2015-2016
108
Chapter Four
Equivalent Frame
Method
Prof. Dr. Mustafa B. Dawood
Dr. Bilal Ismaeel Al-Shraify
2015-2016
109
Equivalent Frame Method:
By equivalent frame method the structure is divided for analysis into
continues frames centered on the column lines and extending longitudinally
and transversely, as shown in Fig(4-1). Each frame is composed of a row of
columns and a broad continues beam. The beam or slab beam, includes the
portion of the slab bounded by panel centerlines on either side of the
columns, together with column line beam or drop panel, if used.
For vertical loading, each floor with its columns may be analyzed
separately, with the columns assumed to be fixed at the floor above and
below
The equivalent column:
In the equivalent frame method of analysis, the columns are considered to
be attached to the continuous slab beam by torsional members that are
transvers to the direction of the span for which moment are being found Fig
(4-2).
From Fig.(4-2), it is clear that the rotational restraint provided at the end of
the slab spanning in the direction ( ) is influenced not only by the flexural
stiffness of the column, but also by the torsional stiffness of the edge beam
AC.
With distributed torque ( ) applied by the slab and resisting torque ( )
provided by the column, the edge-beam section at A and C will rotate to a
greater degree than the section at B.
Prof. Dr. Mustafa B. Dawood
Dr. Bilal Ismaeel Al-Shraify
2015-2016
110
CL Column
interior
equivalent
frame
CLColumn
CL panel
CL Column
CL panel
CL Column
floor
Fig (4-1)
kk
Prof. Dr. Mustafa B. Dawood
Dr. Bilal Ismaeel Al-Shraify
2015-2016
111
Arrangement of live load: ACI-code 13.7.6
1- If the live load does not exceed three-quarters of the dead load
maximum moment may be assumed to occur at all critical sections.
When full factored live load is on the entire slab.
3
. ≤
.
4
2- . > 3/4 .
Use ¾ factored live load.
( 43 L.L)
factored
maximum +ve
moment in (AB) (CD)
A
B
C
D
D.L
factored
( 34 L.L)
factored
maximum -ve moment
at B
A
B
C
D
 Factored moments must not be taken less than those corresponding
to full live load on all panels.
Prof. Dr. Mustafa B. Dawood
Dr. Bilal Ismaeel Al-Shraify
2015-2016
112
The actual column and edge beam are replaced by an equivalent column.
1
=
1
1
+ … … … … … … … . (1)
∑
Where
: Flexural stiffness of equivalent column.
: Flexural stiffness of actual column.
: Flexural stiffness of edge beam.
If no beam frames into the column, a portion of the slab equal to the width
of the column or capital is assumed as the effective beam.
slab
Col.
capital
column
drop
C1
C1
C1
If a beam frames into the column, T-bea4m section or L-beam section is
assumed.
=
9∗
… … … … … … (2)
1−
Where
: modules of elasticity of slab concrete
: Size of rectangular column, capital or bracket in direction
: Cross-sectional constant
=
1 − 0.63
3
… … … … … … … … . . (3)
Where
: Shorter dimension of rectangular part of cross section(
).
: Longer overall dimension of rectangular part of cross-section(
Prof. Dr. Mustafa B. Dawood
).
Dr. Bilal Ismaeel Al-Shraify
2015-2016
113
Moment analysis
With the effective stiffness of the slab beam and the supports, the analysis
of the equivalent frame can proceed by moment distribution.
Negative moment obtained from the analysis apply at the centerlines of
supports. At interior supports, the critical section for (-ve) bending in both
column and middle strip may be taken at the face of column or capital but
not greater than 0.175 from center line of the column, according to ACIcode.
For design purposes, the total strip width is divided into column strip and
adjacent half-middle strips.
The distribution of moments to column and middle strips is done using the
same percentages given in connection with the direct design method.
Prof. Dr. Mustafa B. Dawood
Dr. Bilal Ismaeel Al-Shraify
2015-2016
114
Moment of inertia of slab beam and stiffness:
Moment of inertia used for analysis may be based on the concrete cross
section ( ), neglecting reinforcement, but variation in cross section along
the member axis should be accounted for. The stiffness of the slab strip
could be considered infinite within the bounds of the column or capital.
According to ACI-code 13.7.3
: for slab (from center of the column to the face of the column or capital)
= at the face of column or capital/(1 − / )
: for actual column is assumed to be infinite from the top of the slab to the
bottom of the slab
Accounting for these changes in moments of inertia results in a member, for
analysis, in which the moment of inertia varies in a stepwise manner. The
stiffness factors carryover factors and uniform-load fixed-end moment
factors needed for moment distribution analysis are given in table A.13a for
slab without drop panels and in Table A.13b for slab with drop panels.
. .
= coeff.
=
=
Prof. Dr. Mustafa B. Dawood
ℎ
12
:
( .
. )
Dr. Bilal Ismaeel Al-Shraify
2015-2016
115
Prof. Dr. Mustafa B. Dawood
Dr. Bilal Ismaeel Al-Shraify
2015-2016
116
Prof. Dr. Mustafa B. Dawood
Dr. Bilal Ismaeel Al-Shraify
2015-2016
117
Design the interior panel C of the flat plate floor shown in Fig.( ) by
equivalent frame method, using f = 414
,f = 25
, =
4.8 / ,
= 1.0
/
plus self-weight of slab, floor height=
3.66 , =
= 460
, thickness of the slab ℎ = 216
.
−
ℎ
216
=
∗ 24
1000
= 5.18 /
= 1.2 ∗ (1 + 5.18) + 1.6 ∗ 4.8
= 15.10
/
= 1.2 ∗ 6.18 = 7.40
/
= 1.6 ∗ 4.80 = 7.68 /
3
3
= 4.8 >
= ∗ 6.18
4
4
= 4.6
∴
3
(7.68) = 5.76 /
4
ℎ
6700 ∗ 216
=
=
12
12
= 5626713600
4
4 (6700 ∗ 216 )
=
=
12 ∗ 6700
= 3359232
4 (460 ∗ 460 )
=
12 ∗ 3660
= 4077828
Prof. Dr. Mustafa B. Dawood
6700
6700
A
B
B
C
6700
6700
B
6700
6700
1
15.1 kN/m 2
1
B
2
1
3
B
4
3
B
4
5.76kN/m2
7.4
2
C
B
2
C
5.76kN/m2
3
1
B
7.4 kN/m2
2
C
3
B
4
Dr. Bilal Ismaeel Al-Shraify
2015-2016
118
Equivalent torsional member has width = 460 mm depth=216 mm
216 216 ∗ 460
∗
= 1.088 ∗ 10
460
3
9∗
9 ∗ ∗ 1.088 ∗ 10
=
= 1809119
1−
6700 1 −
= 1 − 0.63 ∗
=
1
1
1
+
∑
1
1
=
2 ∗ 4077828 ∗
=
+
1
2 ∗ 1809119 ∗
= 2506316
= 3359232
Ext. col
.
=
.
=
+
+
3359232
= 0.58
3359232 + 2506316
2506316
=
= 0.42
2506316 + 3359232
=
Int. column
3359232
= 0.36
2506316 + 2 ∗ 3359232
2506316
=
= 0.27
2506316 + 2 ∗ 3359232
.
=
.
Case (1)
. .
=
=
12
+
Prof. Dr. Mustafa B. Dawood
=
15.1 ∗ 6.7 ∗ 6.7
= 377
12
8
− 403 = 163
.
.
Dr. Bilal Ismaeel Al-Shraify
2015-2016
119
Moment in flat plat floor
panel
B
C
B
joint
1
2
2
3
3
4
F.E.M
+377 -377 +377 -377 +377 -377
Case 1
Final moment +171 -441 +403 -403 +441 -171
Span M in c
163
F.E.M
+183 -183 +328 -328 +183 -183
Case 2
Final moment +69 -273 +306 -306 273 -69
Span M in c
187
F.E.M
328 -328 328 -328 +183 -183
Case 3
Final moment +146 -396 +374 -283 +216 -71
Span M in c
164
(2)
+ = 187
.
− = 403
.
(1)
Find −
at face of column
− at 0.23 m from center line of
column
0.23
+ 339
2
∗ 0.23 = 328
.
Control
−403 − 101 ∗
−
–
+
= 358
= 187
.
.
Use ACI-code
75% of −
for column strip
60% of +
for column strip
0.75 ∗ 35 = 268.5
. →
.
358 − 268.5 = 89.5
. → .
0.6 ∗ 187 = 112
. →
.
187 − 112 = 75
. → .
Prof. Dr. Mustafa B. Dawood
Dr. Bilal Ismaeel Al-Shraify
2015-2016
120
Design of flat plat reinforcement
Location
268.5
−
112
+
89.5
Two half
−
M.S
75
+
1- Check (d) for flexural.
2- Check one-way shear.
3- Check punching shear.
Col. S
Prof. Dr. Mustafa B. Dawood
∗ 1000
.
3350
3350
3350
3350
178
178
178
178
Dr. Bilal Ismaeel Al-Shraify