LinkStudPSR Speialit in Punhing Shear Reinforement Deign Manual to EC BS EN 199-1-1:004 Verion 3.1 January 018 LinkStud PSR Limited /o Brook Forging Ltd Doulton Road Cradley Heath Wet Midland B64 5QJ Tel: 08456 58 58 www.linktudpr.om
1. Introdution 4. Lit of Symbol 5 3. Information Required to Deign 7 4. Eentriity Fator β 8 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 5. Effetive Depth of Slab 1 8 9 10 Content 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 13 13 13 14 7. 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 15 15 17 17 8. 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 18 18 18 9. Shear Reinforement Detailing Rule 19 10. Retangular or Square Column Cirular Pattern a) Arranging tud b) Calulating in di erent ondition 1 1 3 LinkStudPSR Limited Page
11. Cirular Column Cirular Pattern a) Arranging tud b) Calulating in di erent ondition 4 4 6 1. Cruiform Pattern a) Arranging rail and tud 7 7 13. Hole in the lab 8 14. Example Calulation Internal Condition 30 a) Cirular pattern quare olumn (internal ondition) 30 Content 15. Example Calulation Edge Condition a) Cirular pattern irular olumn with hole (edge ondition) 3 3 16. Example Calulation External Corner Condition a) Cirular pattern quare orner 36 36 17. Example Calulation Internal Corner Condition a) Cirular pattern quare olumn 38 38 LinkStudPSR Limited Page 3
The LinkStudPSR 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 LinkStudPSR 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 LinkStudPSR hear tud are welded to arrier / paer rail to allow them to be loated orretly and to be upported by the top flexural reinforement. 1. Introdution LinkStudPSR 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 LinkStudPSR pakage. From appliation advie and deign guidane, through propoal drawing, alulation and quotation, to working drawing and ite upport, you an depend on LinkStudPSR for all your Punhing Shear Reinforement need. Kind Regard Dariuz Nowik MS (Eng) Senior Deign Engineer LinkStudPSR Limited LinkStudPSR Limited Page 4
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 199-1-1-004 6.4.4 (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 199-1-1-004, 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 6.4.4 (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. Lit of Symbol LinkStudPSR Limited Page 5
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-199-1-1-004, Fig. 6.19.. Lit of Symbol LinkStudPSR Limited Page 6
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 9.4.1 and 9.4. (1) 3. Information Required to Deign 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. LinkStudPSR Limited Page 7
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. 6.4.3 (6) The reommended value of the eentriity fator β are li ted 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 ly a 4. Eentriity Fator β β = max (β x, β y, β xy ) Where: β x = 1.4-0.5 ( a lx /( + πd)) β y = 1.4-0.5 ( a ly /( 1 + πd)) β xy = 1.5-0.35 (a lx + a ly ) /( 1 + + 3πd) 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 d d orner point =1.75 1d/ =1.15 / + d =1.4 =1.15 / + d =1.4 LinkStudPSR Limited Page 8
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. External orner ondition β = max (β x, β y, β xy ) Where: β x = 1.4-0.5 4 a lx / (π ( + 4d)) β y = 1.4-0.5 4 a ly / (π ( + 4d)) β xy = 1.5-0.35 ( a lx + a ly ) / (3/4 π ( + 4d))) a ly a lx a lx 4. Eentriity Fator β 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 3d/ d orner point = 1.75 /4 ( + 4d) =1.4 =1.15 /4 ( + 4d) =1.4 LinkStudPSR Limited Page 9
) 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 ) 6.4.3 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 / 0.5 1.0.0 3.0 4. Eentriity Fator β k 0.45 0.6 0.7 0.8 W 1 for retangular loaded area i equal: W 1 = 0.5 1 + 1 + 4 d + 16d + π d 1 6.4.3 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 = 1 + 0.6π M Ed /(V Ed ( + 4d)) 6.4.3 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 = 1 + 1.8 (M Edx /(V Ed ( 1 +4d)) + M Edy /(V Ed ( +4d)) ) 6.4.3 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 1 6.4.3 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 / LinkStudPSR Limited Page 10
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 = 0.5 + 1 + 4 1 d + 8d + π d 6.4.3 equation 6.45 W 1 for irular loaded area i equal: W 1 = + 6d + 8d 1.5d 0.5 1 1.5d (+4d)/8 Drawing 6.0a (inluding the irular loaded area) 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 ) 1 u* 1 * 4. Eentriity Fator β 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 0.5 1 1.5d (+4d)/8 Drawing 6.0b (inluding the irular loaded area) u* 1 1.5d 0.5 * 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 ) 6.4.3 equation 6.39 W 1 for a retangular loaded area (external orner ondition) i equal: W 1 = 0.5 1 + 1 + 0.5 + 4d + 0.5d π W 1 for a irular loaded area (external orner ondition) i equal: W 1 = 11 /8 + 9d + 16d LinkStudPSR Limited Page 11
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 T T T T T T T T T T T B Column B B B B B B B B B B Column T1 B1 5. Effetive Depth of Slab T1 T T d y d x B B B1 LinkStudPSR Limited Page 1
a) Perimeter of the Loaded Area Condition for retangular / quare olumn 1 u 0 Internal Edge External orner Internal orner u 0 = 1 + u 0 = B N + B S + u 0 = B E + B S u 0 = 1 + + 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 a lx a lx B N 1 B N B S u 0 0.5 a ly 1 a ly B E u 0 a lx B u S a lx 1 0 a lx u 0 B E B W a lx B N a ly a ly 0.5 B W B N 6. Punhing Shear at the Loaded Area 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) 6.4.5 (3) 6.53 LinkStudPSR Limited Page 13
) Deign value of the maximum punhing hear reitane at the Loaded Area f d = α f k /γ 3.1.6 (3.15) γ = 1.5 α = 1 v Rd.max = 0.3 f d (1 - (f k / 50)) NA to BS EN 199-1-1-004 6.4.5 (3) note, 6..(6) 6.6N Where: f d the value of the deign ompreive trength of onrete. 3.1.6 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 199-1-1-004 3.1.6 (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. 6.4.3 (a) 6. Punhing Shear at the Loaded Area LinkStudPSR Limited Page 14
a) Bai Control Perimeter length The Bai Control Perimeter i loated from the loaded area Internal olumn Edge olumn 1 = 1 + + 4πd = π (4d + ) a lx 1 = 1 + + πd + a lx = π/ (4d + ) + a lx a lx 7. Punhing Shear at the Bai Control Perimeter without Reinforement LinkStudPSR Limited Page 15
External orner olumn Internal orner olumn a lx = 1 + + πd + a lx + a ly = π/4 (4d + ) + a lx + a ly a lx a ly 1 1 = ( 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) a ly a lx a lx a ly a ly 7. Punhing Shear at the Bai Control Perimeter without Reinforement LinkStudPSR Limited Page 16
b) Deign value of the maximum hear tre at the Bai Control Perimeter v Ed 1 = β V Ed / ( d) 6.4.3 (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 ) 6.4.4 (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 6.4.3 (b) 7. Punhing Shear at the Bai Control Perimeter without Reinforement LinkStudPSR Limited Page 17
a) Deign value of the punhing hear reitane at the Bai Control Perimeter f ywd.ef = 50 + 0.5 d f ywd = (f y / 1.15) 6.4.5 (6.5) v Rd. = 0.75 v Rd. + 1.5 (d / r ) A w f ywd.ef (1 / ( d)) v Ed 1 6.4.5 (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 1-0.75 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. 8. Punhing Shear at the Bai Control Perimeter with Reinforement LinkStudPSR Limited Page 18
Shear reinforement hould be detailed in aordane with BS EN 199-1-1:004 6.4.5 (4), 9.4.3. and NA to BS EN 199-1-1-004 6.4.5 (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. 9. Shear Reinforement Detailing Rule 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 LinkStudPSR Limited Page 19
d d 1.5d additional reinforement may be required to omply with paing rule within bai ontrol perimeter d 1.5d u - bai ontrol perimeter 1 u lat hear tud to be loated 1.5d from the u ontrol perimeter out,ef out ontrol perimeter 9. Shear Reinforement Detailing Rule LinkStudPSR Limited Page 0
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 10. Retangular or Square Column Cirular Pattern 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 LinkStudPSR Limited Page 1
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. p ~ 1.5d / (o( /)) p 1 p max 0.75d max 0.75d max 0.75d 0.3d - 0.5d u out 1 r p /(o( /)) lat r 3 Rail A = max 1.5d 3 = max lat Rail B 1.5d 10. Retangular or Square Column Cirular Pattern LinkStudPSR Limited Page
Deign Manual to EC b) Calulating in different ondition p p u out Internal olumn = (u out - ( 1 + ) + 8p )/ n p 1 p a ly p a lx 1 u out p Edge olumn = (u out - 1 - + 4p - a lx )/ n u out p p a ly 10. Retangular or Square Column Cirular Pattern a lx 1 u out a lx 1 External orner olumn = (u out - 1 - + p - a lx - a ly )/ n Internal orner olumn = (u out - 1 - + 6p - a lx - a ly )/ n LinkStudPSR Limited Page 3
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. 11. Cirular Column Cirular Pattern 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. LinkStudPSR Limited Page 4
p 1.5d / (o( /)) p p 1 max 0.75d max 0.75d max 0.75d 0.3d - 0.5d r r lat 3 Rail A = max 1.5d 3 = max lat Rail A 1.5d u out 11. Cirular Column Cirular Pattern LinkStudPSR Limited Page 5
Deign Manual to EC b) Calulating in different ondition p Internal olumn = u out / n u out a lx p Edge olumn = (u out - a lx )/ n u out 11. Cirular Column Cirular Pattern 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 LinkStudPSR Limited Page 6
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/8-0.5 - ) 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 /4-0.75 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. 1. Cruiform Pattern F d d p 1.5d 1.5d F min Deign Manual to EC 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). LinkStudPSR Limited Page 7
. 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)) l Deign Manual to EC 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. 13. Hole in the Slab 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 LinkStudPSR Limited Page 8
. Cae - hole dim l 1 > l u 0 red Thee tud annot be ued in alulation o. Deign Manual to EC (l l ) 1. red l 1 u out Ineffetive l 13. Hole in the Slab 6d u out Effetive u out Effetive LinkStudPSR Limited Page 9
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 = H16 @ 175/ Compreive trength of onrete f k = 30MPa Effetive depth of the lab d x = 35-5 - 16/ d y = 35-5 - 16-16/ 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 + = 350 + 350 = 1400 mm v Ed 0 = β V Ed / (u 0 d) = 1.15 1070 1000 / (1400 84) = 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)) = 0.3 0 (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 = (350+350) + 4 π 84 = 4968.8mm C Rd. = 0.18 / γ = 0.18 / 1.5 = 0.1 k = 1+ (00 / d) = 1 + (00 / 84) = 1.839 < v min = 0.035 k 3/ f 1/ k = 0.035 (1.839) 3/ (30) 1/ = 0.478 MPa v Ed 1 = β V Ed / ( d) = 1.15 1070 1000 / (4968.8 84) = 0.87 MPa 14. Example Calulation Internal Condition Conider reinforement over 350 + 6 84 =.054m width in both diretion from entre of olumn. For ρ l ue b = 1000mm. Uing H16 @ 175/ in both diretion = 1148.9 mm /m T1 & T ρ l = (( A x / (b d x ) A y / (b d y )) = (1148.9 / (1000 9) 1148.9 / (1000 76)) = = 0.00405 < 0.0 v Rd. = C Rd. k (100 ρ l f k ) 1/3 = 0.1 1.839 (100 0.00405 30) 1/3 = 0.507 MPa Chek if v Rd. > v min 0.507 MPa > 0.478 MPa Chek if v Ed 1 < v Rd. 0.87 MPa < 0.507 MPa Shear reinforement required Punhing hear at the bai ontrol perimeter with reinforement u out req. = β V Ed / (v Rd. d) = 1.15 1070 1000 / (0.507 84) = 8541.5 mm 350/ = 175 mm therefore: poition rail entral about the loaded area fae in eah diretion LinkStudPSR Limited Page 30
therefore p = 175 mm Try 1 rail. α = 360 /1 = 30 = u out / 1 = 8541.5 / 1 = 711.8 mm p = /( in(α/) ) = 711.8 / ( in(30 /)) = = 1375.1 mm p 1 = p - p - 1.5d/ (o(α/)) = 1375.1-175 - 1.5 84/ (o(30 /)) = 759.1 mm 0.75d = 0.75 84 = 13 mm 0.5d = 0.5 84 = 14 mm tud paing = 10 mm, ditane to firt tud = 140 mm Stud paing hek Ditane to 3 rd tud r 3 = 175/ o 30 + 140 + 10 = 76.1 mm Ditane to lat tud r lat = 175/ o 30 + 140 + 3 10 = 97.1 mm t 3 = ((r 3 - r 3 o(α)) + (r 3 in(α)) ) = = ((76.1-76.1 o(30 )) + (76.1 in(30 )) ) = 394.5 mm < 1.5d = 46 mm t lat = ((r lat - r lat o(α)) + (r lat in(α)) ) = = ((97.1-97.1 o(30 )) + (97.1 in(30 )) ) = 503. mm < = 568 mm Area of hear reinforement f ywd.ef = 50 + 0.5 d = 50 + 0.5 84 = 31 N/mm f ywd = (f y / 1.15) = 500 / 1.15 = 434.8 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 ) = 0.08 503 10 30 / (1.5 500) = 61.7 mm A w1 = (v Ed 1-0.75 v Rd. ) r / (1.5 f ywd.ef rail no.) A w1 = (0.87-0.75 0.507) 4968.8 10 / (1.5 31.0 1) = 88.8 mm tud dia = 1 mm (A = 113.1 mm ) Provide 1 No 1-4-75-910 (1357. mm ). Spaing: 140/10/10/10/140 48 Stud total 394 503 14. Example Calulation Internal Condition 350 350 140 10 10 10 u out req. LinkStudPSR Limited Page 31 1 No 1-4-75-910 Spaed @ 140/10/10/10/140
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 = H16 @ 150/ Reinforement T = H16 @ 150/ Compreive trength of onrete f k = 30MPa Yield trength of reinforement f y = 500MPa Hole data - a per drawing Effetive depth of the lab d x = 300-30 - 16/ d y = 300-30 - 16-16/ d = (d x + d y )/ = (6 + 46)/ = 6 mm = 46 mm = 54 mm Punhing hear at the loaded area fae a lx = 900 < π/4 ( + 4 d) = π/4 (300 + 4 54) = 1033.6 therefore by interpolation eentriity fator β = 1.188 a lx - / = 900-300/ = 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 300 + 3 35.6-59. = 883.3 mm v Ed 0 = β V Ed / (u 0 d) = 1.188 610 1000 / (883.3 54) = 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)) = 0.3 0 (1 - (30 / 50)) = 5.8 MPa Chek if v Ed 0 v Rd.max 3.3 MPa < 5.8 MPa Aepted. lab edge Punhing hear at the bai ontrol perimeter without reinforement A = ( + 4d) π - Ared = (300 + 4 54) π - 59.8 = 3874.6 mm B = ( + 4d) π/ + a lx - Bred = (300 + 4 54) π/ + 900-61.5 = 3605.7 mm A > B = 3605.7 mm C Rd. = 0.18 / γ = 0.18 / 1.5 = 0.1 k = 1+ (00 / d) = 1 + (00 / 54) = 1.887 < v min = 0.035 k 3/ f 1/ k = 0.035 (1.887) 3/ (30) 1/ = 0.497 MPa v Ed 1 = β V Ed / ( d) = 1.188 610 1000 / (3605.7 54) = 0.791 MPa Conider reinforement over 300 + 6 54 = 184 mm width in both diretion from entre of the loaded area. For ρ l ue b = 1000mm. 900 100 500 300 100 00 59. 35.6 15. Example Calulation Edge Condition LinkStudPSR Limited Page 3
lab edge 500 100 59.8 00.6 100 u = 3874.6 mm 1A lab edge Uing H16 @ 150/ in both diretion = 1340.4 mm /m T1 Uing H16 @ 150/ in both diretion = 1340.4 mm /m T ρ l = ((A x / (b d x ) A y / (b d y )) = (1340.4/ (1000 6) 1340.4/ (1000 46)) = = 0.0058 < 0.0 v Rd. = C Rd. k (100 ρ l f k ) 1/3 = 0.1 1.887 (100 0.0058 30) 1/3 = 0.569 MPa Chek if v Rd. > v min 0.569 MPa > 0.497 MPa Chek if v Ed 1 < v Rd. 0.791 MPa < 0.569 MPa Shear reinforement required 500 1B 100 61.5 00.6 100 u = 3605.7 mm 15. Example Calulation Edge Condition.6 lab edge 900 63.5 1158 1009 u out req. = 4933 mm u out req. LinkStudPSR Limited Page 33
Punhing hear at the bai ontrol perimeter with reinforement u out req. = β V Ed / (v Rd. d) = 1.188 610 1000 / (0.569 54) = 5015 mm 0.5 u out req. = 0.5 5015 = 153.8 > 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 = (5015-900) /6 = 535.8 mm α = 90 / (mid pur in quarter + 1) = 90 /3 = 30 p = /( in (α/)) = 535.8 /( in (30 /)) = 1035.1 mm Angle δ =.6 γ = 6.0 u out req. = u out req. / (1 - δ/ γ ) = 5015 / (1 -.6 /6.0 ) = 5489.1 mm = (u out req. - a lx )/ no.of = (5489.1-900) /6 = 614.8 mm p = /( in (α/)) = 614.8 /( in (30 /)) = 1187.8 mm p = / = 300/ = 150 mm p 1 = p - p - 1.5d /(o (α /)) = 1187.8-150 - 1.5 54 /(o (30 /)) = 643.4 mm 0.5d = 0.5 54 = 17 mm ditane to firt tud = 15 mm p 3 = p 1-1 t tud to olumn dit. = 643.4-15 = 518.4 mm 0.75d = 0.75 54 = 190.5 mm 518.4 /190.5=.7 518.4 /3=17.8 tud paing = 175 mm, 4 tud on a rail. Stud paing hek Rail A - ditane to lat tud r 3 A = 150 + 15 + 175 = 65 mm 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 = 150 + 15 + 3 175 = 800 mm t lat = r lat A in(α/) = 800 in(30/) = 414.1 mm < = 508 mm Area of hear reinforement f ywd.ef = 50 + 0.5 d = 50 + 0.5 54 = 313.5 N/mm f ywd = (f y / 1.15) = 500 / 1.15 = 434.8 N/mm > 313.5 t = 414.1 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 ) = 0.08 414.1 175 30 / (1.5 500) = 4.3 mm A w1 = (v Ed 1-0.75 v Rd. ) r / (1.5 f ywd.ef rail no.) A w1 = (0.791-0.75 0.569) 3605.7 175 / (1.5 313.5 7) = 69.9 mm tud dia = 10 mm (A = 78.5 mm ) 15. Example Calulation Edge Condition LinkStudPSR Limited Page 34
Provide: 13 No 10-4-40-775. Spaing: 15/175/175/175/15 No 10-3-40-600. Spaing: 15/175/175/15 58 Stud in total. 77 80 376 34 414 15 175 175 175 15 15. Example Calulation Edge Condition 374 175 175 137 175 175 175 175 137 u out req. = 5015 mm 80 80 80 LinkStudPSR Limited Page 35
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 = H16 @ 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 = 400-30 - 16/ d y = 400-30 - 16-16/ d = (d x + d y )/ = (36 + 346)/ = 36 mm = 346 mm = 354 mm 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) = 300 mm B E = min. (, + a ly, 1.5d) = min. (300, 300, 531) = 300 mm u 0 = B S + B E = 300 + 300 = 600mm v Ed 0 = β V Ed / (u 0 d) = 1.5 30 1000 / (600 354) =.6 MPa f d = α f k / γ = 1 30 / 1.5 = 0 MPa v Rd.max = 0.3 f d (1 - (f k / 50)) = 0.3 0 (1 - (30 / 50)) = 5.8 MPa Chek if v Ed 0 v Rd.max.6 MPa < 5.8 MPa Aepted. Punhing hear at the bai ontrol perimeter without reinforement = 1 + +π d = 300 + 300 + π 54 = 171.1 mm C Rd. = 0.18 / γ = 0.18/1.5 = 0.1 k = 1+ (00 / d) = 1 + (00/ 354) = 1.75 < v min = 0.035 k 3/ f 1/ k = 0.035 (1.75) 3/ (30) 1/ = 0.444 MPa v Ed 1 = β V Ed / ( d) = 1.5 30 1000 / ( 171.1 354) = 0.79 MPa Conider reinforement over 300 + 3 354 = 136 mm width in both diretion from entre of the loaded area. For ρ l ue b = 1000mm. 300 300 16. Example Calulation External Corner Condition Uing H16 @ 150/ in both diretion = 1340.4 mm /m ρ l = ((A x / (b d x ) A y / (b d y )) = (1340.4 / 36 1340.4 / 346) / 1000 = = 0.00379 < 0.0 v Rd. = C Rd. k (100 ρ l f k ) 1/3 1/3 = 0.1 1.935 (100 0.00586 30) T1 & T = 0.604 MPa Chek if v Rd. > v min 0.473 MPa > 0.444 MPa Chek if v Ed 1 < v Rd. 0.7 9 MPa < 0.473 MPa Shear reinforement required LinkStudPSR Limited Page 36
Punhing hear at the bai ontrol perimeter with reinforement u out req = β V Ed / (v Rd. d) = 1.5 30 1000 / (0.473 354) = 869.4 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) = (869.4-150) / 3 = 856.5 mm p = /( in(α/) ) = 856.5 / ( in(30 /)) = = 1654.6 mm p 1 = p - p - 1.5d/ (o(α/)) = 1654.6-150 - 1.5 354/ (o(30 /)) = 954.8 mm 0.75d = 0.75 354 = 65.5 mm 0.5d = 0.5 354 = 177 mm tud paing = 60 mm, ditane to firt tud = 175 mm Stud paing hek Ditane to 3 rd tud r 3 = 150/ o 30 + 175 + 60 = 868. mm Ditane to lat tud r lat = 150/ o 30 + 175 + 3 60 = 118. mm t 3 = ((r 3 - r 3 o(α)) + (r 3 in(α)) ) = = ((868. - 868. o(30 )) + (868. in(30 )) ) = 449.4 mm < 1.5d = 531 mm t lat = ((r lat - r lat o(α)) + (r lat in(α)) ) = = ((118. - 118. o(30 )) + (118. in(30 )) ) = 584.0 mm < = 708 mm Area of hear reinforement f ywd.ef = 50 + 0.5 d = 50 + 0.5 354 = 338.5 N/mm f ywd = (f y / 1.15) = 500 / 1.15 = 434.8 N/mm > 338.5 t = 584.0 mm, r = 60 mm, rail no. taken a 4 A w.min = 0.08 t r f k / (1.5 f yk ) = 0.08 584.0 60 30 / (1.5 500) = 88.7 mm A w1 = (v Ed 1-0.75 v Rd. ) r / (1.5 f ywd.ef rail no.) A w1 = (0.79-0.75 0.473) 171.1 60 / (1.5 338.5 4) = 95.9 mm tud dia = 1 mm (A = 113.1 mm ) Provide 4 No 1-4-340-1130 Spaing: 175/60/60/60/175 (45.4 mm ) 16 Stud total 300 300 175 60 60 60 175 16. Example Calulation External Corner Condition 449 Rail layout 584 u out req. LinkStudPSR Limited Page 37
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 = H0 @ 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 = 350-5 - 0/ d y = 350-5 - 0-0/ d = (d x + d y )/ = (315 + 95)/ = 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) = 400 mm B W = min. (, + a ly, 1.5d) = min. (400, 400, 457.5) = 400 mm u 0 = 1 + + B N + B W = 400 + 400 + 400 + 400 = 1600mm v Ed 0 = β V Ed / (u 0 d) = 1.75 1100 1000 / (1600 305) =.87 MPa f d = α f k / γ = 1 40 / 1.5 = 6.67 MPa v Rd.max = 0.3 f d (1 - (f k / 50)) = 0.3 6.67 (1 - (40 / 50)) = 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 = 1 + + 3 π d = 400 + 400 + 3 π 305 = 4474.6 mm C Rd. = 0.18 / γ = 0.18/1.5 = 0.1 k = 1+ (00 / d) = 1 + (00/ 305) = 1.81 < v min = 0.035 k 3/ f 1/ k = 0.035 (1.81) 3/ (40) 1/ = 0.539 MPa v Ed 1 = β V Ed / ( d) = 1.75 1100 1000 / ( 4474.6 305) = 1.08 MPa Conider reinforement over 400 + 3 305 = 1315 mm width in bo th diretion from entre of the loaded area. For ρ l ue b = 1000mm. Uing H0 @ 175/ in both diretion = 1795. mm /m T1 & T ρ l = ((A x / (b d x ) A y / (b d y )) = (1795. / 315 1795./ 95) / 1000 = = 0.00589 < 0.0 v Rd. = C Rd. k (100 ρ l f k ) 1/3 1/3 = 0.1 1.810 (100 0.00589 40) = 0.63 MPa 400 400 17. Example Calulation Internal Corner Condition Chek if v Rd. > v min 0.63 MPa > 0.539 MPa Chek if v Ed 1 < v Rd. 1.08 MPa < 0.63 MPa Shear reinforement required LinkStudPSR Limited Page 38
Punhing hear at the bai ontrol perimeter with reinforement u out required = β V Ed / (v Rd. d) = 1.75 1100 1000 / (0.63 305) = 7386.3 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 = (7386.3-150) / 9 = 776.3 mm p = /( in(α/) ) = 776.3 / ( in(30 /)) = = 1499.6 mm p 1 = p - p - 1.5d/ (o(α/)) = 1499.6-00 - 1.5 305/ (o(30 /)) = 86.0 mm 0.75d = 0.75 305 = 8.75 mm 0.5d = 0.5 305 = 15.5 mm tud pain g = 30 mm, ditane to firt tud = 150 mm Stud paing hek Ditane to 3 rd tud r 3 = 00/ o 30 + 150 + 30 = 840.9 mm Ditane to lat tud r lat = 00/ o 30 + 150 + 3 30 = 1070.9 mm t 3 = ((r 3 - r 3 o(α)) + (r 3 in(α)) ) = = ((840.9-840.9 o(30 )) + (840.9 in(30 )) ) = 435.3 mm < 1.5d = 457.5 mm t lat = ((r lat - r lat o(α)) + (r lat in(α)) ) = = ((1070.9-1070.9 o(30 )) + (1070.9 in(30 )) ) = 554.4 mm < = 610 mm Area of hear reinforement f ywd.ef = 50 + 0.5 d = 50 + 0.5 305 = 36.3 N/mm f ywd = (f y / 1.15) = 500 / 1.15 = 434.8 N/mm > 36.3 t = 554.4 mm, r = 30 mm, rail no. taken a 1 A w.min = 0.08 t r f k / (1.5 f yk ) = 0.08 554.4 30 40 / (1.5 500) = 86.0 mm A w1 = (v Ed 1-0.75 v Rd. ) r / (1.5 f ywd.ef rail no.) A w1 = (1.08-0.75 0.63) 4474.6 30 / (1.5 36.3 1) = 98.3 mm tud dia = 1 mm (A = 113.1 mm ) Provide 1 No 1-4-300-990 Spaing: 150/30/30/30/150 (1357. mm ) 48 Stud total 435 554 17. Example Calulation Internal Corner Condition Rail Layout 150 30 30 30 150 400 400 u out req. LinkStudPSR Limited Page 39