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Chapter 10
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SC BN VT LIU 2SC BN VT LIU 2Trn Minh Ti hc Xy dng H ni
B mn Sc bn Vt liuKhoa Xy dng Dn dng & Cng nghip
zx
zy
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Chng 10
Tnh bn kt cu theo ti trng gii hn
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10.1. Cc khi nim chung
10.2. Phng php tnh bn theo ti trnggii hn
10.3. Tnh h thanh chu ko (nn) ng tm
10.4. Tnh thanh chu un thun ty phng10.5. Tnh thanh chu un ngang phng
Tnh bn kt cu theo ti trng gii hn
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10.1. Cc khi nim chung
1. Cc quan im tnh ton kt cu Mi cc nh gi bn u km theo cc quan nim,
cc tiu chun. C hai quan im chnh tnh tonkt cu: quan im tnh theo ng sut cho php vquan im tnh theo ti trng gii hn.
a. Tnh bn theo ng sut cho php
Ch cho php vt liu thanh lm vic trong min nhi, khi mt im bt k hay mt mt ct no thucvt th xutt hin bin dng do (ng sut t ti sch)
=> h b ph hoi iu kin bn:
[ ] 0maxn
= [ ] 0maxn
=
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10.1. Cc khi nim chung
ch u im: n gin, ch cho phpbin dng b ( 0,2%) Nhc im: qu thin v an tonnn lng ph vt liu, cha xem xtn s lm vic ca ton b kt cu
b. Tnh bn theo ti trng gii hn
=> cn c mt phng php khc
khc phc nhc im
- i vi vt liu do, khi xut hin bin dng do mt vi
im (un, xon), mt vi mt ct ngang (h siu tnh) h vncha b ph hoi (vn cn kh nng chu lc)
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F
ch
ch
P
1 2 3
=>nh gi bn ca kt cu cn phi xt n kh nng chu lc ca c h.
=> Cn xt xem hp ng hay khng p ng c cc yu cu t ra vmt chu lc
Trng thi trung gian gia hai trng thi: p ng v khng p ng c
cc yu cu chu lc gi l trng thi gii hn, ti trng tng ng gi l titrng gii hn = > K hiu Fgh Tnh bn theo ti trng gii hn cho php pht sinh bin dng do, h trng thi gii hnkhi bin dng do pht trin ti mc ton kt cu mt khnng chu lc.
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10.2. Phng php tnh bn theo ti trng gii hn
ch
T th ko vt liu do: bindng do >> bin dng n hi
C th quan nim th ch
gm 2 giai on: n hi v do=> th Prandtl
ch
Biu qui c (Prandtl)
iu kin bn :
ghgh
PP Pn
=
u im: tit kim vt liu
Nhc im: cho php bindng ln => khng ph hp chongnh c kh chnh xc
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10.3. Tnh h thanh chu ko (nn) ng tm
Thanh chu ko (nn) ng tm: z=const Khi bt kim no c ng sut php t tich => c tit din u t ti ch.=> Lc dc trn mt ct ngang gi l Nd
d chN A=
1. Thanh n hoc h thanh tnh nh: tnh theo
SCP v TTGH l nh nhau
Khi 1 im no trn mt ct ngang c ng sut t ti ch th theoSCP y l trng thi nguy him. ng thi do =const nn ton bmt ct ngang u t ti ch => bin dng ca thanh l ty => thanhmt kh nng chu lc: TTGH
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10.3. Tnh h thanh chu ko (nn) ng tm
2. H siu tnh: s lng lin kt nhiu hn s lng cn thit khi1 thanh xut hin bin dng do th h vn cn kh nng chu lc,cho n khi h c (n+1) thanh b chy do h mi hon ton mtkh nng chu lc
Phng php gii theo ti trng gii hn
Phng php n hi:
- Xc nh ni lc trong tt c cc thanh => xc nh ng sut
- Ln lt cho (n+1) thanh c tr s ng sut ln nht xuthin chy do (lc dc Nd=ch.A)- Khi thanh th (n+1) thanh b chy do: h TTGH => xcnh ti trng gii hn tng ng
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H gm dm tuyt i cng BCD,chu ti trng phn bu q. Dm clin kt khp ti A v treo bi 2 thanh1 v 2 c cng chiu di v cngEA. Xc nh ti trng cho php theophng php USCP v TTGH, bitch ca vt liu thanh treo
BC D
1 2
a a
q
B C D
N1 N2q
Bi gii
- Gii theo SCP2
2 12 2 0
B M N a N a qa= + =
l1 l2
2 1 2 12 2l l N N = => =
1 2
2 4;
5 5 N qa N qa = =
iu kin bn : [ ]224
5
N qa
A A
= = [ ][ ]5 5
4 4
chdh
Aq
A nA
= =
10.3. Tnh h thanh chu ko (nn) ng tm
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Khi thanh 1 b chy do: N1=Nd=chA=> h TTGH
BC D
Nd=chAqgh Nd=chA
1 2 2gh ch ch N q a A A = =
3
2
chgh
qa
=
BC D
N1Nd=chA
q1
1 2
2 4;
5 5 N qa N qa = =
=> thanh 2 chy do trc => s:
2
1 12 2 0
B chM a N a q a= + =
1 12 2 ch N q a A =
3
2
gh chgh
n na
= =
[ ][ ]5 5
4 4
chdh
Aq
A nA
= =
6
5
gh
dh
q
q
=
- Gii theo TTGH
10.3. Tnh h thanh chu ko (nn) ng tm
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10.3. Tnh h thanh chu ko (nn) ng tm
Phng php ng:- Gi thit (n+1) thanh bt k b chy do => H TTGH gi thit =>Xc nh ti trng gii hn gi thit tng ng Fgh
i
- Gi tr nh nht trong cc ti trng gii hn gi thit l ti trng giihn ca kt cu
Fgh=min{Fghi}
V d: Cho h thanh chu ti trng nh
hnh v. Tm [Fgh] bit A1=A2=A3=A, giihn chy ca vt liu ch, h s an ton n
F
1
2
3300 450
Nhn xt: - Cc thanh 1, 2, 3 u chu ko- H siu tnh bc 1
=> h TTGH khi 2 trong 3 thanh b chy do- Thanh 1 v 2 b chy do => Fgh1
Cc trng thi gii hn gi thit - Thanh 1 v 3 b chy do => Fgh2
- Thanh 2 v 3 b chy do => Fgh3
=> Loi
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10.3. Tnh h thanh chu ko (nn) ng tm
TH1: Thanh 1 v 2 b chy do
Fgh1
3300450
N2=chA
N1=chA
u
0 1 0os30 os30 0
ch ch ghu A Ac F c = + =1 2 3 3
2,153gh ch chF A A
+
=
TH3: Thanh 2 v 3 b chy do
1 300450
Fgh3
N2=chA
N3=chA
v
0 3 0os45 sin 30 0
ch ch gh
v A Ac F = + =
( )3 1 2 2, 41gh ch chF A A = + Fgh=min{Fgh
i}
12,15
gh gh chF F A= =
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10. 4. Tnh dm chu un thun ty
Xt dm chu un thun ty, vt liudm l n hi tuyn tnh
max=ch
min=ch
- Biu ng sut l ng bcnht, iu kin bn:
maxW
xch
x
M =
Khi ti trng tng n gi tr :
max=lminl=ch => dm trng thinguy him
, ,W
x dh ch x dhM =
max=ch
min=ch
Ti trng tip tc tng, min dolan rng dn v min n hi thuhp li
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10. 4. Tnh dm chu un thun ty
Ti trng tng n lc ton b mtct ngang b chy do hon ton, dmmt kh nng chu lc => TRNGTHI GII HN
max=ch
min=ch
trng thi gii hn, ng phn cchgia 2 min ko v nn gi l ng trungho chy do.ng trung ho chy do chia mt ct
ngang lm 2 phn c din tch bng nhau x
x1
min=ch
max=ch
Ak
An
Ak = An
trng thi gii hn, m men un nilc trn mt ct ngang gi l m menun do Mx,d
, ,d ch x d W=
Wx,d m men chng un do
1 1
( ) ( )
,k nA A
x d x xW S S= +
1
( )kA
xS- m men tnh caA
ki vi x
1
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10. 4. Tnh dm chu un thun ty
Khi mt ct ngang c 2 trc i xng th x x1
h
b
2
,6
x dh
bhW =
2
,4
x d
bhW =
D
- Mt ct ngang ch nht
- Mt ct ngang trn
3
,32
x dh
DW
=
3
,6
x d
DW =
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10. 5. Tnh dm chu un ngang phng
Xt dm chu un ngang phng F- Do Mx const => cc mt ctngang c mc chy do khcnhau, khng ging nhau nh unthun ty phng
Mmax
ch
ch
- Khi Mmax=Mx,h: xut hin bindng do u tin ti mp trn vdi ca mt ct ngang im tlc F
- Khi Mmax>Mx,h: bin dng do landn vo trong v ra hai bn dc
theo chiu di dm- Khi Mmax = Mx,d: tit din im t lc b chy do hon ton, trong lccc tit din ln cn cha b chy do hon ton. Min chy do c hnhdng nh hnh v
Do
n hin hi
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10. 5. Tnh dm chu un ngang phng
F
Mmax
ch
ch
Do
n hin hi
Hai phn thanh n hi haipha tri, phi lin kt vi nhau ch mt im. im ni ny ngvai tr nh l khp v gi l
khp do. Khp tht c th xoay t do vc hai pha, v c Mx=0
Khp do ch c th xoaychuyn ng v pha th cng, vc Mx=Mx,d
Mx,d Mx,d- Vi dm tnh nh, khi xut hinkhp do, dm tr thnh c cu=> mt kh nng chu lc =>TTGH => Fgh
- Vi dm siu tnh bc n, h TTGH khi hnh thnh (n+1) khp do.
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???
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