Bi thi tm hiu

Bi tp ln: C hc t

Phn I: u bi

S th t u bi: . N0 . . . . .

Aih1=.. m; Bih2=.. m; Cih3=.. m; Ti ; a x b=.. m; hm =.. m

1)- S mng v cc lp t:

Hnh 1-1- S mng v cc lp t

(Ch hnh v 1-1: Cn ghi y chiu dy cc lp, chiu su chn mng, kch thc mng (bng s) ng vi s liu bi tp ln ca tng ngi)

2)- Cc s liu v nn t:

Nn gm 3 lp t theo th t t trn xung l A; B; C, cc ch tiu c l cho bng sau:

Ch tiu c l cc lp t

S m wTGii Gii Gc maLc H s rng ng vi

lpt nhintrnghn nhohn dost trongdnh c cc cp p lc p=(1(4) (KG/cm2)

tW(%)(g/cm3)ht (Wnh(%)Wd(%)()KG/cm21234

Ai - - - - - - - - - - -

Bi - - - - - - - - - - -

Ci - - - - - - - - - - -

3)- Cc s liu v ti trng:Gm c lc tc dng thng ng Ntc0; Mtc0 t ti cao trnh ct (0,00 do ti trng cng trnh truyn xung, m men tc dng theo 1 phng (y l ti trng tiu chun):

Ntc0=T; Mtc0=.Tm

4)- Cc s liu v mng:

Mng cng, y mng ch nht, c cc cnh a x b; chiu su chn mng hm ; a =m; b =m; hm=m.

Phn II: Yu cu v hng dn (Bi lm)

1)- Xc nh tn t, trng thi ca t. (1)+ Vi t dnh (c gc ma st trong v lc dnh), tn t xc nh thng qua ch s do (. Bng 1

( = Wnh - Wd

(1)

trng thi ca t thng qua st (ch s st) B: Bng 5

(2)

+ Vi t ri: tn t xc nh theo bng 2

+ Tnh h s rng t nhin ca cc lp t:

(3)

Vi cc gi tr (0; (1; (2; (3; (4 v ng cong quan h (( - p)

Lp 1

Lp 2

Lp 3

Hnh 1-2- Biu ng cong (( - p) ca cc lp t

(Ch : Ghi y cc tr s ( trn trc tung ca cc ng cong). Khi v ((-p) t gi tr (4 st trc honh, tng t l trc tung cho d nhn.+ Tnh E1-2 ca tng lp t, cho (=0,8.

2)- Xc nh ng sut tip xc di mng.(0,5)+ Ti tm mng (im O):

(4)

+ Ti A v B:

(5)

trong : Gm + G = F.hm.(tb (c th ly (tb = 2,0 T/m3)

(6)

Hnh 1-3- ng sut tip xc ti im A; O; B.

3)- Tnh p lc tiu chun Rtc ln lp t t mng.(0,5)

Theo cng thc:

(7)

Trong :

A; B; D- cc h s ph thuc gc ma st trong ( ca lp t t mng (bng tra);

b- chiu rng mng;

h- chiu su chn mng;

(t dung trng lp t t mng tr ln;

(d dung trng lp t t mng tr xung; ly trong vng ch ng (c th ly gn ng bng dung trng lp t t mng).c- lc dnh ca lp t t mng;

m1; m2; ktc - cc h s, trong bi tp ny, tm ly bng 1,0.

(Ch : n vi tnh ca gi tr Rtc v lc dnh c).

So snh: p0 (( Rtc c php tnh ton US, bin dng theo cc kt qu ca l thuyt n hi nh cc cng thc nghin cu. (Nu khng tho mn, c th tng kch thc mng).

4)- Tnh ln cho cc im O; A; B theo PP tng cc lp phn t.

a)- Ti im O: (3)

Bc 1: Tnh US gy ln:

pgl = p0 - (t hmBc 2: Chia nn t di y mng thnh cc lp phn t mng vi chiu dy hi = (0,2 ( 0,4)*b (b-chiu rng mng). Ch khi chia lp phi m bo sao cho cc lp phn t nm trn trong lp ln (lp 1; 2; 3). Nn chia vi chiu dy (0,2*b; (0,3*b; (0,4*b theo th t t trn xung (cng gn y mng chia cng mng). (C th chia lp phn t mng hn hng dn trn, vi ch l cng chia mng ln cng chnh xc).

Bc 3: Tnh US (bt ; (z ((gl ) ti y cc lp hi, kt qu ghi bng 1.2

;

Bng 1.2- Bng tnh (z v (bt cho im O

Lp im zbtaba/b2z/bH sk0=zO

t tnh(m)(T/m2)(m)(m)kc4*kc(T/m2)

0 - - - - - - - - -

11 - - - - - - - - -

2 - - - - - - - - -

2 3 - - - - - - - - -

4 - - - - - - - - -

5 - - - - - - - - -

36 - - - - - - - - -

...................

Bc 4: Xc nh phm vi tnh ln H, da vo iu kin ti y lp t phn t tho mn iu kin:

(z ( 0,2*(bt khi E ( 50 KG/cm2; (z ( 0,1*(bt khi E < 50 KG/cm2;

Bc 5: ln ca im O tnh theo cng thc (8) (v ghi bng 1.3)

(8)Bng 1.3- Kt qu tnh ln cho im OLp Lp phnhip1ip2i1i2iSi

t t(m)(KG/cm2)(KG/cm2)(cm)

11 - - - - - -

2 - - - - - -

2 - - - - - - -

- - - - - - -

3 - - - - - - -

- - - - - - -

- - - - - - -

..............

Cng: -

Hnh 1-4- Biu US (z v (bt ti im O

(Ch : Ghi cc tr s ng sut (z y cc lp t phn t; US (bt y cc lp t 1; 2; 3).

b)- Ti im A v B. (5) tnh US ca cc im qua cc trc A; B ta vn dng phng php im gc v nguyn l phn ti trng thnh cc dng c bn, sau cng tc dng ta s c US ti trc qua A v B.

Hnh 1-5- Phn ti trng v din chu ti

V ti trng phn thnh 2 dng: ch nht (phn b u) c cng pglA v tam gic vi cng (pglB - pglA ).

Vi pglA = pm - (t . hm ; pglB = pM - (t . hm .

V din chu ti phn thnh hai din nh bng nhau bin im A; B thnh im gc ca 2 hnh ch nht bng nhau (hnh 1-4), c th:

- US ti A:

(9)

Trong :

;

;

- US ti B:

(10)

Trong :

;

;

Vi ch : a; b chiu di v chiu rng ca mng cho (u bi)Cc kt qu tnh ton trnh by bng 1-4

Bng 1-4- Kt qu tnh US ti trc qua im A

Lp

tim

tnhzi(m)

kc

kA

(8+12)

(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13)

-------------

-------------

-------------

Cc kt qu va tnh cho php v c biu US bn thn v US gy ln, th hin tng t nh trn hnh 1-6 (im A). Lp li bc 4 mc (a) xc nh c phm vi tnh ln, lp bng ghi kt qu tnh ln nh bng 1-3, ta s c kt qu tnh ln ti im A.

Ch : Tnh ln im B cng thc hin nh i vi im A, v cc biu US trn trc i qua B ri tnh ln.

Trong bi tp ln ny phm vi tnh ln H c ly theo gi tr tnh ln im O p dng cho tnh ln im A v B.

Hnh 1-6- Hnh dng cc biu US qua trc A; O; B.

(Ch : Ghi cc tr s ng sut (z y cc lp t phn t; US (bt y cc lp t).

Trnh by: Thuyt minh A4; c th nh trn my vi tnh hoc vit tay; t ba ngoi theo trang 8; Ghi y cc s liu u bi; u bi theo th t c giao.

Hc vin k thut qun s

Vin k thut cng trnh c bit======= ( ======

Bi tp ln

Mn hc: c hc t

S th t u bi: . . . .

Ni dung: Tnh ton ln ca nn t

Gio vin: . . . . . . . . . . . . . .

Hc vin: . . . . . . . . . . . . . .

Lp: . . . . . . . . . . . . . . . . . . . .

h ni .. /20

Phn nh gi

I)- Thang im:

a)-Mc 1: 1,0 im;

b)-Mc 2: 0,5 im;

c)-Mc 3: 0,5 im;

d)-Mc 4: 8 im

- Tnh ln im O: 3,0 im;

- Tnh ln ti im A: 2,5 im;- Tnh ln ti im B: 2,5 im.II)- nh gi:

a)- Sai u bi (cc s liu t nn hoc ti trng hoc kch thc mng): 0 im cho bi tp ln;

b)- Sai phn no tr m ca phn ;

c)- Sai cc s liu v US tip xc; tra bng h s ng sut khng ng, tnh h s rng ( khng ng dn n sai cc kt qu tnh ln-nh gi sai Mc 4: 0 im;

d)- Yu cu trnh by thuyt minh sch s, sng sa; hnh v phi c tn v ni dung; cn ghi ch y cc yu t trn hnh v ng vi bi lm c th ca mnh. Tu theo mc th hin m nh gi tr im theo tng phn.

e)- T ba ghi thng tin nh ch dn; nu thiu hoc sai tr 0,5im.PAGE 6

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