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1. Asl n 16 4 cm2 Manual check z 0 9d 0 9 65 58 5 cm 0 0 5 sin or 0 5 sin ee a o 28 3 0 20 1 fa 250 fapbz 0 20 1 25 250 25 350 585 V1 0 6 1 fe 250 0 6 1 25 250 0 54 VRd max Gcw bw Z Vifeg cot 1 cot 6 1 35 58 5 0 54 1 67 cot 28 3 1 cot 28 3 770 KN Vea 770 100 cm Awan a a aa se ee Ny Z cotO fya 2 58 5 cot 28 3 43 5 m md m12 13 9 cm lt 0 75d 48 75 cm 4A 4 815 Selected 12 at spacing s Page 14 of 17 N RC Expert EC v 2 7 2014 Design of reinforced concrete members to EN 1992 1 1 MPOEKTCOOT User Manual Example 2 Bending design of rectangular cross section Solution with RC Expert Cross section b 30 0 cm h 55 0 cm b f 0 0 cm hr 0 0 cm bi 0 0 cm hr 0 0 cm d 5 0cm d 5 0 cm Internal forces Bending moment Mea 200 0 kN m Axial force Nea 0 0 kN Shear force Veg 0 0 kN Torsional moment Tea 0 0 kKN m Long term load factor KG 75 0 Te 16 0 MPa f ctk 0 05 1 3 MPa Concrete grade C16 20 Tc 10 7 MPa faa 0 9 MPa Main reinforcement grade B500 E 200 0 GPa fyk 500 0 MPa fya 435 0 MPa Input data Strength Reduction Factors Existing Reinforcement Bending with axial force design results Manual verification Mea 20000 Med bd2nf 30 502 1 1 33 0 2817 0 8872 E694 3 5 o0 Es1q 8 92 0 05 43 5kKN cm Mra 20000 Ag
2. Press the Results button in order to start the calculations The following results are provided Asi As2 areas of top and bottom reinforcement Hy Ha reinforcement ratios Os1 Os2 stresses in bottom and top reinforcement x compression zone height Page 6 of 17 N RC Expert EC v 2 7 2014 Design of reinforced concrete members to EN 1992 1 1 MPOEKTCOOT User Manual Theoretical background Og e Concrete tension stress is excluded and all tension is taken by en ee i i reinforcement e Parabolic linear stress strain relationship with maximum value of fog iS Jed used for calculation of concrete design stress e Linear plastic stress strain relationship with maximum value of fyg is Ee 0 Ec Ecu Oh fydalykl used for calculation of steel design stress e Bond exists between steel and concrete e Bernoulli hypothesis for planarity of sections is assumed strain diagram is linear along section height e Section capacity is reached when tensile strain in reinforcement and or compressive strain in concrete reach their ultimate values e Section forces due to external loads are in equilibrium with internal forces resulting from concrete and reinforcement stresses Jya Es Eud Euk Initial imperfections nominal eccentricity and second order effects When the Column checkbox is selected initial imperfections as included as defined in equation 5 2 to EN 199
3. 25730 33 0 asjas cars The following symbols are used in the table above C30 37 33 0 Ecm concrete secant modulus of elasticity fek cube Characteristic cube strength Sek characteristic cylinder strength Sotk 0 05 characteristic tensile strength with 5 probability of failure ted Mec fek Ye design compressive strength feta Act fetk 0 05 Yc design tensile strength Ec compressive strain at maximum stress for parabolic linear stress strain Ecy2 ultimate compressive strain at concrete edge Design values for compressive and tensile strengths in the table are determined for partial safety factor ye 1 5 They still do not include a and ag factors which should be defined additionally Some countries use Occ 0 85 and aet is usually equal to 1 0 You should look for these values in your national annex document Reinforcement steel grades Design Expert include the following steel grades Name Es fya yk an The following symbols are used in the table y ida E design modulus of elasticity fya design yield strength yk characteristic yield strength Page 5 of 17 N RC Expert EC v 2 7 2014 Design of reinforced concrete members to EN 1992 1 1 MPOEKTCOOT User Manual Eyd design ultimate strain Design to Eurocode 2 Bending with axial force design Dew a ee New Open Save Results Help Quit Site Building Section And Materials Bending And Axi
4. N Ac fea normalized axial force A Lgl slenderness ratio i lee radius of gyration fed design value of concrete compressive strength fek characteristic compressive cylinder strength of concrete at 28 days Ac area of cross section A second moment of area of concrete section I second moment of area of reinforcement about the center of area of the cross section Ecd Ecm design value of the modulus of elasticity of concrete yc 1 3 partial safety factor for concrete E design value of the modulus of elasticity of reinforcement Calculation of reinforcement The design procedure is developed for sections with arbitrary shapes An iterative algorithm is applied for calculation of reinforcement The unknown parameters in this procedure are listed below As area of tensile reinforcement As2 area of compressive reinforcement Es1 Strain at bottom reinforcement Ec Strain at the most compressed edge of the concrete or Es Strain at top reinforcement for sections entirely in tension The unknown parameters are obtained by the equilibrium of the internal forces Neg Nc Nsa Ns2 0 1 Meg Mec Ms Ms2 O 2 Meg and Neg are the design values of internal forces due to external loads The resultant forces in concrete and reinforcement and the respective moments about the center of area are calculated as follows X X Ne fod be a Mc Neze ce
5. 10 37 cm st do 0 8872 50 43 5 om 0 2 Page 15 of 17 N RC Expert EC v 2 7 2014 Design of reinforced concrete members to EN 1992 1 1 MPOEKTCOQO User Manual Example 3 Bending with axial force design of symmetrically reinforced section Solution with RC Expert Cross section b 30 0 cm h 50 0 cm ba 0 0 cm ha 0 0 cm b f2 0 0 cm hi 0 0 cm d 5 0 cm d2 5 0 cm Internal forces Moment Meg 119 7 kN m Axial force Nea 199 5 kN Long term load factor KG 75 0 E 31 5 GPa Concrete grade C25 30 fca 16 7 MPa Main reinforcement grade B500 E 200 0 GPa f 500 0 MPa Shear reinforcement grade B500 Eyw 200 0 GPa fyw 500 0 MPa Input data Strength Reduction Factors Existing Reinforcement alpha_cc 0 85 Bottom Bending with axial force design results 2 i2 0 3 one 233 5 MPa Compr zone heightx 92cm FO 7 Manual verification o Mea _ 1197 10 a4 Med bh2n fg 300 5002 11 33 N 199 5 103 te 0 117 Ed banfg 300 500 11 33 Read from fig 15 5 the value of Otot 0 225 bh 300 500 11 33 Med _ 9 995 300 900 1138 a 435 As tot Owi 879 mm 8 8 cm Page 16 of 17 N RC Expert EC v 2 7 2014 Design of reinforced concrete members to EN 1992 1 1 MPOEKTCOOT User Manual Example 4 Interaction curves for bending with axial force Interaction curves for bending with axial force are generated using RC Expert The section
6. at bottom edge reaches zero Case 3 The section is entirely in compression so the strain diagram rotates around point C where the strain always remain amp 2 The location of point C from the top edge is determined by the distance a h 1 Ec2 Ecu2 That is how the number of unknown parameters is reduced due to the condition that the section always should remain in limit state We have to search either for 1 or and obtain the other from the strain diagram If we know amp 1 and e we can calculate the areas of top and bottom reinforcement directly For each Page 9 of 17 2 Z1 N RC Expert EC v 2 7 2014 Design of reinforced concrete members to EN 1992 1 1 MPOEKTCOOT User Manual reinforcement we can use the moment equilibrium around the center of the opposite reinforcement to get the equations Agi Nea z2 Meg Ne z2 2c 0s1 Zs 3 As2 Nea z Mea Ne Z1 Ze 0s2 Zs 4 The distance between centers of top and bottom reinforcement is assumed to be Z Z1 Z2 If the section is entirely in tension the above equations give the final values of both reinforcement areas having N 0 and Os1 Os2 fyd In case of bending with axial force and non symmetric reinforcement we have 2 equations and 3 unknown parameters There are unlimited number of solutions that satisfy equations 1 and 2 Then the values of As and Asz are obtained by solving the optimization problem for minimi
7. axial force design Effective lengths are entered by the user The program is quick and easy to use with rich capabilities and friendly graphical user interface Input data and results are presented in professional looking HTML report for viewing and printing Data input Input data is divided into several pages You can browse among pages consequently by clicking the respective tabs When you finish entering the input data click the Results button If current file is not saved you will be prompted to save Input data is entered in tables and text fields on each page You can move to the next field using the Tab key Moving back to the previous field is performed by the Shift Tab key combination Page 2 of 17 N RC Expert EC v 2 7 2014 Design of reinforced concrete members to EN 1992 1 1 MPOEKTCOOT User Manual Files RC Expert has its own file format and the data for each problem can be saved to a file on the disk Input files have rce extension while the design results are recorded in rce html files New file Click the LJ button to save the current data to a new file A standard new file dialog appears Select or write down file path and name and click Save Open a file Click the gt button A standard file selection dialog appears Select or write down file path and name and click Open Save a file Click the l button A standard save file dialog appears Select or write down file path and name If file al
8. is rectangular with symmetric reinforcement As1 ini As2 ini and d h 0 1 The curves are compared with those provided in the manual on figure 15 5 Rectangular cross section b 25 0cm h 40 0cm d 40cm d 4 0 cm Concrete grade C20 25 fek 20 0 MPa feg 13 3 MPa Ecm 30 GPa Steel grade B500 fyk 500 MPa fyq 435 MPa E 200 GPa Results in absolute coordinates M N As1 0 00 cm2 0 0 Ac 6 55 cm w 0 5 As1 13 1 cm 1 0 Ag4 19 6 cm w 1 5 As1 26 2 cm 2 0 Results in the manual B500 d h 0 100 St ERRANS 2 20 D SNAS E SIRNA ARS 1 80 PRNSRORAIN IAS OAR 1 FRR RRQROQLNH a a ii z EE a TA T T T a T T TOO TT 140 aTa WATA TA TR ANS AN Me TOT TD NASA ARS S 1 00 KS N 0 80 0 60 0 40 0 20 0 60 i 0 80 1 00 5 Nine E f 1620 91 4 1 40 ee 1 60 SLL ae M 1 80 c c 198 21 2 00 ee 220 Nfyalfe Page 17 of 17
9. point in the M N coordinate system By connecting all points consequentially a closed curve is obtained We can put external moments and axial forces as points in the same coordinate system If a point falls outside the interaction curve then section capacity is not sufficient for the applied loads If all points are located inside then the section has sufficient capacity The reinforcement is determined so that the point of the external load forces is located right on the contour of the diagram Page 10 of 17 N RC Expert EC v 2 7 2014 Design of reinforced concrete members to EN 1992 1 1 MPOEKTCOOT User Manual Shear force design Dew a e New Open Save Results Help Quit Site Problem Building Internal Forces Bending moment M Axial force N Shear Force Design Q Shear reinforcement cuts nw 2 Shear reinforcement diameter d 8 v mm Ld Critical crack projection c 0 cm _ Compression struts angle amp 21 8 Shear links andge Si Tensile reinforcement area Asl 2 cm2 Concrete only shear resistance VRd c 60 16 kN Concrete maximum resistance VRd max Shear reinforcement area Asw 226 04 kN 1 40 cm2 m Required shear reinforcement 8 27 0 cm Shear reinforcement ratio w 0 11 Additional tension reinforcement Asl n 2 9 cm2 T 0 kNm V 100 kN Long tem load factor Torsion Design Cover to main reinforcement centers aw 3 5 cm Torsion is distributed among _ Top
10. the reinforcement obtained from bending and shear design Calculations for each rectangular part are performed as described further Concrete cover for torsion dy should be specified from the concrete edge to the center of main reinforcement Shear force is included in calculation of torsional stresses The following checks are performed Requirement for shear reinforcement Teg Tract Vea VRa c gt 1 6 29 Capacity of concrete compressive struts TRd max VA cfed Ak tefsin 2 0 6 30 Tta T is the design torsion force Values for Veg VRa c and Vag max are according to the Shear force design chapter above in this manual The calculation of the reinforcement is performed as follows Page 12 of 17 N RC Expert EC v 2 7 2014 Design of reinforced concrete members to EN 1992 1 1 MPOEKTCOOT User Manual Shear reinforcement Acsyw P fywa cot 0 Main reinforcement Astot P U cot fyd 6 28 Effective section width for torsion tep b h 2 b 2 h fef S 2 dy Effective section area A b tef A tes Perimeter of the effective area Uk 2 b h 2 te P Tga 2 Ax Results are provided separately for each section part Report You can generate a detailed report in HTML format for each problem by clicking the Results Ei button The report is opened in Internet Explorer by default but you can use other web browsers as well Most office programs like MS Word can also edit html files The report is saved
11. 2 1 1 They are accounted as for single element in the form of additional eccentricity eji O Lo 2 where Le effective member length O Oo ah am member inclination 1 200 basic value of inclination a 2 VL height reduction factor L is the geometrical height of the member in meters m i reduction factor for number of members In case of compression second order effects are included using nominal stiffness method The design value of the bending moment is determined by equation 5 30 Mora Meg _ Ed 1 N Nz Mogg M N e is the first order bending moment including initial imperfections The value of eccentricity e must be not be less than e h 30 2 20 mm according to EN 1992 1 1 section 6 1 4 h cross section height N design axial force due to external load Ng critical compressive force It is calculated to eq 5 17 Ng EI Lo EI Ko Egle Ks Es ls nominal stiffness obtained by eq 5 21 Page 7 of 17 N RC Expert EC v 2 7 2014 Design of reinforced concrete members to EN 1992 1 1 MPOEKTCOOT User Manual K 1 factor that accounts for the contribution of the reinforcement Ke ky ko 1 Pet factor that accounts for effects of cracking creep shrinkage etc Pef P to Moegp Moed effective creep factor ki Vf ck 20 factor that depends on concrete grade k n A 170 lt 0 20 factor that depends on axial force and slenderness n
12. N NPOEKTCOOT Software Package Design Expert version 2 7 Structural Design and Detailing to Eurocode RC Expert Design of reinforced concrete members to EN 1992 1 1 User Manual Section And Materials Bending And Axial Force Shear And Torsion Internal Forces Bending moment M 0 0 kN Axial force N 0 0 kNm Long tem load factor Materials Cross Section Dimensions cm E Concrete fcd 13 3 MPa All rights reserved 2014 N RC Expert EC v 2 7 2014 Design of reinforced concrete members to EN 1992 1 1 MPOEKTCOOT User Manual TABLE OF CONTENTS PUDOUE TING Proca eroaan E NE EE ET E AA 2 Data ea EE E T 2 PINGS E EE E E A E E E see 3 IOUT TING T EEE E EA A N EEEN EEA NE E A S E 3 Bjo aE TN NA AE A A A EE A E E E OEE E A 3 Save TUS oor o E A A E 3 Dece COIE a EEE EEEE aaia 3 Merna TOC OS a EE E E 3 Sections and IW ALORS ese ccice case aecavsterencesoaucesevabsanencdtucusenteysesndstonwsusacerneane eases Ee 4 FOSS SOCION srini eves verepestsccdsenevactengbeectuvesnasuseresisiuveasacasvestacdueeshesusecimabesresvectuvebuastetseseveveves 4 IV RCSA T E E E ai vaca se cette sete se A A O E A E E E E N 4 MTO a E E E E E 4 Materials according to EWOCOCG ecdecwiecnscnctnecsannacsedocactedtewaevasenssancononneteesddeesanmaessinumeesennnsaseuseieaunteots 5 Design t Eu urocode 2 ois sce sac atccpcsecnt enc cuesances copauneenen
13. al Force Shear And Torsion Internal Forces Bending moment M 100 kNm 100 kN Long tem load factor Axial force N 200 kN compress 0 kNm Bending And Axial Force S reduction factors C Column 0 cm Existing reinforcement alpha_cc 0 850 Buckling lengths Bottom As1 ini 0 cm2 alpha_ct 1 000 in plane of bending x 1 0 L Hl Top As2 ini 0 cm2 L _ Symmetric reinforcement out plane of bending Loy 10 L El M N Interaction Curve 1500 N kN A C Seismic loads Creep factor 35 Ratio Stress MPa Hi 0 64 1 435 00 u2 0 10 2 435 00 E tBucking y Astt 0 00 Compression zone height x Bending and axial force design completed successfully n 64 The design procedure can handle all kinds of bending and axial force with symmetric and non symmetric reinforcement Axial force can be tensile or compressive Factors acc and act should be defined additionally by user Nominal eccentricity initial imperfections and second order effects can be taken into account Select the Column checkbox and enter member length L and effective length factors for in plane Lo and out of plane Loy bending In case of non symmetric reinforcement you can perform additional reinforcement optimization using the Existing reinforcement option You can enter the value of Ag jni equal to As from previous combination with opposite direction of bending and take its favorable influence upon Ag
14. caled preview of the section Shape can be rectangular T or I To obtain a rectangular section enter flange dimensions to be zeroes Concrete cover is defined as the distance from the edge of concrete to the center of reinforcement Materials Select concrete grade and steel grades for main and shear reinforcement Characteristic and design values for material properties are predefined in tables Concrete compressive and tensile strengths are additionally multiplied by the sustained loading factors Qcc and ag They should be defined separately in the respective fields since they are not included in the table values Material tables You can open material tables by clicking the button A dialog containing both concrete and reinforcement tables appears on screen You can modify values add new rows by clicking the button and remove rows using the button Finally you should press Save to save changes and close the dialog If you want to discard changes just press Exit and you will return to the main window Material tables are common for the whole computer Any changes you make will reflect all Design Expert modules and input files Page 4 of 17 N RC Expert EC v 2 7 2014 Design of reinforced concrete members to EN 1992 1 1 MPOEKTCOOT User Manual Materials according to Eurocode Concrete Design Expert includes the following concrete grades according to EN 1992 1 1 Table 3 1 GPa MPa MPa MPa MPa MPa
15. cetsanectanisiwsacnes seutuneutetsenvenedtaeececteseacenehoutenscstenes 6 Bending with axial force GESIgM ccscsccccscscecsccscsceccccscnceccececscsceccecscnceccccecececeececscesescecscsceseececscesess 6 REVO GET allt aC SIO INO seis etictscur ars cnte artes wncace aus pouausc onaataica N 7 Initial imperfections nominal eccentricity and second Order effects ccccccesssccccesseceeeeseceeeeeeceeeeees 7 CACM UO of reinforcement enneren EE EEE EENE ENa 8 ea Coa E EEE EAEE A 10 Shear force COSION iarcronerie uann ann EE ESEE 11 TOSIN AES IE unra E S 12 REDOT assin a E aeclaeeuoesouesausueselsue 13 EAU S oaa a E E sacs 13 Example 1 Shear force GeSIgin scscecsccscscsccccscscsccccecscsccccecscsceececscnceceececscesescecscsceseececscesesececssess 14 Example 2 Bending design of rectangular cross section sccscscsccscscsceccccscsccccccscsceccccscsceseececscssess 15 Example 3 Bending with axial force design of symmetrically reinforced section sesessesessesessssese 16 Example 4 Interaction curves for bending with axial fOrce csccscscscsscscscecsccscsceccccscnceseccecscscess 17 About the program RC Expert is a software for design of reinforced concrete members with rectangular T and I cross sections to Eurocode EC2 Bending with axial force shear and torsion design are performed Initial imperfections nominal eccentricity and second order effects are taken into account for bending with
16. ement kN m dsw Veg z cot cota sina Area of shear reinforcement cm m Asw 4sw nw fywd Additional tensile force in the main reinforcement Fia 0 5 Veg cot0 cota Veg is the design value of the shear force in the section Additional main reinforcement is provided for the additional tensile force In most programs it is accepted to cover the diagram of the required reinforcement instead of the diagram of the tension force This is the case with Design Expert as well Then the additional area of main reinforcement is calculated as follows As n Fralfyd It should be added to the main reinforcement determined by bending design The quantity of the calculated shear reinforcement depends on the angle of the compressive strut The value of 8 can be defined by user within the limits of 21 8 lt 0 lt 45 1 lt cot lt 2 5 If you select the respective checkbox the custom defined value is used in the calculations Otherwise the program will calculate the angle from the condition Veg max Veg Link spacing is determined according to the selected diameter by the equation s tdyw7 4Asw lt Smax 0 75d 1 cota Torsion design For and T sections all section parts web and flanges can be considered in torsion design as rectangular segments Total torsional moment is distributed among the different parts proportionate to their torsional stiffness Main and shear reinforcement is calculated additional to
17. flange V Web _ Bottom flange Z Reals cm m cm MPa Design check can performed with vertical or inclined shear links no bended bars Additional input data is required Nw number of shear link legs dw shear links diameter 0 angle of compressed concrete strut a angle of shear links Ag area of tensile reinforcement cm with sufficient anchoring after the section of interest Calculations are performed according to the procedure defined in EN 1992 1 1 Shear capacity without shear reinforcement is calculated Vrac Crack 100 pr fek k1Ocp bw d VRd c Z VRd c min Vmin k1 Ccp bw d k 1 4 200 d lt 2 CRa c 0 18 yc Ocp Ned Ac lt 0 2 fea k 0 15 p As bw d lt 0 02 Maximum shear capacity of compressed concrete strut Vod max amp cw bw z V1i fea cotO cota 1 cot 0 eq 6 14 in EN 1992 1 1 If Veg gt VRd max then concrete capacity is not sufficient for the external shear force and must be increased Page 11 of 17 N RC Expert EC v 2 7 2014 Design of reinforced concrete members to EN 1992 1 1 MPOEKTCOOT User Manual z 0 9 d is the lever arm of the internal forces d is the effective section depth v 0 6 1 fek 250 eq 6 6N in EN 1992 1 1 AOw 1 if Ocp 0 1 O p fed if O lt dcp lt 0 25feq 1 25 if 0 25fcd lt Ccp lt 0 5fcd 2 5 1 Ocp fca if 0 5fcd lt Ccp lt fed Distributed load in shear reinforc
18. in a file named data_file_name html The file always comes together with a folder named data_file_name html_files Always keep the file and the folder together otherwise all pictures and formatting will be lost Examples A lot of examples have been solved for verification of the software but only a few are presented in this manual The examples are elaborated according to RC Design Manual Eurocode 2 KIIP Sofia 2011 Prof Konstantin Rusev at al Page 13 of 17 N RC Expert EC v 2 7 2014 Design of reinforced concrete members to EN 1992 1 1 MPOEKTCOOT User Manual Example 1 Shear force design Solution with RC Expert Cross section b 35 0 cm h 70 0 cm d 5 0cm d 5 0 cm Internal forces Shear force Veg 770 0 KN E chi 31 5 GPa Tk 25 0 MPa f ctk 0 05 1 8 MPa Concrete grade C25 30 f Main reinforcement grade B500 E 200 0 GPa yk 500 0 MPa fya 435 0 MPa Shear reinforcement B500 E w 200 0 GPa fyw 500 0 MPa fywa 435 0 MPa Input data Number of shear link legs Nw 2 Diameter of shear links dw 12 0 mm Angle of inclination of shear links a 90 0 deg Main reinforcement As 0 0 cm Shear design results Compression strut angle 0 28 3 deg Concrete only resistance Vrdc 77 2 KN Concrete maximum resistance Vrdmax 770 0 KN Shear reinforcement area A sw 8 2 cm m Required shear reinforcement 12 13 5 Shear reinforcement ratio Mw 0 5 Required main reinforcement
19. oDd h yc x 0 0 Ns1 As1 Os1 Ma Net Z1 Z1 Ye d Page 8 of 17 N RC Expert EC v 2 7 2014 Design of reinforced concrete members to EN 1992 1 1 MPOEKTCOOT User Manual Ns2 s2 0s2 Ms2 Ns2 Z2 29 yc h yc d The compression zone height is obtained by the equation x dE Ec s Concrete stress is obtained from the section strain diagram using equations 3 17 and 3 18 oc Z fea f ee 3a0 lt z lt amp Ec2 O Z fea 34 Ec2 S Ec Z S Ecyz where fea design value of concrete compressive strength n 2 amp 2 2 0 0 y2 3 5 0 for concrete grade lower than C50 60 Reinforcement stress is determined by fya S Osi Es Esi S fya i 1 2 where fyd design yield strength of the reinforcement steel E is the design value of the modulus of elasticity of reinforcement Di Eud Es1 Iterations are performed by variation of strain in tensile reinforcement 1 and compressed concrete edge c SO that the section always remains in limit state The procedure is different depending on the case where strain diagram is located Case Tensile reinforcement strain remains constant s1 yg while strain at top edge varies from Eud in top reinforcement to cu2 at concrete edge Case 2 Concrete strain at top edge remains constant amp cu2 while tensile reinforcement strain varies from amp yq to Ecy2 d4 h At that moment concrete strain
20. ready exists you can overwrite it or select a different name Design code Design Expert is compatible to Eurocodes mainly EN 1992 1 1 and EN 1998 1 1 It is applicable to most countries as far as you can define your own material properties partial safety factors loads and some other important parameters Detailed description of all design methods and formulas used in this program is provided further in this manual Internal forces Design values of internal forces are entered bending moment Mgg axial force Neg shear force Veg and torsional force Teg Negative axial force is compressive and positive is tensile Page 3 of 17 N RC Expert EC v 2 7 2014 Design of reinforced concrete members to EN 1992 1 1 MPOEKTCOOT User Manual Sections and Materials A iri Save File To Disk e RESUS Fie Section And Materials Bending And Axial Force Shear And Torsion Internal Forces Bending moment M 100 kNm Shear force V kN f Long tem load factor Axial force N 200 KN compress Torsional moment T 0 kNm Materials Cross Section Dimensions cm Reinforcement Concrete 20 25 v Mainbars B500 v fcd 13 3 MPa fyd 435 MPa The input data described below is common for all design checks Additional data that is specific for each kind of check is entered separately in the respective page Cross section Cross section dimensions are entered in the respective text fields in the picture Press the button to see a s
21. zing the total reinforcement A 7 A 2 min The solution is found by a special iterative algorithm In case of symmetric reinforcement a unique solution exists At each iteration the area of either top or bottom reinforcement is obtained by formulas 3 or 4 while the other is taken to be equal For arbitrary strain that usually violates the respective equilibrium equation so the iterations continue until we find that value of A 1 As that satisfy both 3 and 4 Then we say that we have a convergence In case of bending we have only tension reinforcement until we reach the condition for sufficient ductility x d 0 45 Then we provide compressive reinforcement that allows moment re distribution in statically indeterminate systems to occur before reaching the ultimate limit state The solution is performed in the following sequence First we assume x d 0 45 and determine the reinforcement by equations 3 and 4 If A 2 gt 0 then the compressive reinforcement is actually needed and this is the final solution Otherwise we take A 0 and iteratively search the value of A necessary to satisfy the equation 3 Interaction diagrams M N interaction diagrams can be generated for the calculated or the initial reinforcement if the latest is defined by user The diagram is obtained by variation of the strain diagram defined by 1 and amp For each iteration the resulting axial force and moment are calculated representing a

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