Đáp án đề thi cuối kỳ học kỳ I năm học 2019-2020 môn Cơ sở công nghệ chế tạo máy - ĐH Sư phạm Kỹ thuật

DAP AN DE T H I C U O l K Y H Q C K Y 1
N A M H Q C 2019-2020
M o n : C O SO C O N G N G H E C H E T A O M A Y
M a m o n h p c : 18 2 F M M T 3 3 0 8 2 5
Cau 1: ( 1.5 diem )

cd 3 cau

Cho so do khoan nhu H in h , hay :

1 ghi ro cac thong so: chieu sail cat t, lu o n g chay dao S, chieu day cat a, chieu rong cat b, goc nghieng
-

chi'nh tp (ve len h in h ) ( 0.2 x 5 = 1 d ie m )
3.

trin h bay dinh nghla : ( 0.5 diem )

chieu day cat a: la khoang djch chuyen cua dao
sau 1 lu on g chay dao Sz, do theo phuong vuong
goc v o i b. o r a=Sz. Sin (9 ) ( 0.25 diem )

chieu rong cat b: la khoang each giua be mat da
gia cong va be mat chua gia cong, do theo
phuong ltro i cat chinh tren mat day.
hoac -'doan chieu dai ltro i cdt chinh tham gia
cdt” o r d iin g cong there b=t/sin((p) ( 0.25 diem )

2. V ie t cong thirc tin h xac dinh m oi quan he giira van toe cat V va so vong quay n ( 0.5 di£m )

V = ? ' D ' \ m/ ph)

1000

Khong co don v j o r sai don \ j - 0.25 diem

S6 hieu: BM3/QT-PDBCL-RDTV


Cau 2: T u ket cau cua m ui khoet, hay trin h bay cac dac diem cua nguyen cong khoet ? ( 0.5
diem)
M u i khoet co lo i Ion hon m u i khoan, nen do cung v u ng cua no cao hon. co kha nang dam bao do chinh
xac va do n h ln bong be mdt cao hon khoan. Khoet con co kha ndng nang cao liro n g chay dao nen nang suat
cao hon khoan ( 0.25 diem )

Sira dirge nhirng sai so ve vi tri tirong quart do khoan 16 hoac 16 due san. ren sin de lai. ( 0 .2 5 d ilm )

Cau 3: Tai sao lai khong nen doa 16 co ranh ? ( 0.5 diem)
Lo co ranh doc lam kha n&ng dinh huong cua dao khong dung va 16 gia cong se bi meo.

Cau 4: G ia cong co Phoi day thu ong gap khi nao ? ( 0.5 diem)
phoi day_thuong gap kh i gia cong vat lieu deo 6 toe do cao, chieu day cat be

Cau 5:
H ien tuo ng co rut phoi la gi ? (

0.5 diem)

K ich th iro c cua phoi tach ra khong giong v o i cac kich thiroc lop cat tu o n g tin g kh i no con nam tren chi
t i e t : Hi?n tu o n g co ru t phoi

Nhdn to anh huang den CO rut phoi:

-

Vat lieu gia cong

-

Che do cat

-

Thong so hinh hoc cua dao
dung dich trom nguoi

Cau 6 cho so do Phay, dao phay tru rang th in g , nhu hinh ve. ( 2 diem )

1.

Hay cho biet, day la phuong phap phay gi ? trin h bay

dac diem cua phuong phap.
la truong hop phay thuan ( 0.5 diem )
dac diem cua phuong phap.
Chieu quay cua dao va chieu tinh tien cua ban
may cung chieu nhau.
lire co xu huong de chi tiet xuong (thanh phan
thang dung Pv) nen cat em. do bong tang
phoi cat tir day den mong : do bong tang

S6 hieu: BM3/QT-PDBCL-RDTV

f
C au 7 Cho so <36 ti$n tru nhu hinh ve ( 1.5 diem )

1. C hiian gia cong tro n g sa do nay la gi ?
Chuan gia cong tro ng so do nay la 2 16 tam. ( 0.25 diem )
2.

Ve hinh, ghi va phan tich d jnh v j dung tro ng so do

(hinh lun g tn iii him. (
.ham then hinh rc. ky hieu he tru e ton Jo)
Ve hinh, ghi ro ky hi#u d jnh v j tren tung m ui tam ( 0.5 diem )
phan tich d jnh vj _ke ten cac bac tir do bj khong che _( 0.5 d ilm )
3.

D ung cir ty 1, v it dieu chinh 2 va bo can mau 3, de tien hanh gia cong cac kich thuoc L I ,
L2, L3, L4. Hay trin h bay each gia cong dat kich thuoc L I va L4 ( 0.5 diem )

L I : gia cong khong can can mau, de ban xe dao tiep xuc true tiep vao cu 2
L4: dung can mau co chieu dai L I- L 4

So hieu: BM3/QT-PDBCL-RDTV


X
Cau 8 Cho chi tie t nhu hinh ve ( 2.5 diem )

Be phay co kich thuac L x h , tren phoi tru D dD x H 8H, da gia cong 2 mat dau.

1- De phay 2 mat phang, trong cung 1 nguyen cong, co kich thuac L x h . T rin h bay each
gia cong (dung may van nang. dang san xuat loat vua, do chinh xac va do nham dat
duac: gia cong tho). Phai N o i ro loai may, dao. V D may phay ngang + dao phay dia 3 mdt ( 0.5
diem ).

(cham theo bai lam cua sv)

2.

V a i each gia cong da trin h bay d can 1. Ve h in h khong co d ird n g ta m - 0 .2 5 d .
-

Ve hinh. xac djnh chuan dinh vi va so bac djnh v i tira ng ung. ( 0.5 diem )

-

G oi ten chi tiet djnh v j da dung. ( 0.5 diem ).
V e cac kich thude cong nghe , ghi ky hieu ( 0.5 diem )
(cham theo bai lam cua sv)

3.

T in h ec(h ) b an g each la p c h u o i k ic h th u d e c o n g nghe.
la p d u g c c h u 6 i bang h in h ve : ( 0.25 d ilm ) _
ve m iii ten 2 dau : chuoi sai.

T in h d u a c g ia tr i ec( h ) = 5 H ( 0.25 d ie m )_ c h i cham khi chuoi dung
(cham theo bai lam cua sv)

S6 hieu: BM3/QT-PBBCL-RDTV