Ket Cau Gach Da go

Chng 2 : Khi Xy Gch

CHNG 3

TNH TON KHI XY KHNG C CT THP

1. NGUYN L TNH TON 1. Phng php tnh theo trng thi gii hn a. Trng thi gii hn 1 (theo kh nng chu lc) - Trng thi ny ng vi lc kt cu khng th chu lc c thm na v bt u b ph hoi hay b mt n nh :T Tgh

T : ni lc trong kt cu do ti trng tnh ton gy ra Tgh : kh nng chu lc b nht ca kt cu - Tnh ton theo trng thi gii hn v kh nng chu lc l cn thit cho mi kt cu b. Trng thi gii hn 2 (theo iu kin s dng bnh thng) - m bo kt cu s dng bnh thng cn hn ch bin dng, m rng khe nt, dao ng ca kt cu. - Kim tra bin dng : f f gh f : bin dng ca kt cu do ti trng tiu chun gy ra fgh : tr s bin dng gii hn cho php ca kt cu - Kim ra vng v khe nt : gh : bin dng ln nht mp chu ko ca cu kin do ti trng tiu chun gy ra gh : bin dng cho php trc khi hnh thnh vt nt i vi kt cu khng cho php nt, vi kt cu cho php nt th l m rng cho php ca khe nt 2. Ti trng tc ng - Ti trng thng xuyn (tnh ti) - Ti trng tm thi (hot ti) - Ti trng c bit Ly theo TCVN 2737-1995 3. Cng tiu chun v cng tnh ton ca khi xy - Cng tiu chun ca khi xy :R =c

Rii =1

n

n

- Cng tnh ton ca khi xy :R= Rc k13

Kt Cu Gch V Gch Ct Thp

Chng 2 : Khi Xy Gch

k : h s an ton xt n cc yu t lm gim cng khi xy k=2 khi xy chu nn k = 2,25 khi xy chu ko xt nh hng ca phng php ch to vt liu, iu kin thi cng v tnh trng lm vic ca kt cu a vo h s iu kin lm vic, m c quy nh trong tiu chun. + Khi kim tra cng ca cc tr v mng tng gia hai ca c din tch tit din ngang 0,3m2 ly m = 0,8 + Khi tnh cu kin c tit din trn khng c li thp xy bng gch thng (khng cong) ly m = 0,6 + Khi kim tra cng khi xy chu nn ca nhng cng trnh cha xy xong ly m = 1,25 + Khi tnh cc khi xy chu nn m ti trng t vo khi khi xy kh cng mt thi gian di ly m = 1,1 + Khi tnh ton khi xy c ct thp :c Ra Ra = .ma ka

cao

ka = 1,1 1,25 i vi thp cn nng ka = 1,5 1,75 i vi thp si ko ngui v thp si cng ma = 0,5 0,9 : h s iu kin lm vic

2. TNH TON KHI XY CHU NN NG TM 1. Cng thc c bn - Cu kin chu nn ng tm l cu kin chu lc nn N trng vi trng tm - CK gch chu nn ng tm : ct, tng... - iu kin cng : N N .mdh .R.F (3.1) N : lc nn do ti trng tnh ton gy ra R : cng chu nn tnh ton ca khi xy Khi din tch tit din ca khi xy F 0,3m2 th R phi c nhn vi h s iu kin lm vic mk = 0,8 mdh : h s xt n nh hng ca ti trng tc dng di hn lm gim KNCL ca khi R xy, khi cnh nh nht ca tit din < 30cm hoc bn knh qun tnh r < 8,7cmmdh = 1 . N dh N

(3.2)

l0 l0 , b r l Khi b 30cm hoc r 8,7cm hoc o 10 b

: h s tra bng ph thuc vo

mdh =114


Chng 2 : Khi Xy Gch

l0 : c xc nh ph thuc vo trng hp tnh ton, lin kt v dng kt cuP P P P

l0 =2H

l0 =H

l 0 =0.7H

l 0 =0.5H

P

P

P

P

P

l0 =1.5H

l 0 =1.25H

: h s un dc

tra bng ph thuc vo v r hay h 0 th Vi VL n hi : = ch (3.3) Khi xy l vt liu n hi do : = th c1,1.R0 th =

.E0 .r 2 l0

2

(3.4) (3.5) th

th = 2 .E.( ) 2) 1,1.R c r 0 th = 2 .E 0 .(1 th c ).( ) 2 = th (1 th c ) l0 1,1.R 1,1.R

r l0

Mt khc : E = E0 (1-

(3.6)

th = 1+

0 th 0 th

(3.7)0 1 + 0

1,1.R c

Chia 2 v cho 1,1Rc ta c : =

(3.8)


15

Chng 2 : Khi Xy Gch

Trong : 0 =

0 th

1,1.R c

(3.9)

r 2 .E0 .( ) 2 l0 T (3.4) v (3.9) 0 = c 1,1.R c Mt khc ta c : E 0 = .R 0 = r r 2 . .( ) 2 9 ( ) 2 1,1 l0 l0

(3.10)0,75

(3.10) tnh cho tit din bt k, vi tit din CN th 0 = Tr s , mdh c xc nh ph thuc vo dng lin ktN N N

b

=1,mdh=1

=1,mdh=1

=1,mdh=1

,mdh

,mdh

,mdh

=1,mdh=1

2. p dng Kim tra KNCL ca ct nh mt nhp xy bng gch t st nung p do mc 100, dng va hn hp mc 25. Ct cao 3.3m, mt u lin kt vi mng c xem nh ngm, mt u lin kt vi dm sn bn trn c xem nh gi di ng. Ct c TDCN 45x60cm, chu Ntt = 36T, trong Ndh = 20T Bi gii : - Tra ph lc 2, ng vi mc gch 100, mc va 50 ta c R = P P 15KG/cm2 - Tra ph lc 1 vi mc va 50 v khi xy gch t st nung p do c = 1000 - Ct c lin kt nh hnh v : lo = 1,5.H = 1,5 x3,3 = 4,95m = 495cm b = l0 495 = = 11 b 45

- Kch thc tit din bxh = 45x60cm Tra bng

= 0,86

V cnh b ca tit din b = 45cm > 30cm mdh = 1 Kh nng chu lc ca ct : Ct khng kh nng chu lc

[ N ] = .mdh .R.F = 0,86 x1x15x(45x60) = 34830Kg = 34.83T < 36T


16

Chng 2 : Khi Xy Gch

3. TNH TON KHI XY CHU NN LCH TM 1. Khi nim chung - Cu kin chu nn lch tm l cu kin chu lc nn N t khng trng trng tm tit dinN N

- lch tm : eo = e01 + eng Vi :e01 =eng

M N

: lch tm ngu nhin eng = 2cm : tng chu lc eng = 1cm : tng t mang eng = 0 khi b > 25cm - Tu theo lch tm e0 ca lc dc m trn tit din c th chu nn hon ton hoc mt phn chu ko. Nu ng sut ko ln hn cng chu ko ca khi xy th trong mch mch va ngang s xut hin khe nt lm thay i chiu cao lm vic ca tit din, lc ny chiu cao lm vic ca tit din l hc. Trong khi xy khng cho php lc N t lch ra khi phm vi tit din ca cu kin m lc N phi t trong phm vi tit din v phi tho : y e0 2cm ng thi m bo s lm vic an ton ca kt cu gch th : e0 0.9 y khi tnh vi t hp c bn 1 e0 0.95 y khi tnh vi t hp c bn 2 Khi chiu dy tng 25cm th : e0 0.8 y khi tnh vi t hp c bn 1 e0 0.85 y khi tnh vi t hp c bn 2 Khi e0 0.7 y : cn phi kim tra v nt


17

Chng 2 : Khi Xy Gch

2. Cng thc c bn - B qua kh nng chu ko ca gch v gch chu ko km, b qua s lm vic ca vng ko - Biu ng sut nn c dng HCN v t ti cng N tnh ton v nn cc b .RN e .medh . .R.Fn - iu kin cng : (3.11) N : lc dc tnh ton e : h s un dc dng trong tnh ton cu kin chu nn lch tm

e =

+ n2

(3.12)x

: h s un dc tnh vi ton b tit din tra bng ph thuc vo h ,

n : h s un dc tnh ring vi phn chu nn' l0 v tra bng ph thuc vo x = x ' l 0 : ph thuc vo dng ca biu momen

e0 y h

l'o

M

medh = 1 .

1.2e0 dh N dh (1 + ) N h

l'o

(13)

e0 dh : lch tm ca ti trng tc dng di hn M e0 dh = dh N dh

: h s hiu chnh xc nh theo cng thc thc nghim e = 1 + 0 1.45 cho tit din ch nht h e = 1 + 0 1.45 cho tit din bt k 2. y

Vi khi xy bng gch rng, tng b tng rng hoc bng thin nhin = 1 - khi xy c cn bng th ngoi lc v ni lc phi nm trn cng mt ng thng tc im t lc phi trng vi trng tm tit din chu nn o Vi tit din ch nht th iu kin c th hin :


18

Chng 2 : Khi Xy Gch

x h = e0 x = h 2e0 2 2 (3.11) v (3.14) : N e. medh ..R(h 2e0 ).b h 2.e0 ).b.h N e .medh . .R.( h 2.e N e .medh . .R.(1 0 ).F h

(3.14)

(3.15)

O N

O N

o Vi tit din ch T khong cch e2 t im t lc N ti trc trung ho v din tch vng nn Fn c xc nh theo cc cng thc : Vng chu nn v pha cnh :e2 = b1 .c (2.e1 c) + (e1 c) 2 b2

Fn = b1 .c + (e1 + e2 c).b2 c Nu e1 : vng nn nm gn trong cnh 2 e1 = e2

Vng chu nn v pha sn :e2 = b2 .d .(2.e1 d ) + (e1 d ) 2 b1

Fn = d .b2 + [c ( y1 + e0 e2 )].b1 d Nu e1 : vng nn nm lt trong phn sn 2 e1 = e2

3. Trng hp lch tm e0 ln - Lch tm ln khi : e0 > 0.7y vi THCB 1 e0 > 0.8y vi THCB 2


19

Chng 2 : Khi Xy Gch N N M=N.e0

k

n

k

n

- iu kin bn : k mn .Rkue 1 M N N F .e = N .( 0 ) = .( 0 1) mn .Rku W F W F F W m .R .F N n ku (16) F .e0 1 W b.h 2 ; F = b.h Vi tit din ch nht : W = 6 m .R .F N n ku (16) (17) 6.e0 1 h

Mt khc : k =

mn : h s iu kin lm vic ca khi xy khi tnh ton hn ch b rng khe nt ca mch va ngang ph thuc vo cp hay thi hn s dng ca cng trnh Cp cng trnh Thi hn s dng mn Cp I 100 nm 1.5 Cp II 50 nm 2 Cp III 25 nm 3

Rku : cng chu ko khi un ca khi xy tra bng 4. V d 1. Kim tra kh nng chu lc ca mt tr gch chu nn lch tm, biu mmen khng i du vi cc s liu sau : Gch t st nung p do mc 100 Va XM mc 50 lch tm theo phng cnh ln e0h = 8cm Kch thc tit din : b x h = 33cm x 45cm l0 = 4.2m Chu lc nn N = 9.5T BI GII - Gch t st nung p do M100, VXM M50 Tra ph lc 1 ta c = 1000, R = 15kg / cm 2


20

Chng 2 : Khi Xy Gch

- V dng va XM nn cng tnh ton gc :R = 15 x0.85 = 12.75kg / cm 2

- Kim tra kh nng chu lc theo (15)N [ N ] = e .medh . .R.F .(1

e =

+ n

2.e0 ) h

2 l 0 430 h = = = 9.33 Tra bng 5.1 vi = 1000; h = 9.33 ta 45 h c = 0.894

V biu mmen cng du nn l0 = l0 = 420cm lch tm e0 = e + eng V b =33cm > 25cm eng = 0 e0 = e = e0 h = 8cm x = h 2.e0 = 45 2.8 = 29cm' l0 420 = = 14.48 tra bng n = 0.789 x 29 0.894 + 0.789 e = = 0.841 2 medh : v b = 33cm > 30 cm ly medh = 1

x =

R : ta c F = 33 x 45 = 1485cm2 = 0.1485m2 < 0.3m2 R = m k .R = 0.8 x12.75 = 10.2kg / cm 2

e0 8 = 1+ = 1.17 < 1.45 4 .5 h 2 x8 [ N ] = 0.841x1x1.17 x10.2 x1485x(1 ) = 9604.9 Kg = 9.6T > 9.5T 45

= 1+

Vy ct khng kh nng chu lc 4. TNH TON KHI XY CHU NN CC B 1. Khi nim - Nn cc b l trng hp ch mt phn tit din chu nn - Trong thc t dm, sn, x g, lanh t, gc ln tng, ct - ng sut trong khi xy c th c cc dng : HCN, hnh thang, tam gic, ng cong tu theo vt liu gc ln khi xy -

Fcb


21

Chng 2 : Khi Xy Gch

- Khng nn k trc tip cc kt cu chu lc ln tng hoc ct gch m cn dng tm m bng b tng ct thp, chiu dy cc tm m l bi s ca chiu dy lp gch v khng nh hn 14cm, trong tm m c li thp, hm lng ct thp theo mi phng a 0.5% - Bn m khng t trc tip ln gch m phi c mt lp va dy 15mm

- Dm, sn k ln b tr th bn m phi t su vo tng 2. Cng thc tnh ton - Kim tra nn cc b theo cng thc : N .d .Rcb .Fcb (3.18) : h s hon chnh biu p lc ca ti trng cc b =1 : biu HCN = 0.5 : biu tam gic d : h s xt n s phn phi li ng sut trong vng chu nn cc b d = 1.5 0.5 : khi xy bng gch, gch b tng c d = 1 : khi xy bng b tng c l rng, b tng t ong cb = .d : tiu chun quy nh vi ti trng cc b phn b khng u, cb khng ph thuc vo dng biu p lc m ch ph thuc vo loi khi xy cb = 0.75 khi k ln tng, ct gch cb = 0.5 khi k ln tng b tng c l rng, b tng t ong Rcb : cng tnh ton ca khi xy khi chu nn cc bRcb = R.3 F .R Fcb

F : din tch tnh ton bao gm Fcb v mt phn xung quanh F c ly nh sau

Fcb

Fcb

Fcb=a.b=a.(2a+bcb)

Fcb=a.b=a.(a+bcb)

A


22

Chng 2 : Khi Xy Gch

Fcb

Fcb=(acb+bcb)(acb+bcb)

Fcb=acb.B

C

Fcb

Fcb=acb.(2a+bcb)

Fcb=bcb.(acb+bcb)

G

: h s ph thuc vo loi khi xy v v tr t ti trng cc b Ti trng t khong gia Ti trng t gc khi xy khi xy ( hnh a, d, e) (hnh b, c) Loi khi xy Tnh vi ti Tnh vi ti Tnh vi ti Tnh vi ti trng cc b trng cc b trng cc b trng cc b v ton b v ton b - Gch t st v gch b 2 2 1 1.2 tng thng - B tng hc, tng b 1.5 2 1 1.2 tng ln - B tng rng, b tng 1.2 1.5 1 1 t ong bng thin nhin


23

Chng 2 : Khi Xy Gch

5. TNH TON KHI XY CHU KO, UN, CT 1. Khi xy chu ko - Khng cho php thit k kt cu gch chu ko theo tit din khng ging m ch c thit k KC gch chu ko theo tit din gingN [ N ] = Rk .F

Rk : cng chu ko tnh ton ca khi xy F : din tch tit din 2. Khi xy chu ct a. Theo tit din khng ging Q ( Rc + 0,8.n. f . 0 ).F - iu kin kim tra : Rc : cng chu ct tnh ton theo tit din khng ging (theo bng ph lc 3) f : h s ma st f = 0.7 vi khi xy bng gch c c quy cch f = 0.3 vi khi xy bng gch rng n : h s vt ti n = 1 vi khi xy bng gch c n = 0.5 vi khi xy bng gch rng 0 : ng sut nn trung bnh c tnh vi ti trng nh nht, vi h s vt ti v ti trng l 0.9 F : din tch chu ct ca khi xy b. Theo tit din c ging - Xy ra khi s hiu thp, cng chu ct ca khi xy do gch quy nhQ Rcg .Fg

Rcg : cng tnh ton chu ct ca gch Fg : din tch chu ct ca gch khng k din tch mch va 3. Khi xy chu un M Rku .W - iu kin v mmen : Rku : cng chu ko khi un ca khi xy (ph lc 3) W : mmen chng un ca tit din Q Rkc .b.z - Kim tra theo lc ct : Rkc : cng tnh ton v ng sut ko chnh ( ph lc 3) b : b rng tit din z : cnh tay n ni lc, vi tit din ch nht z = 2.h/3


24

Chng 4 : Tnh ton khi xy c ct thp

CHNG 4

TNH TON KHI XY C CT THP

1. KHI XY T LI THP NGANG 1. Cu to v tc dng ca li thp - tng kh nng chu lc cho kt cu gch ta c th t ct thp vo trong khi xy dng. - Cch t li thp ngang c dng cho khi xy chc nn ng tm hoc nn lch tm b v thanh mnh nh. - Kt cu chu nn lch tm b eo 0,33 y vi tc dng bt k eo 0,17 h CN - CK c thanh mnh nh khi b 15 - Cu to li thp

d = (3 8)mm c1 , c 2 = (3 12)cm s 40cm (4 5) lp gch phi c 1 li thp

- Xt mt khi xy dng c gch cho trn hnh v. Hm lng ct thp :=Va .100 0 0 Vkfa

- Vi Va : th tch ct thp Va = fa(c1 + c2) Vk = s.c1.c2 : tnh tch khi xy dng ang xt =f a (c1 + c2 ) .100 0 0 s.c1 .c2

- Nu dng li vung c1= c2= c = - min max

2 fa .100 0 0 s.c

min = 0,1 0 0 vi khi xy dng c quy cch

Vi cu kin ng tm max = 50.Phn 1 : Kt Cu Gch V Gch Ct Thp

R Ra

25


lch tm max =

50 R 2l 0 .Ra 1 y

R : l cng tnh ton ca khi xy khng c ct thp Ra: cng tnh ton ca li ct thp ( tra ph lc 9 ) Thng thng = 0,1 1 0 0 - Va : + s hiu va 25 cho mi trng kh ro + s hiu va 50 cho mi trng c m ln - Tiu chun Vit Nam quy nh va cho khi xy c ct thp 50 + Chiu dy mch va ngang ln hn ng li ct thp t nht l 4mm 2. c trng tnh ton ca khi xy - c trng n hi ca khi xy c ct thp a = .Rc c Rak

Rc: cng tiu chun ca khi xy khng c ct thp Vi khi xy bng c quy cch : Rc = 1,5 R hc phng y. Rc = 2 R ( R : cng tnh ton gc cha k h s iu kin lm vic mk) Rcak: cng tiu chun ca khi xy c gia cng bng li ct thpRc ak c 2 R a =R + 100 c

: c trng n hi ca khi xy khng c ct thp

:hm lng ct thp ca li

Rca :li ct thp Rca = 1,1Ra vi thp AI (CI ), AII (CII ) Rca = 1,25Ra si thp ko ngui 3. Tnh ton khi xy chu nn ng tm a. Cng thc tnh ton a. iu kin cng : (1) mdn, , F : ging nh khi xy khng c ct thp nhng c tnh ton vi c trng n hi a . Rak : cng chu nn ng tm ca khi xy c gia cng li ct thp.Rak = R + 2 Ra 2R 100N [N ] = mdn . .Rak .F

(2)

R: cng tnh ton ca khi xy c k n h s iu kin lm vicPhn 1 : Kt Cu Gch V Gch Ct Thp 26


: h s lm gim cng ca khi xy khi dng va

s hiu thp nu va c s hiu 25 = 1 Nu va c s hiu < 25 =R R25

Ra : cng tnh ton ca ct thp trong khi xy b. V d : Kim tra KNCL ca ct gch c kch thc 51x51cm, bng gch t st p do c s hiu 100, va ba ta c s hiu 50 chu Ndh= 36t, Nng.h =13t. Nu khi xy khng KNCL th hy thit k li. lo=420cm Bi gii a.Tra s liu: Gch t st p do c s hiu 100, va ba ta c s hiu 50 Tra bng c = 1000 , R = 15 kg 2cm

b. Kim tra khi xy theo cng thc: N [N ] = mdh . .R.F V cnh ca tit din l 51cm > 30cm mdh = 1 b =l 0 420 = = 8.24 < 10 h 51 Tra bng ta c = 0.91

F = 51x51 = 2600cm 2 < 3000cm 2 m k = 0.8 Cng tnh ton cacm 2 [N ] = 1.0,91.12.2600 = 28,5(T ) N = N dh + N ng .h = 36 + 13 = 49T > [N ] R = mk R = 0,8.15 = 12 kg

khi

xy

Ct khng kh nng chu lc

V gch v va c s hiu tng i cao chn phng n gia cng li ct thp ngang bng thp AI Tra bng ta c Ra = 1500 kg 2 maxcm 12 R = 50. = 50. = 0,4 0 0 1500 Ra

c. Gi thuyt gia cng li ct thp vi hm lng = 0,35 0 0 < maxRak = R + 2 .R a 2.0,35.1500 = 12 + .1 = 22,5 kg 2 cm 100 100

< 2 R = 24 kg

cm 2

d. Kim tra khi xy c gia cng li ct thp theo cng thc:N [N ] = m dh .Rak .F

Tnh a :

Rc a = . c Rak27

Phn 1 : Kt Cu Gch V Gch Ct Thp

Chng 4 : Tnh ton khi xy c ct thpR c = 2 R = 2.15 = 30 kg 2 cm c 2 .R a c Rak = R c + 100

Vi thp AI : Rac = 1,1.Ra = 1,1.1500 = 1650 kg2.0,35.1650 = 41,55 kg 2 cm 100 30 a = 1000. = 722 41,55c Rak = 30 +

cm 2

Tnh :0,75.722 = 7,975 8,24 2 0 7,975 = = = 0,888 1 + 0 1 + 7,975

0 =

0,75 a2 b

=

[N ] = 1.0,888.22,5.2600 = 51948kg = 51,95T > 49 tn Tho mn kh nng chu lc

Li thp ngang gia c dng li vung Gi thit dng 6 c fa= 0,283cm2 s=15cm (thch hp 3cm c 12cm ) 4. Tnh ton khi xy chu nn lch tm a. Cng thc tnh ton tnh ton (3) - Vi e , medh , Fn c xc nh nh trng hp khi xy khng t ct thp chu nn lch tm - Rak : cng tnh ton tng ng ca khi xy t li thp ngang chu nn lch tm Khi s hiu va 25Rak = R + 2.Ra 100 2e0 1 2R y N e me dh.Rak .Fn

Khi s hiu va < 25Rak = R 2.Ra R 2e0 1 2R . 100 R25 y

e0 : lch tm ca lc dc y : khong cch t trong tm tc dng n mp chu nn Ch : khi xc nh cc h s un dc v n phi ly theo im trng n hi ca khi xy t li thp ngang a . a = .Rc c Rak

: c trng n hi ca khi xy khng c ct thp

Rcak : cng tiu chun ca khi xy t li thp ngang28


Chng 4 : Tnh ton khi xy c ct thpc Rak = kR + c 2 .R a 100

k : h s an ton, vi khi xy chu nn Rac : cng tiu chun ca ct thp Rca = 1,1Ra vi thp CI, CII Rac = 1,25Ra si thp b. v d Xc nh kh nng chu lc ca tr c tit din 51 x 64 cm xy bng gch t st p do. Chiu di tnh ton ca ct l0 = 3,2m, lch tm ca lc dc e0=5cm theo pha cnh ln. Mc gch l 100. Va xi mng M75. Trong ct t li thp hn bng si thp 5 (fa = 0,196cm2) khong cch cc thanh c=4cm, khong cch gia li s = 21cm Bi gii e. Tra bng vi gch M100 v va M75 c :R = 17 kg cm 2

, = 1000cm 2

Vi li thp dng si thp 5 : Ra = 2000 kg f. V h = 64cm > 30cm nn medh = 1 g. V dng li vung nn hm lng ct thp=

Va 2f 2.0,196 .100 0 0 = a .100 0 0 = .100 = 0,467 0 0 V cs 4.21

h. Khong cch t trng tm tc dng n mp chu nny = 0,5h = 0,5.40 = 32cm

i. Cng tnh ton tng ngRak = R + 2. Ra 2l 0 1 y 100

Vi F = 51 x 64 = 3264cm2 > 3000cm2 mk =1 R = mk R = 1.17 = 17 kg Rak

cm 2 2000 2.5 kg < 2 R = 34 kg 2 = 17 + 2.0,467. 1 = 29,84 cm 2 cm 100 32 c 2 .R a 100

j. Cng tiu chun ca khi xy c li thpc Rak = KR +

Vi Rac = 1,25Ra = 1,25.2000 = 2500 kg Khi xy chu nn K = 2R = 0,85R = 0,85.17 = 14,45 kg

cm 2

cm 2 2.0,467.2500 c Rak = 2.14,45 + = 52,25 kg 2 cm 100

c trng n hi ca khi xy c ct thp


29


a =

Rc 28,9 = 1000 = 553 c 52,25 Rakcm 2

(Vi R c = KR = 2.14,45 = 28,9 kg mnh : h = h , a )

)

l 0 320 = = 5 tra bng = 0,95 (ph thuc vo h 64

k. Xc nh e : c th xc nh ging nh khi xy khng c ct thp chu nn lch tm hoc ta c th dng ct thp thc nh sau ; e = 1 0 0,06 0 0,2 h h 320 5 = 0,951 0,06. 0,2 64 64 = 0,943 e l

l. Din tch vng nn :

Fn = b(h 2e0 ) = 51(64 2.5) = 2754cm 2 e 5 m. Tnh : = 1 + 0 = 1 + = 1,078 15 hoc r >53) hoc cc tr v tng chu ti trng rung ng - C hai cch t ct thp dc + t bn trong: thp c bo v tt nhng kh thi cng v kh kim tra cht lng, hiu qu chu lc khng cao (do gn trc trung ho ) cch t ny t c dng trong thc t.


30


+ t bn ngoi: ct thp c p bn ngoi khi xy c lp va bo v. Cch t ny d thi cng, ct thp t xa trc trung ho nn chu lc tt - Va: phi c s hiu 25 khi khi xy lm vic trong mi trng kh ro v c m bnh thng, 50 khi khi xy lm vic trong mi trng m t hoc su di t - Ct thp : Ct thp dc: chu nn d 8mm , chu ko d 3mm Ct ai: d = ( 3 8 ) mm khong cch ai u = bi s ca chiu cao hng gch ( k c va ) u 50cm u 15d : khi ct thp t bn ngoi 20d : khi t ct thp bn trong Trong on ni ct thp : u 10d Vi d l ng knh ct dc Hm lng ct thp : 0,1 0 0 vi ct thp chu nn 0,5 0 0 vi ct thp chu ko - Chiu dy lp bo v ct thp dc 10mm i vi tng khi mi trng kh ro 15mm i vi tng khi ngoi tri 20mm i tng khi khi xy di t Vi tr, ct : 20mm khi khi xy mi trng kh ro 25mm khi khi xy ngoi tri 30mm khi khi xy di t - Khi gn n trng thi ph hoi do khi xy c ct thp b/d ln lm khi xy v ct chp chu nn s b tch ri ra, khng lm vic cng nhau na. Do cng ca khi xy khng s dng ht m ch 85 0 0 so vi khi xy khng c ct thp. - Khi khi xy c ct thp chu nn th ta nhn vi h s iu kin lm vic mk = 0,85 - Khi khi xy ch t ct thp chu ko th ly mk = 1 2. Tnh ton chu ko, chu nn ng tm , (4 ) N m dh (0,85 RF + Ra Fa, ) , mdh : xc nh nh trng hp khi xy khng t ct thp R : cng tnh ton chu nn ca khi xy F : din tch tit din ca khi xy , :C. tnh ton ca ct thp trong khi xy=>tra ph lc 9 Ra , : din tch tit din ct thp dc Fa 3. Tnh ton chu ko chu nn lch tm a. Trng hp lch tm b - iu kin xy ra trng hp lch tm bPhn 1 : Kt Cu Gch V Gch Ct Thp 31


-

-

(5) S n , S 0 : mmen tnh ca din tch phn chu nn v ton b din tch tit din i vi trng tm ct thp chu ko (hoc chu nn t) Vi tit din ch nht th cng thc c th vit li : x 0.55h0 x : chiu cao vng nn h0 : chiu cao tnh ton ca tit din Lch tm b c 2 trng hp : ton b tit din chu nn, trn tit din c mt phn nh chu ko C 2 trng hp trn th s ph hoi bt u t vng nn nhiu S ng sut nh hnh vS n 0. 8 S 0N N

M / F M / F

RaFa

D

Ra'Fa'

RaFa' a

D' a ' a

Ra'Fa'

a ' a

Ne e .medh (0.85..R.S 0 + R F Z )' ' ' Ne ' e .medh (0.85..R.S 0 + Ra Fa' Z a )

e : khong cch im t lc N n trng tm ct thp Fa vi tit din ch nht e = 0.5h + e0 a e : khong cch im t lc N n trng tm ct thp Fa' vi tit din ch nht e = h0 a ' e e , medh : xc nh nh i vi khi xy khng c ct thp chu nn lch tm ' S 0 , S 0 : mmen tnh ca ton b tit din ly i vi trng tm ct thp Fa , Fa' Za : khong cch gia 2 trng tm Fa , Fa' Vi tit din ch nht th hai cng thc trn c th vit li' Ne e .medh [0.42..R.b.h02 + Ra Fa' ( h0 a ' )]

Ne ' e .medh [0.42 .R.b.h02 + Ra Fa ( h0 a ' )

Trng hp t ct n th iu kin cng l :Ne e .medh 0.5.R.b.h02


32


b. Trng hp lch tm ln - iu kin xy ra lch tm ln : S n < 0.8S 0 Vi tit din ch nht : x < 0.55h0 - Khi xy ra nn lch tm ln th tit din c 2 min ko nn r rt. S ph hoi bt u t vng ko. - Biu ng sut nh hnh v, xem gn ng ng sut vng nn c dng hnh ch nht t tr s R , b qua kh nng chu ko ca khi xy (13) M / Fa Ne e .medh .(0.85.RS n + Ra' Fa' Z a ) ' ' Y = 0 N e .medh .(0.85.RFn + Ra Fa Ra .Fa ) (14) - Vi tit din HCN th (13) v (14) c th vit li

N

RD

RaFa

Ra'Fa'

M / F Ne .m .[0.85.Rb.x(h 2 ) + R F (h a )] (15) Y = 0 N .m .(0.85.R.b.x + R F R .F ) (16)a e edh 0 ' a ' a ' 0e edh ' a ' a a a

x

- Vi tit din HCN t ct n : (18) Vi 1,25 c xc nh nh i vi trng hp khi xy dng khng c ct thp chu nn lch tm. - iu kin hn ch: + ct thp chu nn t ti tr s gii hn v cng chu nn Ra th iu kin: Z ho a ' vi tc dng x 2a ' c. Vn dng Khi thit k xy dng c ct thp dc ta thng gp hai dng bi ton sau: Bi ton 1: Bit kch thc tit din, s hiu gch, s hiu va, nhm thp, iu kin lin kt, ti trng xc nh Fa v Fa Tra s liu Ra, Ra ' , R, Coi nh s dng ht kh nng chu nn ca ct thp Fa T (6) hoc (8) tnh c Fa hoc t (13) hoc (15) vi gi thit Sn = 0,8 So (hoc x = 0,55ho) tnh c Fa Nu Fa ' 0 : tc ta khng cn ct thp chu nn trng hp t ct n Nu Fa ' > 0 : + Xc dnh chiu cao vng nn x Gi thit ri vo trng hp lch tm ln (13) v ( 15) x + Nu X 0,55ho : lch tm ln T (14) hoc (16) tnh Fa + Nu X > 0,55ho : ri vo lch tm b t (7) hoc (9) tnh FaPhn 1 : Kt Cu Gch V Gch Ct Thp 33x Ne e me dhRbx(ho ). 2 N e me dh(.Rbx RaFa )

(17)


Bi ton 2: Bit kch thc tit din, s hiu gch v va, nhm thp iu kin lin kt, din tch ct thp xc nh kh nng chu lc ca cu kin. - Tm s liu tnh ton - T (14), (16) hoc (18) tnh c chiu cao vng nn X - Nu X 0,55ho : ri vo trng hp lch tm ln kim tra kh nng chu lc theo (13), (15) hoc (17) - Nu X > 0,55ho : ri vo trng hp lch tm b kim tra kh nng chu lc theo (6), (7), (8), (9) hoc (10) d. V d Cho ct gch c tit din 64 x 51cm chiu di tnh ton lo = 8m chu tc dng ca mt lc nn N = 20t t lch tm theo phng cnh ln ca tit din mt on eo = 15cm. Ct xy bng gch t st p do mc 100 dng va xi mng mc 50. Yu cu xc nh din tch ct thp dc Fa v Fa (thp nhm CII) - Tra s liu tnh ton Khi xy dng bng gch t st p do mc 100 dng va xi mng mc 50 tra bng ta c = 1000, R = 15 kg 2cm

Ct thp dc CII c Ra = Ra ' = 2400 kg mnh ca ct: b =

cm 2

l o 800 = = 15,7 > 15 b 51

-

-

Cn phi gia c khi xy bng ct thp dc Vi b = 17,5 tra bng = 0,748 e l Xc nh: e = [1 0 (0,06. 0 0,2)] h h 15 800 = 0,748[1 (0,06. 0,2)] = 0,652 64 64 Cnh b ca tit din b = 51cm > 30cm medh = 1 e 15 = 1+ 0 = 1+ = 1,234 < 1,45 h 64

- Cc thanh thp t vo bn trong ct cch mp ct mt hng gch (chiu rng vin gch l 105mm, chiu dy lp va ngoi 15mm, lp va gia l 10mm) a = a ' = 10,5 + 1,5 + e = eo + = 15 + h a 2 10 = 12,5cm 2

54 12,5 = 34,5cm 2 ho = h a = 64 12,5 = 51,5cm N .e 0,42.R.b.h02 e medh (8) Fa ' = Ra ' (ho a' )


34

Chng 4 : Tnh ton khi xy c ct thp20000.34,5 0,42.1,234.12,75.51(51,5) 2 0,652.1 = = 1,75cm 2 2400(51,5 12,5) Vi R = mk R = 0,85.15 = 12,75 kg 2 cm 2 chn 212 c Fa ' = 2,26cm

- Xc nh x

N .e = e .medh [0,85.R.b.x( ho 0,5 x ) + Ra ' Fa ' ( ho a ' )] 20000.34,5 = 0,652.1[0,85.1,234.12,75.51.x(51,5 0,5 x) + 2400.2,26(51,5 12,5)] 690000 = 0,652[682,05 x(51,5 0,5 x) + 211536]

222,35 x 2 + 22902 x 552078,5 = 0

nhng trng hp x = 38,48cm > 0,55ho = 0,55.51,5 = 28,33cm Lch tm b N .e' e .medh 0,42.R.b.ho 2 (9) Fa ' = Ra (ho a ' ) 20000.4,5 0,42.12,75.51.51,5 2 0,652.1 = =15d s1 s1 s1 s3 s2 s2 s2 s2 s3 >=15d s1 s1 s1

b. B tr bulng kiu c

>=15d

s1

s1

s1

c. B tr inh theo hng thng Hnh:3-8

d. B tr inh kiu c

e. B tr inh kiu xin

B tr lin kt cht tr

Loi cht Bulng Cht tr bng g do tt inh

Bng III-4: Khong cch tiu chun gia cc cht Khong cch S1 Khong cch S2 Khong cch S3 b 10d 6d 7d 4d 5d b 10d 3,5d 3d B tr theo c 3d b 10d 2,5d 3d 2,5d 2,5d

c >=10d c = 10d 15d 25d

4d

3. Cht bn a. Cu to Cht bn thng lm bng cc loi g tt, do c x l k. Cht bn c tc dung lm tng tit din chiu cao. C kh nng chng trt gia cc phn t. Yu cu cu to ca cht bn:l b = 4.5 b

lb + 0.1cm 2 s 2l b = 9 b hr =

S: khong cch gia trc hai cht bn.Phn II : Kt Cu G 27

Chng 3 : Lin kt trong kt cu g

b. Tnh ton Kh nng chu lc ca mt cht bn T c xc nh nh sau: - Theo iu kin p mt ca cht bn: b Tem = 14 l b b b ( daN) vi lb, bb tnh bng n v cm. - Theo iu kin chu un ca cht bn: Tub = 63 b b b ( daN) vi b, bb tnh bng n v cm.b T = min Tem , Tub

[

]

S lng cht bn: n =

Q T

Q: lc ct xut hin trong cu kin. 5. LIN KT CHU KO BNG KIM LOI Loi lin kt ny lm bng kim loi dng lin kt cc thanh ko hoc trc tip chu lc ko thay cho cu kin g chu ko. Thng lm bng thp CT3.1.5d l1 2mm a d a

d

l1 2mm

d

2d

6-:-7d

H nh3.9: Lin kch ke bg kim loa t u o n i

1. inh v vt chu nh Kh nng chng nh ca inh v vt tnh theo cng thc:T = R nh d l1

Rnh: Cng chu nh ca inh khi ma st.R nh = 3.0( R nh kg ) : g kh cm 2 kg = 1.0( 2 ) : g t cm

Yu cu v cu to:

Phn II : Kt Cu G

28

Chng 3 : Lin kt trong kt cu g

l1 2a l1 10d a 4d

Ch : - inh lm vic do ma st nn khi khoan l ri mi ng inh hoc inh chu ti trng ng th inh khng lm vic. - i vi inh vt: phi khoan l trc, dl = d - (12)mm, Rnh = 10 kg/cm2. 2. Bulng sit, thanh cng Dng chng tch trong lin kt chm, lm neo, cc thanh chu ko trong dn, lm thanh cng. Cng thc kim tra:=N mkR Fth

mk= 0.8. t thm cc tm lt c hnh vung cnh a m bo p mt cho g: em =N R em a F

N R em

3. inh a C tc dung chu ko hoc ging cc cu kin trong cc cng trnh ph, tm thi nh dn gio, tr cu g ... inh a thng lm bng thp trn c ng knh d = 1218mm. $6. LIN KT DN L lin kt ph hp vi cng ngh ho trong xy dng, lm cho dng kt cu v kin trc thm phong ph, tit din khng b gim yu, phng p, khi tnh ton xem nh cu kin nguyn.

Phn II : Kt Cu G

29

Chng 1 : M u

CHNG 1I. S lc lch s pht trin ca kt cu gch

M U

Kt cu gch c dng rt sm gn lin vi s hnh thnh v pht trin ca x hi loi ngi, t thi k nguyn thy con ngi bit xp cc khi thnh hang hc . Cch y 8000 nm trc con ngi dng gch khng nung, 5000 6000 nm trc con ngi bit dng c gia cng, 3000 nm trc dng gch nung. Nhng cng trnh ni ting trn th gii c xy bng kt cu gch : Kim t thp Ai Cp: xy cch y trn 5000 nm, cao 146,6m, cnh y di 233m, khong hn 2 triu vin mi vin nng t 2,5 n 50 tn n th n thn ian Hy Lp : xy vo th k th 6 trc cng nguyn, trong n c 125 ct cao 19m Cy Hi ng thnh ph Alcxngri (Ai Cp) : 127m bng (b hng vo nm 1935 do ng t) Vn treo Babilon : xy dng vo th k 15 trc cng nguyn in Pantheon Rm : cao 42.7m Nh th c B, in Kremlanh (Nga), Vn l trng thnh (Trung Quc) Vit Nam cng c mt s cng trnh bng gch nh : Thnh Ty (1397) ti Vnh Lc - Thanh Ha, Thp Bnh Sn (Vnh Phc), Thnh H Ni, Ct C, Cha Thin M ... Ngy nay kt cu gch c dng rng ri. Xy dng cu vm nhp n 90m, vm, v mng cong 2 chiu, mi vm gch nhp 15m, nh, ng khi... II. u nhc im v phm vi s dng ca kt cu gch 1. u im C cng ln, kh vng chc v bn lu, t cn c bo v v tu b Chng chy tt Dng vt liu a phng r tinPhn I : Kt Cu Gch V Gch Ct Thp 1

Chng 1 : M u

Cch m, cch nhit tt

Tit kim thp, xi mng so vi b tng ct thp, khng cn vn khun, thit b thi cng n gin 2. Nhc im Trng lng bn thn ln, khu nh, vn chuyn nhiu Kh nng chu lc khng cao Kh c gii ha thi cng Lc dnh km nn chu ko, chu ti trng ng kt cu d nt

Dng nhiu t => nh hng n t nng nghip Khc phc : ch to ra cc vt liu xp nh, dng gch bng b tng nh, gch rng, khi xy rng, nhiu lp. thun tin cho vic c gii ha thi cng ngi ta dng cch lp ghp cc tm ln. Dng cc kt cu hp l : tng rng, mi vm, v, gch c ct thp...

Phn I : Kt Cu Gch V Gch Ct Thp

2

Chng 2 : Khi Xy Gch

CHNG 2

KHI XY GCH

1. VT LIU DNG TRONG KHI XY GCH I. Gch 1. Phn loi Theo vt liu : Gch t st Gch b tng Gch sillicat Theo phng php ch to : Gch nung Gch khng nung Theo trng lng : Gch nng : 1800kG / m 3 Gch nh : < 1800kG / m 3 Gch rt nh : < 1000kG / m 3 Theo s hiu : Gch c cng thp : 4,7,10,15,25,35,50 Gch c cng trung bnh : 75,100,125,150,200 Gch c cng cao : 250 1000 Kch thc v trng lng vin gch c quy nh ph hp vi sc khe trung bnh ca cng nhn v kch thc tiu chun ca kt cu. Gch t st nung c kch thc tiu chn : 220x105x60 Gch th : 5x10x20, 4x8x19 Gch ng : 10x10x20, 8x8x19 Gch ng 6 l : 10x13.5x22, 8.5x13x20 Gch b tng : 20x20x40, 15x20x40, 10x20x40, 15x20x30 Gch sillicat : 6.5x12x25 2. Cc c trng ca Gch Cng chu nn : xc nh bng th nghim nnRg = Ru = N F M W

Cng chu un : Cng tiu chun : c ly trung bnh ca 5 mu th Cng tiu chun khi nn : R =c g

Rgii =1 5

5

5

Cng tiu chun khi un : R =c u

Ruii =1

53


Chng 2 : Khi Xy Gch

Cng chu ko ca gch th ch bng (5 10)% cng chu nn Mc gch c xc nh trn c s cng trung bnh v cng b nht ca cc mu th khi nn v un Gch mc thp : 7, 10, 25, 35, 50 Gch mc trung bnh : 75, 100, 125, 150, 200 Gch mc cao : 300, 400, 500, 600, 800, 1000 Cng trnh dn dng : 50, 75, 100, 125, 150 Quan h gia ng sut v bin dng ca gch t st gn nh theo quy lut ng thng Vi gch t st p do v gch sillicat : Eg = (12).105 kG/cm2 Vi gch t st p kh v gch sillicat : Eg = (0.20.4).105 kG/cm2 H s bin dng ngang ca gch tng theo cng vi s tng ng sut gch t st nung c h s bin dng ngang 0.03 0.1 II. c khai thc trong t nhin, c th gia cng n mt mc no c hnh dng v kch thc nht nh nng c 1800kG / m3 nh : hoa cng, vi sa thch, bazan, lmit ... nh c < 1800kG / m3 nh : bt, vi v s ... Mc cng cn c vo gii hn cng chu nn, c cc loi mc t 4, 10, 25 n 3000 III. Va 1. Tc dng Lin kt cc vin gch thnh khi xy vng chc Truyn v phn phi ng sut trong khi xy t vin ny n vin khc Lp kn khe h trong khi xy Cch m, cch nhit 2. Yu cu ca va Phi c cng nht nh ng vi tng loi khi xy Phi c tnh bn vng, tnh linh ng, do Phi c kh nng gi nc m bo d xy 3. Phn loi Theo thnh phn : Va xi mng : gm, ct, ximng v nc V khng c cht kt dnh do nn va xi mng kh cng nhanh, c cng khc cao nhng dn kh thi cng Va bata (va tam hp) : gm ct, ximng, vi, t st v nc Loi va ny c tnh do cn thit, thi gian kh cng va phiPhn I : Kt Cu Gch V Gch Ct Thp 4

Chng 2 : Khi Xy Gch

Va khng c ximng : Va vi gm vi, ct, nc Va t st gm ct, t st, nc Va thch cao Theo trng lng : Va nng : c 1500kG / m3 Va nh : c < 1500kG / m3 Ngoi ra cn c th phn loi theo : Va thng thng : va vi, va xi mng, va tam hp Va hon thin : dng trang tr mt ngoi cho cng trnh nh : va ht lu, va trt gai, va mattt Va chu axt : dng trt, lt, p bo v cc b phn ca cng trnh lm vic trong mi trng chu tc dng ca axt (cht kt dnh thng dng l thu tinh lng) Va chu nhit : dng xy, trt cc b phn cng trnh chu nhit, xy l nung, xy bp, ng khi ... (cht kt dnh l ximng pooclng, hoc ximng pooclng ho do, ct liu l bt samt) Va chng thm : l loi va dng trt, lng bao bc cc b phn cng trnh chu nc (thng dng ximng mc cao hoc va c ph gia chng thm) Thnh phn, cp phi v phm vi s dng ca va c quy nh trong quy phm 4. Cc c trng Cng chu nn ca va c xc nh bi cc mu th khi vung cnh 7,07cm trong iu kin tiu chun (t=(1525)0C, thi gian 28 ngy) Cc s hiu va : 0,4, 10, 25, 50, 75, 100, 200 (va s hiu 0 l va mi xy xong, s hiu 2 dng xc nh bin dng ca khi xy bng va vi khi tui di 3 thng, s hiu 4 dng lp m Va xy c mc 25 Cng ca va xi mng v va tam hp c xc nhRt = 1,5t .R28 vi t 50 ngy 14 + t

R28 : cng ca va ngy th 28 Bin dng ca va rt khc nhau ph thuc vo mc va, thnh phn v cp phi, tnh cht ca ti trng. Va nh c bin dng ln hn va nng Bin dng ca va tng ln khi ti trng tc dng di hn tnh t bin Co ngt v t bin tng theo thi gian v ph thuc vo nhiu yu t 5. Chn cp phi va Chn cp phi tc l xc nh cc thnh phn ca va Gi Qx l lng ximng tnh bng kG cho 1m3 ct ht trung bnh v ln khi m t 1 3%Phn I : Kt Cu Gch V Gch Ct Thp 5

Chng 2 : Khi Xy Gch

Qx =

Rv .1000 0,7 Rx

(2.1)

Rv : mc va Rx : mc xi mng Khi dng ct kh lng ximng tng 5%, khi dng ct nh lng ximng cng cn c tng ln Lng vi ti trong va cho 1m3 ct (tnh bng lt) c xc nh theo cng thc : D = 170(1 0,002.Q x ) (2.2) Lng nc dng trn vaN = 1,3 1,6 X

Chn cp phi va theo phng php tra bng o nh mc cp phi vt liu cho 1m3 VXM ct vng, ct c ML > 2, xi mng PC30/PC40 Mc va Thnh phn n hao ph v 25 50 75 100 125 150 - Xi mng - Ct vng kg m3116.01 88.91.19 1.19 213.02 163.02 1.15 1.16 296.03 227.02 1.12 1.13 385.04 279.02 1.09 1.11 462.05 361.04 1.05 1.08 425.04 1.06

M hiuB.121 B.221

2

3

4

5

6

7

o nh mc cp phi vt liu cho 1m3 VXM cat mn, cat co ML = 1.5 2, xi mng PC30/PC40 Thanh Mac va n Ma phn v hiu 25 50 75 100 125 hao ph 230.02 230.02 320.03 410.04 - Xi kg 247.02 320.03 389,04 176.02 176.02 B.122 mngB.222

- Ct mn

m3

2,16 1,18

1,12 1,14

1,09 1,12

1,05 1,09

1,06

2 o nh mc cat mn, PC30/PC40 Thanh Ma phn hiu hao phB.123 B.223

3

4

5

6

cp phi vt liu cho 1m3 VXM cat co ML = 0,7 1,4, xi mng n v kg Mac va 25142,01 108,02

50261,03 200,02

75360,04 278,03

100 359,04

- Xi mng


6

Chng 2 : Khi Xy Gch

- Ct mn

m3

1,13 1,14

1,09 1,11

1,05 1,1

1,04

2

3

4

5

IV. Khi xy gch 1. Cu to Khng c xy trng mch, cc mch va ng phi nm v tr (1/4 1/2) vin gch cho ti trng t bn trn truyn xung cho ton b khi xy Chiu dy mch va ngang khng c qu ln v qu mng i vi khi xy gch : = (8 15)mm i vi khi xy c quy cch : 20 mm Cc vin gch xy phi c ging vo nhau. Cc hng ging nhau t 3-5 hng gch theo chiu cao tng (3 dc 1 ngang hoc 5 dc 1 ngang) Lc tc dng ln khi xy cn phi vung gc vi lp va nm ngang. Cc vin gch trong khi xy cn phi t thnh hng trong 1 mt phng Cc mch va ng phi song song vi mt ngoi ca khi xy v cc mch va ngang phi vung gc vi mt ngoi ca khi xy B rng ca tng xy gch l bi s ca na vin gch Tng 1/2 gch : dy 105mm Tng mt gch : dy 220mm Tng mt gch ri : dy 335mm Tng 2 gch : dy 450mm 2. S lm vic ca vin gch trong khi xy Di tc dng ca ti trng N t ng tm th ng sut trong vin gch kh phc tp : nn lch tm, nn cc b, un, ct ko Nguyn nhn : Do tnh cht khng dng nht ca lp va xy, thnh phn cu va khc nhau, iu kin kh cng ca va trong khi xy khng ging nhau Do bin dng ca va v gch khng ging nhau, khi kh cng va co li nhiu hn gch Kt qu lm cho mch va tch ra thnh nhiu on Do thi cng, ngi cng nhn khng th tri tht bng mch va nn khi t vin gch n khng to nn s nn u Do b mt ca vin gch khng u Do bin dng ca gch v va khng ging nhau. Khi chu lc gch cn tr bin dng ngang ca va Gy ng sut nn trong va v gy ra ng sut ko trong gch. ng sut ko ny cng vi ng sut ko khi vin gch chu un c th vt qu gii hn cng chu ko ca gch gch nt Trong khi xy hc hnh dng cc vin khng c quy cch, ng sut tp trung ln nhng vin nh ra


7

Chng 2 : Khi Xy Gch

3. Cc giai on lm vic ca khi xy chu nnN

Ket Cau Gach Da go

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