Uon Ngang Va Uon Doc Suc Ben Vat Lieu Chuong 12

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Chng 12: Uon ngang va uon doc ong thi 1

Chng 12

UON NGANG VA UON DOC ONG THI

12.1 AC IEM BAI TOAN

Xet mot thanh chu uon bi tac ong ong thi cua lc ngang R va lc nen doc P nh tren H.12.1. Neu chuyen v la ang ke th can phai xet can bang cua thanh tren s o bien dang va momen noi lc se bao gom anh hng cua lc R va P: M(z) = MR + MP = MR + Py(z) (12.1) trong o: MR - momen uon do rieng tai trong ngang gay ra Py(z) - momen uon do lc doc gay ra.

R

P z y(z)

Hnh 12.1 Uon ngang va uon doc ong thi

Bai toan nh vay c goi la uon ngang va uon doc ong thi. ac iem cua bai toan: - Momen M(z) phu thuoc vao o vong y(z) - Momen M(z) phu thuoc phi tuyen vao lc P v o vong y(z) cung phu thuoc vao P. V vay, nguyen ly cong tac dung khong ap dung c cho loai bai toan nay.

12.2 PHNG PHAP CHNH XAC

e tm c momen uon, trc het can thiet lap phng trnh vi phan ng an hoi cua dam chu lc nen P va tai trong ngang.

P P

q(z)

y(z)

q(z)

O

dz

P

Q + dQ

M + dMPM

Q

Hnh 12.2 Thanh chu uon nen


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Xet can bang tren s o bien dang cua phan to thanh dz nh tren H.12.2 0:0 =+= tgPdzQdzMdMMMo chu y rang :

dzdytg =

ta co: QdzdyP

dzdM = (12.2)

lay ao ham hai ve cua (12.2), chu y rang )(zqdzdQ = , ta co phng trnh:

)(22

2

2zq

dzydP

dzMd = (12.3)

the "EIyM = (*) vao (12.3) ta thu c: )(" zqPyEIyIV =+ (12.4) ay la phng trnh vi phan ng an hoi cua dam chu nen uon. Neu biet tai trong tac dung va cac ieu kien bien th co the giai (12.4) e tm ng an hoi, t o suy ra momen uon theo phng trnh (*). Trong thc te, thng co nhieu quy luat tai trong khac nhau tren chieu dai thanh nen viec giai phng trnh (12.4) rat phc tap. V vay, ngi ta thng ap dung phng phap gan ung di ay.

12.3 PHNG PHAP GAN UNG

Xet dam n gian chu tai trong oi xng nh H.12.3. q

f0

a)

q

f

b)

P

lll

Hnh 12.3 ng an hoi oi xng

S o (a) ch chu tai trong ngang, vi o vong gia nhp fo. S o (b) chu ong thi tai trong ngang va tai trong doc, co o vong gia nhp f. Gia thiet ng an hoi co dang hnh sine (giong dang mat on nh), ta co phng trnh ng an hoi trong hai trng hp nh sau:

lzfy oo

= sin ; lzfy = sin

Dang phng trnh nay thoa ieu kien bien 0" == yy tai hai khp. Momen uon noi lc tng ng nh sau:

oooo ylEIlzf

lEIEIyM 2

2

2

2" sin ===


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yl

EIlzf

lEIEIyM 2

2

2

2" sin ===

The cac ket qua nay vao phng trnh (12.1) ta co:

Pyyl

EIyl

EI o += 22

2

2 (12.5)

t o suy ra: 2

2/1

)()(

lEIP

zyzy o =

hay: th

o

PPzyzy

=

1

)()( (12.6)

vi: 22

lEIPth

= la lc ti han cua thanh khi mat on nh trong mat phang

uon. ao ham hai ve cua (12.6) va nhan vi EI ta co:

thPP

zEIyzEIy

=1

)()("0"

hay: th

o

PP

MzM

=1

)( (12.7)

Chu y: - Neu tai khong oi xng nhng cung hng ve mot pha th cac cong thc tren kem chnh xac hn nhng van dung c. - Neu thanh co lien ket hai au khac th van dung c cac cong thc (12.6), (12.7) nhng can xet ti he so lien ket trong cong thc Pth: 2

2

)( lEIPth

= (12.8)

12.4 NG SUAT VA KIEM TRA BEN

ng suat ln nhat c tnh theo cong thc:

)1(max

th

o

PPW

MAP

WM

AP

+=+= (12.9)

V ng suat phu thuoc phi tuyen vao tai trong nen kiem tra ben theo ng suat cho phep khong am bao an toan theo he so n d kien. Trong trng hp nay, ngi ta dung ieu kien an toan theo tai trong nh sau:

oth

o

PnPW

nMA

nP

+)1(

(12.10)

V du 12.1 Tm momen uon va o vong ln nhat cua dam thep ch INo36 chu lc nh tren H.12.4.


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x

q = 2 kN/m

S = 120 kN

4my

Hnh 12.4

Giai. S dung bang tra thep nh hnh, tng ng vi so hieu INo36 va cac ky hieu tren hnh tren, ta co:

A = 61,9 cm2; Ix = 516 cm4; Iy = 13380 cm4; E = 2,1.104 kN/cm2

Tr so ln nhat cua momen uon, o vong do tai trong ngang gay ra tai gia nhp: kNmqlMo 48

4.28

22===

cmEIqly

xo 615,0516.10.1,2

400.10.2.3845.

3845

4

424===

Tr so lc ti han: ( ) ( ) kNl

EIP xth 668400.1516.10.1,2.

2

42

2

2==

=

o vong cua dam, theo cong thc gan ung: cm

PS

yy

th

o 75,0

6681201

615,0

1=

=

= , tang 22% so vi oy

Momen uon ln nhat, theo cong thc gan ung th nhat: kNmSyMM o 9,4075,0.1204 =+=+= Momen uon ln nhat, theo cong thc gan ung th hai: kNm

PS

MM

th

o 87,4

6681201

4

1=

=

= sai so 0,5% so vi cong thc gan ung th

nhat. Gia tr momen trong trng hp uon ngang va doc tang 22,5% so vi momen ch do lc ngang gay ra, tc la thien ve an toan hn.

12.5 THANH CO O CONG BAN AU

1- Anh hng cua o cong ban au

Xet thanh co o cong ban au, chu lc nen P nh tren H.12.5. Gia s ng cong ban au co dang:

lzayo= sin (12.11)


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P z

yo

y1

a

y l/2l/2

Hnh 12.5 Thanh co o cong ban au

Do tac dung cua lc P, thanh b vong them co phng trnh y1(z). o vong toan phan: y = yo + y1 (12.12) Momen uon do lc P gay ra: )( 1yyPPyM o +== (12.13) Phng trnh vi phan o vong them: )( 1''1 yyPMEIy o +== (12.14) the (12.11) vao (12.14) va at:

EIP=2 ta co:

lzayy =+ sin212''1 (12.15)

Nghiem cua phng trnh nay co dang:

lza

l

zBzAy

++= sin11cossin

22

21 (12.16)

Cac ieu kien bien: 00)(00)0(

1

1

====

AlyBy

Do o: lza

lEIP

lza

l

y

=

= sin11sin

1

1

2

2

22

21

hay: lza

kky = sin11 (12.17)

vi: 2

2

lEIP

PPkth

== (12.18)

o vong toan phan:lz

ka

lza

kkayyy o

=

+=+= sin1sin)1(1

hay: th

o

PP

yy

=1

(12.19)

Momen ln nhat gia nhp:

thPP

PaPyM

==1

maxmax (12.20)

Neu ng cong ban au co dang bat ky th co the phan tch thanh chuoi Fourier nh sau: ...2sinsin 21 ++= l

zalzayo (12.21)

the (12.13) vao (12.21) va giai ra 1y ta co:


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++

= ...2sin

2sin

1 221

1 lz

ka

lz

kaky (12.22)

v: 1


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12.6 COT CHU NEN LECH TAM

Xet cot manh chu nen lech tam bi lc P nh tren H.12.8.

lzayo= sin (12.11)

Do tac dung cua lc P, cot b cong va co phng trnh y(z). Momen uon tai mot tiet dien do lc P gay ra: )()}({ zPyPezyePM +=+= (12.23) trong o: e - la o lech tam ban au; y - la o vong cua truc cot. Phng trnh vi phan ng an hoi nh sau:

EIMzy =)('' (12.24)

The (12.23) vao (12.24) va at EIP=2 ta

c:

eyy 22" =+ (12.25)

Nghiem tong quat cua phng trnh nay la

tong cua nghiem thuan nhat va nghiem rieng:

ezBzAy += cossin (12.26)

trong o: A va B - la cac hang so cua nghiem

thuan nhat; e - la nghiem rieng. Cac ieu kien bien: eBy == 0)0(

2

tansin

)cos1(0)( lel

leAly ===

Phng trnh ng an hoi tr thanh:

)1cossin2

(tan += zzley (12.27)

o vong ln nhat tai gia nhp, tc 2lz = la:

)12

cos

1(max == ley (12.29)

(12.28)

Neu e = 0 hoac P = 0 th 0= .

P

y

l

z

P

y(z)

e

e

Hnh 12.8 Cot co o cong ban au


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o th quan he gia P - c cho trong H.12.9. o th nay ch co y ngha khi vat lieu con an hoi, tc la con nho va P < Pth.

Pth

P

e = 0

e = e1

e = e2

e2 > e1

Hnh 12.9 o th quan he gia P - Momen uon ln nhat tai gia nhp c tnh:

2cos

1)( maxmax lEIP

PeyePM =+= (12.30)

Quan he maxM - P cho bi H.12.10. Khi P nho th PeM max , nhng khi P ln th maxM tang rat nhanh.

T cac o th nay ta thay quan he P - va maxM - P phi tuyen. Trong thc te, tnh cot manh chu nen lech tam can thiet phai xet ac iem phi tuyen nay e am bao an toan.

P th

Mmax

PHnh 12.10 Quan he gia Mmax - P

Pe

ng suat cc ai trong thanh:

+=+=

2cos

11 2max

max lEIPr

ecAP

IcM

AP (12.31)

vi: A - dien tch tiet dien thanh; r - ban knh quan tnh

c - khoang cach t truc trung tam en mep xa nhat cua tiet dien. V ng suat phu thuoc phi tuyen vao tai trong nen kiem tra ben theo ng suat cho phep khong am bao an toan theo he so n d kien. Trong


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trng hp nay, ngi ta dung ieu kien an toan theo tai trong nh phng trnh (12.10).

BAI TAP CHNG 12

12.1 Tnh ng suat nen ln nhat theo phng phap gan ung cua dam chu uon ngang va uon doc ong thi cho tren H.12.11.

a)

100

2 m2 m

q = 200 N/m

P = 4 kN

4 m

1

E = 103 kN/cm2

100

1

1 - 1

2 m2 m

q = 3 kN/m

P = 257 kN

4 m

1

Po = 5 kN

1

1 1

2C No20

b)

Hnh 12.11

12.2 Cho dam chu lc nh tren H.12.9. Hay tnh ng suat phap ln nhat va he so an toan n neu [] = 24 kN/cm2 . Tnh o vong ln nhat.

q = 0,5 kN/m

2 mP = 4 kN

E = 103 kN/cm2b)

10 cm10 cm

P1 = 1 kN20 cm

P = 8 kN40 cm

E = 2 x 104 kN/cm2

1 m1 m

a) Hnh 12.12

12.3 Tnh cng o tai trong cho phep tac dung len dam AB nh tren H.12.10, biet he so an toan ve o ben n = 1,6. Dam AB bang thep so 3 co mat cat hnh ong vi ng knh trong d = 6 cm va ng knh ngoai D = 10 cm, vat lieu co [] = 24 kN/cm2, khi tnh bo qua trong lng cua dam.

Kiem tra on nh cua dam neu lay ko = 2. Cho E = 2.104 kN/cm2.

5 m

60o

q

A B

Hnh 12.13

Uon Ngang Va Uon Doc Suc Ben Vat Lieu Chuong 12

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