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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar




                                            Lecture 13
                                            Lecture-13

               Equivalent Frame Method
                                          By: Prof Dr. Qaisar Ali
                                    Civil Engineering Department
                                          NWFP UET Peshawar
                                             drqaisarali@nwfpuet.edu.pk


Prof. Dr. Qaisar Ali            CE 5115 Advance Design of Reinforced Concrete Structures                    1




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar




                        Topics Addressed
               Introduction

               Stiffness of Slab-Beam Member

               Stiffness of Equivalent Column
                       Stiffness of Column

                       Stiffness of Torsional Member

               Examples




Prof. Dr. Qaisar Ali                                                                                        2




                                                                                                                1
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar




                       Topics Addressed
               Moment Distribution Method

               Arrangement of Live Loads

               Critical Sections for Factored Moments

               Moment Redistribution

               Factored Moments in Column and Middle Strips

               Summary



Prof. Dr. Qaisar Ali                                                                                        3




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (ACI 13.7)
                Introduction
                       Consider a 3D structure shown in figure. It is intended to transform this 3D
                       system into 2D system for facilitating analysis. This can be done by using
                       the transformation technique of Equivalent Frame Analysis (ACI 13.7).




Prof. Dr. Qaisar Ali                                                                                        4




                                                                                                                2
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (ACI 13.7)
                Introduction
                       First, a frame is detached from the 3D structure. In the given figure, an
                       interior frame is detached.

                       The width of the frame is same as mentioned in DDM. The length of the
                       frame extends up to full length of 3D system and the frame extends the full
                       height of the building.




Prof. Dr. Qaisar Ali                                                                                        5




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Introduction
                       Interior 3D frame detached from 3D structure.




Prof. Dr. Qaisar Ali                                                                                        6




                                                                                                                3
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Introduction
                       This 3D frame is converted to a 2D frame by taking effect of stiffness of
                       laterally present members (slabs and beams).




Prof. Dr. Qaisar Ali                                                                                        7




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Introduction
                       This 3D frame is converted to a 2D frame by taking effect of stiffness of
                       laterally present members (slabs and beams).




Prof. Dr. Qaisar Ali                                                                                        8




                                                                                                                4

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                Two Way Slab
          Equivalent Frame Method (EFM)
                Introduction
                       This 3D frame is converted to a 2D frame by taking effect of stiffness of
                       laterally present members (slabs and beams).




Prof. Dr. Qaisar Ali                                                                                         9




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                Two Way Slab
          Equivalent Frame Method (EFM)
                Introduction
                       Ksb represents the combined stiffness of slab and longitudinal beam (if any).

                       Kec represents the modified column stiffness. The modification depends on lateral
                       members (slab, beams etc) and presence of column in the storey above.
                                Ksb            Ksb            Ksb             Ksb            Ksb

                       Kec            Kec            Kec            Kec             Kec                Kec
                                Ksb            Ksb            Ksb             Ksb            Ksb

                       Kec            Kec            Kec            Kec             Kec                Kec
                                Ksb            Ksb            Ksb             Ksb            Ksb

                       Kec            Kec            Kec            Kec             Kec                Kec
                                Ksb            Ksb            Ksb             Ksb            Ksb

                       Kec            Kec            Kec            Kec             Kec                Kec


Prof. Dr. Qaisar Ali                                                                                         10




                                                                                                                  5
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                 Two Way Slab
          Equivalent Frame Method (EFM)
                Introduction
                       Therefore, the effect of 3D behavior of a frame is transformed into a 2D frame in terms of
                       these stiffness i.e., Ksb and Kec.

                       Once a 2D frame is obtained, the analysis can be done by any method of 2D frame analysis.
                                 Ksb             Ksb          Ksb             Ksb            Ksb

                       Kec             Kec             Kec          Kec             Kec                Kec
                                 Ksb             Ksb          Ksb             Ksb            Ksb

                       Kec             Kec             Kec          Kec             Kec                Kec
                                 Ksb             Ksb          Ksb             Ksb            Ksb

                       Kec             Kec             Kec          Kec             Kec                Kec
                                 Ksb             Ksb          Ksb             Ksb            Ksb

                       Kec             Kec             Kec          Kec             Kec                Kec


Prof. Dr. Qaisar Ali                                                                                           11




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                 Two Way Slab
          Equivalent Frame Method (EFM)
                Introduction
                        Next the procedures for determination of Ksb and Kec are presented.




                                 Ksb             Ksb          Ksb             Ksb            Ksb

                       Kec             Kec             Kec          Kec             Kec                Kec
                                 Ksb             Ksb          Ksb             Ksb            Ksb

                       Kec             Kec             Kec          Kec             Kec                Kec
                                 Ksb             Ksb          Ksb             Ksb            Ksb

                       Kec             Kec             Kec          Kec             Kec                Kec
                                 Ksb             Ksb          Ksb             Ksb            Ksb

                       Kec             Kec             Kec          Kec             Kec                Kec


Prof. Dr. Qaisar Ali                                                                                          12




                                                                                                                    6
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                       Two Way Slab
          Equivalent Frame Method (EFM)
                 Stiffness of Slab Beam member (Ksb):
                                               ( )
                            The stiffness of slab beam (Ksb = kEIsb/l) consists of combined stiffness of
                            slab and any longitudinal beam present within.

                            For a span, the k factor is a direct function of ratios c1/l1 and c2/l2

                            Tables are available in literature (Nilson and MacGregor) for determination
                            of k for various conditions of slab systems.


                       c1
            l2              c2


                                  l1
Prof. Dr. Qaisar Ali                                                                                        13




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                       Two Way Slab
          Equivalent Frame Method (EFM)
                 Stiffness of Slab Beam member (Ksb):
                                               ( )

                            Determination of k




Prof. Dr. Qaisar Ali                                                                                        14




                                                                                                                 7
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Stiffness of Slab Beam member (Ksb):
                                              ( )

                       Isb determination




Prof. Dr. Qaisar Ali                                                                                        15




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Stiffness of Slab Beam member (Ksb): Values of k for usual
                                              ( )
                cases of structural systems.

            Column             l1     l2    c1/l1        c2/l1       k
            dimension
                 12 × 12        10    10       0.10        0.10          4.182      As evident from the
                                15    15       0.07        0.07          4.05       table, the value of k for
                                20    20       0.05        0.05          4.07       usual cases of structures
                 15 × 15        10    10       0.13        0.13          4.30       is 4.
                                15    15       0.08        0.08          4.06
                                20    20       0.06        0.06          4.04
                 18 × 18        10    10       0.15        0.15          4.403
                                15    15       0.10        0.10          4.182
                                20    20       0.08        0.08          4.06

Prof. Dr. Qaisar Ali                                                                                        16




                                                                                                                 8

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                 Two Way Slab
          Equivalent Frame Method (EFM)
                Stiffness of Equivalent Column (Kec):
                              q                ( )
                        Stiffness of equivalent column consists of stiffness of actual columns
                        {above (if any) and below slab-beam} plus stiffness of torsional members.

                        Mathematically,

                                                                                       nKc × mKt
          1/Kec = 1/nKc + 1/mKt                           OR              Kec =
                                                                                       nKc + mKt

                       Where,
                                  n = 2 for interior storey (for flat plates only)
                                    = 1 for top storey (for flat plates only)
                                  m = 1 for exterior frames (half frame)
                                     = 2 for interior frames (full frame)
                       Note: n will be replaced by ∑ for columns having different stiffness
Prof. Dr. Qaisar Ali                                                                                        17




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                 Two Way Slab
          Equivalent Frame Method (EFM)
                Stiffness of Column (Kc):
                                    ( )
                        General formula of flexural stiffness is given by K = kEI/l

                        Design aids are available from which value of k can be readily obtained for
                        different values of (ta/tb) and (lu/lc).

                        These design aids can be used if moment distribution method is used as
                        method of analysis.




Prof. Dr. Qaisar Ali                                                                                        18




                                                                                                                 9
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Stiffness of Column (Kc):
                                    ( )




Prof. Dr. Qaisar Ali                                                                                        19




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Stiffness of Column (Kc):
                                    ( )

                       Determination of k




Prof. Dr. Qaisar Ali                                                                                        20




                                                                                                                 10
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                 Two Way Slab
          Equivalent Frame Method (EFM)
                Stiffness of Column (Kc):
                                    ( )

                       Determination of k: Values of k for usual cases of structural
                       systems.
                           ta    tb ta/tb     lc      lu       lc/lu   k
                                                                                    As evident from the
                                                                                    table, the value of k for
                            3    3    1.00     10      9.5     1.05     4.52
                                                                                    usual cases of structures
                            4    3    1.33     10      9.4
                                                       94      1.06     4.56
                                                                                    is 5.5.
                            5    3    1.67     10      9.3     1.07     4.60
                            6    3    2.00     10      9.3     1.08     5.20
                            7    3    2.33     10      9.2     1.09     5.39
                            8    3    2.67     10      9.1     1.10     5.42
                            9    3    3.00     10      9.0     1.11     5.46
                            10   3    3.33     10      8.9     1.12     5.5
Prof. Dr. Qaisar Ali                                                                                        21




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                 Two Way Slab
          Equivalent Frame Method (EFM)
                Stiffness of Torsional Member (Kt):
                                              ( )
                       Torsional members (transverse members) provide moment transfer
                       between the slab-beams and the columns.

                       Assumed to have constant cross-section throughout their length.

                       Two conditions of torsional members (given next).




Prof. Dr. Qaisar Ali                                                                                        22




                                                                                                                 11
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Stiffness of Torsional Member (Kt):
                                              ( )
                       Condition (a) – No transverse beams framing into columns




Prof. Dr. Qaisar Ali                                                                                        23




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Stiffness of Torsional member (Kt):
                                              ( )
                       Condition (b) – Transverse beams framing into columns




Prof. Dr. Qaisar Ali                                                                                        24




                                                                                                                 12

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Stiffness of Torsional member (Kt):
                                              ( )
                       Stiffness Determination: The torsional stiffness Kt of the torsional member is
                       given as:




                       If beams frame into the support in the direction of analysis the torsional
                                                                           analysis,
                       stiffness Kt needs to be increased.




   Ecs = modulus of elasticity of slab concrete; Isb = I of slab with beam; Is = I of slab without beam

Prof. Dr. Qaisar Ali                                                                                        25




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Stiffness of Torsional member (Kt):
                                              ( )
                       Cross sectional constant, C:




Prof. Dr. Qaisar Ali                                                                                        26




                                                                                                                 13
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                Two Way Slab
          Equivalent Frame Method (EFM)
                Equivalent Frame
                 q

                       Finally using the flexural stiffness values of the slab-beam
                       and equivalent columns, a 3D frame can be converted to 2D
                       frame.
                                         Ksb                    Ksb                   Ksb

                           Kec                    Kec                    Kec                   Kec
                                         Ksb                    Ksb                   Ksb

                           Kec                    Kec                    Kec                   Kec
                                         Ksb                    Ksb                   Ksb

                           Kec                    Kec                    Kec                   Kec

Prof. Dr. Qaisar Ali                                                                                        27




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                Two Way Slab
          Equivalent Frame Method (EFM):
                Example: Find the equivalent 2D frame for 1st storey of the E-W interior
                frame of fl t plate structure shown b l
                f      f flat l t t t          h    below. Th slab i 10″ thi k and LL i
                                                           The l b is    thick d      is
                144 psf so that ultimate load on slab is 0.3804 ksf. All columns are 14″
                square. Take fc′ = 4 ksi and fy = 60 ksi. Storey height = 10′ (from floor
                top to slab top)
                Data:
                l1 = 25′ (ln = 23.83′)
                l2 = 20′
                Column strip width = 20/4 = 5′




Prof. Dr. Qaisar Ali                                                                                        28




                                                                                                                 14
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 01: 3D frame selection.




                                                                                                   20′




Prof. Dr. Qaisar Ali                                                                                        29




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 01: 3D frame extraction.

                       20′




                                                                                                    10′


                                                                                                    10′
                              25′
                                             25′
                                                                                                    10′
                                                             25′
Prof. Dr. Qaisar Ali                                                          25′                           30




                                                                                                                 15
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                 Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                         Step 02: Extraction of single storey from 3D frame for separate analysis.




                   20′



                               25′
                                            25′
                                                                                               10′
                                                             25′
                                                                              25′



Prof. Dr. Qaisar Ali                                                                                        31




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                 Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                         Step 03a: Slab-beam Stiffness calculation.




                                      Table: Slab beam stiffness (Ksb).
                        l1 and l2 and                      k
                  Spa
                  Span                 c1/l1 c2/l2                I =l h 3/12
                                                                     l /      Ksb=kEIs/l
                           c1     c2                 ( bl A-20) s 2 f
                                                     (table A 20)
                         25' & 20' and
                 A2-B2                 0.05 0.06         4.047      20000       270E
                          14"    14"
                 The remaining spans will have the same values as the geometry is same.




Prof. Dr. Qaisar Ali                                                                                        32




                                                                                                                 16

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                    Two Way Slab
          Equivalent Frame Method (EFM):




Prof. Dr. Qaisar Ali                                                                                         33




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                    Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                            Step 03b: Equivalent column stiffness calculation

                                 (1/Kec = 1/∑Kc +1/Kt)

                                 Calculation of torsional member stiffness (Kt)


                                               Table: Kt calculation.
  Column
               l2      c2       C = ∑ (1 – 0 63x/y)x3y/3 (i 4)
                                           0.63x/y)x     (in               Kt = ∑ 9EcsC/ {l2(1 – c2/l2)3}
  location
     A2       20′ 14" {1 – 0.63 × 10/ 14} × 103 × 14/3 = 2567 2 × [9Ecs×2567/ {20×12 (1–14/ (20×12))3}]=231Ecs


    Note 01: Kt term is multiplied with 2 because two similar torsional members meet at column A2.
    Note 02: Kt values for all other columns will be same as A2 because of similar column
    dimensions.
Prof. Dr. Qaisar Ali                                                                                         34




                                                                                                                  17
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                   Two Way Slab
          Equivalent Frame Method (EFM):                                                    A

                                                                                                            lu
                Solution:
                                                                                            B
                        Step 03b: Equivalent column stiffness calculation

                               (1/Kec = 1/∑Kc +1/Kt)

                               Calculation of column stiffness (Kc)

                                                    Table: ∑Kc calculation.
                                                                                     kAB     CAB
                                                            Ic (in4)
 Column                                                                            (from   (from
                lc     lu = (lc – hf)     lc / lu        for 14″ × 14″    ta/tb                      ΣKc = 2 × kEIc/lc
 location                                                                          table   table
                                                           column
                                                                                   A23)    A23)
               10′                      120/110 =        14 × 143/12 =                             2×(5.09Ecc×3201/ 120)
 A2                        110″                                          5/5 = 1   5.09    0.57
             (120″)                       1.10               3201                                        = 272Ecc


    Note: For flat plates, ∑Kc term is multiplied with 2 for interior storey with similar columns
    above and below. For top storey, the ∑Kc term will be a single value (multiplied by 1)
Prof. Dr. Qaisar Ali                                                                                                     35




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                   Two Way Slab
          Equivalent Frame Method (EFM):




Prof. Dr. Qaisar Ali                                                                                                     36




                                                                                                                              18
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 03b: Equivalent column stiffness calculation

                            (1/Kec = 1/∑Kc +1/Kt)

                            Calculation of column stiffness (Kc)

                                  Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt)

                                  1/Kec = 1/∑Kc +1/Kt = 1/272Ecc + 1/231Ecs

                                  Because the slab and the columns have the same strength
                                  concrete, Ecc = Ecs = Ec.

                                  Therefore, Kec = 124.91Ec

                                  As all columns have similar dimensions and geometric
                                  conditions, the Kec value for all columns will be 124.91Ec
Prof. Dr. Qaisar Ali                                                                                        37




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 04: Equivalent Frame; can be analyzed using any method of analysis




Prof. Dr. Qaisar Ali                                                                                        38




                                                                                                                 19
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                  Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 04: To analyze the frame in SAP, the stiffness values are multiplied by
                       lengths.
                                                                                    Ksblsb = 270×25×12=81000E

                                                                                    Keclec = 124.91×10×12=14989E




                                                                                             10′




Prof. Dr. Qaisar Ali                                                                                               39




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                  Two Way Slab
          Equivalent Frame Method (EFM):
                                                                               Load on frame:
                Solution:                                                      As the horizontal frame element
                       Step 04: SAP results (moment at center).                represents slab beam, load is
                                                                               computed by multiplying slab load
                                                                               with width of frame

                                                                               wul2 = 0.3804 × 20 = 7.608 kip/ft




Prof. Dr. Qaisar Ali                                                                                               40




                                                                                                                        20

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Ductile detailing
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This document discusses ductile detailing of reinforced concrete (RC) frames according to Indian standards. It explains that detailing involves translating the structural design into the final structure through reinforcement drawings. Good detailing ensures reinforcement and concrete interact efficiently. Key aspects of ductile detailing covered include requirements for beams, columns, and beam-column joints to improve ductility and seismic performance. Specific provisions are presented for longitudinal and shear reinforcement in beams and columns, as well as confining reinforcement and lap splices. The importance of cover and stirrup spacing is also discussed.

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This document discusses T-beams, which are more suitable than rectangular beams in reinforced concrete. There are two types of T-beams: monolithic and isolated. It provides notations and code recommendations for T-beams from IS: 456. There are three cases for finding the depth of the neutral axis in a T-beam: when it lies in the flange, in the rib, or at the junction. An example problem is worked through to find the moment of resistance for a given T-beam section using the provided concrete and steel properties.

engineeringscienceenergy
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 04: SAP results (moment at center).




Prof. Dr. Qaisar Ali                                                                                        41




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 04: SAP results (moment at faces).




Prof. Dr. Qaisar Ali                                                                                        42




                                                                                                                 21
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 04: Comparison with SAP 3D model results.



                                                                               Load on model = 144 psf (LL)
                                                                               Slab thickness = 10″
                                                                               Columns = 14″× 14″




Prof. Dr. Qaisar Ali                                                                                          43




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 04: Comparison of SAP 3D model with EFM.




Prof. Dr. Qaisar Ali                                                                                          44




                                                                                                                   22
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM):
                Example: Find the equivalent 2D frame for 1st storey of the E-W interior
                frame of b
                f      f beam supported f
                                    t d frame structure shown b l
                                               t t       h    below. Th slab i 7″
                                                                     The l b is
                thick with LL of 144 psf so that ultimate load on slab is 0.336 ksf. All
                columns are 14″ square. Take fc′ = 4 ksi and fy = 60 ksi. Storey height =
                10′ (from floor top to slab top)
                Data:
                l1 = 25′ (ln = 23.83′)
                l2 = 20′
                Column strip width = 20/4 = 5′




Prof. Dr. Qaisar Ali                                                                                        45




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 01: 3D frame selection.




                                                                                                   20′




Prof. Dr. Qaisar Ali                                                                                        46




                                                                                                                 23
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                 Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                         Step 01: 3D frame extraction.

                       20′




                                                                                                     10′


                                                                                                     10′
                               25′
                                             25′
                                                                                                     10′
                                                             25′
Prof. Dr. Qaisar Ali                                                          25′                           47




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                 Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                         Step 02: Extraction of single storey from 3D frame for separate analysis.




                   20′



                               25′
                                            25′
                                                                                               10′
                                                             25′
                                                                              25′



Prof. Dr. Qaisar Ali                                                                                        48




                                                                                                                 24

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42
4242
42

This document describes research using neural networks to predict the propagation path of plastic hinges in moment resisting frames under seismic loading. Pushover analyses were conducted on various frame configurations to create a database for training a neural network. The neural network takes frame element stiffness values as input and outputs the plastic hinge condition at different nodes. Training results showed the network could accurately predict plastic hinge distribution and collapse mechanisms. Validation on additional frames found reasonable correlation between predicted and actual plastic hinge statuses. The research demonstrates neural networks may provide a useful tool for assessing frame post-elastic behavior and collapse mechanisms at the design stage.

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                    Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                            Step 03a: Slab-beam Stiffness calculation.




                                                    Table: Slab beam stiffness (Ksb).
                                       l1 and l2 and                      k
                                 Span                  c1/l1 c2/l2                  Isb     Ksb=kEIs/l1
                                          c1     c2                 (table A 20)
                                                                           A-20)
                                        25' & 20' and
                                A2-B2                 0.0467 0.058      4.051     25844        349E
                                         14"    14"
                                The remaining spans will have the same values as the geometry is same.




Prof. Dr. Qaisar Ali                                                                                        49




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                    Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                            Step 03b: Equivalent column stiffness calculation

                                 (1/Kec = 1/∑Kc +1/Kt)

                                 Calculation of torsional member stiffness (Kt)


                                               Table: Kt calculation.
  Column
               l2      c2       C = ∑ (1 – 0 63x/y)x3y/3 (i 4)
                                           0.63x/y)x     (in               Kt = ∑ 9EcsC/ {l2(1 – c2/l2)3}
  location
     A2       20′ 14"                       11208                                    3792.63Ecs

     B2       20′ 14"                       12694                                    4295.98Ecs




Prof. Dr. Qaisar Ali                                                                                        50




                                                                                                                 25
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                   Two Way Slab
          Equivalent Frame Method (EFM):                                                   A

                                                                                                            lu
                Solution:
                                                                                           B
                        Step 03b: Equivalent column stiffness calculation

                                (1/Kec = 1/∑Kc +1/Kt)

                                Calculation of column stiffness (Kc)

                                             Table: ∑Kc calculation.
                                                             Ic (in4)
                                                                                           kAB (from
Column location          lc         lu         lc / lu    for 14″ × 14″
                                                              14 14             ta/tb                            Kc
                                                                                          table A23)
                                                            column
                         10′                120/100 =     14 × 143/12 =      16.5/3.5 =
 A2 (bottom)                      100″                                                         7.57         201.9Ecc
                       (120″)                 1.20            3201              4.71
                         10′                120/100 =     14 ×  143/12   =   3.5/16.5=
 A2 (top)                         100″                                                         5.3      141.39Ecc
                       (120″)                 1.20            3201              0.21


                                 ∑Kc = 202Ecc + 141Ecc = 343Ecc
Prof. Dr. Qaisar Ali                                                                                                   51




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                   Two Way Slab
          Equivalent Frame Method (EFM):                                                   A

                                                                                                            lu
                Solution:
                                                                                           B
                        Step 03b: Equivalent column stiffness calculation

                                (1/Kec = 1/∑Kc +1/Kt)

                                Calculation of column stiffness (Kc)

                                             Table: ∑Kc calculation.
                                                             Ic (in4)
                                                                                           kAB (from
Column location          lc         lu         lc / lu    for 14″ × 14″
                                                              14 14             ta/tb                            Kc
                                                                                          table A23)
                                                            column
                         10′                120/100 =     14 × 143/12 =      16.5/3.5 =
 B2 (bottom)                      100″                                                         7.57         201.9Ecc
                       (120″)                 1.20            3201              4.71
                         10′                120/100 =     14 × 143/12 =      3.5/16.5=
 B2 (top)                         100″                                                         5.3      141.39Ecc
                       (120″)                 1.20            3201              0.21


                                 ∑Kc = 202Ecc + 141Ecc = 343Ecc
Prof. Dr. Qaisar Ali                                                                                                   52




                                                                                                                            26
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 03b: Equivalent column stiffness calculation (Column A2)

                            (1/Kec = 1/∑Kc +1/Kt)

                            Calculation of column stiffness (Kc)

                                  Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt)

                                  1/Kec = 1/∑Kc +1/Kt = 1/343Ecc + 1/3792.63Ecs

                                  Because the slab and the columns have the same strength
                                  concrete, Ecc = Ecs = Ec.

                                  Therefore, Kec = 315Ec




Prof. Dr. Qaisar Ali                                                                                        53




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 03b: Equivalent column stiffness calculation (Column B2)

                            (1/Kec = 1/∑Kc +1/Kt)

                            Calculation of column stiffness (Kc)

                                  Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt)

                                  1/Kec = 1/∑Kc +1/Kt = 1/343Ecc + 1/4295.98Ecs

                                  Because the slab and the columns have the same strength
                                  concrete, Ecc = Ecs = Ec.

                                  Therefore, Kec = 318Ec




Prof. Dr. Qaisar Ali                                                                                        54




                                                                                                                 27
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                  Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 04: Equivalent Frame; can be analyzed using any method of analysis




Prof. Dr. Qaisar Ali                                                                                           55




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                  Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 04: To analyze the frame in SAP, the stiffness values are multiplied by
                       lengths.                                                   Ksblsb = 349×25×12=104700E

                                                                                  Keclec = 315×10×12=37800E

                                                                                  Keclec = 318×10×12=38160E




Prof. Dr. Qaisar Ali                                                                                           56




                                                                                                                    28

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                               Two Way Slab
          Equivalent Frame Method (EFM):
                                                                               Load on frame:
                Solution:                                                      As the horizontal frame element
                       Step 04: SAP results (moment at center).                represents slab beam, load is
                                                                               computed by multiplying slab load
                                                                               with width of frame

                                                                               wul2 = 0.336 × 20 = 6.72 kip/ft




Prof. Dr. Qaisar Ali                                                                                             57




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 04: SAP results (moment at center).




Prof. Dr. Qaisar Ali                                                                                             58




                                                                                                                      29
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 04: SAP results (moment at faces).




Prof. Dr. Qaisar Ali                                                                                          59




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 04: Comparison with SAP 3D model results.



                                                                               Load on model = 144 psf (LL)
                                                                               Slab thickness = 7″
                                                                               Columns = 14″× 14″
                                                                               Beams = 14″× 20″




Prof. Dr. Qaisar Ali                                                                                          60




                                                                                                                   30
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM):
                Solution:
                       Step 04: Comparison of beam moments of SAP 3D model with beam
                       moments of EFM by SAP 2D analysis.




Prof. Dr. Qaisar Ali                                                                                        61




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Moment Distribution Method:
                       The original derivation of EFM assumed that moment distribution would be
                       the procedure used to analyze the slabs, and some of the concepts in the
                       method are awkward to adapt to other methods of analysis.

                       In lieu of computer software, moment distribution is a convenient hand
                       calculation method for analyzing partial frames in the Equivalent Frame
                       Method.

                       Once stiffnesses are obtained from EFM, the distribution factors are
                       conveniently calculated.




Prof. Dr. Qaisar Ali                                                                                        62




                                                                                                                 31
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                Two Way Slab
          Equivalent Frame Method (EFM)
                Moment Distribution Method:
                        Distribution Factors:
                                                    Kct
                                    Ksb1
                         1                                      Kt
                                                2                     Ksb2                     lc
                                  l1 Kt
                                                          Kec
                                                                 l1              3
                                                    Kcb
                       K = kEI/l                                                               lc



Prof. Dr. Qaisar Ali                                                                                        63




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                Two Way Slab
          Equivalent Frame Method (EFM)
                Moment Distribution Method:
                        Distribution Factors:

                             Slab Beam Distribution Factors:


                                                                             Ksb1
                       DF (span 2-1) =
                                                                Ksb1 + Ksb2 + Kec

                                                                             Ksb2
                       DF (span 2-3) =
                                                                Ksb1 + Ksb2 + Kec
Prof. Dr. Qaisar Ali                                                                                        64




                                                                                                                 32

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anfisrc jointshear strength
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Moment Distribution Method:
                       Distribution Factors:

                            Equivalent Column Distribution factors:



                                                              Kec
                                 DF =
                                                Ksb1 + Ksb2 + Kec



Prof. Dr. Qaisar Ali                                                                                        65




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Moment Distribution Method:
                       Distribution Factors:

                            These distribution factors are used in analysis.

                            The equivalent frame of example 02 shall now be analyzed using
                            moment distribution method.

                            The comparison with SAP 3D model result for beam moments is also
                            done.
                            done




Prof. Dr. Qaisar Ali                                                                                        66




                                                                                                                 33
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                       Two Way Slab
          Equivalent Frame Method (EFM):
                   Solution:
                           Step 04: Comparison of SAP 3D model with EFM done by Moment
                           distribution method.
            Joint                A                       B                        C                        D                        E
          CarryOver                       0.5034                   0.5034                    0.5034                   0.5034
             DF        0.000 0.301    0.699   0.412     0.177   0.412   0.412    0.177   0.412   0.412    0.177    0.412   0.699    0.301 0.000


                        Slab Column Slab     Slab Column Slab       Slab Column Slab       Slab Column Slab       Slab Column Slab
             FEM       0.000 0.000 399.103 ‐399.103 0.000 399.103 ‐399.103 0.000 399.103 ‐399.103 0.000 399.103 ‐399.103 0.000 0.000
             Bal       0.000 ‐119.955 ‐279.148 0.000
                              119.955 279.148           0.000   0.000   0.000    0.000   0.000   0.000    0.000    0.000 279.148 119.955 0.000
          Carry over                  0.000 ‐140.529            0.000   0.000            0.000   0.000            140.529 0.000
             Bal       0.000 0.000    0.000   57.838 24.854 57.838      0.000    0.000   0.000 ‐57.838 ‐24.854 ‐57.838 0.000        0.000 0.000
          Carry over                 29.117   0.000             0.000   29.117           ‐29.117 0.000             0.000 ‐29.117
             Bal       0.000 ‐8.751 ‐20.365   0.000     0.000   0.000   0.000    0.000   0.000   0.000    0.000    0.000   20.365   8.751 0.000
          Carry over                  0.000   ‐10.252           0.000   0.000            0.000   0.000            10.252   0.000
             Bal       0.000 0.000    0.000   4.220     1.813   4.220   0.000    0.000   0.000   ‐4.220   ‐1.813 ‐4.220    0.000    0.000 0.000
            Total      0.000‐129.395 129.395 ‐488.302 26.810 461.492‐367.695 0.000 367.695‐461.492‐26.810488.302‐129.395129.395 0.000


Prof. Dr. Qaisar Ali                                                                                                                              67




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                       Two Way Slab
          Equivalent Frame Method (EFM):
                   Solution:
                           Step 04: Comparison of SAP 3D model with EFM.




Prof. Dr. Qaisar Ali                                                                                                                              68




                                                                                                                                                       34
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                      Two Way Slab
          Equivalent Frame Method (EFM):
                   Solution of example 02 by Moment Distribution Method:
                                   p       y
                          Step 04: Analysis using Moment distribution method.


             Joint               A                      B                       C                        D                        E
          CarryOver                      0.5034                   0.5034                  0.5034                   0.5034
              DF       0.000 0.474    0.526   0.344    0.313   0.344   0.344   0.313   0.344    0.344   0.313   0.344   0.526    0.474 0.000

                       Slab Column    Slab    Slab Column Slab         Slab Column Slab         Slab Column Slab         Slab    Column Slab

             FEM       0.000 0.000 351.891 ‐351.891 0.000 351.891‐351.891 0.000 351.891‐351.891 0.000 351.891‐351.891 0.000 0.000
             Bal       0.00 ‐166.90 ‐185.00   0.00     0.00    0.00    0.00    0.00    0.00     0.00    0.00    0.00    185.00 166.90 0.00
          Carry over                  0.00    ‐93.13           0.00    0.00            0.00     0.00            93.13    0.00
             Bal       0.00   0.00    0.00    31.99    29.15   31.99   0.00    0.00    0.00     ‐31.99 ‐29.15 ‐31.99     0.00     0.00   0.00
          Carry over                  16.11   0.00             0.00    16.11           ‐16.11   0.00            0.00    ‐16.11
             Bal       0.00   ‐7.64   ‐8.47   0.00     0.00    0.00    0.00    0.00    0.00     0.00    0.00    0.00     8.47     7.64   0.00
          Carry over                  0.00    ‐4.26            0.00    0.00            0.00     0.00            4.26     0.00
             Bal       0.00   0.00    0.00    1.46     1.33    1.46    0.00    0.00    0.00     ‐1.46   ‐1.33   ‐1.46    0.00     0.00   0.00
            Total       0. ‐174.900 174.900 ‐415.961 30.544 385.417‐335.012 0.000 335.012‐385.417‐30.544415.961‐174.900174.900 0.000


Prof. Dr. Qaisar Ali                                                                                                                            69




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                      Two Way Slab
          Equivalent Frame Method (EFM):
                   Solution:
                          Step 04: Comparison of beam moments of SAP 3D model with EFM analysis
                          results obtained by moment distribution method.




Prof. Dr. Qaisar Ali                                                                                                                            70




                                                                                                                                                     35
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Arrangement of Live loads (ACI 13.7.6):
                     g                    (          )

                       When LL ≤ 0.75DL
                            Maximum factored moment when Full factored LL on all spans

                       Other cases
                            Pattern live loading using 0.75(Factored LL) to determine maximum
                            factored moment




Prof. Dr. Qaisar Ali                                                                                        71




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab




Prof. Dr. Qaisar Ali                                                                                        72




                                                                                                                 36

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iosr journal of mechanical and civil engineering (
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Critical section for factored moments (ACI 13.7.7):
                                                      (          )

                       Interior supports
                            Critical section at face of rectilinear support but ≤ 0.175l1 from center of
                            the support

                       Exterior supports
                            At exterior supports with brackets or capitals, the critical section < ½ the
                                          pp                        p     ,
                            projection of bracket or capital beyond face of supporting element.




Prof. Dr. Qaisar Ali                                                                                        73




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab




Prof. Dr. Qaisar Ali                                                                                        74




                                                                                                                 37
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Moment Redistribution (ACI 13.7.7.4):
                                      (            )


                                                                               Mu2
                             Mu1
                                                    Mo

                                                                      Mu3
                                                        ln
               c1/2                                                                              c1/2
                                                         l1

Prof. Dr. Qaisar Ali                                                                                        75




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                               Two Way Slab
          Equivalent Frame Method (EFM)
                Factored moments in column strips and middle strips:
                                               p                 p
                       Same as in the Direct Design Method




Prof. Dr. Qaisar Ali                                                                                        76




                                                                                                                 38
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar



                                Two Way Slab
          Equivalent Frame Method (EFM)
                Summary of Steps required for analysis using EFM
                      y       p    q              y        g
                       Extract the 3D frame from the 3D structure.

                       Extract a storey from 3D frame for gravity load analysis.

                       Identify EF members i.e., slab beam, torsional member and columns.

                       Find stiffness (kEI/l) of each EF member using tables.

                       Assign stiffnesses of each EF member to its corresponding 2D frame member.

                       Analyze the obtained 2D frame using any method of analysis to get longitudinal moments
                       based on center to center span.

                       Distribute slab-beam longitudinal moment laterally using lateral distribution procedures of
                       DDM.




Prof. Dr. Qaisar Ali                                                                                                 77




              Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar




                                            The End




Prof. Dr. Qaisar Ali                                                                                                 78




                                                                                                                          39

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Lecture 12 equivalent frame method

  • 1. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Lecture 13 Lecture-13 Equivalent Frame Method By: Prof Dr. Qaisar Ali Civil Engineering Department NWFP UET Peshawar drqaisarali@nwfpuet.edu.pk Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 1 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Topics Addressed Introduction Stiffness of Slab-Beam Member Stiffness of Equivalent Column Stiffness of Column Stiffness of Torsional Member Examples Prof. Dr. Qaisar Ali 2 1
  • 2. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Topics Addressed Moment Distribution Method Arrangement of Live Loads Critical Sections for Factored Moments Moment Redistribution Factored Moments in Column and Middle Strips Summary Prof. Dr. Qaisar Ali 3 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (ACI 13.7) Introduction Consider a 3D structure shown in figure. It is intended to transform this 3D system into 2D system for facilitating analysis. This can be done by using the transformation technique of Equivalent Frame Analysis (ACI 13.7). Prof. Dr. Qaisar Ali 4 2
  • 3. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (ACI 13.7) Introduction First, a frame is detached from the 3D structure. In the given figure, an interior frame is detached. The width of the frame is same as mentioned in DDM. The length of the frame extends up to full length of 3D system and the frame extends the full height of the building. Prof. Dr. Qaisar Ali 5 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Introduction Interior 3D frame detached from 3D structure. Prof. Dr. Qaisar Ali 6 3
  • 4. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Introduction This 3D frame is converted to a 2D frame by taking effect of stiffness of laterally present members (slabs and beams). Prof. Dr. Qaisar Ali 7 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Introduction This 3D frame is converted to a 2D frame by taking effect of stiffness of laterally present members (slabs and beams). Prof. Dr. Qaisar Ali 8 4
  • 5. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Introduction This 3D frame is converted to a 2D frame by taking effect of stiffness of laterally present members (slabs and beams). Prof. Dr. Qaisar Ali 9 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Introduction Ksb represents the combined stiffness of slab and longitudinal beam (if any). Kec represents the modified column stiffness. The modification depends on lateral members (slab, beams etc) and presence of column in the storey above. Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Prof. Dr. Qaisar Ali 10 5
  • 6. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Introduction Therefore, the effect of 3D behavior of a frame is transformed into a 2D frame in terms of these stiffness i.e., Ksb and Kec. Once a 2D frame is obtained, the analysis can be done by any method of 2D frame analysis. Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Prof. Dr. Qaisar Ali 11 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Introduction Next the procedures for determination of Ksb and Kec are presented. Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Prof. Dr. Qaisar Ali 12 6
  • 7. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Slab Beam member (Ksb): ( ) The stiffness of slab beam (Ksb = kEIsb/l) consists of combined stiffness of slab and any longitudinal beam present within. For a span, the k factor is a direct function of ratios c1/l1 and c2/l2 Tables are available in literature (Nilson and MacGregor) for determination of k for various conditions of slab systems. c1 l2 c2 l1 Prof. Dr. Qaisar Ali 13 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Slab Beam member (Ksb): ( ) Determination of k Prof. Dr. Qaisar Ali 14 7
  • 8. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Slab Beam member (Ksb): ( ) Isb determination Prof. Dr. Qaisar Ali 15 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Slab Beam member (Ksb): Values of k for usual ( ) cases of structural systems. Column l1 l2 c1/l1 c2/l1 k dimension 12 × 12 10 10 0.10 0.10 4.182 As evident from the 15 15 0.07 0.07 4.05 table, the value of k for 20 20 0.05 0.05 4.07 usual cases of structures 15 × 15 10 10 0.13 0.13 4.30 is 4. 15 15 0.08 0.08 4.06 20 20 0.06 0.06 4.04 18 × 18 10 10 0.15 0.15 4.403 15 15 0.10 0.10 4.182 20 20 0.08 0.08 4.06 Prof. Dr. Qaisar Ali 16 8
  • 9. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Equivalent Column (Kec): q ( ) Stiffness of equivalent column consists of stiffness of actual columns {above (if any) and below slab-beam} plus stiffness of torsional members. Mathematically, nKc × mKt 1/Kec = 1/nKc + 1/mKt OR Kec = nKc + mKt Where, n = 2 for interior storey (for flat plates only) = 1 for top storey (for flat plates only) m = 1 for exterior frames (half frame) = 2 for interior frames (full frame) Note: n will be replaced by ∑ for columns having different stiffness Prof. Dr. Qaisar Ali 17 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Column (Kc): ( ) General formula of flexural stiffness is given by K = kEI/l Design aids are available from which value of k can be readily obtained for different values of (ta/tb) and (lu/lc). These design aids can be used if moment distribution method is used as method of analysis. Prof. Dr. Qaisar Ali 18 9
  • 10. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Column (Kc): ( ) Prof. Dr. Qaisar Ali 19 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Column (Kc): ( ) Determination of k Prof. Dr. Qaisar Ali 20 10
  • 11. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Column (Kc): ( ) Determination of k: Values of k for usual cases of structural systems. ta tb ta/tb lc lu lc/lu k As evident from the table, the value of k for 3 3 1.00 10 9.5 1.05 4.52 usual cases of structures 4 3 1.33 10 9.4 94 1.06 4.56 is 5.5. 5 3 1.67 10 9.3 1.07 4.60 6 3 2.00 10 9.3 1.08 5.20 7 3 2.33 10 9.2 1.09 5.39 8 3 2.67 10 9.1 1.10 5.42 9 3 3.00 10 9.0 1.11 5.46 10 3 3.33 10 8.9 1.12 5.5 Prof. Dr. Qaisar Ali 21 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Torsional Member (Kt): ( ) Torsional members (transverse members) provide moment transfer between the slab-beams and the columns. Assumed to have constant cross-section throughout their length. Two conditions of torsional members (given next). Prof. Dr. Qaisar Ali 22 11
  • 12. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Torsional Member (Kt): ( ) Condition (a) – No transverse beams framing into columns Prof. Dr. Qaisar Ali 23 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Torsional member (Kt): ( ) Condition (b) – Transverse beams framing into columns Prof. Dr. Qaisar Ali 24 12
  • 13. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Torsional member (Kt): ( ) Stiffness Determination: The torsional stiffness Kt of the torsional member is given as: If beams frame into the support in the direction of analysis the torsional analysis, stiffness Kt needs to be increased. Ecs = modulus of elasticity of slab concrete; Isb = I of slab with beam; Is = I of slab without beam Prof. Dr. Qaisar Ali 25 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Torsional member (Kt): ( ) Cross sectional constant, C: Prof. Dr. Qaisar Ali 26 13
  • 14. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Equivalent Frame q Finally using the flexural stiffness values of the slab-beam and equivalent columns, a 3D frame can be converted to 2D frame. Ksb Ksb Ksb Kec Kec Kec Kec Ksb Ksb Ksb Kec Kec Kec Kec Ksb Ksb Ksb Kec Kec Kec Kec Prof. Dr. Qaisar Ali 27 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Example: Find the equivalent 2D frame for 1st storey of the E-W interior frame of fl t plate structure shown b l f f flat l t t t h below. Th slab i 10″ thi k and LL i The l b is thick d is 144 psf so that ultimate load on slab is 0.3804 ksf. All columns are 14″ square. Take fc′ = 4 ksi and fy = 60 ksi. Storey height = 10′ (from floor top to slab top) Data: l1 = 25′ (ln = 23.83′) l2 = 20′ Column strip width = 20/4 = 5′ Prof. Dr. Qaisar Ali 28 14
  • 15. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 01: 3D frame selection. 20′ Prof. Dr. Qaisar Ali 29 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 01: 3D frame extraction. 20′ 10′ 10′ 25′ 25′ 10′ 25′ Prof. Dr. Qaisar Ali 25′ 30 15
  • 16. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 02: Extraction of single storey from 3D frame for separate analysis. 20′ 25′ 25′ 10′ 25′ 25′ Prof. Dr. Qaisar Ali 31 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 03a: Slab-beam Stiffness calculation. Table: Slab beam stiffness (Ksb). l1 and l2 and k Spa Span c1/l1 c2/l2 I =l h 3/12 l / Ksb=kEIs/l c1 c2 ( bl A-20) s 2 f (table A 20) 25' & 20' and A2-B2 0.05 0.06 4.047 20000 270E 14" 14" The remaining spans will have the same values as the geometry is same. Prof. Dr. Qaisar Ali 32 16
  • 17. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Prof. Dr. Qaisar Ali 33 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 03b: Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) Calculation of torsional member stiffness (Kt) Table: Kt calculation. Column l2 c2 C = ∑ (1 – 0 63x/y)x3y/3 (i 4) 0.63x/y)x (in Kt = ∑ 9EcsC/ {l2(1 – c2/l2)3} location A2 20′ 14" {1 – 0.63 × 10/ 14} × 103 × 14/3 = 2567 2 × [9Ecs×2567/ {20×12 (1–14/ (20×12))3}]=231Ecs Note 01: Kt term is multiplied with 2 because two similar torsional members meet at column A2. Note 02: Kt values for all other columns will be same as A2 because of similar column dimensions. Prof. Dr. Qaisar Ali 34 17
  • 18. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): A lu Solution: B Step 03b: Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) Calculation of column stiffness (Kc) Table: ∑Kc calculation. kAB CAB Ic (in4) Column (from (from lc lu = (lc – hf) lc / lu for 14″ × 14″ ta/tb ΣKc = 2 × kEIc/lc location table table column A23) A23) 10′ 120/110 = 14 × 143/12 = 2×(5.09Ecc×3201/ 120) A2 110″ 5/5 = 1 5.09 0.57 (120″) 1.10 3201 = 272Ecc Note: For flat plates, ∑Kc term is multiplied with 2 for interior storey with similar columns above and below. For top storey, the ∑Kc term will be a single value (multiplied by 1) Prof. Dr. Qaisar Ali 35 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Prof. Dr. Qaisar Ali 36 18
  • 19. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 03b: Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) Calculation of column stiffness (Kc) Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) 1/Kec = 1/∑Kc +1/Kt = 1/272Ecc + 1/231Ecs Because the slab and the columns have the same strength concrete, Ecc = Ecs = Ec. Therefore, Kec = 124.91Ec As all columns have similar dimensions and geometric conditions, the Kec value for all columns will be 124.91Ec Prof. Dr. Qaisar Ali 37 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Equivalent Frame; can be analyzed using any method of analysis Prof. Dr. Qaisar Ali 38 19
  • 20. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: To analyze the frame in SAP, the stiffness values are multiplied by lengths. Ksblsb = 270×25×12=81000E Keclec = 124.91×10×12=14989E 10′ Prof. Dr. Qaisar Ali 39 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Load on frame: Solution: As the horizontal frame element Step 04: SAP results (moment at center). represents slab beam, load is computed by multiplying slab load with width of frame wul2 = 0.3804 × 20 = 7.608 kip/ft Prof. Dr. Qaisar Ali 40 20
  • 21. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: SAP results (moment at center). Prof. Dr. Qaisar Ali 41 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: SAP results (moment at faces). Prof. Dr. Qaisar Ali 42 21
  • 22. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Comparison with SAP 3D model results. Load on model = 144 psf (LL) Slab thickness = 10″ Columns = 14″× 14″ Prof. Dr. Qaisar Ali 43 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Comparison of SAP 3D model with EFM. Prof. Dr. Qaisar Ali 44 22
  • 23. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Example: Find the equivalent 2D frame for 1st storey of the E-W interior frame of b f f beam supported f t d frame structure shown b l t t h below. Th slab i 7″ The l b is thick with LL of 144 psf so that ultimate load on slab is 0.336 ksf. All columns are 14″ square. Take fc′ = 4 ksi and fy = 60 ksi. Storey height = 10′ (from floor top to slab top) Data: l1 = 25′ (ln = 23.83′) l2 = 20′ Column strip width = 20/4 = 5′ Prof. Dr. Qaisar Ali 45 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 01: 3D frame selection. 20′ Prof. Dr. Qaisar Ali 46 23
  • 24. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 01: 3D frame extraction. 20′ 10′ 10′ 25′ 25′ 10′ 25′ Prof. Dr. Qaisar Ali 25′ 47 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 02: Extraction of single storey from 3D frame for separate analysis. 20′ 25′ 25′ 10′ 25′ 25′ Prof. Dr. Qaisar Ali 48 24
  • 25. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 03a: Slab-beam Stiffness calculation. Table: Slab beam stiffness (Ksb). l1 and l2 and k Span c1/l1 c2/l2 Isb Ksb=kEIs/l1 c1 c2 (table A 20) A-20) 25' & 20' and A2-B2 0.0467 0.058 4.051 25844 349E 14" 14" The remaining spans will have the same values as the geometry is same. Prof. Dr. Qaisar Ali 49 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 03b: Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) Calculation of torsional member stiffness (Kt) Table: Kt calculation. Column l2 c2 C = ∑ (1 – 0 63x/y)x3y/3 (i 4) 0.63x/y)x (in Kt = ∑ 9EcsC/ {l2(1 – c2/l2)3} location A2 20′ 14" 11208 3792.63Ecs B2 20′ 14" 12694 4295.98Ecs Prof. Dr. Qaisar Ali 50 25
  • 26. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): A lu Solution: B Step 03b: Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) Calculation of column stiffness (Kc) Table: ∑Kc calculation. Ic (in4) kAB (from Column location lc lu lc / lu for 14″ × 14″ 14 14 ta/tb Kc table A23) column 10′ 120/100 = 14 × 143/12 = 16.5/3.5 = A2 (bottom) 100″ 7.57 201.9Ecc (120″) 1.20 3201 4.71 10′ 120/100 = 14 × 143/12 = 3.5/16.5= A2 (top) 100″ 5.3 141.39Ecc (120″) 1.20 3201 0.21 ∑Kc = 202Ecc + 141Ecc = 343Ecc Prof. Dr. Qaisar Ali 51 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): A lu Solution: B Step 03b: Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) Calculation of column stiffness (Kc) Table: ∑Kc calculation. Ic (in4) kAB (from Column location lc lu lc / lu for 14″ × 14″ 14 14 ta/tb Kc table A23) column 10′ 120/100 = 14 × 143/12 = 16.5/3.5 = B2 (bottom) 100″ 7.57 201.9Ecc (120″) 1.20 3201 4.71 10′ 120/100 = 14 × 143/12 = 3.5/16.5= B2 (top) 100″ 5.3 141.39Ecc (120″) 1.20 3201 0.21 ∑Kc = 202Ecc + 141Ecc = 343Ecc Prof. Dr. Qaisar Ali 52 26
  • 27. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 03b: Equivalent column stiffness calculation (Column A2) (1/Kec = 1/∑Kc +1/Kt) Calculation of column stiffness (Kc) Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) 1/Kec = 1/∑Kc +1/Kt = 1/343Ecc + 1/3792.63Ecs Because the slab and the columns have the same strength concrete, Ecc = Ecs = Ec. Therefore, Kec = 315Ec Prof. Dr. Qaisar Ali 53 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 03b: Equivalent column stiffness calculation (Column B2) (1/Kec = 1/∑Kc +1/Kt) Calculation of column stiffness (Kc) Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) 1/Kec = 1/∑Kc +1/Kt = 1/343Ecc + 1/4295.98Ecs Because the slab and the columns have the same strength concrete, Ecc = Ecs = Ec. Therefore, Kec = 318Ec Prof. Dr. Qaisar Ali 54 27
  • 28. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Equivalent Frame; can be analyzed using any method of analysis Prof. Dr. Qaisar Ali 55 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: To analyze the frame in SAP, the stiffness values are multiplied by lengths. Ksblsb = 349×25×12=104700E Keclec = 315×10×12=37800E Keclec = 318×10×12=38160E Prof. Dr. Qaisar Ali 56 28
  • 29. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Load on frame: Solution: As the horizontal frame element Step 04: SAP results (moment at center). represents slab beam, load is computed by multiplying slab load with width of frame wul2 = 0.336 × 20 = 6.72 kip/ft Prof. Dr. Qaisar Ali 57 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: SAP results (moment at center). Prof. Dr. Qaisar Ali 58 29
  • 30. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: SAP results (moment at faces). Prof. Dr. Qaisar Ali 59 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Comparison with SAP 3D model results. Load on model = 144 psf (LL) Slab thickness = 7″ Columns = 14″× 14″ Beams = 14″× 20″ Prof. Dr. Qaisar Ali 60 30
  • 31. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Comparison of beam moments of SAP 3D model with beam moments of EFM by SAP 2D analysis. Prof. Dr. Qaisar Ali 61 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Moment Distribution Method: The original derivation of EFM assumed that moment distribution would be the procedure used to analyze the slabs, and some of the concepts in the method are awkward to adapt to other methods of analysis. In lieu of computer software, moment distribution is a convenient hand calculation method for analyzing partial frames in the Equivalent Frame Method. Once stiffnesses are obtained from EFM, the distribution factors are conveniently calculated. Prof. Dr. Qaisar Ali 62 31
  • 32. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Moment Distribution Method: Distribution Factors: Kct Ksb1 1 Kt 2 Ksb2 lc l1 Kt Kec l1 3 Kcb K = kEI/l lc Prof. Dr. Qaisar Ali 63 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Moment Distribution Method: Distribution Factors: Slab Beam Distribution Factors: Ksb1 DF (span 2-1) = Ksb1 + Ksb2 + Kec Ksb2 DF (span 2-3) = Ksb1 + Ksb2 + Kec Prof. Dr. Qaisar Ali 64 32
  • 33. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Moment Distribution Method: Distribution Factors: Equivalent Column Distribution factors: Kec DF = Ksb1 + Ksb2 + Kec Prof. Dr. Qaisar Ali 65 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Moment Distribution Method: Distribution Factors: These distribution factors are used in analysis. The equivalent frame of example 02 shall now be analyzed using moment distribution method. The comparison with SAP 3D model result for beam moments is also done. done Prof. Dr. Qaisar Ali 66 33
  • 34. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Comparison of SAP 3D model with EFM done by Moment distribution method. Joint A B C D E CarryOver 0.5034 0.5034 0.5034 0.5034 DF 0.000 0.301 0.699 0.412 0.177 0.412 0.412 0.177 0.412 0.412 0.177 0.412 0.699 0.301 0.000 Slab Column Slab Slab Column Slab Slab Column Slab Slab Column Slab Slab Column Slab FEM 0.000 0.000 399.103 ‐399.103 0.000 399.103 ‐399.103 0.000 399.103 ‐399.103 0.000 399.103 ‐399.103 0.000 0.000 Bal 0.000 ‐119.955 ‐279.148 0.000 119.955 279.148 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 279.148 119.955 0.000 Carry over 0.000 ‐140.529 0.000 0.000 0.000 0.000 140.529 0.000 Bal 0.000 0.000 0.000 57.838 24.854 57.838 0.000 0.000 0.000 ‐57.838 ‐24.854 ‐57.838 0.000 0.000 0.000 Carry over 29.117 0.000 0.000 29.117 ‐29.117 0.000 0.000 ‐29.117 Bal 0.000 ‐8.751 ‐20.365 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 20.365 8.751 0.000 Carry over 0.000 ‐10.252 0.000 0.000 0.000 0.000 10.252 0.000 Bal 0.000 0.000 0.000 4.220 1.813 4.220 0.000 0.000 0.000 ‐4.220 ‐1.813 ‐4.220 0.000 0.000 0.000 Total 0.000‐129.395 129.395 ‐488.302 26.810 461.492‐367.695 0.000 367.695‐461.492‐26.810488.302‐129.395129.395 0.000 Prof. Dr. Qaisar Ali 67 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Comparison of SAP 3D model with EFM. Prof. Dr. Qaisar Ali 68 34
  • 35. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution of example 02 by Moment Distribution Method: p y Step 04: Analysis using Moment distribution method. Joint A B C D E CarryOver 0.5034 0.5034 0.5034 0.5034 DF 0.000 0.474 0.526 0.344 0.313 0.344 0.344 0.313 0.344 0.344 0.313 0.344 0.526 0.474 0.000 Slab Column Slab Slab Column Slab Slab Column Slab Slab Column Slab Slab Column Slab FEM 0.000 0.000 351.891 ‐351.891 0.000 351.891‐351.891 0.000 351.891‐351.891 0.000 351.891‐351.891 0.000 0.000 Bal 0.00 ‐166.90 ‐185.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 185.00 166.90 0.00 Carry over 0.00 ‐93.13 0.00 0.00 0.00 0.00 93.13 0.00 Bal 0.00 0.00 0.00 31.99 29.15 31.99 0.00 0.00 0.00 ‐31.99 ‐29.15 ‐31.99 0.00 0.00 0.00 Carry over 16.11 0.00 0.00 16.11 ‐16.11 0.00 0.00 ‐16.11 Bal 0.00 ‐7.64 ‐8.47 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 8.47 7.64 0.00 Carry over 0.00 ‐4.26 0.00 0.00 0.00 0.00 4.26 0.00 Bal 0.00 0.00 0.00 1.46 1.33 1.46 0.00 0.00 0.00 ‐1.46 ‐1.33 ‐1.46 0.00 0.00 0.00 Total 0. ‐174.900 174.900 ‐415.961 30.544 385.417‐335.012 0.000 335.012‐385.417‐30.544415.961‐174.900174.900 0.000 Prof. Dr. Qaisar Ali 69 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Comparison of beam moments of SAP 3D model with EFM analysis results obtained by moment distribution method. Prof. Dr. Qaisar Ali 70 35
  • 36. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Arrangement of Live loads (ACI 13.7.6): g ( ) When LL ≤ 0.75DL Maximum factored moment when Full factored LL on all spans Other cases Pattern live loading using 0.75(Factored LL) to determine maximum factored moment Prof. Dr. Qaisar Ali 71 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Prof. Dr. Qaisar Ali 72 36
  • 37. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Critical section for factored moments (ACI 13.7.7): ( ) Interior supports Critical section at face of rectilinear support but ≤ 0.175l1 from center of the support Exterior supports At exterior supports with brackets or capitals, the critical section < ½ the pp p , projection of bracket or capital beyond face of supporting element. Prof. Dr. Qaisar Ali 73 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Prof. Dr. Qaisar Ali 74 37
  • 38. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Moment Redistribution (ACI 13.7.7.4): ( ) Mu2 Mu1 Mo Mu3 ln c1/2 c1/2 l1 Prof. Dr. Qaisar Ali 75 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Factored moments in column strips and middle strips: p p Same as in the Direct Design Method Prof. Dr. Qaisar Ali 76 38
  • 39. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Summary of Steps required for analysis using EFM y p q y g Extract the 3D frame from the 3D structure. Extract a storey from 3D frame for gravity load analysis. Identify EF members i.e., slab beam, torsional member and columns. Find stiffness (kEI/l) of each EF member using tables. Assign stiffnesses of each EF member to its corresponding 2D frame member. Analyze the obtained 2D frame using any method of analysis to get longitudinal moments based on center to center span. Distribute slab-beam longitudinal moment laterally using lateral distribution procedures of DDM. Prof. Dr. Qaisar Ali 77 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar The End Prof. Dr. Qaisar Ali 78 39