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Ain Shams University Faculiy of Engineering Third Year Mechanical Power 26 - June -2004 Timeallowed:3hours Desisn & Prod. Ene. De Machine Construction (2 Exuminer : Asso. Prof.l Hrtlr!- /!y Ey!4 . Stuclents may assume any missing data . Constructional ciruwings should be fully dimensioned with ilnachining si':::bcls antl classes of fits clearly indicated. The sketch shows the outlines of a Double Stage Gear Box utilized to spilt the input power from a diesel engine to drive the cutter and pump of a dredger. The gear unit consists of two gear reductions (a Bevel Gear Drive and a Straight spu]. Gear drive). The diesel engine is rated at 55 KW at 2000 rpm. The pump consurnes 20 KW at 1250 rpm, while the cutter requires a maximum of 30 KW aL 625 rprn. The number of teeth of the pinion bevel gear and the pinion stlaighlgear are-50 and 40 respectively' The module of all gears is 4 mm' You are asked to: 1- Find out the detailed dimensions of all the gears and specity their materials based on static strength, and check the dynamic load for bevel gears only' 2- Find the reactions atlhe bearing suppons tA&Bl a.s shown in the attached sketch and calculate their dynamic loid carrying capacity. The life is expected to be 25000 hr. 3 Fer: the lntprmediate shaft AB draw the bending moment and torque diagrarns, then find the minimum shaft diameter if its material is st 50' - 4- Draw a constructional drawing for the gear unit assembly in at least-two views' . Gears Design Equations: . Pt = ou. b. m . (l-b/L) .K K:Ku.Ky/K- . Velocity factor for gear = 3 / (l+v) r Modified form factor = 0.48 - (3 '85 I Z) . Deformation factor: 0.1 | e I lllEr+ llE*] . Speed factor :2 \t / [l + u] . Scrvice factor -- 1.7 . Lubrication factor : 1.25 . E.r : (21x 1 0a MPa) ----- Ecr : ( 12 x 104 MPa) . P.n:l.ZPd P,,:l.25Pd t I'.n:o.n'ku'b'rn . P',,,=k,u. kn. b. Dn/cos6 k,.b+P, D -p r o 'd-'t' 0.15 t Lr '! ,*Y#ffi'b+q

Final and Answers 2004 Design2 Ahmedawad

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Page 1: Final and Answers 2004 Design2 Ahmedawad

Ain Shams UniversityFaculiy of Engineering

Third Year Mechanical Power26 - June -2004

Timeallowed:3hoursDesisn & Prod. Ene. De

Machine Construction (2

Exuminer : Asso. Prof.l Hrtlr!- /!y Ey!4

. Stuclents may assume any missing data

. Constructional ciruwings should be fully dimensioned with ilnachining si':::bcls

antl classes of fits clearly indicated.

The sketch shows the outlines of a Double Stage Gear Box utilized to spilt the

input power from a diesel engine to drive the cutter and pump of a dredger. The

gear unit consists of two gear reductions (a Bevel Gear Drive and a Straight

spu]. Gear drive). The diesel engine is rated at 55 KW at 2000 rpm. The pump

consurnes 20 KW at 1250 rpm, while the cutter requires a maximum of 30 KW

aL 625 rprn. The number of teeth of the pinion bevel gear and the pinion

stlaighlgear are-50 and 40 respectively' The module of all gears is 4 mm'

You are asked to:

1- Find out the detailed dimensions of all the gears and specity their materials

based on static strength, and check the dynamic load for bevel gears only'

2- Find the reactions atlhe bearing suppons tA&Bl a.s shown in the attached sketch

and calculate their dynamic loid carrying capacity. The life is expected to be

25000 hr.

3 Fer: the lntprmediate shaft AB draw the bending moment and torque diagrarns,

then find the minimum shaft diameter if its material is st 50' -

4- Draw a constructional drawing for the gear unit assembly in at least-two views'

. Gears Design Equations:. Pt = ou. b. m . (l-b/L) .K K:Ku.Ky/K-. Velocity factor for gear = 3 / (l+v)r Modified form factor = 0.48 - (3 '85 I Z)

. Deformation factor: 0.1 | e I lllEr+ llE*]

. Speed factor :2 \t / [l + u]

. Scrvice factor -- 1.7

. Lubrication factor : 1.25

. E.r : (21x 1 0a MPa) ----- Ecr : ( 12 x 104 MPa)

. P.n:l.ZPd P,,:l.25Pdt I'.n:o.n'ku'b'rn. P',,,=k,u. kn. b. Dn/cos6

k,.b+P,D -p r o'd-'t' 0.15 tLr '! ,*Y#ffi'b+q

Page 2: Final and Answers 2004 Design2 Ahmedawad

Third Year Mechanical Power - Machine Construction (2) 126-June-20041

. c : Prqbable error IPmi

Type of mq4llrng-Module C A P

Up to 4.5 50 25 t2

6 60 30 15

7 70 35 t7

Permissible

Strength of some materials(Ultimate and Yield StrCss, MPa)

ccts ll8o l--GG22 l22G I --G,:.16 1260 l-:::Vv !v

GG 30 I 3oo l--GGG40 - l40u lz)uGTW 40 | +oo l22o

l5Cr3l6MnCr5l8 crNi 8

Material

st 34st 37st 42

st 50

St 60

St705s0800760900s00

30Mn537MnSi 5

42CrMo4]6CrNiMo414CrNi Mo630CrNiMo8

Page 3: Final and Answers 2004 Design2 Ahmedawad

Third Year Mechani.ut fqtn.t - Muthinu gont 26-June -2004

Cutter30Kw I625 rPm I

Diesel Engine55 KW

2000 rPm

PumP IzoKW i

1250 rPm

q l',

€€

o(o

l,Ir

-

" --'----G-L.

Page 4: Final and Answers 2004 Design2 Ahmedawad

Bt,ve L e(1r

U, 2ooalZs. = t'6

Dp= 4 xsor-Os = zoo xr.t^.

6f = tori' r

a t ^r /'eo3 -- tqn t.6IL_ = 1ga .l[ = 63

-\\--P = zao -

\rn9 = 32o =

St *qhl- G.o.

2 oo r\"n

')^ -: JZ, (',) Ynm

= 58"

6 3 Si;^, 3263 Stn sg

? €.4r-xx /56.6xZaoo

6-o ;;-E5o

-- <q(as 3z

3' kv=

,k,'rf -

.kv'3

l"f :2865

=

T^=

Zc,P i

-IrrE = o' t46

o.4g - -=3:.?ss7- =

o.4B- i.ss =t5lSo ooo ^

-- ? E65tl.4s6u, x 63.4

^ 3= 339 .s t"lfa

6.4 t4

".454kru

6.\46xo.4t4

:

An,^T-* t ?./l-./-

-r--_o**l f

= t7.45

Z"rg

l) - t25o

-

=')6z-5

'P = 4ox 4 = 46o n,r^Tg= 16oxz=3zo

ha,n

l*r'=,,46o + Lzx\) = 46g -*l*, = {5o - z(t.Zx/t\ = ,(so.4

^nD*, = 32o 1 (z 14) = 32 g rnyr^

Da, ^ 32o _ (zx\.2,4) = 31o.4 mwr

-lob.6 1/|rrl

266. 6 h7^.

D,OP

-/s

- 8o

st-ort.z

Page 5: Final and Answers 2004 Design2 Ahmedawad

oaa

f,r= 6bP = 3o8. 6 Hlao- Ath

-

o.454

-Gg -- Y/b t tl\l-lre

. Str*ght si", G"* ,

V= -E^ 16o x \25o = '{o. 47 ^ls| 6o ooat) = 6c, rnrn

kU= = = = o.ZZZs + lo.hlVlr = o,4B - +5- = o. 384

Kys= o.4g 3.8s = o.4166

Pt = Toooo = 2g6s n,- 10.47=6 -bp^ 6ax4 x ryt- .l

6Ir = ;16s 1-tPa

1fi, = s"l-rrh /

eb, /68x # = /37. ilf^

cL".k ffi 4nnq*i, /,./ n4/lurr/ fuV^, t7.45 ^ls -------------r- €*o, =

.'o n, f7. / //*r/,;n;y ,s @ltre q//. €lfor = eail = 25 /tn

N/n^kJ=

n=

o.tlX o,oZSrZlXlo4= z ag.7'

(zts.1 x63 +zT6s,\ _ 2tosd

Page 6: Final and Answers 2004 Design2 Ahmedawad

*-;'a55- '

clJ = 2865 a 21o56I .t- o'lS't t7.hs

Dlen = o.5 x lloo a o,4t4 x 63 x4ir

'--r'J -. o. ) x i6o,x o,4 s4 x 6z ^ 4

, Beve L Gear ?r,ri, n

Ft = 286s N

F& = 2865 x Tan2o x :;in 32 = 553Fa= zg6s^ hn 2o r cos 32 = gg3

. Stro^qL! Geq.

E= z66s N

F. = 2B6s hnzo = io47 N

t223 4 N

t2.23 KN

= s7, 38 K N s-8e

= 56. o 6 KN s-?u

N=N=

F3

t*3

\irr-ot,

+.-&C4jNo+--

U

\rdE *f -

I

I

,lI

loI tt

"l*1YI!-* lS'

"f5lO-

(n[6-+lol

I--*i,,

^.llqlml\o" l+*:*f *;t;

\^,00vlrn'hlc\" *{,\^lX ----{o Yl*I -:-ldr

c {d+- rn

loatl/I

I

L(

r,--

60!llr\.

qF. -a---(fl i

cO

zz,1 .z rtU 342

ktt.ins

3? 4.zsl(v.*

Ztosd

n 8 "A/ lti Lea,i,t

H 1- 4;"g

Page 7: Final and Answers 2004 Design2 Ahmedawad

AFr-z/\R -- .l lr.l + 3342.5I

'-_!s::t! L

. B.o.in3 tr

Rts ='[

L=

Cn=

s34z N

3.3 kr.,t

382 -----+

: Copa"ity y' Lr..,rny @ tfi.xoJ)

r

s.l t+ l{ai' 244t.3 N

2.4 FN

2Sooox6oX\256 f 8 75 nilllavr rc,Ur.lo6

316/

"G;

Dte1 =

{

Ij

rrr

/8o -

l8o =

/1-x Jj

l(ilfr+xSglt +

_-_______-F J= 23.A

. C"f &'F/ / le^inJ @ (Fr..)---1----a--7-

zA\l

sf--LB= 2A4t \l lgTs =

. Bin/,'n1 Ho^r*t *l C

. B.n/,i, -o*ent a,f DrltJ=[7'r"*.ir' : 3j4 N.w.

G = 382 N.rn

o.6 x6oo = 36a Hfa

rf_e = /8o H&2-

Hq=Jil'1 iqz.l. = ZZ? N.rn

[At p /,

Page 8: Final and Answers 2004 Design2 Ahmedawad