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2012/11/22
1
Chapter 6 Seismic Design of Reinforced Concrete Buildings
Chapter 6
6.1 Introduction
6.2 Earthquake Damage in Reinforce Concrete Buildings
6.3 Structural System and Seismic Grading for Structures
6.4 Seismic design of RC frames
6.5 Seismic design of RC walls
6.6 Detailing
6.7 Dual system
6.8 Case Study
6.1 Introduction
RC structural system
Frame structure
Structural wall
Dual system
Mega structure
钢筋混凝土框架
钢筋混凝土框架梁柱楼板配筋
梁柱节点钢筋连接
2012/11/22
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梁柱节点钢筋连接
梁柱节点钢筋连接
一、现行设计能实现强柱弱梁吗 6.2 Earthquake Damage in Reinforce Concrete Buildings
Sources: China Southwest Architecture Design Institute Co., Ltd
Con’d
现行设计能实现强柱弱梁吗
Sources: China Southwest Architecture Design Institute Co., Ltd
Weak Column
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Con’d
现行设计能实现强柱弱梁吗
Sources: China Southwest Architecture Design Institute Co., Ltd
如何考虑框架、框剪结构中填充墙对整体结构的刚度贡献
Con’d
Sources: China Southwest Architecture Design
Institute Co., Ltd
如何考虑框架、框剪结构中填充墙对整体结构的刚度贡献
Con’d
Sources: China Southwest Architecture Design Institute Co., Ltd
怎样合理设计框架、框剪结构中的填充墙
Con’d
Sources: China Southwest Architecture
Design Institute Co., Ltd
建议提高竖向构件的最低配筋水准
Con’d
Sources: China Southwest Architecture Design Institute Co., Ltd
箍筋设置问题 Damage regarding to stirrups
建筑材料、施工管理问题
Con’d
Sources: China Southwest Architecture Design Institute Co., Ltd
2012/11/22
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十六、重视角柱、加腋梁柱的抗震设计
Photo by Prof. Xiong Haibei, 2008.7
Photo by Prof. Xiong Haibei, 2008.7 Photo by Prof. Xiong Haibei, 2008.7
Key lessons from 5.12 earthquake
From the earthquake vulnerability of RC Frames, we
learnt:
• Vulnerability of corner columns was worse than other exterior and
interior columns
• Column’s failure were more occurred and more severe than beam’s.
• All elements must be detailed so that they can respond to strong
earthquakes in a ductile mode.
• Non-ductile modes such as shear and bond fai lures must be
avoided.
• A high degree of structural redundancy should be provided.
6.3 Structural System and Seismic Grading for Structures
Conceptual Design for RC frame structures
Height
Structural System
Seismic Fortification Intensity Grading
Requirements in Design
Configuration
Calculation
Details
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The appropriate maximum height for R/C buildings (m)
(Table 6.1.1 in GB50011-2010)
Notes: the height in () is the value listed in GB 50011-2008
Structure Types System Seismic Fortification Intensity
6 7 8 (0.2g) 8 (0.3g) 9
Frame System 60 50 (55) 40 (45) 35 24 (25)
Frame-Wall System 130 120 100 80 50
Structural Wall System 140 120 100 80 60
Frame supported Wall System
120 100 80 50 N.A
Frame- Tube System 150 130 100 90 70
Tube in Tube System 180 150 120 100 80
Slab-Column and Wall System
80(40) 70(35) 55(30) 40 N.A
Seismic grading for reinforced concrete buildings
(Table 6.1.2 in GB50011-2010)
Types of structure Seismic fortification intensity
6 7 8 9
Fram structure
Height (m) ≤24 >24 ≤24 >24 ≤24 >24 ≤24
Frames 4th 3rd 3rd 2nd 2nd 1st 1st
Large span frames
3rd 2nd 1st 1st
Wall-Frame
structure
Height (m) ≤60 >60 ≤24 25~60 >60 ≤24 25~60 >60 ≤24 25~60
Frames 4th 3rd 4th 3rd 2nd 3rd 2nd 1st 2nd 1st
Structural walls
3rd 3rd 2nd 2nd 1st 1st
Structural wall
structure
Height (m) ≤80 >80 ≤24 25~80 >80 ≤24 25~80 >80 ≤24 25~60
Structural walls
4th 3rd 4th 3rd 2nd 3rd 2nd 1st 2nd 1st
be continued
Types of structure Seismic fortification intensity
6 7 8 9
Frame -supported
wall structure
Height (m) ≤80 >80 ≤24 25~80
>80 ≤24 25~80
Struc-tural walls
General 4th 3rd 4th 3rd 2nd 3rd 2nd
Streng-thening
3rd 2nd 3rd 2nd 1stI 2nd 1st
Frames that supporting
walls
2nd 2nd 1st 1st
Framed-tube structure
Frame 3rd 2nd 1st 1st
Tube 2nd 2nd 1st 1st
Tube in tube structure
Exterior tube 3rd 2nd 1st 1st
Interior tube 3rd 2nd 1st 1st
Slab-column-wall structure
Height (m) ≤35 >35 ≤35 >35 ≤35 >35
Columns 3rd 2nd 2nd 2nd 1st
Walls 2nd 2nd 2nd 1st 2nd 1st
(Table 6.1.2 in GB50011-2010)
Seismic design flow chart
6.4 Seismic design of RC
frames
Earthquake action and responses
22
Three kinds of Calculation method
• Base Shear Method
• Response Spectrum Method
• Time History Analysys Method
To get the lateral force to every storey of
the building struture, then
To get the shear force of every storey of
the building struture
The Min Value of story shear force
Determination of story shear force
(GB 50011-2010, 5.2.5) The horizontal seismic shear force at each floor level of structure shall be comply with the requirement of the following equations:
n
ij
jeki GV
6 7 8 9
Structures with obvious torsion effect or fundermental period is less than 3.5s
0.008 0.016
(0.024)
0.032
(0.048)
0.064
Structures with fundermental period greater than 5s
0.006 0.012
(0.018)
0.024
(0.036)
0.048
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the shear force of the k-th column in i-th floor
where
Vik is the horizontal shear force induced in column k on i th story;
Dik is relative shear stiffness of column k on i th story;
in
k
ik
ikik V
D
DV
1
(5.7)
n
k
ikD1
is summation of relative shear stiffness of
total columns on i th story.
D Value methods 2
c
h
i12D
Determine the internal forces of
lateral-force-resisting frames :
Determine flexural stiffness for beam and column;
Calculate D value;
Determine shear force for each column;
Determine the position of the point of contra-flexure in column;
Calculate column moments, then derive the beam moments based on
equilibrium of beam to column joint, and distribution of beam stiffness
/hVM ikkl (5.11)
)( /hhVM ikku (5.12)
6.4 Seismic design of RC frames
6.4.1 Principle of seismic design
Three Principles of seismic design
strong shear - weak flexure
strong column - weak beam
strong joints - weak members
M-φ Curve
Lesson from this Diagram
M-φ Curve of a beam
Lesson from this Curve
Ductility and compress ratio
Lesson from these picture
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6.4.2 Shear resistance of members
Members in a Frame:
BEAM, COLUMN, and JOINT
What's different from beam and column?
A member subjected to an axial force of 0.1fc Ag or less can be treated as a beam (Ag = the gross area of the section, fc = design compression strength for concrete).
To dirtermin size of a member's section we should:
),,(0 ccRE fVfbh
• Shear compression ratio : defined as nominal shear stress
divided by design compression strength of concrete (cylinder
strength), or V/fcbh0, to quantify a nominal shear stress across a
beam section
• To ensure that premature diagonal compression failure not
occur before the onset of yielding of shear reinforcement, the
diagonal compression principal stress should be limited.
6.4.2 Shear resistance of members
For Beam:If ratio of span to depth (跨高比) is greater than 2.5,
For Column and Wall:If shear span ratio (剪跨比) is greater than 2,
the design shear force should satisfy the following equation:
ccREccRE
cc
RE
fVfVbh
bhfV
/50.20/
)20.0(1
0
0
6.4.2 Shear resistance of members
)15.0(1
0bhfV cc
RE
For Beam:If ratio of span to depth (跨高比) is NOT greater than 2.5,
For Column and Wall:If shear span ratio (剪跨比) is NOT greater
than 2, the design shear force should satisfy the following equation:
)30.0(1
jjcj
RE
j hbfV
(5.46) For Joint:
6.4.2 Shear resistance of members
To ensure ductile flexural failure and prevent brittle shear failure, a principle of “strong shear—weak flexure” should be followed in seismic design of reinforced concrete beams.
Gb
n
r
b
l
bvb V
l
MMV
6.4.3 strong shear—weak flexure
For beams in grade 1 frames, is 1.3.
For beams in grade 2 frames, is 1.2.
For beams in grade 3 frames, is 1.1. vbvb
vb
Gb
n
r
bua
l
bua Vl
MMV
1.1
For beams in grade 1 and earhtquake intensity 9,the beam
and lintel may not be adjusted,but it should satisfy the
following requirement.
Shear strength of beams
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Considering shear strength reduction in the diagonal plan of
beam under the repeated reversal loading, the shear strength in
diagonal plan should be checked as the follows
)25.142.0(1
00 hs
AfbhfV sv
yvt
RE
b
)1
05.1(
100 h
s
AfbhfV sv
yvt
RE
b
Shear strength of beams
Principle of strong shear and weak flexure
To ensure ductile flexural failure and prevent brittle
shear failure
n
b
c
t
cvc
H
MMV
(5.37)
Shear strength of columns
For columns in grade 1 frames, is 1.5, the others are 1.4
For columns in grade 2 frames, is 1.3. 1.2
For columns in grade 3 frames, is 1.2. 1.1
For columns in grade 4 frames, is 1.1. 1.1
vc
vc
vc
vc
n
t
cua
b
cua
H
MMV
2.1
For columns in grade 1 and earhtquake intensity 9, the
columns may not be adjusted, but it should satisfy the
following requirement.
Shear strength of columns shear resistance of columns
)056.0
1
05.1(
100
Nhs
AfhbfV
c
sv
yvcct
RE
col
)2.01
05.1(
100
Nhs
AfhbfV
c
sv
yvcct
RE
col
Design value of bending moments of members around a
beam-column joint should satisfy the following requirements:
bcc MM
For frame structures in grades 1,2,3 and 4, is 1.7,1.5,1.3 and 1.2;
For other types of frame, is 1.4 in grade 1,1.2 in grade 2,1.1 in grade 3and 4.
c c c c
cc
c
6.4.5 strong column—weak beam
Mc,up
Mc,low
Mb,r
Mb,l
Vc,up
Vb,l
Vb,r
Nc,up
Nc,low
Vc,low Vc,low
Vc,low
Beam
Column
buac MM 2.1
For frame structures in grade 1 and earhtquake intensity 9, the
design value of combined bending moment of menbers
should satisfy the following requirement.
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6.4.5 strong joints-weak members
(1) Design of beam-column joints
The design of beam-column joints is primarily
aimed at:
(i) preserving the integrity of the joint so that the strength
and deformation capacity of the connected beams and
columns can be developed and substantially maintained.
(ii) preventing significant degradation of the joint stiffness
due to cracking of the joint and loss of bond between
concrete and the longitudinal column and beam
reinforcement or anchorage failure of beam reinforcement.
bc
sbo
sb
bjb
jhH
ah
ah
MV
/
/
0
1 (5.41)
bc
sbo
sb
bua
jhH
ah
ah
MV
/
/
0
115.1
(5.42)
(2) Design value of shear force
(b) Shear strength checking of beam-column joint
s
ahAf
b
bNhbfV sb
svjyv
c
j
jjjtj
RE
j05.01.1
1
47.5
(3) Seismic shear strength checking of joint core
s
ahAfhbfV sb
svjyvjjtj
RE
j09.0
1
48.5
6.4.6 Lateral deflection evaluation
• Drift, or inter-story displacement should be limited
(causes non-structural damage and human discomfort
and secondary stress in the main structure)
• To provent the structure collapsed.
(1) Checking elastic floor drifts under frequently accured earthquake,
(2) Checking plastic floor drifts under rare accured earthquake,
A simplified equation to calculate elastic-plastic story drift is the follows:
hu pp (5.50)
epp uu (5.51)
hu ee
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