The ‘Equivalent Frame’ Concept As mentioned in Section 11.1.3, in the case of beam-supported two-way slabs, 100 percent of the gravity loads on the slabs are transmitted to the supporting columns, in bot

Ih longitudinal and transverse directions (see Fig. 11.3(b)). The mechanism of load transfer from slab to columns is achieved by flexure, shear and torsion in the various elements. The slab-beam-column system behaves integrally as a three-dimensional system, with the involvement of all the floors of the building, to resist not only gravity loads, but also lateral loads. However, a rigorous three-dimensional analysis of the structure is complex, and not warranted except in very exceptional structures. Conventionally, when stiff beams are provided along column lines, the slab design is separated from the design of beams (and columns), as in the case of wall-supported slabs. The remaining part of the structure, comprising a three-dimensional skeletal framework of beams and columns, is separated for convenience, into (two-dimensional) plane frames in the longitudinal and transverse direction of the building. As the integrally cast slab also contributes to the strength and stiffness of the beams, the beam members are considered as flanged beams (T-beams, L-beams), with portions of the slab acting as the flanges of these beams; this concept was explained in Chapter 9. However, when the beams are flexible or absent, it is not appropriate to separate the slab design from the beam design.

In using the concept of a plane frame comprising columns and slab-beam members at various floor levels, fundamentally, the slab-beam member should consist of the entire floor member (slab and beam, if any) tributary to a line of columns forming the frame. This is illustrated in Fig. 11.24(a) and (b), which show how a building structure may be considered as a series of ‘equivalent (plane) frames’, each consisting of a row of columns and the portion of the floor system tributary to it. The part of the floor bound by the panel centrelines, on either side of the columns, forms the slab-beam member in this plane frame. Such ‘equivalent frames’ must be considered in both longitudinal and transverse directions, to ensure that load transfer takes place in both directions [Fig. 11.24(a)].

The equivalent frames can now be analysed under both gravity loads and lateral loads using the procedures mentioned in Chapter 9. The primary difference between the frame in Fig. 9.1(b) and the one in Fig. 11.24(b) lies in the width of the slab-beam member and the nature of its connections with the columns. Whereas in the conventional skeletal frame, the full beam is integral with the column, and the rotational restraint offered by the column at the joint is for the entire beam (with both beam and column undergoing the same rotation at the joint), in the ‘equivalent frame’, the column connection is only over part of the slab-beam member width, and hence the flexural restraint offered by the column to the slab-beam member is only partial. Thus, the rotation of the slab-beam member along a transverse section at the column support will vary, and will be equal to the column rotation only in the im For example, for the purpose of gravity load analysis, it is possible to simplify the mes. Accordingly, instead of mediate vicinity of the column. This, in turn, results in torsion in the portion of the slab transverse to the span and passing through the column (i.e., a cross-beam running over the column).

In the elastic analysis of the plane frame in Fig. 9.1(b), it was shown (in Section 9.3) that several approximations can be made, subject to certain limitations. Similar approximations can also be made in the present case.

For example, for the purpose of gravity load analysis, it is possible to simplify the analysis by applying the concept of substitute frames. Accordingly, instead of analysing the full ‘equivalent frame’ [Fig. 11.24(b)], it suffices to analyse separate partial frames [Fig. 11.24(c)], comprising each floor (or roof), along with the columns located immediately above and below. The columns are assumed to be fixed at their far ends [refer Cl. 24.3.1 of the Code]. Such substitute frame analysis is permissible provided the frame geometry (and loading) is relatively symmetrical, so that no significant sway occurs in the actual frame.

Fig. 11.25 Moment variations in a two-way slab

Variations of Moments in a Two-Way Slab Panel Although the horizontal member in the ‘equivalent partial frame’ in Fig. 11.24(c) is modelled as a very wide beam (i.e., slab with or without beam along the column line), it is actually supported on a very limited width. Hence, the outer portions of the member are less stiff than the part along the column line, and the distribution of moment across the width of the member is not uniform — unlike the beam in the conventional plane frame [Fig. 9.1(b)]. The probable variations of moments in a typical panel of the ‘equivalent frame’ are shown in Fig. 11.25. The variation of bending moment in the floor member along the span, under gravity loads is sketched in Fig. 11.25(b). Such a variation — with ‘negative moments’ near the supports and ‘positive moments’ in the neighbourhood of the midspan — is typical in any beam subject to uniformly distributed loads. In Fig. 11.25(b), Mab denotes the total ‘negative’ moment in the slab-beam member along the support line AB (extending over the full width of the panel), and Mef denotes the total ‘positive moment’ along the middle line EF of the panel. These moments are distributed across the width of the panel nonuniformly, as sketched in Fig. 11.25(c). The actual variation along AB or EF (marked by the solid line in Fig. 11.25(c) depends on several factors, such as the span ratio l2/l1, relative stiffness of beam (if any) along column lines, torsional stiffness of transverse beams (if any), etc. The actual moment variation is very difficult to predict exactly, and hence suitable approximations need to be made. This is generally achieved by dividing the slab panel into a column strip (along the column line) and two half-middle strips [Fig. 11.25(a)], and by suitably apportioning the total moment (Mab or Mef) to these strips with the assumption that the moment within each strip is uniform. This is indicated by the broken lines in Fig. 11.25(c), and is also clearly shown in Fig. 11.25(d). When beams are provided along the column line, the beam portion is relatively stiffer than the slab and resists a major share of the moment at the section. In this case, the moment has to be apportioned between the beam part and the slab part of the slab-beam member as indicated in Fig. 11.25(e). The calculations involved in the design procedure are given in the next section, which follows the unified procedure of design for all types of column-supported slabs — with or without beams (i.e., including flat slabs).

11.4 DESIGN OF COLUMN-SUPPORTED SLABS (WITH / WITHOUT BEAMS) UNDER GRAVITY LOADS 11.4.1 Code Procedures Based on the Equivalent Frame Concept Two-way slabs supported on columns include flat plates [Fig. 1.12], flat slabs [Fig. 1.13], waffle (ribbed) slabs [Fig. 1.11], and solid slabs with beams along the column lines [Fig. 1.10(b)]. Such slabs may be designed by any procedure which satisfies the basic conditions of equilibrium and geometrical compatibility, and the Code requirements of strength and serviceability. Specific design procedures have been laid out in l. 31) for the design of ‘flat slabs’, which are defined, according to the the Code (C Code (Cl. 31.1) as follows:

The equivalent frame method is recommended by ACI 318-02. It is given in Subsection 31.5, IS:456 - 2000. This method is briefly covered in this section for flat plates and flat slabs.

In the Direct Design Method, the statical moment, M0 , is calculated for each panel. This moment is then divided between positive and negative moment regions using arbitrary moment coefficients and the positive moments are adjusted to reflect pattern loadings. In the Elastic Frame Method, all of this is accomplished by frame analyses

The slab system is represented by a series of two dimensional equivalent frames for each spanning direction. An equivalent frame along a column line is a slice of the building bound by the centre-lines of the bays adjacent to the column line.

The use of frame analyses to analyse slabs was first proposed by Peabody in 1948 and a method of slab analysis referred to as “design by elastic analysis” was incorporated in the 1956 and 1963 editions of the ACI code. In the late 1940s Siess and Newmark studied the application of moment distribution analyses to two-way slabs on stiff beams.

Following extensive research on two-way slabs carried out at the University of Illinois, Corley and Jirsa resented a more refined method of frame analysis for slabs. This has been incorporated in the 1971 and subsequent ACI codes and in A23.3. Corley and Jirsa considered only gravity loads. Studies of the use of frame analyses for laterally loaded column-slab structures14–5 led to treatment of this problem in the

1983 and subsequent ACI codes. The Equivalent Frame Method in the ACI code corresponds to the Elastic Frame Method in A23.3.

The Elastic Frame Method is intended for use in analyzing moments in any practical building frame. Its scope is thus wider than the Direct Design Method, which is subject to the limitations presented in Sec. 13–6 (A23.3 Cl. 13.10.1). It works best for frames that can also be designed by the Direct Design Method, however.

The width of the equivalent frame is divided into a column strip and two middle strips. The column strip (CS) is the central half of the equivalent frame. Each middle strip (MS) consists of the remaining portions of two adjacent equivalent frames. The following figure shows the division in to strips along one direction. The direction under investigation is shown by the double headed arrow.

Figure 9-3.3 Equivalent frame along Column Line 2

In the above figure, l1

= span of the equivalent frame in a bay

l2

= width of the equivalent frame. This is the tributary width for calculating the loads.

The following figure shows a typical elevation of an equivalent frame.

Figure 9-3.4 Elevation of an equivalent frame

The analysis is done for each typical equivalent frame. An equivalent frame is modelled by slab-beam members and equivalent columns. The equivalent frame is analysed for gravity load and lateral load (if required), by computer or simplified hand calculations. Next, the negative and positive moments at the critical sections of the slab-beam members are distributed along the transverse direction. This provides the design moments per unit width of a slab.

If the analysis is restricted to gravity loads, each floor of the equivalent frame can be analysed separately with the columns assumed to be fixed at their remote ends, as shown in the following figure. The pattern loading is applied to calculate the moments for the critical load cases.

Torsional Members and Equivalent Columns

When the beam-and-column frame shown in Fig. 14–8a is loaded, the ends of the column and beam undergo equal rotations where they meet at the joint. If the flexural stiffness, K = M/θ , is known for the two members, it is possible to calculate the joint rotations and the end moments in the members. Similarly, in the case shown in Fig. 14–8b, the ends

14–2 Elastic Frame Analysis of Slab Systems for Vertical Loads

Figure 14–11 Division of edge members for calculation of C.

The constant C in Eq. 14–6 is calculated by subdividing the cross section into rectangles and carrying out the following summation

C=∑ (1−0.63xy¿)

x3 y3

¿ 14-7

where x is the shorter side of a rectangle and y is the longer side. The subdivision of the cross section of the torsional members is illustrated in Fig. 14–11. Several possible combinations of rectangles may have to be tried to get the maximum value of C. To do this, the wide rectangles should be made as large as possible in their smaller dimension. Thus the rectangles chosen in Fig. 14–11b will give larger values of C than those shown in Fig. 14–11a.

When using a moment distribution analysis, the frame analysis is carried out for a frame with slabs having stiffnesses, Ks , equivalent columns having stiffnesses, Kec , and possibly beams parallel to the slabs with stiffnesses, Kb .

Figure 9-3.5 Simplified model of an equivalent frame The steps of analysis of a two-way slab are as follows.

1. Determine the factored negative ( M u−¿¿

) and positive moment (M u+¿¿

demands at the critical

sections in a slab-beam member from the analysis of an equivalent frame. The values of M u−¿¿

are calculated at the faces of the columns. The values of¿¿ are calculated at the spans. The following sketch shows a typical moment diagram in a level of an equivalent frame due to gravity loads.

i.Figure 9-3.6 Typical moment diagram due to gravity loads

2. Distribute M u−¿¿ to the CS and the MS. These components are represented as M u

−¿¿CS

and

M u−¿¿

MS,respectively. Distribute M u

+¿¿ the CS and the MS. These components are represented

as M u+¿¿

CS and M u

+¿¿MS,

respectively.

Figure 9-3.7 Distribution of moments to column strip and middle strips

3. If there is a beam in the column line in the spanning direction, distribute each of M u−¿¿

CS and

M u+¿¿

CS between the beam and rest of the CS

Figure 9-3.8 Distribution of moments to beam, column strip and middle strips

4. Add the moments M u−¿¿

MS and M u

+¿¿MS

for the two portions of the MS (from adjacent equivalent

frames). 5. Calculate the design moments per unit width of the CS and MS

Calculation of Kt , Kc , and Kec for an Edge Column and an Interior ColumnThe 200 mm thick flat plate shown in Fig. 14–4 is attached to 300mm × 600mm columns oriented with the 300 mm dimension perpendicular to the edge, as shown in Fig. 14–4. The storeyto- storey height is 2700 mm. The slab and columns are 30 MPa concrete. Compute Kt , _Kc , and Kec for the connections between the slab strip along line 2 and columns A2 and B2.

1. Compute the values for the exterior column, A2.

Calculation of Kt , Kc , and Kec for an Edge Column and an Interior Column Calculated by

Sheet No

Checked by Job No

Client Date

Compute the values for the exterior column, A2Define the cross section of the torsional members. According to A23.3 Cl. 13.9.2.7, the attached torsional member at the exterior column corresponds to condition (a) in Fig. 14–10 as shown in Fig. 14–12a. Here x = 200 mm and y = 300 mm.

Compute C.

C=∑ (1−0.63xy¿)

x3 y3

¿

C=∑ (1−0.63200300

¿)2003300

3¿

= 464x106 mm4

Compute Kt

K t = ∑ 9 Ecs C

lt ¿¿¿

where the summation refers to the beams on either side of line 2 and lt refers to the length of the beams on each side of line 2. Since both beams are similar:

K t = ∑ 9 Ecs 464 x106

5500¿¿¿ = 2.15x106Ecs N•mm

Compute Kc for the edge columns. The height centre-to-centre of the floor slabs is 2700mm. The distribution of stiffnesses along the column is similar to Fig. 14–7a. The edge columns are bent about an axis parallel to the edge of the slab.

Ic = 600 x 3003

12 = 1.35 × 109 mm4

For this column the overall height l = 2700 mm, the unsupported clear height, lu = 2500 mm and l/lu = 1.080. The distance from the centre-line of the slab to the top of the column at the top surface of the slab, ta , is 100 mm as is the corresponding distance, tb, at the bottom of the column. Interpolating in Table A–22 for l/lu = 1.080 and ta/tb = 1.0 gives

K c= 4.86I c Ecc

lc

and the carryover factor is −0.56. Because there are two columns, one above and one below the floor, with the same stiffness

X= 200mmY = 300mm

C2 = 600mmlt = 5500mm

∑ K c = 2(4.86 Ecc x 1.35 x 109

2700) = 4.86x106 Ecc Nmm

Compute the equivalent column stiffness Kec for the edge column connection

1K cc

= 1

∑ K c

+ 1K t

= 1

4.86 x106 Ecc

+ 1

2.15 x106 Ecs

The slab and the columns have the same concrete strength, so Ecs = Ecc = Ec . Therefore, Kec = 1.49 × 106Ec N•mm. Note that Kec is only 31% of _Kc . This illustrates the large reduction in effective stiffness due to the lack of a torsionally stiff member at the edge of the slab along line 1.

Compute the values at the interior column B2. The torsional member at column B2 also has a section corresponding to condition (a) in Fig. 14–10 with x = 200 mm and y = 600 mm as shown in Fig. 14–12b. Thus C = 1.26×109 mm4 and Kt = 4.89×106Ecs N•mm. In the slab strip along line 2, the columns are bent about their strong axes and have Ic = 5.40 ×109 mm4. Again Kec = 3.91 ×106Ec N•mm

∑ K c = 2(4.86I c Ecc

lc

) = 19.84x106 Ecc Nmm

Kec = 3.91×106Ec N•mm

It is important to note that unless Kt is very large, Kec will be much smaller than ∑ K c

Fig 14.14 Torsional members at B1 and B2—Example 14–4.b) Compute C. To compute C, divide the torsional member into rectangles to maximize C as shown in Fig. 14–14a:C = (1-0.63x450/450)x4503x450/3 + (1-0.63x180/270)x1803x270/3 = 5.36x109mm4

Compute Kt

Kt = ∑ 9CEcs

lt ¿¿¿

For span A1–B1, _t = 5775 mm, while for span B1–C1, _t = 7000 mm. Thus

x = 200 mmy = 600 mmIc = 5.40 ×109 mm4