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by: Rowan Spinka DVM


Rowan Spinka DVM
GPA 3.62


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Class Notes
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This 49 page Class Notes was uploaded by Rowan Spinka DVM on Thursday October 15, 2015. The Class Notes belongs to MAE 589W at North Carolina State University taught by Staff in Fall. Since its upload, it has received 30 views. For similar materials see /class/224032/mae-589w-north-carolina-state-university in Aerospace Engineering (AE) at North Carolina State University.

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Date Created: 10/15/15
NC STATE UNIVERSITY Internal uomousuon Ingnes MAE 589V Spring 2009 Engine Performance and Testing Dr Tiegang Fang 3184 Broughton Hall Mechanical and Aerospace Engineering Department North Carolina State University NC STATE UNVVERSITY Engine Full L erformance Three characteristics curves showing the brake torque brake power and brake specific fuel consumption emmgnszmwza kW w 22 D1005 rounufml u 395 w 32a 1 m am so 7n so so Iu n u zm nitIII NC STATE UNWEHSITY Example Full Load Performance Po weri39ecn Piusquot 4045H Diesel Engine Specifications PERFORMANCE DATA Rated Speed Intermittent 73 hp 129 kW if 2400 rpm Peak Pgwgr Power Bulge a 173 11p 129 kW it 3400 1pm 0 34 it 2400 rpm 476 lbft 645 Nan i745 ISOU 1pm 26 Q 500 1pm Peak Torque Torque Rise l i RATED BIIP is the power rating for variable speed and load applications when full power is required intermittently CONTINUOUS BI IP is the power rating for applications operating under a constant load and speed for long periods oftime HEAVY DUTY see application ratingside nitions engine perfonmanee curves POWER OUTPUT is withing or 5 at standard SAE J 1995 and ISO 3046 CERTIFICATIONS EPA Tier 3 EU STAGE III A AND CARE NC STATE UNWEHSITY Example Full Load Performance PERFORMANCE CURVE 4T6 Ih 645 N m 379 Ilait Him 173 hp 129 kW Brake Power hp le 60 45 Fuel Consumption PM 3quot 4045H Diesel Engine Specifications Poweri39ech 00 1000 1200 1400 1600 mm 2090 220 2400 Engine Speed a rpm 540 732 450 524 330 515 040 242 035 219 Torque lbft N mJ Fuel lbihpI1r gikWh NC STATE UNIVERSITV Engine Performance Maps Add contours of constant BSFC and constant power to the load speed graph Load is commonly denoted by torque or BIVIEP Engine Torque mN Specific Fuel TURW PURWCnnsumpTiun gr kWh constant power curves optimum efficiency curve NC STATE UNIVERSITV Engine Performance Maps cont d Help to describe the effect of loadspeed variations Engine Torque mm Specific Fuel TDkW g kw uneump rinn gr kWh There is one loadspeed comMn banw BSFCreacmnga n nknuni upfimum NC STATE UNIVERSITV Optimizing Engine Performance To make the engine work at lower BSFC and higher efficiency region Engine Torque mm Specific Fuel TUkW g kw uneump rinn gr kWh For constant speedBSFC decreases with increasing load duetornechanmal efficiency loss and theninmeasesup tothequload condMonsdueto mmrhdMgwa poorcombus on Engine Speed rpm NC STATE UNIVERSITV Optimizing Engine Performance cont d Engine Torque mm Specific Fuel TDkW g kw ungump rinn gr kWh Ior constant load BSFC decreases with increasing seed duetoheat ossandthen inmeasesuptothe rnaxknunispeed condMonsdueto friction loss upfimum NC STATE UNIVERSITV Optimizing Engine Performance cont d Optimum efficiency curve can only be achieved by using Continuously Variable Transmission CVT technique Engine Specific Fuel Ter que mew 90km Consump rinn mN ngWh npfimum Map Example Englne 5pm rpm Fig 14 Spam Fuel Consumpnon ach Map New Beeue LSITDI 10 Map Example Engine Torque vs Throttle Engmu Tnlqvu Map mama Engmnquul llh Engine Testing Parameters to be measured Power brake indicated Torque Speed Fuel consumption Air consumption lncylinder pressure for combustion analysis Other parameters such as pressures temperatures etc Specific equipment is required for different parameters P P SquotPPNT Engine Test Cell Control Room 12 NC STATE UNIVERSITY Power Torqu and speed Measurement Power Torque and Speed are measured by dynamometers dynos Three types of dynos are commonly used 1 Fluid or Hydraulic dynamometers water brakes 2 Eddy current dynamometers 3 Electric dynamometers Type 1 and 2 can only absorb power from the engine Type 3 electric can absorb power from the engine or deliver power to engine which can be used to motor the engine to measure the friction power mm m Water 39mm measures pm mumquot b slam mm rind to turcusnm ucs Water brake dyno NC STATE UNIVERSITY Power Torque and Speed Measurement cont d lady nmm gnu mm am i39m7 Eddy current dyno AC electric dynos are newer and response faster 7 7 14 NC STATE UNIVERSITY Fuel Consumption Measurement Measurement includes two types Mass or Volume Volume measured using rotameter Float in a tapered tube Reading affected by viscosity and temperature of fuel Volume measured using turbine flow meter Volume measured by positive displacement flow meters Advantage independent of viscosity ROtameter I LII Ull IU flow I m meter Positive displacement flow meters 15 NC STATE H N lVEHS ITV Fuel Consumption Measurement cont d Mass steady state batch ow measurement with scale and beaker Mass Coriolis flow measurement Fluid passes through a vibrating tube causing the tube to twist the amount of twisting is proportional to the mass flow 7 Coriolis mass ow meter NC STATE UNIVERSITY Air Consumption Measurement Most common calibrated orifice or nozzle Pressure drop across orifice or nozzle increases with flow rate Pressure differential measured with pressure transducer Normally calibrated by manufacturer under controlled conditions Orifice or nozzle cause slight air restriction rip 17 NC STATE UNIVERSITY Engine Combustion Data Acquisition Pressurevolume conditions in combustion chamber easily measured Twochannel shaft encoder synchronize the timing Top Dead Center TDC pulse Crank angle pulses at lt025 degree intervals Piezoelectri transducer for measuring incylinder pressure Sensor slight recessed Effect of connecting passage Special coatings on the surface to limit radiant energy transfer Water cooled to prevent damage at high temperatures Can be used to calculate energy release rate 18 Enginv Test Facility Reference Data and Alarms RealTime Test Cell Controller HlIhSIEEIU Network Operator Interface Typical engine test facility 19 or vB 100 h WW 7 Do t 7 94 C 39 2117K map ijxzzxL My 0 1DD Ma 759 72i tli 4300 7 7 0 HIV xzyzo gt 60 gt 0 63 1 IQa f 0 f55 on a A 773 0 v d M c wou l x HOWXO m 28 m 144 9 33 6 p 7 le W 17 7739s Z 7 vi i M 1 0048 0 7 0000 I 7 W 13 77 175 Chapter 8 2D Elasticity Theory Useful for inplane loads with no bendingextension coupling eg B 0 9 M 5 Z R 81 Elasticity Equations for plane stress ie SW oxz 022 0 Anisotropic ElasticitV Two general classes of problems 1 Plane Stress 2 Quasi 3D no zdependence Will use cornpleX variable techniques Lekhnitskii s method I Basic eguations 2D Plane Stress A StraInDISplacement 6y 2 A2l 2 8x 176 6 strains 3 displacements gt strains must have some conditions if real displacements are expected B Compatibility 32ny 32 Ex 62 Ey My 6y2 6x2 6 equations 39 A26 52 l 2 E 6x832 2 62 6x 8y 62 C Equilibrium sly3 Fi 0 A28 2F 2 O D Constitutive Hooke s law 6139 52 Old E Reduction of equation from 3D to 2D l Plane Stress Conditions a O C C 6 0 aZ x2 yz Z 2 O TyZ O sz Z Y Z ze y GyZ 177 gt Compatibility reduces to one equation 32ny 32 Ex 62 Ey axay ayz axz 3 Equilibrium without body forces reduces to two equations 5 55 hr xy0 1 5x 5y a 6 zo 5x 5y For equation 1 let ox 35F and TW 35F 2 2 6F aF0 For equation 2 let 6y 2 0 and rxy x 62G 62G 6My 6my aFaG Butt 1 gt X 35 ax 8y 178 Thus there must be some function I such that andGz iegjz 5y 5y 5x 5y 5y 5x 2 2 2 mpg 64gt M o 1 ayz yaxz W My We have reduced three components of stress down to one stress function 4 Constitutive equations becomes 2 2 1 V 1 v Ex6xxy6 Ex Ex y Ex 322 Ex 6x2 ny 1 ny 52 1524 EyZ E GxE Gy E 2 E 2 x x 8y yax T a2 ny ny xy nyaxay 5 Plug these strains into the remaining compatibility equation 179 82 1 824 i i82 vxy82l 8x8y ny 8x8y 5y2 Ex yZ Ex 3x2 2 2 2 8 nya 16 8x2 Ex 8y2 Ey 8x2 or i l 2ny 844 1 84d Ex 8y4 ny Ex 8x28y2 E 8x4 We have reduced the problem to one 4th order differential equation 6 Strains can be used to calculate displacements using the straindisplacement relations Therefore the complete solution is known if I can be found F Complex Variable Solution Technique Review of Complex Variables 3quot 7 5 1 Description of part Z M g v 2 xaya9 orzxiyaib 7 i9 0 x or z re WM Real axis 1 r a2 b22 9 tan1 180 2 Square Root l 9 irz 2 3 Trig relations IiGZI ie i9 i9 cosez sin62 2 21 z rcose l39sin 6 l l 22 ir2cosgiisin 2 2 4 Conjugates z x iy E x iy f2 22 22 x iy2 2x 1y f2 22 22 x iy2 2x 1y 2 Z 2x 2 Real number 181 f Z f Z Rea1fZ 2Refz 1 De ne 2 new operators i2 3 and i 2 621 ay lax 522 ay 26x where LL and M 2 are two of the four roots of the characteristic equation E E 4 1 2V12H2 1O G12 E2 Example i 6O2S Ex 2 143 gtlt106 vxy 0306 6 6 Ey 673gtlt10 ny 400gtlt10 Characteristic equation 4 i 2vxy uz zO ny Ey 2 biw1b2 4ac 2a 182 b 2 i ZVW 20306 0255 ny 400 cz z z 0212 Ey 6 73 02 0127 i 0196 0127ii0443 Square root of a complex number w waib cos i sin 0 where r a2 122 2 0461 1 g 0443 a 0127 0tan 7400 7400 2 cos 2 izsm 2 183 i 0542 ii 0409 wrong M 0542 139 0409 or H or H or H but u12aiB and uzz aiB 1 0542 i 0409 H2 W iw Other possibilities include H1 21131 Hz 13982 31 7 32 both real and possible H1 113 Hz 1B 3 is real and possible Then we have 0 621 622 62 822 4 2 Integrating we nd I Flx my F2 x uzy 184 F1XH1yF2xH2y or l F1ltzlgtF2zzgtElt21gtElt22gtl where 21 xu1y and 22 x uzy 3 Comments We now have the stress function F in a convenient form We now it is a function of two analytic harmonic functions of the variables 21 and 22 The fact that these functions are analytic greatly simpli ed the analysis and let us use some of the methods that are well established in complex analysis eg conformal mapping 4 De ne new functions 13121 E 13222 2 1 and 2 are similar Z1 dZ2 This step is not necessary but keeps with the notation of Lekhnitsiu 2 2 2 5 Recall stresses ox 6 1 6y 6 1 rxy a F ay ax My 2 2 2 2 0x H1C131H2qgt2 H1 131 Hchzl 0y CD391CD3925i5392 185 Txy H1qu H2 D392 mii E252 6 Other equations i Displacements u 191131 12 1 9151 172532 61y 62 V c11 131 92 67151 61252 61x C4 2 2 where Pk511Hk 512 Sl6Hk 61k Sunk 522 526Hk k 1 2 C1 C2 C4 are rigid body rotations and displacements ii Boundary Tractions Tloynj Txcos96xxsin91xy Ty s1n91xycosoyy 131 132 51 52 i I T de constant top inside HICDI H2sz E161 H262 ITde constant bottom outside 186 7 Comments Once C131 and 132 are known average stress strain displacement or traction can be found throughout the body 11 Elliptic Hole in an In nite Plate Useful for two practical cases 4 4 amp 0 0 a A A 39 l l C9 1 i r 7 l MM t t J l t Plate with hole cracked plate stress cone stress singularity 187 Case 1 Can use average laminate properties Ex Ey etc or can analyze a unidirectional laminate or laminae Case 2 Practical only for a UD laminate or lamina even for these there are restrictions A Form of the Stress Functions CDk Solution requires superposition of two stress elds r 7quot T t l g l gt 4 4 6 Uniform Stress State 4quot 9 09 P z 1 roe K 17700 4 J F a k 2k constant Iii rel 7 28quot t J L L a 65 PL Hole in 00 plate CIDO39k 2k constant CD39k C13094219 CDIIEOQk 188 H139cZk gk B Far Field BC s i Let Hl39coo 0 Then 60 p1u12g1uig2 H12 1H 2 63192 g1g2 1 2 Tgcoy P12 Higi Hzgz E1 1 Hzgz 3 equations and 4 unknowns gt we are free to choose a fourth equation ii Arbitrarily choose Im g1 Im g2 0r 0g1g2 1 2 iii Solve 4 X 4 to get Pl HI2P2 1izH1 H2 kzlaz gk 2 2 2Hk HI l3k iV Comments The arbitrary choice at ii above means that the stress functions gk are not unique The stresses however are unique 189 3 BC s at Hole Opening Tractions are zero x T20 szo TyZO Recall 131132 51622Jl fv0ds H1 Di 2 E151 E252 If0d5 Also reca112CD39k2 chzk gk or bk 2 Hkzk gkzk NOWI H1 8121H2 8222171 H2 8222 0 1 H1H1 8121 H2H2 8222 H1U71 H2U72 g2220 2 4 Solution of i and E 00 i Assume H ff 2k 2 20 2139 Laurent Series 00 190 for H39 ooO 099020 nl23 I 00 XS 2 Hkltzkgt Z n 1121 Zk ii Plug this into 3 and 4 Integrate H 39 first The result is an in nite set of simultaneous equations for ocm No solution eXists with this form of stress function Try a conformal map zk gt pk iii Choose the simplest map first On boarder p k cs unit circle 2 2 2 2 z i 1 2 a u b Recall o a lukb iV Stress function H k 2k becomes H k pk Hl39czkaH WampDkzjjkpk k de apk 62k 62k apk pk where 2 2 2 2 aZk zk a ukb 191 7 nl r 00 ak 1 WW Hkltpkgtz n21 pk or k k k k a a a Hkpka1 ln pk 2 3 24 3 pk pk pk at boundary pk c k k k Hkoa1klnoa2 a3 2 3m 5 c c Vi Plug into eqs 1 and 2 and compare like powers of o ak aPM 1912 ibP12HI P1 k 1 2 2 2Hk HI 3 k k all other 612 are zero ie equations are uncoupled 5 Final Solution 2 2kP1 P2HI P12HkHZk k12 2 2 Zak HI I 3 k 192 ap2ul 1912 ibp12u 191 a iukb 2 2 2 2 2 k HI ZkiJZk Hkb Cl 2 r H Hm 191 mm 2 P12 Hk I 2Hk HI 64mm 1912 ibp12u 192 a iukb 1 2Hk HI Zkiv V 111 Stress Concentration at the Edge of a Circular Hole 0y Plot of along x was N J HM Peak value is called the stress concentration A Hole Description a b 1 unit radius zxiy zkxuky 193 for stress at hole edge look aty O x a 1 Z 1 Zk 1 B Stress function CD39k For 630sz 63601320 2 qyk p2 HI 1P2 I mi M2 2HkHk w c Stress Concentration oyz j 5i5 P2 HIHZ 6 y1lH1H2 note for isotropic materials 1 Hz 21 G y1lll P2 1 ie p1 p12 0 D Numeric Examples UniDirectional Quasi Isotropic x GPa 7 y GPa 145 G xy Gm 35 Vyx 034 iH1H2 15513 2 Mluz 02197 1 0V 806 3 192 34 2 390 h 12 12 194 i 45 1286 1286 3702 0837 08206 10 182 195 F Loading in xdirection for 00P1 P2P120 G x1iu1u2 at x0 y1 P1 fl 1 KTis given incorrectly in the book as equation 639 pg 209 Should be l 2 E KT1 VLT L ET GLT E L is modulus in loading direction 196 l EL 2 EL lH1H2 2 VLT ET GLT This is a good check on ul and uz for your homework G Design Considerations Classic Industrial Case UD with hole loaded in ber direction Premature failure due to high stress conc Traditional approach is to add to thickness i Non composites engineer add more 00 layers ii Composites engineer add 45 layers to reduce KT This is a classic example of how composite materials can be optimized for given structural components 197 RICHARD K CORYELL 2 NOVEMBER 1987 MAE 589 MECHANICS OF COMPOSITE STRUCTURES PROGRAM TO CALCUALTE THE STRESS CONCENTRATION FACTORS FOR A ONE INCH HOLE IN VARIOUS FOURPLY LAMINATES 0 10 30 45 60 90 TH EXgtlt 195000 178865 55018 16265 11219 12000 106 ny 12000 11819 11219 16265 55018 195000 106 nygtlt 04400 09776 38867 50356 388867 04400 106 ny 02700 07062 16022 08483 03267 00166 U1 000 000 125 092 056 000 0611 0971 0811 0401 0371 0151 U2 000 000 125 092 O56 000 659i 3991 0811 0401 0371 163i oxPl 82000 59664 26248 17915 17337 27861 198 Stress Concentration for a Hole in a Symmetric Laminate 5 V E Id x ew is 50 50 70 no 90 Theta 199 200 NJ Zn 7 7M 3M 70 Io vsj J39 xquot 739 0 at N f 1E7quot n w 39 0 n 5616 W5 1 39 aIAfaay a f0 fa rm m gg w awx Jaa wad q l H 11 20 5 Yf 39 551 r39 5114 39 04K 4A 1 may mow7 AMAVK Q yd may 5 a r 54 7A 4 r C v M Jaw55 ram Sc c mmb j 45 09 a 201 IV Failure Analysis of Stress Concentrations Stress at 5c 3 is much greater for the large hole than it is for the smaller hole even though both cases have the same stress concentration factor Since the composite with the larger hole has more of its cross section highly stressed it will fail sooner than the composite with the smaller hole This is known as the hole size effect Whitney and Naismer 1974 proposed two methods which help predict failure There is some question as to the accuracy of the methods but the concepts are valid A Point Stress Failure Criterion Failure occurs when the laminate yield stress is reached at a critical distance from the hole edge


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