3 Lateral Load Distribution

Earthquake Resistant Design of structures
GTU # 3732007
Earthquake Resistant
Design of structures
Civil Engineering Department

Darshan Institute of Engineering & Technology, Rajkot
dipak.jivani@darshan.ac.in
9409530079
Section - 3
Components of EQ Acceleration
 Two Horizontal components of Acceleration
 One Vertical Component of Acceleration
 Rotations
Inertia force due to vertical acceleration Inertia force due to Horizontal acceleration
Prof. Dipak Jivani 3

Floor Diaphragm
 Diaphragm: It is a horizontal, or nearly horizontal system, which transmits lateral forces to the vertical
resisting elements, for example, reinforced concrete floors and horizontal bracing systems.
 Types of Diaphragm
 Rigid Diaphragm
 Flexible Diaphragm
Rigid Diaphragm Flexible Diaphragm

Center of Mass - CM
 The point through which the resultant of the masses of a system acts. This point corresponds
to the centre of gravity of masses of system.
 Uniform distribution of mass then, CM=Geometrical center
 Uneven distribution of mass then CM ≠ Geometrical center

Mass of part I is 1200 kg/m2, while that of the other two parts
is 1000 kg/m2.
Let origin be at point A, and the coordinates of the centre of

mass be at (X, Y)
𝑚𝑖 𝑥𝑖
𝑋=
𝑚𝑖
𝑚𝑖 𝑦𝑖
𝑌=
𝑚𝑖
Hence, coordinates of centre of mass are (9.76, 4.1) m.

Centre of Stiffness
 The point through which the resultant of the restoring forces of a system acts.
 Geometrical center of stiffnesses of lateral load resisting elements
𝑘𝑖 𝑥𝑖
𝑋=
𝑘𝑖
𝑘𝑖 𝑦𝑖
𝑌=
𝑘𝑖

In the X-direction
frames located at uniform spacing. Hence, the y
coordinate of centre of stiffness is located symmetrically,
Y = 5.0 m from the left bottom corner.
In the Y-direction,
Let the lateral stiffness of each transverse frame = k,
Hence, coordinates of centre of stiffness are (8.75, 5.0) m.

Torsionally coupled and Uncoupled system
Torsionally Uncoupled system Torsionally coupled system
CM CS CM
CS
Center of mass and center of stiffness coincide Center of mass and center of stiffness is not coinciding

Torsionally coupled system
CM CM
CS CS
1) If CM & CS will not coincide then EQ force produces translation + rotation of floor
2) Rotation of floor produce torsion in vertical elements.

Lateral Load Distribution -Example
Consider a simple one-storey building having two shear
walls in each direction. All four walls are in M25 grade
concrete, 200 thick and 4 m long. Storey height is 4.5 m.
Design shear force on the building is 100 kN in either
direction.
Compute design lateral forces on different shear walls
using the torsion provisions of 2016 edition of IS 1893
(Part 1).

Center of Mass
Centre of mass (CM) will be the geometric centre of the floor slab,
Hence, Xm = 8.0m and Ym = 4.0m.
Eccentricity
e = 2m
Center of Stiffness
𝑘𝑖 𝑥𝑖 𝑘𝑥0 + (𝑘𝑥12) CR CM
𝑋= = =6𝑚
𝑘𝑖 2𝑘 (6.0, 4.0) (8.0, 4.0)
𝑘𝑖 𝑦𝑖 𝑘𝑥0 + (𝑘𝑥8)
𝑌= = =4𝑚
𝑘𝑖 2𝑘
Centre of rigidity (CR) will be at (6.0, 4.0).

Design Eccentricity
Eccentricity along X-Direction e = 0.0 m Eccentricity along Y-Direction = 2.0 m
Design eccentricity: Design eccentricity:

𝑒𝑑 = (1.5×0.0) + (0.05×8) = 0.4 m 𝑒𝑑 = (1.5 × 2.0) + (0.05 × 16) = 3.8 m
and and
𝑒𝑑 = (0.0)−(0.05×8) = −0.4 m 𝑒𝑑 = (2.0) − (0.05 × 16) = 1.2 m

EQ Force in X-direction
Lateral forces in the walls due to translation
For Design eccentricity ey = ± 0.4 m
CR
Lateral Load = 100 kN
(6.0, 4.0)
Stiffness in Y Lateral forces Lateral force due to Total

Stiffness in torsional moment
Direction due to translation Force
Wall X Direction rx ry Kx ry Ky rx 𝑘 𝐹2
Ky F=
Kx 𝐹1 = 𝐹 𝑘𝑥 𝑟𝑦 + 𝑘𝑦 𝑟𝑥
𝑘 =
𝑘 𝑟2
(𝐹 × 𝑒) F1+F2
A 0 k -6 0 0 -6k 0 2.31 2.31

B 0 k +6 0 0 6k 0 2.31 2.31
C k 0 0 +4 4k 0 50 1.54 51.54
D k 0 0 -4 -4k 0 50 1.54 51.54
2k 2k
𝐼𝑝 = 𝑘𝑟 2 = 𝑘𝑥 𝑟𝑦2 + 𝑘𝑦 𝑟𝑥2 = 36 + 36 + 16 + 16 𝑘 = 104 𝑘

EQ Force in Y-direction Lateral forces in the walls due to translation:
For Design eccentricity ex = 3.8 m CR

Lateral Load = 100 kN (6.0, 4.0)
Lateral forces Lateral force due to

Stiffness in X Total
Stiffness in due to torsional moment
Direction Force
Wall X Direction rx ry Kx ry Ky rx translation 𝐹2
Ky 𝑘 F=
Kx 𝑘𝑥 𝑟𝑦 + 𝑘𝑦 𝑟𝑥
𝐹1 = 𝐹 = (𝐹 × 𝑒) F1+F2
𝑘 𝑘 𝑟2
A 0 k -6 0 0 -6k 50 -21.92 28.08
B 0 k +6 0 0 6k 50 + 21.92 71.92
C k 0 0 +4 4k 0 0 − 14.62 -14.62
D k 0 0 -4 -4k 0 0 + 14.62 14.62
2k 2k
𝐼𝑝 = 𝑘𝑟 2 = 𝑘𝑥 𝑟𝑦2 + 𝑘𝑦 𝑟𝑥2 = 36 + 36 + 16 + 16 𝑘 = 104 𝑘

EQ Force in Y-direction Lateral forces in the walls due to translation:
For Design eccentricity ex = 1.2 m

Lateral Load = 100 kN
Lateral forces Lateral force due to

Stiffness in X Total
Stiffness in due to torsional moment
Direction Force
Wall X Direction rx ry Kx ry Ky rx translation 𝐹
Ky 𝑘 F=
Kx 𝑘𝑥 𝑟𝑦 + 𝑘𝑦 𝑟𝑥
𝐹1 = 𝐹 = (𝐹 × 𝑒) F1+F2
𝑘 𝑘 𝑟2
A 0 k -6 0 0 -6k 50 - 6.93 43.07
B 0 k +6 0 0 6k 50 + 6.93 56.93
C k 0 0 +4 4k 0 0 - 4.62 - 4.62
D k 0 0 -4 -4k 0 0 4.62 4.62
2k 2k
𝐼𝑝 = 𝑘𝑟 2 = 𝑘𝑥 𝑟𝑦2 + 𝑘𝑦 𝑟𝑥2 = 36 + 36 + 16 + 16 𝑘 = 104 𝑘

51.54 kN 14.62 kN
43.07 kN 71.92 kN
2.31 kN 2.31 kN
51.54 kN 14.62 kN

Flow of Inertia Forces to Foundation
 The lateral inertia forces are
transferred by the floor slab to the
walls or columns, to the foundations,
and finally to the soil system
underneath.
 The load resisting system must be of

closed loops, enable to transfer all the
forces acting either vertically or
horizontally to the ground

Lateral Load Resisting Systems
 To efficiently transfer lateral load from floor diaphragm to foundation.
 Moment Resisting Frame
 Shear wall
 Braced frame
 Combination of above
Moment Resisting Frame Shear wall Braced frame

Moment Resisting Frame
• Columns and Girders joined by moment resisting connections

• Lateral stiffness of the frame depends on the flexural stiffness of the
beams, columns, and connections.
• Economical for buildings up to about 10-15 stories.
• Well suited for reinforced concrete construction due to the inherent
continuity in the joints.
• Gravity loads also resisted by frame action.
• This system is generally preferred by architects because they are
relatively un-obtrusive compared with shear walls or braced frame

• Rigid jointed frame,
• Stiffness of column is important.
Deflected shape due To Deflected shape due To Flexural Flexural Deformation Of

Flexural Deformation Of Deformation Of Column And Beam And Column & Axial
Column Beams Deformation Of Column

Shear wall
• wall elements designed to take vertical as well as inplane
horizontal (lateral) forces
• Concrete buildings
• Wood buildings
• Masonry buildings
• resist lateral forces by shear deformation
• stiffer buildings
• Generally constructed with concrete, Shear walls have high
in-plane stiffness and strength.
• Well suited for tall buildings up to about 35 stories.
• Can be used around elevator and/or stair cores.

• Act as a deep cantilever,
• Concrete wall for elevators, stairs and service well.
H / L <= 1 H/L>3
Shear Deformation Bending Deformation

Reinforced concrete shear walls
– an excellent structural system
Layout and symmetry

Shear wall – Frame (Dual system) combination
• This system consists of shear wall (or braced frame) and moment resisting frame
• The two systems are designed to resist the total design force in proportion to their lateral stiffness
considering the interaction of the dual system at all floor levels
• The moment resisting frames are designed to independently resist at least 25% of design seismic base shear

Ensure compatibility of deformation of walls and frames

Braced Frames
• Braced Frames are basically vertical truss systems.

• Almost exclusively steel or timber.
• Highly efficient use of material since forces are primarily axial. Creates a laterally stiff building with
relatively little additional material.
• Good for buildings of any height.
• May be internal or external.
• Braces used to resist lateral loads
• steel or concrete
• Damage can occur when braces buckle
• Stiffer than pure frame

Section - 3
Methods of Seismic Analysis

Code Based Procedure for Lateral Load
 Equivalent Static Force Method
 Dynamic Analysis
 Response Spectrum Method
 Time History Analysis

Equivalent static load procedure
 The equivalent lateral force for an earthquake is a unique concept used in earthquake engineering. The
concept is attractive because it converts a dynamic analysis into partly dynamic and partly static
analyses for finding the maximum displacement (or stresses) induced in the structure due to earthquake
excitation.
 For seismic resistant design of structures, only these maximum stresses are of interest, not the time
history of stresses.
 The equivalent lateral force for an earthquake is defined as a set of lateral static forces which will
produce the same peak response of the structure as that obtained by the dynamic analysis of the
structure under the same earthquake.
 This equivalence is restricted only to a single mode of vibration of the structure

Basis of Equivalent Lateral Force (Static Force) Procedure
Vb = m a
Vb = (W/g) a
Vb = W (a/g)
Vb = W Ah
Ah = Basic horizontal seismic coefficient

Vb = Base shear
W = Total weight of the structure
a = Acceleration induced at the base during earthquake
g = Acceleration due to gravity
Ah is modified to consider the following effects.

• Natural period
• Damping
• Modal shapes
• Types of structure and place(zone)
• Subsoil conditions
• Importance of the structure
Assumptions
• Assume that structure is rigid.

• Assume perfect fixity between structure and foundation.
• During ground motion every point on the structure experience same accelerations
• Dominant effect of earthquake is equivalent to horizontal force of varying magnitude over the height.

Response Spectrum Method
 Here Period and mode shapes of the structure are obtained using free vibration analysis not
from Empirical formula
 (Sa / g ) is obtained from the same response chart for all the modes separately
 Distribution of forces at various story's is carried out using mode shape, Participation Factors
etc.
 Response quantities ( BM, SF etc. ) are combined using CQC Complete Quadratic combination

Lateral forces are found by superimposition of the
Forces resulting from each mode

The advantages of RSA, compared with time-history analysis
 The size of the problem is reduced to finding only the maximum response of a limited number of modes of the
structure, rather than calculating the entire time history of responses during the earthquake.
 The use of smoothed envelope spectra makes the analysis independent of the characteristics of a particular
earthquake record.
 RSA can very often be useful as a preliminary analysis, to check the reasonableness of results produced by time-
history analyses.

Disadvantages of RSA
 RSA is essentially linear and can make only approximate allowance for nonlinear behavior.
 The results are in terms of peak response only, with a loss of information on frequency content, phase and number
of damaging cycles, which have important consequences for low-cycle fatigue effects.

Time History Analysis
 Obtain the design parameters by giving the actual Earthquake excitation

 To perform such an analysis, a representative earthquake time history is required for a
structure being evaluated.
 The time-history method is applicable to both elastic and inelastic analysis.

Choice Of Method For Multistoried Building

Section - 3
Example
Example - 2

Example - 2

Example - 2

Example - 2

Example - 2
Wi.hi2 Wi.hi2
Floor Wi hi Qi = VB. Vi
Wi.hi2 Wi.hi2
Lvl. (kN) (m) Wi.hi2 (kN)
(kN)
Roof (7) 4277.77 24.5 2567731.44 0.2943 1319.35 1319.35
6 5523.84 21.0 2436013.44 0.2792 1251.65 2571.00
5 5523.84 17.5 1691676.00 0.1938 868.80 3439.80
4 5523.84 14.0 1082672.64 0.1241 556.34 3996.14
3 5523.84 10.5 609003.36 0.0698 312.91 4309.05
2 5523.84 7.0 270668.16 0.031 138.97 4448.02
1 5523.84 3.5 67667.04 0.0077 34.52 4483.00
Wi.hi2 = Qi = 4483 kN
8725432.08

Example - 2

Section - 3
 For three storey building frame lumped mass and storey stiffness are shown in figure. Using
response spectrum method find out lateral force at storey level.
𝑚3 = 2000 kg
𝑘3 = 5𝑥105 𝑁/𝑚
𝑚2 = 3000 kg
𝑘2 = 5𝑥105 𝑁/𝑚
𝑚1 = 3000 kg
𝑘1 = 5𝑥105 𝑁/𝑚

𝑚1 0 0 3000 0 0
𝑀𝑎𝑠𝑠 𝑚𝑎𝑡𝑟𝑖𝑥 = 𝑚 = 0 𝑚2 0 = 0 3000 0 𝑘𝑔
0 0 𝑚3 0 0 2000
𝑘1 + 𝑘2 −𝑘2 0 10 × 105 −5 × 105 0

𝑆𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠 𝑚𝑎𝑡𝑟𝑖𝑥 = 𝑘 = −𝑘2 𝑘2 + 𝑘 3 −𝑘3 = −5 × 105 10 × 105 −5 × 105 𝑁/𝑚
0 −𝑘3 𝑘3 0 −5 × 105 5 × 105

eigen values and eigen vectors
𝑘 − 𝜔2 𝑚 = 0
10 × 105 −5 × 105 0 3000 0 0

2
−5 × 105 10 × 105 −5 × 105 − 𝜔 0 3000 0 =0
0 −5 × 105 5 × 105 0 0 2000
10 −5 0 0.03 0 0
2
−5 10 −5 − 𝜔 0 0.03 0 =0
0 −5 5 0 0 0.02
10 − 0.03𝜔2 −5 0
−5 10 − 0.03𝜔2 −5 =0
0 −5 5 − 0.02𝜔2
Solving above equation,

Natural Frequencies
𝜔12 = 575.79 𝜔1 = 6.33 𝑟𝑎𝑑/𝑠𝑒𝑐

𝜔22 = 40.09 𝜔2 = 17.34 𝑟𝑎𝑑/𝑠𝑒𝑐
𝜔32 = 300.77 𝜔3 = 24.00 𝑟𝑎𝑑/𝑠𝑒𝑐
Time Periods (T)
2𝜋 2𝜋
𝑇1 = = = 0.992 𝑠𝑒𝑐
𝜔1 6.33
2𝜋 2𝜋
𝑇2 = = = 0.362 𝑠𝑒𝑐
𝜔2 17.34
2𝜋 2𝜋
𝑇3 = = = 0.262 𝑠𝑒𝑐
𝜔3 24

Mode Shape -1
𝐹𝑜𝑟 𝜔1 = 6.33 𝑟𝑎𝑑/𝑠𝑒𝑐
𝑘 − 𝜔2 𝑚 ∅ = 0 ∅31 = 2.095
10 − 0.03𝜔2 −5 0 ∅11 0
−5 10 − 0.03𝜔2 −5 ∅21 = 0
0 −5 5 − 0.02𝜔2 ∅31 0
10 − 0.03 × 6.332 −5 0 ∅11 0 ∅21 = 1.759

−5 10 − 0.03 × 6.332 −5 ∅21 = 0
0 −5 5 − 0.02 × 6.332 ∅31 0
8.798 −5 0 ∅11 0
−5 8.798 −5 ∅21 = 0 ∅11 = 1
0 −5 4.198 ∅31 0
8.798 ∅11 − 5∅21 = 0

−5 ∅11 + 8.798 ∅21 − 5∅31 = 0
−5 ∅21 + 4.198 ∅31 = 0 Mode Shape -1
1
Considing ∅11 =1 and by solving above equation ∅1 = 1.759
2.095
Mode Shape -2
𝐹𝑜𝑟 𝜔2 = 17.34 𝑟𝑎𝑑/𝑠𝑒𝑐 ∅32 = −0.961
𝑘 − 𝜔2 𝑚 ∅ = 0
10 − 0.03𝜔2 −5 0 ∅12 0
−5 10 − 0.03𝜔2 −5 ∅22 = 0
0 −5 5 − 0.02𝜔2 ∅32 0
∅22 = 0.196
10 − 0.03 × 17.34 2
−5 0 ∅12 0
−5 10 − 0.03 × 17.342 −5 ∅22 = 0
0 −5 5 − 0.02 × 17.342 ∅32 0
0.979 −5 0 ∅12 0
−5 0.979 −5 ∅22 = 0 ∅12 = 1
0 −5 −1.02 ∅32 0
0.979 ∅12 − 5∅22 = 0

−5 ∅12 + 0.979 ∅22 − 5∅32 = 0
−5 ∅22 − 1.02 ∅32 = 0 Mode Shape -2
1
Considing ∅12 =1 and by solving above equation ∅2 = 0.196
−0.961
Mode Shape - 3
𝐹𝑜𝑟 𝜔3 = 24 𝑟𝑎𝑑/𝑠𝑒𝑐
𝑘 − 𝜔2 𝑚 ∅ = 0 ∅33 = 1.12
10 − 0.03𝜔2 −5 0 ∅11 0
−5 10 − 0.03𝜔2 −5 ∅21 = 0
0 −5 5 − 0.02𝜔2 ∅31 0
10 − 0.03 × 242 −5 0 ∅13 0

∅23 = 0 ∅23 = −1.456
−5 10 − 0.03 × 242 −5
0 −5 5 − 0.02 × 242 ∅33 0
−7.28 −5 0 ∅13 0
−5 −7.28 −5 ∅23 = 0 ∅13 = 1
0 −5 −6.52 ∅33 0
7.28 ∅13 − 5∅23 = 0

−5 ∅13 + 7.28 ∅23 − 5∅33 = 0
−5 ∅23 − 6.52∅33 = 0
Mode Shape - 3
1
Considing ∅13 =1 and by solving above equation ∅3 = −1.456
1.12
Modal Participation Factor Calculation
Storey Seismic weight Mode-1

level Wi (kg) ∅𝑖𝑘 𝑊𝑖 ∅𝑖𝑘 𝑊𝑖 ∅2𝑖𝑘
3 2000 2.095 4190 8778.05
2 3000 1.759 5277 9282.24
1 3000 1.000 3000 3000
𝑊𝑖 ∅𝑖𝑘 = 12467 𝑊𝑖 ∅2𝑖𝑘 = 21060.29
Modal Mass 12467 2
𝑊𝑖 ∅𝑖𝑘 2 𝑀1 = = 7380 𝑘𝑔
21060.29
𝑀𝑖 =
𝑔 𝑊𝑖 ∅2𝑖𝑘
% of Total Weight 7380
= × 100 = 92.25 %
8000
Modal Participation Factor 12467
𝑊𝑖 ∅𝑖𝑘 𝑃1 = = 0.592
21060.29
𝑃𝑖 =
𝑊𝑖 ∅2𝑖𝑘

3 2000 -0.961 -1922 1847.042
2 3000 0.196 588 115.248
1 3000 1.000 3000 3000
𝑊𝑖 ∅𝑖𝑘 = 1666 𝑊𝑖 ∅2𝑖𝑘 = 4962.29
Modal Mass 1666 2
𝑊𝑖 ∅𝑖𝑘 2 𝑀2 = = 559.33 𝑘𝑔
4962.29
𝑀𝑖 =
% of Total Weight 559.33
= × 100 = 6.99 %
8000
𝑊𝑖 ∅𝑖𝑘 𝑃2 = = 0.336
4962.29
𝑃𝑖 =

3 2000 1.12 2240 2508.8
2 3000 -1.456 -4368 6359.808
1 3000 1.000 3000 3000
𝑊𝑖 ∅𝑖𝑘 = 872 𝑊𝑖 ∅2𝑖𝑘 = 11868.61
Modal Mass 872 2
𝑊𝑖 ∅𝑖𝑘 2 𝑀3 = = 64 𝑘𝑔
11868.61
𝑀𝑖 =
% of Total Weight 64
= × 100 = 0.8 %
8000
𝑊𝑖 ∅𝑖𝑘 𝑃3 = = 0.0734
11868.61
𝑃𝑖 =

Design Horizontal seismic coefficient
𝑍 = 𝑍𝑜𝑛𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 = 0.36 𝑓𝑜𝑟 𝑧𝑜𝑛𝑒 𝑉

𝐼 = 𝐼𝑚𝑝𝑜𝑟𝑡𝑎𝑛𝑐𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 = 1.5
𝑅 = 𝑅𝑒𝑠𝑝𝑜𝑛𝑠𝑒 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 = 5 𝑓𝑜𝑟 𝑆𝑀𝑅𝐹
Mode 1 Mode 2 Mode 3

Time period T (sec) 0.992 0.632 0.262
𝑆𝑎 1.008 2.50 2.50
𝑔
𝑧 𝐼 𝑆𝑎 0.36 1.5 0.36 1.5 0.36 1.5
𝐴ℎ = = 1.008 = 1.008 = 1.008
2𝑅 𝑔 2 5 2 5 2 5
𝐴ℎ1 = 0.0544 𝐴ℎ2 = 0.135 𝐴ℎ1 = 0.135

Lateral Load Calculation

level Wi (N) =mi x g ∅𝑖𝑘 𝐴𝑘 𝑃𝑘 Design Lateral Force Storey Shear
𝑄𝑖𝑘 = 𝐴𝑘 ∅𝑖𝑘 𝑃𝑘 𝑊𝑖 𝑉𝑖𝑘 (N)
(N)
3 2000x9.81 = 19620 2.095 0.0544 0.592 𝑄31 = 1323.74 𝑉31 = 1323.74
2 3000x9.81 = 29430 1.759 0.0544 0.592 𝑄21 = 1667.15 𝑉21 = 2990.89
1 3000x9.81 = 29430 1.000 0.0544 0.592 𝑄11 = 947.78 𝑉11 = 3938.67

(N)
3 2000x9.81 = 19620 2.095 0.135 0.336 𝑄32 = −855.25 𝑉32 = −855.25
2 3000x9.81 = 29430 1.759 0.135 0.336 𝑄22 = 261.65 𝑉22 = −593.6
1 3000x9.81 = 29430 1.000 0.135 0.336 𝑄12 = 1334.95 𝑉12 = 741.35

(N)
3 2000x9.81 = 19620 2.095 0.135 0.0734 𝑄33 = 217.74 𝑉33 = 217.74
2 3000x9.81 = 29430 1.759 0.135 0.0734 𝑄23 = −424.6 𝑉23 = −206.86
1 3000x9.81 = 29430 1.000 0.135 0.0734 𝑄13 = 291.62 𝑉13 = 84.76

𝑉31 = 1323.74 𝑁 𝑉32 = −855.32 𝑁 𝑉33 = 217.74 𝑁
𝑄31 = 1323.74 𝑁 𝑄32 = −855.25 𝑁 𝑄33 = 217.74 𝑁

+ − +
𝑉23 =
−206.86 𝑁
𝑄22 = 261.55𝑁 𝑉22 = 𝑄21 = −424.6𝑁
𝑄21 = 1667.15𝑁 𝑉21 = 2990.89 𝑁 −593.6 𝑁
− −
+ 𝑉12 = 𝑉13 =
741.35 𝑁 84.76 𝑁
𝑄11 = 947.78 𝑁 𝑄12 = 1334.95 𝑁 𝑄31 = 291.62 𝑁
𝑉11 = 3938.67 𝑁
+ +
+
𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝐹𝑜𝑟𝑐𝑒 𝑆𝑡𝑜𝑟𝑒𝑦 𝑠ℎ𝑒𝑎𝑟 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝐹𝑜𝑟𝑐𝑒 𝑆𝑡𝑜𝑟𝑒𝑦 𝑠ℎ𝑒𝑎𝑟 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝐹𝑜𝑟𝑐𝑒 𝑆𝑡𝑜𝑟𝑒𝑦 𝑠ℎ𝑒𝑎𝑟
𝐌𝐨𝐝𝐞 𝟏 𝐌𝐨𝐝𝐞 𝟐 𝐌𝐨𝐝𝐞 𝟑

Complete Quadratic combination method (CQC)
𝜔1 = 6.33 𝑟𝑎𝑑/𝑠𝑒𝑐
𝜔2 = 17.34 𝑟𝑎𝑑/𝑠𝑒𝑐
𝜔3 = 24.00 𝑟𝑎𝑑/𝑠𝑒𝑐
𝜔1 𝜔2 𝜔3
𝛽11 𝛽21 𝛽31 𝜔1 𝜔1 𝜔1 1 2.73 3.79
𝛽32 = 𝜔1 𝜔2 𝜔2 𝜔3
𝛽 = 𝛽12 𝛽22 𝜔2 𝜔2 = 0.365 1 1.384
𝛽13 𝛽23 𝛽33 𝜔1 𝜔2 𝜔3 0.263 0.722 1
𝜔3 𝜔3 𝜔3
Damping ratio ζ = 5% =0.05
𝜌11 𝜌21 𝜌31 1 0.008 0.0039

𝜌 = 𝜌12 𝜌22 𝜌32 = 0.008 1 0.0846
𝜌13 𝜌23 𝜌33 0.0039 0.0846 1

𝜌𝟏𝟏 𝜌𝟐𝟏 𝜌𝟑𝟏 V31
V3 = V31 V32 V33 𝜌𝟏𝟐 𝜌𝟐𝟐 𝜌𝟑𝟐 V32
𝜌𝟏𝟑 𝜌𝟐𝟑 𝜌𝟑𝟑 V33
1 0.008 0.0039 1323.74

𝑉3 = 1323.74 −855.25 217.74 0.008 1 0.0846 −855.25
0.0039 0.0846 1 217.74
V3 = 1576.015 𝑁
𝜌𝟏𝟏 𝜌𝟐𝟏 𝜌𝟑𝟏 V21

V2 = V21 V22 V23 𝜌𝟏𝟐 𝜌𝟐𝟐 𝜌𝟑𝟐 V22
𝜌𝟏𝟑 𝜌𝟐𝟑 𝜌𝟑𝟑 V23
1 0.008 0.0039 2990.89

𝑉2 = 2990.89 −593.6 −206.86 0.008 1 0.0846 −593.6
0.0039 0.0846 1 −206.86
V2 = 3054.209 𝑁
𝜌𝟏𝟏 𝜌𝟐𝟏 𝜌𝟑𝟏 V11
V1 = V11 V12 V13 𝜌𝟏𝟐 𝜌𝟐𝟐 𝜌𝟑𝟐 V12
𝜌𝟏𝟑 𝜌𝟐𝟑 𝜌𝟑𝟑 V13
1 0.008 0.0039 3938.67

𝑉1 = 3938.67 741.35 84.76 0.008 1 0.0846 741.35
0.0039 0.0846 1 84.76
V1 = 4016.184 𝑁

SRSS method
V3 = V31 2 + V32 2 + V33 2
V3 = 1323.74 2 + −855.25 2 + 217.74 2
V3 = 1590.96 𝑁
V2 = V21 2 + V22 2 + V23 2
V2 = 2990.89 2 + −593.6 2 + −206.86 2
V2 = 3056.23 𝑁
V1 = V11 2 + V12 2 + V13 2
V1 = 3938.67 2 + 741.35 2 + 84.76 2
V1 = 4008.72 𝑁

ABS method
V3 = V31 + V32 + V33 = 1323.74 + 855.25 + 217.74 = 2396.73 N
V2 = V21 + V22 + V23 = 2990.89 + 593.6 + 206.86 = 3791.35 N
V1 = V11 + V12 + V13 = 3938.67 + 741.35 + 84.76 = 4767.78 N

𝑉3 = 1576.015 𝑁
𝑄3 = 1576.015 𝑁
+ Comparison of Storey shear
Storey CQC (N) SRSS (N) ABS (N)
𝑉2 = 3054.209 𝑁 3 1576.015 1590.96 2396.73

𝑄2 = 1478.19 𝑁
2 3054.209 3056.23 3791.35
+
1 4016.184 4008.72 4767.78
𝑄1 = 961.975 𝑁
𝑉1 = 4016.184 𝑁
+
𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝐹𝑜𝑟𝑐𝑒 𝑆𝑡𝑜𝑟𝑒𝑦 𝑠ℎ𝑒𝑎𝑟
𝐂𝐐𝐂 𝐌𝐞𝐭𝐡𝐨𝐝

 Looping
Thank you…!
Next Topic

3 Lateral Load Distribution

Uploaded by

Copyright:

Available Formats

3 Lateral Load Distribution

Uploaded by

Document Information

Original Description:

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

3 Lateral Load Distribution

Uploaded by

Copyright:

Available Formats

Earthquake Resistant Design of structures

Civil Engineering Department

Prof. Dipak Jivani 3

Rigid Diaphragm Flexible Diaphragm

 Uniform distribution of mass then, CM=Geometrical center

 Uneven distribution of mass then CM ≠ Geometrical center

Prof. Dipak Jivani 5

Let origin be at point A, and the coordinates of the centre of

Hence, coordinates of centre of mass are (9.76, 4.1) m.

Prof. Dipak Jivani 6

Prof. Dipak Jivani 7

Hence, coordinates of centre of stiffness are (8.75, 5.0) m.

Prof. Dipak Jivani 8

Torsionally Uncoupled system Torsionally coupled system

Prof. Dipak Jivani 9

Prof. Dipak Jivani 10

Prof. Dipak Jivani 13

Centre of rigidity (CR) will be at (6.0, 4.0).

Prof. Dipak Jivani 14

Eccentricity along X-Direction e = 0.0 m Eccentricity along Y-Direction = 2.0 m

Design eccentricity: Design eccentricity:

Prof. Dipak Jivani 15

Stiffness in Y Lateral forces Lateral force due to Total

A 0 k -6 0 0 -6k 0 2.31 2.31

𝐼𝑝 = 𝑘𝑟 2 = 𝑘𝑥 𝑟𝑦2 + 𝑘𝑦 𝑟𝑥2 = 36 + 36 + 16 + 16 𝑘 = 104 𝑘

Prof. Dipak Jivani 16

For Design eccentricity ex = 3.8 m CR

Lateral forces Lateral force due to

𝐼𝑝 = 𝑘𝑟 2 = 𝑘𝑥 𝑟𝑦2 + 𝑘𝑦 𝑟𝑥2 = 36 + 36 + 16 + 16 𝑘 = 104 𝑘

For Design eccentricity ex = 1.2 m

Lateral forces Lateral force due to

𝐼𝑝 = 𝑘𝑟 2 = 𝑘𝑥 𝑟𝑦2 + 𝑘𝑦 𝑟𝑥2 = 36 + 36 + 16 + 16 𝑘 = 104 𝑘

Prof. Dipak Jivani 18

Prof. Dipak Jivani 19

 The load resisting system must be of

Prof. Dipak Jivani 20

Moment Resisting Frame Shear wall Braced frame

• Columns and Girders joined by moment resisting connections

Prof. Dipak Jivani 22

Deflected shape due To Deflected shape due To Flexural Flexural Deformation Of

Prof. Dipak Jivani 23

Prof. Dipak Jivani 24

Shear Deformation Bending Deformation

Layout and symmetry

Prof. Dipak Jivani 26

Prof. Dipak Jivani 27

Prof. Dipak Jivani 28

• Braced Frames are basically vertical truss systems.

Prof. Dipak Jivani 29

Prof. Dipak Jivani 33

 Equivalent Static Force Method

Prof. Dipak Jivani 34

 This equivalence is restricted only to a single mode of vibration of the structure

Prof. Dipak Jivani 35

Ah = Basic horizontal seismic coefficient

Ah is modified to consider the following effects.

• Assume that structure is rigid.

Prof. Dipak Jivani 37

Prof. Dipak Jivani 38