14
A23.3 presents two parallel methods for calculating moments in two-way slab systems.
These are the Direct Design Method presented in Chap. 13 and the Elastic Frame Method presented in this chapter. In addition, A23.3 Cl. 13.6.1 allows other methods if they meet certain requirements. The relationship between the Direct Design Method and the Elastic
Frame Method is explained in Sec. 13–6 of this book. In the Direct Design Method, the statical moment, M
0
, 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 use of frame analyses to analyse slabs was first proposed by Peabody
14
–
1 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
14
–
2 , 14
–
3 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
14 – 4 presented 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 structures
14
–
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.
This chapter builds on the basic knowledge of the behaviour and design of slabs in flexure and shear presented in Chap. 13.

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
Figure 14–8
Frame action and twisting of edge member.
14–2 Elastic Frame Analysis of Slab Systems for Vertical Loads
705

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 .
63 x y x 3
3 y
(14–7;
A23.3 Eq. 13–14) 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
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, K s
, equivalent columns having stiffnesses, possibly beams parallel to the slabs with stiffnesses, K b
.
K ec
, and
EXAMPLE 14–3 Calculation of K t
, K c
, and K ec for an Edge Column and an Interior Column
The 200 mm thick flat plate shown in Fig. 14–4 is attached to 300 mm
×
600 mm 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
K ec for the connections between the slab strip along line 2 and columns A 2 and B 2.
K t
, K c
, and
1. Compute the values for the exterior column,
A
2 .
(a) Define 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.
(b) Compute
C
.
C
=
1
−
0 .
63 x y x 3
3 y
(14–7;
A23.3 Eq. 13–14)
=
1
−
0 .
63
200 mm
300 mm
( 200 mm ) 3
3
×
300 mm
=
464
×
10
6 mm
4
(c) Compute
K t .
K t
=
9 E cs
C t
( 1
− c
2
/ t
) 3
(14–6;
A23.3 Eq. 13–13)
14–2 Elastic Frame Analysis of Slab Systems for Vertical Loads
709

y = 300 mm y = 600 mm x = 200 mm
Figure 14–12
Attached torsional members—
Example 14–3.
710 where the summation refers to the beams on either side of line 2 and t beams on each side of line 2. Since both beams are similar: refers to the length of the
K t
=
9 E cs
N/mm
2 ×
464
×
10
6 mm
4
5500 mm ( 1
−
600 mm / 5500 mm ) 3
=
2 .
15
×
10
6 E cs
N • mm
Note that K t is in N • mm for E cs
(d) Compute
K c in MPa because the value 2 .
15
×
10 6 has the units of C/
− mm 3 for the edge columns.
The height centre-to-centre of the floor slabs is 2700
.
mm. 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.
I c
=
600
×
300
3
=
1 .
35
×
10
9 mm
4
12
For this column the overall height and / u
=
2700 mm, the unsupported clear height, u
=
2500 mm
=
1 .
080. The distance from the centre-line of the slab to the top of the column at the top surface of the slab, t a
, is 100 mm as is the corresponding distance, column. Interpolating in Table A–22 for / u
=
1 .
080 and t a
/t b t b
, at the bottom of the
=
1 .
0 gives
K c
=
4 .
86 E cc
I c c and the carryover factor is
−
0 .
56. Because there are two columns, one above and one below the floor, with the same stiffness
K c
=
2
4 .
86 E cc
N/mm
2 ×
1 .
35
×
10
9 mm
4
2700 mm
=
4 .
86
×
10
6 E cc
N • mm
(e) Compute the equivalent column stiffness
K ec for the edge column connection.
K
1 ec
=
1
K c
+
1
K t
=
1
4 .
86
×
10 6 E cc
+
1
2 .
15
×
10 6 E cs
The slab and the columns have the same concrete strength, so
K ec
=
1 .
49
×
10 6 E c
N • mm.
Note that K ec is only 31% of K
E cs
=
E cc
=
E c
. Therefore, c
. 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.
2. Compute the values at the interior column
B 2 .
The torsional member at column B 2 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
×
10
9 mm
4 and K t
=
4 .
89
×
10
6 strip along line 2, the columns are bent about their strong axes and have I
E c cs
N • mm. In the slab
=
5 .
40
×
10
9 mm
4
.
Again
K c
=
2
4 .
86 E cc
I c c
=
19 .
4
×
10
6
E cc
N • mm
K ec
=
3 .
91
×
10
6 E c
N • mm
It is important to note that unless K t is very large, K ec will be much smaller than K c
.
䊏
Chapter 14 Two-Way Slabs: Elastic Frame Method

450 mm
450 mm 270 mm
180 mm
270 mm 450 mm 270 mm
180 mm
270 mm
Figure 14–14
Torsional members at and
B 1
B 2—Example 14–4.
712
(b) Compute
C
.
To compute C , divide the torsional member into rectangles to maximize C as shown in Fig. 14–14a:
C
=
1
−
0 .
63
450
450
=
5 .
36
×
10
9 mm
4
450
3 ×
450
3
+
1
−
0 .
63
180
270
180
3 ×
270
3
(c) Compute
K t .
K t
= t
9 E cs
C
( 1
− c
2
/ t
) 3
(14–6;
A23.3 Eq. 13–13)
For span A 1– B 1, t
=
5775 mm, while for span B 1– C 1, t
=
7000 mm. Thus
K t
=
9 E cs
×
5 .
36
×
10 9
5775 ( 1
−
450 / 5775 ) 3
+
9 E cs
×
5 .
36
×
10 9
7000 ( 1
−
450 / 7000 ) 3
=
19 .
1
×
10
6
E cs
N • mm
Because the span for which moments are being determined contains a beam between the columns parallel to the span, K t is multiplied by the ratio of the moment of inertia of a cross section including the beam stem,
Example 14–2, I sb
I sb
(Fig. 14–6b), to the moment of inertia,
=
9 .
41
×
10
9 mm
4
.
I s
, of the slab alone. From
I s
=
6388
×
180 3
12
=
3 .
10
×
10
9 mm
4
Therefore, I sb
/I s
=
3 .
03 and K t
=
3 .
03
×
19 .
1
×
10 6 E cs
=
57 .
9
×
10 6 E cs
N • mm.
Chapter 14 Two-Way Slabs: Elastic Frame Method

the centroid of the shear perimeter in Fig. 14–15c gives the moment to be transferred as
M sj
−
V cant
(e c
+ e sc
)
−
V slab e sc
−
M f
=
0
Replacing V cant e c by M c gives
M f
=
M sj
−
M c
−
(V slab
+
V cant
)e sc
The connection is then designed for (V slab
+
V cant
) and M f
.
(14–8)
EXAMPLE 14–5 Design of a Flat-Plate Floor Using the Elastic Frame Method
Figure 14–4 shows a plan of a flat-plate floor without spandrel beams. Design this floor using the
Elastic Frame Method. Use 30 MPa concrete for the columns and slab and Grade 400 reinforcement.
The storey-to-storey height is 2.7 m. The floor supports its own dead load plus 1.0 kPa for partitions and finishes and a live load of 2.1 kPa.
This is the same slab that was designed in Example 13–7. Only those parts of the design that differ from Example 13–7 are discussed here. The step numbers in the design process are the same as in Example 13–7.
1. Select the design method.
Although the slab satisfies the requirements for the use of the Direct
Design Method, it has been decided to base the design on moments calculated using the Elastic
Frame Method.
2. Select the thickness.
The selection of thickness is based on Table 13–1 in Sec. 13–6 and also on providing adequate shear strength. Based on the calculations in Example 13–7, a 200 mm slab will be used.
3. Compute the moments in the equivalent frame along column line 2.
(Examples 13–7 and
14–5 differ in this step.) The strip of slab along column line 2 acts as a rigid frame spanning between columns A 2, B 2, C 2, and D 2. For the purposes of analysis, the columns above and below the slabs will be assumed fixed at their far ends.
(a) Determine moment distribution coefficients for slab beams.
From Example 14–1:
Span A 2 – B 2 :
K
A 2
– B 2
=
4 .
11 EI
1
1 where
1
=
5500 mm and
I
1
=
5500
×
200 3
12
=
3 .
67
×
10
9 mm
4
K
A 2
– B 2
=
4 .
11 E c
×
3 .
67
×
10
9
5500
=
2 .
74
×
10
6 E c
N • mm
COF
A 2
–
B 2
= −
0 .
509
Similarly, K
B 2
– A 2
=
2 .
74
×
10
6 E c
N • mm and COF
B 2
– A 2
= −
0 .
509.
Fixed end moments
=
0 .
085 w f 2 a
2
1
Span B 2 – C 2 :
K
B 2 – C 2
=
4 .
10 E c
I
1
1
=
4 .
10 E c
×
3 .
67
×
10
9
6000
=
2 .
51
×
10
6 E c
N • mm
COF
B 2
–
C 2
= −
0 .
507 fixed end moments
=
0 .
084 w f 2 a
2
1
Span C 2 – D 2 : Same as A 2 – B 2 .
14–2 Elastic Frame Analysis of Slab Systems for Vertical Loads
717

0
0
(b) Determine the moment distribution coefficients for the equivalent columns.
From Example 14–3:
Column A 2: K ec
=
1 .
49
×
10
6 E c
N • mm , COF
= −
0 .
56
Column B 2: K ec
=
3 .
91
×
10
6
E c
N • mm , COF
= −
0 .
56
(c) Compute the distribution factors.
The distribution factors are computed in the usual manner, thus:
DF
A 2 – B 2
=
K
K
A 2
–
B 2
A 2
– B 2
+
K ecA 2
=
2 .
74
×
10
6 E c
2 .
74
×
10 6 E c
+
1 .
49
×
10 6 E c
=
0 .
648
DF column A 2
=
1
−
0 .
648
=
0 .
352
The distribution factors and carryover factors are shown in Fig. 14–18. The cantilever members projecting outward at joints A 2 and D 2 refer to the slab that extends outside the column to support the wall.
(d) Select the loading cases and compute the fixed-end moments.
Since less than three-fourths of w
D w
L
=
2 .
1 kPa is
=
5 .
8 MPa, only the case of uniform live load on each panel need be considered (A23.3 Cl. 13.9.4.2). We will assume the live load of 2.1 kPa already includes any reduction for tributary areas.
w f
=
1 .
25
×
( 0 .
2
×
24
+
1 .
0 )
+
1 .
5
×
2 .
1
=
10 .
4 kPa
For the slab beam in question,
2 a
=
5500 mm.
Span A 2 – B 2 :
M
=
0 .
085 w f 2 a
2
1
=
0 .
085
×
10 .
4
×
5 .
5
×
5 .
5
2 =
147 .
1 kN • m
Note that the fixed-end moment M is based on the centre-to-centre span,
1
, rather than on the clear span, n
, used in the Direct Design Method.
Span B 2 – C 2 :
M
=
0 .
084
×
10 .
4
×
5 .
5
×
6 .
0
2 =
173 .
0 kN • m
0
A2
Stiffness
×
10
6
Distribution factor to B2–A2
B2
0.299
0.648
2.74
0.509
0.274
2.51
0.507
Carryover factor
C2
0.274
0.299
2.74
0.509
D2
0.648
0
0
0
Figure 14–18
Stiffness, carryover, and distribution factors—Example 14–5.
718
Chapter 14 Two-Way Slabs: Elastic Frame Method

720

14.1
12.6 kN
m
23.7
25.2
25.2 kN
m
14.1
Figure 14–20
Moments in column
Example 14–5.
A 2—
(b) Compute the shear on the section and the moment about the centroid of the section.
A free-body diagram of the joint is shown in Fig. 14–21b. The shear forces and bending moments computed using the Elastic Frame Analysis act at the centre-line of the column, as shown in
Fig. 14–21b. It is necessary to compute statically equivalent forces acting at the centroid of the critical shear perimeter, axis force acting through axis
X – X in Fig. 14–21b. From vertical force equilibrium, the shear
X – X is 38 .
5
+
133 .
8
=
172 .
3 kN. When the shear forces act through the centroid of the shear perimeter the corresponding moment is (Eq. 14–8)
M f
=
57 .
5
−
7 .
1
−
( 38 .
5
+
133 .
8 )
×
0 .
097
=
33 .
7 kN • m
The rest of the combined moment and shear transfer check proceeds as in Example 13–7.
11, 12, and 13.
Check the shear at the interior and corner columns and design the torsional reinforcement.
These steps are similar to the corresponding steps in Example 13–7.
䊏
The Elastic Frame Method was derived assuming that the structural analysis would be carried out using the moment distribution method. Following the procedures illustrated in
Example 14–5, the fixed-end moments, stiffnesses, and equivalent column stiffnesses are computed for use in such an analysis. If a standard frame analysis program based on the stiffness method is to be used, the frame must be specially modelled to get answers that agree with those obtained using the Elastic Frame Method.
The slab-column frame shown in Fig. 14–22a can be modelled for a computer analysis as shown in Fig. 14–22b. For this very simple frame, a total of 27 joints and 26 members are needed, compared to 9 joints and 8 members if the variation in stiffness is ignored. In addition, it is necessary for the analyst to reduce the stiffness of each column to account for the flexibility caused by the attached torsional member. Instead of carrying out a detailed
722
Chapter 14 Two-Way Slabs: Elastic Frame Method

3.
Superimpose the results from steps 1 and 2.
This is a complex series of steps. It is hoped that by the next code revision, the Elastic
Frame Method will be restated in a clearer and a more computer-compatible form.
A23.3 presents two parallel methods for calculating moments in a two-way slab: the Direct
Design Method, presented in Chap. 13, and the Elastic Frame Method, presented in Chap. 14, which corresponds to the Equivalent Frame Method in the ACI code. In the Elastic Frame
Method, the two-way slab system is idealized as two plane frame structures consisting of columns and slabs, one in each orthogonal direction.
Section 14–2 reviews the modelling of the beams and columns in the idealized elastic frame and the derivation of the stiffnesses of the frame members. Calculation of distribution factors, carryover factors and fixed-end moments that allow nonprismatic members to be analysed using the moment distribution method are presented. The column stiffnesses require modification because, at a joint, the average rotation of the beam and slab is greater than the rotation of the end of the column. The column stiffness is therefore computed for an equivalent column consisting of the actual column and an attached torsional member on one or both sides of the joint.
Other aspects concerning the use of the Elastic Frame Method to analyze slab systems for vertical loads are presented in Sec. 14–2, including the live load patterns to be considered in the structural analysis, the computation of moments at the faces of supports, and the calculation of the moment at the centroid of the shear perimeter of a column. To illustrate the difference between the Direct Design Method, presented in Chap. 13, and the Elastic
Frame Method, the long example from Sec. 13–10 is recomputed at the end of Sec. 14–2.
The Elastic Frame Method was derived assuming that the moment distribution method would be used to compute the moments in frames. Section 14–3 presents some of the considerations necessary for modelling a frame for analysis by a standard computer program to obtain results that are consistent with those obtained using the Elastic Frame Method.
In practice, computer analysis programs that are specifically written for the analysis of reinforced concrete slab systems are often used.
Section 14–4 presents the use of the Elastic Frame Method to analyse slabs in sway frames subjected to lateral loads. A procedure is described where the stiffnesses used in the analysis for gravity loads are modified to obtain values appropriate for the analysis for lateral loads.
PROBLEMS
14–1 Design the north-south strips in the slab shown in Fig. P13–6 using the Elastic Frame Method.
Loadings, dimensions, and material strengths are as given in Problem 13–6.
14–2 Repeat Problem 14–1 for the east-west strips in
Fig. P13–6.
14–3 Compute the moments in the east-west strips along lines A and B in the slab shown in Fig. 14–5. The floor supports its own weight, superimposed dead loads of 0.2 kPa for ceiling and mechanical fixtures and 1.2 kPa for future partitions, plus a live load of
4.8 kPa. The exterior wall weighs 5 kN/m and is supported by the edge beam. The storeys above and below the floor in question are 3.6 m and 4.2
m high, respectively. Lateral loads are resisted by an elevator shaft. Use 25 MPa concrete and Grade
400 reinforcement.
726
Chapter 14 Two-Way Slabs: Elastic Frame Method