Lecture-08 Gravity Load Analysis of RC Structures By: Prof Dr. Qaisar Ali Civil Engineering Department UET Peshawar www.drqaisarali.com 1 Contents Analysis Approaches Point of Inflection Method Equivalent Frame Method Case Study Limit Analysis Plastic Analysis References 2 1
Analysis Approaches Analysis Approaches Elastic Analysis Based on elastic properties of structure. Presents the behavior of the structure within certain limitations. Plastic Analysis Based on inelastic capacities of structure, presents the behavior of the structure more accurately. Exact FEM, Slope Deflection, Moment Distribution, Stiffness methods, Equivalent Frame etc. Approximate ACI Coefficients, Direct Design, Point of Inflection, Portal Frame Methods etc Non-Linear pushover Analysis etc. 3 Analysis Approaches The approximate analysis methods such as ACI Coefficients and Direct Design Method have been discussed in detail in earlier lectures. In this lecture, another approximate method known as Point of Inflection Method will be briefly discussed. The exact analysis methods such as Slope Deflection, Moment Distribution and Stiffness method etc. have already been studied. The Equivalent Frame Analysis method will be discussed in detail in this lecture. 4 2
Point of Inflection Method In this method, points of inflection are located on the frame and the members are assumed separate determinate members at point of inflection. The individual members can be analyzed by statics as shown next. 5 Point of Inflection Method 1. Indeterminate frame 2. Point of inflection located over frame by inspection 3. Indeterminate frame broken into determinate members 4.Analysis of determinate membersby statics 5. Determinate member s BMD combined to form final approximate BMD of indeterminate structure 6 3
Introduction Consider a 3D structure shown in figure. It is intended to transform this 3D system to 2D system for facilitating analysis. This is done using the transformation technique of. 7 Introduction The width of the frame is same as 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. 8 4
Introduction This 3D frame is converted to a 2D frame by taking the effect of stiffnesses of laterally present members (slabs and beams). In EFA method, the horizontal members of the converted 2D frame are called slab-beam members and the vertical members are called equivalent columns. 9 Introduction The modified stiffnesses K sb and K ec are calculated and assigned to slab-beam and equivalent columns as shown in the figure. Ksb represents the combined stiffness of slab and longitudinal beam and Kec represents the modified column stiffness. 10 5
Stiffness of Slab Beam member (K sb ): The stiffness of slab beam (K sb = kei sb /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 c 1 /l 1 and c 2 /l 2 Tables are available for determination of k for various conditions of slab systems. l 2 c 1 c 2 l 1 Plan view of a floor of width l 2 which is to be converted to slab beam line element 11 Stiffness of Slab Beam member (K sb ): Determination of k 12 6
Stiffness of Slab Beam member (K sb ): I sb determination 13 Stiffness of Equivalent Column (K ec ): Stiffness of equivalent column consists of stiffness of actual columns {above and below the slab} plus stiffness of torsional members. Mathematically, 1/K ec = 1/ K c + 1/ K t OR K ec = K c K t K c + K t Where, K c = sum of flexural stiffnesses of columns above and below the slab. K t = Torsional stiffness of attached torsional members 14 7
Stiffness of Column (K c ): 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 (t a /t b ) and (l u /l c ). These design aids can be used if moment distribution method is used as method of analysis. 15 Stiffness of Column (K c ): 16 8
Stiffness of Column (K c ): Determination of k 17 Stiffness of Torsional Member (K t ): Torsional members (transverse members) provide moment transfer between the slab-beams and the columns. Assumed to have constant cross-section throughout their length. 18 9
Stiffness of Torsional Member (K t ): Stiffness Determination: The torsional stiffness K t of the torsional member is given as: If beams frame into the support in the direction of analysis, the torsional stiffness K t needs to be increased. E cs = modulus of elasticity of slab concrete; I sb = I of slab with beam; I s = I of slab without beam = l 2 h 3 /12 19 Stiffness of Torsional Member (K t ): Cross sectional constant, C: 20 10
Moment Distribution Method: The original derivation of EFM assumed that moment distribution would be the procedure used to analyze the slabs. In lieu of computer software, moment distribution is a convenient hand calculation method for analyzing partial frames in the Equivalent Frame Method. Next slides discuss the application of Moment Distribution Method to complete the analysis using EFM. 21 Moment Distribution Method: Distribution Factors: Slab Beam Distribution Factors: DF (span 2-1) = K sb1 K sb1 + K sb2 + K ec DF (span 2-3) = K sb2 K sb1 + K sb2 + K ec 22 11
Moment Distribution Method: Distribution Factors: Equivalent Column Distribution Factor: DF = K ec K sb1 + K sb2 + K ec 23 Moment Distribution Method: Distribution of unbalanced moment to columns: Portion of unbalanced moment to upper column = K ct K cb + K ct Portion of unbalanced moment to lower column = K cb K cb + K ct 24 12
Arrangement of Live loads (ACI 8.11.1) ACI 8.11.1 states that when the loading pattern is known, the equivalent frame shall be analyzed for that load. 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 moment. 25 26 13
Summary of Steps required for analysis using EFM 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 Moment Distribution 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. Slab analysis can be done using DDM. 27 Analysis by EFM Now 3D model will be converted into equivalent 2D model using EFM. 25 ft 25 ft 25 ft 25 ft 20 ft Beam 20 ft 20 ft 28 14
Analysis by EFM Sizes: Sizes will be as in 3D line model. Slab: Slab thickness, h f = 7 Columns: All columns are 14 14 as before. Beams: All beams are 14 wide and 20 deep. 29 Analysis by EFM Step a: 3D frame selection 25 ft 25 ft 25 ft 25 ft 20 20 ft 20 ft 20 ft 30 15
Analysis by EFM Step b: 3D frame extraction 20 10.5 25 25 25 25 10.5 10.5 31 Analysis by EFM Step c: Extraction of single storey from 3D frame for separate analysis 20 According to ACI 8.11.2.5, It shall be permitted to assume that the far ends of columns built integrally with the structure are considered to be fixed for gravity load analysis. 25 25 25 25 10.5 32 16
Analysis by EFM Step d (i): Slab-beam Stiffness calculation A B C D E Table: Slab beam stiffness (K sb ). l Span 1 and l 2 and c c 1 c 1 /l 1 c 2 /l 2 k I sb K sb =kei s /l 1 2 25' & 20' and AB 0.0467 0.058 4.051 25844 349E 14" 14" The remaining spans will have the same values as the geometry is same. Table A-20 (Reinforced concrete: Mechanics and Design, 3 rd Ed) 33 Analysis by EFM Step d (ii): Equivalent column stiffness calculation (1/K ec = 1/ K c +1/K t ) C2 C1 C1 C1 C2 Calculation of torsional member stiffness (K t ) Column location Table: K t calculation. l 2 c 2 C = (1 0.63x/y)x 3 y/3 (in 4 ) K t = 9E cs C/ {l 2 (1 c 2 /l 2 ) 3 } C2 20 14" 11208 3792.63E cs C1 20 14" 12694 4295.98E cs 34 17
Analysis by EFM A Step d (ii): Equivalent column stiffness calculation (1/K ec = 1/ K c +1/K t ) B l u Calculation of column stiffness (K c ) for Column C2 Table: K c calculation. Column location l c l u l c / l u for 14 14 I c (in 4 ) column C2 (bottom) 10.5 126/106 14 14 106 /12 = (126 ) 1.20 3201 C2 (top) 10.5 (126 ) 106 126/106 1.20 14 14 3 /12 = 3201 t a /t b k AB K c 16.5/3.5 = 4.71 3.5/16.5= 0.21 7.57 192E cc 5.3 135E cc K c = 192E cc + 135E cc = 327E cc 35 Analysis by EFM 36 18
Analysis by EFM Step d (ii): Equivalent column stiffness calculation (1/K ec = 1/ K c +1/K t ) Calculation of column stiffness (K c ) for Column C1 Table: K c calculation. Column location l c l u l c / l u for 14 14 I c (in 4 ) column C1 (bottom) 10.5 126/106 14 14 106 /12 = (126 ) 1.20 3201 C1 (top) 10.5 (126 ) 106 126/106 1.20 14 14 3 /12 = 3201 t a /t b k AB K c 16.5/3.5 = 4.71 7.57 192E cc 3.5/16.5= 0.21 5.3 135E cc K c = 192E cc + 135E cc = 327E cc 37 Analysis by EFM Step d (ii): Equivalent column stiffness calculation (Column C2) (1/K ec = 1/ K c +1/K t ) Calculation of column stiffness (K c ) Equivalent column stiffness calculation (1/K ec = 1/ K c +1/K t ) 1/K ec = 1/ K c +1/K t = 1/327E cc + 1/3792.63E cs Because the slab and the columns have the same strength concrete, E cc = E cs = E c. Therefore, K ec = 301E c 38 19
Analysis by EFM Step d (ii): Equivalent column stiffness calculation (Column C1) (1/K ec = 1/ K c +1/K t ) Calculation of column stiffness (K c ) Equivalent column stiffness calculation (1/K ec = 1/ K c +1/K t ) 1/K ec = 1/ K c +1/K t = 1/327E cc + 1/4295.98E cs Because the slab and the columns have the same strength concrete, E cc = E cs = E c. Therefore, K ec = 303E c 39 Analysis by EFM Step e: Equivalent Frame; can be analyzed using any method of analysis. 40 20
Analysis by EFM As the ground storey is same as 1 st one, therefore the stiffness calculated shall also be assigned to ground storey. For the top storey, the slab beam stiffness will be same as lower stories. However the equivalent stiffness of the top storey column is computed next. 41 Analysis by EFM Step d (ii): Equivalent column stiffness calculation (1/K ec = 1/ K c +1/K t ) Calculation of column stiffness (K c ) Table: K c calculation. Column location l c l u l c / l u for 14 14 I c (in 4 ) column C2 (bottom) 10.5 (126 ) 100 126/106 1.20 14 14 3 /12 = 3201 t a /t b k AB K c 16.5/3.5 = 4.71 7.57 192E cc K c = 192E cc Similarly for interior column, K c = 192E cc 42 21
Analysis by EFM Step d (ii): Equivalent column stiffness calculation (Column C2) (1/K ec = 1/ K c +1/K t ) Calculation of column stiffness (K c ) Equivalent column stiffness calculation (1/K ec = 1/ K c +1/K t ) 1/K ec = 1/ K c +1/K t = 1/192E cc + 1/3792.63E cs Because the slab and the columns have the same strength concrete, E cc = E cs = E c. Therefore, K ec = 182E c 43 Analysis by EFM Step d (ii): Equivalent column stiffness calculation (Column C1) (1/K ec = 1/ K c +1/K t ) Calculation of column stiffness (K c ) Equivalent column stiffness calculation (1/K ec = 1/ K c +1/K t ) 1/K ec = 1/ K c +1/K t = 1/192E cc + 1/4295.98E cs Because the slab and the columns have the same strength concrete, E cc = E cs = E c. Therefore, K ec = 183E c 44 22
Analysis by EFM Step e: Equivalent Frame 45 Analysis by EFM Step f: Analysis using moment Distribution Method Load on frame for Bending Analysis: As the horizontal frame element represents slab beam, load is computed by multiplying slab load with width of frame w DL = 0.0875 20 = 1.75 kip/ft = 0.144 20 = 2.88 kip/ft w LL 46 23
Analysis by EFM Step f: Analysis results for dead load (interior storey) using moment distribution method. 47 Analysis by EFM Step f: Analysis results for dead load (top storey) using moment distribution method. 48 24
Analysis by EFM Step f: Analysis results for dead load (Values at centerline). 49 Analysis by EFM Step f: Analysis results for dead load Distribution of moments to slab and beam for top storey. Table: E-W Interior frame analysis (Top Storey). Length 25-0 (Ext) 25-0 (Int) Longitudinal moment section Column Strip Longitudinal Moment %age moments (LM) factor (Graph A4) Column Strip slab Moment (CSSM) = 0.15CSM Column strip Beam Moment (BM)= 0.85CSM Middle Strip slab Moment E - 34 0.93 5 27 2.38 + 67 0.8 8 46 13.4 I - 111 0.8 13 75 22.2-105 0.8 13 71 21 + 42 0.8 5 29 8.4-85 0.8 10 58 17 50 25
Analysis by EFM Step f: Analysis results for dead load Distribution of moments to slab and beam for interior storey. Length 25-0 (Ext) 25-0 (Int) Table: E-W Interior frame analysis (Interior Storey). Longitudinal moment section Column Strip Longitudinal Moment %age moments (LM) factor (Graph A4) Column Strip slab Moment (CSSM) = 0.15CSM Column strip Beam Moment (BM)= 0.85CSM Middle Strip slab Moment E - 45 0.93 6 36 3.15 + 64 0.8 8 44 12.8 I - 109 0.8 13 74 21.8-101 0.8 12 69 20.2 + 43 0.8 5 29 8.6-87 0.8 10 59 17.4 51 Analysis by EFM Step f: Analysis results for dead load Analysis of columns for DL (factors for moment distribution) The computed unbalanced longitudinal moments shall be transferred to columns and shall be distributed to top and bottom columns as follows: Portion of unbalanced moment to upper column = Portion of unbalanced moment to lower column = K ct K cb + K ct K cb K cb + K ct 52 26
Analysis by EFM Step f: Analysis results for dead load Analysis of columns for DL (factors for moment distribution) 34 6 45 8 45 8 K cb3 /(K cb3 =1.00 K ct2 /(K ct2 +K cb2 )=0.41 K cb2 /(K ct2 +K cb2 )=0.59 K ct1 /(K ct1 +K cb1 )=0.41 K cb1 /(K ct1 +K cb1 )=0.59 K cb3 =192E K ct2 =135E K cb2 =192E K ct1 =135E K cb1 =192E C2 C1 C1 C1 C2 53 Analysis by EFM Step f: Analysis results for dead load Analysis of columns for DL 34 1.00=34 6 1.00=6 K cb3 /(K cb3 =1.00 K cb3 =192E 45 0.41=18.45 45 0.59=27 8 0.59=4.7 8 0.41=3.3 K ct2 /(K ct2 +K cb2 )=0.41 K cb2 /(K ct2 +K cb2 )=0.59 K ct2 =135E K cb2 =192E 45 0.41=18.45 45 0.59=27 8 0.41=3.3 8 0.59=4.7 K ct1 /(K ct1 +K cb1 )=0.41 K cb1 /(K ct1 +K cb1 )=0.59 K ct1 =135E K cb1 =192E C2 C1 C1 C1 C2 54 27
Analysis by EFM Step f: Analysis results for live load 55 Analysis by EFM Step f: Analysis results for live load Distribution of moments to slab and beam Length 25-0 (Ext) 25-0 (Int) Table: E-W Interior frame analysis (Top Storey). Longitudinal moment section Column Strip Longitudinal Moment %age moments (LM) factor (Graph A4) Column Strip slab Moment (CSSM) = 0.15CSM Column strip Beam Moment (BM)= 0.85CSM Middle Strip slab Moment E - 56 0.93 7.8 44 3.92 + 111 0.8 13.3 75 22.2 I - 183 0.8 22.0 124 36.6-172 0.8 20.6 117 34.4 + 69 0.8 8.3 47 13.8-140 0.8 16.8 95 28 56 28
Analysis by EFM Step f: Analysis results for live load Distribution of moments to slab and beam Length 25-0 (Ext) 25-0 (Int) Table: E-W Interior frame analysis (Interior Storey). Longitudinal moment section Column Strip Longitudinal Moment %age moments (LM) factor (Graph A4) Column Strip slab Moment (CSSM) = 0.15CSM Column strip Beam Moment (BM)= 0.85CSM Middle Strip slab Moment E - 73 0.93 10.2 58 5.11 + 105 0.8 12.6 71 21 I - 179 0.8 21.5 122 35.8-166 0.8 19.9 113 33.2 + 70 0.8 8.4 48 14-144 0.8 17.3 98 28.8 57 Analysis by EFM Step f: Analysis results for live load Analysis of columns for LL (factors for moment distribution) Interior column moment is due to pattern load. The computed unbalanced longitudinal moments shall be transferred to columns and shall be distributed to top and bottom columns as follows: Portion of unbalanced moment to upper column = Portion of unbalanced moment to lower column = K ct K cb + K ct K cb K cb + K ct 58 29
Analysis by EFM Step f: Analysis results for live load Analysis of columns for LL (factors for moment distribution) 56 151 73 151 43 151 K cb3 /(K cb3 =1.00 K ct2 /(K ct2 +K cb2 )=0.41 K cb2 /(K ct2 +K cb2 )=0.59 K ct1 /(K ct1 +K cb1 )=0.41 K cb1 /(K ct1 +K cb1 )=0.59 K cb3 =192E K ct2 =135E K cb2 =192E K ct1 =135E K cb1 =192E C2 C1 C1 C1 C2 59 Analysis by EFM 56 1.00=56 Step f: Analysis results for live load Analysis of columns for LL 151 1.00=151 K cb3 /(K cb3 =1.00 K cb3 =192E 73 0.41=30.00 151 0.41=62 73 0.59=43 151 0.59=89 K ct2 /(K ct2 +K cb2 )=0.41 K cb2 /(K ct2 +K cb2 )=0.59 K ct2 =135E K cb2 =192E 73 0.41=30.00 151 0.41=62 73 0.59=43 151 0.59=89 K ct1 /(K ct1 +K cb1 )=0.41 K cb1 /(K ct1 +K cb1 )=0.59 K ct1 =135E K cb1 =192E C2 C1 C1 C1 C2 60 30
Case Study: Comparison of the Results of Equivalent Frame Method, ACI Coefficient Method, Direct Design Method & SAP 2D Model with respect to SAP2000 3D Line Model 61 Case Study Dead Load Bending Moment in Beam 54 17.98 19.2 30 23 54 17.98 32 39 30 62 56.24 46 40 44 62 56.24 54 38 40 87 69.07 65 62 60 87 69.07 78 60 60 66 64.14 62 58 57 66 64.14 72 57 56 46 34.53 29 30 28 46 34.53 30.5 30 29 66 64.14 48 52 50 66 64.14 60 52 50 ACI Coefficient Method Direct Design Method Equivalent Frame Method by SAP 2D model SAP 2D Model SAP 3D Model C2 C1 C1 C1 C2 62 31
Case Study Dead Load Bending Moment Column 54 20 34 49 42 20 6 6.0 6.9 5.4 ACI Coefficient Method Direct Design Method Equivalent Frame Method by SAP 2D model SAP 2D Model SAP 3D Model 27 10 18.45 38 35 27 10 18.45 38 34 27 10 27 33 31 27 10 27 23 19 10 3 4.7 2.6 3.0 10 3 4.7 2.4 2.6 10 3 3.3 4.2 3.7 10 3 3.3 4.2 4.0 C2 C1 C1 C1 C2 63 Case Study Live Load Bending Moment in Beam 79 26 33 49 30 79 26 47 63 40 90 79 75 67 62 90 79 71 63 56 125 97 112 101 82 125 97 109 99 82 94 90 126 96 77 94 90 99 92 75 64 48 47 49 39 64 48 48 47 39 94 90 95 86 68 94 90 85 86 68 ACI Coefficient Method Direct Design Method Equivalent Frame Method by SAP 2D model SAP 2D Model SAP 3D Model C2 C1 C1 C1 C2 64 32
Case Study Live Load Moment in Columns 78.7 33 56 81 59 86 133 151 68 45 ACI Coefficient Method Direct Design Method Equivalent Frame Method by SAP 2D model SAP 2D Model SAP 3D Model 39.35 16.5 30 62 48 39.35 16.5 30 62 47 39.35 16.5 43 55 43 39.35 16.5 43 37 27 43 66.5 62 55 39 43 66.5 62 54 38 43 66.5 89 49 35 43 66.5 89 31 21 C2 C1 C1 C1 C2 65 Limit Analysis Introduction Most RC structures are designed using following approach: Moments, shears, and axial forces in RC structures are found by elastic theory. The actual proportioning of members is done by strength methods, in which inelastic section and member response is considered. Although this design approach is safe and conservative but is inconsistent to total analysis-design process. 66 33
Limit Analysis Redistribution of Moments A frame normally will not fail when the nominal moment capacity of just one critical section is reached: A plastic hinge will form at that section Large rotation at constant resisting moment will occur. Load transfer to other locations (having more capacity) along the span will occur. On further increase in load, additional plastic hinges may form at other locations along the span. As a result, structure will collapse, but only after a significant redistribution of moments. 67 Limit Analysis Redistribution of Moments Full use of the plastic capacity of reinforced concrete beams and frames requires an extensive analysis of all possible mechanisms and an investigation of rotation requirements and capacities at all proposed hinge locations. On the other hand, a restricted amount of redistribution of elastic moments can safely be made without complete analysis, yet may be sufficient to obtain most of the advantages of limit analysis. 68 34
Limit Analysis Negative Moments Redistribution in Continuous Flexural Members (ACI 6.6.5) A limited amount of redistribution is permitted by ACI Code 6.6.5. depending upon a rough measure of available ductility, without explicit calculation of rotation requirements and capacities. 69 Limit Analysis Negative Moments Redistribution in Continuous Flexural Members (ACI 6.6.5) The net tensile strain in the extreme tension steel at nominal strength ε t given in eq. below, is used as an indicator of rotation capacity. Accordingly, ACI Code 6.6.5 provides as follows: Except where approximate values for moments are used, it shall be permitted to increase or decrease negative moments calculated by elastic theory at supports of continuous flexural members for any assumed loading arrangement by not more than 1000ε t percent, with a maximum of 20 percent. ε t = ε u (d c)/c) 70 35
Limit Analysis Negative Moments Redistribution in Continuous Flexural Members (ACI 6.6.5) d A s ε u c ε t = ε u (d c)/c) ε u = 0.003 As example, for given A s if: d = 16.5 ; c = 4 ε t = 0.009 ε t 1000ε t = 9 % < 20 % 71 Limit Analysis Negative Moments Redistribution in Continuous Flexural Members (ACI 6.6.5) Graphical representation of ACI code provision 72 36
Limit Analysis Negative Moments Redistribution in Continuous Flexural Members (ACI 6.6.5) 6.6.5.1 The modified negative moments shall be used for calculating moments at sections within the spans. 6.6.5.1 Redistribution of negative moments shall be made only when ε t is equal to or greater than 0.0075 at the section at which moment is reduced. 73 Limit Analysis Negative Moments Redistribution (Example) For the beam shown, find moment redistribution. 74 37
Limit Analysis Negative Moments Redistribution (Example) Solution: To obtain maximum moments at all critical design sections. it is necessary to consider three alternative loadings. M max+, ext M max+, int M max-, int It will be assumed that 20 % adjustment of support moment is permitted throughout. 75 Limit Analysis Decrease in exterior positive moment: Negative moment increased 20 %, which results in decrease in M max+,ext from 109 to 101 Decrease in interior positive moment: Negative moment increased 20 %, which results in decrease in M max+,int from 72 to 57 Decrease in interior negative moment: Negative moment decreased 20 %, without exterior positive and interior positive moments exceeding M max+,ext and M max+,int respectively. 82+0.2 82 = 98 78+0.2 78 = 93 M max+, ext M max+, int M max-, int 134 0.2 134 = 107 76 38
Limit Analysis Conclusion on Redistribution of Moments It can be seen, then, that the net result is a reduction in design moments over the entire beam. This modification of moments does not mean a reduction in safety factor below that implied in code safety provisions; rather, it means a reduction of the excess strength that would otherwise be present in the structure because of the actual redistribution of moments that would occur before failure. It reflects the fact that the maximum design moments are obtained from alternative load patterns, which could not exist concurrently. The end result is a more realistic appraisal of the actual collapse load of the indeterminate structure. 77 Plastic Analysis Non-Linear Static (Pushover) Analysis Points on the structure whose performance (when it yields, cracks or fails) is required to be monitored are selected. The structure is pushed at the top. Top Drift (D) and corresponding base-shear (V) is calculated and plotted on V-Δ curve. Points of interest PUSH D Structure is further pushed in steps and V-Δ curve is plotted. Also performance of the selected points is monitored and marked on the V-Δ curve. Therefore a single chart that shows the performance of the whole structure (or separate charts for all points of interests) is obtained. These charts can be used to identify points where strengthening of structure is required (i.e points that fail or start to fail in the start of the curve. V V (base shear) Beam Crack Beam yield Cracking in Column Failure D 78 39
References ACI 318-14 PCA Notes on ACI 318 RC Design Teaching Aids by PCA Example by Dr. Qaisar Ali and Engr. Umer 79 The End 80 40