c im chung ca kt cu thp nh cng nghip

PAGE 22

Chng 2: kt cu thp nh nhp ln2.1 phm vi s dng v cc c im ca kt cu nh nhp ln

Nh nhp ln l nh c khong cch ct theo phng ngang ln (thng thng 40m) nhm mc ch hn ch s lng ct bn trong nh. Kt cu nh nhp ln c dng trong cc cng trnh dn dng v cng nghip. Trong cng trnh dn dng nh rp ht, nh trin lm, sn vn ng, nh ga, nh thi u, ch Trong cng trnh cng nghip nh gara t, hngga my bay, xng ng tu, xng lp rp my bay

Kt cu nh nhp ln c nhng c im sau: Cng trnh nhp ln khng phi l nhng cng trnh xy dng hng lot m l cc cng trnh n chic, yu cu v kin trc rt cao ph hp vi tnh nng ca cng trnh . Kch thc ca cng trnh thay i trong phm vi rt rng. Nhp ca nh cng nghip thay i t 50(100m, Xng lp rp my bay c nhp 100(120m, chiu cao 8 (10m, Xng ng tu c nhp 20 ( 60m, chiu cao 30( 40m. Hnh dng phong ph v phc v nhu cu ring nn rt kh c m un in hnh. Do vy phi trit in hnh ho cc cu kin khi thit k.

Kt cu nh nhp ln ch yu chu ti trng do trng lng bn thn v ca tm lp. Vic gim trng lng ca kt cu l nhim v chnh ca ngi thit k. gim trng lng bn thn ngi ta s dng cc cch sau: s dng vt liu thp cng cao, dng hp kim nhm, dng vt liu lp mi nh nh tm tn mng, cht do, vi bt , dng kt cu ng sut trc, h khng gian hoc dng mi dy.Kt cu chu lc chnh ca nh nhp ln bao gm cc dng sau: Kt cu phng chu lc: h dm, h khung v h vm Kt cu kiu dm v khung thch hp vi mt bng hnh ch nht

H vm c hnh dng kin trc p, tit kim vt liu, hp l khi s dng vt nhp 80m Kt cu khng gian chu lc: H thanh khng gian, cupn, mi dy

H thanh khng gian c cu to t cc dn phng an cho nhau, do chu lc theo cc phng nn tit kim vt liu.

Mi cupn c s dng khi nh c mt bng hnh trn hoc a gic.

H mi dy c s dng khi nhp rt ln, em li hiu qu cao.

2 Nh nhp ln vi kt cu phng chu lc

1. Kt cu kiu dm, dn

Kt cu kiu dm dn c dng trong cng trnh cng cng nh rp ht, nh vn ho, cng trnh th thao c mt bng hnh ch nht. Nhp ca kt cu kiu dm dn l 40 ( 90m, thng thng ngi ta s dng dn nhiu hn dm.

Kt cu kiu dm s dng tng i t trong nh nhp ln, nhp ca chng khong 35( 40m. Kt cu kiu dm c u im: sn xut n gin, d bo dng

Kt cu dm ca mi sn trt bng chu u xy dng nm 1969, mt bng 100x73m k 4 ct cao 12m; dm chnh dng hp 700x4300, dm ph theo phng ngang nh.

Kt cu kiu dn:

Kt cu sn trt bng ngh thut

1. H ging dc theo mt phng cnh trn ca dn; 2. Dn chnh; 3. Ct

Hnga my bay nhp 60m La chn hnh dng dn ph thuc vo yu cu s dng, yu cu kin trc cng nh cc yu cu khc (thng gi, chiu sng ). Hnh dng ca dn c th l dn cnh song song, dn hnh thang, dn a gic, dn tam gic hoc dn cnh cung.

Dn cnh song song (hnh a, b): dng cho nh mi c dc nh.

S n gin hoc lin tc L ( 60m.

Rt nhiu nt ging nhau, chiu di cc thanh bng bng nhau. Chiu cao dn h=(1/8 ( 1/14)L vi dn n gin, h=(1/10 ( 1/18)L vi dn lin tc (gim i 15(20% so vi dn n gin). Dn hnh thang (hnh c, d):

dc cnh trn ph hp vi yu cu dc mi.

i = 1/12 ( 1/15. Chiu cao gia dn h=(1/8 ( 1/12)L vi dn n gin, h=(1/10 ( 1/15)L vi dn lin tc. Dn a gic (hnh e):

Cnh trn gy khc.

Tit kim vt liu, ch to phc tp.

Dng cho nh 1 nhp c L = 60 ( 90m

Dn cnh cung (hnh i, k):

Ni lc trong cc thanh phn b hp l.

Vt nhp ln L = 60 ( 100m. Dn cnh cung c dng 2 parabn (hnh k) cho tit din thanh cnh di v cnh trn bng nhau, n nh ca dn tng do trng tm dn h thp.

Dn tam gic (hnh g, h):

Dng khi ti trng nh v dc i = 1/5 ( 1/7.

Cu to t hai na dn lin kt vi nhau bng thanh cng dng khi L=40(50m; Chiu cao dn h=(1/6 ( 1/9)L

Cu to t hai dn cnh song song c thanh cng v thanh ng dng khi L ( 90m, chiu cao dn h=(1/6 ( 1/10)L, chiu cao dn cnh song song h=(1/12 ( 1/20)L.

Vic la chn h thanh bng ca dn ph thuc vo hnh dng dn, vo ti trng tc dng v ph thuc vo cc kt cu khc lin kt vo dn. H thanh bng c la chn sao cho trng lng dn v cng ch to t nht. Trong nh nhp ln hay s dng h thanh bng tam gic c b sung thanh ng, h thanh bng xin. Gc hp l ca thanh bng vi thanh cnh trng h thanh bng tam gic l 45o v trong h thanh bng xin l 35o. Do nhp ln, chiu cao dn ln, chiu di thanh bng ln, gim chiu di tnh ton trong mt phng ca thanh bng v thanh cnh di, ngi ta s dng h thanh chng ph (hnh d, e) vic ny tuy lm tng cng ch to nhng s gim trng lng dn. Trong dn cnh cung, ni lc ca cc thanh bng khng ln, vic dng h thanh bng cho nhau (lm vic chu ko) s tit kim vt liu hn so vi h thanh bng tam gic.

Dn nhp ln cn phi tnh ton vng ca dn do tnh ti v hot ti tiu chun gy ra. vng cho php l 1/250 L. gim vng ca dn, ngi ta cu to vng xy dng ca dn. vng xy dng ca dn ly bng tng vng do tnh ti tiu chun v do mt na hot ti tiu chun gy ra.

Tit din thanh dn c la chn sao cho d cu to nt, d lin kt vi kt cu khc. Khi la chn tit din thanh dn cn ch : Chiu cao tit din thanh dn khng vt qu (1/8 (1/10) chiu di thanh gim ng sut ph do cng ca nt.

Khi ni lc thanh cnh thay i nhiu th cn thanh i tit din thanh, c gng ch thanh i tit din l nt khuych i dn.

lch tm khi thay i tit din khng vt qu 1,5% chiu cao tit din ch H, ch thp v ch T, khng vt qu 4% cho tit din ch I v cc tit din kn. Nu lch tm ln hn phi k n trong tnh ton.

Tit din thanh dn c th dng cc dng sau:

Tnh ton dn nhp ln tin hnh nh vi dn thng.

2. Kt cu khung

Dng ph mi c nhp 40 m - 150 m

Thng c 2 loi: khung c v khung rng Khung c: Tit din ct v x l dng c, dng cho nhp 50 m - 60 m, thng s dng khung hai khp, t thanh cng gim bt lc x ngang ca mng. Gim cng ch to v chuyn ch. Khung hai khp c thanh cng di

1. X ngang; 2. Thanh cng; 3. Cu trc treo

gim m men un ca x do ti trng thng ng, thit k sao cho cng n v ca x v ca ct xp x nhau, t l chiu cao ca tit din x v nhip 1/30 - 1/40 Khung rng: Tit din x v ct l dng rng. Dng cho nhp t 100 m- 150 m, s kt cu c th l khung hai khp hoc khung khng khp.

u khuyt im ca kt cu khung:

- Trng lng b hn kt cu dn dm cng nhp.

- cng ln hn kt cu dn dm cng nhp.

- Chiu cao x ngang nh hn kt cu dn dm cng nhp.

- Chiu cao tit din ct ln.

- Chu nh hng ca nhit v mng ln khng u.

2.1. Kt cu khung c nh nhp ln

Nhp thng gp: 40 ( 100m (hp l: L = 40 ( 50m).

Ch to v vn chuyn n gin.

Thng dng dng 2 khp chn.

Gim lc x ngang mng dng thanh cng t mt di nn.

Chiu cao tit din x ngang: h = (1/30 ( 1/40)L

Lin kt khp chn ct dng gi u.

2.2. Kt cu khung rng nh nhp ln

Nhp thng gp: L = 100 ( 150m.

Dng 2 khp: chn hoc nt lin kt x vi ct.

Dng khng khp (tng cng), ct ngm vi mng v vi x ngang.

X ngang: h = (1/12 ( 1/20)L

Chiu cao ct: hc = d (khong cch nt dn x ngang)

Bin php gim mmen un cho x ca khung hai khp: to lch tm ti chn ct.

2.3. c im cu to nt khung v tnh ton khung nhp ln

c im cu to: ti gc khung (nt) c ng sut tp trung:

- Khung c: c b sung sn gia cng.

- Khung rng: thm bn p v sn.

Tnh ton: M hnh ho cc cu kin bng phn t thanh, t vo trc ca cc cu kin.

+ Khung c: cc thanh lin kt theo s ngm cng nt, khp hoc ngm chn ct. + Khung rng: to thnh h thanh nh dn, c tnh ton nh dn c k n bin dng ca tt c cc thanh;Phng php tm ni lc:

+ Lc, chuyn v... nh mn c hc kt cu

+ Chng trnh phn mm tnh ton kt cu Sap...

Tnh ton v kim tra cc cu kin ca ct v x theo cng thc ca cu kin chu nn lch tm.

3. kt cu vm

3.1. Gii thiu v kt cu vm

Kt cu mi vm sn vn ng

Kt cu mi vm nhiu nhp ga tu ho Phm vi s dng:

- Trin lm. - Cung vn ho. - B bi, nh thi u, ...

Cc kch thc chnh ca vm: nhp L, mi tn vng f . f - ph thuc vo iu kin s dng, kin trc, kinh t. T s li nht f/L = 1/5 - 1/6 . Khi f tng s gim c mmen v lc dc trong vm. Do cc iu kin v kin trc, ti a c th ly f/L = 1/2 - 1/5. c im: lc x ngang ln do phi to kt cu chu lc x ngang nh dy cng; khung chu x ngang.

Cc kiu vm:

Vm 2 khp: l loi dng rt ph bin, tit kim c vt liu, t l vng chu mmen b l t, mmen phn b tng i u dn ti tit din chn c hp l, t chu nh hng ca nhit v ln gi ta.

Vm 3 khp: mmen phn b khng u dn ti lng ph v vt liu, h kt cu l tnh nh, lp dng kh khn do phi to nt khp, khng chu nh hng ca nhit v ln gi ta. Ni lc chn vm ln.

Vm khng khp: l h siu tnh bc 3, ni lc nh dn ti tit kim c vt liu, chu nh hng ln ca nhit v ln gi ta.

Biu m men ca 3 loi khi chu ti phn b u.

1. Biu mmen i vi vm 3 khp; 2. biu mmen vm hai khp;

3. biu mmen vm khng khp- La chn trc vm

1. ng gi thit ban u; 2. ng cong do gi;3. ng trung bnh; 4. trc vm thit k3.2. c im cu to kt cu vm

T l hp l: f/L=1/5(1/6

Vm c: tit din (, 2 cnh song song, t hp hn.

; h khng ln hn 2m.

- Nhp thng gp: L=50 ( 60m

- Ch to thnh tng on vn chuyn 6 ( 9m.

Vm rng: h =

- Tit din thn vm rng.

Lin kt khp bn cu to n gin, s dng khi phn lc gi khng ln. Cu to gm con ln mt tr c lin kt trc tip vi chn vm, tht di l thp bn, c lin kt vi mng. Con ln t p vo tht di v c gi c nh bng bulng neo b tr theo trc vm khng ngn cn xoay ca chn vm.

Lin kt khp bn c tnh theo iu kin p mt ca con ln vo tht di. Chiu dy con ln c tnh theo iu kin chu un. Lin kt khp ci dung khi phn lc gi ln. Cu to gm hai mt v tr cng tip xc vi nhau, bulng neo gn c nh ci di vi mng. tng cng cho chn vm, ti v tr truyn lc, chn vm c gia cng bng cc sn cng. Lin kt khp u dng khi phn lc gi rt ln. Cu to ca khp u bao gm tht trn v di, gia hai tht c b tr thanh tr c. Vm c gn vo tht trn bng bu lng, tht di rng hn tht trn bo m iu kin ng sut truyn vo mng nh hon cng chu nn ca mng. Tht di lin kt vi mng bng bulng neo trnh trng hp khi gi bc gy ko cho chn vm.

Khp nh vm dng khp bn hoc khp u, khi vm nh c th dng khp nh dng tm hoc bu lng.

Cu to lin kt dng tm gm hai bn thp t dc theo trc vm (khng lm cn tr s xoay ca tit din) cho php truyn c lc dc, cc tm thp c m rng to lin kt vi h ging. 3.3. Tnh ton vm

- Ti trng sau:

+ ti trng ng do trng lng bn thn vm v cc lp mi;

+ ti trng gi;

+ Nhit ;

+ s dch chuyn ca gi (trng hp nu c th sy ra)

- Phng trnh ng cong ca vm c th l ng parabol, cung trn, ng cong hnh elp

+ Phng trnh ng cong ca vm parabol c dng sau:

- Ni lc:

Mx= Md - H.y; Nx= Qd.sin( + H.cos(; Qx= Qd.cos( - H.sin(

H: lc x ngang; y: tung trc vm (ymax = f)

(: gc tip tuyn (trc vm vi phng ngang)

Md, Qd: m men v lc ct ca dm n gin cng nhp.

Vi vm 2 khp gii bng cc phng php c hc kt cu hc, tnh vi h siu tnh mt bc X1 = H = ;

(11 chuyn v ca h c bn theo hng lc tc dng X1 = 1

(1p chuyn v theo hng X1 di tc dng ca ti trng

Vm rng: ;

; - ni lc trong thanh th i ca vm do lc n v v ti trng gy ra trong h c bn.Ai , li - din tch tit din v chiu di thanh th i; n - s thanh ca vm

a) Vi vm hnh parabol, hai khp chu ti phn b u trn ton nhp :

Lc x ngang: ; Lc dc: ; ( - gc nghing vi tip tuyn ca cung vm;

b) Vi vm hnh parabol, hai khp khi chu lc tp trung: M s c tnh bng hiu M ca dm lin kt hai u khp chu ti tp trung v M sinh ra do lc x ngang H c tnh bng cng thc sau:

a,b - khong cch t lc tp trung P ti gi tri v gi phi vm

c) Vi vm hnh parabol, hai khp chu ti phn b u trn mt na nhp:

Lc x ngang: ; Phn lc ng: ;

Mx= Md - H.y; Nx= Qd.sin( + H.cos(; Qx= Qd.cos( - H.sin(Md, Qd: m men v lc ct ca dm n gin cng nhp chu ti trng phn b u trn mt na nhp.d) Vm hai khp vi ng sut do s thay i u ca nhit

t - chnh lch nhit ; h - chiu cao tit din vm ; e) Vi vm khng khp chu ti trng tp trung Pg) Vi vm rng cnh song song: Ni lc trong cc thanh ca vm c th xc nhbng cch phn mmen, lc dc cho cc thanh cnh, lc ct cho cc thanh bng chu sau khi c Mx; Nx; Qx

Xc nh lc dc trong thanh cnh:

Trong thanh bng xin:

Trong thanh bng ng:

h - khong cch trng tm ca hai thanh cnh;

a - khong cch t trng tm tit din n trng tm thanh

cnh i din;

(, ( - c m t nh trn hnh v

Tnh ton tit din:

- Vm c: tnh theo iu kin chu un.

- Vm rng: cc thanh c tnh nh thanh dn thngn nh tng th ca vm theo phng ngoi mt phng un c m bo bng cc h ging ngang, cc thanh chng dc v h x g. Khong cch gia cc im c kt khng b tr vt qu t 16-20 ln chiu rng cnh vm.

Lc ti hn v mt n nh trong mt phng lm vic ca vm c chu tc dng ca lc dc vi nh hng ca mmen l khng ng k c xc nh nh sau:

S - chiu di na vm;

- cng vm ti 1/4 nhp;

( - h s chiu di tnh ton, k n cong ca vm, ph thuc s vm v t s f/l

iu kin n nh ca vm

3. gii thiu kt cu mi khng gian ca nh nhp ln

c im: l dng mi c kt cu m trc ca cc b phn chu lc khng cng nm trong mt mt phng v truyn lc theo c hai phng, ni lc c dn u trn mt mi, nn mi nh hn v c hnh dng kin trc p hn so vi kt cu phng.

Gm hai loi: h li thanh khng gian phng v h li thanh khng gian dng v1. H li thanh khng gian phng hai lpDng cho cc cng trnh nhp nh (l< 30 m), nhp va( l = 30-60m) hoc nhp ln L > 60 m.

a) Cc dng s b tr h thanh H mi c to bi cc dn phng giao nhau, t theo hai hng: trc giao (a), hoc cho (b); t theo ba hng (c,d). Tu theo cch b tr m cc thanh cnh hp vi nhau to nn mng li hnh vung, tam gic hoc lc gic.

Hnh 1.1 S mi gm cc dn thng ng giao nhau

a), b) - b tr cc dn theo hai hng; c), d) - b tr cc dn theo ba hng

H mi ghp bi cc n nguyn nh hnh dng hnh chp 4 mt, 5 mt hoc 7 mt. Cc cch ghp ny to nn cc dn t cho trong mi.

Hinh 1.2 S mi ghp bi cc n nguyn hnh thp

a), b) - t cc n nguyn hnh chp 5 mt; c) - t cc n nguyn hnh chp 4 mt; d) - t cc n nguyn hnh chp 7 mt.

b) La chn s b tr thanh: ph thuc dng mt bng mi, c nhp, s b tr gi k, cu to nt lin kt gia cc thanh, dng tit din cc thanh...

Mi c cc li hnh vung ( H. 1.1, a), c cc n nguyn hnh chp 5 mt (H. 1.2, a,b) dng hp l khi mt bng mi l hnh vung, hoc mi ch nht khi t s 2 cnh < 1: 0,8 khi s lm vic ca mi theo hai hng l gn nh nhau.

Mi c mt bng hnh ch nht khi t s 2 cnh > 1: 0,8 nn dng mi gm cc dn t cho gc 45o so vi chu vi (H. 1.1,b,c), vi cc n nguyn chp 4 mt (H. 1.2,c).

Mi c cc thanh cnh to nn cc li hnh vung (H. 1.1,a),vi cc n nguyn hnh chp 5 mt (H. 1.2,a,b) hoc hnh chp 7 mt (H. 1.2,d) c th b bin hnh nn khng chu c mmen xon. V vy khi cu to mi c con sn cn b tr sao cho phn con sn ch chu un ngang.

Mi c cc cnh to nn hnh tam gic (H. 1.1,c), t cc n nguyn hnh chp 4 mt (H. 1.2, c) to nn h li khng gian c tnh bt bin hnh v cng tng, v vy thch hp cho cc dng mt bng hnh dng phc tp v c cc b phn lm vic dng con sn. Tuy nhin phc tp hn khi ch to v dng lp HHnh 1.3. Mt s phng n b tr gi ta cho mt bng hnh vung v tam gic

a) - gc, b), d) - con sn gc; c), e) - theo chu vi; g), h) - dng consn xung quanh; i) - thnh h nhiu nhp (vi gi ta bn trong)

c) B tr gi ta

c im ca h mi li khng gian v phng din kt cu l gi ta c th b tr ti v tr bt k, tuy nhin chng c b tr theo cc nguyn tc sau:

+ Theo yu cu b tr kin trc;

+ m bo tt nht tnh lm vic khng gian ca kt cu, sao cho ni lc cng phn b u trong mt mi cng tt;

+ Trong cc mt bng i xng nn b tr gi i xng.

Khi b tr gi ta gc v to con sn gc (H. 1.3,a,b), cc thanh khu vc gi ta s c ni lc ln, ni lc phn b khng u trn mt mi. khc phc iu ny c th phn b phn lc gi ln mt s thanh nh cc phng n trn (H. 1.4), (H. 1.5).

Hnh 1.4. Cc phng n m rng gi ta

a)- gi thng thng; b) - dng thm thanh chng xin; c) - t thm thanh cng; d) dng gi dng khng gian.

Hnh 1.5. Th d v cch m rng gi ta

a) - dng thanh chng xin; b) - dng thm thanh cng.

B tr nhiu gi ta theo chu vi (H. 1.3, c, e) khi bc ct nh (hoc trn cc dm ging u ct - khi bc ct ln) cho ni lc phn b kh u trong mt mi.

B tr gi ta to nn vng con sn chung quanh mi vi nhp con sn lc = (0,2 - 0,25)l (H.1.3, g, h) lm gim ni lc cho vng gia tm v vy c hiu qu kinh t hn.

B tr dng nhiu nhp (H. 1.3, i) mmen gi lm gim ni lc cc thanh nhp, tuy nhin khng gian s dng gim v chi ph ct tng.

ni lc trong h phn b u hn c th dng cc phng n b tr gi lin hp khc nhau: dng thm h dy vng (H. 1.6, a, b), h dy treo (H. 1.6, c, d), to h dn gi (H. 1.6, e) hoc vm (H. 1.6, h).

Hnh 1.6. Cc phng n kt cu gi lin hp

a),b)- dng h dy vng; c), d)- h dy treo; e)- to h dn gi; h) - gi ta ln vm.

d) Kch thc hnh hc ca mi, tit din thanh v vt liu lm thanh

- Kch thc hnh hc ca mi

+ Nhp l ca mi c ln bt k tu theo yu cu b tr kin trc;

+ Chiu cao ca dn h = (1/15 - 1/30) l ;

+ Gc nging ca cc thanh xin so vi phng nm ngang ( = 400 - 500 ;

+ Chiu di cc thanh: Thng thng chiu di cc thanh dn a = 1,2 - 3 m.

- Tit din thanh

+ Tit din hnh ng l loi dng nhiu hn c, so vi tit din t hp t thp gc u cnh v nu cc thanh c mnh nh nhau th tit din thp ng tit kim c khong 15 % trng lng thp;

+ Tit din l thp gc n hoc t hp;

+ Tit din l nh hnh mng dp ngui ( vung, ch nht ch ( ).

+ Chiu dy ca thp lm thanh khng c nh hn 2,5 mm; thp ng dng nh nht l 48x2,5 mm; thp gc nh nht l L 50x3 mm v L 63x40x3 mm.

Cc loi tit din trn c th ly theo cc bng thp hnh c sn.

mnh gii hn ca cc thanh

+ i vi thanh chu nn ly theo bng 1

Bng 1. mnh gii hn ca cc thanh nn

Cc thanh mnh gii hn

1.Thanh cnh, thanh xin, thanh ng nhn trc tip phn lc gi ta bng thp ng hoc t hp t 2 thp gc

2.Cc thanh khc:

- bng thp ng, thp gc n hoc t hp t 2 thp gc

- bng thp gc n dng lin kt bulng

180- 60(210-60(220- 40(

Ghi ch: ( = N/((Af); N- lc nn trong thanh; A- din tch tit din nguyn ca thanh; f- cng tnh ton ca thp; ( ( 0,5 - h s un dc ca thanh.

+ i vi cc thanh chu ko ly theo bng 2

Bng 2. mnh gii hn ca cc thanh ko

Cc thanh mnh gii hn khi chu ti trng

ng trc tiptnhcu trc

Thanh cnh, thanh xin, thanh ng nhn trc tip phnlc gi ta

Cc thanh khc250

350400

400250

300

Khi thit k s b, sau khi xc nh c chiu di cc thanh, da theo mnh gii hn ( nn ly nh hn mt cht so vi gi tr gii hn) c th nh trc tit din cc thanh.

- Vt liu lm thanh.

Cc thanh ca kt cu mi li khng gian c lm bng thp cc bon thng thng, c gii hn chy t 240 n 290 N/ mm2.

e) Nt lin kt gia cc thanh

- Nt l cc bn m c hnh dng khc nhau: Loi ny thng dng lin kt cc thanh dn bng thp gc, thp hnh, cc nh hnh mng khc nhau. Chng c cu to kh n gin. Cc thanh lin kt vi bn m bng bulng hoc ng hn. Bn m c to thnh bng cch dp hoc hn cc bn ri vi nhau to nn hnh dng cn thit theo s hi t ca cc thanh.

Hnh 1.7. Bn m c sn xut bng cch dp

1. bn m; 2. l dp lm nh v thanh khi lp

Hnh 1.8. Nt dn t cc bn thp hn

a) Ch thp ct mng ; b) Ch thp hn

.Mt s qui nh i vi nt lin kt dng bn m

+ Thp dng lm bn m ly cng loi vi thp ca thanh dn;

+ Chiu dy ca bn m phi c dy ln hn dy ca cc thanh t nht l 2 mm; chiu dy ti thiu ca bn m l 6 mm;

+ ng trc ca cc thanh phi giao nhau ti mt im, nu khng phi k n mmen lch tm;

+ Lin kt cc thanh vi bn m dng bulng cng cao hoc ng hn gc.

+ Ti bn m dng lin kt hn, khong cch gia thanh cnh vi thanh bng, u cc thanh cnh v thanh bng vi mp ca bn m khng c nh hn 20 mm (H. 1.9);

Hnh 1.9. Khong cch ti thiu gia cc thanh trong nt dn

- Nt cu ( dng Mero- ca c)

Nt cu dng Mero dng lin kt cc thanh dn bng thp ng. Loi ny ang c dng ph bin ta hin nay.

+ Cc chi tit ca nt cu bulng gm: cu thp c l ren, bulng cng cao, ng lng (ng vai tr ca cu), vt ch (hoc cht) bt vo rnh ca ng lng, v u cn hoc tm bt u ng ( H.1.10).

Hnh 1.10. Cu thp lin kt vi bulng

- Vt liu thp cho cc lin kt

+ Cu thp dng thp CT45 hoc 14Mn2Si1 c cng tnh ton f = 3650 daN/cm2;

+ Bulng cng cao dng thp c lp bn 8.8; 10.9; 12.9 hoc thp hp kim 40Cr;

- ng knh cn thit ca cu thp c xc nh theo cc iu kin sau:

+ Theo iu kin 2 bu lng khng chm nhau trong cu: + din tch tip xc gia ng lng v mt cu (2 ng lng ca 2 thanh k nhau khng chm nhau)

- Xc nh kh nng chu lc ko ca mt bulng cng cao c tnh theo cng thc sau:

[N]blc (An ftb

Trong :

[N]blc - kh nng chu lc ko thit k;

( - h s k n nh hng ca ng knh bulng n lc ko thit k, khi ng knh bung d < 30 mm, ( = 1,0; khi ng knh bulng d 30 mm ly (= 0,93;

ftb = 0,7fub- cng chu ko tnh ton ca bulng sau khi gia cng nhit, fub l cng ko t tiu chun ca thp lm bulng, An - din tch tit din tnh ca bulng (tr gim yu do ren) mm2 ,

i vi cc thanh ch lm vic chu nn (trong mi t hp ti trng) th ng knh cn thit ca bulng c th gim i, khi din tch tit din ca ng lng phi c kim tra theo iu kin chu nn v lm vic p mt ca ng lng vi mt phng ct vt ca cu theo lc nn tnh ton ca thanh. u thanh dn c th c lin kt vi u cn (H. 1.12,a) hoc tm bt u (mt bch - H. 1.12,b). Khi , bn ca ng hn ni u cn (hoc mt bch) vi thanh dn v bn ca tit din ngang bt k ca u cn phi bng bn ca thanh dn. B rng bh ca ca ng hn ly theo chiu dy ca thanh dn t 2- 5 mm. Chiu dy ca tm bt u phi ly theo kh nng chu lc thc t (th nghim), khi chiu dy ca thnh ng t ( 4 mm th chiu dy ca tm bt u tt 1/5 ng knh ngoi ca ng thp.

Hnh 1.12. Chi tit u cn v ng bt u thanh dn.

a) u cn; b) tm bt u ( mt bch).

- Nt cu rng, hn

Cu rng rut c hn bi 2 bn cu, c 2 loi: c sn v khng c sn bn trong (H. 1.13, a,b).

Hnh 1.13. Nt cu rng bng thp

a) - cu rng khng c sn; b) - cu rng c sn

Cu rng thng dng khi lin kt cc thanh thp ng vi cu bng ng hn. Vt liu thp lm cu l thp cc bon thp thng thng, thp cc bon cn nng dng trong xy dng, hoc thp 16Mn.

Khi xc nh ng knh ngoi ca cu rng, khong cch a gia mt ngoi ca 2 thanh dn cnh nhau a 10 mm (H. 1.14). Theo iu kin ny c th s b xc nh ng knh cu theo cng thc:

D = ( d1 + 2a + d2 )/ Trong :

d1, d2 - ng knh ngoi ca 2 ng thp cnh nhau to thnh gc ;

- gc to bi ng thp cnh nhau (rad).

Hnh 1.14. Xc nh ng knh ngoi ca cu.

g) Nt gi

Nt gi phi cu to sao cho kt cu lm vic ng s tnh, truyn lc tin cy v c cu to n gin.

i vi gi chu lc nn c th dng mt s s cu to nh sau:

1. Gi lc nn dng tm gi phng (H. 1.15, a, b), loi ny dng cho dn li c nhp nh.

Hnh 1.15. Gi khi nhp nh.

a) - khi thanh dn l thp hnh; b) - khi thanh dn l thp ng.

2. Gi chu nn c mt tht cong (H. 1.16, a, b), dng cho dn li c nhp trung bnh.

Hnh 1.16. Gi chu nn c mt tht cong

a) khi bulng nh v t trn trc trng tm v khng c tr thp ni vi cu gia;

b) khi bulng nh v t khng ng trng tm v c tr thp gia.

3. Gi chu nn c 2 tht cong (H. 1.17), dng cho loi dn li nhp ln.

Hnh 1.17. Gi chu nn c 2 tht cong

4. Gi chu nn bn cu (H. 1.18) dng cho dn nhp ln c nhiu gi .

Hnh 1.18. Gi chu nn bn cu.

h) Tnh ton dn li thanh khng gian dng phng

Xc nh ti trng

- Ti trng tc dng ln kt cu dn li gm: ti trng thng xuyn (trng lng bn thn dn, cc lp lp, cc lp cch m, cch nhit...), ti trng tm thi (hot ti mi, ti trng gi...), ti trng khi thi cng, dng lp... Tt c cc ti trng v t hp ti trng phi tun theo cc qui nh ca TCVN 2737 - 1995. Ti trng v tc ng - Tiu chun thit k .

- Ti trng bn thn dn c th xc nh c sau khi s b chn tit din cc thanh dn.

- Khi tnh a ti trng thnh lc tp trung t ti cc nt.

vng khng ch:

- dn mi nh : l2 / 200;

- dn sn nh : l2 /300.

Vi l2 l nhp tnh ton (thng l cnh ngn ca mi).

Tnh ton ni lc trong cc thanh dn nh phn mm my tnh.

Khi tnh ni lc cc thanh a ti trng v thnh cc lc tp trung t ti cc nt, gi tr cc lc tp trung tnh theo din chu ti ca cc nt.

xc nh ni lc trong cc thanh dn, vng ca kt cu c th dng cc chng trnh my tnh thng mi ho nh: SAP-90; SAP- 2000; STAAD III... Kt qu tnh ton bng my tnh c th coi nh chnh xc v dng kim tra bn, n nh ca cc thanh dn.

3.2. H thanh khng gian dng v

V mt lp l h thanh khng gian mt lp, c mt ngoi cong theo mt chiu, dng cho cc cng trnh c mt bng hnh ch nht nhip n 90m. Yu t chnh quyt nh s lm vic ca v l kt cu gi v cu to h li thanh.

Nu h gi c b tr lin tc dc theo v tr L, trng hp ny v tr lm vic ging nh vm vi nhp B, b rng ca thn vm c ly ph thuc cu to h li dc theo thn v;

Nu v tr c ta ln h gi (ct) ti bn gc ca v, trng hp ny v tr lm vic theo s dm chu un vi nhp l L; H li hnh thoi c cu to n gin, c s lng ln nht cc chi tit v cu kin ging nhau, thng thng c s dng cho v tr lm vic theo kiu h vm. Trong trng hp ny gc ( hp l c ly t (45o-60o). Ngc li, trng hp khi v tr lm vic theo s dm vi nhp L th yu cu ( < 30o.H li hnh thoi v sn ngang s dng hp l hn cho cc trng hp v tr lm vic theo kiu vm, cc sn ngang to thnh cc vm cung lm tng ccng cho vm v gim c chi ph vt liu, cc sn ngang ng vai tr ca h ging.H li hnh thoi c sn dc s dng hp l trong trng hp lm vic ca v tr theo kiu dm chu un, trong sn dc lm tng thm cng cho v.

H li c c sn dc v ngang s dng hp l i vi cc v tr c kch thc L/B y = Mx / H Lc ko trong dy ti v tr x:

- Mmax; Qx ti v tr x c xc nh nh dm n gin.

- Lc ko ln nht trong dy:

V: phn lc ng gi ta.

H: Phn lc ngang gi ta:

Vi ti trng phn b u, chiu di dy ld c xc nh theo:

( (

D: c trng ca ti trng (tra bng 2.2 SGK).

4.3. Kt cu mi dy hai lp

Lp vng xung l lp chu lc

Lp vng ln l dy cng (lp dy n nh)

Cc thanh chng chu nn (c khi ko nh s sau)

H chu lc c 2 chiu

4.4. Kt cu mi dy trc giao (mi yn nga)

c hnh thnh t 2 lp dy vung gc nhau:

Dy vng xung l dy ch - chu lc.

Dy vng ln l dy cng.

Ta, lin kt, neo vo gi cng l vnh (hoc dm bin)

Dy lun lun cng vi bt k ti trng tc dng do mi lun n nh.

4.5. kt cu hn hp dy v thanh cng

p ng nhu cu khng gian rng.

X c treo bng dy vo ct tr.

Gim m men x bng nhiu lp dy.

Thng dng cho nh trin lm, hng ga my bay...

a) Lin kt khp bn; b) lin kt khp u; c) lin kt tm; d) lin kt khp bu lng

Lin kt khp bn; b) lin kt khp ci; c) lin kt khp u

1. con ln; 2. thanh tr c; 3. tht trn v di

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