MH 12 SUC BEN VAT LIEU_2.pdf

5

, nm 2012

Mn hc: Sc bn vt liu

M mn hc : MH10

V tr, tnh cht, ngha v vai tr ca mn hc

- V tr: + Sc bn vt liu l mn hc k thut c s c b tr sau khi hc sinh hc cc mn: C l thuyt v Vt liu kim loi. + Sc bn vt liu cung cp kin thc cho cc mn chi tit my v k thut chuyn mn ca ngnh. - Tnh cht: + Sc bn vt liu l mn khoa hc kt hp cht ch gia l thuyt v thc nghim. + L mn hc thuc cc mn hc, m-un k thut c s bt buc. - ngha Gip sinh vin nm c cc kin thc c bn, c s trong k thut v vn dng tnh ton trong thc t - Vai tr

L mn l thuyt c s cho cc mn chuyn ngnh nn c vai tr quan trng trong chng trnh o to ngh ct gt kim loi.

Mc tiu ca mn hc:

- Trnh by c cc khi nim c bn ca mn hc nh: bin dng, ni lc, ng sut, bn, cng, n nh ca chi tit my.

- Phn tch c ngha ca cc i lng c trng cho tnh cht c hc ca vt liu.

- Xc nh c cc phng php a chi tit t kt cu thc v s tnh v phn tch c thnh cc loi bin dng c bn.

- V c cc biu ni lc v xc nh c mt ct nguy him trn chi tit.

- Vn dng c cc iu kin bn, iu kin cng, iu kin n nh gii ba bi ton c bn ca mn sc bn vt liu.

- C thc trch nhim, ch ng hc tp.

B GIAO THNG VN TI

TRNG CAO NG NGH GTVT TRUNG NG II

..............*&*..............

GIO TRNH

Tn mn hc: Sc bn vt liu M MN HOC: MH12

NGH: HN TRNH TRUNG CP NGH/CAO NG NGH

(LU HNH NI B)

r

F

Hi phng, nm 2011

6

LI NI U

Ngy nay khoa hc k thut pht trin nh v bo, cc ngnh k thut

chim mt v tr tng i quan trng trong nn kinh t .V vy vic o to

nhn lc cho cc ngnh k thut ng vai tr quan trng to ra ngun nhn

lc c nng lc phc v cho nn kinh t ang pht trin ca nc ta.

Sc bn vt liu l mn khoa hc bn thc nghim thuc khi k thut c ging dy trong cc trng cao ng, i hc k thut. N cung cp nhng kin thc cn thit v nhng tc dng c hc gi quyt cc vn

thc t trong vic thit k ch tao, tnh bn cho chi tit, n l mn hc c s cho rt nhiu cc mn hc chuyn ngnh thuc kh k thut.

Gio trnh Sc bn vt liu c xy dng trn c s nhng gio trnh c ging dy trong cc trng k thut kt hp vi kinh nghim ging dy ca nhng gio vin trong ngnh. Gio trnh c bin son cho

ph hp vi c im ca sinh vin trng cao ng ngh.

Gio trnh Sc bn vt liu c bin son ngn gn, d hiu, b sung nhiu kin thc mi, ni dung cp ti nhng kin thc c bn v tnh

ton bn, cng, s n nh ca chi tit. p ng c nhng tnh cht c trng ca ngh c kh.

Trong khi bin son gio trnh tc gi c nhiu c gng nhng khng trnh khi nhng khim khuyt. Rt mong nhn c s ng gp kin t bn c.

Cu trc chung ca gio trnh c 10 chng

Chng I: Nhng khi nim chung. Chng II : Ko v nn ng tm.

Chng III: Ct - Dp Chng IV: c trng c hc ca hnh phng. Chng VI: Un ngang phng. Chng V: Xon thun ty.

Chng VII: Thanh chu lc phc tp. Chng VIII: n nh ca thanh thng chu nn ng tm.

Chng IX: Tnh bn ca thanh thng chu ng sut thay i.

Tc gi

7

MC LC

mc Li ni u Mc lc

Trang

3

4

Chng I: Nhng khi nim chung.

1. Gii thiu lch s mn hc. 7 2. Nhim v v i tng nghin cu ca mn hc 8 3. Cc gi thuyt c bn v vt liu. 9

4. Ngoi lc, ni lc, phng php mt ct v ng sut. 10 5. Cc loi bin dng c bn. 15 Chng II : Ko v nn ng tm.

1. Khi nim v ko - nn ng tm 17

2.Ni lc 17

3.ng sut v bin dng. 20 4. c trng c hc ca vt liu. 24 5. Tnh ton v ko (nn) ng tm. 27 Chng III: Ct - Dp

1. Khi nim v Ct 36

2. Khi nim v Dp 39 Chng IV: c trng c hc ca hnh phng.

1. Khi nim v mmen tnh. 43 2. Khi nim v mmen qun tnh. 45 3. Bn knh qun tnh. 49

Chng V: Xon thun ty.

1.Khi nim v xon thun ty. 52 2. ng sut v bin dng trong thanh mt ct trn chu xon 54 3. Tnh ton v xon thun ty. 57

Chng VI: Un ngang phng. 1. Khi nim v un ngang phng. 61 2. Ni lc v biu ni lc. 61 3. ng sut trong dm chu un ngang phng. 64

4. Tnh ton v un ngang phng. 67 5. Chuyn v ca dm chu un. 69

Chng VII: Thanh chu lc phc tp.

1. Khi nim thanh chu lc phc tp. 71 2. Un xin. 72

3. Un ngang phng v ko (nn) ng thi. 78 4. Un v xon ng thi. 83 Chng VIII: n nh ca thanh thng chu nn ng tm.

1.Khi nim v n nh, lc ti hn v ng sut ti hn. 87 2. Cng thc tnh lc ti hn, ng sut ti hn theo Euler. 89 3. Cng thc tnh lc ti hn v ng

sut ti hn theo Iasinki.

89

4. Tnh ton v n nh 90 Chng IX: Tnh bn ca thanh thng chu ng sut

8

thay i. 1. Khi nim v thanh chu ng sut thay i. 93 2. Hin tng mi ca vt liu. 93 3. Chu trnh v c trng chu trnh ng sut. 93 4. Gii hn mi. 94

5. Cc nhn t nh hng n gii hn mi, cc bin php khc phc.

95

6. Tnh bn theo h s an ton. 97

9

CHNG I:

NHNG KHI NIM M U

M chng: CHI Gii thiu:

Nhng khi nim m u c ngha rt quan trng trong qu trnh nghin cu, tnh ton ca mn hc. Nhng khi nim ny gip sinh vin hiu c nhng cm t v quy c k hiu thng c s dng trong mn hc. Mc tiu:

+ Trnh by c nhim v v i tng nghin cu ca mn hc.

+ Trnh by c cc khi nim: Vt rn thc, ngoi lc, ni lc,

phng php mt ct, ng sut, cc bin dng c bn.

+ C thc trch nhim, ch ng hc tp.

1. Gii thiu lch s mn hc.

T th k 18 con ngi c nhng cng trnh c xem l s khi u ca mn hc

Nm 1729 Buyphinghe a ra dng quan h phi tuyn gia nhs sut v bin dng. Sau nm 1768 Hc a ra quy lut c bn v vt th n hi vi dng tuyn tnh ng thi ng c nhng cng trnh :

- L thuyt ton hc v un ca thanh n hi ca le v Becnuli.

- Tnh n nh ca le

- Dao ng ngang ca thanh n hi

- Nghin cu v l thuyt lc n hi ca khng kh(Lmnxp)

Cui th k 18 u th k 19 nh bc hc ngi Php Navie xut pht t quan im v lc tng tc gia cc phn t ca Niu tn xut ra l thuyt n hi ri rc. Nm 1822 Csi a ra khi nim v trng thi ng sut ti mt im v vit cc phng trnh cn bng cng vi cc biu thc biu din s tng quan gia ng sut v bin dng cho vt th ng hng. Ta c th kt lun rng Navi, Csi v Ostrogratxki, Potxng l nhng ngi t nn mng cho l thuyt n hi ton hc.

Vo cui th k 19 nhu cu v pht trin cng nghip thi thc cc nh khoa hc tm cch tnh ton nhanh chng nhng bi ton trong thc t do pht sinh ra ngnh l thuyt n hi ng dng v l thuyt v sc bn vt liu.

10

Vo cui th k 19 v sang u th k 20 ngnh c hc vt rn bin dng pht trin v cng rng ln.

2. NHIM V V I TNG CA C HCVT RN BIN DNG

Mc tiu:

+Trnh by c nhim v v i tng nghin cu ca mn hc.

2.1. Nhim v

Sc bn vt liu l mn khoa hc nghin cu thc nghim, kh nng chu lc v bin dng ca vt th ra phng php tnh sao cho cc vt th bn, cng, n nh v tit kim vt liu. - bn: l kh nng chu lc ln nht ca chi tit sao cho chi tit khng b ph hng. - cng: l kh nng chu lc ln nht ca chi tit sao cho bin dng khng qu ln lm nh hng n iu kin lm vic bnh thng - n nh: L kh nng chu lc ln nht ca chi tit sao cho chi tit khng b thay i hnh dng hnh hc trong qu trnh lm vic bnh thng Sc bn vt liu ra phng php tnh ton ,lp nn cc biu thc ton hc tha mn iu kin bn, iu kin cng v iu kin n nh .Xut pht t Sc bn vt liu ch yu gii quyt 3 dng bi ton c bn: + Bi ton kim tra bn + Bi ton xc nh kch thc hp l + Bi ton xc nh ti trng cho hp l

2.2. i tng nghin cu

- i tng nghin cu ca b mn sc bn vt liu l vt rn thc - Vt rn thc l vt rn khi c tc dng ca ngoi lc s xy ra bin dng v c th b ph hng - Vt rn thc c phn lm 3 dng c bn: + Vt th dng khi: Vt th c kch thc theo ba phng ln tng ng nhau. (Hnh 1-1a) + Vt th dng thanh: Vt th c kch thc mt phng ln hn rt nhiu so vi phng cn li. (Hnh 1-1b) + Vt th dng tm: L vt th m kch thc hai phng ln hn rt nhiu so vi phng cn li, phng c kch thc b gi l b dy. (Hnh 1-1c)

c,

a, b,

Hnh 1-1

11

Sc bn vt liu trong chng trnh ch yu nghin cu v vt th dng thanh thng

Phn loi theo tit din: - Hnh ch nht

- Hnh vung - Hnh trn

i tng nghin cu ca mn hc l vt rn thc (tc l vt rn bin dng)

3. CC GI THIT C BN V VT LIU

Mc tiu:

+ Trnh by c cc khi nim: Vt rn thc, ngoi lc, ni lc,

phng php mt ct, ng sut, cc bin dng c bn.

3.1. Cc gi thit v s lin tc, ng tnh v ng hng

*S lin tc: Cc phn t vt liu mi ni trong vt th phn b u v lin tc .Tc l gia chng khng c khe h coi vt th khng c khuyt tt. *S ng tnh: Cc phn t vt liu tt c mi ni trong vt th c cng tnh cht *S ng hng: Kh nng chu lc ca cc phn t vt liu trong vt th theo mi hng u nh nhau

3.2. Vt liu c tnh n hi hon ton

- Tnh n hi l kh nng tr v trng thi ban u khi vt c bin dng do tc dng ca ngoi lc - n hi hon ton l di tc dng ca ngoi lc (ngoi lc phi nm trong gii hn n hi ca vt) vt th b bin dng, khi thi tc dng lc vt th tr li y nguyn trng thi ban u (tc l khng c bin dng d)

Hnh 1-3.

Gi thuyt ny ch r sc bn vt liu ch nghin cu bi ton trong

P

Mt ct ngang

Hnh 1-2

12

giai on n hi. Ngoi min n hi bi ton s c nghin cu trong

mt mn hc khc l l thuyt do.

3.3. Gi thit v quan h bc nht gia lc tc dng v bin dng

Khi lc tc dng cn nm trong gii hn n hi ca vt th bin dng ca vt c quan h bc nht vi lc tc dng gy nn bin dng . * Th nghim th ko vt liu do: Khi lc tc dng cn nm trong gii hn n hi (0 Ptl)ca vt liu. Bin dng l on ON. Trong gii hn ny ta thy lc tng nhanh cn bin dng tng rt chm.Quan h gia lc v bin dng l ng cong OA. Do cong ca OA rt nh nn ta c th coi n l ng thng . Quan h gia lc v bin dng l quan h bc nht . Kt lun : Tt c cc loi vt liu l i tng nghin cu trong mn sc bn th n phi tha mn cc gi thit trn.

3.4. Nguyn l c lp tc dng ca lc

a) Nguyn l: Tc dng ca h lc ln vt bng tng cc lc thnh phn tc dng ln vt

Tc l : Nu mt h chu tc dng ng thi ca nhiu yu t th c

th kho st h di tc dng ca tng yu t ring r ri cng cc kt

qu li (hnh1-5).

b. ngha: Mt bi ton phc tp c phn tch thnh cc bi ton n gin v kt qu ca bi ton bng tng cc bi ton n gin

Nu vt liu lm vic ngoi min n hi th nguyn l trn

khng c p dng v sai s m. Cc yu t tc dng ln h c th

bao gm c goi lc ln cc tc nhn khc nh nhit , p sut, v.v...

4. NGOI LC NI LC - NG SUT H S AN TON

4.1. Ngoi lc

P

O

P tl P c

A B

N l

Hnh 1-4

Hnh 1-5

13

4.1.1. nh ngha

Ngoi lc l nhng lc hoc m men lc t vt th khc hoc t mi trng xung quanh tc dng ln vt th kho st Ngoi lc c hai loi: Ti trng(lc) tc dng v phn lc lin kt

4.1.2. Phn loi

a.Phn loi ngoi lc:

nh ngha: L ngoi lc tc dng ln vt th m im t, phng, chiu, tr s bit trc. + Phn loi theo hnh thc tc dng: - Ti trng tp trung: L nhng lc hoc ngu lc tc dng ln vt trn mt din tch rt nh, coi nh tc dng ti mt im. - Ti trng phn b:

. Ti trng phn b ng (Hnh 1-5):

Ti trng tc dng ln vt th theo mt ng. Q = q .l

Trong : Q : L ln ca h lc phn b q : Lc n v l : di ca on thng m h lc phn b . Ti trng phn b m (Hnh 1-6): Ti trng tc dng ln vt th trn mt mt no . Q = q .S Trong : Q : L ln ca h lc phn b q : Lc n v S : din tch m h lc phn b

. Ti trng phn b khi (Hnh 1-7): Ti trng tc dng lin tc trn mt khi.

Q = q .V Trong : Q : L ln ca h lc phn b q : Lc n v V : th tch m h lc phn b Hnh 1-8

+ Theo mc tc dng: - Ti trng tnh: L ti trng tc dng ln vt th c tr s tng dn t 0 n gi tr xc nh ri sau khng thay i na. Ti trng tnh thng gp nh: trng lng, v cc phn lc - Ti trng ng: L ti trng c tr s, phng, chiu hoc im t lin tc thay i theo thi gian v lm cho vt th chuyn ng c gia tc.

b. Phn lc lin kt

q

q

q

Hnh 1-6

Hnh 1-7

14

nh ngha: Phn lc lin kt l lc , mmen do vt gy lin kt gy ra chng li chuyn ng hay xu hng chuyn ng ca vt kho st.

*Mt s lin kt phng thng s dng: - Lin kt gi di ng : y l mt loi lin kt n, trong mt phng nch hn ch mt dch chuyn thng. Cc lin kt thc t nh bi lng

cu, con ln di ng, v.v... Khi s ho u a v dng gi ny. Gi c

mt thnh phn phn lc lin kt Y

Hnh 1-9. Lin kt gi di ng

- Lin kt gi c nh: L loi lin kt hn ch hai dch chuyn thng (trong khng gian hai chiu) v 3 dch chuyn thng (trong khng gian ba

chiu). V d: nh cc con ln c nh di cc nhp cu, cc bi chn

trong my cng c, v.v... K hiu gi c nh ch ra trn hnh 1-10.Gi c hai thnh phn phn lc lin kt Y, Z

Hnh 1-10. Lin kt gi c nh

- Lin kt ngm: L loi lin kt hn ch hon ton su bc t do ca h. V d lin kt gia chn ct v mt t, lin kt gia cc dm hnh lang

vi tng nh, v.v... k hiu ngm ch ra trn hnh 1-11. Lin kt ngm c ba thnh phn phn lc lin kt Y, Z, M

Hnh 1-11. Lin kt ngm

c. Phn loi ti trong.

Ti trng c phn thnh ti trng tnh v ti trng ng. + Ti trng tnh l ti trng m gi tr ca n tng dn t khng

n mt tr s xc nh trong qu trnh gia tc chuyn ng ca cc cht im l khng ng k v c th b qua.

P Y

P

Z

Y

M

P Y

Z

15

+ Ti trng ng l ti trng tc dng ln h lm cho cc cht im ca h chuyn ng c gia tc hoc c xut hin lc qun tnh.

- Ti trng ng m tr s thay i rt nhanh trong mt khong thi gian nh c gi l ti trng va chm.

- Ti trng m phng chiu, ln bit cn im t. Thay i c gi l ti trng di ng.

V d: Trng lng m khi chy tc dng ln cu. - Ti trng bin thin tun hon theo thi gian l ti trng gy nn dao ng.

4.2. Ni lc

- Ni lc l lc do chnh bn thn vt sinh ra chng li bin dng khi c ngoi lc tc dng. - Ni lc l phn tng ln ca lc lin kt phn t ca vt liu khi c ngoi lc tc dng. - Khng c ngoi lc tc dng th khng c ni lc. Khi ngoi lc tng th ni lc cng tng theo nhng ni lc ch tng ti mt gii hn nht nh, nu ngoi lc c tip tc tng m ni lc khng tng c na th lin kt phn t b ph v hay vt liu b ph hng.

4.3. Cch xc nh ni lc ( Phng php mt ct )

Xt thanh thng chu tc dng ca h lc cn bng nh trn (hnh 1-12a).

dng phng php mt ct :

Hnh 1-12.

Pn Q

Tri

P 1

P 2

P 3

P 4

P 5

P 6

P 7 Phi

Pn

z Tri

P 5

P 6

P 7

R'

x

y

R Q y

N Z

Q x

M x M z

My

a,

b,

16

- Tng tng dng mt mt phng (Q) vung gc vi trc thanh, ct thanh lm hai phn .Gi li mt phn bt k kho st (gi s gi li phn tri) - Xt cn bng cho phn tri(hnh 1-11b) . phn tri cn bng th phi c lc sinh ra cn bng vi cc lc tc dng ln phn tri. chnh l ni lc sinh ra trn mt ct ngang ca phn tri, ta hp cc ni lc c vc

t hp lc l R . - Phn kho st cn bng di tc dng ca ngoi lc v ni lc

RtriPi )(

( ni FFFFtriP ...)( 765 ) Lp h trc ta Oxyz c gc ta O trng vi tm mt ct ngang, cc trc Ox, Oy nm trong mt phng cha ct ngang ca thanh, trc Oz trng vi trc thanh.

Di chuyn R bng phng php di lc song song v tm O ta

c mt vc t lc 'R v mmen M

* Chiu vc t lc 'R v m men M ln h trc ta Oxyz ta c6 thnh phn ni lc Nz, Qx, Qy, Mx, My, Mz gi l 6 thnh phn ni lc trn ton b mt ct ngang ang kho st, mi thnh phn ni lc c mt tn ring

- Thnh phn Nz gi l Lc dc c phng vung gc vi mt ct ngang

)(TriPN izZ - Thnh phn Qx , Qy gi l lc ct hay lc ngang c phng vung gc vi trc ca thanh

)(

)(

triPQ

triPQ

iyy

ixx

- Thnh phn Mz : Mmen xon quanh trc Oz

)( izz PmM

tri

- Thnh phn Mx , My : Mmen un quanh trc Ox, Oy

)(

)(

iyy

ixx

PmM

PmM

tri

4.4. ng sut

4.4.1. nh ngha

ng sut l gi tr ca ni lc sinh ra trn mt n v din tch mt ct * Nu ni lc phn b u: ng sut = Din tch mt ct

Ni lc

17

* Nu ni lc phn b khng u: Cn phi tm c quy lut phn b, xc nh c vng pht sinh ln nht sau xc nh ng sut ln nht trong mt ct tnh ton.

Hnh 1-13. ng sut trn mt ct ngang

* n v ca ng sut: N/m2, kN/m

2, MN/m

2.

4.4.2. Phn loi ng sut

Da vo 2 phng c bn ca ni lc, ng sut c phn thnh hai thnh phn l: ng sut php v ng sut tip - ng sut php: K hiu : Khi ni lc c phng vung gc vi mt ct ngang ta c ng sut sinh ra trn mt ct ngang l ng sut php - ng sut tip: K hiu : Khi ni lc c phng tip tuyn (trng) vi mt ct ngang ta c ng sut sinh ra trn mt ct ngang l ng sut tip

5. CC LOI BIN DNG C BN

Ngoi lc tc dng ln chi tit vi nhiu hnh thc khc nhau th cc bin dng cng s khc nhau. Trong k thut kho st 4 loi bin dng c bn sau: Ko- nn ng tm, ct-dp, xon, un - Ko- nn ng tm: Nu mt thanh thng chu tc dng ca cc lc c phng trng vi trc thanh th thanh chu Ko - Nn ng tm - Ct Dp:

+ Ct: Nu tc dng vo thanh hai lc song song, ngc chiu, cng ln v t hai mt phng ct st gn nhau th thanh s xy ra hin tng ct.

+ Dp: Dp l hin tng nn cc b xy ra trn mt din tch truyn lc tng i nh ca hai chi tit p vo nhau. - Xon: Nu tc dng vo thanh cc ngu lc hay cc mmen c chiu quay ngc nhau v c mt phng tc dng trng vi cc mt ct trong thanh.

- Un: Nu ngoi lc tc dng l lc tp trung, lc phn b, ngu lc nm trong mt phng i xng cha trc ca thanh.

r

21F m

18

CU HI N TP

1. Cc gi thuyt c bn v vt liu, gi thuyt v tnh lin tc, ng cht v ng hng, gi thuyt v vt liu n hi tuyt i, gi thuyt v tng quan gia bin dng v lc, nguyn l c lp tc dng.? 2. nh ngha ngoi lc, ni lc, v ng sut? Phn loi ng sut? 3. Nu phng php mt ct xc nh ni lc ? 4. Cc loi bin dng c bn?

19

CHNG II: KO - NN NG TM

M chng: CHII

Bin dng ko v nn chng ta gp rt nhiu trong thc t c bit l trong cc chi tit my v cc cu kin ca cng trnh.V d: Dy cp ko vt, ng khi ca cc nh my, cc thanh trong kt cu dn...tt c cc chi tit trn u chu ko hoc nn.

Mc tiu

+ Trnh by c khi nim thanh chu ko - nn ng tm + Phn tch c khi nim lc dc. + V c biu lc dc, biu ng sut trn mt ct ngang. + Tnh c ng sut v bin dng trong thanh. + p dng thnh tho ba bi ton c bn theo iu kin bn. + C thc trch nhim, ch ng hc tp.

1. NH NGHA

Mc tiu - Trnh by c khi nim thanh chu ko - nn ng tm

+ nh ngha: Khi mt thanh thng chu tc dng ca cc lc c phng trng vi trc thanh th thanh chu Ko - Nn ng tm.

V d:

Ko ng tm Nn ng tm

Hnh 2-1

- Thanh chu ko ng tm: Ngoi lc hng t trong thanh ra ngoi. - Thanh chu nn ng tm: Ngoi lc hng t ngoi vo trong thanh. Thanh chu nn ng tm l trng hp ngc li ca thanh chu ko ng tm do trong qu trnh nghin cu chng ta ch nghin cu thanh chu ko ng tm cn thanh chu nn th ngc li.

2. NI LC BIU NI LC

Mc tiu + Phn tch c khi nim lc dc. + V c biu lc dc trn mt ct ngang.

2.1. Phng php mt ct xc nh ni lc.

* Xt mt thanh thng chu ko ng tm trng thi cn bng (Hnh 2-2a). Xc nh ni lc trong thanh ?

k P P k P n P n

20

- Tng tng dng mt mt phng (Q) vung gc vi trc thanh ct thanh lm hai phn, gi li phn A kho st. Theo phng php

mt ct th phn A cn bng di tc dng ca ngoi lc kP

v ni lc.

Gi ni lc trn phn A l zN

th ta phi xc nh

zN

(Hnh 2-2b).

- Phn A cn bng nn kP

v zN

l hai lc cn bng: ( kP

,

zN

)~0

Vy da vo kP

xc nh zN

:

Kt lun: - Ni lc trong thanh chu ko (nn) ng tm ch c mt thnh phn dc theo trc thanh, ta gi l lc dc.

- K hiu: zN

C + Phng: Trng vi trc ca thanh. + Chiu: Ngc chiu vi ngoi lc tc dng.

+ Tr s: NZ = kP

+ im t: Ti tm mt ct.

* Quy c du:

+ Ni lc hng t trong mt ct ra th mang du dng (thanh chu ko) + Ni lc hng t ngoi vo trong mt ct mang du m(thanh chu nn)

2.2. Biu ni lc.

2.2.1. nh ngha.

Biu ni lc l th biu din s bin thin ca ni lc dc theo trc thanh.

2.2.2. Cc bc v biu ni lc.

- Bc 1: Xc nh phn lc lin kt (nu cn) - Bc 2: Chia on cho thanh, da trn c s im t ca lc tng

Q

A

P k P k

A P k N z

B

Hnh 2-2

a,

b,

21

ng vi mt im ,hai im lin tip l mt on. - Bc 3: Xc nh ni lc trong tng on + Dng phng php mt ct ,ct thanh lm hai phn ,gi li mt phn kho st + t ni lc vo mt ct (gi nh ni lc dng v hng ra ngoi mt ct) + Vit phng trnh cn bng v gii cc phng trnh - Bc 4: V biu ni lc.

+ K ng thng song song vi trc thanh gi l ng khng. + K cc on thng song song vi nhau v vung gc vi ng khng + in du ,in gi tr ni lc

* V d 2.1:

Cho thanh AC chu tc dng ca cc lc dc trc P1=10 kN; P2= 30kN. V biu ni lc cho thanh AB ?

Bi lm

+ Xc nh phn lc: Phng trnh cn bng : ZA +P1 - P2 = 0

ZA = P2 - P1 =30 -10 =20 kN + Chia on cho thanh: Chia thanh lm 2 phn AB, BC + Xc nh ni lc trn tng on: - Xt on AB :Dng mt ct (1-1), ct thanh,lm hai phn, gi li phn tri kho st Ta c phng trnh cn bng NZ

1-1 + ZA= 0

NZ1-1

=-ZA= -20 KN

Vy on AB chu nn, ni lc mang du m, -Xt on BC: Dng mt ct (2-2), ct thanh lm hai phn ,gi li phn phi kho st Ta c phng trnh cn bng

P 1 P 2 A B C

Hnh 2-3

P 1 P 2

1

1

2

2

A B

C

P 1 N z 2-2

N z 1-1 X A

1

1

2

2

N Z

10KN

X A

20KN

Hnh 2-4

22

NZ2-2

- P1= 0 NZ

2-2 = P1= 10 KN

Vy on BC chu ko +V biu ni lc.( Hnh 2-4 )

Nhn xt biu ni lc: Nhn vo biu ni lc thy on AC l on nguy him nht.

3. NG SUT V BIN DNG CA THANH

Mc tiu + V c biu ng sut trn mt ct ngang. + Tnh c ng sut v bin dng trong thanh.

3.1. ng sut

3.1.1.Th nghim

Xt thanh thng c tit din hnh ch nht chu ko ng tm - Trc khi cho thanh chu ko + K ln mt ngoi ca thanh cc on thng song song vi trc thanh, cc on thng ny c trng cho cc th dc v k cc on thng vung gc vi trc thanh, cc on thng ny c trng cho cc mt ct ngang. To thnh mt li vung

Hnh 2-5

- Sau khi cho thanh chu ko: Li vung bin thnh li ch nht

Hnh 2-6

Lm nhiu ln th nghim ta u thu c kt qu nh trn - Nhn xt: + Cc th dc: Vn thng, vn song song vi nhau v song song vi trc thanh. Cc th dc b gin di ra ,khong cch gia chng b thu hp li nhng chng vn c chiu di bng nhau iu ny chng t cc th dc bin dng ging nhau. + Cc mt ct ngang: Khong cch gia chng tng ln ,tit din mt ct b thu hp li nhng cc mt ct vn phng v vn vung gc

l F

l1 F1

P P

23

vi trc thanh .iu ny chng t cc mt ct ngang c bin dng ging nhau

+ Chiu di ca thanh thay i mt on 1l l l ( l : bin dng di

tuyt i)

+ Tit din mt ct ngang co li 1F F F

+ Ta thy: F

24

Trong : + NZ: Ni lc (lc dc) + F: Din tch mt ct ngang + E: Mdun n hi ca vt liu (tra bng)

3.2.2. nh lut Hc

Trong giai on n hi, ng sut php t l vi bin dng di t i

..

Z ZZ Z Z Z

NlE

l E F E

Trong : +

Z : Bin dng di t i

+ EF: cng chng ko (nn)

3.2.3. nh lut Poat xng

Bin dng theo phng ngang t l thun vi bin dng dc t i

x y Z

Trong :+ ,x y : Bin dng theo phng ngang

+ : H s poat xng (0<

25

+ Xc nh phn lc lin kt : Phng trnh cn bng : PA - P1 + P2 P3 =

PA= P1- P2 + P3 = 3050 +60

PA = 40 kN + Chia on cho thanh: Chia thanh lm 3 on l : AC, CD ,DB - Xt onBD:Dng mt ct(1-1), ct thanh , gi li phn phi kho st Ta c phng trnh cn bng NZ

1-1- P1 = 0

NZ1-1

= P1 =30 kN Vy on DB chu ko, ni lc NZ

1-1 mang du dng

-Xt onCD:Dng mt ct(2-2), ct thanh, gi li phn phi kho st Ta c phng trnh cn bng NZ

3-3 + P2 P1 = 0

NZ3-3

= - P2+ P1= -50 + 30

NZ3-3

= -20 kN Vy on CD chu nn, chiu NZ

2-2 c chiu ngc li.

-Xt on AC:Dng mt ct(3-3), ct thanh ,gi li phn tri kho st Ta c phng trnh cn bng NZ

3-3 - PA = 0

NZ3-3

= PA= 40 kN Vy on AC chu ko, Ni lc NZ

3-3 mang du dng

Biu ni lc Nz ( hnh 2-9)

*Bin dng di tuyt i ca thanh AB

p dng cng thc:

33

3

33

22

2

22

11

1

11

.

.

.

.

.

.

.

.

FE

lN

FE

lN

FE

lN

FE

lNl ZZZ

ii

i

ii

Z

V thanh c tit din khng i nn F = F1= F2 = F3 ;

E = E1= E2 = E3 = 2.104 kN/cm

2

cml 014,08.10.2

60.4080.2050.304

Hnh 2- 9

20KN

P 1 P 2 P 3

l 1 l 2 l 3

1

1

2

2

3

3

P 1 N Z

1

P 1 P 2

2

2

1

P A A C D B

1-1

P A A

3

3

N Z 2-2

N Z 3-3

N Z

30KN

40KN

26

4. C TRNG C HC CA VT LIU

Mc tiu + Trnh by c th nghim th ko vt liu do v th nghim th ko vt liu dn

+ Phn tch c qu trnh bin dng ca mu

4.1. Th nghim ko nn vt liu do.

tin hnh th nghim trc tin phi c cc mu th nghim theo tiu chun tng nc. Trn hnh 2 - 10 nhng mu th trn v dt

c dng Vit Nam.

Phn thanh c chiu di l0 gi l phn lm vic ca mu. Thit b to lc ko mu trong cc mu th nghim c th l cc thit b c kh hoc thy lc . Hnh 2- 10 d l s nguyn l ca my th nghim c thit b thy lc. Nh p lc du

trong tr A tng ln t t m pt tong c nng ln v to ra lc ko trong

mu B. Lc ko mu B c th c xc nh bi gi tr o trn ng h o p lc C. Sau khi kp cht mu vo my ngi ta cho lc ko mu tng chmchm t gi tr 0. Qu trnh bin dng ca mu c my v thnh biu ( - ) trn hnh 2-11.Ta thy chiu di mu tng dn, chiu ngang mu hp dn cho n khi lc ko P t cc i P0 th c m ch no trn mu b tht hn li(ng vi im D). Sau thanh tip tc b dn di trong khi lc ko

Hnh 2-11

O

ch h t

B A B

C

D M E

Hnh 2- 10

C 32 23 d=20

20 25 220 25 20 l0=10d=200

40 30

50 5 220 5 50 200

10

32

l0=6d=36

d=6

a,

b,

c,

d,

B

A

27

gim dn v n mt gi tr P no (ng vi m M). Th mu b t ti ch tht. Quan h f i vi thp CT3 ca Nga c biu lc ko

nh trn hnh 2-11 vi cc im c trng A, B, C, D, M. Bin dng ko ca vt liu gm 3 giai on chnh sau:

* Giai on 1:

Vt liu c tnh n hi tuyt i, quan h f l quan h

tuyn tnh c biu din bi on thng OA: .E (2-1) Gia on ny gi l giai on t l, quan h (2-1) c gi l nh lut Hc khi ko, E l m un n hi dc ca vt liu. Gi P tl l gi tr ln nht ca lc ko trong giai on ny v F0 l din tch mt ct ngang ban u ca mu th, ta c

0F

Ptltl

ng sut tl l gii hn t l (i vi thp

CT3 22 /210/21 cmMNcmkNtl )

dc ca on OA xc nh bng m un n hi E

Etg

K t trn im A biu khng cn quan h tuyn tnh na m t nh lut Hc mt hiu lc. rt gn im A trn on cung ny ca biu c mt im B, ng sut ng vi im B k hiu h v c

gi l gii hn n hi. Trong cc tiu chuaanrr k thut h c xc

nh khi m cc bin dng d ca mu th t c 0,05% v vit

05.0 h

* Giai on 2:

Bin dng ng bin vi lc cht t ri tng r rt trong khi lc khng tng c na (on nm ngang k t C). Ta gi giai on ny

laf giai on chy do. Lc ko ng vi giai on ny c k hiu ch

ng sut : 0F

Pchch

- ch gi l gii hn chy (i vi thp CT322 /240/24 cmMNcmkNch )

Trong giai on chy, nu quan st mt mu lm bng thp t cc bon c mi nhn, ta thy nhng vt gn nghing trn trc thanh mt gc 45

0. chnh l nhng vt gy ra do s trt gia cc tinh th

vt liu do ng sut tip cc i gy ra. Nhng vt trt ny gi l ng Liuder-Trernov.

* Giai on 3:

28

Sau khi kt thc giai on chy do,trong vt liu li xut hin kh nng t cng c . C th l bin dng ch tng nu lc ko mu tng.Biu trong giai on ny l mt ng cong trn.

ng sut ng vi im D cao nht trong giai on ny l gii hn bn :

0F

PBB

i vi thp CT3 22 /400/40 cmMNcmkNB

Sau khi t gii hn bn th c mt ch trong mu th b tht li v t bin dng tip tc tng nghch bin vi lc cho n khi mu b t ng vi im M.

4.2. Th nghim ko vt liu dn

a. Biu

Vt liu dn chu ko rt km nn b ph hng t ngt ngay khi gin cn rt nh. Nhn biu ng 4 hnh 2-12 Ta thy khng c giai on t l v giai on chy do, biu l mt dng ng cong ngay khi ng sut cn rt nh. Tuy vy trong gii hn lm vic, thong thng i vt liu dn vn c th p dng nh lut Hc, vi vt liu dn ta ch c gii hn bn:

0F

PBB

Nu em so snh vi vt liu do th gii hn ny rt nh Khi b nn vt liu cng b ph hy ngay khi bin dng cn b, nhng gii hn bn cn c tr s ln hn nhiu so vi khi ko.

b. Biu lP

Trn hnh 2-13 cho ta tng quan gia lc tc

dng P v bin dng l khi ko vt liu dn. Tr s lc ko ng vi lcmub ph hu (im A) gi l

PB cc loi vt liu dnb ph hu t ngt bin dng

cn rt nh, chngt kh nng chu ko ca vt liu n l rt km. Dng ca ng cong tu thuc vo bn cht ca vt liuth nghim. Nhng loi vt liu dn nh gang xm,thp c t l cc bon cao, , thu tinh, v.v... khi b ph hu bin dng ca chng

thng khngvt qu 2.5%, trong trng hp biu thngc thay bng ng thng (ng nt t trn hnh 2-13).

O

A

D

C

4

A

C

2

CT3

3

CT3

B tl

tl

1

ch

ch

B

B

Hnh 2-12

Hnh 2-13

29

5. TNH TON V KO NN NG TM

Mc tiu + Trnh by c khi nim ng sut nguy him, ng sut cho php v h s an ton + Xc nh c iu kin bn

+ p dng tnh ton c ba bi ton c bn theo iu kin bn

5.1. Khi nim v ng sut cho php v h s an ton.

5.1.1. ng sut nguy him v ng sut cho php

- ng sut nguy him : l gi tr ng sut nh nht m tng ng vi

n vt liu xem nh b ph hng, k hiu l: ,o o

- ng sut cho php: l gi tr ng sut ln nht m tng ng vi n

vt liu cn lm vic c. Nu vt qu gi tr vt liu xem nh b ph hng. K hiu l: : ng sut php cho php : ng sut tip cho php * Gi tr ca ng sut cho php c tra bng trong s tay k thut .

5.1.2. H s an ton : n > 1

H s an ton l h s d tr bn ty theo iu kin lm vic ca chi tit , n ph thuc vo cc yu t sau:

- Vt liu - iu kin lm vic v thi gian lm vic - Trnh cng ngh - Mc quan trng ca chi tit.

T cc yu t nh hng trn ngi ta tng hp c h s an ton n

;0n

n

0

Trong : n - l h s an ton

5.2. iu kin bn v cc bi ton c bn

5.2.1. iu kin bn

iu kin cn v thanh chu ko nn ng tm m bo bn l ng sut sinh ra trn mt ct ngang ca thanh phi nh hn hoc bng ng sut cho php.

nkZ ,

5.2.2. Cc bi ton c bn

5.2.2.1. Kim tra bn

30

T iu kin bn ta c cng thc kim tra bn:

nkZZF

N, Thanh m bo bn

* Cc bc gii bi ton - Bc 1: V biu ni lc((vi thanh c tit din thay i th phi v biu ng sut)) - Bc 2:Xc nh im nguy him(im c ng sut ln nht trong thanh). Tnh ng sut sinh ra ti im nguy him - Bc 3: p dng iu kin bn kim tra bn

+ Nu maxZ : Thanh bn

+ Nu maxZ : Thanh khng bn Bi 1: Cho thanh AB chu ko ng tm (hnh v). Din tch mt ct

ngang: F=20(cm2), P=120KN, = 10(KN/cm2).

Kim tra bn cho thanh ? Bi lm

- Dng mt ct (1-1) ct thanh lm 2 phn , gi li phn phi kho st, Ta c phng trnh cn bng

NZ = P = 120(KN) - ng sut sinhh ra trn mt ct (1-1) ca thanh l

)/(620

120 2cmKNF

NZZ

p dng iu kin bn ta thy : Z Kt lun : Thanh m bo bn Bi 2: Cho thanh AC c tit din trn ng knh d = 2 cm, chu tc dng ca cc lc dc trc P1= 10kN, P2= 30kN .( Hnh 2-15)

a, V biu ni lc cho thanh AB ?

b, Kim tra bn cho thanh AC ? Bit nk , = 10 kN/cm2

Bi lm

a, Biu ni lc ca thanh AC: ( Hnh 2-16)

P 1 P 2 A B

C

Hnh 2-14

Hnh 2-15

1

1

1

1

P N

Z

P

31

b, Kim tra bn cho thanh AC Nhn vo biu ni lc ta thy on AB l nguy him nht .Vy ta ch cn kim tra bn cho on AB. Nu on AB m bo bn th thanh AC m bo bn p dng cng thc kim tra bn :

nkZ

ZF

N,

11

- Ni lc Nz = 20KN

- Din tch mt ct ngang : 222

14,34

2.14,3

4

.cm

dF

ng sut sinh ra trn mt ct ngang ca on AB l

211 /37,614,3

20cmKNZ

So snh ta thy 2,211 /10/37,6 cmKNcmKN nkZ

Kt lun : Thanh AC bn. Bi tp 3 : Cho thanh thng AD, tit din trn ng knh d = 2cm, chu tc dng ca lc dc trc P1= 30kN, P2= 50kN, P3=60kN.

-V biu ni lc cho thanh AD? - Kim tra bn ca thanh ? Bit = 10 KN/cm

2

Bi lm

a. V biu ni lc Hnh 2 - 18

P 1 P 2 P 3 B A D C

Hnh 2-17

P 1 P 2

1

1

2

2

A B

C

N Z

10kN

d X A

20kN

Hnh 2-16

32

b. Kim tra bn Nhn vo biu ta thy ni lc trn on AC l ln nht, do

AC l on nguy him nht . Nn ta ch cn kim tra bn cho on AC. Nu on AC m bo bn th thanh cng m bo bn . iu kin bn :

Tac: + KNNZ 4033

+ Din tch mt ct ngang ca thanh l :

F=2 23,14.2

4 4

d =3,14 (cm

2)

ng sut sinh ra trn mt ct ngang ca on AC :

So snh ta thy : 2,2 /10/7,12 cmKNcmKN nkZ

Kt lun : Thanh khng bn

Bi tp 4 : Thanh AE c tit din trn c cc ng knh tng ng l d1= 6cm, d2= 4cm, d3= 8cm (hnh 2-19), thanh chu tc dng ca cc lc dc trc P1=100 KN, P2=60 KN, P3= 140 KN.

a. V biu ni lc cho thanh AE ? b. V biu ng sut cho thanh AE? c. Tnh bn cho thanh AE ? Bit k,n = 10 KN/cm

2

F

NZZ

33

max

)/(7,1214,3

40

14,3

233

max cmKNN Z

Z

P 1 P 2 P 3 A B D E d 1 d 2 d C

Hnh 2-19

Hnh 2-18

P 2 P 3 P 1 P A A C D B

N Z

30kN

20kN

40kN

33

a.V biu ni lc:

+Bc1:Xc nh phn lc lin kt :

Phng trnh cn bng : PA - P1 + P2 P3 = 0 PA= P1- P2+ P3= 100 -60 +140

PA = 180 kN + Bc 2: Chia on cho thanh: Chia thanh lm 3 on l : AB, BD ,DE

+ Bc 3: Xc nh ni lc cho tng on -Xt on DE: Dng mt ct (1-1), ct thanh ,gi li phn tri kho st Ta c phng trnh cn bng NZ

1-1 P1 = 0

NZ1-1

= P1= 100 kN

Vy on DE chu ko,chiu NZ1-1

c chiu nh hnh v - Xt on BD:Dng mt ct (2-2), ct thanh,gi li phn phi kho st Ta c phng trnh cn bng NZ

3-3 + P2 P1 = 0

NZ3-3

=-P2+P1=-60+100= 40 kN

Vy on BD chu nn, chiu NZ2-2

c chiu nh hnh v. - Xt on AB:Dng mt ct (3-3),ct thanh, gi li phn phi kho st Ta c phng trnh cn bng NZ

3-3- PA = 0

NZ3-3

= PA =180 kN

Vy on AB chu ko, ni lc NZ

3-3 mang du dng, c chiu

hnh v + Bc 4: V biu ni lc Nz ( hnh 2-20)

* Trong bi ton ny tnh ton cho thanh c tit din thay i. V vy khng th da vo biu ni lc xc nh im nguy him.

180KN

3

3

2

2

P 1 P 2 P 3 A

B D E d 1 d 2 d 3 C

1

1

3

3

A

P A

P A N z 3-3

2

2

P 1 P 2 D E C N z 2-2

P 1 E

1

1

N z 1-1

N Z

100KN 40KN

Z

0

7,96KN/cm2 1,41

12,73

6,37

0

Hnh 2-20

34

Nh chng ta bit im nguy him l im c gi tr ng sut ln nht, do tm c im nguy him cn phi v biu ng sut. b. V biu ng sut:

Din tch mt ct ngang ca cc on AB, BC, CE tng ng l:

222

11 56,12

4

4.14,3

4

.cm

dF

222

22 14,3

4

2.14,3

4

.cm

dF

222

33 26,28

4

6.14,3

4

.cm

dF

+ ng sut sinh ra trn mt ct ca on DE l:

2

1

11

/96,756,12

100cmkN

F

NZDEZ

+ ng sut sinh ra trn mt ct ca on CD l:

2

3

22

/41,126,28

40cmkN

F

NZCDZ

+ ng sut sinh ra trn mt ct ca on BC l:

2

2

22

/73,1214,3

40cmkN

F

NZBCZ

+ ng sut sinh ra trn mt ct ca on AB l:

2

3

33

/37,626,28

180cmkN

F

NZDEZ

Biu ng sut: z (Hnh 2-20) * Nhn xt biu ng sut: Nhn vo biu ng sut ta thy

on BC nguy him nht. c. Kim tra bn cho thanh AE:

Nhn vo biu ta thy ng sut sinh ra trn mt ct ngang ca on BC l ln nht, do on BC l on nguy him nht . Nn ta ch cn kim tra bn cho on BC. Nu on BC m bo bn th thanh cng m bo bn . * p dng iu kin bn :

Ta c: 2

max /73,12 cmkNZBC

Z

2, /10 cmkNnk

So snh ta thy : 2,2

max /10/73,12 cmkNcmkN nkZ

* Kt lun : Thanh AE khng bn

5.2.2.2. Xc nh kch thc mt ct ngang hp l

Ni dung: Cho lc tc dng v ng sut cho php. Xc nh kch

maxZ

35

thc mt ct hp l sao cho thanh m bo bn ? T iu kin bn ta c

maxZNF

= F

F : l din tch mt ct cho php * Cc bc gii bi ton

- Bc 1: V biu ni lc - Bc 2: Xc nh ng sut ln nht trong thanh - Bc 3: Xc nh din tch mt ct t iu kin bn

* Ch : Chn mt kch thc F hp l trong khong :

FFFF %5

Bi 5: Cho thanh thng AD, tit din trn ng knh d, chu tc dng ca lc dc trc P1=30 KN, P2=50 KN, P3=60 KN .Bit k,n = 10 KN/cm

2 (hnh v). Xc nh ng knh hp l cho thanh ?

Bi lm Ta thy on AC l nguy him nht vy ta phi xc nh ng

knh hp l cho thanh v tr AC (Nu cn ta cng phi tnh ng knh hp l cho on CB v DB) Ta c:

KNNZ 4033

T iu kin bn ta c Din tch mt ct ngang ca thanh l

)(410

40 233

cmN

F Z

,

M c 4

. 2dF

)(26,2

14,3

4.44

4

. 2cmd

d

Chn ng knh d = 2,3 (cm)

5.2.2.3. Xc nh lc tc dng hp l

Ni dung: Cho kch thc mt ct ngang, ng sut cho php, phng, chiu, im t ca ti trng nhng cha bit tr s. Yu cu xc nh tr s ca ti trong sao cho thanh bn. T iu kin bn ta c :

PFPN nkZ ,.)( * Cc bc gii bi ton

- Bc 1: V biu ni lc - Bc 2: Xc nh ti trng t iu kin bn

* Ti trng tc dng hp l chn trong khong : PPPP %5 Bi 6: Cho thanh AB c din tch mt ct ngang F=8cm

2,

=10(KN/cm2). Xc nh lc tc dng hp l thanh bn ? Bi lm

36

T iu kin bn ta c

nkZ

ZF

P

F

N,

KNFP nk 8010.8. , * Ti trng tc dng hp l chn trong khong : PPPP %5

Vy chn P = 78 KN

CU HI N TP

1.Nu cc nh ngha thanh chu ko - nn ng tm, quy c du ni lc Nz trong thanh chu ko - nn ng tm ? 2.Trnh by phng php v biu ni lc trong thanh chu ko - nn ng tm ? 3.Nu cc nh ngha,vit biu thc tnh ng sut sinh ra trn mt ct ngang ca thanh chu ko - nn ng tm ? 4. Biu thc tnh bin dng di ca thanh ,cc nh lut Hc , nh lut Poat-xng

5. Vit biu thc tnh bin dng di ca thanh AB ? 6. Vit iu kin bn v cng thc kim tra bn cho thanh chu ko nn ng tm ?

BI TP

Bi 1: Cho thanh AB c tit din trn ng knh D = 4cm chu tc dng ca cc lc dc trc P1= 40KN ; P2= 70KN ; P3= 50KN. a) V biu ni lc cho thanh AB ?

b) Kim tra bn cho thanh AB ? Bit 2, /10 cmKNnk

Bi 2: Trc AB c ng knh khng i d = 4 cm , chu lc nh hnh v Bit : F1= 60 kN, F2 = 40 kN F3 = 50 kN , E = 2.10

4 kN/cm

2

a,V biu ni lc cho thanh ? b, Tnh bin dng di tuyt i cho trc AB?

c, Kim tra iu kin bn cho thanh ? Bit []k,n = 10 kN/cm2

P 1 P 2 P 3 A

C D B

Hnh 2-20

37

Bi 3: Trc AD c ng knh khng i d1= 4cm, d2= 6cm chu lc nh

hnh2-22. Bit : F1= 100 kN; F2 = 80 kN ; F3 = 40 kN

a,V biu ni lc cho thanh ? b, V biu ng sut cho thanh ? c, Kim tra iu kin bn cho thanh ? Bit []k,n = 10 KN/cm

2

Hnh 2-21

P 1 P 2 P 3 A B D d 1 d 2 C

Hnh 2-22

P 1 P 2 P 3 A

C D B

2m 1m 1m

38

CHNG III: CT V DP

M chng: CHI Gii thiu: Mc tiu:

+ Trnh by c khi nim v ct, dp. + Gii c ba bi ton c bn ca sc bn v ct v dp theo iu kin bn. + C thc trch nhim, ch ng hc tp.

1. HIN TNG CT

Mc tiu: + Trnh by c khi nim v ct. + Gii c ba bi ton c bn ca sc bn v ct theo iu kin bn.

1.1. nh ngha v ct, ni lc, ng sut v bin dng ct.

1.1.1. nh ngha v ct

Khi tc dng vo thanh hai lc song song, ngc chiu, cng ln v t hai mt phng ct st gn nhau th thanh s xy ra hin tng ct.

1.1.2. Ni lc: Qx , Qy

Dng phng php mt ct ta tng tng ct thanhlm hai phn. Gi li phn I kho st. Phn tri s cn bng di tc dng ca ngoi lc v ni lc sinh ra trn mt ct ngang thuc phn tri. Ni lc ny l Qx hoc Qy c : K hiu: Qx hoc Qy c

- Phng: Tip tuyn vi mt ct - Chiu: Ngc chiu ngoi lc tc dng - im t: Thuc mt ct - Tr s: Q = P

1.1.3. Bin dng

Xt mt thanh thng c tit din hnh ch nht chu tc dng ca hai lc P song song, ngc chiu, cng tr s v nm trong hai mt ct st gn nhau qua AB,CD. - Trc khi cho thanh chu ct : K ln mt ngoi ca thanh cc on thng song

I Q

P

II I

P

P

Hnh 3-1

39

song vi trc thanh, cc on thng ny c trng cho cc th dc v k cc on thng vung gc vi trc thanh, cc on thng ny c trng cho cc mt ct ngang v t hai mt phng ct st gn nhau. - Sau khi cho thanh chu ct : Bin dng ch xy ra phn vt liu c gii hn bi hai mt phng ct cha cc lc ct (hnh 3-3)

Nhn xt: - Mt ct cha lc tnh khng xy ra bin dng - Mt ct cha lc ng c xy ra bin dng : B dch chuyn thnh mt ct cha B`C`. - Qua nhiu ln lm th nghim ngi ta xc nh c rng BB`= CC`= S, kch thc mt ct ngang khng thay i, cc mt ct ngang vn phng v vn song song vi nhau. iu ny chng t bin dng ca cc phn t vt liu trong cng mt mt ct l ging nhau . - Cc th dc b trt so vi phng ban u mt gc , nhng vn song song vi nhau :gi l gc trt ca cc th dc

T hnh v ta c :

l

Stg

Kt lun : Bin dng trong thanh chu ct l bin dng trt ca vt liu .

1.1.4. ng sut

- Ni lc phn b u v c phng tip tuyn vi mt ct nn ng sut cng c phng tip tuyn vi mt ct hay ng sut trong thanh chu ct l ng sut tip, k hiu l C

C

xC

F

Q ,

C

y

CF

Q

n v : N/m2, kN/cm

2

1.2. nh lut Hoohs v ct. Trong giai on n hi, ng sut tip t l vi bin dng gc ca vt liu.

.G Trong :

A

B'

C' D

l

P c

C

B

P c

Hnh 3-3

l

C D

A B

Hnh 3-2

Hnh 3-4

I Q

P

40

: Bin dng gc

G : M dun n hi trt Thp: G= 8.10

6N/cm

2

1.3. Tnh ton cho ct

1.3.1. iu kin bn

Thanh chu ct tha mn iu kin bn khi ng sut ln nht pht sinh trong thanh phi nh hn ng sut cho php.

c c


a. Kim tra bn

T iu kin bn ta c cng thc kim tra bn:

Trong thc t thng gp trng hp chu lc theo phng trc Oy, rt t trng hp chu lc theo phng trc Ox.

+ Trng hp lc tc dng theo phng trc Oy

cc

y

cF

Q m bo bn

+ Trng hp lc tc dng theo phng trc Ox

cc

xc

F

Q m bo bn

V d 1 : Cho thanh AB tit din trn c ng knh d = 4cm. Thanh chu tc dng ca lc P = 120KN .

Kim tra bn cho thanh AB: Bit 2/8 cmKNc Bi lm

Dng mt ct ngang ct thanh lm 2 phn , gi li mt phn phi kho st Ta c phng trnh cn bng

Qy = P =120(KN) + Din tch mt ct ngang ca thanh l:

222

56,124

4.14,3

4

.cm

dFc

- ng sut sinhh ra trn mt ct ngang ca thanh l

)/(55,956,12

120 2cmKNF

Q

c

y

c

iu kin bn : cc So snh ta thy : 22 /8/55,9 cmKNcmKN cc

Kt lun : Thanh khng m bo bn

b. Xc nh kch thc mt ct ngang hp l

41

T iu kin bn ta c

FQ

Fc

y

c

* Mt ct hp l chn trong khong : FFFF %.5 V d 2: Cho thanh AB tit din trn c ng knh d . Thanh chu tc dng ca lc P = 120KN .Xc nh ng knh hp l cho thanh AB. Bit 2/8 cmKNc

Bi lm

T iu kin bn ta c

FQ

Fc

y

c

2158

120cmFc

M din tch mt ct ngang l : 4

. 2dFc

Ta c : cmdd

37,414,3

4.1515

4

. 2

ng knh hp l chn trong khong : FFFF %.5 Chn d = 4.5 cm

c. Xc nh lc tc dng hp l

T iu kin bn ta c

PFP ccc . * Ti trng tc dng hp l chn trong khong : PPPP %5 V d 3: Cho thanh AB tit din trn c ng knh d . Thanh chu tc dng ca lc P = 120KN .Xc nh ng knh hp l cho thanh AB.

Bit 2/8 cmKNc Bi lm T iu kin bn ta c

PFP ccc . + Din tch mt ct ngang ca thanh l:

2

22

56,124

4.14,3

4

.cm

dFc

KNPc 48,1008.56,12

*Ti trng tc dng hp l chn trong khong:

PPPP %5

Chn P = 98 KN

2. HIN TNG DP

Mc tiu: +Trnh by c khi nim v dp. + Gii c ba bi ton c bn ca sc bn v dp theo iu kin bn.

10

8

30

Hnh 3-5

42

2.1. nh ngha

Dp l hin tng nn cc b xy ra trn mt din tch truyn lc tng i nh ca hai chi tit p vo nhau.

V d: inh tn trong mi ni chu ct, thnh l ca tm ghp p vo thn inh gy hin tng dp, ng thi thn inh cng b ct ngang,

2.2. Ni lc:

Ni lc sinh ra trn mt ct chu dp K hiu :Pd c phng vung gc vi mt ct chu dp

2.3. Bin dng

Xt thanh thng chu dp (hnh 3-7a) Sau khi tc dng vo thanh h lc phn b Pd (hnh 3-7b) Nhn xt :

- Bin dng ch xy ra phn chu lc - Khong cch gia cc th dc trong din tch chu dp b thu hp li (b nn)

Kt lun :Bin dng trong thanh chu dp l bin dng nn ca vt liu

2.4. ng sut

Ni lc c phng vung gc vi mt ct chu dp,nn ng sut sinh ra trn mt ct chu dp l ng sut php K hiu : d

d

dd

F

P (N/m

2, kN/cm

2)

2.5. Tnh ton cho thanh chu dp

2.5.1. iu kin bn

Thanh chu ct tha mn iu kin bn khi ng sut ln nht pht sinh trong thanh phi nh hn ng sut cho php.

dd

P P

P P

Hnh 3-6

P d

a,

Hnh 3-7

b,

43


a. Kim tra bn

Cng thc kim tra bn :

dd

dd

F

P m bo bn


T iu kin bn ta c

dd

dd F

PF


T iu kin bn ta c

ddd FP .

Bi tp 1:

Mi ghp inh tn gm c 8 inh, ng knh thn inh d = 0,8cm chu tc dng lc P = 30kN (Hnh 3-12),C chiu dy cc tm 1= 3cm, 2= 4cm. Kim tra bn ct ,dp cho inh tn ?

Bit: =8 kN/cm2. Bi lm

* Kim tra bn ct

- p dng cng thc kim tra bn ca thanh chu ct

cc cc

P

F

- ng sut sinh ra trn din tch mt ct ca 8 inh tn l:

ccc

P

F

- Bit PC = 40KN - Din tch mt ct ca mt inh tn

222

1 5,04

8,0.14,3

4

.cm

dF

- Din tch mt ct 8 inh tn FC = 8.F1 = 8.0,5 = 4 (cm

2)

2/5,74

30cmkNC

- So snh vi ng sut cho php, ta thy: cc

P P

P

Hnh 3-12

44

inh tn m bo bn ct.

* Kim tra bn Dp

p dng iu kin bn ca thanh chu dp

dd dd

P

F

C: Pd = 40KN

Din tch b mt chu dp ca 8 inh tn Fd = n. 1.d = 8.3.0,8 = 19,2 (cm

2)

Vy ng sut sinh ra trn b mt chu dpca 8 inh tn

2/6,12,19

30cmkNd

So snh vi ng sut cho php ta c: d d Vy inh tn m bo bn dp.

+ Kt lun: inh tn m bo bn

CU HI N TP

1.nh ngha thanh chu ct ? 2.Vit cc biu thc ca cc bi ton tnh ton cho thanh chu ct? 3.nh ngha thanh chu dp ? 4.Vit cc biu thc ca cc bi ton tnh ton cho thanh chu dp?

BI TP

Bi 1: Mi ghp inh tn gm c n = 8 inh, ng knh thn inh

d=0,8cm, chu tc dng lc ngang P = 60kN, bit : =8kN/cm2.

2/10 cmkNd . C chiu dy cc tm ghp l 1= 2cm, 2= 3cm Kim tra bn ct v bn dp cho inh tn ?

Bi 2: Mi ghp inh tn gm c n = 4 inh, c ng knh thn inh l

d, chu tc dng lc ngang P = 80KN, bit : =10kN/cm2. C chiu dy cc tm ghp l 1= 2cm, 2= 3cm Tnh ng knh hp l cho inh khi chu ct v dp?

P

P

P

P

1 2

Hnh 3-19

Hnh 3-20

2

P

P

P

P 1

45

CHNG IV:

C TRNG HNH HC CA MT CT

M chng: CHIV Cc c trng hnh hc xc nh cho cc mt ct ngang ca thanh chu lc, n c xc nh cho tng mt ct c th tng ng vi hnh dng ca mt ct. Cc c trng hnh hc ny c s dng tnh ton cho thanh chu un, xon v thanh chu lc phc tp. Mc tiu:

+ Trnh by c khi nim v momen tnh v cc momen qun tnh.

+ Xc nh c trng tm ca hnh phng. + V c h trc qun tnh chnh trung tm. + Tnh c momen qun tnh chnh trung tm. + C thc trch nhim, ch ng hc tp.

1. M MEN TNH

Mc tiu: + Trnh by c khi nim v mmen tnh + Vit c cng thc xc nh ta trng tm

+ Xc nh c ta trng tm Xt mt mt ct c din tch F v h trc

xoy ta ly xung quanh im M(x, y) mt phn t c din tch dF

1.1. nh ngha : K hiu Sx ; Sy

Mmen tnh ca mt ct F ly i vi trc

Ox,Oy c nh ngha bng biu thc sau:

Mmen tnh ca mt ct F ly i vi trc Ox

.xF

S y dF (chiu di)3

Mmen tnh ca mt ct F ly i vi trc Oy

.yF

S x dF (chiu di)3

Tr s ca mmen tnh c th (+); (-) hoc khng Nu mmen tnh ca mt ct F ly i vi mt trc no bng

khng th trc trung tm gi l trng tm ca mt ct.

O

F

M

x

y

y

x

Hnh 4-1

46

1.2. Cng thc xc nh ta trng tm

- Gi s xoy l h trc trung tm th khi :

+ Sx = S y = 0 ; Xc= a, Yc= b

-> SX = YC.F ; SY = XC.F

Cng thc xc nh trng tm

,X YC C

S SY X

F F

b. Ta trng tm C ( xC, yC ) ca hnh phng phc tp (ghp bi nhiu hnh n gin)

i

iCi

n

nCnCC

CF

Fx

FFF

FxFxFxx

.

....

.......

21

2211

i

iCi

n

nCnCC

CF

Fy

FFF

FyFyFyy

.

....

.......

21

2211

Trong : (xCi , yCi ):L ta trng tm cu hnh phng th i Fi : Din tch ca hnh th i V d 4-1 : Xc nh ta trng tm ca tm phng (Hnh 4 -3) - t h trc ta Oxy vo tm phng

- Gi C(xc,yc) l trng tm ca c tm phng - Chia hnh phng lm 2 phn : 1, 2 - Ta trng tm ca cc tm phng tng ng l : C1(x1,y1) ; C2(x2, y2)

x1=2m; y1 =3m ; x2= 5m, y2= 2m

Din tch cc tm phng tng ng l S1= 24m

2 , S2= 9m

2

Thay vo cng thc ta c :

mSS

SxSxxC

11

31

924

9.524.2..

21

2211

;

mSS

SySyyC

11

30

924

9.224.3..

21

2211

Trng tm ca c tm phng trn l C

11

30,

11

31

V d 4-2 :

Xc nh ta trng tm ca tm phng(Hnh 4 -4) - t h trc ta Oxy vo tm phng - Gi C(xc,yc) l trng tm ca c tm phng - Chia hnh phng lm 2 phn : 1, 2

- Ta trng tm ca cc tm phng tng

Y c

X c O 1

O

F

x

y

X

Y

C

Hnh 4-2

Hnh 4 -3

7m

6m

4m

y

x O

C 1 C 2

C

x C

(1) (2)

yC

47

ng l : C1(x1,y1) ; C2(x2, y2)

x1=20cm; y1 =20cm; x2= 20cm; y2= 8,5cm

- Din tch cc tm phng tng ng l

S1= 1800cm2 , S2= 628cm

2

- Din tch c tm phng:

S= S1 - S2 = 1800 - 628=1172cm2

Thay vo cng thc ta c :

cmSS

SxSxxC 20

1172

628.201800.20..

21

2211

;

cmSS

SySyyC 16,26

1172

628.5,81800.20..

21

2211

Trng tm ca c tm phng C (20; 26,16)

2. KHI NIM V MMEN QUN TNH.

Mc tiu: + Trnh by c khi nim cc mmen qun tnh trc, mmen qun tnh c cc, mmen qun tnh ly tm. + Xc nh c c trng hnh hc ca mt s mt ct n gin + V c h trc qun tnh chnh trung tm.

+ Tnh c momen qun tnh chnh trung tm. 2.1. nh ngha.

2.1.1. Mmen qun tnh trc : K hiu Jx ; Jy

Mmen qun tnh ca mt ct F ly i vi trc Ox, Oy

2.xF

J y dF (chiu di)4

2.yF

J x dF (chiu di)4

Tr s mmen qun tnh lun dng

2.1.2. Mmen qun tnh c cc : (mmen qun tnh i vi mt im).

K hiu Jp Mmen qun tnh c cc ca mt ct F ly i vi gc ta

,c nh ngha bng cng thc sau :

2.F

J dF (chiu di)4

2 2( ). x yF

J x y dF J J (chiu di)4

Hnh 4 -4

60cm 40cm

60cm

y

x O

C

(1)

(2)

y C

x C

48

iu ny chng t mmen qun tnh c cc bao gi cng c tr s dng.

2.1.3. Mmen qun tnh ly tm : K hiu Jxy

Mmen qun tnh ca mt ct F ly i vi h trc xoy

. .xyF

J x y dF (chiu di)4

Tr s ca mmen qun tnh ly tm c th (+), (-) hoc khng

+ Mmen qun tnh chnh trung tm:

Khi mmen qun tnh ly tm ca mt ct ly i vi h trc no

bng khng th h trc gi l h trc qun tnh chnh trung tm.

Nu mt ct c mt trc i xng th trc i xng chnh l

mt trc ca h trc qun tnh chnh trung tm.

- H trc qun tnh chnh l h trc c m men qun tnh ly tm bng khng ( Jxy= 0 ) - H trc qun tnh chnh trung tm l h trc c gc ta trng vi trng tm mt ct (Jxy= Sx = Sy = 0)

- M men qun tnh trc ca h trc qun tnh chnh gi l m men qun tnh chnh - Khi mmen tnh ca din tch F i vi mt trc no bng khng, th trc c gi trc trung tm. Giao im ca hai trc trung tm c gi l trung tam ca mt ct.

Mmen qun tnh trc ca h trc qun tnh chnh trung tm gi l mmen qun tnh chnh trung tm

* Ch : Nu mt ct c mt trc i xng th h trc c to bi mt trc i xng v trc vung gc vi n chnh l h trc qun tnh chnh

2.3. c trng hnh hc ca mt s mt ct n gin

2.3.1. Mt ct hnh ch nht

- M men tnh : Sx=Sy=0

- M men qun tnh ly tm : Jxy=0

- M men qun tnh trc : 12

. 3hbJ x

12

. 3bhJ y

h

b

O x

y

49

- Bn knh qun tnh : 2 3

xx

J hi

F

2 3

y

y

J bi

F

2.3.2. Mt ct hnh tam gic :

- M men qun tnh trc : Jx= 3

12

bh

Jy= 3

12

hb

2.3.3. Mt ct ngang l hnh trn c



- M men qun tnh c cc : 4

40,132

DJ D

- M men qun tnh trc :

4

40,052 64

x y

J DJ J D

- Bn knh qun tnh:

4

2

4

64 4

x

x y

J D Di i

F D

2.3.4. Mt ct ngang l hnh trn rng



- M men qun tnh c cc :

4

4 4 41 0,1 132

DJ D

- M men qun tnh trc :

4 40,05 12

x y

JJ J D

y

x b

h

O

y

x D

Hnh 4-5

Hnh 4-6

Hnh 4-7

y

d

x

O D

Hnh 4-8

50

- Bn knh qun tnh : 214

x

x y

J Di i

F

Trong : d

D

2.3.5 Mt ct ngang ca cc thp nh hnh

Tra bng c trng hnh hc ca cc mt ct trong cc s tay k

thut

2.4. Cng thc chuyn trc song song. (nh l tnh tin trc)

Trong thc t ta thng gp cc chi tit, b phn cng trnh m

tit din mt ct ngang c ghp bi nhiu tit din n gin, to

kh nng chu lc tt nht. Tit kim nht. iu ny yu cu chng ta

phi bit cch tnh cc loi mmen qun tnh khi bit mmen qun tnh

ca nhng hnh n gin.

- Cho mt ct F vi cc c trng hnh hc ca h trc xoy coi nh bit. ( Hnh 4-7)

- Xc nh c trng hnh hc ca mt

ct F i vi XO1Y c to thnh khi ta

tnh tin trc 0x i mt khong b v 0y i

mt khong a

X x a

Y y b

2.4.1. Mmen tnh

Mmen tnh ca mt ct F ly i vi trc O1X, O1Y

SX= . ( ) . .x xF F F

Y dF y b dF S b dF S b F

Tng t: SY= .yS a F

Hnh 4-9

b

a

Y

X

Y

O 1

X

x

y

y

x

dF

O

Hnh 4-10

51

2.4.2. Mmen qun tnh

Mmen qun tnh ca mt ct F ly i vi trc O1X, O1Y

JX =2 2 2 2 2( ) ( 2 ) 2

x x

F F F

Y dF y b dF y yb b dF J bS b F

Tng t: JY=Jy + 2aSy + a2F

2.4.3. Mmen qun tnh li tm

- Mmen qun tnh li tm ca mt ct F ly i vi h trc XO1Y

( )( ) . .XY xy y x

F F

J XYdF x a y b dF J b S a S abF

- Gi s xoy l h trc trung tm (Hnh 4-11), th khi :

+ Sx = S y = 0 ; Xc= a, Yc= b

Ta c: SX = YC.F ; SY = XC.F

JX = Jx + b2F

JY= Jy + a2F

Cng thc xc nh trng tm

,X YC CS S

Y XF F

+ Mmen tnh ca mt mt ct ly i vi mt trc th bng din tch ca mt ct nhn vi khong cch t trng tm mt ct y ly i vi trc.

JX = Jx+(YC)2F

JY = Jy+(XC)2F

JXY = Jxy+XCYCF

3. BN KNH QUN TNH : K hiu ix ; iy

,yx

x y

JJi i

F F (chiu di)

V d 4.3: Xc nh m men qun tnh chnh trung tm ca hnh phng (Hnh 4 -12)?

Lp h trc xOy, Oy l trc i xng nn l trc qun tnh chnh trung tm.

* Xc nh trng tm ca hnh phng: C xC = 0

Y c

X c O 1

O

F

x

y

X

Y

C

Hnh 4-11

52

cmSS

SySyyC 2,2

128

12.18.4..

21

2211

Lp h trc qun tnh chnh trung tm x0Cy0

* Xc nh m men qun tnh chnh trung tm Jx0, Jy0

p dng cng thc:

21

xoxoxo JJJ

Ta c:

15

3768.8,1

12

4.2.

12

. 3

11

31 FCC

hbJ xo

5

9212.2,1

12

2.6.

12

. 3

22

32 FCC

hbJ xo

15

652

5

92

15

37621 xoxoxo JJJ

Tng t ta c

3

116

12

6.2

12

2.4 3321 yoyoyo JJJ

M men qun tnh chnh trung tm l:

Jx0= 15

652 , Jy0=

3

116

CU HI N TP

1. nh ngha mmen tnh,vit cng xc nh ta trng tm ca hnh phng ?

2. nh ngha cc mmen qun tnh trc, mmen qun tnh c cc, mmen qun tnh ly tm?

3. Vit cc cng thc xc nh cc c trng hnh hc ca mt s mt ct n gin ?

4. nh ngha h trc qun tnh chnh trung tm, m men qun tnh chnh trung tm ?

5. Vit cc cng thc chuyn trc song song ?

6cm

2cm

4cm

2cm

x 0

x

y y 0

C 1

C C 2

Hnh 4 -12

53

BI TP

Bi 1: Xc nh m men qun tnh chnh trung tm ca hnh phng (Hnh 4 -13)?

Bi 2: Xc nh m men qun tnh chnh trung tm ca hnh phng

(Hnh 4 -14)?

4cm

14cm 8cm

8cm

Hnh 4 -13

6cm 2cm 6cm

6cm 4cm 8cm

Hnh 4 -14

54

CHNG V : XON THUN TY THANH TRN

M chng: CHV

Bin dng xon thun ty thanh trn chng ta gp rt nhiu trong thc t c bit l trong cc chi tit my dng trc.

V d: Mi khoan khi ang khoan, trc vt, trc bnh li, cha vn....

Mc tiu: + Trnh by c khi nim v xon thun ty, bin dng trong xon. + V c biu momen xon ni lc, phn tch v tnh c ng sut trn mt ct. + Tnh c bin dng trong thanh chu xon. + Tnh thnh tho ba bi ton c bn ca sc bn theo iu kin bn v iu kin cng. + C thc trch nhim, ch ng hc tp.

1. NI LC V BIU NI LC

Mc tiu: + Trnh by c khi nim v xon thun ty + V c biu mmen xon ni lc

1.1. nh ngha

Thanh chu xon thun ty l thanh m ngoi lc tc dng l cc ngu lc hay cc mmen c chiu quay ngc nhau v c mt phng tc dng trng vi cc mt ct trong thanh.

V d: Mi khoan, trc ng c, trc hp gim tc

1.2. Ni lc v biu m men xon ni lc

1.2.1: Ni lc

Xt thanh thng c tit din trn chu tc dng ca cc m men nh hnh v (Hnh 5-1)

Dng phng php mt ct xc nh ni lc.

Ta xc nh c m men xon ni lc Mz c: - Phng: Trng vi mt ct ngang ca thanh - Tr s: Bng tng i s ca mmen ngoi lc tc dng

Hnh 5-1

m m

m M Z

55

*Quy c du

Mmen xon ni lc : K hiu : Mz

+ Nhn t bn ngoi vo mt ct thy mmen Mz quay cng chiu kim ng h th Mz mang du dng. + Nhn t bn ngoi vo mt ct thy mmen Mz quay ngc chiu kim ng h th Mz mang du m.

- n v: N.m, KN.m,

1.2.2. Biu ni lc

a. Cc bc v biu ni lc.

- Bc 1: Xc nh phn lc lin kt (nu cn) - Bc 2: Chia on cho thanh, da trn c s v tr tc dng ca

mmen tng ng vi mt im ,hai im lin tip l mt on. - Bc 3: Xc nh ni lc trong tng on

+ Dng phng php mt ct ,ct thanh lm hai phn ,gi li mt phn kho st + t ni lc vo mt ct (gi nh ni lc Mz dng ) + Vit phng trnh cn bng v gii cc phng trnh gi tr ca ni lc - Bc 4: V biu ni lc.


*V d 1: Cho thanh chu xon thun ty nh trn ( hnh 5-2) : m1 = 20KNm, m2= 60KNm. V biu ni lc cho thanh ?

Bi lm

- B1: Xc nh phn lc lin kt (Hnh5-3) Ta c phng trnh cn bng

021 mmmm Az mKNmmmA .40206012

- B2: Chia on cho thanh: AB, BC - B3: Xc nh ni lc trn tng on + Xt on AB:Ct AB bi mt phng (1-1), xt cn bng phn bn phi, ta c:

m 1

A B C

m 2

Hnh 5-2

56

01211 mmM Z

KNmmM Z 2030501211

+ Xt on BC: Ct BC bi mt ct (2-2), xt cn bng phn bn phi, ta c:

0122 mM Z

KNmM Z 30122

- B4: V biu ni lc (Hnh5-3) *Nhn xt : Nhn vo biu ta thy on AB l nguy him nht

2. NG SUT V BIN DNG TRONG THANH TRN CHU XON

Mc tiu: + Trnh by c bin dng trong thanh chu xon thun ty. + Phn tch v tnh c ng sut sinh ra trn mt ct ngang. + Tnh c bin dng trong thanh chu xon.

2.1. Bin dng

Xt mt thanh thng c tit din trn, chiu di l l, bn knh R. + Trc khi cho thanh chu xon (Hnh 5- 4)

- K ln mt ngoi ca thanh cc ng thng song song vi trc thanh, cc ng thng ny c trng cho cc th dc.

- K cc ng trn vung gc vi trc ca thanh, cc ng ny c trng cho cc mt ct ngang.

Tc dng vo thanh m men xon m lm cho thanh chu xon. + Sau khi cho thanh chu xon (Hnh 5- 5)

l a

Hnh 5- 4

40KN.m

2 1

m 1

A B C

2 1 m 2 m A

m A 1

A

1

2

C

m 1

2

M z 1-1

M z 2-2

20KN.m

0 Mz

Hnh 5-3

57

Nhn xt : - Cc th dc:

+ Cc th dc b lch i so vi ban u mt gc gi l , khi xt

cc th dc trn cng mt tr ngoi cng th cc gc bng nhau. + on thng biu th cho th dc l tp hp ca v s im do

khi on thng b lch i mt gc th cc im trn on thng ny cng b dch chuyn, thy c iu ny chng ta xt ba im nh trn hnh: A, B, C. im A khng b dch chuyn, im B dch chuyn n v tr B, im C dch chuyn n v tr C. Vy im B v C u b trt trn chu vi ngoi cng ca mt ct ngang m c th l trt trn cc cung trn c ln khc nhau. Nh vy khi th dc b lch i mt

gc th ch c im A v tr lin kt ngm cng l khng b dch chuyn cn cc im cn li trn th dc u b trt trn cc cung trn c kch thc khc nhau v cc cung ny u chn gc v vy gc c gi l gc trt.

Ta c gi thit bin dng trong lng thanh ging bin dng bn

ngoi thanh vy th cc th dc trong thanh cng b lch i mt gc

v cc th dc trn cng mt mt tr th gc bng nhau vy khi ta xt

cc th dc trn cc mt tr khc nhau th gc c bng nhau khng, chng ta cng tm hiu. Trn mt tr ny th ng mu trng vn th hin th dc khi chua bin dng v ng mu vng th hin th dc khi bin dng.

Trong cng l mt th dc trng vi trc thanh. Vy ta thy th dc trng vi trc thanh khng b bin dng vy gc =0. Cng tin ra

mt tr ngoi cng th gc cng tng dn v mt tr ngoi cng th

gc t gia tr ln nht l max

Ta c: max

0

- Cc mt ct ngang: + Khong cch gia cc mt ct ngang khng i, chiu di thanh

khng i -> thanh khng b bin dng dc theo trc thanh. + Hnh dng ca cc mt ct ngang vn l hnh trn, vn phng v

vn vung gc vi trc thanh.

Hnh 5-5

m

l a

58

+ Kch thc ca mt ct ngang ca thanh trc v sau bin dng vn khng i v bng F.

+ Xt mt bn knh bt k trn mt ct ngang ta thy khi thanh chu xon bn knh ny vn thng nhng b xoay i mt gc quanh tm chng t mt ct ngang cng b xoay i mt gc quanh tm. V gc

xoay ny ta gi l . Quan st trn hnh ta thy cc mt ct ngang khc

nhau th gc c ln khc nhau v max

0 .

Khi nghin cu v bin dng ca cc th dc v mt ct ngang ta u thy khi thanh chu xon cc phn t vt liu u b trt trn cc cung trn quanh tm do bin dng trong thanh chu xon thun ty l bin dng trt.

Vy bin dng trt ny phn b nh th no trn mt ct ngang? Kt lun: Bin dng l bin dng trt. Bin dng phn b khng

u trn mt ct.

2.2. ng sut

2.2.1. Biu phn b ng sut trn mt ct ngang

Theo nh lut Hc c: .G Trong :

- G l mun n hi trt ca vt liu G = const - l bin dng trt ca vt liu + Quy lut phn b ng sut:

- Khi R=0 = 0 x = 0

- Khi R tng tng x tng

- Khi Rmax max x max Biu phn b ng sut trn mt ct ngang (Hnh 5-6) Nhn xt biu :

- ng sut tng dn t tm mt ct n bn knh ln nht ca mt ct v t gi tr ln nht khi bn knh ln nht.

- ng sut c gi tr thay i t max0 x

2.2.2. ng sut ln nht trn mt ct ngang

* ng sut ln nht c xc nh bng cng thc:

p

Z

W

Mmax

Trong : - Mz: M men xon ni lc (N.cm ; KN.m ,) - Wp: Mmen chng xon ca mt ct ngang ca thanh (chiu

di3)

Hnh 5-6

B A

O

max

59

R

JW

p

p

+ Vi mt ct ngang ca thanh c tit din trn c :

334

.02,016

.

.32

2..D

D

D

D

R

JW

p

p

+ Vi mt ct ngang ngang ca thanh c tit din trn rng

D

dD

DWp

;1..2,0

32

1.. 4343

Trong : - D l ng knh ngoi - d l ng knh trong

3. IU KIN BN V CC BI TON C BN

Mc tiu: + Tnh thnh tho ba bi ton c bn ca sc bn theo iu kin bn v iu kin cng.

3.1. iu kin bn v ba bi ton c bn.

3.1.1. iu kin bn iu kin cn v thanh chu xon thun ty m bo bn

l ng sut sinh ra trn mt ct ngang ca thanh phi nh hn hoc bng ng sut cho php.

x max Nu chi tit m bo iu kin trn n s m bo bn khi chu lc. 3.1.2. Ba bi ton c bn

a. Kim tra bn

T iu kin bn ta c cng thc kim tra bn :

xp

z

W

M max

- Tm ng sut ln nht - So snh ng sut ln nht vi ng sut cho php - Kt lun

+ Nu max X thanh bn + Nu max X thanh khng bn


T iu kin bn ta c

xz

p

MW

60

+ Vi mt ct ngang ca thanh c tit din trn c :

Ta xc nh c ng knh hp l ca thanh l

DMD

x

z 32,0.

* Ch : Nn chn ng knh hp l ca thanh trong khong:

DDDD %5


( )X X XM m W BI TP V D:

Bi 1 : Cho thanh chu xon thun ty nh trn hnh v: m1=30kNcm , m2=60kNcm ; m3 = 50kN.cm V biu m men xon ni lc cho thanh ?

Bi lm

- Bc 1: Xc nh phn lc lin kt Ta c phng trnh cn bng

0321 mmmmm AA 306050213 mmmmA

mkNmA .40

- Bc2: Chia on cho thanh: AB, BC, CD

- Bc3: Xc nh ni lc trn tng on

+ Xt on AB: Dng mt ct (1-1) ct thanh, xt cn bng phn bn phi, ta c:

011 AZ mM

kNmmM AZ 4011

+ Xt on BC: Dng mt ct (2-2) ct thanh, xt cn bng phn bn phi, ta c:

m 3

C B A

m 2 m 1

D

Hnh 5-7

40kN.m

3

3

2-2 M z

1-1 M z

2

m 2

C

2

1

A

1 m A

m A m 3 1 2

C B A

m 2

1 2

m 1

m 1

D

D

3

3 m 1

D

3-3 M z

90kN.m

30kN.m

M z

Hnh 5-8

61

02122 mmMZ

kNmmmMZ 9060302122

+ Xt on CD: Dng mt ct (3-3) ct thanh, xt cn bng phn bn phi, ta c:

0133 mM Z

kNmmM z 30133

- Bc4: V biu ni lc (Hnh 5-8) *Nhn xt : Nhn vo biu ta thy on BC l nguy him nht

3.2. iu kin cng v ba bi ton c bn.

3.2.1. iu kin cng

L iu kin sao cho: max []

- max l gc xon t i ln nht tnh c (n v: Rad/m).

- [] l gc xon t i cho php thng cho mRad / , (nu cho l m/0 th i ra Rad/m vi 360

0 =2. rad)

- Trng hp thanh ch c mt mmen xon ngoi lc v tit din khng i:

p

z

JG

M

.max

Trng hp thanh c nhiu on, mi on c ni lc Mzi v

cng GJpi khc nhau th ta phi tnh i trn tng on: pi

zii

JG

M

. sau

tm max kim ta theo iu kin cng.

3.2.2. Ba bi ton c bn.

+ Bi ton kim tra cng + Bi ton xc nh kch thc hp l theo iu kin cng + Bi ton xc nh ti trng cho hp l theo iu kin cng

CU HI NG TP

1. Trnh by cc nh ngha thanh chu xon nu quy c du ni lc Mz v cc bc v biu ni lc Mz ca thanh chu xon thun ty ?

2. V biu phn b ng sut trn mt ct ngang ca thanh chu xon thun ty?

3. Vit cng thc tnh ng sut ln nht sinh ra trn mt ct ngang ca thanh chu xon thun ty? Gii thch cc k hiu?

4. Vit cc cng thc tnh ton ca cc bi ton c bn tnh theo iu kin bn cho thanh chu xon thun ty ?

62

5. Vit cc cng thc tnh ton ca cc bi ton c bn tnh theo iu kin cng cho thanh chu xon thun ty ?

BI TP

Bi 1 : Cho thanh chu xon thun ty nh trn hnh v: m1=30KNm, m2=50KNm

V biu m men xon ni lc cho thanhAC ?

Bi 2: Trc chu xon thun ty c tit din trn ng knh d= 4cm nh trn (hnh 5-9) :Trc chu tc dng ca cc mmen m1=80KNm, m2=50KNcm ; m3 = 60KN.cm a.V biu ni lc cho thanh ?

b. Kim ta bn cho thanh AD? Bit 2/10 cmKNx

Bi 3: Trc AB c tit din trn c cc ng knh tng ng l d1= 4cm , d2= 6cm ; (hnh 5-10), thanh chu tc dng ca cc lc dc trc m1=50kN.cm ; m2=160 kN.cm

a. V biu ni lc cho thanh AD ? b. V biu ng sut cho thanh AD?

c. Tnh bn cho thanh AD ? Bit ]x = 10 kN/cm2

m 2

C B A

m 1

Hnh 5-8

m 3 m 2 m 1

A B D C

Hnh 5-9

A D d 1 d 2

m 1 m 2

Hnh 5-10

63

CHNG VI : UN NGANG PHNG THANH THNG

M chng: CHVI

Bin dng un ngang phng thanh thng chng ta gp rt nhiu trong thc t c bit l trong cc chi tit my, cc dm chu ti thng ng. V d : Thanh dm ca kt cu mi, dm chu ti thng ng trong kt cu dn....

Mc tiu: + Trnh by c khi nim v un ngang phng. + V c biu ni lc trong thanh chu un ngang phng. + p dng thnh tho ba bi ton c bn theo iu kin bn v ng sut php +Tnh c vng v gc xoay ca mt s dm chu un n gin. + C thc trch nhim, ch ng hc tp.

1. KHI NIM V UN NGANG PHNG

Mc tiu:

+ Trnh by c khi nim v un ngang phng.

- Khi c ngoi lc tc dng, trc ca thanh b cong i ngi ta ni thanh chu un. - Nu trc thanh b cong nhng vn nm trong mt phng thng ng th thanh b un ngang phng - Ngoi lc: lc tp trung, lc phn b, ngu lcnm trong mt phng ti trng ca thanh - Mt phng ti trng ca thanh l mt phng cha ti trng ca thanh v trc ca thanh . - Khi ngoi lc tc l cc ngu lc hoc mmen lc c mt phng tc dng trng vi mt phng ti trng ca thanh th thanh chu un phng thun ty.

2. NI LC V BIU NI LC

Mc tiu: + Trnh by c quy c du v ni lc Qy v MX trong thanh

chu un ngang phng. + V c biu ni lc trong thanh chu un ngang phng.

2.1. Ni lc

- Thanh un phng c hai thnh phn ni lc l lc ct Qy v mmen un ni lc MX - Thanh un phng thun ty c mt v ch mt thnh phn ni lc l mmen un ni lc MX

64

- Quy c du: + Lc ct Q mang du (+) khi php tuyn ngoi ca mt ct

quay 900 theo chiu kim ng h n trng vi vc t lc Qy v ngc

li Qy mang du m + Mmen un c du (+) nu ni lc lm cho th pha di ca

dm b dn ra, th trn ca dm b co li

2.2. Biu ni lc

Cc bc v biu ni lc. - Bc 1: Xc nh phn lc lin kt (nu cn) - Bc 2: Chia on cho thanh, da trn c s im t ca lc

tng ng vi mt im ,hai im lin tip l mt on. - Bc 3: Xc nh ni lc trong tng on

+ Dng phng php mt ct ,ct thanh lm hai phn ,gi li mt phn kho st + t ni lc vo mt ct (gi nh ni lc dng) + Vit phng trnh cn bng v gii cc phng trnh - Bc 4: V biu ni lc.


V d 6.1: Cho dm AC di a=1m, chu tc dng lc un P=100N. V biu ni lc Qy, Mx cho dm AC ?

Bi lm - Bc1: Xc nh phn lc lin kt

a a

P

B A

C

Hnh 6-2

Phn phi Phn tri

Qy>0

Mx>0 Qy>0

Mx>0

Hnh 6-1

65

0

0

100( )

. .2 0

10050( )

2 2 2

A

A C

A C

A C

C A C

X X

Y Y Y P

Y Y P N

m P a Y a

P PY Y Y N

* Chia thanh lm 2 on: AB ,BC

+ Xt on AB: Dng mt ct (1-1) ct thanh, mt ct (1-1) tin t A n B, tc l

(1

0 z a ).

Xt cn bng phn tri, ta c:

* 01 Ay YQF

NYQ A 501

* 0. 11 zYM Ax

11 .zYM Ax

- Khi z1=0Mx1=0(Nm)

- Khi z1=a=1m Mx1=50(Nm)

+ Xt on BC: Ct BC bi mt ct (2-2) cch gc mt khong z2 (

10 z a ). Xt cn bng phn

phi, ta c:

* 02 Cy YQF NYQ C 502

* 0. 22 zYM Cx 22 .zYM Cx

- Khi z2=0 Mx2=0(Nm)

- Khi z2=a=1m Mx2=50(Nm)

V d 6.2: Cho thanh AB chu tc

dng ca lc phn b q=10(N/m),

chiu di thanh l=10(m). (Hnh 6-4)

V biu ni lc cho dm AB ?

Y A X A

Yc

X A

Y A Yc

Mx

Q

Mx 2

Q 2 z 2 2

2

C

z 1

Mx 1

Q 1

1

1

a a

P

B A C

50N

50Nm

50N

Hnh 6-3

l

A B

q

Hnh 6-4

66

Bi lm

- Bc1: Xc nh phn lc lin kt

0

. 0

. 100( )

. . . 02

. 10.1050( )

2 2

50( )

A

A B

A B

A B

B

A B

X X

Y Y Y q l

Y Y q l N

lm q l Y l

q lY N

Y Y N

- Bc2: Chia on v xc nh ni lc - Bc3: + Xt onAB: Ct AB bi mt ct (1-1) cch gc A mt khong

z (0 z l ).

Xt cn bng phn phi, ta c

* 0.1 zqYQF By

0.1 zqYQ B

zQ 10501

- Khi z = 0 Q1 = -50N

- Khi z = l =10m Q1 = 50N

* 02

... z

zqzYM Bx 2

..2z

qzYM Bx

2.5.50 zzM x

- Khi z = 0 Mx =0 (Nm)

- Khi z = l/2 = 5m Mx =125(Nm)

- Khi z = l =1m Mx = 0(Nm)

-Bc 4: V biu ni lc (Hnh 6-5)

3. NG SUT TRONG DM CHU UN

3.1. Th nghim:

Xt thanh c tit din hnh ch nht,

Hnh 6-5

q Y B

B A

q

l

B

1

1 z O

Q

Mx

Mx

Y B Y A

X A

50N

50N

l/2

Q

125Nm

67

thun tin cho vic quan st bin dng ta xt trng thi trc v sau khi chu un + Trc khi cho thanh chu un(Hnh 6-6a) K ln mt ngoi ca thanh cc ng thng song song vi trc thanh: c trng cho cc th dc, v k cc ng thng vung gc vi trc: c trng cho cc mt ct ngang

Tc dng ngoi lc l hai mmen un, thanh chu un thun ty (hai ngu lc nm trong mt phng i xng cha trc thanh) + Sau khi cho thanh chu un (Hnh 6-6b,c)

* Nhn xt - Trc thanh b cong i so vi ban u, trc thanh b bin dng - Cc mt ct ngang: Khi c ngoi lc tc dng cc mt ct ngang trong thanh vn phng v vn vung gc vi tip tuyn ca trc thanh v khng cn song song vi nhau na - Cc th dc: + Trc bin dng cc th dc thng, song song vi nhau v song song vi trc thanh + Xt 7 th dc, cc th dc trong thanh b un cong ng dng vi trc v chiu di ca chng c s thay i lin tc t th dc b co li ti th dc b dn di nh vy t th dc th nht n th dc th 7 s c mt th dc c chiu di khng thay i, th gi l th trung ha. + Tp hp tt c nhng th c chiu di khng i trong thanh s to thnh mt lp th gi l lp th trung ha (mt s ti liu gi l mt trung ha) + Giao tuyn ca lp th trung ha vi mt ct ngang gi l trc trung ha

2

m m

1

4 5

3

1 2

3 4

5

Th trung ha

a,

b,

D

C B

A

m m

Mt trung ha

Trc trung ha

c,

Hnh 6-6

68

+ Nu mt ct l mt ct i xng th trc trung ha s trng vi trc i xng trn mt ct. * Kt lun: Bin dng ca vt liu trong thanh chu un l s kt hp ca bin dng ko v bin dng nn

3.2. ng sut

- ng sut trong thanh chu un l ng sut php

- K hiu: u - ng sut khng u trn mt ct (bin dng khng u) + Vt liu trng trc trung ha khng b bin dng ko-nn, cng xa trc trung ha ng sut cng tng v xa trc trung ha nht th ng sut l ln nht Xt phn chu ko: Theo nh lut Hc ta c .E

M : E = const; l

ll 2

+ Khi y=0 =0 k=0

+ Khi y tng tngk tng

+ Khi ymax max = max (biu din trn hnh v) Biu ng sut trn mt ct ngang ca thanh chu un + Ti trc trung ha c y = 0 = 0

+ Ti th ngoi cng c y = 2

h

c max min

,

+ ng sut ln nht khi y ln nht : ymax= h/2 Tnh ng sut ln nht trn mt ct ngang bng cng thc:

Mxmax

Trong : max

x

u

JW

y (m men chng un ca mt ct i vi trc trung

ha x, n v l m3). Ph thuc vo tit din mt ct ngang.

+ Mt ct ngang hnh ch nht c:

Jx=3 2 2

,12 6 6

x y

bh bh b hW W

+ Mt ct ngang hnh trn c:

A'

A

O

Nn

Ko

y

x O

Ym

ax

min

max

x Hnh 6-7

69

Jx=4 3

30,164 32

x y

d dW W d

Wx cng ln kh nng chng un cng cao v max cng gim. * Vi cc mt ct c hnh dng phc tp th m men chng un c th tra bng trong s tay k thut.

Bng hng lot th nghim v l thuyt

n hi chng minh: mt ct ngang ca

dm chu un ngang phng khng hon ton

phng v vung gc v trc thanh nh un

thun tu, nhng s bin dng ca mt ct

ngang d l khng ng k v c th b qua.

V vy, ngi ta vn dng cng thc ng sut phpca un thun tu.

Cng thc tnh ng sut ln nht trn mt ct ngang ca dm chu un Mx

max

Trong : max l ng sut ln nht trn mt ct ngang ca dm chu un Wx l m men chng un Mx l m men un ni lc

4. TNH TON CHO THANH CHU UN PHNG

Mc tiu: + Vit c iu kin bn v cc bi ton c bn + p dng thnh tho ba bi ton c bn theo iu kin bn v ng sut php

4.1. iu kin bn v ng sut php v ba bi ton c bn

4.1.1. iu kin bn v ng sut php

iu kin cn v thanh chu un m bo bn l ng sut sinh ra trn mt ct ngang ca thanh phi nh hn hoc bng ng sut cho php.

u max

4.1.2. Ba bi ton c bn

a. Kim tra bn

Cng thc kim tra bn

uMx

max

Hnh 6-8

70

- Tm ng sut ln nht - So snh ng sut ln nht vi ng sut cho php - Kt lun

+ Nu u thanh bn + Nu u thanh khng bn


ux

x

MW


uxu WM .

4.2. Ton p dng.

Bi 1: Cho mt dm mt ct trn chu chu tc dng lc un P=100kN , chiu di dm 2m, mt ct c D =20cm; dm c ta trn 2 gi nh hnh 6-7.

- V biu ni lc Qy, Mx cho dm AC - Hy kim tra bn ca dm g theo ng sut php bit: 2/100 mMN

Bi lm

p dng cng thc kim tra bn:

ux

x

W

M max

- M men chng un l:

c D= 20cm= 0,2m

WX = 0,1.D3= 0,1. 0,2

3 =

8.10-4

(m3)

- M men un ni lc l:

MNmkNmM x3

max 10.5050

- ng sut ln nht trn mt ct ngang ti im nguy him (B) ca dm chu un

24

3

max /5,6210.8

10.50mMN

W

M

x

x

So snh ta thy 22max /100/5,62 mMNmMN

Kt lun : Dm AC m bo bn.

Y A X A

Yc

X A Y A Yc

Mx

Q

Mx 2

Q 2 z 2 2

2

C

z 1

Mx 1

Q 1

1

1

a a

P

B A C

50kN

50kNm

50kN

Hnh 6-9

71

5. CHUYN V CA DM CHU UN.

Mc tiu:

- Tnh c chuyn v ca dm chu un ngang phng

Xt mt thanh thng chu un ngang phng. Sau khi chu un, trc thanh b cong i, ng cong ca thanh gi l ng n hi (Hnh 6-8)

Phng trnh ca ng n hi

y = y(z)

Xt chuyn v ca im K ta thy sau bin dng im K di chuyn thnh im K`. Khi chuyn v va im K l :

K = KK`

- Phn KK` thnh 2 thnh phn u v v u: l thnh phn nm ngang v: l thnh phn thng ng L thuyt n hi chng minh c rng : u l v cng b bc cao so vi v cho nn khi kho st ta thng b qua u. Vy ta c :

K v(z)

v(z): l vng T hnh v ta c th suy ra :

K v(z) y(z)

Xt mt ct ngang i qua im K trc v sau bin dng to nn mt gc (z)

(z) : l chuyn v gc (gc xoay)

- Qua K` k tip tuyn vi ng n hi to vi trc z mt gc l

T ta c: tg (bi v rt nh)

Vy : tg y (`z) v (`z)

Kt lun: o hm ca vng bng gc xoay.

CU HI N TP

1. Khi nim v un ngang phng ? Ni lc v biu ni lc ca thanh chu un ngang phng ?

z

K` y (z)

u

K

z

P

v

ng n hi

Hnh 6-10

72

2. Vit cng thc tnh ng sut trong dm chu un ?Gii thch cc i lng trong cng thc? 4. Vit iu kin bn v ng sut php v cng thc tnh ton ca ba bi ton c bn trong thanh chu un ngang phng?

BI TP

Bi 1: Cho mt dm mt ct ch nht cnh (axb) =(4x6)cm chu tc dng lc un P=90kN , chiu di dm a= 1m, dm c nm ngang bi 2 gi nh hnh 6-11. - V biu ni lc Qy, Mx cho dm AC - Hy kim tra bn ca dm g theo ng sut php bit: 2/100 mMN

Bi 2: Cho mt dm AB mt ct ch nht cnh (axb) =(2x3)cm chu tc dng h lc phn b q=20kN/m , chiu di l =4 m, dm c ta trn 2 gi nh hnh 6-12. - V biu ni lc Qy, Mx cho dm AC

- Hy kim tra bn ca dm g theo ng sut php bit: 2/100 mMN

Bi 3: Cho mt dm AB mt ct ch nht cnh (axb) =(3x4)cm chu tc dng ca m men m =120kN.cm , chiu di a =2 m, dm c ta trn 2 gi nh hnh 6-13. - V biu ni lc Qy, Mx cho dm AC

- Hy kim tra bn ca dm g theo ng sut php bit: 2/100 mMN

Hnh 6-12

Hnh 6-11

2a

C

P

a B

A

B A

q

l

m

B A C

a a

Hnh 6-13

73

CHNG VII: THANH CHU LC PHC TP

M chng: CHVII

Trong thc t nhng chi tit my chu cc hnh thc bin dng c bn thng gp rt t m ch yu l cc chi tit chu ng thi mt lc t hai hnh thc bin dng tr ln. Chi tit chu ng thi mt lc t hai hnh thc bin dng tr ln gi l chi tit chu lc phc tp. V d: Un xin, un ng thi xon, un ng thi vi ko nn ng tm.....

Mc tiu:

+ Trnh by c c cc khi nim v un xin, un ng thi vi ko nn ng tm, ko nn lch tm, un ng thi xon..

+ V c s tnh tng qut v s tnh tng loi bin dng c bn t thc t.

+ Xc nh c mt ct nguy him v p dng c iu kin bn gii ba bi ton c bn ca sc bn .

+ C thc trch nhim, ch ng hc tp.

1. KHI NIM THANH CHU LC PHC TP.

Mc tiu:

+ Trnh by c c cc khi nim v chu lc phc tp v phng php nghin cu nhng chi tit my hay c cu my chu lc phc tp.

1.1. Khi nim.

Cc chng trc chng ta ch kho st thanh chu lc n gin nh: Ko- nn ng tm, ct ,dp, xon thun ty v un ngang phng.

Trong thc t c nhng chi tit my hay c cu my chu lc phc tp.V d mt trc truyn va chu xon va chu un....

1.2. Phng php nghin cu

* Ko nn ng tm :

- Ni lc: Lc dc Nz

- ng sut php phn b u: F

N zz

* Xon thun ty:

- Ni lc: M men xon ni lc Mz

- ng sut tip ln nht phn b theo quan h bc nht vi bn

knh: p

z

W

Mmax

74

* Un ngang phng:

+ Un quanh trc x:

- Ni lc: Qy, Mx

- ng sut php ln nht phn b theo quan h bc nht vi y:

x

x

W

Mmax

+ Un quanh trc y:

- Ni lc: Qx, My

- ng sut php ln nht phn b theo quan h bc nht vi x:

y

y

W

Mmax

Bi tp kt hp:

Mx + My c un xin

Mx + My + Nz c un xin + ko (nn)

Mx + My + Mz c un xin + xon

gii quyt cc bi ton trn chng ta s p dng nguyn l cng tc dng: ng sut hay bin dng do nhiu yu t (ngoi la, nhit , ln ca gi...) gy ra ng thi trnmt thanh bng tng ng sut hay bin dng do tng yu t gy nn. Nguyn l ny ch c p dng khi vt liu lm vic trong min n hi v bin dng ca thanh l nh.

2. UN XIN.

Mc tiu:

+ Trnh by c c cc khi nim v un xin. + V c s tnh tng qut v s tnh un xin + Xc nh c mt ct nguy him v p dng c iu kin

bn gii ba bi ton c bn ca sc bn .

2.1. nh ngha.

Du hiu ni lc: Thanh chu un xin l thanh chu lc sao cho trn mi mt ct ngang ca n c hai thnh phn ni lc l m men un Mx , My nm trong cc mt phng qun tnh chnh trung tm ca mt ct ngang. (Hnh 7-1a)

Mt phng qun tnh chnh trung tm ca mt ct ngang l mt phng c to bi mt trc qun tnh chnh trung tm ca mt ct ngang v trc ca thanh

Mx thuc mt phng z0y (0y l trc qun tnh chnh trung tm).

75

My thuc mt phng z0x. (0x l trc qun tnh chnh trung tm).

Hp hai m men Mx , My ta c mt m men Mu :

22yxu MMM

Mx: ng ti trng l y

My : ng ti trng l x

Mu : ng ti trng hnh7-1b

2.2. ng sut.

Ta gi gc l gc gia trc x v ng ti trng, > 0 khi chiu quay t trc x n ng ti trng l thun chiu kim ng h. ( Hnh 7-1b)

Ta c quan h:

cos.

sin.

MM

MM

y

x

H s gc ca ng ti trng

y

x

M

Mtg

Ta c Mx gy nn ng sut php c gi tr l :

yJ

M

x

xMx

z .

My gy nn ng sut php c gi tr l :

xJ

M

y

yMy

z .

Mx

My

a,

x

My z

Mu

Mx O

y

ng ti trng

b,

Hnh 7-1

76

p dng nguyn l cng tc dng, ta c th coi ng sut ti mt im no trn mt ct ngang c ta x, y l tng ng sut do tng

m men un Mx , My gy ra mt cch ring l.

xJ

My

J

M

y

y

x

xz .. (7.1)

Biu thc (7.1) l cng thc tng qut tnh ng sut php cho un xin.

Khi s dng cng thc trn ta phi ch n du ca x, y v ca Mx , My. trnh nhm ln ta thng dng cng thc k thut sau:

xJ

My

J

M

y

y

x

x

z ..

Ta ly du cng hay tr trc mi s hng ty theo cc m men un Mx , My gy ra ng sut ko hay nn im ang xt.

Du (+) tng ng vi im chu ko

Du (-)tng ng vi im chu nn

Du +, - c xc nh theo tng vng (Hnh 7-2)

2.3. iu kin bn v ba bi ton c bn.

2.3.1. iu kin bn

* Vi vt liu do : nk nn trong hai gi tr ng sut ta chn ng sut no c tr s tuyt i ln nht kim tra:

minmax ,max

* Vi vt liu dn: nk

Nn ta phi kim tra bn cho c im chu ko ln nht v chu nn ln nht.

+ k max

Mx

My z

y

x

z y

x

Hnh 7-2

77

+ n min

2.3.2. Ba bi ton c bn.

a. Bi ton kim tra bn.

T iu kin bn ta c cng thc kim tra bn

* Vi vt liu do :

nky

y

x

x

W

M

W

M,minmax

* Vi

Documents

MH 12 SUC BEN VAT LIEU_2.pdf