Like this document? Why not share!

# Lecture 12 equivalent frame method

## by alialhussainawi on Nov 01, 2011

• 6,868 views

### Views

Total Views
6,868
Views on SlideShare
6,868
Embed Views
0

Likes
2
371
0

No embeds

## Lecture 12 equivalent frame methodDocument Transcript

• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Lecture 13 Lecture-13 Equivalent Frame Method By: Prof Dr. Qaisar Ali Civil Engineering Department NWFP UET Peshawar drqaisarali@nwfpuet.edu.pkProf. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 1 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Topics Addressed Introduction Stiffness of Slab-Beam Member Stiffness of Equivalent Column Stiffness of Column Stiffness of Torsional Member ExamplesProf. Dr. Qaisar Ali 2 1
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Topics Addressed Moment Distribution Method Arrangement of Live Loads Critical Sections for Factored Moments Moment Redistribution Factored Moments in Column and Middle Strips SummaryProf. Dr. Qaisar Ali 3 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (ACI 13.7) Introduction Consider a 3D structure shown in figure. It is intended to transform this 3D system into 2D system for facilitating analysis. This can be done by using the transformation technique of Equivalent Frame Analysis (ACI 13.7).Prof. Dr. Qaisar Ali 4 2
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (ACI 13.7) Introduction First, a frame is detached from the 3D structure. In the given figure, an interior frame is detached. The width of the frame is same as mentioned in DDM. The length of the frame extends up to full length of 3D system and the frame extends the full height of the building.Prof. Dr. Qaisar Ali 5 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Introduction Interior 3D frame detached from 3D structure.Prof. Dr. Qaisar Ali 6 3
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Introduction This 3D frame is converted to a 2D frame by taking effect of stiffness of laterally present members (slabs and beams).Prof. Dr. Qaisar Ali 7 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Introduction This 3D frame is converted to a 2D frame by taking effect of stiffness of laterally present members (slabs and beams).Prof. Dr. Qaisar Ali 8 4
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Introduction This 3D frame is converted to a 2D frame by taking effect of stiffness of laterally present members (slabs and beams).Prof. Dr. Qaisar Ali 9 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Introduction Ksb represents the combined stiffness of slab and longitudinal beam (if any). Kec represents the modified column stiffness. The modification depends on lateral members (slab, beams etc) and presence of column in the storey above. Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec KecProf. Dr. Qaisar Ali 10 5
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Introduction Therefore, the effect of 3D behavior of a frame is transformed into a 2D frame in terms of these stiffness i.e., Ksb and Kec. Once a 2D frame is obtained, the analysis can be done by any method of 2D frame analysis. Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec KecProf. Dr. Qaisar Ali 11 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Introduction Next the procedures for determination of Ksb and Kec are presented. Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec Kec Ksb Ksb Ksb Ksb Ksb Kec Kec Kec Kec Kec KecProf. Dr. Qaisar Ali 12 6
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Slab Beam member (Ksb): ( ) The stiffness of slab beam (Ksb = kEIsb/l) consists of combined stiffness of slab and any longitudinal beam present within. For a span, the k factor is a direct function of ratios c1/l1 and c2/l2 Tables are available in literature (Nilson and MacGregor) for determination of k for various conditions of slab systems. c1 l2 c2 l1Prof. Dr. Qaisar Ali 13 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Slab Beam member (Ksb): ( ) Determination of kProf. Dr. Qaisar Ali 14 7
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Slab Beam member (Ksb): ( ) Isb determinationProf. Dr. Qaisar Ali 15 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Slab Beam member (Ksb): Values of k for usual ( ) cases of structural systems. Column l1 l2 c1/l1 c2/l1 k dimension 12 × 12 10 10 0.10 0.10 4.182 As evident from the 15 15 0.07 0.07 4.05 table, the value of k for 20 20 0.05 0.05 4.07 usual cases of structures 15 × 15 10 10 0.13 0.13 4.30 is 4. 15 15 0.08 0.08 4.06 20 20 0.06 0.06 4.04 18 × 18 10 10 0.15 0.15 4.403 15 15 0.10 0.10 4.182 20 20 0.08 0.08 4.06Prof. Dr. Qaisar Ali 16 8
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Equivalent Column (Kec): q ( ) Stiffness of equivalent column consists of stiffness of actual columns {above (if any) and below slab-beam} plus stiffness of torsional members. Mathematically, nKc × mKt 1/Kec = 1/nKc + 1/mKt OR Kec = nKc + mKt Where, n = 2 for interior storey (for flat plates only) = 1 for top storey (for flat plates only) m = 1 for exterior frames (half frame) = 2 for interior frames (full frame) Note: n will be replaced by ∑ for columns having different stiffnessProf. Dr. Qaisar Ali 17 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Column (Kc): ( ) General formula of flexural stiffness is given by K = kEI/l Design aids are available from which value of k can be readily obtained for different values of (ta/tb) and (lu/lc). These design aids can be used if moment distribution method is used as method of analysis.Prof. Dr. Qaisar Ali 18 9
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Column (Kc): ( )Prof. Dr. Qaisar Ali 19 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Column (Kc): ( ) Determination of kProf. Dr. Qaisar Ali 20 10
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Column (Kc): ( ) Determination of k: Values of k for usual cases of structural systems. ta tb ta/tb lc lu lc/lu k As evident from the table, the value of k for 3 3 1.00 10 9.5 1.05 4.52 usual cases of structures 4 3 1.33 10 9.4 94 1.06 4.56 is 5.5. 5 3 1.67 10 9.3 1.07 4.60 6 3 2.00 10 9.3 1.08 5.20 7 3 2.33 10 9.2 1.09 5.39 8 3 2.67 10 9.1 1.10 5.42 9 3 3.00 10 9.0 1.11 5.46 10 3 3.33 10 8.9 1.12 5.5Prof. Dr. Qaisar Ali 21 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Torsional Member (Kt): ( ) Torsional members (transverse members) provide moment transfer between the slab-beams and the columns. Assumed to have constant cross-section throughout their length. Two conditions of torsional members (given next).Prof. Dr. Qaisar Ali 22 11
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Torsional Member (Kt): ( ) Condition (a) – No transverse beams framing into columnsProf. Dr. Qaisar Ali 23 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Torsional member (Kt): ( ) Condition (b) – Transverse beams framing into columnsProf. Dr. Qaisar Ali 24 12
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Torsional member (Kt): ( ) Stiffness Determination: The torsional stiffness Kt of the torsional member is given as: If beams frame into the support in the direction of analysis the torsional analysis, stiffness Kt needs to be increased. Ecs = modulus of elasticity of slab concrete; Isb = I of slab with beam; Is = I of slab without beamProf. Dr. Qaisar Ali 25 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Stiffness of Torsional member (Kt): ( ) Cross sectional constant, C:Prof. Dr. Qaisar Ali 26 13
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Equivalent Frame q Finally using the flexural stiffness values of the slab-beam and equivalent columns, a 3D frame can be converted to 2D frame. Ksb Ksb Ksb Kec Kec Kec Kec Ksb Ksb Ksb Kec Kec Kec Kec Ksb Ksb Ksb Kec Kec Kec KecProf. Dr. Qaisar Ali 27 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Example: Find the equivalent 2D frame for 1st storey of the E-W interior frame of fl t plate structure shown b l f f flat l t t t h below. Th slab i 10″ thi k and LL i The l b is thick d is 144 psf so that ultimate load on slab is 0.3804 ksf. All columns are 14″ square. Take fc′ = 4 ksi and fy = 60 ksi. Storey height = 10′ (from floor top to slab top) Data: l1 = 25′ (ln = 23.83′) l2 = 20′ Column strip width = 20/4 = 5′Prof. Dr. Qaisar Ali 28 14
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 01: 3D frame selection. 20′Prof. Dr. Qaisar Ali 29 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 01: 3D frame extraction. 20′ 10′ 10′ 25′ 25′ 10′ 25′Prof. Dr. Qaisar Ali 25′ 30 15
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 02: Extraction of single storey from 3D frame for separate analysis. 20′ 25′ 25′ 10′ 25′ 25′Prof. Dr. Qaisar Ali 31 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 03a: Slab-beam Stiffness calculation. Table: Slab beam stiffness (Ksb). l1 and l2 and k Spa Span c1/l1 c2/l2 I =l h 3/12 l / Ksb=kEIs/l c1 c2 ( bl A-20) s 2 f (table A 20) 25 & 20 and A2-B2 0.05 0.06 4.047 20000 270E 14" 14" The remaining spans will have the same values as the geometry is same.Prof. Dr. Qaisar Ali 32 16
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM):Prof. Dr. Qaisar Ali 33 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 03b: Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) Calculation of torsional member stiffness (Kt) Table: Kt calculation. Column l2 c2 C = ∑ (1 – 0 63x/y)x3y/3 (i 4) 0.63x/y)x (in Kt = ∑ 9EcsC/ {l2(1 – c2/l2)3} location A2 20′ 14" {1 – 0.63 × 10/ 14} × 103 × 14/3 = 2567 2 × [9Ecs×2567/ {20×12 (1–14/ (20×12))3}]=231Ecs Note 01: Kt term is multiplied with 2 because two similar torsional members meet at column A2. Note 02: Kt values for all other columns will be same as A2 because of similar column dimensions.Prof. Dr. Qaisar Ali 34 17
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): A lu Solution: B Step 03b: Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) Calculation of column stiffness (Kc) Table: ∑Kc calculation. kAB CAB Ic (in4) Column (from (from lc lu = (lc – hf) lc / lu for 14″ × 14″ ta/tb ΣKc = 2 × kEIc/lc location table table column A23) A23) 10′ 120/110 = 14 × 143/12 = 2×(5.09Ecc×3201/ 120) A2 110″ 5/5 = 1 5.09 0.57 (120″) 1.10 3201 = 272Ecc Note: For flat plates, ∑Kc term is multiplied with 2 for interior storey with similar columns above and below. For top storey, the ∑Kc term will be a single value (multiplied by 1)Prof. Dr. Qaisar Ali 35 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM):Prof. Dr. Qaisar Ali 36 18
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 03b: Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) Calculation of column stiffness (Kc) Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) 1/Kec = 1/∑Kc +1/Kt = 1/272Ecc + 1/231Ecs Because the slab and the columns have the same strength concrete, Ecc = Ecs = Ec. Therefore, Kec = 124.91Ec As all columns have similar dimensions and geometric conditions, the Kec value for all columns will be 124.91EcProf. Dr. Qaisar Ali 37 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Equivalent Frame; can be analyzed using any method of analysisProf. Dr. Qaisar Ali 38 19
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: To analyze the frame in SAP, the stiffness values are multiplied by lengths. Ksblsb = 270×25×12=81000E Keclec = 124.91×10×12=14989E 10′Prof. Dr. Qaisar Ali 39 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Load on frame: Solution: As the horizontal frame element Step 04: SAP results (moment at center). represents slab beam, load is computed by multiplying slab load with width of frame wul2 = 0.3804 × 20 = 7.608 kip/ftProf. Dr. Qaisar Ali 40 20
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: SAP results (moment at center).Prof. Dr. Qaisar Ali 41 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: SAP results (moment at faces).Prof. Dr. Qaisar Ali 42 21
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Comparison with SAP 3D model results. Load on model = 144 psf (LL) Slab thickness = 10″ Columns = 14″× 14″Prof. Dr. Qaisar Ali 43 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Comparison of SAP 3D model with EFM.Prof. Dr. Qaisar Ali 44 22
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Example: Find the equivalent 2D frame for 1st storey of the E-W interior frame of b f f beam supported f t d frame structure shown b l t t h below. Th slab i 7″ The l b is thick with LL of 144 psf so that ultimate load on slab is 0.336 ksf. All columns are 14″ square. Take fc′ = 4 ksi and fy = 60 ksi. Storey height = 10′ (from floor top to slab top) Data: l1 = 25′ (ln = 23.83′) l2 = 20′ Column strip width = 20/4 = 5′Prof. Dr. Qaisar Ali 45 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 01: 3D frame selection. 20′Prof. Dr. Qaisar Ali 46 23
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 01: 3D frame extraction. 20′ 10′ 10′ 25′ 25′ 10′ 25′Prof. Dr. Qaisar Ali 25′ 47 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 02: Extraction of single storey from 3D frame for separate analysis. 20′ 25′ 25′ 10′ 25′ 25′Prof. Dr. Qaisar Ali 48 24
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 03a: Slab-beam Stiffness calculation. Table: Slab beam stiffness (Ksb). l1 and l2 and k Span c1/l1 c2/l2 Isb Ksb=kEIs/l1 c1 c2 (table A 20) A-20) 25 & 20 and A2-B2 0.0467 0.058 4.051 25844 349E 14" 14" The remaining spans will have the same values as the geometry is same.Prof. Dr. Qaisar Ali 49 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 03b: Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) Calculation of torsional member stiffness (Kt) Table: Kt calculation. Column l2 c2 C = ∑ (1 – 0 63x/y)x3y/3 (i 4) 0.63x/y)x (in Kt = ∑ 9EcsC/ {l2(1 – c2/l2)3} location A2 20′ 14" 11208 3792.63Ecs B2 20′ 14" 12694 4295.98EcsProf. Dr. Qaisar Ali 50 25
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): A lu Solution: B Step 03b: Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) Calculation of column stiffness (Kc) Table: ∑Kc calculation. Ic (in4) kAB (fromColumn location lc lu lc / lu for 14″ × 14″ 14 14 ta/tb Kc table A23) column 10′ 120/100 = 14 × 143/12 = 16.5/3.5 = A2 (bottom) 100″ 7.57 201.9Ecc (120″) 1.20 3201 4.71 10′ 120/100 = 14 × 143/12 = 3.5/16.5= A2 (top) 100″ 5.3 141.39Ecc (120″) 1.20 3201 0.21 ∑Kc = 202Ecc + 141Ecc = 343EccProf. Dr. Qaisar Ali 51 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): A lu Solution: B Step 03b: Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) Calculation of column stiffness (Kc) Table: ∑Kc calculation. Ic (in4) kAB (fromColumn location lc lu lc / lu for 14″ × 14″ 14 14 ta/tb Kc table A23) column 10′ 120/100 = 14 × 143/12 = 16.5/3.5 = B2 (bottom) 100″ 7.57 201.9Ecc (120″) 1.20 3201 4.71 10′ 120/100 = 14 × 143/12 = 3.5/16.5= B2 (top) 100″ 5.3 141.39Ecc (120″) 1.20 3201 0.21 ∑Kc = 202Ecc + 141Ecc = 343EccProf. Dr. Qaisar Ali 52 26
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 03b: Equivalent column stiffness calculation (Column A2) (1/Kec = 1/∑Kc +1/Kt) Calculation of column stiffness (Kc) Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) 1/Kec = 1/∑Kc +1/Kt = 1/343Ecc + 1/3792.63Ecs Because the slab and the columns have the same strength concrete, Ecc = Ecs = Ec. Therefore, Kec = 315EcProf. Dr. Qaisar Ali 53 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 03b: Equivalent column stiffness calculation (Column B2) (1/Kec = 1/∑Kc +1/Kt) Calculation of column stiffness (Kc) Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt) 1/Kec = 1/∑Kc +1/Kt = 1/343Ecc + 1/4295.98Ecs Because the slab and the columns have the same strength concrete, Ecc = Ecs = Ec. Therefore, Kec = 318EcProf. Dr. Qaisar Ali 54 27
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Equivalent Frame; can be analyzed using any method of analysisProf. Dr. Qaisar Ali 55 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: To analyze the frame in SAP, the stiffness values are multiplied by lengths. Ksblsb = 349×25×12=104700E Keclec = 315×10×12=37800E Keclec = 318×10×12=38160EProf. Dr. Qaisar Ali 56 28
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Load on frame: Solution: As the horizontal frame element Step 04: SAP results (moment at center). represents slab beam, load is computed by multiplying slab load with width of frame wul2 = 0.336 × 20 = 6.72 kip/ftProf. Dr. Qaisar Ali 57 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: SAP results (moment at center).Prof. Dr. Qaisar Ali 58 29
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: SAP results (moment at faces).Prof. Dr. Qaisar Ali 59 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Comparison with SAP 3D model results. Load on model = 144 psf (LL) Slab thickness = 7″ Columns = 14″× 14″ Beams = 14″× 20″Prof. Dr. Qaisar Ali 60 30
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Comparison of beam moments of SAP 3D model with beam moments of EFM by SAP 2D analysis.Prof. Dr. Qaisar Ali 61 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Moment Distribution Method: The original derivation of EFM assumed that moment distribution would be the procedure used to analyze the slabs, and some of the concepts in the method are awkward to adapt to other methods of analysis. In lieu of computer software, moment distribution is a convenient hand calculation method for analyzing partial frames in the Equivalent Frame Method. Once stiffnesses are obtained from EFM, the distribution factors are conveniently calculated.Prof. Dr. Qaisar Ali 62 31
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Moment Distribution Method: Distribution Factors: Kct Ksb1 1 Kt 2 Ksb2 lc l1 Kt Kec l1 3 Kcb K = kEI/l lcProf. Dr. Qaisar Ali 63 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Moment Distribution Method: Distribution Factors: Slab Beam Distribution Factors: Ksb1 DF (span 2-1) = Ksb1 + Ksb2 + Kec Ksb2 DF (span 2-3) = Ksb1 + Ksb2 + KecProf. Dr. Qaisar Ali 64 32
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Moment Distribution Method: Distribution Factors: Equivalent Column Distribution factors: Kec DF = Ksb1 + Ksb2 + KecProf. Dr. Qaisar Ali 65 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Moment Distribution Method: Distribution Factors: These distribution factors are used in analysis. The equivalent frame of example 02 shall now be analyzed using moment distribution method. The comparison with SAP 3D model result for beam moments is also done. doneProf. Dr. Qaisar Ali 66 33
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Comparison of SAP 3D model with EFM done by Moment distribution method. Joint A B C D E CarryOver 0.5034 0.5034 0.5034 0.5034 DF 0.000 0.301 0.699 0.412 0.177 0.412 0.412 0.177 0.412 0.412 0.177 0.412 0.699 0.301 0.000 Slab Column Slab Slab Column Slab Slab Column Slab Slab Column Slab Slab Column Slab FEM 0.000 0.000 399.103 ‐399.103 0.000 399.103 ‐399.103 0.000 399.103 ‐399.103 0.000 399.103 ‐399.103 0.000 0.000 Bal 0.000 ‐119.955 ‐279.148 0.000 119.955 279.148 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 279.148 119.955 0.000 Carry over 0.000 ‐140.529 0.000 0.000 0.000 0.000 140.529 0.000 Bal 0.000 0.000 0.000 57.838 24.854 57.838 0.000 0.000 0.000 ‐57.838 ‐24.854 ‐57.838 0.000 0.000 0.000 Carry over 29.117 0.000 0.000 29.117 ‐29.117 0.000 0.000 ‐29.117 Bal 0.000 ‐8.751 ‐20.365 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 20.365 8.751 0.000 Carry over 0.000 ‐10.252 0.000 0.000 0.000 0.000 10.252 0.000 Bal 0.000 0.000 0.000 4.220 1.813 4.220 0.000 0.000 0.000 ‐4.220 ‐1.813 ‐4.220 0.000 0.000 0.000 Total 0.000‐129.395 129.395 ‐488.302 26.810 461.492‐367.695 0.000 367.695‐461.492‐26.810488.302‐129.395129.395 0.000Prof. Dr. Qaisar Ali 67 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Comparison of SAP 3D model with EFM.Prof. Dr. Qaisar Ali 68 34
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution of example 02 by Moment Distribution Method: p y Step 04: Analysis using Moment distribution method. Joint A B C D E CarryOver 0.5034 0.5034 0.5034 0.5034 DF 0.000 0.474 0.526 0.344 0.313 0.344 0.344 0.313 0.344 0.344 0.313 0.344 0.526 0.474 0.000 Slab Column Slab Slab Column Slab Slab Column Slab Slab Column Slab Slab Column Slab FEM 0.000 0.000 351.891 ‐351.891 0.000 351.891‐351.891 0.000 351.891‐351.891 0.000 351.891‐351.891 0.000 0.000 Bal 0.00 ‐166.90 ‐185.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 185.00 166.90 0.00 Carry over 0.00 ‐93.13 0.00 0.00 0.00 0.00 93.13 0.00 Bal 0.00 0.00 0.00 31.99 29.15 31.99 0.00 0.00 0.00 ‐31.99 ‐29.15 ‐31.99 0.00 0.00 0.00 Carry over 16.11 0.00 0.00 16.11 ‐16.11 0.00 0.00 ‐16.11 Bal 0.00 ‐7.64 ‐8.47 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 8.47 7.64 0.00 Carry over 0.00 ‐4.26 0.00 0.00 0.00 0.00 4.26 0.00 Bal 0.00 0.00 0.00 1.46 1.33 1.46 0.00 0.00 0.00 ‐1.46 ‐1.33 ‐1.46 0.00 0.00 0.00 Total 0. ‐174.900 174.900 ‐415.961 30.544 385.417‐335.012 0.000 335.012‐385.417‐30.544415.961‐174.900174.900 0.000Prof. Dr. Qaisar Ali 69 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM): Solution: Step 04: Comparison of beam moments of SAP 3D model with EFM analysis results obtained by moment distribution method.Prof. Dr. Qaisar Ali 70 35
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Arrangement of Live loads (ACI 13.7.6): g ( ) When LL ≤ 0.75DL Maximum factored moment when Full factored LL on all spans Other cases Pattern live loading using 0.75(Factored LL) to determine maximum factored momentProf. Dr. Qaisar Ali 71 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way SlabProf. Dr. Qaisar Ali 72 36
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Critical section for factored moments (ACI 13.7.7): ( ) Interior supports Critical section at face of rectilinear support but ≤ 0.175l1 from center of the support Exterior supports At exterior supports with brackets or capitals, the critical section < ½ the pp p , projection of bracket or capital beyond face of supporting element.Prof. Dr. Qaisar Ali 73 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way SlabProf. Dr. Qaisar Ali 74 37
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Moment Redistribution (ACI 13.7.7.4): ( ) Mu2 Mu1 Mo Mu3 ln c1/2 c1/2 l1Prof. Dr. Qaisar Ali 75 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Factored moments in column strips and middle strips: p p Same as in the Direct Design MethodProf. Dr. Qaisar Ali 76 38
• Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar Two Way Slab Equivalent Frame Method (EFM) Summary of Steps required for analysis using EFM y p q y g Extract the 3D frame from the 3D structure. Extract a storey from 3D frame for gravity load analysis. Identify EF members i.e., slab beam, torsional member and columns. Find stiffness (kEI/l) of each EF member using tables. Assign stiffnesses of each EF member to its corresponding 2D frame member. Analyze the obtained 2D frame using any method of analysis to get longitudinal moments based on center to center span. Distribute slab-beam longitudinal moment laterally using lateral distribution procedures of DDM.Prof. Dr. Qaisar Ali 77 Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar The EndProf. Dr. Qaisar Ali 78 39