technical guideline

2www.namcong.com

NI DUNG

Namcong Engineering CorporationTp. H Ch Minh: Tel : (84)-8-3547 2679 Fax: (84)-8-3547 2763Tp. Nng: Tel : (84)-5-116 260 805 Fax: (84)-5-116 260 806Email: [email protected]: www.namcong.com

LIN H

LI NI U

B-TNG D NG LC

NGUYN L LM VIC

PHM VI NG DNG

HIU QU K THUT

HIU QU KINH T

C TNH K THUT CA VT LIU

CP

U NEO SNG

U NEO CHT

VA

CC TIU CH TH NGHIM CA VT LIU

TH NGHIM NEO/ NM

TH NGHIM CP D NG LC

TH NGHIM VA

NGUYN TC C BN TRONG THIT K D NG LC

QUY TRNH THIT K

KCH THC TIT DIN S B

VT LIU

CC BC TNH TON CU KIN B-TNG D NG LC

KIM TRA, TNH TON CC TRNG THI: SLS, ULS

PHNG N THIT K SN TRN NN T N NH By Brian M. Juedes

CHI TIT CU TO

CC BC THI CNG CP D NG LC

CC BC THI CNG CP D NG LC CHO DM SN TRN NN T (tham kho)

LU QUY TRNH THI CNG CP D NG LC

DANH SCH CC D N THI CNG CP D NG LC

3

4 - 6

4

4

5

6

7 - 16

7

8 - 12

13-16

16

16-17

16

17

17

18-38

18

18

18

19

20-38

38-40

41-43

44-46

47

48

49

Trang

3 www.namcong.com

Tri qua qu trnh xy dng v pht trin, Namcong khng ngng hon thin b my qun l, nng cao trnh chuyn mn ngh nghip nhm to nn mt bn sc cho mt Namcong nng ng v nhit huyt.

Trong , i ng nhn s ca Namcong lun mit mi lm vic, nghin cu v sng to nhm to nn nn tng vng chc v kin thc v c trng vn ha cng ty, vi mc tiu l to dng c mt h thng dch v ti u nhm phc v khch hng ngy cng tt hn.

Thnh qu lao ng khng ch c khng nh qua uy tn trn th trng D ng lc, vi hn 100 d n v ang thi cng, m cn c th hin qua nhng ti sn tr tu qu bu c c kt qua tng nm thng lm vic, nghin cu v ci tin.

C th cho cun ti liu ny, vi Phin bn th 5 c tn gi: Technical Guideline Post-tensioning - Namcongs system, chnh l s k tha, c kt v pht huy xng ng ca nhng phin bn trc, vi nhng ch dn c th cho cc bc thit k da trn nhng tiu chun c p dng trn ton th gii trong lnh vc D ng lc.

Phin bn ln ny c bin son vi mc tiu s l mt ti liu tham kho hu ch cho nhng ai ang tm hiu v cng ngh d ng lc, cho cc k s thit k v cc k s thi cng ang lm trong lnh vc D ng lc hin nay.

Vi mt tn ch gin d: Khch hng xng ng nhn c dch v tt nht, Namcong hy vng s lun gi vng mt nim tin chin lc trong lng khch hng v s a Namcong i xa hn trong cc dch v ca Namcong.

LI NI U 3

4 - 6

4

4

5

6

7 - 16

7

8 - 12

13-16

16

16-17

16

17

17

18-38

18

18

18

19

20-38

38-40

41-43

44-46

47

48

49

Trang

4www.namcong.com

B-TNG D NG LC

NGUYN L LM VIC

Ko cng thp (hoc cp) cng cao nm trong b-tng nhm to ra lc khng vi ti s dng (lc ng sut trc).

B-tng d ng lc

Phng php cng trc (dng cho cu kin c sn)

Phng php cng sau (dng cho cu kin b-tng ton khi)

Cng sau khng bm va

Hnh 1: Cu kin b-tng d ng lc

Biu 1: H thng b-tng D ng lc

PHM VI NG DNG

Sn phng d ng lc Sn d ng lc c m ct

Sn d ng lc c dm dt

Hinh 2: Cc loi sn d ng lc

C ba kiu sn d ng lc thng c s dng: sn phng, sn c m ct v sn dm dt.Sn phng d ng lc thng c ng dng cho bc ct t 7-10m, gip thi cng nhanh, cu kin c kch thc nh, kim sot c vng.Sn d ng lc c m ct c ng dng cho sn c bc ct t 9-12m. Loi sn ny kim sot chng thng tt. Sn d ng lc c dm dt c ng dng cho cng trnh c bc ct t 12-20m v c th ln hn.Cng ngh D ng lc mang li hiu qu ti u khi ng dng cho sn c bc ct t 7m tr ln.

Lc cng

Ti s dngLc ng sut trc

Lc cng

Cng sau c bm va

Sn d ng lc c ng dng vo tt c cc cng trnh xy dng nh: chung c, vn phng, khch sn, nh hng, nh xng cng nghip, silo, cu ng...

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HIU QU K THUT

Lm nn cu kin b-tng. Lm gim ng sut ko; hn ch vng v vt nt trong cu kin b-tng. Tng bc ct.

Gim chiu cao tit din. Gim trng lng bn thn. Gim khi lng thp gia cng trong mt ct b-tng.

B-TNG D NG LC

Chiu cao thng thy

Chiu cao thng thy

Hnh 3: Dm sn b-tng ct thp thng thng

Chiu cao thng thy

Chiu cao thng thy

Chiu cao tit kim

Hnh 4: Sn b-tng ct thp d ng lc

Hnh 5: Dm sn b-tng ct thp thng thng Hnh 6: Sn b-tng ct thp d ng lc

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Chi ph phn th

Sn phng

d ng lc

7m 9m 12m Span column

Dm sn

b-tng ct thp

Sn phng

b-tng ct thp

Thi cng nhanh v n gin.

Hon thin nhanh, gim chi ph hon thin.

Gim ti a chiu cao ca mi tng.

B tr linh hot cho mt bng s dng.

Thay i cng nng s dng mt cch d dng.

Cng ngh d ng lc s mang hiu qu cao v chi ph u t.

i vi nhng cng trnh c bc ct t 7m tr ln, gii php kt cu d ng lc p dng cho h dm sn s tit kim t 5%-20% chi ph u t phn th so vi kt cu b-tng ct thp thng thng.

c bit l nhng cng trnh c tng din tch sn ln hn hn 4,000m2 th hiu qu kinh t khi p dng cng ngh d ng lc s cng c pht huy.

Biu 2: Chi ph phn th

B-TNG D NG LC

HIU QU KINH T

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Lp bo v

V nha Cp

Cp

Hnh 7: Cp c v bc, khng bm va Hnh 8 : Cp khng c v bc, c bm va


CP

Hnh 9: Cp khng bm va Hnh 10: Cp c bm va

Bng 1: c tnh k thut ca cp d ng lc

Tiu chunASTM A416

Grade 270 KsiBS 5896

Grade 1860 MpaASTM A416

Grade 270 KsiBS 5896

Grade 1860 Mpa

n v tnhLow-relaxation

strand7-wire super

strandLow-relaxation

strand7-wire super

strand

ng knh danh ngha mm 12.70 (0.5") 12.90 (0.5") 15.24 (0.6") 15.70 (0.6")

Din tch ngang mm2 98.71 100.00 140.00 150.00

Khi lng kg/m 0.775 0.785 1.102 1.180

Gii hn chy N/mm2 1,6701) 1,6401) 1,6701) 1,6401)

Gii hn bn N/mm2 1,860 1,860 1,860 1,860

Lc ko t kN 183.7 186.0 260.7 279.0

Chng ng sut2) sau 1,000 h 0.7 x gii hn bn fpk

max. 2.5 % max. 2.5 % max. 2.5 % max. 2.5 %

Chng ng sut2) sau 1,000 h 0.8 x gii hn bn fpk

max. 3.5 % max. 4.5 % max. 3.5 % max. 4.5 %

Module n hi Gpa

1) yield measured at 1% extension under load

2) applicable for relaxation class 2 according to BS 5896: or low relaxation complying with ASTM A416, respectively

195 5% (195 10 Gpa)

Chng loiTiu ch k thut

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Bng 2: c tnh k thut u neo dp

A

R

E

C

D B

F

G

Type A (mm) B (mm) C (mm) D (mm) E (mm) F (mm) G (mm) R (mm)

ST13F- 3 169 94 135 152 45 50 107 110

ST13F- 4 265 94 216 248 45 50 172 110

ST13F- 5 265 94 216 248 45 50 210 110

ST15F- 4 265 94 216 248 45 50 191 110

ng lun cp dp


U NEO SNG

u neo sng phi m bo chu c 92% lc ko t ca si cp (theo tiu chun BS 4447) hoc 95% lc ko t ca si cp (theo tiu chun ACI 318:2002).

Hnh 11: u neo sng

Hnh 12: Chi tit u neo dp (xem bng 2)

minimum

Ghi ch: 13 - 3: ng knh si cp - s lng si cp.

(20mm)

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Diameter of fixing holes

Grouting hole M27x2

Anchorbearing plate Strand

Corrugated duct

Spiralreinforcement

Working anchor plate Wedge

Hnh 13: Chi tit u neo trn (xem bng 3)

Bng 3: c tnh k thut ca u neo trn

Anchor head (mm)

A x B C x D x E Hole Dist Hole Dia F G H Wnd

ST15R-1 46 x 48 80 x 80 x 14 / / 70 6 30 4

ST13R-1 42 x 45 70 x 70 x 14 / / 70 6 30 4

ST15R-2 83 x 48 115 x 90 x 80 85 M10 130 8 40 4

ST13R-2 75 x 46 115 x 90 x 80 85 M10 130 8 40 4

ST15R-3 85 x 48 130 x 100 x 90 90 M10 130 8 40 4

ST13R-3 80 x 46 130 x 100 x 90 90 M10 130 8 40 4

ST15R-4 95 x 48 140 x 110 x 95 100 M10 140 10 50 4

ST13R-4 88 x 46 130 x 100 x 90 90 M10 130 8 40 4

ST15R-5 105 x 48 150 x 120 x 100 110 M10 160 10 50 4

ST13R-5 98 x 46 140 x 100 x 95 100 M10 140 10 50 4

ST15R-6 125 x 48 165 x 130 x 115 120 M10 180 12 50 4

ST13R-6 115 x 48 150 x 120 x 100 110 M10 160 10 50 4

ST15R-7 125 x 50 175 x 140 x 120 125 M10 180 12 50 4

ST13R-7 115 x 48 150 x 120 x 100 110 M10 160 10 50 4

ST15R-8 135 x 51 185 x 150 x 130 135 M10 200 14 50 5

ST13R-8 125 x 50 165 x 130 x 115 120 M10 180 12 50 4

ST15R-9 145 x 52 200 x 160 x 135 150 M10 200 14 50 5

ST13R-9 135 x 50 185 x 150 x 120 135 M10 180 12 50 4

ST15R-10 155 x 54 210 x 180 x 150 160 M10 220 14 50 5.5

ST13R-10 145 x 52 200 x 160 x 135 150 M10 200 14 50 5

ST15R-11 165 x 55 220 x 200 x 155 170 M10 220 14 50 5.5

ST13R-11 155 x 52 210 x 180 x 150 160 M10 220 14 60 5.5

ST15R-12 165 x 58 230 x 200 x 155 170 M10 240 14 60 5.5

ST13R-12 155 x 52 210 x 180 x 150 160 M10 220 14 60 5.5

ST15R-13 165 x 60 240 x 200 x 155 170 M10 240 14 60 5.5

ST13R-13 155 x 52 210 x 180 x 150 160 M10 220 14 60 5.5

ST15R-14 175 x 62 250 x 210 x 160 185 M10 260 16 60 6

ST13R-14 165 x 55 220 x 200 x 155 170 M10 220 14 60 5.5

ST15R-15 185 x 63 260 x 220 x 160 190 M10 260 16 60 6

ST13R-15 175 x 60 240 x 200 x 155 170 M10 240 14 60 5.5

ST15R-16 195 x 63 265 x 240 x 160 200 M10 280 16 60 6

ST13R-16 185 x 62 240 x 200 x 155 185 M10 260 16 60 6

ST15R-17 195 x 66 270 x 240 x 160 200 M10 280 16 60 6

ST13R-17 185 x 62 250 x 210 x 155 185 M10 260 16 60 6

ST15R-18 205 x 68 280 x 240 x 170 210 M10 300 16 60 6

ST13R-18 185 x 63 250 x 210 x 155 185 M10 260 16 60 6

ST15R-19 205 x 70 290 x 240 x 170 210 M10 300 16 60 6

ST13R-19 185 x 63 250 x 210 x 155 185 M10 260 16 60 6

ST15R-20/21/22 225 x 75 300 x 260 x 185 220 M10 320 18 60 6.5

ST13R-20/21/22 210 x 65 280 x 240 x 170 210 M10 280 16 60 6

ST15R-23/24 240 x 80 315 x 280 x 195 230 M10 320 18 60 6.5

ST13R-23/24 225 x 72 290 x 270 x 180 210 M10 300 16 60 6

ST15R-25/26/27 245 x 82 320 x 290 x 195 230 M10 320 18 60 6.5

ST13R-25/26/27 230 x 76 300 x 280 x 185 220 M10 320 18 60 6

ST15R-28~31 260 x 88 340 x 300 x 205 250 M10 350 20 60 6.5

ST13R-28~31 240 x 85 320 x 290 x 195 230 M10 320 18 60 6.5

23,24

20,21,22

25,26,27

28~31

Anchorage TypeNumber of

strands

14

13

12

11

10

19

18

17

16

15

4

5

6

9

8

7

Anchor Bearing Plate (mm) Spiral reinforcement (mm)

1

2

3



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Bng 4: c tnh k thut ca u neo cp n

ST6 - 1

ST5 - 1

C (mm)

39

48

A (mm)

125

145

B (mm)

57

90

Type


Hnh 14: Chi tit u neo cp n (xem bng 4)

Hnh 15: Khong cch gia cc u neo cp dp (xem bng 5)

Bng 5: Bng tra khong cch gia u neo cp dp

Hnh 16: Kch thc hc ko cp ca u neo cp dp (xem bng 6)

Anchorage Type A (mm) B (mm) C (mm) D (mm) E (mm)

ST15 (13)F -2 300 200 230 130 100

ST15 (13)F -3 340 240 230 130 100

ST15 (13)F -4 390 280 250 140 110

ST15 (13)F -5 440 320 260 140 120

Bng 6: Bng tra kch thc hc ko cp ca u neo cp dp

Ghi ch: Cc thng s trn c th thay i ty thuc vo nh sn xut. 15 - 3: ng knh si cp - s lng si cp.

C(mm) D(mm) E(mm) F(mm) C(mm) D(mm) E(mm) F(mm)

ST15 (13)F -2 135 x 65 (130 x 60) 155 (150) 80 (75) 95 (90) 55 (50) 150 (145) 75 (70) 90 (85) 50 (45)

ST15 (13)F -3 180 x 70 (152 x 60) 205 (190) 90 (75) 120 (110) 60 (50) 200 (185) 85 (70) 110 (105) 55 (50)

ST15 (13)F -4 220 x 70 (190 x 60) 250 (230) 100 (85) 150 (135) 70 (55) 240 (225) 90 (80) 135 (130) 60 (55)

ST15 (13)F -5 255 x 70 (243 x 65) 300 (270) 110 (85) 175 (160) 75 (60) 280 (265) 95 (80) 160 (155) 70 (65)

Note: The specifications and the dimension of the anchor bearing plates hereinabove can be designed and manufactured to meet the users

requirements.

Anchorage TypeAnchor Bearing

Plate A x B (mm)

Concrete Strength

fcu = 40 Mpafcu = 30 Mpa


AB

D

C

E

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Hnh 17: Hc ko cp ca u neo cp b dp

Hnh 18: Kch thc hc ko cp ca u neo trn (xem bng 7)

Bng 7: Bng tra kch thc hc ko cp ca u neo trn

Anchorage Type A (mm) B (mm) C (mm) Anchorage Type A (mm) B (mm) C (mm)

ST15 R - 3, 4, 5 380 250 120 ST13 R - 3, 4, 5 380 250 120

ST15 R - 6, 7 450 310 130 ST13 R - 6, 7 380 250 120

ST15 R - 8, 9 540 380 130 ST13 R - 8, 9 450 310 130

ST15 R - 10, 11, 12, 13 540 380 140 ST13 R - 10, 11, 12, 13 450 310 130

ST15 R - 14, 15 580 420 140 ST13 R - 14, 15 500 360 140

ST15 R - 16, 18, 19, 21, 22 600 450 150 ST13 R - 16, 18, 19, 21, 22 540 380 140

ST15 R - 24, 25, 27 680 550 190 ST13 R - 24, 25, 27 600 450 150

ST15 R - 31 700 550 200 ST13 R - 31 640 530 170

ST15 R - 37 780 650 210 ST13 R - 37 680 550 190

ST15 R - 43 800 650 220 ST13 R - 43 680 650 190

ST15 R - 55 900 750 250 ST13 R - 55 780 650 210

15 - 8: diameter of strand - No. of strandsGhi ch: 15 - 3: ng knh si cp - s lng si cp.

B

AC

B A

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Hnh 19: Khong cch ti thiu gia cc u neo trn (xem bng 8)

Bng 8: Bng tra khong cch ti thiu gia cc u neo trn

Ghi ch: Kch thc trn c th thay i ph thuc vo cng cp, iu kin thc t ct thp cng trng, ti trng

tnh ton. 15 - 3: ng knh si cp - s lng si cp.

a(mm) b(mm) a(mm) b(mm) a(mm) b(mm)

ST15 (13)R -2 115 (115) 149 (149) 97 (97) 138 (138) 92 (92) 126 (126) 86 (86)

ST15 (13)R -3 130 (130) 169 (169) 110 (110) 156 (156) 104 (104) 143 (143) 97 (97)

ST15 (13)R -4 140 (130) 182 (169) 119 (110) 168 (156) 112 (104) 154 (143) 105 (97)

ST15 (13)R -5 150 (140) 195 (182) 127 (119) 180 (168) 120 (112) 165 (154) 112 (105)

ST15 (13)R -6 165 (150) 214 (195) 140 (127) 198 (180) 132 (120) 181 (165) 123 (112)

ST15 (13)R -7 175 (150) 227 (195) 148 (127) 210 (180) 140 (120) 192 (165) 131 (112)

ST15 (13)R -8 185 (165) 240 (214) 157 (140) 222 (198) 148 (132) 203 (181) 138 (123)

ST15 (13)R -9 200 (185) 260 (240) 170 (157) 240 (222) 160 (148) 220 (203) 150 (138)

ST15 (13)R -10 210 (200) 273 (260) 178 (170) 252 (240) 168 (160) 231 (220) 157 (150)

ST15 (13)R -11 220 (210) 286 (273) 187 (178) 264 (252) 176 (168) 242 (231) 165 (157)

ST15 (13)R -12 230 (210) 299 (273) 195 (178) 276 (252) 184 (168) 253 (231) 172 (157)

ST15 (13)R -13 240 (210) 312 (273) 204 (178) 288 (252) 192 (168) 264 (231) 180 (157)

ST15 (13)R -14 250 (220) 325 (286) 212 (187) 300 (264) 200 (176) 275 (242) 187 (165)

ST15 (13)R -15 260 (240) 338 (312) 221 (204) 312 (288) 208 (192) 286 (264) 195 (180)

ST15 (13)R -16 265 (265) 344 (344) 225 (225) 318 (318) 212 (212) 291 (291) 198 (198)

ST15 (13)R -17 270 (265) 351 (344) 229 (225) 324 (318) 216 (212) 297 (291) 202 (198)

ST15 (13)R -18 280 (265) 364 (344) 238 (225) 336 (318) 224 (212) 308 (291) 210 (198)

ST15 (13)R -19 290 (265) 377 (344) 246 (225) 348 (318) 232 (212) 319 (291) 217 (198)

ST15 (13)R -20 300 (280) 390 (364) 255 (238) 360 (336) 240 (224) 330 (308) 225 (210)

ST15 (13)R -21 300 (280) 390 (364) 259 (238) 336 (336) 244 (224) 335 (308) 228 (210)

ST15 (13)R -22 300 (280) 403 (364) 263 (238) 372 (336) 248 (224) 341 (308) 232 (210)

ST15 (13)R -23 315 (290) 409 (377) 267 (246) 378 (348) 252 (232) 346 (319) 236 (217)

ST15 (13)R -24 315 (290) 413 (377) 270 (246) 381 (348) 254 (232) 349 (319) 238 (217)

ST15 (13)R -25 320 (300) 416 (390) 272 (255) 384 (360) 256 (240) 352 (330) 240 (225)

ST15 (13)R -26 320 (300) 422 (390) 276 (255) 390 (360) 260 (240) 357 (330) 243 (225)

ST15 (13)R -27 320 (300) 429 (390) 280 (255) 396 (360) 264 (240) 363 (330) 247 (225)

ST15 (13)R -31 340 (320) 468 (416) 306 (272) 432 (384) 288 (256) 396 (352) 270 (240)

ST15 (13)R -37 390 (360) 533 (468) 348 (306) 492 (432) 328 (288) 451 (396) 307 (270)

ST15 (13)R -43 440 (390) 585 (507) 382 (331) 540 (468) 360 (312) 495 (429) 337 (292)

ST15 (13)R -55 520 (440) 676 (572) 442 (374) 624 (528) 416 (352) 572 (484) 390 (330)

Note: The dimension in the table is calculated based on 2000 Mpa strands strength. The one of 1860 Mpa can be shortened

properly, please adjust, according to the duct aperture diameter, the strength of the strands and the practical layout condition of the

anchored steel bars, after calculating the local bearing strength based on JT J023 and GB20010 design standard.

Anchorage TypeAnchor Bearing

Plate

Concrete Strength

fcu = 30 Mpa fcu = 40 Mpa fcu = 50 Mpa

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U NEO CHT

tnh chiu di u neo cht khi thiu s liu k thut, c th tnh chiu di ny trong iu kin lc ko cng ca cp t 75% lc ko t (theo tiu chun BS 8110 : 4.10.3). lt =

fci Cng b-tng khi ko cng ng knh danh ngha ca si cpKt H s ph thuc vo loi cp v c chn nh sau: a. Si thp trn: Kt = 600 b. Si thp c g (chiu cao ca sng 0.15): Kt = 400 c. Cp 7 si: Kt = 240

Ktfci

(Mpa), (Mpa)fci

Hnh 20: u neo cht

Bng 9: c tnh k thut u neo cht loi H cho b cp trn, vi lc ko cng ca cp t 80% lc ko t

ST15 (13)-3H ST15 (13)-4H ST15 (13)-5H ST15 (13)-6,7H ST15 (13)-8,9H ST15 (13)-12H

A 190 (130) 190 (150) 200 (160) 210 (170) 210 (220) 330 (270)

B 90 (70) 210 (170) 220 (180) 230 (190) 310 (250) 390 (310)

C (min) 1000 (750) 1000 (750) 1000 (750) 1100 (850) 1200 (850) 1300 (850)

ST15 (13)-19H ST15 (13)-27H ST15 (13)-31H ST15 (13)-37H ST15 (13)-43H ST15 (13)-55H

A 390 (310) 450 (410) 510 (430) 510 (430) 550 (560) 620 (560)

B 470 (390) 520 (430) 570 (470) 690 (570) 750 (580) 850 (680)

C (min) 1300 (950) 1700 (1150) 1700 (1150) 2000 (1680) 2500 (1680) 2500 (1980)

Specification

Size

Specification

Size


Hnh 21: Chi tit cu to u neo cht cho b cp trn (xem bng 9)

C B

u c hnhThp nh vCpng ghen c np

Tension ring

A

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ng ghen c np Cp Thp nh v u c hnh

a

a

a

a

B

A

Type A (mm) B (mm) a (mm)

ST5F min.300 min.750 min.75

ST6F min.300 min.1000 min.75

ng va Tension ring

ng lun cp

u c hnh

Thp nh v


Hnh 22: Chi tit cu to u neo cht cho b cp dp loi H (xem bng 10)

Bng 10: Chi tit cu to u neo cht cho b cp dp (loi H) vi lc ko cng ca cp t 80% lc ko t

NG LUN CP

Hnh 23: u neo cht loi H

ng lun cp, van bm va v cc ng ni phi cng gi nguyn hnh dng trong qu trnh thi cng:

ng km trn: dy ti thiu 0.23mm - 0.3mm.

ng km gp np: dy ti thiu 0.23mm - 0.3mm.ng nha cng: dy ti thiu 2.0mm.

Din tch mt ct ngang ca ng lun cp phi ln hn hoc bng 2 ln din tch mt ct ngang ca tt c cc si cp chim ch (theo tiu chun ACI 318: 18.17.3).

Hnh 24: ng lun cp trn

ng vang lun cp

u c hnh

Thp nh v

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Kch thc ng lun cp trn

Bng 11: Kch thc ng lun cp trn

Hnh 25: Kch thc ng lun cp trn (xem bng 11)

I.D (mm) O.D (mm)

ST13R -1 25 30 6

ST13R -3 50 57 9

ST13R -4 50 57 9

ST13R -7 60 67 9

ST13R -12 80 87 11

ST13R -19 90 97 12

ST13R -22 100 107 12

ST13R -31 120 127 14

ST13R -37 115 121 22

ST13R -43 130 137 24

ST13R -55 147 153 23

Tendon Unit EccentricityDuct diameter

ng lun cp

lch tm(Eccentricity)

Bng 12: lch tm ca b cp trn

Hnh 26: lch tm ca b cp trn (xem bng 12)

O.D(mm)

I.D(mm)

O.D(mm)

I.D(mm)

ST15R - 1 33 27 ST15R - 14 107 100

ST13R - 1 30 25 ST13R - 14 97 90

ST15R - 2 50 45 ST15R - 15 107 100

ST13R - 2 50 45 ST13R - 15 97 90

ST15R - 3 55 50 ST15R - 16 117 110

ST13R - 3 55 50 ST13R - 16 97 90

ST15R - 4 60 55 ST15R - 18 117 110

ST13R - 4 55 50 ST13R - 18 97 90

ST15R - 5 60 55 ST15R - 19 117 110

ST13R - 5 60 55 ST13R - 19 97 90

ST15R - 6,7 75 70 ST15R - 21 127 120

ST13R - 6,7 60 55 ST13R - 21 107 100

ST15R - 8,9 85 80 ST15R - 22 127 120

ST13R - 8,9 75 70 ST13R - 22 107 100

ST15R - 10 97 90 ST15R - 24 127 120

ST13R - 10 85 80 ST13R - 24 117 110

ST15R - 11 97 90 ST15R - 25 127 120

ST13R - 11 85 80 ST13R - 25 117 110

ST15R - 12 97 90 ST15R - 27 127 120

ST13R - 12 85 80 ST13R - 27 117 110

ST15R - 13 97 90 ST15R - 31 137 130

ST13R - 13 87 80 ST13R - 31 127 120

Corrugated duct dimension Corrugated duct dimension

Sheathing Sheathing

Type Type



I.D.

O.D.

Tendon Type A B

ST13F - 5 70 75 20

ST13F - 4 60 70 20

ST13F - 3 50 55 20

ST15F - 5 85 90 25

ST15F - 4 70 75 25

ST15F - 3 60 65 25

ST15F - 2 45 50 25

lng: >10 giy v < 25 giy

r nc: 0.0% v < 0.3%

co ngt: -1% v +5%

Cng : > 27MPa sau 07 ngy (khi lp phng 40x40x160mm)

T l nc / Xi mng: 0.35

c tnh ca va

17 www.namcong.com

CC TIU CH TH NGHIM CA VT LIU

Cp d ng lc c th nghim trn mi 20 tn cp vi mt t mu 3 si, c ly t cun cp ch nh bi T vn gim st hoc Ch u t.

Hnh nh th nghim cp thc t

1. chy Kim tra chy ca va bng phu hnh cn (BS EN 445-2007). Vic o c c thc hin ti cng trng, khong 4 pht sau khi trn va, thi gian chy ca va phi t 10-25 giy. Th nghim ny c tin hnh cho mi m trn ti cng trng.

2. Cng chu nn Khun c mu (khi lp phng 40x40x160mm) c y li bng tm knh sau khi y va. Mi ln th nghim cng chu nn cho 03 mu. Cng nn ca mu va sau 28 ngy ti thiu phi t 30 Mpa hoc 27 Mpa ti thi im 07 ngy trong trng hp xut nh gi cng nn ti thi im 28 ngy da vo cng nn ti thi im 07 ngy.

Th nghim ny c thc hin mi sn hoc cu kin ring bit.

Hnh nh th nghim va thc t

TH NGHIM CP D NG LC (ASTM A416)

TH NGHIM VA

Tham kho bng 1: c tnh k thut ca Cp d ng lc.

Cc ch tiu k thut c kim tra nh sau:

7-wire strand 0.5 ASTM A416 BS 5896

Gii hn chy 1670 Mpa 1640 Mpa

Lc ko t ti thiu 183.7 kN 186.0 kN gin di tng i 3.5% 3.5%

18www.namcong.com

- BS 8110-1997 Part 1, 2, 3; BS 5400- Eurocode

- ACI 318- AS 3600- TCVN 2737


QUY TRNH THIT K

VT LIU

British Standard

Tham chiu BS 8500-1-2006 Tham kho Bng 7, BS EN 206-1-2000: 4.3

Eurocode

Tham chiu Mc e, BS EN 206-1

fy = 390 Mpa hoc 490 Mpa f

y = 390 Mpa hoc 490 Mpa

Tham chiu Bng 1: c tnh k thut ca cp D ng lc (ASTM A416, BS 5896)

B-tngCu kin cng sau

THP S DNG

CP Loi 0.5~12.7mm (12.9mm) khng c v bc cho cp c bm va v c v bc cho cp khng bm va. Loi 0.6~15.2mm (15.7mm) khng c v bc cho cp c bm va v c v bc cho cp khng bm va.

KCH THC TIT DIN S B

}Vi L: chiu di nhp h: chiu dy snGhi ch: Kch thc sau cng phi tha mn cc iu kin ca tiu chun p dng.

Span / 5

Span / 15

Cc phng n snKch thc tit

din s b

Chiu dy sn:

L/50 L/40

1. Sn phng d ng lc

2. Sn d ng lc c m ct

h h

> span / 3

Kch thc m

ct: L/3

Chiu cao dm

dt:

L/20 L/35

Chiu cao dm

cao: L/15 L/20

3. Sn d ng lc c dm dt

4. Sn d ng lc c dm cao

* CC TIU CHUN THIT K

C P DNG

Tham kho

KCH THC TIT DIN S B:

DM, SN M CT...

THNG TIN BAN U

BN V KIN TRC, M&E

TIU CHUN THIT K

TNH TON CP, B TR CP

LP M HNH

H S T HP

KIM TRA CC TRNG THI:

SLS, ULS

THUYT MINH, D TON

KIM TRA KT QU TNH TON

TRIN KHAI SHOP DRAWING

TI TRNG

VT LIU THIT K: B-TNG,

CP, CT THP

SLS:

GIAI ON TRANSFER:

KIM TRA NG SUT, VNG

GIAI ON SERVICE:

KIM TRA VNG, NG SUT,

V VT NT

ULS:

KIM TRA / TNH TON KHNG

UN, CHNG CT, CHNG XON,

GIA CNG U NEO

19 www.namcong.com


CC BC TNH TON CU KIN B-TNG D NG LC

British Standard Eurocode

1. H s t hp1.1. T hp ti kim tra vng.a1 1.0SW + 1.0SDL + 1.0WL + 1.0LL + ncPTa2 1.0SW + 1.0SDL + 1.0WL + 2LL + ncPT a3 1.0SW + 1.0SDL + 1.0WL + 2LL + ncPTa5 1.0SW + 1.0PT 1.2. T hp ti kim tra ng sut Transfer: 1.0SW + 1.0PTSLS: 1.0SW + 1.0SDL + 1.0WL + 1.0LL + ncPT

ULS: 1.4SW + 1.4SDL + 1.4WL + 1.6LL + ncPT*Trong SW : Ti trng bn thn SDL: Ti hon thin LL : Hot ti WL : Ti tng PT : Ti ng lc trcPT* : Phn m men th cp do ti ng lc trcnc = 0.82, theo Namcong2 = 0.25 (vn phng); hoc 2 = 0.75 (kho bi)2. Tnh ton s lng cp trn 1 di tnh ton:2.1. Theo phng php cn bng moment

1. H s t hp1.1. T hp ti kim tra vng.a1 1.0SW + 1.0SDL + 1.0WL + 1.0LL + ncPTa2 1.0SW + 1.0SDL + 1.0WL + 2LL + ncPTa3 1.0SW + 1.0SDL + 1.0WL + 2LL + ncPTa5 1.0SW + 1.0PT1.2. T hp ti kim tra ng sutTransfer: 1.0SW + 1.0PTSLS: 1.0SW + 1.0SDL + 1.0WL + 1.0LL + ncPT (t hp ton phn) 1.0SW + 1.0SDL + 1.0WL + 2LL + ncPT (t hp di hn)ULS: 1.35SW + 1.35SDL + 1.35WL + 1.5LL + ncPT*Trong SW : Ti trng bn thn SDL: Ti hon thinLL : Hot tiWL : Ti tng PT : Ti ng lc trcPT* : Phn m men th cp do ti ng lc trcnc = 0.82, theo Namcong2: xem bng A1.1 BS EN 1990:20022. Tnh ton s lng cp trn 1 di tnh ton: 2.1. Theo phng php cn bng moment

PHN BS

, =

, =

a = kl

1

r

1

=

1

=

:

:

1

=

1

=

= 1.15,

1 +

a = (a a) + a

a = Kl

1

r

1

r +

1

r

250

a = a = (a a) + a

1

500; 20

1

350; 20

a = a a

a = f,

,()

(1 + ) a()

=3

1 + 2

= ( )

3( )

=

:

:

= +

2

:

b

d: Aps.

= 0.45x

Ti trng, moment di ang xt

PHN BS

, =

, =

a = kl

1

r

1

=

1

=

:

:

1

=

1

=

= 1.15,

1 +

a = (a a) + a

a = Kl

1

r

1

r +

1

r

250

a = a = (a a) + a

1

500; 20

1

350; 20

a = a a

a = f,

,()

(1 + ) a()

=3

1 + 2

= ( )

3( )

=

:

:

= +

2

:

b

d: Aps.

= 0.45x

PHN BS

, =

, =

a = kl

1

r

1

=

1

=

:

:

1

=

1

=

= 1.15,

1 +

a = (a a) + a

a = Kl

1

r

1

r +

1

r

250

a = a = (a a) + a

1

500; 20

1

350; 20

a = a a

a = f,

,()

(1 + ) a()

=3

1 + 2

= ( )

3( )

=

:

:

= +

2

:

b

d: Aps.

= 0.45x

PHN BS

, =

, =

a = kl

1

r

1

=

1

=

:

:

1

=

1

=

= 1.15,

1 +

a = (a a) + a

a = Kl

1

r

1

r +

1

r

250

a = a = (a a) + a

1

500; 20

1

350; 20

a = a a

a = f,

,()

(1 + ) a()

=3

1 + 2

= ( )

3( )

=

:

:

= +

2

:

b

d: Aps.

= 0.45x

Ti trng, moment di ang xt

B

rng

d

i an

g x

t

Q/m

M1

M2

M3

P Pcentroida1 a3

a2

W

B

rng

d

i an

g x

t

Q/m

M1

M2

M3

P Pcentroida1 a3

a2

W

n : s lng si cp si b

M : m men tnh tonPav

: lc ng sut trc

trung bnh ca si cp

n : s lng si cp si b

M : m men tnh tonPav

: lc ng sut trc

trung bnh ca si cp

20www.namcong.com

Vi:

wc l vng ln ca cu kin cha chu ti.w1 l vng tc thi ca cu kin khi chu ti trng di hn (1) theo biu thc t (6.14a) n (6.16b) - BS EN 1990-2002.w2 l vng gia tng ca cu kin trng thi di hn(2)

khi chu ti trng di hn (1).

w3 l vng gia tng do cc tc ng thay i(3) theo biu thc (6.14a) n (6.16b) - BS EN 1990:2002.wtot l tng vng ca w1, w2, w3.wmaxtng vng cn li c tnh n vng ln ca cu kin.(1) quasi-permament loads, (2) long-term state, (3) variable actions.1.2.1.1. Cc iu kin chung 1) wmax hoc wtot L/250. 2) vng sau thi cng L/500 khi chu ti trng di hn. L: chiu di nhp hoc cng xn.

1.2.1.2. ng sut gii hn trong ct thp 310/s = 500/fyk.(As,req/As,prov) [s]SLS

1. TrNG THI GII HN S DNG (SLS): 1.1. Giai on truyn lc (Transfer stage) 1.1.1. Kim tra ng sut ng sut nn khng vt qu 0.4 fci ( BS 8110-1-1997:4.3.5.1 ) ng sut ko khng vt qu 0.36 fcui ( BS 8110-1- 1997: 4.3.5.2) Nu ng sut vt qu gii hn cho php, ly ng sut d tnh ton thp tng cng.1.1.2. Kim tra vngTham kho BS 8110-1-1997:3.4.6.1.2. Giai on lm vic bnh thng (Service stage)Tham kho BS 8110-2-1985:3.7, BS 8110-1-1997: 3.4.6.1.2.1. Kim tra vng:1.2.1.1. Cc bc tnh vng: a) vng tc thi do tng ti trng gy ra a1: T hp ti gy ra vng tc thi: 1.0SW + 1.0SDL + 1.0LL + ncPT

vng :

Vi :

Bn knh cong tc thi do tng ti trng gy ra.

MTL: Moment do tng ti trng ti thi im tc thi.Ec : M un n hi ca b-tng ti thi im tc thi.Ii : Moment qun tnh (ph thuc vo vt nt trn mt ct tit din). b) vng tc thi do ti di hn gy ra a2: T hp ti di hn gy ra vng tc thi: 1.0SW + 1.0SDL + ncPT + 2LL 2 = 0.25 (vn phng); hoc 2 = 0.75 (kho bi) vng :

Vi :

Bn knh cong tc thi do ti di hn gy ra.MPL: Moment do tng ti trng di hn gy ra.Ec : M un n hi ca b-tng.



British Standard Eurocode

1.1. Giai on truyn lc (Transfer stage) 1.1.1. Kim tra ng sut ng sut nn khng vt qu 0.6 fck(t) (BS EN 1992-1-1-2004) ng sut ko = fctm(t) (BS EN 1992-1-1-2004) Nu ng sut vt qu gii hn cho php, ly ng sut d tnh ton thp tng cng.1.1.2. Kim tra vngTham kho BS EN 1990-2002.1.2. Giai on lm vic bnh thng (Service stage)1.2.1. Kim tra vng

PHN BS

, =

, =

a = kl

1

r

1

=

1

=

:

:

1

=

1

=

= 1.15,

1 +

a = (a a) + a

a = Kl

1

r

1

r +

1

r

250

a = a = (a a) + a

1

500; 20

1

350; 20

a = a a

a = f,

,()

(1 + ) a()

=3

1 + 2

= ( )

3( )

=

:

:

= +

2

:

b

d: Aps.

= 0.45x

a = kl1

r

1

=

1. TrNG THI GII HN S DNG (SLS):

wc

wmax

w1w2

w3

wtot

Hnh A1.1 - M t vnga = Kl

1

r

a = Kl1

r

a = Kl1

r

a = Kl1

r

21 www.namcong.com


Vi:s l ng sut trong ct thp chu ko ti gia nhp (ti gi ta i vi cngxn) di tc dng ca ti trng tnh ton trng thi gii hn s dng (SLS).As,req l din tch ct thp theo yu cu tnh ton ti tit din ny i vi trng thi gii hn bn (ULS).As,provl din tch ct thp b tr ti tit din ny.

1.2.1.3. Cc bc tnh ton vngCc bc tnh vng ging nh trong BS 8110, nhng kt qu w1 hoc w2 th tnh theo ti trng di hn. w0 = w(Gk + Pm), trng thi tc thi. w1 = w(Gk + 2.Qk + Pm), trng thi tc thi ca ti trng di hn (1).

wa = w1 + w3 = w(Gk + Qk + Pm), trng thi tc thi ca ton b ti trng c tnh n cc tc ng thay i(3).

wb = w2 + w1 = w(Gk + 2.Qk + Pm), trng thi di hn(2) ca ti trng di hn (1).

Ghi ch: w0 = w(Gk; Pm; Ecm) w1 = w(Gk; Qk; Pm; Ecm) wa = w(Gk; Qk; Pm; Ecm) wb= w(Gk; Qk; Pm; Ecm/(1+ )), = (,t0) l h s t bin.

V vy:

w3 = wa w1 w2 = wb w1

T kt qu trn, chng ta so snh cc gi tr vng vi cc gi tr vng gii hn:

wmax

= w1 + w2 + w3 L/250, theo khon 4 - mc 7.4.1 trong

BS EN 1992-1-1-2004.

wafter construction

= wmax

w0 L/500, theo khon 5 mc 7.4.1

trong BS EN 1992-1-1-2004.

Ii: Moment qun tnh (ph thuc vo vt nt trn mt ct tit din). c) vng trng thi di hn do ti di hn gy ra a3: T hp ti di hn gy ra vng di hn SW + SDL + ncPT + 2LL 2 = 0.25 (vn phng); hoc 2 = 0.75 (kho bi) vng :

Vi: : Bn knh do ti di hn gy ra.

E3 = : M un n hi ca b-tng.

MPL : Moment do ti trng di hn gy ra.Ii : Moment qun tnh (ph thuc vo vt nt trn mt ct tit din.3 : H s t bin, theo iu kin mi trng.

1.2.1.2. vng sau cng ca cu kin:

1

=

PHN BS

, =

, =

a = kl

1

r

1

=

1

=

:

:

1

=

1

=

= 1.15,

1 +

a = (a a) + a

a = Kl

1

r

1

r +

1

r

250

a = a = (a a) + a

1

500; 20

1

350; 20

a = a a

a = f,

,()

(1 + ) a()

=3

1 + 2

= ( )

3( )

=

:

:

= +

2

:

b

d: Aps.

= 0.45x

K: h s tra theo bng 3.1 BS 8110-2-1985.l : chiu di nhp, hoc chiu di cng xn.

1.2.1.3. Kim tra vng cu kin theo BS 8110-1-1997:3.4.6a) iu kin 1:

L: chiu di nhp, hoc chiu di cng xn.

Vi :

b) iu kin 2:

PHN BS

, =

, =

a = kl

1

r

1

=

1

=

:

:

1

=

1

=

= 1.15,

1 +

a = (a a) + a

a = Kl

1

r

1

r +

1

r

250

a = a = (a a) + a

1

500; 20

1

350; 20

a = a a

a = f,

,()

(1 + ) a()

=3

1 + 2

= ( )

3( )

=

:

:

= +

2

:

b

d: Aps.

= 0.45x

PHN BS

, =

, =

a = kl

1

r

1

=

1

=

:

:

1

=

1

=

= 1.15,

1 +

a = (a a) + a

a = Kl

1

r

1

r +

1

r

250

a = a = (a a) + a

1

500; 20

1

350; 20

a = a a

a = f,

,()

(1 + ) a()

=3

1 + 2

= ( )

3( )

=

:

:

= +

2

:

b

d: Aps.

= 0.45x

1

500; 20

1

350; 20

i vi cu kin vt liu gin,

d nt, gy.

i vi cu kin cc loi vt liu

khng d b ph hoi.

a = Kl1

r

=

1.15,1 +


22www.namcong.com


1.2.2. Kim tra ng sut Gii hn ng sut nn: 0.45 f

ck i vi t hp di hn

(quasi-permanent) v 0.6 fck

i vi t hp ton phn (characteristic) theo tiu chun BS EN 1992-1-2004. Gii hn ng sut ko: f

ctm theo tiu chun

BS EN 1992-1-1-2004.

1.2.3. Tnh ton vt nt Tham kho BS EN 1992-1-1-2004: 7.3. 1.2.3.1. Tnh ton b rng vt nt a) Bc 1 Tnh ton m un n hi ca b-tng. E

c,eff = E

cm / (1 + (,t

0))

(,t0) l h s t bin.

e = E

s / E

c,ef

b) Bc 2 Tnh ton chiu cao ca vng chu nn: tm x

Cn bng m men vi trng tm l v tr t lc nn trong b-tng.V ta c: x = (((

e.)2 + 2 e.)1/2 - e.)

Vi: = A

s / (b.d)

b : chiu rng ca tit din.d : chiu cao tnh ton t mp ca vng ct thp thng. c) Bc 3 tnh ton ng sut ca thp chu ko: tm

s

d) Bc 4 tnh ton chch lch gia bin dng trong b-tng v ct thp: tm (

sm

cm).

Vi:

1.2.2. Kim tra ng sut ng sut nn khng c vt qu 0.4 f

cu

(BS 8110 - 1997: 4.3.4.2). ng sut ko khng vt qu: a (gi tr a ni suy trong bng 4.2 ng vi kt cu loi 3 v vt nt 0.2mm sau nhn vi h s ni suy c trong bng 4.3 BS 8110-1-1997).

1.2.3. Tnh ton vt nt Tham kho BS 8110-2-1985: 3.8/ BS 8007-1987.1.2.3.1. Tnh b rng vt nt

PHN BS

, =

, =

a = kl

1

r

1

=

1

=

:

:

1

=

1

=

= 1.15,

1 +

a = (a a) + a

a = Kl

1

r

1

r +

1

r

250

a = a = (a a) + a

1

500; 20

1

350; 20

a = a a

a = f,

,()

(1 + ) a()

=3

1 + 2

= ( )

3( )

=

:

:

= +

2

:

b

d: Aps.

= 0.45x

: Bin dng xut hin cng vi vt nt. : Bin dng ti v tr dng xt, b qua s nh hng b-tng trong vng ko.

a) Khi vt nt gii hn 0.2mm.

PHN BS

, =

, =

a = kl

1

r

1

=

1

=

:

:

1

=

1

=

= 1.15,

1 +

a = (a a) + a

a = Kl

1

r

1

r +

1

r

250

a = a = (a a) + a

1

500; 20

1

350; 20

a = a a

a = f,

,()

(1 + ) a()

=3

1 + 2

= ( )

3( )

=

:

:

= +

2

:

b

d: Aps.

= 0.45x

PHN BS

, =

, =

a = kl

1

r

1

=

1

=

:

:

1

=

1

=

= 1.15,

1 +

a = (a a) + a

a = Kl

1

r

1

r +

1

r

250

a = a = (a a) + a

1

500; 20

1

350; 20

a = a a

a = f,

,()

(1 + ) a()

=3

1 + 2

= ( )

3( )

=

:

:

= +

2

:

b

d: Aps.

= 0.45x

PHN BS

, =

, =

a = kl

1

r

1

=

1

=

:

:

1

=

1

=

= 1.15,

1 +

a = (a a) + a

a = Kl

1

r

1

r +

1

r

250

a = a = (a a) + a

1

500; 20

1

350; 20

a = a a

a = f,

,()

(1 + ) a()

=3

1 + 2

= ( )

3( )

=

:

:

= +

2

:

b

d: Aps.

= 0.45x

tham kho BS 8007-1987; B.4 Eq.2; Appendix B.b) Khi vt nt gii hn 0.1mm.

= 1.5( )(

)

3( )

tham kho BS 8007-1987; B.4 Eq.3 Appendix B.

: Bin dng trung bnh ti v tr ang xt.

PHN BS

, =

, =

a = kl

1

r

1

=

1

=

:

:

1

=

1

=

= 1.15,

1 +

a = (a a) + a

a = Kl

1

r

1

r +

1

r

250

a = a = (a a) + a

1

500; 20

1

350; 20

a = a a

a = f,

,()

(1 + ) a()

=3

1 + 2

= ( )

3( )

=

:

:

= +

2

:

b

d: Aps.

= 0.45x

PHN BS

, =

, =

a = kl

1

r

1

=

1

=

:

:

1

=

1

=

= 1.15,

1 +

a = (a a) + a

a = Kl

1

r

1

r +

1

r

250

a = a = (a a) + a

1

500; 20

1

350; 20

a = a a

a = f,

,()

(1 + ) a()

=3

1 + 2

= ( )

3( )

=

:

:

= +

2

:

b

d: Aps.

= 0.45x

tham kho BS 8110-2-1985; Eq.13:3.8.

: khong cch t mt b-tng chu nn n im tnh b rng vt nt.x : chiu cao trc trung ha.h : chiu cao cu kin.

= 1.5( )(

)

3( )

h

b

x

150

150 150

Ap

Centroidalaxis

dp

= ,

,()

= (1 + ) ()

=3

1 + 2

h d

x

As

b

Neutral axis

c

s1

fc

fs * As

fs

z

0.5*fc*b*x

Section Strain Stresses/force

S

C min

acr DIA


23 www.namcong.com

AsAp

d

XAc

cu3

X

fcd

Fc

FS

Z

s

e) Bc 5 tnh ton khong cch ln nht gia cc vt nt: tm s

r,max

f) Bc 6 tnh ton b rng vt nt: g) Bc 7 so snh b rng vt nt vi cc gii hn.

2. TrNG THI GII HN BN (ULS): 2.1. Tnh ton kh nng chu un2.1.1. Kh nng chu un theo ULSTham kho cc k hiu trong BS EN 1992-1-1-2004: 1.6Ta c, F

c = x . f

cd . b = (./

c).x.b.f

ck = 2 (d z).b.f

cd

Fb + F

s = A

p.f

pd + A

s.f

yd

V, M = Fc.z; z = (d .x/2); F

c = M / z

M/z = Ap.f

pd + A

s.f

yd

As = (M/z A

p.f

pd) / f

yd

Vy, M = 2.b.fcd

(d z).z = (2/ Yc)b.d2.f

ck(z/d-(z/d)2)

M/( b.d2.fck

) = (2/ Yc).(z/d-(z/d)2

Gi: K = M / ( b.d2.fck

);

= (2/ c) = 1.333, i vi f

ck 50 Mpa, = 1.0

K/=z/d - (z/d)2

(z/d)2 - z/d + K/ = 0 K /4 (z/d) = 0.5 + (0.25 K/)1/2, bi v:


s : Bc ln nht ca 2 thanh thp trong vng chu ko.c

min: Lp b-tng bo v thp.

Es: M un n hi thp.

As: Din tch thp.

bt : B rng tit din i qua trc ca thp chu nn.

Ghi ch: trc trung ha c xc nh nh sau:Trong trng hp mt ct chu ko, gi tr (h-x) trong cng thc tnh

r c xc nh bng cch ni suy gia

cc iu kin gii hn: Trc trung ha ti mt chu nn ln nht: (h-x) = h. Trc ko: (h-x) = 2h.

Nu < 0: Tit din khng b nt.

Strain diagram

Strainc, s

s

Cra

ck

c

sm

cm

LengthLe Le

S

wk = s . (sm cm)sr,max includes fos of 1.7

cs cult

PHN BS

, =

, =

a = kl

1

r

1

=

1

=

:

:

1

=

1

=

= 1.15,

1 +

a = (a a) + a

a = Kl

1

r

1

r +

1

r

250

a = a = (a a) + a

1

500; 20

1

350; 20

a = a a

a = f,

,()

(1 + ) a()

=3

1 + 2

= ( )

3( )

=

:

:

= +

2

:

b

d: Aps.

= 0.45x

wk = s . (sm cm)

2. TrNG THI GII HN BN (ULS): 2.1. Tnh thp khng un (tham kho BS 8110-1-1997:4.3.7)

K hiu:

Ap

: Din tch cp trong vng b-tng chu ko.A

s : Din tch thp.

b : B rng tit din.d : Chiu cao tnh t mt cu kin n tm din tch thp A

ps.

dn=0.45x : Chiu cao tnh t trng tm ca vng nn.

fpb

: ng sut ko thit k ca cp (khng cn tra bng, c th tnh theo cng thc tham kho trong Namcongs Handbook Part A 2015).x : Chiu cao trc trung ha.M

u : M men khng tit din.

Mf : M men do ti ngoi (P) gy ra.

=

+

2


Figure 3.5: Rectangular stress distribution

24www.namcong.com


Khi cnh tay n t n gii hn thp hn, x = d/2, tit din c th chu c moment ti a bi tit din ct thp n.V vy,

(z/d)min

= 1 /2.(x/d) = 1 /4 (z/d)

min = 1 0.8/4 = 0.8, for f

ck 50 Mpa, = 0.8

Klim

= 0.213

Nu K Klim

, ch cn gia cng ct thp chu ko (ct n):

As = (M/z A

p.f

pd) / f

yd

Nu K > Klim

, yu cu phi c gia cng chu ko v gia

cng chu nn (ct kp):

Gia cng chu nn nh sau:

x = d/2, for K = Klim

.

z/d = 0.8, for fck

50 Mpa, = 0.8

As = (M K

lim.b.d2.f

ck) / (f

yd.(d d))

Tnh ton fpd

:

p = (d

p x).

cu3

s = (d

s x).

cu3

ud ;

ud da vo tiu chun thit k.

pd

= p(0)

+ p

fpd

= f(pd

)

p =

p.E

p, if

pd (f

p.eff/

s) /E

p

p = f

pd(

pd) f

p.eff/

s, if

pd > (f

p.eff/

s) /E

p

Trong bt k trng hp no, p 500 Mpa.

Ch thch: fp.eff

tham kho trong Namcongs Handbook: Part B 2015.

CCH NG X CA CU KIN TRNG THI ULS (CLASS 3)

iu kin gii hn: Mf M

u

Theo iu kin cn bng lc th:

T= C

T = fpb Aps

Mu = (d - dn ) . fpb . Aps

Tnh: : Din tch cp v thp trong tit din ang xt. f

pb: xc nh da vo bng 14:

t : tra bng 14.

; vi m = 1.15 tra bng 14.

x

C =0.67f

T = f A

= M d.

C = (M 0.45x) f A

A = A + Af

f

f

=f

f= f

T T

;TT

= = f

M = 0.45x = 0.45 f

T T

;

v mi n(0.8 f ; 5a )

v =V

bM

[V] = 0.1 bMf

[V] = 1 0.55f

f vbM + d

V

[V] = max (V , [V ]mi n)

0.5[V] < < [V] + 0.4bM

A = 0.4 bs

0.87f

V [V] + 0.4bM

A = (V [V]s)

0.87fM

v mi n0.8f; 5a

v [v]

[v] = v + v

v =d.{dd/()}

(dd/)

.

T

v = 0.6

.

v =

[]

= area of s teel i n cons i MereM s ecti on.

= area of s tranM i n cons i MereM s ecti on.

= (A + A).

= length of cons i MereM s ecti on.

= Mepth of cons i MereM s ecti on.

: Reaction of column drop.

: axial stress

[v] < < 1.6[v]

As i n ([])

d.T

1.6[v] < < 2[v]

0,9x b

x

C =0.67f

T = f A

= M d.

C = (M 0.45x) f A

A = A + Af

f

f

=f

f= f

T T

;TT

= = f

M = 0.45x = 0.45 f

T T

;

v mi n(0.8 f ; 5a )

v =V

bM

[V] = 0.1 bMf

[V] = 1 0.55f

f vbM + d

V

[V] = max (V , [V ]mi n)

0.5[V] < < [V] + 0.4bM

A = 0.4 bs

0.87f

V [V] + 0.4bM

A = (V [V]s)

0.87fM

v mi n0.8f; 5a

v [v]

[v] = v + v

v =d.{dd/()}

(dd/)

.

T

v = 0.6

.

v =

[]



= (A + A).




: axial stress

[v] < < 1.6[v]

As i n ([])

d.T

1.6[v] < < 2[v]

x

C =0.67f

T = f A

= M d.

C = (M 0.45x) f A

A = A + Af

f

f

=f

f= f

T T

;TT

= = f

M = 0.45x = 0.45 f

T T

;

v mi n(0.8 f ; 5a )

v =V

bM

[V] = 0.1 bMf

[V] = 1 0.55f

f vbM + d

V

[V] = max (V , [V ]mi n)

0.5[V] < < [V] + 0.4bM

A = 0.4 bs

0.87f

V [V] + 0.4bM

A = (V [V]s)

0.87fM

v mi n0.8f; 5a

v [v]

[v] = v + v

v =d.{dd/()}

(dd/)

.

T

v = 0.6

.

v =

[]



= (A + A).




: axial stress

[v] < < 1.6[v]

As i n ([])

d.T

1.6[v] < < 2[v]

x

C =0.67f

T = f A

= M d.

C = (M 0.45x) f A

A = A + Af

f

f

=f

f= f

T T

;TT

= = f

M = 0.45x = 0.45 f

T T

;

v mi n(0.8 f ; 5a )

v =V

bM

[V] = 0.1 bMf

[V] = 1 0.55f

f vbM + d

V

[V] = max (V , [V ]mi n)

0.5[V] < < [V] + 0.4bM

A = 0.4 bs

0.87f

V [V] + 0.4bM

A = (V [V]s)

0.87fM

v mi n0.8f; 5a

v [v]

[v] = v + v

v =d.{dd/()}

(dd/)

.

T

v = 0.6

.

v =

[]



= (A + A).




: axial stress

[v] < < 1.6[v]

As i n ([])

d.T

1.6[v] < < 2[v]

x

C =0.67f

T = f A

= M d.

C = (M 0.45x) f A

A = A + Af

f

f

=f

f= f

T T

;TT

= = f

M = 0.45x = 0.45 f

T T

;

v mi n(0.8 f ; 5a )

v =V

bM

[V] = 0.1 bMf

[V] = 1 0.55f

f vbM + d

V

[V] = max (V , [V ]mi n)

0.5[V] < < [V] + 0.4bM

A = 0.4 bs

0.87f

V [V] + 0.4bM

A = (V [V]s)

0.87fM

v mi n0.8f; 5a

v [v]

[v] = v + v

v =d.{dd/()}

(dd/)

.

T

v = 0.6

.

v =

[]



= (A + A).




: axial stress

[v] < < 1.6[v]

As i n ([])

d.T

1.6[v] < < 2[v]

x

C =0.67f

T = f A

= M d.

C = (M 0.45x) f A

A = A + Af

f

f

=f

f= f

T T

;TT

= = f

M = 0.45x = 0.45 f

T T

;

v mi n(0.8 f ; 5a )

v =V

bM

[V] = 0.1 bMf

[V] = 1 0.55f

f vbM + d

V

[V] = max (V , [V ]mi n)

0.5[V] < < [V] + 0.4bM

A = 0.4 bs

0.87f

V [V] + 0.4bM

A = (V [V]s)

0.87fM

v mi n0.8f; 5a

v [v]

[v] = v + v

v =d.{dd/()}

(dd/)

.

T

v = 0.6

.

v =

[]



= (A + A).




: axial stress

[v] < < 1.6[v]

As i n ([])

d.T

1.6[v] < < 2[v]

P

Dd

APS

W

D -

x x

x

C =0.67f

T = f A

= M d.

C = (M 0.45x) f A

A = A + Af

f

f

=f

f= f

T T

;TT

= = f

M = 0.45x = 0.45 f

T T

;

v mi n(0.8 f ; 5a )

v =V

bM

[V] = 0.1 bMf

[V] = 1 0.55f

f vbM + d

V

[V] = max (V , [V ]mi n)

0.5[V] < < [V] + 0.4bM

A = 0.4 bs

0.87f

V [V] + 0.4bM

A = (V [V]s)

0.87fM

v mi n0.8f; 5a

v [v]

[v] = v + v

v =d.{dd/()}

(dd/)

.

T

v = 0.6

.

v =

[]



= (A + A).




: axial stress

[v] < < 1.6[v]

As i n ([])

d.T

1.6[v] < < 2[v]

x

C =0.67f

T = f A

= M d.

C = (M 0.45x) f A

A = A + Af

f

f

=f

f= f

T T

;TT

= = f

M = 0.45x = 0.45 f

T T

;

v mi n(0.8 f ; 5a )

v =V

bM

[V] = 0.1 bMf

[V] = 1 0.55f

f vbM + d

V

[V] = max (V , [V ]mi n)

0.5[V] < < [V] + 0.4bM

A = 0.4 bs

0.87f

V [V] + 0.4bM

A = (V [V]s)

0.87fM

v mi n0.8f; 5a

v [v]

[v] = v + v

v =d.{dd/()}

(dd/)

.

T

v = 0.6

.

v =

[]



= (A + A).




: axial stress

[v] < < 1.6[v]

As i n ([])

d.T

1.6[v] < < 2[v]

0.67

C 0.9

x/2

0.9

x0.0035

xT

MT CT TIT DIN NG SUT BIN DNG

d

d

b

As

Ap As

NeutralAxis

x = d/2

0.0035

s

st

X Fc

Fs

Fp + Fs

Z

fcd

(a) Mt ct (b) Bin dng (c) ng sut


25 www.namcong.com


2.2. Tnh thp chng ct cho dm v chng thng cho sn2.2.1. Chng ct cho dmV

Ed l lc ct tnh ton ti tit din ang xt do ti trng

bn ngoi v ng sut trc (bm dnh hoc khng bm dnh) gy ra.V

Rd,c l kh nng chu ct tnh ton ca cu kin khng c

ct thp chu ct.V

Rd,s l gi tr tnh ton ca lc ct m ct thp chu ct

khi chy do c th duy tr c.

VRd

l kh nng chu ct ca cu kin c ct thp chu ct.V

Rd,max l gi tr tnh ton ca lc ct ln nht m cu kin

c th duy tr c n gii hn b ph hoi khi nn. Vi: V

Rd = V

Rd,s(+ V

ccd + V

td)

Tham kho thm cc k hiu trong BS EN 1992-2004:1.6.

Bng 14. Bng tra h s t l gia ng sut trong cp v cng thit k.

Ghi ch: xem BS 8110-1-1997: 4.3.7.3 2.2. Tnh thp chng ct cho dm v chng thng cho sn2.2.1. Chng ct cho dm:A

sv din tch tit din ca hai nhnh ct ai.

bv b rng cu kin, i vi tit din ch T, I hay L l

chiu rng sn.d khong cch t th chu nn ln nht n trng tm vng ct thp (A

ps+A

s) trong vng ko.

dt chiu cao t th nn ln nht n ct thp dc hoc

trng tm cc thanh cng, (ly gi tr ln hn).f

pe ng sut trc tnh ton thit k trong thanh cng

sau khi xy ra tt c cc tn tht ng sut, phi ly nh hn 0.6f

pu.

Ghi ch: Khi din tch thp trong vng chu ko ny gm cc thanh cng v ct thp, gi tr ca f

pe c th ly bng gi tr

xc nh bng cch chia lc ng sut trc tnh ton cho din tch tng ng ca thanh cng bng(A

ps + A

sf

y/f

pu)

fyv

bn c trng ca ct thp, khng c ly ln hn 500 N/mm2.

fA

fbd

f/f x/d

f/f f/f

0.6 0.5 0.4 0.6 0.5 0.4 0.05 1.00 1.00 1.00 0.12 0.12 0.12 0.10 1.00 1.00 1.00 0.23 0.23 0.23 0.15 0.95 0.92 0.89 0.33 0.32 0.31 0.20 0.87 0.84 0.82 0.41 0.40 0.38 0.25 0.82 0.84 0.76 0.48 0.46 0.45 0.30 0.78 0.75 0.72 0.55 0.53 0.51 0.35 0.75 0.72 0.70 0.62 0.59 0.57 0.40 0.73 0.70 0.66 0.69 0.66 0.62 0.45 0.71 0.68 0.62 0.75 0.72 0.66 0.50 0.70 0.65 0.59 0.82 0.76 0.69

Vccd

Vtd


26www.namcong.com


2.2.1.1. Quy trnh thit ka) Trng hp 1: Nu V

Ed V

Rd,c, khng cn tnh thp chu ct. Khi khng c

yu cu ct thp chu ct trn c s tnh ton lc ct, phi b tr ct thp chu ct ti thiu theo 9.2.2 trong BS EN 1992-1-1-2004.V,VEd 0.5bwd fcdVi: = 0.6[1 f

ck/250], (fck tnh bng Mpa)

b) Trng hp 2:

Trong cc vng c VEd

> VRd,c

, , phi b tr ct thp chu

ct VEd

VRd

.

Tng ca lc ct tnh ton l:

VEd

(- Vccd

- Vtd

) VRd

VRd,max

2.2.1.2. Biu thc v tnh tonTham kho thm k hiu v cng thc trong Namcongs handbook Part B 2015.a) Trng hp 1:

VRd,c

= [CRd,c

k (100I f

ck)1/3 + k1 cp] bwd, [tnh bng N]

VRd,c,min

= (vmin

+ k1 cp)bwd, [tnh bng N]

Vi:

fck

tnh bng Mpa.

k = 1 + (200/d)1/2 2.0 vi d tnh bng mm.

I = A

sl / b

wd 0.02

Asl l din tch ct thp chu ko (l

bd + d).

bw

l chiu rng nh nht ca tit din trong vng chu

ko [mm].

cp

= NEd

/Ac < 0.2 f

cd (Mpa)N

Ed l lc dc trc trn tit din ngang do ti trng gy ra

hoc ng sut trc [tnh bng N] gy ra (NEdt

>t0 i

vi lc nn). nh hng ca cc bin dng lm thay

i lc dc NEd

th c th b qua.

Ac l din tch tit din b-tng [mm2].

k1 = 0.15

CRd,c

= 0.18/c

vmin

= 0.035 k3/2 fck

1/2

M0 moment gy ra ng sut = 0 trong cu kin.

sv bc ai dc theo cu kin.

vc ng sut ca b-tng theo bng 3.8 BS 8110-1-1997

vi As c thay th bng (A

ps + A

s), trong A

ps v

As l din tch cp v thp trong vng chu ko.

V & M gi tr lc ct v m men un ca tit din do ti cc hn gy ra.V

c lc khng ct gii hn thit k b-tng.

Vcr

lc khng ct gii hn thit k ca tit din b nt do un.2.2.2. Biu thc v tnh ton theo tiu chun BS 8110-1-1997: 4.3.8.ng sut ct thit k cc hn:

Nu (v) khng tha, th thay i tit din ca dm.

Lc ct gii hn ti tit din dm b nt do un:

Vi :

fpe

: sut ko trc ca cp sau khi tnh n mt lc,

fpe

0.6 fpu

.

fpu

: Cng cp d ng lc.

x

C =0.67f

T = f A

= M d.

C = (M 0.45x) f A

A = A + Af

f

f

=f

f= f

T T

;TT

= = f

M = 0.45x = 0.45 f

T T

;

v mi n(0.8 f ; 5a )

v =V

bM

[V] = 0.1 bMf

[V] = 1 0.55f

f vbM + d

V

[V] = max (V , [V ]mi n)

0.5[V] < < [V] + 0.4bM

A = 0.4 bs

0.87f

V [V] + 0.4bM

A = (V [V]s)

0.87fM

v mi n0.8f; 5a

v [v]

[v] = v + v

v =d.{dd/()}

(dd/)

.

T

v = 0.6

.

v =

[]



= (A + A).




: axial stress

[v] < < 1.6[v]

As i n ([])

d.T

1.6[v] < < 2[v]

xem (BS 8110-1-1997: 4.3.8.2) (1)

x

C =0.67f

T = f A

= M d.

C = (M 0.45x) f A

A = A + Af

f

f

=f

f= f

T T

;TT

= = f

M = 0.45x = 0.45 f

T T

;

v mi n(0.8 f ; 5a )

v =V

bM

[V] = 0.1 bMf

[V] = 1 0.55f

f vbM + d

V

[V] = max (V , [V ]mi n)

0.5[V] < < [V] + 0.4bM

A = 0.4 bs

0.87f

V [V] + 0.4bM

A = (V [V]s)

0.87fM

v mi n0.8f; 5a

v [v]

[v] = v + v

v =d.{dd/()}

(dd/)

.

T

v = 0.6

.

v =

[]



= (A + A).




: axial stress

[v] < < 1.6[v]

As i n ([])

d.T

1.6[v] < < 2[v]

v min(0.8f ; 5MPa )

x

C =0.67f

T = f A

= M d.

C = (M 0.45x) f A

A = A + Af

f

f

=f

f= f

T T

;TT

= = f

M = 0.45x = 0.45 f

T T

;

v mi n(0.8 f ; 5a )

v =V

bM

[V] = 0.1 bMf

[V] = 1 0.55f

f vbM + d

V

[V] = max (V , [V ]mi n)

0.5[V] < < [V] + 0.4bM

A = 0.4 bs

0.87f

V [V] + 0.4bM

A = (V [V]s)

0.87fM

v mi n0.8f; 5a

v [v]

[v] = v + v

v =d.{dd/()}

(dd/)

.

T

v = 0.6

.

v =

[]



= (A + A).




: axial stress

[v] < < 1.6[v]

As i n ([])

d.T

1.6[v] < < 2[v]

x

C =0.67f

T = f A

= M d.

C = (M 0.45x) f A

A = A + Af

f

f

=f

f= f

T T

;TT

= = f

M = 0.45x = 0.45 f

T T

;

v mi n(0.8 f ; 5a )

v =V

bM

[V] = 0.1 bMf

[V] = 1 0.55f

f vbM + d

V

[V] = max (V , [V ]mi n)

0.5[V] < < [V] + 0.4bM

A = 0.4 bs

0.87f

V [V] + 0.4bM

A = (V [V]s)

0.87fM

v mi n0.8f; 5a

v [v]

[v] = v + v

v =d.{dd/()}

(dd/)

.

T

v = 0.6

.

v =

[]



= (A + A).




: axial stress

[v] < < 1.6[v]

As i n ([])

d.T

1.6[v] < < 2[v]

(BS 8110-1-1997: 4.3.8.7)

(BS 8110-1-1997: 4.3.8.7)

[V] = max([V],[V]min)

A = (V [V])s0.87fd

A = 0.4bs

0.87f


A = (V [V])s0.87fd

A = 0.4bs

0.87f


A = (V [V])s0.87fd

A = 0.4bs

0.87f


27 www.namcong.com


b) Trng hp 2:

Gc phi c gii hn: 1 cot 2.5 i vi cc cu kin c ct thp chu ct thng ng:

VRd,s

= Asw

z fywd

cot / s (a)

VRd,max

= cw

bw

z 1 f

cd / (cot + tan) =

cw b

w z

1 f

cd sin

cos, t gi tr ln nht khi cot = 1

VRd,max

= cw

bw

z 1 f

cd / 2

Vi: A

sw l din tch tit din ca ct thp chu ct.

s l khong cch ct thp ai. f

ywd l gii hn chy tnh ton ca ct thp chu ct.

z l cnh tay n, z = 0.9d.

1 l h s gim cng i vi b-tng c vt nt

khi ct.

cw l h s tnh n trng thi ng sut ca b-tng

chu nn.


28www.namcong.com


V:

Gi tr 1 c th c tnh nh sau:

Nu 0.8fyk

fywd

fyd

= fyk

/S:

1 = = 0.6[1 f

ck/250], (f

ck tnh bng Mpa)

Nu fywd

< 0.8fyk

, v trong biu thc (a), (b) fywd

c thay

bi 0.8 fywd

:

1 = 0.6 vi f

ck 60 Mpa

v1 = 0.9 f

ck /200 vi f

ck > 60 Mpa

Gi tr cw

c tnh nh sau

cw = 1 i vi kt cu khng ng sut trc

cw

= 1 + cp

/fcd

vi 0 < cp

0.25fcd

cw

= 1.25 vi 0.25fcd

< cp

0.5fcd

cw

= 2.5(1 - cp

/fcd

) vi 0.5fcd

< cp

< fcd

Din tch tit din ngang ln nht ca ct thp chu ct

i vi cot = 1:A

sw,max f

ywd / (b

w s)

cw 1 fcd / 2

i vi cu kin c ct thp nghing chu ct, kh nng chu ct l gi tr nh hn gia hai kt qu sau:V

Rd,s = A

sw z f

ywd (cot + cot) sin / s (b)

VRd,max

= cw

bw

z 1 f

cd (cot + cot) / (1 + cot2),

t gi tr ln nht khi cot = 1 V

Rd,max =

cw b

w z 1 fcd (1 + cot) / 2

Ct thp chu ct tnh ton ln nht, vi cot = 1:A

sw,max f

ywd / (b

w s)

cw 1 fcd / (2 sin)

Ghi ch quan trng 1, cho bw: Khi bn cnh c cha ng lun cp c bm va vi ng knh > b

w/8, gi tr b

w c thay bi b

w,nom

bw,nom

= bw

0.5 v khi bw

/8, bw,nom

= bw

i vi ng lun cp khng bm va, ng bng nha c bm va v thanh cng khng bm dnh, chiu dy danh ngha ca sn l: b

w,nom = b

w 1.2

Ghi ch quan trng 2 cho lc ko b sung do lc ct VEd gy ra:Lc ko pht sinh, F

td, c to ra do lc ct V

Ed tc ng

ln ct thp dc.

Ftd

= 0.5 VEd

(cot cot),

Ftd

c gi tr cc i khi cot =1.

Ftd

= 0.5 VEd

(1 cot)


29 www.namcong.com

2.2.3. Tnh ton chng thng cho sn: tham chiu BS 8110-1-1997: 3.7.62.2.3.1. Tnh thp chng thng sn Kim tra iu kin chc thng ti mt ct C1:

Nu iu kin (1) khng tha th phi thay i tit din cu kin ti nt ct. Nu iu kin (1) tha =>tip tc kim tra iu kin 2. v [v] (2) Kh nng chu ct c kim tra u tin ti chu vi cch mt vng cht ti mt khong bng 1.5d. [v]=vc+ vp

Vi:[v] ng sut khng ct ca cu kinvc ng sut khng ct ca b-tng, tra bng 3.8 BS 8110-1:1997.vp ng sut khng ct do lc dc trc cu kin to ra.v ng sut gy ct ti chu vi ang xt.As din tch thp qua chu vi ang xt.Ap din tch cp qua chu vi ang xt.Asp= (As + Ap).bv ly cnh ca chu vi c tnh khng ct nh nht.d chiu cao lm vic ca sn.Vt Phn lc ti u ct.P Lc dc trc ca cu kin (k c lc nn ca cp).

Ast = Ftd / fyd, l ct thp chu ko b sung.

2.2.2. Tnh ton chng thng cho sn2.2.2.1. Cc bc thit k chng thng


(1)

x

C =0.67f

T = f A

= M d.

C = (M 0.45x) f A

A = A + Af

f

f

=f

f= f

T T

;TT

= = f

M = 0.45x = 0.45 f

T T

;

v mi n(0.8 f ; 5a )

v =V

bM

[V] = 0.1 bMf

[V] = 1 0.55f

f vbM + d

V

[V] = max (V , [V ]mi n)

0.5[V] < < [V] + 0.4bM

A = 0.4 bs

0.87f

V [V] + 0.4bM

A = (V [V]s)

0.87fM

v mi n0.8f; 5a

v [v]

[v] = v + v

v =d.{dd/()}

(dd/)

.

T

v = 0.6

.

v =

[]



= (A + A).




: axial stress

[v] < < 1.6[v]

As i n ([])

d.T

1.6[v] < < 2[v]

x

C =0.67f

T = f A

= M d.

C = (M 0.45x) f A

A = A + Af

f

f

=f

f= f

T T

;TT

= = f

M = 0.45x = 0.45 f

T T

;

v mi n(0.8 f ; 5a )

v =V

bM

[V] = 0.1 bMf

[V] = 1 0.55f

f vbM + d

V

[V] = max (V , [V ]mi n)

0.5[V] < < [V] + 0.4bM

A = 0.4 bs

0.87f

V [V] + 0.4bM

A = (V [V]s)

0.87fM

v mi n0.8f; 5a

v [v]

[v] = v + v

v =d.{dd/()}

(dd/)

.

T

v = 0.6

.

v =

[]



= (A + A).




: axial stress

[v] < < 1.6[v]

As i n ([])

d.T

1.6[v] < < 2[v]

BT U

Tnh ton gi tr (theo hnh 7 hoc Biu thc t (6.38) n (6.46) trong BS EN 1992-1-1-2004)

Tnh gi tr VEd,max (ng sut ct thit k ti mt ct):

VEd,max = VEd/(u1deff)trong u1 l chu vi ct

deff = (dy + dz )/ 2(dy v dz l chiu cao hu hiu ca cc phng

vung gc)

Tnh gi tr VRd,max

Nu VEd,max VRd,max Thit k li sn

Tnh gi tr VEd, (ng sut ct thit k):VEd,max = VEd/(u1deff)

trong u1 l chiu di ca chu vi ang tnh ton

Tnh ton kh nng thng ca b-tng (khng c ct thp chng ct), VRD,c

trong 1 = (ty tz)0.5

(ty, tz l t l ct thp theo hai phng. T l ct thp c tnh cho nhng cy thp c chiu di i ra khi vng 3d tnh t mt ct c tnh c on

neo thp vo b-tng)

Yes

No

min(0.8; 5)

v =.{/()}

(/)

.

Cornercolumn

Veff = 1.25Vt(see 3.7.6.3)

Edgecolumn

use equation 26 or, for approximately equal spans

Veff = 1.4Vt (see 3.7.6.3)

Veff = 1.25Vt(see 3.7.6.3)

Internalcolumn

use equation 25 or, for approximately equal spans

Veff = 1.15Vt (see 3.7.6.2)


30www.namcong.com


Nu VEd > VRd,c ?Khng cn ct

thp chng thng

Tnh din tch ct thp chng thng trn chu vi:

ASW = (VEd - 0.75VRd,c)Sru1/(1.5 fywd,ef)(khong cch gia cc chu vi b tr thp chng

thng)

fywd,ef = 250 + 0.25 deff fywd

Tnh ton di chu vi ngoi cng khng c yu cu ct thp chng ct:uout,ef = VEd/(VRd,c d)

Xc nh mt bng ct thp chng thng

Yes

No

Khi v > 2[v] v h thng ct thp c b tr nng cao kh nng chu ct, cn phi chng minh bng cc chng c thit k.Asvsin: khng ly nh hn 0.4ud/0.87fyv (0.4 c n v l Mpa )fyv : Cng thp chng thng.Asv: Tng din tch thp chng thng. : Gc nghing gia thp chng thng v mt phng sn.

ng sut ct phi c kim tra trn cc chu vi k tip nhau vi khong cch 0.75d cho n khi tm c chu vi khng c yu cu ct thp chu ct.

Chu vi tnh ton ti a tnh t mt ct khng c vt qu 3.75d. Nu khng tha th phi thay i tit din cu kin ti nt ct.

As i n 5(0.7v [v])uM

0.87f

v 2[v]

v 2[v]

l = K

f

Where f the concrete s trength at trans f er.

nomi nal Mi ameter of the tenMon

K

= (; )

=

=

;

;

:

v =2

h h

h3

v + v v

v =V

bM

v

v

v v + 0.4

v v + 0.4

As

=

0.8xyf

A =As

.f

f. (x + y)

u mi n x,y2

, 200

u 300

(5).

Nu v khng tha iu kin iu kin (2) cn tnh ton b tr thp chng thng:

(3).

(4).

[v] < < 1.6[v]

Asin ([])

.

1.6[v] < < 2[v]

[v] < 1.6[v] 1.6[v] < 2[v]

[v] < 1.6[v] 1.6[v] < 2[v]

(

5(

Punching shear layout


31 www.namcong.com


2.3. Tnh ton gia cng u neo sng, u neo cht2.3.1. Vi u neo cht (BS EN 1992-1-1-2004: 8.10.10.2.2)Truyn ng sut trc(1) Khi ko cng si cp ng sut ca b-tng gia si cp c gi nh c gi tr khng i fbpt, trong :fbpt = p1 1 fctd (t)Trong : p1 l h s tnh n dng si cp v tnh trng bm dnh lc bung si cp.

p1 = 2.7 i vi cc si thp c kha. p1 = 3.2 i vi cp c 3 v 7 si. 1 = 1.0 i vi nhng iu kin bm dnh tt. = 0.7 ngc li, ngoi tr khi c th iu chnh thnh cc gi tr cao hn lin quan n cc iu kin

c bit trong khi thi cng.

fctd l gi tr cng chu ko tnh ton ca b-tng ti thi im bung si cp; fctd (t) = ct.0.7 fctm (t) / c.Ghi ch: Ph thuc vo T chc ph chun k thut Chu u, c th s dng cc gi tr

p1 i vi cc dng thanh

cng khc vi cc dng nu trn.(2) Gi tr c bn ca chiu di truyn lpt:lpt = 1 2 pm0/ fbpt Trong :

1 = 1.0 khi bung t t. = 1.25 khi bung t ngt.2 = 0.25 i vi si cp c tit din trn. = 0.19 i vi cp 3 v 7 si. l ng knh danh ngha ca si cp.

pm0 l ng sut trong thanh cng ngay sau khi bung.(3) Ph thuc vo tnh hung thit k, gi tr tnh ton ca

chiu di truyn phi ly theo gi: lpt1 = 0.8 lpt

2.3. Tnh ton gia cng u neo sng, u neo cht: Tham kho BS 8110-1-1997: 4.10.32.3.1. Chiu di u neo cht:Khi cha c cc s liu thc nghim, c th s dng phng trnh sau y tnh ton chiu di truyn ng sut trc l

t, i vi lc cng ng sut trc ban u bng

75% bn c trng ca thanh cng khi cc u cu kin c dm cht.

Vi fci - cp bn b-tng trng thi Transfer. - ng knh danh ngha ca cp.Kt = h s ph thuc vo loi cp (K = 240).


0.87f

v 2[v]

v 2[v]

l = K

f



K

= (; )

=

=

;

;

:

v =2

h h

h3

v + v v

v =V

bM

v

v

v v + 0.4

v v + 0.4

As

=

0.8xyf

A =As

.f

f. (x + y)

u mi n x,y2

, 200

u 300

2.3.2. i vi u neo snga) ng sut sau tm neo c kim tra nh sau:

Khong cch ti thiu gia ng trc ca neo vi mp (edge) ca b-tng khng c nh hn gi tr quy nh trong T chc ph chun k thut chu u. Gi tr ti thiu da trn cng ca b-tng vo thi im ko cng. Ct thp cn thit ngn n hoc v vng u neo c xc nh theo lng tr b-tng ch nht, c gi l quy tc lng tr chnh, v tr pha sau u neo.


2.3.2. Thp gia cng u neo sng: Tham kho BS8110-1-1997

VERTICAL: Y0 = min (Yov ; Yov)HORIZONTAL: Y0 = min (Yov ; Yov)

anchor 1

anchor 2

anchor 3 anchor 4

anchor 5

Yo

h2Yo

h2Y

oh1

Yoh1

Yo

h3Yo

h3

Oh

Ov

Yov4 Yov4

HORIZONTAL

LINE OF ACTION

VE

RT

ICA

L

LIN

E O

F A

CT

ION

32www.namcong.com

Gia cng theo phng ng.

Vi: Xem bng 15

Po: Lc ko ca b cp


Tit din ca lng tr lin kt vi mi u neo c gi l khi ch nht lin kt. Khi ch nht lin kt c cng tm v trc i xng vi tm neo (c hai trc i xng) v phi tha mn:

Trong :

Pmax = Ap . p,max (Ap l din tch tit din ca cp; p,max l

ng sut cc i ln cp = min {k1 . fpk; k2 . fp0.1k}.Ghi ch: Gi tr ca k1 v k2 c th tm thy trong ph lc

Quc gia. Gi tr ngh l k1= 0.8 v k2= 0.9.c .c l kch thc ca khi ch nht lin kt.fck (t) l cng b-tng trong thi im ko cng.

Khi ch nht lin kt phi c t l xp x vi tm neo.

iu kin ny c tha khi c/a v c/ a khng ln hn:

trong a v a l kch thc ca khi ch nht

nh nht bao gm c tm neo.

Nhng khi ch nht lin kt vi neo trong cng mt tit din phi lun nm trong khi b-tng, khng c chng cho.

Primary regularisation prism(4) th hin xp x khi lng m trong cc ng sut thay i t gi tr rt cao ngay sau tm neo n gi tr hp l i vi b-tng theo lc nn n trc (uniaxial). Trc ca lng tr c ly theo trc ca cp, y chnh l khi ch nht lin kt, cn chiu cao pha sau u neo c ly gi tr 1.2 max (c .c). Khi lng tr lin kt vi nhng u neo khc nhau c th chng cho (iu ny xy ra khi cc b cp khng song song) nhng vn phi nm trong b-tng.(4) Quy tc lng tr theo ngi dch.b) Ct thp chng v v n trong b-tng, trong mi lng tr (nh ngha trong (1)) khng c nh hn:

Trong :

Pmax l lc ti a ln cp theo 5.10.2.1 trong EN 1992-1-1-2004 v fyd l cng thit k ca thp gia cng.Ct thp phi c phn b mi hng theo khi lng tr.

.

0.6. ()

1.25 .

.

= 0.15

, with , 1.20

xPo

2Ypo

bursting stress

Anchorage Zone Po / 2Yo

2Yo

Figure: Bursting stresses in rectangular beam subjected to an axial symmetric force.

2Y = 2min(Y; Y )

2Y = a

F = Pf

;

where f

;

seetable15

P: tendonjackingforce

2Y = 2min(Y; Y )

2Y = a

F = Pf

;

where f

;

seetable15

P: tendonjackingforce

Bng 15: Bng tra lc ko v u neo sng.

Ghi ch: xem bng 4.7 - BS 8110-1-1997

0.2 0.3 0.4 0.5 0.6 0.7

0.23 0.23 0.2 0.17 0.14 0.11

Y

Y

FP

Y

Y

FP


33 www.namcong.com


Din tch ca ct thp b mt ti b mt cht ti khng

nh hn 0.03 theo mi hng.

2.4. Tnh ton thp chng xon2.4. Tnh ton thp chng xon(Tham kho BS 8110-2-1985: 2.4.2, Reinforce Concrete Structures of Pretoria University)

Xc nh thp dc v thp chng ct ca dm.

As : Thp dc chu un.Asv: Thp ai chng ct.ng sut ct do moment xon (T) gy ra.

2.4.1. Kim tra iu kin chng xon ca dm

v + vt vtu (BS 8110-2-1985:2.4.5)

v = : ng sut do lc ct ti tit din gy ra.

vt : ng sut do moment xon gy ra.

vtu: ng sut gii hn (tra bng 16).

= 0.15

, with , 1.20

- ng tm

- Cnh ngoi ca tit

din ngang tnh ton,

chu vi u

- Lp bo v

Hnh 6.11: Cc ch thch v nh ngha s dng trong mc 6.3

2.4.1. Cc yu t hnh hc

a = 300 mm; b = 500 mm

A = 150000 mm2 (din tch ca mt ct tit din)

u = 1600 mm (chu vi ca mt ct tit din)

t = A/u = 94 mm

Ak= (500 - 94) . (300 - 94) = 83636 mm2

T hp ng sut trong thnh dm [EC 2-2 Fig. 6.104].

Xon ng sut ct T hp


0.87f

v 2[v]

v 2[v]

l = K

f



K

= (; )

=

=

;

;

:

v =2

h h

h3

v + v v

v =V

bM

v

v

v v + 0.4

v v + 0.4

As

=

0.8xyf

A =As

.f

f. (x + y)

u mi n x,y2

, 200

u 300


0.87f

v 2[v]

v 2[v]

l = K

f



K

= (; )

=

=

;

;

:

v =2

h h

h3

v + v v

v =V

bM

v

v

v v + 0.4

v v + 0.4

As

=

0.8xyf

A =As

.f

f. (x + y)

u mi n x,y2

, 200

u 300

Cp bn b-tng (fcu)

vt,min vtu

MPa N/mm2 N/mm2

25

30

35

40

0.33

0.37

0.385

0.4

4

4.38

4.69

5

Bng 16: Bng gi tr vt,min v vtu

Ghi ch: tham kho bng 2.3 - BS 8110-2-1985: 2.4.6

Diagonal cracking continuting along all the four sides

Both logitudinal bars and links are intersecting the cracks: they both work in equilibrating torsion

resistant hollow section

Rectangular section subjected to shear and torsion

t

a

b

tt

hmax

X1

hmin Y1

Y1

X1


34www.namcong.com

2.4.2. Tnh thp chng xon.

Thp dc chng xon phi c b tr u chu vi trong ct ai. Khong cch gia hai thanh thp khng c nh hn 300mm; v phi c t nht 4 thanh, mi thanh mi gc ai.

Thp dc chng xon c th t cng v tr vi thp chng un bng cch dng thp c ng knh ln hn ng knh ca thp chng un.

Thp chng xon nn c neo mt on t nht bng cnh ln nht ca tit din ang xt.


2.4.2. Gii thch:VEd,i(T) lc ct do xon thnh th i.VEd,i(V) lc ct do tc ng ca lc ct vo thnh th i A l tng din tch ca tit din trong phm vi chu vi ngoi, bao gm c cc din tch l rng.u l chu vi ngoi ca tit din ngang.tef,i l dy hu hiu ca thnh th i. Trong trng hp tit din l hnh ch nht th tef,I = teff = A/u.zi l chiu di cnh ca thnh th i. Trng hp tit din hnh ch nht th z

i (= h hoc = b) l chiu di

cnh bn ca din tch ch nht Ak.

Ak l din tch c bao bi ng tm ca cc thnh lin kt vi nhau, bao gm c cc din tch l rng bn trong (xem hnh 6.11).uk l chu vi ngoi ca din tch Ak.q lu lng lc ct do moment xon, (in N/m).q = t,i . tef,it,i l ng sut ct do xon trong thnh th i.

2.4.3. Quy trnh thit k:i vi tit din ch nht, v = 450 ~ cot = 1:b, h l di cc bn ca tit din Ak.

Moment xon trung tm ca tit din Ak:

TEd = 2qh.(b/2) + 2.qb.(h/2)= 2.q.bh = 2.q.Ak

a) Chng xon trong ct thp ai:

q = TEd / (2.Ak)

VEd,i(T) = q . zi = TEd.zi / (2.Ak)

VEd,i(T) = VRd,s = Asw zi fywd cot / s

TEd = TRd,s = (2.Ak) . Asw fywd cot / s

TEd = TRd,s = (2.Ak) . Asw fywd /s, for =450~cot = 1

vt < vt,min vt > vt,min

v vc + 0.4

v > vc + 0.4

B tr hm lng

thp chng ct ti

thiu (cu to);

khng cn thp

chng xon.

Tnh lng thp

chng ct ; khng

cn thp chng

xon.

Tnh ton thp

chng xon, khng

c nh hn hm

lng thp cu to

chu ct.

Tnh ton ct thp

chng ct v chng

xon.

Bng 17: iu kin tnh ct thp chu ct v xon

Ghi ch: tham kho bng 2.4 - BS 8110-2-1985 : 2.4.6


0.87f

v 2[v]

v 2[v]

l = K

f



K

= (; )

=

=

;

;

:

v =2

h h

h3

v + v v

v =V

bM

v

v

v v + 0.4

v v + 0.4

As

=

0.8xyf

A =As

.f

f. (x + y)

u mi n x,y2

, 200

u 300

q

qb

qh

qb

qh

S

h

h

450

Asw . fyd


35 www.namcong.com


tnh ton chng ct cho tit din ch nhtTheo phng ng:

V*Ed = VEd,i(V) + VEd,i(T) < VEd,max Trong : VEd,i(V) = VEd / 2, v VEd l lc ct. V*Ed VRd,s Theo phng ngang: VEd,i(T) < VEd,max VEd,i(T) VRd,sTrong : z trong biu thc ca VEd,max v VRd,s l cng gi tr nh zi

b) Chng xon trong ct thp dc Asl:

Lc ko c th c cn bng theo phng ngang:

H = 2qh + 2qb = q(2h + 2b) = q . uk = Asl . fyd / uk q = Asl.fyd / ukTEd = 2. q . Ak q = TEd / 2Ak Asl.fyd / uk = TEd / 2Ak, vi = 450 ~ cot = 1 TEd = TRd,Als = (2. Ak). Asl . fyd / uk (c)Tng qut: TEd=TRd,Als=(2. Ak) . Asl.fyd .cot / u (d)Lu cho cp d ng lc c bm va:

Asl.fyd trong biu thc (c) v (d) c thay th bi Asl.fyd + Ap.pTrong : p 500 Mpa, l ng sut tng ln trong cp d ng lc c bm va.c) Kh nng chu lc ln nht ca cu kin chu xon v ct

phi tha mn iu kin sau:

TEd / TRd,max + VEd,(T) / VRd,max 1.0Trong

TEd l moment xon tnh ton.VEd,(T) l lc ngang tnh ton, vi zi=b trong din tch Ak.TRd,max = 2..cw. fcd . Ak. tef,i. sin . cos.VRd,max kh nng chu ct tnh ton ln nht (xem li mc 2.2.1 ca ti liu ny)

h

450

450

Nc Nc qh

H = qh


36www.namcong.com


d) i vi cc tit din c hnh dng gn ch nht, ch yu cu b tr ct thp ti thiu (xem BS EN 1992-1-1-2004: 9.2.1.1), min l tha mn iu kin sau:

TEd / TRd,c + VEd,(T) / VRd,c 1.0

Trong :

TRd,c = (2.Ak).fctd.teff VRd,c = [CRd,c k (100I fck)1/3 + k1 cp] bwd

PHNG N THIT K SN TRN NN T N NH By Brian M. Juedes

Chng 2.0 ca PTI Design Manual, phin bn 3, a ra quy trnh thit k cho sn trn nn t n nh. V bn

cht, quy trnh ch yu cu mt lc ng trc ti thiu l

0.05A sau khi mt ton b ng sut (losses), bao gm c

ma st trn nn t (sub-grade friction). Khng c iu

kin quy nh v kh nng chu lc cho php ca t,

cng ca t hay ti trng chu vi (perimeter load). Cng

khng c iu kin quy nh v moment, lc ct hay

cng ca sn. Bi vit ny th hin mt phng n thit

k cho sn dy ng nht trn nn t n nh. Quy trnh

ny gii thch v kh nng chu lc cho php ca t;

cng ca t v ti trng chu vi; v tnh ton lc ct, mo-

ment v vng thit k.

CC T KHA

Kh nng chu lc cho php ca t; cc u neo; lch

tm; nn t h tr (ground-supported); sn d ng lc

cng sau; cng ca t; t n nh; sn dy ng

nht.

GII THIU

PTI Design of Post-Tensioned Slabs-on-Ground, phin

bn th 3 phn ln vit v nn t m rng. C mt s

phn cp v t nn v nn t n nh. Chng 2.0

trnh by v cc thng tin tng qut v xut quy trnh

thit k cho sn trn nn t n nh.

Thng tin tng qut tham kho trn BR

technical guideline

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