Design Manual to EC2 BS EN :2004

Transcription

1 Deign Manual to EC BS EN :004 Verion 3.0 Marh 011 = Limited Gloueter Houe Anon Court Buine Centre Dyon Way Stafford, ST18 0GB Tel/Fax: SPECIALISTS [Type text] IN PUNCHING SHEAR REINFORCEMENT

2 Content 1. Introdution 4. Lit of Symbol 5 3. Information Required to Deign 7 4. Eentriity Fator β a) Beta fator for regular and quare olumn and antilever lab b) Beta fator for irular olumn and antilever lab ) Calulating eentriity fator β with bending moment preent Effetive Depth of Slab 1 6. Punhing Shear at the Loaded Area a) Perimeter of the Loaded Area b) Deign value of the hear tre at the Loaded Area ) Deign value of the maximum punhing hear reitane at the Loaded Area Punhing Shear at the Bai Control Perimeter without reinforement a) Bai Control Perimeter length b) Deign value of the maximum hear tre at the Bai Control Perimeter ) Punhing hear reitane at the Bai Control Perimeter Punhing Shear at the Bai Control Perimeter with reinforement a) Deign value of the punhing hear reitane at the Bai Control Perimeter b) The area of hear reinforement Shear Reinforement Detailing Rule Retangular or Square Column Cirular Pattern a) Arranging tud b) Calulating in different ondition Limited 011 Page

3 Content 11. Cirular Column Cirular Pattern a) Arranging tud b) Calulating in different ondition Cruiform Pattern a) Arranging rail and tud Hole in the lab Example Calulation Internal Condition a) Cirular pattern quare olumn (internal ondition) Example Calulation Edge Condition a) Cirular pattern irular olumn with hole (edge ondition) Example Calulation External Corner Condition a) Cirular pattern quare orner Example Calulation Internal Corner Condition a) Cirular pattern quare olumn Limited 011 Page 3

4 1. Introdution The Sytem offer utomer a fat, eay and extremely ot effetive method of providing Punhing Shear Reinforement around olumn and pile within flat lab and pot-tenioned lab, at lab to hearwall juntion, beam to olumn juntion and within footing and foundation lab. The Sytem omprie hort length of arbon teel deformed bar reinforement with end anhorage provided by enlarged, hot forged head at both end, giving a ro-etional area ratio of 9:1. Thee tud head anhor eurely in the lab, eliminating lippage and providing greater reitane to punhing hear. The double-headed hear tud are welded to arrier / paer rail to allow them to be loated orretly and to be upported by the top flexural reinforement. i a tehnologially advaned and proven ytem - a fully-teted, fullyaredited, fully-traeable Punhing Shear Reinforement Sytem approved by CARES for ue in reinfored onrete lab deigned in aordane with EC deign tandard. Through our total fou on Punhing Shear Reinforement we have beome expert in our field, with unparalleled experiene in the deign of PSR heme and a thorough knowledge of the intriaie and omplexitie of the EC deign tandard. We are pleaed to be able to offer you thi expertie a a ornertone of the pakage. From appliation advie and deign guidane, through propoal drawing, alulation and quotation, to working drawing and ite upport, you an depend on for all your Punhing Shear Reinforement need. Kind Regard Dariuz Nowik MS (Eng) Senior Deign Engineer Limited Limited 011 Page 4

5 . Lit of Symbol Greek Symbol α - angle between rail in quarter α - oeffiient for long term effet β - eentriity fator γ - partial fator for material for ULS δ - when a hole i preent δ indiate an angle between the tangent line of the dead zone σ p - the normal onrete tree in the ritial etion σ x, σ y - the normal onrete tree in the ritial etion in x- or y- diretion ρ l - tenion reinforement ratio ρ lx, ρ ly - ratio of tenion reinforement in both diretion Latin Symbol A - area of onrete aording to the definition of N ed a lx, a ly - ditane to lab edge in x- and y- diretion A w - area of one perimeter of hear reinforement around the loaded area A w min - minimum area of one perimeter of hear reinforement around the loaded area A w1 - area of one hear tud A w1 min - minimum area of one hear tud A x, A y - area of T1 and T main reinforement per width of the loaded area + 3d eah fae b - onidered width of the lab B - effetive part of the perimeter of the quare or retangular loaded area faing: B N - north, B E - eat, B S - outh, B W - wet. B C - effetive part of the perimeter of the irular loaded area faing: B CN - north, B CE - eat, B CS - outh, B CW - wet. - diameter of the irular olumn 1, - quare/ retangular olumn dimenion C Rd. NA to BS EN (1) d - effetive depth of the lab d x d y - effetive depth of the reinforement in two orthogonal diretion f d - deign ompreive trength of onrete f k - harateriti ompreive ylinder trength of onrete at 8 day (BS EN , table 3.1) f yk - harateriti tenile trength of the reinforement f ywd - deign trength of the punhing reinforement f ywd.ef - effetive deign trength of the punhing reinforement h - lab depth k, k 1 - fator (1) k - oeffiient dependent on the ratio between the olumn dimenion 1 and l 1, l - hole dimenion M Ed - bending moment n - number of rail with the firt tud at a maximum of 0.5 d from the loaded area fae N ed,x N ed.y - longitudinal fore aro the full bay for internal olumn and the longitudinal fore aro the ontrol etion for edge olumn n - number of egment around u out on a irular layout pattern n t - number of tud per rail Limited 011 Page 5

6 . Lit of Symbol p - ditane from the entre point of a irular layout pattern to u out perimeter p 1 - ditane from the loaded area fae to the lat tud on the rail p - ditane from the loaded area fae to the entre point of a irular layout pattern r 3 - ditane to the 3 rd tud r lat - ditane to the lat tud - length of equal egment around u out on a irular layout pattern 3 - maximum tangential paing inide the bai ontrol perimeter (uually between 3 rd tud) lat - maximum tangential paing between lat tud outide the bai ontrol perimeter t - tangential tud paing r - radial tud paing t t, t b - top over, bottom over u 0 - loaded area perimeter u 0 red - part of the loaded area perimeter within the "dead zone" - bai ontrol perimeter red - part of the bai ontrol perimeter within the "dead zone" u out - ontrol perimeter beyond whih hear reinforement i not required u out - extended u out ontrol perimeter whih take aount of the preene of a hole u out.ef - u out for ruiform pattern when the ditane between the lat tud i greater then V Ed - deign value of the hear fore v Ed 0 - deign value of the hear tre at the loaded area fae v Ed 1 - deign value of the hear tre at the bai ontrol perimeter v min - minimum onrete hear tre reitane v Rd. - deign value of the punhing hear reitane of a lab without hear reinforement at the bai ontrol perimeter v Rd. - deign value of the punhing hear reitane of a lab with hear reinforement at the bai ontrol perimeter v Rd.max - deign value of the maximum punhing hear reitane at the loaded area fae W 1 - orrepond to a plati ditribution of the hear tre a deribed in BS EN , Fig Limited 011 Page 6

7 3. Information Required to Deign The following information i required to deign hear reinforement: V Ed - deign value of the hear load the hape and ize of the loaded area f k - the harateriti ompreive ylinder trength of the onrete the tenion reinforement diameter and paing in both diretion within 3d from the loaded area fae the thikne of the lab, top and bottom over to the main reinforement the ditane to the lab edge in both diretion the loation and ize of any hole() within 6d from the loaded area fae the loation of any tep in the lab or any hange to the lab thikne We aume that: the thikne of the lab i equal or greater than 00mm the load given have been fatored with EC fator the load given do not inlude the load from the olumn above the onrete lab ha not been made uing lightweight aggregate the main reinforement bar are plaed aordingly to the detailing rule deribed in and 9.4. (1) In order to deign uing the proper value of the hear tre, we reommend engineer provide u with value and diretion of the bending moment or the value of eentriity fator β a alulated by the Projet Engineer if thee value are not provided the approximate value of β will be ued. Limited 011 Page 7

8 4. Eentriity Fator β For truture where lateral tability doe not depend on frame ation between the lab and the olumn, and where the adjaent pan do not differ in length by more than 5%, approximate value for β may be ued (6) d d / + d a ly The reommended value of the eentriity fator β are lited below: (a per figure 6.1N) Internal olumn β = 1.15 Edge olumn β = 1.4 External orner olumn β = 1.5 Internal orner olumn β = 1.75 (the value found by interpolation) a) Beta fator for retangular and quare olumn and antilever lab If the lab edge doe not line up with the loaded area fae, the following rule may apply: Edge ondition If a lx ( + πd)/ than β = 1.15, If a ly = 0 than β = 1.4. Beta fator for a lx between the above value may be found by interpolation. External orner ondition a lx 1 β = max (β x, β y, β xy ) Where: β x = ( a lx /( + πd)) β y = ( a ly /( 1 + πd)) β xy = (a lx + a ly ) /( πd) a lx 1 Internal orner ondition If: a lx / + πd, or a ly 1 / + πd, or a lx + a ly πd, and a lx > 0 and a ly > 0 than β = 1.15 If: a lx 0, and a lx - 1, and a ly 0, and a ly - than β = 1.75 If: a lx 0, and a ly - -, or a ly 0, and a lx - 1 -, than β = 1.4 Beta fator for a lx, a ly between the above value may be found by interpolation. lab edge loation =1.15 (example) =1.15 d orner point =1.75 1d/ =1.15 =1.4 =1.15 / + d =1.4 Limited 011 Page 8

9 4. Eentriity Fator β 3d/ /4 ( + 4d) b) Beta fator for irular olumn and antilever lab Similar rule apply for the irular loaded area. See the detail below: Edge ondition If a lx π/4 ( + 4 d) than β = 1.15, If a lx / than β = 1.4. Beta fator for a lx between the above value may be found by interpolation. a lx External orner ondition β = max (β x, β y, β xy ) Where: β x = a lx / (π ( + 4d)) β y = a ly / (π ( + 4d)) β xy = (a lx + a ly ) / (3/4 π ( + 4d))) a ly a lx Internal orner ondition a lx If: a lx π/4 ( + 4d), or a ly π/4 ( + 4d), or a lx + a ly π/4 ( + 4d), and a lx > 0 and a ly > 0 than β = 1.15 If: a lx /, and a lx - /, and a ly /, and a ly - / than β = 1.75 If: a lx /, and a lx - /, and a ly - - /, or a ly /, and a ly - /, and a lx - - /, than β = 1.4 Beta fator for a lx, a ly between the above value may be found by interpolation. a ly lab edge loation =1.3 (example) =1.15 =1.15 d orner point = 1.75 =1.4 =1.15 /4 ( + 4d) =1.4 Limited 011 Page 9

10 4. Eentriity Fator β ) Calulating eentriity fator β with bending moment preent The Eentriity Fator β may be alulated by taking exiting bending moment into aount. The expreion below give a more aurate value of β fator. Internal ondition The Eentriity Fator hould be alulated a follow: β = 1 + k M Ed /(V Ed W 1 ) equation 6.39 Where: M Ed - bending moment V Ed - hear fore W 1 - orrepond to a hear ditribution (um of multipliation of bai ontrol perimeter length and the ditane from gravity entre to the axi about whih the moment at). - bai ontrol perimeter k - oeffiient dependent on the ratio between the olumn dim. 1 and (ee tab.6.1) 1 / k W 1 for retangular loaded area i equal: W 1 = d + 16d + π d equation 6.41 W 1 for irular loaded area i equal: W 1 = ( + 4d) Therefore the Eentriity Fator for irular loaded area i equal: β = π M Ed /(V Ed ( + 4d)) equation 6.4 In ae where moment in both diretion for retangular loaded area are preent, the Eentriity Fator hould be determined uing the expreion: β = (M Edx /(V Ed ( 1 +4d)) + M Edy /(V Ed ( +4d)) ) equation 6.43 Where: M Edx - bending moment about x axi (parallel to 1 ) M Edy - bending moment about y axi (parallel to ) For a irular loaded area 1 = Edge ondition Where the eentriity in both orthogonal diretion i preent, the Eentriity Fator hould be determined uing the expreion: β = / * + k e par /W equation 6.44 Where: * - redued bai ontrol perimeter (drawing 6.0a) e par - eentriity from the moment perpendiular to the lab edge e par = M Ed /V Ed k - may be determined from tab. 6.1 with the ratio 1 / replaed by 1 / Limited 011 Page 10

11 4. Eentriity Fator β W 1 - i alulated for the bai ontrol perimeter about the axi perpendiular to the lab edge W 1 for retangular loaded area i equal: W 1 = d + 8d + π d equation 6.45 W 1 for irular loaded area i equal: W 1 = + 6d + 8d 1.5d d (+4d)/8 u* 1 * Drawing 6.0a (inluding the irular loaded area) 1 If the bending moment about the axi parallel to the lab edge exit, eentriity i toward the interior and there i no other bending moment, then the punhing fore i uniform along the redued ontrol perimeter * (ee drawing 6.0a). Fator β i equal: β = 1 + k M Ed * /(V Ed W 1 ) Where: k - may be determined from tab. 6.1 with the ratio 1 / replaed by 1 / W 1 - i alulated for the redued ontrol perimeter about the axi parallel to the lab edge loated in the entroid of the redued ontrol perimeter. Corner ondition In orner ondition, when eentriity i toward the interior of the lab, the Eentriity Fator may be onidered a: * β = / 1.5d d (+4d)/8 Drawing 6.0b (inluding the irular loaded area) u* d * 1.5d (+4d)/8 1 If the eentriity i toward the exterior, expreion 6.39 applie. β = 1 + k M Ed /(V Ed W 1 ) equation 6.39 W 1 for a retangular loaded area (external orner ondition) i equal: W 1 = d + 0.5d π W 1 for a irular loaded area (external orner ondition) i equal: W 1 = 11 /8 + 9d + 16d Limited 011 Page 11

12 d y d x 5. Effetive Depth of Slab The effetive depth of the lab i aumed to be ontant and may be taken a: d = (d x + d y )/ 6.4. (1) equation 6.3 where: d x = h - t t - T1/ d y = h - t t - T1 - T/ t t - top over Column T1 T T T T T T T T T T T B B B B B B B B B B B B1 Column T1 T T B B B1 Limited 011 Page 1

13 6. Punhing Shear at the Loaded Area a) Perimeter of the Loaded Area Condition for retangular / quare olumn 1 a lx 1 B N 1 a ly B N a ly B E B W u 0 B S u 0 u 0 a lx B u S a lx 1 0 Internal Edge External orner Internal orner u 0 = 1 + u 0 = B N + B S + u 0 = B E + B S u 0 = B N + B W Where: B N, B S = min. ( 1, 1 + a lx, 1.5 d) B E, B W = min. (, + a ly, 1.5 d) Condition for irular olumn B N a lx a lx B N 0.5 a ly u 0 a ly 0.5 a lx B E B W u 0 u 0 u 0 B S B S 0.5 Internal Edge External orner Internal orner u 0 = π u 0 = π/4 + B N + B S u 0 = B E + B S u 0 = π/ + B N + B W Where: B N, B S = min. (0.5 π, 0.15 π + a lx, 1.5 d) B E, B W = min. (0.5 π, 0.15 π + a ly, 1.5 d) Note: if any hole within 6d from the fae of the loaded area i preent, the loaded area perimeter hould be redued (ee 6.4. (3) EC and etion 13) Note: if the lab edge i offet at leat from the loaded area fae then it preene hould be ignored when alulating u 0. b) Deign value of the hear tre at the Loaded Area v Ed 0 = β V Ed /(u 0 d) (3) 6.53 Limited 011 Page 13

14 6. Punhing Shear at the Loaded Area ) Deign value of the maximum punhing hear reitane at the Loaded Area f d = α f k /γ (3.15) γ = 1.5 α = 1 v Rd.max = 0.3 f d (1 - (f k / 50)) NA to BS EN (3) note, 6..(6) 6.6N Where: f d the value of the deign ompreive trength of onrete f k - the harateriti ompreive ylinder trength of onrete at 8 day. table 3.1 α the oeffiient for long term effet NA to BS EN (1)P γ the partial fator for material for ULS.4..4 table.1n Pleae note that f d i limited to the trength of C50/60, unle otherwie proven. If v Ed 0 > v Rd.max - the lab depth or the olumn ize mut be inreaed (a) Limited 011 Page 14

15 7. Punhing Shear at the Bai Control Perimeter without Reinforement a) Bai Control Perimeter length The Bai Control Perimeter i loated from the loaded area 1 Internal olumn = πd = π (4d + ) a lx 1 a lx Edge olumn = πd + a lx = π/ (4d + ) + a lx Limited 011 Page 15

16 7. Punhing Shear at the Bai Control Perimeter without Reinforement External orner olumn = πd + a lx + a ly = π/4 (4d + ) + a lx + a ly 1 Internal orner olumn a ly a lx a ly a lx 1 a lx a ly a lx a ly = ( 1 + ) + 3πd + a lx + a ly = 3/4 π (4d + ) + a lx + a ly Note: if any hole within 6d from the fae of the loaded area i preent, the loaded area perimeter hould be redued (ee 6.4. (3) EC and etion 13) Limited 011 Page 16

17 7. Punhing Shear at the Bai Control Perimeter without Reinforement b) Deign value of the maximum hear tre at the Bai Control Perimeter v Ed 1 = β V Ed / ( d) (3), (6.38) ) Punhing Shear Reitane at the Bai Control Perimeter k = 1+ (00/d) v min = 0.035k 3/ 1/ f k 6.. (1), (6.3N) C Rd. = 0.18/ γ k 1 = 0.1 σ x = N ed.y / A x σ y = N ed.y / A y σ p = (σ x + σ y ) / ρ lx = A x / (bd x ) ρ ly = A y / (bd y ) ρ l = (ρ lx ρ ly ) 0.0 v Rd. = C Rd. k (100 ρ l f k ) 1/3 + k 1 σ p (v min + k 1 σ p ) (1), (6.47) Where: σ x, σ y - the normal onrete tre in the ritial etion in x- and y- diretion (MPa, poitive if ompreion) N ed.x, N ed.y - the longitudinal fore aro the full bay for internal olumn and the longitudinal fore aro the ontrol etion for edge olumn. The fore may be from a load or pretreing ation. A - the area of onrete aording to the definition of N ed ρ lx, ρ ly the ratio of tenion reinforement in both diretion b the width of the loaded area + 3d eah ide ρ l the tenion reinforement ratio If v Ed 1 < v Rd. - punhing hear reinforement i not required (b) Limited 011 Page 17

18 8. Punhing Shear at the Bai Control Perimeter with Reinforement a) Deign value of the punhing hear reitane at the Bai Control Perimeter f ywd.ef = d f ywd = (f y / 1.15) (6.5) v Rd. = 0.75 v Rd (d / r ) A w f ywd.ef (1 / ( d)) v Ed (6.5) Where: r - radial tud paing f ywd.ef - the effetive deign trength of the punhing reinforement. A w - area of one perimeter of the hear reinforement around the loaded area. b) The area of the hear reinforement A w1.min = 0.08 ( r t ) (f k ) / 1.5 f yk A w1 = (v Ed v Rd. ) r / (1.5 f ywd.ef n) A w1.min Where: t - tangential tud paing n - number of rail with the firt tud at a maximum of 0.5 d from the loaded area fae. A w1 - area of one hear tud. A w1 min the minimum area of one hear tud. f yk - the harateriti tenile trength of reinforement. Limited 011 Page 18

19 9. Shear Reinforement Detailing Rule Shear reinforement hould be detailed in aordane with BS EN : (4), and NA to BS EN (4) The firt tud hould be plaed between 0.3d and 0.5d from the loaded area fae. For ruiform pattern the reommended ditane would be 0.35d. There hould be a minimum of two perimeter of reinforement. The radial paing of the hear reinforement ( r ) hould not exeed 0.75d. The tangential paing of the hear reinforement ( t ) hould not exeed 1.5d within the Bai Control Perimeter. The tangential paing of the irular pattern hear reinforement outide the Bai Control Perimeter hould not exeed. In the ae of ruiform pattern, tangential paing an go beyond but thi will effet u out by leaving gap in the perimeter (ee the drawing on page 0). u out (or u out.ef ) hould be alulated uing the following formula: u out = β V Ed / (v Rd. d) The outermot perimeter of hear reinforement hould be plaed at a ditane not greater than kd within u out (or u out.ef ) Where: k = 1.5 unle the perimeter u out (or u out.ef ) i loated loer than 3d from the loaded area fae. In thi ae the hear reinforement hould be plaed in the zone 0.3d to 1.5d from the loaded area fae. The hape of the perimeter u out (or u out.ef ) will depend on the arrangement of the hear reinforement and on paing limitation. max max 1.5d 1.5d max 0.75d 0.3d - 0.5d lat hear tud to be loated 1.5d from the u ontrol perimeter out u 1 - bai ontrol perimeter at from the loaded area u out ontrol perimeter Limited 011 Page 19

20 9. Shear Reinforement Detailing Rule d 1.5d d u - bai ontrol perimeter 1 d 1.5d lat hear tud to be loated 1.5d from the u ontrol perimeter out u out,ef ontrol perimeter additional reinforement may be required to omply with paing rule within bai ontrol perimeter Limited 011 Page 0

21 10. Retangular or Square Column Cirular Pattern a) Arranging tud When u out i alulated, the perimeter an be drawn around the loaded area and the zone inide an be reinfored with hear tud. Care hould be taken to enure the hear reinforement detailing rule deribed in etion 9 are implemented. The number of rail mut be aumed in order to loate u out. The entre line of eah orner rail mut be in line with the pivot point loated p from the loaded area. p i half of the horter ide of the loaded area ide but not more than 0.75d. The u out perimeter i reated with egment of line offet by 1.5d from the lat perimeter of the hear tud. The length of thee egment are not idential with the one in the middle of the quarter being the longet. Thi inreae an be ignored on the bai that the larger perimeter will have an inreaed load apaity and i therefore onidered to be a wort-ae enario. The angle between the rail in quarter (α) equate to: α = 90 / n Where: n the number of egment around u out on a irular layout pattern For equation to alulate in different ondition, ee Setion 10 (b). In order to loate the u out perimeter, p mut be alulated: p = / ( in(α/)) Therefore p 1 the ditane from the loaded area fae to lat tud equate to: p 1 = p - p - 1.5d / (o(α/)) When p 1 i worked out, the number of tud (n t )on the rail an eaily be alulated uing the following formula: n t > (p 1-0.5d)/ 0.75d The outome hould be rounded up to the nearet integer value. The next tep i to hek the maximum tangential paing between the tud on the rail in the two following ae - inide bai ontrol perimeter hek if 3 1.5d - outide bai ontrol perimeter hek if lat Where: 3 the maximum tangential paing inide the bai ontrol perimeter (uually between the 3 rd tud) lat the maximum tangential paing between the tud outide the bai ontrol perimeter To hek tangential paing, the ditane from the pivot point to the lat tud inide the bai ontrol perimeter (uually the third) and the ditane from the pivot point to the lat tud outide the bai ontrol perimeter mut be alulated. r 3 = p / o(α*) + ditane to firt tud + tud paing r lat = p / o(α*) + ditane to firt tud + (number of tud on rail - 1) tud paing Where: r 3 the ditane to the 3 rd tud r lat the ditane to the lat tud α* - angle might be multiplied with integer value depending on number of rail Limited 011 Page 1

22 10. Retangular or Square Column Cirular Pattern The two rail (A and B) with the longet r mut be hoen ( r for rail A and B are equal on the example below but thi i not a rule). The paing for both ae (inide and outide the bai ontrol perimeter) hould be alulated uing the following formula: t = ((r A + r B o(α)) + (r B in(α)) ) When the tangential paing exeed the maximum value, the number of rail hould be inreaed or intermediate paer rail hould be ued. u out ~ 1.5d / (o( /)) Rail A p max 0.75d max 0.75d lat = max 3 = max 1.5d p 1 max 0.75d r lat r 3 Rail B p 0.3d - 0.5d p /(o( /)) 1.5d 1 Limited 011 Page

23 10. Retangular or Square Column Cirular Pattern p b) Calulating in different ondition u out u out p p p p p 1 a lx 1 Internal olumn = (u out - ( 1 + ) + 8p )/ n Edge olumn = (u out p - a lx )/ n u out p a ly p a ly a lx 1 u out a lx 1 External orner olumn = (u out p - a lx - a ly )/ n Internal orner olumn = (u out p - a lx - a ly )/ n Limited 011 Page 3

24 11. Cirular Column Cirular Pattern a) Arranging tud When u out i alulated, the perimeter an be drawn around the loaded area and the zone inide an be reinfored with hear tud. Care hould be taken to enure the hear reinforement detailing rule deribed in etion 9 are implemented. The number of rail mut be aumed in order to loate u out. The entre line of eah rail mut be in line with entre point of the loaded area. p i half of the loaded area diameter. The u out perimeter for irular loaded area i reated with egment of line offet by 1.5d from the lat perimeter of the hear tud. Segment perpendiular to the radiu of the loaded area are alled. The length of thee egment i equal for the irular loaded area. The angle between the rail (α) equate to: α = 90 / n Where: n the total number of ''. The number of ' i defined by the 90 angle of eah quarter being plit into equal egment by the plaing rail. For equation to alulate in different ondition, ee Setion 11 (b). In order to loate the u out perimeter, p mut be alulated: p = / ( in(α/)) Therefore p 1 the ditane from the loaded area fae to lat tud equate to: p 1 = p - p - 1.5d / (o(α/)) When p 1 i worked out, the number of tud (n t )on the rail an eaily be alulated uing the following formula: n t > (p 1-0.5d)/ 0.75d The outome hould be rounded up to the nearet integer value. The next tep i to hek the maximum tangential paing between the tud on the rail in the following two ae - inide bai ontrol perimeter hek if 3 1.5d - outide bai ontrol perimeter hek if lat Where: 3 the maximum tangential paing inide the bai ontrol perimeter (uually between the 3 rd tud) lat the maximum tangential paing between the tud outide the bai ontrol perimeter To hek the tangential paing, the ditane from the entre point of the loaded area to the lat tud inide the bai ontrol perimeter (uually third) and the ditane from the entre point of the loaded area to the lat tud outide the bai ontrol perimeter mut be alulated. r 3 = p + the ditane to firt tud + tud paing r lat = p + the ditane to firt tud + (number of tud on rail - 1) tud paing Where: r 3 the ditane to the 3 rd tud r lat the ditane to the lat tud The paing for both ae (inide and outide the bai ontrol perimeter) hould be alulated uing the following formula: t = ((r + r o(α)) + (r in(α)) ) When tangential paing exeed the maximum value, the number of rail hould be inreaed or intermediate paer rail hould be ued. Limited 011 Page 4

25 11. Cirular Column Cirular Pattern 1.5d / (o( /)) Rail A max 0.75d lat p p 1 max 0.75d lat 3 = max 1.5d = max max 0.75d r 3 r Rail A p 0.3d - 0.5d 1.5d u out Limited 011 Page 5

26 11. Cirular Column Cirular Pattern b) Calulating in different ondition u out u out a lx p p Internal olumn = u out / n Edge olumn = (u out - a lx )/ n a lx u out a ly u out p a lx a ly p External orner olumn = (u out - a lx - a ly )/ n Internal orner olumn = (u out - a lx - a ly )/ n Limited 011 Page 6

27 F p min 1.5d 1. Cruiform Pattern The deign to EC uing the ruiform pattern look imilar to the irular pattern deign. The whole deign methodology deribed in etion 3 to 9 i valid for ruiform pattern deign. a) Arranging rail and tud The key differene appear in the rail arrangement. A the ruiform pattern ha no diagonal rail, the proedure of loating u out ef look lightly different. With the length of u out ef we an loate the u out ef perimeter. The minimum ditane between the u out ef perimeter and the entre point of the loaded area an be alulated uing the following expreion: p min = u out ef /8 - d (3π/ ) The number of tud i given by the formula below (the value hould be rounded up to the nearet integer): n t = (p min - 1() / - ditane to firt tud - 1.5d)/0.75d + 1 The minimum ditane (F) between the firt and lat rail on eah ide of the loaded area may be alulated uing the following expreion: F = u out ef / π d - The Ditane F hould be infilled with rail taking into aount the 1.5d paing rule between rail. The drawing below illutrate the point above. F 1.5d d d 1 u out ef When all the rail are et, the final tep i to alulate the required area of tud. The equation are idential to thoe for irular pattern deign (ee etion 8b for detail). Pleae note that only the rail with the firt tud at 0.3 d to 0.5 d from the loaded area fae and the lat tud at 1.5 d from the u out ef perimeter (hown in green above) an be taken to aount when alulating the area of the tud (i.e. in the drawing above only 8 rail an be taken to aount). Limited 011 Page 7

28 . l 13. Hole in the Slab In ae where the loaded area i ituated near a hole, "if the hortet ditane between the perimeter of the loaded area and the edge of the hole doe not exeed 6d, that part of the ontrol perimeter ontained between two tangent drawn to the outline of the hole from the entre of the loaded area i onidered to be ineffetive". (6.4. (3)) Beaue part of the u out ontrol perimeter beome ineffetive (u out Ineffetive ), the additional ontrol perimeter (u out ) mut be found, the effetive part of whih will be equal to u out o: u out = u out Effetive + u out Ineffetive In order to loate the u out ontrol perimeter we have to aume that the ontrol perimeter exit between the tangent line a well. Therefore: u out = u out / (1-δ/360) The part of the new ontrol perimeter whih i loated outide the dead zone (u out ) hould equal the required u out ontrol perimeter. The tud within the "dead zone" annot be ued in alulation whih might aue an inreae in the tud diameter required. Rail around the "dead zone" an be ut or moved. Alternatively, additional rail might be plaed to uit the hole and to omply with the paing rule. The drawing below how how to draw tangent line. Pleae note that part of the perimeter (u 0 red, red, u out Ineffetive ) loated within the "dead zone" are ineffetive and hould be ubtrated from the total length of the perimeter. Cae 1 - hole dim l 1 l Thee tud annot be ued in alulation u 0 red o red u out Ineffetive l 1 6d u out Effetive u out Effetive Limited 011 Page 8

29 . 13. Hole in the Slab Cae - hole dim l 1 > l Thee tud annot be ued in alulation u 0 red o (l. 1 l ) l. red l 1 u out Ineffetive 6d u out Effetive u out Effetive Limited 011 Page 9

30 14. Example Calulation Internal Condition a) Cirular Pattern - quare olumn (internal ondition) Data Slab depth h = 35 mm Column dimenion: 1 = 350 mm, = 350 mm Load V Ed = 1070 kn Cover = 5 mm (top and bottom) Reinforement T1 & T = 175/ Compreive trength of onrete f k = 30MPa Effetive depth of the lab d x = / d y = / d = (d x + d y )/ = (9 + 76)/ = 9 mm = 76 mm = 84 mm Punhing hear at the loaded area fae Eentriity fator β for internal olumn = 1.15 u 0 = 1 + = = 1400 mm v Ed 0 = β V Ed / (u 0 d) = / ( ) = 3.09 MPa f d = α f k / γ = 1 30 / 1.5 = 0 MPa v Rd.max = 0.3 f d (1 - (f k / 50)) = (1 - (30 / 50)) = 5.8 MPa Chek if v Ed 0 v Rd.max 3.09 MPa < 5.8 MPa Aepted. Punhing hear at the bai ontrol perimeter without reinforement = ( 1 + ) + 4π d = ( ) + 4 π 84 = mm C Rd. = 0.18 / γ = 0.18 / 1.5 = 0.1 k = 1+ (00 / d) = 1 + (00 / 84) = < v min = k 3/ f 1/ k = (1.839) 3/ (30) 1/ = MPa v Ed 1 = β V Ed / ( d) = / ( ) = 0.87 MPa Conider reinforement over =.054m width in both diretion from entre of olumn. For ρ l ue b = 1000mm. Uing 175/ in both diretion = mm /m T1 & T ρ l = ((A x / (b d x ) A y / (b d y )) = ( / (1000 9) / ( )) = = < 0.0 v Rd. = C Rd. k (100 ρ l f k ) 1/3 = ( ) 1/3 = MPa Chek if v Rd. > v min MPa > MPa Chek if v Ed 1 < v Rd MPa < MPa Shear reinforement required Punhing hear at the bai ontrol perimeter with reinforement u out req. = β V Ed / (v Rd. d) = / ( ) = mm 350/ = 175 mm therefore: poition rail entral about the loaded area fae in eah diretion Limited 011 Page 30

31 Example Calulation Internal Condition therefore p = 175 mm Try 1 rail. α = 360 /1 = 30 = u out / 1 = / 1 = mm p = /( in(α/)) = / ( in(30 /)) = = mm p 1 = p - p - 1.5d/ (o(α/)) = / (o(30 /)) = mm 0.75d = = 13 mm 0.5d = = 14 mm tud paing = 10 mm, ditane to firt tud = 140 mm Stud paing hek Ditane to 3 rd tud r 3 = 175/ o = 76.1 mm Ditane to lat tud r lat = 175/ o = 97.1 mm t 3 = ((r 3 - r 3 o(α)) + (r 3 in(α)) ) = = (( o(30 )) + (76.1 in(30 )) ) = mm < 1.5d = 46 mm t lat = ((r lat - r lat o(α)) + (r lat in(α)) ) = = (( o(30 )) + (97.1 in(30 )) ) = 503. mm < = 568 mm Area of hear reinforement f ywd.ef = d = = 31 N/mm f ywd = (f y / 1.15) = 500 / 1.15 = N/mm > 31 t = 503 mm, r = 10 mm, rail no. taken a 1 A w.min = 0.08 t r f k / (1.5 f yk ) = / ( ) = 61.7 mm A w1 = (v Ed v Rd. ) r / (1.5 f ywd.ef rail no.) A w1 = ( ) / ( ) = 88.8 mm tud dia = 1 mm (A = mm ) Provide 1 No (1357. mm ). Spaing: 140/10/10/10/ Stud total u out req. Limited 011 Page 31 1 No /10/10/10/140

32 Example Calulation Edge Condition a) Cirular Pattern - irular olumn with hole (edge ondition) Data Slab depth h = 300 mm Slab edge offet a lx = 900 mm Column dia = 300 mm Load V Ed = 610 kn Cover = 30 mm (top and bottom) Reinforement T1 = 150/ Reinforement T = 150/ Compreive trength of onrete f k = 30MPa Yield trength of reinforement f y = 500MPa Hole data - a per drawing lab edge Effetive depth of the lab d x = / d y = / d = (d x + d y )/ = (6 + 46)/ 900 = 6 mm = 46 mm = 54 mm Punhing hear at the loaded area fae a lx = 900 < π/4 ( + 4 d) = π/4 ( ) = therefore by interpolation eentriity fator β = a lx - / = / = 750 > = 508 B N,B S,B W = 0.5 π = 0.5 π 300 = 35.6 mm Angle δ = ar tan ((100)/(500)) =.6 u 0 red =.6/360 π 300 = 59. mm u 0 = 0.5 π + B N + B S + B W - u 0 red = 0.5 π = mm v Ed 0 = β V Ed / (u 0 d) = / ( ) = 3.3 MPa f d = α f k / γ = 1 30 / 1.5 = 0 MPa v Rd.max = 0.3 f d (1 - (f k / 50)) = (1 - (30 / 50)) = 5.8 MPa Chek if v Ed 0 v Rd.max 3.3 MPa < 5.8 MPa Aepted. Punhing hear at the bai ontrol perimeter without reinforement A = ( + 4d) π - Ared = ( ) π = mm B = ( + 4d) π/ + a lx - Bred = ( ) π/ = mm A > B = mm C Rd. = 0.18 / γ = 0.18 / 1.5 = 0.1 k = 1+ (00 / d) = 1 + (00 / 54) = < v min = k 3/ f 1/ k = (1.887) 3/ (30) 1/ = MPa v Ed 1 = β V Ed / ( d) = / ( ) = MPa Conider reinforement over = 184 mm width in both diretion from entre of the loaded area. For ρ l ue b = 1000mm. Limited 011 Page 3

33 Example Calulation Edge Condition lab edge lab edge u = mm 1A u = mm 1B Uing 150/ in both diretion = mm /m T1 Uing 150/ in both diretion = mm /m T ρ l = ((A x / (b d x ) A y / (b d y )) = (1340.4/ (1000 6) / ( )) = = < 0.0 v Rd. = C Rd. k (100 ρ l f k ) 1/3 = ( ) 1/3 = MPa Chek if v Rd. > v min MPa > MPa Chek if v Ed 1 < v Rd MPa < MPa Shear reinforement required.6 lab edge u out req. = 4933 mm u out req. Limited 011 Page 33

34 15. Example Calulation Edge Condition Punhing hear at the bai ontrol perimeter with reinforement u out req. = β V Ed / (v Rd. d) = / ( ) = 5015 mm 0.5 u out req. = = > a lx = 900 mm u out req. ha U hape. Try mid pur in quarter no.of = (+1) = 6 = (u out req. - a lx )/ no.of = ( ) /6 = mm α = 90 / (mid pur in quarter + 1) = 90 /3 = 30 p = /( in (α/)) = /( in (30 /)) = mm Angle δ =.6 γ = 6.0 u out req. = u out req. / (1 - δ/ γ ) = 5015 / (1 -.6 /6.0 ) = mm = (u out req. - a lx )/ no.of = ( ) /6 = mm p = /( in (α/)) = /( in (30 /)) = mm p = / = 300/ = 150 mm p 1 = p - p - 1.5d /(o (α /)) = /(o (30 /)) = mm 0.5d = = 17 mm ditane to firt tud = 15 mm p 3 = p 1-1 t tud to olumn dit. = = mm 0.75d = = mm /190.5= /3=17.8 tud paing = 175 mm, 4 tud on a rail. Stud paing hek Rail A - ditane to lat tud r 3 A = t 3 = r 3 A in(α/) = 65 in(30/) = 33.5 mm < 1.5d = 381 mm Rail A - ditane to lat tud r lat A = t lat = r lat A in(α/) = 800 in(30/) = mm < = 508 mm = 65 mm = 800 mm Area of hear reinforement f ywd.ef = d = = N/mm f ywd = (f y / 1.15) = 500 / 1.15 = N/mm > t = mm, r = 175 mm, rail no. taken a 7 (for rail within perimeter) A w.min = 0.08 t r f k / (1.5 f yk ) = / ( ) = 4.3 mm A w1 = (v Ed v Rd. ) r / (1.5 f ywd.ef rail no.) A w1 = ( ) / ( ) = 69.9 mm tud dia = 10 mm (A = 78.5 mm ) Limited 011 Page 34

35 Example Calulation Edge Condition Provide: 13 No Spaing: 15/175/175/175/15 No Spaing: 15/175/175/15 58 Stud in total u out req. = 5015 mm Limited 011 Page 35

36 16. Example Calulation External Corner Condition 300 a) Cirular Pattern - quare olumn Data Slab depth h = 400 mm Column dimenion: 1 =300 mm, =300 mm Load V Ed = 30 kn Cover = 30mm (top and bottom) Reinforement T1 & T = 150/ Compreive trength of onrete f k = 30MPa Ditane to lab edge a lx = 0 mm, a ly = 0 mm Effetive depth of the lab d x = / d y = / d = (d x + d y )/ = ( )/ Punhing hear at the loaded area fae a lx = 0, a ly = 0 eentriity fator β = 1.5 B S = min. ( 1, 1 + a lx, 1.5d) = min. (300, 300, 531) B E = min. (, + a ly, 1.5d) = min. (300, 300, 531) u 0 = B S + B E = v Ed 0 = β V Ed / (u 0 d) = / ( ) f d = α f k / γ = 1 30 / 1.5 v Rd.max = 0.3 f d (1 - (f k / 50)) = (1 - (30 / 50)) Chek if v Ed 0 v Rd.max.6 MPa < 5.8 MPa Aepted. 300 = 36 mm = 346 mm = 354 mm = 300 mm = 300 mm = 600mm =.6 MPa = 0 MPa = 5.8 MPa Punhing hear at the bai ontrol perimeter without reinforement = 1 + +π d = π 54 = mm C Rd. = 0.18 / γ = 0.18/1.5 = 0.1 k = 1+ (00 / d) = 1 + (00/ 354) = 1.75 < v min = k 3/ f 1/ k = (1.75) 3/ (30) 1/ = MPa v Ed 1 = β V Ed / ( d) = / ( ) = 0.79 MPa Conider reinforement over = 136 mm width in both diretion from entre of the loaded area. For ρ l ue b = 1000mm. Uing 150/ in both diretion = mm /m ρ l = ((A x / (b d x ) A y / (b d y )) = ( / / 346) / 1000 = = < 0.0 v Rd. = C Rd. k (100 ρ l f k ) 1/3 = ( ) 1/3 T1 & T = MPa Chek if v Rd. > v min MPa > MPa Chek if v Ed 1 < v Rd MPa < MPa Shear reinforement required Limited 011 Page 36

37 Example Calulation External Corner Condition Punhing hear at the bai ontrol perimeter with reinforement u out req = β V Ed / (v Rd. d) = / ( ) = mm 300/ = 150 mm therefore: poition rail entral about the loaded area fae in eah diretion. Therefore p = 150 mm Try 4 rail. α = 90 / (4-1) = 30 = (u out - p ) / (4-1) = ( ) / 3 = mm p = /( in(α/)) = / ( in(30 /)) = = mm p 1 = p - p - 1.5d/ (o(α/)) = / (o(30 /)) = mm 0.75d = = 65.5 mm 0.5d = = 177 mm tud paing = 60 mm, ditane to firt tud = 175 mm Stud paing hek Ditane to 3 rd tud r 3 = 150/ o = 868. mm Ditane to lat tud r lat = 150/ o = 118. mm t 3 = ((r 3 - r 3 o(α)) + (r 3 in(α)) ) = = (( o(30 )) + (868. in(30 )) ) = mm < 1.5d = 531 mm t lat = ((r lat - r lat o(α)) + (r lat in(α)) ) = = (( o(30 )) + (118. in(30 )) ) = mm < = 708 mm Area of hear reinforement f ywd.ef = d = = N/mm f ywd = (f y / 1.15) = 500 / 1.15 = N/mm > t = mm, r = 60 mm, rail no. taken a 4 A w.min = 0.08 t r f k / (1.5 f yk ) = / ( ) = 88.7 mm A w1 = (v Ed v Rd. ) r / (1.5 f ywd.ef rail no.) A w1 = ( ) / ( ) = 95.9 mm tud dia = 1 mm (A = mm ) Provide 4 No Spaing: 175/60/60/60/175 (45.4 mm ) 16 Stud total Rail layout 584 u out req. Limited 011 Page 37 4 No

38 17. Example Calulation Internal Corner Condition 400 a) Cirular Pattern - quare olumn Data Slab depth h = 350 mm Column dimenion: 1 =400 mm, =400 mm Load V Ed = 1100 kn Cover = 5 mm (top and bottom) Reinforement T1 & T = 175/ Compreive trength of onrete f k = 40MPa Ditane to lab edge a lx = 0 mm, a ly = 0 mm Effetive depth of the lab d x = / d y = / d = (d x + d y )/ = ( )/ 400 = 315 mm = 95 mm = 305 mm Punhing hear at the loaded area fae a lx = 0, a ly = 0 eentriity fator β = 1.75 B N = min. ( 1, 1 + a lx, 1.5d) = min. (400, 400, 457.5) B W = min. (, + a ly, 1.5d) = min. (400, 400, 457.5) u 0 = B N + B W = v Ed 0 = β V Ed / (u 0 d) = / ( ) f d = α f k / γ = 1 40 / 1.5 v Rd.max = 0.3 f d (1 - (f k / 50)) = (1 - (40 / 50)) = 400 mm = 400 mm = 1600mm =.87 MPa = 6.67 MPa = 6.7 MPa Chek if v Ed 0 v Rd.max.87 MPa < 6.7 MPa Aepted. Punhing hear at the bai ontrol perimeter without reinforement = π d = π 305 = mm C Rd. = 0.18 / γ = 0.18/1.5 = 0.1 k = 1+ (00 / d) = 1 + (00/ 305) = 1.81 < v min = k 3/ f 1/ k = (1.81) 3/ (40) 1/ = MPa v Ed 1 = β V Ed / ( d) = / ( ) = 1.08 MPa Conider reinforement over = 1315 mm width in both diretion from entre of the loaded area. For ρ l ue b = 1000mm. Uing 175/ in both diretion = mm /m T1 & T ρ l = ((A x / (b d x ) A y / (b d y )) = (1795. / / 95) / 1000 = = < 0.0 v Rd. = C Rd. k (100 ρ l f k ) 1/3 = ( ) 1/3 = 0.63 MPa Chek if v Rd. > v min 0.63 MPa > MPa Chek if v Ed 1 < v Rd MPa < 0.63 MPa Shear reinforement required Limited 011 Page 38

39 Example Calulation Internal Corner Condition Punhing hear at the bai ontrol perimeter with reinforement u out required = β V Ed / (v Rd. d) = / ( ) = mm 400/ = 00 mm therefore: poition rail entral about the loaded area fae in eah diretion. Therefore p = 00 mm Try 4 rail in quarter. α = 90 / (4-1) = 30 = (u out - p ) /9 = ( ) / 9 = mm p = /( in(α/)) = / ( in(30 /)) = = mm p 1 = p - p - 1.5d/ (o(α/)) = / (o(30 /)) = 86.0 mm 0.75d = = 8.75 mm 0.5d = = 15.5 mm tud paing = 30 mm, ditane to firt tud = 150 mm Stud paing hek Ditane to 3 rd tud r 3 = 00/ o = mm Ditane to lat tud r lat = 00/ o = mm t 3 = ((r 3 - r 3 o(α)) + (r 3 in(α)) ) = = (( o(30 )) + (840.9 in(30 )) ) = mm < 1.5d = mm t lat = ((r lat - r lat o(α)) + (r lat in(α)) ) = = (( o(30 )) + ( in(30 )) ) = mm < = 610 mm Area of hear reinforement f ywd.ef = d = = 36.3 N/mm f ywd = (f y / 1.15) = 500 / 1.15 = N/mm > 36.3 t = mm, r = 30 mm, rail no. taken a 1 A w.min = 0.08 t r f k / (1.5 f yk ) = / ( ) = 86.0 mm A w1 = (v Ed v Rd. ) r / (1.5 f ywd.ef rail no.) A w1 = ( ) / ( ) = 98.3 mm tud dia = 1 mm (A = mm ) Provide 1 No Spaing: 150/30/30/30/150 (1357. mm ) 48 Stud total 554 Rail Layout u out req. 1 No /30/30/30/150 Limited 011 Page Stud total