Seismic Analysis of a Multi Storey Plane Frame using Static & Dynamic Methods

Seismic Analysis of a Multi Storey Plane Frame using Static & Dynamic Methods

Comparison of Equivalent Static Force & Response Spectrum on Frame with & without Infills

Siva Kiran Kollimarla Department of Civil Engineering R.V.R&J.C college of engineering Guntur-A.P, INDIA

Chelli Jagadeesh babu R.V.R&J.C College of Engineering Guntur-A.P, INDIA

Chadalawada Jagan Mohan Department of Civil Engineering R.V.R&J.C college of engineering Guntur-A.P, INDIA

M Venkateswara Rao

1. V. R & J. C college of engineering

   Guntur-A.P, INDIA

   AbstractIn the context of seismic analysis and design of structures, in earthquake engineering, a verity of methods are available. Standard codes provide provision for certain methods for the analysis of wide range of structures of engineering interest. In this paper an attempt is made to present the provisions of IS 1893:2002, part-1 for the analysis of structures and its suitability.

   Keywords Lumped mass; infills; CQC; seismic analysis;

   1. INTRODUCTION

      In the context of seismic analysis and design of structures, a variety of methods are available for seismic analysis of structures. In this paper an attempt has made to present static and dynamic methods of seismic analysis as per Indian standard code IS 1893:2002 (Part 1). The basic concepts and codal provisions of Equivalent static force method and Response spectrum method as per IS 1893:2002 (Part 1) are explained in detail.

      1. Equivalent static analysis (ESA)

         This approach defines a series of forces acting on a building to represent the effect of earthquake ground motion, typically defined by a seismic design response spectrum. It assumes that the building responds in its fundamental mode. For this to be true, the building must be low rise and must not twist significantly when the ground moves. The response is read from design response spectrum, given the natural frequency of the building. The applicability of this method is extended in many building codes by applying factors to account for higher buildings with some higher modes and for low levels of twisting. To account for effects due to yielding of the structure, many codes apply modification factors that reduce the design forces.de-composing as a means of reducing degrees of freedom in the analysis.

      2. Response spectrum analysis (RSM)

         This approach permits the multiple modes of response of a building to be taken in to account (in the

         frequency domain). This is required in many building codes for all except for very simple or very complex structures. The response of a structure can be defined as a combination of many special shapes (modes) that in a vibrating string correspond to the harmonics. Computer analysis can be used to determine these modes for a structure. For each mode, a response is read from the design spectrum, based on modal mass, and they are then combined to provide an estimate of the total response of the structure. In this we have to calculate the magnitude of forces in all directions i.e. X, Y & Z and then see the effects on the building. Combination methods include the following.

         1. Absolute values are added together (AVS)

         2. Square root of sum of the squares (SRSS)

         3. Complete Quadratic Combinations (CQC).

      A method that is an improvement on SRSS for closely shaped modes.

      The results of a response spectrum analysis using the response spectrum from a ground motion is typically different from that which would be calculated from linear dynamic analysis using that ground motion directly, since phase information is lost in the process of generating the response spectrum.

      In cases where structures are either too irregular, too tall are of significance to a community in disaster response, the response spectrum approach is no longer appropriate and more complex analysis is often required such as non-linear static analysis or dynamic analysis.

   2. STEP BY STEP PROCEDURE FOR STATIC AND DYNAMIC METHODS

      1. Equivalent static force method

         1. Calculation of Lumped masses to various floor levels The earthquake forces shall be calculated for the full dead load plus the percentage of imposed load as given in Table-8 of IS

            1893 (Part 1):2002. The imposed load on roof is assumed to be zero. The lumped mass of each floor are worked out as follows:

            1. Mass of each floor is calculated as

               Mass of columns +Mass of beams in longitudinal and transverse direction of that floor + Mass of slab+ Imposed load of that floor if permissible.

            2. Roof

               Mass of infill + Mass of columns + Mass of beams in longitudinal and transverse direction of that floor + Mass of slab + Imposed load of that floor if permissible

               5) Calculation of moment due to design lateral forces

               n

               n

               M Fihi i1

      2. Response Spectrum Method

         1. Determination of Eigen values and Eigen vectors

            The mass and stiffness matrices at each floor level are calculated in S.I units and then modeled in the form of matrix as

            • 50% of imposed load, if imposed load is greater than 3 kN/m2

              M1 0

              0 M

              M 2

              0 0

              0

              0

              0

              0 0

              0

              0

              0

              M3 0

              M

              M

              0 4

              1. Seismic weight of building

      The seismic weight of each floor is its full dead load plus appropriate amount of imposed load, as specified in Clause

      k1 k 2

      k

      K 2

      • k 2

        k 2 k 3

        0 0

        3

        3

        k 0

        7.3.1 and 7.3.2 of IS 1893 (Part 1):2002. Any weight supported in between stories shall be distributed to the floors above and below in inverse proportion to its distance from the floors.

        0 k 3 k 3 k 4

        0

        0

        0

        0

        4

        4

        k

        12EI

        k 4

        k

        k

        4

        1. Determination of Fundamental Natural Period

           Column stiffness of storey,

           k c

           L3

           The approximate fundamental natural period of a vibration (Ta), in seconds, of a moment resisting frame building without brick infill panels may be estimated by the empirical expression (Clause 7.6.1)

           The stiffness of infill is determined by modeling the infill as an equivalent diagonal strut

           1

           E I h 4

           h

           f c

           Ta = 0.075xh0.75

           For RC frame building

           2 2E m tsin2

           1

           Ta = 0.085xh0.75 For Steel frame building

           E I l 4

           l f b

           Emtsin2

           Where h is the height of the building, in meters.

           If the structure considered with infills (Clause 7.6.2)

           Ta = 0.09H/D

           Width of the strut Where

           W= 1 2 2

           2 h l

        2. Determination of Design Base Shear

           Design seismic base shear, (Clause 7.5.3)

           VB = AhW

        3. Vertical Distribution of Base Shear

           The design base shear (VB) computed shall be distributed along the height of the building as per the expression, (Clause 7.7.1)

           Wihi 2

           Ef = Youngs modulus of the concrete

           Em = Youngs modulus of the masonry infill h = Height of infill wall

           l = length of wall

           t = Thickness of wall Ic=Moment of inertia of columns Ib =Moment of inertia of beam

           A=cross sectional area of diagonal stiffness

           <>Qi VB

           n

           n

           Wih i 2 i1

           ld =Diagonal length of strut = p + l2

           Therefore, stiffness of infill is

           Where

           Qi = Design lateral forces at floor i Wi= Seismic weights of the floor i

           hi = Height of the floor i measured from base and n = Number of stories

           k d

           AEm

           ld

           cos 2

           2

           2

           Taking k/m = n

           The design horizontal seismic coefficient Ah for various modes are,

           A Z I Sa1

           2K 2m

           k2 0 0

           p 2 R g

           K 2M

      • k2

      2K 2m

      k3 0

      0 k3

      2K 2m

      k4

      For rocky, or hard soil sites

      0

      0 k4

      k 0.575 2m

      1 15T;

      S a 2.5;

      g

      0.00

      0.10

      T 0.10

      T 0.40

      The solution for the above equation is given as

      1.00/T;

      0.40

      T 4.0

      Eigen values

      2

      For medium soil sites

      2

      2

      1 0 0 0

      2

      2

      2 0 2 0 0

      1 15T;

      S

      0.00 T 0.10

      4

      4

      0 0 3 0

      a 2.5;

      0.10 T 0.55

      Eigen vectors

      0 0

      11

      0

      12

      2

      13

      14

      g

      For soft soil sites

      1.36/T;

      0.55 T 4.0

      1 15T;

      S

      0.00 T 0.10

      1 2 3 4 21 22

      23 24 .

      a 2.5;

      0.10 T 0.67

      31

      32

      33

      34

      g 1.67/T;

      0.67 T 4.0

      41

      42

      43

      44

      1. Determination of storey shear in each mode

         Natural frequency in various modes

         The peak shear force will be obtained by

         n

         1 0 0 0

         0

         0

         0

         0

         2 0

         Vik

         Qik ji1

         Natural time period

         0 0

         0

         0

         0

         T1 0

         0

         0

         T T2

         0 0

         0

         0

         0

         3 0

         0 4

         0 0

         0

         0

         0

         T3 0

         T

         T

         0 4

      2. Determination of storey shear force due to all modes

      The peak storey shear force (Vi) in storey i due to all modes considering is obtained by combining those due to each mode in accordance with modal combination SRSS (Square Root of Sum of Squares) or CQC (Complete Quadratic Combination) methods.

      1. Maximum Absolute Value Sum (AVS)

         2) Determination of Modal Participation factor

         4

         4

         Wiik

         4

         4

         P i1

         n

         n

         Vi vik

         k1

         3.15

         2

         2

         Wi (ik )

         i1

         1. Determination of model mass

            4 2

      2. Square Root of Sum of Squares (SRSS)

         If the building does not have closely spaced modes, the peak response quantity () due to all modes considered shall be obtained as,

         r

         Wii1

         ( )2

         k

         k

         i

         i

         M i1

         k

         k

         k1

         g4 W ( )2

         Where

         i

         i1

         i1

         k = Absolute value of quantity in mode k, and r is the

         1. Determination of lateral force on each floor

         The design lateral force (Qik) at floor i in mode k is given by,

         Qik (Ak Pk ik Wk)

         numbers of modes being considered.

      3. Complete Quadratic Combination (CQC)

      r r 8 2 1 ij 1.5

      Seismic weight of building

      = Seismic weight of all floors = M13+.2M52+M3+M4

      Where

      iij j i1 j1

      ij =

      1 ij

      2 4 2ij 1 ij2

      = 64.46+64.46+64.46+37.09 = 2303.4.276

      1. Determination of Fundamental Natural Period

         r = Number of modes being considered

         ij = Cross modal coefficient

         i = Response quantity in mode i (including sign)

         j = Response quantity in mode j (including sign)

         = Modal damping ration (in fraction)

         The approximate fundamental natural period of a vibration (Ta), in seconds, of a moment resisting frame building without brick infill panels may be estimated by the empirical expression

         T = 0.075xh0.75

         = Frequency ratio / a

         ij j i

         i = Circular frequency in ith mode, and j = Circular frequency in jth mode.

         7) Determination of lateral forces at each storey

         The design lateral forces Froof and Fi, at roof and at ith floor, are calculated as,

         = 0.075×140.75 = 0.5423 s

         Where h is the height of the building, in meters.

      2. Determination of Design Base Shear

         Design seismic base shear, VB = AhW

         Froof = Vroof, and Fi = Vi Vi+1

         Z I S

         0.24 1

         Ah = a

         = 1.842 0.04343.27

         H: Calculation of moment due to design lateral forces

         2 R g 2 5

         n

         n

         M Fihi

         For Ta

         Sa 1

         = 0.5423 => = = 1.842, for rock site from Fig 2

         i1

         g Ta

   3. ANALYSIS OF MULTI STOREY PLANE FRAME

1. Equivalent static force method

   1. Frame Without infills

      1. Calculation of Lumped masses to various floor

         levels

         of IS 1893 (part 1): 2002

         Design seismic base shear, VB = 0.0443 x (230.43×9.81) = 99.933 kN

         1. Vertical Distribution of Base Shear

            Qi VB

            Qi VB

            W h

            W h

            2

            1 1

            W1h 2 W2p2 W3p2 W4h42

            Roof

            = Mass of infill + Mass of columns + Mass of beams in longitudinal and transverse direction of that floor + Mass of

            =99.93

            1

            632.25×3.52

            2 2 2 2

            slab + Imposed load of that floor if permissible

            ={((0.25x10x(3.5/2)+0.15x15x(3.5/2))20}+{(0.25x10x0.40+0. 25x15x0.35)25}+{0.10x5x10x25}+{(0.25×0.45x(0.35/2)x3)x

            25}

            632.25×3.5

            =4.307kN

            Similarly,

            632.25×7

            632.25×10.5

            363×14

            = 363.82 kN (weight) = 37.087 ton (mass) 3rd, 2nd, 1st Floors

            ={((0.25x10x3.5)+(0.15x15x3.5))20}+{(0.25x10x0.40+0.25x 15×0.35)25}+{0.10x5x10x25}+

            {0.25×0.45×3.5x3x25}+{5x10x3.5×0.5}

            = 632.43 kN (weight) = 64.45 ton (mass)

            • 50% of imposed load, if imposed load is greater than 3 kN/m2

            Q2 = 0.1724 x 99.933 = 17.23 kN Q3 = 0.3872 x 99.933 = 38.768kN Q4 = 0.3967 x 99.933 = 39.654kN

            Maximum overturning moment=1097.9kN-m

            1. Vertical Distribution of Base Shear

               Qi VB

               W1h 2

               1

               1

               W1h 2 W2p2 W3p2 W4h42

               1

               =135.6

               632.25×3.52

               632.25×3.52 632.25×72 632.25×10.52 363×142

               Fig.1. Distribution of lateral forces and shear by Equivalent static force method on frame without infill

   2. Frame With infills

      1. Calculation of Lumped masses to various floor levels

         Roof

         Mass of infill + Mass of columns + Mass of beams in longitudinal and transverse direction of that floor + Mass of slab + Imposed load of that floor if permissible

         = 363.82 kN (weight) = 37.087 ton (mass) 3rd, 2nd, 1st Floors

         ={((0.25x10x3.5)+(0.15x15x3.5))20}+{(0.25x10x0.40+0.25x 15×0.35)25}+{0.10x5x10x25}+

         {0.25×0.45×3.5x3x25}+{5x10x3.5×0.5}

         = 632.43 kN (weight) = 64.45 ton (mass)

         • 50% of imposed load, if imposed load is greater than 3 kN/m2

           Seismic weight of building

           = Seismic weight of all loors = M1+M2+M3+M4

           = 64.46+64.46+64.46+37.09 = 230.47

      2. Determination of Fundamental Natural Period

         Ta = 0.09H/DD

         = 0.09×14/1010= 0.3984 sec

      3. Determination of Design Bash Shear

   = 5.8kN

   Similarly,

   Q2 = 0.1724 x 135.63 = 23.215kN Q3 = 0.3872 x 135.63 = 52.234kN Q4 = 0.3967 x 135.63 = 53.429kN

   Maximum overturning moment=1479.3kN-m

   Fig.2. Distributions of lateral forces and shear by Equivalent static force method on frame with in fills

2. Response Spectrum Method

1. Frame without stiffness of infills

   1. Determination of Eigen values and Eigen vectors

      The mass and stiffness matrices at each floor level are calculated in S.I units and then modeled in the form of matrix as

      Ah =

      Z I Sa

      = 0.24 1 x2.5 0.06

      M1 0

      0 0

      2 R g 2 5

      Sa 1

      0

      0

      M

      0

      M2 0

      0

      0

      0 M3 0

      For Ta = 0.3984 => g

      = = 2.5, for rock site from Figure 2

      Ta

      0

      0

      M

      M

      0

      0

      0

      0

      4

      of IS 1893 (part 1): 2002

      64.46 0 0 0

      0

      0

      0

      0

      0

      0

      Design seismic base shear, VB = 0.06 x (230.43×9.81) = 134.68kN

      M 103 0

      0

      64.46

      0

      0

      64.46

      0 kg

      0

      37.09

      k1 k 2

      k2

      K

      • k 2

      k 2 k3

      0 0

      k3 0

   2. Modal participation factor

      4

      14.40

      0 k3

      k3 k4

      k 4

      Wiik

      0 0 k k

      i1

      (W11k W22k ….. W44k) 4.30

      4 4

      P

      4 W1(1k)2 W2(2k)2 ….. W4(4k)2 1.95

      2

      2

      Column stiffness of storey,

      Wi(ik)

      i1

      0.68

      3 0.25×0.453

      12EI

      12x22360x10

      12

   3. Determination of model mass

      k L3

      11880.78 kN/m 3.53

      [9.81(64.45(0.0328) … 37.08(0.0872))]2

      Total lateral stiffness of each storey,

      k1= k2= k3= k4= 3×11880.78 = 35642.36 kN/m

      M1

      9.81[9.81(64.45(0.0328)2 … 37.08(0.0872)2 )]

      207.60

      Similarly, M2 = 18.54, M3 = 3.82, M4 = 0.47

      71284.72 3564.36 0 0

      3564.36 71284.72 3564.36

      K

      0

      0

      kN/m

      Modal contributions of various modes

      0

      0

      0

      0

      0 3564.36 71284.72 35642.36

      M1 207.60

      35642.36 35642.36

      For mode 1,

      For mode 2,

      M

      M 2

      230.43

      18.54

      0.90 90%

      0.0804 8.04%

      Eigen values

      81	0	0	0
      0	657	0	0
      0	0	1475	0
      0	0	0	206

      81	0	0	0
      0	657	0	0
      0	0	1475	0
      0	0	0	206

      2

      For mode 3,

      M

      M 3

      M

      M

      230.43

      3.82

      230.43

      0.47

      0.0165 1.65%

      5

      For mode 4, 4

      M

      230.43

      0.0020 0.20%

      Eigen vectors

      0.0328

      0.0795

      0.0808

      0.0397

      1. Determination of lateral force on each floor

      The design lateral force (Qik) at floor i in mode k is given by,

      0.0608 0.0644 0.0540 0.0690

      Q A P W

      1 2 3 4 .

      ik k ik k i

      0.0798 0.0273 0.0448 0.0799

      0.0872 0.0865 0.0839 0.0696

      The design horizontal seismic coefficient Ah for various modes are,

      Natural frequency in various modes

      A Z I Sa1 0.24 11.433 0.0343

      9 0 0 0

      p 2 R g 2 5

      0 25.63 0 0

      Z I S

      0.24 1

      A p

      a2

      2.5 0.060

      0 0

      38.41 0

      2 R g 2 5

      0

      0

      0

      Natural time period

      0 45.44

      Similarly,

      Ap 0.060, Ah4 0.060.

      0.6977 0 0 0

      T 0 0

      0.2450

      0

      0 0

      0.1636 0

      For rocky, or hard soil sites

      0

      0

      0

      0

      0

      0

      0.1383

      1 15T;

      S

      0.00 T 0.10

      a 2.5;

      0.10 T 0.40

      g 1.00/T;

      0.40 T 4.0

      For

      T 0.6978 Sa1 1.433

      1. Determination of storey shear force due to all modes

         1 g

         S Maximum Absolute Value Sum (AVS)

         For

         T2 0.2450 a2 2.5

         g

         V1 70.056 36.51416.572 5.824

         128.986

         For

         T 0.1636 Sa3 2.5

         n V2 59.781 6.927 27.63215.925

         110.282

         3 g Vi vik

         k1

         V3 40.738 42.1271.867 21.796

         106.544

         S

         For

         T4 0.1382 a4 2.5

         g

         V415.720 27.207 26.385 21.878

         91.207

         Design lateral force

         Square Root of Sum of Squares (SRSS)

         V1 = [(V11)2 + (V12)2 + (V13)2 + (V14)2]1/2

         (Ap

         P1 11

         W1 )

         = [(70.056)2 + (-36.514)2 + (-16.572)2 + (5.824)2]1/2

         Qi1

         Qi1

         (Ap P1 21 W2 )

         (Ap P1 31 W3 )

         = 80.930kN

         Similarly,

         V2 = 68.110kN

         (A

         (A

         h4

         P1 41

         )

         )

         W4

         V3 = 62.553kN V4 = 46.507kN

         ((0.0343)(14.40)(0.0328)(64.45 9.81))

         ((0.0343)(14.40)(0.0608)(64.45 9.81))

         (10.275)

         (19.043)

         Complete Quadratic Combination (CQC)

         kN

         V1 =

         ((0.0343)(14.40)(0.0798)(64.45 9.81)) (25.018)

         ((0.0343)(14.40)(0.0872)(37.08 9.81)) (15.720)

         1 0.0073

         0.0031

         0.0023 70.056

         0.0073 1

         0.0559

         0.0278 36.51

         [70.056 – 36.514 – 16.572 + 5.824]

         0.0031

         0.0559

         1 0.2597 16.57

         Similarly,

         43.44

         35.199

         44.204

         29.499

         21.749

         37.722

         V1 = [80.70] V2 =

         0.0023

         0.0278

         0.2597

         1 5.82

         Qi2

         , Qi3

         , Qi2

         14.920

         24.517

         43.675

         1 0.0073

         0.0031

         0.0023 59.78

         27.207

         26.385

         21.878

         0.0073 1

         0.0559

         0.0278 6.92

         [59.781+ 6.927 27.632 -15.925]

         0.0031

         0.0559 1

         0.2597 27.63

         1. Determination of storey shear in each mode

            0.0023

            0.0278

            0.2597

            1 15.92

            The storey shear forces for the first mode is,

            V2 = [66.61]

            (Q11 Q21 Q31 Q41)

            70.056

            V3 =

            n (Q21 Q31 Q41)

            59.925

            Vi1 Qi1

            (Q Q )

            kN

            .7

            1 0.0073

            0.0031

            0.003 40.738

            ji1

            31 41

            40

            38

            (Q )

            .7

            [40.738 + 42.127 -1.867 + 21.796]0.0073 1

            0.0559

            0.027842.127

            41

            15

            20

            0.0031

            0.0559 1

            0.25971.867

            V12

            36.514

            V13

            16.572

            0.0023

            0.0278

            0.2597

            1 21.796

            V22

            6.927

            V

            27.632

            V3 = [62.95]

            Vi2

            , Vi3

            23 ,

            V32 42.127 V33 1.867

            V4 =

            V42

            27.207

            V43

            26.385

            1 0.0073

            0.0031

            0.0023 15.720

            0.0073 1

            0.0559

            0.0278 27.207

            V

            5.824

            [15.720 27.207 – 26.385 – 21.878]

            14 0.0031

            0.0559 1

            0.2597 26.385

            V V24 15.925

            0.0023

            0.0278

            0.2597

            1 21.878

            i4 V34

            21.796

            V = [48.48]

            4

            V44

            21.878

            1. Determination of Lateral Forces at Each Storey

               Maximum Absolute Value Sum (AVS) Froof = V4 = 91.207 kN

               Ffloor 3 = F3 = V3 V4 = 106.544 91.207 = 15.337 kN

               Ffloor 2 = F2 = V2 V3 =110.2815 106.544 = 3.738 kN Ffloor 1 = F1 = V1 V2 = 128.986 110.282 = 18.705 kN

               Square Root of Sum of Squares (SRSS) Froof = F4 = V4 = 46.508kN

               Ffloor 3 = F3 = V3 V4 = 62.562 46.508 = 16.055kN

               Ffloor 2 = F2 = V2 V3 = 68.119 62.562 = 5.56kN Ffloor 1 = F1 = V1 V2 = 80.941 68.119 = 12.823kN

               Complete Quadratic Combination (CQC) Froof = F4 = V4 = 48.492kN

               Ffloor 3 = F3 = V3 V4 = 62.957 48.492 = 14.465kN Ffloor 2 = F2 = V2 V3 = 66.625 62.957 = 3.67kN Ffloor 1 = F1 = V1 V2 = 80.714 66.625 = 14.089kN

            2. Calculation of moment due to design lateral forces

      Absolute Value Sum (AVS) Overturning moment = 1529.6 kN-m

      Square Root of Sum of Squares (SRSS) Overturning moment = 903.456 kN-m

      Complete Quadratic Combination (CQC) Overturning moment = 905.758 kN-m

      Fig.4. Distributions of lateral forces and shear by SRSS method on frame without infills

      Fig.5. Distribution of lateral forces and shear by CQC method on frame without infills

2. Frame with the stiffness of Infills

   1. Determination of Eigen values and Eigen vectors

M1 0

0

0

M M 2

0 0

0

0

0

0 0

0 0

M 3 0

0 M

0 M

4

64.46 0 0 0

Fig.3. Distribution of lateral forces and shear by Absolute value sum method on frame without infills

M 103 0

64.46 0

0 kg

Column stiffness of storey, Eigen values

12EI

12x22360x103 0.001893

1443 0 0 0

k

L3

3.53

11846.758 kN/m

2 0

0

11707

0

0 0

26247 0

Ef = Youngs modulus of the concrete = 22360 N/m2

Em = Youngs modulus of the masonry infill = 13800 N/m2 h = Height of infill wall = 3.5 m

Eigen vectors

0

0

0

0

0.0010

0

0.0025

36746

0.0025

0.0013

l = length of wall = 5 m

t = Thickness of wall = 250 mm

1 2 3 4

0.0019

0.0025

0.0028

0.0020

0.0009

0.0027

0.0017

0.0014

0.0027

0.0022

.

0.0025 0.0022

c

c

I =Moment of inertia of columns = 1 x (0.25 x 0.453)

12

Natural frequency in various modes

= 0.001893 m4

I =Moment of inertia of beam = 1 x (0.25 x 0.403)

37.99

b 12

= 0.001333 m4

W = 1x J2 + 2 = 0.7885

0

0

9

9

0 3

2 h 1

A=cross sectional area of diagonal stiffness

= W*t = 0.7885 x 0.25 = 0.1972 m2

Natural time period

0.1654 0 0 0

0

0

0

0

0

0

0

0

h

22360 x 0.001893 x 3.5

1/4

0.611m

0.03

28

2 2 x 13800 x 0.25 x sin 2 x 35

22360 x 0.001333 x 5.0 1/4

1. Determination of model participation factor

   4

   l 1.45 m

   Wiik

   13800 x 0.25 x sin 2 x 35

   Therefore, stiffness of infill is

   P

   i1

   4

   4

   Wi (ik ) 2 i1

   14.40

   AEm cos 2

   0.1972 x 13800 x 106

   0.8192

   (W11k W2 2k ….. W4 4k )

   4.30

   ld 6.103

   W1 (1k ) 2 W2 ( 2k ) 2 ….. W4

   ( 4k

   ) 2

   1.95

   299086.078 N/m

2. Determination of model mass

   4 2

   0.68

   For the frame with two bays there are two struts participating

   Wii1

   in one direction, total lateral stiffness of each storey

   M1

   i1

   4

   k = k = k = k = 3×11846.758 + 2×299086078

   gWi (i1 ) 2

   1 2 3 4

   i1

   = 633712.430 kN/m

   1.2686

   9

   9

   0.6343

   K 10

   0.6343

   1.2686

   0

   0.6343

   0

   0 N/m

   [9.81(64.45(0.0328) … 37.08(0.0872))]2

   M1

   9.81[9.81(64.45(0.0328) 2 … 37.08(0.0872) 2 )]

   0 0.6343

   0

   0

   0

   0

   1.2686

   0.6343

   0.6343 0.6343

   Similarly, M2 = 18.54, M3 = 3.82, M4 = 0.47.

   Modal contributions of various modes

   For mode 1,

   M1 207.60 0.90 90%

   M 230.43

   For mode 2,

   M 2

   M

   M3

   18.54

   230.43

   <>3.82

   0.0804 8.04%

   Similarly,

   32.512

   26.343

   27.972

   18.667

   12.980

   22.512

   For mode 3,

   0.0165 1.65%

   M 230.43

   Qi2

   , Qi3

   , Qi2

   11.166

   20.361

   15.514

   16.696

   26.065

   13.057

   For mode 4,

   M 4 0.47 0.0020 0.20%

   M 230.43

   1. Determination of storey shear in each mode

3. Determination of lateral force on each floor

The design lateral force (Qik) at floor i in mode k is given by

The peak shear force

n

n

Vik Qik

Q A P W

ji1

ik k ik k i

The design horizontal seismic coefficients Ah for various modes are,

The storey shear forces for the first mode is,

A Z I Sa1 0.24 1 2.5 0.060

(Q11 Q21 Q31 Q41) 70.056

p 2 R g 2 5

n (Q Q Q ) 59.925

Vi1 Qi1

21 31

41

kN

A Z I Sa2 0.24 1 1.871 0.045

ji1

(Q31 Q41)

40.738

p 2 R g 2 5

(Q )

0

Similarly,

Ap 0.037, Ah4 0.035

Similarly,

41

15.72

For rocky, or hard soil sites

V12

27.327

V13

10.487

V22

5.184

V23

17.485

Vi2

, Vi3

,

1 15T;

0.00 T 0.10

V32

31.528

V33

1.182

Sa

V

20.361

V

16.696

2.5;

0.10 T 0.40

42

43

g 1.00/T;

0.40 T 4.0

V14

3.476

V

9.5042

Vi2

24

For

T 0.1655 Sa1 2.5

V34

13.008

1 g V

13.057

For

T 0.0581 Sa2 1 15T 1.871

44

For For

2 g

T 0.0388 Sa3 1 15T 1.582

3 g

4

4

T 0.0382 Sa4 1 15T 1.492

1. Determination of storey shear force due to all modes

   Maximum Absolute Value Sum (AVS)

   V1 122.194 27.327 10.487 3.476

   g n

   Design lateral force

   Vi

   vik

   V2 104.272 5.184 17.485 9.5042

   V 71.057 31.528 1.182 13.008

   k1

   3

   (Ap P1 11 W1 )

   (Ap P1 21 W2 )

   V4 27.419 20.36116.696 13.057

   163.508

   Qi1

   136.463

   (Ap P1 31 W3 )

   kN

   (A P W )

   116.791

   h4 1 41 4

   ((0.060)(14.40)(0.0328)(64.45 9.81))

   ((0.060)(14.40)(0.0608)(64.45 9.81))

   (17.922)

   (33.215)

   77.544

   Square Root of Sum of Squares (SRSS)

   kN

   ((0.060)(14.40)(0.0798)(64.45 9.81)) (43.637) r

   ((0.060)(14.40)(0.0872)(64.45

   (k )2

   9.81)) (27.419)

   k1

   V1 = [(V11)2 + (V12)2 + (V13)2 + (V14)2]1/2

   = [(122.194)2+(-27.327)2+(-10.487)2+(3.475)2]1/2

   = 125.699 kN

   Similarly,

   V2 = 106.281kN V3 = 78.827kN V4 = 40.196kN

   Complete Quadratic Combination (CQC)

   Square Root of Sum of Squares (SRSS)

   Froof = F4 = V4 = 40.202kN

   Ffloor 3 = F3 = V3 V4 = 78.840 40.202 = 38.638kN Ffloor 2 = F2 = V2 V3 = 106.297 78.840 = 27.457kN Ffloor 1 = F1 = V1 V2 = 125.716 106.297 = 19.422kN

   Complete Quadratic Combination (CQC) Froof = F4 = V4 = 44.990 – 0 = 44.990kN

   r r Ffloor 3 = F3 = V3 V4 = 79.136 44.990 = 38.146kN

   iij j i1 j1

   Ffloor 2 = F2 = V2 V3 = 105.995 79.136 = 26.859kN Ffloor 1 = F1 = V1 V2 = 125.533 105.995 = 19.538kN

   1 0.0073

   0.0031

   0.0023 122.272

   0.0073 1

   0.0559

   0.0278 27.327

   h) Calculation of moment due to design lateral forces

   V1 =

   [122.194 27.327 10.487 3.475]

   0.0031 0.0559 1 0.2597 10.487 n

   i i

   V1 = [125.512]

   0.0023 0.0278 0.2597 1 3.475

   M

   Absolute Value Sum (AVS)

   F h

   i1

   1 0.0073

   0.0073 1

   0.0031

   0.0559

   0.0023104.272

   0.0278 5.184

   Overturning moment=1730.1kN-m

   Square Root of Sum of Squares (SRSS)

   [104.272 5.184 17.485 9.504]

   V = 0.0031

   0.0559 1

   0.2597 17.485

   Overturning moment=1228.7kN-m

   2 0.0023

   0.0278 0.2597

   1 9.504

   V2 = [105.977]

   Complete Quadratic Combination (CQC) Overturning moment=1230.8kN-m

   1 0.0073

   0.0031

   0.0023 71.057

   0.0073 1

   0.0559

   0.0278 31.528

   [71.057 31.528 1.182 13.008]

   V3 =

   0.0031

   0.0559 1

   0.2597 1.182

   0.0023 0.0278 0.2597 1 13.008

   V3 = [79.125]

   1 0.0073

   0.0031

   0.0023 27.419

   0.0073 1

   0.0559

   0.0278 20.362

   [27.419 20.362 16.696 13.057]

   V4 =

   0.0031

   0.0559 1

   0.259716.696

   0.0023

   0.0278 0.2597

   1 13.057

   V4 = [40.964]

   Fig.6. Distribution of lateral forces and shear by AVS method on frame with infills

2. Determination of Lateral Forces at Each Storey

Maximum Absolute Value Sum (AVS) Froof = F4 = V4 = 77.544 kN

Ffloor 3 = F3 = V3 V4 =116.791 77.544 = 39.247 kN

Ffloor 2 = F2 = V2 V3 =136.463 116.791 = 19.672kN Ffloor 1 = F1 = V1 V2 =163.508 136.463 = 27.045kN

Fig.7. Distribution of lateral forces and shear by SRSS method on frame with infills

Fig.8. Distribution of lateral forces and shear by CQC method on frame with infills

CONCLUSION

In case of Equivalent Static Force method, the design base shear value is found maximum for frame with infills over the frame without infills with the seismic weight being unchanged. The increase in base shear is due to natural period of the structure according to IS 1890:2002. This method is suitable for structures with low to medium range heights.

In case of Response spectrum method, Complete Quadratic Combination (CQC) provides, reasonable accuracy over AVS and SRSS methods due to provision for modal contribution. This method is suitable for almost all kinds of engineering interest.

REFERENCES

IS 1893 (part1): 2002, Criteria for earthquake Resistant Design of Structures-Part1: general Provisions and Buildings

, Bureau of Indian Standards, New Delhi.

Pankaj Agarwal, Manish Shrikhande (2007),Earthquake Resistant Design Of Structures Prentice-Hall of India Limited Private Limited, New Delhi.

T.K.Data(2010), Seismic Analysis of Structures , John Wiley &Sons (Asia) Pte Ltd.