IS 1893:2002 CRITERIA FOR EARTHQUAKE RESISTANT DESIGN OF STRUCTURESPART1 GENERAL PROVISIONS AND BUILDINGS The Code is now split into five parts Part 1 - General provisions and buildingsPart 2 - Liquid retaining tanks - Elevated and ground supportedPart 3 - Bridges and retaining wallsPart 4 - Industrial structures including stack like structuresPart 5 - Dams and embankments Part 1 contains provisions that are general in nature and applicable to allstructures. 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 isupgraded to Zone II. Killari area is enhanced to Zone III. Bellary isolated zoneis removed. East coast is enhanced to Zone III and connected with Zone III ofGodavari 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 offounding strata viz. Hard, Medium and Soft. 4. The empirical formula for calculating fundamental natural period T=0.1nfor moment resisting frames without bracing or shear walls is replaced withTa=0.075h0.075 for RC framed buildings. This formula applies to bare framese.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 beexperienced by the structure during the probable maximum earthquake, if it wereto remain elastic. Then response reduction due to ductile deformation orfrictional energy dissipation in the cracks is applied via `response reductionfactor' 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 Vα0 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 VZ 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 31.3III 0.04W 0.064W 0.040W 0.067W 0.0 4.7IV 0.05W 0.080W 0.060W 0.100W 20.0 25.0V 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 betaresults 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 maximumstorey 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 resistingelement > 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 thestorey 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 lateralstorey shear force calculated as before. Non-Parallel Systems:Earthquake effects about the two orthogonal axes must be combined: a. Ex ± 0.3Eyb. 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 : DeadL : LiveE : EarthquakeX : along X axisY : along Y axisP : Positive eccentricityN : Negative eccentricity This results in 25 actual combinations but can be reduced rationally. Limits of static analysis------------------------- Zone II Zone III Zone IV Zone VRegular buildings <90 m <90 m <40 m <40 mIrregular 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  3Foundation depth mm       1500   Slab cover mm   25   Frame type  OMRFGround storey height mm 3000  Beam cover mm 50  Soil type  MediumTypical 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 = 60Walls 2.25*20 = 45 6.25*20 = 125 6.25*20 = 125Columns 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/mMu+ = 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 kNTriangular live load from slab = 125/4 = 31.25 kNTotal = 112.50 kN  Simply supported moment :Dead = 50*5/8 + 31.25*5/6 = 57.29 kN-mLive = 31.25*5/6 = 26.04 kN-mTotal = 83.33 kN-m

Fixed end moment : Dead = 50*5/12 + 31.25*5*5/48 = 37.11 kN-mLive = 31.25*5*5/48 = 16.28 kN-mTotal = 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, effectivewidth of flange is taken as lo/12+bw+3*Df where lo is the distance between pointsof contraflexure, assumed as 0.7 times beam span and Df is the flange depth.  Gross Moment of Inertia of Beams: Overall depth = 500 mmWeb width = 300 mmFlange depth = 150 mmFlange 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 * KUMc = ------------------------- (KU+KL+0.5*KB/2)

where:

MF = Fixed end beam moment = 80.08 kN-mKU = I/L of upper column = 0.000675 / 3 = 0.000225 m3KL = I/L of lower column = 0.000675 / 3 = 0.000225 m3KB = I/L of beam = 0.005290 / 5 = 0.001058 m3

80.08 * 0.000225Mc = ------------------------- = 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.mmMu(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 kNShear capacity of stirrups = 0.87*415*0.289*450/1000 = 46.95 kNTotal 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 60.0Walls above 45.0 125/2 = 62.5 125/2 = 62.5Walls below 125/2 = 62.5 125/2 = 62.5 0.0Columns above 0.0 27/2 = 13.5 27/2 = 13.5Columns below 27/2 = 13.5 27/2 = 13.5 13.5/2 = 6.75Live 0.0 0.5*125 = 62.5 0.0 ---- ----- ------ wi (kN) 371.0 414.5 142.75  Time Period Ta h = 7.5 md = 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*2.5Ah = ------ = ------------ = 0.0667 2R 2*3.0 where Z Zone factor = 0.16I Importance factor = 1.0R Response reduction factor = 3.0 for OMRFSa/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 arelocated at (2.5,2.5) meters with respect to origin at lower left corner. Thus

xcm = ycm = xsc = ysc = 2.5 m

Static eccentricity esiFrames along x-axis esiy = (xcm-xsc) = (2.5-2.5) = 0.00 mFrames along y-axis esix = (ycm-ysc) = (2.5-2.5) = 0.00 m

Accidental eccentricity .05biFrames along x-axis = 0.05*biy = 0.05 * 5 = 0.25 mFrames along y-axis = 0.05*bix = 0.05 * 5 = 0.25 m

Design eccentricity ediFrames along x-axis = ediy = esiy+0.05*biy = 0 + 0.25 = 0.25 mFrames 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

whereVxi,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=1/4xi,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 levelsFrames along x-axis = δxi = 1 + ediy*yi,k/rk

2 = 1 + 2.5*0.25/12.5 = 1.05Frames along y-axis = δyi = 1 + edix*xi,k/rk

2 = 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 ColumnsDue to dead loads (322+352)/4 = 168.50 kNDue to live loads ( 25+125)/4 = 37.50 kNDue to earthquake 22.29 kN Unfactored Moment in ColumnsDue to dead loads 18.4*37.11/80.08 = 8.53 kN-mDue to live loads 18.4*16.28/80.08 = 3.74 kN-mDue to earthquake 22.97 kN-m Unfactored Shear in ColumnsDue to dead loads 2*8.53/3 = 5.69 kNDue to live loads 2*3.74/3 = 2.49 kNDue to earthquake 61.24/4 = 15.31 kN Pu = 1.2 * (168.5 + 37.5 + 22.29*1.05) = 275.29 kNMux = 1.2 * (8.53 + 3.74) = 14.72 kN-mMuy = 1.2 * (8.53 + 3.74 + 22.97*1.05) = 36.39 kN-mVux = 1.2 * (5.69 + 2.49 + 61.24*1.05/4) = 24.10 kNVuy = 1.2 * (5.69 + 2.49) = 9.82 kN Check for As = 1080 sq.mm  p 1080 * 100--- = -------------- = 0.048fck 300 * 300 * 25

Puz = 0.45fc.b.D + 0.75fy.As  = 0.45*25*300*300 + 0.75*415*1080 = 1348650 N Pu 275290--- = -------- = 0.204Puz 1348650 αn = 1 + (.204 - .2) / 0.6 = 1.0067 Refer Chart 45 of SP-16  Pu 275290

------- = ---------- = 0.122fck.b.D 25*300*300  Mux1-------- = 0.09fck.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-mDue to live loads 36.8*46.875/(168.75*1.5) = 6.82 kN-mDue 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-mDL-E -1.2*17.72 - 1.2*6.82 + 1.2*1.05*39.35 = 20.13 kN-mD+E -1.5*17.72 - 1.5*1.05*39.35 = -88.56 kN-mD-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 kNDue to live loads 31.25/2 = 15.63 kNDue 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 kNE 1.5*1.05*15.74 = 24.79 kN Design for hogging moment = -88.56 kN-mAs = 513 sq.mmMu = 0.87*415*513*450*(1-513*415/300*450*25)/1000000 = 78.1 kN-m Design for sagging moment = 46.03 kN-mAs = 277 sq.mmMu = 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 ofsymmetric as well as asymmetric buildings with rigid diaphragms. The treatmentof 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, ata 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 toaccount 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 twomodes 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 factorcan 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 inthe direction opposite to the static eccentricity which is what the second part ofthe 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 thecenter of rigidity at a given floor. Centers of rigidity are points on each floor ofa 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 appliedseismic loads along x and y axes are analyzed.

2. Free body diagram of each floor is taken along with storey shears vi+1,j andvi,j above and below that floor respectively where subscript i refers to storey andsubscript 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 cannow 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 massfrom 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 appliedseismic loads along x and y axes are analyzed.

2. Free body diagram of the substructure from roof down to the level being consideredis taken along with shears Vxi,j and Vyi,j at the cut where subscript i refersto 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 inan eccentrically loaded footing, magnification factors on all forces and momentsobtained 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 iδyi = 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 axisediy = Maximum additive design eccentricity at level i along y axis

xsci = x ordinate of shear center at level i = ΣVyi,j*xi,j / ΣVyi,jysci = y ordinate of shear center at level i = ΣVxi,j*yi,j / ΣVxi,j

xi,j = x ordinate of j-th column at level iyi,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 fromcenter 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 theeffect 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 levelsrespectively. 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 ordinateof 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 locatedat 2.5 m for all of the components. Density of reinforced concrete is taken as25 kN/m3 for this example.

                        Roof Floor         Ground       xcg m ---------------- ----------------- ---------------- -----Slab+finishes           280.00             200.00              0.00  4.000Beams                   116.25             116.25             93.00  4.161Walls above              69.75  193.75/2 =  96.88  193.75/2 = 96.88  4.161Walls below  193.75/2 = 96.88  193.75/2 =  96.88              0.00  4.161Columns above             0.00   40.50/2 =  20.25   40.50/2 = 20.25  4.333Columns below 40.50/2 = 20.25   40.50/2 =  20.25   20.25/2 = 10.13  4.333Live                      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 arecalculated as :

xcm = Σwj*xcgj/Σwjycm = Σwj*ycgj/Σwj

where subscript j refers to each component of seismic weight from roof downwardsto 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 mysc = 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 - xscRoof   : 4.090 - 4.333 = -0.243 mFloor  : 4.094 - 4.333 = -0.239 mGround : 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 mAlong 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 mFloor  : -1.5*0.239 - 0.40 = -0.759 mGround : -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 mFloor  : -0.239 + 0.40 = 0.161 mGround : -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,rk

2 = {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=0.960Floor  : 1+(0-4.333)*(-0.759)/17.14 = 1.192 or 1+(0-4.333)*0.161/17.14=0.959Ground : 1+(0-4.333)*(-0.738)/17.14 = 1.186 or 1+(0-4.333)*0.175/17.14=0.956Maximum: 1.193  Frame along grid 2Roof   : 1+(5-4.333)*(-0.765)/17.14 = 0.970 or 1+(5-4.333)*0.157/17.14=1.006Floor  : 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=1.007Maximum: 1.007  Frame along grid 3Roof   : 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=1.038Maximum: 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/rk

2 = 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 actionsfound 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.