and chord sectionprofile. 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 thatdifferent brace angle and overlap values may change how the joint is calculated and you should refer tothe Tata Design of Welded Joints literature before tackling any joint to make sure you are using thecorrect method and formulae.The force combination of the braces can be seen at 5.6.2 Figure 51(c) (asa (from left to right) compression-compression-tension) of our Design Of Welded Joints literature.

Reference

45° 50°

e = - 19.5

N2,Ed = - 650kNN1 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


37/56


37

GEOMETRIC CALCULATIONS

For Eccentricity; Brace 1 in relation to Brace 3;

( ) 2h

sin

sinsing

sin2

b

sin2

be 0

31

31

3

3

1

11 −⎥

⎦

⎤⎢⎣

⎡

θ+θθ×θ

×⎥⎦

⎤⎢⎣

⎡+

θ×+

θ×=

Brace 1 to 3 overlap, λov = 50%;

where;i

iov

sin100

bg

θ×

×λ=

( )1

45sin100

0.900.50g1 −×°×

×=

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

( ) ( ) ( ) ( ) ( )

( )

mm0.50

2

0.200

9045sin

90sin45sin6.63

90sin2

0.100

45sin2

0.90e1

−=

−⎥⎦

⎤⎢⎣

⎡

°+°°×°

×⎥⎦

⎤⎢⎣

⎡−+

°×+

°×=

For Eccentricity, Brace 2 in relation to Brace 3;

( ) 2h

sin

sinsing

sin2

b

sin2

be 0

32

32

3

3

2

22 −⎥

⎦

⎤⎢⎣

⎡θ+θθ×θ

×⎥⎦

⎤⎢⎣

⎡+

θ×+

θ×=

Brace 2 to 3 overlap, λov = 50%;

where;2

2ov2

sin100

bg

θ×

×λ=

( )°××

=50sin100

1000.50g2

mm3.65g2

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

( ) ( ) ( ) ( ) ( )

( )

mm4.40

2

0.200

9050sin

90sin50sin3.65

90sin2

0.100

50sin2

0.100e2

−=

−⎥⎦

⎤⎢⎣

⎡

°+°°×°

×⎥⎦

⎤⎢⎣

⎡−+

°×+

°×=

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/f y0)] = 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


38/56


38

33ε = 33 [√(235/f y0)] = 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 ∴PASSh2/t2 = 100.0/6.3 = 15.9 ∴PASS

0.25 ≤ bi/b0 ≤ 1.0 b1/b0 = 90.0/200.0 = 0.45 ∴PASSb2/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 ∴PASS0.5 ≤ hi/ bi ≤ 2.0 h1/b1 = 90.0/90.0 = 1.0 ∴PASS

h2/b2 = 100.0/100.0 = 1.0 ∴PASSh3/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 ≤ +50e1 = -50 mm ∴PASSe2 = -40.4 mm ∴PASS

λov ≥ 25% λov,1 = |g| sin θ1 /h1 x 100%

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

30° ≤ θi ≤ 90° θ1 = 45° ∴PASSθ2 = 50° ∴PASSθ3 = 90° ∴PASS

bi/b j ≥ 0.75 b1/b3 = 90.0/100.0 = 0.9 ∴PASSb2/b3 = 100.0/100.0 = 1.0 ∴PASS

Brace 1 & 3 as overlap K-Joint

Brace Effective Width Failure for 50% ≤λov


39/56


39

mm4.71b0.90but4.71b

0.90bbut0.900.5355

3.6355

0.100

3.610b

1,ov,e1,ov,e

1,ov,e1,ov,e

=∴≤=

≤×××

××

=

where;

11,eff 111y

00y

0

0

1,eff bbbutbtf

tf

b

t10

b ≤××

×

×

×

=

90bbut900.5355

10355

200

1010b 1,eff 1,eff ≤××

××

×=

mm90b90bbutmm90b 1,eff 1,eff 1,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;)

( )yiiyj j

Rd,iRd, jf A

f ANN

×

××=

221

223

mm1670cm7.16 A

mm2320cm2.23 A

==

==

( )

( )3551670

3552320570N Rd,3 ×

××=

=Rd,3N 792kN > 150kN ∴PASS

Brace 2 & 3 as overlap K-Joint

Brace Effective Width Failure for 50% ≤λov


40/56


40

mm63b0.100but63b

0.100bbut0.1003.6355

3.6355

0.100

3.610b

2,ov,e2,ov,e

2,ov,e2,ov,e

=∴≤=

≤×××

××

=

where;

22,eff 222y

00y

0

02,eff bbbutb

tf

tf

b

t10b ≤×

××

××

=

100bbut1003.6355

10355

200

1010b 2,eff 2,eff ≤××

××

×=

mm4.79b100bbutmm4.79b 2,eff 2,eff 2,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

yj jRd,iRd, j

f A

f ANN

×

××=

222

223

mm2320cm2.23 A

mm2320cm2.23 A

==

==

( )( )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 thenature of the joint, it can be treated as two overlap K-Joints. This means Brace 3 (the vertical overlappedbrace) is checked twice in relation to Brace 1 and Brace 2 and must pass on both of these checks. It isimportant to check the joint configuration and refer to our Design of Welded Joints literature to make sure thecorrect method and formulae is used.

Reference

5.3.2.2

5.3.3 (K- & N-Overlap Sect)

(Tata TechnicalGuide TST02)


41/56

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 mmt0 = 10.0 mm t1 = 5.0 mm t2 = 6.3 mm t3 = 6.3 mm

γM5 = 1.0 A1 = 13.2cm2 A2 = 18.9cm

2 A3 = 18.9cm

2

This example is to show how to calculate a gap KT-Joint with CHS chord and braces. Therecommendations for an all-CHS gap KT-Joint are in EN1993-1-8 Table 7.6. The method of checking aKT-Joint depends on the direction of the brace forces, gap or overlap and chord section profile. Thisexample 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 forchord face failure and punching shear failure for each brace. Please be aware that different brace angleand gap values may change how the joint is calculated and you should refer to the Tata Design ofWelded Joints literature before tackling any joint to make sure you are using the correct method andformulae.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 = 52mmg1,3 = 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 = 21.2mme1 = 46.8mm


42/56


42

Geometric Calculations


( ) 2d

sin

sinsing

sin2

d

sin2

de 0

31

31

3

3

1

11 −⎥

⎦

⎤⎢⎣

⎡

θ+θθ×θ

×⎥⎦

⎤⎢⎣

⎡+

θ×+

θ×=

Brace 1 to 3 gap, g1 = 30mm;

( ) ( )

( ) ( )

( )

mm8.46

2

7.193

9045sin

90sin45sin0.30

90sin2

6.101

45sin2

9.88e1

=

−⎥⎦

⎤⎢⎣

⎡

°+°

°×°×⎥

⎦

⎤⎢⎣

⎡+

°×

+

°×

=


( ) 2d

sin

sinsing

sin2

d

sin2

de 0

32

32

3

3

2

22 −⎥

⎦

⎤⎢⎣

⎡

θ+θθ×θ

×⎥⎦

⎤⎢⎣

⎡+

θ×+

θ×=

Brace 2 to 3 gap, g2 = 52mm;

( ) ( )

( ) ( )

( )

mm2.21

2

7.193

9030sin

90sin30sin0.52

90sin2

6.101

30sin2

6.101e

2

=

−⎥⎦

⎤

⎢⎣

⎡

°+°

°×°×

⎥⎦

⎤

⎢⎣

⎡+

°×+

°×=


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/f y0)]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 ∴PASSd2/t2 = 101.6/6.3 = 16.13 ∴PASSd3/t3 = 101.6/6.3 = 16.13 ∴PASS

di/ti ≤ 70ε2(Class 1 or 2 for compression bracings)

70ε2 = 70 [√(235/f y0)]2 = 70√ (235/355)2 = 46.34

d1/t1= 17.78 ∴PASSd3/t3 = 16.13 ∴PASS

0.2 ≤ di/d0 ≤ 1.0 d1/d0 = 88.9/193.7 = 0.46 ∴PASSd2/d0 = 101.6/193.7 = 0.52 ∴PASSd3/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 ≤ +48.4e1 = 46.8 mm ∴PASSe2 = 21.2 mm ∴PASS

Reference

5.1.1 Figure 27

(EN1993-1-1)Table 5.2

(EN1993-1-1)Table 5.2


43/56


43

g ≥ t1 + t2 but ≤ 12t0 for g1,3 ≥ t1 + t3 = 5.0 + 6.3 = 11.3 mm but ≤ 12t0g2,3 ≥ t2 + t3 = 6.3 + 6.3 = 12.6 mm but ≤ 12t012t0 = 12 x 10.0 = 120 mm

g1,3 = 30 mm ∴PASSg2,3 = 52 mm ∴PASS

30° ≤ θi ≤ 90° θ1 = 45° ∴PASS

θ2 = 30° ∴PASSθ3 = 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

sin

tf kkN γ⎟⎟

⎠

⎞⎜⎜⎝

⎛

×

++×+×

θ

×××=

Gap/lap factor, kg;

Using Formulae:

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

−+

γ+γ=

33.1t/g5.0exp1

024.01k

0

2.12.0

g

where;

69.90.102

7.193

t2

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

69.9024.0169.9k

2.12.0

g

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 graph5.1.2.2 Fig. 29


44/56


44

PASSkN369kN538N Rd,1 ∴>=



0,op,el

Ed,0,op

0,ip,el

Ed,0,ip

0

Ed,pEd,p

W

M

W

M

A

N++=σ

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

2

2Ed,pN/mm7.88

107.57

1000512=

×

×=σ


7.88

f n

0y

Ed,pp =⎟

⎠

⎞⎜⎝

⎛ =

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ σ=



kp = 1 - 0.3 np (1 + np) but ≤ 1.0( ) 0.1but25.0125.03.01kp ≤+×−=

91.0kp =

from graph;

kp = 0.91

0.1/7.1933

6.1016.1019.882.108.1

45sin

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


45/56


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 mmd1 = 88.9 mmd2 = 101.6 mmd3 = 101.6 mm

Therefore check for Chord Punching Shear;

5M

i2

i10

0yRd,i /

sin2

sin1dt

3

f N γ

θ×

θ+××π××=

For Brace 1;

0.1/45sin2

45sin19.880.10

3

355N

2Rd,1 °×

°+××π××=


For Brace 2;

0.1/30sin2

30sin16.1010.10

3

355N

2Rd,2 °×

°+××π××=


For Brace 3;

0.1/90sin2

90sin16.1010.10

3

355N

2Rd,3 °×

°+××π××=


Summary

Here, the validity limits are combined to include all three braces. The eccentricity values use both brace 1 and2 (diagonals) in relation to brace 3 (the vertical brace). It is important to check the joint configuration and referto 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


46/56

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’ overlappingBrace ‘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 mmt0 = 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 cm

2 A3 = 18.9 cm

2

f u1 = 470 N/mm2 f u2 = 470 N/mm

2 f u3 = 470 N/mm

2

This example is to show how to calculate an overlap KT-Joint with CHS chord and braces. Please beaware that there are no recommendations for an overlap KT- Joint in EN1993-1-8 or elsewhere as faras Tata Steel is aware. This is a Tata Steel method and is included in the Tata Steel Design of WeldedJoints literature. It is a theoretical method which is not supported by research or tests. The method ofchecking a KT-Joint depends on the direction of the brace forces, gap or overlap and chord sectionprofile. 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 forchord face failure. Please be aware that different overlap values may change how the joint is calculatedand you should refer to the Tata Design of Welded Joints literature before tackling any joint to makesure you are using the correct method and formulae.The force combination of the braces can be seenat 5.6.2 Figure 51(c) (as a (from left to right) compression-compression-tension) of our Design OfWelded Joints literature. Due to one of the braces falling within the limits (overlapping percentage) forthe localised brace-to-chord shear check, this calculation has also been completed.

Reference

‘f u‘ values takenfrom ProductStandardEN10210-1Table A.3 asrequired by UKNational 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 = -60.4mme1 = -73.2mm


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47

Geometric Calculations


( ) 2d

sin

sinsing

sin2

d

sin2

de 0

31

31

3

3

1

11 −⎥

⎦

⎤⎢⎣

⎡

θ+θθ×θ

×⎥⎦

⎤⎢⎣

⎡+θ×

+θ×

=

Brace 1 to 3 overlap, λov = 71.6%;

where;1

1ov

sin100

dg

θ×

×λ=

( )1

45sin100

9.886.71g1 −×°××=

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

( ) ( ) ( ) ( ) ( )

( )

mm2.73

2

7.193

9045sin

90sin45sin0.90

90sin2

6.101

45sin2

9.88e1

−=

−⎥⎦

⎤⎢⎣

⎡

°+°

°×°×⎥⎦

⎤⎢⎣

⎡−+

°×+°×

=


( ) 2d

sin

sinsing

sin2

d

sin2

de 0

32

32

3

3

2

22 −⎥

⎦

⎤⎢⎣

⎡θ+θθ×θ×⎥

⎦

⎤⎢⎣

⎡ +θ×

+θ×

=

Brace 2 to 3 overlap, λov = 65.2%;

where;2

2ov2

sin100

dg

θ×

×λ=

( )°××

=50sin100

6.1012.65g2

mm5.86g2 −=

(gap is negative as it’s an overlap)

( ) ( ) ( ) ( ) ( )

( )

mm4.60

2

7.193

9050sin

90sin50sin5.86

90sin2

6.101

50sin2

6.101e2

−=

−⎥⎦

⎤⎢⎣

⎡

°+°°×°

×⎥⎦

⎤⎢⎣

⎡−+

°×+°×

=


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/f y0)]2 = 70√ (235/355)2 = 46.34d0/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


48/56


48

di/ti ≤ 50 d1/t1 = 88.9/5.0 = 17.78 ∴PASSd2/t2 = 101.6/6.3 = 16.13 ∴PASSd3/t3 = 101.6/6.3 = 16.13 ∴PASS

di/ti ≤ 70ε2(Class 1 or 2 for compression bracings)

70ε2 = 70 [√(235/f y0)]2 = 70√ (235/355)2 = 46.34

d1/t1= 17.78 ∴PASSd3/t3 = 16.13 ∴PASS

0.2 ≤ di/d0 ≤ 1.0 d1/b0 = 88.9/193.7 = 0.46 ∴PASSd2/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 ≤ +48.4e1 = -73.2 mm ∴PASSe2 = -60.4 mm ∴PASS

λov ≥ 25% λov,i = |g| sin θi /di x 100%λov,1 = 90.0 x sin 45°/88.9 x 100%λov,1 = 71.6% ∴PASSλov,2 = 86.5 x sin 50°/101.6 x 100%λov,2 = 65.2% ∴PASS

30° ≤ θi ≤ 90° θ1 = 45° ∴PASSθ2 = 50° ∴PASSθ3 = 90° ∴PASS

Chord Face Failure (deformation);

Compression brace (1);

5M0

1

1

200ypg

Rd,1 /d

d2.108.1

sin

tf kkN 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

024.01k

0

2.12.0

g

where;

69.90.102

7.193

t2

d

0

0 =×==γ

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

Use the smallest overlap for (g)


49/56


49





0,op,el

Ed,0,op

0,ip,el

Ed,0,ip

0

Ed,pEd,p

W

M

W

M

A

N++=σ

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

2

2Ed,pN/mm7.88

107.57

1000512=

×

×=σ


7.88

f n

0y

Ed,pp =⎟

⎠

⎞⎜⎝

⎛ =⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ σ=



kp = 1 - 0.3 np (1 + np) but ≤ 1.0( ) 0.1but25.0125.03.01kp ≤+×−=

91.0kp =

from graph;

kp = 0.91

0.1/7.193

9.882.108.1

45sin

0.1035591.015.2N

2

Rd,1 ⎟ ⎠

⎞⎜⎝

⎛ ×+×

°×××

=

Tension Brace (2);

Rd,12

1Rd,2 N

sin

sinN ×

θ

θ=

63750sin

45sinN 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

69.9024.0169.9k

2.12.0

g

15.2kg =

From graph;

g/t0 = -86.5/10.0 = -8.7

kg = 2.15


50/56


50

Localised Brace-to-Chord Shear Check for overlapped Jo ints;

Although this failure mode check was included as a corrigendum in section 7.1.2 (6) of EN 1993-1-8, noequations have been included in tables 7.2, 7.10, 7.21 or 7.24. In their absence, the following criteria foradequacy of the shear connection may be used. They are based on research by CIDECT but expressed usingEN1993-1-8 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%


51/56


51

where Brace Effective Width (overlapped brace);

j j,eff j jyj

00y

0

0 j,eff ddbutd

tf

tf

d

t12d ≤×

×

××=

6.101dbut6.1013.6355

0.10355

7.193

0.1012d 3,eff 3,eff ≤××

××

×=

6.101dbut100d 3,eff 3,eff ≤=

mm100d 3,eff =

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

( )2cweldeds/toehiddenIf 1c1c

s2s

1s

==

=

Therefore;

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

........50sin

3.61006.1012100

2.65100

3

470

45sin

0.59.889.882100

6.71100

3

470

4 °

×⎥⎦

⎤⎢⎣

⎡+××⎟

⎠

⎞⎜⎝

⎛ −

×+°

×⎥⎦

⎤⎢⎣

⎡+××⎟

⎠

⎞⎜⎝

⎛ −

×⎜⎜⎝

⎛ ×π

( )0.1

0.1

90sin

3.610011001

3

470×⎟ ⎠

⎞°××+×

×+

( ) 0.13419073809712674674

kN575 ×++×π≤

( )9897454

kN575 ×π≤

Summary

Here, the validity limits are combined to include all three braces. The eccentricity values use both brace 1 and2 (diagonal) in relation to brace 3 (the vertical overlapped brace). The localised brace-to-chord shear checkhas 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 overlapwithin the limits for the shear check. The brace-to-chord shear check equation sees the addition of thehorizontal 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 jointconfiguration and refer to our Design of Welded Joints literature to make sure the correct method andformulae 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



52/56

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 mmh0 = 311.0 mm b1 = 100.0 mm b2 = 100.0 mmtw = 9.0 mm t1 = 6.3 mm t2 = 6.3 mmtf = 14.0 mmdw = h0 – 2 x (tf – r 0) = 265.2 mmr 0 = 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/f y0) = 38 √(235/355) = 30.92

dw/tw = 265.2/9 = 29.5 ∴PASSdw ≤ 400 dw = 265.2 ∴PASSChord: Flange;

cf /tf ≤ 10ε (Class 1 or 2 compression) (EN1993-1-110ε = 10 √(235/f y0) = 10 √(235/355) = 8.1 Table 5.2)cf = (b0 – 2r 0 - 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 ∴PASSbi/ti ≤ 35 bi/ti = 100/6.3 = 15.87 ∴PASShi/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 kNN1,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 10210UB CHORD: ADVANCE S355 to EN 10025


53/56


53

g ≥ t1 + t2 = 6.3 + 6.3 = 12.6 Referenceg = 165.8 (zero eccentricity) ∴PASS30° ≤ Ө1 ≤ 90° Ө1 = 45° ∴PASS30° ≤ Ө2 ≤ 90° Ө2 = 45° ∴PASS

Chord Web Yielding;

5Mi

i,ww0yRd,1 /

sin

btf N γ

θ

××=

5.7.4

where… ( ) ( )0f i0f i

ii,w r t10t2butr t5

sinhb ++≤++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/

45sin

6.2410.9355N Rd,1 =

°

××=

and N2,Rd = N1,Rd as same brace and angle dimensions

1092 > 150 kN ∴PASS

Chord Shear ;

5M

i

0v0yRd,i /

sin3

Af N 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

0.143

8.16541

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


54/56


54

Reference

∴ 0.1/45sin3

9.3114355N Rd,1

°×

×=

kN903N Rd,1 =


903kN > 150 kN ∴PASS

Bracing Effective Width;

5Mi,eff iyiRd,i /ptf 2N 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 (f y0 / f yi) 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


Chord Axial Force Resistance in gap;

( ) 5MRd,0,pl

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

V

V1f Af A AN γ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −××+×−=

5.7.4

where… ( )2Ed,21Ed,1Ed,0 sinN,sinNmaxV θ×θ×= 5.7.4

( )°×−°×= 45sin150,45sin150maxV Ed,0


55/56


55

Vo,Ed = 106.1kN Reference

where…( ) ( )

0.1

3/3558.31413/f AV

0M

0y0,vRd,0,pl

×=

γ=

5.7.4

Vpl,0,Rd = 643.9kN

( ) 0.1/9.6431.106

13558.31413558.31416120N 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


56/56

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While care has been taken to ensure that the

information contained in this brochure is

accurate, neither Tata Steel Europe Limited, nor

its subsidiaries, accept responsibility or liability

for errors or for information which is found to be

misleading.

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Tata Steel Europe Limited

Tata Steel

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