NAGY GYÖRGY Tamás Assoc. Prof, PhD
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1 NAGY GYÖRGY Tamás Assoc. Prof, PhD E mail: tamas.nagy gyorgy@upt.ro Tel: Web: Office: A219
2 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 =?
3 Support L calc l rez L calc =
4 Support L calc l rez L calc = 3.00m
5 Diagrams: M + V M Ed = V Ed = = = = P/2 P/2 L calc = 3.00m
6 Diagrams: M + V M Ed = V Ed = = = = P/2 P/2 L calc = 3.00m M Ed V Ed
7 Diagrams: M + V M Ed = V Ed = = = = P/2 P/2 L calc = 3.00m M Ed V Ed
8 Diagrams: M + V M Ed = V Ed = = = = 3,5t 3,5t L calc = 3.00m M Ed V Ed
9 Diagrams: M + V M Ed = V Ed = 35 knm 35 kn = = = 35kN 35kN L calc = 3.00m M Ed 35kNm V Ed 35kN
10 !!! 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 = mm for beams 6 14 mm for slabs V Ed 35kN
11 d = h - d s d s = c nom + s /2 s = 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.)
12 Δ, ;, ; 10 bond durability
13 Δ, ;, ; 10 bond,
14 Δ, ;, ; 10 bond, where = mm for beams 6 14 mm for slabs 20 mm longitudinal 8 mm transversal
15 Δ, ;, ; 10 durability
16 Δ, ;, ; 10 durability
17
18 Δ, ;, ; 10 durability,
19 LONGITUDINAL REINFORCEMENT Δ TRANSVERSAL REINF. (stirrup), ;, ; ; 10 ; 10, 20 mm Δ 10 (A.N.),,, 22, 22, 20 8 ; 10 ; 10, 10 mm Δ 10 (A.N.),,, 28, 28, 30, OK!!!,,
20 2 d = h d s d s = c nom + s /2 s =20 mm d s = c nom + s /2 =
21 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 =
22 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
23 2 f cd f ck c
24 2 f cd f ck = 13,33 N/mm 2 c
25
26 / f yd f yk s
27 / f yd f yk = 435 N/mm 2 s
28 /
29 / = 0,617
30 = 0, / = 0,617
31 = 0, / = 0,617
32 112
33 112 = 0,306
34 112 = 0,306 = 365 mm 2 REINFORCEMENT PROPOSAL?
35
36 112 = 0,306 = 365 mm
37 2 16 d s = c nom + s /2 = d = h d s =, 0,26, 0,0013, 0,04
38 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!
39 ,,,
40
41
42 min 58 2 ; 30 29
43
44 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
45 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
46 , ,,
47 2,25 N/mm 2, 4 =773 mm ,, =773 mm
48 , max,100 1/3 1 1, 0,18/ ,15 / 0,2
49 , max,100 1/3 1 1, 0,18/ ,15 / 0,2 =0
50 , max,100 1/ A sl =???
51 , max,100 1/ A sl =??? - is the area of the tensile reinforcement, which extends (lbd + d) beyond the section considered
52 , max,100 1/ A sl = 216 = 402 mm 2 - is the area of the tensile reinforcement, which extends (lbd + d) beyond the section considered
53 , max,100 1/ ,035 3/2 1/2 =,
54 , max,100 1/ ,035 3/2 1/2 = 0.403, kn
55 24.03 kn, max,100 1/ kn 0,035 3/2 1/2 = 0.403, kn
56 ,
57 , SHEAR REINFORCEMENT IS REQUIRED min, ;,
58 , SHEAR REINFORCEMENT IS REQUIRED min, ;,
59 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, (A.N.) z 0.9d =
60 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, (A.N.) z 0.9d = 236 mm
61 CAPACITY OF THE COMPRESSED (CONCRETE) CHORD (V Rd,max ), / Choosing 45 90,
62 CAPACITY OF THE COMPRESSED (CONCRETE) CHORD (V Rd,max ), / Choosing 45 90, kn
63 CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ), Choosing - the cross sectional area of the shear reinforcement or and the spacing of the stirrups From condition,
64 CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ) Diametre of stirrups are imposed () A sw =na s (n= no shear arms!!!) s (spacing)
65 CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ) Diametre of stirrups are imposed () A sw =na s (n= no shear arms!!!) 26? 28? s (spacing)
66 CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ) Diametre of stirrups are imposed () A sw =na s (n= no shear arms!!!) s (spacing) mm 2
67 CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ) mm Diametre of stirrups are imposed () A sw =na s (n= no shear arms!!!) s (spacing) mm 2
68 CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ) mm 2 Reinforcement ratio for shear reinforcement, 0,08 / >(?),
69 CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ) mm 2 Reinforcement ratio for shear reinforcement, 0,08 / 1.439x x10 3 >(?),
70 CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ) mm mm 2 s =
71 CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ) mm mm 2 s = 166 mm
72 CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ), 0,75 1 Choosing s = 150 mm, s,
73 CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ), 0, mm Choosing s = 150 mm, s,
74 CAPACITY OF SHEAR REINFORCEMENT (V Rd,s ), 0, mm Choosing s = 150 mm, kn s,
75 DUCTILITY CONDITION FOR SHEAR REINFORCEMENT,, For a crack at 45 (ctg = 1) 1 1 0,5
76 DUCTILITY CONDITION FOR SHEAR REINFORCEMENT 0,5? <?
77 DUCTILITY CONDITION FOR SHEAR REINFORCEMENT 0, N/mm N/mm 2 OK!
78 DUCTILITY CONDITION FOR SHEAR REINFORCEMENT 0, N/mm N/mm 2 OK!
79 2 ND VARIANT!!! DESIGN OF LONGITUDINAL REINFORCEMENT AND SHEAR REINFORCEMENT USING STIRRUPS AND INCLINED BARS
80 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
81 2,25 N/mm 2, 4 =676 mm ,, =676 mm
82 , max,100 1/3 1 1, 0,18/ ,15 / 0,2 =0
83 , max,100 1/3 1 1 A sl = 214 = 308 mm s 0.02
84 , max,100 1/ ,035 3/2 1/2 = 0.401, kn
85 , SHEAR REINFORCEMENT IS REQUIRED min, ;,
86 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, (A.N.) z 0.9d = 236 mm
87 CAPACITY OF THE COMPRESSED (CONCRETE) CHORD (V Rd,max ), 1 Choosing = 45, kn = 45
88 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
89 SHEAR CAPACITY OF ELEMENTS REINFORCED WITH INCLINED BARS Maximum distance between inclined bars, 0, mm Choosing s = 300 mm, 74.4 kn
90 SHEAR CAPACITY OF ELEMENTS REINFORCED WITH INCLINED BARS,, For an inclined bar at 45 and compressed chord of 45 : 0,5 1
91 SHEAR CAPACITY OF ELEMENTS REINFORCED WITH INCLINED BARS 0, N/mm N/mm 2
92 SHEAR CAPACITY ALONG THE INCLINED BAR (V Rd,s ) = = = P/2 P/2 L V Ed 5cm 26 cm V Ed V Rd
93 SHEAR CAPACITY AFTER THE INCLINED BAR (V Rd,s ) Se impune diametrul () etrierului A sw =na s (n= nr ramuri!!!) s (pasul) mm 2
94 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
95 SITUATION WITH DISTRIBUTED LOADS p = 23,33 kn/m, 0, mm L Imposing 6 /19 cm V Rd V Ed V Ed =27,78kN V Ed,??? s
96 SITUATION WITH DISTRIBUTED LOADS p = 23,33 kn/m, 0, mm L Imposing 6 /19 cm V Rd V Ed V ed =27,78kN V Ed,, s 5cm 26 cm
97
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