[Corus] SHS Joint Worked Examples

Job No: Sheet: 1 of 4 Made by: CPM Date: 24/02/05 Checked by: Date: Client: Job Title: SHS Joints Worked Example Subject: CHS Gap K-joint, Axial Capacity

Corus Tubes PO Box 101, Weldon Road Corby, Northants NN17 5UA Tel: +44 (0) 1536 402121 Fax: +44 (0) 1536 404049 www.corusgroup.com

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

References are to Corus publication ‘Design of SHS welded joints’

-500 kN 400 kN

CHS 219.1x12.5All material EN 10210 S355J2H

CHS 139.7x5.0 CHS 114.3x3.6

45º45º 40

-1000 kN -1636 kN

Parameter limits

d0/t0 ≤ 50 d0/t0 = 219.1/12.5 = 17.53 ∴PASS di/ti ≤ 50 d1/t1 = 139.7/5 = 27.94 ∴PASS d2/t2 = 114.3/3.6 = 31.75 ∴PASS di/d0 ≥ 0.2 d1/d0 = 139.7/219.1 = 0.64 ∴PASS d2/d0 = 114.3/219.1 = 0.52 ∴PASS g ≥ t1 + t2 40 ≥ 5 + 3.6 = 8.6 mm ∴PASS 30° ≤ θi ≤ 90° θ1 = 45° ∴PASS θ2 = 45° ∴PASS -0.55 d0 ≤ e ≤ +0.25 d0 e = 0 mm -0.55 x 219.1 ≤ e ≤ +0.25 x 219.1 -120.5 ≤ e ≤ +54.8 ∴PASS

5.1.1 Table 6 (page 20)

3.1

(page 9)




Chord face deformation Note: Brace 1 usually designated compression and brace 2 tension.

Compression brace (1):

( ) ( )p0

1

1

200y

Rd,1 nfgfdd2.108.1

sintf

N

+

θ=

where, Gap/lap function f(g) Using formulae: Using graph:

( ) ( )

−+

γ+γ=

33.1t/g5.0exp1024.01gf

0

2.12.0

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

764.85.122

1.219t2

d

0

0 =×

==γ

( ) ( )

−×+

×+=

33.15.12/405.0exp1764.8024.01764.8gf

2.12.0

( ) 761.1gf =

from graph; f(g) = 1.761 where; g/to = 40/12.5 = 3.2 do/to = 219.1/12.5 do/to = 17.528

5.1.3 (page 21)

Formulae Graph

5.1.2(page

20)

7.1(page

43)

Graph5.1.2

Fig. 22(page

21)




CHS chord end load function f(np) Using formulae: Using graph:

0.1 butf

3.0f

3.01)n(f2

0y

p

0y

pp ≤

σ−

σ+=

where; CHS chord least compressive stress σp;

0,el

20,op

20,ip

0

0,pp W

MMAN +

−=σ

Note: Moment part is additive to compressive stress which is negative, hence minus sign for moments. For CHS chords use least compressive chord stress.

2p 101.8110001000

××−

=σ

2

p mm/N30.123−=σ

2

p 3553.1233.0

3553.1233.01)n(f

−−

−+=

0.1860.0)n(f p ≤=


0,el

20,op

20,ip

0

0,pp W

MMAN +

−=σ


2p 101.8110001000

××−

=σ

2

p mm/N30.123−=σ CHS chord stress ratio;

347.0355

30.123f 0y

p −=−

=σ

from graph; f(np) = 0.860

Formulae Graph

5.1.2(page

20)

5.1.2 (page

20)

5.1.2 (page

20)

5.1.2 (page

20)

5.1.2 Graph

Fig. 21 (page

21)




860.0761.11.2197.1392.108.1

45sin5.12355N

2

Rd,1 ×

+

°×

=

kN986N Rd,1 =

Tension brace (2):

98645sin45sinN

sinsinN Rd,1

2

1Rd,2 ×

°°

=θθ

=

kN986N Rd,2 =

Chord punching shear (valid when di ≤ d0 - 2t0)

i2

ii00yRd,i sin2

sin13

dtfN

θθ+

×π

=

Brace (1):

°°+

××π××

=45sin245sin1

37.1395.12355N 2Rd,1

kN1919N Rd,1 =

Brace (2):

°°+

××π××

=45sin245sin1

33.1145.12355N 2Rd,2

kN1570N Rd,2 =

Joint strength dictated by chord face deformation for both bracings; Brace 1 joint capacity, kN986N Rd,1 = Brace 2 joint capacity, kN986N Rd,2 =

5.1.3 (page 21)

5.1.3 (page 22)

Job No: Sheet: 1 of 4 Made by: CPM Date: 24/02/05 Checked by: Date: Client: Job Title: SHS Joints Worked Example Subject: CHS Overlap K-joint, Axial Capacity




-500 kN

-1000 kN CHS 219.1x12.5All material EN 10210 S355J2H

CHS 139.7x5.0 400 kN

-1636 kN

CHS 114.3x3.6

45º45º -45

e = -42.2

Parameter limits d0/t0 ≤ 50 d0/t0 = 219.1/12.5 = 17.53 ∴PASS di/ti ≤ 50 d1/t1 = 139.7/5 = 27.94 ∴PASS d2/t2 = 114.3/3.6 = 31.75 ∴PASS di/d0 ≥ 0.2 d1/d0 = 139.7/219.1 = 0.64 ∴PASS d2/d0 = 114.3/219.1 = 0.52 ∴PASS 25% ≤ Overlap ≤ 100% Overlap = g sin θi /di x 100% Overlap = 45 sin 45° / 114.3 x 100% Overlap = 27.8% 25% ≤ 27.8% ≤ 100% ∴PASS 30° ≤ θi ≤ 90° θ1 = 45° ∴PASS θ2 = 45° ∴PASS -0.55 d0 ≤ e ≤ +0.25 d0 e = -42.2 mm -0.55 x 219.1 ≤ e ≤ +0.25 x 219.1 -120.5 ≤ e ≤ +54.8 ∴PASS

5.1.1 Table 6 (page 20)

7.1 Fig. 31 (page 43,44)

3.1 (page 9)




Chord face deformation Note: Brace 1 usually designated compression and brace 2 tension.


( ) ( )p0

1

1

200y

Rd,1 nfgfdd2.108.1

sintf

N

+

θ=

5.1.3 (page 21) where, Gap/lap function f(g) Using formulae: Using graph:

( ) ( )

−+

γ+γ=

33.1t/g5.0exp1024.01gf

0

2.12.0

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

764.85.122

1.219t2

d

0

0 =×

==γ

( ) ( )( )

−−×+

×+=

33.15.12/455.0exp1764.8024.01764.8gf

2.12.0

( ) 024.2gf =

from graph; f(g) = 2.024 where; g/to = -45/12.5 = -3.6 do/to = 219.1/12.5 do/to = 17.528

Formulae Graph

5.1.2(page

20)

7.1(page

43)

Graph5.1.2

Fig. 22(page

21)




CHS chord end load function f(np) Using formulae: Using graph:

0.1 butf

3.0f

3.01)n(f2

0y

p

0y

pp ≤

σ−

σ+=


0,el

20,op

20,ip

0

0,pp W

MMAN +

−=σ


2p 101.8110001000

××−

=σ

2

p mm/N30.123−=σ

2

p 3553.1233.0

3553.1233.01)n(f

−−

−+=

0.1860.0)n(f p ≤=


0,el

20,op

20,ip

0

0,pp W

MMAN +

−=σ


2p 101.8110001000

××−

=σ

2

p mm/N30.123−=σ CHS chord stress ratio;

347.0355

30.123f 0y

p −=−

=σ

from graph; f(np) = 0.860

Formulae Graph

5.1.2(page

20)

5.1.2 (page

20)

5.1.2 (page

20)

5.1.2 (page

20)

5.1.2 Graph

Fig. 21 (page

21)




860.0024.21.2197.1392.108.1

45sin5.12355N

2

Rd,1 ×

+

°×

=

kN1134N Rd,1 =

Tension brace (2):

113445sin45sinN

sinsinN Rd,1

2

1Rd,2 ×

°°

=θθ

=

kN1134N Rd,2 =

Chord punching shear check not required for overlapping bracings. Joint strength dictated by chord face deformation for both bracings; Brace 1 joint capacity, kN1134N Rd,1 = Brace 2 joint capacity, kN1134N Rd,2 =

5.1.3 (page 21)

Job No: Sheet: 1 of 6 Made by: CPM Date: 24/02/05 Checked by: Date: Client: Job Title: SHS Joints Worked Example Subject: RHS Gap K-joint, Axial Capacity




-650 kN

RHS 200 x 200 x 10All material EN 10210 S355J2H

SHS 120 x 120 x 5

45º

5

45º

650 kN

40

-1000 kN -1920 kN

Parameter limits

b0/t0 ≤ 35 b0/t0 = 200/10 = 20 ∴PASS h0/t0 ≤ 35 h0/t0 = 200/10 = 20 ∴PASS bi/ti ; hi/ti ≤ 35 and ≤ 34.5√(275/fyi) = 30.4 compression brace b1/t1 = 120/5 = 24 ∴PASS h1/t1 = 120/5 = 24 ∴PASS bi/ti ; hi/ti ≤ 35 tension brace b2/t2 = 120/5 = 24 ∴PASS h2/t2 = 120/5 = 24 ∴PASS bi/b0 ≥ 0.35 and ≥ 0.1+0.01 b0/t0 = 0.3 b1/b0 = 120/200 = 0.6 ∴PASS b2/b0 = 120/200 = 0.6 ∴PASS g ≥ t1 + t2 40 ≥ 5 + 5 = 10 mm ∴PASS 0.5(b0-(b1+b2)/2) ≤ g ≤ 1.5(b0-(b1+b2)/2) g = 40 mm 0.5(200-(120+120)/2) ≤ g ≤ 1.5(200-(120+120)/2) 40 ≤ g ≤ 120 mm ∴PASS 30° ≤ θi ≤ 90° θ1 = 45° ∴PASS θ2 = 45° ∴PASS -0.55 h0 ≤ e ≤ +0.25 h0 e = 5 mm -0.55 x 200 ≤ e ≤ +0.25 x 200 -110 ≤ e ≤ +50 ∴PASS

5.2.1 Table 7 (page 25)

2.1 (page 4)

3.1 (page 9)




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


( )nfb4

hbhbtb

sintf3.6

N0

2211

0

0

1

200y

Rd,1

+++θ

=

where, RHS chord end load function f(n) Using formulae: Using graph:

( ) 0.1butf4.03.1nf0y

0 ≤βσ

+=

where;

6.02002120120

b2bb

0

21 =×

+=

+=β

RHS chord most compressive stress σ0;

0,op,el

0,op

0,ip,el

0,ip

0

00 W

M

W

M

AN

++=σ

Note: Moment part is additive to compressive stress which is negative, hence minus sign for moments. For RHS chords use most compressive chord stress.

20 109.7410001920

××−

=σ

2

0 mm/N34.256−=σ

( ) ( ) 0.16.035534.2564.03.1nf ≤

×−×

+=

( ) 0.1819.0nf ≤=

where; RHS chord most compressive stress σ0;

0,op,el

0,op

0,ip,el

0,ip

0

00 W

M

W

M

AN

++=σ


20 109.7410001920

××−

=σ

2

0 mm/N34.256−=σ RHS chord stress ratio;

722.0355

34.256f 0y

0 −=−

=σ

from graph;

for 6.0200120

bb

0

i ==

f(n) = 0.819

5.2.3 (page 28)

Formulae Graph

5.2.2(page

25)

7.1(page

43)

5.2.2 (page

25)

5.2.2 (page

25)

5.2.2 (page

25)

5.2.2 Graph

Fig. 23 (page

26)




819.02004

12012012012010200

45sin103553.6N

2

Rd,1

×+++

×°××

=

kN695N Rd,1 =

Chord shear check – brace 1

i

v0yRd,i sin3

AfN

θ=

where

212.0

1034041

1

t3g41

1

5.0

2

2

5.0

20

2 =

××

+=

+=α

( ) ( ) 2

000v mm442410200212.02002tbh2A =×+×=α+=

°×

=45sin3

4424355N Rd,1

kN1282N Rd,1 =

5.2.3 (page 28)

5.2.2 (page 27)

5.2.2 (page 27)




Bracing Effective width – brace 1

( )effiiiiyiRd,i bbt4h2tfN ++−=

where iiiyi

00y

0

0eff bbutb

tftf

bt10b ≤××=

mm120butmm1201205355

10355200

1010beff ≤=×××

××

=

( )1201205412025355N Rd,1 ++×−××=

kN817N Rd,1 =

Chord punching shear – brace 1 (valid when β ≤ 1 - 2t0/b0)

++

θθ= epi

i

i

i

00yRd,i bb

sinh2

sin3

tfN

where ii0

0ep bbutb

bt10b ≤×=

120but120200

1010bep ≤××

=

mm60bep =

++

°×

°×

= 6012045sin1202

45sin310355N Rd,i

kN1506N Rd,1 =

Summary - brace 1 Joint strength for brace 1 dictated by chord face deformation. ∴ Brace 1 joint capacity, kN695N Rd,1 =

5.2.3 (page 28)

5.2.2 (page 26)

5.2.3 (page 28)

5.2.2 (page 26)




Brace 2 Repeat brace capacity formulae for brace 2. Note: As both braces are of same geometry, brace 2 capacity will be the same: Chord face deformation, kN695N Rd,2 = Chord shear check, kN1282N Rd,2 = Bracing effective width, kN817N Rd,2 = Punching shear, kN1506N Rd,2 = ∴ Brace 2 joint capacity, kN695N Rd,2 =




Chord axial load resistance in gap Chord: Axial load in gap (valid when V ≥ 0.5 Vp)

( )[ ]2

pv00yRd,gap,0 1V/V2AAfN −−= where 2

0 mm7490A = 2

v mm4424A = kN619.45945sin650sinNV ii =°=θ=

kN740.9063

44243553

AfV v0yp =×==

−×−=

2

Rd,gap,0 1740.906619.459244247490355N

kN2659N Rd,gap,0 =

-650 650

-1000 -1000 -460 -460-1920

-1000 -460-1460

Axial load in gap (kN)

kN1460N gap,0 =

2659 > 1460 ∴ Passes axial load in gap

5.2.3 (page 28)

Job No: Sheet: 1 of 3 Made by: CPM Date: 01/11/04 Checked by: Date: Client: Job Title: SHS Joints Worked Example Subject: RHS Overlap K-joint, Axial Capacity




-600 kN

-1000 kN -1849 kNSHS 200 x 200 x 10All material EN 10210 S355J2H

SHS 120 x 120 x 5 600 kN

45º45º 70 i j

-50.1

Parameter limits

b0/t0 ≤ 40 b0/t0 = 200/10 = 20 ∴PASS h0/t0 ≤ 40 h0/t0 = 200/10 = 20 ∴PASS bi/ti ; hi/ti ≤ 30.4√(275/fyi) = 26.8 compression brace b1/t1 = 120/5 = 24 ∴PASS h1/t1 = 120/5 = 24 ∴PASS bi/ti ; hi/ti ≤ 35 tension brace b2/t2 = 120/5 = 24 ∴PASS h2/t2 = 120/5 = 24 ∴PASS bi/b0 ≥ 0.25 b1/b0 = 120/200 = 0.6 ∴PASS b2/b0 = 120/200 = 0.6 ∴PASS bi/bj ≥ 0.75 bi/bj = 120/120 = 1.0 ∴PASS 25% ≤ overlap ≤ 100% Overlap = g sin θi /hi x 100% Overlap = 70 sin 45° / 120 x 100% Overlap = 41.2% 25% ≤ 27.8% ≤ 100% ∴PASS 30° ≤ θi ≤ 90° θ1 = 45° ∴PASS θ2 = 45° ∴PASS -0.55 h0 ≤ e ≤ +0.25 h0 e = -50.1 mm -0.55 x 200 ≤ e ≤ +0.25 x 200 -110 ≤ e ≤ +50 ∴PASS

5.2.1 Table 7 (page 25)

7.1 Fig. 31 (page 43,44)

5.2.1 Table 7

(page 25)

2.1 (page 4)

3.1

(page 9)




Bracing Effective width Note: Only the overlapping brace (i) need be checked. The capacity of the overlapped brace (j) is

based on on an efficiency ratio to that of the overlapping brace. Overlapping brace, i (2): For 25% ≤ Ov < 50%

( )

++−

= eoveffiiv

iyiRd,i bbt4h250OtfN

where, iiiyi

00y

0

0eff bbutb

tftf

bt10b ≤××=

120but1205355

10355200

1010beff ≤×××

××

=

mm120beff =

iiiyi

jyj

j

jeov bbutb

tftf

bt10

b ≤××=

120but12053555355

120510beov ≤×

××

××

=

mm50beov =

( )

++×−×

×= 50120541202

502.415355N Rd,i

kN624N Rd,i =

5.2.3 (page 28)

5.2.2 (page 26)

5.2.2 (page 26)




Overlapped brace, j (1):

yii

yjjRd,iRd,j fA

fANN =

3557.223557.22624N Rd,j ×

××=

kN624N Rd,j =

5.2.3 (page 28)

Job No: Sheet: 1 of 4 Made by: CPM Date: 01/11/04 Checked by: Date: Client: Job Title: SHS Joints Worked Example Subject: RHS T-joint, Moment In-Plane Capacity




RHS 150 x 150 x 10All material EN 10210 S355J2H

RHS 150 x 150 x 8

90º

54 kNm -19.2 kN

35.8 kNm-136 kN

18.2 kNm-350 kN

For simplicity only the moment in plane capacity calculations are shown as other examples cover axial capacity. Axial and moment out of plane capacity would need to be calculated and included in the interaction check.

Parameter limits b0/t0 ≤ 35 b0/t0 = 150/10 = 15 ∴PASS h0/t0 ≤ 35 h0/t0 = 150/10 = 15 ∴PASS bi/ti ; hi/ti ≤ 35 and ≤ 34.5√(275/fyi) = 30.4 compression brace b1/t1 = 150/8 = 24 ∴PASS h1/t1 = 150/8 = 24 ∴PASS bi/b0 ≥ 0.25 b1/b0 = 150/200 = 0.75 ∴PASS 30° ≤ θi ≤ 90° θ1 = 45° ∴PASS

5.2.1 Table 7 (page 25)

2.1 (page 4)




Chord face deformation (Valid when β ≤ 0.85)

0.1150150

bb

0

1 ===β ∴Check not required

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

RHS chord end load function f(n) Using formulae: Using graph:

( ) 0.1butf4.03.1nf0y

0 ≤βσ

+=


0,op,el

0,op

0,ip,el

0,ip

0

00 W

MWM

AN

++=σ


320 10236100010008.35

109.541000136

×××−

+××−

=σ

2

0 mm/N47.176−=σ

( ) ( ) 0.10.135547.1764.03.1nf ≤

×−×

+=

( ) 0.1)n(f0.1but101.1nf =∴≤=


0,op,el

0,op

0,ip,el

0,ip

0

00 W

MWM

AN

++=σ


320 10236100010008.35

109.541000136

×××−

+××−

=σ

2

0 mm/N47.176−=σ RHS chord stress ratio;

497.0355

47.176f 0y

0 −=−

=σ

from graph;

for 0.1150150

bb

0

i ==

f(n) = 1.101

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

7.1

(page 43)

Formulae Graph

5.2.2(page

25)

5.2.2 (page

25)

5.2.2 (page

25)

5.2.2 (page

25)

5.2.2 Graph

Fig. 23 (page

26)




Chord side wall crushing (Valid when 0.85 < β ≤ 1.0)

( )2

010ykRd,ip t5htf5.0M += where, 0yyk ff = for T-joints

( )2Rd,ip 105150103555.0M ×+××=

kNm71M Rd,ip =

Bracing Effective width (Valid when 0.85 < β ≤ 1.0)

−−= 111

1

eff1,pl1yRd,ip thb

bb1WfM

where, 1111y

00y

0

0eff bbutb

tftf

bt10b ≤××=

mm150butmm1251508355

10355150

1010beff ≤=×××

××

=

××

−−×= 8150150

150125110237355M 3

Rd,ip

kNm49.73M Rd,ip =

5.2.5.1 (page 28)

5.2.5.1 (page 28)

5.2.2 (page 26)




Summary – Moment in plane Joint moment in-plane capacity dictated by chord side wall failure. Brace joint capacity, kNm71M Rd,ip =

Interaction formula (When moments are present an interaction check is required)

RHS Chord, Check 0.1MM

MM

NN

Rd,op

op

Rd,ip

ip

Rd

≤++

CHS Chord, Check 0.1M

M

MM

NN

Rd,op

op2

Rd,ip

ip

Rd

≤+

+

Note: For CHS chords the in-plane moment term is squared

2.4 (page 7)

2.4 (page 7)

Documents

[Corus] SHS Joint Worked Examples