Bai Giang Sbvl2
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Transcript of Bai Giang Sbvl2
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Sc bn vt liu 2(SBVL2)
Mechenics of Materials 2
SSccbnbnvvttliliuu22(SBVL2)(SBVL2)
MechenicsMechenicsof Materials 2of Materials 2
TrTrngng ii hhcc HHngng cc
KhoaKhoa KTCNKTCN BB mnmn KK thuthutt cngcng trtrnhnh
GV.GV. LL ThThThanhThanh TmTm
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Ti liu tham kho
Bt buc: Phm Ngc Khnh - Sc bnvt liu, NXB Xy dng 2002.
Tham kho:
- Nguyn Xun Lu, Binh Nghi, Vnh Lai - Sc bn vt liu tp 2, NXBGiao thng vn ti 2005.
- L Ngc Hn g - Sc bn vt liu, NXBXy dng 2002
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Thanh chu lc phc tp
Khi nim chung
Thanh chu un xin
Thanh chu un v ko (nn) ng thi
Thanh chu ko (nn) lch tm. Li mtct ngang
Thanh chu un v xonng thi
Thanh chu lc tng qut
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Thanh chu lc phc tp
nh ngha: Thanh chu lc phc tp khi trn ccmt ct ngang c tc dngng thi ca nhiuthnh phn ni lcnh: lc dc Nz, m menun Mx,My v m men xon Mz
Cch tnh thanh chu lc phc tp: p dngnguyn l cng tc dng
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Cc bi ton chu lc phc tp
Bi ton un xin: Khi trn cc mt ctngang tn ting thi m men un Mx vm men un My
Bi ton un v ko (nn) ng thi: Khitrn cc mt ct ngang tn ting thilc dc Nz v cc m men un Mx, My
Bi ton un v xonng thi: Khi trn
cc mt ct ngang tn ting thi mmen xon Mz v cc m men un Mx, My
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Bi ton un xin
Mt thanhc gi l un xin khi trncc mt ct ngang ca nng thi tn tihai thnh phn ni lc l m men un Mx
v m me un My
x
yz
M
M
x
y
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Bi ton un xin
Nhn dng bi ton:Thanh chu un xin khi ti trng (l cc lc tp trunghoc phn b) cphng vung gc vi trc dm, cttrc dmnhng khng trng vi trc qun tnh chnhtrung tm ca mt ct
Pq
y
x
P q
x
P
P
1
2
P1
y
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Bi ton un xin
Quyc du:
- Mx > 0 khicng th y > 0
- My > 0 khicng th x > 0
x
yz
M
M
x
y
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Bi ton un xin
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Bi ton un xin
ng sut tiim A(x,y) trn mt ct ngang cathanh chu un xin:
(1)
Hoc:
(2)
x
J
My
J
M
y
y
x
xz ..
xJ
My
J
M
y
y
x
x
z
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Bi ton un xin
Biu din cc min chu ko, nn trn mtct do Mx, My gy ra
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Bi ton un xin
V d: Tit din ch nht bxh=20x40 cm2 chu un xin, cho Mx=8 kNm, My=5 kNm.Tnhng sut php tiim B (nh hnh v)
Ta im B: xB=+10 cm
yB=-20 cm
Tnh theo cng thc (1):
Tnh theo cng thc (2):
Mx gy ko nhngim nmdi Ox v gy nn
nhngim trn Ox
My gy ko pha tri Oy v gy nn pha phi OyTa thy tiim B Mx gy nn, My gy ko
)/)(10(
12
20.40
500)20.(
12
40.20
800 2
33 cmkNBz
)2
33 /)(10(
12
20.40
500)20.(
12
40.20
800cmkNBz
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Bi ton un xin
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ng trung ha vB phn bS
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ng trung ha vB phn bS
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Bi ton un xin
Biu ng sut trn mt ct ngangc v nh sau:- Ko ding trung ha ra khi mt ct
- V ng vung gc ving trung ha lmng chun.
- ng sut ti ccim trnng thng song song ving trunghac biu din bng mton thng c gc trnng chun
v cphng nm trnng thng song song.T biu ng sut trn mt ct ngang ta
tmc ccim nguy him trong min
chu ko v nn. Nhngim ny l nhng
im nm xang trung ha nht.
i vi mt ct ngang hnh ch nht, ch I,im nguy him lun lun l cc gc
(im B v C)
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Cc tr s ng sut cc i vcc tiu
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Bi ton un xin
iu kin bn ca thanh chu un xin:- Vi vt liu gin:
- Vi vt liu do:
1. Bi ton kim tra bn (Bton thmnh)
2. Bi ton chn kchthc mt ct (Bton thit k)
3. Bi ton chn ti trng cho php (Bton s dng)
nn
kk
max
max
max
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V d
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Bi ton un xin
V d: Mt dm tit din ch T chu lcnh hnh v.- V biu ni lc
- Xcnhng trung ho ti tit din ngm
- Tnhng sutmax, minCho q=4 kN/m; P=qL; L=2 m; a=5 cm
Ccctrng ca tit din: yo = 7a/4, Jx = 109a4/6; Jy = 34a4/6.
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Bi ton un xin
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Bi ton un xin
Phng trnhng trung ho:Ti tit din ngm:
Chiu Mx, My biu dinnh hnh v, cc m men unuc dudng
Ta c:
Biu din tit din bng hnh
phng theo t l, t ptrnh trnc th v chnh xcng trungho
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Bi ton un xin
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Bi ton un v ko (nn) ng thi
Mt thanh chu un v ko nnng thikhi trn mt ct ngang ca n xut hincc thnh phn: lc dc Nz v m menun M
x, M
y
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Bi ton un v ko (nn) ng thi
Nhn dng bi ton:Thanh s b un v ko nnng thi khi ti trng tcng vo thanh l cc lc tp trung P v cc lc phnb q cphng ct trc thanhnhng khng vung gcvi trc thanh, hoc cc lc cphng song song vi
trc thanhnhng t lch tm mt ct
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Bi ton un v ko (nn) ng thi
ng sut: (1)-ng sut do Nz ly (+) khi lc dc l lc
ko-ng sut do M
x, M
ylynh sau: Nunh
hng trc y,xdng v pha gy koca Mx, My th ly theo du ca y v x
nhhng htrc x, y
khi dng cng thc (1)(2)
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Cng thc k thut:
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ng trung ha B phn b S
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ng sut php cc tr
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ng sut php cc tr
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K bn
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Bi ton un v ko (nn) ng thi
Thanh chu ko hay nn lch tm:Thanh chu ko hay nn lch tm l thanh chulc sao cho trn mi mt ct ngang ca thanhchc thnh phn lc dc N cngphng
nhng t lch mton e so vi trc z cathanh
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Bi ton un v ko (nn) ng thi
ng sut:
t:
Ta c:
(ix, iy: bn knh qun tnh ca mt ct ngangi vi cc tr cx,y)
F; 22 y
yx
x
Ji
F
Ji
)
..
1( 22 y
K
x
K
z i
xx
i
yy
F
N
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Bi ton un v ko (nn) ng thi
S phn b ng sut trn mt ct ngang:
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Bi ton un v ko (nn) ng thi
Ta thyphng trnh biu dinng sut ti mtim trnmi mt ct ngang l mt hm 2 bin theo x v y, y lphng trnh biu din mt mt phng. Ta gi mt phngny l mt phngng sut.
Giao tuyn ca mtng sut vi mt ct ngang l qu tchnhngim cng c gi tr ng sut php bng khng.Giao tuyn chnh lng trung ha.
Phng trnh ca n l:
trong ko (nn) lch tm:
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Bi ton un v ko (nn) ng thi
Nhn xt:
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Bi ton un v ko (nn) ng thi
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Bi ton un v ko (nn) ng thi
Cch xcnh li tit din:- Gi s ng trung ho tip xc vi mt
cnh tit din
- Vitphng trnhng trung ho, t suy ra to imt lc K tng ng
p dngtng t vi cc cnh cn li, niv tr ccimt lc, tac li tit din
Ch : Li tit din lun la gic li
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Bi ton un v ko (nn) ng thi
iu kin bn:Gi A(xA,yA) v B(xB,yB) l 2 im xang trungho nht v min chu ko v chu nn
K bn ca thanh:
B
y
y
B
x
xz
B
Ay
y
Ax
xz
A
xJ
My
J
M
F
N
xJ
M
yJ
M
F
N
min
max
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Bi ton un v ko (nn) ng thi
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nnh ca thanh chu nnng tm
Khi nim chungBi ton Euler xcnh lc ti hn
ng sut ti hn, gii hn p dng cng
thc Eulernnh ca thanh ngoi gii hnn hi
Phng php thc hnh tnhnnh
Thanh chu un ngang v un dcngthi
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nnh ca thanh chu nnng tm
nh ngha:Gi s c mt thanh di
v mnh, udi b
ngm, u trn chu
mt lc nnng tm P,khi lc P nh hn
mt gii hn nhtnh
th thanh thng, khi
thanh chu nn thun tu
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nnh ca thanh chu nnng tm
Khi mtnnh ti trng lnhn ti trngti hn, bin dng ca h tng rt nhanh.
V d: thanh chu nnnh hnh v ta tnhc :
Khi P = 1,010 Pth th f = 9%l
Khi P = 1,015 Pth th f = 22%l
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nnh ca thanh chu nnng tm
V vy, khi thit k ngoi vicm bo an tonv mt bn v cng, cn phim boiu kinnnh: ti trng tcng nh hn titrng ti hn.
k : h s an ton v mtnnh
V vy gii bi tonnnh, vicc bn l
xcnh ti trng ti hn Pth
od
th
k
PP
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Bi ton Euler xcnh lc ti hn
Cc TH thanh c dng lin kt khc:(Cng thc Euler)
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ng sut ti hn
f
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Gii hn p dng cng thc Euler
f
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Gii hn p dng cng thc Euler
V d:- i vi thp CT3: E= 2,1.107N/cm2;tl=21.000N/cm2
- i vi thp CT5: E= 2,1.107N/cm2;tl=30.000N/cm2
o = 81- i vi gang: o = 80
- i vi g: E= 1,1.106N/cm2;tl=1100N/cm2 o = 100
- i vi g thng: E= 9.105
N/cm2
;tl=1600N/cm2
o =75
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nnh ca thanh ngoi gii hnn hi
ngha: CT Euler chp dngc khivt liun hi.
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nnh ca thanh ngoi gii hnn hi
CT thc nghim Iasinski: xut da trn nhius liu thc nghim, ph thuc vo mnh cathanh.
- Thanh c mnh va :
a,b,c l cc hng s ph thuc vt liu, x bng thcnghim
Thngthng c = 0 nnngi tathng coi:
00
5,3
0
114124
1200
19,4
310003450077600
2930
Thp CT3Thp CT5Gang xm
G
c(N/cm2)b(N/cm2)a(N/cm2)Vt liu
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nnh ca thanh ngoi gii hnn hi
Thanh c mnh b :Lc ny thanhkhng mtnnh mtn trng thi phhoi ca vt liu
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nnh ca thanh ngoi gii hnn hi
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V d
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V d
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Phng php thc hnh tnhnnh
Tuy nhin, nu thanh c mt ct ngang b gim yu ccb (v d l khot bt bulng hocinh tn) th sgim yu ch nhhngn bn mnhhngkhngng k n tnhnnh ca thanh.
V vy trongtrng hp ny, khi kim tra theoiu kinbn ta phi dng din tch thc Fth ca mt ct ngang bgim yu
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i vi bi tonnnh cng c 3 bi ton:
Trong 2 bi ton trn, v tit din thanh bitnn c th suy ra h s theo trnh t:
(tra bng)
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Vic tm F phi lmng dn. Trnh t nh sau:
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Chn hnh dng mt ct hp l v vt liu
- Vi nhng thanh c mnh va v nh:+ 1< < o: th= a-b+c
2 (a,b,c ph thucvt liu)
+