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