Earthquake Engineering

Authored by : Indrajit Chowdhury Page 1 8/30/2006 Head of Department Civil & Structural Engg. Petrofac International Ltd. Sharjah United Arab Emirates

Analytical and Design Concepts for Earthquake Engineering Introduction: - In this chapter we will deal with some fundamental concepts pertaining to earthquake engineering. On completion of this chapter you should have an understanding of: - Why earthquake happens in nature The essential engineering parameters, which affect the geo-technical and structural

aspect of a system under earthquake. Basic concepts of dynamic analysis as applied to Earthquake engineering

pertaining to buildings, and different types of industrial and infra-structural systems like chimney, frame foundation, retaining wall, water tank, RCC and earth dams etc.

Have an understanding of different provisions of IS 1893 (2002) code as well as UBC 1997 and IBC 2000 the two American earthquake codes used almost internationally for projects in different part of the world.

Before reading this chapter we however feel you should go through the following chapter as a pre-requisite. 1. Chapter 4 Basic concepts in structural dynamics 2. Chapter 5 Basic concepts in soil dynamics 3. Also have some fundamental awareness of how earthquake can affect a structure-

foundation system. Earthquake is perhaps the most complex natural phenomenon which human being is trying to understand, combat and harness from the early history of mankind. In spite of scientific study of the subject for last 100 years or more it is felt that we are still in the infancy of our knowledge in this subject. For data affecting this phenomenon are so vast and varying and also from different branch of science, we at best can arrive at a simplified model of the problem amenable to human perception and try to arrive at a solution which would in all probability survive this natures assault with some limited damage if ever the structure is faced with such vagary. The basic objective of an earthquake resistant design is not to make the structure fool proof but to limit its damage to the extent of minimizing the loss of human life and property. Though earthquake is a global phenomenon, yet there are some countries in the world which has been severely affected by earthquake leading to significant loss of human life and properties- like USA, Japan, Turkey India, New Zealand etc, while there are countries whose geological characteristics are seismically considered inert like United Kingdom, Gulf countries like Oman, Kuwait, UAE, Qatar etc which has no significant history of earthquake. Based on the above it is evident that there are countries where significant research and investigation has been carried out to develop a procedure for earthquake resistant design of structures. Countries like USA, Japan, India Mexico etc have contributed significantly on this issue. In USA and Japan considerable research has been carried out in University of California Berkley, California Institute of Technology, University of North Carolina, University of Tokyo to name a few.


In India significant work has been carried out in University of Roorkee- Earthquake Engineering Research Center, Indian Institute of Technology Kanpur to understand the seismicity of the region and develop a unified code for earthquake resistant design of structures and foundations. Why does Earthquake happen in Nature? The topic itself can be subject of a complete book. A detailed discussion on this is beyond the scope of this work, however as civil engineers designing structures, which can withstand such calamity -some fundamental understanding on this issue is essential. Earth Crust Earth Core Molten Magma Figure-1 Earth with its central core As shown in the above figure (1) the earth constitutes of a central core, consisting of molten magma, which is undergoing continuous upheaval. While the outer core, which has solidified in million of years forms the outer earth crust. The inner magma (the molten core) is continuously creating a pressure on the outer core and trying to come out by seeking some weaknesses in the earth crust. Whenever it can come out it generates what is known as a volcanic eruption. When it cannot it tries to push the crust upward thus creating folds and faults resulting in a source which stores a significant amount of potential/strain energy. As by law of nature all system in course of time try to achieve minimum state of energy, these storehouse of potential energy keep on releasing their stored strain energy as kinetic energy generating waves on the surface of the outer crust which is known as earthquake commonly. It is said that the Himalayan mountain range is one such formation due to pressure of the inner magma. The deformation, which the earth crust underwent due to formation of the mountain range, is still being adjusted naturally. It is for this reason, areas in its close proximity like Assam, Nepal and portions of south China is often subjected to severe earthquake. There is also a phenomenon called seismotectonic movement otherwise known as continental drift, which also generates earthquake at certain location of the earth. According to this the outer crust of earth is made up of undistorted plates of lithosphere. These plates are in differential motion, and at places they move away from each other where new plate are added from the interior of the earth while in places they collide with each other.


All major earthquakes which mark the active zones of the earth closely follows the plate boundaries and has been found to be a function of the movements of these plates1. Human interference can also sometimes modifies stresses in the earth surface to trigger minor or even moderate earthquakes. In many mining areas tremors and shocks results due to underground explosion in mines which can cause damages to structures on the ground. One of the classic case of man made earthquake was the Koyna Dam incident in 1967 in India when pounding of large amount of water behind the dam resulted in an earthquake causing extensive damage to surrounding2. Essential difference between systems subjected to Earthquake and Vibrations from Machine In earlier chapters we had discussed in detail response of machine foundation under dynamic loading. In machine foundations the unbalanced force from the machine gets transmitted to the ground via structure/foundation to the soil media. In such cases normally a limited part of the soil is affected significantly. Moreover the strain range induced in the soil is usually limited to low strain range (usually 10-3%). However in case of earthquake the phenomenon is quite different. In this case when an earthquake shock generates due to rupture of a fault within the earth surface it generates waves within the soil, which induces much larger strain in the soil (10-2 to10-1%) for a major earthquake.

Time History response Velocity

-0.3

-0.2

-0.1

0

0.1

0.2

0

1.14

2.28

3.42

4.56 5.7

6.84

7.98

9.12

10.3

11.4

12.5

13.7

14.8 16

17.1

18.2

19.4

Time steps

Velo

city

v2(

m/s

ec)

Figure 2: -Typical propagation of earthquake waves through soil Shown above is a typical propagation of waves through soil medium and is usually a combination of specifically four types of waves namely

1) P waves (body waves) 2) 2) S waves (body waves) 3) Rayleigh Waves (surface waves)

1 For further information reader may refer to the monograph New Zealand Adrift by Stevens G.R.; A.H &A.W.Reed Wellington 1980. 2 Anil K Chopra and Partha Chakrabarti The Koyna Earthquake and damage to Koyna Dam Bulletin of Seismological Soceity of America 63 No-2 381-97 (1973)


4) Love waves (surface waves) The primary or P waves are the fastest traveling of all the waves and generally produces longitudinal compression and extension within a soil media. These waves can travel both through soil and water and are the first one to arrive at a site. However soil being relatively more resistant to compression and dilation effect its impact on ground distortion is minimal. The S waves, also otherwise known as secondary or shear waves usually causes shear deformation in the medium through which it propagates. The S waves can usually propagate through soil only 3.The soil being weak in resisting shear deformation it travels at a much slower speed through the ground then primary waves and are found to cause maximum damage to the ground surface. The Rayleigh waves are surface waves which are found to produce ripples on the surface of the ground 4.These waves produce both horizontal and vertical movement of the earth surface as the waves travel away from the source. Love waves are similar to S waves and produces transverse shear deformation to the ground. As described above all these waves combine together to produce shock waves from which an engineer extracts the value of the maximum ground acceleration (amax) which is the major parameter which governs his design. Thus based on above it is apparent the mechanics of earthquake is opposite to dynamics of machine foundation in the sense that here the forces are transmitted from soil to the structure. It is the shock within the ground which excites the structure and induces the inertial force in the system. Effect of earthquake on soil foundation system :- Having explained in the earlier section that the primary source of disturbance is in the soil itself, it is important to assess and know what could be the effects of an earthquake on the soil on which a structure is built. For it should be understood that irrespective of how well an earthquake resistant design is carried out for a structure if the ground supporting it fails, the structure will invariably undergo significant damage and which at times could even be catastrophic5. The major effect on soil affected by an earthquake can be classified as follows :-

1) Liquefaction of soil 2) Settlement of foundation due to deep seated liquefaction failure 3) Reduction of bearing capacity 4) Ground Subsidence 5) Land Slides

Of all the phenomenon defined above liquefaction is perhaps the most important factor which has caused catastrophe in many previous earthquakes, and unfortunately gets very little attention from structural engineers in a design office 6.Thus it is important to understand what is the phenomenon and what are methods available to assess and mitigate it. What is Liquefaction? Conceptually speaking liquefaction is very much akin to giving a rapid squeeze to a sponge ball saturated with water. When the squeeze is applied, we observe that the water stored inside the sponge comes out and the sponge feels lighter as the water comes out. For soil sample (specially when it is cohesionless) the shear strength is given by the expression

3 Since liquid have no shear resistance it cannot travel through water. 4 This is very much similar to ripples produced by a pebble dropped in a pond. 5 Nigata Earthquake in Japan 1964 was one of the primary example where a number of structures underwent significant damages due to ground subsidence and liquefaction of soil 6 Especially in India where in previous earthquakes significant damage has been recorded due to this phenomenon.


tan)( us =

Here s= shear strength of the soil = Overburden pressure of the soil sample u= insitu pore pressure within the soil sample = angle of internal friction of the soil sample When earthquake force acts on the soil sample it produces a rapid shock on the body, by virtue of which there is a sudden increase in pore pressure, which cannot dissipate readily. When the force of earthquake is significantly high(M>=6.5) which also results in ground shaking for a good amount of time the pore pressure increment becomes such that it equals the overburden pressure and the soil looses its shear strength altogether(i.e. s=0) and starts flowing like a liquid. This phenomenon is otherwise known as liquefaction of soil. When such phenomenon is observed during an earthquake soil collapses completely and sand boils are observed in the ground. Even c- soils losses significant part of its strength resulting in bearing capacity failures of foundation and or significant settlement. Liquefaction of soil has been observed in a number of earthquakes throughout the world like Nigata in Japan(1964), Kobe in Japan(1995), Dhubri and Koyna(1967) Earthquakes in India. From above discussion it is obvious that non-plastic cohesionless soil under saturated condition are most susceptible to earthquake. As SPT value has been extensively used to define the static engineering strength of cohesionless soil consistently it was but natural that researchers tried to co-relate SPT values of cohesion less sandy soil to liquefaction potential of soil samples to earthquake shocks. Pioneering research work was done in this area by Seed, Idriss and Tokimatsu who correlated the observed SPT values to Cyclic resistance ratio which is one of the major parameters used to define the liquefaction potential of a soil sample. We will talk more about this later, first let us see how liquefaction is measured for a particular soil sample. The susceptibility of a soil sample undergoing liquefaction is measured by a term called liquefaction potential, which is measured as a factor of safety against cyclic resistance ratio to cyclic stress ratio. Mathematically speaking it is defined as :-

0.1=CSRCRRFS (10.1)

Here FS.= Factor of safety against Liquefaction CSR= Cyclic stress ratio CRR=Cyclic resistance ratio In other words ( based on the above formulation) if the factor of safety is less than or equal to 1.0 the soil has very good possibility of undergoing liquefaction under an earthquake, however if the value is greater than 1.0 the possibility of soil failure due to liquefaction is remote. Thus it is obvious that we need to first understand what does CSR and CRR stand for. During earthquake the soil under influence of earthquake will be subjected to repetitive shear stress( known as cyclic shear stress) and is estimated by the expression

dv

v

v

av rg

aCSR

==

'65.0

'max

(10.2)


Here amax=Maximum acceleration at the ground surface v= Total overburden pressure at the design depth v' = Effective overburden pressure at the design depth

g= Acceleration due to gravity =dr Stress reduction factor which varies with depth and is given by zrd 000765.00.1 = for mz 15.9 (10.3a)

zrd 0267.0174.1 = for mzm 2315.9 (10.3b) zrd 008.0744.0 = for mzm 3023 (10.3c)

5.0=dr for mz 30 (10.3d) For ease of electronic computation dr may also be expressed by the expression7 ( )( )25150

5150

00121000062050572904177000010017530040520411300001

z.z.z.z..z.z.z..r ..

..

d ++++= (10.4)

The maximum acceleration of the ground ( maxa ) is another factor, which needs careful evaluation. For practical design office purpose one of the expression used to evaluate amax is

( ) gDa M 8.0320.0max 10184.0 = (10.5) Here =maxa Maximum ground acceleration M= Expected Moment Magnitude of earthquake D= Maximum epicentral distance in km. It may be noted that if more reliable observed earthquake data is available for the site (predicting ground acceleration more accurately) it may well be used in lieu of the above formula. Having calculated the cyclic stress ratio based on the above expressions it is essential to evaluate the cyclic resistance ratio(CRR) of the in-situ soil. It is evident that the CRR value of the soil sample will depend on its in-situ strength. Since Laboratory testing can be carried out under a better controlled environment, one of the plausible method which has been tried is to collect in-situ undisturbed soil sample for evaluation of the parameter CRR in the laboratory. However, one of the major difficulty encountered in this is that generally the in-situ stress state cannot be established in the laboratory, and specimens of granular soil retrieved with typical drilling techniques are far too disturbed to yield any meaningful results. Only through very specialized sampling techniques such as ground freezing sufficiently undisturbed sample can be obtained, which again becomes a prohibitively costly affair for all but most critical projects. It is for this co-relating the CRR value with field observed test data is still the sate of the art practice.

7 This formula was proposed by Thomas F Blake (Fugro-West Inc, Ventura Califormia)


Co-Relation between CRR and SPT value:- For calculation of CRR based on observed SPT value(No), as a first step the observed SPT value is subjected to certain corrections as expressed hereafter ( ) ( )( )( )( )SRBENo CCCCCNN )(601 = (10.6)

Here =oN Measured SPT value at the site

=NC Is a correction factor for overburden pressure =EC Is a correction factor for Hammer energy ratio =BC Is a correction factor for borehole diameter =RC Is a correction factor for rod length =SC Is a correction factor for sampler with or without liners ( )601N = Corrected SPT value with 60% hammer efficiency.

The correction factors for various equipment parameters are as shown hereafter in Table-1 Factor Equipment Parameter Term Correction factor Overburden pressure Independent of Equipment NC

v

aP'

Energy Ratio Safety Hammer Doughnut Hammer

EC 0.6 to1.17 0.45 to 1.0

Rod length 3 to 4 m 4 to 6 m 6 to10 m 10 to 30 m > 30m

RC 0.75 0.85 0.95 1.0 >1.0

Bore Hole Diameter 65 to 115 mm 150 mm 200 mm

BC 1.0 1.05 1.15

Sampling Method Standard Sampler Sampler without Liners

SC 1.0 1.2

Table-1 Correction factors to observed SPT values Here Pa= Atmospheric pressure or 100kPa(100kN/m2)

v' = Effective overburden pressure at depth of the standard penetration sample Having established the design SPT value ( )601N the cyclic resistance ratio(CRR) is given by the expression for clean sands(i.e.< 5% contents) as

432

32

1 hygyfyeydycybyaCRR ++++

+++= where (10.7)


a=0.048, b=0.004721,c=0.0006136, d=-1.673X10-5,e=-0.1248,f=0.009578,g=-0.0003285, h=3.714X10-6 and y= ( )601N . The above equation is valid for ( )601N less than 30.For clean granular soil for N>30 are far too dense to liquefy and are generally classed as non-liquefiable. Another expression, which is used for clean sand base for computation of CRR is8

[ ] 2001

45)(1050

135)(

)(341

2601

601

6015.7 +++= N

NN

CRR (10.8)

Here =5.7CRR The Cyclic resistance ratio at Earthquake Magnitude of 7.5 Influence of fine contents on CRR value:- While developing the original expression Seed et al.(1985) noted an apparent increase of CRR value with an with increased fine contents. Whether this can be attributed to an increase in resistance or decrease in penetration resistance is not clear. However to cater to this it has been recommended correction to SPT values for the influence of fine contents. Other grain characteristics like Plasticity index (PI) may also affect the liquefaction resistance as well , however is not so well defined till date. Hence corrections based solely on fine contents is used and should be mellowed with judgment and caution. I.M.Idriss and R.B.Seed proposed corrections of ( )601N to an equivalent clean sand value ( ) CSN 601 given by ( ) 601601 )(NN CS += (10.9) where and are determined from the following relationships as shown in Table-2 Sl NO Values of and Fine content

1 0= For %5FC 2

= 2

19076.1FCe

%35%5 FC 3 0.5= %35FC 4 0.1= For %5FC 5

+=

100099.0

5.1FC %35%5 FC

6 2.1= %35FC Table-2 :- Modification factor to SPT value based on Fine contents The above equations can now be used for routine liquefaction resistance calculation for soil subjected to SPT at field. Effect of Earthquake Magnitude on Liquefaction Resistance :- The original study of the liquefaction potential was based on a magnitude of 7.5 earthquake. To evaluate the potential at earthquake of other magnitude correction factors were proposed that allows induced stress ratios for other magnitudes be adjusted to correspond to a magnitude of 7.5 by dividing the stress ratios by the factors as shown hereafter Table-3

8 After Alan.F.Rauch at the University of Texas 1998.


Sl No Earthquake Magnitude Magnitude scaling

Factor 1 5.25 1.5 2 6 1.32 3 6.75 1.13 4 7.5 1.0 5 8.5 0.89

Table-3 :- Magnitude scaling factor as proposed by Seed and Idriss The magnitude scaling factor as proposed above based on recent research is now believed to be very conservative for moderate size earthquake. A new set of MSF has now been proposed by Idriss where the MSF is defined as function of Moment Magnitude and is given by

56.2

24.210M

MSF = (10.10) We furnish below in Taqble-4 data furnished by other researchers on the MSF value varying with earthquake magnitude:- Sl No

Magnitude Seed And Idriss(original)

Idriss(Later) Arango Ambreseys Andrus & Stokoe

1 5.5 1.43 2.2 3 2.2 2.86 2.8 2 6 1.32 1.76 2 1.65 2.2 2.1 3 6.5 1.19 1.44 1.6 1.4 1.69 1.6 4 7 1.08 1.19 1.25 1.1 1.3 1.25 5 7.5 1 1 1 1 1 1 6 8 0.94 0.84 0.75 0.85 0.67 0.8 7 8.5 0.89 0.72 0.44 0.65 Table-4:- Magnitude scaling factor as proposed by various investigators The factor of safety against Liquefaction can now be expressed as

MSFCSR

CRRFS

= 5.7 (10.11) Where CRR 7.5 = Cyclic resistance ratio for a magnitude 7.5 earthquake Whatever has been discussed previously will now be further clarified by a suitable problem, which covers the whole gamut of the above conditions


Example 10.1 Density of soil =18/m3 GWL 1.0 m Density of soil =20kN/m3 2.0m Average SPT Value=13 Saturated Density of soil=19.6 kN/m3 6.0m Figure-3:- Soil Profile of a site with typical soil properties. As shown in the above figure 3 is a site soil profile which consists of 3.0meter of silty clay underlain by 6 meter of sand whose average SPT value is 13.The ground water table is observed to be at a level of 1.0 meter below Ground level. The dry density of the of the silty clay is 18 KN/m3, while that in saturated condition is 20 KN/m3.The saturated density of sand is 19.6kN/m3.Sieve analysis shows the sand to have Fines content as 15%.Find the liquefaction potential when the site is considered to be 150 kM away from the epicentre having an earthquake Moment magnitude of 6.5?The SPT test was carried out by standard sampler with safety hammer & having rod length of 6.0 meter. The diameter of the bore hole was 150mm Solution Considering ( ) gDa M 8.0320.0max 10184.0 = Here M=6.5 and D=150 kM which gives

( ) gga 4017.015010184.0 8.05.6320.0max == Effective vertical stress at center of the sand layer is

8.8636.92200.118' =++=v KN/m2 The gross vertical pressure at center of the sand layer

8.11636.192200.118 =++=v KN/m2 The depth below ground where liquefaction potential is calculated is 1+2+3=6m Thus z=6.0 m


The corrected SPT value is given by ( ) ( )( )( )( )SRBENo CCCCCNN )(601 = Here 13=oN

8.86100=NC , 0.1=EC , 05.1=BC , 85.0=RC , 0.1=SC

Thus ( ) 45.120.185.005.10.18.86

10013601 ==N For FC=15% we have

= 219076.1FCe

or 498.2215

19076.1 ==

e

and

+=

100099.0

5.1FC

or 048.110001599.0

5.1

=

+=

Thus corrected SPT value is given by ( ) 601601 )(NN CS += Or, ( ) 5.1545.12048.1498.2601 =+=CSN

Considering [ ] 2001

45)(1050

135)(

)(341

2601

601

6015.7 +++= N

NN

CRR

We have [ ] 16511.02001

455.151050

1355.15

5.15341

25.7 =+++=CRR The Magnitude scaling factor is given by

56.2

24.210M

MSF =

or, 44.15.6

1056.2

24.2

==MSF

Thus MSFCSR

CRRFS

= 5.7

Or 0.16798.044.13497.016511.0


Thus as the factor of safety being less than 1.0 the soil has a high chance of liquefaction during an earthquake. Correlation between CRR and CPT value:- Other than SPT, cone penetration test(CPT) is also used in field for evaluation of geo-technical engineering parameters. As such investigators have also tried to co-relate the CPT value with CRR for evaluation of liquefaction potential. One of the advantages with CPT being that since it is a continuous process, thin layers of soil that one can miss by SPT will not be missed in this case. As stated earlier, equation (10.2) is used to determine the CSR value .The CRR value is indirectly co-related to CPT by developing relationship between CPT and SPT value. As per Seed and Idriss

Ntoqc 5 4= for clean sand and Ntoqc 4.5 5.3= for silty sand

Here =cq Is the observed CPT value in Mpa Once an equivalent SPT value is obtained from the observed cq rest of the procedure remains same as stated earlier. Murthy et al has given following relationship which can also be used to obtain equivalent SPT values from the observed cone penetration values.

State of sand Dr N qc(Mpa) Very loose


Liquefaction of clay :- Normally clay is a substance which is deemed non liquefiable. However based on experience of earthquake in China it is now established that there are certain types of clay, which under shaking do undergo liquefaction. As a rule of thumb, a clay sample will be deemed liquefiable provided all of the following criteria as mentioned below are complied with :-

Weight of soil particles finer then 0.005 mm is less then 15% of the dry weight of the soil.

The liquid limit(LL) of the soil is less then 35%. The moisture content of the soil is less than 0.9 times the liquid limit of soil.

Clayey soil meeting not all of the above criteria are usually considered non liquefiable. Settlement of foundation due to liquefaction failure:- We had stated in our earlier section of liquefaction that during earthquake due to shock, there is a sudden increase in pore pressure that cannot dissipate immediately resulting in lose of shear strength of soil. However, in course of time, this pore pressure do dissipate away towards the surface resulting in volumetric deformation of the ground. Considering the above phenomenon and heterogeneous nature of soil the soil may undergo differential settlement which could be critical for building foundations and underground lifelines. A technique to estimate the ground settlement has been proposed by Ishihara and Yoshimine wherein they developed a chart based on which the post liquefaction volumetric strain is co-related to the FS value(CRR/CSR) and the SPT value as shown here after. Insert Ishihara & Yoshimine curve. Figure-4 Ishihara and Yoshmine Curve for computation of volunmetric strain Based on above curve once we know the FS and SPT value the volumetric strain is read off from the curve and settlement is obtained based on multiplying this strain with the depth of the soil. The above is now further elaborated by a problem hereafter Example 10.2 For the soil sample as described in example 10.1 estimate the settlement of the sandy layer considering all other boundary conditions remaining identical. Solution:- From previous example we have seen

0.16798.044.13497.01651.0


Reduction of Bearing Capacity of soil Normally it is believed that earthquake has marginal effect on the bearing capacity of soil. As a matter of fact it is often a common practice and advised in many codes to increase the allowable

bearing capacity by 25 to 3133 %.

The reason for this can be attributed to the fact that normally when we find the bearing capacity of soil, we find out the ultimate bearing capacity of soil. Dividing it by a factor of safety we arrive at the allowable bearing capacity of the soil. This is mostly as per the general shear failure theory of the soil whence, either Terzaghi, Meyerhof or Brinch Hansens formula is used. However in many cases and especially for cohesive soil, it is the settlement -which governs the design bearing capacity of soil. Thus during an earthquake which is considered once in a lifetime phenomenon on the structure, a lowering of the factor of safety on the bearing capacity is usually deemed acceptable, and hence allowable bearing capacity is increased. However it should be made clear that such increment is valid for a particular case of when the foundation is resting on

Crystalline rocks having no horizontal fragments or laminations Dense compacted sand having SPT value > 30 Stiff to very stiff clay with nominal plastic flow.

If the soil is otherwise made of fragmented rock, loose sand or soft plastic clay sensitive to vibration this increased bearing capacity value should not be used.9 In such cases there could be significant reduction in strength when the foundation can undergo either a local shear failure( when the foundation punches through the overlying soil due to liquefaction of the bottom layer) or undergo a general shear failure when there is significant change in the soil property for which the bearing capacity factors Nc,Nq, and N undergoes reduction resulting in a reduced bearing capacity. Punching Shear Failure of soil:- To understand how local shear failure can occur let us consider the soil profile as shown in figure 4 P Figure-5:- Soil Profile of a site with foundation resting on top layer on non-liquefiable soil

9 Unfortunately many design engineers hardly give consideration to this and believes this increase of bearing capacity of foundation almost a sacrosanct issue.


As shown in figure 4 let us consider a case of a foundation resting on top layer of shallow clayey soil which is non liquefiable, underlain by a layer of loose sand susceptible to liquefaction. It is apparent from figure 4 that depth of the layer below the footing to the top of liquefiable sand layer is quite less and it might so happen that if the bottom layer looses its strength and the foundation is subjected to heavy load from superstructure the foundation may punch through this thin layer of soil and collapse causing serious damage to the super-structure. Similar to a column punching through a RCC footing here the whole foundation punches through the soil along the vertical dotted line to collapse. To prevent this happening we calculate a factor of safety (FS) expressed as

PZLB

FS f)(2 += for isolated footing

and P

ZFS f

2= for strip footing. Here B= Width of foundation in meter L= Length of foundation in meter Z= Depth of soil layer from bottom of footing to the top of liquefiable soil =f Shear strength of un-liquefiable layer of soil in kN/m2. If the top layer of non liquefiable soil is cohesive in nature( clay) then the shear strength is given by

uf S= where =uS Un-drained shear strength of the soil. For c soil (undrained shear strength parameters) the shear strength is given by:-

tanhf c += Here =h Horizontal total stress in kN/m2. For cohesive soil this is often assumed as v5.0 For an un-liquefiable soil layer of cohesionless soil the shear strength is given by

'tan''tan' 0 vhf k== Here =h' Effective horizontal stress in kN/m2 and is equal to the coefficient of passive pressure at rest times the vertical effective stress v'

=' Effective angle of friction of cohesionless soil. We now show the application of the above based on a suitable problem as shown hereafter. Example 10.3 P =650kN 3m Layer-1 Non-Liquefiable plastic clay 9m Layer-2 Liquefiable sandy layer


Figure-6:- Soil Profile of a site with foundation resting on top layer on non-liquefiable soil As shown in the figure 5 a footing of size 3mX2m is place on a stiff clayey silt layer of undrained shear strength Su=50 kN/m2 and =10o.The footing has maximum load of 650kN on it (including its own weight).The clay layer is underlain by a layer of loose sand 9.0 meter deep which susceptible to liquefaction. Find the factor of safety of the foundation under punching shear failure. The foundation is resting at depth of 1.5 meter below ground level. Density of soil of top layer is 20 KN/m3 Solution:- As per the problem

5.15.10.3 ==Z meter.

305.120 ==v kN/m2 Thus 15305.0 ==h kN/m2.

64.5210tan1550 =+=f kN/m2 Thus resistive force = 6.78964.525.1)23(2)(2 =+=+ fZLB kN

Thus 214.1650

6.789 ==FS . Considering the uncertainty in soil FS=1.2 could be a low value General Shear failure capacity reduction due to liquefaction :- This phenomenon is generally observed in case of the soil supporting the foundation is a stiff clay layer underlain by sandy layer susceptible to liquefaction. The ultimate bearing capacity of foundation based on general shear failure theory is given by Terzaghis equation as

BNqNcNq sqcult 21++=

The first term ccN gives the strength of the soil due to its cohesive property. The second term depicts the effect of overburden soil which goes on to increase the bearing capacity of the soil

and the last term BNs21 gives the frictional strength of the soil where the term N is a

function of the friction angle . For clayey soil as 0= , gives 0=N and 1=qN ,For spread footing considering the aspect ratio(B/L) correction we thus have

fcult DLBcNq +

+= 3.01 which is modified to

fcuult DLBNSq +

+= 3.01 .

For shallow foundation near the ground as the second term has minimal effect, for all practical purpose we can consider the equation to be

+=LBNSq cuult 3.01

For the bottom layer of liquefiable soil there is obviously a reduction in value of Nc and this is usually function of the ratio of Z/B as follows


BZ c

N

0 0 0.25 0.7

0.5 1.3 1.0 2.5 1.5 3.8 5.5

Table 7- Reduction in value of Nc for Z/B ratio Here B= Width of the foundation Z= Height of soil from bottom of foundation to the top of liquefiable soil. Example 10.3 For the example problem cited in example 10.3 find the reduced bearing capacity of the foundation considering the top layer of soil as stiff clay of undrained shear strength of 50kN/m2.All other parameters remains same as the earlier problem.Consider Nc=5.5 for clayey soil. Solution:- Under unliquefied state the ultimate bearing capacity is given by

fcuult DLBNSq +

+= 3.01

For =0 Nc=5.5 which gives

3605.120323.015.550 =+

+=ultq kN/m2

Considering foundation size as 2mX3m we have

216032360 ==ultQ kN

32.36502160 ==FS

When the bottom soil is liquefied considering

75.00.25.1 ==

BZ

referring to table 7 reduced Nc value =1.9

Thus 1445.120323.019.150 =+

+=ultq kN/m2

Hence 86432144 ==ultQ kN.

Thus 3.1650864 ==FS which is low and should preferably be about 1.5


Ground Subsidence due to earthquake During an earthquake of major magnitude there are many cases of ground subsidence and land slides which has wrecked havoc on many structures and specially underground services which may get severely damaged due to this. In the Sanfrancisco Bay earthquake(1906) major source of damage was fire which broke out as an aftermath of the earthquake and could not be contained as most of the underground water pipe lines were severely damaged due to ground subsidence and became non-functional. The major reason for this subsidence is again deep seated liquefaction for which the soils starts to flow and due to differential or non uniform flow can split apart a structure built on it. Roads and pavements have been observed to undergo extensive damage due to subsidence and was a major observation in Chi Chi Earthquake in Taiwan(1999).When the slope of the ground is less or equal to 6% the flow of soil is generally defined as a lateral displacement of soil. When this slope is more than 6% the same is know as a land slide. A number of researches have been carried out to develop a mathematical model, which would effectively predict the subsidence of the ground during a major earthquake. However, parametric functions being so many in numbers and uncertain that there is yet a model which can stated as unconditionally applicable. The most used mathematical model for practical engineering purpose is one empirical model developed by Bartlet and Youd developed based on historical data collected from 6 earthquakes in USA and two in Japan. The proposed to expressions one for sites near steep banks with a free face, the other with sites having gently sloped terrain. For free faced condition :-

153483.06572.00133.09275.01782.13658.16 LogTLogWRLogRMLogDH +++=1515 509224.0)100(5270.4 DFLog +

For sloped terrain condition :- 153483.04293.00133.09275.01782.17870.15 LogTLogSRLogRMLogDH +++=

1515 509224.0)100(5270.4 DFLog + Here DH=Estimated average ground displacement in meters

=1550D Average mean grain size of the liquefiable layer M=Moment magnitude of the earthquake R=Epicentral distance in kM F15=Average fine content(passing ASTM 200 sieve) for the liquefiable layer in% T15 is the cumulative thickness of the saturated granular layer in meters having blow count


Example 10.4 30meter Density of soil =20kN/m3 Average SPT Value=13 Saturated Density of soil=19.6 kN/m3 6.0m Figure-7:- Soil Profile of a site with typical soil properties. As shown in the above figure 6 is a site soil profile which consists of 3.0meter of clay underlain by 6 meter of sand whose average SPT value is 13 which is susceptible to earthquke. The site consists of a canal flowing across as shown in the above figure. The density of the clay is 20 KN/m3.The saturated density of sand is 19.6kN/m3.Sieve analysis shows the sand to have Fines content as 15%.The average grain size diameter of the sand layer is 0.032 .A power house is to be in built on this site located at distance of 30 meter from the canal bank. The site is considered to be 50 kM away from the epicentre having an earthquake Moment magnitude of 6.75.Find the estimated movement of soil with this free face condition. Solution:- Here R=50 kM M=6.75 W=H/L=3/30=0.1=10% T=6 meter D50=0.32 Considering

153483.06572.00133.09275.01782.13658.16 LogTLogWRLogRMLogDH +++=1515 509224.0)100(5270.4 DFLog + we have

63483.0106572.0500133.0509275.075.61782.13658.16 LogLogLogLogDH +++=

032.09224.0)10100(5270.4 + Log 1751.1295.0846.8271.06572.0665.0576.195285.73658.16 =++++=HLogD

which gives DH=0.068 meters Considering uncertainties this value can vary from half to double thus estimated value is 0.04m to 0.134 m


Effect of earthquake on structures : From above discussion it is obvious that earthquake has a profound influence soil, and since a structure is built on this soil- it do also affects its response. Potential energy stored in earth faults are released due to its rupture and generates kinetic energy in form of stress waves in soil which propagates as P and S waves on the surface of the earth and induces acceleration on structures and foundations built on the surface of the earth. Thus as per Newtons law of motion the structure is subjected to force based on its inertial mass, which it has to resist based on its stiffness and ensure that the stresses and deformations induced in the structure and foundation are within the safe limits. The above in essence is the basic philosophy of earthquake resistant design. The analytical methods adapted for earthquake analysis for different class of structures and foundations may be classified into following category :-

Seismic coefficient method or equivalent static method Response spectrum Method or Psuedo Static analysis Dynamic analysis which is further subdivided into:-

o Modal analysis o Time History analysis.

We as a first step would study in general the basic principles underlying the above methods and finally see their application to different class of structures and foundations like buildings, tall chimneys, elevated water tank, retaining walls earth dams etc. Seismic Coefficient Method:- This is an approach where the earthquake force is treated as an equivalent static force based on the zonal classification of a country10. Though earthquake force in essence is dynamic in nature based on the potential occurrence of earthquake in a particular zone, the soil condition, the type of foundation code recommended a certain percentage of weight of the structure which would it expect to resist as lateral force. It should be noted that this method is now obsolete in terms of latest code IS-1893 2002 and may only be used with caution just to get an idea about the extent of force it may generate in a particular zone for a particular type of structure, and that too only for cases where large number of human life is not endangered- either due to direct or indirect effect of earthquake. Based on the seismic zoning, soil foundation system, importance factor etc we derive a factor

h ,which is given by 0 Ih = where

=A coefficient depending on the soil foundation system as given in Table 9 I= Importance factor as furnished in Table 10

=0 Basic horizontal seismic coefficient as given in Table 8

10 It is presumed the reader has a copy of the earthquake code like IS-1893(1984 and 2002) at hand for cross reference.


Zone Classification

Seismic Coefficient(0)

V 0.08 IV 0.05 III 0.04 II 0.02 I 0.01

Table-8:- Basic seismic coefficient as per IS-1893 1984 Type of soil constituting the foundation

Pile passing through any soil but resting on rock

Piles on any other soil

Raft foundations

Combined or Isolated RCC Foundation with tie beams

Isolated Fdn without tie beams

Well Foundations

Rock or hard soil

1.0 Not applicable

1.0 1.0 1.0 1.0

Medium soil

1.0 1.0 1.0 1.0 1.2 1.2

Soft soil 1.0 1.2 1.0 1.2 1.5 1.5 Table-9:- Soil Foundation factor for various soil foundation system as per IS-1893-1984 Type of structure Importance factor (I) Dams(all types) 3.0 Containers of inflammable or poisonous gases or liquids.

2.0

Important service and community structures such as hospitals, water towers and tanks, schools important bridges, important power houses, monumental structures, emergency buildings like telephone exchange fire bridge , large assembly buildings like structures like cinemas, assembly halls and subway stations.

1.5

All others 1.0 Table-10:- Value of importance factor I as per IS-1893 1984


Based on above having derived the value of h, the base shear acting at the soil foundation level is given by V=KChW for multistoried frames or buildings and V= hW for all other type of structures Where V= Base shear on the structure due to a given earthquake K= A factor known as the performance factor of the frame C=A coefficient defining flexibility of a structure with the increase in number of storeys depending on fundamental time period. The value of performance factor K for different type of framing is as given in table 11 below :- Structural Framing System Value of

performance factor K

Moment resistance frame with appropriate ductility details as given in Is-4326

1.0

Frame as above with RC shear walls or steel bracing members designed for ductility

1.0

Frame as in figure 1a with either steel bracing members or plain or nominally reinforced concrete infill panels

1.3

Frame as in 1a in combination with masonry infill

1.6

Reinforced concrete framed buildings(Not covered by 1 or 2 above)

1.6

Table-11:- Value of performance factor I as per IS-1893 1984 The value of flexibility factor C versus time is as given in Figure 8 as presented below

Flexibility factor C

0

0.2

0.4

0.6

0.8

1

1.2

0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3

Time period in seconds

Flex

ilbili

ty fa

ctor

C

C


Figure-8:- Value of flexibility factor C as per IS-1893 1984 For calculation of time period ,code has furnished some empirical formulas from which it may be found out as follows:-

For moment resisting frame without bracings or shear walls for resisting the lateral loads

nT 1.0= Here n= number of storeys including basement storeys.

For all others

dHT 09.0=

Here H= Total height of the main structure in meters and d= Maximum base dimension of building in meters in direction parallel to the applied seismic force. The above formulations are only valid for buildings which are regular in shape and have regular distribution of mass or stiffness both in horizontal and /or vertical plane. The value of 0 @ 0.08 has been obtained for zone V based on observations of earthquake occurrence in that zone however the values for other has been reduced proportionally, the basis of this reduction has never been very explicit. Though the above method has now been made obsolete in the recent code(IS-1893-2002) but still remains in practice in design offices to estimate preliminarily the magnitude of earthquake force before a more detailed analysis is carried out. Example 10.5 4.0 4.0 6.0 6.0 6.0 6.0 Figure 9:- Plan view of the frame An RCC building having frame layout is as shown in figure 4 above. The transverse cross section of the frame is a shown in figure 3 below. Given the following loading and geometric dimensions of the various


structural members calculate the base on the building as per seismic coefficient method IS-1893(1984) considering zone IV.Consider soil foundation system as of medium stiffness. EL 116.4 EL 112.8 EL 109.2 EL 105.6 EL 102.0 Tie beam all round EL 100.0

4.0 4.0 Figure 10:- Transverse elevation of the frame Loadings:-

Live load on roof = 2kN/m2 Live load on other floors= 4kN/m2 Parapet wall on roof = 1.5 meter all round Internal Partition walls = 1 kN/m2 Floor finish = 1.5 kN/m2 Cement plaster on ceiling = 50 mm

Geometric Properties( Dimensions in mm) :-

Column size = 300X 600 Beam size in transverse direction = 300X 450 Beam size in longitudinal direction = 300X 600 Average thickness of water proofing on roof= 75 mm All external walls 250 mm thick

Material Properties :-

Density of concrete = 25 kN/m3 Density of brick = 20 kN/m3 Density cement plaster = 24 kN/m3 Grade of concrete = M25

Seismic zone properties :-


Seismic zone = Zone IV Soil type Medium stiff Foundation type= Isolated footings with tie beams at 1.0 meter below Ground level

Consider no live load on roof and 50% reduction in live load for other floors during earthquake Solution :- Calculation of roof load(El 116.4)- Assume slab thickness =125 mm Wt of slab = kN 60025824125.0 = Live Load on roof = kN3848240.2 = Parapet wall (1.5 m high) = ( ) kN 48020824225.05.1 =+ (considering 250 mm thk) Water proofing on roof= kN 6.34524824075.0 = Cement plaster on ceiling = kN 2402582405.0 = Wt of long beam = ( ) kN 5.25625324125.06.03.0 = Wt of short beam = ( ) kN 5.972558125.0450.03.0 = Wt of columns = kN 5.12125158.16.03.0 = Total load on roof = 600+384+480+346+240+257+98+122=2527 kN Calculation of load on other floors(El 112.8 109.2 and 105.6) - Wt of slab = kN 60025824125.0 = Live Load on floor = kN7688240.4 = Wt of partition wall = kN1928240.1 = Load from external brick wall = ( ) ( ) kN 1016201625.0275.06.3204825.0475.06.3 =+ Cement plaster on ceiling = kN 2402582405.0 = Flooring on slab = kN 2888245.1 = Wt of long beam = ( ) kN 5.25625324125.06.03.0 = Wt of short beam = ( ) kN 5.972558125.0450.03.0 = Wt of columns = kN 24325156.36.03.0 = Total load on each floor = 600+768+192+1016+240+288+257+98+243=3702 kN


Calculation of load on ground floor (El 102.6)- Load from external brick wall = ( ) ( ) kN 1016201625.0275.06.3204825.0475.06.3 =+ Wt of long beam = ( ) kN 5.25625324125.06.03.0 = Wt of short beam = ( ) kN 5.972558125.0450.03.0 = Wt of columns = kN 18925158.26.03.0 = Total load on ground floor = 1016+257+189+98=1560 kN Total Load to be considered for earthquake :- Load at roof level =2527-384=2143 kN ( Considering no live load on roof during earthquake) Load at EL 112.8 =3692-768+0.5X768=3308 kN( Considering 50% live load on each floor during earthquake). Load at El 109.2= 3308 kN (Same as other floor) Load at El 104.6= 1560 kN Total Weight = 2143+3X3308+1560=13627 kN Calculation of Seismic Coefficient :- As stated in theory above 0Iah = For Seismic zone IV 05.00 = For medium stiff soil with isolated foundations connected by tie beam b=1.0 For normal residential building importance factor I=1.0 Thus 05.005.00.10.1 ==ha Considering nT 1.0= where n= number of storeys we have T=0.5 secs based on which as IS-1893 1984 flexibility factor C=-.075 Considering Moment resistant frame with ductile detailing K=1.0 Thus WKCV hb = Or kN 51101.5111361805.075.00.1 ==bV Thus total base shear acting on building for an earthquake force acting in either transverse or longitudinal direction is =511 KN11. 11 This is strictly not correct for we will see later that time period will vary in both direction based on its stiffness and mass thus earthquake force will also vary accordingly. Moreover the force calculated herein is the total force acting on the building considered as stick model. How the force is distributed in each frame in plan as well as on each floor( vertically) we will see at a later stage.


Response spectrum method:- This method has undergone almost a radical change compared to what is furnished in IS-1893 2002 and that what was furnished in IS-1893 1984. In previous code(1984 version) it was observed that base shear developed based on seismic coefficient method and that by response spectrum method were almost matching or were very close for 5% damping in the system. However with the present version (2002) this force is almost double the previous version. This we believe would significantly enhance the project cost of all projects to come in future. Response Spectrum Method as per 1984 version:- Though the 1984 version has been made obsolete however for historical reason and also for comparison with the present code we present below the steps followed in this method. The 1984 version gave a set of curves representing the values Sa/g versus different time period in seconds for different level of damping. The sets of curves are as shown in figure 11. INSERT I984 CURVE OF IS-1893 Figure11:- Response Curve as per IS-1893 1984 The above curve is actually based on the curves generated by Housner based on his observations and average spectrum obtained using four earthquake time histories. Based on the response spectra curve as furnished in figure 11 for a particular time period of a structure, the corresponding Sa/g is obtained for a particular damping ratio. Based on the zonal demarcation like I, II, III, IV etc. code gives a values of response spectrum factor F012 based on which the coefficient of horizontal seismic force is given by

gSaIFh 0 =

Here and I are as already defined factors in the seismic coefficient method and the factor F0 is as defined in Table 12 below.

12 This is exactly 5 times the value of 0 as given for seismic coefficient method.


Zone Classification Seismic Zone factor (F0)

V 0.40 IV 0.25 III 0.20 II 0.10 I 0.05

Table-12:- Value of Seismic zone factor F0 as per IS-1893 1984 Once the value of h is known the rest of the procedure remains same as that for seismic coefficient method. It may be noted that here that the time period may either be obtained based on formulations as given in code or may be found out based on a detailed dynamic analysis and forces then obtained based on modal response technique13. We now explain the above procedure based on a suitable numerical problem. Example 10.6 For the building cited in example 10.1 find the base shear as per response spectrum technique based on IS-1893-1984. Consider the site to be zone 4 with medium stiff soil. Consider 5% damping ratio for the structure. Referring to figure-9 the time period of the building is given by

nT 1.0= = 0.5 sec. For 0.5 sec and 5 % damping the Sa/g obtained from the curve as shown in Figure 11 is Sa/g =0.16 As stated previously in example 10.1 =1.0 and I=1.0 and F0=0.25 as per Table 12 Thus

gSaIFh 0 =

Or =h 1.0X1.0X0.25X0.16=0.04 As shown in example 10.1 total weight of the structure W=13627 kN For T=0.5 sec C=0.75 K=1.0. Thus considering V=KChW we have V=1.0X0.75X0.04X13627=408.81=409 kN Response Spectrum Method as per IS-1893 2002:- 13 This we are going to study in detail subsequently.


As stated at the outset the method has undergone a drastic modification with respect to the present code. In lieu of the soil foundation factor () considered in the earlier code, the latest version now defines the Sa/g curve for different type of soil starting with rock to soft soil. Sa/g curve for various type of soil as per IS-1893 (2003) is shown in Figure 12 below for 5% damping.

Spectral Response as per IS-1893 2002

0

0.5

1

1.5

2

2.5

30

0.3

0.7

1.0

1.4

1.8

2.1

2.5

2.8

3.2

3.6

3.9

Time Period(secs)

Spec

tral

Acc

eler

atio

nC

oeffi

cien

t(Sa/

g)

Sa/g(Hardsoil/Rock)Sa/g(Mediumsoil)Sa/g(Soft soil)

Figure 12:- Response Spectrum Curve Sa/g as per IS-1893(2002). Moreover as computer analysis has almost become a daily routine work in day to day design office practice-where it is preferable to have digital data of Sa/g for computer input, the code now defines the Sa/g curve by direct formulas enabling one to furnish numerical input for earthquake analysis by computer. The formulas suggested by code for various types of soil as per Clause 6.4.4 of the code for 5% damping ratio are as follows:- Type of soil Value of Sa/g Range Rock or hard soil 1+15T 0.0


Spectral Acceleration Soft soil

0

0.5

1

1.5

2

2.5

3

0

0.3

0.6

0.9

1.2

1.5

1.8

2.1

2.4

2.7 3

3.3

3.6

3.9

Time period(secs)

Spec

tral

acc

eler

atio

n co

effic

ient

s(Sa

/g)

Sa/g(5%)Sa/g(7%)Sa/g(10%)Sa/g(15%)Sa/g(20%)Sa/g(25%)Sa/g(30%)

Figure 13:- Response Spectrum Curve Sa/g for soft soil as per IS-1893(2002). Damping Ratio(%)

0 2 5 7 10 15 20 25 30

Factors 3.2 1.4 1.0 0.9 0.8 0.7 0.6 0.55 0.5 Table- 13:- Multiplying factors for obtaining values for other damping as per IS-1893 (2002) The country unlike previously that was classified into 5 zones( zone I to V) in the present code zone I has completely been deleted and the zones now constitute of zone II to V only. The zone factors to be considered as per the present code is as presented in Table 14 below. Seismic Zone II III IV V Seismic intensity

Low Moderate Severe Very severe

Z 0.1 0.16 0.24 0.36 Table- 14:- Seismic Zone factor as per IS-1893 (2002) The importance factor I has remain unchanged and as such the factors furnished earlier in Table 10 still holds good. To bring it in line with international practice followed by other countries14, the code has now introduced a new factor R which is known as the response reduction factor and also called the ductility factor in many literature. This is the property of a body to dissipate energy by means of its ductile behaviour and may be generated means of special detailing15. The R factor for buildings constituting of different types of frames like Ordinary moment resistant frames or special moment resistant frame etc whether it has shear wall etc has furnished 14 Specially UBC 1997, and NEHRP as followed in USA. 15 We will discuss more about this later.


different values. The value of the response reduction factor R for different types of structural system as defined in IS-1893 2002 is furnished in Table 15 below. Sl No Lateral Load Resistant System R 1 Ordinary moment resistant frame 3.0 2 Special Moment resisting frame specially detailed to provide

ductile behaviour 5.0

3 Steel Frame with:- 3a Concentric Bracing 4.0 3b Eccentric Bracing 5.0 4 Special moment resistant frame with ductile detailing 5.0 Buildings with shear walls 5 Load bearing Masonry wall buildings 5a Un-reinforced 1.5 5b Reinforced with horizontal RC band 2.5 5c Reinforced with horizontal RC band and vertical bars at

corners of rooms and jamb openings 3.0

6 Ordinary reinforced concrete shear walls 3.0 7 Ductile shear walls 4.0 Buildings with Dual systems Ordinary shear wall with OMRF 3.0 Ordinary shear wall with SMRF 4.0 Ductile shear wall with OMRF 4.5 Ductile shear wall with SMRF 5.0 Table- 15:- Response reduction factor R as per IS-1893 (2002) Based on the above data the design horizontal seismic coefficient Ah for a structure is determined by the expression :-

RgZIS

A ah 2= and the base shear is furnished by the expression

WAV h=

The empirical relation furnished by time period has also undergone some modifications. As per the latest code the approximate fundamental time period in seconds for a moment resistant frame without brick infill panels may be estimated by the empirical expression:

75.0075.0 hTa = for RC frame building 75.0085.0 hTa = for steel buildings

where h= height of the building For all other buildings including moment resistant frame buildings with brick infill panels is estimated from the formula


dhT 09.0=

Where h=height of the building and d=base dimension of the building at plinth level in meter along the direction of the lateral force. Based on above we now solve a numerical problem to illustrate how base shear is obtained as per latest IS-1893. Example 10.7 For the building cited in example 10.1 find the base shear as per response spectrum technique based on IS-1893-2002. Consider the site to be zone 4 with medium stiff soil. Consider 5% damping ratio for the structure. Referring to figure-9 the time period of the building is given by

dhT 09.0=

Here h=16.4 meter and d=8m in transverse direction and d=24m in long direction thus

sec 5218.08

4.1609.0 ==T in short direction and

sec 3012.024

4.1609.0 ==T in long direction Thus based on the response spectrum curve Sa/g=2.50 for both short and long direction As per IS-1893 2002 for Zone IV Z=0.24 Considering SMRF with ductile detailing as per Table 9 R=5.0

Thus Rg

ZISA ah 2

=

Or 06.052

5.2124.0 ==hA

As shown in example 10.1 total weight of the structure W=13627 kN Thus considering V=AhW we have V=0.06X13627=817.6 kN. Thus based on the above three examples if we compare the base shear for the given building we have as follows :-


Sl No Code Method Base Shear(kN) Remarks 1 IS-1893-1984 Seismic

Coefficient Method

511

2 IS-1893-1984 Response spectrum Method

409

3 IS-1893-2002 Do 818 Table- 16:- Comparison of Base shear as per IS-1893 (1984) and IS-1893 2002 Dynamic analysis under earthquake loading:- To understand the basic concept we start with system having single degree of freedom and subsequently extend this to system having multi-degree of freedom. Y ut X ug Figure-14:- Single bay portal subjected to Earthquake force As shown in figure 14 a single bay portal subjected to an earthquake force for which the body moves through a distance ug at base and undergoes additional deformation of u at top. We had shown earlier that under time dependent force the equation of motion is given by

0=++ kuucum &&& where m= mass of the system c=damping of the system(usually represented by a dash pot) k= Stiffness of the system =uuu ,, &&& Acceleration, Velocity and displacement vectors. As during the motion the body undergoes a rigid body motion in terms of ug it does not affect the stiffness and damping of the system, which are affected by ut only. Thus the above equation may represented as

0)( =+++ tttg kuucuum &&&


From which we arrive at the expression

gttt umkuucum &&&&& =++ or ettt Fkuucum =++ &&& Where Fe= The earthquake force induced on the system and is equal to the mass of the body times ground acceleration due to earthquake. How do we evaluate the earthquake force ? Before we proceed further to analyse the above equation of equilibrium, it is essential to understand the nature and characteristics of earthquake force and how do we evaluate it. The earthquake force in essence is a transient force and acts on a body for a small instant of time. In terms of Newtonian mechanics this can also be termed as an impulsive force acting on a body. According to the basic law of physics an impulse force is expressed as

= dttFF )( The above expression means a force F which is a function of time is acting upon a body for a very small duration of time dt and is normally defined as an impulse.

As dtdvmF = we can write this as

mdvFdt = . Thus if an impulse force F , is acting on a body ,it will result in a sudden change in its velocity without significant change in its displacement. For spring mass system under free vibration we had seen earlier that the displacement is given by

tBtAx nn cossin += , where A and B are integration constants and their magnitudes depend on the boundary condition. For boundary conditions at t=0, velocity =v0 and displacement x=x0 the above expression can be written as

txtvx nnn

cossin 00 += where

mk

n = Thus for the spring mass initially at rest and acted upon by an impulse force is given by

tm

Fx nn

sin=

When considering damping for the system the free vibration equation is written as

)1sin( 2 += tAex ntn Considering the impulse load the above can modified to


tem

Fx nt

n

n 2

21sin

1

= where is the damping ratio of the system The above is know as Duhamel integral and is effectively used for evaluation of earthquake. While considering earthquake the above expression can be further reduced to the expression

texx nt

n

n 2

21sin

1=

&&

Under earthquake the shock induced on the ground is generally represented by a response spectra or a velocity spectra. Moreover as we are interested in the peak value( or maximum force in the system) the above integral can effectively used to obtain the peak velocity from which maximum displacement and acceleration are obtained subsequently a shown here after. We had seen earlier that equation of motion for the portal structure under earthquake is given by the expression:-

ettt Fkuucum =++ &&& Dividing each tem by m we have

mF

umku

mcu ettt =++ &&&

or gtntnt uuuu &&&&& =++ 22 Since the force is impulsive in nature acting for a duration of time (say) the displacement

tu can be represented by

dteuu n

tt

g

n

tn )(1sin)(

11 2)(

02

= &&

Differentiating the above we have

dtdteuu nnnn

tt

g

n

tn )](1cos1)(1sin[)(

11 222)(

02

+

= &&&

Considering deuC nt

tg

n 2

01 1cos)( = && and

deuC nt

tg

n 2

02 1sin)( = && the velocity can be expressed as

[ ] [ ]{ }tCCtCCeu nntt n 222122212 1cos11sin11 ++= & or )1sin(

122

22

12

+

=

tCCeu n

t

t

n

&


The velocity spectrum or the peak velocity is given by the maximum value of the above

Or max

22

2121

CCeuSt

gv

n +==

& when the maximum displacement is given by

n

vd

SS = and

2n

ad

SS =

where Sa is the acceleration spectrum .

Thus the maximum force the system may experience is given by Fmax= dn Sm

2 It is obvious that that for response spectrum analysis the value Sa is function of the time period or natural frequency of the system which is given by the expression

mk= and

2=T . Certain type of structures can very well be modelled as systems with single degree of freedom and the base force can be found out as follows:- Eample 10.8 As shown in the figure 14 below an air cooler of weight 450 KN is supported on a structure as shown . Determine the force on the system calculating time period based on dynamic analysis. Consider the soil is medium stiff and the site is in zone III .Consider 5% damping for the structure. For beams and columns section properties are as follows Ixx=1268.6cm4 Iyy=568 cm4 and A=78 cm2 , Area of the bracing members = 12 cm2, Esteel=2X106 Kg/cm2.Density of column material=78.5kN/m3 What will be the force on the frame based formulation as given in code? 6500 6000 3000 Figure-14 :- Structure supporting an Air cooler


Solution :- For earthquake force in transverse direction:-

Stiffness of each column is given by 312

LEIK =

Here 454 102686.16.1268 mcmI ==

28 kN/m 102=E L=6.5 meter

Thus ( ) kN/m 86.1105.6102686.110212

3

58

==

K

Considering four columns =

==4

1kN/m 46.44386.1104

iiK

Weight of the air cooler = 450 kN

Thus mass of air cooler = /msec-kN 87.4581.9

450 2=

Mass of each column = /msec-kN 4057.081.9

5.65.78108.7 23 =

Considering 1/3rd weight of column contributing to top mass of 4 column

/msec-kN 5409.03

44057.0 24

1==

=iim

Weight of top beam = ( ) kN 11108.75.78126 3 =+

Mass of beam = 1213.181.9

11 = kN-sec2/m Thus total mass = /msec-kN 532.471283.15409.087.45 2=++

Considering KmT 2= we have

sec. 057.246.443

532.472 == T for which as per IS-1893(2002) Sa/g=0.661

Considering Rg

ZISA ah 2

= here Z=0.16 for zone III, I-1.0 Sa/g=0.661 and R=3.0 we have

0176.032

661.00.116.0 ==hA

Thus kN. 22.881.9532.470176.0 ==hV


For earthquake in longitudinal direction (i.e. in the direction of the braced bay) Stiffness of per column ( considered hinged at base)=

= ( ) KN/m 135.61068.5101.233

3

68

3 ==

LEI

6000 Stiffness of bracing = 2cosL

AE

o99.5775.36tan 1 ==

6500

Stiffness of each bracing= kN/m 1089399.57cos5.6

101.2102.1 283 =

Thus total stiffness of the frame in longitudinal direction = kN/m 43624410893134 =+

Considering KmT 2= we have

sec. 2074.043624

532.472 == T for which as per IS-1893(2002) Sa/g=2.5

Considering Rg

ZISA ah 2

= here Z=0.16 for zone III, I-1.0 Sa/g=2.5 and R=4.0( for concentric bracing) we have

05.042

5.20.116.0 ==hA

Thus kN. 31.2381.9532.4705.0 ==hV in longitudinal direction As per code for steel frame 75.0085.0 hTa = in transeverse direction Or sec346.05.6085.0 75.0 ==aT for which the value Sa/g=2.5 Thus Considering

RgZISA ah 2

= here Z=0.16 for zone III, I-1.0 Sa/g=2.5 and R=3.0 we have

066.032

5.20.116.0 ==hA

Thus maximum force on the frame =31.08 kN which is 3.78 times the force obtained by dynamic analysis. Earthquake Analysis of systems with Multi-degree of freedom:- Before we delve into the detailed dynamic analysis of systems with multi-degree of freedom under earthquake force( based on modal analysis or time history response), we deal with a


particular technique often used in practical engineering design where for many buildings effect of fundamental time period is most pre-dominant. In such cases higher mode participation vis--vis its effect being insignificant are ignored without causing any significant errors. Analysis based on assumed shape function :- This is a technique in which a multi-degree freedom system is converted into an equivalent system having mass and stiffness of that of a single degree of freedom based on an assumed shape function to find out the time period of a system. To start with let us consider a stick model of a system having multi-degree of freedom as shown hereafter. Mn Kn M3 Displaced Shape(1st Mode) K3 M2 K2 M1 K1 Figure-16 A stick model having multi-degree of freedom The kinetic energy of the system is given by

( ) 21

),(21

= = t

tzymtTn

ii (5)

We consider here )()(),( tztzy = where

=)(z Admissible shape function which satisfies the boundary condition of system =)(t Generalized co-ordinate

( )

= ===

n

kkkn

n

jj

n

ii tztzmtT

111)()()()(

21 &&

or

= == =

n

ikji

n

j

n

kkj zzmtttT

11 1)()()()(

21)( && from which we conclude that generalized

mass of the system is given by,

= =

n

ikji zzmM

1)()(*

Thus for fundamental mode for j=k we have


= =

n

iii zmM

1

2 )(* Similarly potential energy is given by

( ) [ ]21

),(21 tzyktV

n

ii =

=

Here = Difference in displacement between two adjacent level or, ( )

=

===

n

kkkn

n

jj

n

ii tztzktV

111)()()()(

21

or,

= == =

n

ikji

n

j

n

kkj zzktttV

11 1)()()()(

21)(

Thus for fundamental mode for j=k we have

= =

n

iii zkK

1

2 )(*

Now knowing KmT 2= we have for this generalized case

*

*

2*KMT =

From the above mathematical derivation it is obvious that if we know what could be the assumed shape function correctly it is possible to arrive at the fundamental time period of the system correctly. Based on the aspect ratio(H/D) Naeem16 has proposed the following shape functions which may be considered for buildings modeled as stick having multi-degrees of freedom. Here H=Height of the building D= Width of building in direction of the earthquake force considered. Sl No H/D Shape function

1 5.1/


We will now solve the previous building problem( vide example 10.5) to see how base shear results differ with what we have calculated earlier Example10.9:- Refer the problem as shown in example10.5 calculate the time period of the building based on assumed shape function method and calculate the base shear in both transverse and longitudinal direction and find out the base shear based on IS-1893-2002.Consider all other boundary conditions remains same as was defined in the previous problem EL 116.4 EL 112.8

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