11
By Denis Mitchell
and Patrick Paultre
Seismic Design
11.1
11.2
11.3
11.4
11.4.1
11.4.2
11.4.2.1
11.4.2.2
11.4.3
11.4.4
11.4.4.1
11.4.4.2
11.4.4.3
11.4.4.4
11.4.4.5
11.4.4.6
11.4.4.7
11.4.5
11.4.5.1
11.4.5.2
11.4.5.3
11.4.5.4
11.4.5.5
11.4.5.6
11.4.6
11.4.6.1
11.4.6.2
11.4.6.3
11.4.6.4
11.5
11.5.1
11.5.2
11.5.3
11.5.3.1
Introduction..................................................................................... 11–3
Seismic Design Considerations ..................................................... 11–5
Loading Cases ............................................................................... 11–5
Design of a Six-Storey Ductile Moment-Resisting
Frame Building ............................................................................... 11–6
Description of Building and Loads.................................................. 11–6
Determination of Design Forces..................................................... 11–7
Gravity Loading .............................................................................. 11–7
Seismic Loading ............................................................................. 11–8
Deflections, Drift Ratios and Torsional Sensitivity........................ 11–12
Design of Ductile Beam................................................................ 11–12
Determination of Design Moments............................................... 11–13
Moment Redistribution and Moment Envelopes .......................... 11–13
Design of Flexural Reinforcement at Critical Sections ................. 11–14
Design of Transverse Reinforcement in Beams .......................... 11–17
Checking Extent of Plastic Hinging .............................................. 11–19
Bar Cut-offs .................................................................................. 11–19
Splice Details................................................................................ 11–19
Design of Interior Ductile Column ................................................ 11–21
Column End-Actions from Analysis .............................................. 11–21
Factored Axial Loads and Moments............................................. 11–22
Preliminary Selection of Column Reinforcement.......................... 11–22
"Strong Column - Weak Beam" Requirement .............................. 11–23
Design of Transverse Reinforcement in Column ......................... 11–24
Splice Details................................................................................ 11–28
Design of Interior Beam-Column Joint ......................................... 11–28
Determination of Factored Forces in Joint ................................... 11–28
Check Factored Shear Resistance of Joint.................................. 11–30
Transverse Reinforcement Required in Joint............................... 11–30
Bond of Beam Bars ...................................................................... 11–31
Analysis of a Ductile Core-Wall Structure .................................... 11–31
Description of Building and Loads................................................ 11–31
Analysis Assumptions .................................................................. 11–32
Seismic Loading ........................................................................... 11–32
Minimum Lateral Earthquake Force ............................................. 11–33
11–2
Seismic Design
11.5.3.2
11.5.3.3
11.5.3.4
11.5.3.5
11.5.3.6
11.5.4
11.5.5
11.5.5.1
11.5.5.2
11.5.5.3
11.5.6
11.5.6.1
11.5.6.2
11.5.6.3
11.5.6.4
11.5.6.5
11.5.6.6
Accidental Torsion........................................................................ 11–35
Degree of Coupling ...................................................................... 11–35
Check on Structural Irregularity.................................................... 11–36
Dynamic Analysis ......................................................................... 11–36
Deflections and Drift Ratios.......................................................... 11–40
Design Forces .............................................................................. 11–40
Design of Coupling Beams........................................................... 11–45
Design Forces for Coupling Beams ............................................. 11–45
Design and Detailing of Coupling Beams..................................... 11–46
Ductility of Coupling Beams ......................................................... 11–47
Design of Ductile Walls ................................................................ 11–47
Design Forces in E-W Direction ................................................... 11–47
Design Forces in N-S Direction .................................................... 11–49
Design of Base of Wall for Flexure and Axial Load...................... 11–49
Ductility of Walls ........................................................................... 11–53
Checking Wall Thickness for Stability (Clause 21.6.3) ................ 11–53
“Buckling Prevention” Ties for Concentrated Reinforcement
(Clause 21.6.6.9).......................................................................... 11–54
11.5.6.7 Design for Shear at Base of Walls (Clause 21.6.9) ..................... 11–54
11.5.6.8 Checking Sliding Shear Resistance at Construction Joints
(Clause 21.6.9.4).......................................................................... 11–56
11.5.6.9 Determination of Plastic Hinge Region (Clause 21.6.2) ............... 11–57
11.5.6.10 Changes in Horizontal Distributed Reinforcement Over the
Height of the Walls (Clause 21.6.5) ............................................. 11–57
11.5.6.11 Changes in Vertical Distributed Reinforcement Over the
Height of the Walls (Clause 21.6.5) ............................................. 11–58
11.5.6.12 Changes in Concentrated Vertical Reinforcement Over the
Height of the Walls (Clause 21.6.6) ............................................. 11–58
11.5.7
Frame Members Not Considered Part of the SFRS .................... 11–58
11.5.7.1 Slab-Column Connections (Clause 21.12.3) ................................ 11–58
11.5.7.2 Check on Design And Detailing of Columns (Clause 21.12)........ 11–59
11.5.8
Comparisons With the Design Using the 1994 CSA Standard .... 11–59
11.5.9
References ................................................................................... 11–59
CAC Concrete Design Handbook
11.1
11–3
Introduction
The 2005 National Building Code of Canada (NBCC) gives the minimum lateral earthquake
force for the equivalent static force procedure as:
V =
S (Ta ) M v I E W S (2.0 )M v I E W
≥
Rd Ro
Rd Ro
and for a Seismic Force Resisting System (SFRS) with an R d equal to or greater than 1.5 ,
2 S (0.2)IEW
V need not be taken greater than
3 Rd Ro
where:
S (Ta ) =
Mv
IE
Ta
=
=
=
W
=
Rd
=
Ro
=
design spectral response acceleration, expressed as a ratio to gravitational
acceleration for a period of Ta
factor to account for higher mode effect on base shear
earthquake importance factor for the structure
fundamental period of vibration of the building in seconds in the direction under
consideration
dead load, plus 25% of the design snow load, plus 60% of storage load and the full
contents of any tanks. Minimum partition load need not exceed 0.5 kPa
ductility-related force modification factor that reflects the capability of a structure to
dissipate energy through inelastic behaviour
overstrength-related force modification factor that accounts for the dependable
portion of reserve strength in a structure.
The designer chooses the type of SFRS, with the corresponding force modification factors,
R d and R o . The values of R d and R o are a function of the type of lateral load resisting system
and the manner in which the structural members are designed and detailed. Table 11.1 provides
a guide for the required design and detailing provisions of CSA Standard A23.3 associated with
the corresponding factors, R d and R o .
11–4
Seismic Design
Table 11.1 Design and Detailing Provisions Required for Different Reinforced Concrete
Structural Systems and Corresponding R d and R o Factors
Type of SFRS
Rd
Ro
Ductile
moment
resisting
frames
4.0
1.7
Moderately
ductile
moment
resisting
frames
Ductile
coupled walls
2.5
1.4
4.0
1.7
Ductile
partially
coupled walls
3.5
1.7
3.5
1.6
Moderately
ductile
shearwalls
Conventional
construction:
Moment
resisting
frames
2.0
1.4
1.5
1.3
Conventional
construction:
Shearwalls
1.5
1.3
Ductile
shearwalls
Summary of design and detailing requirements in CSA
A23.3-04
Beams capable of flexural hinging with shear failure and
bar buckling avoided. Beams and columns must satisfy
ductile detailing requirements. Columns properly
confined and stronger than beams. Joints properly
confined and stronger than beams.
Beams and columns must satisfy detailing requirements
for moderate ductility. Beams and columns to have
minimum shear strengths. Joints must satisfy moderate
ductility detailing requirements and must be capable of
transmitting shears from beam hinging.
At least 66% of base overturning moment resisted by
wall system must be carried by axial tension and
compression in coupled walls. Coupling beams to have
ductile detailing and be capable of flexural hinging or
resist loads with diagonal reinforcement (shear failure
and bar buckling avoided). Walls must have minimum
resistance to permit attainment of nominal strength in
coupling beams and minimum ductility level.
Coupling beams to have ductile detailing and be
capable of flexural hinging or resist loads with diagonal
reinforcement (shear failure and bar buckling avoided).
Walls must have minimum resistance to permit
attainment of nominal strength in coupling beams and
minimum ductility level.
Walls must be capable of flexural yielding without local
instability, shear failure or bar buckling. Walls must
satisfy ductile detailing and ductility requirements.
Walls must satisfy detailing and ductility requirements
for moderate ductility. Walls must have minimum shear
strength.
Beams and columns must have factored resistances
greater than or equal to factored loads. Columns and
beams must satisfy minimum detailing requirements for
conventional construction. Closely spaced hoops
required in columns unless factored resistance of
columns greater than factored resistance of beams or if
R d R o = 1.0 .
Walls must have factored resistances greater than or
equal to factored loads. Factored shear resistance must
exceed shear corresponding to factored flexural
resistance or shear corresponding to R d R o = 1.0 .
Walls must satisfy minimum detailing requirements for
conventional construction.
Other
SFRS(s)
1.0
1.0
CAC Concrete Design Handbook
11.2
11–5
Seismic Design Considerations
Seismic design is concerned not only with providing the required strength but also with
providing minimum levels of ductility and choosing appropriate structural systems. These goals
may be achieved by:
(i)
choosing structural systems which are as symmetrical as possible in plan and as
uniform as possible in elevation (minimizing structural irregularities);
(ii) designing the primary lateral load resisting structural components so that desirable
energy dissipating systems will form (e.g., "weak-beam strong-column");
(iii) detailing the energy dissipating regions of the primary lateral load resisting components
to ensure that substantial inelastic deformations can be achieved without significant
loss of strength, and
(iv) ensuring that secondary members which are not part of the lateral load resisting system
can maintain their gravity load carrying capacity as they undergo the required lateral
deformations.
In the design of ductile members it is necessary to determine the hierarchy of strengths of
different members. To ensure that certain hierarchy of strengths are achieved the CSA Standard
defines "probable", "nominal" and "factored" resistances. Table 11.2 summarizes the various
types of flexural resistances used in the CSA Standard and suggests approximate relationships
between these resistances.
Table 11.2 Factored, Nominal and Probable Moment Resistances
Type of flexural
resistance
Calculated
using
Where used
M r = factored
resistance
φ c = 0.65
φ s = 0.85
All members must
satisfy M r ≥ M f
M n = nominal
resistance
φ c = 1.0
φ s = 1.0
To ensure columns
stronger than beams
M p = probable
φ c = 1.0
φ s = 1.0
resistance
Approximate
relationships for
flexure
M n ≈ 1.2M r
M p ≈ 1.47M r
fs = 1.25f y
Note: the relationship between M n and M r for the case of flexure and axial load depends on the
level of axial load
11.3
Loading Cases
For loading combinations including earthquake, the factored load combinations shall include:
Principal loads:
1.0D + 1.0E
and either of the following combinations of principal and companion loads:
1) For storage occupancies, equipment areas and service rooms:
1.0D + 1.0E + 1.0L + 0.25S
2) For other occupancies:
1.0D + 1.0E + 0.5L + 0.25S
11–6
Seismic Design
11.4
11.4.1
Design of a Six-Storey Ductile Moment-Resisting Frame
Building
Description of Building and Loads
The six-storey reinforced concrete frame building shown in Fig. 11.1 is located in Vancouver
and is to be designed as a ductile moment resisting frame structure. The six-storey reinforced
concrete office building has 7 - 6 m bays in the N-S direction and 3 bays in the E-W direction which
consist of 2 - 9 m office bays and a central 6 m corridor bay.
The interior columns are all 500 x 500 mm while the exterior columns are 450 x 450 mm.
The one-way slab floor system consists of a slab 110 mm thick spanning in the E-W direction
supported by beams in the N-S direction. The secondary beams supporting the slab are 300 mm
wide x 350 mm deep (from top of slab to bottom of beam). The beams of both the N-S and E-W
frames are 400 mm wide x 600 mm deep for the first three storeys and 400 x 550 mm for the top
three storeys.
Material Properties
Concrete: normal density concrete with f c′ = 30 MPa
Reinforcement: f y = 400 MPa
Live loads
Floor live loads:
2
2.4 kN/m on typical office floors
2
4.8 kN/m on 6 m wide corridor bay
Roof load
2
2.2 kN/m snow load, accounting for parapets and equipment projections
2
1.6 kN/m mechanical services loading in 6 m wide strip over corridor bay
Dead loads
3
self-weight of reinforced concrete members calculated as 24 kN/m
2
1.0 kN/m partition loading on all floors
2
0.5 kN/m mechanical services loading on all floors
2
0.5 kN/m roofing
Wind loading
2
1.84 kN/m net lateral pressure for top 4 storeys
2
1.75 kN/m net lateral pressure for bottom 2 storeys
The fire-resistance rating of the building is assumed to be 1 hour.
CAC Concrete Design Handbook
11–7
Fig. 11.1 Six-storey structure located in Vancouver
11.4.2
Determination of Design Forces
11.4.2.1 Gravity Loading
To determine the member forces, the structure was analyzed using ETABS. To make
allowances for cracking, member stiffnesses were assumed to be 0.4 of the gross I for all beams
as required by CSA A23.3. To account for the influence of the axial load level on the column
stiffnesses, average estimated cracked moments of inertia of 0.6 and 0.7 of I g were used for the
columns in the top three storeys and bottom three storeys, respectively (Clause 21.2.5.2). The
analysis models and the gravity loading are illustrated in Fig. 11.2.
11–8
Seismic Design
To illustrate the requirements for the design of a ductile moment-resisting frame,
components of a typical interior E-W frame will be designed in the following examples.
Fig. 11.2 Unfactored loading cases considered in design of typical interior frame
11.4.2.2 Seismic Loading
Minimum Lateral Earthquake Force
The structure is located in Vancouver and is founded on very dense soil and soft rock.
Therefore the site classification is “C” and the acceleration-based and velocity-based site coefficients
are Fa = 1.0 and Fv = 1.0 , respectively.
The seismic response factor, S, is dependent on the fundamental period, T, of the structure.
The 5% damped spectral response accelerations, Sa (0.2) , Sa (0.5 ) , Sa (1.0 ) and Sa (2.0 ) are 0.94,
0.64, 0.33 and 0.17, respectively. The design spectral response accelerations are given by the
product of the site coefficients and S a as shown in Fig. 11.3.
Figure 11.3 Design spectral response acceleration
The empirical fundamental lateral period, Ta , for concrete moment frames is given by:
Ta = 0.075hn3 / 4 = 0.075 × 21.9 3 / 4 = 0.759 s
CAC Concrete Design Handbook
11–9
The calculated period for this structure, using the computer program ETABS is 1.35 s. The
value of the fundamental lateral period cannot be taken greater than 1.5 × 0.759 = 1.139 , and
hence use Ta = 1.14 s. The corresponding value of S (Ta ) is 0.308 (see Fig. 11.3).
For this office building, the earthquake importance factor, I E = 1.0 . The values of M v and
J depend on the ratio of Sa (0.2) / Sa (2.0 ) = 0.94 / 0.17 = 5.53 and the value of Ta . For this case
M v = 1.0 and J = 1.0 .
For this ductile moment resisting frame structure R d = 4.0 and R o = 1.7 . Hence the seismic
base shear, V, is:
V =
S (Ta ) Mv IEW 0.308 × 1.0 × 1.0 × W
=
= 0.0453W
Rd Ro
4.0 × 1.7
Vmin =
S (2.0 ) Mv I EW 0.17 × 1.0 × 1.0 × W
=
= 0.025W
Rd Ro
4.0 × 1.7
Vmax =
2 S (0.2)IEW
2 0.94 × 1.0 × W
=
= 0.092W
3 Rd Ro
3
4.0 × 1.7
For this structure, W = 44,765 kN.
Hence, V = 0.0453W = 0.0453 × 44,765 = 2026 kN.
The
portion
of
V concentrated
at
the
top
of
the
building
is
kN, but need not be taken greater
Ft = 0.07Ta V = 0.07 × 1.139 × 2026.1 = 161.5
than 0.25V = 0.25 × 2026.1 = 506.5 kN.
The calculations of the seismic lateral forces at each floor level are summarized in Table
11.3.
Table 11.3 Lateral Load Calculations for Each Floor Level
Floor
6 roof
5
4
3
2
1
0
Total
hx ,
m
21.90
18.25
14.60
10.95
7.30
3.65
0.00
-
Wx ,
kN
7457
7365
7365
7526
7526
7526
44,765
hxW x ,
kN·m
163,308
134,411
107,529
82,410
54,940
27,470
570,068
Fx ,
kN
696
440
352
270
180
90
0
2,026
Vx ,
kN
0
696
1,135
1,487
1,757
1,936
2,026
-
Tx ,
kN·m
2953
1866
1493
1144
763
381
8601
Accidental Torsion
The 3-D model shown in Fig. 11.4 was used to calculate accidental torsional effects by
applying the lateral forces Fx (see Table 11.3) at an accidental eccentricity of ±0.1D nx , where
Dnx is the plan dimension of the building at level x, perpendicular to the direction of seismic
loading. This gives an accidental torsional eccentricity of 4.245 m, from the centre of mass (same
as centre of rigidity) for loading in the E-W direction. The resulting floor torques, T x , are given in
Table 11.3.
11–10
Seismic Design
Dynamic Analysis
This symmetrical structure has no structural irregularities in the vertical or horizontal
directions and in addition is not sensitive to torsion (see Section 11.4.3). Therefore, In accordance
with NBCC a dynamic analysis is not required. However, a dynamic analysis was carried to
determine the lateral period of vibration (see above). This dynamic analysis was also used to
determine the design forces for the members and to estimate the lateral displacements. The
purpose of carrying out a dynamic analysis in this example is to illustrate the approach required
and to obtain a more realistic design force distribution.
The first step is to determine Ve from a linear dynamic analysis. The design base shear
Vd is obtained from:
Vd =
Ve
IE
Rd Ro
However for this regular structure, Vd shall not be taken less than 0.8V .
All forces and deflections obtained from the linear dynamic analysis are scaled by the factor
Vd / Ve to obtain the design values. However, in order to obtain realistic values of anticipated
deflections and drifts, the design values need to be multiplied by R d R o / I E .
Fig. 11.4 shows the 3-D model for the dynamic analysis, using ETABS. In the analysis, rigid
end offsets were used to simulate the dimensions of the joints and rigid diaphragms were
assumed. The total mass for each floor was concentrated at the centre of mass (coincident with
the centre of rigidity for this structure). To account for sway effects (P-Delta) the ETABS program
option, accounting for second order effects by the addition of the so-called geometric stiffness,
which is a function of the compression forces in the columns from gravity loads, was used. These
compressive forces were obtained from the consistent loading case of 1.0D + 0.5L + 0.25S , with
live load reduction factors.
The first three lateral modes in the E-W direction are shown in Fig. 11.5, together with the
associated periods of vibration and the modal participating mass ratios. Note that the sum of
these ratios is 96.9% of the total mass and hence exceeds the minimum required ratio of 90% of
the total mass (NBCC). Spectral modal superposition, using SRSS for the first three modes in the
E-W direction was used to determine all forces and deformations.
CAC Concrete Design Handbook
11–11
Figure 11.4 3-D Model used for dynamic analysis
Mode 1
T = 1.349
MPMR = 0.82
Mode 2
T = 0.453
MPMR = 0.11
Mode 3
T = 0.250
MPMR = 0.04
Figure 11.5 Mode shapes, corresponding lateral periods of vibration and modal
participating mass ratios
The base shear determined by dynamic analysis is Ve = 10690 kN.
Therefore:
10690
1.0 = 1572 kN
4.0 × 1.7
However for this regular building Vd shall not be taken less than 0.8V = 0.8 × 2026 .1 = 1620.9 kN.
Hence, all forces and deflections obtained from the dynamic analysis shall be multiplied by
Vd / Ve = 1620 .9 / 10690 = 0.152 .
Vd =
11–12
11.4.3
Seismic Design
Deflections, Drift Ratios and Torsional Sensitivity
The deflections obtained from the dynamic analysis need to be multiplied by the factor 0.152
to account for the total anticipated displacements, including the inelastic effects. It is necessary to
multiply these deflections by the factor R d R o / I E to obtain the design values. Note that the
deflections obtained from dynamic analysis include P-Delta effects. The deflections arising from
accidental torsional eccentricity shall be added to the deflections from the dynamic analysis.
To determine if the structure is sensitive to torsion, the value of B x is determined from the
maximum and average displacements of the structure at level x in the E-W and N-S directions.
The maximum value, B , of the B x values is at the first floor level for loading in the E-W direction
(an average displacement of 5.1 mm and a maximum displacement of 6.8 mm), giving:
δ
6.8
B = max =
= 1.35
5.1
δ ave
Because B is less than 1.7, the structure is not sensitive to torsion.
The maximum interstorey drift ratio occurs in Frames 1 and 8 in the second storey for the EW direction of loading. From the dynamic analysis the maximum interstorey drift ratio is 0.0015
and the interstorey drift ratio from accidental torsion at this level is 0.0007, for a maximum
interstorey drift ratio of 0.0022. Therefore the anticipated interstorey drift ratio, including inelastic
effects is 0.0022 × 4.0 × 1.7 / 1.0 = 0.0146 . This anticipated maximum interstorey drift ratio is less
than the NBCC limit of 0.025.
11.4.4
Design of Ductile Beam
To illustrate the procedures involved in designing a beam in a ductile moment-resisting
frame, a typical first storey interior beam will be designed below. For illustration purposes frame 2
will be designed. This frame, although it has a smaller torsional shear than frame 1, will require
more reinforcement than frame 1 because it carries larger dead and live loads. The details of the
beam and column framing are given in Fig. 11.6.
Fig. 11.6 Typical beam and column framing
CAC Concrete Design Handbook
11–13
11.4.4.1 Determination of Design Moments
The moments in the beams resulting from dead load, D, live load, L, and earthquake loading,
E, as determined from frame analyses, are in Fig. 11.7. Note that the moments are given at the
face of the columns. Since most of the gravity loading in beams AB, BC and CD is introduced at
the locations of the secondary beams, the small uniformly distributed loading has been
approximated by additional concentrated loads at the secondary beam locations.
Fig. 11.7 Loading cases on typical interior beam at second floor level
Table 11.4 gives the unfactored moments at critical locations and also gives the factored moment
combinations which need to be considered.
11.4.4.2 Moment Redistribution and Moment Envelopes
Instead of designing each of the critical sections for the maximum factored moments given in
Table 11.4, moment redistribution will be used to reduce some of the maximum design moments.
Since the beams in a ductile moment-resisting frame structure are designed and detailed to
exhibit considerable ductility the maximum redistribution of 20%, permitted by Clause 9.2.4, will be
used.
While it is possible to redistribute the earthquake moments, care must be exercised to
ensure that the total column shears in any one storey remain unchanged after redistribution. A
simpler approach is to redistribute only the dead load and live load moments.
In order to reduce the magnitude of the negative moments at location BA, the dead and live
load support moments are reduced at this location by the maximum permitted amount (20%). The
positive moments in span AB are increased by the appropriate amounts. The resulting moments
are summarized in Table 11.5.
11–14
Seismic Design
Table 11.4 Moments at Critical Locations (kN·m) Before Redistribution
D
L
Accidental Torsion
E without accidental
torsion
E with accidental
torsion
1.25D+1.5L
1.0D+1.0E
1.0D-1.0E
1.0D+0.5L+1.0E
1.0D+0.5L-1.0E
AB
-167
-66
± 30
± 98
a
+119
+50
± 11
± 36
b
+105
+45
±9
± 30
BA
-206
-90
± 28
± 92
BC
-104
-79
± 41
± 131
c
+29
+41
± 15
± 48
± 128
± 47
± 39
± 120
± 172
± 63
-308
-39
-295
-72
-328
+224
+166
+72
+191
+97
+199
+144
+66
+167
+88
-393
-86
-326
-131
-371
-249
+68
-276
+29
-316
+98
+92
-34
+112
-13
Note: controlling load combinations shown in bold
Table 11.5 Moments at Critical Locations (kN·m) after Redistribution
D
L
Accidental Torsion
E without accidental
torsion
E with accidental
torsion
1.25D+1.5L
1.0D+1.0E
1.0D-1.0E
1.0D+0.5L+1.0E
1.0D+0.5L-1.0E
AB
-167
-66
± 30
± 98
a
+132
+56
± 11
± 36
b
+133
+57
±9
± 30
BA
-165
-72
± 28
± 92
BC
-104
-79
± 41
± 131
c
+29
+41
± 15
± 48
± 128
± 47
± 39
± 120
± 172
± 63
-308
-39
-295
-72
-328
+249
+180
+85
+208
+113
+252
+172
+94
+201
+122
-314
-45
-284
-81
-320
-249
+68
-276
+29
-316
+98
+92
-34
+112
-13
Note: controlling load combinations shown in bold
It is noted that, after redistribution, earthquake loading governs at all negative moment
sections at the second floor level.
11.4.4.3 Design of Flexural Reinforcement at Critical Sections
Top bars at column faces
In deciding on the appropriate top reinforcement, note that Clause 21.5.5.6 limits the
diameter, db, passing through the joint to l j / 24 for this normal density concrete structure and
uncoated bars. Thus for this case the maximum diameter of beam bars passing through the
interior columns is 500 / 24 = 21 mm. Hence the maximum beam bar size is 20M.
At column A, a factored moment resistance of at least 328 kN·m is required. Assuming a
flexural lever arm of 0.75h = 0.75 × 0.600 = 0.450 m, the required area of top bars would be
2
328 × 1000 / (0.450 × 0.85 × 400 ) = 2144 mm . If it is assumed that slab reinforcement within a
distance of 3hf from the sides of the beam is effective, then 4–10M bars in the slab are effective.
CAC Concrete Design Handbook
11–15
It is assumed that these 10M bars in the flange are effective under reversed cyclic loading even
though there is no anti-buckling reinforcement for these bars. Note that larger bars may not be
2
effective. The additional reinforcement required is then 1744 mm . Note that it is unwise to be too
conservative when designing the top reinforcement since beam shears, joint shears, column
moments and column shears are all increased if the flexural capacity at the end of the beam is
increased.
Let us try an arrangement of 6–20M bars as shown in Fig. 11.8. Keeping in mind that the
positive moment resistance of the beam needs to be at least one-half of the negative moment
resistance, try using 4–20M bars on the bottom of the beam.
Fig. 11.8 Beam cross section near column face
Accounting for the presence of the compression reinforcement, the depth of compression, c,
is found to be 92 mm and the factored negative moment resistance is 359 kN·m. Hence the
moment capacity is satisfactory.
The required minimum top and bottom reinforcement, As ,min , from Clause 21.3.2.1 is:
2
As ,min = 1.4bw d / f y = 1.4 × 400 × 527 / 400 = 738 mm ≤ 1200 mm
2
O.K.
The maximum reinforcement permitted is:
2
0.025bw d = 0.025 × 400 × 527 = 5270 mm ≥ 2200 mm
2
O.K.
Note that in choosing the arrangement of the beam bars at column faces the following
factors must be considered:
(a) the need to restrain the longitudinal bars from buckling by providing lateral restraint in the
form of hoops and ties.
(b) the need to pass the beam bars through the column cage, and
(c) the need to provide adequate space between top bars to permit placement and vibration of
concrete.
Since the magnitudes of the negative moment resistances required at column faces AB, BA
and BC are all about the same, we will use the same reinforcing arrangement at these three
locations.
11–16
Seismic Design
Bottom bars for positive moment regions
Span BC:
For span BC, the effective compressive flange width is 1600 mm (see Clause 10.3.3). For
2
4–20M bars ( As = 1200 mm ), M r at the column face, accounting for the large amount of top
reinforcement, is 229 kN·m which is larger than one half of 359 kN·m (i.e., M r at column face
2
where As = 2200 mm ).
As 4–20M bottom bars are provided at column faces AB, BA and BC, use 4–20M bars in
span BC. The positive moment resistance M r is 229 kN·m in the midspan regions of span BC.
As 229 kN·m exceeds 112 kN·m (Table 11.5), 4-20M bars will be satisfactory.
Span AB:
For span AB, the effective compressive flange width is 2200 mm (see Clause 10.3.3). For
M r ≥ 252 kN·m (Table 11.5) try 6- 20M bottom bars. Neglecting the top reinforcement, the depth
of the equivalent rectangular stress block is 18 mm and M r = 315 kN·m (see Fig. 11.9).
Accounting for the large amount of top reinforcement, the positive moment M r at the column face
is 246 kN·m.
Fig. 11.9 Positive and negative moment capacities of beam
CAC Concrete Design Handbook
11–17
11.4.4.4 Design of Transverse Reinforcement in Beams
Shear requirements
Determine the shears corresponding to the development of flexural hinging at both ends of
the beam. For the chosen reinforcement at the beam ends the probable moment resistances are:
(i)
−
Probable negative moment resistance, M pr
−
Using a strain compatibility approach to calculate M pr
results in a depth of compression,
−
c = 95 mm, a stress block depth of 85 mm and M pr
= 528 kN·m (see Fig. 11.9). Note that the
probable moment resistance of the beams can be estimated by multiplying M r by the ratio
1.25 / 0.85 = 1.47 . In this case M pr would be 359 × 1.47 = 528 kN·m. This simple approach is
sufficiently accurate for design purposes.
(ii)
+
Probable positive moment resistance, M pr
The factored moment resistance at the ends in span BC is 229 kN·m and hence we can
+
estimate the probable moment resistance as M pr
= 1.47 × 229 = 337 kN·m. Similarly, the
probable moment resistance at the ends of span AB is 1.47 × 246 = 362 kN·m.
(iii) Determine factored shears
The shear diagrams shown in Fig. 11.10 are drawn for lateral forces acting in the west
direction. For lateral forces acting in the east direction the shear diagrams will be "mirror images"
of those shown (e.g., the shear at B would be 234 kN).
Fig. 11.10 Determinaton of shears corresponding to flexural hinging
11–18
Seismic Design
Using θ = 45° and β = 0 gives (Clause 21.3.4.2):
Vr = φs Av f y d v / s
At the column faces the transverse reinforcement consists of 4–10M legs, hence
2
Av = 400 mm .
At the ends A and B the required spacing for shear is:
s=
0.85 × 400 × 400 × 0.9 × 527
= 276 mm
234 × 1000
At the ends B and C the required spacing for shear is:
s=
234
× 276 = 247 mm.
262
2
If 2-legged 10M stirrups are used as transverse reinforcement ( Av = 200 mm ) the required
spacings in the middle regions of beams AB and BC are 344 mm and 228 mm, respectively.
Other shear design requirements:
(i)
Maximum shear (Clause 11.3.3)
Vr ,max = 0.25φ c f c′ bw d v = 0.25 × 0.65 × 30 × 400 × 0.9 × 527 = 925 kN
(ii)
Minimum amount of stirrups (Clause 11.2.8.2):
for 4 stirrup legs:
s≤
Av f y
0.06 f c′ bw
=
400 x 400
0.06 30 x 400
= 1217 mm
for 2 stirrup legs:
s ≤ 608 mm
(iii) Spacing limits (Clause 11.3.8.3):
Since V f < 0.125 φ c f c′ bw d v = 0.125 x 0.65 X 30 x 400 x 0.9 × 527 = 462 kN
Then s max = 600 mm or 0.7d v = 0.7 × 0.9 × 527 = 332 mm.
Note that near the ends of the beams the stirrup spacing required for shear cannot exceed
276 and 247 mm for spans AB and BC, respectively.
"Anti-buckling” requirements (Clause 21.3.3.2)
Hoops to prevent buckling of longitudinal bars are required over a length of 2d from the face
of the columns. The spacing of the hoops shall not exceed:
(i)
d / 4 = 527 / 4 = 132 mm
(ii)
8d bl = 8 × 20 = 160 mm
(iii)
24d bh = 24 × 10 = 240 mm
(iv) 300 mm
Note that the 4-legged arrangement of transverse reinforcement satisfies Clause 21.3.3.3.
Hence use a spacing of 130 mm for 4-legged hoops over a length of at least
2d = 2 × 527 = 1054 mm.
CAC Concrete Design Handbook
11–19
11.4.4.5 Checking Extent of Plastic Hinging
The moment diagrams corresponding to plastic hinging at both ends of beams AB and BC
are shown in Fig. 11.10. Hinging can spread over a distance of about 3.43 m from the face of
column B in beam AB. Since the earthquake loading can reverse, provide 4-legged hoop
reinforcement spaced at 130 mm as shown in Fig. 11.11. The bottom 4-20M bars can only be
spliced near midspan of the beam (Clause 21.3.2.3). Therefore, to satisfy the maximum spacing
requirements of Clause 21.3.2.3 for regions of lap splices, provide 2-legged hoops spaced at 100
mm in the middle region of beam AB (see Fig. 11.11). For span BC, provide 4-legged hoops at a
spacing of 130 mm over a distance of 2d (1054 mm) from the column faces. Outside of these
regions, the provision of 2-legged hoops at a spacing of 100 mm satisfies both the shear
requirements and the confinement requirements for lap splices (see Fig. 11.11).
11.4.4.6 Bar Cut-offs
The locations of bar cut-offs are determined from the moment diagrams corresponding to the
formation of plastic hinges at the ends of the beams. The theoretical cut-off location is located at
a distance of 1.51 m from the face of the column (see Fig. 11.10). From Clause 12.10.4 it is
required to provide an embedment length beyond the theoretical cut-off point of at least d or 12db.
Hence the minimum length required is 1510 + d = 1510 + 527 = 2037 mm. Continue the 6–20M
top bars a distance of 3 m from the column face such that the bars are terminated in a region of
lower shear. For span BC, extend the 6-20M top bars a distance of 2.0 m from the face of the
column.
11.4.4.7 Splice Details
Flexural reinforcement cannot be spliced within a distance of 2d from the column face nor
within a distance d from of a potential plastic hinge location (Clause 21.3.2.3). In evaluating cutoff locations, d was taken as 527 mm.
In determining locations of bar cut-offs and splices we will consider the moment diagram
corresponding to the formation of hinges at the ends of the beams (see Fig. 11.10). The splices
for the top bars will be located in a region of the beam where the bars are predicted to remain in
compression. However, as it is required to have a minimum negative and positive moment
resistance at the face of the joint (Clause 21.3.2.2) the splice length will be calculated as for a
tension splice.
(a)
Splicing of the 2–20M "continuous" top bars
The required minimum moment capacity along the length of the beam (Clause 21.3.2.2) is
0.25 × 359 = 90 kN·m. M r for 2-20M top bars is 175 kN·m. Hence for the classification of
tension lap splices in accordance with Clause 12.15.2 (As provided)/ As required) is
175 / 90 = 1.94 . Hence Class B splices are required. The development length l d for these top
bars from Table 12-1 is:
fy
400
l d = 0.45k 1k 2 k 3 k 4
d b = 0.45 × 1.3 × 1 × 1 × 0.8 ×
× 20 = 684 mm
f c′
30
Thus the splice length is 1.3 × 684 = 890 mm.
(b) Splicing of the 4–20M "continuous" bars
ld
The development length for these bottom bars is:
fy
400
= 0.45k 1k 2 k 3 k 4
d b = 0.45 × 1 × 1 × 1 × 0.8 ×
× 20 = 526 mm
f c′
30
11–20
Seismic Design
Thus the splice length is 1.3 × 526 = 684 mm.
The details of the reinforcement in the beam are illustrated in Fig. 11.11.
Fig. 11.11 Reinforcement details in beams
CAC Concrete Design Handbook
11.4.5
11–21
Design of Interior Ductile Column
To illustrate the procedures involved in designing a column in a ductile moment resisting
frame a typical first storey interior column in Frame 2 or 7 of the building described in Section 11.4
will be designed. The column that will be designed is shown in Fig. 11.12.
Fig. 11.12 Beam-column framing
11.4.5.1 Column End-Actions from Analysis
The column end actions obtained from analysis are summarized in Table 11.6. The
earthquake forces given are those for lateral seismic forces acting in the E-W direction. For the
live load values, pattern loading and live load reduction values were considered.
Table 11.6 Column End Actions
nd
2 floor bottom
of column
st
1 floor top of
column
st
1 floor bottom of
column
PD
kN
1598
PL
kN
486
PE
kN
±106
MD
kN·m
+41
ML
kN·m
+28
ME
kN·m
±147
VD
kN
+20
VL
kN
+13
VE
kN
±92
1704
556
±137
-53
-37
±112
+28
+19
±94
1704
556
±137
+32
+22
±183
+28
+19
±94
11–22
Seismic Design
11.4.5.2 Factored Axial Loads and Moments
The column must be designed to resist the appropriate combinations of axial load and
moment. From Table 11.6, it is evident that the factored moments at the base of the column will
be larger than that at the top.
For the base of the column at the ground floor level the factored axial load and moment
combinations are given in Table 11.7.
Table 11.7 Factored Axial Load and Moments at Column Bases
Pf
kN
Mf
kN·m
Case 1
1.25D
+1.5L
Case 2
1.0D
+1.0E
Case 3
1.0D
-1.0E
Case 4
1.0D+0.5L
+1.0E
Case 5
1.0D+0.5L
-1.0E
-2964
-1841
-1567
-2119
-1845
+73
+216
-151
+226
-140
Note that for this member Ag fc′ / 10 = (500 × 500 ) × 30 / 10 = 750 kN. As Pf exceeds this
value, the requirements of Clause 21.4 apply (Clause 21.4.1.1).
11.4.5.3 Preliminary Selection of Column Reinforcement
In selecting the column bars, recall that the diameter of these bars must satisfy the
requirements that d b ≤ l j / 24 = 600 / 24 = 25 mm (Clause 21.5.5.6) for this normal density
concrete and for uncoated bars. Hence the maximum bar size is 25M. Try using 8-25M bars as
shown in Fig. 11.13.
Fig. 11.13 Column reinforcement details
2
For this arrangement of reinforcement Ast = 8 × 500 = 4000 mm . From Clause 21.4.3.1
2
the minimum area of longitudinal steel is 0.01 × 500 × 500 = 2500 mm and the maximum area of
longitudinal steel outside of lap splice regions (assuming lap splicing with an equal area of steel)
2
is 0.03 × 500 × 500 = 7500 mm . Hence this steel arrangement satisfies these requirements.
CAC Concrete Design Handbook
11–23
Checking Column Capacity
The axial load-moment interaction diagram for the chosen column section is shown in Fig.
11.14. It can be seen that the column has adequate capacity to resist the various combinations of
Pf and M f which occur at the base of the column.
Fig. 11.14 Pr - M r interaction diagram for column section
Although there is a considerable excess of moment capacity in the column, this additional
capacity is needed at the top of the column in order to ensure that the columns are stronger than
the beams (see below).
11.4.5.4 "Strong Column - Weak Beam" Requirement
The flexural capacity of the columns must exceed the flexural capacity of the beams so that
∑M
nc
≥
∑M
pb
Hence it is necessary to first determine the probable resistances of the beams framing into the
column.
11–24
(a)
Seismic Design
−
Probable negative moment resistance, M pb
Note
that
the
probable
resistance,
−
M pb
,
can
be
approximated
as
1.47M r = 1.47 × 359 = 528 kN·m. This simple approach is sufficiently accurate for design
purposes as can be seen from Fig. 11.9.
(b)
+
Nominal positive moment resistance, M pb
+
= 1.47 × 246 = 362 kN·m.
M pb
(c)
Determination of
To determine
∑M
∑M
nc
nc
for a particular loading case we need to calculate the nominal moment
resistance of the column above and below the beam-column joint. The lowest flexural resistance
will occur at either the highest or lowest axial load, that is, load cases 3 and 4 need to be
investigated (load case 1 does not involve lateral load). The axial load corresponding to cases 3
and 4 are given in Fig. 11.15 along with the column nominal moment resistances corresponding to
these axial loads (from the P-M interaction diagram).
Fig. 11.15 Capacity design of columns and factored loads on column
Thus the requirement that
∑M
nc
≥
∑M
pb
is satisfied. Note that for simplicity the above
calculations have neglected the influence of the beam and column shears acting at the joint faces.
11.4.5.5 Design of Transverse Reinforcement in Column
Shear requirements
The column must have a factored shear resistance, Vr, which exceeds the column shear
corresponding to the probable moment resistance in the beams and which exceeds the shear
CAC Concrete Design Handbook
11–25
forces due to factored loads (Clause 21.4.5.1). From Table 11.6, load case 4 (1.0D + 0.5L + 1.0E )
gives the maximum factored shear of 132 kN.
The moment at the top of the column corresponding to the development of the probable
moment resistances of the beams may be estimated from:
Kc
1 / 3.05
+
−
M c = M pr
+ M pr
×
= (362 + 528 ) ×
= 445 kN·m
2 × 1 / 3.05
Kc
(
)
∑
However since we are designing a ground storey column a different approach is required at
the base of this column. It is assumed that the column frames into a substructure that is
considerably stronger and stiffer than the column and hence the possibility of hinging at the
column base must be accounted for. To ensure adequate column shear capacity, it is necessary
to determine the maximum probable moment resistance corresponding to all axial loads.
Because the axial loads for all the seismic load cases are close to the balanced axial load level,
the moment at the base of the column will be taken as the probable moment resistance
corresponding to the balanced loading conditions (i.e., the highest probable moment resistance
possible). The calculations involved in determining this moment resistance are summarized in
Fig. 11.16.
Fig. 11.16 Determination of factored shear strength in ground storey column
11–26
Seismic Design
The column actions which will correspond to the formation of hinges in the beams at the top
of the column and the formation of a hinge at the base of the column are shown in Fig. 11.16.
From Clause 21.4.5.2 the shear carried by the concrete is determined with values of β and
θ taken from Clause 11 but limited to a maximum of 0.10 and a minimum of 45°, respectively.
For this column containing greater than minimum amounts of transverse reinforcement and
subjected to axial compression, Clause 11.3.6.3 applies, but the limits for β and θ given above
control. The shear resistance attributed to the concrete, assuming that d v = 0.72h , is:
Vc = φ c λβ f c′ bw d v = 0.65 × 0.10 × 30 × 500 × 0.72 × 500 = 64.1 kN
The required Vs is equal to 359 − 64.1 = 294.9 kN. Using the transverse reinforcement
arrangement shown in Fig. 11.13 with square and diamond shaped hoops, the effective area of
2
shear reinforcement is Av = 2 + 2 cos 45 o × 100 = 3.41× 100 = 341 mm . Hence, the required
stirrup spacing can be found from Equation (11-7) as:
(
s=
φ s Av f y d v cot θ
Vs
=
)
0.85 × 341 × 400 × 0.72 × 500 × cot 45 o
= 142 mm
294.9 × 1000
Since Vf of 359 kN is less than 0.125λφ c fc′bw d v = 439 kN, then from Clause 11.3.8.1, the
maximum spacing of the shear reinforcement is the smaller of 0.7 × 0.72 × 500 = 252 mm or
600 mm.
In order to satisfy the minimum shear reinforcement requirements of Clause 11.2.8.2, the
maximum spacing of the 10M stirrups is:
s=
Av f y
0.06 f c′ bw
=
341 × 400
0.06 × 30 × 500
= 830 mm
Therefore for shear a spacing of 142 mm controls.
(b) Confinement requirements
Since the column under consideration is at the base of the structure confinement
reinforcement must be provided over the full height of the column (Clause 21.4.4.6).
From Clause 21.4.4.2, the total cross-sectional area of rectangular hoop reinforcement
depends on the following factors:
kn =
nl
8
=
= 1.33
nl − 2 8 − 2
(
)
(
)
Po = α1fc′ Ag − Ast + fy Ast = 0.805 × 30 500 2 − 4000 + 400 × 4000 = 7541 kN
k p = Pf / Po = 2119 / 7541 = 0.281
Hence, the total area of confinement reinforcement is:
Ash = 0.2k n k p
Ag f c′
500 2
30
shc = 0.2 × 1.33 × 0.281 ×
×
× 420s = 3.34s
2
Ach f yh
400
420
but not less than:
Ash = 0.09shc
2
f c′
30
= 0.09 × 420 ×
s = 2.84s
f yh
400
For Ash = 341 mm , s = 341 / 3.34 = 102 mm.
CAC Concrete Design Handbook
11–27
From Clause 21.4.4.3 the spacing of the hoops shall not exceed:
(i)
h / 4 = 500 / 4 = 125 mm
(ii)
6d b = 6 × 25 = 150 mm
(iii)
s x = 100 + (350 − hx ) / 3 = 100 + (350 − 187.5 ) / 3 = 154 mm
Hence use 10M hoops at 100 mm centres as shown in Fig. 11.17. The chosen arrangement
of hoops and longitudinal reinforcement also satisfies Clauses 21.4.4.4 and 7.6.5. The details of
the first-storey column reinforcement are given in Fig. 11.17.
Fig. 11.17 Details of reinforcement in first-storey column
11–28
Seismic Design
11.4.5.6 Splice Details
We will splice the column bars at mid-height of the column with tension lap splices in
accordance with Clause 21.4.3.2. The development length, l d , can be found from Table 12-1 as:
l d = 0.45k 1k 2 k 3 k 4
fy
f c′
d b = 0.45 × 1 × 1 × 1 × 1 ×
400
30
× 25 = 822 mm
Provide a lap length of 1.3l d = 1.3 × 822 = 1068 mm (see Fig. 11.17).
11.4.6
Design of Interior Beam-Column Joint
To illustrate the procedures involved in designing a beam-column joint in a ductile momentresisting frame, an interior joint in the structure described in Section 11.4 will be designed. A
description of the joint details is given in Fig. 11.18.
Fig. 11.18 Geometry of interior beam-column joint
11.4.6.1 Determination of Factored Forces in Joint
In accordance with Clause 21.5.1.2, assume that the tensile force in the beam reinforcement
is 1.25 As f y .
CAC Concrete Design Handbook
11–29
To estimate the corresponding shear, Vcol , in the column above the joint, assume that
flexural hinging occurs in the beams at the first and second storey levels. The calculations are
summarized in Fig. 11.19.
Fig. 11.19 Determination of factored shear resistance in joint
11–30
Seismic Design
11.4.6.2 Check Factored Shear Resistance of Joint
Since four equal depth beams frame into the joint and each covers more than 3/4 of each
face of the joint, the joint is considered to be externally confined (Clause 21.5.4.1). Hence the
factored shear resistance of the joint is taken as:
V r = 2.2λφ c f c′ A j = 2.2 × 0.65 30 × 500 × 500 = 1958 kN
As the design shear in the joint of 1408 kN is less than 1958 kN, the shear resistance of the
joint is adequate.
11.4.6.3 Transverse Reinforcement Required in Joint
As the joint is framed by four equal depth beams which provide confinement, only one-half of
the confinement steel required for the column is required through the joint (Clause 21.5.2.2). The
spacing required for confinement in the joint is therefore 200 mm. However the spacing limits of
Clause 21.4.4.3 control ( s max = h / 4 = 125 mm). Hence provide 3 sets of 10M hoops between
the flexural bars in the beams as shown in Fig. 11.20.
Fig. 11.20 Details of joint reinforcement
CAC Concrete Design Handbook
11–31
11.4.6.4 Bond of beam Bars
As the beam bars pass through the joint their bond characteristics are checked by the
requirement in Clause 21.5.5.6 that the bar diameters be not greater than
l j / 24 = 500 / 24 = 21 mm (normal density concrete and uncoated bars). Since this exceeds the
actual bar diameter of 20 mm this requirement is met.
11.5
11.5.1
Analysis of a Ductile Core-Wall Structure
Description of Building and Loads
The twelve-storey reinforced concrete building shown in Fig. 11.21 is located in Montreal and is
founded on stiff soil.
Fig. 11.21 Plan and elevation of twelve-storey office building
11–32
Seismic Design
The twelve-storey reinforced concrete office building has a centrally located elevator core.
Each floor consists of a 200 mm thick flat plate with 6 m interior spans and 5.5 m end spans. The
columns are all 550 x 550 mm and the thickness of the core wall components 400 mm. The 400
mm thick wall thickness was initially chosen such that it exceeds l u / 14 = 4650 / 14 = 332 mm
(Clause 21.6.3). This value is checked in Section 11.5.6.5. The core wall measures 6.4 m by 8.4
m, outside to outside of the walls. Two 400 mm wide x 900 mm deep coupling beams connect the
two C-shaped walls at the ceiling level of each floor. The core walls extend one storey above the
th
th
roof at the 12 floor level forming an elevator penthouse at the 13 floor level. The slab has a 100
mm overhang.
Material Properties
Concrete: normal density concrete with fc′ = 30 MPa
Reinforcement: f y = 400 MPa
Gravity and Wind Loadings
Floor live load:
Roof load:
2
2.4 kN/m on typical office floors
2
4.8 kN/m on 12 m by 12 m corridor area around core
2
2.2 kN/m full snow load
2
1.6 kN/m mechanical services loading in 6 m wide strip over corridor bay
3
Dead loads:
self-weight of members calculated at 24 kN/m
2
1.0 kN/m partition loading on all floors
2
0.5 kN/m ceiling and mechanical services loading on all floors
2
0.5 kN/m roofing
Wind loading:
varies from 1.1 to 1.37 kN/m net lateral pressure over the height of the building
2
The building is to be designed with a fire-resistance rating of 2 hours.
11.5.2
Analysis Assumptions
To determine the forces in the walls and the coupling beams and the periods of vibration, the
three-dimensional core wall system was analyzed using ETABS. To make allowances for
cracking, member stiffnesses were based on effective properties equal to 0.25I g for the moment
of inertia and 0.45 Ag for the shear area for all diagonally reinforced coupling beams as required
by Clause 21.2.5.2.1. The walls were modeled with an effective flexural stiffness of 0.7EI g and an
effective axial stiffness of 0.7EAg , determined as a function of the axial loading at the base of the
walls (see Clause 21.2.5.2.1).
11.5.3
Seismic loading
For the force modification factors, Rd and Ro , we will assume that the core-wall system will
take 100% of the lateral loads as allowed by the NBCC. In the N-S direction we will design and
detail the walls as ductile shear walls and hence Rd = 3.5 and Ro = 1.6 . In the E-W direction we
will design and detail the coupling beams and walls as a ductile coupled wall system and hence
Rd = 4.0 and Ro = 1.7 . In order for the E-W direction to qualify as a ductile coupled wall system
we must check the degree of coupling as determined by analysis of the structure.
CAC Concrete Design Handbook
11–33
11.5.3.1 Minimum Lateral Earthquake Force
The structure is located in Montreal and is founded on stiff soil. Therefore the site
classification is “D”. The acceleration-based site coefficient Fa = 1.124 and the velocity-based site
coefficient Fv = 1.360 . The seismic response factor, S (Ta ) , is dependent on the fundamental
period, Ta , of the structure. The 5% damped spectral response accelerations, Sa (T ) , for Montreal
are given in Table 11.8. Table 11.8 also gives the design spectral response accelerations, S (T ) ,
obtained from the product of the site coefficients and S a as shown in Fig. 11.22.
Table 11.8 Spectral response accelerations and design spectral response accelerations
T ≤ 0.2
T = 0.5
T = 1.0
T = 2.0
T ≥ 4.0
Sa (T )
0.69
0.34
0.14
0.048
0.024
S (T )
0.776
0.462
0.190
0.065
0.033
Figure 11.22 Design spectral response acceleration
The empirical fundamental lateral period, Ta , for this shear wall structure in the N-S and E-W
directions, is given by:
Ta = 0.05hn3 / 4 = 0.05 × 45.0 3 / 4 = 0.869 s
N-S Direction
The calculated period for this structure in the N-S direction, using the computer program
ETABS, is 1.83 s. Note that a 3-D model including the walls, the slabs and the columns was also
analysed and resulted in a period of 1.75 s. Because this period is within 15% of the periods of the
walls alone, then a period of 1.83 s was used.
The value of the fundamental lateral period cannot be taken greater than 2 × 0.869 = 1.74 and
hence use Ta = 1.74 s. From linear interpolation, S (Ta ) = 0.0981 (see Fig. 11.22).
The values of M v and J depend on the ratio of Sa (0.2) / Sa (2.0 ) = 0.69 / 0.048 = 14.4 and the
value of Ta . It is necessary to interpolate the value of S (Ta ) Mv and the value of J between periods of
1.0 and 2.0 s and 0.5 and 2.0 s, respectively. This interpolation results in S (Ta ) Mv = 0.170 ,
Mv = 1.736 and J = 0.505 .
11–34
Seismic Design
This office building has an earthquake importance factor, I E = 1.0 . For this ductile shear wall
Rd = 3.5 and Ro = 1.6 .
Hence the seismic base shear, V, is:
V =
S (Ta ) Mv IEW 0.0981× 1.736 × 1.0 × W
=
= 0.0304W
Rd Ro
3.5 × 1.6
Vmin =
S (2.0 ) Mv I EW 0.0653 × 2.5 × 1.0 × W
=
= 0.0291W
Rd Ro
3.5 × 1.6
Vmax =
2 S (0.2)IEW
2 0.7756 × 1.0 × W
=
= 0.0923W
3 Rd Ro
3
3.5 × 1.6
For this structure, W = 90590 kN.
Hence V = 0.0304W = 0.0304 × 90590 = 2756 kN.
The
portion
of
V concentrated
at
the
top
of
the
building
is
Ft = 0.07TaV = 0.07 × 1.737 × 2755.6 = 335 kN, but need not be taken greater than
0.25V = 0.25 × 2755.6 = 688.9 kN.
The calculations of the seismic lateral forces at each floor level are summarized in Table
11.9.
E-W Direction
The calculated period for this structure in the E-W direction, using the computer program
ETABS is 1.72 s. It is noted that this period may be used because the period for the full 3-D
structure (walls, slabs and columns) is within 15% of this value. The value of the fundamental
lateral period cannot be taken greater than 2 × 0.869 = 1.74 and hence use Ta = 1.72 s. From
linear interpolation, S (Ta ) = 0.101 (see Fig. 11.22).
It is necessary to interpolate the value of S (Ta ) Mv and the value of J between periods of 1.0
and 2.0 s and 0.5 and 2.0 s, respectively. This interpolation results in
S (Ta ) Mv = 0.110 ,
Mv = 1.093 and J = 0.757 .
For the ductile coupled wall system in the E-W direction Rd = 4.0 and Ro = 1.7 .
Hence the seismic base shear, V, is:
V =
S (Ta )Mv IEW 0.101× 1.093 × 1.0 × W
= 0.0162W
=
Rd Ro
4.0 × 1.7
Vmin =
S (2.0 ) Mv I EW 0.065 × 1.2 × 1.0 × W
= 0.0115W
=
Rd Ro
4.0 × 1.7
Vmax =
2 S (0.2)IEW
2 0.776 × 1.0 × W
=
= 0.0760W
3 R d Ro
3
4.0 × 1.7
For this structure W = 90590 kN.
Hence V = 0.0162W = 0.0162 × 90590 = 1466 kN.
The
portion
of
V
concentrated
at
the
Ft = 0.07TaV = 0.07 × 1.72 × 1466 = 176.2 kN, but need
0.25V = 0.25 × 1466 = 366.5 kN.
top
of
the
not be taken
building
is
greater than
CAC Concrete Design Handbook
11–35
The calculations of the seismic lateral forces at each floor level using the equivalent static
force procedure are summarized in Table 11.9. The weight of the penthouse has been included
at the roof level.
Table 11.9 Lateral Load Calculations for Each Floor Level
Floor
N-S
E-W
hi ,
m
Wi ,
kN
hi Wi ,
kN·m
Fx
Tx
Fx
Tx
12
45.00
8154
366930
727.1
2163
385.1
1146
11
41.35
7467
308761
329.8
981
175.8
523
10
37.70
7467
281506
300.7
895
160.3
477
9
34.05
7467
254251
271.6
808
144.7
431
8
30.40
7467
226997
242.5
721
129.2
384
7
26.75
7467
199742
213.4
635
113.7
338
6
23.10
7467
172488
184.3
548
98.2
292
5
19.45
7467
145233
155.2
462
82.7
246
4
15.80
7467
117979
126.0
375
67.2
200
3
12.15
7467
90724
96.9
288
51.7
154
2
8.50
7467
63470
67.8
202
36.1
108
1
4.85
7766
37665
40.2
120
21.4
64
90590
2265745
2756
8198
1466
4362
0
Total
11.5.3.2 Accidental Torsion
The 3-D model shown in Fig. 11.23 was used to calculate accidental torsional effects by
applying the lateral forces Fx (see Table 11.9) at an accidental eccentricity of ±0.1Dnx , where
Dnx is the plan dimension of the building at level x, perpendicular to the direction of seismic
loading. This gives an accidental torsional eccentricity of 2.975 m, from the centre of mass (same
as centre of rigidity) for loading in the N-S and E-W directions. The values of Tx are given in
Table 11.9.
The structure was analysed with a 3-D model of the core-wall structure for both wind and
seismic loading, with and without eccentricity. In these analyses the participation of the flat plate
and columns was neglected.
11.5.3.3 Degree of Coupling
In the calculations of the base shear it was assumed that there was sufficient coupling of the
walls in the E-W direction to qualify this wall system as a ductile coupled wall system rather than a
partially coupled wall system. To check the degree of coupling by the wall system in the E-W
direction, the base overturning moment resisted by axial tension and compression forces in the
walls (resulting from shear in the coupling beams), divided by the total base overturning moment
is determined. Although the design forces were obtained from dynamic analysis, it is not
11–36
Seismic Design
appropriate to use these values to determine the degree of coupling because the values obtained
from modal combination (e.g., SRSS or CQC) does not satisfy static equilibrium. The degree of
coupling was determined using static analysis with the Fx forces from the equivalent static force
procedure, giving:
Tl
5513 × 6.5
=
= 0.74
M1 + M 2 + Tl 2 × 6400 + 5513 × 6.5
where
T
l
= axial tension and compression acting at centroid of coupled walls
= distance between centroids of coupled walls, equal to 6.5 m for this example
The degree of coupling is 74%, which is exceeds the minimum limit for ductile coupled walls of
66%. Hence Rd = 4.0 and Ro = 1.7 , as assumed above.
11.5.3.4 Check on Structural Irregularity
To determine if the structure is sensitive to torsion, the values of B need to be determined at
all levels from the maximum and average displacements of the structure at in the E-W and N-S
directions. The maximum value, B (determined at the extreme points of the structure), in the N-S
direction occurs in the first storey, with a displacement due to accidental torsion of 1.04 mm and a
displacement due to Fx of 1.30 mm. Hence:
B=
δ max 1.04 + 1.30
=
= 1.80
1.30
δ ave
Because B is greater than 1.7, the structure is sensitive to torsion and hence is designated as
irregular. The maximum value of B in the E-W direction occurs in the first storey and is 1.66.
Note that a 3-D analysis of the structure, including the columns and slabs as well as the
actual mass distributions indicates that the first and fourth modes of vibration are torsional with
periods of 1.89 and 0.54 s, respectively. This confirms that the structure is indeed torsionally
sensitive.
This design example illustrates the steps necessary to design this common type of structure,
that is torsionally sensitive.
11.5.3.5 Dynamic Analysis
The NBCC requires that the Dynamic Analysis Procedure be used except that the Equivalent
Static Force Procedure may be used for structures that meet any one of the three conditions in
parts (a), (b) and (c) of Clause 4.1.8.7. For this building, the term I E FaSa (0.2 ) is greater than 0.35
and hence the condition in part (a) is not satisfied. The presence of the structural irregularity due
to torsion sensitivity means that the Equivalent Static Force Procedure cannot be used (part (b) of
4.1.8.7). Part (c) of 4.1.8.7 is also not satisfied. Accordingly, the Equivalent Static Force
Procedure is not permitted as an alternative to the Dynamic Analysis Procedure for this example
building. The first step is to determine Ve from a linear dynamic analysis. The design base shear
Vd is obtained from:
Vd =
Ve
IE
Rd Ro
CAC Concrete Design Handbook
11–37
Because this is an irregular structure, that requires dynamic analysis (NBCC 4.1.8.7), Vd shall not
be taken less than 1.0V rather than 0.8V , permitted for a regular structure.
All forces and deflections obtained from the linear dynamic analysis are scaled by the factor
Vd / Ve to obtain the design values. However, in order to obtain realistic values of anticipated
deflections and drifts, the design values need to be multiplied by R d R o / I E .
Fig. 11.23 shows the 3-D ETABS model that considers only the core wall system (SFRS) and
is used for the dynamic analysis. The second model used is the entire structure including the
frame members not considered part of the SFRS (columns and the slabs) to check the ductility
and strength of these members subjected to seismically induced deformations. The total mass for
each floor was concentrated at the centre of mass (same as centre of rigidity for this example)
and rigid diaphragms were assumed at each floor level. Sway effects (P-Delta) were included
using the ETABS program option. For this analysis, compressive loads on the walls were obtained
from the consistent loading case of 1.0D + 0.5L + 0.25S , with live load reduction factors.
The first three lateral modes in the N-S and E-W directions are shown in Fig. 11.24, together
with the associated periods of vibration and the modal participating mass ratios. Note that the sum
of these ratios is 94.9% and 93.8% of the total mass in the N-S and E-W directions, respectively.
These ratios exceed the minimum required ratio of 90% of the total mass (NBCC). Spectral modal
superposition, using SRSS for the first three modes in the both directions was used to determine
all forces and deformations.
The base shear in the N-S direction determined by dynamic analysis is Ve = 14159 kN.
Therefore:
Vd = 14159 ×
1.0
= 2528 kN
3.5 × 1.6
However for this irregular building Vd shall not be taken less than V = 2755.6 kN. Hence, all
forces and deflections obtained from the dynamic analysis shall be multiplied
byVd / Ve = 2755.6 / 14159 = 0.195 in the N-S direction.
The base shear in the E-W direction determined by dynamic analysis is Ve = 11021 kN.
Therefore:
Vd = 11021 ×
1.0
= 1621 kN
4.0 × 1.7
However for this irregular building Vd shall not be taken less than V = 1466.1kN. Hence, all
forces and deflections obtained from the dynamic analysis shall be multiplied by
Vd / Ve = 1621 / 11021 = 0.147 in the E-W direction.
11–38
Seismic Design
Figure 11.23 3-D Model used for dynamic analysis
CAC Concrete Design Handbook
Mode 1
T = 1.830
MPMR = 0.67
11–39
Mode 2
T = 0.339
MPMR = 0.22
Mode 3
T = 0.143
MPMR = 0.07
Mode 2
T = 0.435
MPMR = 0.17
Mode 3
T = 0.199
MPMR = 0.05
(a) N-S direction
Mode 1
T = 1.717
MPMR = 0.72
(b) E-W direction
Figure 11.24 Mode shapes, corresponding lateral periods of vibration and modal
participating mass ratios in the N-S and E-W directions
11–40
Seismic Design
11.5.3.6 Deflections and Drift Ratios
The deflections obtained from the dynamic analysis are multiplied by the factor 0.195 in the
N-S direction and 0.147 in the E-W direction. To account for the total anticipated displacements,
including the inelastic effects it is necessary to multiply these deflections by the factor R d R o / I E
to obtain the design values. Note that the factor
Vd Rd Ro
is equal to 1.0 unless Vd is controlled
Ve I E
by the value of V , as in the N-S direction in this example (see 11.5.3.5). The deflections obtained
from dynamic analysis include P-Delta effects. The deflections arising from accidental torsional
eccentricity are added to the deflections from the dynamic analysis.
The maximum total interstorey drift in the N-S direction occurs in the eighth storey. From the
dynamic analysis the interstorey drift in this storey is 0.00060 and the interstorey drift from
accidental torsion at this level is 0.00041, for a maximum interstorey drift of 0.0010. Therefore the
anticipated interstorey drift, including inelastic effects, is 0.0010 × 3.5 × 1.6 / 1.0 = 0.0056 .
Similarly the maximum anticipated interstorey drift in the E-W direction is 0.0047. These
anticipated maximum interstorey drift ratios are less than the NBCC limit of 0.025.
11.5.4
Design Forces
The results from the 3-D analyses for both seismic and wind loading are summarized in
Tables 11.10, 11.11, 11.12 and 11.13. It is noted that for wind loading, the case with eccentric
wind loading does not govern. Table 11.10 gives the forces from seismic loading analysis in the
N-S direction, without and with accidental torsion effects.
Table 11.11 gives the forces from seismic loading analysis in the E-W direction, without
accidental torsion effects.
Table 11.12 gives the forces from accidental torsion due to seismic loading analysis in the EW direction. It is noted that accidental torsion is resisted by shear flow around the components of
the C-shaped walls and by shear in the coupling beams. The accidental torsion does not create
any global moments, axial loads or shears in the C-shaped walls, but results in local moments,
axial loads and shears in the component parts, AB, BC and CD (see Fig. 11.27).
CAC Concrete Design Handbook
11–41
Table 11.10 Results of Seismic Loading Analyses (1.0E) in N-S direction for one wall,
including accidental torsion
Storey
13 top
13 bot
12 top
12 bot
11 top
11 bot
10 top
10 bot
9 top
9 bot
8 top
8 bot
7 top
7 bot
6 top
6 bot
5 top
5 bot
4 top
4 bot
3 top
3 bot
2 top
2 bot
1 top
1 bot
Wall moment
without
torsion,
kN·m
Wall moment
with
torsion,
kN·m
Wall shear
without
torsion,
kN
Wall shear
with
torsion,
kN
0
0
0
1484
1484
3689
3689
6022
6022
8037
8037
9504
9504
10391
10391
10831
10831
11097
11097
11647
11647
13033
13033
15567
15567
20396
492
907
1503
2611
3297
4839
5623
7040
7922
8797
9767
9890
10931
10494
11375
11548
11212
12582
11514
14090
13107
16685
15859
20803
20210
28332
0
0
406
406
605
605
647
647
589
589
518
518
515
515
596
596
738
738
923
923
1119
1119
1283
1283
1378
1378
114
114
509
509
787
787
898
898
902
902
886
886
934
934
1062
1062
1249
1249
1478
1478
1719
1719
1943
1943
2057
2057
11–42
Seismic Design
Table 11.11 Results of Seismic Loading Analyses (1.0E) in E-W direction (Coupled Wall)
for one wall
Storey
13 top
13 bot
12 top
12 bot
11 top
11 bot
10 top
10 bot
9 top
9 bot
8 top
8 bot
7 top
7 bot
6 top
6 bot
5 top
5 bot
4 top
4 bot
3 top
3 bot
2 top
2 bot
1 top
1 bot
Wall
moment,
kN·m
Wall
axial
load,
kN
Wall
shear,
kN
Coupling
beam
shear
without
torsion,
kN
Coupling
beam
shear with
torsion,
kN
452
452
1078
309
1086
618
945
1135
1019
1515
1280
1720
1527
1815
1708
1859
1826
1860
1820
1889
1597
2205
1303
3090
1957
5577
139
139
332
332
596
596
913
913
1251
1251
1583
1583
1895
1895
2188
2188
2469
2469
2753
2753
3047
3047
3341
3341
3588
3588
0
0
252
252
390
390
441
441
438
438
430
430
440
440
466
466
508
508
581
581
678
678
764
764
810
810
69.6
113.0
96.3
149.0
132.2
192.8
159.5
228.9
173.6
251.6
179.3
265.1
184.6
276.7
194.6
290.7
210.2
307.4
228.9
323.3
242.9
329.8
237.3
310.4
191.8
244.3
CAC Concrete Design Handbook
11–43
Table 11.12 Local forces due to accidental torsion (1.0E) in E-W direction (Coupled Wall
Direction) in different components of C-shaped wall
Storey
13 top
13 bot
12 top
12 bot
11 top
11 bot
10 top
10 bot
9 top
9 bot
8 top
8 bot
7 top
7 bot
6 top
6 bot
5 top
5 bot
4 top
4 bot
3 top
3 bot
2 top
2 bot
1 top
1 bot
Wall component AB or CD ( ± )
3.2 m long segments
Wall component BC ( ± )
6.4 m long segment
Moment,
kN·m
Axial
load,
kN
Shear,
kN
Moment,
kN·m
Axial
load,
kN
Shear,
kN
90.4
-44.4
115.0
-86.8
142.0
-113.5
168.2
-140.2
191.0
-165.9
209.9
-189.6
224.2
-210.6
233.2
-228.4
235.9
-242.4
231.0
-251.8
217.2
-256.7
189.4
-243.9
221.8
-361.0
-58.0
-58.0
-94.9
-94.9
-106.4
-106.4
-105.9
-105.9
-95.3
-95.3
-75.6
-75.6
-47.1
-47.1
-9.0
-9.0
40.1
40.1
102.9
102.9
183.4
183.4
286.1
286.1
457.3
457.3
-36.9
-36.9
-55.3
-55.3
-70.0
-70.0
-84.5
-84.5
-97.8
-97.8
-109.4
-109.4
-119.1
-119.1
-126.5
-126.5
-131.0
-131.0
-132.3
-132.3
-129.8
-129.8
-118.7
-118.7
-120.2
-120.2
-81.1
131.3
232.9
30.2
330.6
-26.6
398.0
-93.7
442.9
-169.1
468.1
-252.1
475.1
-343.8
464.1
-446.1
433.9
-562.7
382.1
-699.3
304.9
-861.8
190.8
-1089.6
248.5
-1493.1
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
58.2
58.2
-55.6
-55.6
-97.9
-97.9
-134.7
-134.7
-167.7
-167.7
-197.3
-197.3
-224.4
-224.4
-249.4
-249.4
-273.0
-273.0
-296.3
-296.3
-319.7
-319.7
-350.8
-350.8
-359.1
-359.1
11–44
Seismic Design
Table 11.13 Results of Wind Loading Analyses (1.4W)
Storey
13 top
13 bot
12 top
12 bot
11 top
11 bot
10 top
10 bot
9 top
9 bot
8 top
8 bot
7 top
7 bot
6 top
6 bot
5 top
5 bot
4 top
4 bot
3 top
3 bot
2 top
2 bot
1 top
1 bot
N-S
Direction
E-W Direction
(Coupled Wall Direction)
Wall
moment,
kN·m
Wall
moment,
kN·m
Wall axial
load,
kN
Coupling beam
shear,
kN
0
0
-74
122
47
577
503
1287
1212
2263
2190
3498
3427
4990
4923
6734
6672
8726
8670
10962
10915
13433
13396
16161
16136
20087
159
159
444
178
633
72
735
-84
800
-281
827
-508
819
-767
770
-1060
665
-1402
473
-1823
130
-2385
-496
-3233
-1669
-5599
49
49
119
119
241
241
426
426
679
679
1001
1001
1391
1391
1845
1845
2358
2358
2918
2918
3504
3504
4073
4073
4546
4546
24.5
38.9
64.4
95.7
128.8
162.5
195.5
227.0
255.5
278.3
290.5
281.5
233.3
The design forces for both seismic and wind loading are given in Fig. 11.25. The distribution
of wall moments for wind loading is typical for a coupled wall system. The distribution of wall
moments for seismic loading was obtained from modal combinations (SRSS) and therefore the
moments obtained are absolute values.
CAC Concrete Design Handbook
11–45
Fig. 11.25 Results of analyses for seismic and wind loading
11.5.5
Design of Coupling Beams
11.5.5.1 Design Forces for Coupling Beams
The maximum coupling beam shear due to factored wind loading is 290.5 kN and due to
earthquake effects (including accidental torsion) is 329.8 kN. Hence the seismic loading case
governs the design of the coupling beams. These maximum shears occur in the coupling beams
in the third storey (see Tables 11.11 and 11.13).
In designing the coupling beams we can account for redistribution of moments between the
beams and hence the factored shear resistance required for earthquake loading would be the
sum of the shears in the coupling beams divided by the number of coupling beams. This would
require a factored shear resistance of 3283 / 13 = 252.5 kN for seismic loading.
The coupling beams have a depth of 900 mm and a clear span of 2000 mm and hence
satisfy the dimensional limitation that the depth must not be greater than twice the clear span
(Clause 21.6.8.5). The ductile coupling beams must be designed with diagonal reinforcement,
rather than longitudinal bars and vertical hoops, because the clear span of each beam is not equal
to or greater than four times the effective depth (see Clauses 21.6.8.6 and 21.3.1.1).
Since the design torsions arise only from accidental torsional eccentricity, which can act in
either direction, the same coupling design forces and beam details will be used on the north and
south sides of the core wall.
11–46
Seismic Design
11.5.5.2 Design and detailing of coupling beams
Fig. 11.26 shows the details of the diagonal reinforcement in a coupling beam. From the
geometry of the reinforcement, the angle α between the centroidal axis of one set of diagonal
bars and the horizontal is 19.4û.
Fig. 11.26 Coupling beam reinforcing details
If 4-20M bars are provided in each set of diagonal bars then the factored shear resistance is:
Vr = 2φs As fy sin α = 2 × 0.85 × 4 × 300 × 400 × sin 19.4 = 271.0 kN
If this reinforcement is placed in the lower storeys, then a moment redistribution of 17.8%
would be required in the coupling beam with maximum moment due to seismic loading and 6.7%
redistribution would be required for wind loading. Clause 9.2.4 permits a moment redistribution up
to 20%.
The diagonal reinforcement must have closely spaced hoops as required in Clause 21.6.8.7
with a maximum spacing given by the smaller of:
(a) 6d bl = 6 × 20 = 120 mm
(b)
24d bh = 24 × 10 = 240 mm
(c)
100 mm
CAC Concrete Design Handbook
11–47
Hence use 10M hoops spaced at 100 mm.
The diagonal reinforcement must extend into the wall at each end a minimum distance of
1.5l d = 1.5 × 526 = 789 mm. Hence use an embedment length of 800 mm. In addition to the
diagonal reinforcement, provide minimum transverse and longitudinal reinforcement, as shown in
Fig. 11.26, but only in the clear span of the beam.
Although a designer might choose to use the same coupling beams details, with 4-20M
diagonal bars, over the full height of this building, reducing the amount of reinforcement in the
coupling beams near the top of the structure is advantageous whenever possible. This approach
is illustrated below.
If the reinforcement in the coupling beams is reduced to 4-15M near the top of the structure,
then the factored resistance per beam is:
Vr = 2φs As fy sin α = 2 × 0.85 × 4 × 200 × 400 × sin 19.4 = 180.7 kN.
The beams with 4-15M bars would require a hoop spacing of 90 mm.
Provide coupling beams with 4-20M bars for storeys 1 to 11 and beams with 4-15M bars for
the top two coupling beams. This arrangement gives a total shear capacity of the 13 beams of:
∑V
r
= 11× 271.0 + 2 × 180.7 = 3342 kN
This shear resistance of the thirteen coupling beams exceeds the required shear resistance of
3283 kN.
Hence provide diagonally reinforced coupling beams with 4-20M for the first 11 storeys and
beams with 4-15M diagonal bars for the top two coupling beams.
11.5.5.3 Ductility of coupling beams
The inelastic rotational capacity of the coupling beams must be greater than the rotational
demand. The rotational demand is given by (Clause 21.6.8.4):
 ∆ f Rd Ro
 hw
θ id = 
 l cg  0.019 × 4.0 × 1.7  6.5

= 0.0093
=

45.0
 2.0

 lu
The rotational capacity of diagonally reinforced coupling beams (Clause 21.6.8.4) is 0.04.
Hence the coupling beams have sufficient ductility.
11.5.6
Design of Ductile Walls
11.5.6.1 Design Forces in E-W Direction
The design forces in the walls are determined from a capacity design approach. It is desired
that the walls be strong enough such that flexural hinging occurs in the coupling beams, which are
the primary energy dissipators of the structural system. In order to determine the design forces in
the walls we will consider a “push-over” type of loading on the coupled wall system. For this
analysis it is required that the factored resistance of the walls be at least equal to the moment
corresponding to the development of flexural hinging in the beams.
The “push-over” approach used for this twelve-storey coupled wall structure provides a
conservative design approach to the requirements of Clause 21.6.8.12. For taller coupled wall
structures the provisions of Clause 21.6.8.12 should be used.
The nominal shear resistance, Vn , for the coupling beam containing 4-20M diagonal
reinforcing bars is:
Vn = 2φs As f y × sin α = 2 × 1.0 × 1200 × 400 × sin 19.4 = 318.9 kN
Similarly the nominal resistance for the beams containing the 4-15M bars is 212.6 kN.
11–48
Seismic Design
In order to satisfy the capacity design requirement, the factored wall moments will be
increased at each level x by the factor γ bx , determined as:
γ bx =
∑V
∑V
n
f
where:
∑V
n
= sum of the shears corresponding to the nominal flexural resistance of coupling
f
beams above level x
= sum of factored shears above level x
∑V
The beam overstrength factors, cumulative dead and live loads and the factored axial loads
and factored moments in each wall multiplied by γ bx are shown on Table 11.14. The values of
∑V
f
are taken as the wall axial loads given in Table 11.14, since the axial load wall is the sum of
the shears in the coupling beams.
Table 11.14 Coupling Beam Overstrength Factors and Dead and Live Loads per Wall
Storey
13 top
13 bot
12 top
12 bot
11 top
11 bot
10 top
10 bot
9 top
9 bot
8 top
8 bot
7 top
7 bot
6 top
6 bot
5 top
5 bot
4 top
4 bot
3 top
3 bot
2 top
2 bot
1 top
1 bot
γ bx
Vn ,
kN
212.6
Vf ,
kN
113.0
1.88
212.6
149.0
1.62
318.9
192.8
1.64
318.9
228.9
1.55
318.9
251.6
1.48
318.9
265.1
1.42
318.9
276.7
1.37
318.9
290.7
1.32
318.9
307.4
1.28
318.9
323.3
1.24
318.9
329.8
1.21
318.9
310.4
1.19
318.9
244.3
1.20
PD ,
kN
-945
-945
-1701
-1701
-2457
-2457
-3213
-3213
-3969
-3969
-4725
-4725
-5481
-5481
-6237
-6237
-6993
-6993
7749
7749
8506
8506
9262
9262
9797
9797
PL ,
kN
0
0
-196
-196
-394
-394
-513
-513
-621
-621
-725
-725
-825
-825
-924
-924
-1020
-1020
1115
1115
1208
1208
1301
1301
1393
1393
Pn ,
kN
± 425
± 425
± 850
± 850
± 1488
± 1488
± 2126
± 2126
± 2764
± 2764
± 3402
± 3402
± 4039
± 4039
± 4677
± 4677
± 5315
± 5315
± 5953
± 5953
± 6591
± 6591
± 7228
± 7228
± 7866
± 7866
γ bx M1 ,
kN·m
851
851
1750
501
1777
1012
1469
1765
1506
2239
1814
2438
2088
2483
2259
2459
2338
2382
2259
2344
1929
2664
1550
3676
2345
6682
CAC Concrete Design Handbook
11–49
The required moment capacity at the base of each wall, given in Table 11.14, is 6682 kN·m
for seismic loading without accidental torsion. It is noted that the required moment at the base of
each wall due to wind (see Table 11.13) is 5599 kN·m. Hence the seismic capacity design
requirements control the design.
From Table 11.14, the minimum axial force at the base of the "tension wall" corresponding to
development of flexural hinges in the coupling beams is:
Ps + Pn = −9797 + 7866 = −1931 kN, that is a compressive load.
The maximum axial force at the base of the "compression wall" corresponding to
development of flexural hinges in the coupling beams is:
Ps + Pn = −9797 − 0.5 × 1393 − 7866 = −18360 kN.
To account for the local forces (see Table 11.12) from accidental torsion a simplified
approach will be taken. Although the axial loads and moments on parts AB and CD have opposite
signs it will be assumed that the C-shaped wall is subjected to an additional axial load of twice the
axial tension or compression acting on segments AB or CD and an additional moment of twice the
moment acting in these segments. For example, at the base of the structure the additional axial
load for design is 2 × ±457 = ±914 kN and the additional moment is 2 × ±361 = ±722 kN·m.
11.5.6.2 Design Forces in N-S Direction
From Tables 11.10 and 11.13, the base moment in each wall due to lateral seismic loading is
28332 kN·m (including accidental torsion) and the base moment due to factored wind loading is
20087 kN·m, respectively. Hence, for flexural design in the N-S direction, the earthquake loading
controls.
The
corresponding
axial
load
for
the
two
walls
is
2(PD + 0.5PL ) = 2(− 9797 − 0.5 × 1393 ) = 2 × −10494 = −20988 kN.
11.5.6.3 Design of base of wall for flexure and axial load
Preliminary choice of vertical reinforcement
(a)
Minimum area of concentrated reinforcement (Clause 21.6.6.4)
In the 3.2 m long walls in the E-W direction:
As = 0.0015 bw l w = 0.0015 × 400 × 3200 = 1920 mm
2
2
Therefore try 4–25M bars ( As = 2000 mm ) as concentrated reinforcement at one end of the 3.2
m long wall (segments AB and CD).
In the 6.4 m long wall (segment BC) in the N-S direction
As = 0.0015 bw l w = 0.0015 × 400 × 6400 = 3840 mm
2
For the case of flanged walls, concentrated reinforcement at the ends of the effective flanges
may supply up to one half of the required minimum wall web concentrated reinforcement (Clause
21.6.6.5). Hence the required concentrated reinforcement at the web-flange intersection is
2
0.5 × 3840 = 1920 mm . Hence try 4-25M at the intersection of the two walls.
Outside the plastic hinge regions, only two-thirds of this area of the concentrated
reinforcement is required.
Provide a clear cover for the hoops of 40 mm, resulting in a clear cover of 50 mm for the
main vertical reinforcement in the wall, as required for a two-hour fire-resistance rating.
In regions of plastic hinging, the uniformly distributed horizontal reinforcement must be
anchored within the region of concentrated reinforcement to develop 1.25f y (Clause 21.6.5.5).
11–50
Seismic Design
The development length required for the 10M bars, using the simplified equation in Clause
12.2.3 is:
1.25fy
500
l d = 0.45k1k 2 k 3 k 4
d b = 0.45 × 1× 1× 1× 0.8
× 10 = 329 mm
fc′
30
The length provided in the region of concentrated reinforcement (see Fig. 11.27) is 320 mm.
With the significant cover provided on the bars and the additional confinement provided in this
region of concentrated reinforcement, the development length can be shown to be less than 320
mm using Eq. 12-1 in Clause 12.2.2. Hence, the hoop configuration shown in Fig. 11.27 is
adequate.
(b) Maximum area of concentrated reinforcement (Clause 21.6.4.3)
Clause 21.6.4.3 limits the reinforcement ratio, including regions with lap splices, to 0.06.
With the layout of the 4-25M bars at the ends of the flanges and at the web-flange junctions, as
shown in Fig. 11.27, the percentage of steel equals (4 × 500 ) / (400 × 400 ) = 0.0125 . This
arrangement allows for lap splicing of the reinforcement without exceeding the limit of 0.06.
Fig. 11.27 Details of reinforcement in walls.
At the base of the walls the spacing of the horizontal distributed reinforcement must be decreased
to 120 mm in component BC and to 150 mm in components AB and CD (see Section 11.5.6.7).
(c)
Maximum bar diameters (Clause 21.6.4.4)
In the 400 mm thick walls, the maximum diameter of reinforcement is 400 / 10 = 40 mm.
CAC Concrete Design Handbook
11–51
(d) Distributed reinforcement (Clauses 21.6.5)
In the plastic hinge region, the spacing of the distributed reinforcement must not exceed
300 mm in each direction. Outside of this region, the maximum spacing is 450 mm. The
distributed reinforcement ratio must be greater than 0.0025 in each direction.
In the 400 mm thick wall elements, the maximum spacing, assuming two curtains of 10M
reinforcing bars is (2 × 100 ) / (0.0025 × 400 ) = 200 mm. Hence, at the base of the walls, use 2
curtains of 10M bars at 200 mm spacing in the horizontal and vertical directions.
In the regions of plastic hinging, two curtains of reinforcement must be provided in the N-S
direction if the design shear in one wall exceeds (Clause 21.6.5.3):
0.18λφc fc′ Acv = 0.18 × 0.65 × 30 × 400 × 6400 = 1641 kN.
Two curtains of reinforcement must be provided in the E-W direction if the design shear in
one wall exceeds:
0.18λφc fc′ Acv = 0.18 × 0.65 × 30 × 800 × 3200 = 1641 kN.
Calculation of M r at base of walls
The factored moment resistances for different loading cases were determined using the
stress block factors of Clause 10.1.7, strain compatibility and a maximum concrete compressive
strain of 0.0035. Table 11.15 summarizes the results.
Table 11.15 Predicted Factored Moment Resistances and Depths of Compression per Wall
at the base (Global wall forces)
Load Case
E-W (1 “C-shaped” wall)
“tension”
1.0D + 1.0E
wall
“compression”
wall
Nf
kN
Mf
kN·m
Nr
kN
Mr
kN·m
c
mm
-1931
6682
-1931
15058
493
1.0D + 1.0E
-17663
6682
-17663
21773
226
1.0D + 0.5L + 1.0E
-18360
6682
-18360
22280
233
1.0D + 1.0E
-9797
28332
-9797
49058
290
1.0D + 0.5L + 1.0E
-10494
28332
-10494
51097
303
N-S (per wall)
The required moment capacity in the E-W direction from analysis is 6682 kN·m per wall. As
can be seen from Table 11.15, the “tension” wall has a factored moment resistance of
15058 kN·m, while the compression wall has a factored moment resistance of 22280 kN·m.
Hence the factored flexural resistance exceeds the required factored moment. Figure 11.28
illustrates the moment resistances for the walls in the E-W direction.
It is noted that the total forces in the E-W tension wall, including global (Table 11.15) and
local (Table 11.12) forces at its base is:
N f = −1931 + 914 = −1017 kN and M f = 6682 + 722 = 7405 kN·m.
For this level of axial load the factored moment resistance is 13385 kN·m. Hence the wall
strength is adequate for both global and local forces. It can be shown, in a similar way, that the
factored resistance of the compression wall is adequate for global and local forces.
Clause 21.6.8.8 requires that the walls at each end of a coupling beam be designed so that
the factored moment resistance of the wall about its centroid, calculated using axial loads Ps + Pn ,
11–52
Seismic Design
exceeds the moment corresponding to the nominal resistance of the coupling beam. This
requirement will be satisfied at all the different levels of the structure because the “push-over”
analysis summarized in Table 11.14 already considers the attainment of the nominal resistances
of the coupling beams over the entire height of the structure. Hence all of the design moments
and axial forces correspond to these conditions.
In the N-S direction, the factored moment resistance for both load cases significantly
exceeds the required moment (see Table 11.15 and Fig. 11.29).
Fig. 11.28 Factored moment resistances of ductile coupled walls (E-W direction)
Fig. 11.29 Factored moment resistance of ductile shear walls (N-S direction)
CAC Concrete Design Handbook
11–53
11.5.6.4 Ductility of walls
E-W Direction - Ductile Coupled Walls (Clause 21.6.8)
For ductile coupled walls, the inelastic rotational demand θ id is taken as:
θ id =
∆ f Ro Rd 0.019 × 1.7 × 4.0
=
= 0.0029 ≥ 0.004
hw
45.0
The inelastic rotational capacity θ ic , taking l w as the length of the coupled wall system
(Clause 21.6.8.3) and assuming ε cu = 0.0035 is:
ε l
  0.0035 × 8.4

− 0.002  = 0.0278 ≤ 0.025
θ ic =  cu w − 0.002  = 
c
2
×
2
0
.
493


 
Because the rotational capacity of 0.025 exceeds the rotational demand of 0.004, sufficient
ductility is provided.
N-S Direction - Ductile Shear Walls (Clause 21.6.7)
For ductile shear walls, the ratio of the nominal flexural resistance to the factored flexural
resistance, M f , at the base is 1.95 and hence the inelastic rotational demand θ id is taken as:
θ id =
(∆ f Ro Rd
− ∆ f γ w ) (0.028 × 1.6 × 3.5 − 0.028 × 1.95 )
=
= 0.0024 ≥ 0.004
l 
6.4 


 hw − w 
 45.0 −

2 
2 


The inelastic rotational capacity θ ic , assuming ε cu = 0.0035 is:
 ε cu l w
  0.0035 × 6.4

− 0.002  = 
− 0.002  = 0.0350 ≤ 0.025

 2c
  2 × 0.303
θ ic = 
Because the rotational capacity of 0.025 exceeds the rotational demand of 0.004, sufficient
ductility is provided.
Confinement of concentrated reinforcement (Clause 21.6.7.3)
Because the inelastic rotational capacities of the walls in the E-W and N-S directions were
determined using ε cu = 0.0035 , it is not necessary to check the confinement requirements of
Clause 21.6.7.4 for the concentrated reinforcement in the walls.
11.5.6.5 Checking wall thickness for stability (Clause 21.6.3)
Clause 21.6.3 requires a wall thickness of l u / 10 in those parts of a wall that, under factored
vertical and lateral loads, are more than half way from the neutral axis to the compression face of
the wall section.
The 3200 mm long portions of the E-W "tension wall" may be considered as simple
rectangular wall elements as shown in Fig. 11.28 with a neutral axis depth of 493 mm. Since the
neutral axis depth is less than
4bw = 4 × 400 = 1600 mm and is less than
0.3l w = 0.3 × 3200 = 960 mm, the l u / 10 limit need not apply (see Clause 21.6.3.4). According
to Clause 21.6.3.2 the wall thickness in the plastic hinge region must not be less
than l u / 14 = 4650 / 14 = 332 mm. Hence the wall thickness of 400 mm is adequate.
11–54
Seismic Design
For the 6400 mm long portion of the E-W "compression wall", it is noted that
c / 2 = 233 / 2 = 117 mm, which is less than the wall thickness, and furthermore, the wall is
laterally supported at its ends by the 3200 mm wall portions. Hence, this portion of the wall need
not have a thickness of l u / 10 (see Clause 21.6.3.5).
For stability considerations for the wall loaded in the N-S direction, the value of
c / 2 = 303 / 2 = 152 mm is smaller than the 400 mm thickness of the flanges. Therefore the
400 mm wall dimension is adequate and the width of the flange of 3200 mm, greatly exceeds
l u / 5 (see Clause 21.6.3.5).
Therefore all of the stability requirements are satisfied.
11.5.6.6 “Buckling prevention” ties for concentrated reinforcement
(Clause 21.6.6.9)
The concentrated reinforcement should have buckling prevention ties in accordance with
Clause 7.6 and the ties must be detailed as hoops (Clause 21.6.6.9). In plastic hinge regions, the
hoop spacing shall not exceed:
(i)
6d bl = 6 × 25 = 150 mm
(ii)
24d bh = 24 × 10 = 240 mm
(iii) one-half the wall thickness = 400 / 2 = 200 mm.
Hence provide 10M hoops at a spacing of 150 mm as shown in Fig. 11.27.
11.5.6.7 Design for shear at base of walls (Clause 21.6.9)
The walls must be designed to resist the shear corresponding to the formation of plastic
hinges at their bases (Clause 21.6.9.1).
Determine probable moment resistances of walls
In order to determine the probable moment resistances of the walls, axial load-moment
calculations were carried out with φc = φs = 1.0 and using an equivalent "yield" strength of steel
of 1.25fy . In the E-W direction, we only need to determine the probable moment resistance of the
"compression wall" subjected to an axial load corresponding to 1.0E + 1.0D + 0.5L , since it results
in the larger resistance. From calculations, the probable moment resistance of the wall, M pw , is
26235 kN·m in the E-W direction and 60679 kN·m per wall in the N-S direction.
It is assumed that earthquake loading causes plastic hinging at the base of the walls.
Assuming that the ratio of the shear to moment at the base of a wall remains constant as the
moment increases to the probable resistance, the shear at the base as the wall develops a plastic
hinge will be:
M pw
V =
× Vf
Mf
N-S direction:
V =
M pw
Mf
× Vf =
60679
× 2057 = 4406 kN
28332
In calculating the shear capacity of the wall, the effective shear depth d v is taken as
0.9d = 0.9 × 6200 = 5580 mm, but need not be taken less than 0.8l w = 0.8 × 6400 = 5120 mm
(Clause 21.6.9.3).
CAC Concrete Design Handbook
11–55
In the region of expected plastic hinging, at the base of the wall, the inelastic rotational
demand θ id is less than 0.005 and hence the factored shear demand cannot exceed:
0.15φc fc′bw d v = 0.15 × 0.65 × 30 × 400 × 5120 = 5990 kN
The factored shear demand of 4406 kN is less than this upper limit.
Because the inelastic rotational demand θ id is less than 0.005, the factored shear resistance
is calculated using β = 0.18 . The axial load on the “tension” wall is -10494 kN (compression). This
axial load is less than: 0.1fc′ Ag = 0.1× 30 × (3200 × 800 + 5600 × 400 ) = 14400 kN.
Hence θ is taken as 45° (Clause 21.6.9.6). The factored shear resistance (Clauses 11.3.4
and 11.3.5), assuming pairs of 10M bars at 120 mm spacing is:
Vr = φ c β f c′ bw d v +
φ s Av f y d v cot θ
s
= 0.65 × 0.18 30 × 400 × 5580 +
0.85 × 200 × 400 × 5580 cot 45 o
120
= 1430 + 3162 = 4592 kN
Hence the shear resistance is adequate with pairs of 10M horizontal bars spaced at 120 mm.
It is noted that the accidental torsion causes shear in segments AB and CD. This shear, due
to accidental torsion, must be considered in design, accounting for the fact that the critical flange
will be in tension and have large flexural cracks due to the attainment of plastic hinging for loading
in the N-S direction. A method to check the shear resistance of these segments, accounting for
large cracks and possible redistribution of shear resisting torsion, is described in the calculations
below for loading in the E-W direction.
E-W direction:
In the E-W direction it is necessary to first determine the shear from seismic loading and
include accidental torsion effects. The shear at the base without accidental torsion is
810 / 2 = 405 kN in segment AB. The analysis including accidental torsion indicates a shear force
of 120 kN on wall segments AB and CD (see Table 11.12). In accordance with Clause 21.6.8.13 it
may be necessary to redistribute the shear force of 359 kN in the segment BC because under
plastic hinging this wall segment will experience large tensile strains and cracks. If the torsion
arising from this 359 kN force is redistributed to wall segments AB and CD, then an additional
shear force will be necessary. This additional shear force is:
∆VAB = VBC ×
8.0
8.0
= 359 ×
= 479 kN in two segments
6.0
6.0
Hence the total shear in segment AB is 405 + 120 + 479 / 2 = 765 kN.
In lieu of redistributing the torsional shear from wall segment BC, a more detailed analysis
will be conducted (Clause 21.6.8.13). Wall component BC is lightly loaded in shear but must resist
the shear with large cracks in this tension wall. In order to determine the shear resistance under
these conditions the “dowel action” resistance will be determined, considering the resistance of
the reinforcement only. From Paulay and Priestley (1992) the dowel resistance can be taken as:
Vr = 0.25φ s Av f y = 0.25 × 0.85 × 27 × 200 × 400 = 459 kN.
Because this resistance exceeds the shear force of 359 kN, redistribution of the shear from the
tension wall BC to wall AB is not necessary. Hence the total shear in segment AB is
405 + 120 = 525 kN.
11–56
Seismic Design
The design shear for segment AB, including torsional effects is:
V =
M pw
Mf
× Vf =
26235
× 525 = 1860 kN
6682 + 722
For segment CD, the shear from accidental torsion acts in the opposite direction from that in
segment AB. Hence the total shear for segment CD is 405 − 120 = 285 kN. This results in a
design shear for this segment of:
V =
M pw
Mf
× Vf =
26235
× 285 = 1010 kN.
6682 + 722
Segments AB and CD must both be designed for the larger shear of 1860 kN because the
accidental torsion can reverse.
In calculating the shear capacity of the wall, in accordance with Clause 21.6.9.3, we will
assume an effective shear depth, dv = 0.8l w = 0.8 × 3200 = 2560 mm.
In the region of expected plastic hinging, at the base of the wall, the inelastic rotational
demand θ id is less than 0.005 and hence the factored shear demand in one segment cannot
exceed:
0.15φc fc′bw dv = 0.15 × 0.65 × 30 × 400 × 2560 = 2995 kN
The factored shear demand of 1860 kN is less than this upper limit.
Because the inelastic rotational demand θ id is less than 0.005, the factored shear resistance
is calculated using β = 0.18 . The axial load on the tension wall is -1017 kN. This axial load is less
than: 0.1fc′ Ag = 0.1× 30 × (3200 × 800 + 5600 × 400 ) = 14400 kN.
Hence θ is taken as 45° (Clause 21.6.9.6). The factored shear resistance (Clauses 11.3.4
and 11.3.5) for wall segment AB at the base with pairs of 10M bars at a spacing of 150 mm is:
Vr = φ c β f c′ bw d v +
φ s Av f y d v cot θ
s
= 0.65 × 0.18 30 × 400 × 2560 +
0.85 × 200 × 400 × 2560cot 45 o
150
= 656 + 1161 = 1817 kN
Segment AB is overstressed by only 2%, while segment BC is very lightly loaded and hence
a portion can be redistributed to this segment. Hence the shear strength in the E-W direction is
adequate with 2-10M bars at 150 mm spacing.
Extend the 10M horizontal reinforcement into the confined core of the region of concentrated
reinforcement as close to the outside surface of the walls as cover will permit (see Fig 11.27 and
Section 11.5.6.3).
11.5.6.8 Checking sliding shear resistance at construction joints
(Clause 21.6.9.4)
In accordance with Clause 21.6.9.4, we must check the sliding shear resistance of the
construction joints. Since the vertical uniformly distributed reinforcement is constant over the
height of the walls, the most critical situation is at the base of the walls.
CAC Concrete Design Handbook
11–57
N-S direction:
In the N-S direction, the required shear strength is 4406 kN per wall. If the construction joint
is intentionally roughened, the factored shear stress resistance from Clauses 11.5.1 and 11.5.2 is:


N  
v r = φc  c + µ  ρv fy +


Ag  




9797 × 1000


= 0.65 0.50 + 1.0 0.0025 × 400 +

800 × 3200 + 400 × 5600  


= 2.302 MPa
Hence, the sliding shear resistance is: 2.302 × Acv = 2.302 × 400 × 6400 = 5893 kN
Since the sliding shear resistance exceeds the shear corresponding to plastic hinging, sliding
shear will be prevented.
E-W direction:
In the E-W direction, the required shear strength of segment AB is 1860 kN. It will be
assumed that the net compressive axial load is acting on the segments AB and CD (the segment
BC is in tension). If the construction joint is intentionally roughened, the factored shear stress
resistance from Clause 11.5 is:


N  
v r = φc  c + µ ρv f y +


Ag  




1931 × 1000  

= 0.65 0.50 + 1.0 0.0025 × 400 +

800 × 3200  


= 1.465 MPa
Hence, the sliding shear resistance of segment AB is:
1.465 × Acv = 1.465 × 400 × 3200 = 1875 kN.
The sliding shear resistance is adequate.
11.5.6.9 Determination of plastic hinge region (Clause 21.6.2)
As the wall cross sectional dimensions remain constant over the 48.65 m height of the wall
and provided that the main flexural reinforcement is appropriately curtailed, only one plastic hinge
region will form, near the base of the walls. The height over which plastic hinging could take place
from the base of the wall is governed by the longer, N-S, wall (Clause 21.6.2.2) and is taken as
1.5 × 6.4 = 9.6 m. Therefore detail the first three storeys as plastic hinge regions.
11.5.6.10 Changes in horizontal distributed reinforcement over the height
of the walls (Clause 21.6.5)
The maximum spacing of the 2–10M horizontal bars, outside of the plastic hinge region is
200 mm, since the minimum reinforcement ratio of 0.0025 must be satisfied. Therefore use 2 10M bars at 200 mm spacing above the plastic hinge region.
11–58
Seismic Design
11.5.6.11 Changes in Vertical distributed reinforcement over the height of
the walls (Clause 21.6.5)
Once again, the minimum reinforcement ratio of 0.0025 governs the selection of vertical
distributed reinforcement. Hence use 2-10M bars at 200 mm spacing over the entire height of the
wall.
11.5.6.12 Changes in concentrated vertical reinforcement over the height
of the walls (Clause 21.6.6)
The minimum area of concentrated reinforcement which can be used outside the plastic
hinge region is 0.001bw l w (Clause 21.6.6.3). At one end of the 3200 mm long wall, the minimum
2
amount of concentrated reinforcement is 0.001× 400 × 3200 = 1280 mm . Similarly, the minimum
amount of concentrated reinforcement required at the intersection of the wall components is
2
2560 − 1280 = 1280 mm (Clause 21.6.6.5).
Note that in deciding on the changes to the concentrated reinforcement over the height of the
structure, it is necessary to ensure that the factored moment resistance at each floor level is
sufficient to develop the plastic hinging at the base of the structure. This check should be made
with due consideration for the effect of lap splices in the reinforcement. An example of the
calculations necessary to ensure adequate flexural strength is given in Reference 1.
11.5.7
Frame Members Not Considered Part of the SFRS
Clause 21.12 provides design requirements for members that are not considered part of the
seismic force resisting system. The shear walls and coupled wall system were designed to take
100% of the seismic loading effects. The slabs and the columns must be checked to determine if
the levels of ductility and strength of these important vertical load carrying members are sufficient.
11.5.7.1 Slab-column connections (Clause 21.12.3)
The reduction factor, RE , on two-way slab shear stress is a function of the interstorey drift.
th
The maximum drifts at the extremities of the structure including torsional effects is 0.00565 (8
th
storey) in the N-S direction and 0.0047 (8 storey) in the E-W direction. Hence the reduction
factor is:
 0.005 

RE = 
 δi 
0.85
 0.005 
=

 0.00565 
0.85
= 0.901
Interior slab-column connection:
For an interior slab-column connection, the gravity load two-way shear stress (excluding
shear
from
unbalanced
loading
and
determined
using
the
seismic
load
combinations (1.0D + 0.5L ) can be determined for a first interior columns location as follows:
(
)
Vf = [1.0(4.8 + 1.5 ) + 0.5(2.4 )]× 5.75 × 6.0 − 0.712 = 255 kN
A clear cover of 25 mm and 15M top bars are assumed for the slab.
The corresponding shear stress is:
vf =
Vf
255 × 1000
=
= 0.561 MPa
bo d 4 × (550 + 160 ) × 160
CAC Concrete Design Handbook
11–59
The limiting shear stress obtained by multiplying RE by the two-way shear stress for this
square column from Clause 13.3.4.1 is:
v c = 0.38φc fc′ RE = 0.38 × 0.65 30 × 0.901 = 1.219 MPa
Hence, no shear reinforcement is required.
Corner slab-column connection:
Applying the same principles to a corner slab-column connection gives a factored shear of:
(
)
Vf = [1.0(4.8 + 1.5 ) + 0.5(2.4 )] × 3.125 2 − 0.73 2 = 69.2 kN
vf =
Vf
69.2 × 1000
=
= 0.296 MPa
bo d 2 × (100 + 550 + 160 / 2) × 160
The limiting shear stress obtained by multiplying RE by the one-way shear stress for this
corner column from Clause 13.3.6 and 11.3.6.2 is:
v c = 0.21φc fc′ RE = 0.21× 0.65 30 × 0.901 = 0.674 MPa
Hence, no shear reinforcement is required.
11.5.7.2 Check on design and detailing of columns (Clause 21.12)
Clauses 21.12.1 to 21.12.2 provide detailed requirements for the columns which are not
considered part of the seismic force resisting system. Minimum design and detailing requirements
must be applied or the columns must be analyzed to determine if the factored moments in the
columns exceed their nominal resistances when the structure is deformed laterally to the design
displacements.
11.5.8
Comparisons with the Design Using the 1994 CSA Standard
The structure designed in this chapter is the same structure designed in Reference 2, except
that the structure designed in this chapter was for a foundation on soil of site Class D (stiff soil),
whereas the structure in Reference 2 was for the same structure founded on rock. In addition,
900 mm deep diagonally reinforced coupling beams were used, instead of 600 mm deep coupling
beams with conventional reinforcement. It is noted that, for this structure, the design force levels
using the 2005 NBCC are somewhat lower than those using the 1995 NBCC.
11.5.9
References
1.
Mitchell, D. and Collins, M.P., "Chapter 11 - Seismic Design", Concrete Design Handbook,
Canadian Portland Cement Association, Ottawa, 1985, pp. 11-1 – 11-31.
2.
Mitchell, D., Paultre, P. and Collins, M.P., "Chapter 11 - Seismic Design", Concrete Design
Handbook, Cement Association of Canada, Ottawa, 1995, pp. 11-1 – 11-33.
3.
Paulay, T., Priestley, M.J.N., “Seismic Design of Reinforced Concrete and Masonry
Buildings”, John Wiley and Sons, NY, 1992, 744p.
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Seismic Design