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8/3/2019 8 CHS Joints 2011
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Hollow sections in structural applications Course
Wardenier part 8 CHS Joints page
Wardenier Tubular Structure Course - 2011
Tubular Structures
Circular Hollow Section Joints
em. Prof.dr.ir. J. Wardenier
Delft University of Technology
National University of Singapore
Wardenier Tubular Structure Course - 2011
Now many Applications
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Hollow sections in structural applications Course
Wardenier part 8 CHS Joints page
Wardenier Tubular Structure Course - 2011
Basic Types of welded Joints
Wardenier Tubular Structure Course - 2011
Symbols used (e.g. for K Joints)d
b
h
t
e
g
01
2
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Hollow sections in structural applications Course
Wardenier part 8 CHS Joints page
Wardenier Tubular Structure Course - 2011
Joint parameters
0
1
d
d=
0
21
2d
dd +=
for T and X
for K-joint
0
02t
d=
0
'tgg =
00
'y
op
fANn
=
0
0
yfn =
Wardenier Tubular Structure Course - 2011
Failure modes (basic)
Chord punching shear
Chord plastification
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Wardenier part 8 CHS Joints page
Wardenier Tubular Structure Course - 2011
Further possible failure modes (additional)
Chord shear failure
Brace effective width
Local Buckling in chord or brace
Wardenier Tubular Structure Course - 2011
Analytical models
Ring model
Punching shear model
Shear failure of the chord
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Wardenier part 8 CHS Joints page
Wardenier Tubular Structure Course - 2011
Ring model (elastic stress distribution)
1,Ed
joint joint
N1
N1
1,Ed
joint joint
N1
N1
Wardenier Tubular Structure Course - 2011
Ring model (assume line loads)
Be
2,5 to 3d0
11 sin
2
N
11 sin
2N
N1
N1
c1d11
1 sin2
Ne
11
B2
sinN
Be
2,5 to 3d0
11 sin
2
N
11 sin
2N
N1
N1
c1d11
1 sin2
Ne
11
B2
sinN
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Wardenier part 8 CHS Joints page
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Ring model (plastic hinges)
0y20p ft
4
1m =
Wardenier Tubular Structure Course - 2011
Ring model
=
2
dc
2
d
2
sinNBm2 11011ep
1
0y20
1
0e1
sin
ft
)c1(
d/B2N
=
C0
Qf= f(n)f
1
0y20
1
01 Q
sin
ft
)c1(
cN
=
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Wardenier part 8 CHS Joints page
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Ring model
some effects have been neglected, e.g.:
- influence axial and shear force on mp
and
- chord stress effect
Wardenier Tubular Structure Course - 2011
Effect chord loading (be careful different def. in codes)
Function f(n) = Qfbased on n (IIW 2009)
(In Eurocode based on n)
0,0
0,2
0,4
0,6
0,8
1,0
1,2
-1,0 -0,8 -0,6 -0,4 -0,2 0,0 0,2 0,4 0,6 0,8 1,0
2 = 63,5
2 = 25,4
2 = 63,5
2 = 50,8
2 = 25,4
N1usin1
/
(fy0
t02Q
u)
N0/ Npl,0
N0
N2
N0p
A
A
1
N1
2N0
N2
N0p
A
A
1
N1
2
p011gap,0
2
1i
p0ii0
NcosNN
NcosNN
+=
+==
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Wardenier part 8 CHS Joints page
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Punching shear model
N1sin 1 = d1t0. f(ellips )vp with vp = 0,58fy0
Assumption:
at failure about uniform
punching shear yield
stress distribution
(check by experiments)
validity !
Wardenier Tubular Structure Course - 2011
f(ellips )
Punching shear model
12
10y011
sin2
sin1ftd58,0N
+=
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Chord shear model
Based on yield criterion of Huber-Hencky-Von Mises
Ngap,0Ngap,0
Wardenier Tubular Structure Course - 2011
Chord shear model
)f58,0(A2
3
fAV 0y0
0y
v0,pl
==
Npl,0 = A0 fy0 = (d0 - t0) t0 fy0
The shear load capacity is given by:
The axial load capacity is given by:
)f58,0(A2
3
fAV 0y0
0y
v0,pl
==
The shear load capacity is given by:
The axial load capacity is given by:
)f58,0(A2
3
fAV 0y0
0y
v0,pl
==
The shear load capacity is given by:
Npl,0 = A0 fy0 = (d0 - t0) t0 fy0
The axial load capacity is given by:
)f58,0(A2
3
fAV 0y0
0y
v0,pl
==
The shear load capacity is given by:
)f58,0(A2
3
fAV 0y0
0y
v0,pl
==
The shear load capacity is given by:
The axial load capacity is given by:
)f58,0(A2
3
fAV 0y0
0y
v0,pl
==
The shear load capacity is given by:
Npl,0 = A0 fy0 = (d0 - t0) t0 fy0
The axial load capacity is given by:
)f58,0(A2
3
fAV 0y0
0y
v0,pl
==
The shear load capacity is given by:
Ngap,0
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Chord shear model
0,1N
N
V
sinN2
0,pl
0,gap
2
0,pl
ii
+
2
v0y
ii
0y00,gap Af58,0
sinN
1fAN
)f58,0(A2
3
fAV 0y0
0y
v0,pl
==
Npl,0 = A0 fy0 = (d0 - t0) t0 fy0
The shear load capacity is given by:
The axial load capacity is given by:
or
2
v0y
ii
0y00,gap Af58,0
sinN
1fAN
or
0,1N
N
V
sinN2
0,pl
0,gap
2
0,pl
ii
+
2
v0y
ii
0y00,gap Af58,0
sinN
1fAN
or
Ngap,0
Wardenier Tubular Structure Course - 2011
Experiments
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Wardenier part 8 CHS Joints page
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Finite element simulation
0
100
200
300
400
500
0 20 40 60 80 100
Ovalisation [mm]
Load[kN]
EX-03 - Collar, Compression
Experiment = 0.54
Numerical 2= 50.6
Calibration with experiments
(Choo/Van de Vegte)
Wardenier Tubular Structure Course - 2011
Verification with experiments
For example:
X-joints
analysis for
IIW (2009)
0
2
4
6
8
0,0 0,2 0,4 0,6 0,8 1,0
f(N1u
)Xj
oints
+
7,01
1
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Wardenier part 8 CHS Joints page
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Evaluation to design rules
Analytical models
ExperimentsStrength formulae
Scatter in test results
scatter in mechanical properties
tolerances in dimensions
tolerances in fabrication
failure mode
Design strength
N1,Rd =Nk/m
Wardenier Tubular Structure Course - 2011
Strength formulae for CHS joints(IIW 2009 = Draft ISO = CIDECT)
or
f
1
200y
Rd,1 Qsin
tf)'g(f)(f)(fN
=
or
i
2
0y0fuRdi,
sintfQQN =
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Wardenier part 8 CHS Joints page
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Criterion
to be checked
Axially loaded joints with CHS
Braces and Chord
Design strength:
chord plastification
Design strength:
chord punching shear
(only for di d0 2t0)
Limit states criteria IIW (2009)
i
20y0
fuRdi,sin
tfQQN =
12
101y0Rd1,
sin2
sin1tdf0,58=N
+
i
20y0
fuRdi,sin
tfQQN =
12
101y0Rd1,
sin2
sin1tdf0,58=N
+
Criterion
to be checked
Axially loaded joints with CHS
Braces and Chord
Design strength:
chord plastification
Design strength:
chord punching shear
(only for di d0 2t0)
i
20y0
fuRdi,sin
tfQQN =
12
101y0Rd1,
sin2
sin1tdf0,58=N
+
Wardenier Tubular Structure Course - 2011
Function Qu
Notes: 1) When cos1 >, check shear
( )8.612.6Q 0.22u +=
]
)tg(1.2
1[1)8(165.1Q
0.8
0
0.31.6u
+
++=
1
N1
t1
d1
N1
d0 t0
0.15u
0.7 -1
12.6Q
+=
1
N1
t1
d1d
0 t0
2
N2
t2
d2N1
1
t1
d1g
+e
d0
N0
X joints 1)
T and Y joints
K gap joints
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Function Qf
Compression tension
T, Y, X joints C1 = 0,45 0,25 C1 = 0,20
K gap joints C1 = 0,25
Qfequations X, T and K gap joints (IIW 2009)
0,pl
0
0,pl
0
M
M
N
Nn +=( ) 1Cf |n|1Q =
Wardenier Tubular Structure Course - 2011
Range of validity based on:
- test evidence
- validity of formulae
- limitation of failure criteria (e.g. avoid local buckling)
- deformation- deformation capacity
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Comparison Qu for T, X and K gap joints for
2=25 (IIW, 2008)
0
10
20
30
40
0 0.2 0.4 0.6 0.8 1
Qu
K gap jointg'=2
K gap jointg'=infinite
T joint
X joint
g'=g/t0
Relation Qu between joint capacities
9
Comparison Qu for T, X and K gap joints for
2=25 (IIW, 2008)
0
10
20
30
40
0 0.2 0.4 0.6 0.8 1
Qu
K gap jointg'=2
K gap jointg'=infinite
T joint
X joint
g'=g/t0
9
Wardenier Tubular Structure Course - 2011
Overlap joints IIW (2009)
Brace effective width (1)
Chord M-N yield (2)
Brace shear (3)
brace i = overlapping member; brace j = overlapped member
di
d0
dj
ti
t0
Ni
Nop
Nj
No
ij
4.8
4.11
4.10
4.11
4.10
tj
brace i =overlapping member; brace j = overlapped member
brace i = overlapping member; brace j = overlapped member
di
d0
dj
ti
t0
Ni
Nop
Nj
No
ij
4.8
4.11
4.10
4.11
4.10
tj
brace i = overlapping member; brace j = overlapped member
(2) (2)
(1)
(3)
N0
brace i = overlapping member; brace j = overlapped member
di
d0
dj
ti
t0
Ni
Nop
Nj
No
ij
4.8
4.11
4.10
4.11
4.10
tj
brace i =overlapping member; brace j = overlapped member
brace i = overlapping member; brace j = overlapped member
di
d0
dj
ti
t0
Ni
Nop
Nj
No
ij
4.8
4.11
4.10
4.11
4.10
tj
brace i = overlapping member; brace j = overlapped member
(2) (2)
(1)
(3)
brace i = overlapping member; brace j = overlapped member
di
d0
dj
ti
t0
Ni
Nop
Nj
No
ij
4.8
4.11
4.10
4.11
4.10
tj
brace i = overlapping member; brace j = overlapped member
di
d0
dj
ti
t0
Ni
Nop
Nj
No
ij
4.8
4.11
4.10
4.11
4.10
tj
brace i =overlapping member; brace j = overlapped member
brace i = overlapping member; brace j = overlapped member
di
d0
dj
ti
t0
Ni
Nop
Nj
No
ij
4.8
4.11
4.10
4.11
4.10
tj
brace i = overlapping member; brace j = overlapped member
(2) (2)
(1)
(3)
N0
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Relation CHS vs RHS overlap Joints
IIW (2009)
- criteria for RHS overlap joints multiplied by /4
(ratio between the cross sectional areas for d=b)
- all b and h dimensions in formulae replaced by d.
Wardenier Tubular Structure Course - 2011
Simplification
Ce in graph
i
f
iyi
0y0
e
yii
Rdi,
sin
Q
tf
tfC
fA
Neff. ==
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Design graphs for X joints
X joint efficiency
0,0
0,1
0,2
0,3
0,4
0,5
0,6
0,7
0,8
0,9
1,0
0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1
effciencyCX
2=10
2=15
2=20
2=30
2=40
1
f
11y
00yX
1y1
Rd,1
s in
Q
tf
tfC
fA
N
=
Wardenier Tubular Structure Course - 2011
Design graph for T joints
T joint efficiency
0,0
0,10,2
0,3
0,4
0,5
0,6
0,7
0,8
0,9
1,0
0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1
effciencyCT
2=10
2=15
2=20
2=30
2=40
2=50
1
f
11y
00yT
1y1
Rd,1
s in
Q
tf
tfC
fA
N
=
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Design graphs for K gap joints
Compare the strength for various d0/t0 ratios and various gaps
1
f
11y
00yK
1y1
*1
sin
Q
tf
tfC
fA
N
=
K gap joint efficiency g'=2
0,00,1
0,2
0,3
0,4
0,5
0,6
0,7
0,8
0,9
1,0
0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1
effciencyCK
2=10
2=15
2=20
2=30
2=40
2=50
i
21
i
f
iyi
00yK
yii
Rd,i
d2
dd
sin
Q
tf
tfC
fA
N +
=
1
f
11y
00yK
1y1
*1
sin
Q
tf
tfC
fA
N
=
K gap joint efficiency g'=10
0,0
0,1
0,2
0,3
0,4
0,5
0,6
0,7
0,8
0,9
1,0
0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1
effciencyCK
2=10
2=15
2=20
2=30
2=40
2=50
i
21
i
f
iyi
00yK
yii
Rd,i
d2
dd
sin
Q
tf
tfC
fA
N +
=
Wardenier Tubular Structure Course - 2011
Joints related to basic types (IIW 2009)
N2 N1N2 N11N2N
X
X
KK
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Joints related to basic types IIW (2009)
h1
d0
t0
N1
N1
1t1
d0
b1
t0
N1
N1
h
1
d0
1
t0
t
N1
N1
h1
d0
t 0
N1
b1
N1
1
f20y01 Qtf)(f)(f)(fN =
Wardenier Tubular Structure Course - 2011
Multi-planar Joints
Storm Surge Barrier near Hook of Holland, The Netherlands
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Multiplanar joints
Geometrical effectGeometrical effect
Loading effectLoading effect
Effect depends on: - type of joint (T, X, K, CHS, RHS)
- type of loading (axial, b.i.p., b.o.p.)
Effect for strength, stiffness and deformation capacity
Wardenier Tubular Structure Course - 2011
Example:XX joint of CHS
multiplanar effects
N2
= N1
N2 = 0.6 N1
N2 = 0.0
N2 = -0.6 N1
Uniplanar joint = 16.0 = 0.60
2 = 50.8
1111 / d0
N1
/(fy
o*to)2
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Moment joints
Bending in plane
Bending out of plane
In frames and Vierendeel girders
(see beam to column joints)
In principle the
same philosophy
Wardenier Tubular Structure Course - 2011
Interaction between N, Mip and Mop
0,1M
M
M
M
N
N
Rd,i,op
Ed,i,op
2
Rd,i,ip
Ed,i,ip
Rd,i
Ed,i +
+
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Special items
Concrete filling of the chord
complete
or
only between tubulars e.g. between pile and leg ina jacket
Punching shear check
Wardenier Tubular Structure Course - 2011
Joints with a Can
Can length for X joints L > 2d0
llcancan/d/d00
f(Nf(N))
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Semi-flattened end joints
T and X-joint:
replace d1 by d1,min.
K-joint with gap:
replace di by
0,5(di +di,min.)
Wardenier Tubular Structure Course - 2011
Slotted Gusset Plate Joints
Exposed roof structure
of
Toronto Convention Centre
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Slotted plate joints
Lw 1,3d An
Lw
An=A
g
a)
crack
slot
GussetPlate b)
CHS
crack
CHSLw
Wardenier Tubular Structure Course - 2011
Tee end to CHS Joints
N1,Rd = 2fy1 t1 (tw + 5tp) A1 fy1
N1,Rd
= 2fyw
tw
(t1
+ 2.5tp
+ s)
2fyw tw (t1 + 5tp)
2.51
tp
N1
d1
tw
5tp+ t
w
s s
t1
2.51
tp
N1
d1
tw
5tp+ t
w
s s
2.51
tp
N1
d1
tw
5tp+ t
w
s s
t1
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The end ofthe lectures
on
CHS joints
Wardenier Tubular Structure Course - 2011