Chuyen de Tinh Cot Btct Chiu Nen Lech Tam Xien

CHUYEN E NGHIEN CU ANH GIA PHNG PHAP TNH CAU KIEN BTCT CHU NEN

LECH TAM XIEN

(1) Phep bien oi tng ng tiet dien chu nen lech tam xien thanh tiet dien chu nen lech tam

phang.

(2) He so quy oi dien tch thep hai au tiet dien trong cach tnh gan ung ra dien tch thep bo

tr theo chu vi cua ca tiet dien.

Page 239

CHUYEN E: NGHIEN CU ANH GIA PHNG PHAP TNH CAU

KIEN BTCT CHU NEN LECH TAM XIEN

TOM TAT

Trong noi dung cua chuyen e nay, sinh vien se phan tch lam ro s lam viec cua

cau kien chu nen (keo) lech tam xien. Tren c s ly thuyet o sinh vien s dung cac

ngon ng lap trnh Visual Basic va VBA trong excel e viet nen phan mem s dung

mat cong tng tac e tnh toan cot thep va kiem tra kha nang chu lc. Ben canh o

sinh vien cung xay dng mot phan mem tnh cau kien chu nen xien theo phng phap

gan ung (1)

. Hai phan mem nay co the import d lieu t ETABS t o lam cho viec

thiet ke c tien li hn. T viec so sanh ket qua tnh cua hai phan mem sinh vien

sinh vien e ngh s dung he so k(2)

=0.4 trong tnh toan cau kien chu nen xien theo

phng phap gan ung.

T khoa: cau kien chu nen xien, ngon ng lap trnh VB, mat cong tng tac, quy

hai phng ve mot phng.

A. TONG QUAN VE CAU KIEN CHU NEN (KEO) LECH TAM XIEN

Nen lech tam xien la nen lech tam ma mat phang uon khong nam trong mat

phang oi xng cua tiet dien.

oi vi tiet dien tron th khong xay ra nen lech tam xien.


LECH TAM XIEN

(3) capacity factor

Page 240

Hnh CD.1 S o noi lc tiet dien chu nen lech tam xien

Cot chu nen lech tam xien thng gap trong cac khung khi xet s lam viec cua

cot ong thi chu uon theo hai phng.

Tiet dien ch nhat chu nen lech tam xien th cot thep thng at theo chu vi va

oi xng qua hai truc. Trng hp Mx~My th nen lam tiet dien vuong.

S lam viec cua cau kien chu nen (keo) lech tam xien

Vi cau kien lam bang vat lieu ong chat va ang hng th khi chu nen lech

tam xien co the dung phng phap cong tac dung e tnh ng suat:

yx

x y

MM Nx y

J J F

ieu kien ben la han che ng suat khong c vt qua ng suat cho phep

hoac cng o tnh toan cua vat lieu.

oi vi ket cau be tong cot thep th vat lieu khong phai la ong nhat ang hng

cung nh la an hoi tuyen tnh. Nen viec xac nh ng suat cua tiet dien khong the

dung phng phap cong tac dung ma phai xet tac dung ong thi cua N, Mx, My.

Khi chu nen lech tam xien tuy theo tng quan gia tr cua N, Mx, My ma tiet

dien vung be tong chu nen co the nam vao 1 trong 4 trng hp sau (xem hnh CD.2).


LECH TAM XIEN

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Page 241

Hnh CD.2 cac dang cua vung nen

en trang thai gii han th ng suat trong be tong c xem nh la phan bo eu

va at en gia tr Rb. ng suat trong cot thep xa truc trung hoa co the at en Rs

(keo) hoac Rsc (nen), trong khi o cot thep gan truc trung hoa co ng suat be hn.

Gia thiet tnh toan

Tiet dien phang.

Be tong vung keo c bo qua.

ng suat trong vung be tong chu nen la eu.

Cac phng trnh tnh toan tiet dien chu nen xien

Lay hp cua lc doc va momen theo 2 phng ta c cac phng trnh sau:

x b b b si si

y b b b si si

b b si si

M R A y y

M R A x x

N R A A

x

xo

x

xo

x

xo

x

xo


LECH TAM XIEN

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Hnh CD.3 S o noi lc va bieu o ng suat tren tiet dien thang goc vi truc doc

cau kien

Tuy theo quan iem tnh toan ma cac nc khac nhau a ra cach tnh toan ng

suat trong tng thanh thep i khac nhau.

Tnh ng suat cot thep theo quan iem

ng suat

Tieu chuan TCXDVN 356-2006 a ra

cong thc thc nghiem e xac nh s:

,( 1)

11.1

sc u

i

i

Trong o

su = 400Mpa

= 0.85-0.008Rb

Aho1

ho5

ho6

s s

s s

s s

s s

s s

s s

bRb

y

x

ho5

ho5ho5xo

A'A

D' A'

A

x

a'

h

h

Hnh CD.4 ng sut trong

ct thp i v i

trong ct thp i v i


LECH TAM XIEN

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Page 243

i

oi

x

h

sc i sR R

Tnh ng suat cot thep theo quan iem bien

dang.

Xuat phat t bien dang cua be tong tai mep vung

nen a c quy nh, dung gia thiet tiet dien phang,

khi biet v tr truc trung hoa (biet xo) va v tr cua thanh

cot thep (hoi) se tnh c bien dang cua no la i (xem

hnh CD.4).

i = coi

x

xh

0

0

Khi i T th i = Rs

i < T th i = iRs , vi T = s

s

E

R

Vi cot thep chu keo th ieu kien la x h0i

i vi cot thep chu nen: iu kin la x hoi

He so theo tieu chuan ACI 318 th bang 0.60.85, theo tieu chuan BS 8110 th

bang 0.9, theo EUROCODE 2 th 0.8, con trong TCXDVN khong co noi en b

bang bao nhieu tuy nhien chung ta co the tm c he so theo TCXDVN thong qua

cong thc tnh ng suat cot thep tren.

AAAA

x

h

h

h

h

A A'

Hnh CD.5 ng sut trong

ct thp i c tnh theo

bin dng i.


LECH TAM XIEN

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S tng ong gia hai quan iem

Cong thc xac nh ng suat thanh thep trong TCVN 356-2005:

,( 1)

11.1

sc u

i

i

Tai v tr truc trung hoa th ng suat bang 0 i= hay x = hoi

Cot thep chu keo khi i >0 x < hoi

Cot thep chu nen khi i >0 x < hoi

Nh vay 2 quan niem tnh tren co tnh tng ong, he so trong quan iem bien

dang co chung y ngha vi he so trong TCVN 356-2005.

Y ngha cua he so hay o la vung be tong chu nen trong tnh toan khong lay

la vung gii han bi ng trung hoa ma c lay giam i vi he so hay

B. PHNG PHAP TNH TOAN CAU KIEN CHU NEN

Phng phap dung bieu o tng tac

nynxnP - M -M

Mt cong tung tc

ng bao

ti trngMx0

My0

nMt phng PnP

(b)

(c)

(a)

y

xM

M

P


LECH TAM XIEN

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HnhCD.6 Mat cong tng tac Pn-Mnx-Mny

Nguyen tac tnh thep cua phng phap nay la th dan ham lng thep sao cho

kha nang mat cong tng tac bao ben ngoai iem lc can tnh toan cot thep.

Phng phap gan ung

Thc hin theo trnh t sau:

Xet tiet dien cnh Cx, Cy

ieu kien ap dung: 0.5 2x

y

c

c

Tiet dien chu lc nen N , moment uon xM , yM , o lech tam ngau nhien axe , aye .

Sau khi xet uon doc theo hai phng, tnh c he so x , y . Moment gia tang:

1 1;x x x y y yM M M M

Tuy theo tng quan gia hai gia tr 1xM , 1yM vi kch thc cac canh ma a ve

mot trong hai mo hnh tnh toan (theo phng x hoac y )

Mo hnh Theo phng x Theo phng y

ieu kien

y

y

x

x

c

M

c

M 11

x

x

y

y

c

M

c

M11

K hieu

xh c ; yb c

1 1xM M ; 2 1yM M

0.2a ax aye e e

xb c ; yh c

1 1yM M ; 2 1xM M

0.2a ay axe e e


LECH TAM XIEN

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Xac nh s bo vung be tong chu nen:

1.b

Nx

R b

Nu x1 ho:1

0

0

0.61

xm

h, ngc li th m0=0.4

Momen tng ng :

1 0 2

hM M m M

b

o lech tam: e1=M/N; eo = max(e1,ea)

Da vao o lech tam e va t so x1 e phan biet cac trng hp tnh toan sau

Trng hp 1: 2ax1< Rho => lech tam ln loai 1. Dien tch thep c

xac nh theo cong thc sau:

Ast = ZkR

xhbxRNe

sc

b2

0

Trng hp 2: x1 lech tam ln loai 2. Dien tch thep c xac

nh theo cong thc sau:

Ast = ZkR

ZeN

ZkR

Ne

ss

)('

Trng hp 3: x1> Rho => lech tam be. Gia tr x1 phai c xac nh lai,

x1 la nghiem cua phng trnh bac 3 sau:

x3 + a2x

2 + a1x + a0 = 0


LECH TAM XIEN

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Trong o

a2 = -(2+ R)h0

a1 = ZhhbR

NeRR

b

0

2

0 122

a0 = bR

hZeN

b

RR 0)1(2

Giai phng trnh bac 3 tren c gia tr cua x. Lu y khi giai ra neu x>ho

th lay s=-Rsc roi giai he 2 phng trnh can bang e tm lai gia tr cua x.

Dien tch cot thep c xac nh theo cong thc sau:

Ast= zkR

xhbxRNe

sc

b )2

( 0

C. XAY DNG MAT BIEU O TNG TAC

Khai niem ve bieu o tng tac

Vi 1 tiet dien cho trc chu nen (keo) lech tam, kha nang chu lc c bieu

dien thanh mot ng tng tac. o la ng cong the hien theo hai truc trong mat

phang Oxy. Truc ng Oy the hien gia tr lc nen Pn, truc ngang Ox the hien momen

Mn.

Tren ng cong tng tac Pn-Mn, ng tia the hien o lech tam e=Mn/Pn.

Truc ng Oy the hien kha nang chu nen ung tam Po (momen uon bang 0) cua cot.

Truc ngang Ox the hien kha nang chu momen uon Mo (lc doc truc bang 0).

Thanh TuyenHighlight


LECH TAM XIEN

(3) capacity factor

Page 248

x

y (P )

(M )ee=

0

P

M

- ng t ia e=M /Pn n

Hnh CD.7 ng cong tng tac Pn-Mn

Mat bieu o tng tac

Vi nen lech tam xien th kha nang chu lc c bieu dien thanh mat bieu o

tng tac. o la mot mat cong the hien theo ba truc Oxyz. Truc ng Oz the hien gia

tr lc nen (keo). Cac truc Ox va Oy the hien momen Mx, My. Moi iem tren mat

bieu o c xac nh bi toa o x, y, z the hien noi lc tng ng. K hieu C, Dx, Dy

la giao iem cac truc vi mat bieu o. ng net gach OkDkxDky la giao tuyen cua

mot mat phang ngang (song song vi mat xOy) vi mat phang toa o va mat cua bieu

o tng tac. ng cong CDk D la giao tuyen cua mat phang cha truc Oz vi mat

bieu o.

C

kxD

Dky

kO

O

DyD

Dx

z(max)N

Nzz

y

x

Mx0

nP

x

yM

M

Hnh CD.8 Mat bieu o tng tac


LECH TAM XIEN

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Page 249

Nguyen tac xay dng bieu o tng tac

Gia thiet ng gii han vung be tong chu nen cua be tong t o xac nh cac

gia tr Nz, Mx, My.

T s o bo tr thep, ng knh cot thep ta xac nh c toa o cua tng thanh

thep, ng suat trong cot thep si, xac nh c lc tac dung len moi thanh cot thep.

Biet cac thong so vung nen la kch kch thc vung nen, ng suat trong be tong Rb,

xac nh c trong tam vung nen, hp lc tac dung len vung nen. T o gia tr lc

doc Nz c xac nh bang cach lay hp lc theo phng z, Mx va My c xac nh

c bang viec lay momen cua cac lc trong cot thep va be tong vung nen vi truc x

va truc y.

ng gii han vung be tong chu nen

Hnh CD.9 ng giihan vung be tong chu nen

ng gii han vung be tong chu nen c xac nh da vao hai thong so va

x.

Trng hp ac biet ng vi ->0 hay -> /2 tng ng vi nen lech tam phang.


LECH TAM XIEN

(3) capacity factor

Page 250

Xac nh dien tch va toa o trong tam vung be tong chu nen.

Nh a trnh bay tren th gia thiet la ng suat trong be tong vung nen la eu v

vay toa o trong tam vung be tong chu nen chnh la toa o trong tam hnh hoc cua

vung be tong chu nen.

Trng hp I: vung nen tam giac

Hnh CD.10 ng gii han vung nen trng hp I

Phng trnh ng trung hoa la y =-(tan x+c

Phng trnh ng thang qua A, B la: y = h/2

Phng trnh ng thang qua B, C la: x = b/2

Toa o iem E la nghiem cua he phng trnh sau:

(tan )

/ 2

y x c

x h

Toa o iem F la nghiem cua he phng trnh sau:

(tan )

/ 2

y x c

x b


LECH TAM XIEN

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Page 251

Dien tch vung be tong chu nen: Ab=S(EBF)

Toa o trong tam vung be tong chu nen:

3

3

B E FGAb

B E FGAb

x x xx

y y yy

Trng hp II: vung nen hnh thang loai II

Hnh CD.11 ng gii han vung nen trng hp II


Phng trnh ng thang qua A, D la: x=-b/2

Phng trnh ng thang qua B, C la: x=b/2


(tan )

/ 2

y x c

x b


(tan )

/ 2

y x c

x b

Dien tch vung be tong chu nen: Ab=S(ABFG)-S(EFG)


LECH TAM XIEN

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Page 252


( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

G ABFG ABFG G EFG EFG

GAb

b

G ABFG ABFG G EFG EFG

GAb

b

x S x Sx

A

y S y Sy

A

Trng hp III: vung nen hnh thang loai II



Phng trnh ng thang qua A, B la: y=h/2

Phng trnh ng thang qua D, C la: y=-h/2


(tan )

/ 2

y x c

y h


(tan )

/ 2

y x c

y h

Dien tch vung be tong chu nen: Ab=S(EBCG)-S(EFG)


LECH TAM XIEN

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Page 253


( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

G EBCG EBCG G EFG EFG

GAb

b


GAb

b

x S x Sx

A

y S y Sy

A

Trng hp IV: vung nen hnh ngu giac



Phng trnh ng thang qua A, D la: x=-b/2

Phng trnh ng thang qua D, C la: y=-h/2


(tan )

/ 2

y x c

x b


(tan )

/ 2

y x c

y h

Dien tch vung be tong chu nen: Ab=S(ABCD)-S(EFD)


LECH TAM XIEN

(3) capacity factor

Page 254


( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

G ABCD ABCD G EFD EFD

GAb

b


GAb

b

x S x Sx

A

y S y Sy

A

Xac nh ng suat cot thep

Theo TCXDVN 356-2005 th ng suat cot thep c xac nh nh sau:

1

1,11

,

i

usc

si mang dau am khi chu nen, dng khi chu keo.

Khi tnh ra si>Rsi th lay si=Rsi, khi tnh ra si


LECH TAM XIEN

(3) capacity factor

Page 255

Xac nh mat bieu o tng tac

Hnh CD.14 S o tnh tiet dien chu nen xien

Lay hp lc theo phng ng Oz va lay momen theo 2 truc Ox, Oy ta c cac

phng trnh sau:

x b b b si si

y b b b si si

b b si si

M R A y y

M R A x x

N R A A

Nh vay ng vi moi cap gia tr q va c ta co c 1 iem xac nh tren he toa o

(Oxyz) tng ng vi he (OMxMyN).

Khi ta thay oi gia tr t trong khoang (0, /2) va c trong khoang ( -tan b / 2 -

h / 2; -tan b / 2 - h / 2 ) ta c tap hp cac iem tao thanh mot mat cong tng tac.

Aho1

ho5

ho6

s s

s s

s s

s s

s s

s s

bRb

y

x

ho5

ho5ho5xo


LECH TAM XIEN

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Page 256

Hnh CD.15 mat cong tng tac

S dung bieu o tng tac e tnh toan va kiem tra kha nang chu lc

Nguyen tac tnh toan

Vi moi tiet dien chu nen lech tam vi kch thc va bo tr cot thep a xac inh,

ta se xay dng c mot mat bieu o tng tac tng ng.

e kiem tra kha nang chu lc cua tiet dien, ta se tnh c N=N,

Mx=Neeoy, My=Neeox.

Neu iem (N; Mx; My) nam trong bieu o tng tac th tiet die n o u kha

nang chu lc.


LECH TAM XIEN

(3) capacity factor

Page 257

Neu iem (N; Mx; My) khong nam trong bieu o tng tac th ta se thay oi

cot thep hoac tiet dien, luc nay se co mot mat tng tac mi. Qua trnh giai lap ket

thuc khi iem lc nam tren bieu o tng tac. Lng thep va tiet dien tng ng la ket

qua can thiet ke.

Phng phap xac nh

Bc 1: Gia s ham lng thep, t o ve mot mat cong tng tac.

Bc 2: Tm ng cong tng tac tren o cac iem co Mx/My bang nhau va

bang t so Mx/My cua noi lc ang can tnh thep.

Viet mot phng trnh mat phang co phng trnh Mx/My=(Mx/My)can tnh thep

Tm giao iem cua mat phang nay vi mat cong cua bieu o tng tac 3D.

Tap hp cac ng cac giao iem nay cho ta c 1 ng cong tng tac can

tm.

Hnh CD.16 ng cong tng tac co Mx/My=(Mx/My) can tnh thep

-5000.00

-4000.00

-3000.00

-2000.00

-1000.00

0.00

1000.00

2000.00

3000.00

4000.00

5000.00

6000.00

0 200 400 600 800 1000 1200

N(KN)

M(KNm)


LECH TAM XIEN

(3) capacity factor

Page 258

Bc 3: Tm he so kha nang chu lc

nh ngha he so chu lc: he so chu lc la t so CR(3)

=AP

AC

Hnh CD.17 Xac nh he so kha nang chu lc

Bc 4: Kiem tra kha nang chu lc

Co cac trng hp xay cua he so kha nang chu lc:

- p < 1 : lng thep dung e ve bieu o tng tac lam cho cot d kha nang chu

lc.

- p =1 : lng thep dung e ve bieu o tng tac la va u kha nang chu lc.

- p > 1 : lng thep dung e ve bieu o tng tac lam cho cot khong u nang chu

lc.

Bc 5: Tnh lap e tm lng thep lam cho he so CF.


LECH TAM XIEN

Page 259

D. VIET CHNG TRNH VE MAT CONG TNG TAC

Viec xay dng bieu o tng tac yeu cau khoi lng cong viec rat ln. V vay

chung ta can s giup cua may tnh e xay dng mat cong tng tac.

S o khoi ve mat cong tng tac (xem hnh CD.18)

Muc tieu cua chng trnh

ng dung e thiet ke cau kien chu keo nen lech tam xien (cot, vach).

Co the import d lieu c xuat ra t ETABS t o lam giam khoi lng

cong viec cua ngi thiet ke.

Co the tnh toan theo tieu chuan 356-2005-VIET NAM, ACI, EURO CODE.

Giao dien chng trnh CCRD1 (cho phep nhap va tnh toan 1 d lieu bat k)

Hnh CD.18 Giao dien chng trnh CCRD1 viet bang ngon ng Visual Basic


LECH TAM XIEN

Page 260

Giao dien chng trnh CCRD1 (viet bang VBA)

Hnh CD.19 Giao dien chng trnh CCRD1 viet bang VBA trong excel

Giao dien chng trnh CCRD1 (cho phep import d lieu t ETABS)

Hnh CD.20 Giao dien chng trnh CCRD2 viet bang VBA trong exce


LECH TAM XIEN

Page 261

Hnh CD.21 s o khoi lap trnh ve bieu o tng tac

Cac bc ve bieu o

tng tac 3D

Nhap d lieu (tiet dien, noi

lc, so thanh thep bo tr tren

tng canh)

Cho gia tr thay oi t 0 en

90 (bc thay oi cang nho

cang tot)

= 1

Thay oi gia

tr cua x

x=x1

Tm v tr trong tam

vung be tong chu

nen

Tm ng suat trong

cac thanh thep

T cac phng

trnh can bang tm

c cac gia tr

Mx, My, N

ve c 1 iem

trong he toa o

(MxMyN)

x=x2

...

ve c 1 iem

trong he toa o

(MxMyN)

x=xi

...

ve c 1 iem

trong he toa o

(MxMyN)

= 2

...

= i

...

Ve c 1 ng

cong gom nhieu

iem tng ng

vi cac gia tr

khac nhau cua x


LECH TAM XIEN

Page 262

E. KIEM TRA O TIN CAY CUA PHAN MEM

e kiem tra o tin cay cua chng trnh nay th sinh vien so sanh ket qua tnh

cua phan mem cua mnh vi cac phan mem tnh lech tam xien cua nc ngoai. e

tang o tin cay th sinh vien chon so sanh vi 2 phan mem rat manh ve ket cau la

ETABS 9.7 va CSICol 8 (ay la phan mem chuyen ve tnh cot nen xien).

V cac phan mem nc ngoai khong co tnh theo tieu chuan Viet Nam nen e

quy ve cung he quy chieu th chung ta can bien oi vat lieu tng ng ve cung mot

he quy chieu.

Kiem tra vi chng trnh CSICol 8

S dung tieu chuan BS 8110-97 e kiem tra

Vat lieu B25 co Rb=CD.5 Mpa tng ng vi fcu = 1.5CD.5/0.67 =32.5 MPa

Cot thep CII co Rs=280 Mpa tng ng vi fy = 1.05280 = 294(Mpa)

Mx=My=60KN Tiet dien 300600 (mm2)

N=400KN Cot thep 6d20 (xem hnh CD.20)

Lp be tong bao ve a=3 (cm)

Hnh CD.22 Tiet dien tnh toan

Chon ve giao tuyen cua mat phang thang ng i qua truc Oz va iem tng ng

vi bo ba noi lc can kiem tra vi mat cong tng tac e so sanh.


LECH TAM XIEN

Page 263

Ket qua t phan mem CSICol 8

Hnh CD.23 ket qua ve bieu o tng tac cua phan

mem CSICol 8

CD.24 Bang ket qua toa o cac iem thuoc bd

tng tac t phan mem CSICol 8


LECH TAM XIEN

Page 264

Ket qua t phan mem CCRD1 cua sinh vien lap

Hnh CD.25 ket qua ve bieu o tng tac cua phan mem CCRD1

e so sanh c trc quan hn th sinh vien lay so lieu output cua 2 chng trnh

roi ve tren cung mot he toa o.

Hnh CD.26 So sanh b tng tac cua hai chng trnh.

-1000

-500

0

500

1000

1500

2000

2500

3000

3500

-50 0 50 100 150 200

N(KN)

M(KNm)

csicol

ccrd1


LECH TAM XIEN

Page 265

Kiem tra vi phan mem ETABS

Xet khung nh hnh ve H CD.26

Cot C1 co tiet dien va cach bo tr

thep giong nh tiet dien dung trong bai

kiem tra vi phan mem CSICol 8 (xem hnh

CD.22).

Khai bao tiet dien cot C1 trong

ETABS la reinforcement to be check.

ng knh thep chon s bo la d20.

Tien hanh thay oi ng knh thep

trong cot C1 trong ETABS en khi nao he so kha nang chu lc CR~1.

Sau khi tien hanh thay oi en khi d=33mm ng vi dien tch thep =52.2 (cm2)

th c he so CR=0.93.

Hnh CD.28 Bang ket qua ETABS gom noi lc va he so kha nang chu lc CR

ng vi d=33mm

Hnh CD.27 S o

ket cau chon e tnh

e kiem tra


LECH TAM XIEN

Page 266

Lay gia tr noi lc nh trong hnh CD.28 e nhap vao chng trnh CCRD1, va

chnh lai he so CR =0.935 e tm dien tch cot thep.

Ket qua cua chng trnh CCRD1: As=49.2 cm xem hnh CD.29 ben di)

Hnh CD.29 Ket qua tnh thep cua chng trnh CCRD1 tng ng vi noi lc tnh

toan cho cot C1 ng vi he so kha nang chu lc bang 0.935

Nh vay ket qua tnh chenh lech 5.7 (%)

Nhan xet: Qua hai v du tren cung nh nhieu v du khac sinh vien a kiem tra th

ket qua tnh t chng trnh CCRD1 cua mnh gan trung khp vi ket qua cua hai

chng trnh tnh ket cau ln cua cong ty nc ngoai CSI la CSICol l 8 va ETABS 9.7.

Nh vay ket qua tnh t chng trnh CCRD1 cua sinh vien la ang tin cay.


LECH TAM XIEN

Page 267

F. LAP LAP TRNH TNH TOAN COT CHU NEN LECH TAM THEO

PHNG PHAP GAN UNG

S o khoi (xem hnh CD.26 en CD.28)

NHP S LIU

BT U

CHNG TRNH CHNH

Nhp s liu kch thc

tit din: Cx, Cy

(I)

M hnh theo

phng Mx

25.0y

x

C

C

zxxy

zyyx

NeM

NeM

01

01

x

y

y

x

C

M

C

M 11

M hnh theo

phng My

ng sai

h = Cy; b = CxM1 = Mx1M2 = My1ea = eax+0.2eay

h = Cx; b = CyM1 = My1M2 = Mx1ea = eay+0.2eax

x1 hng sai

h

xm 10

7,01 3,00m

yx

y

y

y

x

xx

a

C

l

C

l

h

ea

hee

hle

N

Me

b

hMmMM

,max;288,0

;288,0

;2

;30

;600

max;;

00

0

00

01201

Chuyn qua nn lch tm phng

(IV)

I

NHP S LIU

- Chiu di tnh ton ct: l0- M men 2 trc: Mx, My- Lc nn ng tm tc dng ln ct: Nz- Chiu dy lp b tng bo v t tm ct thp n

mp ct gn nht: a

- Mc b tng: Mbt , xc inh c cng tnh

ton ca b tng: Rb v m un n hi ca b

tng Eb- Nhm ct thp: Nct , xc nh c cng tnh

ton ca ct thp chu ko: Rs , cng tnh ton

ca ct thp chu nn: Rsc v m un n hi ca

ct thp Es

KT THC

NHP S LIU

Tnh y (II)

Tnh x (III)

bR

Nx

ahZ

ahh

b

z

usc

SR

R

1

,

0

1,111

2

ng

sai

gt = 15% = 0.015

(V)

2

sgt

gt

e0 = e1 + ea (KC tnh nh)

hoc e0 = max(e1,ea) (KC siu tnh)

Hnh CD.26 Phan 1- S o khoi

lap trnh tnh theo pp gan ung


LECH TAM XIEN

Page 268

(IV)

> 104

Ct mt n nh

x1 > Rh0

Tnh theo trng hp

nn lch tm b

ZkR

xhbxRNe

Asc

b

st

20

Tnh theo trng hp

nn lch tm ln

XUT KT Qu

- Ct thp b tr theo chu vi:, Ast- Hm lng ct thp:, s

KT THC

CHNG TRNH

ng

ng sai

sai

0bh

Asts

sai01,0

gt

gts

smin

s = minAst = sbh0

(V)

bR

hZeNa

ZhhbR

Nea

ha

b

RR

RR

b

R

00

0201

02

12

)1(22

2

ngsai

ng

x1 a

ZkR

xhbxRNe

Asc

b

st

20

ng

ZkR

ZeNA

s

ast

sai

Gii phng trnh bc 3

x3 + a2x2 + a1x + a0 = 0

3

2;

2

34

327

2

3

3

1

3

1

3

2

120

32

22

1

azx

DAz

KD

KA

K

aaa

a

aa

x = x1

x h0

sai

ng

02 8

2

1e

bR

Nhhx

b

Hnh CD.27 Phan 2- S o khoi

lap trnh tnh theo pp gan ung


LECH TAM XIEN

Page 269

y

yC

l

288,0

0

y = 1

14yng sai

x

y

x

MM

Ch

Cb

cr

s

l

bcr

dhdhl

dhdh

p

ep

e

a

b

b

s

gts

N

N

ISI

l

EN

NyM

yNM

hyNNMM

S

h

e

hle

N

Me

Rh

l

E

E

ah

bhI

ahhbh

I

1

1

4,6

1

5,0;1};;{

1,0

1,0

11,0;1

),max(

30;

600max;

01,001,05,0

2

;12

20

min0

01

0min

2

0

0

3

II

TNH y

KT THC TNH y

x

xC

l

288,0

0

x = 1

14x

ng sai

III

TNH x

KT THC TNH x

Tnh

y

y

x

y

MM

Ch

Cb

cr

s

l

bcr

dhdhl

dhdh

p

ep

e

a

b

b

s

gts

N

N

ISI

l

EN

NyM

yNM

hyNNMM

S

h

e

hle

N

Me

Rh

l

E

E

ah

bhI

ahhbh

I

1

1

4,6

1

5,0;1};;{

1,0

1,0

11,0;1

),max(

30;

600max;

01,001,05,0

2

;12

20

min0

01

0min

2

0

0

3

Tnh

x

e0y = e1 + ea (KC tnh nh)

hoc e0y = max(e1,ea) (KC siu tnh)

e0x = e1 + ea (KC tnh nh)

hoc e0x = max(e1,ea) (KC siu tnh)

Hnh CD.28 Phan 3- S o khoi lap trnh tnh toan theo phng phap gan ung


LECH TAM XIEN

Page 270

G. ANH GIA PHNG PHAP TNH GAN UNG

Trong phan nay sinh vien se khao sat 2 phan sau:

- Tm he so k hp ly

- Phan tch sai so cua phng phap tnh gan ung theo moi quan he cua bo

ba noi lc (N,Mx,My).

Tm he so k hp ly trong cong thc tnh dien tch cot thep cua phng phap

gan ung.

Trong sach cua thay NGUYEN NH CONG, trong cac cong thc xac nh dien

tch cot thep th quy nh lay he so k=0,4. Con theo mot e tai luan van thac s cua

NGUYEN PHAN C HUNG th Thac s tm c he so k hp ly la 0.45.

e xac nh lai mot lan na he so nao la hp ly th trong e tai nay sinh vien

tien hanh tnh toan va so sanh s chenh lech dien tch thep trong phng phap gan

ung so vi phng phap dung bieu o tng tac ng vi hai he so k lan lt bang 0.4

va 0.45

Cac bc thc hien

Chon mot tiet dien vi mot lng cot thep nao o e tnh toan.

Ve bieu o tng tac cho tiet dien a chon.

Ve giao tuyen cua mat phang (c) thang ng i qua truc Oz va cha nhng

iem co Mx/My bang mot gia tr chon trc vi mat cong tng tac cua tiet dien a

chon. Lay toa o cua cac iem thuoc giao tuyen e tnh thep theo phng phap gan

ung.


LECH TAM XIEN

Page 271

Chon vat lieu: Be tong B25

Cot thep CII

Tiet dien tnh toan: 300600 (mm2

).

Cot thep 12d22 bo tr nh hnh ve CD.29

Chieu cao tang nha; 3.5 (m).

Be day lp be tong bao ve tnh en trong tam cot

thep: 4 (cm). Hnh CD.29 tiet dien tnh thep

Ket qua ve 4 giao tuyen cua cac mat phang vi mat cong tng tac

Hnh CD.30 cac bieu o tng tac ng vi cac mat cat khac nhau cho tiet dien tren

CHUYEN E NGHIEN CU ANH GIA PHNG PHAP TNH CAU KIEN BTCT CHU NEN LECH TAM XIEN

(4) Lc doc nen khi mang dau -, lc doc la keo khi mang dau +; (5) Bo qua cac trng hp keo lech tam

Page 272

Bang 1: Bang tnh toan cot thep theo phng phap gan ung ng vi Mx/My = 2 Be tong B15; Cot thep CII

B

rng

Chiu

cao

Chiu

cao tng

s tt

tren

cnh h

s tt cnh

b

k thep

Lp be

tong bao

ve

Lc dc(4)

Momen M-

22

(My)

Momen M-

33

(Mx)

Dien tch

thep ung

PP gan ung vi k=0.4 PP gan ung vi k=0.45

b (m) Cy h (m) Cx htang(m) nh nb d(mm) a(cm) N(KN) M2(KNm) M3(KNm) As(cm2) As(cm2) (%) As(cm2) (%)

0.30 0.60 3.50 4.00 4.00 22 4.00 1277.25 0.00 0.00 45.62 57.02 -(5)

50.68 -

0.30 0.60 3.50 4.00 4.00 22 4.00 1269.42 1.09 2.19 45.62 56.67 - 50.37 -

0.30 0.60 3.50 4.00 4.00 22 4.00 1269.42 1.09 2.19 45.62 56.67 - 50.37 -

0.30 0.60 3.50 4.00 4.00 22 4.00 1083.84 22.78 45.56 45.62 48.39 - 43.01 -

0.30 0.60 3.50 4.00 4.00 22 4.00 917.47 39.94 79.89 45.62 40.96 - 36.41 -

0.30 0.60 3.50 4.00 4.00 22 4.00 627.71 63.15 126.30 45.62 28.02 - 24.91 -

0.30 0.60 3.50 4.00 4.00 22 4.00 329.33 82.39 164.79 45.62 CD.7 - 13.07 -

0.30 0.60 3.50 4.00 4.00 22 4.00 36.86 94.49 188.98 45.62 1.65 - 1.46 -

0.30 0.60 3.50 4.00 4.00 22 4.00 -250.73 103.33 206.67 45.62 68.83 33.73 61.25 25.53

0.30 0.60 3.50 4.00 4.00 22 4.00 -547.87 108.15 216.29 45.62 57.47 20.63 51.22 10.94

0.30 0.60 3.50 4.00 4.00 22 4.00 -849.50 108.13 216.26 45.62 50.78 10.17 45.32 -0.65

0.30 0.60 3.50 4.00 4.00 22 4.00 -1169.86 102.55 205.11 45.62 52.47 13.06 46.87 2.68

0.30 0.60 3.50 4.00 4.00 22 4.00 -1534.22 90.54 181.07 45.62 56.8 19.69 50.75 10.12

0.30 0.60 3.50 4.00 4.00 22 4.00 -1841.88 76.04 152.09 45.62 60.83 25.01 54.34 16.05

0.30 0.60 3.50 4.00 4.00 22 4.00 -2123.98 58.71 117.42 45.62 62.22 26.69 55.55 17.88

0.30 0.60 3.50 4.00 4.00 22 4.00 -2349.72 42.88 85.76 45.62 62.19 26.65 55.48 17.78

0.30 0.60 3.50 4.00 4.00 22 4.00 -2519.25 29.43 58.86 45.62 61.22 25.49 54.57 16.41

0.30 0.60 3.50 4.00 4.00 22 4.00 -2636.84 18.91 37.81 45.62 59.82 23.74 53.27 CD.37

0.30 0.60 3.50 4.00 4.00 22 4.00 -2720.13 10.35 20.70 45.62 58.12 21.51 51.72 11.80

0.30 0.60 3.50 4.00 4.00 22 4.00 -2775.94 4.05 8.11 45.62 56.58 19.38 50.32 9.35

0.30 0.60 3.50 4.00 4.00 22 4.00 -2799.42 1.09 2.19 45.62 55.72 18.13 49.54 7.92

0.30 0.60 3.50 4.00 4.00 22 4.00 -2807.25 0.00 0.00 45.62 55.36 17.60 49.21 7.30

Mx/My=5 Trung bnh 21.53 11.96



Page 273

Bang 2: Bang tnh toan cot thep theo phng phap gan ung ng vi Mx/My = 1 Be tong B15; Cot thep CII

B

rng

Chiu

cao

Chiu

cao tng

s tt

tren

cnh h

s tt cnh

b

k thep

Lp be

tong bao

ve

Lc dc(4)

Momen M-

22

(My)

Momen M-

33

(Mx)

Dien tch

thep ung



0.30 0.60 3.50 4.00 4.00 22 4.00 1277.25 0.00 1277.25 0.00 57.02 -(5)

50.68 -

0.30 0.60 3.50 4.00 4.00 22 4.00 1252.64 3.55 1252.64 3.55 55.92 - 49.71 -

0.30 0.60 3.50 4.00 4.00 22 4.00 1252.64 3.55 1252.64 3.55 55.92 - 49.71 -

0.30 0.60 3.50 4.00 4.00 22 4.00 1012.21 32.53 1012.21 32.53 45.19 - 40.17 -

0.30 0.60 3.50 4.00 4.00 22 4.00 639.19 73.49 639.19 73.49 28.54 - 25.36 -

0.30 0.60 3.50 4.00 4.00 22 4.00 505.51 84.33 505.51 84.33 22.57 - 20.06 -

0.30 0.60 3.50 4.00 4.00 22 4.00 225.01 106.31 225.01 106.31 10.05 - 8.93 -

0.30 0.60 3.50 4.00 4.00 22 4.00 -18.21 122.90 -18.21 122.90 73.87 38.25 65.66 30.53

0.30 0.60 3.50 4.00 4.00 22 4.00 -287.32 133.52 -287.32 133.52 65.96 30.84 58.71 22.30

0.30 0.60 3.50 4.00 4.00 22 4.00 -562.15 138.34 -562.15 138.34 57.51 20.68 51.29 11.06

0.30 0.60 3.50 4.00 4.00 22 4.00 -850.92 139.35 -850.92 139.35 54.08 15.65 48.3 5.56

0.30 0.60 3.50 4.00 4.00 22 4.00 -1161.78 130.84 -1161.78 130.84 57.06 20.06 51 10.56

0.30 0.60 3.50 4.00 4.00 22 4.00 -1494.60 115.97 -1494.60 115.97 60.73 24.89 54.29 15.98

0.30 0.60 3.50 4.00 4.00 22 4.00 -1776.74 97.32 -1776.74 97.32 62.77 27.33 56.1 18.69

0.30 0.60 3.50 4.00 4.00 22 4.00 -2052.95 75.87 -2052.95 75.87 63.22 27.85 56.48 19.24

0.30 0.60 3.50 4.00 4.00 22 4.00 -2281.78 56.58 -2281.78 56.58 62.85 27.42 56.11 18.70

0.30 0.60 3.50 4.00 4.00 22 4.00 -2443.46 41.42 -2443.46 41.42 61.79 26.18 55.11 17.23

0.30 0.60 3.50 4.00 4.00 22 4.00 -2580.24 27.50 -2580.24 27.50 60.31 24.36 53.75 15.13

0.30 0.60 3.50 4.00 4.00 22 4.00 -2685.41 15.83 -2685.41 15.83 58.66 22.24 52.22 12.65

0.30 0.60 3.50 4.00 4.00 22 4.00 -2738.32 9.53 -2738.32 9.53 57.57 20.76 51.23 10.96

0.30 0.60 3.50 4.00 4.00 22 4.00 -2782.64 3.55 -2782.64 3.55 56.24 18.89 50.02 8.80

0.30 0.60 3.50 4.00 4.00 22 4.00 -2807.25 0.00 -2807.25 0.00 55.36 17.60 49.21 7.30

Mx/My=5 Trung bnh 24.20 CD.98



Page 274

Bang 3: Bang tnh toan cot thep theo phng phap gan ung ng vi Mx/My = 0.35 Be tong B15; Cot thep CII

B

rng

Chiu

cao

Chiu

cao tng

s tt

tren

cnh h

s tt cnh

b

k thep

Lp be

tong bao

ve

Lc dc(4)

Momen M-

22

(My)

Momen M-

33

(Mx)

Dien tch

thep ung



0.30 0.60 3.50 4.00 4.00 22 4.00 1277.25 0.00 0.00 45.62 57.02 -(5)

50.68 -

0.30 0.60 3.50 4.00 4.00 22 4.00 1224.64 7.58 2.65 45.62 54.67 - 48.6 -

0.30 0.60 3.50 4.00 4.00 22 4.00 1224.64 7.58 2.65 45.62 54.67 - 48.6 -

0.30 0.60 3.50 4.00 4.00 22 4.00 893.47 45.89 16.06 45.62 39.89 - 35.45 -

0.30 0.60 3.50 4.00 4.00 22 4.00 398.61 100.42 35.15 45.62 17.8 - 15.82 -

0.30 0.60 3.50 4.00 4.00 22 4.00 298.91 110.65 38.73 45.62 13.34 - 11.86 -

0.30 0.60 3.50 4.00 4.00 22 4.00 56.95 129.59 45.36 45.62 2.54 - 2.26 -

0.30 0.60 3.50 4.00 4.00 22 4.00 -174.03 145.32 50.86 45.62 61.32 25.61 54.56 16.39

0.30 0.60 3.50 4.00 4.00 22 4.00 -372.72 152.48 53.37 45.62 55.67 18.06 49.6 8.03

0.30 0.60 3.50 4.00 4.00 22 4.00 -604.15 156.72 54.85 45.62 51.86 12.04 46.3 1.48

0.30 0.60 3.50 4.00 4.00 22 4.00 -836.98 156.87 54.90 45.62 50.82 10.24 45.43 -0.41

0.30 0.60 3.50 4.00 4.00 22 4.00 -1060.95 151.79 53.13 45.62 54.98 17.03 49.17 7.23

0.30 0.60 3.50 4.00 4.00 22 4.00 -1303.90 141.35 49.47 45.62 58.45 21.96 52.27 12.73

0.30 0.60 3.50 4.00 4.00 22 4.00 -1611.33 118.13 41.35 45.62 60.15 24.16 53.8 15.21

0.30 0.60 3.50 4.00 4.00 22 4.00 -1912.06 93.40 32.69 45.62 61.22 25.49 54.73 16.65

0.30 0.60 3.50 4.00 4.00 22 4.00 -2123.72 74.31 26.01 45.62 61.25 25.53 54.73 16.65

0.30 0.60 3.50 4.00 4.00 22 4.00 -2311.68 56.38 19.73 45.62 60.85 25.04 54.33 16.04

0.30 0.60 3.50 4.00 4.00 22 4.00 -2482.84 39.12 13.69 45.62 60.06 24.05 53.57 CD.85

0.30 0.60 3.50 4.00 4.00 22 4.00 -2627.67 23.42 8.20 45.62 58.88 22.53 52.46 13.05

0.30 0.60 3.50 4.00 4.00 22 4.00 -2698.58 CD.94 5.23 45.62 57.91 21.23 51.55 11.51

0.30 0.60 3.50 4.00 4.00 22 4.00 -2754.64 7.58 2.65 45.62 56.79 19.68 50.52 9.71

0.30 0.60 3.50 4.00 4.00 22 4.00 -2807.25 0.00 0.00 45.62 55.36 17.60 49.21 7.30

Mx/My=5 Trung bnh 20.68 11.10


LECH TAM XIEN

(6) He so theo e ngh trong tai lieu cua thay NGUYEN NH CONG

(7) He so theo e ngh trong e tai luan van cua Thac sy NGUYEN PHAN C HUNG

Page 275

Khai niem vung an toan va vung nguy hiem tren bieu o tng tac

T ket qua phan tch tren, sinh vien phan biet cac vung an toan va nguy hiem

khi tnh bang phng phap gan ung ng vi hai he so k=0.4(6)

va k=0.45(7)

.

Hnh CD.31 Bieu o tng tac P=const

Hnh CD.32 Bieu o tng tac Mx/My=const

Ghi chu: (1) Phng phap gan ung (2) Bieu o tng tac chnh xac


LECH TAM XIEN

Page 276

Nhan xet ket qua tnh

o chenh lech trung bnh cua dien tch thep khi khi tnh theo phng phap gan

ung s dung he so k=0.4 so vi phng phap chnh xac la: 22%.

o chenh lech trung bnh cua dien tch thep khi khi tnh theo phng phap gan

ung s dung he so k=0.45 so vi phng phap chnh xac la: 12%. Tuy nhien co mot

so v tr th thien ve nguy hiem.

Mc o an toan cua ket qua tnh thep theo phng phap gan ung la phu thuoc

vao tng quan gia cac thanh phan trong bo ba (N, Mx, My). Cac iem cang nam

gan ve pha hai phang MxOP hay MyOP. Khi s dung he so k=0.4 th ket qua tnh thep

cua cac iem nay se gan vi nghiem chnh xac, con khi s dung he so k=0.45 th ket

qua tnh thep co the thien ve nguy hiem.

H. KET LUAN

Sau khi kiem tra ket qua tnh toan vi cac phan mem ln ve ket cau nh CSICol

8 va ETABS chng to rang phan mem sinh vien lap ra la ang tin cay.

Trong tnh toan cau kien chu nen lech tam nen viet chng trnh ve bieu o

tng tac e co the tnh toan chnh xac dien tch cot thep -> mang lai hieu qua kinh te.

Mc o an toan cua phng phap gan ung khi tnh tiet dien co o lech tam rat

be hay rat ln se nho hn so vi trng o lech tam trung gian.

Ket qua tnh thep theo phng phap gan ung ng vi he so k=0.4 th thien ve an

toan vi o sai sai trung bnh la 22%, con khi tnh ng vi he so k=0.45 th tuy thuoc

vao tng quan cua bo ba noi lc (N,Mx,My) ma co the ket qua tnh thep se thien ve

an toan hay khong an toan. V vay trong thiet ke neu khong the s dung bieu o tng

tac e tnh thep th nen chon he so k=0.4 trong phng phap tnh toan gan ung e

am bao an toan trong moi trng hp noi lc.


LECH TAM XIEN

Page 277

I. OAN CODE CHNG TRNH THIET KE COT BTCT CHU NEN

LECH TAM XIEN

Public Sub tinhthep()

Textbox12.Text = "": Textbox11.Text = "": Textbox13.Text = "": TextboxCD.Text = ""

Dim k As Integer, dk As Double, hamluongtrungbinh As Double, hamluongminimum As

Double, hamluongmax As Double, fs1 As Double, fs2 As Double, capratio1 As Double, capratio2 As

Double, capratiotb As Double

Call nhapdulieu

hamluongmax = Combobox4.Text / 100

hamluongminimum = hamluongmin(landa)

fs1 = b * h * hamluongminimum / (2 * (n1 + n2 - 2))

fs = fs1

Call bdtt

capratio1 = capratio

If capratio1 > 0.9 Then

fs2 = b * h * hamluongmax / (2 * (n1 + n2 - 2))

fs = fs2

Call bdtt

capratio2 = capratio

If capratio2 < 0.9 Then

Do

hamluongtrungbinh = (hamluongmax + hamluongminimum) / 2

fs = b * h * hamluongtrungbinh / (2 * (n1 + n2 - 2))

Call bdtt

capratiotb = capratio

If capratiotb > 0.9 Then

hamluongminimum = hamluongtrungbinh

Else

hamluongmax = hamluongtrungbinh

End If

k = k + 1

Loop Until k = 10

Else


LECH TAM XIEN

Page 278

MsgBox ("Ham luong thep vuot qua ham luong cho phep! Hay tang tiet dien")

End If

End If

dk = Math.Round(Sqr(1000000 * Math.Abs(fs) * 4 / 3.14), 2)

Textbox13.Text = Math.Round(2 * (n1 + n2 - 2) * fs / b / h * 100, 2) & "%"

Textbox12.Text = 2 * (n1 + n2 - 2) & " d"

TextboxCD.Text = Fix(dk) + 1

Textbox11.Text = Math.Round(2 * (n1 + n2 - 2) * fs * 10 ^ 4, 2)

ProgressBar1.Visible = False

End Sub

'ham tinh khoang cach tu 1 diem den truc trung hoa

Private Function kc(ByVal x As Double, ByVal Y As Double) As Double

kc = (Math.Abs(dth(x, Y))) / Sqr(hesoa ^ 2 + 1)

End Function

'ham tinh ung suat trong cac thanh thep

Private Function us(ByVal x As Double, ByVal Y As Double) As Double

Dim xnen As Double, omega As Double, xi As Double

xnen = kc(b / 2, h / 2)

omega = 0.85 - 0.008 * rb / 1000

xi = xnen / (xnen - dth(x, Y) / Sqr(hesoa ^ 2 + 1))

If xi = 0 Then xi = 0.0001

us = -400000 * (omega / xi - 1) / (1 - omega / 1.1)

If us > rphs Then us = rphs

If us < -rs Then us = -rs

End Function

Private Sub gdkc()

'tim giao diem cua truc trung hoa va cac canh cua tiet dien

hdgd(1) = -(c + h / 2) / hesoa

tdgd(1) = h / 2

hdgd(2) = -(c + (-h / 2)) / hesoa

tdgd(2) = -h / 2

tdgd(3) = -(c + hesoa * (-b / 2))


LECH TAM XIEN

Page 279

hdgd(3) = -b / 2

tdgd(4) = -(c + hesoa * (b / 2))

hdgd(4) = b / 2

End Sub

Private Sub toadodiem()

'lap ma tran toa do cac dinh tiet dien va vi tri cot thep

hd(1) = -b / 2: hd(2) = b / 2: hd(3) = b / 2: hd(4) = -b / 2

td(1) = h / 2: td(2) = h / 2: td(3) = -h / 2: td(4) = -h / 2

For i = 5 To 4 + n1

hd(i) = -b / 2 + a / 100 + (i - 5) * (b - 2 * a / 100) / (n1 - 1)

td(i) = h / 2 - a / 100

Next

For i = 5 + n1 To 4 + n1 + n2 - 1

hd(i) = b / 2 - a / 100

td(i) = h / 2 - a / 100 - (i - 5 - n1 + 1) * (h - 2 * a / 100) / (n2 - 1)

Next

For i = 4 + n1 + n2 To 4 + n2 + 2 * n1 - 2

hd(i) = b / 2 - a / 100 - (i - 4 - n1 - n2 + 1) * (b - 2 * a / 100) / (n1 - 1)

td(i) = td(3) + a / 100

Next

For i = 3 + n2 + 2 * n1 To 4 + 2 * (n1 + n2 - 2)

hd(i) = hd(5)

td(i) = -h / 2 + a / 100 + (i - 3 - n2 - 2 * n1 + 1) * (h - 2 * a / 100) / (n2 - 1)

Next

End Sub

Private Function dth(ByVal x As Double, ByVal Y As Double) As Double

dth = hesoa * x + Y + c

End Function

'ham tinh dien tich vung be tong chiu nen

Private Function ab(ByVal x As Double) As Double

If x = hesoa * b / 2 + h / 2 Then

ab = b * h

Else


LECH TAM XIEN

Page 280

Call gdkc

If hdgd(1) = b / 2 Then

'tinh dien tich tam gia CDD'

dt1 = 0.5 * b * (tdgd(3) - tdgd(4))

'tinh dien tich hcn 12DD'

dt2 = b * (h / 2 - tdgd(4))

Else

'tinh dien tich tam gia CBB'

dt1 = 0.5 * (hdgd(2) + b / 2) * (tdgd(3) + h / 2)

'tinh dien tich hcn 1234

dt2 = b * h

End If

Else

If hdgd(2) >= b / 2 Then

'tinh dien tich tam gia ADD'

dt1 = 0.5 * (b / 2 - hdgd(1)) * (h / 2 - tdgd(4))

'tinh dien tich hcn A2DD'

dt2 = (b / 2 - hdgd(1)) * (h / 2 - tdgd(4))

Else

'tinh dien tich tam gia ABB'

dt1 = 0.5 * (hdgd(2) - hdgd(1)) * h

'tinh dien tich hcn A2BB'

dt2 = (b / 2 - hdgd(1)) * h

End If

End If

ab = dt2 - dt1

End If

End If

End Function

'ham tinh hoanh do trong tam vung be tong chiu nen

Private Function xgab(ByVal x As Double) As Double

Dim x1 As Double, x2 As Double

If x


LECH TAM XIEN

Page 281

If x >= hesoa * b / 2 + h / 2 Then

xgab = 0

Else

Call gdkc



dt1 = 0.5 * b * (tdgd(3) - tdgd(4))


dt2 = b * (h / 2 - tdgd(4))

'tinh toa do trong tam tam gia CDD' G1

x1 = 1 / 3 * (-b / 2)

'tinh toa do trong tam hcn 12DD'

x2 = 0

Else


dt1 = 0.5 * (hdgd(2) + b / 2) * (tdgd(3) + h / 2)


dt2 = b * h

'tinh toa do trong tam tam gia CBB' G2

x1 = 1 / 3 * (-b + hdgd(2))

'tinh toa do trong tam hcn 1234

x2 = 0

End If

Else



dt1 = 0.5 * (b / 2 - hdgd(1)) * (h / 2 - tdgd(4))


dt2 = (b / 2 - hdgd(1)) * (h / 2 - tdgd(4))


x1 = 1 / 3 * (b / 2 + 2 * hdgd(1))

'tinh toa do trong tam hcn A2DD'

x2 = (b / 2 + hdgd(1)) / 2

Else


dt1 = 0.5 * (hdgd(2) - hdgd(1)) * h


LECH TAM XIEN

Page 282


dt2 = (b / 2 - hdgd(1)) * h


x1 = 1 / 3 * (2 * hdgd(1) + hdgd(2))

'tinh toa do trong tam hcn A2BB'

x2 = (b / 2 + hdgd(1)) / 2

End If

End If

xgab = (dt2 * x2 - dt1 * x1) / (dt2 - dt1)

End If

End If

End Function

'ham tinh tung do trong tam vung be tong chiu nen

Private Function ygab(ByVal x As Double) As Double

Dim y1 As Double, y2 As Double

If x = hesoa * b / 2 + h / 2 Then

ygab = 0

Else

Call gdkc



dt1 = 0.5 * b * (tdgd(3) - tdgd(4))


dt2 = b * (h / 2 - tdgd(4))


y1 = 1 / 3 * (tdgd(3) + 2 * tdgd(4))

'tinh toa do trong tam hcn 12DD'

y2 = 0.5 * (h / 2 + tdgd(4))

Else


dt1 = 0.5 * (hdgd(2) + b / 2) * (tdgd(3) + h / 2)



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dt2 = b * h


y1 = 1 / 3 * (tdgd(3) - h)

'tinh toa do trong tam hcn 1234

y2 = 0

End If

Else



dt1 = 0.5 * (b / 2 - hdgd(1)) * (h / 2 - tdgd(4))


dt2 = (b / 2 - hdgd(1)) * (h / 2 - tdgd(4))


y1 = 1 / 3 * (h / 2 + 2 * tdgd(4))

'tinh toa do trong tam hcn A2DD'

y2 = (h / 2 + tdgd(4)) / 2

Else


dt1 = 0.5 * (hdgd(2) - hdgd(1)) * h


dt2 = (b / 2 - hdgd(1)) * h


y1 = 1 / 3 * (-h / 2)

'tinh toa do trong tam hcn A2BB'

y2 = 0

End If

End If

ygab = (dt2 * y2 - dt1 * y1) / (dt2 - dt1)

End If

End If

End Function

'lap ma tran tra dac trung vat lieu

Private Sub vatlieu()

betong(1, 1) = "b12.5": betong(2, 1) = "b15": betong(3, 1) = "b20": betong(4, 1) = "b25":

betong(5, 1) = "b30": betong(6, 1) = "b35": betong(7, 1) = "b40": betong(8, 1) = "b45": betong(9, 1) =

"b50": betong(10, 1) = "b55": betong(11, 1) = "b60"


LECH TAM XIEN

Page 284

betong(1, 2) = 7.5 * 10 ^ 3: betong(2, 2) = 8.5 * 10 ^ 3: betong(3, 2) = 11.5 * 10 ^ 3:

betong(4, 2) = CD.5 * 10 ^ 3: betong(5, 2) = 17 * 10 ^ 3: betong(6, 2) = 19.5 * 10 ^ 3: betong(7, 2) = 22

* 10 ^ 3: betong(8, 2) = 25 * 10 ^ 3: betong(9, 2) = 27.5 * 10 ^ 3: betong(10, 2) = 30 * 10 ^ 3:

betong(11, 2) = 33 * 10 ^ 3

betong(1, 3) = 21 * 10 ^ 6: betong(2, 3) = 23 * 10 ^ 6: betong(3, 3) = 27 * 10 ^ 6: betong(4,

3) = 30 * 10 ^ 6: betong(5, 3) = 32.5 * 10 ^ 6: betong(6, 3) = 34.5 * 10 ^ 6: betong(7, 3) = 36 * 10 ^ 6:

betong(8, 3) = 37.5 * 10 ^ 6: betong(9, 3) = 39 * 10 ^ 6: betong(10, 3) = 39.5 * 10 ^ 6: betong(11, 3) =

40 * 10 ^ 6

thep(1, 1) = "ci,ai": thep(2, 1) = "cii,aii": thep(3, 1) = "aiii(d6,8)": thep(4, 1) =

"ciii,aiii(d10,40)": thep(5, 1) = "civ,aiv": thep(6, 1) = "av": thep(7, 1) = "avi"

thep(1, 2) = 225 * 10 ^ 3: thep(2, 2) = 280 * 10 ^ 3: thep(3, 2) = 355 * 10 ^ 3: thep(4, 2) =

365 * 10 ^ 3: thep(5, 2) = 510 * 10 ^ 3: thep(6, 2) = 680 * 10 ^ 3: thep(7, 2) = 815 * 10 ^ 3


365 * 10 ^ 3: thep(5, 3) = 450 * 10 ^ 3: thep(6, 3) = 500 * 10 ^ 3: thep(7, 3) = 500 * 10 ^ 3


200 * 10 ^ 6: thep(5, 4) = 190 * 10 ^ 6: thep(6, 4) = 190 * 10 ^ 6: thep(7, 4) = 190 * 10 ^ 6

End Sub

Private Function hamluongmin(ByVal x)

If x


LECH TAM XIEN

Page 285

momenquantinhthep = 0

If x = h Then

momenquantinh = b * h ^ 3 / 12

deltae = max(ey / h, 0.5 - 0.01 * lo / h - 0.01 * rb / 1000)

For i = 5 To 4 + 2 * (n1 + n2 - 2)

momenquantinhthep = fs * td(i) ^ 2 + momenquantinhthep

Next

Else

momenquantinh = h * b ^ 3 / 12

deltae = max(ex / b, 0.5 - 0.01 * lo / b - 0.01 * rb / 1000)

For i = 5 To 4 + 2 * (n1 + n2 - 2)

momenquantinhthep = fs * hd(i) ^ 2 + momenquantinhthep

Next

End If

s = 0.11 / (0.1 + deltae) + 0.1

nth = 6.4 / lo ^ 2 * (momenquantinh * eb * s / 2 + e * momenquantinhthep)

hesouondoc = 1 / (1 - n / nth)

End Function

Private Function max(ByVal x, ByVal Y) As Double

If x >= Y Then

max = x

Else

max = Y

End If

End Function

Private Sub giaihept(ByVal x1, ByVal y1, ByVal z1, ByVal x2, ByVal y2, ByVal z2)

Dim a(3, 3) As Double, b(3) As Double, l(3, 3) As Double, u(3, 3) As Double, Y(3) As

Double

a(1, 1) = y2 - y1

a(1, 2) = x1 - x2

a(1, 3) = 0

a(2, 1) = 0

a(2, 2) = z2 - z1

'If Math.Abs(a(2, 2)) < 0.00001 Then a(2, 2) = 1 / 10000

a(2, 3) = y1 - y2

a(3, 1) = my


LECH TAM XIEN

Page 286

a(3, 2) = -mx

a(3, 3) = 0

b(1) = x1 * (y2 - y1) - y1 * (x2 - x1)

b(2) = y1 * (z2 - z1) - z1 * (y2 - y1)

b(3) = 0

If Math.Abs(a(2, 2)) < 0.00001 Then

x(3) = z2

x(2) = -a(3, 1) * b(1) / (a(1, 1) * a(3, 2) - a(3, 1) * a(1, 2))

x(1) = a(3, 2) * b(1) / (a(1, 1) * a(3, 2) - a(3, 1) * a(1, 2))

Else

For i = 1 To 3

u(1, i) = a(1, i)

Next

For i = 1 To 3

l(i, 1) = a(i, 1) / u(1, 1)

Next

For i = 2 To 3

For j = 2 To 3

If j < i Then

u(i, j) = 0

Else

u(i, j) = a(i, j)

For k = 1 To i - 1

u(i, j) = u(i, j) - l(i, k) * u(k, j)

Next

If u(i, j) = 0 Then u(i, j) = 1 / 100

End If

If j > i Then

l(i, j) = 0

Else

l(i, j) = a(i, j) / u(j, j)

For k = 1 To j - 1

l(i, j) = l(i, j) - l(i, k) * u(k, j) / u(j, j)

Next

If l(i, j) = 0 Then l(i, j) = 1 / 1000000


LECH TAM XIEN

Page 287

End If

Next

Next

Y(1) = b(1)

Y(2) = (b(2) - l(2, 1) * Y(1)) / l(2, 2)

Y(3) = (b(3) - l(3, 1) * Y(1) - l(3, 2) * Y(2)) / l(3, 3)

x(3) = Y(3) / u(3, 3)

x(2) = (Y(2) - u(2, 3) * x(3)) / u(2, 2)

x(1) = (Y(1) - u(1, 3) * x(3) - u(1, 2) * x(2)) / u(1, 1)

End If

End Sub

Private Sub nhapdulieu()

b = Textbox1.Text 'chieu rong b(m)

h = Textbox2.Text 'chieu cao h(m)

n1 = Textbox6.Text 'so thanh thep tren canh h

n2 = Textbox5.Text 'so thanh thep tren canh b

a = Textbox4.Text 'lop be tong bao ve

mx = Textbox9.Text

If mx = 0 Then mx = 1 / 1000

my = Textbox8.Text

If my = 0 Then my = 1 / 1000

n = Textbox7.Text 'luc doc

htang = Textbox3.Text 'chieu cao tang nha"

If n = 0 Then n = 0.1

khbt = Combobox1.Text 'cuong do be tong

khthep = Combobox2.Text 'cuong do thep

Call vatlieu

'tra cuong do be tong

For i = 1 To 11

If UCase(betong(i, 1)) = khbt Then rb = betong(i, 2): eb = betong(i, 3): Exit For

Next

'tra cuong do thep

For i = 1 To 7

If UCase(thep(i, 1)) = khthep Then rs = thep(i, 2): rphs = thep(i, 3): e = thep(i, 4): Exit

For

Next

'do toan khoi: chieu dai tinh toan


LECH TAM XIEN

Page 288

lo = 0.7 * htang

'do manh

landa = lo / b

End Sub

Public Sub bdtt()

Dim teta1 As Double, teta As Double, tongus1 As Double, tongus2x As Double, tongus2y

As Double, deltac As Integer

Dim k As Integer, l As Integer, x1 As Double, y1 As Double, z1 As Double, x2 As Double,

y2 As Double, z2 As Double, mxycheck As Double

Call nhapdulieu

'do lech tam

If n > 0 Then

ex = Math.Abs(my / n)

ey = Math.Abs(mx / n)

'do lech tam ngau nhien

eax = max(htang / 600, b / 30): eay = max(htang / 600, h / 30)

'do lech tam tong

ex = max(ex, eax)

ey = max(ey, eay)

mx = n * ey

my = n * ex

mxcheck = mx * hesouondoc(h)

mycheck = my * hesouondoc(b)

Else

mxcheck = mx

mycheck = my

End If

'lap ma tran toa do cac dinh tiet dien va vi tri cua cot thep

Call toadodiem

mxcheck = mx * hesouondoc(h)

mycheck = my * hesouondoc(b)

If Checkbox1.Value = False Then my = 1: mx = 0.001 * HScrollbar1.Value

teta1 = -1.57079632 / 10

k = 0

'tim toa do 3d cua cac diem tren duong cong tuong tac


LECH TAM XIEN

Page 289

Do

k = k + 1

teta1 = teta1 + 1.570796327 / 10

teta = (1.570796327 * Sin((1.570796327 * Sin(1.570796327 * Sin(teta1)))))

hesoa = Abs(Tan(teta))

c = 0

l = 0

For deltac = 1 To 21

l = l + 1

c = -hesoa * b / 2 - h / 2 + (deltac - 1) * (hesoa * b + h) / 20

Call gdkc

tongus1 = 0

For i = 5 To 4 + 2 * (n1 + n2 - 2)

tongus1 = us(hd(i), td(i)) + tongus1

Next

tongus2x = 0

For i = 5 To 4 + 2 * (n1 + n2 - 2)

tongus2x = (us(hd(i), td(i))) * td(i) + tongus2x

Next

tongus2y = 0

For i = 5 To 4 + 2 * (n1 + n2 - 2)

tongus2y = (us(hd(i), td(i))) * hd(i) + tongus2y

Next

mx3d(k, l) = rb * ab(c) * ygab(c) + tongus2x * fs

my3d(k, l) = rb * ab(c) * xgab(c) + tongus2y * fs

n3d(k, l) = ab(c) * rb + fs * tongus1

Next

Loop Until teta1 > 1.5707963 Or k = 10

'tim giao diem cua mat phang my*x-mx*y=0 voi cac mat phang tuong tac

For l = 2 To 20

For k = 1 To 11

If Math.Abs(mx3d(k, l) / my3d(k, l)) < mx / my Then x1 = mx3d(k - 1, l): y1 = my3d(k

- 1, l): z1 = n3d(k - 1, l): x2 = mx3d(k, l): y2 = my3d(k, l): z2 = n3d(k, l): Call giaihept(x1, y1, z1, x2,

y2, z2): Exit For

Next


LECH TAM XIEN

Page 290

hoanhdox(l) = x(1): hoanhdoy(l) = x(2): tungdo(l) = x(3): hoanhdo(l) = Sqr((hoanhdox(l))

^ 2 + (hoanhdoy(l)) ^ 2)

Next

hoanhdo(0) = Sqr((mx3d(1, 1)) ^ 2 + (my3d(1, 1)) ^ 2): tungdo(0) = n3d(1, 1)

hoanhdo(1) = hoanhdo(2): tungdo(1) = tungdo(2)

hoanhdo(21) = 0: tungdo(21) = rs * fs * 2 * (n1 + n2 - 2) + rb * b * h

' tinh ti so kha nang chiu luc cua diem (Mx,My,N)

tungdomin = 0

For i = 0 To 38

If tungdomin > tungdo(i) Then tungdomin = tungdo(i)

Next

tungdomax = 0

For i = 0 To 38

If tungdomax < tungdo(i) Then tungdomax = tungdo(i)

Next

If n > tungdomax Or n < tungdomin Then

capratio = 100

Else

For l = 1 To 21

hesogoc = n / Sqr(mxcheck ^ 2 + mycheck ^ 2)

If tungdo(l) / hoanhdo(l) > hesogoc Then

tuso = hoanhdo(l) * (tungdo(l) - tungdo(l - 1)) - tungdo(l) * (hoanhdo(l) - hoanhdo(l -

1))

mauso = tungdo(l) - tungdo(l - 1) - (hoanhdo(l) - hoanhdo(l - 1)) * hesogoc

hoanhdocheck = tuso / mauso

tungdocheck = hesogoc * hoanhdocheck

'chieu dai doan tu goc toa do dien diem luc can kiem tra

oa = Sqr(n ^ 2 + mxcheck ^ 2 + mycheck ^ 2)

'chieu dai doan tu goc toa do den diem thuoc duong toi han

op = Sqr(hoanhdocheck ^ 2 + tungdocheck ^ 2)

capratio = Math.Round(oa / op, 2): Textbox16.Text = capratio

Exit For

End If

Next

End If

End Sub


LECH TAM XIEN

Page 291

Private Sub vebdtt()

Call bdtt

Dim hoanhdomax As Double

Dim checkx(1) As Double, checky(1) As Double

checkx(0) = Sqr(mxcheck ^ 2 + mycheck ^ 2): checky(0) = n: checkx(1) = Sqr(mxcheck ^ 2

+ mycheck ^ 2): checky(1) = n

hoanhdomax = 0

For i = 0 To 38

If hoanhdomax < hoanhdo(i) Then hoanhdomax = hoanhdo(i)

If hoanhdomax < Sqr(mxcheck ^ 2 + mycheck ^ 2) Then hoanhdomax = Sqr(mxcheck ^ 2

+ mycheck ^ 2)

Next

If tungdomin > n Then tungdomin = n

If tungdomax < n Then tungdomax = n

With AxTChart1.Axis.Bottom

.Maximum = hoanhdomax + 100

.Minimum = hoanhdo(0) - 50

End With

With AxTChart1.Axis.Left

.Maximum = tungdomax + 200

.Minimum = tungdomin - 200

End With

AxTChart1.Series(0).AddArray 22, tungdo, hoanhdo

If Checkbox1.Value = 1 Then

AxTChart1.Series(1).AddArray 2, checky, checkx

End If

End Sub

Private Sub Button5_Click()

Dim a(100) As String

Dim bd As Integer

Dim x(1) As String

Dim b() As String

'FileOpen(1, filenhap, OpenMode.Input)

'For i = 1 To 1000


LECH TAM XIEN

Page 292

'Input(1, a(i))

'If a(i) = "COLUMN FORCES" Then bd = i: Exit For

'Next

'For j = bd + 3 To 1000

'Input(1, a(j))

' b = Split(a(j), " ")

'Debug.Print(b(0), b(5), b(6))

'Next

'FileClose (1)

End Sub

Private Sub Button2_Click_1()

'OpenFileDialog1.ShowDialog()

'filenhap = OpenFileDialog1.FileName

'TextBox10.Text = filenhap

End Sub


'ProgressBar1.Visible = True

Call tinhthep

End Sub

Private Sub HScrollBar1_Scroll()

Dim dk As Double

dk = Combobox3.Text ' duong kinh cot thep

fs = 3.1416 * (dk / 1000) ^ 2 / 4

Call vebdtt

End Sub

Private Sub form_load()

Dim dk As Double

Checkbox1.Value = 1


fs = 3.1416 * (dk / 1000) ^ 2 / 4

Call vebdtt

ProgressBar1.Visible = False

End Sub



LECH TAM XIEN

Page 293

Dim dk As Double


fs = 3.1416 * (dk / 1000) ^ 2 / 4

Call vebdtt

End Sub


If MsgBox("Ban co muon dong that khong", vbYesNo, "chu y") = vbYes Then End

End Sub

Private Sub TextBox1_click()

Textbox1.SetFocus

End Sub

Private Sub TextBox2_MouseClick()

Textbox2.Refresh

End Sub


'TextBox3.SelectAll()

End Sub


End Sub


End Sub


End Sub


End Sub


End Sub


End Sub


Dim cx As Double, cy As Double, ea As Double, m As Double, m1 As Double, m2 As

Double, mo As Double, e1 As Double, eo As Double, landax As Double, landay As Double, gamae As

Double

Dim phi As Double, omega As Double, xir As Double, it As Double, mxtemp As Double,

mytemp As Double


LECH TAM XIEN

Page 294

Call nhapdulieu

Call toadodiem

mxtemp = mx: mytemp = my

fs = 20 * 10 ^ -4

cx = h

cy = b

For i = 1 To 10

fs = fs / (2 * (n1 + n2 - 2))

omega = 0.85 - 0.008 * rb / 1000: xir = omega / (1 + rs / 400000 * (1 - omega / 1.1))

h = cx

b = cy

If n > 0 Then

ex = Math.Abs(mytemp / n)

ey = Math.Abs(mxtemp / n)

mx = n * ey * hesouondoc(h)

my = n * ex * hesouondoc(b)

End If

If mx / h > my / b Then

h = cx: b = cy

m1 = mx: m2 = my

ea = eay + 0.2 * eax

Else

b = cx: h = cy

m1 = my: m2 = mx

ea = eax + 0.2 * eay

End If

If n / rb / b


LECH TAM XIEN

Page 295

If landa

Chuyen de Tinh Cot Btct Chiu Nen Lech Tam Xien

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