7
Examiner: Prof.r Hesh am Aly Sonbol You are asked to: ! r-":r.-0-!r.:.:··:;:~c ::::<1 number of-the V-belts based on the motorcapacity .: _ :..: 120 Marks] 2- Find the helical gearsmaterial based on static strength only _ [20 Marks] 3- Calculate the bearing capacity for the output shaft for the gear unit if the expected life. is 20000 hr. [20 Marks] 4- (beck the strength of the 40 rnm diameter output gear unit shaft if its material is St 42 [15 Marks] 5- Draw a constructional drawing for the gear unit assembly : [30 Marks] iliestion 2: [20 Marks] A bevel gear assembly as shown in Figure (2) working under the following con-ditions: .. Bearings i\: and 13:60xl30x31x34 (C=6IKN), Rearing C: 70~170x5l (C=132KN). The thrust arising from the straight bevel gear of diameter 80 111111 and rotating at 1200 rpm is 6KN" Spcci ficaiions of threaded end of shaft: M48x4, core diameter=4).1 mrn, height of nut = 38111111 .. T: True F: Falls . IJ__: AJI ron~0>Larings. are roiling bearings bUI not all rolling bearings are !oller bearings. ('I".o_r..£.L_J-· ! 2 ! 1'1...: s!ll)Wn system should be lubricated using grease. (T or F). ! ~-.~ .,n·;;~illg~ "A" and "B" are termed: cylindrical~~lf-a_ligning· or tapercd-=--' ---=~ ! 4 ! The detachable bearing is "A" or "13" or "C" or all of them. f·:) ;~The ,~dia,-loaion the shaft is carried by: three bearings or t-w-o-b-e-a-ri·-n-gs-or-oJ~ bearing~ ~=__] ,6 ! The axial load on the shaft is carried bv: bcarinz "A" or "B" or "C". _._ ._- ---- .--- ._-----_ ..__-_ p ._--- ._----. -------- ----, :~7__I Fk;JJ~IHs."C' i~ fix~d frolll its lour side~ and t~~s ShO_L~ld be the locating bearing. (T <?r_F)_ .._ I ! X The threads of [he threaded part of the shaft arc subjected to an axial torce equal 6KN. (T or F) I ~J? ..j Tile lhr~L~t car:ied by each bearing.= 2 KN. (T o·;!f"6)mment . . --- ... =~! ! 10 Calculate the thrust actina on each bolt of the 4 bolts fixing the cover f) to the housing. I 4•• • __ _ _ _ .__ _ •__ Gear Center distance 200 !TIm Pressure angle _.:. ~_Qo _ Belt center distance 400 mm Tooth width 20 rom Motor output speed 1800 rpm Gear/Pinion teeth 150/50 Hell speed ratio- 3:·1 Smaller pulley diam. : )00 rnrn Coefficient offrietion 0.2 The iJarticulars of the drive arc as follows: Question 1: [105 Marks] The attached sketch, Figure (1), shows the layout of a Helical Spur Gear Reducer which it has to be in't~\II::~d up-side for a special application in factory. Power is supplied to the unit through an electric motor or:, K wand a V-bclt drive. The output of the unit is finally delivered to a conveyor through a system of fla: belt. ,c ••••••••••••••••••••••••••••••••••••••••••••• c •• ~ ••• a •••••••••••••••••••. Machine Construction and Design of Machine Equipment ---- ----.--------~~----------------------~--~---------------- __ • Students may assume any missing data. • Constructional drawings should he fully dimensioned with machining symbols and classes offits clearly indicated. Time: 3 Hours Maximum mark = (125) AIN SHAMS UNIVERSITY FACULTY OF ENGINEERING DESIGN & PRODUCTION ENG!NEERING DEPARTMENT Third Year Mecltqnical, Power Section - , .

Final and Answers 2010 2011 Design2 Ahmedawad

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Page 1: Final and Answers 2010 2011 Design2 Ahmedawad

Examiner: Prof.r Hesh am Aly Sonbol

You are asked to:! r-":r.-0-!r.:.:··:;:~c::::<1 number of-the V-belts based on the motorcapacity .: _ :.. : 120 Marks]2- Find the helical gearsmaterial based on static strength only _ [20 Marks]3- Calculate the bearing capacity for the output shaft for the gear unit if the expected life.

is 20000 hr. [20 Marks]4- (beck the strength of the 40 rnm diameter output gear unit shaft if its material is St 42 [15 Marks]5- Draw a constructional drawing for the gear unit assembly : [30 Marks]

iliestion 2: [20 Marks]A bevel gear assembly as shown in Figure (2) working under the following con-ditions: ..Bearings i\: and 13: 60xl30x31x34 (C=6IKN), Rearing C: 70~170x5l (C=132KN). The thrust arisingfrom the straight bevel gear of diameter 80 111111 and rotating at 1200 rpm is 6KN"Spcci ficaiions of threaded end of shaft: M48x4, core diameter=4).1 mrn, height of nut = 38111111..T: True F: Falls .IJ__: AJI ron~0>Larings. are roiling bearings bUI not all rolling bearings are !oller bearings. ('I".o_r..£.L_J-·! 2 ! 1'1...: s!ll)Wn system should be lubricated using grease. (T or F). !~-.~.,n·;;~illg~"A" and "B" are termed: cylindrical~~lf-a_ligning· or tapercd-=--' ---=~! 4 ! The detachable bearing is "A" or "13" or "C" or all of them.f·:) ;~The,~dia,-loaion the shaft is carried by: three bearings or t-w-o-b-e-a-ri·-n-gs-or-oJ~bearing~ ~=__],6 ! The axial load on the shaft is carried bv: bcarinz "A" or "B" or "C".• _._ ._- ---- .--- ._-----_ ..__ -_ p ._--- ._----. -------- - ---,:~7__I Fk;JJ~IHs."C' i~ fix~d frolll its lour side~ and t~~s ShO_L~ldbe the locating bearing. (T <?r_F)_ .._ I! X The threads of [he threaded part of the shaft arc subjected to an axial torce equal 6KN. (T or F) I

~J?..j Tile lhr~L~tcar:ied by each bearing.= 2 KN. (T o·;!f"6)mment . . --- ...=~!! 10 Calculate the thrust actina on each bolt of the 4 bolts fixing the cover f) to the housing. I• 4 •• • __ _ _ _ .__ _ • __ •

Gear Center distance 200 !TImPressure angle _.:. ~_Qo _Belt center distance 400 mmTooth width 20 rom

Motor output speed 1800 rpmGear/Pinion teeth 150/50Hell speed ratio- 3:·1Smaller pulley diam. : ) 00 rnrnCoefficient offrietion 0.2

The iJarticulars of the drive arc as follows:

Question 1: [105 Marks]The attached sketch, Figure (1), shows the layout of a Helical Spur Gear Reducer which it has to bein't~\II::~d up-side for a special application in factory. Power is supplied to the unit through an electricmotor or:,Kwand a V-bclt drive. The output of the unit is finally delivered to a conveyor through asystem of fla: belt.

,c ••••••••••••••••••••••••••••••••••••••••••••• c •• ~ ••• a •••••••••••••••••••.

Machine Construction andDesign of Machine Equipment

---- ----.--------~~----------------------~--~----------------__• Students may assume any missing data.• Constructional drawings should he fully dimensioned with machining symbols and

classes offits clearly indicated.

Time: 3 HoursMaximum mark = (125)

AIN SHAMS UNIVERSITYFACULTY OF ENGINEERING

DESIGN & PRODUCTION ENG!NEERING DEPARTMENTThird Year Mecltqnical, Power Section

- ,.

Page 2: Final and Answers 2010 2011 Design2 Ahmedawad

@

I Figure 21

Motor

\ Pinjon

Flat belt Pu

Figu-re 1"',,..

Motor

v- Belti.-> .~

· il'facltille Design-January -2011- Third Year Mechanical, Power Section [Page ..

Page 3: Final and Answers 2010 2011 Design2 Ahmedawad

,------: 3/4

• \i-Belts Design Equations:o ::.~180v :i: 2 sin"' (D-d i 2C]r, UfO-~.:-::e'F1

:':''0,.>' : Permissible (maximumierror, [urn]:-t :-i-!-- T2--'J-r4r5-r6-,fl-f-r9-I-iO-~12-115'2oJls~~~QTJ].:;~{tfr]~j~:-_e~·~CIfJ=~oQ1J21~~1~3T]fIIc[t44I~~Il~Iltt_2Jl i-4_L0_Il~IJlJ_lJ~~ Y."

c : Probable crrorlum]- -- --- - _._-_._-----_._-----------,: Module '; Commercial i Accurate: Precision I

.: Uplo4..5 -:- 50--' 25 : 12 J"~---6---1-- 60 30' 15 IL . __ , -L- __ --l

i I 70 35' 17 ,t ~ ~------+.------~! £; j 80 ; 40 j 20 I~-------r___ ,--,I 12 ,. 100 j 50 I 25 ,r---:"-:---l---.,---, ---I~-----, ., ,. ,:,.) , I/.O , 60.. .)0..________________________ ~ J

.. Coefficient of friction' between bells and pulley = 0.2.• Bearing axial factor (y) = 1

2A =-~~. b . cos y " P,

• P: =- (J'e.b.rn.I<.. K·"Kc .. K:!K~.~;:_... \1(Jdificd form Iactor> O.4~ - [2.85 j Z:']• Deformation factor :.--:150 N!mm• Lubrication factor _'"0 1.25• Veloc ity factor =--: 3i(3~v)• S~['\"i(:..;:factor ~o 1.2<II Sp~:c,-!tacror= 2 u ! [I ;-u]o l:.s~:-o. (21 xlO" MPa) ECl = (12 xl04 MPa)• 1\. :'_'aell . kv . b . 111$ P - 1 1)- r' - 'I' k b J) i C(I,2 .., .\ - .- c - \.w > \. n . . p . S;,

e Pr:: P: Ian a sec r.. p~=-: P,tan 1

Useful Data

.Machine Design-January -2011- Third YearMechanical, Power Section [Page: 3/4/

1-..1

Page 4: Final and Answers 2010 2011 Design2 Ahmedawad

Load stress factor MPaMaterial BHN KwSt I St 1500/1500 0.29--- ..

0.562500/15002000/2000 0.52500/2500 0:-95"'~-- --- ._-_ .3000/2500 1.14-3500/2500

..1.38-.

4000/3000 1.8- 4000/3500 2.26000 14500 3.6~.6000 16000 5.5--_ "'

SUCI 1500/1800 0.420GO 11800 0.82500/1600 1.5

'"CII CI iROO /1800 1.6

Material in MPaType UTS Yield BHN· -'GG18 150·220 _ .._- .1700._-

23C-280--GG2G ---- 2100-GG30 . --- .:_

250-300 ---- 2200--~!?38 380 180 --.GS60 600 360 --..-St42

- . - - -420-500 220 1300-Si60-- . -

600·700 300 1800SI70 70G-850 340 2200

t-c15 --.- -50'O~550 300 1400'-en 5QJ-600 300 1550---'C35 600-720 370 1700'e60 -

750-900 490 240018CrNi8 650-1400 - 1900-6000

f--O-'.1700-300037MnSi5 600-1050 ---

Strength of some materials(Ultimate, Yield Stress and BHN), MPa

L _17 "-' .-.)

20 (ll

23.5.526:50211.7030.9131.65

13.25J 6.~)3

504.4 !<i8.832

-n.s)

I n ! 17 I 10 - 1 2S 'l3~ ..I ~~~ I.s 6 8 10 "''''1,- I 4' I - "''''','!1-i 0.0'181. 0.937 0.074 6.140 0.272 : 05 I5 I 0.736' 0.8'10 1.104 1.7G6 2. IL.', . .-

f 4 0:035 0.074 0.140 0.27.2 ~0.5451 O.?-57 I 1.398 1:-693 2.208 3.459 5+16 .

+r0~OSOI 0.'1.10 i 0.20~-. ---0!105 0:-810 1.39& I 2.061 2.502 3.238 5.152 &.0%-8 J 0.063 0.140 0.265 0.530 1.030 I-!'~~OI 2.723 3.238 4.195 6.771 10.30 I

J1i·J:.9-74 r.O-t62 0,316- -O-=<)4()_ 1.25.1 i 2.2&2 ~3.312 3.901 5.078'- '8·.1~0 12.51'1i2 0.081 0.184' 0.353 0.736 1.472 I 2.650 3.827 4,490 5.888 9.421 14.72_.14 o.os: 0.191 0.383 0.810 1.619 2.944 4:269 5.152 6.6211 10.60 16.191)6Io.081_ -

1.7'66 3.165 I 4.6370.199 0.412 0.883 5.520 7.213 11.56 17.6618 0.074 0.191 0.412 0.883 1.914 3.386 4.931 5.888 7.654 12.22 . 19.14:J.-..-

• 20 0.059 0.177 0.397 0.957 1.987 3.533 5.078 6.035 7.875 12.59 19.87 I22 0.037 0.155 0.361 . . 0.883 1.987 3.533 5.}52 6.109" _ltO?? 177)' 19.871

24 - 0.110 0.309 0.810 1.914 3.459 5.005 6.035 7.581 '12.51 19,14 J..~ - 0.059 0.221 0.736 L840· 3.312 4.784 5.741 7.434 11.85 18.40 i'-.F+~.=--r-2-._132 I 0.66 1.693 3:~.1814.416 5.226 6.845 10.8~1·1.6.9::; II 30.' -. - - - 1.472 3.650 3.754 4.563 5.888 9~~68..~72 i_--". ~---.! .... .;

Belt width, mm

Transmitted Power per Belt, IKwj ..~

;~

: ~_1-m=5 6 TI]Jo 13 L 17 . '20~2. I 25 .32 I 40 I 50 IU!-l _l_ (~_D-L6 _~_i_I_1~.4 ( 16_ ..20 25-r''l2 ILdn!_l__ll. ; 32 i _45 I 63 '. 90 I 125J~ .2J2J]~q 355 _soo---- 710 '1

lVlinimum Pullev dianH~tcr CUm), urrn:

Machine Design-January -2011- Third Year Mechanicfli, Power Section (Page: 41• o. ~ ··0

.s: ."..;~>-,...-, .. ~~~.- ..

Page 5: Final and Answers 2010 2011 Design2 Ahmedawad

x z , 20r.:jj)= o.4S- 2-55 =: 0.44

I 7~·6

;'0 95 b = 6'b, o·48~ )f._ o· 4 it1-2..5 x.."2.

k l.( '0 '2,35._ O,/(i'{-.J I'l = 0 .il 0 _ -- - "i i

:::J '2..2~·5

150,Zf ::;_-- =.9 tc:"~ 7..8·qs

-\:,:;::::'20

'3x.\o"o :::: 95 b N.: '3 .~Jj

,..." 7('1.."100)('600 - I. I. Llfr. == _~-:..:......::;;.,=.....:::.. '3.;j1-1. . rY1 S6'0 oco

,~t :::

f'l"\n-:' \.75

'1)\ = 100

" 'Su:e i3

N=

B: k2.= ~ -= 2.22J..j5

/'-i. ::: _ '3 x. \. '2 := 5.1; (;, := bo ' "6 03 x: 0.q8 x 2·2.. 2

C . -7 _ J V I0 0 - I 5c~ c> If-e. 'IO! "2.. -= (53 - . 0

1<6 :::

c

:.:=16&.5

Page 6: Final and Answers 2010 2011 Design2 Ahmedawad

~...

c= 6 ::: 758 73~24o z: 41149 N:: it1 kN

fCO-=rJ]1ill Tt;'c

5etxf('nj (3

Pet::: KPr:: 7587

.B etun'n9 AF,. .::,~_?f' -r .'C ?0. -

H'j

= -20 S 7 -t- 5"3 it = 2 5 g \

CA:: 253 \ 3~~ = 16 jol N.::. /6., kN

- -- -400'5<:>2o(!lo,,60x.2~c> = 2tto

/06

I ',/ - 'R.~= \0bg'~;~_4):' (!;1~nc32Y __

75S7 -N ' -

i '

12.1\= ~ (154 + 66S);- (2364 -47~t= 2.057 N

fr::. 956. t-aY120. Sec 28·95 :; 30i N

fa.=- ~5b. ta.n29.QS -= 53~ N,£"\0..+ ~ett

- AI e - o·2-~ 'A_

_b_~e -es_ -

Page 7: Final and Answers 2010 2011 Design2 Ahmedawad

..

. ~O( s1

'Bf.,d(tg Ha~e.,,:i :::.H -== Z83680 N.""""

'Tot'lu.e . = T::: 1+3sit 0 N· ,....w\

37::: ({; y, 10 (2.83 .62)2.-t- (1~3.S4) 7..~ (Ao):'

:::: 2..5.'3 Mf6._

'. ~'j ~- '~~ H p~f--'fP.l:I~ ~. ·55 \(A_

streSs ;'1' lieoJed shalt

~i r;J Se_t;t"Oil .d: bear-IJ f3