Phụ lục chung cho đề thi CTM

Phụ lục đề thi trắc nghiệm Chi tiết máy:  .

Phụ lục đề thi trắc nghiệm Chi tiết máy:





2

H

M

H

q

Z



)]

1 ( ) 1 ( [

2

2 2 1

2 1 2

2 1







E E

E E

Z M

s

s

lim

gh

]











1

m

N





E L

N

N

2 c

x 0,7

2

d k

T





u

u

u 0,7

4

d k

M W

M



   ;   22  

u c

] [

6 2 k/n









b

F b

M

] [ 7 , 0 ) 6 / (

7 ,













l k

F b

b l k

M d F

M

] [ 7 , 0 7 ,





l k

F b l k

M d F

1

3 , 1

4

d f i

F s z

k







]

[

/

2

k

1  

i

F

] [

3 , 1 4

k

F k V

3 1

1 (5,2 6,4) T

a

d d d d a L

4 2

2

2 1 2 2

1  





























1

1 2

; 1

;

1

0 2









f

f t f

t f

f

t

e

e F F e

F F e

e

F

2 cos

2F0 

F r  ;  18001

0 2 1

2 1



F F

F F

F









b v F

d t

b v F

d

C C C

k F C

C C v

k P b

]

.[

]

.[

1000

0





z l u

d

C C C

C

P

k

P

z

]

[

0

1



a

d d d d a L

4 2

2

2 1 2 2

1  



































1 2

2 2 1 2

2 2

4

1

d d d

d L d

d

L

z1 = 29 – 2.u d = p/sin( /z)

a

p z z p

a z z p

l

2

2 2

2 1 2 2









 











 ]

[ /

.k k k k P0

P

P t  z n day 

Số dẫy xích 1 2 3 4

kday 1 1,7 2,5 3







cos 2

cos )

(

w tw

a

m z





 2tg

) inv inv

)(

t

z z

) cos / tg

(

arctg  

t  ; tw arccos(a.cost/a w) ; b = arctg(costw.tg)

ba=2bd/(u1) ; lH = K..bw/cosb ;  = 0,5.dw.sinw

1

.(

cos

2

1 m u

a

 cos 2

) (z1 z2 m





















 



2 1

1 1 2 , 3 88 , 1

z

m

b w

sin





 























3

1

4

;

tw

b H

Z



 2 sin

cos 2

2 2 1

1/ 2 /

t T d T d

F   ; F a F ttgw ; F r F ttgtwF ttgnw cosw

w nw w w nw

t

n

d

T F

F







2 cos



1 i i

3 1 i

' H1 Hi

HE HN 60c T T n t 60.c.n.L T T t /t

HE H0 HL

m N N

xH v R HL

0

lim

s

K

H

H



2 H

H 1 a

w ( 1) [ ] u

K T u

K a

ba 







bd

2 1

F 1

m [ F]

F

z

Y Y K T K m











3

2 H

H 1 2

R

e

] [ ) 1 (

1





u K K

K T u

K

R

be be





2

2 1

2 2

2

e

)

tg(

tg











 ; tg=z1/q ;

x q

z d

z m w w

2

1

1







19100

1

1

t m n z q

] [

170

2

H 2 2

w

q

K T z

q z

a      ; aw = (q+z2+2.x)m/2

Fa1 = Ft2 = 2.T2/d2 ; Ft1 = Fa2 = Fa1.tg( ) = Ft2.tg( ) ;

 











cos

tg cos t2 r2

F F F

 









cos

cos

cos

t2

n

n

F

1

H 2 2 H

d

K T

cos

4 , 1

F 2

2

F Fv Fβ 2



m d b

Y K K T

j

j j j j j

j

d

t d t b

d

W

2

) (

16

1 1

3

0







;

j

j j j j j j

d

t d t b d W

2

) (

32

1 1





) (

2

1

t

h

l

d

T

l b d

T l b

F

2 



 ; L = 60.10-6.n.Lh ; Q = (X.V.Fr + Y.Fa).kd.kt ;

p = R/ld  [p] ; pv = R.n/(19100.l)  [pv] ; hmin = (D-d)/2 - e = (1-)/2  s.(Rz1+Rz2)

] [ 4 /

2

2

0

c c

d

D

z

T



2 0

d v c d

l d D z

T

1 , 0

0 3

0

u c

u

z D d

l T

3

4

2

4







c

c

] [

6 ,



c F k

) (

8

)

min max

d G x F

F

x n









 ]

[

2

p

F

d

H

h

a





 2

2

.l

s

EJ

F a





