1
Cu 1: (3 im) Cho lp ghp d = D = 85S7/h6.a) Tra bng tm sai lch
gii hn ca trc v l.
(1 im)b) V s lp ghp v cho bit mi ghp thuc loi no?
(1 im)c) Tnh cc gi tr h hoc di, dung sai lp ghp.
(1 im)Cu 2: (2 im) Cho lp ghp d = D = 75.a) Tra kch thc b, h,
t1, t2 ca then bng v dung sai lp ghp then bng, bit mi ghp then bng
l mi ghp bnh thng.
(1 im)b) V bn v lp then bng.
(1 im)Cu 3: (3 im) Cho chi tit nh hnh v:
a) Tra bng tm dung sai tr ca mt A v mt B.
(1 im)b) Tra bng tm dung sai ng trc ca mt A v mt B
(1 im) c) Ghi k hiu dung sai hnh dng v v tr trn bn v.
(1 im)Cu 4: (1 im)V hnh panme ang o kch thc 37,81 mm.
Cu 5: (1 im)
Dng b cn mu 83 ming chn cn mu kim tra kch thc gii hn ln nht ca
rnh 85H7.
Ngy 22 thng 06 nm2011Khoa C kh
Ging vin son
..
T Ngc Thin
P N 1
Cu 1: (3 im) Cho lp ghp d = D = 85S7/h6.
d) Tra bng tm sai lch gii hn ca trc v l.
(1 im)i vi l 85S7: Tra bng 1.19 trang 26 (Sch BTDSLG): ES= -58m;
EI=-93mi vi trc 85h6:Tra bng 1.29 trang 41 (Sch BTDSLG): es= 0 m;
ei = -22 m
e) V s c tnh lp ghp v cho bit mi ghp thuc loi no?(1 im)
T s trn cho thy y l lp ghp c di v min dung sai trc nm trn min
dung sai l.
f) Tnh cc gi tr h hoc di, dung sai lp ghp.
(1 im) di ln nht: Nmax = dmax Dmin = es EI
= 0 (- 93) = 93 m
di nh nht: Nmin = dmin - Dmax = ei - ES
= -22 ( - 58 ) = 36 m
di trung bnh: Nm = (Nmax + Nmin )/ 2 = (93 + 36)/ 2 = 64,5 m
Dung sai lp ghp:
TLG = N max - Nmin = 93 36 = 57 m.
Th li: TLG = TD + T d
m Td = es ei = 0 ( - 22) = 22 m TD = ES EI = -58 (-93) = 35
m
nn TLG = 22 + 35 = 57 m.Cu 2: (2 im) Cho lp ghp vi d = D =
75.
c) Tra kch thc b, h, t1, t2 ca then bng v dung sai lp ghp then
bng, bit mi ghp then bng l mi ghp bnh thng.
(1 im) Kch thc then v rnh then bng :
Tra bng 4.1 trang 115 (BT DSLG), ta c:
b = 20 mm.
h = 12 mm
t1 = 7,5 mm
t2 = 4,9 mm
Dung sai lp ghp then bng :
Tra bng 4.4 trang 118 (BT DSLG) ,ta c:
Mi ghp ca then vi rnh trn trc: N9/h9
Mi ghp ca then vi rnh trn bc: JS9/h9
d) V bn v lp then bng.
(1 im)
Cu 3: (3 im) Cho chi tit nh hnh v:
c) Tra bng tm dung sai tr ca mt A v mt B.
(1 im)Ta c: TA = es ei = 0,018 0,002 = 0,016mm
TB = es ei =0,019 0,001 = 0,018mm
Tra bng 1.4 trang 4 (BTDSLG), ta c:
Cp chnh xc ca mt A l IT6
Cp chnh xc ca mt B l IT7
Tra bng 2.11 trang 78 (BTDSLG), ta c:
i vi mt A: Ttr = 8m. i vi mt B: Ttr = 8 m.d) Tra bng tm dung sai
ng trc ca mt A v mt B
(1 im)Tra bng 2.21 trang 91 (BTDSLG),
Tt = 20 m. c) Ghi k hiu dung sai hnh dng v v tr trn bn v.
(1 im)
Cu 4: (1 im)
V hnh panme ang o kch thc 37,81 mm.
Cu 5: (1 im)
Dng b cn mu 83 ming chn cn mu kim tra kch thc gii hn ln nht ca
rnh 85H7.
Tra bng 1.14 trang 18 (BTDSLG), ta c:
ES=0,035 mm; EI = 0
Dmax = DN + ES = 85 + 0,035 = 85,035mm
Kch thc cn kim tra : 85,035 mm
Chn cn mu th nht: 1,005 mm, kch thc cn li: 84,03 mm
Chn cn mu th hai: 1,03 mm, kch thc cn li: 83 mm
Chn cn mu th ba: 3 mm, kch thc cn li: 80 mm
Chn cn mu th t: 80 mm
Vy cn t nht 4 cn mu kim tra kch thc trn.Ngy 26 thng 07
nm2011
Ngi son
T Ngc Thin
2:
Cu 1: (3 im) Cho lp ghp d = D = 125M8/h9
g) Tra bng tm sai lch gii hn ca trc v l.
(1 im)h) V s lp ghp v cho bit mi ghp thuc loi no?
(1 im)i) Tnh cc gi tr h hoc di, dung sai lp ghp.
(1 im)Cu 2: (2 im) Cho lp ghp vi d = D = 120.
e) Tra kch thc b, h, t1, t2 ca then bng v dung sai lp ghp then
bng, bit mi ghp then bng l mi ghp bnh thng.
(1 im)f) V bn v lp then bng.
(1 im)Cu 3: (3 im) Cho chi tit nh hnh v:
e) Xc nh dung sai trn ca mt A v mt B.
(1 im)f) Xc nh dung sai o hng knh ca mt A v mt B(1 im) c) Ghi k
hiu dung sai trn bn v.
(1 im)Cu 4: (1 im)
V hnh panme ang o kch thc 46,73 mm.
Cu 5: (1 im)
Dng b cn mu 83 ming chn cn mu kim tra kch thc gii hn ln nht ca
rnh 125H6.
Ngy 22 thng 06 nm2011
Ging vin son
T Ngc Thin
P N 2:
Cu 1: (3 im) Cho lp ghp d = D = 125M8/h9
j) Tra bng tm sai lch gii hn ca trc v l.
(1 im)i vi l 125M8: Tra bng 1.16 trang 23 (Sch BTDSLG): ES= 8m;
EI= - 55mi vi trc 125h9: Tra bng 1.29 trang 41 (Sch BTDSLG): es= 0
m; ei = -100m
k) V s lp ghp v cho bit mi ghp thuc loi no?
(1 im)
T s trn cho thy y l lp ghp trung gian v min dung sai trc nm xen
k min dung sai l.
l) Tnh cc gi tr h hoc di, dung sai lp ghp.
(1 im) di ln nht: Nmax = dmax Dmin = es EI
=0 ( - 55) = 55 m
h ln nht: Smax = Dmax dmin = ES ei
= 8 (-100) = 108 m
V Smax > Nmax nn ta tnh h trung bnh:
Sm = (Smax - Nmax)/2 = (108 - 55)/2 = 26,5 m
Dung sai lp ghp: TLG = Nmax + Smax = 108 + 55 = 163 mCu 2: (2
im) Cho lp ghp vi d = D = 120.
g) Tra kch thc b, h, t1, t2 ca then bng v dung sai lp ghp then
bng, bit mi ghp then bng l mi ghp bnh thng.
(1 im) Kch thc then v rnh then bng :
Tra bng 4.1 trang 115 (BT DSLG), ta c:
b = 32 mm.
h = 18 mm
t1 = 11 mm
t2 = 7,4 mm
Dung sai lp ghp then bng :
Tra bng 4.4 trang 118 (BT DSLG) ,ta c:
Mi ghp ca then vi rnh trn trc: N9/h9
Mi ghp ca then vi rnh trn bc: JS9/h9
h) V bn v lp then bng.
(1 im)
Cu 3: (3 im) Cho chi tit nh hnh v:
g) Xc nh dung sai trn ca mt A v mt B.
(1 im)Ta c: TA = es ei = 0,018 0,002 = 0,016mm
TB = es ei =0,019 0,001 = 0,018mm
Tra bng 1.4 trang 4 (BTDSLG), ta c:
Cp chnh xc ca mt A l IT6
Cp chnh xc ca mt B l IT7
Tra bng 2.11 trang 78 (BTDSLG), ta c:
i vi mt A: Ttrn = 8m. i vi mt B: Ttrn = 8 m.h) Xc nh dung sai o
hng knh ca mt A v mt B(1 im)Tra bng 2.21 trang 91 (BTDSLG),
Thk = 20 m.
c) Ghi k hiu dung sai trn bn v.
(1 im)
Cu 4: (1 im)
V hnh panme ang o kch thc 46,73 mm.
Cu 5: (1 im)
Dng b cn mu 83 ming chn cn mu kim tra kch thc gii hn ln nht ca
rnh 125H6.
Tra bng 1.14 trang 18 (BTDSLG), ta c:
ES=0,025 mm; EI = 0
Dmax = DN + ES = 125 + 0,025 = 125,025mm
Kch thc cn kim tra : 125,025 mm
Chn cn mu th nht: 1,005 mm, kch thc cn li: 124,02 mm
Chn cn mu th hai: 1,02 mm, kch thc cn li: 123 mm
Chn cn mu th ba: 3 mm, kch thc cn li: 120 mm
Chn cn mu th t: 20 mm, kch thc cn li: 100 mm
Chn cn mu th nm: 100 mm.
Vy cn t nht 5 cn mu kim tra kch thc trn.Ngy 26 thng 07
nm2011
Ngi son
T Ngc ThinTRNG CKT CAO THNG
KHOA C KH
THI KT THC HC PHN
CAO NG C IN T 2010
MN: DUNG SAI K THUT 0 (90 pht)
(Sinh vin c php s dng ti liu)
THI KT THC HC PHN
CAO NG C IN T 2010
MN: DUNG SAI K THUT 0 (90 pht)
(Sinh vin c php s dng ti liu)
TRNG CKT CAO THNG
KHOA C KH
THI KT THC HC PHN
CAO NG C IN T 2010
MN: DUNG SAI K THUT 0 (90 pht)
(Sinh vin c php s dng ti liu)
TRNG CKT CAO THNG
KHOA C KH
THI KT THC HC PHN
CAO NG C IN T 2010
MN: DUNG SAI K THUT 0 (90 pht)
(Sinh vin c php s dng ti liu)
TRNG CKT CAO THNG
KHOA C KH
