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SỨc Bền Vật Liệu.TRANSCRIPT
AI HOC QUOC GIA THANH PHO HO CH MINH
Sc ben vat lieu
GV: Le Hoang Tuan
Chng 1
CAC KHAI NIEM C BAN
1.1 Khai niem ve mon hoc Sc ben vat lieu- NHIEM VU CUA MON SC BEN VAT LIEU
Sc ben vat lieu (SBVL) la mon hoc ky thuat c s, nghien cu tnh chat chu lc cua vat lieu e e ra cac phng phap tnh ve o ben, o cng va o on nh cua cac bo phan cong trnh hay chi tiet may - goi chung la vat the - chu cac tac ong khac nhau nh tai trong, s thay oi cua nhiet o va che tao khong chnh xac, nham thoa man yeu cau an toan va tiet kiem vat lieu.
Nh vay, muc ch cua mon hoc nay la xay dng cac khai niem va phng phap tnh, e tnh toan cac vat the chu tac dung cua cac tac ong ben ngoai.
Vat the lam viec c an toan th phai:
- Thoa ieu kien ben ngha la khong b pha hoai (nt gay, sup o).
- Thoa ieu kien cng ngha la bien dang va chuyen v nam trong mot gii han cho phep.
- Thoa ieu kien on nh ngha la bao toan hnh thc bien dang ban au.
e am bao s tin cay cua cac phng phap tnh, mon hoc ket hp chat che gia nghien cu thc nghiem va suy luan ly thuyet. Nghien cu thc nghiem nham phat hien ra tnh chat ng x cua cac vat lieu va cac dang chu lc khac nhau, lam c s e xuat cac gia thiet n gian hn e xay dng ly thuyet. Cung v vay, ly thuyet SBVL mang tnh gan ung va neu qua trnh suy dien cang nhieu th ket qua tnh cang co kha nang sai lech nhieu hn. Trong nhieu trng hp, ngi ta phai lam th nghiem tren mo hnh cong trnh thu nho trc khi xay dng hoac th tai cong trnh trc khi a vao s dung.
Thong thng, khi kch thc cua vat the ln hn th kha nang chu lc cung tang va do o o an toan cung c nang cao; tuy nhien, vat lieu phai dung nhieu hn nen nang ne va ton kem hn. Kien thc cua mon SBVL giup giai quyet hp ly mau thuan gia yeu cau an toan va tiet kiem vat lieu.
1.2 Hnh dang vat the
Cac vat the c s dung trong ky thuat c chia ra ba loai c ban:
Khoi: la nhng vat the co kch thc theo ba phng tng ng. Th du: e ap, mong may...
Tam va vo: la nhng vat the
mong co kch thc theo mot
phng rat nho so vi hai phng
con lai; tam co dang phang, vo co
dang cong.
Th du: san nha, mai vo, vo noi hi, vo may bay
Thanh: la nhng vat the hnh dang dai co kch thc theo mot phng rat ln so vi hai phng con lai. ay la loai vat the c dung rong rai trong thc te, nh thanh cua dan cau, cot ien, truc may va c nghien cu chu yeu trong SBVL. Thanh c thay bang truc thanh va mat cat ngang vuong goc vi truc thanh (H.1.3).
Tuy theo truc thanh thang, cong,
gay khuc (phang hay khong gian)
ma goi la thanh thang, thanh cong
hay khung (phang hay khong gian)
nh tren H.1.4.
1.3 NGOAI LC. LIEN KET VA PHAN LC LIEN KET
1.3.1 Ngoai lc
Ngoai lc la lc tac ong t moi trng hoac vat the ben ngoai len vat the ang xet. ay la loai tac ong quan trong va thng gap trong thc te. Ngoai lc c phan loai theo nhieu cach khac nhau.
+ Theo tnh chat chu ong va b ong
Ngoai lc c phan ra tai trong va phan lc.
-Tai trong : la nhng lc chu ong, ngha la co the biet trc ve v tr, phng va o ln. Tai trong la "au vao" cua bai toan, thng c quy nh bi cac quy pham thiet ke hoac c tnh toan theo kch thc cua vat the.
- Phan lc : la nhng lc thu ong (phu thuoc vao tai trong), phat sinh tai v tr lien ket vat the ang xet vi cac vat the xung quanh no.
Th du ve tai trong va phan lc c
minh hoa tren H.1.5. Trong lng xe
va cau la tai trong, lc tac dung
goi la phan lc.
+ Theo hnh thc phan bo
Ngoai lc c phan ra lc tap trung va lc phan bo.
Lc tap trung: la lc tac dung tai mot iem cua vat the. Trong thc te, khi dien tch truyen lc be th ngi ta coi nh lc truyen qua mot iem e n gian hoa s phan tch. Th du, trong lng mot chiec xe o to truyen xuong mat cau c thay bang cac lc tap trung at tai trong tam cua dien tch tiep xuc gia cac banh xe va mat cau, hoac phan lc tai mat tiep xuc cua goi ta cung c thay bang lc tap trung nh H.1.5.
Lc phan bo: la lc tac dung tren mot dien tch, mot the tch hoac mot ng cua vat the. Lc trong trng la mot Th du cua lc phan bo the tch v no tac ong len moi iem cua trong vat the. Cng o cua lc phan bo the tch co th nguyen la lc/the tch, hay [F/L3]. Ap lc nc len be cha hay ap lc at len tng chan la minh hoa cho lc phan bo dien tch. Cng o lc phan bo dien tch co th nguyen la lc/dien tch, hay [F/L2]. Khi lc phan bo tren mot dai hep th ngi ta thay lc phan bo dien tch bang lc phan bo ng vi cng o lc co th nguyen la lc/chieu dai, hay [F/L]. Th du trong lng mot thanh c thay bang lc phan bo ng tren truc thanh. Lc phan bo ng la loai lc phan bo thng gap trong bai toan SBVL. Cac loai lc phan bo c minh hoa tren H.1.6.
+ Theo tnh chat tac dung
Ngoai lc c phan ra lc tnh va lc ong. Lc tnh la lc bien oi cham hoac khong thay oi theo thi gian, v vay gay ra gia toc chuyen ong rat be co the bo qua khi xet can bang. Ap lc at len tng chan, trong lng cua cong trnh la cac lc tnh Lc ong la lc thay oi nhanh theo thi gian, gay ra chuyen ong co gia toc ln. Th du, rung ong do mot ong c gay ra, va cham cua bua xuong au coc, chuyen ong cua oan xe la tren cau la cac trng hp lc ong Vi lc ong th can xet en s tham gia cua lc quan tnh trong phng trnh can bang tnh hoc.
Trong SBVL, ca hai loai lc nay eu c xet ti.
1.3.2 Lien ket va phan lc lien ket
1.3.2.1 Cac loai lien ket va phan lc lien ket:
Mot thanh muon duy tr hnh dang, v tr ban au khi chu tac ong cua ngoai lc th no phai c lien ket vi vat the khac hoac vi at. Tuy theo tnh chat ngan can chuyen ong ma ngi ta a ra cac s o lien ket, thng gap la goi ta di ong, goi co nh hay ngam nh tren H.1.7.
Goi di ong (H.1.7a) ch ngan can mot chuyen ong thang va phat sinh mot phan lc R theo phng cua lien ket
Goi co nh (H.1.7b) ngan can chuyen v thang theo phng bat k va phat sinh phan lc R cung theo phng o. Phan lc R thng c phan tch ra hai thanh phan V va H
Ngam (H.1.7c) ngan can bat k chuyen v thang nao va chuyen v xoay. Phan lc thng c phan tch ra ba thanh phan V, H va M.
1.3.2.2 Cach xac nh phan lc:
Cac thanh phan phan lc c xac nh t ieu kien can bang tnh hoc.
Bai toan phang co ba phng trnh can bang oc lap, c thiet lap cac dang khac nhau nh sau:
1. (x, y khong song song)
2. (A, B, C khong thang hang)
3. (AB khong vuong goc vi x)
Bai toan khong gian co sau phng trnh can bang oc lap, thng co dang:
1.4 Cac dang chu lc va bien dang c ban CHUYEN V
1.4.1Bien dang cua vat the:
Trong thc te, s chu lc cua mot thanh co the phan tch ra cac dang chu lc c ban gom keo (nen), xoan, cat va uon nh c minh hoa tren H.1.8. Truc thanh khi chu keo hoac nen se dan dai hoac co ngan; khi chu uon se b cong i nh H.1.8e, con thanh chu xoan th truc thanh van thang nhng ng sinh tren be mat tr thanh ng xoan tru. Khi chu cat, hai phan cua thanh co xu hng trt oi vi nhau. cac chng sau, cac dang chu lc c ban nay se lan lt c nghien cu.
1.4.2 Bien dang cua phan to: Neu tng tng tach mot phan to hnh hop t mot thanh chu lc th s bien dang cua no trong trng hp tong quat co the phan tch ra hai thanh phan c ban, gom bien dang dai va bien dang goc nh tren H.1.9.
Phan to tren H.1.9a ch thay oi chieu dai, khong thay oi goc. Chieu dai dx ban au cua phan to b dan dai (hay co ngan) mot lng (dx. Bien dang dai tng oi theo phng x, k hieu la , c nh ngha bi t so (dx va dx:
Phan to tren H.1.9b ch co thay oi goc, khong thay oi chieu dai. o thay oi cua goc vuong ban au goi la bien dang goc hay goc trt, ky hieu la (.
1.4.3 Chuyen v:
Khi vat the b bien dang, cac iem trong vat the noi chung b thay oi v tr. o chuyen di t v tr cu cua iem A sang v tr mi A c goi la chuyen v dai. Goc hp bi v tr cua mot oan thang AC trc va trong khi bien dang AC cua vat the c goi la chuyen v goc (H.1.10).
1.5 Cac gia thiet
Khi giai bai toan SBVL, ngi ta chap nhan mot so gia thiet nham n gian hoa van e nhng co gang am bao s chnh xac can thiet phu hp vi yeu cau thc te. Cac gia thiet nay lien quan en s o hnh hoc cua vat the, tnh chat cua vat lieu va tnh chat bien dang, chuyen v cua vat the.
1.5.1 Gia thiet ve s o tnh
Khi tnh toan, ngi ta thay vat the thc bang s o tnh. Th du, thanh chu tai trong ban than tren H.1.11a c thay bang s o tren H.1.11b.
Trong trng hp nay, vat the thanh c thay bang truc thanh, tiet dien thanh thay bang cac ac trng hnh hoc (se c nghien cu trong chng sau), lien ket ta c thay bang cac goi co nh va di ong, trong lng ban than thay bang lc phan bo eu.
1.5.2 Gia thiet ve vat lieu
Vat lieu c coi la lien tuc, ong nhat, ang hng va an hoi tuyen tnh.( Ta tng tng lay mot phan to bao quanh mot iem trong vat the. Neu cho phan to be tuy y ma van cha vat lieu th ta noi vat lieu lien tuc tai iem o. Gia thiet ve s lien tuc cua vat lieu lam c s e xay dng khai niem ng suat va bien dang tai mot iem, cho phep s dung cac phep tnh cua toan giai tch nh gii han, vi phan, tch phan... Vat lieu lien tuc la mo hnh toan hoc cua vat lieu that, co cac ac trng c hoc giong nh cac ac trng v mo (xac nh tren mot the tch vat lieu u ln) tng ng cua vat lieu that. Trong thc te, ngay ca vi vat lieu c coi la hoan hao nhat nh kim loai th cung co cau truc vi mo (chang han, t mc o mang tinh the tr i) khong lien tuc theo ngha toan hoc. Gia thiet nay giup cho SBVL tranh c viec khao sat cau truc vi mo cua vat lieu that, la viec rat phc tap, tham ch khong lam c.
( Vat lieu ong nhat ngha la tnh chat c hoc tai moi iem trong vat the la nh nhau, vat lieu ang hng ngha la tnh chat c hoc tai mot iem theo cac phng eu giong nhau. Tnh chat c hoc c ac trng bi cac hang so vat lieu nh mo un an hoi, he so bien dang hong, gii han an hoi... Thc ra, cau truc vi mo cua vat lieu that khong hoan toan ong nhat va ang hng, nhng s sap xep cua chung thng la ngau nhien theo moi hng, nen neu vat the co kch thc u ln th gia thiet tren noi chung chap nhan c.
( Mot vat the that se co thay oi hnh dang di tac dung cua ngoai lc. Tnh chat
an hoi cua vat the la kha nang khoi phuc
lai hnh dang ban au cua no khi
ngoai lc thoi tac dung. Neu quan he gia
ngoai lc va bien dang la bac nhat, th vat lieu c goi la an hoi tuyen tnh nh minh hoa tren H.1.12.
. Gia thiet vat lieu an hoi tuyen tnh lam giam bt s phc tap cua bai toan SBVL.
1.5.3 Gia thiet ve bien dang va chuyen v
Khi chu tac ong ben ngoai, vat the co bien dang va chuyen v be so vi kch thc ban au cua vat . V vay, co the khao sat s can bang cua vat the hoac cac bo phan cua no tren hnh dang ban au.
Gia thiet nay xuat phat t ieu kien cng cua cac vat the (nh cong trnh, cac bo phan may moc...) c s dung trong thc te ky thuat. ieu kien cng oi hoi bien dang va chuyen v ln nhat trong vat the phai nam trong mot gii han tng oi nho.
He qua:
Khi vat the co chuyen v be va vat lieu an hoi tuyen tnh th co the ap dung nguyen ly cong tac dung nh sau:
Mot ai lng do nhieu nguyen nhan ong thi gay ra se bang tong ai lng o do tac ong cua cac nguyen nhan rieng le.
Th du: xet thanh chu uon nh tren H.1.13.
Chuyen v ( tai au thanh do lc P1 va P2 gay ra co the phan tch nh sau:
Nguyen ly cong tac dung bien bai toan phc tap thanh cac bai toan n gian nen de giai quyet hn. V vay, no thng c s dung trong SBVL.
Chng 2
LY THUYET NOI LC
2.1 KHAI NIEM VE NOI LC - PHNG PHAP KHAO SAT - NG SUAT
1- Khai niem ve noi lc:
Xet mot vat the chu tac dung cua ngoai lc va trang thai can bang (H.2.1). Trc khi tac dung lc, gia cac phan t cua vat the luon co cac lc tng tac gi cho vat the co hnh dang nhat nh. Di tac dung cua ngoai lc, cac phan t cua vat the co the dch lai gan nhau hoac tach xa nhau. Khi o, lc tng tac gia cac phan t cua vat the phai thay oi e chong lai cac dch chuyen nay. S thay oi cua lc tng tac gia cac phan t trong vat the c goi la noi lc.
Mot vat the khong chu tac ong nao t ben ngoai th c goi la vat the trang thai t nhien va noi lc cua no c coi la bang khong.
2-Phng phap khao sat noi lc: Phng phap mat cat
Xet lai vat the can bang va 1 iem C trong vat the (H.2.1),.
Tng tng mot mat phang ( cat qua C va chia vat the thanh hai phan A va B; hai phan nay se tac ong lan nhau bang he lc phan bo tren dien tch mat tiep xuc theo nh luat lc va phan lc.
Neu tach rieng phan A th he lc tac ong t phan B vao no phai can bang vi ngoai lc ban au (H.2.2).
Xet mot phan to dien tch (A bao quanh iem khao sat C tren mat cat ( co phng phap tuyen v. Goi la vector noi lc tac dung tren (A. Ta nh ngha ng suat toan phan tai iem khao sat la:
Th nguyen cua ng suat la [lc]/[chieu dai]2 (N/m2, N/cm2).
ng suat toan phan p co the phan ra hai thanh phan:
+ Thanh phan ng suat phap (v co phng phap tuyen cua mat phang (
+ Thanh phan ng suat tiep (v nam trong mat phang ( ( H.2.3 ).
Cac ai lng nay lien he vi nhau theo bieu thc:
(2.1)
ng suat la mot ai lng c hoc ac trng cho mc o chu ng cua vat lieu tai mot iem; ng suat vt qua mot gii han nao o th vat lieu b pha hoai. Do o, viec xac nh ng suat la c s e anh gia o ben cua vat lieu, va chnh la mot noi dung quan trong cua mon SBVL.
2.2 CAC THANH PHAN NOI LC - CACH XAC NH
1- Cac thanh phan noi lc:
Nh a biet, oi tng khao sat cua SBVL la nhng chi tiet dang thanh, ac trng bi mat cat ngang (hay con goi la tiet dien) va truc thanh.
Goi hp lc cua cac noi lc phan bo tren mat cat ngang cua thanh la R. R co iem at va phng chieu cha biet .
Di R ve trong tam O cua mat cat ngang ( co phng bat ky
at mot he truc toa o Descartes vuong goc ngay tai trong tam mat cat ngang, Oxyz, vi truc z trung phap tuyen cua mat cat, con hai truc x, y nam trong mat cat ngang.
Khi o, co the phan tch R ra ba thanh phan theo ba truc:
+ Nz, theo phng truc z ( mat cat ngang) goi la lc doc
+ Qx theo phng truc x (nam trong mat cat ngang) goi la lc cat.
+ Qy theo phng truc y (nam trong mat cat ngang) goi la lc cat.
Momen M cung c phan ra ba thanh phan :
+ Momen Mx quay quanh truc x goi la momen uon .
+ Momen My quay quanh truc y goi la momen uon .
+ Momen Mz quay quanh truc z goi la momen xoan.
Sau thanh phan nay c goi la cac thanh phan noi lc tren mat cat ngang (H.2.4).
2- Cach xac nh:
Sau thanh phan noi lc tren mot mat cat ngang c xac nh t sau phng trnh can bang oc lap cua phan vat the c tach ra, tren o co tac dung cua ngoai lc ban au PI va cac noi lc.
Cac phng trnh can bang hnh chieu cac lc tren cac truc toa o:
(2.2)
trong o: Pix, Piy, Piz - la hnh chieu cua lc Pi xuong cac truc x, y, z.
Cac phng trnh can bang momen oi vi cac truc toa o ta co:
(2.3)
vi:mx(Pi), my(Pi), mz(Pi) - cac momen cua cac lc Pi oi vi cac truc x,y, z.
3-Lien he gia noi lc va ng suat:
Cac thanh phan noi lc lien he vi cac thanh phan ng suat nh sau:
- Lc doc la tong cac ng suat phap
- Lc cat la tong cac ng suat tiep cung phng vi no
- Momen uon la tong cac momen gay ra bi cac ng suat oi vi truc x hoac y
- Momen xoan la tong cac momen cua cac ng suat tiep oi vi truc z2-3 BAI TOAN PHANG:
Trng hp bai toan phang ( ngoai lc nam trong mot mat phang ( th du mat phang yz)), ch co ba thanh phan noi lc nam trong mat phang yz : Nz, Qy, Mx.
( Qui c dau (H.2.5)
- Lc doc Nz ( 0 khi gay keo oan thanh ang xet (co chieu hng ra ngoai mat cat)
- Lc cat Qy ( 0 khi lam quay oan thanh ang xet theo chieu kim ong ho.
- Momen uon Mx ( 0 khi cang th di ( th y dng ).
( Cach xac nh:
Dung 3 phng trnh can bang tnh hoc khi xet can bang phan A) hay phan B)
Th du 2.1 Xac nh cac tr so noi lc tai mat cat 1-1 cua thanh AB, vi :
q = 10 kN/m; a = 1m; Mo = 2qa2. ( H.2.6)
Giai. Tnh phan lc: Giai phong cac lien ket va thay vao o bang cac phan lc lien ket VA, HA, VB.
Viet cac phng trnh can bang tnh hoc khi xet can bang thanh AB
( HA = 0; ;
Tnh noi lc: Mat cat 1-1 chia thanh lam hai phan.
Xet s can bang cua phan ben trai (H.2.6) :
Neu xet can bang cua phan phai ta cung tm c cac ket qua nh tren.
2.4 BIEU O NOI LC ( BAI TOAN PHANG )
1. nh ngha: Thng cac noi lc tren cac mat cat ngang cua mot thanh khong giong nhau.
Bieu o noi lc (BNL) la o th bieu dien s bien thien cua cac noi lc theo v tr cua cac mat cat ngang.
Nh vao BNL co the xac nh v tr mat cat co noi lc ln nhat va tr so noi lc ay.
2. Cach ve BNL- Phng phap giai tch:
e ve bieu o noi lc ta tnh noi lc tren mat cat cat ngang mot v tr bat ky co hoanh o z so vi mot goc hoanh o nao o ma ta chon trc. Mat cat ngang chia thanh ra thanh 2 phan. Xet s can bang cua mot phan (trai, hay phai) , viet bieu thc giai tch cua noi lc theo z..
Ve ng bieu dien tren he truc toa o co truc hoanh song song vi truc thanh (con goi la ng chuan), tung o cua bieu o noi lc se c dien ta bi cac oan thang vuong goc cac ng chuan.
Th du 2.2- Ve BNL cua dam mut tha (H.2.7)
Giai
Xet mat cat ngang 1-1 co hoanh o z so vi goc A, ta co ( 0 ( z ( l )
Bieu thc giai tch cua lc cat
va momen uon tai mat cat 1-1
c xac nh t viec xet can bang
phan phai cua thanh:
Cho z bien thien t 0 en l, ta se c bieu o noi lc nh tren H.2.7.
Qui c:+Bieu o lc cat Qy tung o dng ve pha tren truc hoanh.
+Bieu o momen uon Mx tung o dng ve pha di truc hoanh.
(Tung o cua bieu o momen luon ve pha th cang cua thanh).
Th du 2.3 Ve BNL cua dam n gian chu tai phan bo eu q (H.2.8a).
Giai
Phan lc: Bo cac lien ket tai A va B, thay bang cac phan lc ( H.2.8a).
(Z = 0 ( HA =0.
Do oi xng (
Noi lc: Chon truc hoanh nh tren H.2.8b. Xet mat cat ngang 1-1 tai K co hoanh o la z, ( 0 ( z ( l ). Mat cat chia thanh lam hai phan.
Xet can bang cua phan ben trai AK (H.2.8b)
T cac phng trnh can bang ta suy ra:
Qy la ham bac nhat theo z, Mx la ham bac 2 theo z.
Cho z bien thien t 0 en l ta ve c cac bieu o noi lc (H2.8).
Cu the: +Khi z=0 ( Qy = ql/2 , Mx = 0
+Khi z=l ( Qy = -ql/2 , Mx = 0
+Tm Mx, cc tr bang cach cho ao ham dMx / dz =0,
dMx / dz =0 (
Qua cac BNL, ta nhan thay:
Lc cat Qy co gia tr ln nhat mat cat sat goi ta,
Momen uon Mx co gia tr cc ai gia dam.
Th du 2.4 Ve BNL cua dam n gian chu lc tap trung P ( H.2.9a) .
Giai
Phan lc: Cac thanh phan phan lc tai cac goi ta la:
;;
Noi lc : V tai trong co phng vuong goc vi truc thanh nen lc doc Nz tren moi mat cat ngang co tr so bang khong.
Phan oan thanh: V tnh lien tuc cua cac ham so giai tch bieu dien cac noi lc nen phai tnh noi lc trong tng oan cua thanh; trong moi oan phai khong co s thay oi ot ngot cua ngoai lc .
( oan AC- Xet mat cat 1-1 tai iem K1 trong oan AC va cach goc A mot oan z, ( 0 ( z ( a ).
Khao sat can bang cua phan ben trai ta c cac bieu thc giai tch cua noi lc:
(a)
( oan CB- Xet mat cat 2-2 tai iem K2 Trong oan CB cach goc A mot oan z , ( a ( z ( l ). Tnh noi lc tren mat cat 2-2 bang cach xet phan ben phai (oan K2B). Ta c:
(b)(b)
T (a) va (b) de dang ve c cac bieu o noi lc nh H.2.9d,e.
Trng hp ac biet : Neu a=b= L/2, khi o momen cc ai xay ra tai gia dam va co gia tr: Mmax = PL/4
Th du 2.5 Ve BNL cua dam n gian chu tac dung cua momen tap trung Mo (H.2.10a.)
Giai
Phan lc: Xet can bang cua toan dam ABC ( cac phan lc lien ket tai A va B la:
; , chieu phan lc nh H.2.10a.
Noi lc:
oan AC: Dung mat cat 1-1 cach goc A mot oan z1 ;(0 ( z1 ( a ).Xet can bang cua oan AK1 ben trai mat cat K1 ( cac noi lc nh sau
(c)
oan CB: Dung mat cat 2-2 trong oan CB cach goc A mot oan z2 vi (a ( z2 ( l ) . Xet can bang cua phan ben phai K2B ( cac bieu thc noi lc tren mat cat 2-2 la:
(d)
BNL c ve t cac bieu thc (c), (d) cua noi lc trong hai oan (H.2.10d-e).
Trng hp ac biet: Momen tap trung Mo at tai mat cat sat goi ta A (H.2.11).
Qy va Mx se c xac nh bi (d) ng vi
a = 0. BNL ve nh H.2.11
Cac nhan xet :
- Ni nao co lc tap trung, bieu o lc cat ni o co bc nhay. Tr so cua bc nhay bang tr so lc tap trung. Chieu bc nhay theo chieu lc tap trung neu ta ve t trai sang phai
- Ni nao co momen tap trung, bieu o momen uon ni o co bc nhay. Tr so cua bc nhay bang tr so momen tap trung. Chieu bc nhay theo chieu momen tap trung neu ta ve t trai sang phai
Kiem chng cac nhan xet :
Khao sat oan (z bao quanh mot iem K co tac dung lc tap trung P0 , momen tap trung M0 ( H.2.12b).
Viet cac phng trnh can bang ( (Y = 0 ( Q1 + P0 Q2 = 0 ( Q2 Q1 = P0 (i)
(M/K = 0 ( M1 +M0 - M2 + Q1 - Q2 =0
Bo qua vo cung be bac mot Q1 , Q2 , ( M2 - M1 = M0 (ii)
Bieu thc (i) a kiem chng nhan xet ve bc nhay cua bieu o lc cat.
Bieu thc (ii) a kiem chng nhan xet ve bc nhay cua bieu o momen.
2.4. LIEN HE VI PHAN GIA NOI LC VA TAI TRONG PHAN BO TRONG THANH THANG
Xet mot thanh chu tai trong bat ky (H.2.13a). Tai trong tac dung tren thanh nay la lc phan bo theo chieu dai co cng o q(z) co chieu dng hng len (H.2.13b).
Khao sat oan thanh vi phan dz, gii han bi hai mat cat 1-1 va 2-2 (H.2.13b). Noi lc tren mat cat 1-1 la Qy va Mx. Noi lc tren mat cat 2-2 so vi 1-1 a thay oi mot lng vi phan va tr thanh Qy + dQy; Mx + dMx . V dz la rat be nen co the xem tai trong la phan bo eu tren oan dz.
Viet cac phng trnh can bang:
1-Tong hnh chieu cac lc theo phng ng
(Y = 0 ( Qy + q(z)dz (Qy + dQy) = 0
(
(2.4)
ao ham cua lc cat bang cng o cua lc phan bo vuong goc vi truc thanh.
2- Tong momen cua cac lc oi vi trong tam mat cat 2-2 ta c:
(M/o2 = 0 (
Bo qua lng vo cung be bac hai (
(2.5)
ao ham cua momen uon tai mot mat cat bang lc cat tai mat cat o
T (2.4) va (2.5) (
(2.6)
ngha la: ao ham bac hai cua momen uon tai mot iem chnh la bang cng o cua tai trong phan bo tai iem o.
Th du 2.6 Ve BNL cho dam n gian AB chu tac dung cua tai phan bo bac nhat nh H.2.14.
Giai
Phan lc: Giai phong lien ket, at cac phan lc tng ng cac goi ta, xet can bang cua toan thanh,
(X =0 ( HA = 0,
Noi lc: Cng o cua lc phan bo mat cat 1-1 cach goc A mot oan z cho bi: q(z)= q0
Dung mat cat 1-1 va xet s can bang cua phan ben trai (H.2.14b).
(Y = 0 ( (e)
(M/o1 = 0 ( (g)
T (e) va (g) ta ve c bieu o lc cat va momen cho dam a cho. Cac bieu o nay co tnh chat nh sau:
Bieu o lc cat Qy co dang bac 2. Tai v tr z = 0, q(z) = 0 nen ay bieu o Qy at cc tr: (Qy)z = 0 = Qmax =
Bieu o momen uon Mx co dang bac 3. Tai v tr ; Qy = 0. Vay tai ay Mx at cc tr:
Th du 2.7 Ve BNL cho dam chu lc tong quat (H.2.15)
Giai
Phan lc: Giai phong lien ket, xet can bang toan thanh, suy ra phan lc lien ket tai A va C la:
HA = 0 ,VA = 2qa; VC = 2qa
Noi lc:
* oan AB: Mat cat 1-1, goc A (0 ( z ( a), xet can bang phan trai
* oan BC: Mat cat 2-2, goc A (a ( z ( 2a) va xet can bang phan trai:
* oan CD: Mat cat 3-3, goc A, (2a ( z ( 3a) xet can bang phan phai:
(2a ( z ( 3a)
Bieu o momen va lc cat ve nh H.2.15.
Th du 2.8 Ve bieu o noi lc trong khung chu tai trong nh tren H.2.16.
Giai
Tnh phan lc lien ket
Xet s can bang cua toan khung di tac dung cua tai trong ngoai va cac phan lc lien ket ta suy ra:
(Ngang = 0 ( HA = 0
(ng = 0 ( VA + VD= 0 ( VD =( ung chieu a chon )
Vay chieu that cua VA ngc vi chieu a chon
Ve bieu o noi lc
oan AB: dung mat cat 1-1 va xet can bang oan AK1 ta c:
(0 ( z1 ( a)
oan BC: dung mat cat 2-2 va xet can bang oan ABK2 ta c:
(0 ( z2 ( a)
oan CD: dung mat cat 3-3 va xet can bang DK3
(0 ( z3 ( a)
Kiem tra s can bang nut
oi vi khung, co the kiem tra ket qua bang viec xet can bang cac nut. Neu tach nut ra khoi he th ta phai at vao nut cac ngoai lc tap trung (neu co) va cac noi lc tai cac mat cat, gia tr cua chung c lay t bieu o va ve.
Sau khi at cac lc tren, neu tnh ung cac noi lc cac nut th nut se can bang, ngha la cac phng trnh can bang c thoa man. Ngc lai, neu cac phng trnh khong thoa man th cac noi lc tnh sai.
Cu the oi vi khung ang xet, ta tach nut B va at vao o momen tap trung qa2 va cac thanh phan noi lc tren cac oan thanh ngang va ng nh H.2.16d:
- Tai mat cat tren thanh ngang co lc doc +qa hng ra ngoai mat cat, lc cat co chieu hng len va momen gay cang th di.
- Tai mat cat tren thanh ng co lc doc hng ra ngoai mat cat (hng xuong) lc cat +qa hng t phai sang trai va momen gay ra cang th trong khung nen chieu quay co mui ten hng ra ngoai.
Ta de dang thay cac phng trnh can bang thoa man:
( X = 0 ; ( Y = 0 ; ( M/B = 0
Tng t, tach nut C va at vao o lc tap trung qa hng t trai sang phai va cac thanh phan noi lc tren cac oan thanh ngang va ng nh H.2.16d.
- Tai mat cat tren thanh ngang co lc doc +qa hng ra ngoai mat cat, lc cat co khuynh hng lam quay phan oan thanh ang xet ngc chieu kim ong ho nen co chieu hng xuong, con momen th bang khong.
- Tai mat cat tren thanh thang ng ton tai lc doc co chieu hung vao mat cat (hng len) va khong co lc cat cung nh momen.
Ta de dang thay rang cac phng trnh can bang c thoa man:
; ;
Vay cac nut B va C eu can bang ngha la cac he noi lc tai cac nut ung.
Th du 2.9 Ve BNL trong thanh cong (H.2.17)
Giai
Cat thanh tai tiet dien 1-1, xac nh bi goc ( (0 ( ( ( 90o), xet can bang cua phan tren di tac dung cua cac ngoai lc va cac thanh phan noi lc at theo chieu dng quy c nh H.2.17b.
Phng trnh can bang hnh chieu cac lc theo phng phap tuyen vi mat cat cho: N = 2Psin( Pcos( = P(2sin( cos() (a)
Phng trnh can bang hnh chieu cac lc theo phng ng knh cho:
Q = 2Pcos( + Psin( = P(2cos( + sin() (b)
Phng trnh can bang cua cac momen cac lc oi vi trong tam mat cat dan en:
M = 2PRsin( PR(1 cos() = PR(2sin( + 1 cos() (c)
Cho ( mot vai tr so ac biet va tnh cac tr so noi lc tng ng, ta ve c bieu o.
Lc cat at cc tr khi , ngha la khi:
-2sin( + cos( = 0 ( tg( = 0,5 ( ( = (o = 26o56
sin(o = 0,4472 ; cos(o = 0,8944
Ta co bang noi lc sau:(0(o45o900
N
Q
M P2 P00
2,236 P- PR0,7 P2,12 P-1,7 PR2 P+P-3PR
Khi ve can chu y at cac tung o theo phng vuong goc vi truc thanh, tc la theo phng ban knh nh tren H.2.17c,d,e.
CACH VE BIEU O NHANH
2.5.1 Phng phap ve tng iem
Da tren cac lien he vi phan, ta nh dang cac BNL tuy theo dang tai trong a cho va t o ta xac nh so iem can thiet e ve bieu o.
Tren 1 oan thanh
+ q =0 ( Q = hang so, M = bac nhat.
+ q = hang ( Q = bac nhat, M = bac hai.
.
+ Neu bieu o co dang hang so , ch can xac nh mot iem bat ky.
+ Neu bieu o co dang bac nhat , can tnh noi lc tai hai iem au va cuoi oan thanh.
+ Neu bieu o co dang bac hai tr len th can ba gia tr tai iem au, iem cuoi va tai ni co cc tr, neu khong co cc tr th can biet chieu loi lom cua bieu o theo dau cua ao ham bac hai. oan thanh co lc phan bo q hng xuong se am, nen be lom cua bieu o momen hng len. Ngc lai, neu q hng len se dng nen be lom cua bieu o momen hng xuong. Tom lai, ng cong momen hng lay lc phan bo q.Th du 2.10: Ve BNL trong dam cho tren H.2.18 (phng phap ve iem)
Giai.
Phan lc lien ket
Noi lc
oan AB: q=0( Qy = hang so,
Mx = bac nhat.
Trong trng hp nay Qy la hang so bang khong v QA(AB) = 0.
( Mx trong oan nay se la hang so
MA (AB) = MB (BA) = Mo = -qa2
oan BD: q= hang ( Qy = bac 1,
Mx = bac 2.
Tai B:
Tai D:
Bieu o Qy trong oan nay khong co v tr nao =0 ( bieu o Mx khong co cc tr.
Ch can noi hai gia tr momen tai B va D bang ng cong bac hai co be lom sao cho hng lay lc q.
oan DC: q= hang ( Qy = bac 1, Mx = bac 2.
Tai D:
;
Tai C:
Bieu o Qy trong oan nay khong co v tr nao =0 ( bieu o Mx khong co cc tr.
Ch can noi hai gia tr momen tai D va C bang ng cong bac hai co be lom sao cho hng lay lc q.
Cac bieu o lc cat Qy va momen Mx lan lt c ve tren H.2.18b,c.
2.5.2 Cach ap dung nguyen ly cong tac dung
Khi thanh chu tac dung nhieu loai tai trong, ta co the ve bieu o noi lc trong thanh do tng tai trong rieng le gay ra roi cong ai so lai e c ket qua cuoi cung.
Th du 10. Ve bieu o mo men trong dam nh H.2.18a bang cach cong bieu o.
Giai. Tai trong tren thanh c chia thanh hai trng hp c ban:
+ Hnh 2.18b bieu dien mo men do lc tap trung P gay ra
+ Hnh 2.18c bieu dien mo men do lc phan bo eu q gay ra
Hnh 2.18dbieu dien mo men tong hp can tm, cac tung o bang tong ai so cac tung o tai cac tiet dien tng ng tren H.2.18b,c.
BAI TAP CHNG 2
2.1. Ve bieu o noi lc cua cac dam cho tren H.2.1.
2.2. Khong can tnh ra phan lc, ve BNL cua cac dam cho tren H.2.2.
2.3. Ve bieu o noi lc nh tren H.2.3.
2.4. Ve bieu o noi lc cua dam tnh nh nh tren H.2.4.
2.5. Ve bieu o noi lc cho he khung sau (H.2.5).
2.6. Ve bieu o lc doc, momen uon, momen xoan cho thanh khong gian (H.2.6).
Chng 3
THANH CHU KEO - NEN UNG TAM
3.1 Khai niem
( nh ngha: Thanh c goi la chu keo hay nen ung tam khi tren moi mat cat ngang cua thanh ch co mot thanh phan noi lc la lc doc Nz.Nz ( 0 khi hng ra ngoai mat cat- Keo
Nz ( 0 khi hng vao trong mat cat- Nen
ay la trng hp chu lc n gian nhat. Ta gap trng hp nay khi thanh chu 2 lc bang nhau va trai chieu hai au doc truc thanh .
Thanh chu keo ung tam (H.3.2a) hay chu nen ung tam (H.3.2b).
(Thc te : co the gap cac cau kien chu keo hay nen ung tam nh: day cap trong can cau (H.3.3a), ong khoi (H.3.3b), cac thanh trong dan (H.3.3c).
3.2 ng suat tren mat cat ngang
Xet thanh thang chu keo (nen) ung tam (H.3.4a)
Cac mat cat ngang CC va DD trc khi chu lc cach nhau oan dz va vuong goc truc thanh. Cac th doc trong oan CD (nh GH) bang nhau (H.3.4b).
Khi thanh chu keo (nen), noi lc tren mat cat ngang DD hay bat ky mat cat ngang khac la Nz = P (H.3.4c) thanh se dan ra, mat cat DD di chuyen doc truc thanh z so vi mat cat CC mot oan be (dz (H.3.4b).
Quan sat cac th doc trong oan CD (nh GH), bien dang eu bang HH va khong oi, mat cat ngang trong suot qua trnh bien dang van phang va vuong goc vi truc thanh, ieu nay cho thay cac iem tren mat cat ngang ch co ng suat phap (z khong oi (H.3.4d).
Quan he gia ng suat va noi lc: (
V (z = const (
hay:
(3.1)
vi: A - dien tch mat cat ngang cua thanh.
3.3 Bien dang cua thanh chu keo (nen) ung tam
1- Bien dang doc
Bien dang doc truc z cua oan dai dz chnh la (dz (H.3.4b).
( Bien dang dai tng oi cua oan dz la: (a)
Theo nh luat Hooke, ta co:
(b)
trong o: E - la hang so ty le, c goi la mo un an hoi khi keo (nen), no phu thuoc vao vat lieu va co th nguyen , n v N/m2.
Bang 3.1 Tr so E cua mot so vat lieu.
Vat lieuE (kN/cm2)(
Thep (0,15 ( 0,20)%C
Thep lo xo
Thep niken
Gang xam
ong
ong thau
Nhom
Go doc th
Cao su2 x 1042,2 x 104
1,9 x 104
1,15 x 1041,2 x 104(1,0 (1,2)104(0,7 ( 0,8)104(0,08 ( 0,12)104
0,80,25 ( 0,33
0,25 ( 0,33
0,25 ( 0,33
0,23 ( 0,27
0,31 ( 0,34
0,31 ( 0,34
0,32 ( 0,36
0,47
T (a) ( (dz, sau o the (b) vao , (Bien dang dai doc truc cua oan dz la:
(c)
( Bien dang dai (dan dai khi thanh chu keo, co ngan khi thanh chu nen) cua mot oan thanh co chieu dai L la:
;
(3.2)
Trng hp E khong oi, A la hang so va Nz cung khong oi tren suot chieu dai L cua thanh, ta c:
;
(3.3)
Neu thanh gom nhieu oan chieu dai Li va tren moi oan Nz, E, A khong oi th ta co:
(3.3)
Tch so EA goi la o cng khi keo hay nen ung tam cua thanh.
oi khi, ngi ta con dung o cng tng oi
2- Bien dang ngang
Theo phng z doc truc thanh co bien dang doc, theo phng ngang x, y vuong goc z thanh cung co bien dang ngang (H.3.4d).
Neu goi (x va (y la bien dang dai tng oi theo hai phng x va y, th ta co quan he sau: (3.4)
trong o: ( = (0 ( 0,5 tuy vat lieu) - he so Poisson, la hang so ( bang 3.1)
Dau () trong (3.4) ch rang bien dang doc va ngang ngc nhau.
Th du 3.1 Ve bieu o doc Nz , tnh ng suat va bien dang dai toan phan thanh tren H.3.5a Biet E = 2.104 kN/cm2; F1 = 10 cm2; F2 = 20 cm2.
Giai. Dung phng phap mat cat ve bieu o Nz (H.3.5b)
( ng suat tren mat cat ngang moi oan la:
;
;
Bien dang doc toan phan chnh la bien dang dai tuyet oi cua thanh. Cong thc (3.3) ap dung cho bon oan cua thanh.
(L= = 0,005 cm
Bien dang doc mang dau + ch ro thanh b dai ra.
3.4 ac trng c hoc cua vat lieu
1. Khai niem
Van e: Can phai so sanh ng suat , bien dang trong vat lieu cua thanh khi chu lc vi ng suat bien dang cua vat lieu cung loai a biet.
Can th nghiem keo, nen ung tam e tm hieu tnh chat chu lc va qua trnh bien dang t luc bat au chu lc en luc pha hong cua cac loai vat lieu khac nhau.
Phan loai vat lieu: Can c vao bien dang va s pha hong, kha nang chu keo, nen khac nhau, ngi ta phan vat lieu thanh hai loai c ban:
Vat lieu deo la vat lieu b pha hoai khi bien dang kha ln (thep, ong)
Vat lieu don la vat lieu b pha hoai khi bien dang con nho (gang, a, betong
Nh vay, ta co bon th nghiem c ban sau:
2. Th nghiem keo vat lieu deo (thep)
1- Mau th nghiem
Theo tieu chuan TCVN 197 - 85 (H.3.6)
Chieu dai Lo th nghiem la oan thanh ng knh do, dien tch Ao
2- Th nghiem
Tang lc keo t 0 en khi mau t, vi bo phan ve bieu o cua may keo, ta nhan c o th quan he gia lc keo P va bien dang dai (L cua mau (H.3.7a) . Hnh dang mau sau khi t nh H.3.7b.
3- Phan tch ket qua
Qua trnh chu lc cua vat lieu co the chia lam ba giai oan.
OA: giai oan an hoi, quan he P - (L la bac nhat. Lc ln nhat trong giai oan nay la lc t le Ptl, ng suat tng ng trong mau la
Gii han t le
(3.5)
AD: giai oan chay, lc khong tang nhng bien dang tang lien tuc. Lc keo tng ng la lc chay Pch , ng suat trong mau la
Gii han chay
(3.6)
DBC: giai oan cung co (tai ben), quan he P - (L la ng cong. Lc ln nhat la lc ben PB , ng suat ln nhat la
Gii han ben
(3.7)
Goi chieu dai mau sau khi t (H.3.7b) la L1 va dien tch mat cat ngang ni t la A1 ,ta co cac nh ngha ac trng cho tnh deo cua vat lieu nh sau:
Bien dang dai tng oi (tnh bang phan tram):
( =
(3.8)
o that ty oi (tnh bang phan tram):
( = % (3.9)
4- Bieu o ( - ( (bieu o quy c)
T bieu o P - (L de dang suy ra bieu o tng quan gia ng suat va bien dang dai tng oi .
Bieu o nay co hnh dang giong nh bieu o P - (L (H.3.8). Tren bieu o ch ro va ca moun an hoi: = tan(.
Neu ke en s giam dien tch mat cat ngang ta se co bieu o quan he gia va ng suat thc (ng net t).
3. Th nghiem keo vat lieu don (gang)
Bieu o keo vat lieu don co dang ng cong (H.3.9). Vat lieu khong co gii han ty le va gii han chay ma ch co gii han ben.
(3.10)
Tuy vay, ngi ta cung quy c mot gii han an hoi nao o va xem o th quan he lc keo va bien dang la ng thang (ng quy c).
4. Nen vat lieu deo
Mau nen vat lieu deo (va don) (H.3.10a).
Bieu o nen vat lieu deo (H.3.10b). Ch xac nh c gii han ty le
va gii han chay
khong xac nh c gii han ben do s phnh ngang cua mau lam cho dien tch mat cat ngang mau lien tuc tang len.
Sau th nghiem, mau co dang hnh trong (H.3.10c).
5. Nen vat lieu don
Bieu o quan he P - (L tng t bieu o keo vat lieu don (H.3.9).
Ch xac nh c gii han ben tng ng vi lc nen pha hong Pb. Mau th nghiem b v ot ngot, co dang hnh non (H.3.10d).
Cac th nghiem keo va nen cac vat lieu deo va don, cho thay :
Gii han chay cua vat lieu deo khi keo va nen nh nhau.
Vat lieu don co gii han ben khi keo be hn nhieu so vi gii han ben khi nen. Th du, vi gang xam gii han ben khi keo la 2,5 kN/cm2 , con gii han ben khi nen co the at en 10 kN/cm2.
3.6 The nang bien dang an hoi (TNBDH)
1- Khai niem
Xet thanh chu keo lam viec trong giai oan an hoi (H.3.11a).
Lc keo tang dan t 0 en gia tr P, trong qua trnh tang lc, thanh dan ra t t en gia tr (L.
Sau khi at en gia tr P, ta bo lc i, thanh se an hoi hoan toan.
Ngi ta noi cong W cua ngoai lc phat sinh trong qua trnh di chuyen a chuyen hoa thanh the nang bien dang an hoi U tch luy trong thanh va chnh the nang nay lam cho thanh an hoi sau khi khong tac dung lc.
2- Tnh the nang bien dang an hoi
H.3.11b bieu dien quan he gia lc keo P va bien dang (L.
( Tnh cong cua lc P tren chuyen di (L.
Cho P tang len dP, bien dang doc thanh tang len d(L.
( cong cua ngoai lc dW do lc (P + dP) la:
dW = (P + dP)d(L = Pd(L + dPd(L
Bo qua lng be bac cao dPd(L ( dW = Pd(L
Cong nay bieu dien bang dien tch hnh ch nhat gach cheo (H.3.11b).
( cong cua lc keo P khi tang t 0 en P c bieu th tren o th bang dien tch tam giac OAC.
W =
( Cong nay bien thanh the nang bien dang an hoi U:
U = W =
Thay (L = ( U = (3.11)
( Goi u la the nang bien dang an hoi rieng (the nang tch luy trong mot n v the tch), ta co: u =
vi: V = AL va ( u = (3.12)
( Xet oan thanh co chieu dai dz co noi lc Nz (H.3.12):
TNBDH: dU =
( TNBDH cua oan thanh dai L, co noi lc Nz la:
U =
Khi trong oan thanh khong oi, ta co:
U =
(3.13)
Vi nhieu oan dai Li : U=(Ui= ( (3.13)
TNBDH thng dung e tnh chuyen v cua he thanh.
Th du 3.2 Tnh chuyen v ng cua iem at lc .
Cho E = 20000kN/cm2;
L = 2m; P =300kN; ( = 30o ; A= 10 cm2 (H.3.13a)
Giai
Xac nh noi lc
Tach mat A (H.3.13b), dung hai phng trnh hnh chieu:
(X = 0 ( NAB = NAC = N
(Y = 0 ( 2Ncos( = P N =
Chuyen v ng cua iem A
a) Phng phap tnh theo bien dang hnh hoc
Goi (AB, (AC lan lt la bien dang dai cua thanh AB, AC tng ng, cac iem bien dang nay bieu dien bi oan AI, AK (H.3.13b). T I, K ke hai ng vuong goc vi AB va AC, chung cat nhau A, AA chnh la o di chuyen cua iem A.
Trng hp tren, v NAB = NAC nen (AB = (AC va A nam tren ng thang ng ke t A, ( AA chnh la chuyen v can tm.
Xet tam giac AIA ta co:
AAcos( = AI ( AA = =
AA = = = 0,4cm
b) Phng phap dung the nang bien dang an hoi
Ta co: W = U
(*)
trong o:
Cong ngoai lc: W = P.AA
The nang bien dang an hoi cua he:
U = +
a co:
NAB = NAC = N; LAB = LAC = L
(
U = 2
The vao (*) ta c: P.AA = 2
(
AA =
EMBED Equation.3 = = 0,4 cm
3.7 ng suat cho phep - He so an toan - Ba bai toan c ban
( ng suat nguy hiem, ky hieu , la tr so ng suat ma ng vi no vat lieu c xem la b pha hoai. Vat lieu deo
Vat lieu don .
( ng suat cho phep: Khi che tao, vat lieu thng khong ong chat , trong qua trnh s dung tai trong tac dung co the vt qua tai trong thiet ke, ieu kien lam viec cua ket cau hay chi tiet cha c xem xet ay u, cac gia thiet khi tnh toan cha ung vi s lam viec cua ket cau.
V the ta khong tnh toan theo . Chung ta phai chon mot he so an toan n ( 1 e xac nh ng suat cho phep
(3.15)
va dung tr so e tnh toan.
( He so an toan n do nha nc hay hoi ong ky thuat cua nha may quy nh.
e chon he so an toan c chnh xac, nhieu khi ngi ta phai chon nhieu he so theo rieng tng nguyen nhan dan en s khong an toan cua cong trnh hay chi tiet may, co the neu ra:
- He so ke en o ong chat cua vat lieu
- He so ke en s vt qua tai trong thiet ke
- He so ke en s lam viec tam thi hay lau dai
( ieu kien ben : e am bao s lam viec an toan ve o ben khi thanh chu keo (nen) ung tam, ng suat trong thanh phai thoa man ieu kien ben la:
(3.16)
( Ba bai toan c ban: T ieu kien ben, ta co ba bai toan c ban:
Kiem tra ben: Kiem tra xem ng suat trong thanh co thoa ieu kien ben hay khong?
Chon kch thc mat cat ngang: ay la bai toan thiet ke, ta phai nh kch thc mat cat ngang cua thanh sao cho am bao ieu kien ben.
T (3.16) (
nh tai trong cho phep: t (3.16) de dang xac nh c noi lc ln nhat co the at c cua thanh la:
hay :
T co the tm c tr so cho phep cua tai trong tac dung len cong trnh hay chi tiet may.Th du 3.3
Cho ket cau gom hai thanh chu lc nh H.3.14a.
a) Kiem tra ben thanh AB.
b) nh so hieu thep V dung cho thanh BC. Biet: sin( =;
kN/cm2 Giai.
( Noi lc: Tach mat B, (H.3.14b). Dung cac phng trnh can bang:
(Y = 0 ( NAB sin( ( P = 0
( NAB = = 52 kN (keo)
(X = 0 ( ( NBC ( NAB cos( = 0
( NBC = ( 48 kN (nen)
( Kiem tra ben thanh AB
Ta co:
= 13,8 kN/cm2
nen thanh AB am bao o ben.
( Chon so hieu thep cho thanh BC
Ta phai co: ABC ( = = 3,43 cm2
Tra bang thep nh hnh (phu luc 1), chon c thep dung cho BC la:
2V 25 ( 25 ( 4 co: A = 2 ( 1,86 cm2 = 3,72 cm2Th du 3.4 Cho ket cau nh H.3.15a
nh tai trong cho phep [P ] theo ieu kien ben cua cac thanh 1, 2, 3.
Cho biet [( ] = 16 kN/cm2,A1= 2 cm2, A2= 1 cm2, A3 = 2 cm2.
Giai.
( Noi lc trong cac thanh. Cat cac thanh 1, 2, 3, co lap he nh H.3.15b.
Viet cac phng trnh can bang:
(X = 0 ( N2 cos45o + N3 = 0
(Y = 0 ( P + N1 + N2 sin45o = 0
(M/A = 0 ( P2a + N1a = 0
( N1 = 2P; N2 = P (nen); N3 = P
( nh [P ]: Viet ieu kien ben cua cac thanh 1, 2, 3:
= ( ( P ( = = 16 kN
= ( ( P ( = = 11,3 kN
= ( ( P ( A3 = 16.2 = 32 kN
So sanh ta c [P] = 11,3 kN.
3.8 BAI TOAN SIEU TNH
nh ngha. Bai toan sieu tnh la bai toan ma ch vi cac phng trnh can bang tnh hoc se khong u e giai c tat ca cac phan lc hay noi lc trong he.
Cach giai. Can tm them cac phng trnh dien ta ieu kien bien dang (phng trnh bien dang, phng trnh hnh hoc) cua he sao cho cong so phng trnh nay vi cac phng trnh can bang tnh hoc va u bang so an so phan lc, noi lc can tm.
Th du 3.5 Xet thanh chu lc nh H.3.16a.
hai ngam co hai phan lc VA va VB.
Phng trnh can bang:
-VA - VB + P = 0 (a)
Phng trnh nay co hai an, muon giai c can tm them phng trnh ieu kien bien dang cua thanh.
Tng tng bo ngam B va thay bang phan lc VB (H.3.16b).
ieu kien bien dang cua he la:
(L = (BA = (BC + (CA = 0 (b)
Goi NBC va NCA la noi lc trong cac oan BC va CA , ta se c:
(L = + = 0(c)
vi: NBC = VB; NCA = VB - P; (c) tr thanh:
= 0
( VB =
Sau khi tnh c phan lc VB, bai toan tr thanh bai toan tnh nh bnh thng.
Th du 3.6 He gom ba thanh, treo lc P (H.3.17a). Tnh noi lc trong cac thanh treo.
Giai.
( Phng trnh can bang: Tach mat A (H.3.17b)
Ta co hai phng trnh can bang:
(X =NAB sin( +NAD sin( = 0 (a)
(Y = P +NAB cos( + NAC + NAD cos( = 0 (b)
Hai phng trnh, 3 an so noi lc ,can them mot phng trnh bien dang.
( Phng trnh bien dang: Xet he thanh sau khi chu lc.
V oi xng nen iem A di chuyen thang ng en A.
T A ke ng AI va AK lan lt vuong goc vi AB va AD. V bien dang la nho nen goc ABA va ADA la vo cung be va goc BAC va DAC van gi gia tr (, (
Trong tam giac AIB:
v goc (ABA) be ( cos ( 1 ( BI ( BA.
( IA = (AB : o dan dai cua AB; KA = (AD : o dan dai cua AD.
Ngoai ra, AA = (AC : o dan dai cua AC.
Xet tam giac AIA va AKA, ta co : IA = KA = AAcos( (c)
thay: IA =(AB = ; KA = (AD = ; AA = (AC =
vao (c) roi vao (a) va (b) ta c:
NAB = NAD =
NAC =
BAI TAP chng 3
3.1 Ve bieu o lc doc, tnh ng suat trong moi oan. Tnh bien dang toan phan cua thanh.
Cho P = 10 kN; A = 5 cm2; E = 2.104 kN/ cm2 (H.3.1)
3.2 Cho he nh H.3.2. Hay:a) Tnh noi lc trong thanh AC, t o tm ng knh mat cat ngang thanh AC sao cho thanh am bao o ben
b) Gia s thanh AB cng tuyet oi, hay tnh chuyen v cua iem A.
Cho: P = 10 kN ; q = 10 kN/m ; = 16 kN/cm2 ; E = 2.104 kN/cm2
3.3 Cho khoi tuyet oi cng at tren ba thanh 1, 2, 3 trong lng khoi Q = 40 kN; lc ngang q =1kN/m; = 16kN/cm2 (H.3.3)
a) Xac nh noi lc trong cac thanh
b) Chon so hieu mat cat ngang cua cac thanh theo ieu kien ben biet rang ca ba thanh 1, 2, 3 eu cau tao bi hai thep goc eu canh ghep vi nhau.
3.4 Tnh chuyen v ng cua iem at lc P. Cho E = 2.104 kN/cm2 ,
( He thanh H.3.4c co cac thanh AB va GC tuyet oi cng (H.3.4)).
3.5 Ve bieu o lc doc Nz ; Cho P = 10 kN; E = hang (H.3.5).
3.6 nh noi lc trong cac thanh treo, P = 10 kN; q = 10 kN/m; E = hang (H.3.6).
3.7 Xac nh [(] sao cho ng suat trong cac thanh BD va CG khong vt qua ng suat cho phep [(], gia thiet thanh AB tuyet oi cng va cac thanh khac eu co cung loai vat lieu vi moun an hoi E (H3.7).
3.8 Tnh noi lc trong cac thanh treo hoac chong khi nhiet o trong cac thanh tang len (to. Cac dam AB xem nh tuyet oi cng.
Ap dung bang so, cho Eth = 2.104 kN/cm2, (th = 125.107
Eth = 1.104 kN/cm2, ( = 165.107; a = 1m (H.3.8)
Chng 4
TRANG THAI NG SUAT
4.1 Khai niem VE TRANG THAI NG SUAT tai mot iem
4.1.1 TRANG THAI NG SUAT (TTS)
Xet mot iem K trong mot vat the can bang va cac mat cat qua K, tren cac mat cat ay co cac ng suat phap ( va ng suat tiep (. Cac ng suat nay thay oi tuy v tr mat cat (H.4.1).
nh ngha TTS: TTS tai mot iem la tap hp tat ca nhng ng suat tren cac mat i qua iem ay.
TTS tai mot iem ac trng cho mc o chu lc cua vat the tai iem o. Nghien cu TTS la tm ac iem va lien he gia cac ng suat ( , (, xac nh ng suat ln nhat, nho nhat e tnh toan o ben hay giai thch, oan biet dang pha hong cua vat the chu lc.
4.1.2 Bieu dien TTS tai mot iem
Tng tng tach mot phan to hnh hop vo cung be bao quanh iem K. Cac mat phan to song song vi cac truc toa o (H 4.2).
Tren cac mat cua phan to se co chn thanh phan ng suat:
+Ba ng suat phap: (x , (y , (z
+Sau ng suat tiep. (xy , (yx , (xz , (zx ,
(yz , (zy ,
ng suat phap ( co 1 ch so ch phng phap tuyen mat co ( .
ng suat tiep ( co hai ch so: Ch so th nhat ch phng phap tuyen cua mat cat co (, ch so th hai ch phng cua (.
4.1.3 nh luat oi ng cua ng suat tiep
Tren hai mat vuong goc, neu mat nay co ng suat tiep hng vao canh ( hng ra khoi canh ) th mat kia cung co ng suat tiep hng vao canh ( hng ra khoi canh ), tr so hai ng suat bang nhau ( H.4.3)
((xy ( = ((yx (; ((xz(=((zx( ; ((yz ( =((zy ( (4.1)
TTS tai mot iem con 6 thanh phan ng suat4.1.4 Mat chnh, phng chnh va ng suat chnh. Phan loai TTS
Ly thuyet an hoi a chng minh rang tai mot iem bat ky cua vat the chu lc luon tm c mot phan to hnh hop vuong goc ma tren cac mat ch co ng suat phap, ma khong co ng suat tiep (H4.4a).
Nhng mat o goi la mat chnh.
Phap tuyen cua mat chnh goi la phng chnh.
ng suat phap tren mat chnh goi la ng suat chnh va ky hieu la:
(1 , (2 va (3. Quy c: (1 > (2 > (3.
Th du :
(1 = 200 N/cm2;
(2 = (400 N/cm2;
(3 = (500 N/cm2 Phan loai TTS :
- TTS khoi : Ba ng suat chnh khac khong (H.4.4a).
- TTS phang: Hai ng suat chnh khac khong (H.4.4b).
- TTS n: Mot ng suat chnh khac khong (H.4.4c).
TTS khoi va TTS phang goi la TTS phc tap.4.2 TTS TRONG BAI TOAN phang- PHNG PHAP GIAI TCH
4.2.1 Cach bieu dien Quy oc dau
Cach bieu dien:
Xet mot phan to (H.4.5a). ng suat tren mat vuong goc vi truc z bang khong va mat nay la mot mat chnh v co ng suat tiep bang khong.
e de hnh dung, ta bieu dien phan to ang xet bang hnh chieu cua toan phan to len mat phang Kxy (H.4.5b).
Quy c dau: + ( ( 0 khi gay keo ( hng ra ngoai mat cat)
+ ( ( 0 khi lam cho phan to quay thuan kim ong ho
Hnh 4.5b bieu dien cac ng suat ( 0
4.2.2 ng suat tren mat cat nghieng bat ky
Van e: Xac nh ng suat tren mat cat nghieng song song vi truc z va co phap tuyen u tao vi truc x mot goc ( (( > 0 khi quay ngc chieu kim ong ho ke t truc x) (H.4.6a). Gia thiet a biet ng suat (x, (y va (xy.
( Tnh (u va (uv : Tng tng cat phan to bang mat cat nghieng a neu, mat cat chia phan to ra lam hai phan, xet can bang cua mot phan phan to (H.4.6b).
Tren mat nghieng co ng suat (u va (uv , chung c xac nh t phng trnh can bang tnh hoc.
* (U=0 (
* (V=0 (
Ke en: ((xy ( = ((yx (; dx = ds sin( ; dy = ds cos(,
(
(4.2a)
(4.2b)
( Tnh (v : Xet mat nghieng co phap tuyen v, vuong goc mat co phap tuyen u (H.4.7). Thay the ( bang (( + 90() vao (4.2a) ,
( ng suat phap tac dung tren mat co phap tuyen v:
(4.3)
Tong (4.2a) va (4.3), (
(4.4)
Bieu thc tren cho thay, tong ng suat phap tac dung tren hai mat vuong goc cua phan to ng suat phang tai mot iem la hang so va khong phu thuoc vao goc (.
o la Bat Bien Th Nhat cua ng suat phap
Th du 4.1 Thanh co dien tch 5 cm2, chu keo vi lc P = 40 kN. Xac nh ng suat tren mat cat nghieng mot goc 30o vi mat cat ngang (H.4.8).
Giai
ng suat phap tren mat cat ngang (Chng 3)
Tach phan to hnh hop bao iem K nam tren mat cat ngang.
Ta co:
Mat cat nghieng co phap tuyen hp vi truc vi truc x (truc thanh). mot goc 30o .
T (4.2) (
4.2.3 ng suat chnh - Phng chnh - ng suat phap cc tr
1- ng suat chnh - phng chnh
Ngoai mat chnh la mat a biet vuong goc vi truc z, hai mat chnh con lai la nhng mat song song vi truc z (v phai vuong goc vi mat chnh a co).
Mat chnh la mat co ng suat tiep = 0 ( Tm hai mat chnh con lai bang cach cho =0
Neu goi la goc cua truc x hp vi phng chnh th ieu kien e tm phng chnh la: =0 (
( Phng trnh xac nh (0 :
(4.5)
(4.5) cho thay co hai gia tr (0 sai biet nhau 90(. V vay, co hai mat chnh vuong goc vi nhau va song song vi truc z. Tren moi mat chnh co mot ng suat chnh tac dung.
Hai ng suat chnh nay cung la ng suat phap cc tr (ky hieu la (max hay (min ) bi v
giong vi (4.5)
Gia tr ng suat chnh hay ng suat phap cc tr co the tnh c bang cach the ngc tr so cua ( trong (4.5) vao (4.2a).
e y rang:
(
(4.6)
Ta lai thay ( max + ( min = ( 1 + ( 3 = ( x + ( y
Th du 4.2 Tm ng suat chnh va phng chnh cua TTS (H.4.10a). n v cua ng suat la kN/cm2.
Giai
Theo quy c dau, ta co:
EMBED Equation.3 Phng chnh xac nh t (4.5):
(
(
(i)
Co 2 phng chnh ( 2 mat chnh) vuong goc nhau
Cac ng suat chnh c xac nh t (4.6):
(ii)
e xac nh mat chnh nao t (i) co ng suat chnh (ii) tac dung, ta dung (4.2b), chang han vi , ta co:
Vay : (1 = 4,41 kN/cm2 ng vi goc nghieng ,
(2 = 1,58 kN/cm2 tac dung tren mat co .
Cac mat va ng suat chnh bieu dien tren phan to H.4.10b.
2- ng suat tiep cc tr
Tm ng suat tiep cc tr va mat nghieng tren o co ng suat tiep cc tr bang cach cho
(4.7)
(
(4.7)
So sanh (4.7) vi (4.5) ( (4.8)
( hay (
Mat co ng suat tiep cc tr hp vi nhng mat chnh mot goc 45(.
The (4.8) vao (4.2b), ta c :
(4.9)
4.2.4 Cac trng hp ac biet
1- TTS phang ac biet
Phan to tren H.4.12 co:
T (4.6) (
(4.10)
Phan to co 2 ng suat chnh ( se gap trng hp thanh chu uon ).
2- TTS trt thuan tuy (H.4.13) ay, ;Thay vao (4.6)
( hay (4.11)
Hai phng chnh c xac nh theo (4.5):
( (4.12)
Nhng phng chnh xien goc 45o vi truc x va y.
3- Trng hp phan to chnh (H.4.14)
Phan to chnh ch co ( 1 , ( 3 ,( = 0;
Thay vao (4.9), ta c: (4.13)
4.3 TTS TRONG BAI TOAN PHANG- PHNG PHAP O TH
1- Vong tron Mohr ng suat
Cong thc xac nh ng suat tren mat cat nghieng (4.2) co the bieu dien di dang hnh hoc bang vong tron Mohr. e ve vong tron Mohr, ta sap xep lai (4.2) nh sau:
(4.14)
(4.14)
Bnh phng ca hai ve cua hai ang thc tren roi cong lai, ta c:
(4.15)
at:
(4.16)
(4.15) thanh: (4.17)
Trong he truc toa o, vi truc hoanh ( va truc tung (, (4.17) la phng trnh cua mot ng tron co tam nam tren truc hoanh vi hoanh o la c va co ban knh R . Nh vay, cac gia tr ng suat phap va ng suat tiep tren tat ca cac mat song song vi truc z cua phan to eu bieu th bang toa o nhng iem tren vong tron. Ta goi vong tron bieu th TTS cua phan to la vong tron ng suat hay vong tron Mohr ng suat cua phan to.
Cach ve vong tron: (H.4.16):
- nh he truc toa o : truc hoanh ( // truc x, truc tung ( // truc y cua phan to va hng len tren.
-Tren truc ( nh iem E((x, 0) va iem F((y, 0)
Tam C la trung iem cua EF
- nh iem cc P ((y, (xy ) .
- Vong tron tam C, qua P la vong tron Mohr can ve
Chng minh: + C la trung iem cua EF (
Trong tam giac vuong CPF:
Do o (
2- ng suat tren mat cat nghieng
Dung vong tron Mohr e tm ng suat tren mat cat nghieng cua phan to co phap tuyen u hp vi truc x mot goc (.
Cach tm (u ; (uv
Ve vong tron Mohr nh H.4.17.
T cc P ve tia Pu // vi phng u cat vong tron tai iem M.
Hoanh o cua M = (u ; Tung o cua M = (uvChng minh:
Ky hieu 2(1 la goc (CA,CD), 2( la goc (CD,CM).
Hnh 4.17 cho:
nhng:
nen:
Tng t, ta co:
Ta nhan lai c phng trnh (4.2)
3- nh ng suat chnh- phng chnh- ng suat phap cc tr
Tren vong tron ng suat ( H.4.17)
iem A co hoanh o ln nhat, tung o = 0( (max = ; ( =0
Tia PA bieu dien mot phng chnh.
iem B co hoanh o nho nhat, tung o = 0( (min = ; ( =0
Tia PB bieu dien phng chnh th hai.
4- nh ng suat tiep cc tr
Tren vong tron (H.4.17): hai iem I va J la nhng iem co tung o ( ln va nho nhat. Do o, tia PI va PJ xac nh phap tuyen cua nhng mat tren o co ng suat tiep cc ai va cc tieu. Nhng mat nay tao vi nhng mat chnh mot goc 45o.
ng suat tiep cc tr co tr so bang ban knh ng tron.
ng suat phap tren mat co ng suat tiep cc tr co gia tr bang hoanh o iem C, tc la gia tr trung bnh cua ng suat phap:
5- Cac trng hp ac biet
- TTS phang ac biet Phan to co hai ng suat chnh 1 va 3 (H.4.18).
- TTS trt thuan tuy
Phan to co 2 ng suat chnh:
Cac phng chnh xien goc 45o vi truc x va y (H.4.19)
- TTS chnh ( H.4.20)
Th du 4.3 Phan to TTS phang (H.4.21),cac ng suat tnh theo kN/cm2. Dung vong tron Mohr, xac nh:
a) ng suat tren mat cat nghieng
b) ng suat chnh va phng chnh
c) ng suat tiep cc tr.
Giai.
Theo quy c ta co:
(Tam vong tron C.
( Cc P(1, + 4). T P ve tia song song vi truc u cat vong tron Mohr tai M. Toa o iem M bieu th ng suat tren mat cat nghieng vi :
(Hoanh o A va B bieu th ng suat chnh co gia tr bang:
Hai phng chnh xac nh bi goc (o:
(Tung o I va J co gia tr bang ng suat tiep cc tr:
Cac ng suat nay tac dung len cac mat, tng ng vi cac goc nghieng:
4.3 BIEU DIEN HNH HOC TTS khoi
( Tong quat, TTS tai mot iem la TTS khoi (H.4.22).
( Xet nhng mat // mot phng chnh ( th du phng III) , ng suat chnh (3 khong anh hng en (, ( tren cac mat nay (H.4.23). ( co the nghien cu ng suat tren nhng mat nay tng t TTS phang.
Ve vong tron ng suat bieu dien cac ng suat tren mat nghieng nay (vong tron so 3 tren H.4.24) .
T vong tron nay, ta thay tren nhng mat song song vi phng chnh III co mat co ng suat tiep cc ai (ky hieu (max,3)
( Tng t, oi vi nhng mat song song vi phng chnh th I va th II, ta cung ve c cac vong tron ng suat (Vong tron so 1 va vong tron so 2) (H.4.24).
( Ly thuyet an hoi a chng minh rang gia tr cua ( va ( tren mot mat bat ky cua mot phan to trong TTS khoi co the bieu th bang toa o cua mot iem nam trong mien gach cheo ( H.4.24 ).
( Qua hnh ve, ng suat tiep ln nhat cua phan to bieu th bang ban knh cua vong tron ln nhat, (H.4.24).
(4.18)
4.4 Lien he ng suat va bien dang
4.4.1 nh luat Hooke tong quat
1- Lien he ng suat phap va bien dang dai
(TTS n: trong chng 3, a co:
nh luat Hooke lien he gia ng suat phap va bien dang dai : (4.19)
- bien dang dai tng oi theo phng
Theo phng vuong goc vi cung co bien dang dai tng oi ngc dau vi : (4.20)
( TTS khoi: vi cac ng suat chnh theo ba phng chnh I, II, III (H.4.25). Tm bien dang dai tng oi theo phng I .
Bien dang dai theo phng I do gay ra:
Bien dang dai theo phng I do gay ra:
Bien dang dai theo phng I do gay ra:
Bien dang dai tng oi theo phng I do ca ba ng suat sinh ra se la tong cua ba bien dang tren:
(4.21)
Tng t, bien dang dai tng oi theo hai phng chnh II , III con lai:
(4.22)
(4.23)
( TTS tong quat: Ly thuyet an hoi a chng minh oi vi vat lieu an hoi ang hng, ( ch sinh ra bien dang dai ma khong sinh ra bien dang goc , ( ch sinh ra bien dang goc ma khong sinh ra bien dang dai.
( Trong trng hp phan to TTS tong quat, van co
(4.24)
2-Lien he gia ng suat tiep va bien dang goc
( nh luat Hooke ve trt)
Phan to TTS trt thuan tuy (H.4.26). Bien dang goc (goc trt) ( bieu th o thay oi goc vuong.
nh luat Hooke ve trt:
(4.25)
trong o: G - la moun an hoi trt. Th nguyen cua G la [lc/(chieu dai)2] va n v thng dung la N/m2 hay MN/m2.
Lien he gia E, ( va G nh sau: (4.26)
4.4.2 nh luat Hooke khoi
Tnh o bien oi the tch cua mot phan to hnh hop co cac canh bang da1, da2 va da3 .
The tch cua phan to trc bien dang la:
Sau bien dang, phan to co the tch la:
Goi bien dang the tch tng oi la (, ta co:
(4.27)
The (4.21)(4.22),(4.23) vao (4.27) (
(4.28)
at tong ng suat phap la:
(4.28) thanh:
(4.29)
cong thc (4.29) c goi la nh luat Hooke khoi bieu th quan he tuyen tnh gia bien dang the tch tng oi va tong ng suat phap.
Nhan xet:(T (4.29), neu vat lieu co he so Poisson ( = 0,5 ( cao su), th luon luon bang khong tc la the tch khong oi di tac dung cua ngoai lc.
( Cong thc tren cho thay phu thuoc vao tong ng suat phap ch khong phu thuoc vao rieng tng ng suat phap. Nh vay, neu cung vi phan to ay ta thay cac ng suat chnh bang mot ng suat trung bnh tb co gia tr bang trung bnh cong cua ba ng suat chnh noi tren:
th bien dang the tch tng oi cua phan to tren van khong thay oi.
That vay, vi nhng ng suat chnh la (tb , bien dang the tch bang:
Ket qua tren co y ngha nh sau: vi phan to ban au la hnh lap phng, trong hai trng hp tren ta thay the tch phan to eu bien oi nh nhau.
- Tuy nhien, trong trng hp au khi cac ng suat chnh khac nhau, phan to va bien oi the tch va bien oi hnh dang tc la tr thanh phan to hnh hop ch nhat sau khi bien dang.
- Con trong trng hp th hai, khi thay cac ng suat chnh bang ng suat trung bnh, phan to ch bien oi ve the tch ma khong bien oi hnh dang, ngha la sau khi bien dang phan to van gi hnh lap phng.
- Ve mat ly luan, co the phan phan to TTUS khoi chu cac ng suat chnh (1 , (2 , (3 thanh 2 phan to (H. 4.28). Phan to b) ch bien oi the tch, phan to c) ch bien oi hnh dang.
4.5 The nang bien dang an hoi
( chng 3, phan to TTS n (thanh b keo hoac nen):
The nang bien dang an hoi rieng (4.30)
( Trong TTS khoi, s dung nguyen ly oc lap tac dung, ta co the nang bien dang an hoi rieng bang:
(4.31)
thay theo nh luat Hooke trong (4.21) - (4.23) vao , (
hay
(4.32)
Ta co the phan tch the nang bien dang an hoi u thanh hai thanh phan:
-Thanh phan lam oi the tch goi la the nang bien oi the tch utt-Thanh phan lam oi hnh dang goi la the nang bien oi hnh dang uhdTa co:
u = utt + uhd
e tnh the nang bien oi hnh dang, ta thay cac ng suat va bang ng suat ((1 -(tb ), ((2 -(tb ), ((3 -(tb ), tac dung len cac mat phan to.
The vao (4.32) ta co the nang bien oi hnh dang bang:
hay :
(4.33)
( TTS n , thay 0 vao (4.32) va (4.33), ta c the nang rieng va the nang bien oi hnh dang nh sau:
(4.34)BAI TAP chng 4
4.1 Tm gia tr ng suat phap va ng suat tiep tren mat cat AB cua phan to nh tren H.4.1 bang phng phap giai tch va o th. n v ng suat tnh bang kN/cm2.
4.2 Tren hai mat tao vi nhau mot goc ( = 60o va i qua mot iem TTS phang co cac ng suat nh tren H.4.2. Hay tnh cac ng suat chnh tai iem o, ng suat phap (u va bien dang tng oi (u theo phng u. Cho: E = 2.10 kN/cm2; (= 0,3.4.3 Tren mat cat m - n i qua mot iem trong vat the TTS phang co ng suat toan phan p = 3000 N/cm2, ng suat nay co phng tao thanh goc 60o vi mat cat. Tren mat vuong goc vi mat cat o ch co ng suat tiep (H.4.3).
Tnh ng suat phap va ng suat tiep tren mat cat hp vi mat cat m - n mot goc 45o. Tnh ng suat phap ln nhat tai iem o.
4.4 Tai mot iem tren be mat cua vat the, ng suat tac dung len phan to nghieng mot goc 30o vi truc x co tr so va hng nh tren H.4.30.
a) Xac nh ng suat chnh va phng chnh.
b) Xac nh ng suat tiep cc tr va ng suat phap tren be mat co ng suat tiep cc tr. Bieu dien cac ng suat o tren H.4.4.
4.5 Mot tam mong co kch thc nh tren H.4.5 chu tac dung cua ng suat keo ( = 30 kN/cm2 theo phng chieu dai cua tam va ng suat tiep ( = 15 kN/cm2.
a) Xac nh ng suat phap theo phng ng cheo mn va phng vuong goc vi ng cheo
b) Tnh bien dang dai tuyet oi cua ng cheo mn.
Cho E = 2.104 kN/cm2, (= 0,3.
4.6 Mot tam thep mong hnh ch nhat chu ng suat phap phan bo eu (x va (y nh tren H.4.6. Cac tam ien tr A va B c gan len tam theo hai phng x va y cho cac so o nh sau: (x = 4,8.104 va (y = 1,3.104. Tnh (x va (y, biet E = 2.104 kN/cm2; (= 0,3.
4.7 Tai mot iem tren mat vat the chu lc, ngi ta gan cac tam ien tr A, B, C e o bien dang ty oi theo cac phng Om, On va Ou (H.4.7). Cac so o thu c:
Xac nh ng suat chnh, phng chnh tai iem o.
Cho : E = 2.104 kN/cm2 ; (= 0,3.
4.8 Tai iem A cua mot dam cau co gan hai tenxomet e o bien dang theo phng nam ngang va phng thang ng (H.4.8). Khi xe chay qua cau, ngi ta o c: (x = 0,0004; (y = 0,00012.Tnh ng suat phap theo phng doc va phng thang ng cua dam. Cho biet E = 2.104 kN/cm2; (= 0,3.
4.9 Co mot phan to hnh hop co cac canh: a = 2cm;
b = 4 cm; c = 2 cm, chu tac dung cua cac lc P1, P2 tren bon mat cua phan to (xem H.4.9). Cho : P1 = 60 kN; P2 = 120 kN; E = 2.104 kN/cm2; (= 0,3.
a) Xac nh cac bien dang dai (a, (b, (c cua cac canh a, b, c va bien oi the tch cua phan to hnh hop.
b) Muon bien oi the tch (V = 0 th phai at them lc phap tuyen P3 bang bao nhieu vao hai mat con lai?
Tnh (max trong trng hp nay.
4.10 Mot khoi hnh hop lam bang thep co kch thc cho tren H.4.10, c at gia hai tam cng tuyet oi, chu lc nen P = 250 kN. Tnh lc tac dung tng ho gia mat tiep xuc cua hnh hop vi cac tam cng. Cho (= 0,3.
.
4.11 Mot khoi lap phng bang be tong at va kht ranh cua vat the A chu ap suat phan bo eu mat tren P = 1 kN/cm2 (H.4.11).
Xac nh ap lc nen vao vach ranh va o bien dang the tch tuyet oi. Cho canh a = 5 cm; E = 8.102 kN/cm2; (= 0,36.
. Vat the A coi nh cng tuyet oi.
4.12 Mot tam thep kch thc a ( b ( c at gia hai tam tuyet oi cng, hai tam nay c lien ket vi nhau bang bon thanh nh H.4.12. Khi tam thep chu ap lc p phan bo tren hai mat ben th ng suat keo cua thanh la bao nhieu? Tnh ng suat chnh trong tam thep. Cho Etam = Ethanh va dien tch F cua thanh.
Chng 5
LY THUYET BEN
5.1 KHAI NIEM VE LYTHUYET BEN
( ieu kien ben thanh chu keo hoac nen ung tam ( chng 3),
( TTS n) :
trong o,
;
ng suat nguy hiem (0 co c t nhng th nghiem keo (nen) ung tam:
- oi vi vat lieu deo la gii han chay ch
- oi vi vat lieu don la gii han ben b.
( e viet ieu kien ben mot iem cua vat the TTS phc tap (phang hay khoi), can phai co ket qua th nghiem pha hong nhng mau th TTS tng t. Viec thc hien nhng th nghiem nh the rat kho khan v:
- ng suat nguy hiem phu thuoc vao o ln cua cac ng suat chnh va phu thuoc vao t le gia nhng ng suat nay. Do o phai thc hien mot so lng rat ln cac th nghiem mi ap ng c t le gia cac ng suat chnh co the gap trong thc te
- Th nghiem keo, nen theo ba chieu can nhng thiet b phc tap, khong pho bien rong rai nh th nghiem keo nen mot chieu
V vay, khong the can c vao th nghiem trc tiep ma phai da tren cac gia thiet ve nguyen nhan gay ra pha hong cua vat lieu hay con goi la nhng thuyet ben e anh gia o ben cua vat lieu.
nh ngha :Thuyet ben la nhng gia thuyet ve nguyen nhan pha hoai cua vat lieu, nh o anh gia c o ben cua vat lieu moi TTS khi ch biet o ben cua vat lieu TTS n ( do th nghiem keo, nen ung tam).
Ngha la, vi phan to TTS bat ky co cac ng suat chnh 3, ta phai tm ng suat tnh theo thuyet ben la mot ham cua 3 roi so sanh vi hay TTS n.
( ieu kien ben cua vat lieu co the bieu dien di dang tong quat nh sau: ( hay )
(t , (t c goi la ng suat tnh hay ng suat tng ng. Van e la phai xac nh ham f hay la tm c thuyet ben tng ng.
5.2 CAC THUYET BEN (TB) C BAN
1- Thuyet ben ng suat phap ln nhat (TB 1)
( Nguyen nhan vat lieu b pha hong la do ng suat phap ln nhat cua phan to TTS phc tap at en ng suat nguy hiem TTS n.
( Neu ky hieu:
(1 , (2 , (3 : ng suat chnh cua TTS phc tap
0khay 0n - ng suat nguy hiem ve keo va nen
n - he so an toan
( ieu kien ben theo TB 1:
(5.1a)
(5.1b)
trong o: t1 - la ng suat tnh hay ng suat tng ng theo TB 1
( u khuyet iem: TB 1, trong nhieu trng hp, khong phu hp vi thc te. Th du trong th nghiem mau th chu ap lc giong nhau theo ba phng (ap lc thuy tnh), du ap lc ln, vat lieu hau nh khong b pha hoai. Nhng theo TB 1 th vat lieu se b pha hong khi ap lc at ti gii han ben cua trng hp nen theo mot phng.
TB 1 khong ke en anh hng cua cac ng suat khac cho nen TB nay ch ung oi vi TTS n.
2- Thuyet ben bien dang dai tng oi ln nhat (TB 2)
( Nguyen nhan vat lieu b pha hong la do bien dang dai tng oi ln nhat cua phan to TTS phc tap at en bien dang dai tng oi ln nhat trang thai nguy hiem cua phan to TTS n.
( Goi 1 : bien dang dai tng oi ln nhat cua phan to TTS phc tap
0k : bien dang dai tng oi trang thai nguy hiem cua phan to b keo theo mot phng ( TTS n).
Theo nh luat Hooke, ta co:
(a)
(b)
Ket hp (a) va (b), ke en he so an toan n
( ieu kien ben theo TB 2:
(c)
hay
(5.2a)
oi vi trng hp bien dang co ngan, ta co
(5.2b)
( u khuyet iem: TB bien dang dai tng oi tien bo hn so vi TB ng suat phap v co ke en anh hng cua ca ba ng suat chnh. Thc nghiem cho thay TB nay ch phu hp vi vat lieu don va ngay nay t c dung trong thc te.3- Thuyet ben ng suat tiep ln nhat (TB 3)( Nguyen nhan vat lieu b pha hong la do ng suat tiep ln nhat cua phan to TTS phc tap at en ng suat tiep ln nhat trang thai nguy hiem cua phan to TTS n.
( Goi: max - ng suat tiep ln nhat cua phan to TTS phc tap ;
0k - ng suat tiep ln nhat trang thai nguy hiem cua phan to b keo theo mot phng ( TTS n).
n He so an toan
( ieu kien ben theo TB 3:
(d)
trong o, theo (4.18), chng 4, ta co:
(e)
(e) vao (d), (
( ieu kien ben theo TB 3:
(5.3)
( u khuyet iem: TB ng suat tiep ln nhat phu hp vi thc nghiem hn nhieu so vi hai TB 1 va TB 2 . Tuy khong ke ti anh hng cua ng suat chnh 2 song TB nay to ra kha thch hp vi vat lieu deo va ngay nay c s dung nhieu trong tnh toan c kh va xay dng. No cung phu hp vi ket qua mau th chu ap lc theo ba phng.
4- Thuyet ben the nang bien oi hnh dang (TB 4)( Nguyen nhan vat lieu b pha hong la do the nang bien oi hnh dang cua phan to TTS phc tap at en the nang bien oi hnh dang trang thai nguy hiem cua phan to TTS n.
( Goi: uhd - The nang bien oi hnh dang cua phan to TTS phc tap
(uhd)o - The nang bien oi hnh dang trang thai nguy hiem cua phan to b keo theo mot phng ( TTS n).
n He so an toan
( ieu kien e phan to TTS phc tap khong b pha hong la ben theo TB 4 la:
uhd ( (uhd)o
(g)
Theo 4.5 ,chng 4, ta a co:
(h)
The (h) vao (g) , lay can bac hai cua hai ve , ke en he so an toan n
( ieu kien ben theo TB 4:
hay la:
(5.4)
trong o: t4 - la ng suat tng ng theo thuyet ben th t.
( u khuyet iem: TB the nang bien oi hnh dang c dung pho bien trong ky thuat v kha phu hp vi vat lieu deo. Ngay nay c s dung nhieu trong tnh toan c kh va xay dng .
CAC KET QUA AC BIET:
1- TTS phang ac biet (H.5.3):
Cac ng suat chnh :
Theo TB ng suat tiep (5.3):
(5.5)
Theo TB the nang bien oi hnh dang (5.4):
hay:
(5.6)
2- TTS trt thuan tuy (H.5.4):
Cac ng suat chnh :
Theo TB ng suat tiep:
hay:
(5.7)
Theo TB the nang bien oi hnh dang:
hay:
(5.8)
5- Thuyet ben ve cac TTS gii han (TB 5 hay la TB Mohr)
TB Mohr c xay dng tren c s cac ket qua thc nghiem, khac vi cac TB trc xay dng tren c s cac gia thuyet.
chng 4, ta a biet mot TTS khoi vi ba ng suat chnh va co the bieu dien bang ba vong tron Morh 1, 2 va 3 vi ng knh tng ng la va nh Hnh.4.22. Neu vat lieu trang thai nguy hiem th nhng vong tron tng ng vi TTS nguy hiem c goi la nhng vong tron Mohr gii han. Thc nghiem cho thay, ng suat phap t anh hng en s pha hoai cua vat lieu nen ta ch e y en vong tron Mohr ln nhat goi la vong tron chnh xac nh bi ng knh Tien hanh th nghiem cho cac TTS khac nhau va tm trang thai gii han tng ng cua chung, tren mat phang toa o ta ve c mot ho cac ng tron chnh gii han nh H.5.5. Neu ve ng bao nhng vong tron o ta se thu c mot ng cong gii han, ng cong nay cat truc hoanh iem tng ng vi trang thai co ba ng suat chnh la ng suat keo co gia tr bang nhau. Gia thiet rang ng bao la duy nhat oi vi moi loai vat lieu, ta nhan thay neu TTS nao bieu th bang mot vong tron chnh nam trong ng bao th vat lieu am bao ben, vong tron chnh tiep xuc vi ng bao th TTS o gii han ben con neu vong tron chnh cat qua ng bao th vat lieu b pha hong.
Viec phai thc hien mot so lng ln cac th nghiem e xac nh cac vong tron gii han va ve chnh xac ng cong gii han la khong n gian.V vay, ngi ta thng ve gan ung ng bao bang cach da tren c s hai vong tron gii han keo va nen theo mot phng vi ng knh tng ng la [k va n. ay, e cho tien ta thay the cac ng suat nguy hiem va n bang ky hieu ng suat cho phep [k va [n tc la a co ke ti he so an toan. ng bao c thay the bang ng thang tiep xuc vi hai vong tron gii han nh tren H.5.6.
Xet mot TTS khoi co vong tron Mohr ln nhat va 3 tiep xuc vi ng bao, nam gii han ve o ben. Tren H.5.7, vong tron nay c ve bang ng net t. Sau ay, ta thiet lap lien he gia nhng ng suat chnh va 3 vi cac ng suat cho phep [k va [n. T hnh ve ta co ty le thc:
Thay the cac tr so:
vao ty le thc tren, ta nhan c ieu kien gii han:
hoac:
Nh vay, ieu kien ben theo TB Mohr (TB 5) c viet la:
(5.9a)
vi he so:
(5.9b)
Tuy bo qua anh hng cua ng suat chnh (2 va n gian hoa ng cong gii han thanh ng thang, thuyet ben Mohr co u iem hn nhng thuyet ben tren v no khong da vao gia thuyet nao ma can c trc tiep vao trang thai gii han cua vat lieu. Thc te cho thay TB nay phu hp vi vat lieu don, tuy nhien no cho ket qua chnh xac ch khi vong tron gii han cua TTS ang xet nam trong khoang hai vong tron gii han keo va nen.
5.3 VIEC AP DUNG CAC TB
Tren ay la nhng TB c dung tng oi pho bien. Viec ap dung TB nay hay TB khac e giai quyet bai toan cu the phu thuoc vao loai vat lieu s dung va TTS cua iem kiem tra.
oi vi TTS n, ngi ta dung TB 1 e kiem tra o ben.
oi vi TTS phc tap, neu la vat lieu don, ngi ta thng dung TB 5 (TB Mohr) hay TB 2, neu la vat lieu deo ngi ta dung TB 3 hay TB 4.
Hien nay, co nhieu TB mi c xay dng, tong quat hn va phu hp hn vi ket qua thc nghiem. Tuy vay, nhng TB nay cung co nhng nhc iem nhat nh nen cha c s dung rong rai.
Th du: Kiem tra ben phan to vat the TTS khoi nh tren H.5.8. ng suat cho theo kN/cm2. Cho biet: .
Giai.
Chon he toa o nh tren H.5.8.
Theo quy c ta co:
(x = -5 kN/cm2 , (y = 6 kN/cm2 , (zy = -(yz = 4 kN/cm2 (z =0 , (xz = (zx =(yx = (xy =0
Mat vuong goc vi truc x la mat chnh vi ng suat chnh . Hai ng suat chnh con lai nam trong mat phang vuong goc vi ng suat chnh a cho va co gia tr bang:
Do o:
Theo TB ng suat tiep:
Theo TB the nang bien oi hnh dang:
Nh vay, theo ca hai TB phan to nay am bao ben.
BAI TAP chng 5
5.1 Khi nen vat lieu theo ba phng cung vi tr so ng suat phap (H.5.1), ngi ta thay vat lieu khong b pha hoai. Hay kiem tra ben oi vi phan to tren bang TB ng suat tiep ln nhat va TB the nang bien oi hnh dang ln nhat.
5.2 Dung TB ng suat tiep ln nhat e tnh ap lc p ln nhat tac dung tren khoi thep tren H.5.2. Khoi thep o c at kht vao trong khoi thep ln.
Cho E = 2.107 N/cm2; ( = 0,28; [( ] = 16 kN/cm2.
5.3 Cho TTS nh H.5.3. Tnh ng suat tng ng (ve trai cua cong thc kiem tra ben) theo TB the nang bien oi hnh dang va TB Mohr. Cho (ok/(on = 0,25.
5.4 Cho TTS tai mot iem cua vat the chu lc nh H.5.4:
(1 = 20 kN/cm2; (2 = 40 kN/cm2; (3 = 80 kN/cm2Kiem tra o ben theo TB 3 va TB 4.
Biet [(] = 120 kN/cm2.
5.5 Mot tru tron bang thep ((= 0,3) at kht gia hai tng cng nh H.5.5. Phan gia cua tru chu ap lc p phan bo eu. Tnh ng suat tng ng theo TB 4 phan gia va phan au cua hnh tru.Chng 6
AC TRNG HNH HOC CUA MAT CAT NGANG
6.1 Khai niem
chng 3, khi tnh o ben cua thanh chu keo (nen) ung tam, ta thay ng suat trong thanh ch phu thuoc vao o ln cua dien tch mat cat ngang A (mat cat F, dien tch A). Trong nhng trng hp khac, nh thanh chu uon, xoan th ng suat trong thanh khong ch phu thuoc vao dien tch F ma con phu thuoc vao hnh dang, cach bo tr mat cat ngha la vao nhng yeu to khac ma ngi ta goi chung la ac trng hnh hoc cua mat cat ngang.
Xet thanh chu uon trong hai trng hp mat cat at khac nhau nh tren H.6.1. Bang trc giac, de dang nhan thay trng hp a), thanh chu lc tot hn trng hp b), tuy rang trong trong hai trng hp dien tch cua mat cat ngang thanh van nh nhau. Nh vay, kha nang chu lc cua thanh con phu thuoc vao cach sap at va v tr mat cat ngang oi vi phng tac dung cua lc.
Cho nen, ngoai dien tch A can phai nghien cu cac ac trng hnh hoc khac cua mat cat ngang e tnh toan o ben, o cng, o on nh va thiet ke mat cat cua thanh cho hp ly.
6.2 Momen tnh - Trong tam
Xet mot hnh phang bieu dien mat cat ngang A ( mat cat A ) nh tren H.6.2. Lap mot he toa o vuong goc Oxy trong mat phang cua mat cat. M(x,y) la mot iem bat ky tren hnh. Lay chung quanh M mot dien tch vi phan dA
( Momen tnh cua mat cat A oi vi truc x (hay y) la tch phan:
(6.1)
v x, y co the am hoac dngnen momen tnh co the co tr so am hoac dng.
Th nguyen cua momen tnh la [(chieu dai)3].
( Truc trung tam x0 ,y0 l truc co momen tnh cua mat cat A oi vi cc truc o bang khong.
( Trong tam C la giao iem cua hai truc trung tam.
( Momen tnh oi vi mot truc i qua trong tam bang khong.
( Cach xac nh trong tam C cua mat cat A:
Dng he truc song song vi he truc xOy ban au (H.6.2). Ta co
Thay vao (6.1), (
v truc la truc trung tam nen ,(
(6.2)
Tng t, ta co:
(6.3)
T (6.2) va (6.3), (
(6.4)
Ket luan: Toa o trong tam c xac nh trong he truc xOy ban au theo momen tnh Sx , Sy va dien tch A theo (6.4).
Ngc lai, neu biet trc toa o trong tam, co the s dung (6.2), (6.3) e xac nh cac momen tnh.
Nhan xet 1:
Mat cat co truc oi xng, trong tam nam tren truc nay v momen tnh oi vi truc oi xng bang khong (H.6.3a,b).
Mat cat co hai truc oi xng, trong tam nam giao iem hai truc oi xng (H.6.3c).
Thc te, co the gap nhng mat cat ngang co hnh dang phc tap c ghep t nhieu hnh n gian.
Tnh chat: momen tnh cua hnh phc tap bang tong momen tnh cua cac hnh n gian.
Vi nhng hnh n gian nh ch nhat, tron, tam giac hoac mat cat cac loai thep nh hnh I, U, V, L ta a biet trc (hoac co the tra theo cac bang trong phan phu luc ) dien tch, v tr trong tam (De dang tnh c momen tnh cua hnh phc tap gom n hnh n gian:
(6.5)
trong o: - dien tch va toa o trong tam cua hnh n gian th i, n - so hnh n gian.
( Toa o trong tam cua mot hnh phc tap trong he toa o xy.
; (6.6)
Th du 6-1 Xac nh trong tam mat cat ch L ch gom hai hnh ch nhat nh tren H.6.4. Toa o trong tam C cua hnh tren la:
EMBED Equation.3 Th du 6.2 Mot mat cat thanh ghep, gom thep ch ( so hieu No55, thep ch [ so hieu No27, va thep ch nhat 15 ( 1,2 cm (H.6.5). Xac nh trong tam C cua mat cat.
Giai.
Tra bang ((OCT 8239-89) ( so lieu sau:
- oi vi thep ch ( No55:
h2 = 55 cm
t = 1,65 cm
A2 = 118 cm2- oi vi thep ch [ No27:
h3 = 27 cm
A3 = 35,2 cm2
z3 = 2,47 cm
- Hnh ch nhat:
A1 = 15 cm x 1,2 cm = 18 cm2Chon he truc toa o xy qua goc C2 ( toa o trong tam cua ba hnh tren la:
; ;
Dien tch va momen tnh cua toan mat cat la:
A = A1 + A2 + A3 = 18 + 118 + 35,2 = 171,2 cm2
v y la truc oi xng, trong tam C se nam tren truc nay.
( Toa o iem C la:
Dau () cho thay trong tam C nam pha di truc x.
Chu y rang, truc x co the chon tuy y song th du nay ta at truc x i qua trong tam C2 cua mat cat ch ( cho tien tnh toan.
6.3 momen quan tnh- HE TRUC QUAN TNH CHNH TRUNG TAM
1- Momen quan tnh (MMQT)
(Momen quan tnh oc cc
( MMQT oi vi iem) cua mat cat A oi vi iem O c nh ngha la bieu thc tch phan:
(6.7)
vi :- khoang cach t iem M en goc toa o O,
(Momen quan tnh oi vi truc y va x cua mat cat A c nh ngha:
(6.8)
(Momen quan tnh ly tam cua mat cat A oi vi he truc x,y c nh ngha:
(6.9)
T nh ngha cac momen quan tnh, ta nhan thay:
- MMQT co th nguyen la [chieu dai]4- Ix , Iy , Ip ( 0
- MMQT ly tam Ixy co the dng, am hoac bang khong.
- V nen
(6.10)
MMQT oc cc bang tong MMQToi vi hai truc vuong goc x, y co goc tai iem cc:
Theo nh ngha cua MMQT , ta cung co:
Tnh chat: Momen quan tnh cua mot hnh phc tap bang tong momen quan tnh cua tng hnh n gian.
2- He truc quan tnh chnh trung tam (QTCTT)
Mot he truc toa o co MMQT ly tam oi vi he truc o bang khong c goi la he truc quan tnh chnh.
He truc quan tnh chnh i qua trong tam mat cat c goi la he truc quan tnh chnh trung tam.
oi vi he truc nay, ta co:
Tnh chat: Khi mat cat A co mot truc oi xng th bat ky he truc vuong goc nao cha truc oi xng eu la he truc quan tnh chnh cua mat cat.
Chng minh: xet mat cat A co truc oi xng la y nh tren H.6.7. Ta luon tm c nhng cap vi phan dien tch oi xng e:
Nhan xet 2:
Bat ky truc nao vuong goc vi truc oi xng i qua trong tam cung hp vi truc oi xng thanh mot he truc QTCTT.
MMQT oi vi truc chnh trung tam c goi la momen quan tnh chnh trung tam cua mat cat A.
6.4 momen quan tnh CUA CACHNH N GIAN
1- Hnh ch nhat
nh momen quan tnh chnh trung tam cua hnh ch nhat b ( h (H.6.8).
He co hai truc oi xng x, y cung la he truc QTCTT.
e tnh Ix, lay dien tch vi phan dA la mot dai be rong b, be day dy, khoang cach en truc la y. Ta co:
(6.11)
Tng t, oi vai tro cua x va y, b va h, ta c:
(6.12)
2- Hnh tam giac
Tnh MMQT hnh tam giac oi vi truc x i qua ay (H.6.9).
Dien tch dA la dai vi phan song song vi ay, co chieu day la dy, khoang cach en truc x la y va co be rong c tnh nh sau:
( (6.13)
3- Hnh tron - Hnh vanh khan
Tnh MMQT cua hnh tron oi vi truc x (hay y) la ng knh.
He truc (x,y) cung la he truc chnh trung tam
Trc tien tm momen quan tnh oc cc oi vi trong tam O.
Xet vong tron ban knh R H.6.10a. Lay phan to dien tch dA dang mot vanh tron manh ban knh va be day . Nh vay,
Momen quan tnh oc cc cua toan bo hnh tron:
Do oi xng, ta co:
Theo (6.10), ta co:
(
(6.14)
Theo tnh chat cua MMQT oi vi truc a biet muc 6.3,
( MMQT cua mat cat hnh tron rong (hnh vanh khan) (H.6.10b) la hieu MMQT cua hai hnh tron ng knh D va d:
(6.15)
Va (6.16)
Vi
6.5 CONG THC CHUYEN TRUC SONG SONG
Biet cac ac Trng Hnh Hoc cua mat cat A trong he truc toa o Oxy.
Xac nh MMQT mat cat nay trong he truc OXY song song vi he truc a cho (H.6.11).
Goi a va b la toa o cua O trong he toa o OXY.
Ta co :
( Theo nh ngha:
(6.17a)
tng t: (6.17b)
( oi vi momen quan tnh ly tam:
(6.18)
( Neu he truc Oxy la he truc trung tam cua hnh A th cac cong thc tren co dang: (6.19)
( Neu he truc Oxy la he truc chnh trung tam cua hnh A th cac cong thc tren co dang:
(6.20)
Cong thc (6.21) thng c s dung e tnh cac momen quan tnh chnh trung tam cua mot hnh phc tap khi a biet momen quan tnh chnh trung tam cua tng hnh n gian.
T cong thc nay, ta nhan thay: trong tat ca cac truc song song th momen quan tnh oi vi truc trung tam luon co gia tr nho nhat.
Th du 6.2 Tnh momen quan tnh oi vi truc BB i qua ay cua hnh ch nhat (H.6.8).
Giai.
Dung cong thc chuyen truc song song e tnh JBB:
IBB = Ix +
Th du 6.3 nh MMQT chnh trung tam cua mat cat (H.6.12).
Giai
nh trong tam C: Chon he truc x,y nh hnh (truc y la truc oi xng, C nam tren truc y)
(Tnh MMQT oi vi he truc chnh trung tam IX,IY IX =Ix1+ Ix2 +Ix3 = 4352 cm4 Vi Ix1=+; Jx2=Jx3 = +;
Iy =Iyln- Iynho = - = 14336 cm4 Th du 6.4 Tnh MMQT chnh trung tam cua mat cat th du 6.2 (H.6.6).
Giai. Lay lai ket qua th du 6.2:
A1 = 18 cm2; A2 = 118 cm2; A3 = 35,2 cm2;
A = A1 +A2 +A3 = 171,2 cm2y1 = 28,1cm; y3 = 29,97cm; yC = 3,2 cm
MMQT cua ba hnh I, II, III oi vi cac truc trung tam cua chung nh sau:
I
Tra bang, ta co: ; ; ;
Dung cong thc chuyen truc song song e tnh momen quan tnh oi vi truc X i qua C, cua tng hnh I, II , III:
I
I
I
MMQTtnh chnh trung tam IX cua toan mat cat:
IX = + + = 100068,73 cm4MMQT tnh chnh trung tam IY:
IY = + + =
EMBED Equation.3 6.6 Cong thc XOAY TRUC CUA MMQT- XAC NH HE TRUC QUAN TNH CHNH (HTQTC)
1- Cong thc xoay truc:
Biet cac MMQT Ix , Iy , Ixy cua A trong he truc toa o Oxy.
Tnh cac MMQT Iu , Iv , Iuv cua A oi vi he truc mi Ouv ; He truc Ouv hnh thanh t viec xoay he truc Oxy mot goc ( ngc kim ong ho ( H.6.13)Toa o cua iem trong he truc mi va he toa o cu c lien he nh sau:
Theo nh ngha, cac MMQT oi vi truc u, v la:
; ;
( Tnh Ju
(a)
S dung cac cong thc lng giac:
(a) tr thanh:
(6.21)
( Tnh Iv : Tng t nh tnh MMQT Iu, ta c momen quan tnh Iv (hoac bang cach the trc tiep bang trong phng trnh (6.21)):
(6.22) ( Tnh Iuv :
(b)
(6.23)
2- He truc quan tnh chnh- Cach xac nh
( He truc QTC: Theo nh ngha muc 6.3, he truc quan tnh chnh la he truc co MMQT ly tam bang khong.
e xac nh he truc nay, cho Iuv = 0 (
(6.24)
trong o: - la goc xac nh truc quan tnh chnh.
Phng trnh (6.24) luon co hai nghiem 2 sai khac nhau mot goc . ( co hai nghiem ( sai biet nhau mot goc , ngha la luon tm c hai truc chnh vuong goc vi nhau.
( MMQT cc tr : e tm goc ( sao cho momen quan tnh co tr so ln nhat hoac nho nhat, lay ao ham cua Iu theo ( va cho bang khong:
(c)
De thay nghiem ( cua (c) cung la nghiem cua (6.24).
Nh vay oi vi he truc chnh vuong goc, momen quan tnh co gia tr ln nhat va nho nhat, goi la momen quan tnh chnh.
The ngc lai 2( t (6.24) vao (6.21) va (6.22), ta c tr so cac momen quan tnh chnh:
(6.25)
va:
(6.26)
( Cach xac nh he truc QTCTT cua mot hnh phang bat ky
Trong trng hp tong quat, khi dien tch A khong co truc oi xng, he truc QTCTT c xac nh theo trnh t nh sau:
- Chon he truc Oxy bat ky ban au. Xac nh trong tam cua hnh trong he truc nay
- Chuyen truc song song ve trong tam cua hnh. Tnh cac momen quan tnh oi vi he truc trung tam
- Xoay truc e tm truc quan tnh chnh i qua trong tam.
Viec xac nh he truc QTCTT cung nh tnh toan cac momen quan tnh chnh la rat can thiet trong viec tnh toan ng suat, chuyen v cua thanh chu uon, xoan ma ta se nghien cu cac chng sau.
6.7- VONG TRON MOHR QUAN TNHVe mat toan hoc, ta nhan thay co s tng ong gia phng trnh chuyen oi momen quan tnh va phng trnh chuyen oi ng suat vi:
- Iu tng ng vi
-Iuv tng ng vi
-Ix tng ng vi
-Iy tng ng vi
-Ixy tng ng vi .
V vay, neu dung mot he truc toa o vi truc hoanh bieu dien Iu va truc tung bieu dien th oi vi mot he truc u,v ta se co , se c bieu dien bang mot iem tren mot ng tron goi la vong tron Mohr quan tnh (H.6.14). Tng t nh vong tron Mohr ng suat, ta tm thay cac phng cua truc quan tnh chnh la PA va PB, neu phng cua truc hoanh song song vi truc Ox tren hnh phang F
Th du 6.5 Xac nh he truc quan tnh chnh trung tam va momen quan tnh chnh trung tam cua mat cat ch Z tren H.6.15, vi cac kch thc:
h = 400 mm; b = 175 mm; t = 28 mm
Giai.
Chia mat cat Z lam ba hnh ch nhat:
- Hnh I va III (nam ngang) be day t, be rong b t
- Hnh II be day t, chieu cao h.
Tuy khong co truc oi xng, song mat cat lai co iem oi xng (hay tam oi xng), trung vi trong tam hnh II, sao cho moi ng thang tren mat cat i qua iem C2 eu oi xng oi vi iem nay. De thay trong tam mat cat C trung vi tam oi xng.
Chon he truc toa o xy e tnh toan, i qua trong tam C. Momen quan tnh oi vi cac truc x, y la:
Ix = I+ 2
Iy = I+ 2
Momen quan tnh ly tam oi vi he truc xy la:
Ixy = I+ 2
e y rang, momen quan tnh cua hai hnh I va III bang nhau. Momen quan tnh ly tam cua hai hnh tren eu am v nam goc toa o th hai va th t.
Phng cua he truc quan tnh chnh trung tam so vi truc Ox:
va 217osuy ra: va
the gia tr cua (o vao (6.21), tnh c tr so momen quan tnh
Iu = 47944cm4 va Iv = 3381 cm4.
Nh vay, momen quan tnh chnh trung tam va phng he truc quan tnh chnh trung tam tng ng la:
IX = 47994 c