cÁc ĐỀ thi chi tiẾt mÁy chÍnh thỨc -...
TRANSCRIPT
-
CC THI CHI TIT MY CHNH THC 2002-2008 9
CC THI CHI TIT MY CHNH THC
2002-2008
THI NM 2002
Cu 1 Cho hp gim tc nh hnh v (hnh 1.1)
x
x
x
Z1
Z2
Z3
Z4
xIII
II
I A
Ba
a
Hnh 1.1
1.1 a) Phn tch lc n khp (im t lc ti A v B) ca cc b truyn
trong HGT nu trn vi chiu quay bt k ca trc dn I v hng
rng cho nh hnh v.
b) Vit cc biu thc tnh gi tr ca cc lc n khp theo Ti, ni, di,
v i vi i = 1,2.
c) Nu nhn xt v vit biu thc xc nh tng lc dc trc tc dng
trn trc II.
1.2 Khi thay i chiu quay trc dn I v hng rng th chiu ca lc n
khp c thay i khng? V sao?
Hy chn hng rng trn cc bnh rng sao cho hp l nht? Vit
biu thc xc nh lc dc trc tc dng trn trc II.
1.3 Khi chiu ca cc lc ny thay i th sc bn ca trc v ln c b
nh hng khng? V sao?
-
10 Chi tit my v DTH trong Chi tit my
Khi tnh sc bn trc nn tnh cho trng hp no nu chiu quay
thay i?
1.4 V sao li xut hin lc ph trong khp ni. Cch xc nh lc ny v
gii thch ti sao li chn chiu lc thay i khi tnh trc v chn ln.
Cu 2
2.1 Nu nguyn tc chung chn vt liu v rn mt rng khi thit k
bnh rng v cc cp bnh rng khc nhau trong mt HGT.
2.2 Phn tch cc ch tiu tnh ton thit k b truyn ai dt. m bo
tui th ca b truyn ai khi tnh ai theo kh nng ko cn lu
nhng vn g?
2.3 Cho trc chu ti nh hnh v (trc trung gian hnh 1.1).
a) V dng biu m men un v m men xon ca trc.
b) Ti sao khi thit k trc cn phi tnh trc theo h s an ton?
c) Nu cc gii php khi trc khng m bo h s an ton (s < [s]).
Cu 3 Cho kt cu c kch thc nh hnh v (hnh 1.2), bit: a1 = 300mm;
a2 = 240mm; b1 = 450mm; b2 = 390mm; s = 20mm; L = 350mm; R = 8000N.
R
2 3
41
0
l
a1
a2
b1
b2
s
Hnh 1.2
Dng 4 bu lng nh s 1, 2, 3 v 4. Bu lng bng thp C30 c
[k] = 240MPa; h s ma st gia gi v nn b tng l f = 0,15;
ng sut dp cho php ca nn b tng l [d] = 1,80MPa v h s an
ton khi xit cht vi ti trng tnh l k = 1,5.
-
CC THI CHI TIT MY CHNH THC 2002-2008 11
3.1 a) Xc nh ng knh cn thit ca bu lng cho 2 phng n: bu
lng lp khng khe h v bu lng lp c khe h.
b) Trong ren kp cht nn dng loi ren g? V sao?
3.2 Kim tra bn dp ca nn xi mng. Nu d < [d] th gii quyt ra
sao?
3.3 Xc nh ti trng ln nht tc dng ln mi ghp nu s dng bu lng
lp c khe h.
3.4 a) Khi ti trng ngoi thay i t Rmin n Rmax < R th c cn tnh li
kch thc bu lng khng? Ti sao?
b) Trong trng hp no cn kim tra bu lng theo bn mi?
c) Nu cc gii php chng hin tng t tho lng trong mi ghp ren.
Ghi ch
a) C th s dng cng thc sau tnh ng knh ngoi ren bu lng
(khi khng c cc bng tra cc thng s ng knh ren): d = d1
+2.h vi h = 0,54p. p l bc ren (vi bu lng c ng knh n
30mm, c th ly p theo dy sau: 2,5; 2,0; 1,5; 0,75 v 0,5mm).
b) Dy tiu chun cc ng knh bu lng: (dy 1): M8p; M10p;
M12p; M16p; M18p; M20p; M30p).
THI NM 2003
Cu 1
1.1 Chng minh rng trong truyn ng trc vt, ngoi trt bin dng
cn c trt dc ren vi vn tc trt ln.
1.2 nh hng ca trt dc ren n kh nng lm vic v dng hng ca
truyn ng trc vt.
1.3 Nu cc gii php trong thit k v s dng gim ma st, mi mn
rng bnh vt.
Cu 2
2.1 Cho h dn ng nh hnh v (hnh 1.3), trong cp bnh rng Z1/Z2 t trong hp kn v cp Z3/Z4 t bn ngoi (khng c bi trn).
-
12 Chi tit my v DTH trong Chi tit my
Qx
x
x
x
Z1
Z2Z3
Z4
ng co
HGT
Tang
Hnh 1.3
a) Cc cp bnh rng trn c thit k theo ch tiu sc bn no? V
sao? Chng t rng cc ng sut tip v ng sut un u thay i
c chu k?
b) Vit cng thc v nu ngha ca cc h s ZH v Z trong cng
thc kim nghim rng theo bn tip xc, t suy ra gii php
ci thin sc bn ca bnh rng.
c) Nu ngha v phn tch cc yu t nh hng ca h s KH v
KF trong cng thc tnh bnh rng theo bn tip xc v theo
bn un. Nu cc gii php ti trng phn b u trn chiu di
ng tip xc.
2.2 Mt khch hng mang n mt bnh tr rng thng b hng (do
mn; trc r b mt hay do rng b gy) v ngh thit k li
ch to bnh rng mi thay th bnh rng b hng trn.
a) ch to bnh rng mi thay th bnh rng b hng th cn phi
xc nh nhng thng s no ca bnh rng.
b) xc nh cc thng s , cn phi o nhng kch thc no ca
bnh rng?
Cu 3 Cho kt cu tay quay to lc nh hnh 1.4. Bit lc dc trc tc
ng ln vt me l Fa = 75.000N; vt c s u mi z = 1; bc ren p = 8;
cc ng knh d = 55mm v d2 = 51mm v ren thang c =300. May, tay
n v bu lng u bng thp c ch =220MPa; h s ma st gia vt bng
thp v ai c bng ng l f = 0,15. Khng kim tra lc xit; ti trng
ngoi khng i v chn [n] = 6.
-
CC THI CHI TIT MY CHNH THC 2002-2008 13
Fa
A
B
B
Theo A
30
1250
1000100
10
0
h1
h2
Hnh 1.4
3.1 Chng minh rng truyn ng trc vt me ai c c kh nng t
hm. Hiu sut truyn ng = 0,3.
3.2 Xc nh t s truyn qui c.
3.3 Xc nh lc tc dng ln tay n quay.
3.4 Xc nh ng knh bu lng ghp tay n vi may theo 2 phng n
lp khng khe h v lp c khe h vi chiu dy h1 = h2 = 8mm.
Nu nhn xt v chn phng n s dng? V sao?
Ghi ch
a) C th s dng cng thc sau tnh ng knh ngoi ren bu lng
(khi khng c cc bng tra cc thng s ng knh ren): d = d1
+2.h vi h = 0,54p. p l bc ren (vi bu lng c ng knh n
30mm, c th ly p theo dy sau: 2,5; 2,0; 1,5; 0,75 v 0,5mm ).
b) Dy tiu chun ng knh bu lng (dy 1): M8p; M10p; M12p;
M16p; M18p; M20p; M30p).
THI NM 2004
Cu 1
1.1. Cc dng tip xc v cc thng s c trng v iu kin lm vic ca
cc b mt i tip gp trong tnh ton chi tit my.
-
14 Chi tit my v DTH trong Chi tit my
1.2. Vit biu thc tnh i lng c trng v iu kin chu ti khi hai
mt tip xc nhau v nu gii hn s dng cng thc .
1.3. a) Ti sao ma st v mn trong a tr li nh hn trong bi.
b) Chng minh rng khi vng trong quay th tui th ca ln ln hn
khi vng ngoi quay (minh ha bng hnh v tn s thay i ng
sut trn vng ).
Yu t ny c xt n khi tnh nh th no?
Cu 2
2.1 a) Cc loi ti trng tc dng trn chi tit my (nh ngha). Trong tnh
ton cn phn bit nhng loi ti trng g? Cho th d minh ha.
b) Tnh ng sut cho php ng vi cc trng hp: chi tit my chu
ng sut khng i, ng sut thay i n nh v khng n nh.
ngha ca h s tui th KN.
2.2 Vit v gii thch ngha cc i lng trong cng thc tnh h s tui
th KHL v KFL khi xc nh ng sut cho php trong truyn ng
bnh rng. Cc gi tr gii hn ca KHL v KFL.
Cu 3 Cho kt cu nh hnh v (Hnh 1.5)
0
a
3
2
1
a
k
do
1
T
2 3
6 5 4
a
a
S
R
h
Hnh 1.5
-
CC THI CHI TIT MY CHNH THC 2002-2008 15
3.1. ng 1 c hn vi tm 2 c chiu dy s1= 8mm bng mi hn c cnh
hn k nh hnh v; hn tay, dng que hn E42. Vt liu ng v gi bng
thp CT3 c ch= 225MPa, h s an ton khi xc nh ng sut cho php
s =1,5. Chu ngoi lc R= 40000N v mmen T = 1,75.106Nmm. ng c
ng knh ngoi d = 100mm, ng knh trong d0= 68mm v chiu
cao h = 200mm
a) Xc nh kch thc cnh hn k
b) - C th thay kt cu mi hn cho bng mi hn ch K c khng?
- V kt cu mi hn ch K v tnh kim nghim mi hn (hn gip mi )
- C nn thay th mi hn cho bng mi hn ch K khng? v sao?
c) Trnh by phng php tnh mi hn khi ti trng thay i t
RMin= 0 n RMax= R v T = const
Xc nh h s gim ng sut cho php v ng sut cho php, bit h
s tp trung ng sut kt = 2,5.
3.2 Chi tit 2 c ghp ni vi thanh thp ch U (N0 = 28 ) c chiu dy
s2 = 6mm bng mi ghp 6 bu lng c s nh hnh v v kch
thc a =100mm.
a) Xc nh ng knh bu lng (dng bu lng lp c khe h) kp
cht tm 2 vi gi vi iu kin: H s an ton khi xc nh lc
xit cht k = 1,5, bu lng bng thp 45 c ch = 350MPa, khng
kim tra lc xit do chn h s an ton khi xc nh ng sut
cho php s = 2,5, h s gim ti = 0,25 v h s ma st f = 0,15.
(khng cn tnh chnh xc li h s gim ti ).
b) Trnh by phng php tnh bu lng khi:
- Ti trng R thay i t RMin = 0 n RMax= R v T = const.
- Ti trng R = const v T thay i t TMin= 0 n TMax= T.
Ch thch
Kich thc cc yu t ca mi ghp bu lng c chn theo ng knh ngoi ca ren nh sau:
ng knh nh ren M12 M16 M20 M24
Kch thc trong d1 (mm) 10,106 13,385 17,294 20,752
Chiu cao ai c H(mm) 10 14 16 19
Kch thc cha vn S(mm) 19 24 30 36
Chiu dy m h0(mm) 3 3,5 4,5 5,5
-
16 Chi tit my v DTH trong Chi tit my
THI NM 2005
Cu 1
1.1. a) Trnh by nhng nguyn tc v chn vt liu khi tnh ton thit k
chi tit my.
b) Nhng nguyn tc th hin trong vic chn vt liu ch to trc
vt v bnh vt nh th no?
V sao vt liu vnh rng bnh vt c chn ph thuc vo vn tc trt.
1.2 a) Nu cc c im tnh ton thit k chi tit my.
b) V sao phi tin hnh thit k chi tit my theo hai bc: Tnh thit
k v tnh kim nghim.
c) Ly 3 v d cho ba chi tit khc nhau gii thch thm v c im ny.
d) Vi cc kch thc v thng s thu c trong bc tnh thit k,
khi kim nghim khng t yu cu cn x l nh th no?
e) Ly 3 v d cho 3 loi chi tit khc nhau v nu cc gii php c th
s dng nu trong bc tnh kim nghim khng t yu cu.
Cu 2 Cho kt cu mi ghp hn (hnh 1.6a), hn chi tit 1 (thanh thp
L90909, din tch mt ct A = 14cm2) hn vi tm 2 c chiu dy = 12mm.
Vt liu ca 2 chi tit bng thp CT3 c []k = 160MPa, hn bng tay vi
que hn 42A.
l1
l2
e1h1
G
1
2
x x
h2
F
T
Hnh 1.6a
F
l1
l2
T
e1
Gh1
x x
h2
Hnh 1.6b
-
CC THI CHI TIT MY CHNH THC 2002-2008 17
2.1 Xc nh chiu di mi hn dc l1 v l2 khi mi ghp chu ti trng
dc trc F(N).
Bit khong cch e1 = 25,1mm, cnh hn k v [].
2.2 Xc nh kch thc cnh hn k khi mi ghp ng thi chu ti
trng F = 6.104 N v T = 3,5.106 Nmm; chiu di cnh hn l1 = 155mm
v l2 = 60mm, khong cch t trng tm tit din mi hn n ng
tm trc x-x l h1 = 27mm v h2 = 67mm v ng sut cho php ca
mi hn [] = 105MPa.
2.3 Nu thay i kt cu mi hn nh phng n hnh 1.6b. Theo bn
phng n no hp l hn, v sao? (khng cn tnh c th).
Cu 3 Cho b truyn ng trc vt (hnh 1.7a). Bit n1 = 930 vng/pht; u = 20;
m = 6,3mm; q = 12,5 v z1 = 2, mmen trn trc bnh vt T2 = 300000Nmm.
dm
80
1
2
n2
2
1 Hnh 1.7a Hnh 1.7b
3.1 a) Xc nh tr s (ch ly phn nguyn ca gi tr lc) v chiu cc
lc n khp xut hin trn b mt ren trc vt v rng bnh vt theo
s hnh 1.7a (b qua lc ma st).
b) Khi trc vt quay theo chiu ngc li, cc lc ny c thay i
khng? V sao?
3.2 a) Xc nh di lp vnh rng bnh vt vi thn khi truyn ti
trng trn, bit:
- Vnh bnh vt bng ng thanh lp vi thn bnh vt bng gang
(E2 E1= 105MPa; 1 2 = 0,3).
- Cc kch thc b mt ghp cho trn hnh 1.7b
- B mt ghp c gia cng vi nhm b mt Rz1= 3,2m v Rz2 =
6,3m, f = 0,05 v h s an ton k =1,7.
b) di mi ghp c xc nh khi bnh vt quay theo chiu no?
V sao?
-
18 Chi tit my v DTH trong Chi tit my
THI NM 2006
Cu 1
1.1 Cc thng s v yu cu c bn ca truyn ng cng sut l g
(truyn ng c kh)?
1.2 a) Vit cng thc tnh lc cng trn cc nhnh ai: F1; F2; Vit cng
thc xc nh mi quan h ca Ft v F0 vi gc m 1, h s ma st
f (b qua lc cng do lc ly tm). T suy ra gii php nng cao
kh nng ti b truyn ai.
b) V sao ai c th b ph hng do mi.Vit cng thc tnh ng sut
max v min (b qua ng sut do lc ly tm gy ra). Nu cc yu t
nh hng n tui th ca b truyn ai.
1.3 a) Nu v phn tch nguyn nhn dn n t s truyn thay i trong
b truyn ai, xch v bnh rng.
b) nh hng ca t s truyn thay i n kh nng lm vic ca b
truyn v cc gii php khc phc (nu c).
Cu 2 Cho s hp gim tc Cn Tr ( hnh 1.8).
II III
I
l1 l2
l3
A
B
Khp ni
a xch
Hnh 1.8
2.1 a) t lc n khp ti cc im A v B.
b) Vit biu thc v tnh gi tr (ch ly phn nguyn) cc lc n khp
trn cp bnh cn rng thng bit: T1 = 150000Nmm; dm1 = 150mm;
1 = 130.
2.2 C bao nhiu phng n b tr gi trc cho trc c lp bnh rng
cn dn. V cc s b tr gi trc nu.
-
CC THI CHI TIT MY CHNH THC 2002-2008 19
2.3 Cho bit: l1 = 50mm; l2 = 120mm v l3 = 200mm (xem s hnh
1.8); u vo lp khp ni vng n hi c D0 = 100mm (ng knh
qua tm cht vng n hi).
a) Xc nh gi tr Fkmin v Fkmax (ch ly phn nguyn) v xc nh
phng chiu ca lc Fk (lc do khp gy ra).
b) Tnh cc phn lc gi ta trc vo HGT cho s b tr nh hnh 2
vi Fkmax (ch tnh cho phng n khi Fkmax ngc chiu vi Ft).
c) Xc nh ti trng qui c Q cho lp theo s trn hnh 1.8 ( s
ch O) bit: s dng a cn c = 260 v V = 1; K = Kt = 1.(Ghi
ch: Nu Fa / VFr > e, ly X = 0,40 v Y= 0,45cotg)
2.4 a) Tnh s b ng knh trc vo HGT, bit [] = 20MPa.
b) Chn ng knh cc on trc c lp khp ni, ln v bnh rng,
bit ng knh trc ng c in dc = 42mm. ng knh cc
on trc cn tha mn nhng yu cu g? (Bit dy tiu chun
ng knh thn trc:..30; 32; 34; 36; 38; 40; 42; 45...).
c) V kt cu trc cho phng n chn (ch v hnh dng m khng
cn t l).
2.5 S dng mi ghp then bng c kch thc 12x8xl truyn m men
T1= 150000Nmm. Bit ng knh trc d = 38mm; chiu su rnh then
trn trc v trn bc: t1 = 5mm v t2 = 3,3mm; vt liu then c [d] =
100MPa v []= 40MPa. ( Bit chiu di tiu chun ca then bng:..28;
32; 36; 40; 45; 50; 56; 63; 70...).
Xc nh chiu di then v chiu di may bnh rng cn lp trn trc.
2.6 Nu thay cp bnh rng cn thng bng cp bnh cn rng cong th nn
chn hng rng nh th no? V sao?
THI NM 2007
Cu 1
1.1 a) Th no l trc trn, trc bc. Trong thc t loi trc no c s
dng nhiu hn, ly 2 v d mi loi minh ha.
b) Nu v phn tch u nhc im ca trc trn v trc bc
c) Nu cc gii php khc phc nhng nhc im ca trc bc
trong thit k kt cu nng cao tnh cng ngh trong ch to trc
v cc chi tit lp trn trc.
-
20 Chi tit my v DTH trong Chi tit my
1.2 a) Vit biu thc tnh gi tr ca bin ng sut v ng sut trung bnh
khi trc quay 1 chiu, bit Mu = 150000Nmm; T = 750000Nmm;
trc c ng knh d = 45mm; then c kch thc b = 14mm v
t1 = 5,5mm.
V th v s thay i ca ng sut un (u) v ng sut xon ()
trong mt chu trnh thay i ng sut.
b) Vit cng thc tnh kim nghim trc theo h s an ton, bit S v S.
- Nu cc gii php khi S < [S].
- Trng hp trc quay 2 chiu, tnh ton trc theo bn mi c g
thay i so vi trng hp trc quay mt chiu? V sao?
Cu 2
2.1 a) Nu cc c im chnh ca ni trc cht, ni trc b v ni trc
n hi l g?
b) Nu nguyn nhn dn n s xut hin lc hng tm tc dng ln
trc khi s dng ni trc? Cc gii php khc phc nu c.
c) Cch xc nh tr s v phng chiu ca lc Fk trong tnh ton trc
v chn . Ly v d minh ha iu va nu l ng.
2.2 Cho ni trc a c kt cu nh hnh 1.9, bit m men xon tc dng
ln trc T = 1400Nm; s bu lng z = 6; ng knh qua tm bu lng
D0 = 260mm; Chiu dy a h1 = 10mm v h2 = 12mm. Bu lng lm
bng thp C30 c ch= = 260MPa tng ng vi [k] = 100MPa; [c] =
125MPa v [d] = 240MPa; h s ma st trn b mt tip xc f = 0,15
v h s an ton s = 1,5.
P.n B
h1 h2
D0
d
Km
P.n A
dm
I
Hnh 1.9
-
CC THI CHI TIT MY CHNH THC 2002-2008 21
a) Xc nh chiu dy cnh hn km hn a vo may , bit: dm =
180mm. H s tp trung ti trng theo chiu di cnh hn l 3 (ch
c 1/3 chiu di cnh hn chu lc). Ti trng tnh, hn tay, dng
que hn E42, a bng thp c [k] = 157MPa.
b) Xc nh ng knh bu lng lp 2 na ni trc theo 2 phng n:
- Lp c khe h (Phng n A).
- Lp khng khe h (Phng n B).
c) Da vo kt qu tnh ton v kch thc ni trc cho quyt
nh phng n s dng cho hp l, v sao?
d) Ngoi phng n cho trong hnh v, cn c th dng cc phng
n no khc m bo ng tm gia 2 u trc ni.
e) Nu s dng 3 bu lng tinh va nh tm va truyn lc v 3
bu lng c khe h lp xen k nhau ni 2 na khp trc. Theo
bn th:
- C th thc hin c khng? V sao?
- Nu thc hin c th nu phng php xc nh ng knh bu
lng cho mi ghp hn hp ni trn. Gi thit l a c cng
v chiu dy ln m bo bn ct v bn dp ca bu lng;
Cc bu lng c cng ng knh.
Ghi ch: Dy tiu chun ren h mt:
d M8; M10; M12; M16; M20; M24; M27;
d1 6,647; 8,376; 10,106; 13,835; 17,924; 20,752); 23,752;
THI NM 2008
Cu 1
1.1 Th no l bnh rng tr tng ng vi bnh rng cn. Xc nh
cc thng s ca bnh tr tng ng khi thay th qua tit din trung
bnh. Bit cc thng s ca cp bnh rng cn Zi, me, b v i.
1.2 Nu thay th qua tit din khc (v d qua tit din y ln vi me) th
kt qu tnh ton sc bn c thay i khng? V sao? Xc nh cc
thng s ca bnh rng tr tng ng qua tit din y ln.
1.3 Chng minh rng ng sut tip xc hoc ng sut un khng thay i
trn chiu di rng v v vy c th tnh sc bn rng bt c tit din
no. V n gin, ta tnh qua tit din trung bnh ca rng.
-
22 Chi tit my v DTH trong Chi tit my
1.4 H s 0,85 trong cc cng thc tnh sc bn bnh cn rng thng c
ngha g? V sao?
Cu 2
2.1 Cng dng ca HGT trong h thng dn ng my.
2.2 Nu nhng u nhc im ca HGT khai trin thng. Cc gii php
khc phc khi thit k cc chi tit (bnh rng, trc...) trong HGT
khai trin thng.
2.3 C bao nhiu phng n b tr bnh rng trong s HGT 2 cp dng
khai trin hoc ng trc nu s dng cc bnh rng tr thng v
nghing. Hy nu cc phng n b tr c th. Trong nhng trng
hp no nn s dng cc phng n b tr tng ng. Cn ch g
khi chn hng rng trong trng hp s dng ton rng nghing (cp
nhanh v cp chm u rng nghing)?
2.4 Khi thit k bnh rng trong HGT, gi tr m un ca cc cp bnh
rng nn chn nh th no? V sao?
Cu 3 Chi tit 1 c c nh bng vng kp 2 trn thanh tr trn 3 (ng
knh d = 60 mm) nh vo hai bulng 4 (hnh 1.10).
Hnh 1.10
Thanh tr trn c c nh vi thn my 5 bng mi hn gc,
ng knh hnh tr ti mi hn l D = 80mm. Cho bit l1 = 400 mm,
l2 = 200 mm.
H s ma st gia trc v vng kp f = 0,20, h s an ton khi xit
cht vi ti trng tnh k = 1,3. Vng kp mm hnh dng b mt tip
xc c dng tr v p lc p phn b u trn b mt tip xc. Ti trng
khng i tc dng F = 3000N, b qua khi lng ca chi tit.
-
CC THI CHI TIT MY CHNH THC 2002-2008 23
3.1 Xc nh lc xit cn thit V trn mi bulng ?
3.2 Xc nh ng knh bulng nu vt liu bulng l thp c cp bn
8.8 (gii hn chy ch = 640 MPa, h s an ton s = 2,5).
3.3 Xc nh chiu dy cnh hn km hn trc 3 vo thn 5. H s tp
trung ti trng theo chiu di cnh hn l 3 (ch c 1/3 chiu di cnh
hn chu lc). Ti trng tnh, hn tay, dng que hn E42, thn v trc
u bng thp c [k] = 157MPa.
3.4 Khi ti trng ngoi thay i c cn tnh li ng knh bu lng v mi
ghp hn khng? V sao?
Bit dy tiu chun ren h mt:
d M8; M10; M12; M16; M20; M24; M27;
d1 6,647; 8,376; 10,106; 13,835; 17,924; 20,752); 23,752;