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ANALY SIS OF EARTHQUAK E GROUND MOTIONS – I: STATIC ANALYSIS ANALYSIS
ANALYSISOF EARTH EARTHQUAK QUAKE E GROUND MOTIONS MOTI ONS : STAT TATIIC ANALYSIS
Requirements of Earthquake Resistant Design Seismic Coefficient Method Torsion due to Eccentricities
Storey Drift Calculations Appendages Appendages
Numerical Examples
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Requirements of Earthqua Earthquake Resistant Design
Earthqua ke can cause dam age not only on account of the shaking results from them but also due t o o th the r chain ef fe fec ts ts like lan ds dsli de des , floods, floods, fires fires and disrupti disruption on to communi communicati cation. on. Therefore it is impo rtant to take n ecessary preca utions in the planning an d design of structures so that they are safe against such secondar secondary y effects effects also. also.
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6.1.1 .1T TheCharacteristi cteristics cs (inte (intensity, durati tion, on, etc) of Seismic Ground Ground Vibrations Vi brations expected – Depends on
Magnitude of EQ Depth of Focus
Distance from Epicenter
Characteristics of the Path through which seismic waves travel Soil strata on which the structure stands
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6.1.2 Response of Structures to Ground Vibrations is a Function of --
6.1.3 Design Approach: Structure should
Nature of Foundation Soil
Materials, Forms, Size and Mode of Construction of Structures Duration and Characteristics of Ground Motions
IS 1893(Part I): 2002 specif ies design forces for structures standing on Rocks or soils which do not settle, liquefy or slide due to loss of strength during ground vibrations.
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However – Ductility, arising from inelastic material behavior and detailing. Over-strength, arisin g f rom the additional reserve strength in structures over and above the design strength, are relied upon to account for this difference in actualand design lateral forces. The specified EQ loads are based upon post-elastic energy dissipation in the structure and hence the provisions of this standard for design, detailing and construction shall be satisfied even for structures and memb ers fo r which l oad combinations w hi ch do not contain the EQ eff ect indicate larger demands than combinationincluding EQ. 7
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Possess at least a minimum strength to withstand minor EQs (
6.1.7 Additions to the Existing Structures
Structurally Independent – shall be designed as NEW structure Structurally Dependent – shall be designed as Entire conforms to the seismic force resistance requirements for new structure (1) Addition shall comply with the requirements for new structures. (2) Addition shall not increase forces in any structural elements by 5 % unl ess the capac ity of the el eme nt subject to increased force is still in compliance with this standard, and (3) Addition shall not decrease the seismic resistance of any structural element unless >= required 8
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6.3.2 Design Horizontal EQ Load
7.5 Design Lateral Force:
When the Lateral Load resisting elements are oriented along orthogonal horizontal directions the structure shall be designedfor the effects due to full Design EQLoad in one horizontal direction at time.
When the lateral load resisting elements are not oriented along the orthogonal horizontal directions, the structure shall be designed for the effects due to full design EQ load in one horizontal direction plus 30 % of the design EQ load in the other direction
The Design Lateral Force shall first be computed for the building as a whole. This Design Lateral Force shall then be distributed to the various floor levels. The overall design seismic force thus obtained at each floor level, shall then be distributed to individual Lateral Load Resisting elements depending on the floor diaphragm action.
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7.5.3 Design Seismic BaseShear
7.6 Fundamental Natural Period (Tn)
Total Design Lateral Force or Design Sei smic Base Shear (Vb) along any principal direction shall be determined by Vb = Ah * W Where, Ah = Design Horizontal acceleration spectrum using the fundamental natural Period Ta , as per the considered direction of vibration
W = Seismic weight of the building.
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The Approximate Fundamental Natural Period of Vibration (Tn), in seconds, of a Moment-Resisting Frame Building Without Brick Infill Panels may be estimated by the empirical expression Tn = 0.075 * h0.75 for RC Frame Building = 0.085 * h0.75 for Steel Frame Building
Where, h – Height of Building in m, excludes the basement storeys, where basement walls are connected with the groundfloor deck or fitted between the building columns.
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7.6.2 Tn For all Other Buildings
6. 4 Design Spectrum
For Moment-Resisting Frame Buildings with the infill panels
Tn = 0.09 * h / (√d) Where, h – height of building
d – Base dimension of the building at the Plinth Level, in m, along the considered direction of the Lateral Force
6.4.2 The Design Horizontal Seismic Coeffici ent Ah for a structure Ah = (Z * I * Sa) / (2 * R * g) = (Z/2)*(I/R)*(Sa/g)
Provided that for any structure with T <= 0.1 s, the value of Ah will not be taken less than (Z/2) w hatever be the value of (I/R)
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7.3 Design Imposed Loads for EarthquakeCalculation
Where, Z = Zone Factor –
II,
III,
IV,
V
Low, Moderate, Severe, Very Severe 0.10, 0.16,
0.24,
Up-to and including 3.0 kN/m^2 – 25 % Above 3.0 kN/m^2 – 50 %
0.36
for the Maximum Considered Earthquake (MCE) and Service Life of Structure in the Zone. i.e. DBE = MCE/2 I – Importance Factor R – Response Reduction Factor Sa/g – Average Response Acceleration Coefficient as per Soil Condition
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7.3.1 The EQ Force shall be calculated for the Full DL + the percentage of Imposed Load
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7.3.2 For Calculating the Design Seismic Forces of the Structure, the Imposed Load on Roof NEED not be considered. 7.3.3 Percentage of Imposed Loads shall also be used for “Whole Frame Loaded” condition in the Load Combinations where the Gravity Loads are combined with the EQ Loads, No reduction as per IS 875 for 16 numberof storeys.
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7.4 Seismic Weight
7.7 Distribution of Design Force
7.4.1 Seismic Weight of Floors –
DL + % of LL
the Weight of Columns and Walls in any storey shall be equally distributed to the floors above and below the storey.
7.4.2 Seismic Weight of Building –
Sum of the Seismic Wt of all the f loors.
Qi = (Vb) * (Wi * hi^2)/(∑ Wi * h i^2) Qi – Design Lateral Force at Floor I, Wi – Seismic Weight of Floor I, hi – Height of Floor I measured from Base, n – number of Storeys in the building is the number of levels at which the masses are located
7.4.3 Any Weight supported in between the storeys shall be distributed to the floors above and below in Inverse proportion to its distance from the floors.
7.7.1 Vertical Distribution of Base Shear to Different Floor Levels –
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7.9 Torsion
7.11Deformations
7.9.1 Provision shall be made in all buildings for Increase in SF on the Lateral Force Resisting Elements resulting from the Horizontal Torsional Moment arising due to eccentricity between the Centre of Mass and Centre of Rigidity. However, the Negative Torsional Shear shall be neglected. 7.9.2 Design Eccentricity ( whichever gives the more severe effect) e di – static eccentricty, distance between CM and CR
7.11.1 Storey Drift Limitation The storey drift in any storey due to minimum specified design lateral force with partial factor of 1.0 shall not exceed 0.004 times the storey height. No drift limit for single storey bui lding w hich has been designed to accommodatestorey drift.
Bi – Floor Plan dimension perpendicular to the direction of Force. edi 1.5 * esi 0.05 bi
OR
esi - 0.05 bi 19
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7.12.2 Cantilever Projections
7.11.2 Def ormation structures
Compatibility
of
7.12.2.1 Vertical Projections
Non-Seism ic
For buildings located in seismic zones IV and V, it shall be ensured that the structural components that are not part of the seismic Force Resisting System in the direction under consideration, do not lose their vertical load-carrying capacity under the induced moments resulting from storey deformations equal to R times the storey displacements calculated.
Tower, Tanks, Parapets, Smoke Stacks ( Chimneys) and Other Vertical cantilever projections attached to buildings and projecting above the roof shall be designed and checked for stability for FIVE times the design horizontal seismic coefficient Ah. In the analysis of the building, the weight of these projecting elements will be lumped with the roof weight.
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Calculation of Design Seismic Force by Static Analysis
7.12.2.2 Horizontal Projections
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All Horizontal Projections like cornices and balconies shall be designed and checked for stability for FIVE times the design Vertical coefficient
Problem Statement:
7.12.2.3 The increased Design Forces are ONLY for designing the projecting parts and their connections with the main structures. For the design of the main structure, such increase NEED not be considered.
Consider G+3 Storey building. The building is located in seismic zone III. The soil conditi ons are medium stiff. The R.C. frames are infilled with brick masonry. The lumped weight due to dead loads is 10 KN/m2 on floors and on roof. The floors carry live load of 5 KN/m2 on floors and roof.
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PLAN AND ELEVATION OF A BUIL DING
Cases considered for the analysis 3.0 3.0 3.0
(a) Only Building.
5 @ 5m c/c
(b) Building + Vertical Projection --- Water Tank placed concentrically
PLAN (c) Building + Vertical Projection --- Water Tank placed eccentrically. 3.0
(d) Analysis of cantilever portions --- Vertical & horizontal projections. 3.0 3.0 3.0 5 @ 5m c/c
ELEVATION 25
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Case a): Analysis of G + 3 BUILDING FundamentalPeriod:
Design Parameters:
EQ in X-direction: For seismic zone III, z= 0.16 (Table 2 of IS:1893) T Importance factor I = 1.0
= 0.09h/√d = 0.09 X 1 2/√25
= 0 .216 s ec.
Response Reduction factor R = 5 Ah = (Z / 2) X (I/R) X (Sa/g) = (0.16/2) X (1/5) X 2.5 = 0.04 Seismic Weights:
Floors:
Vb Roof:
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Floor area= 25 X 9 = 225 m
W1 = W2 = W3 = 225 X (10+0.5 X 5)
EQ in Y-direction: T = 0.09h/√d = 0. 09 X 1 2/√9 = 0 .3 6 s e c
W4 = 225 x 10 = 2250 KN
Ah = (Z / 2) X (I/R) X (Sa/g) = (0.16/2) X (1/5) X 2.5 = 0.04
= 3750 KN
Vb
Total Seismic weight of the structure, W = ∑Wi = 3 X 3750+2250 = 13500 KN 27
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= Ah X W = 0 .04 X 13500 = 540 K N.
= Ah X W = 0 .04 X 13500 = 540 K N.
SeismicForce is samein both directions. 28
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VERTICAL PROJECTION (WATER TANK) CONCENTRICALLY
Lateral force distribution along height by Static Analysis
Storey hi L eve l W i (K N) (m) W i hi2
Wihi/(∑Wihi
2
Lateral Force at ith Level for EL in X & Y-direction (KN)
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2250
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324
0.40678
219.66
3
3750
9
303.75
0.381356
205.93
2
3750
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135
0.169492
91.52
1
3750
3
33.75
0.042373
22.88
796.5
1
540
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Case b): Building + Vertical Projection --- Water Tank placed concentrically
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Seismic weight on roof: w4 = DL + wt. of tank wt. of tank :
DATA:
a) Roof slab = 0.120 x 5.2 x 3.2 x 25 =49.92 = 50 KN
DL on floors = 10 KN/m2 , LL on floors = 5 kN/m2 seismic zone III
b)Walls -0.2 x 16 x 2 x 25 = 160 KN
z = 0.16 ; I = 1 : R = 5 c)Bottom slab = 0.2 x 5.2 x 3.2 x 25 = 83.2KN
Seismic Weight calculations: 4)Beams = 0.6 x 0.3 x (5 x 2 + 3 x 2) x 25 = 72 KN L = 9m
;
b = 25m
; floor area = 225 sq.m 5)Staging 4-columns 4 x 0.3 x 0.3 x 2.1 = 18.9 = 19 KN
LL >= 5KN/sqm , 50% LL need to be considered on all floors except roof. Total wt. of tank = 384.2 KN ; Total wt. on roof = 2634.2 KN Seismic weight on floors : Total wt.of structure = 11071.7 KN w1 = 225(10 + 0.5 x 5) = 2812.5 KN = w2 = w3 31
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VERTICAL PROJECTION (WATER TANK) AT ECCENTRICITY
Lateral force distri bution along height by static method.
3.0 3.0
Storey Le vel
Wi (KN)
hi, m
4
2634.0
12
Wi hi 2 x 1000
Wihi / ∑Wihi2
Lateral force at ith level for EQx and EQy (KN)
3.0 5 @ 5m c/c
PLAN
379.3
0.520
229.0
2.0 2.1
3
2812.5
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227.8
0.310
2
2812.5
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101.3
0.140
61.1
1
2812.5
3
25.3
0.034
15.3
∑=
733.7
1
137.5 3.0 3.0
442.9
3.0 3.0 5 @ 5m c/c
ELEVATION
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Case c): Building + Vertical Projection --- Water Tank placed eccentrically. The center of mass (CM) is calculated as follows
The lumped wt. of tank on roof causes a shift in center of mass of
wt. of tank = 384.2 KN = 25.613 KN/m2 roof level whereas the centre of rigidity remains at the geometrical
mass of tank = 2561.33 Kg/m2
center of roof. Thus an eccentricity in both directions is induced which X= 11.55 m,
Y= 4.782 m
creates torsion & increases the shear. CM (11.56 , 4.73) , CR (12.5, 4.5) EQ in X direction:EQ.in x direction :
T = 0.09h/√d = 0.09 x 12√25 = 0.216 Sec. Calculated eccentricity esi = 12.5 - 11.56 = 0.94 m Ah = (Z / 2) X( I / R) X (Sa/g) = 0.04
Design eccentricity edi = 1.5 x esi + 0.05b
OR
Design base shear = Ah x W = 442.868 KN edi = esi - 0.05b
whichever gives more severe effect
EQ in Y – direction:- Ah = 0.04, Design base shear = Ah x W = 442.868 KN
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For, edi = 1.5 x 0.94+ 0.05 x 25 = 1.41 + 1.25 = 2.66 m OR Ist column line, Vx = 8.195 KN edi = 0.94 -0.05 x 25 = 0.94 - 1.25 = -0.31m
2nd column line, Vx = 2.730 KN
edi = 2.66 m
3rd column line, Vx = - 2.731 KN
EQX = 228.95 KN
4th column line, Vx = - 8.195 KN
Torsional moment T =(EQX) x (edi) = 609.007 KNm Additional shear due to torsion is given as
The -ve values are to be neglected while +ve shear to be added i.e. force is
Vx = (T / Ip) x (y) x (Kxx)
not to be deducted as per IS – 1893 – 2002.
Ip = ( Kx )x Y2 + (Ky) x X2 , Kx = 12E I / L3 , I = 0.000675 m4 ,
Similar calculations follow for Y direction .
E = 25 X 106 KN/m2 , Kx = 7500 KN/m, (Kx) x Y2 = 1923750, ( Ky) x X2 = 13125000, Ip = 15048750.
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Case d): Design of Projection (Tank- Column)
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Case e): Design cons iderations for hori zontal projections
Design of container Column: Wt. of container = 384.2 KN ;
Design of horizontal projections for vertical earthquake acceleration.
Wt.of water = 30 x 10 = 300 KN.
Vertical acceleration = 10/3 (Ah) = 0.133.
Total load = 384.2 + 300 = 684.2 KN The projection is to be designed for 5 times Ah
Consider a horizontal balcony 0f 1m x 1m with a parapet wall of ht = 1m
--- IS 1893 – 2002
M – 20 concrete ; Fe – 415 Steel
= 5 x Ah = 5 x 0.04 = 0.2
L / d =7
Vb = Ah x W = 0.2 x 684.2 = 136.84 KN.
M.F.= 1.3
d = 1000/7 x 1.3 = 109.89mm :
Moment = 2.1 x (136.84/4) = 71.841 KNm & Axial load = 684.2 / 4 = 171.05 KN
D = 109.89 + 20 = 129.89 mm =130 mm (say)
The column is to be checked for,
d = 110 mm .
Axial load = 166.32 KN Moment = 139.713 KNm. along both directions. 39
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For the seismic acceleration in vertical direction the stability is checked
Loading calculation :
Av = (2/3) Ah also these projections are designed for five times
1) DL = 25 x 0.130 + 1.5 = 4.75 kn/m2 Assume LL = 5 KN / m2
the horizontal accn.hence Av = (10 / 3) x Ah = (10 / 3) x 0.04 = 0.133
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Total load = 9.75 KN/m
The overturning moment is obtained for a critical or worst combination as
ultimate load = 1.5 x 9.75 = 14.635 KN/m2
Mo = w1 (1 + Av)L / 2 + w2 (1+ Av)L
Wt. of parapet wall = 1 x 0.1 x 25 = 2.5 KN w1 = 1 x 1 x 0.130 x 25 + LL; w2 = 2.5 KN ( Point load of the Parapet wall)
Design moment = WuL2 + Wu x L = 14.635 x 1 x 1/ 2 + 2.5 x 1 = 9.82 KN m Section design ; for given material
Ru = 2.76
dreq =√.(9.82 x 106 / 2.76 x 1000) = 59.65 mm < provided. -- o.k.
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