Software Package. Design Expert version 2.0. RC Expert. Design of reinforced concrete elements. User Manual

Transcription

1 Software Package Deign Expert verion.0 RC Expert Uer Manual All right reerved 011

2 TABLE OF CONTENTS ABOUT THE PROGRAM... 3 ENTERING DATA... 3 FILES... 3 INPUT DATA... 3 Deign code... 3 Load... 3 Cro Section... 3 Material... 4 DESIGN TO BULGARIAN CODE (NPBSTBK)... 4 Bending with axial force... 4 Shear... 6 Torion... 6 Crack width and deflection... 7 DESIGN TO EUROCODE... 7 Bending with axial force... 7 Shear... 9 Torion CALCULATION REPORT EXAMPLES TO BULGARIAN CODE NPBSTBK... 1 Example 1. Deign of reinforced concrete beam... 1 Example. Deign of reinforced concrete column EXAMPLES TO EUROCODE Example 3. Shear force deign Example 4. Deign of rectangular ection for bending Example 5. Ultimate bending capacity of rectangular ection Example 6. Bending deign of doubly reinforced eciton Example 7. Ultimate bending capacity of doubly reinforced ection Example 8. Ultimate bending capacity of ection with flange... 0 Example 9. Bending deign of ection with flange... 1 Example 10. Ultimate bending capacity of ection with flange... Example 11. Capacity curve for bending and axial force... 3 Page of 3

3 About the program RC Expert i created for deign of reinforced concrete element with rectangular, T and I ection according to Bulgarian code and Eurocode. Calculation are performed for bending, compreion or tenion, hear force, torion, crack and deflection (lat i to Bulgarian code only). Second order effect are included for column if effective length are entered. Software i quick and eay to work, with rich functionality and friendly uer interface. Input data and deign reult can be printed in a profeional html report. Entering data Input data and reult are divided into everal page: You can witch between eparate page by clicking the correponding tab. Enter input data and click "Reult" to view the report. If file i not aved, you will be prompted to do that. Input data in each page i filled in text field or table. You can move to the next field with left click or Tab key. With Shift+Tab key combination you can go back to the previou field. File Input data for each problem i aved in a file with extenion *. tb. Deign output i written to a *. tb.html file in HTML format. You can open a file by the Open button. You can ave a file by the Save button. If a file i aved for firt time, a tandard dialog appear where you hould elect file path and name. Otherwie, file i aved uing current filename. You can change filename with the "New" command. Input data remain unchanged. To enter multiple element in one eion: input the firt, calculate and draw, click Save, click New, input the econd, calculate and draw, click Save and o on. Input data Following input data i common for all deign check. There i additional data for different kind of check that i decribed in the repective chapter. Deign code You can elect between two deign code Bulgarian "NPBStBK" and "Eurocode ". Other deign code may be developed in future verion. Load Enter ultimate value for internal force in the ection bending moment M Ed, axial force N Ed, hear force Q (V Ed ) and torion T Ed. Axial force i negative for compreion and poitive + for tenion. Cro Section Enter ection dimenion directly into the picture on the right ide of the window. You can ee caled preview of the ection by clicking the button. Section can be rectangular, T or I haped. Enter zeroe for dimenion of flange for rectangular ection. Concrete cover d1,d (a, a ) i meaured from concrete edge to center of area of main reinforcement. Page 3 of 3

4 Material Select concrete grade and teel grade for main and hear reinforcement. You can view detailed table with deign propertie of concrete and teel by clicking the button. You can alo modify material propertie or add new material. Deign to Bulgarian code (NPBStBK) Bending with axial force Input of trength reduction factor b and i required additionally. To include econd order effect check the Column box and enter element length L and buckling length in plane of bending L ox and out of plane L oy. You can enter initial compreive reinforcement A,ini that will be favorable for calculation of tenion reinforcement. Compreive reinforcement may come from other load combination with oppoite moment or may be required by deign code. Option Seimic load include an additional factor of 0.85 for calculation of limit compreion zone ratio R and an additional check N/(A R bn ) < 0.5 according to eimic code. Seimic factor k, are not included automatically. You hould enter them in field b and by multiplying them to the repective trength reduction factor. Deign can handle all cae of combined bending and tenion (+) or compreion (-). Reinforcement can be either ymmetrical or unymmetrical. Second order effect are included by the factor, calculated according to NPBStBK, equation (19): = 1/(1 N/N cr ). Buckling force i calculated by equation (68): N cr 6.4E b l 0 I b 0.11 l 0.1 e 0.1 E I E b Creep factor i determined by the formula l = 1+M G /M. Ratio KG = M G /M i the ratio of bending due to dead and permanent load to total bending. It i entered next to load a Long term load factor. Buckling force N cr depend on econd moment of inertia of reinforcement I t and the reinforcement i till unknown. Several iteration are required in order to obtain I t for the final reinforcement that i deigned. Calculation are performed by clicking the Reult button. Following value are calculated: A, A area of bottom and top reinforcement;, repective reinforcement ratio;, tree at center of bottom and top reinforcement; x A,tot height of compreion zone. total column reinforcement for out of plane bucking. Deign i performed by numerical procedure for arbitrary ection with general loading. Although thi verion upport only rectangular, T and I ection it will be eay to add new hape in the future (e.g. circular, ring etc.). Deign i baed on the following aumption: Concrete tre i completely neglected in tenion zone and all tenion i carried by reinforcement. Concrete compreion tre diagram i rectangular with contant value R b. Unknown value are: A tenile reinforcement area; A compreive reinforcement area; x compreion zone height. They are determined from the condition for balance of internal force: Page 4 of 3

5 N + N b + N + N = 0 (1) M + M b + M + M = 0 () M and N are ection load. Internal force for concrete and reinforcement and reulting moment in ection center of are calculated a follow: N b = A b R b ; M b = N b (h y c x/); N = A ; M = N (y c a) N = A ; M = N (h y c a ); and are tree in bottom and top reinforcement repectively: R c i = 1 y i / x 1 /1.1 c,u R c y i = h 0 за y i = a за Formula for tree are baed on equation (77), = R b. Solution i found by an iterative algorithm a follow: Initial value for reinforcement are aumed to be minimum allowed value by code. Iteration are repeated until equation () i atified a follow: Compreive zone height x i determined by equation (1). Thi i alo an iterative proce: Firt interval {0; h} i calculated for x = 0, x = h/ and x = h. One of the two ubinterval {0; h/} and {h/; h} i elected where equation (1) ha different ign in both end. Calculation are repeated and continued the ame way until N + N b + N + N <, where i the acceptable error. Then the condition M + M b + M + M < i verified and if atified, calculation are finihed. If it i not atified then value of N and N are calculated by equation for balance of moment about center of bottom and top reinforcement, repectively: N = (M M b + N (h y c a ))/z N = (M M b N (y c a))/z Moment about center of bottom and top reinforcement due to concrete tre are a follow: M b = N b z b M b = N b (z z b) where z b i the ditance between the center of compreion zone and the bottom reinforcement and z i the ditance between center of bottom and top reinforcement. Neceary area of bottom and top enforcement are calculated from the repective force: A = N /R при N > 0; A = N /R c при N < 0; In cae of bending where = x/h 0 < R or in cae of compreion/ tenion R 1 1 cu 1.1 combined with bending, area of compreion reinforcement i calculated: A = N /R if N > 0; A = N /R c if N < 0; Iteration i repeated. Additional check are performed for out of plane buckling due to axial force N < m (R b A + R c A,tot ). Buckling factor i determined from Table 33 in Bulgarian code NPBStBK uing effective length L oy. Total reinforcement area A,tot i calculated a a reult. Page 5 of 3

6 Shear Shear deign i performed for element with hear link and traight bar (no inclined bar). The following additional data i required: n w number of hear link cut; d hear link diameter; c critical crack projection. Concrete capacity without hear reinforcement i alo provided. Ultimate concrete tre i verified for combined hear and torion. Influence of axial force N i taken into account. Compreion flange i neglected. The following value are calculated a a reult: Q b,min = b3 R bt b h 0 concrete capacity without hear reinforcement Q max = 0.3 b1 w R b b h 0 ultimate concrete capacity. If Q > Q max, ection ize or concrete grade hould be increaed. A w area of hear reinforcement for one meter and one cut; w ratio of hear reinforcement. Shear reinforcement i calculated a follow: The value of M b = b R bt b h 0 i found. Projection of critical crack i h 0 c 0 = M b /Q.5 h. Equation (9) from NPBStBK i ued where q w i replaced by formula (94а) for Q = Q b,w If value of c 0 i entered then c intead c 0 i ued. Shear capacity of concrete i: Q b = M b / c 0 > Q b,min equation (88) from NPBStBK. Load carried by hear reinforcement (kn/m) i: q w = (Q Q b )/c 0 Q b,min /( h 0 ) Area of hear reinforcement (cm /m) i: A w = q w 100 /(n w 0.8 R w ) Deign tenion reitance of hear reinforcement (without reduction) i noted a R w. Factor ued in the above equation are calculated according to the deign code a follow: w = E w /E b n w A w /(b 100) 1.3 b1 = R b; b = 1.5; b3 = 0.6; b4 = 1.5 = 1 + f + n 1.5 n = 0.1 N / (R bt A w0 ) - за натиск (N < 0) (0. N / (R bt A w0 ) 0.8) - за опън (N > 0) The factor for Т ection i neglected conervatively - f = 0 Torion All ection part are included into calculation for I and T ection. Total torional moment i ditributed among eparate part proportionate to their torional tiffne. Additional main and hear reinforcement are calculated to thoe by other analyi. Shear tre i calculated including hear force: = Q / A w0 +1. T / W t, where A w0 = b w h 0 Requirement for hear reinforcement: Concrete capacity check: > 0.6 R bt < 0.3 R b Shear reinforcement: A w,t = p 100 /(0.9 R w ) Page 6 of 3

7 Main reinforcement: A,tot = p U ef /(0.9 R ); p=t/( A ef ); b ef = b (a w ; h ef = h a w ; U ef = (b ef +h ef ); A ef = b ef h ef Concrete cover a w i defined to the center of main reinforcement. Crack width and deflection The program calculate moment of cracking M crc, crack width a crc and deflection for imply upported or cantilever beam. Deflection i compared to the admiible value in deign code. In addition, curvature and ratio of nonlinear to linear tiffne are calculated. Curvature-moment diagram i provided a well. Linear curvature i drawn in blue and nonlinear i in red. Required input data include beam type (imply upported or cantilever), beam length L, nominal ditributed load q n or maximal moment M n. Main reinforcement hould be entered a actual number and diameter of bottom and top bar in the element. There i no option for different diameter on one ide. Environment condition for the tructure are pecified. For calculation of bending due to permanent and long-term load, Long term load ratio i ued a defined in Internal force ection. In verion.0 calculation are to Bulgarian code only. Deign to Eurocode Bending with axial force Deign can handle all cae of combined bending with tenion (+) or compreion (-). Reinforcement can be either ymmetrical or unymmetrical. Following aumption are ued in the deign: - concrete tenile tre i ignored and all tenion i carried by the reinforcement; - tre-train relationhip for concrete i paraboliclinear with maximum value of f cd ; - tre-train relationhip for teel i linear-platic with maximum value of f yd ; for 0 c c - - there i friction between concrete and reinforcement; - plane ection remain plane after deformation (Bernoulli cae); - train in reinforcement and concrete i limited and ection capacity i reached when limit train i achieved; for c c cu - c = f cd - internal force in concrete and reinforcement hould balance ection force due to external load. Page 7 of 3

8 f yd = E y f yd Unknown value are A 1 tenile reinforcement area; A compreive reinforcement area; 1 train in tenion reinforcement; c train in compreed concrete edge or tenion train in top reinforcement for ection entirely in tenion. Strain i limited to: cu c 0 for concrete and c yd for reinforcement. For ection entirely in compreion train diagram hould rotate around point C. Strain in point C i c and ditance of point C to top edge i equal to a C = h (1 c / cu ). Unknown value are determined by the condition for balance of internal force: N Ed + N c + N 1 + N = 0 (1) M Ed + M c + M 1 + M = 0 () where M Ed and N Ed are ection load. Internal force for concrete and reinforcement and reulting moment in ection center are calculated a follow: x x N c = c(z) b(z) dz; 0 N 1 = A 1 1; N = A ; M c = N c ( z c + h y c x); M 1 = N 1 (y c d 1 ); M = N (h y c d ); Solution i found by an iterative algorithm a follow: z c = c(z) b(z) z dz/n c Initial value for reinforcement are aumed to be minimum allowed value by code. Iteration are repeated until equation () i atified a follow: Strain 1 и c are calculated by condition (1) uing iterative algorithm until N Ed + N c + N 1 + N <, where i the acceptable error. Then the condition M Ed + M c + M 1 + M < i verified and if atified, calculation are finihed. If it i not atified then value of N 1 and N are calculated by equation for balance of moment about center of bottom and top reinforcement, repectively: N 1 = (M Ed M c + N Ed (h y c d ))/z N = (M Ed M c1 N Ed (y c d 1 ))/z Moment about center of bottom and top reinforcement due to concrete tre are a follow: M c1 = N c (h x d 1 + z c ) M c = N c (x z c ) where z i the ditance between center of bottom and top reinforcement. Neceary area of bottom and top enforcement are calculated from the repective force: 0 Page 8 of 3

9 A 1 = N 1 /f yd if N > 0; A 1 = N 1 /f yd if N 1 < 0; In cae of bending where K = M /(b h 0 f ck ) > K`= or in cae of compreion/ tenion combined with bending, area of compreion reinforcement i calculated: A = N /f yd at N > 0; A = N /f yd at N < 0; Iteration i repeated. Second order effect are included for member in compreion. Bending moment due to econd order effect are calculated by the equation: M Ed = M 0Ed /(1 N/N B ) (Eq from EC) Firt order bending moment M 0Ed = M + N e i include initial imperfection. e i = i L ox / M 0Ed N e 0 (Eq. 5. from EC) e 0 = h/30 0 mm i = 0 h m; 0 = 1/00; h = /(L/100) ½ ; m = 1; N B = EI/L ox EI = K c E cd I c +K I (Eq from EC) (Eq. 5.1 from EC) K = 1; K c = k 1 k /(1 + ef ); k 1 = (f ck /0) ½ ; k = n /170; n = N ed /(A c f cd ); = L ox /i; i = (I c /A c ) ½ Effective creep factor i calculated by the formula: ef = (,t0) M 0Eqp /M 0Ed. Ratio M 0Eqp /M 0Ed i entered by the uer a Long term load factor. Creep factor (,t 0 ) hould be alo defined by uer according to Figure 3.1 in EC. Buckling force N B depend on econd moment of inertia of reinforcement I and the reinforcement i unknown at thi tage. Several iteration are required in order to obtain I for the final reinforcement that i deigned. Calculation are performed by clicking the Reult button. Following value are calculated: A 1, A 1, area of bottom and top reinforcement; repective reinforcement ratio; 1, tree at center of bottom and top reinforcement; x height of compreion zone. The program provide M-M capacity curve for the ection with the calculated reinforcement. You can et initial reinforcement and obtain the capacity curve for it. Diagram i calculated by changing train 1 and c within pecified limit. Ultimate ection reitance i calculated for each poition and i plotted in M-N coordinate ytem. Diagram i a cloed curve. If we plot external moment and axial force a a point in the ame ytem we can compare it to the ection capacity. If the point i inide the capacity curve then ection i OK for the load. If the point i outide the capacity curve then ection will fail. Reinforcement i calculated o that capacity curve goe through the point of external load. Shear Shear deign i performed for element with vertical or inclined hear link and traight main bar (no inclined bar). The following additional data i required: n w number of hear link cut; d hear link diameter; c critical crack projection. Page 9 of 3

10 angle of compreion trut; angle between link and a vertical line; A exiting tenion reinforcement (cm ) with ufficient anchorage behind the ection of interet. The following value are calculated to Eurocode : V Rd,c = (C Rd,c k (100 l f ck ) k 1 cp ) b h 0 concrete capacity without hear reinforcement (Eq. 6..a) V Rd,c V Rd,c,min = (v min +k 1 cp ) b h 0 (Eq. 6..b) k = 1+(00/h 0 ) ½ ; cp = N Ed /A c < 0, f cd ; C Rd,c = 0,18/ c ; k 1 = 0.15; v min =0,035 k 3/ f ck ½ ; l = A /(b h 0 ) 0.00; V Rd,max = c b z 1 f cd (cot +cot)/(1+cot ) ultimate concrete capacity in the compreion trut (Eq from ЕС). If V Ed > V Rd,max, ection dimenion or concrete grade hould be increaed. z = 0.9 h 0 i lever arm of internal force; 1 = 0.6 (1 f ck /50) - (Eq. 6.6N от ЕС). Load, carried by hear reinforcement (kn/m): q w = V Ed / (z (cot +cot) in) Area of hear reinforcement (cm /m): A w = q w /(n w f yw ) Additional tenion force in main reinforcement: F td = 0.5 V Ed (cot cot) Additional main reinforcement to take thi force: A lн = F td /f yd V Ed = V i the hear load entered by the uer. If angle i entered to be 0, the pecified value i ued in calculation. If angle i = 0, it i calculated by the condition V Rd,max = V Ed, providing that (1 cot.5). Link pacing i calculated by the formula: = A w /( d /4) < max = 0.75 h 0 (1 + cot). Torion All ection part are included into calculation for I and T ection. Total torional moment i ditributed among eparate part proportionate to their torional tiffne. Additional main and hear reinforcement are calculated to thoe by other analyi. Deign of each rectangular part of the ection i decribed below. Concrete cover a w i defined to the center of main reinforcement. The following check are performed for combined torion with hear force: Concrete capacity without hear reinforcement: T Ed /T Rd,c + V Ed /V Rd,c > 1 (6.9) Ultimate concrete capacity in compreion trut:t Ed /T Rd,max + V Ed /V Rd,max < 1 (6.31) Ultimate torion without hear reinforcement: T Rd,c = A k t ef fctd Ultimate torion for the concrete ection: T Rd,max = c f cd A k t ef in( ) (6.30) T Ed = T i the torional moment entered by the uer. Value for V Ed V Rd,c V Rd,max are calculated a decribed in chapter "Shear" above. Reinforcement i calculated a follow: Page 10 of 3

11 Shear reinforcement: A w = P / (f yw cot) Main reinforcement: A tot = P U k cot / f yd (6.8) Effective thickne: t ef = b h / ( b + h) t ef i taken not le than a w Effective cro ection area: A k = (b - t ef ) (h - t ef ) Perimeter of effective area: U k = (b + h t ef ) Reult are provided eparately for each part of the ection. Calculation report P = T Ed / ( A k ) Profeional html report can be generated for each problem by clicking the "Reult" button. Report i diplayed in Internet Explorer by default but other web brower can be ued. Mot office program like MS Word can edit html file. Report filename i data_file_name.html. It come together with a folder data_file_name.html_file. Alway keep together report file with the folder. Otherwie picture and formatting will be lot. Page 11 of 3

12 Example to Bulgarian code NPBStBK Example 1. Deign of reinforced concrete beam Deign with RC Expert Cro ection b = 5,0 cm h = 50,0 cm b f = 0,0 cm h f = 0,0 cm b`f = 0,0 cm h`f = 0,0 cm a = 3,5 cm a` = 3,5 cm Section Load Bending moment - M = 16,0 kn.m Axial force- N = 0,0 kn Shear Force - Q = 108,0 kn Torional moment - T = 0,0 kn.m Long term load factor - KG = 90,0 % Material Concrete grade B0 E b = 7,5 MPa R b = 11,5 MPa R bt= 0,9 MPa Main reinforcement grade AIII E = 00,0 MPa R = 375,0 MPa R c= 375,0 MPa Shear reinforcement grade AI E w= 00,0 MPa R w= 180,0 MPa Bending And Axial Force Deign Strength Reduction Factor Exiting Reinforcement Concrete b = 1,0 Bottom A,ini = 0,0 cm Reinforcement = 1,0 Top A`,ini = 0,0 cm Shear Force Deign Shear link cut - n w = Shear link diameter - d = 8,0 mm Compreion trut angle - c = 0,0 cm Deflection And Crack Width Beam type - imply upported, beam length L = 600,0 cm Ditributed load Bending moment Axial force q n = 30,0 kn/m M n = 135,0 kn.m N n = 0,0 kn Exiting Reinforcement Tenile 4N0, A = 1,57 cm Compreive N1, A` =.6 cm Environment condition - contant air moiture < 70% - b =,00, = 0,15 Page 1 of 3

13 Reult Bending With Axial Force Deign Reinforcement Area Reinf. ratio Reinf. tre Bottom A = 10,9 cm = 0,9 % = 375,0 MPa Top A`= 0,0 cm `= 0,0 % `= -375,0 MPa Out of plane A,tot=0,0 cm Compr. zone height x = 14,0 cm Bending with axial force deign completed uccefully! Shear Force Deign Concrete only reitance - Q b,min = 6,8 kn Concrete ultimate reitance - Q max = 374,3 kn Shear reinforcement area - A w = 1,9 cm /m Required hear reinforcement - Ф8/5,0 cm Shear reinforcement ratio - w = 0, % Shear force deign completed uccefully! Deflection And Crack Width Crack opening moment - M crc = 18,43 kn.m Crack width - a crc = 0,3 mm Deflection - D max =,68 cm< 3 cm Curvature for moment Mn - 1/r = 7, /m Stiffne for moment Mn- B/EI = 0,6 Deflection and crack width deign completed uccefully! Example. Deign of reinforced concrete column Deign with RCExpert Cro ection b = 5,0 cm h = 50,0 cm b f = 0,0 cm h f = 0,0 cm b`f = 0,0 cm h`f = 0,0 cm a = 3,5 cm a` = 3,5 cm Section Load Bending moment - M = 400,0 kn.m Axial force- N = -1000,0 kn Shear Force - Q = 0,0 kn Torional moment - T = 0,0 kn.m Long term load factor - KG = 75,0 % Page 13 of 3

14 Material Concrete grade B35 E b = 33,0 MPa R b = 19,5 MPa R bt= 1,3 MPa Main reinforcement grade AIII E = 00,0 MPa R = 375,0 MPa R c= 375,0 MPa Shear reinforcement grade AI E w= 00,0 MPa R w= 180,0 MPa Bending And Axial Force Deign сила Strength Reduction Factor Exiting Reinforcement Concrete b = 1,0 Bottom A,ini = 0,0 cm Reinforcement = 1,0 Top A`,ini = 0,0 cm Column Length Effective length Symmetrical reinforcement L = 600,0 cm In plane of bending - L ox = 1,0 *L Out plane - L oy = 1,0 *L Reult Bending With Axial Force Deign MI + MII = 456,5 kn.m Reinforcement Area Reinf. ratio Reinf. tre Bottom A = 18,7 cm = 1,6 % = 375,0 MPa Top A` = 18,7 cm `= 1,6 % `= -375,0 MPa Out of plane A,tot = 5,0 cm Compr. zone height x = 1,1 cm Bending with axial force deign completed uccefully! Example to Eurocode Example 3. Shear force deign Deign with RC Expert Cro ection b = 0,0 cm h = 80,0 cm b f1 = 0,0 cm h f1 = 0,0 cm b f = 0,0 cm h f = 0,0 cm d 1 = 5,0 cm d = 5,0 cm Section Load Bending moment - M Ed = 0,0 kn.m Axial force- N Ed = 0,0 kn Shear Force - V Ed = 600,0 kn Torional moment - T Ed = 0,0 kn.m Long term load factor - K G = 75,0 % Page 14 of 3

15 Material C = 1,50 S = 1,15 Concrete grade C30/37 E c = 3,0 MPa f ck = 30,0 MPa f ctk,0.05=,0 MPa Main reinforcement grade S500 E y = 00,0 MPa f yk = 500,0 MPa f yd = 435,0 MPa Shear reinforcement grade S450 E yw= 00,0 MPa f ywk= 450,0 MPa f ywd= 391,0 MPa Shear Force Deign Shear link cut - n w = Shear link diameter - d = 1 mm Compreion trut angle - = 8,7 deg Shear link angle - = 90,0 deg Tenile reinforcement - A l = 0,0 cm Reult - Shear Force Deign Concrete only reitance - Concrete ultimate reitance - V Rd,c = 80,0 kn V Rd,max = 600,0 kn Shear reinforcement area - A w = 6, cm /m Required hear reinforcement - Ф1/17,5 cm Shear reinforcement ratio - w = 0,6 % Additional tenile reinforcement - A lн = 1,6 cm > A l Manual verification V Ed = 600 kn; f ck = 30 MPa; f cd = 0 MPa; f ywd = 391 MPa h 0 = h a = 750 mm; z = 0.9 h 0 = = 675 mm = 1 = 0.6 (1 f ck /50) = 0.58 Calculation of compreive trut angle θ for V Rd,max = V Ed : 1 VEd θ arcin arcin 8.7 > 1.8 ; cot = 1.8 cw v fcd bw z Calculation of required hear reinforcement: A w z f VEd cotθ ywd Aw For one cut: n w mm / mm cm / m; Required link are 1/175 mm Additional tenile force: ΔF td = 0.5 V Ed cotθ = = 546 kn Additional tenile reinforcement for that force: A lн = ΔF td /f yd =546/43.5 = 1,6 cm Page 15 of 3

16 Example 4. Deign of rectangular ection for bending Example 4 to 10 are developed uing the book "Reinforced concrete deign to Eurocode " - Bill Moley, John Bungey, Ray Hule, 007 (Example ). Formula are obtained auming cc = Concrete deign reitance i f cd = cc f ck / C. Calculation by the program are performed with parabolic-linear tre-train relationhip. Manual verification i performed with rectangular tre ditribution and height of compreion zone equal to = 0.8 x. Deign with RC Expert Cro ection b = 6,0 cm h = 48,0 cm b f1 = 0,0 cm h f1 = 0,0 cm b f = 0,0 cm h f = 0,0 cm d 1 = 4,0 cm Section Load Bending moment - Axial force- Shear Force - Torional moment - d = 4,0 cm M Ed = 185,0 kn.m N Ed = 0,0 kn V Ed = 0,0 kn T Ed = 0,0 kn.m Long term load factor - K G = 75,0 % Material C = 1,50; cc =0,85 S = 1,15 Concrete grade C5/30 E c = 31,5 MPa f ck = 5,0 MPa f ctk,0.05= 1,8 MPa f cd = 14,1 MPa f ctd= 1,0 MPa Main reinforcement grade S500 E y = 00,0 MPa f yk = 500,0 MPa f yd = 434,8 MPa Shear reinforcement grade S0 E yw= 00,0 MPa f ywk= 0,0 MPa f ywd= 191,3 MPa Input Data for Bending And Axial Force Deign Exiting Reinforcement Bottom A,ini = 0,0 cm Top A`,ini = 0,0 cm Reult for Bending With Axial Force Deign Reinforcement Area Reinf. ratio Reinf. tre Bottom A = 11,3 cm = 1,0 % = 434,8 MPa Top A` = 0,0 cm `= 0,0 % `= -434,8 MPa Out of plane A,tot = 0,0 cm Compr. zone height x = 16,3 cm Manual verification (Example 4.1) Ultimate bending moment i given M Ed = 185 kn.m. Find required reinforcement area A. M K b d f ck Compreive reinforcement i not required K z d mm A M 0.87 f yk mm z Page 16 of 3

17 Example 5. Ultimate bending capacity of rectangular ection Deign with RC Expert Cro ection b = 6,0 cm h = 56,0 cm b f1 = 0,0 cm h f1 = 0,0 cm b f = 0,0 cm h f = 0,0 cm d 1 = 4,0 cm Section Load Bending moment - Axial force- Shear Force - Torional moment - d = 4,0 cm M Ed = 84,0 kn.m N Ed = 0,0 kn V Ed = 0,0 kn T Ed = 0,0 kn.m Long term load factor - K G = 75,0 % Material C = 1,50; cc =0,85 S = 1,15 Concrete grade C5/30 E c = 31,5 MPa f ck = 5,0 MPa f ctk,0.05= 1,8 MPa f cd = 16,7 MPa f ctd= 1, MPa Main reinforcement grade S500 E y = 00,0 MPa f yk = 500,0 MPa f yd = 435,0 MPa Shear reinforcement grade S50 E yw= 00,0 MPa f ywk= 50,0 MPa f ywd= 17,0 MPa Input Data for Bending And Axial Force Deign Exiting Reinforcement Bottom A,ini = 14,7 cm Top A`,ini = 0,0 cm Reult for Bending With Axial Force Deign Reinforcement Area Reinf. ratio Reinf. tre Bottom A = 15,0 cm = 1,1 % = 435,0 MPa Top A` = 0,0 cm `= 0,0 % `= -435,0 MPa Out of plane A,tot = 0,0 cm Compr. zone height x = 17,7 cm Manual verification (Example 4.) Reinforcement area i given A = 1470 mm. Find ultimate bending reitance M Rd. F fck b 0.87 fyk A cc F t mm x 188mm M F t z 0.87 f yk 150 Ad kN/m Page 17 of 3

18 Example 6. Bending deign of doubly reinforced eciton Deign with RC Expert Cro ection b = 6,0 cm h = 50,0 cm b f1 = 0,0 cm h f1 = 0,0 cm b f = 0,0 cm h f = 0,0 cm d 1 = 6,0 cm d = 5,0 cm Section Load Bending moment - M Ed = 85,0 kn.m Axial force- N Ed = 0,0 kn Shear Force - V Ed = 0,0 kn Torional moment - T Ed = 0,0 kn.m Long term load factor - K G = 75,0 % Material C = 1,50; cc =0,85 S = 1,15 Concrete grade C5/30 E c = 31,5 MPa f ck = 5,0 MPa f ctk,0.05= 1,8 MPa f cd = 14,1 MPa f ctd= 1,0 MPa Main reinforcement grade S500 E y = 00,0 MPa f yk = 500,0 MPa f yd = 435,0 MPa Shear reinforcement grade S0 E yw= 00,0 MPa f ywk= 0,0 MPa f ywd= 191,0 MPa Input Data for Bending And Axial Force Deign Exiting Reinforcement Bottom A,ini = 0,0 cm Top A`,ini = 0,0 cm Reult for Bending With Axial Force Deign Reinforcement Area Reinf. ratio Reinf. tre Bottom A = 18,0 cm = 1,6 % = 435,0 MPa Top A` = 4,3 cm `= 0,4 % `= -435,0 MPa Out of plane A,tot = 0,0 cm Compr. zone height x = 0,4 cm Manual verification (Example 4.3) Deign bending moment i given M Ed = 85 kn.m. Find required reinforcement A and A'. M K b d f d' d A ' ck Compreive reinforcement i required K K 0.87 f bal yk fck b d d d' mm A yk Kbal fck b d 0.87 f z bal A ' mm Page 18 of 3

19 Example 7. Ultimate bending capacity of doubly reinforced ection Deign with RC Expert Cro ection b = 8,0 cm h = 56,0 cm b f1 = 0,0 cm h f1 = 0,0 cm b f = 0,0 cm h f = 0,0 cm d 1 = 5,0 cm Section Load Bending moment - Axial force- Shear Force - Torional moment - d = 5,0 cm M Ed = 443,0 kn.m N Ed = 0,0 kn V Ed = 0,0 kn T Ed = 0,0 kn.m Long term load factor - K G = 75,0 % Material C = 1,50; cc =0,85 S = 1,15 Concrete grade C5/30 E c = 31,5 MPa f ck = 5,0 MPa f ctk,0.05= 1,8 MPa f cd = 14,1 MPa f ctd= 1,0 MPa Main reinforcement grade S500 E y = 00,0 MPa f yk = 500,0 MPa f yd = 435,0 MPa Shear reinforcement grade S0 E yw= 00,0 MPa f ywk= 0,0 MPa Input Data for Bending And Axial Force Deign Exiting Reinforcement Bottom A,ini = 4,1 cm Top A`,ini = 6,3 cm Reult for Bending With Axial Force Deign Reinforcement Area Reinf. ratio Reinf. tre Bottom A = 4,1 cm = 1,7 % = 435,0 MPa Top A` = 6,9 cm `= 0,5 % `= -435,0 MPa Out of plane A,tot = 0,0 cm Compr. zone height x = 4,4 cm Manual verification (Example 4.4) f ywd= 191,0 MPa Reinforcement value are given A =68 mm and A' =410 mm. Find ection capacity moment M Rd. F t F cc 0.87 f x d yk F f c 0.87 f A f b 0.87 f A ' A A ck b yk ck 195mm 44 d' x 5 M F cc d F c yk ' x 0.8 d d' f b d 0.87 f A d d' ck yk ' 44mm kN/m Page 19 of 3

20 Example 8. Ultimate bending capacity of ection with flange Deign with RC Expert Cro ection b = 6,0 cm h = 46,0 cm b f1 = 0,0 cm h f1 = 0,0 cm b f = 80,0 cm h f = 15,0 cm d 1 = 4,0 cm d = 4,0 cm Section Load Bending moment - M Ed = 49,0 kn.m Axial force- N Ed = 0,0 kn Shear Force - V Ed = 0,0 kn Torional moment - T Ed = 0,0 kn.m Long term load factor - K G = 75,0 % Material C = 1,50; cc =0,85 S = 1,15 Concrete grade C5/30 E c = 31,5 MPa f ck = 5,0 MPa f ctk,0.05= 1,8 MPa f cd = 14,1 MPa f ctd= 1,0 MPa Main reinforcement grade S500 E y = 00,0 MPa f yk = 500,0 MPa f yd = 435,0 MPa Shear reinforcement grade S0 E yw= 00,0 MPa f ywk= 0,0 MPa f ywd= 191,0 MPa Input Data for Bending And Axial Force Deign Exiting Reinforcement Bottom A,ini = 14,7 cm Top A`,ini = 0,0 cm Reult for Bending With Axial Force Deign Reinforcement Area Reinf. ratio Reinf. tre Bottom A = 14,7 cm = 1,4 % = 435,0 MPa Top A` = 0,0 cm `= 0,0 % `= -34,3 MPa Out of plane A,tot = 0,0 cm Compr. zone height x = 8,0 cm Manual verification (Example 4.5) Reinforcement area i given A = 1470 mm. Find ection bending capacity M Rd. F fck bf 0.87 fyk A cc F t mm h f 150mm 56 x 70mm z d 40 39mm 8 M F cc z f ck b f z kN/m Page 0 of 3

21 Example 9. Bending deign of ection with flange Deign with RC Expert Cro ection b = 0,0 cm h = 40,0 cm b f1 = 0,0 cm h f1 = 0,0 cm b f = 40,0 cm h f = 10,0 cm d 1 = 5,0 cm d = 5,0 cm Section Load Bending moment - M Ed = 180,0 kn.m Axial force- N Ed = 0,0 kn Shear Force - V Ed = 0,0 kn Torional moment - T Ed = 0,0 kn.m Long term load factor - K G = 75,0 % Material C = 1,50; cc =0,85 S = 1,15 Concrete grade C5/30 E c = 31,5 MPa f ck = 5,0 MPa f ctk,0.05= 1,8 MPa f cd = 14,1 MPa f ctd= 1,0 MPa Main reinforcement grade S500 E y = 00,0 MPa f yk = 500,0 MPa f yd = 434,8 MPa Shear reinforcement grade S0 E yw= 00,0 MPa f ywk= 0,0 MPa f ywd= 191,3 MPa Input Data for Bending And Axial Force Deign Exiting Reinforcement Bottom A,ini = 0,0 cm Top A`,ini = 0,0 cm Reult for Bending With Axial Force Deign Reinforcement Area Reinf. ratio Reinf. tre Bottom A = 14,5 cm =,1 % = 434,8 MPa Top A` = 0,0 cm `= 0,0 % `= -434,8 MPa Out of plane A,tot = 0,0 cm Compr. zone height x = 13,4 cm Manual verification (Example 4.6) Deign moment i given M Ed = 180 kn.m. Find required reinforcement area A. h 100 w 6 w h f 180 Mf Fcwz fckbwwz w50 10 w w50 10 w 500w w 15mm Mf Fcfz f 6 1 Mf 0.567fckbfhf d kN/m 180kN/m x F h f 0.8 w mm 0.41d t Fcf Fcw 0.87fykA 0.567fckbfhf 0.567fckbww A A 140mm Page 1 of 3

22 Example 10. Ultimate bending capacity of ection with flange Deign with RC Expert Cro ection b = 30,0 cm h = 60,0 cm b f1 = 0,0 cm h f1 = 0,0 cm b f = 45,0 cm h f = 15,0 cm d 1 = 5,0 cm d = 5,0 cm Section Load Bending moment - M Ed = 519,0 kn.m Axial force- N Ed = 0,0 kn Shear Force - V Ed = 0,0 kn Torional moment - T Ed = 0,0 kn.m Long term load factor - K G = 75,0 % Material C = 1,50; cc =0,85 S = 1,15 Concrete grade C5/30 E c = 31,5 MPa f ck = 5,0 MPa f ctk,0.05= 1,8 MPa f cd = 14,1 MPa f ctd= 1,0 MPa Main reinforcement grade S500 E y = 00,0 MPa f yk = 500,0 MPa f yd = 435,0 MPa Shear reinforcement grade S0 E yw= 00,0 MPa f ywk= 0,0 MPa f ywd= 191,0 MPa Input Data for Bending And Axial Force Deign Exiting Reinforcement Bottom A,ini = 5,9 cm Top A`,ini = 0,0 cm Reult for Bending With Axial Force Deign Reinforceme nt Area Reinf. ratio Reinf. tre Bottom A = 7,1 cm = 1,6 % = 435,0 MPa Top A` = 0,0 cm `= 0,0 % `= -435,0 MPa Out of plane A,tot = 0,0 cm Compr. zone height x = 1,1 cm Manual verification (Example 4.7) Reinforcement area i giver A = 59 mm. Find ection bending capacity M Rd. F cf 0.567f b h ck f f kN F F F x t cw cw 0.87f yk A 0.567f t cf ck b w 3 118kN 3 h f F F mm 0.43d kN F cw M F cf hf d F cw d hf kN/m Page of 3

23 0,0 0,1 0, 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0 RC Expert v.0/011 Example 11. Capacity curve for bending and axial force Draw capacity curve for bending and axial force for the given ection uing RC Expert. Reinforcement i ymmetrical A 1 =A and d 1 /h=0.1. Cro ection - rectangular b = 5,0 cm d 1 = 4,0 cm h = 40,0 cm d = 4,0 cm C = 1,50 S = 1,15 Material Concrete grade C0/5 E c = 30,0 MPa Main reinforcement grade S500 E y = 00 MPa Shear reinforcement grade S0 E yw= 00 MPa f ck = 0,0 MPa f cd = 13,3 MPa f yk = 500 MPa f yd = 435 MPa f ywk= 0 MPa f ywd= 191 MPa Abolute coordinate M N A 1 = 0.00 cm, =0.0 A 1 = 6.55 cm, =0.5 Relative coordinate -3,0 -,5 =.0 = 1.5 -,0 = 1.0-1,5 A 1 = 13.1 cm, =1.0 A 1 = 19.6 cm, =1.5 = N ed /(bhf cd ) -1,0-0,5 0,0 = 0.5 = 0.0 0,5 A 1 = 6. cm, =.0 1,0 1,5 = M ed /(bh f cd ),0 = (A 1 +A )/bh f yd /f cd Diagram are calculated for yu =10, cu =3.5, c =.0. Same diagram calculated for yu = 5 and c =. are given on fig In the book "Reinforced concrete NPBSK- ЕС", K. Rouev, 008. Page 3 of 3