2015-2016 chapter four equivalent frame method › eprints › publication_7_4417_382.pdfcolumn and...

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Prof. Dr. Mustafa B. Dawood Dr. Bilal Ismaeel Al-Shraify

Chapter Four Equivalent Frame

Method

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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.

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Fig (4-1)

kk

Column

panel

panel

floor

Column Column

interiorequivalent

frame

CLCL

CL

CL

CL

ColumnCL

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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.

. ≤34

.

2- . > 3/4 .

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.

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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.

C1

Col.capital C1

drop

C1

column

slab

If a beam frames into the column, T-bea4m section or L-beam section is assumed.

=9 ∗

1−… … … … … … (2)

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( ).

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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 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.

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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.

=

=ℎ

12: ( . . )

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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

1000 ∗ 24= 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 /

= 4.8 >34

=34∗ 6.18

= 4.6

∴34

(7.68) = 5.76 /

=ℎ

12 =6700 ∗ 216

12= 5626713600

=4

=4 (6700 ∗ 216 )

12 ∗ 6700= 3359232

=4 (460 ∗ 460 )

12 ∗ 3660= 4077828

B

15.1 kN/m

7.4

C B

B C B

2

5.76kN/m2

5.76kN/m2 7.4kN/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

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Equivalent torsional member has width = 460 mm depth=216 mm

= 1 − 0.63 ∗216460 ∗

216 ∗ 4603 = 1.088 ∗ 10

=9 ∗

1−=

9 ∗ ∗ 1.088 ∗ 10

6700 1−= 1809119

1=

1∑ +

1

1=

12 ∗ 4077828 ∗

+1

2 ∗ 1809119 ∗

= 2506316 = 3359232

Ext. col

. =+

=3359232

3359232 + 2506316= 0.58

. = + =2506316

2506316 + 3359232 = 0.42

Int. column

. =3359232

2506316 + 2 ∗ 3359232 = 0.36

. =2506316

2506316 + 2 ∗ 3359232 = 0.27

Case (1)

. . = 12 =15.1 ∗ 6.7 ∗ 6.7

12 = 377 .

+ = 8 − 403 = 163 .

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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

+ = 187 . (2) − = 403 . (1)

Find − at face of column

− at 0.23 m from center line of column

−403− 101 ∗0.23

2+ 339

∗ 0.23 = 328 . − Control

– = 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 . → .

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Design of flat plat reinforcement

Location . ∗ 1000

Col. S − 268.5 3350 178 + 112 3350 178 Two half

M.S − 89.5 3350 178 + 75 3350 178

1- Check (d) for flexural. 2- Check one-way shear. 3- Check punching shear.

2015-2016 chapter four equivalent frame method › eprints › publication_7_4417_382.pdfcolumn and...

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