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2015-2016 108
Prof. Dr. Mustafa B. Dawood Dr. Bilal Ismaeel Al-Shraify
Chapter Four Equivalent Frame
Method
2015-2016 109
Prof. Dr. Mustafa B. Dawood Dr. Bilal Ismaeel Al-Shraify
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 (퐿 ) 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 (푚 ) applied by the slab and resisting torque (푀 ) provided by the column, the edge-beam section at A and C will rotate to a greater degree than the section at B.
2015-2016 110
Prof. Dr. Mustafa B. Dawood Dr. Bilal Ismaeel Al-Shraify
Fig (4-1)
kk
Column
panel
panel
floor
Column Column
interiorequivalent
frame
CLCL
CL
CL
CL
ColumnCL
2015-2016 111
Prof. Dr. Mustafa B. Dawood Dr. Bilal Ismaeel Al-Shraify
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.
퐿. 퐿 ≤34퐷. 퐿
2- 퐿. 퐿 > 3/4 퐷. 퐿
Use ¾ factored live load.
A
D.Lfactored
(34 L.L)
factored
B C D
maximum +vemoment in (AB) (CD)
A
(34 L.L)
factored
B C D
maximum -ve momentat B
Factored moments must not be taken less than those corresponding
to full live load on all panels.
2015-2016 112
Prof. Dr. Mustafa B. Dawood Dr. Bilal Ismaeel Al-Shraify
The actual column and edge beam are replaced by an equivalent column.
1푘
=1∑푘
+1푘
… … … … … … … . (1)
Where
푘 : Flexural stiffness of equivalent column.
푘 : Flexural stiffness of actual column.
푘 : 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.
C1
Col.capital C1
drop
C1
column
slab
If a beam frames into the column, T-bea4m section or L-beam section is assumed.
푘 =9 ∗ 퐸 퐶
푙 1−… … … … … … (2)
Where
퐸 : modules of elasticity of slab concrete
퐶 : Size of rectangular column, capital or bracket in direction 푙
퐶: Cross-sectional constant
퐶 = 1 − 0.63푥푦푥 푦
3 … … … … … … … … . . (3)
Where
푥: Shorter dimension of rectangular part of cross section(푚푚).
푦: Longer overall dimension of rectangular part of cross-section(푚푚).
2015-2016 113
Prof. Dr. Mustafa B. Dawood Dr. Bilal Ismaeel Al-Shraify
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 0.175푙 from center line of the column, according to ACI-code.
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.
2015-2016 114
Prof. Dr. Mustafa B. Dawood Dr. Bilal Ismaeel Al-Shraify
Moment of inertia of slab beam and stiffness:
Moment of inertia used for analysis may be based on the concrete cross section (퐼 ), 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
퐼 : for slab (from center of the column to the face of the column or capital) = 퐼 at the face of column or capital/(1− 퐶 /푙 )
퐼 : 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.
퐹.퐸.푀 = coeff.푤 푙 푙
푠푡푖푓푓푛푒푠푠 표푓 푠푙푎푏 =푘퐸퐼푙
퐼 =퐼 ℎ12
푙 : 푠푝푎푛 (퐶. 푡표.퐶)
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
Prof. Dr. Mustafa B. Dawood Dr. Bilal Ismaeel Al-Shraify
Design the interior panel C of the flat plate floor shown in Fig.( ) by equivalent frame method, using f푦 = 414 푀푃푎,f푐 = 25 푀푃푎,푤 =4.8 푘푁/푚 ,푤 = 1.0 푘푁/푚 plus self-weight of slab, floor height=3.66 푚,퐶 = 퐶 = 460푚푚, thickness of the slab ℎ = 216 푚푚.
푆푒푙푓 − 푤푒푖푔ℎ푡 표푓 푠푙푎푏
=216
1000 ∗ 24= 5.18 푘푁/푚
푤 = 1.2 ∗ (1 + 5.18) + 1.6 ∗ 4.8= 15.10 푘푁/푚
푤 = 1.2 ∗ 6.18 = 7.40 푘푁/푚 푤 = 1.6 ∗ 4.80 = 7.68 푘푁/푚
푤 = 4.8 >34푤 =
34∗ 6.18
= 4.6
∴ 푢푠푒34
(7.68) = 5.76 푘푁/푚
퐼 =푙 ℎ12 =
6700 ∗ 21612
= 5626713600
푘 =4퐸 퐼푙 =
4퐸 (6700 ∗ 216 )12 ∗ 6700
= 3359232 퐸
푘 =4퐸 (460 ∗ 460 )
12 ∗ 3660= 4077828 퐸
B
15.1 kN/m
7.4
C B
B C B
2
5.76kN/m2
5.76kN/m27.4kN/m2
1
2
3
B C B1 2 3 4
1 2 3 4
1 2 3 4
A B
B C B
6700 6700 6700
6700
6700
6700
2015-2016 118
Prof. Dr. Mustafa B. Dawood Dr. Bilal Ismaeel Al-Shraify
Equivalent torsional member has width = 460 mm depth=216 mm
퐶 = 1 − 0.63 ∗216460 ∗
216 ∗ 4603 = 1.088 ∗ 10 푚푚
푘 =9 ∗ 퐸 퐶
푙 1−=
9 ∗ 퐸 ∗ 1.088 ∗ 10
6700 1−= 1809119 퐸
1푘 =
1∑푘 +
1푘
1푘
=1
2 ∗ 4077828 ∗ 퐸+
12 ∗ 1809119 ∗ 퐸
푘 = 2506316 퐸 푘 = 3359232 퐸
Ext. col
퐷.퐹 =푘
푘 + 푘=
3359232퐸3359232퐸 + 2506316퐸
= 0.58
퐷.퐹 =푘
푘 + 푘 =2506316
2506316 + 3359232 = 0.42
Int. column
퐷.퐹 =3359232
2506316 + 2 ∗ 3359232 = 0.36
퐷.퐹 =2506316
2506316 + 2 ∗ 3359232 = 0.27
Case (1)
퐹.퐸.푀 =푤 푙 푙
12 =15.1 ∗ 6.7 ∗ 6.7
12 = 377 푘푁.푚
+푀 푖푛 푠푝푎푛 퐶 =푤 푙 푙
8 − 403 = 163 푘푁.푚
2015-2016 119
Prof. Dr. Mustafa B. Dawood Dr. Bilal Ismaeel Al-Shraify
Moment in flat plat floor
panel B C B joint 1 2 2 3 3 4
Case 1 F.E.M +377 -377 +377 -377 +377 -377 Final moment +171 -441 +403 -403 +441 -171
Span M in c 163
Case 2 F.E.M +183 -183 +328 -328 +183 -183 Final moment +69 -273 +306 -306 273 -69
Span M in c 187
Case 3 F.E.M 328 -328 328 -328 +183 -183 Final moment +146 -396 +374 -283 +216 -71
Span M in c 164 +푀 = 187 푘푁.푚 푐푎푠푒 (2) −푀 = 403 푘푁.푚 푐푎푠푒 (1)
Find −푀 at face of column
−푀 at 0.23 m from center line of column
−403− 101 ∗0.23
2+ 339
∗ 0.23 = 328 푘푁.푚 −푀 푎푡 퐵 Control
–푀 = 358 푘푁.푚 +푀 = 187 푘푁.푚
Use ACI-code
75% of −푀 for column strip
60% of +푀 for column strip
0.75 ∗ 35 = 268.5 푘푁.푚 → 퐶표푙.푆 358 − 268.5 = 89.5 푘푁.푚 → 푀. 푆 0.6 ∗ 187 = 112 푘푁.푚 → 퐶표푙. 푆 187 − 112 = 75 푘푁.푚 → 푀. 푆
2015-2016 120
Prof. Dr. Mustafa B. Dawood Dr. Bilal Ismaeel Al-Shraify
Design of flat plat reinforcement
Location 푀 푘푁.푚 푏 푚푚 푑 푚푚 푀 ∗ 1000
푏 휌 퐴
Col. S −푀 268.5 3350 178 +푀 112 3350 178
Two half M.S
−푀 89.5 3350 178 +푀 75 3350 178
1- Check (d) for flexural. 2- Check one-way shear. 3- Check punching shear.