NAGY GYÖRGY Tamás Assoc. Prof, PhD E mail: tamas.nagy gyorgy@upt.ro Tel: +40 256 403 935 Web: http://www.ct.upt.ro/users/tamasnagygyorgy/index.htm Office: A219
SIMPLE SUPORTED BEAM b = 15 cm h = 30 cm L = 3,30 m l rez = 30 cm P = 7 tones Concrete C20/25 Steel S500 Exposure class XC1 = = = P/2 P/2 L A s =? A sw =? s w =?
Support L calc l rez L calc =
Support L calc l rez L calc = 3.00m
Diagrams: M + V M Ed = V Ed = = = = P/2 P/2 L calc = 3.00m
Diagrams: M + V M Ed = V Ed = = = = P/2 P/2 L calc = 3.00m M Ed V Ed
Diagrams: M + V M Ed = V Ed = = = = P/2 P/2 L calc = 3.00m M Ed V Ed
Diagrams: M + V M Ed = V Ed = = = = 3,5t 3,5t L calc = 3.00m M Ed V Ed
Diagrams: M + V M Ed = V Ed = 35 knm 35 kn = = = 35kN 35kN L calc = 3.00m M Ed 35kNm V Ed 35kN
!!! 112 2 d =??? f cd =??? 100 = = = 35kN 35kN L calc = 3.00m M Ed 35kNm where d = h - d s d s = c nom + s /2 s = 14 25 mm for beams 6 14 mm for slabs V Ed 35kN
d = h - d s d s = c nom + s /2 s = 14 25 mm for beams 6 14 mm for slabs long d s C nom,long etr C nom,etr Δ = 5mm for slabs (A.N.) = 10 mm rest of the elements (A.N.)
Δ, ;, ; 10 bond durability
Δ, ;, ; 10 bond,
Δ, ;, ; 10 bond, where = 14 25 mm for beams 6 14 mm for slabs 20 mm longitudinal 8 mm transversal
Δ, ;, ; 10 durability
Δ, ;, ; 10 durability
Δ, ;, ; 10 durability,
LONGITUDINAL REINFORCEMENT Δ TRANSVERSAL REINF. (stirrup), ;, ; 10 20 ; 10 ; 10, 20 mm Δ 10 (A.N.),,, 22, 22, 20 8 ; 10 ; 10, 10 mm Δ 10 (A.N.),,, 28, 28, 30, OK!!!,,
2 d = h d s d s = c nom + s /2 s =20 mm d s = c nom + s /2 =
2 d = h d s d s = c nom + s /2 s =20 mm d s = c nom + s /2 = 40 mm d = h d s =
2 d = h d s d s = c nom + s /2 s =20 mm d s = c nom + s /2 = 40 mm d = h d s = 260 mm
2 f cd f ck c
2 f cd f ck = 13,33 N/mm 2 c
2 0.259
2 0.259 0.8 1 0.4 3.5 3.5 1000 / f yd f yk s
2 0.259 0.8 1 0.4 3.5 3.5 1000 / f yd f yk = 435 N/mm 2 s
2 0.259 0.8 1 0.4 3.5 3.5 1000 /
2 0.259 0.8 1 0.4 3.5 3.5 1000 / = 0,617
2 0.259 0.8 1 0.4 = 0,372 3.5 3.5 1000 / = 0,617
2 0.259 0.8 1 0.4 = 0,372 3.5 3.5 1000 / = 0,617
112
112 = 0,306
112 = 0,306 = 365 mm 2 REINFORCEMENT PROPOSAL?
112 = 0,306 = 365 mm 2 2 16 3 14
2 16 d s = c nom + s /2 = d = h d s =, 0,26, 0,0013, 0,04
2 16 = 402 mm 2 d s = c nom + s /2 = 38 mm d = h d s = 262 mm, 0,26 =45mm 2 ok!, 0,0013 =51mm 2 ok!, 0,04 = 1800 mm 2 ok!
1 2 3 4 5,,, 4 2.25 1 2
min 58 2 ; 30 29
2.25 design value of the ultimate bond stress where: η 1 η 2 f ctd is a coefficient related to the quality of the bond condition and the position of the bar during concreting (see Figure 8.2): = 1,0 when good conditions are obtained and = 0,7 for all other cases and for bars in structural elements built with slip forms, unless it can be shown that good bond conditions exist - is related to the bar diameter: = 1,0 for φ 32 mm = (132 φ)/100 for φ > 32 mm f ctk0,05 c
2.25 design value of the ultimate bond stress where: η 1 η 2 f ctd is a coefficient related to the quality of the bond condition and the position of the bar during concreting (see Figure 8.2): = 1,0 when good conditions are obtained and = 0,7 for all other cases and for bars in structural elements built with slip forms, unless it can be shown that good bond conditions exist - is related to the bar diameter: = 1,0 for φ 32 mm = (132 φ)/100 for φ > 32 mm f ctk0,05 c 2,25 N/mm 2 =1,00N/mm 2
, 4 1 2 3 4 5,,
2,25 N/mm 2, 4 =773 mm 1 2 3 4 5,, =773 mm
, max,100 1/3 1 1, 0,18/ 1 200 2 1 0,15 / 0,2
, max,100 1/3 1 1, 0,18/ 0.12 1 200 2 1.877 1 0,15 / 0,2 =0
, max,100 1/3 1 1 0.02 A sl =???
, max,100 1/3 1 1 0.02 A sl =??? - is the area of the tensile reinforcement, which extends (lbd + d) beyond the section considered
, max,100 1/3 1 1 0.02 0.01 A sl = 216 = 402 mm 2 - is the area of the tensile reinforcement, which extends (lbd + d) beyond the section considered
, max,100 1/3 1 1 0,035 3/2 1/2 =,
, max,100 1/3 1 1 0,035 3/2 1/2 = 0.403, 24.03 kn
24.03 kn, max,100 1/3 1 1 15.72 kn 0,035 3/2 1/2 = 0.403, 24.03 kn
, 24.03 35
, 24.03 35 SHEAR REINFORCEMENT IS REQUIRED min, ;,
, 24.03 35 SHEAR REINFORCEMENT IS REQUIRED min, ;,
CAPACITY OF THE COMPRESSED (CONCRETE) CHORD (V Rd,max ), / cw 1 - is a coefficient taking account of the state of the stress in the compression chord cw =1 for RC elements cw >1 for PC elements - is a strength reduction factor for concrete cracked in shear 1 0,61 250 (A.N.) z 0.9d =
CAPACITY OF THE COMPRESSED (CONCRETE) CHORD (V Rd,max ), / cw 1 - is a coefficient taking account of the state of the stress in the compression chord cw =1 for RC elements cw >1 for PC elements - is a strength reduction factor for concrete cracked in shear 1 0,61 250 0.552 (A.N.) z 0.9d = 236 mm
CAPACITY OF THE COMPRESSED (CONCRETE) CHORD (V Rd,max ), / Choosing 45 90,
CAPACITY OF THE COMPRESSED (CONCRETE) CHORD (V Rd,max ), / Choosing 45 90, 130.16 kn
CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ), Choosing - the cross sectional area of the shear reinforcement or and the spacing of the stirrups 45 90 From condition,
CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ) Diametre of stirrups are imposed () A sw =na s (n= no shear arms!!!) s (spacing)
CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ) Diametre of stirrups are imposed () A sw =na s (n= no shear arms!!!) 26? 28? s (spacing)
CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ) Diametre of stirrups are imposed () A sw =na s (n= no shear arms!!!) s (spacing) 26 56.5 mm 2
CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ) 0.341 mm Diametre of stirrups are imposed () A sw =na s (n= no shear arms!!!) s (spacing) 26 56.5 mm 2
CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ) 2656.5 mm 2 Reinforcement ratio for shear reinforcement, 0,08 / >(?),
CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ) 2656.5 mm 2 Reinforcement ratio for shear reinforcement, 0,08 / 1.439x10 3 0.715x10 3 >(?),
CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ) 0.341 mm 26 56.5 mm 2 s =
CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ) 0.341 mm 26 56.5 mm 2 s = 166 mm
CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ), 0,75 1 Choosing s = 150 mm, s,
CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ), 0,75 1 197 mm Choosing s = 150 mm, s,
CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ), 0,75 1 197 mm Choosing s = 150 mm, 38.65 kn s,
DUCTILITY CONDITION FOR SHEAR REINFORCEMENT,, For a crack at 45 (ctg = 1) 1 1 0,5
DUCTILITY CONDITION FOR SHEAR REINFORCEMENT 0,5? <?
DUCTILITY CONDITION FOR SHEAR REINFORCEMENT 0,5 1.09 N/mm 2 3.68 N/mm 2 OK!
DUCTILITY CONDITION FOR SHEAR REINFORCEMENT 0,5 1.09 N/mm 2 3.68 N/mm 2 OK!
2 ND VARIANT!!! DESIGN OF LONGITUDINAL REINFORCEMENT AND SHEAR REINFORCEMENT USING STIRRUPS AND INCLINED BARS
2 ND VARIANT = 367 mm 2 s 3 14 = 462 mm2 d s = c nom + s /2 = 38 mm d = h d s = 262 mm
2,25 N/mm 2, 4 =676 mm 1 2 3 4 5,, =676 mm
, max,100 1/3 1 1, 0,18/ 0.12 1 200 2 1.874 1 0,15 / 0,2 =0
, max,100 1/3 1 1 A sl = 214 = 308 mm 2 0.0078 s 0.02
, max,100 1/3 1 1 0,035 3/2 1/2 = 0.401, 22.11 kn
, 22.11 35 SHEAR REINFORCEMENT IS REQUIRED min, ;,
CAPACITY OF THE COMPRESSED (CONCRETE) CHORD (V Rd,max ), 1 cw 1 - is a coefficient taking account of the state of the stress in the compression chord cw =1 for RC elements cw >1 for PC elements - is a strength reduction factor for concrete cracked in shear 1 0,61 250 0.552 (A.N.) z 0.9d = 236 mm
CAPACITY OF THE COMPRESSED (CONCRETE) CHORD (V Rd,max ), 1 Choosing = 45, 260.32 kn = 45
SHEAR CAPACITY OF ELEMENTS REINFORCED WITH INCLINED BARS, Choosing A sw s - aria of inclined bar distance between inclined bars s = 45 = 45 A sw = 114 = 154 mm 2
SHEAR CAPACITY OF ELEMENTS REINFORCED WITH INCLINED BARS Maximum distance between inclined bars, 0,61 314 mm Choosing s = 300 mm, 74.4 kn
SHEAR CAPACITY OF ELEMENTS REINFORCED WITH INCLINED BARS,, For an inclined bar at 45 and compressed chord of 45 : 0,5 1
SHEAR CAPACITY OF ELEMENTS REINFORCED WITH INCLINED BARS 0,5 1 1.49 N/mm 2 5.20 N/mm 2
SHEAR CAPACITY ALONG THE INCLINED BAR (V Rd,s ) = = = P/2 P/2 L V Ed 5cm 26 cm V Ed V Rd
SHEAR CAPACITY AFTER THE INCLINED BAR (V Rd,s ) Se impune diametrul () etrierului A sw =na s (n= nr ramuri!!!) s (pasul) 26 56.5 mm 2
SHEAR CAPACITY AFTER THE INCLINED BAR (V Rd,s ) = = = P/2 P/2 L V Ed V Ed V Rd CALCUL = IDEM PARTEA 1! V Ed = IDEM
SITUATION WITH DISTRIBUTED LOADS p = 23,33 kn/m, 0,751 197 mm L Imposing 6 /19 cm V Rd V Ed V Ed =27,78kN V Ed,??? s
SITUATION WITH DISTRIBUTED LOADS p = 23,33 kn/m, 0,751 197 mm L Imposing 6 /19 cm V Rd V Ed V ed =27,78kN V Ed,, s 5cm 26 cm