IS 1893
IS 1893:2002 CRITERIA FOR EARTHQUAKE RESISTANT DESIGN OF STRUCTURES
PART1 GENERAL PROVISIONS AND BUILDINGS
The Code is now split into five parts
Part 1 - General provisions and buildings
Part 2 - Liquid retaining tanks - Elevated and ground supported
Part 3 - Bridges and retaining walls
Part 4 - Industrial structures including stack like structures
Part 5 - Dams and embankments
Part 1 contains provisions that are general in nature and applicable to all
structures. Also, it contains provisions that are specific to buildings only.
The important changes as compared to IS:1893-1984 are as follows:
1. Seismic zone map is revised with only four seismic zones. Zone I is
upgraded to Zone II. Killari area is enhanced to Zone III. Bellary isolated zone
is removed. East coast is enhanced to Zone III and connected with Zone III of
Godavari Graben area.
2. Seismic zone factor is changed reflecting more realistic value of peak
ground acceleration.
3. Response acceleration spectra are now specified for three types of
founding strata viz. Hard, Medium and Soft.
4. The empirical formula for calculating fundamental natural period T=0.1n
for moment resisting frames without bracing or shear walls is replaced with
Ta=0.075h0.075 for RC framed buildings. This formula applies to bare frames
e.g. in industrial plant buildings. The formula for framed buildings with
in-filled masonry walls is Ta = 0.09h/d0.5 where h and d are the height and
base dimension of the building along the considered direction of earthquake.
5. Revised procedure first calculates the actual force that may be
experienced by the structure during the probable maximum earthquake, if it were
to remain elastic. Then response reduction due to ductile deformation or
frictional energy dissipation in the cracks is applied via `response reduction
factor' R in place of the earlier performance factor K. The list of building
systems and the corresponding values of R is more exhaustive.
The code procedures for calculating base shear VB are summarized below:
IS:1893-1984
------------
VB = K.C.αh.W
where
K = Performance factor 1.0 for SMRF (IS:4326 detail) and 1.6 for OMRF (IS:456
detail)
C = Fundamental time period dependant coefficient
= 1.0 for T <= 0.35 sec and 0.5/T(2/3) for T > 0.35 sec
αh = β.I.α0
β = Soil-foundation system dependant coefficient
= 1.2 for isolated footings without tie beams in medium soils, piles in soft
soils, combined or isolated footings with tie beams in soft soils
= 1.5 isolated footings without tie beams in soft soils
I = Importance factor
= 1.5 for hospitals, schools, cinema halls, monumental structures, telephone
exchanges, radio, fire, railway power stations and 1.0 for others
α0 = Zone dependant design seismic coefficient
Zone II III IV V
α0 0.02 0.04 0.05 0.08
W = Seismic weight of building
= Dead load + appropriate amount of live load
= Dead load + 25% for LL up to 3 kN/sq.m
50% for LL > 3 kN/sq.m
0% for LL on roof
IS:1893-2002
------------
VB = Ah.W
[Z/2].[Sa/g]
Ah = ------------
[R/I]
Z = Zero period acceleration value for the Maximum Considered Earthquake
Zone II III IV V
Z 0.10 0.16 0.24 0.36
Sa/g = Spectral acceleration coefficient for Hard, Medium or Soft soil, 5% damping
= 2.5 for T <= 0.40 and 1.00/T for T > 0.40 (Hard: GP,GW,SP,SW,SC with N>30)
= 2.5 for T <= 0.55 and 1.36/T for T > 0.55 (Medium: All with 10<N<30 SP with N>15)
= 2.5 for T <= 0.67 and 1.67/T for T > 0.67 (Soft: All except SP with N<10)
I = Importance factor
= 1.5 for hospitals, schools, cinema halls, monumental structures, telephone
exchanges, television, radio, fire, railway power stations and 1.0 for
others
R = Response reduction factor
Ordinary RC Moment Resisting Frame (OMRF) 3
Special RC Moment Resisting Frame (SMRF) 5
Ordinary RC Shear Walls 3
Ductile RC Shear Walls 3
Dual Systems with frames carrying >25% of VB
Ordinary RC Shear Walls with OMRF 3
Ordinary RC Shear Walls with OMRF 4
Ductile Shear Walls with OMRF 4.5
Ductile Shear Walls with SMRF 5
W = Seismic weight of building
= Dead load + appropriate amount of live load
= Dead load + 25% for LL up to 3 kN/sq.m
50% for LL > 3 kN/sq.m
0% for LL on roof
Comparative Values of Maximum Base Shear
----------------------------------------
IS:1893-1984 IS:1893-2002 Percent Increase
Zone VB(SMRF) VB(OMRF) VB(SMRF) VB(OMRF) SMRF OMRF
II 0.02W 0.032W 0.025W 0.042W 25.0 31.3
III 0.04W 0.064W 0.040W 0.067W 0.0 4.7
IV 0.05W 0.080W 0.060W 0.100W 20.0 25.0
V 0.08W 0.128W 0.090W 0.150W 12.5 17.2
Elimination of C results in higher force up to 11%, 25% and 43% for hard,
medium and soft soils respectively in the peak region and elimination of beta
results in lower force for soft soils to the extent of 20%.
6. Accidental torsion is introduced:
edi = 1.5 esi + 0.05 bi or
= esi - 0.05 bi whichever governs
edi = Design eccentricity at floor i
esi = Static eccentricity at floor i defined as the distance between center
of mass and center of rigidity
bi = Plan dimension of floor i perpendicular to the direction of force
7. Definition and treatment of irregularities is elaborated:
In Plan
-------
i) Torsional : If floor diaphragms are rigid in their own plane and maximum
storey drift at one end is > 1.2*average storey drift
ii) Re-entrant corners : if projection beyond re-entrant corner is > 15% of
plan dimension in that direction
iii) Diaphragm discontinuity: if open areas > 50% of gross enclosed area or
change in effective diaphragm stiffness from one storey to next > 50%
iv) Out-of-plane offsets: discontinuities in lateral load resisting paths
v) Non-parallel systems
In Elevation
------------
i) Soft-storey: Lateral stiffness < 70% of that in in the storey above or
< 80% of the average lateral stiffness of three storeys above
ii) Mass: seismic weight of any storey except roof < 200% of adjacent storeys
iii) Geometic: horizontal dimension of a lateral force resisting element >
150% of that in adjacent storey
iv) In plane discontinuity: In plane offset of a lateral force resisting
element > length of that element
v) Weak-storey having lateral strength < 80% of that in the storey above
Additional requirements for some of the irregularities are specified:
Soft Storey:
The columns and beams of the soft storey are to be designed for 2.5 times the
storey shears and moments calculated under seismic loads besides the columns
designed and detailed for the calculated storey shears and moments, shear walls
placed symmetrical in both directions of the building as far away from the center
of the building as feasbible to be designed exclusively for 1.5 times the lateral
storey shear force calculated as before.
Non-Parallel Systems:
Earthquake effects about the two orthogonal axes must be combined:
a. Ex ± 0.3Ey
b. Ey ± 0.3Ex
8. More load combinations are required:
Basic Combination Expanded Combinations
1) 1.5(D+L)
2) 1.2(D+L±E)
1.2(D+L+EXP), 1.2(D+L+EXN), 1.2(D+L-EXP), 1.2(D+L-EXN),
1.2(D+L+EYP), 1.2(D+L+EYN), 1.2(D+L-EYP), 1.2(D+L-EYN)
3) 1.5(D±E)
1.5(D+EXP), 1.5(D+EXN), 1.5(D-EXP), 1.5(D-EXN),
1.5(D+EYP), 1.5(D+EYN), 1.5(D-EYP), 1.5(D-EYN)
4) 0.9D±1.5E
0.9D+1.5EXP, 0.9D+1.5EXN, 0.9D-1.5EXP, 0.9D-1.5EXN,
0.9D+1.5EYP, 0.9D+1.5EYN, 0.9D-1.5EYP, 0.9D-1.5EYN
where
D : Dead
L : Live
E : Earthquake
X : along X axis
Y : along Y axis
P : Positive eccentricity
N : Negative eccentricity
This results in 25 actual combinations but can be reduced rationally.
Limits of static analysis
-------------------------
Zone II Zone III Zone IV Zone V
Regular buildings <90 m <90 m <40 m <40 m
Irregular buildings <40 m <40 m <12 m <12 m
Example
The data for this example is as follows:
Storeys 2 Concrete grade 25 Seismic zone 3
Foundation depth mm 1500 Slab cover mm 25 Frame type OMRF
Ground storey height mm 3000 Beam cover mm 50 Soil type Medium
Typical storey height mm 3000 Column cover mm 50 Importance factor 1.0
----------------------------------------------------------------------
Column Data Beam Data Slab Data
----------------------------------------------------------------------
Mark Cx Cy Mark bw D Load Mark t Loads
mm mm mm mm kN/m mm kN/sq.m
Dead Live
----------------------------------------------------------------------
C1 300 300 B1 300 500 6.25 S1 150 1.25 5.00
B1R 300 500 2.25 S1R 150 3.25 1.00
B1G 300 400 6.25
----------------------------------------------------------------------
Loads (kN):
Roof Floor Ground
----------------------- ---------------------- ----------------------
Slab 0.15*25*25 = 93.75 0.15*25*25 = 93.75 = 0
Finish 3.25*25 = 81.25 1.25*25 = 31.25 = 0
Beams 0.3*0.5*25*20 = 75 0.3*0.5*25*20 = 75 0.3*0.4*25*20 = 60
Walls 2.25*20 = 45 6.25*20 = 125 6.25*20 = 125
Columns 0.3*0.3*3*25*4= 27 0.3*0.3*3*25*4= 27 0.3*0.3*1.5*25*4 = 13.5
(below) --- --- -----
Dead 322 352 198.5
Live 1*25 = 25 5*25 = 125 = 0
Analysis & Design of Floor Slab A1B2 :
Dead load (0.15*25+1.25) = 5 kN/m2
Live load = 5 kN/m2
Factored Load 1.5*5+1.5*5 = 15 kN/m2
Aspect ratio = 5.0 / 5.0 = 1.0
M+ = 0.056*15*52 = 21.0 kN-m/m
As+ = 504 sq.mm/m
Mu+ = 0.87*415*504*(150-25)*(1-(504*415/1000*125*25)/106 = 21.2 kN-m/m (OK)
Gravity Analysis & Design of Beam A1B1 :
Direct dead UDL on beam = (75+125)/4 = 50.00 kN
Triangular dead load from slab = 125/4 = 31.25 kN
Triangular live load from slab = 125/4 = 31.25 kN
Total = 112.50 kN
Simply supported moment :
Dead = 50*5/8 + 31.25*5/6 = 57.29 kN-m
Live = 31.25*5/6 = 26.04 kN-m
Total = 83.33 kN-m
Fixed end moment :
Dead = 50*5/12 + 31.25*5*5/48 = 37.11 kN-m
Live = 31.25*5*5/48 = 16.28 kN-m
Total = 53.39 kN-m
Factored fixed end moment = 1.5*37.11 + 1.5*16.28 = 80.08 kN-m
Substitute Frame Properties:
Gross properties of sections will be used in the analysis. For beams, effective
width of flange is taken as lo/12+bw+3*Df where lo is the distance between points
of contraflexure, assumed as 0.7 times beam span and Df is the flange depth.
Gross Moment of Inertia of Beams:
Overall depth = 500 mm
Web width = 300 mm
Flange depth = 150 mm
Flange width = 0.7*5000/12 + 300 + 3*150 = 1042 mm
Depth of CG from top = 300*500*250 + (1042-300)*150*75
------------------------------- = 176 mm
300*500 + (1042-300)*150
Icg =
0.3*0.53/12 + 0.3*0.5*(0.25-0.176)2 + (1.042-0.3)*0.153/12 +
(1.042-0.3)*0.15*(0.176 -0.075)2 = 0.00529 m4
Gross Moment of Inertia of Columns = 0.3*0.33/12 = 0.000675 m4
Moments in upper and lower columns of a single bay symmetrical frame are
MF * KU
Mc = -------------------------
(KU+KL+0.5*KB/2)
where:
MF = Fixed end beam moment = 80.08 kN-m
KU = I/L of upper column = 0.000675 / 3 = 0.000225 m3
KL = I/L of lower column = 0.000675 / 3 = 0.000225 m3
KB = I/L of beam = 0.005290 / 5 = 0.001058 m3
80.08 * 0.000225
Mc = ------------------------- = 18.4 kN-m
(2*0.000225+0.001058/2)
Factored shear in columns = 18.4*2/3 = 12.27 kN
Factored moment at beam ends = 2*18.4 = 36.80 kN-m
Factored moment at mid-span = 1.5*83.33 - 36.8 = 88.20 kN-m
Factored shear at beam ends = 1.5*112.5 / 2 = 84.38 kN
Reinforcement in beam:
Longitudinal steel
As(min) = (0.85/415)*300*(500-50) = 277 sq.mm
Mu(min) = 0.87*415*277*450*(1-277*415/300*450*25)/1000000 = 43.47 kN-m
Required top steel at beam supports < 277 sq.mm
As(mid) = 553 sq.mm
Mu(mid) = 0.87*415*553*450*(1-553*415/1042*450*25)/1000000 = 88.1 kN-m (OK)
Stirrups
Minimum Asv/sv = 0.4*300/415 = 0.289
Shear capacity of concrete = 0.36*300*450/1000 = 48.60 kN
Shear capacity of stirrups = 0.87*415*0.289*450/1000 = 46.95 kN
Total shear capacity = 95.55 kN < 84.38 kN (OK)
Required sv/Asv > 1/0.289 = 3.46
Seismic Analysis
Seismic Weight
Roof Floor Ground
------------ ------------- --------------
Slab+finishes 175.0 125.0 0.0
Beams 75.0 75.0 60.0
Walls above 45.0 125/2 = 62.5 125/2 = 62.5
Walls below 125/2 = 62.5 125/2 = 62.5 0.0
Columns above 0.0 27/2 = 13.5 27/2 = 13.5
Columns below 27/2 = 13.5 27/2 = 13.5 13.5/2 = 6.75
Live 0.0 0.5*125 = 62.5 0.0
---- ----- ------
wi (kN) 371.0 414.5 142.75
Time Period Ta
h = 7.5 m
d = 5.0 m
Ta = 0.09 * h / sqrt(d)
= 0.09 * 7.5 / sqrt(5) = 0.302 second
Design horizontal seismic coefficient Ah
ZISa/g 0.16*1.0*2.5
Ah = ------ = ------------ = 0.0667
2R 2*3.0
where
Z Zone factor = 0.16
I Importance factor = 1.0
R Response reduction factor = 3.0 for OMRF
Sa/g Response acceleration coefficient = 2.5
Base Shear VB
VB = Ah * Σwi = 0.0667*(371.0 + 414.5 + 142.75) = 61.91 kN
Vertical Distribution of Base Shear Qi
Level wi hi wi*hi2 Qi
4 371.00 7.5 20868.8 61.91*20868.8/29583.6 = 43.67 kN
3 414.50 4.5 8393.6 61.91* 8393.6/29583.6 = 17.57 kN
2 142.75 1.5 321.2 61.91* 321.2/29583.6 = 0.67 kN
1 0.00 0.0 0.0 0.00 kN
-------- -----
29583.6 61.91 kN
Seismic Loads and Bending Moments
Storey Shear Column Moment Beam Shear Column Axial
kN kN-m kN kN
2 43.67 43.67*3/2*4=16.38 16.38*2/5 = 6.55 = 6.55
1 61.24 61.24*3/2*4=22.97 (16.38+22.97)*2/5 = 15.74 (6.55+15.74) = 22.29
Effect of Eccentricity in Plan
Due to biaxial symmetry of loads and layout, center of mass and shear center are
located at (2.5,2.5) meters with respect to origin at lower left corner. Thus
xcm = ycm = xsc = ysc = 2.5 m
Static eccentricity esi
Frames along x-axis esiy = (xcm-xsc) = (2.5-2.5) = 0.00 m
Frames along y-axis esix = (ycm-ysc) = (2.5-2.5) = 0.00 m
Accidental eccentricity .05bi
Frames along x-axis = 0.05*biy = 0.05 * 5 = 0.25 m
Frames along y-axis = 0.05*bix = 0.05 * 5 = 0.25 m
Design eccentricity edi
Frames along x-axis = ediy = esiy+0.05*biy = 0 + 0.25 = 0.25 m
Frames along y-axis = edix = esix+0.05*bix = 0 + 0.25 = 0.25 m
Radius of Gyration of Strength
rk = (ΣVxi,j*(ysci-yi,j)2/ΣVxi,j + ΣVyi,j*(xsci-xi,j)2/ΣVyi,j)1/2
where
Vxi,j/ΣVxi,j=fraction of storey shear along x direction at level i in column j=1/4
Vyi,j/ΣVyi,j=fraction of storey shear along y direction at level i in column j=1/4
xi,j =x ordinate of j-th column at level i = 0 and 5 m
yi,j =y ordinate of j-th column at level i = 0 and 5 m
rk2 = 2*(2.5-0)2/4 + 2*(2.5-5)2/4 + 2*(2.5-0)2/4 + 2*(2.5-5)2/4 = 12.5 m2
Magnification Factor at all levels
Frames along x-axis = δxi = 1 + ediy*yi,k/rk2 = 1 + 2.5*0.25/12.5 = 1.05
Frames along y-axis = δyi = 1 + edix*xi,k/rk2 = 1 + 2.5*0.25/12.5 = 1.05
Analysis and Design of Columns of First Storey for D + L + Ex
Unfactored Axial Load in Columns
Due to dead loads (322+352)/4 = 168.50 kN
Due to live loads ( 25+125)/4 = 37.50 kN
Due to earthquake 22.29 kN
Unfactored Moment in Columns
Due to dead loads 18.4*37.11/80.08 = 8.53 kN-m
Due to live loads 18.4*16.28/80.08 = 3.74 kN-m
Due to earthquake 22.97 kN-m
Unfactored Shear in Columns
Due to dead loads 2*8.53/3 = 5.69 kN
Due to live loads 2*3.74/3 = 2.49 kN
Due to earthquake 61.24/4 = 15.31 kN
Pu = 1.2 * (168.5 + 37.5 + 22.29*1.05) = 275.29 kN
Mux = 1.2 * (8.53 + 3.74) = 14.72 kN-m
Muy = 1.2 * (8.53 + 3.74 + 22.97*1.05) = 36.39 kN-m
Vux = 1.2 * (5.69 + 2.49 + 61.24*1.05/4) = 24.10 kN
Vuy = 1.2 * (5.69 + 2.49) = 9.82 kN
Check for As = 1080 sq.mm
p 1080 * 100
--- = -------------- = 0.048
fck 300 * 300 * 25
Puz = 0.45fc.b.D + 0.75fy.As
= 0.45*25*300*300 + 0.75*415*1080 = 1348650 N
Pu 275290
--- = -------- = 0.204
Puz 1348650
αn = 1 + (.204 - .2) / 0.6 = 1.0067
Refer Chart 45 of SP-16
Pu 275290
------- = ---------- = 0.122
fck.b.D 25*300*300
Mux1
-------- = 0.09
fck.b.D2
Mux1 = Muy1 = 0.09*25*300*3002 = 60750000 N-mm
(Mux/Mux1)αn+(Muy/Muy1)αn < 1
(14.72/60.75)1.0067 + (36.39/60.75)1.0067 = 0.241 + 0.598 = 0.837 < 1 (OK)
Column-beam Intersection at top of first storey:
Unfactored Moments at Beam Ends
Due to dead loads 36.8*121.875/(168.75*1.5) = 17.72 kN-m
Due to live loads 36.8*46.875/(168.75*1.5) = 6.82 kN-m
Due to earthquake 16.38 + 22.97 = 39.35 kN-m
Combinations for moment
DL+E -1.2*17.72 - 1.2*6.82 - 1.2*1.05*39.35 = -79.03 kN-m
DL-E -1.2*17.72 - 1.2*6.82 + 1.2*1.05*39.35 = 20.13 kN-m
D+E -1.5*17.72 - 1.5*1.05*39.35 = -88.56 kN-m
D-E -0.9*17.72 + 1.5*1.05*39.35 = 46.03 kN-m
Unfactored Shears at Beam Ends
Due to dead loads 81.25/2 = 40.63 kN
Due to live loads 31.25/2 = 15.63 kN
Due to earthquake 2*39.35/5 = 15.74 kN
Combinations for shear
DL+E 1.2*40.63 + 1.2*15.63 + 1.2*1.05*15.74 = 87.34 kN
E 1.5*1.05*15.74 = 24.79 kN
Design for hogging moment = -88.56 kN-m
As = 513 sq.mm
Mu = 0.87*415*513*450*(1-513*415/300*450*25)/1000000 = 78.1 kN-m
Design for sagging moment = 46.03 kN-m
As = 277 sq.mm
Mu = 0.87*415*277*450*(1-277*415/1042*450*25)/1000000 = 44.6 kN-m (OK)
Minimum Shear capacity of beam = 95.55 kN (OK)
TORSIONAL PROVISIONS IN IS:1893(2002)
New clauses were introduced in the revised Indian seismic code for torsion of
symmetric as well as asymmetric buildings with rigid diaphragms. The treatment
of torsional provisions is elaborated here along with a solved example.
Clause 7.9.2 of IS:1893(2000) reads as follows:
The design eccentricity, edi to be used at floor i shall be taken as:
edi = 1.5esi + 0.05bi or
= esi - 0.05bi
whichever of these gives the more severe effect in the shear of any frame
where
esi = Static eccentricity at floor i defined as the distance between centre of mass
and centre of rigidity
bi = Floor plan dimension of floor i, perpendicular to the direction of force
Under seismic loads, structures experience lateral forces acting, in general, at
a design eccentricity edi with respect to a neutral point, such that deflections on
the side towards edi are higher than those on the other side of the neutral point.
The sides towards and away from edi are known as the flexible and stiff sides
respectively.
Multiplier 1.5 on esi in the first equation is the dynamic amplification factor to
account for possible coupling of torsional and lateral modes of vibration and
depends on the ratio of frequencies in the two modes. When frequencies in the two
modes are far apart, dynamic amplification factor is 1.0 as in the second equation.
Accidental eccentricity due to possible variations of live load, stiffness and
ground motion along the width of building is given by 0.05bi. Obviously, this factor
can take positive or negative value.
Two cases are possible:
1. Lateral-torsional mode coupling occurs and accidental eccentricity is in the
same direction as the static eccentricity which is reflected by the first equation.
In general, this is the governing case for members on the flexible side.
2. Lateral-torsional mode coupling does not occur and the static eccentricity is in
the direction opposite to the static eccentricity which is what the second part of
the equation implies. In general, this is the governing case for members on the
stiff side.
Seismic force acting at the code specified design eccentricity results in torques
at various floor levels. There are two approaches to account for this effect.
1. Floor Torques about Centers of Rigidity:
Static eccentricity is defined as the distance between the center of mass and the
center of rigidity at a given floor. Centers of rigidity are points on each floor of
a multistoreyed building such that lateral loads applied through them do not cause
rotation of any of the floors [1].
In order to locate centers of rigidity, the following procedure is adopted:
1. The structural models constrained to deflect only in the direction of applied
seismic loads along x and y axes are analyzed.
2. Free body diagram of each floor is taken along with storey shears vi+1,j and
vi,j above and below that floor respectively where subscript i refers to storey and
subscript j refers to shear resisting element of that storey.
3. The point of intersection of resultants of net storey shears (vi,j - vi+1,j)
along the orthogonal axes is the center of rigidity for the storey.
A pair of design eccentricities and the resulting floor torques at each storey can
now be calculated. These floor torques are applied to a three-dimensional frame
model taking due care of the fact that 3D frame analysis accounts for static
eccentricity 1.0 esi automatically.
2. Storey Torsion about Shear Center:
Static eccentricity is defined as the distance between the center of cumulative mass
from roof down to the level under consideration and shear center at that level.
In order to locate shear center, the following procedure is adopted:
1. The structural models constrained to deflect only in the direction of applied
seismic loads along x and y axes are analyzed.
2. Free body diagram of the substructure from roof down to the level being considered
is taken along with shears Vxi,j and Vyi,j at the cut where subscript i refers
to level and subscript j refers to shear resisting element at that level.
3. The point of intersection of resultants of shears Vxi,j and Vyi,j defines shear
center at that level.
A pair of design eccentricities at the level under consideration can now be
calculated.
Analogous to the quantity (1+6e/d) commonly used to calculate maximum pressure in
an eccentrically loaded footing, magnification factors on all forces and moments
obtained in step 1 above are given by:
δxi = 1 + ediy*yi,k/rk2
δyi = 1 + edix*xi,k/rk2
where
δxi = Magnification factor for frames in x direction at level i
δyi = Magnification factor for frames in y direction at level i
Vxi,j = Shear along x direction at level i in column j
Vyi,j = Shear along y direction at level i in column j
edix = Maximum additive design eccentricity at level i along x axis
ediy = Maximum additive design eccentricity at level i along y axis
xsci = x ordinate of shear center at level i = ΣVyi,j*xi,j / ΣVyi,j
ysci = y ordinate of shear center at level i = ΣVxi,j*yi,j / ΣVxi,j
xi,j = x ordinate of j-th column at level i
yi,j = y ordinate of j-th column at level i
rk = Radius of gyration of stiffness
= (ΣVxi,j*(ysci-yi,j)2/ΣVxi,j + ΣVyi,j*(xsci-xi,j)2/ΣVyi,j)1/2
It must be emphasized that, in general, location of shear center is different from
center of rigidity. Tso [2] has shown equivalence of the two procedures given above.
The second approach is computationally simpler and will be used to illustrate the
effect of torsion on the two storeyed building shown below consisting of ground,
first floor and roof levels.
The data for this example is as follows:
----------------------------------------------------------------------
Column Data Beam Data Slab Data
----------------------------------------------------------------------
Mark Cx Cy Mark bw D Wall Mark t Loads
mm mm mm mm Load mm Dead Live
kN/m kN/sq.m
----------------------------------------------------------------------
C1 300 300 B1 300 500 6.25 S1 150 1.25 5.00
B1R 300 500 2.25 S1R 150 3.25 1.00
B1G 300 400 6.25
----------------------------------------------------------------------
Here, suffixes R and G on beam and slab marks refer to roof and ground levels
respectively. Storey heights are 3.0 m and foundation depth is 1.5 m below ground.
Seismic Weight wi
Seismic weight at a particular level consists of:
- Dead loads of slab and beams including finishes at the level
- Proportional dead loads of walls and columns above and below
- Appropriate amount of live loads at the level as per code
The following table gives seismic weights at various levels along with x ordinate
of center of gravity, xcg measured from lower left corner which will be used later
for calculating center of mass. Due to symmetry in vertical direction, ycg is located
at 2.5 m for all of the components. Density of reinforced concrete is taken as
25 kN/m3 for this example.
Roof Floor Ground xcg m
---------------- ----------------- ---------------- -----
Slab+finishes 280.00 200.00 0.00 4.000
Beams 116.25 116.25 93.00 4.161
Walls above 69.75 193.75/2 = 96.88 193.75/2 = 96.88 4.161
Walls below 193.75/2 = 96.88 193.75/2 = 96.88 0.00 4.161
Columns above 0.00 40.50/2 = 20.25 40.50/2 = 20.25 4.333
Columns below 40.50/2 = 20.25 40.50/2 = 20.25 20.25/2 = 10.13 4.333
Live 0.00 0.5*200 = 100.00 0.00 4.000
------ ------ ------
wi (kN) 583.13 650.51 220.26
Center of Mass
Centers of mass at various levels with respect to origin at lower left corner are
calculated as :
xcm = Σwj*xcgj/Σwj
ycm = Σwj*ycgj/Σwj
where subscript j refers to each component of seismic weight from roof downwards
to the level under consideration.
xcm at Roof Level
=(280*4+(116.25+69.75+96.88)*4.161+20.25*4.333)/583.13
=4.090 m
xcm at First Floor Level
=(4.090*583.13+200*4+(116.25+2*96.88)*4.161+2*20.25*4.333+100*4)/(583.13+650.51)
=4.094 m
xcm at Ground Level
=((583.13+650.51)*4.094+(93+96.88)*4.161+(20.25+10.13)*4.333)/(583.13+650.51+220.26)
=4.108 m
ycm = 2.5 m at all levels
Seismic Analyses on Constrained Models
In lieu of analyses, it will be assumed here that all columns share storey shears
equally at all levels.
Shear Center
Following the assumption of storey shears being shared equally by all columns,
values of Vxi,j/ΣVxi,j and Vyi,j/ΣVyi,j are all equal to 1/6. Thus at all levels,
xsc = ΣVyi,j*xi,j/ΣVyi,j = 2*(0+5+8)/6 = 4.333 m
ysc = ΣVxi,j*yi,j/ΣVxi,j = 3*(0+5)/6 = 2.500 m
Static Eccentricity esi
The distance between center of mass and shear center gives static eccentricity
along each of the axes at various levels.
Along x-axis: esix = xcm - xsc
Roof : 4.090 - 4.333 = -0.243 m
Floor : 4.094 - 4.333 = -0.239 m
Ground : 4.108 - 4.333 = -0.225 m
Along y-axis: esiy = ycm - ysc = 2.5 - 2.5 = 0 m (at all levels)
Accidental Eccentricity bi
Along x-axis = 0.05*bix = 0.05 * 8 = 0.40 m
Along y-axis = 0.05*biy = 0.05 * 5 = 0.25 m
Maximum Design Eccentricity edimax
Algebraic addition of static and accidental eccentricities gives maximum value of
design eccentricity.
Along x-axis = 1.5*esix + 0.05*bix
Roof : -1.5*0.243 - 0.40 = -0.765 m
Floor : -1.5*0.239 - 0.40 = -0.759 m
Ground : -1.5*0.225 - 0.40 = -0.738 m
Along y-axis = 1.5*esiy + 0.05*biy = 0 + 0.25 = 0.25 m
Minimum Design Eccentricity edimin
Algebraic subtraction of accidental eccentricity from static eccentricity gives
minimum value of design eccentricity.
Along x-axis = ediminx = esix - 0.05*bix
Roof : -0.243 + 0.40 = 0.157 m
Floor : -0.239 + 0.40 = 0.161 m
Ground : -0.225 + 0.40 = 0.175 m
Along y-axis = ediminy = esiy - 0.05*biy = 0 - 0.25 = -0.25 m
Radius of Gyration of Strength
rk = (ΣVxi,j*(ysci-yi,j)2/ΣVxi,j + ΣVyi,j*(xsci-xi,j)2/ΣVyi,j)1/2
The quantities Vxi,j/ΣVxi,j and Vyi,j/ΣVyi,j being equal to 1/6 at all levels,
rk2 = {3*(2.5-0)2+3*(2.5-5)2+2*(4.333-0)2+2*(4.333-5)2+2*(4.333-8)2}/6 = 17.14 m2
Magnification Factors
δxi = 1 + ediy*yi,k/rk2
Frame along grid 1
Roof : 1+(0-4.333)*(-0.765)/17.14 = 1.193 or 1+(0-4.333)*0.157/17.14=0.960
Floor : 1+(0-4.333)*(-0.759)/17.14 = 1.192 or 1+(0-4.333)*0.161/17.14=0.959
Ground : 1+(0-4.333)*(-0.738)/17.14 = 1.186 or 1+(0-4.333)*0.175/17.14=0.956
Maximum: 1.193
Frame along grid 2
Roof : 1+(5-4.333)*(-0.765)/17.14 = 0.970 or 1+(5-4.333)*0.157/17.14=1.006
Floor : 1+(5-4.333)*(-0.759)/17.14 = 0.971 or 1+(5-4.333)*0.161/17.14=1.006
Ground : 1+(5-4.333)*(-0.738)/17.14 = 0.971 or 1+(5-4.333)*0.175/17.14=1.007
Maximum: 1.007
Frame along grid 3
Roof : 1+(8-4.333)*(-0.765)/17.14 = 0.836 or 1+(8-4.333)*0.157/17.14=1.034
Floor : 1+(8-4.333)*(-0.759)/17.14 = 0.838 or 1+(8-4.333)*0.161/17.14=1.034
Ground : 1+(8-4.333)*(-0.738)/17.14 = 0.842 or 1+(8-4.333)*0.175/17.14=1.038
Maximum: 1.038
As expected, frame along grid 1 which is farthest from shear center and on the
flexible side experiences maximum magnification factor.
δyi = 1 + edix*xi,k/rk2
= 1 + 0.25*2.5/17.14 = 1.037 for frames along A and B at all levels.
Storey-wise magnification factors or, conservatively, the maximum value of
magnification factor for each frame in x and y direction is applied to all actions
found from seismic analysis of constrained model.
REFERENCES
1. Jain SK and Murty SVR, "Proposed Changes in Indian Seismic Code,IS:1893
(Part 1) 2002". IITK-GSDMA Project on Building Codes.
2. Tso WK, "Static Eccentricity Concept for Torsional Moment Estimations".
Journal of Structural Engineering, ASCE, Vol. 116, No. 5 pp 1199-1212.
80.08 * 0.000225
Mc = ------------------------- = 18.4 kN-m
(2*0.000225+0.001058/2)
Factored shear in columns = 18.4*2/3 = 12.27 kN
Factored moment at beam ends = 2*18.4 = 36.80 kN-m
Factored moment at mid-span = 1.5*83.33 - 36.8 = 88.20 kN-m
Factored shear at beam ends = 1.5*112.5 / 2 = 84.38 kN
Reinforcement in beam:
Longitudinal steel
As(min) = (0.85/415)*300*(500-50) = 277 sq.mm
Mu(min) = 0.87*415*277*450*(1-277*415/300*450*25)/1000000 = 43.47 kN-m
Required top steel at beam supports < 277 sq.mm
As(mid) = 553 sq.mm
Mu(mid) = 0.87*415*553*450*(1-553*415/1042*450*25)/1000000 = 88.1 kN-m (OK)
Stirrups
Minimum Asv/sv = 0.4*300/415 = 0.289
Shear capacity of concrete = 0.36*300*450/1000 = 48.60 kN
Shear capacity of stirrups = 0.87*415*0.289*450/1000 = 46.95 kN
Total shear capacity = 95.55 kN < 84.38 kN (OK)
Required sv/Asv > 1/0.289 = 3.46
Seismic Analysis
Seismic Weight
Roof Floor Ground
------------ ------------- --------------
Slab+finishes 175.0 125.0 0.0
Beams 75.0 75.0 60.0
Walls above 45.0 125/2 = 62.5 125/2 = 62.5
Walls below 125/2 = 62.5 125/2 = 62.5 0.0
Columns above 0.0 27/2 = 13.5 27/2 = 13.5
Columns below 27/2 = 13.5 27/2 = 13.5 13.5/2 = 6.75
Live 0.0 0.5*125 = 62.5 0.0
---- ----- ------
wi (kN) 371.0 414.5 142.75
Time Period Ta
h = 7.5 m
d = 5.0 m
Ta = 0.09 * h / sqrt(d)
= 0.09 * 7.5 / sqrt(5) = 0.302 second
Design horizontal seismic coefficient Ah
ZISa/g 0.16*1.0*2.5
Ah = ------ = ------------ = 0.0667
2R 2*3.0
where
Z Zone factor = 0.16
I Importance factor = 1.0
R Response reduction factor = 3.0 for OMRF
Sa/g Response acceleration coefficient = 2.5
Base Shear VB
VB = Ah * Σwi = 0.0667*(371.0 + 414.5 + 142.75) = 61.91 kN
Vertical Distribution of Base Shear Qi
Level wi hi wi*hi2 Qi
4 371.00 7.5 20868.8 61.91*20868.8/29583.6 = 43.67 kN
3 414.50 4.5 8393.6 61.91* 8393.6/29583.6 = 17.57 kN
2 142.75 1.5 321.2 61.91* 321.2/29583.6 = 0.67 kN
1 0.00 0.0 0.0 0.00 kN
-------- -----
29583.6 61.91 kN
Seismic Loads and Bending Moments
Storey Shear Column Moment Beam Shear Column Axial
kN kN-m kN kN
2 43.67 43.67*3/2*4=16.38 16.38*2/5 = 6.55 = 6.55
1 61.24 61.24*3/2*4=22.97 (16.38+22.97)*2/5 = 15.74 (6.55+15.74) = 22.29
Effect of Eccentricity in Plan
Due to biaxial symmetry of loads and layout, center of mass and shear center are
located at (2.5,2.5) meters with respect to origin at lower left corner. Thus
xcm = ycm = xsc = ysc = 2.5 m
Static eccentricity esi
Frames along x-axis esiy = (xcm-xsc) = (2.5-2.5) = 0.00 m
Frames along y-axis esix = (ycm-ysc) = (2.5-2.5) = 0.00 m
Accidental eccentricity .05bi
Frames along x-axis = 0.05*biy = 0.05 * 5 = 0.25 m
Frames along y-axis = 0.05*bix = 0.05 * 5 = 0.25 m
Design eccentricity edi
Frames along x-axis = ediy = esiy+0.05*biy = 0 + 0.25 = 0.25 m
Frames along y-axis = edix = esix+0.05*bix = 0 + 0.25 = 0.25 m
Radius of Gyration of Strength
rk = (ΣVxi,j*(ysci-yi,j)2/ΣVxi,j + ΣVyi,j*(xsci-xi,j)2/ΣVyi,j)1/2
where
Vxi,j/ΣVxi,j=fraction of storey shear along x direction at level i in column j=1/4
Vyi,j/ΣVyi,j=fraction of storey shear along y direction at level i in column j=1/4
xi,j =x ordinate of j-th column at level i = 0 and 5 m
yi,j =y ordinate of j-th column at level i = 0 and 5 m
rk2 = 2*(2.5-0)2/4 + 2*(2.5-5)2/4 + 2*(2.5-0)2/4 + 2*(2.5-5)2/4 = 12.5 m2
Magnification Factor at all levels
Frames along x-axis = δxi = 1 + ediy*yi,k/rk2 = 1 + 2.5*0.25/12.5 = 1.05
Frames along y-axis = δyi = 1 + edix*xi,k/rk2 = 1 + 2.5*0.25/12.5 = 1.05
Analysis and Design of Columns of First Storey for D + L + Ex
Unfactored Axial Load in Columns
Due to dead loads (322+352)/4 = 168.50 kN
Due to live loads ( 25+125)/4 = 37.50 kN
Due to earthquake 22.29 kN
Unfactored Moment in Columns
Due to dead loads 18.4*37.11/80.08 = 8.53 kN-m
Due to live loads 18.4*16.28/80.08 = 3.74 kN-m
Due to earthquake 22.97 kN-m
Unfactored Shear in Columns
Due to dead loads 2*8.53/3 = 5.69 kN
Due to live loads 2*3.74/3 = 2.49 kN
Due to earthquake 61.24/4 = 15.31 kN
Pu = 1.2 * (168.5 + 37.5 + 22.29*1.05) = 275.29 kN
Mux = 1.2 * (8.53 + 3.74) = 14.72 kN-m
Muy = 1.2 * (8.53 + 3.74 + 22.97*1.05) = 36.39 kN-m
Vux = 1.2 * (5.69 + 2.49 + 61.24*1.05/4) = 24.10 kN
Vuy = 1.2 * (5.69 + 2.49) = 9.82 kN
Check for As = 1080 sq.mm
p 1080 * 100
--- = -------------- = 0.048
fck 300 * 300 * 25
Puz = 0.45fc.b.D + 0.75fy.As
= 0.45*25*300*300 + 0.75*415*1080 = 1348650 N
Pu 275290
--- = -------- = 0.204
Puz 1348650
αn = 1 + (.204 - .2) / 0.6 = 1.0067
Refer Chart 45 of SP-16
Pu 275290
------- = ---------- = 0.122
fck.b.D 25*300*300
Mux1
-------- = 0.09
fck.b.D2
Mux1 = Muy1 = 0.09*25*300*3002 = 60750000 N-mm
(Mux/Mux1)αn+(Muy/Muy1)αn < 1
(14.72/60.75)1.0067 + (36.39/60.75)1.0067 = 0.241 + 0.598 = 0.837 < 1 (OK)
Column-beam Intersection at top of first storey:
Unfactored Moments at Beam Ends
Due to dead loads 36.8*121.875/(168.75*1.5) = 17.72 kN-m
Due to live loads 36.8*46.875/(168.75*1.5) = 6.82 kN-m
Due to earthquake 16.38 + 22.97 = 39.35 kN-m
Combinations for moment
DL+E -1.2*17.72 - 1.2*6.82 - 1.2*1.05*39.35 = -79.03 kN-m
DL-E -1.2*17.72 - 1.2*6.82 + 1.2*1.05*39.35 = 20.13 kN-m
D+E -1.5*17.72 - 1.5*1.05*39.35 = -88.56 kN-m
D-E -0.9*17.72 + 1.5*1.05*39.35 = 46.03 kN-m
Unfactored Shears at Beam Ends
Due to dead loads 81.25/2 = 40.63 kN
Due to live loads 31.25/2 = 15.63 kN
Due to earthquake 2*39.35/5 = 15.74 kN
Combinations for shear
DL+E 1.2*40.63 + 1.2*15.63 + 1.2*1.05*15.74 = 87.34 kN
E 1.5*1.05*15.74 = 24.79 kN
Design for hogging moment = -88.56 kN-m
As = 513 sq.mm
Mu = 0.87*415*513*450*(1-513*415/300*450*25)/1000000 = 78.1 kN-m
Design for sagging moment = 46.03 kN-m
As = 277 sq.mm
Mu = 0.87*415*277*450*(1-277*415/1042*450*25)/1000000 = 44.6 kN-m (OK)
Minimum Shear capacity of beam = 95.55 kN (OK)
TORSIONAL PROVISIONS IN IS:1893(2002)
New clauses were introduced in the revised Indian seismic code for torsion of
symmetric as well as asymmetric buildings with rigid diaphragms. The treatment
of torsional provisions is elaborated here along with a solved example.
Clause 7.9.2 of IS:1893(2000) reads as follows:
The design eccentricity, edi to be used at floor i shall be taken as:
edi = 1.5esi + 0.05bi or
= esi - 0.05bi
whichever of these gives the more severe effect in the shear of any frame
where
esi = Static eccentricity at floor i defined as the distance between centre of mass
and centre of rigidity
bi = Floor plan dimension of floor i, perpendicular to the direction of force
Under seismic loads, structures experience lateral forces acting, in general, at
a design eccentricity edi with respect to a neutral point, such that deflections on
the side towards edi are higher than those on the other side of the neutral point.
The sides towards and away from edi are known as the flexible and stiff sides
respectively.
Multiplier 1.5 on esi in the first equation is the dynamic amplification factor to
account for possible coupling of torsional and lateral modes of vibration and
depends on the ratio of frequencies in the two modes. When frequencies in the two
modes are far apart, dynamic amplification factor is 1.0 as in the second equation.
Accidental eccentricity due to possible variations of live load, stiffness and
ground motion along the width of building is given by 0.05bi. Obviously, this factor
can take positive or negative value.
Two cases are possible:
1. Lateral-torsional mode coupling occurs and accidental eccentricity is in the
same direction as the static eccentricity which is reflected by the first equation.
In general, this is the governing case for members on the flexible side.
2. Lateral-torsional mode coupling does not occur and the static eccentricity is in
the direction opposite to the static eccentricity which is what the second part of
the equation implies. In general, this is the governing case for members on the
stiff side.
Seismic force acting at the code specified design eccentricity results in torques
at various floor levels. There are two approaches to account for this effect.
1. Floor Torques about Centers of Rigidity:
Static eccentricity is defined as the distance between the center of mass and the
center of rigidity at a given floor. Centers of rigidity are points on each floor of
a multistoreyed building such that lateral loads applied through them do not cause
rotation of any of the floors [1].
In order to locate centers of rigidity, the following procedure is adopted:
1. The structural models constrained to deflect only in the direction of applied
seismic loads along x and y axes are analyzed.
2. Free body diagram of each floor is taken along with storey shears vi+1,j and
vi,j above and below that floor respectively where subscript i refers to storey and
subscript j refers to shear resisting element of that storey.
3. The point of intersection of resultants of net storey shears (vi,j - vi+1,j)
along the orthogonal axes is the center of rigidity for the storey.
A pair of design eccentricities and the resulting floor torques at each storey can
now be calculated. These floor torques are applied to a three-dimensional frame
model taking due care of the fact that 3D frame analysis accounts for static
eccentricity 1.0 esi automatically.
2. Storey Torsion about Shear Center:
Static eccentricity is defined as the distance between the center of cumulative mass
from roof down to the level under consideration and shear center at that level.
In order to locate shear center, the following procedure is adopted:
1. The structural models constrained to deflect only in the direction of applied
seismic loads along x and y axes are analyzed.
2. Free body diagram of the substructure from roof down to the level being considered
is taken along with shears Vxi,j and Vyi,j at the cut where subscript i refers
to level and subscript j refers to shear resisting element at that level.
3. The point of intersection of resultants of shears Vxi,j and Vyi,j defines shear
center at that level.
A pair of design eccentricities at the level under consideration can now be
calculated.
Analogous to the quantity (1+6e/d) commonly used to calculate maximum pressure in
an eccentrically loaded footing, magnification factors on all forces and moments
obtained in step 1 above are given by:
δxi = 1 + ediy*yi,k/rk2
δyi = 1 + edix*xi,k/rk2
where
δxi = Magnification factor for frames in x direction at level i
δyi = Magnification factor for frames in y direction at level i
Vxi,j = Shear along x direction at level i in column j
Vyi,j = Shear along y direction at level i in column j
edix = Maximum additive design eccentricity at level i along x axis
ediy = Maximum additive design eccentricity at level i along y axis
xsci = x ordinate of shear center at level i = ΣVyi,j*xi,j / ΣVyi,j
ysci = y ordinate of shear center at level i = ΣVxi,j*yi,j / ΣVxi,j
xi,j = x ordinate of j-th column at level i
yi,j = y ordinate of j-th column at level i
rk = Radius of gyration of stiffness
= (ΣVxi,j*(ysci-yi,j)2/ΣVxi,j + ΣVyi,j*(xsci-xi,j)2/ΣVyi,j)1/2
It must be emphasized that, in general, location of shear center is different from
center of rigidity. Tso [2] has shown equivalence of the two procedures given above.
The second approach is computationally simpler and will be used to illustrate the
effect of torsion on the two storeyed building shown below consisting of ground,
first floor and roof levels.
No comments:
Post a Comment