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ANALYSIS OF EARTHQUAKE GROUND MOTIONS : STATIC ANALYSIS

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ANALYSIS OF EARTHQUAKE GROUND MOTIONS – I: STATIC ANALYSIS

Requirements of Earthquake Resistant Design

Seismic Coefficient Method Torsion due to Eccentricities Storey Drift Calculations Appendages Numerical Examples

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Requirements of Earthquake Resistant Design

Earthquake can cause damage not only onaccount of the shaking results from them butalso due to other chain effects like landslides,floods, fires and disruption to communication.

Therefore it is important to take necessaryprecautions in the planning and design ofstructures so that they are safe against suchsecondary effects also.

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6.1.1 The Characteristics (intensity, duration, etc) of Seismic Ground

Vibrations 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 ---

Nature of Foundation Soil Materials, Forms, Size and Mode of Construction of

Structures Duration and Characteristics of Ground Motions

IS 1893(Part I): 2002 specifies 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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6.1.3 Design Approach:Structure should - Possess at least a minimum strength to withstand minor

EQs (<DBE) which occur frequently without damage

Resist moderate EQs (DBE) without significant structuraldamage though some non-structural damage may occurand

Aim that structures withstand a major EQ (MCE) withoutcollapse.

Actual forces that appear on structures during EQ aremuch greater than the design forces specified in thisstandard

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However – Ductility, arising from inelastic materialbehavior and detailing.

Over-strength, arising from the additional reservestrength in structures over and above the designstrength, are relied upon to account for this difference inactual and design lateral forces.

The specified EQ loads are based upon post-elasticenergy dissipation in the structure and hence theprovisions of this standard for design, detailing andconstruction shall be satisfied even for structures andmembers for which load combinations which do notcontain the EQ effect indicate larger demands thancombination including EQ.

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6.1.7 Additions to the Existing Structures Structurally Independent – shall be designed as NEW

structure

Structurally Dependent – shall be designed as Entireconforms to the seismic force resistance requirementsfor new structure

(1) Addition shall comply with the requirements for newstructures.(2) Addition shall not increase forces in any structuralelements by 5 % unless the capacity of the elementsubject to increased force is still in compliance with thisstandard, and(3) Addition shall not decrease the seismic resistance ofany structural element unless >= required

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6.3.2 Design Horizontal EQ Load

When the Lateral Load resisting elements are orientedalong orthogonal horizontal directions the structure shallbe designed for the effects due to full Design EQ Load inone horizontal direction at time.

When the lateral load resisting elements are not orientedalong the orthogonal horizontal directions, the structureshall be designed for the effects due to full design EQload in one horizontal direction plus 30 % of the designEQ load in the other direction

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7.5 Design Lateral Force:

The Design Lateral Force shall first be computed for thebuilding as a whole.

This Design Lateral Force shall then be distributed to thevarious floor levels.

The overall design seismic force thus obtained at eachfloor level, shall then be distributed to individual LateralLoad Resisting elements depending on the floordiaphragm action.

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7.5.3 Design Seismic Base Shear

Total Design Lateral Force or Design Seismic BaseShear (Vb) along any principal direction shall bedetermined by

Vb = Ah * WWhere,Ah = Design Horizontal acceleration spectrum using the

fundamental natural Period Ta , as per the considereddirection of vibration

W = Seismic weight of the building.

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7.6 Fundamental Natural Period (Tn)

The Approximate Fundamental Natural Period ofVibration (Tn), in seconds, of a Moment-Resisting FrameBuilding Without Brick Infill Panels may be estimated bythe 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 basementstoreys, where basement walls are connected with theground floor deck or fitted between the building columns.

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7.6.2 Tn For all Other Buildings

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

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6. 4 Design Spectrum

6.4.2 The Design Horizontal Seismic Coefficient 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) whatever be the value of (I/R)

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Where, Z = Zone Factor –

II, III, IV, VLow, Moderate, Severe, Very Severe0.10, 0.16, 0.24, 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 16

7.3 Design Imposed Loads for Earthquake Calculation

7.3.1 The EQ Force shall be calculated for the Full DL + the percentage of Imposed Load

Up-to and including 3.0 kN/m^2 – 25 %Above 3.0 kN/m^2 – 50 %

7.3.2 For Calculating the Design Seismic Forces of theStructure, the Imposed Load on Roof NEED not beconsidered.

7.3.3 Percentage of Imposed Loads shall also be usedfor “Whole Frame Loaded” condition in the LoadCombinations where the Gravity Loads are combinedwith the EQ Loads, No reduction as per IS 875 fornumber of storeys.

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7.4 Seismic Weight 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 floors.

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.

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7.7 Distribution of Design Force 7.7.1 Vertical Distribution of Base Shear to Different

Floor Levels –

Qi = (Vb) * (Wi * hi^2)/(∑ Wi * hi^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

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7.9 Torsion 7.9.1 Provision shall be made in all buildings for

Increase in SF on the Lateral Force ResistingElements resulting from the Horizontal TorsionalMoment arising due to eccentricity between the Centreof Mass and Centre of Rigidity. However, the NegativeTorsional Shear shall be neglected.

7.9.2 Design Eccentricity ( whichever gives the moresevere effect) e di – static eccentricty, distancebetween CM and CR

Bi – Floor Plan dimension perpendicular to the direction of Force.

b 0.05- e OR b 0.05 e* 1.5 e i sii si di 20

7.11 Deformations 7.11.1 Storey Drift Limitation

The storey drift in any storey due to minimum specifieddesign lateral force with partial factor of 1.0 shall notexceed 0.004 times the storey height. No drift limit forsingle storey building which has been designed toaccommodate storey drift.

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7.11.2 Deformation Compatibility of Non-Seismicstructures

For buildings located in seismic zones IV and V, it shallbe ensured that the structural components that are notpart of the seismic Force Resisting System in thedirection under consideration, do not lose their verticalload-carrying capacity under the induced momentsresulting from storey deformations equal to R times thestorey displacements calculated.

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7.12.2 Cantilever Projections

7.12.2.1 Vertical Projections

Tower, Tanks, Parapets, Smoke Stacks ( Chimneys) andOther Vertical cantilever projections attached to buildingsand projecting above the roof shall be designed andchecked for stability for FIVE times the design horizontalseismic coefficient Ah. In the analysis of the building, theweight of these projecting elements will be lumped withthe roof weight.

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7.12.2.2 Horizontal Projections

All Horizontal Projections like cornices and balconiesshall be designed and checked for stability for FIVEtimes the design Vertical coefficient

7.12.2.3 The increased Design Forces are ONLY fordesigning the projecting parts and their connections withthe main structures. For the design of the main structure,such increase NEED not be considered.

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Calculation of Design Seismic Force by Static Analysis

Problem Statement:

Consider G+3 Storey building. The building is located in seismic zone III. The soil conditions 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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(a) Only Building.

(b) Building + Vertical Projection --- Water Tank placed concentrically

(c) Building + Vertical Projection --- Water Tank placed eccentrically.

(d) Analysis of cantilever portions --- Vertical & horizontal projections.

Cases considered for the analysis

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3.0

3.0

3.0

5 @ 5m c/c

3.0

3.0

3.0

3.0

5 @ 5m c/c

PLAN

ELEVATION

PLAN AND ELEVATION OF A BUILDING

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Case a): Analysis of G + 3 BUILDING

Design Parameters:

For seismic zone III, z= 0.16 (Table 2 of IS:1893)

Importance factor I = 1.0 Response Reduction factor R = 5

Seismic Weights:

Floors: Roof:

Floor area= 25 X 9 = 225 m2 W4 = 225 x 10

W1 = W2 = W3 = 225 X (10+0.5 X 5) = 2250 KN

= 3750 KN

Total Seismic weight of the structure, W = ∑Wi = 3 X 3750+2250 = 13500 KN

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Fundamental Period:EQ in X-direction:

T = 0.09h/√d = 0.09 X 12/√25 = 0.216 sec.

Ah = (Z / 2) X (I/R) X (Sa/g) = (0.16/2) X (1/5) X 2.5 = 0.04

Vb = Ah X W = 0.04 X 13500 = 540 KN.

EQ in Y-direction:

T = 0.09h/√d = 0.09 X 12/√9 = 0.36 sec

Ah = (Z / 2) X (I/R) X (Sa/g) = (0.16/2) X (1/5) X 2.5 = 0.04

Vb = Ah X W = 0.04 X 13500 = 540 KN.

Seismic Force is same in both directions.

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Lateral force distribution along height by Static Analysis

Storey Level Wi (KN)

hi (m) Wi hi2 Wihi/(∑Wihi2

Lateral Force at ith Level for EL in X & Y-direction

(KN)

4 2250 12 324 0.40678 219.66

3 3750 9 303.75 0.381356 205.93

2 3750 6 135 0.169492 91.52

1 3750 3 33.75 0.042373 22.88

796.5 1 540

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VERTICAL PROJECTION (WATER TANK) CONCENTRICALLY

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Case b): Building + Vertical Projection --- Water Tank placed concentrically

DATA :

DL on floors = 10 KN/m2 , LL on floors = 5 kN/m2

seismic zone III z = 0.16 ; I = 1 : R = 5

Seismic Weight calculations:

L = 9m ; b = 25m ; floor area = 225 sq.m

LL >= 5KN/sqm , 50% LL need to be considered on all floors except roof.

Seismic weight on floors :

w1 = 225(10 + 0.5 x 5) = 2812.5 KN = w2 = w3

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Seismic weight on roof:w4 = DL + wt. of tank

wt. of tank :

a) Roof slab = 0.120 x 5.2 x 3.2 x 25 =49.92 = 50 KN

b)Walls - 0.2 x 16 x 2 x 25 = 160 KN

c)Bottom slab = 0.2 x 5.2 x 3.2 x 25 = 83.2KN

4)Beams = 0.6 x 0.3 x (5 x 2 + 3 x 2) x 25 = 72 KN

5)Staging 4-columns 4 x 0.3 x 0.3 x 2.1 = 18.9 = 19 KN

Total wt. of tank = 384.2 KN ; Total wt. on roof = 2634.2 KN

Total wt.of structure = 11071.7 KN

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Lateral force distribution along height by static method.

Storey Level

Wi (KN)

hi, m

Wi hi2

x 1000Wihi /∑Wihi2

Lateral force at ithlevel for EQx and EQy (KN)

4 2634.0 12 379.3 0.520 229.0

3 2812.5 9 227.8 0.310 137.52 2812.5 6 101.3 0.140 61.1

1 2812.5 3 25.3 0.034 15.3∑= 733.7 1 442.9

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VERTICAL PROJECTION (WATER TANK) AT ECCENTRICITY

5 @ 5m c/c

5 @ 5m c/c

3.0

3.0

3.0

3.0

3.0

3.0

3.0

PLAN

ELEVATION

2.12.0

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The lumped wt. of tank on roof causes a shift in center of mass of

roof level whereas the centre of rigidity remains at the geometrical

center of roof. Thus an eccentricity in both directions is induced which

creates torsion & increases the shear.

EQ in X direction:-

T = 0.09h/√d = 0.09 x 12√25 = 0.216 Sec.

Ah = (Z / 2) X( I / R) X (Sa/g) = 0.04

Design base shear = Ah x W = 442.868 KN

EQ in Y – direction:- Ah = 0.04,

Design base shear = Ah x W = 442.868 KN

Case c): Building + Vertical Projection --- Water Tank placed eccentrically.

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The center of mass (CM) is calculated as follows

wt. of tank = 384.2 KN = 25.613 KN/m2

mass of tank = 2561.33 Kg/m2

X= 11.55 m, Y= 4.782 m

CM (11.56 , 4.73) , CR (12.5, 4.5)

EQ.in x direction :

Calculated eccentricity esi = 12.5 - 11.56 = 0.94 m

Design eccentricity edi = 1.5 x esi + 0.05b OR

edi = esi - 0.05b whichever gives more severe effect

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edi = 1.5 x 0.94+ 0.05 x 25 = 1.41 + 1.25 = 2.66 m OR

edi = 0.94 - 0.05 x 25 = 0.94 - 1.25 = -0.31m

edi = 2.66 m

EQX = 228.95 KN

Torsional moment T =(EQX) x (edi) = 609.007 KNm

Additional shear due to torsion is given as Vx = (T / Ip) x (y) x (Kxx)

Ip = ( Kx )x Y2 + (Ky) x X2 , Kx = 12E I / L3 , I = 0.000675 m4 ,

E = 25 X 106 KN/m2 , Kx = 7500 KN/m,

(Kx) x Y2 = 1923750, ( Ky) x X2 = 13125000, Ip = 15048750.

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For,

Ist column line, Vx = 8.195 KN

2nd column line, Vx = 2.730 KN

3rd column line, Vx = - 2.731 KN

4th column line, Vx = - 8.195 KN

The -ve values are to be neglected while +ve shear to be added i.e. force is

not to be deducted as per IS – 1893 – 2002.

Similar calculations follow for Y direction .

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Case d): Design of Projection (Tank- Column)

Design of container Column:

Wt. of container = 384.2 KN ; Wt.of water = 30 x 10 = 300 KN.

Total load = 384.2 + 300 = 684.2 KN

The projection is to be designed for 5 times Ah --- IS 1893 – 2002

= 5 x Ah = 5 x 0.04 = 0.2

Vb = Ah x W = 0.2 x 684.2 = 136.84 KN.

Moment = 2.1 x (136.84/4) = 71.841 KNm & Axial load = 684.2 / 4 = 171.05 KN

The column is to be checked for,

Axial load = 166.32 KN Moment = 139.713 KNm. along both directions.

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Case e): Design considerations for horizontal projections

Design of horizontal projections for vertical earthquake acceleration.

Vertical acceleration = 10/3 (Ah) = 0.133.

Consider a horizontal balcony 0f 1m x 1m with a parapet wall of ht = 1m

M – 20 concrete ; Fe – 415 Steel

L / d =7 M.F.= 1.3

d = 1000/7 x 1.3 = 109.89mm :

D = 109.89 + 20 = 129.89 mm =130 mm (say)

d = 110 mm .

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Loading calculation :

1) DL = 25 x 0.130 + 1.5 = 4.75 kn/m2

Assume LL = 5 KN / m2

Total load = 9.75 KN/m2

ultimate load = 1.5 x 9.75 = 14.635 KN/m2

Wt. of parapet wall = 1 x 0.1 x 25 = 2.5 KN

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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For the seismic acceleration in vertical direction the stability is checked

Av = (2/3) Ah also these projections are designed for five times

the horizontal accn.hence Av = (10 / 3) x Ah = (10 / 3) x 0.04 = 0.133

The overturning moment is obtained for a critical or worst combination as

Mo = w1 (1 + Av)L / 2 + w2 (1+ Av)L

w1 = 1 x 1 x 0.130 x 25 + LL;w2 = 2.5 KN ( Point load of the Parapet wall)