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Chng 5

THANH CHU UN PHNG

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NI DUNG

5.1. Khi nim chung

5.2. Un thun ty thanh thng

5.3. Un ngang phng thanh thng

SB1 nghin cu ng sut, bin dng,chuyn v trong thanh di tc dng ca

cc trng hp chu lc c bn

Chng 2;Ko (nn)ng tm

Chng 5:Xon

UN

University of Architechture

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5.1. Khi nim chung (1)

Thanh chu un: khi c tc dng ca ngoi lc trcthanh thay i cong

Dm: thanh chu un


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Gii hn nghin cu: Dm vi mt ct ngang c t nht 1 trci xng (ch I, T, ch nht, trn,); mt phng ti trng trngmt phng i xng ca dm => Un phng

Mt phng ti trng: mt phng cha ti trng v trc thanh

Mt phng qun tnh chnh trung tm: mt phng cha trc

thanh v 1 trc qun tnh chnh trung tm ca mt ct ngang.

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Phn loi un phng Un thun ty phng

Un ngang phng

V d: thanh chu unphng Trn on BC: Mx0, Qy=0

=> Un thun ty phng

Trn on AB,CD: Mx0,Qy0

=> Un ngang phng

F

F

FaFa

M x

Q y

F F

z

V VA D= F

= Fa b a

B C DA


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5.2. Un thun ty phng (1)

Un thun ty phng


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1. nh ngha: Thanh gi l chu un thun tu nutrn cc mt ct ngang ca n ch tn ti thnhphn ng lc l mmen un Mx (hoc My) nm

trong mt phng qun tnh chnh trung tm.

Ti trng gy un: nm trong mt phng i qua trc thanhv vung gc vi trc thanh

2. Cc gi thit v bin dng ca thanh

a. Th nghim


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Vch trn b mt ngoi ca thanh H nhng ng thng // trc thanh

=> th dc

H nhng ng thng vung gcvi trc thanh => mt ct ngang

Cho thanh chu un thun ty phngQUAN ST

Cc ng thng // trc thanh =>ng cong // trc, khong cch giacc ng cong k nhau khng i

Cc ng thng vung gc vi trcthanh => vn thng v vung gc vitrc thanh

Cc th pha trn b co (chu nn),cc th di b dn (chu ko)

th dcmt ct ngang

M M


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Bin dng ca thanh chu un


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GI THITa. Gi thit mt ct ngang phng:

mt ct ngang trc bin dngl phng v vung gc vi trcthanh th sau bin dng vnphng v vung gc vi trc

b. Gi thit v cc th dc: trongqu trnh bin dng cc lp vtliu dc trc khng c tcdng tng h vi nhau

Vt liu lm vic trong min nhi

Tn ti lp trung ho: gm ccth dc khng b dn cngkhng b co.

ng trung ha: Giao tuyn calp trung ho vi mt ct

ngang

M M

Lp trung hong

trung ho

ngtrung ho


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3.ng sut trn mt ct nganga. Bin dng di ca th dcc khong cch y n thtrung ho

Xt vi phn chiu di ca thdc dz = cd. Sau bin dngcd c di l cd.

Bin dng di t i:

th trung ho

c d

a b

c d

d

dz

12

1 2

1 2

1 2

y

ya b

' 'z

y d ddz c d cd ydz cd d

z

y

bn knh cong ca th trung ho


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b. Quan h ng sut - bin dngXt mt ct ngang bt k, phn

t din tch dA cha im K.

Tch phn t lp phng cha

im K.T gt 1: gc vung khng thay

i => t=0T gt 2: sx=sy=0=> Trn mt ct ngang ch tn

ti duy nht ng sut php szTheo nh lut Hooke

y

z

x

dA s

x

y

z

K

K szsz

z zEs zy

Es

1????


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c. Cng thc tnh ng sut phpTi trng gy un nm trong mt

phng yOz v vung gc vitrc thanh nn: Nz=My=0 v

Mx0. Ta c: y

z

x

dA s

x

y

z

K

Mx

0z zA A

EN dA ydA s

0xA

yd A S ng trung ho i qua trng tm

ca mt ct ngang

0y zA A

EM x dA xydA s

0xyA

xydA I H trc Oxy l h trc

qun tnh chnh trung tm


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Mx>0: cng th diMx thun tin ta thng dng

cng thc tnh ton

y

z

x

dA s

x

y

z

K

Mx

2

x z x

A A

E EM y dA y d A I

s

1 x

x

M

EI

EIx cng ca dm chu un

Mx m men un ni lc bn knh cong ca th trung ho

z

y

E

s

x

zx

My

Is

Thay biu thc ca bn knh congvo biu thc xc nh ng sut php

y tung im cn tnh ng sut

x

z

x

My

I s

thuc vng ko

thuc vng nn


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d. Biu ng sut php Cc im cng xaTH th tr tuyt i ng sut php

cng ln Cc im nm trnTH th c sz=0

Cc im nm trn ng thng // TH th csz=const=> Biu din s bin thin ca ng sut php theo

chiu cao mt ct ngang Biu ng sut php l ng thng i qua gc

to => v c biu ch cn tnh ng sutphp ti im bt k trn mt ct ngang

nh du (+) biu din phn ng sut ko v du(-) biu din phn ng sut nn


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Mt ct ngang c hai trc ixng

max2

x x

x x

M Mh

I Ws

min2

x x

x x

M MhI W

s

max mins

/ 2x

x IWh

- m men chng un ca mt ct ngang

x

y

smin

smax

h/2

h/2

2

6x

bhW

330,1

/ 2 32

xx

I DW D

D

Hnh ch nht: Hnh trn:

Hnh vnh khn: 3

4 3 41 0,1 1

/ 2 32

xx

I DW D

D

d

D

vi

z

Mx


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Mt ct ngang c 1 trci xng

x

y

t

h

b smin

smax

ynmax

ykmax

max max

x xk

k

x x

M My

I Ws

min max

x xn

n

x x

M MyI W

max

k xx k

IW

y

max

n xx n

IW

y

ykmax - khong cch xa TH nht thuc vng chu ko

ynmax - khong cch xa TH nht thuc vng chu nn

z

Mx


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4. iu kin bnDm lm bng vt liu do

Dm bng vt liu gin

Ba bi ton c bn

Kim tra iu kin bn:

Xc nh kch thc ca mt ct ngang:

Xc nh ti trng cho php:

max minmax , s s max min;k ns s s

maxx

x

M

Ws s

x

x

MW s

x xM W


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Mt ct ngang c hnh dng hp l: Kh nng chu lc ca dm ln nht

Tit kim vt liu nht

Dm bng vt liu dn: mt ct ngang hp l khi ng

thi tha mn

Dm bng vt liu do:

max maxx k

kx

My

Is s

min maxx n

nx

My

Is s

max

max

(*)

k

k

n

n

y

y

ss

Mt ct ngang khngi xng qua trc x

tho mn k (*)

k n

smax

max

1

k

n

y

y

Mt ct ngang chai trc i xng


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tit kim vt liuT biu ng sut, cng xa TH ng sut cng ln

=> a vt liu ra xa TH

x

y

x

y y

x

y

x

Vt liu dn Vt liu do


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5.3. Un ngang phng (1)

1. nh nghaThanh gi l chu un ngangphng nu trn cc mt ct

ngang ca n ng thi c cpng lc l mmen un Mx, lcct Qy nm trong mt phngqun tnh chnh trung tm.

Gi thit mt ct ngang phng

khng cn ng


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Bin dng thanh chu un ngang phng


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Hai thnh phn ng lcMx => ng sut phpQy => ng sut tip

ng sut phpTrong Mx l mmen un ni lc trn mt ct ngang Ix l mmen qun tnh ca mt ct ngang i

vi trc qun tnh chnh trung tm Ox y l tung ca im tnh ng sutGhi ch: Mx > 0 khi lm cng th div Mx < 0

khi lm cng thtrn.

s

x

zx

M

yI


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2. ng sut tip:Vi mt ct ngang dng hnh ch nht hp b

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5.3. Un ngang phng (*)

Tch phn t gii hnbi: Hai mt ct ngang

cch nhau dz

Mt ct dc // v cch

Oz khong y

y

dz

z

y

x

z

F

x

z y

dzy

x

y

Mt ct ngang


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5.3. Un ngang phng (*)

Xt cn bng phn t

dz y

x

y

szph

sztr

tzy s

zph

sztr

M+dMMQ

Q+dQ

0Z zyt


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Cng thc Zuravxki

t c

y x

zy

x c

Q S

I b

h

b=b

y

THx

y

Ac

c

Qy l lc ct theo phng y ti mt ct ngang.

Ix l mmen qun tnh ca mt ct ngang i vi trc x.

bc chiu rng ca mt ct ngang ti im tnh ng sut

l phn din tch b ct (l phn din tch gii hn bi chiu rng mt ct

ngang ti im tnh ng sut v mp ngoi ca mt ct ngang).

l m men tnh ca phn din tch b ctc

xS

CA


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ng sut tip trn mt ctngang ch nht

bc=b; Ix=bh3/12;

SxC=yC.AC => x

y

h

b=

y

bc

tmax

AC

2 212 2 2 2 4

c

xh h b hS y y b y

2 22 2

3 3

12 6

.. 2 4 4

y y

zy

Q Qb h h

y ybh b bht

02zy

h

y t

ax

30

2

y

m

Qy

bht


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ng sut tip trn mt ctngang ch I

Trong bng thp nh hnh, cho cc kch thc h, b. s,

d, Ix, Sx m men tnh ca tit din. Ta c

- Phn bng: bc=d y

xh

s

b

y

d

tmax

t 1

t 1

21 .

2

C

x xS S d y

21 .2

y

zy x

x

QS d y

I d

max0 =>y x

x

Q SyI d

t

2

1

1=>

2 2 2

y

x

x

Qh hy s S d s

I dt


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Phn b ng sut tip trn tit din ch I


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4. iu kin bnXt tit din ch nht chu un ngang phng. Biu ng sut trn tit din:

K, N - trng thi ng sut nC- trng thi ng sut trt thun tyB- trng thi ng sut phng c bit

x

y

N

K

C

B

Mx

z

maxt

mins

maxs

h/2

h/2

maxsmaxs

minsmins

maxt

maxt

ss

t

tB

B

B

B


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Kim tra bn cho trng thi ng sut nMt ctn g a n g n g u y him: mt ct c m men un ln

nht (vt liu do: tr tuyt i ca m men ln nht,vt liu gin: m men m v m men dng ln

nht)Vt liu do:

Vt liu gin:

max minmax , s s

max min;k ns s s University of Architechture

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Kim tra bn cho trng thi ng sut trt thun tyMt ct nguy him: Mt ct c tr tuyt i Qy ln nht

Vt liu do:

Vt liu gin: Dng thuyt bn Mohr

axmmax t t

0n

tt - nu dng thc nghim tm t0

2

st - nu dng thuyt bn 3

3

st - nu dng thuyt bn 4

max

1

kt t

k

n

s s


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Kim tra bn cho trng thi ng sut phng c bitMt ct ngang nguy him: c tr tuyt i Mx v Qy

cng ln

im kim tra: im c ng sut php v ng sut tip

cng ln (im tip gip gia lng v vi mt ctngang ch I)

Dm bng vt liu do:

Dm bng vt liu gin:

s s t2 2t z zy

( ) 4( ) (TB3)

s s t2 2

t z zy( ) 3( ) (TB4)

2 21 1

42 2

z z zy k

s s t s


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BI TP V d 5.1 (1)

BI:

Cho dm mt ct ngang ch nht chu ti trng nh hnh v

V biu cc thnh phn ng lc

Xc nh ng sut php cc i ti mt ct ngang nguy him

Kim tra iu kin bn cho dm, bit [s]=1,5 kN/cm2


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V d 5.1 (2)

BI GII:

Gii phng lin kt v xc nh cc phn lc

0 : 40kN 14kNy B B DF M R R

Dng phng php mt ct

Mt ct 1 - 1

1 1

1 1 1

0 20 kN 0 20kN

0 20kN 0m 0 0

yF Q Q

M M M

Mt ct 2 - 2

2 2

2 2 2

0 20 kN 0 20kN

0 20kN 2.5m 0 50kN m

yF Q Q

M M M

Mt ct 3 3

Mt ct 4 4

Mt ct 5 5

Mt ct 6 6

3 3

4 4

5 5

6 6

26kN 50kN m

26kN 28kN m

14kN 28kN m

14kN 0

Q M

Q M

Q M

Q M


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V d 5.1 (3)

T biu cc thnh phn ng lc tathy, mt ct ngang nguy him ti B c:

max max26kN 50kN mBQ M M

Q

kN

M

kNm

ng sut php cc i

221 1

6 6

6 3

3

max

max 6 3

0.080m 0.250m833.33 10 m

50 10 N m

833.33 10 m

x

x

W b h

M

Ws

6 2

max 60.0 10 Pa=60MPa=6kN/cms

Ta thy: smax > [s] => Dm khng thomn iu kin bn


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V d 5.2 (1)

Cho dm mt ct ngang thp ch I chu ti trng nh hnhv. Bit ng sut cho php ca thp [s]=16 kN/cm2. Hychn s hiu mt ct ngang thp theo iu kin bn ngsut php ca dm.


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V d 5.2 (2)

Xc nh phn lc lin kt ti A v D

5m 60kN 1.5m 50kN 4m 0

58.0kN

58.0 kN 60 kN 50 kN=0

52.0kN

A D

D

y A

A

M V

V

F V

V

V biu lc ct v m men un ni lc.52.0kN

60kN

8kN

A A

B A q

B

Q V

Q Q S

Q

M men ln nht ti:Q = 0 => z = 2,6 m.

max( ) 67.6kNmQM S AE

Q

kN

58

66Mmax

M

kNm


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V d 5.2 (3)

iu kin bn theo ng sut php cadm:

max

min 2

6 3 3

67.6kN m

16kN/cm

422.5 10 m 422.5 m

MW

c

s

max max

max x

x

M MW

Ws s

s

Chn s hiu thp t bng tra

3xW [mm ]

I 27 371

I 27a 407

I 30 472I 30a 518

I 33 597

Thep

30I


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CC VN SINH VIN CN NM V

Phn bit trng thi trt thun ty v thun ty

Cc cng thc tnh ng sut php v

sut tip

Cc cng thc tnh ng sut cc trKim tra cu kin theo iu kin bn ch

liu do v gin

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SBVL1 - Ch5 - Uon Phang (2012)

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