khang uon cua dam be tong cot thep theo tieu chuan thiet ke cau 6ubng bo TCVN 11823-5:2017 voi nhieu truong hop cot thep thanh chju keo va nen

(1)

KHOA HOC-CONG NGHE

So 10/2018

SLTC khang uon cua dam be tong cot thep theo tieu chuan thiet ke cau 6ubng bo TCVN 11823-5:2017 voi nhieu truong hop cot thep thanh chju keo va nen

• TS. NGUYEN DUY L | £ M - TrUdng Dgi hoc SUpham Ky thuat TP Hd Chi Minh

TOM TAT: Tieu chuan thiet ke cau duong bo TCVN 11823-5:2017 mpi ban hanh dupc bien soan dua tren tieu chuan thiet ke cau AASHTO, LRFD Bridge Design va tieu chuan cu 22TCN 272-05 cung dya tren AASHTO 98, Hoa Ky. Trong ca hai tieu chuan vua neu co trinh bay each xac dmh sdc khang danh dinh cua dam be tong cot thep thuong, sue khang nay gi^ dinh thep chiu keo va neu co cot thep chiu nen deu Pat toi trang thai chay khi pha hoai Oieu nay khong dung cho tat ca cac trudng hop bo tri thep chiu keo va nen voi cac ham luong nhieu it khac nhau. Vi v§y, bai bao trinh bay ket qua xac d|nh sue khang danh dinh cua dam be tong cot thep thuong voi nhteu truong hpp bo tri cot thep thanh khac nhau Ket qua nghien cdu giup ky su thiet ke cau, dac biet la cac sinh vien ung dung de xac dinh sue khang uon phu hop.

Tllf KHOA: Dam be tong cot thep, mo-men cyc hgn, phan tich mgt cSt ngang, sdc khang uon ABSTRACT. The new Highway Bridge Design Specification TCVN 11823-5 2017 was built based on the AASHTO, LRFD Bridge Design Specifications of America as well as the old Specification22 TCN 272-05 based on AASHTO 98 In these specifications, the nominal moment resistances of reinforced concrete beam was provided with tensile steel bar, or with compressive steel bar, both were in plastic state. It is clear that the such provision has not been enough for many cases. Therefore, the paper performed on moment resistance of reinforcement concrete beams with various cases of tensile and compressive steel reinforcing bar The study is highly expected to help bridge engineers and students compute moment resistance of beam properly

KEYWORDS: Reinforced concrete beam, ultimate moment, section analysis, bending resistance

I.DATVANOE

Dam la mot trong nhUng ket cau pho bien nhat trong ePng trinh xay dung ndi chung va eong trinh cau duPng ndi rieng. Hien tai, sUc khang uon eua dam be tong cot thep thudng xae dinh theo ePng thdc cua tieu ehuan thiet kecau mdi ban hanhTCVN 11823-5:2017 [1] (dUa tren nen LRFD Bridge Design Specifications of America [2]) va ngay ea tieu ehuan thiet ke trudc day 22TCN 272-05 [3] (dUa tren nen AASHTO 98 [4]), edn thieu eho nhieu trudng hop Ung xd eho thep ehiu nen va keo. Dieu nay thUe sU gay khP khan cho ngUPi sCrdung, hay sddung se eho ket qua khPng phu hpp. Vi du khi bd tri ham luong thep chju nen bang thep ehiu keo, ehieu day vung nen eua be tong se bang khdng khi dCing ePng thUe (26) hay (30) trong TCVN 11823-5:2017, hoac eong thdc 5.7.3.1.1-4 trong TCN 272- 05. Bai bao bo sung sdc khang uon eua dam be tdng cot thep thudng eo tiet dien chd nhat bo tri thep ehiu keo va nen vPi cac ham lupng nhieu it khac nhau.

2. PHAN TiCH Li THUYET 2.1. Md hinh vat lieu

MP hinh vat lieu eua be tong thudng va edt thep thanh dUOe the hien P Hinh 2.1a, b. Theo bieu do Hinh 2.1a, be tPng xem nhu b\ pha hoai tai bien dang tuong doi la 0.003, dng xCf keo true tiep cua be tong la tuyen tinh va pha hoai giPn, cUPng dp keo rat be so vdi cudng do nen. Theo mP hinh d Hinh 2.1b: Quan he Ung suat - bien dang cua thep trong nen va keo true tiep la tuong t u nhau: Tuyen tinh khi bien dang nho hon bien dang ehay, bieu do di ngang khi bien dang vUpt qua diem ehay.

2.2. COsd phan tich

Mot so gia thuyet eo ban khi phan tieh dam:

- Be tong thuPng la vat lieu ddng nhat va d i n g hudng;

- Mat cat dam phling trUPe va trong khi uon;

- Dp bien dang trong cot thep thanh bang dp bien dang betPng;

- Kha nang ehiu keo eiia be tong thudng la nho so vdi edt thep ed the bo qua.

79

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KHOA H O C - C d N G N G H |

Stress

Strain

a) - Ong xucua be tong thudng

Elastic modulus. E.

b) - Ung xdcua thep thanh Hinh 2.1: Md hinh hda quan hi dng suat • bien dang doc true

cua thep thanh [5]

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Hinh 2.2: Phan bo ung suat bien dang va ndi luc tren tidt dien Xet mpt dam gian don chlu mp-men duong eP tiet dien ehUnhat vdi chieu cao h, be rpng b va thep thanh chju keo. Hinh 2.2 (c) the hien viec don gian hda khdi Ung suat nen cua be tong nhu mpt hinh chU nhat. Vi tri true trung hPa [NA] duac xac dmh bdi cPng thUe c = ^ , dp cong eua

80

dam ((!>) va kha nang chiu mP-men danh djnh {MJ dupe rut ra t d eae phuong trinh can b^ng tTnh hoe eua e^e lUc va mp- men tren mat eat ngang nhUsau (cong thdc 1-3):

Ic,=l7; (1) K=TTryr.^Z^, yo (2)

<^ = £j(d-c) = e^/c ^0.003/c (3) Trong dd:y^., y^^ - Lan lUpt la khoang each tU true

trung hoa tdi lUe keo tUong dUong T. va lUc nen tUOng duong C; O - Do cong tUong Ung vdi mP-men A/_, trong khi e - bien dang tai dinh dam,

, (a) (b) (c)

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/ EB 0 ooisc, g^ErC 0.005 f,=400MPa -* Ev=fy/E>=400y200000=0,003 a)-M$t cat nen kiem soat nen; b) - M0t cit chuyen tidp,

c) - M0t cat kiem soat keo Hinh 2.3: Dieu kien ve bien dang keo eua thep

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Beam 1 Seam 2 Hinh 2.4: Phan bo bien dang vi dng suat trong mat cit

cd cot ehiu kep vi ehiu nen 2.3. Ket quS phan tich

Hinh 2.2 cung cap sU phan bo Ung suat, bien dang va luc trong thep thanh tren mat cat ngang dam bao gom e^ thep thanh ehiu keo va ehiu nen. Bien dang nen gidi han eua be tong thudng tai dinh dam, e^^, dupe de nghj bang 0.003 khi pha huy. Khi bien dang n l n dat den gidi han 0.003, bien dang tai edt thep ehiu keo ed the xay ra 3 trudng hop dupc mP tadH/n/)2.4nhusau (ACI 318-14[6]):

- Mat c^t kiem soat nen: Bien dang thep chiu keo kha nhp khi dam pha huy (e^^ < e).

- Mat c3t ehuyen tiep: Bien dang thep chju keo d mdc vUa p h l i khi dam pha buy (e ,<£,,< 0.005).

- Mat e l t kiem soat keo: Bien dang thep chju keo kha Idn khi dam pha huy (£ > 0.005).

(3)

KHOA H O C - C 6 N G N G H £

So 10/2018

Do mp-dun dan hoi cua thep thudng bang E^ - 200000MPa, nen neu eUdng ddchay eua thep./; =400MPa bien dang chay se la s^^ -,/;/£ ^ 0.002. Thep n^n Ung vdi gidi ban nen 0.003 tren dinh eo the xay ra hai trUdng hop:

Thep nen d trang thai dan hoi neu E^ < e\ {it xiy ra vi thep nen thudng gan mep tren ed bien dang 0.003 Idn hon e), thep nen d trang thai chay neu E\ > e^ (thudng xay ra).

Mpt dam gian don bo tri ca thep chju keo va thep chju nen ed the dupe phan tieh thanh hai dam theo nhu mo ta P Hinh 2.4: Beam 1 cP thep chiu nen va keo can bSng, Beam 2 ehi co thep chiu keo can bang vpi be tPng chju nen. Nham tranh pha hoai ddn, mat c i t dam se dugc thiet ke P trang thai ehuyen tiep hay kiem soat keo. Co 3 trUdng hpp bd tri edt thep thanh nhU sau:

- Mat cat ehi co edt thep thanh ehiu keo (hoac eo them cot thep chiu nen nhung rat nhd so vdi edt thep chiu keo);

- Mat cat vda ep edt thep thanh chju keo, vda cP edt thep ehju nen p trang thai bien dang deo;

- Mat eat vUa eP edt thep thanh ehiu keo, vda cP eot thep ehiu nen d trang thai bien dang dan hoi.

Bang 2.1. Cdng thiie tinh khi nang chju uon M^

va vi tn true trung hda e khi be tdng dat tdi nen cue han (CI)Thep nen va thep keo deu d trang thai chay

0 85/;/?|6 0&5lh

* Kiem tra bien dang thep nen

£[=0003{c-d-)/c>€\

* Kiem tra bien dang thep keo

£^ =0.003id-c)/c>£^

Sdc khang uon cho trUdng hpp nay (tong mo-men ddi vdi trpng tam nen be tdng):

M„ = AJ^(d-a/2)-AJ^Xd'-a/2)

Trong dd:A^J^, - Lan lUpt la dien tich va cUPng dp chay eua thep keo.

(C2) Thep nen d trang thai chSy, thep keo lam viec Strang thai dan hoi (thUcrng xay ra khi thep keo qua nhieu)

Q-?'5fAP,c)b + Aj^ =^,£,e, =.4,£,o.003((/-t)/t, c la nghiem phuong trinh bae 2 (lay nghiem dUOng):

(0 85X^,i)c-+(.4;y, +/i,^,0.003)c-4£,0 003(/-0

* Kiem tra bien dang thep nens, ={i.OQ'i{c-d')lc>e^

* Kiem tra bien dang thep keo^, =G.m3{d-c)lc<E^.

Sdc khang uon eho trudng hpp nay (tong mp-men doi vdi trpng tam nen be tPng):

M^ = A^E^s^(d-a/2)-AJ^(d -a/2) Trong do: £^, £, - Lan lupt la mP-dun dan hdi va bien dang cCia thep keo.

(C3) Thep pen 6 trang thai dan hoi, thep keo lam viec dtrang thai chSy (thUcmg xay ra khi thep nen nhieu)

<=.0.S5l(/i,c)b-i- A,E,O.O0Hc-<i-)/ c = AJ^.

c la nghiem phuong trinh bac 2 (lay nghiem dUOng):

(0.85_/; ^|6)c' + (.^; £,0.003-^,/^)c-^;£, 0.003c/'=0

•" Kiem tra bien dang thep nen s[ =0 002(c-d')/c<s[

* Kiem tra bien dang thep keo e^ = 0.003(<i -c)lc>e^

SUc khang uon cho truPng hup nay (tong md-men ddi vdi trong tam nen be tPng);

M,^ÂJSd-al2)-AÊÊSd'-al2)

Trong dd:E^, s^, E^ - Lan lUpt la mo-dun dan hoi, bien dang eua thep keo, bien dang eua thep nen.

(C4) Thep nen d trang thai dan hoi, thep keo lam viec d trang dan hoi (thUo^ng xSy ra khi thep keo nhieu, thep nen co ldp bito v^ day)

Q.8SfX/],c)f> + A,EX^A^E,E,

o 0 85_/;'(/?,c)ft + A'^E^0 003(c-d')lc = A^E,0.003{a'-c)lc c la nghiem phuong trinh bac 2 (lay nghiem duong):

(0,g5/>,6Jc^ +{A^ + A^ )£,0 003c - ( . ^ y + /!,(/)£, 0-003 = 0

• Kiem tra bien dang thep nenE] = 0.003(c-d')/c<Sy

*Kiem tra bien dang thep keo£, =0-003(<^-c)/f<£, SUc khang uon eho truPng hop nay (tong mP men ddi vdi trpng tam nen be tdng):

M„ = A^E^£^{d -a / 2)- A[E^E[{d' -a / 2) Trong dd: E^, e^, s^ - Lan lUpt la mo-dun dan hdi, bien dang eua thep keo, bien dang cua thep nen.

(C5) Thep nen co ham lUpng I6n hon hay b^ng thep keo (thUcfng xay ra khi thep nen nhieu), tinh nhUC3 Khi tinh toan so bp theo trudng hop (Cl) cd ket qua c < 0, dieu nay ed nghia thep nen kha nhieu nen thudng d trang thai dan hoi. Vi vay, ta cd the tim sUc khang uon danh djnh nhutrudng hop (C3).

2.4. Cac vi du minh hoa cho dam tiet dien chuT nhatthdrdirdi chju keo

2.4.1. Vidul

Xet dam gian don tiet dien chO nhat cd chieu cao h - 45em (d - 40cm), rpng b - 20em, tai day dam bo tri thep thanh dudng kinh 25mm, 4thanh iA_= 19.63 em^), thep nen 02 thanh dUdng kinh 12 {A\ - 2.26 cm^), Idp bao ve thep nen d'=5em. CuPng dp chay cua thep thanh /;. - / - 400MPa; be tong thuong eo eUdng do nen/^ = 30MPa.

Tinh sdc khang uon danh djnh.

Dau tien gia thiet thep keo va nen deu ehay, tinh c theo trudng hpp (Cl): c = 16.3cm, ^9, ^ 0.84, e^,= e.^ 0.002.

* Kiem tra bien dang thep keo:

s, = 0 Q03{d-c)/c = 0.00208 > £,:Dung nhUgia thiet.

* Kiem tra bien dang thep nen:

£, = 0 003(c-(/•)/f = 0 00436> £[:Dung nhugia thiet.

SUc khang uon danh dinh theo (Cl):

M,,-262.29 kN.m.

2.4.2 Vidu2

Nhu v i d u l nhung neu thep thd dudi 6 thanh dudng kinh 25 {A^ = 29.45em^), thep nen van 02 thanh dUdng kinh 12 (/i'-2.26cm^).Tinh sde khang uon danh djnh.

Dau tien gia thiet thep keo va nen deu ehay, tinh c theo trudng hup (Cl);e^25.52em,^i=0.84,c =£-^^0.002.

f;=0,003(c-a")/c = 0.002413 >£;:DungnhUgia thiet.

81

(4)

KHOA H O C - C d N G NGH$

£, =o.003(d-c)/c-0.001703 <£,:Dan hdi, saigiSthiet.

Vay, p h l i tinh theo trudng hpp (C2) (thep nen ehay, thep keo dan hoi):

c = 24.40 cm, ^, = 0.84, e, = d^, = 0.002.

* Kiem tra bien dang th^p nen:

£\ = 0 . 0 0 3 ( c - t / V c - 0.002385 > e,: Chly, Ok.

£, =0.003{d-c)/c = 0.001919 <£.:Dan hdi: Ok.

Sdc khang uon danh djnh theo (C2}:

M=341.59kN.m.

2.4.3.Vidu3

Nhuvidu 1 nhUng neu thepthddudi 3 thanh dudng kinh 25 (A^= 14.73em=), th^p nen van hai thanh dudng kinh 12 (^^^ = 2.26cm^).Tinh sdc khang uon danh djnh.

Dau tien giS thiet thep keo va nen deu ch^y, tinh e theo trudng hpp (Cl): e - 11.7em, ji^ = 0.84, e = £^, = 0.002.

£;=0.003(c-<j'')/c-oooi7i8<£,,:Dan hoi, sai giS thiet.

£, - 0 003((/-c)/c-0.007258>s; Chay, dung gi^ thiet, Vay, ph^i tinh theo truPng hpp {C3) (thep nen dan hdi, thep keo chay):

c = 11.97 em, /?, = 0.84, E^ = e'^, = 0.002.

f;-O.003(c-<:;Vc-0.001747 <£;:ean hoi. Ok.

£, = O.003(t/ -c)lc^ 0,007028 > E;. Chay: Ok.

SUc khang uon danh dinh theo (C3):

A/-206.17kN.m.

2.4.4.Vidu4

Neu thep thd dudi 6 thanh dudng kinh 25 (A^ = 14.73em^), thep nen van hai thanh dudng kinh 12 {A^ = 2.26em=), IPp b^o v& thep nen d' = 9em, Tinh sdc khang uon danh dinh.

Dau tien gia thiet thep keo va nen deu ch^y, tinh c theo trudng hop (Cl):c^25.52cm,^,^0.84,e,=£'^, = 0.002.

£ > 0 003(c-^')/c-0.001942<£;,:Dan hoi, sai gi^thi^t.

£, =0.003(fi/-<:)/c = 0.00l703<£„:Oan hoi, sal gia thiet.

Vay phSi tinh theo trUPng hpp {C4) (thep nen dan hoi, thep keo dan hoi):

e = 24.43em, ^, - 0.84, e^, = e'^ = 0.002.

£;=0.003(c-<^')/c = 0.001895<£^:Danh6i,Ok.

£, = 0 003{t/-c)/c-0.001913 <£,:Dan hoi: Ok.

Sdc khang uon danh djnh theo (C4):

A / - 3 3 6 . 7 4 kN.m.

2.4.5 Vidu 5

Neu thep thd dudi hai thanh dudng kinh 25 [A^

= 9.82em^), thep nen hai thanh duPng kinh 25 (A = 9.82cm').Tinh sUc khang uon danh dinh.

Dau tien gia thiet thep keo va nen deu chay, tinh c theo trUdng hpp (Cl): e < Oem, p^ = 0.84, c,= e' = 0.002.

Vay, ph^i trnh theo trUPng hpp (C3) (thep nen dan hdi, thep keo chay):

c = 6.32cm, i = 0.84, £,:^E' =0.002.

£ , = 0 . 0 0 3 ( c - t / V c = 0.000628<£^:Dan hoi. Ok.

£•, = 0.003(c/ - c ) / c = 0.01598 > s;. ChSy: Ok.

Sdc khang uon danh dinh theo (C3}:

JW = 143.80 kN.m.

3 . K ^ T L U A N

Bai bao eung cap thdng tin hUu feh ve each xac dmh sdc khang uon cua dam be tong cot thep thudng vdi nhieu trUdng hpp bo tri edt thep chju keo va nen cp ham lupng nhieu it khac nhau. Dieu nay giup nguPi ky suthiet ke ap dung dung, thuan tien, phu hop tieu ehuan TCVN 11823-5:2017.

Tai lieu tham khdo

[1]. TCVN 11823-5:2017, Thiet Kecau dUdng bo. tap 5:

K^tcaubetPng.

[2]. AASHTO {2012), LRFD Bridge Design Specifications of America.

[3]. 22 TCN 272-05 (2005), Tiiu chudn thiet kecau.

[4]. AASHJOIRW (1998), Bridge Design Specifications of America, Second Edition, SI Edition.

[5]. Ju H, Lee DH, Cho HC, Kim KS, Yoon S and Seo SY (2014), Application of Hydrophilie SHanol-Based Chemical Grout for Strengthening Damaged Reinforced Concrete Flexural Members, Materials 2014, 7,4823-4844;

doi;10.3390/ma7064823.

[6]. ACI 318-14 (2014), Building code requirements for structural concrete.

Ngay nhan bdi: 15/8/2018 Ngay ch^p nhin dang: 4/9/2018 NgUci phdn bign: TS. Le Bd Khanh

TS.TranVQTir

khang uon cua dam be tong cot thep theo tieu chuan thiet ke cau 6ubng bo TCVN 11823-5:2017 voi nhieu truong hop cot thep thanh chju keo va nen

So 10/2018

SLTC khang uon cua dam be tong cot thep theo tieu chuan thiet ke cau 6ubng bo TCVN 11823-5:2017 voi nhieu truong hop cot thep thanh chju keo va nen

79

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