Machine Design - Computer Action Team
Machine DesignBolt Selections and Design Dimensions of standard threads (UNF/UNC) Strength specifications (grades) of bolts.Clamping forcesFeThe bolt force iskbFb Fi Fekb kcWhere Kb and KC are the bolt and the clamping material stiffness and Fi isthe initial bolt tensioning. Calculating Kb and Kc are relatively difficult andexam problems often give you theses stiffnesses or their ratio.
The clamping force iskcFc Fi Fekb kcFbFiFcFeRecommended initial tension (for reusable bolts)Fi (0.75 to 0.90) SpAtWhere Sp is the proof strength and At is the tensile area ofthe bolt.Recommended tightening torque (based on power screwformulas):T 0.20 FidWhere d is the nominal bolt size.2
Design of bolts in tensionFb At SpWhere At is the tensile area.Example M1aGiven: Two plates are bolted with initial clamping force of2250 lbs. The bolt stiffness is twice the clamping materialstiffness.Find: External separating load that would reduce theclamping force to 225 lbs. Find the bolt force at thisexternal load.SolutionFc Fi kcFekb k c225 2250 1Fe1 2Fe 6075 lbsFb Fi kb2Fe 2250 ( )(6075) 6300 lbskb k c3Example M1bSelect a bolt that would withstand 6300 lbs load in directtension. Apply a factor of safety of 2.5. Use a bolt withSAE strength grade of 2 (which has a proof strength of 55ksi).3
Fb ,design 6300(2.5) 15750Fb ,design S p AtAt (15750) At 0.286 in 255000A ¾” 10-UNC bolt has a tensile area of 0.336 squareinches.Bolts under shear loadingExample M1cA 1”-12 UNF steel bolt of SAE grade 5 is under directdouble shear loading. The coefficient of friction betweenmating surfaces is 0.4. The bolt is tightened to its full proofstrength. Tensile area is 0.663 in2. Proof strength is 85kpsi, and yield strength is 92 kpsia) What shear force would the friction carry?b) What shear load can the bolt withstand w/o yielding ifthe friction between clamped members is completelylost? Base the calculation on the thread root area.Approximate Answers: a) 22500 lbs, b) 70740 lbs4
Design of Bolt Groups in Bending Assume bolted frame is rigid.Use geometry to determine bolt elongations.Assume load distribution proportional to elongations.Assume shear loads carried by friction.LFDdExample M3Consider the bracket shown above. Assume the bracket isrigid and the shear loads are carried by friction. Thebracket is bolted by four bolts. The following is known:F 5400 lbs, L 40 inches, D 12 inches, d 4 inches. Findappropriate UNC bolt specifications for bolts of 120 Ksiproof strength using a factor of safety of 4.Answer: ¾”- 10UNC5
Design of Bolt Groups in Torsion Assume bolted frame is rigid.Use geometry to determine bolt distortion.Assume torque distribution proportional to distortions.Assume bracket rotates around the bolt group C.G.Ignore direct shear stress if its magnitude is small.Assume friction is lostUsually the bolt shank areas are used for analysis ofstresses.Example M4LThe bolts are ½”-13UNC. The distance between bolts is1.25”. The load is 2700 lbs and L 8”. Find the shearstress on each bolt.Answer: 44250 psi6
Design of Bolts in Fatigue LoadingThe factor of safety against fatigue failure of a bolt orscrew is:Sn a aWhere S a Se Su i Seand i is the stress due to initialSe SutensionExample: A M16*2 SAE grade 8.8 bolt is subject to acyclic stress. The minimum nominal stress in the tensilearea is calculated to be 400 Mpa (for initial tension with noexternal load) and maximum nominal stress is 500 Mpa(for maximum external load). Determine the factor ofsafety guarding against eventual fatigue failure for this bolt.Fully corrected endurance limit, including thread effects, is129 Mpa. The ultimate strength of the bolt material is 830Mpa. max min500 400 50 MPa22129(830) 400(129)Sa 57.8129 83057.8n 1.1550 a 7
Gear GeometryKinematic model of a gear setdpnpngdgNpPressure LineNgTerminologyDiametral pitch (or just pitch) P : determines the size of thetooth. All standard pairs of meshing gears have the samepitch.NP d dp Ndm NPp Teeth per inchInches per toothmm per toothP is pitch, p is circular pitch and m is the module.8
I) Regular Gear Trains (External gears)n1N2 n2N112N1 and N2 are the number of teeth in each gear, and n1 andn2 are the gear speed in rpm or similar units.Internal gearsn1 N 2 n2 N1129
II) Epicyclic (Planetary) Gear Trains2ARM1Planetary gear trains have two degrees of freedom – Theyrequire two inputs.Note: When Arm is held stationary, or with respect to theArm, the gears behave like regular gear trains:n2/A : the rpm of 2 with respect to Armn1/A : the rpm of 1 seen standing on Arm2110
Planetary gear trains can be solved by the following tworelationships. (two equations in three unknowns)1) Relative angular velocity formula:n2 n A n2 / A n1 n A n1 / A2) Regular gear train formula with Arm stationaryn2 / AN1 n1 / AN23The Toy Gearbox Sun gear N2 24 Planet gear N3 18 Ring Gear N2 2 N3 60Find Arm speed (assume n2 100 cw)24n2 / A n2 nA 100 nA n4 / A n4 nA0 nAn2 / A n2 / A n3 / AN3 N 4 ( ) 2.5n4 / A n3 / A n4 / AN 2 N3100 nA 2.5( nA ) nA 28.5 cw11
Problem #M5: Gear kinematicsThe figure shows an planetary gear train. The number ofteeth on each gear is as follows:N2 20N5 16N4 30The input is Gear 2 and its speed is 250 rpm clockwise.Gear 6 is fixed. Determine the speed of the arm and thespeed of Gear 4. The drawing is not to scale.d5 d6 d2 d4 and assuming all P are the same we getN5 N6 N2 N4 and N6 34 teethn2 / A n2 n A 250 n A n6 / A n6 n A0 nAn 2 / A n 2 / A n 4 / A n5 / AN6N4 (1)( )n 6 / A n 4 / A n5 / A n 6 / AN2N512
Substituting for the number of teeth on each gearn2 / A3034 ( )( ) 3.187n6 / A20 16n A 114.3Alson4 / A n4 n An4 ( 114.3) n2 / A n2 n A 250 ( 114.3)n4 / AN20 ( 2 ) n2 / AN430From above:n4 -357.1 rpm13
Kinematics of Automobile DifferentialConsidering the Right Wheel, Left Wheel, the RingGear and the Drive Shaft.nRW nLW 2nRG14
Gear Force AnalysisFn FtdFrFn : Normal forceFt : Torque-producing tangential forceFr : Radial force.When n is in rpm and d is in inches:33000(hp)Ft Vandand V nd12Fr Ft tan In SI units:dWatts T and T Ft ( )215
Helical gearsGeometric parametersPn : Normal pitchP : Plane of rotation pitch : Helix angle n: Normal pressure angle : Plane of rotation pressure angleN : Number of teethd: pitch diameterpn and p : circular pitchespa : axial pitchGeometric relationships:NP and P Pn cos( )dtan( n )tan( ) cos( )Pp Pn pn andpa ptan( )Helical gear forcesFt From PowerFr Ft tan( )Fa Ft tan( )When shaft axes are parallel, the helix hands of the two gears mustbe opposite of each other.16
Straight Bevel gearsGeometric Parameters(Pinion)Pinionddavgdp: pitch diameterdavg,p: average diameterb: Face width p Pitch cone angle bBevel gear forces33000(hp)Ft Vavgwhere Vavg nd avg12andd avg d b sin( )Fr Ft tan( ) cos( )Fa Ft tan( ) sin( )These forces are for pinion and act through the toothmidpoint. Forces acting on the gear are the same butact on opposite directions.17
Worm Gear KinematicsdwdgThe velocity ratio of a worm gear set is determined by thenumber of teeth in gear and the number of worm threads(not the ratio of the pitch diameters). g NwNwgNw Number of threads (single thread 1, double thread 2, etc)The worm’s lead isL pa NWThe worm’s axial pitch pa must be the same as the gear’splane of rotation circular pitch p.The worm’s lead angle is the same as the gear’s helixangle he gear and worm must have the same hand.18
ExampleFor a speed reduction of 30 fold and a double threadedworm, what should be the number of teeth on a matchingworm gear.Ng (2) (30) 60 teethThe geometric relation for finding worm lead angletan( ) L d wWorm Gear ForcesThe forces in a worm gearset when the worm is driving isFgr FwrFgt FwaFwtFga FwtFwaFwrThe Fwt is obtained from the motor hp and rpm as before.The other forces are:cos( n ) cos( ) f sin( )Fwa Fwtcos( n ) sin( ) f cos( )19
The worm and gear radial forces are:sin( n )Fwr Fwtcos( n ) sin( ) f cos( )The worm gearset efficiency is:e cos( n ) f tan( )cos( n ) f cot( )Where f is the coefficient of friction. Condition for selflocking when worm is the driverf cos( n ) tan( )Note: In a RH worm, the teeth spiral away as they turn in aCW direction when observed along the worm axis. Whenthe worm in turning in CW direction, the teeth sweeptoward the observer seen along the axis of the worm(imagine a regular bolt and nut).RH Worm20
Bearing Reaction ForcesBearingAny gear or pulleyFrFaFtFFaTotal thrust load on bearings is FaFor the radial reaction forces for spur gears (no axialforces) combine the radial and tangential forces into F:F Fr2 Ft 221
Flat BeltsFlat belts have two configurations: OpenD d)2CD d D 2 sin 1 ()2C d 2 sin 1 (Closed (Crossed) d D 2 sin 1 (D d)2CWhereC: Center-to-center distanceD,d: Diameters of larger and smaller rims : Angle of wrap around pulley22
Slippage Relationship(True only at the verge of slippage)Tight sideF1F1 e F2 is in radians.DriverF2Slack sideTransmitted Hp isHp ( F1 F2 )V33000Where F1 and F2 are in lbs and V is in ft/min.Initial TensionBelts are tensioned to a specified value of Fi. When thebelt is not transmitting torque:F1 F2 FiAs the belt start transmitting power,F1 Fi FF2 Fi - FThe force imbalance continues until the slippage limit isreached.23
Problem M7A 10”-wide flat belt is used with a driving pulley ofdiameter 16” and a driven pulley of rim diameter 36” in anopen configuration. The center distance between the twopulleys is 15 feet. The friction coefficient between the beltand the pulley is 0.80. The belt speed is required to be3600 ft/min. The belts are initially tensioned to 544 lbs.Determine the following. (answers are in parentheses)a) Belt engagement angle on the smaller pulley (3.03radians).b) Force in belt in the tight side just before slippage.(1000 lbs).c) Maximum transmitted Hp. (99.4 hp)Formula for V-beltsP1 Pc e / sin P2 Pc wherePc m ' 2 r 2m’ Mass per unit lengthr Pulley radius24
Disk Brakes and ClutchesdDTorque capacity under “Uniform Wear” condition perfriction surface (when brake pads are not new)T fp a d8(D 2 d 2 )Wheref: Coefficient of frictionpa: Maximum pressure on brake padd,D: Inner and outer pad diametersTorque capacity under “Uniform Pressure” conditions perfriction surface (when brake pads are new)T fp a12( D3 d 3 )25
Maximum clamping forces to develop full torqueFor Uniform WearF p a d2(D d )For Uniform PressureF pa4( D2 d 2 )Example M8Given: A multi-plate disk clutchd 0.5”D 6”Pmax 100 psiCoefficient of friction 0.1Power transmitted 15 hp at 1500 rpmFind: Minimum number of friction surfaces requiredAnswer: N 2 (uniform pressure)N 9 (uniform wear)26
Energy Dissipation in Clutches and BrakesThe time it takes for two rotational inertia to reach the samespeed after engagement through a clutch is:I1I2I 1 I 2 ( 1 2 )t T ( I1 I 2 )whereT: Common transmitted torque : angular speed in rad/secThe total energy dissipated during clutching (braking) is:I1I 2 ( 1 2 ) 2E 2( I1 I 2 )If the answer is needed in BTU, divide the energy in in-lbby 9336.27
Problem M9A brake with braking torque capacity of 230 ft-lb brings arotational inertia I1 to rest from 1800 rpm in 8 seconds.Determine the rotational inertia. Also, determine the energydissipated by the brake.I1I2Solution hints:Convert rpm to rad/sec: 1 188 rad/secNote that 2 0Find the ratio (I1I2/I1 I2) using time and torque 9.79Note that I2 is infinitely large I1 9.79 slugs-ftFind energy from equation 173000 ft-lb28
SpringsCoverage: Helical compression springs in static loadingTerminology: d: Wire diameterD: Mean coil diameterC: Spring index (D/d)Nt: Total # of coilsN: Num. of active coilsp: Coil pitchLf: Free length N*pLs: Solid lengthEnd detail and number of active coils:PlainLsNtLf(Nt 1)dNpN dPlain &GroundNtdN 1p(N 1)Square(Nt 1)dN 2pN 3dSquare &GroundNtdN 2pN 2Note: Spring geometry, especially the end-condition relationships, are notexact. Other books may have slightly different relationships.29
Spring Rate of Helical Springs (compression/extension)d 4GdGk 338D N 8 NCwhere : N is the number of active coilsG: shear modulus E/2(1 )G 11.5*106 psi for steelsShear stress in helical springs for static loading max8FD Ks3 d1where K s 1 2C and C is the spring index.Shear strength in springs 0.45S u Ferrous without presetting 0.65S u Ferrous with presettingNote: it is common in practice (but not academia) to specify strength as“Allowable Stress”. Allowable stress is defined as the strength (yield orshear strength) divided by the factor of safety.30
Spring Surge Frequencyfn 1 kg2 WaWhere g is the gravitational acceleration and Wa is theweight of the active coils:1Wa 2 d 2 DN 4with being the specific gravity of spring material. Forsteel springs when d and D are in inches:13900dfn HZ2NDExample M10Consider a helical compression spring with the followinginformation (not all are necessarily needed):Ends: Squared and groundSpring is not presetMaterial: Music wire (steel) with Sut 283 ksid .055 inches and D 0.48 inchesLf 1.36 inches and Nt 10Find the following. Answers are given in parentheses.Spring constant, K (14.87 lb/in)Length at minimum working load of 5 lbs (1.02”)Length at maximum load of 10 lbs (0.69”)Solid length (0.55”)Load corresponding to solid length (12.04 lbs)Clash allowance (LFmax – LS) (0.137”)Shear stress at solid length (93496 psi)Surge frequency of the spring (415 Hz)31
Design of WeldsWelds in parallel loading and transverse loadingP3”PORWeld GeometryThroat: tLeg : hAnalysis Convention Critical stresses are due to shear stresses in throat areaof the weld in both parallel and transverse loading. For convex welds, t 0.707h is used. The shear strength in weld analysis is taken as 30% ofthe weld ultimate strength.32
Analysis Methodology Under combined loading, different stresses arecalculated and combined as vectors.Stresses based on weld leg (h)Direct tension/compression: F 1.41 Lt LhDirect shear: 1.41VLhBending: Mc Mc IIutTorsion:Tr J utFormulas for Iu, and Ju are attached for different weldshapes.33
34
35
Problem M11a -Welds subject to direct shearTwo steel plates welded and are under a direct shear load P.The weld length is 3 inches on each side of the plate andthe weld leg is 0.375 inches. What maximum load can beapplied if the factor of safety is 2 against yielding? Theweld material is E60.3”PSolution (of M11a)L 2d 61.41V1.41P 0.6284 PLh6(.375)psiThe design strength of the weld material in shear is:Sys 0.3 Sut 0.3(60) 18 ksiUsing a factor of safety of 2, the allowable shear stress is: all 18/2 9 ksiEquating stress and strength.6284P 9000 P 14322 lbs36
Problem M11b – Welds subject to torsionA round steel bar is welded to a rigid surface with a ¼ “fillet weld all around. The bar’s outer diameter is 4.5”.Determine the critical shear stresses in the weld when thebar is subjected to a 20,000 lb-in pure torque.J u 2 (r 3 ) 2 (4.5 / 2)3 71.57 in 31J t (71.57)( )(0.707) 12.6 in 44 Tc (20000)(2.25) 3571 psiJt12.6Problem M11c – Welds subject to bendingSolve the previous problem with a bending moment of35000 lb-in acting on the welds instead of the torsion load.I u r 3 (2.25) 3 35.78 in 3I t 35.78(0.25)(0.707) 6.32 in 4 Mc (35000)(2.25) 12460 psiIt6.3237
Problem M11d – Welds subject to combined loadsIf the shear strength (Sys) in the weld is 27800 psi, what isthe factor of safety against yielding when both stresses inprevious two problems are acting on the bar. 12 22 35712 124602 12961 psiFS 27800/12961 2.1438
Ball and Roller BearingsTerminology Rated or catalog Capacity, C10 : Radial load for a lifeof 1000,000 cycles (or other L10) and 90% reliability. Application or design radial load Fr. Application or design life LLoad/Life relationships (Palmgren formula)F1L10.333 F2 L02.333This means if we double the load, the life of the ballbearing would be reduced by a factor of 10. This formulais for ball bearings. For roller bearings use 0.3 as theexponent.Example: By what factor the radial load capacity of aroller bearing has to be increased if the bearing is to lasttwice as long as its catalog rating.0.3 F1 L2 F 2 1.23 F2 11.23 1 F2 L1 0.3Example: Given:02 series Deep Groove ball bearing,Radial load is 4 KN,Application factor KD 1.2Design life 540 million cycles95% reliabilityFind: Suitable bearing catalog rating based on 106 cycle L10 life.39
Solution:Life multiplier due to reliabilityx1 0.65 (at 95%) - See reliability multiplier belowAdjusted design life:LD 540/0.65 830.77 million cyclesForce multiplier due to life being different from 106 cyclesK1 (830.77).333 9.38Adjusted Design LoadFDA 4 (1.2) (9.38) 45 KNSelection: 60 mm bore with 47.5 KN capacity.40
Problem M12 – BearingsAn angular contact 02-series ball bearing is required to runfor 50000 hours at 480 rpm. The design radial load is 610lbs and the application factor (load multiplier) is 1.4. For areliability of 90%, what is the required capacity of thisbearing? Answer: 42.9 KN (capacities are in SI units) If the required reliability is different than 90%, apply areliability factor to the life. (See Juvinall).Kr Reliability factor90% Kr 195% Kr 0.6550% : Kr 599% : Kr 0.20 If there is substantial thrust or axial loading, then anequivalent radial load should be used. For radial ballbearings (Ft is thrust load):Ignore Ft and use Fe Fr if Ft 35% FrUse Fe 1.18 Ft if Ft 10 FrIf Ft 35%Fr but less than 10 Fr use FtFe Fr 1 1.115( .35) Fr When a bearing system includes several bearings (nbearings), and the reliability of the entire system isgiven (Rsys), then the reliability of each bearing mustbe:RD n Rsys41
KeysSquare keysw d/4 (d is shaft diameter)Length LShear stresses at torque TFSetting F 2T/d and balancing the force and stress2T d ( L) d4 8Td 2LTorque capacity for this key in shear is obtained by setting stress toits yielding limiting value:Ld 2 (0.58 S y )8TmaxFor round pins in double shear 4T Dd 2DdThe torque capacity isT Dd 24(0.58S y )Other key types and splines are all treated similarly: Equating theshear area to the force created by the transmitted torque.42
Power ScrewsLOADCollarTorque required to raise a loadFd m L fd m Ff c d c TR 2 d m fL 2F: Load , dm Screw mean diameterL: Screw lead NW * pf: Thread coefficient of frictionfc : Collar coefficient of frictiondc: Mean collar diameterTorque required to lower the loadFd m L fd m Ff c d c TL 2 d m fL 2Note: All formulas are for power screws with squarethreads which are the most common type.43
Condition for self locking:Lf tan( ) d mEfficiency of power screws (includes collar losses)e FL2 TRProblem M13 – power screwsA single thread power screw is carrying a load of 12500lbs. The mean screw diameter is 1 inch and the screwpitch is 0.25 (4 threads per inch). The mean collardiameter is 1.5 inch. The coefficient of friction of boththreads and collar are 0.1. The thread shape is square.Find a) Major diameter of the screw, b) Torque requiredto lift the load, c) Minimum f to make the screw selflocking if fc 0, d) power screw efficiencyAnswers a)1.125 in , b)2070 lb-in, c) 0.08, d)24%44
worm, what should be the number of teeth on a matching worm gear. N g (2) (30) 60 teeth The geometric relation for finding worm lead angle d w L S tan(O) Worm Gear Forces The forces in a worm gearset when the worm is driving is F gr F wr F gt F wa F ga F wt The F wt is obtained from
team xl team 2. t050710-f xl team 3. t050907-f xl team xl team 4. t050912-f xl team xl team 5. t050825-f xl team xl team 6. t050903-f xl team. 2 7. t050914-f xl team xl team 8. t061018-f xl team 9. t061105-f xl team name xl team 10. t060717-f xl team xl team 11. t070921-f xl team xl team xl team 12. t061116-f xl team. 3 13. 020904-f name/# xl .
decoration machine mortar machine paster machine plater machine wall machinery putzmeister plastering machine mortar spraying machine india ez renda automatic rendering machine price wall painting machine price machine manufacturers in china mail concrete mixer machines cement mixture machine wall finishing machine .
Round 3 Game 1 Game 2 Game 3 Game 4 Team 1 Team 7 Team 8 Team 2 Team 6 Team 5 Team 4 Team 3 Continuing the method, which team plays Team 7 in Round 4? Team . 14 Infection Model This is a simple example of how people in a community might become infected with a disease. O
a) Plain milling machine b) Universal milling machine c) Omniversal milling machine d) Vertical milling machine 2. Table type milling machine 3. Planer type milling machine 4. Special type milling machine 5.2.1 Column and knee type milling machine The column of a column and knee
1.Shaper Machine, 2. Lathe Machine, 3. Drilling Machine, 4. Grinding Machine Table 1 shows the summary of all man machine chart and Table 2 shows the man machine chart for shaping machine and same way we have prepared the man machine chart for other machines. Table 3 Man Machine Charts Summary
SP3 : Technologies de traitement SP4 : Outil global d'aide à la décision Action 6 PCB OPTITRI Action 7 PCB ECODEPOT Action 8 STAB PCB Action 9 PCB SEDICA Action 10 FUNGI EAT PCB Action 12 BIODECHLOR PCB Action 13 DESTHER PCB Action 14 PLATPIL PCB Action 15 SEDIRHONE PCB / / / / SP3.1 : dragage et criblage SP3.2 : confinement SP3.3 : absorption
CAD/CAM Computer-Aided Design/Computer-Aided Manufacturing CADD Computer-Aided Design and Drafting CADDS Computer-Aided Design and Drafting System CADE Computer-Aided Design Equipment CADEX Computer Adjunct Data Evaluator-X CADIS Communication Architecture for Distributed Interactive Simulation CADMAT Computer-Aided Design, Manufacture and Test
team members throughout the contest with each team member losing at least 10 pounds, the team name will be placed in a drawing for one of three grand prizes. The First Team drawn will win 250 each, The Second Team drawn, 500 per team member, and the Third Team drawn will win 1,000 per team member. Team CAVHCS is already off to a great start. The
