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Nonlinear Finite Element Analysis of Steel Frame Connected
with External Diaphragm Connection
Sui WEINING1 Wang ZHANFEI1 Liu LIN AND1 Minoru YAMANARI 2
1 School of Civil Engineering, Shenyang Jianzhu University (Shenyang, 110168 China)
E-mail:[email protected] 2 Associate Professor, Graduate School of Science and Technology, Kumamoto University,
(Kumamoto 860-8555, Japan) E-mail: [email protected]
In this paper, in order to investigate the effects of various geometric parameters on the bearing capacity and rigidity of the joint, analytical models are established through the finite element method. Several models of semi-rigid joint with external stiffener rings are analyzed and results are presented for compar-ison and discussion. It is found that the thickness of column, the width and height of beam as well as the width and height of external stiffener rings have obvious effect on the principal stress distribution and the joint damage. Through a single parameter variation and the curve fitting of the effect on the bearing ca-pacity and rigidity according to the finite element simulation results, the writers propose formulas including ultimate flexural capacity, yielding flexural capacity, initial flexural stiffness and yielding flexural stiffness of the joint. In addition, the reliability of the calculation formula is assessed.
Key Words : External stiffener rings, Semi-rigid joint, Nonlinear finite element analysis, Bearing ca-
pacity, Flexural stiffness
1. INTRODUCTION
Rigid or pin joints are usually designed and utilized in high-rise framework structure. However, in practical engineering applications, most of framework joints are semi-rigid. Rigid and pin joints have many weaknesses, for example, the brittle failure of rigid joint is found in Northridge earthquake in American and Kobe earth-quake in Japan; Though the pin joint increase relative story displacements of high-rise framework building. In order to fulfill bearing capacity and ductility of the joint, researchers designed the semi-rigid joint. The ad-vantages of semi-rigid connection performance include high bearing capacity and better ductility [1-2] . For semi-rigid joint, the main studies includes Frye and Morris Yee and Melchers Kishi and Chen Hsieh and Xu Colson and Louveau Ang and Morris Kishi and Chen , Foley and Vinnakota [3-15]. Domestic research began in in 1990s and mainly focused on structure analysis [16-21]. However, there is a little related research on the analysis of the semi-rigid joint with external diaphragm connection at present. Paper [22]points out that: bearing capacity and rigidity of the joints increase in linear with the increasing of the width of external dia-phragm, and the influence is great. This paper studies bearing capacity and rigidity of the joints through non-linear FEA methods and design formula is deduced for the widly use of external diaphragm connections. 2. FINITE ELEMENT MODEL
Nonlinear finite element analysis of a semi-rigid steel joint with external diaphragm was performed using analysis program ABAQUS , the bi-linear hardening model is adopted, 3D shell elements of the type S4R are used to discretize the part in this work. The details of beam-column joints with external diaphragm connection are shown in Fig.1. Their dimensions are giving in Table 1. In these analyses, the effect of welding residual stress is not taken into consideration.
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bf
D External stiffener rings
Beam
Column
tw hf
hstctf
Fig.1 Joint detail Fig.2 Refined mesh of joint
Table 1 Nominal dimensions and material property of model
H
(mm)
L
(mm)
D
(mm)
Steel fy
( N/mm2)
E
(Gpa)
1600 1350 400 Q235 235 206 0.3 (1) Description of Model
Table 2 lists the parameters for the analysis models, it include: depth of beam (hf) , flange width at top and bottom of beam(bf) , flange thicknesses at top and bottom of beam (tf), web thickness (tw), thickness of col-umn(tc), width(hs) and thickness(ts) of external diaphragm connection.
N
L
H
Displacement loading
Column
Beam
Fig.3 Boundary and load conditions
For the material nonlinearity, an elastoplastic constitutive law with isotropic strain hardening based on the Von Mises yield criterion is adopted. The constitutive model is integrated by means of the Euler algorithm and the geometrical nonlinearity is taken into consideration. In order to save time as well as ensure accuracy, the panel zone is meshed densely and the rest of the model is meshed sparsely as shown in Fig.2. The results shown that the proposed method has high accuracy and short computing time.
(2) Boundary and Load Conditions
The boundary conditions of FEA model same with the experiment one [21], the details are depicted in Fig.3. The columns are both pined, constraint the three direction displacement in the bottom and constraint the other direction displacement except the axial displacement on the top. In order to prevent the out-plane instability, constraints the out-plane displacement with steel cantilever beam.
(3) Calibration of Model with Experimental Evidence
Experimental studies had been conducted by researchers Sui et.al[22]for the boundary conditions and unit type of this model. The analysis result and the experimental data are satisfactory agree.
3. FINITE ELEMENT ANALYTICAL RESULT Calculation parameters:
0y
L
P LK , M P L , is an angle of joints which deducts the angles of connection
and panel, Py is load when joints yield, KL0 is initial bending rigidity (the relationship of bending moment and
tc
(mm)
bf
(mm)
tf
(mm)
tw
(mm)
hf
(mm)
hs
(mm)
ts
(mm)
6~20 140~240 6~22 6~22 160~300 25~150 6~22
Table 2 Geometric parameter scope of joints
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angle when initial loading is applied), KL1 is yield bending rigidity(rigidity when joints yield with the increase of load).M is moment of joints, Mu is ultimate moment(maximum bending carrying capacity), My is yield moment(bending carrying capacity when joints yield). They are shown in Fig.4.
KL0
KL1
MM u
M y
y u Fig.4 Calculation parameters
Fig 5 is the deformation and stress distribution when the column thickness tc is changed from 6mm to 20mm and the other parameters are kept constant. When the thickness of column is increased gradually from 6 mm to 20 mm, the biggest stress position changed from panel zoon to external diaphargm, and the shear deformation of panel zoon decreases gradually. In fig.5(a), panel zoon emerges pronounced shear deformation and out-plane buckling happens to the panel zone. While in Fig.5(b), for tc =20mm, those phenomenons disappear but obvious warp appears in the external diaphragm and flanges.
(a) tc=6mm (b) tc=20mm (a) tc=6mm (b) tc=20mm
Fig.5 Joints deformation and stress Fig.6 Stress flow on the jointdistribution cloud picture Learn form Fig.6, the beam stress transferred to column through external diaphragm and panel zoon, and the
stress concentrate on the panel zoon when tc is 6mm and concentrate on external diaphragm and flanges when tc is 20mm. Therefore, tc can improve column relative rigidity.
-8 -6 -4 -2 2 4 6 8
-4-3-2-1
1234
0
tc=8mm
××110055
tc=16mmtc=18mmtc=20mm
/kN
·mm
M
/rad
tc=6mm
tc=10mmtc=12mmtc=14mm
××1100--44
2 4 6 8
1
2
3
4
0
××110055
/kN
·mm
tc=6mmtc=8mmtc=10mmtc=12mmtc=14mmtc=16mmtc=18mmtc=20mm
M
××1100--44
/rad (a) Bending strength curve (b) Rigidity curve
Fig.7 The influence of tc for the joint Fig.7. shows that Mu of the joint grew about 10% and there is obvious increase trend for My ,Kl0 and Kl1
when the parameter tc increased per 2mm. Flange width: bf changed from 140mm to 240mm while the other parameters are unchanged. Fig.8 is the
stress distribution and defrmation of panel zone. stress concentration phenomenon is evident on the external diaphragm and shear stress of panel zoon increased obviously with the increase of parameter bf.
(a) bf=140mm (b) bf=240mm (a) bf=140mm (b) bf=240mm Fig.8 Joints deformation and stress Fig.9 Stress flow on the jointdistribution cloud picture
Fig.9 shows stress flow on the joint with the influence of bf , stress flow concentrates when bf wider(240mm) or narrower(140mm), it is homogenized relatively when bf is 180mm.
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-8 -6 -4 -2 2 4 6 8
-4-3-2-1
1234
0
bf=160mm
××110055
bf=240mm/k
N·m
mM
/rad
bf=140mm
bf=180mmbf=200mmbf=220mm
××1100--44
2 4 6 8
1
2
3
4
0
××110055
/kN
·mm
bf=140mmbf=160mm
bf=180mm
bf=200mm
bf=220mm
bf=240mm
M
××1100--44
/rad (a) Bending strength curve (b) Rigidity curve
Fig.10 The influence of bf for the joint Fig.10. shows that Mu of the joint increased about 10% and there is obvious increase trend for My and Kl0, but
Kl1 changes a little when bf is increased each 20mm. Effect of flange and external diaphragm thicknesses: External diaphragm thickness should not be designed less
than the flange . The external diaphragm thickness studied in this paper is equal to flange thickness. The influence for the joint is investigated when they change simultaneously.
(a) tf=6mm (b) tf=22mm (a) tf=6mm (b) tf=22mm
Fig.11 Joints deformation and stress Fig.12 Stress flow on the jointdistribution cloud picture Fig.11 shows that shear deformations of panel zoon are increased and with the increase of thicknesses pa-
rameter tf and ts, obvious buckling has happened in column wall. At the same time, the trend of principal stress flowing to the column wall and connection zoon is increased obviously. So, it is proved that the increase of their thicknesses can improve Mu My andKl0, but do not improve Kl1 from Fig.13.
-8 -6 -4 -2 2 4 6 8
-4-3-2-1
1234
0
ts=8mm
××110055
ts=16mm
/kN
·mm
M
/rad
ts=6mm
ts=10mmts=12mmts=14mm
××1100--44
2 4 6 8
1
2
3
4
0
××110055
/kN
·mm
ts=6mmts=8mmts=10mmts=12mm
M
××1100--44
/rad
ts=14mmts=16mm
(a) Bending strength curve (b) Rigidity curve
Fig.13 The influence of ts and tf for the joint Web thickness: tw is changed from 6mm to 22mm, the joint deformation and stress distribution is shown in
Fig.14. In addition, transference of principal stress does not have apparent effect from Fig.15.
(a) tw=6mm (b) tw=22mm (a) tw=6mm (b) tw=22mm Fig.14 Joints deformation and stress distribution Fig.15 Stress flow on the joint
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-8 -6 -4 -2 2 4 6 8
-4-3-2-1
1234
0
tw=8mm
××110055
tw=16mmtw=18mmtw=20mm
/kN
·mm
M
/rad
tw=6mm
tw=10mmtw=12mmtw=14mm
××1100--44
tw=22mm 2 4 6 8
1
2
3
4
0
××110055
/kN
·mm
tw=6mmtw=8mmtw=10mmtw=12mmtw=14mmtw=16mmtw=18mmtw=20mm
M
××1100--44
/rad
tw=22mm
(a) Bending strength curve (b) Rigidity curve
Fig.16 The influence of tw for the joint From Fig.16, it is found that there is no influence for M and K by changing tw, because of all of curves are
overlapped. Depth of the beam: Fig.17 shows that shear area of connection zoon is increased in the progress of when hf is
increased from 160mm to 300mm. And stress flow principal is concentrated when connection zoon area is smaller, but stress stream is homogeneous when hf is 300mm from Fig.18.
(a) hf=160mm (b) hf=300mm (a) hf=160mm (b) hf=300mm Fig.17 Joints deformation and stress Fig.18 Stress flow on the jointdistribution cloud picture
-8 -6 -4 -2 2 4 6 8
-4-3-2-1
1234
0
hf=160mm
××110055
hf=240mm
hhf=160mmhhf=190mm
/kN
·mm
M
/rad
hf=140mm
hf=180mmhf=200mmhf=220mm
××1100--44
hf=260mmhf=280mmhf=300mm 2 4 6 8
1
2
3
4
0
××110055
/kN
·mm
hf=140mmhf=160mmhf=180mmhf=200mmhf=220mmhf=240mm
hhf=160mmhhf=190mm
M
××1100--44
/rad
hf=280mmhf=260mm
hf=300mm
(a) Bending strength curve (b) Rigidity curve
Fig.19 The influence of hf for the joint
Learn from Fig.19, M and K of joint have linear increase with the increase of hf changed from 160mm to 300mm.
The effect of external diaphragm width: Shear deformation of panel zoon increase gradually while the shear deformation and stress of external diaphragm decrease, they finally disappear with the change of hs from 25mm to 150mm shown in Fig.20. Principal stress stream is concentrated when hs is 25mm and homogeneous when it is 150mm for Fig.21.
(a) hs=25mm (b) hs=150mm (a) hs=25mm (b) hs=150mm Fig.20 Joints deformation and stress Fig.21 Stress flow on the jointdistribution cloud picture
Fig.22 shows that grate influence have been made by hs for M when it changes from 25mm to 100mm,
however the influence decreases gradually beyond 100mm. Moreover there are a lot of effects for KL0 and a little for KL1.
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In conclusion, great influence could be made by the change of tc, bf, hf, tf(ts), hs for M and KL0 ,at the same time tc and hf have effect for KL1 .
-8 -6 -4 -2 2 4 6 8
-4-3-2-1
1234
0
hs=50mm
×105
hs=150mm/k
N·m
mM
/rad
hs=25mm
hs=75mmhs=100mmhs=125mm
×10-4
2 4 6 8
1
2
3
4
0
×105
/kN
·mm
hs=25mm
hs=50mm
hs=150mm
hs=75mm
hs=125mm
hs=100mm
M
×10-4
/rad (a) Bending strength curve (b) Rigidity curve
Fig.22 The influence of hs for the joint
3. SINGLE PARAMETER ANALYSIS
Nondimensional Parameters: D/tc=X1, D/bf= X2, D/tf= X3, D/tw= X4, D/hf= X5, D/hs= X6. Establish the rela-tionship curves of M/ yR2L, KL0R3/EIL2 of nondimensional parameters, after that through the curve fitting to get the fitting formula of each curve respectively. The aim is to get M and K nonlinear regression formula.
2 4 6 8
0.5
1
1.5
2
2.5
0X1
×10-2
Mu/ yR2L=0.219X1
-0.763
My/ yR2L=0.047X1
-0.497
×10
Mu analysis result
My analysis result ------regression curve
2 4 6 8
1
2
3
4
0X1
KL0R3/EIL
2=1.44X1
1.49
KL1R3/EIL
2=0.015X1
2.286
×103
KL0 analysis result
KL1 analysis result
-------regression curve
×10
1 2 3 4
0.5
1
1.5
2
2.5
0X2
×10-2
Mu/ yR2L=0.023X2
-0.727
My/ yR2L=0.014X2
-0.907
Mu analysis result
My analysis result ------regression curve
1 2 3 4
1
2
3
4
0X2
KL0R3/EIL
2=4783X2
-1.49
×103
KL0 analysis result
-------regression curve
2 4 6 8
0.5
1
1.5
2
2.5
0X4
×10-2
Mu/ yR2L=0.076X4
-0.547
My/ yR2L=0.024X4
-0.305
×10
Muanalysis result
My analysis result ------regression curve
2 4 6 8
1
2
3
4
0X4
KL0R3/EIL
2=440000X 4
-1.59
×103
KL0 analysis result
-------regression curve
×10
-130-
1 2 3 4
0.5
1
1.5
2
2.5
0X5
×10-2
Mu/ yR2L=0.015X5
-1.03
My/ yR2L=0.01X5
-0.953
Mu analysis result
My analysis result
------regression curve
1 2 3 4
1
2
3
4
0X5
KL0R3/EIL
2=1502X5
-0.9
×103
KL0 analysis result
-------regression curve
KL1R3/EIL
2=225.4X5
-2.28
KL1 analysis result
0.5 1 1.5 2
0.5
1
1.5
2
2.5
0X6
×10-2
Mu/ yR2L=0.018X6
-0.233
My/ yR2L=0.009X6
-0.171
×10
Mu analysis result
My analysis result
------regression curve
0.5 1 1.5 2
1
2
3
4
0X6
KL0R3/EIL
2=4077X6
-0.712
×103
KL0 analysis result
-------regression curve
×10
Fig.23 Nonlinear regression curve According to the changing curve of each single parameter in Fig.23, nonlinear regression analysis is conducted to
determine the My ,Mu ,KL0 and KL1 formula: 2 0.763 0.727 0.547 1.03 0.233
y 1 2 4 5 66.23uM R L X X X X X (1)
2 497 907 05 0.953 711 2 4 5 60.55y yM R L X X X X X (2)
2 1.49 3 1.49 1.51 0.9 0.7120 1 2 4 5 614458LK EIL X R X X X X (3)
2 2.286 3 2.281 1 50.028LK EIL X R X (4)
4. DISPERSES OF FORMULA
The results of formula calculation compared with FEA are shown in Fig.24, the break line is the 95% con-
fidence boundary level, most of the data are in this level. In order to calculate accessibly, the power function form is adopted in the calculation formula, and the formula has very strong practicability. The calculation formula of the semi-rigid joint with external diaphragm is proposed.
5. CONCLUSIONS
The geometry factors of the joint have great influence for M and K: tc, bf, hf, tf(ts), hs have great influence for M and KL0; tc and hf have effect for KL1 . The calculation formula of M and K is obtained through nonlinear regression formula of the results of FEA.
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0.5 1 1.5 2 2.5
0.5
1
1.5
2
2.5
0Mcy/ yR
2L
×10-2
×10-2
My/
yR2 L
0.5 1 1.5 2 2.5
0.5
1
1.5
2
2.5
0Mcu/ yR
2L
×10-2
×10-2
Mu/
yR2 L
1 2 3
1
2
3
0KcL0R
3/EIL2
×103
×103
KL
0R3 /E
IL2
1 2 3
1
2
3
0KcL1R
3/EIL2
×102
×102
KL
1R3 /E
IL2
Fig.24 Comparision of the formulus and the analysis results
ACKNOWLEDGMENT:This study is supported by the National Natural Science Foundation of China (se-rial number: 51178275) and Fund Project of Liaoning Province Education Office (serial number: L2010452), these supports are gratefully acknowledged.
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