ADAPT-BUILDER FLOOR-Pro Module VERIFICATION

Transcription

1 Verification_Title.p65 MNL410 STRUCTURAL CONCRETE SOFTWARE SYSTEM ADAPT-BUILDER FLOOR-Pro Module VERIFICATION Dr. Bijan O. Aalami Professor Emeritus, San Fancisco State University Structural Engineer Redwood City, California Copyright 2003 Consulting Company Member POST-TENSIONING INSTITUTE web site Woodside Road, Suite 220, Redwood City, California, 94061, USA, Tel: (650) Fax (650)

2 ADAPT ADAPT-FLOOR PRO PREFACE This manual presents a number of typical problems, along with the verification of their solutions, using ADAPT-FLOOR Pro. The objective of the manual is to expose the accuracy of the typical solutions from the program, and also to illustrate the interpretation of the results and a method of their validation. The input data of each example is on the program s CD along with an electronic file of this manual in PDF format. i

3 ADAPT ADAPT-FLOOR PRO Verification_Man_TOC.doc LIST OF CONTENTS B_EX1: CANTILEVER BEAM FOR REBAR VERIFICATION...1 B_EX2: SIMPLY SUPPORTED T-SECTION...4 B_EX3: VERIFICATION OF HYPERSTATIC (SECONDARY) ACTIONS...9 B_EX4: OFFSET FEATURE OF FLOOR-PRO...11 B_EX5: SIMPLY SUPPORTED PLATE UNDER UNIFORM LOADING...15 B_EX6: SIMPLY SUPPORTED QUARTER PLATE UNDER UNIFORM LOADING...20 B_EX7: FIXED PLATE UNDER UNIFORM LOADING...24 B_EX8: A SIMPLE FLOOR SYSTEM FOR REBAR VERIFICATION...29 B_EX9: VERIFICATION OF HYPERSTATIC ACTIONS IN FLOOR SYSTEMS...32 ii

4 B_EX1 Verification_Man.doc 12/30/03 CANTILEVER BEAM FOR REBAR VERIFICATION DESCRIPTION This example verifies the structural analysis and reinforcement calculation of a cantilever Beam. The reinforcement is calculated using ACI code. The Beam is of a rectangular cross section. The values obtained from beam theory are compared with the values obtained from the FLOOR-Pro program. The two solutions are in full agreement. STRUCTURE The dimensions and loading of the cantilever Beam of uniform cross-section are shown in Fig. B_EX1-1. Other particulars of the specimen are: 28 days concrete strength f c = 35 MPa Applied load is a concentrated dead load at the cantilever tip, vertical loading at the tip = 5 kn Selfweight is disregarded Width and height of cross section are = 400x600 mm Yield strength of rebar = 400 MPa Modulus of elasticity of rebar = MPa The modulus of elasticity of concrete = MPa Length of cantilever Beam = 5 m Top and bottom cover to rebar = 40 mm The Beam is designed at eight sections along its length. The first cross section is 50 mm away from the fixed end and the 8 th cross section is 50 mm away from the tip. All the actions are calculated for a strength condition with load factor as follows: U = 1.2*D Where, U is the factored action and D is the dead load. The deflection is not factored (serviceability condition). 1

5 RESULTS The bending moment and rebar diagram are shown in Figs. B_EX1-2 and B_EX1-3, respectively. The bending moment and shear force at a section 50 mm away from the fixed end and tip deflection calculated by the program are shown in Table B_EX1-1. Hand calculated values, using simple beam theory, are also entered in the same table for comparison. The agreement is very good. FIGURE B_EX1-1 FIGURE B_EX1-2 FACTORED MOMENT DIAGRAM 2

6 FIGURE B_EX1-3 REINFORCEMENT REQUIREMENT FOR BENDING VERIFICATION For the verification, the cross section at 50 mm away from the fixed end is selected. Applied vertical load (V) = 5000 N Length of cantilever (L) = 5000 mm Moment of inertia (I)= 400*600 3 /12 = 7.20E9 mm 4 a) Factored bending moment (Mu) = 1.2*V*(L-50) =1.2*5000*( ) = 29,700,000 Nmm = knm (ADAPT-FLOOR Pro 29.7 knm, OK) b) Factored shear force (Vu) = 1.2*V = 1.2 *5000 = 6000 N or 6kN (ADAPT-FLOOR Pro 6 kn, OK) c) Tip deflection (d) = V*L 3 /3E c I = 5000*(5000) 3 /(3* *7.20E9) = 0.97 mm (ADAPT-FLOOR Pro 0.97 mm, OK) TABLE B_EX1-1: COMPARISON OF RESULTS Items FLOOR-Pro Theory Moment at a section 50 mm away from fixed end knm knm Shear force at support 6 kn 6 kn Tip deflection 0.97 mm 0.97 mm Rebar at a section 50 mm away from fixed end 151 mm 2 *150 mm 2 *With ADAPT-PULT 3

7 B_EX2 A and B DESCRIPTION SIMPLY SUPPORTED T-SECTION This example verifies the structural analysis and reinforcement calculation of a simply supported posttensioned flanged Beam with a T-section. The reinforcement is calculated using ACI code. The values obtained from beam theory are compared with the values obtained from the FLOOR-Pro program. The agreement is found to be very good. STRUCTURE The dimensions and loading of the simply supported posttensioned Beam are shown in Fig. B_EX2-1. The Tendon consists of 16 strands, each assumed to have a constant force of k along its length. Other particulars of the structure are: 28 days concrete strength f c = 4000 psi Dead load (including selfweight) = 1.61 k/ft Live load = 0.51 k/ft Length of the span = 62 ft Area of each strand = in 2 Poisson s ratio = 0.2 Top and bottom cover to rebar = 2 in Yield strength of rebar = 60 ksi Modulus of elasticity of rebar = ksi Modulus of elasticity of concrete = 3834 ksi The following two cases are compared with theoretical values. B_EX2A: Beam is supported at its ends only. B_EX2B: In addition to the supports at the two ends, the two free edges of the flange in direction of the span are clamped against rotation about the free edge. This case simulates the condition of a flanged Beam from an array of identical parallel Beams all loaded uniformly to the same value. The end supports are placed at the centroid of the entire section which is in from the bottom fiber. The Beam is designed at 21 equally spaced sections along its length. The first cross section is 0.6 in away from one end and 21 st cross-section is 0.6 in away from the other end. All the actions except stresses are calculated for strength 4

8 combination as follows: U service = 1*DL + 1*LL + 1*PT U strength = 1.2*DL + 1.6*LL + 1*HYP Where HYP is the hyperstatic (secondary) moment due to prestressing. In this case, HYP is zero, since the structure is statically determinate. The stresses are verified for serviceability combination, the reinforcement for strength combination. RESULTS The bending moment and rebar diagram for case B_EX2A are shown in Figs. B_EX2-2 and B_EX2-3, respectively. For both cases, the max bending moment, max shear force, top and bottom stresses and rebar at the mid-section calculated by the program are shown in Table B_EX2-1. These are also compared with the theory and they match very well. FIGURE B_EX2-1 5

9 0 Moment Diagrams Project: General name / Support Line 2 / Load Case: Strength 1.20 x Selfweight x Dead load x Live load x Hyperstatic Moment Drawn on Tension Side Moment [k-ft] Span 1 FIGURE B_EX2-2: BENDING MOMENT DIAGRAM FOR STRENGTH COMBINATION Rebar Diagrams Project: General name / Support Line 2 / Load Case: Strength 1.20 x Selfweight x Dead load x Live load x Hyperstatic 0.00 Rebar Required Top Rebar Required Bottom Rebar [in²] Span 1 FIGURE B_ EX2-3 REBAR DIAGRAM 6

10 Items TABLE B_EX2-1: NUMERICAL RESULTS FLOOR-Pro (B_EX2A) FLOOR-Pro (B_EX2B) Theory Total moment at midspan (Mu) k-ft k-ft k-ft Moment due to PT at midspan k-ft k-ft k-ft Max shear force (Vu) k k k Top stress at ksi ksi ksi midspan Bottom stress midspan 0.29 ksi 0.29 ksi 0.29 ksi Rebar at midspan 1.46 in in 2 *1.34in 2 *From ADAPT-PULT program HAND VERIFICATION The values at midspan are verified. Load due to dead and live (w) = = 2.12 k/ft Span (L) = 62 ft Eccentricity of the strand at midspan (e) = in Moment of inertia (I) = (204*5 3 /12)+(204*5* )+(14*25 3 /12)+(14*25* ) = in 4 Area of cross-section (A) = 204*5+14*25 = 1370 in 2 Distance to the top fiber from Distance to the bottom fiber from (y t )= 6.33 in (y b )= in a) Bending moment: Bending moment due to dead load=w D L 2 /8 = 1.61*(62) 2 /8 = k-ft Bending moment due to live load=w L L 2 /8 = 0.51*(62) 2 /8 = k-ft Bending moment for strength combination (M u ) = 1.2* * = k-ft (ADAPT-FLOOR-Pro k-ft, OK) Prestressing moment = 16*P*e = 16*30.13*20.91/12 = k-ft (ADAPT-FLOOR-Pro k-ft, OK) 7

11 Bending moment for service combination (M) = = k-ft b) Top stress f top = P/A + M*y t /I = 482.2/ *12*6.33/ = 0.52 ksi compression (ADAPT-FLOOR Pro 0.52 ksi, OK) c) Bottom stress f bot = P/A - M*y b /I = 482.2/ *12*23.67/ = 0.29 ksi tension (ADAPT-FLOOR Pro 0.29 ksi, OK) 8

12 B_EX3 VERIFICATION OF HYPERSTATIC (SECONDARY) ACTIONS DESCRIPTION This example verifies the correct extraction of hyperstatic (secondary) actions (moment and shear) due to prestressing in statically indeterminate structures. The secondary actions are used explicitly in the strength calculation of prestressed members. STRUCTURE The elevation and dimensions of the two-span Beam for the verification is shown in Fig. B_EX3-1. The Beam is posttensioned with a single Tendon having a constant force of 36 k along its length. Other particulars of the model are: Concrete strength f c = 4000 psi; Width and height of the section = 12 in x 24 in; Tendon profile: simple parabola; Length of each span = 30 ft and Loading: post-tensioning only. RESULTS The hyperstatic reactions on the Beam are shown in Fig. B_EX3-2. These are obtained from the Beam s hyperstatic shear illustrated in Fig. B_EX3-3. The hyperstatic actions are in self-equilibrium, as indicated in Fig. B_EX3-2. For verification, the hyperstatic actions calculated using an independent computer program ADAPT-ABI are listed in Table B_EX3-1 along with the values obtained from FLOOR-Pro for comparison. The agreement is excellent. ITEMS ADAPT-FLOOR ADAPT-ABI Pro Vertical reaction at A 0.60 k k Vertical reaction at B 1.20 k k Vertical reaction at C 0.60 k k 9

13 FIGURE B_EX3-1: TWO-SPAN BEAM WITH SIMPLE PARABOLIC TENDON (A) (B) (C) 0.60 k 1.20 k 0.60 k FIGURE B_EX3-2: HYPERSTATIC ACTIONS Forces Diagrams Project: General name / Support Line 5 / Load Case: Hyperstatic_com 1.00 x Hyperstatic Clockwise Shear Postive Force [k] Span 1 Span 2 FIGURE B_EX3-3 HYPERSTATIC SHEAR DUE TO PRESTRESSING (End support shear = 0.60 k; Center support = 1.2 k) Equilibrium check: Sum of vertical forces = = 0.0 k Sum of moments about point A = 0.60* *30 = 0.0 k-ft 10

14 B_EX4 A to C OFFSET FEATURE OF FLOOR-PRO DESCRIPTION This example verifies the correct implementation of the offset feature in the ADAPT-Floor Pro program. The offset feature is used in several types of applications. One is to shift the natural node of a member to another location. Another example is to have the ability to handle multilayer members, such as in composite members. The concept of offset in ADAPT-Floor Pro is based on a rigid connection between the natural node and its offset. The deformations follow the general premise of the program for plane sections remain plane. The values obtained from the FLOOR-Pro program are compared with beam theory. The agreement is found to be excellent. STRUCTURE A cantilever Beam with general dimensions, construction and loading as shown in Fig. B_EX4-1 is used. Other particulars of the model are: Span = 40 m; Overall dimensions of the cross-section: Width and depth = 400 x 1000 mm; Concrete strength f c = 35 MPa and Modulus of elasticity of concrete = MPa. Applied loading at the tip: Vertical load = 12 kn; Moment = 8 knm and Axial = 1400 kn. Three variations of this Beam are analyzed, each having a different make up of its cross-section, but with the same overall dimensions. These are: (A): One layer: (B): Two layers: = 400 x 1000 mm Upper layer = 400 x 400 mm Lower layer = 400 x 600 mm (C): Three layers: Upper layer = 400 x 200 mm Middle layer = 400 x 200 mm Lower layer = 400 x 600 mm The calculated values are verified for a section 50 mm from 11

15 the support. Moment, shear, axial and deflection are all calculated for service condition; therefore applied loads are not factored. RESULTS The actions and deflection obtained from the program along with the hand-calculated values are listed in Table B_EX4-1. The agreement is good for all cases. FIGURE B_EX4-1 12

16 FIGURE B_EX4-2 Items TABLE B_EX4-1: Comparison of the results FLOOR-Pro (A) FLOOR-Pro (B) FLOOR-Pro (C) Theory Moment at 50 mm from fixed end knm knm knm knm Shear Force at fixed end 12 kn 12 kn 12 kn 12 kn Axial force 1400 kn 1400 kn 1400 kn 1400 kn Tip deflection mm mm mm mm VERIFICATION Values are verified for a section 50 mm away from the fixed end. Moment of inertia (I)= 400* /12 = 3.333E10 mm 4 Modulus of elasticity of concrete (E c )=29910 MPa a) Bending moment = V*(L-50)+ M = 12*( )+ 8 = 487.4E6 Nmm or knm (ADAPT-Floor Pro knm, OK) b) Shear force = V = 12 KN (ADAPT-Floor Pro 12 kn, OK) c) Axial force = P 13

17 = 1400 = 1400 kn (ADAPT-Floor Pro 1400 kn, OK) d) Tip deflection (d) = V*L 3 /3E c I + ML 2 /2E c I = 12000*(40000) 3 /(3*29910*3.333E10)+ 8E6*(40000) 2 /(2*29910* 3.333E10) = mm (ADAPT-Floor Pro mm, OK) 14

18 B_EX5 SIMPLY SUPPORTED PLATE UNDER UNIFORM LOADING DESCRIPTION This example verifies the shell element formulation in ADAPT-FLOOR Pro. The values obtained from the program are compared with those in the literature. The agreement is found to be good. STRUCTURE The structure selected is a simply supported square Slab under uniform loading as shown in Fig. B_EX5-1. Other particulars of the Slab are: The length and width of the Slab = 10m x 10m Slab thickness = 400 mm Concrete strength f c = 35 MPa Poisson s ratio = 0.3 Shear modulus = MPa Modulus of elasticity of concrete = MPa Applied dead loading over the entire plate = 10 kn/m 2. RESULTS The vertical deflection, Myy moment intensity, top stress and principal stress contours are shown in Fig. B_EX5-2 to B_EX5-5, respectively. The bending moment, top/bottom stresses, principal stresses and central deflection calculated by the program are listed in Table B_EX5-1. These are also compared with the theoretical values given in reference listed at the end of this example. The agreement is good. 15

19 Y FIGURE B_EX5-1: PLAN OF SIMPLY SUPPORTED SLAB X FIGURE B_ EX5-2: CONTOUR OF VERTICAL DEFLECTION 16

20 FIGURE B_ EX5-3: CONTOUR OF MOMENT Myy INTENSITY FIGURE B_ EX5-4: CONTOUR OF TOP FIBER STRESS ALONG X-X 17

21 FIGURE B_ EX5-5: CONTOUR OF MAXIMUM PRINCIPAL TOP FIBER STRESSES TABLE B_EX5-1: Comparison of the results Items FLOOR-Pro Ref.1 Maximum vertical deflection 2.31 mm 2.31 mm Mxx or Myy at midspan kn 47.9 kn Top stress at center along X-X Bot stress at center along Y-Y MPa 1.79 MPa MPa 1.79 MPa Principal max stress at center Principal min stress at center 1.79 MPa 1.79 MPa 1.79 MPa 1.79 MPa Twisting moment at corner kn 32.5 kn HAND VERIFICATION Applied uniform vertical load (q) = 10 kn/m 2 = 0.01 N/mm 2 Thickness of the plate (h) = 400 mm Poisson s ratio (ν) = 0.3 Modulus of elasticity of concrete (E c ) = MPa Length of the square plate (a) = 10 m = 10,000 mm From the reference at the end of this example: a) Max deflection = *qa 4 /D Where, D = E c h 3 /12(1-ν 2 ) 18

22 = 29910*400 3 /12( ) = 1.75E11 Nmm Deflection = *0.01* /1.75E11 = 2.31 mm (ADAPT 2.31 mm, OK) b) Max M yy = *qa 2 (from Reference 1) = *0.01* = 47,900 N = 47.9 kn (ADAPT 47.6 kn, OK) c) Top stress at center = 6*M yy /h 2 = 6*47.9*1000/400 2 = 1.79 MPa (ADAPT 1.79 MPa, OK) d) Principal max stress at center = the same theoretical value (ADAPT 1.79 MPa, OK) e) Twisting moment at corner = *qa 2 = *0.01* = 32,500 N = 32.5 kn (ADAPT 31.9 kn, OK) REFERENCE 1- Tomoshenko, S.P. and Woinowsky-Krieger, S., Theory of Plates and Shells, 2 nd Edition, Engineering Societies Monographs, McGraw-Hill Book Company,

23 B_EX6 SIMPLY SUPPORTED QUARTER PLATE UNDER UNIFORM LOADING DESCRIPTION This example verifies the application of restraints to shell elements used in ADAPT-FLOOR Pro. In addition, the example illustrates the application of symmetry in modeling of a structure, and hence the reduction of computational efforts. The values obtained from the program are compared with those given in the literature. Good agreement is obtained. STRUCTURE Taking advantage of the biaxial symmetry of the quarter Slab of Example B_EX5A, this example considers a quarter of the square plate as shown in Fig. B_EX6-1. The two symmetrical edges of the Slab are each restrained against rotation about their line of symmetry, but are otherwise free. The rotational fixity simulates the line of symmetry. Other particulars of the full Slab are as listed below. In the current example, only one-quarter of the plan dimensions and loading apply: The length and width of the Slab = 10 m x 10 m Slab thickness = 400 mm Concrete strength f c = 35MPa Poisson s ratio = 0.3 Shear modulus = MPa Modulus of elasticity of concrete = MPa Applied dead loading over the entire plate = 10 kn/m 2 RESULTS The solution obtained by the program is shown in Figs. B_EX6-2 through B_EX6-4. The principal values of the solution are listed in Table EX6-1 along with the corresponding values from Example B_EX5. The two solutions agree. 20

24 Y FIGURE B_EX6-1 SIMPLY SUPPORTED QUARTER PLATE X FIGURE B_ EX6-2 CONTOUR OF VERTICAL DEFLECTION 21

25 FIGURE B_ EX6-3 CONTOUR OF MOMENT Myy FIGURE B_ EX6-4 CONTOUR OF TOP FIBER STRESS ALONG X-X 22

26 TABLE B_EX6-1: Comparison of the results Items FLOOR-Pro FLOOR-Pro Example B_EX5 Maximum vertical deflection 2.31 mm 2.31 mm Mxx or Myy at midspan kn kn Top stress at center along X-X Bot stress at center along Y-Y MPa 1.79 MPa MPa 1.79 MPa Principal max stress at center Principal min stress at center 1.79 MPa 1.79 MPa 1.79 MPa 1.79 MPa Twisting moment at corner kn kn 23

27 B_EX7 FIXED PLATE UNDER UNIFORM LOADING DESCRIPTION This example verifies the application of fixed boundary conditions (restraints) used for the shell element formulation in ADAPT-FLOOR Pro. The values obtained from the program are compared with the literature. The agreement is good. STRUCTURE The structure is a square Slab of uniform thickness fixed along all its four edges against rotation and vertical displacement. (Fig. B_EX7-1). The Slab is under a uniformly distributed load over its entire surface. Other particulars of the Slab are: The length and width of the Slab = 10m x 10m Slab thickness = 400 mm Concrete strength f c = 35MPa Poisson s ratio = 0.3 Shear modulus = MPa Modulus of elasticity of concrete = MPa Applied dead loading over the entire plate = 10 kn/m 2 RESULTS The calculated displacement and actions are shown in Figs. B_EX7-2 to B_EX7-5. The principal values of the problems are extracted from the solution and listed in TABLE B_EX7-1 along with the corresponding values calculated using the reference at the end of the example. The agreement between the two solutions is good. 24

28 Y X FIGURE B_EX7-1 SQUARE PLATE WITH FIXED BOUNDARIES FIGURE B_ EX7-2 CONTOUR OF VERTICAL DEFLECTION 25

29 FIGURE B_ EX7-3 CONTOUR OF Myy FIGURE B_ EX7-4 CONTOUR OF TOP STRESS ALONG X-X 26

30 FIGURE B_ EX7-5 CONTOUR OF TOP FIBER MAXIMUM PRINCIPAL STRESS TABLE B_EX7-1: Comparison of the results Items FLOOR-Pro Ref. 1 Maximum vertical deflection 0.72 mm 0.72 mm Mxx or Myy at midspan kn 23.1 kn Top stress at center along X-X Bot stress at center along Y-Y MPa 0.86 MPa MPa 0.86 MPa Principal max stress at center 0.86 MPa 0.86 MPa HAND VERIFICATION Applied uniform vertical load (q) = 10 kn/m 2 Thickness of the plate (h) = 400 mm Poisson s ratio (ν) = 0.3 Modulus of elasticity of concrete (E c ) = MPa Length of the square plate (a) = 10 m From the reference at the end of this example: a) Max deflection = *qa 4 /D Where, D = E c h 3 /12(1-ν 2 ) = 29910*400 3 /12( ) = 1.75E11 Nmm 27

31 Deflection = *0.01* /1.75E11 = 0.72 mm (ADAPT 0.72 mm, OK) b) Max M yy = *qa 2 = *0.01* = 23.1 kn (ADAPT 22.8 kn, OK) c) Top stress at center = 6*M yy /h 2 =6*23.1*1000/400 2 = 0.86 MPa (ADAPT 0.86 MPa, OK) d) Principal max stress at center due to symmetry of the problem is the same as problem = 0.86 MPa (ADAPT 0.86 MPa) REFERENCE 1- Tomoshenko, S.P. and Woinowsky-Krieger, S., Theory of Plates and Shells, 2 nd Edition, Engineering Societies Monographs, McGraw-Hill Book Company,

32 B_EX8 A SIMPLE FLOOR SYSTEM FOR REBAR VERIFICATION DESCRIPTION This example verifies the structural analysis and reinforcement calculation of a simple floor system. The reinforcement is calculated using ACI-318 code. The calculated reinforcement from ADAPT-FLOOR Pro is compared with an independent program ADAPT-PULT. Good agreement is obtained. STRUCTURE The example is a rectangular Slab resting on a Wall and two Columns as shown in Fig. B_EX8-1. Other particulars of the specimen are: Slab thickness uniform = 14 in Concrete strength f c = 4000 psi Modulus of elasticity of concrete = 3834 ksi Shear Modulus = 1598 MPa Poisson s ratio = 0.2 Modulus of elasticity of rebar = ksi Yield strength of rebar = 60 ksi Top and bottom cover to rebar = 2 in Selfweight of 150 pcf Dead load of 0.02 k/sf Live load of 0.04 k/sf The Slab is considered to have two support lines along the X-direction, each passing over one of the Columns and the tip of the Wall (see Fig. B_EX8-1). RESULTS The bending moment and reinforcement for the design strip highlighted in Fig. B_EX8-2 are shown in Figs. B_EX8-3 and B_EX8-4. The rebar for a central design section of the design strip marked in Fig. B_EX8-2 is calculated using ADAPT-PULT. The results are given in Table B_EX8-1. The agreement is good. 29

33 FIGURE B_EX8-1 PLAN OF THE FLOOR Rebar for this section is verified with ADAPT-PULT FIGURE B_EX8-2 DESIGN STRIPS AND DESIGN SECTIONS IN X-DIRECTION 30

34 Moment Diagrams Project: General name / Support Line 4 / Load Case: Strength 1.20 x Selfweight x Dead load x Live load x Hyperstatic Moment Drawn on Tension Side Moment [k-ft] Span 1 Span 2 Span 3 FIGURE B_EX8-3 BENDING MOMENT OF THE LOWER HALF DESIGN STRIP IN X-DIRECTION Rebar Diagrams Project: General name / Support Line 4 / Load Case: Strength 1.20 x Selfweight x Dead load x Live load x Hyperstatic Rebar Required Top Rebar Required Bottom Rebar [in²] Span 1 Span 2 FIGURE B_EX8-4 REBAR DIAGRAMS TABLE B_EX8-1: Comparison of the results Items FLOOR-Pro ADAPT-PULT Moment on midspan (M u ) k-ft k-ft Rebar 7.66 in in 2 31

35 B_EX9 VERIFICATION OF HYPERSTATIC ACTIONS IN FLOOR SYSTEMS DESCRIPTION This example verifies the correct extraction of hyperstatic (secondary) actions (moment and shear) in floor systems due to prestressing. The hyperstatic actions are used in the strength design of prestressed members. STRUCTURE The floor system selected is a uniform thickness concrete Slab resting on eight square perimeter Columns and a central round Column (Fig. B_EX9-1). Other features of the Slab include a Drop Panel over the central Column and an Opening. The Tendon layout is displayed in Fig. B_EX9-2. The Tendons are banded along the gridlines 1,2,3 and are distributed uniformly in the transverse direction. All Columns are released against rotation at the connection to the Slab except the central Column (Column 5). Other particulars of the model are: Concrete strength f c = 27 MPa Modulus of elasticity of concrete = MPa Poisson s ratio = 0.2 Shear Modulus = MPa Effective stress of Tendon, fse = 1400 MPa Strand area = 98 mm 2 The loading is limited to prestressing forces. RESULTS The reactions at the supports are due to prestressing only (Fig. B_EX9-3). These are the hyperstatic actions that are shown in Hyperstatic reactions. The following verifies that they are in self-equilibrium. 32

36 FIGURE B_EX9-1 PLAN OF STRUCTURE 33

37 FIGURE B_EX9-2 TENDON LAYOUT FIGURE B_EX9-3 VIEW OF TENDONS 34

38 FIGURE B_EX9-4 PLAN OF SLAB AND COLUMN DESIGNATION 35

39 LOWER COLUMNS CPrestressing ID Label Fz Fr Fs Mrr Mss Mzz kn kn kn kn-m kn-m kn-m 1 Column Column Column Column Column Column Column Column Column Equilibrium Check: Sum of vertical forces = = kn Vertical forces are in equilibrium. Sum of moments along A = ( )* ( )* = 0 knm Mxx moments are in equilibrium. 36