Reactions Shear Moment: Deflection End Slope: Resctions: Shear: Moment Deflection: End Slope: Elements of Sections Rectangle (aus through center) AE bh c= he 1 bhi Ze Ve bhiys Rectangle (xis through base) A=bh ez Ve Bhs, ze ve> bhi/s ven ars. 2S aexsoasst cb Circe ons trough cote) a Aandi = momentca Ve atn2 fod on bridge must be made to ‘ull of 5000 bs, Find the diameter assuming a factor of safety of 5 and © osein Jn{ans) 0, 07310 . 61,800 bs (ans) 1. 66, s00Ibs stress area Se (nt) 42,600 () (0.75) (0.625) = 61,850 bs “what force PIs requred to punch 9 1/2 in fe on 23/8 in thick plate ifthe ultimate shear th on the plate & 42000 ps? (ME Bo. Oct 24940 }bs 24, 620Ibs 24960 Ib¢ 0. 24,740 (ans) Stress «Area wx rt 2,000 (x) (8/2) 78) 4,740 Ibs 42.5 in ameter by 2 eng journal bearing {sto carry» 5500 Io load at 3600 cpm using SAE 40 use oil at 200°F through 2 single hole at 25 psi compute the bearing pressure. (ME Bd, Oct 5) ‘2100p! (ans) C1000 pst 8. 5C0 psi D. 950ps = SB = soaps: 5. Aloucnal bearing with diameter of 76.2 mm s subjected to 2 land of 4900 N while rotating at 200 rpm. If its cooficient of fretion ts 002 and the L/D = 25, find its projected area in. mm! (Me 8 Apr 96) A 12,080, 13,050 8 18165, D. 14,516 ans) Solution: y= 2s L= 025) L = 7622) = 1905 mm Area = DxL = 76.2%1905 = 14516 mm? 6. What modulus of elasticity in tension is Tequired to obtain a unit deformation of 0.00105 tim from 3 toad producing a unit tonsite stress 0f 49,000 psi? (ME Bd. Oct 95) [A 42.300x 10" psi C. 43.101 10 psi a1 207108 psi_D. 41.905. 10 oi (ans) Solution ee a SAMO 8 43.905 10% psi 7. The shaft whase torque varies fram 2000 to {500 n-ts fas Anche inclameter and 60,000, pl veld strength, Compute for the shaft mean average stress. (ME Bd Apr 97) ‘A 6036 psifans) C S162pst 8, 8810s) B. S550ps1 Solution $1200 = 9954 5) sl «2054 Spin = Ee = HO oi st Sun = SSS = SIE 35 pt How many Sf inc les an be pune tne oben Waste pie nade of He 1010 Seah 1/26 ch tek wing» force of 5 ‘Ths unmate song or shew 016 a Haciret sate. (ME 8. Apr 7)no, othoes = si = 5.22 sey5 holes 9. Determine the minimum mean diameter of « taper pin for use to fx 3 lever to a shat, fit sto ‘transmit a maximum torque of 700 ins. The shaft diameter Is 2 inches and the material allowable stesss 15000 psi. Use factor of safety ‘of 2. (ME Bd. 0ct97), 700mm. 5:2mm 8 7.2mm D. 62mm (ans) Solution Torque = Fr orita © tee Ber uot 15000 = 72 0 = a26sin = 519mm 20, A link has load factor of 0.8, the surface factor is 0.92 and the endurance stength is 28000 psi. Compute the alternating stress ofthe link fits subjected toa reversing lad, Assume a factor of safety of 3. (ME 84. 0297), A, 8350 ©, 9333/ans) 8. 10920 . 7260 Solution: stress Bott. 9535.33 pst 11. The diameter of a brass rod Is 6 mm. What force wil stretch it by 0.28 of length. Ese = 9x10" Pa | 5080. (ans) _ 9050N 8, 5060 . 6050, Solution: ae Solution 2002 = aera F = 5089.4 Tensile strength 51500 ps! 42, Ifthe uitimate shear strength of stee pate 542,000 ibn’, wnat force is necessary to punch 42.075 Inch clameter hole in 2 0.625 in thick plate? [ME 6d. Oct 2006) 6.101325 tm = 51,500 psi x 1a pe 54.98 Nien? ‘63,00 bs eternine the extension ofthe wit 8 68,0601bs 0, Apr 2008) 438. The yield strength ofa stivtural see 61850 bs (ANS) Acamm — C. 04mm members 38,000 fi’. The tense stress is 65,0015 O3mmlans) 0. 0.2mm E240 off, (ME Bd, Aye 2008) Soliton ‘A, Whats the equivalent of tho yield strength in : wpa? Force = stress xarea Aasomea ——¢. 100 Ma = Ss(nat) 8.35020. 250.MPa(ans) 4 42,000 (0.750.625) = 61,850 lbs Solution: 101325 APa 43, Compute the induced/compressve stress, in IF tere ean cones altn8 W uray SURSEOOPI rare Pa, of 2 steel solid shafting of SO mm diameter ee Co ineed and 800 mm in length that is subjected 10 an ave ceca e 248.10 Pa os i: placed atthe free end ofthe same Increase of temperature by 80 deg C. (ME Be, Coe races ‘su 2000) ges 8, Thettensle stress can be expressed as ‘A, 136,530 kPa (ANS) . 83 pa © 57 Mpa (ans) Ait eee quslto: (ME B6.Apr 2004) sabi mes Seo Aas ©. 8/3 . 162,256 kPa P Se . 112,187 kPa Selution 101325 kPa Pa Solution puto: S.= 8240 psix TAT pai To00KPa B fox 2 defection of uniformly loaded 1 induced thermal stress = cE(T;—Ts! 96.79 Pa cantilever Bam hea oofentof thermal expansion of steel Es The fctor of safety n tension (Englsh sites 610° = 2 siptn srmreternineend alo, Sa modulus of elasticity of steel = 30x 10* psi % A743 ee ; wat 3. 495 . 8.37 fans) tetss aoe) = seer HEL note) Pe ‘3 Solution: S = 6610°(300 109 248) « 28512 + 23512 si (IO1S25KFa) _ ase szoirn Hast) 14, A rectangular metal bar has a width of 10 J, # carton stoo! having Brine Harchess ‘mm and can support maximum compressive linker 100 should have ultimate tensle . Whats the computed safety facto of tension stress of 20 Mpa determine the minimum Brg loser to (ME Ba Apr 2004) sng St system of calculation? breadth of the bar loaded with a force of 3 kN f 1C00 N/mm C100 N/mm? A 349 ©. 495 IME Bd. Apr 2004) “BON jans) 0, 200N/mm? 8 4.38|ans} 0 939 A 1Smm(ans)— 6 13mm 8. 21mm D.BmmSolution E. ifthe tensile stress is only 7,200 loin? what willbe the new factar of safety in tension? Ag © 5 ans) 8.10 07 Soluton: 2 5s pox 35:00 720 19. A.Linch diameter S0festel roe carries a tense load of 15,000 bf. The elongation is (0.258 , the modulus of elasticity is 2.9 x 10” Ibffin’, (WE Bd. Ape 2008), ‘A. The unstretchod length of the rod can be computed using the equation ‘A none ofthese C. Lo =BEA/F (ans) 8 io=e/t Solution: Ft Elongation, = MY 46 4 oe 8. The computed unstretched lengths. A 9122391. 239933 n(ons) 3132329, 32993in soto 1 4 Ab ono? a oas{"\iPbou0!) 7 aSoME = Le239913% C. The totalength ofthe rod willbe A Y70042in —C 240,071 in (ans) 8 @0.7I0in 0. 730242in Solution abey+l = 0158 + 239.913 = 240,071 in ©. Ifthe modulus of elastcty is 200 Gpa, the unstretched length af the rod will became A. 8155m © 459913 m B.6097m(ans) D. 501.244m Solution: 147 pw = 200x10%kpae 7B O33 pa = 28,015,544 psi BE r 154 5 JOPbo orsstaciot) oe 153000 L = 2400410 284.0 in “TOO cm L= 240,04 in 8097 m E. The new computed length of rod will be A 5.65 in ©, 240.2960 (ans) 8. 688.025in —D. a565in Solution: aleyot =0.0040132 +5097 = 6.101 m = 240.196 in 20. What isthe decrease in lateral dimension due to ang axial tensile force of $0,000 Ibs in 2 bar 3 inches on each side and 6 feet long? the material is steel (€=3.0x 10" psi,=03). (ME Bd, Oct. 2003) [A 5.56x10" inch 8, 185% 10" inch ans) © 234% 10° inch D. 167% 10* inch Solution stress +555 voi = 8 Strain = 1.85185 x 10° nin solution Mods of East, € = 100 5 M78 ee? 1035 he ann 4, 358,772p8 22, Determine the maximum deflection (in mm) ‘fa $0 x 249 mm rectangular aluminum beam 6 rmters long subjected toa load acting downward at midpoint of 800 N. (Figure 6 attached sheet 4/3), (MEB6. Ap 2003) 8. Whats the unit deformation irauced? A 53.50 eed Dovoaa fans). 0.00041 8, 303 ans) . 0.4100 , o.8ico, : solution Solutio a Uniesetoaaton,at= > @ (00510247 36.6 10% mt set Peay £ E& 30x 10" psi 206,786 MPa 0008125 n/n i 261-3) TG wrists modulus or easucny? FA 23590000 al (on) € 54,9600 05 LO ene ee 6206, 7862108 [5.762107 ne Ys 001S112m «54 mm 00 4. th Dor = 24,691,358 psi D, ifthe load applied is 5,000 ke, what would be 23, Determine the polar section madulus, 2p E [i') of shaft delivering 10 Hp at 1S0rem. The diameter ofthe shalt 4° © and alowsble 4 Shear sre of 6,000 psi, (ME Bd Apr 2008) a as8Gn) C70 2, S516 on a eas 025 wy a. 0 ker ee } rs F What woul be the now modus of Aasticy ro “fot pple oad of 5.0002 Penis 4 A. 54321,000psifas) © 92340,000 ps 6 8 21,300,000 pi 24,900,000 ps124 Compute forth polar section modulus of a” hollow shaft with an internal iameter of 2°03 and OD of 3"2. The hatte 48” long. (ME Bd ‘pr 2003) 35 a7 B.a75\ans) 0, 425 Solution 25, Formula for maximum shear of beam as shown on Figure 8 on the attached sheet 1/3, (0m 8a, apr 2003), A ae ce ioe a ‘Therefore, the maximum shear P/2. 26, Find the polar section medulus in cubic inch ‘ofa shaft with diameter 3 inches. (ME Bd. ct 2002), A 265(ns) —c, 445 3355 2258 Solution ab? _ 2G) 2p = 7 2 A va he? 26. Astee train raise 400 meters long, in March tis at-30°C and 40°C in July, What isthe change in length in mm? K for steel =11.7 30" mic, (ME Bd. Oct 2002) Aad mm ©. 328mm (ans) 5. 303mm . 503 mm Solution = LAT = (23.7x10°9(400)40- #03275 m =327.6mm 0) 27. Find the polar moment of inertia of square 31" bya", A 20int 8. 25in* (ans) © 3010 5, 38int Solution: For square section with side 2 J=25io¢ 28, Compute for the loadin kt on 3 em ameter, 100. long rod its maximum ‘longation exceed 0.12 cm. (ME Bd, Apr 2002) AW78(ans) 148 8. 196 D. 2R7 solution 5=0,12em=0,0012 m t= 100em=1m Loum a= 710.03) 000707 ms? S for stool 10% 10" ps = 30x 10% 101825 207 10° koa WT A aE Fi) ‘TlanerOR-207,000000) Fea76kN 0012 = 29, find the polar section modulus af ahollow shaft with OD = 6" and ID3", (MEd py 2002) A2897in? C4545 in? 8. 39:76in' (ans) 0, 51.98in* Solution: 0185 in wast sth Shpe: phi= 25, isin OMin(ans} —¢ jn steel 27/8 inches in i length supporting divest 0.0185 fans) 0, 0.0166 Hor steel = 30% 10° psi (60.000)5 12) Foro he equation of te defection in | Determine the defection of 4 Binches be Leam in Figure No ifload P= 30001 ana IME 8d, Oc 2002) oidin Badin ol steel shalt toad L=25(22) =-3001n, eons (eM Lig int 000300)? <9(303108 17067) 0.109 in ‘34, Determine theloadin kN on 225 mm ‘lameter 11200 mm long steel shaft ts maximum alnngation nxcned 1m (ome 64. oct 2000) A 85 (ans) € 103 5, 128 D. 93 Soluvon: He ae é for steel» 30,000,000 ps = 206,786 x10" kPa i 20001 * eoasPor boars 8) 45. Compute for th induced/compressive stress Kpa, ofa steel slid shafting of 50mm ‘ameter and 890 mm in lenet that is subjected toan increase of temperature by S0Geg, Use for stzel = 0.0000139 m/m*C. {E= 30,000,000 psi= 206,768 Mpa) (ME Bd Oct 2000) A, 181,816 ©. 218,182 8. 242816 1, 196,880 (ens) solution stress =Ke0¢ .0000119|(208 785. 10/80)36. Compute forthe polar section modulus of an ‘SRE 1060 shafting having a dlameter of 3 inches. Use afactor of safety of2 and design stress at 000 psi. (ME Ba. Oct 2000) AAT G42 acre 1. 53 (ans) Solution 837. Maximum moment formula for beam as shown on Figure 1 on the attached Sheet 1/2, (oe Bd, oct 2000), ‘A PLans) ae Reactions: Shear . Moment: End Slope: 38, A rizontal beam 16 e longi subjected to.@ load oF $00 Ib located ta its center The ‘dimension ofthe beam is 2 4 inches Fespectivly and its unit weights 100 bf. Find its fleural stress (ME 8d, Apr 1999) ‘A 11,696.34 Ibfin? ans) 15,677.2 yin? 8, 10.2332 ibn? , 15,388. bj? oem = Meson = 10500 8600 25006) - 5200. “520% oment = #88 250(8) «5.200 200 Fb = 62400 = dstance from the farthest fiber tothe neutral aK =2 in on? BB fora rectangular 70s section 1160634 thin? 39, A stool rod 30 mmin diameter anc 860 mm longhas an sllowable elongation not to exceed 1.5 mm, find the allowable load in kN (ME Ba Apr 1999) A278 (as) 8. 208 © 316N , 2366 mB an[ients4)o oro.) ke isnot ven, use €= 20% 0s for ste! = 30.20% psi= 206.785 Mipa (0800) © (osoP bo re5u0°) Wat isthe bending momnentin feb atthe ied end of a 20 truss with a uniform weight 200 and a concent‘ated vertical load at eee end of 1200 ibs? (ME Bd, Oct 1998) A 35,200 c 1500 1810 1. 17,000 fb ans) lution: fending Woment + Laonoio) =1200 (20) + Loot 0x10) = 17,000 fb 2, A rectangular meta! bar has 2 width of 10 I> and can support a madmum compressive yess of 20 Mpa; determine the minimum {Me 81. ep1 2002) A. iSmm(ans) ©. 13mm | 2mm ©. 13mm |A constant force of 150 Nl & applied eitialy to & wheel of diameter 140 mm. ening the work done, In joules, in 12ENGINEERING MATERIALS ne Important Properties: iitleness - tendency to fracture without “sppreciale deformation. Bucatry BBirmanent Awrson. thet property. that_permits deformation belore fracture in Hasvcty ~ ability of a material to-be deformed fn to return to its original shape. Hardness - resistance to indentation. Machinaaiiey = material can be cut relative ease with which 2 Imatteabitey susceptillty to extreme ‘deformation in roling ané hammering Plasticity = ability of a metal 0 be deformed ‘onsiderabiy without rupture fines « abily to resist deformation ‘Toughness - ability to withstand shock toad without breaking, Important Facts about Engineering Materials: Stoel is the post prevalent enaingering metel due ta the abundance of iran or, low cost, simpity of prouction and bigh performance. [Metals ae the most commonly used materials in entineering dasign Metallurgy le the study that encompasses the procutement and production of metals Estractivesmetallrgy is the study which cGvers the refinement of pure metals from their ores. {com (Fa) is obtained from oxides (F203) ‘the earth and suoge mixed with the ron ones 48 - 8 ype Blast furnace isthe place in which the process to ‘edlce ron oxides to pure ron Coke is nal that hasbeen previously burned. Bessemer and oxygen processes are used to reduce the carbon content ana eurify the iron Classifications of Pain carbon steels: '. Lov carbon steel (less than 0.3% carbon) ore used for wire, stuctural shapes, and. strow machine parts, 8B, Megium-carbon steel (0.30 ~ 0.70% carbon] ‘re used for axles, gears and similar parts Fequicing medium to high hardness and high stronath . High-carbon steel (0.70 ~ 1.8% carbon} are ‘sed for dri, crting tools, and knives Lowealloy steals are stoels containing lass than 8% total alloying elements and have higher sMrength of plain carkon. Types of lowaloy stoels are structural steel, highsstrensth stel and ultra high strength steel. Highsalloy steels contain more tan sx rot alloying elements, Stainless Steels a stel formed by the ation of ehremium, Cassicatinsof tales steels: fs Ferile stiles steels contain more than 12 zr percent chromium and contan ro ickel {par ofthe NI 400 sees). Tune blades te mmarofectured from fer stains tess Martens stainless steels are heat eaable and contain no nickel and primary in Bgher farbon contents (pat of the IS 400 sens) surgieal insuments are. manufactured from imate stainless tel Austenitic stanes steal, contains 22% tickles an ooyingslamont ang eSmfony used fo genenl covraake farses Base ompaston 1s 8 sce 340 2% sremium oF Austenitic stangay sees 76 ‘weldable, non-magnetic, hardenable and tars be Dalshed ta irr fh.SSS ‘BY JAS TORDILLC Heat treatment Pract: Steet Pin carbon fonesng = eats shoe the rantomatan "Caton oe range ssvaly1900%0 2350, andcanmngaowy tenga i to soften the matal and increase ease in Goroh aa machining Nickel ae Nike hromium 3 Hardening ~ heating above the transformation "Heat and conron ris 030 temperature and qvenchingusually in ol, Molyodenure Rios oe fortrepupose of ceasngtnehadnes” Ye fn Molybdenum-chomiumnic amaling «testing to some 10% aboetheMhosensmaner nm 3H utansformation range with subsequent Molybdenum:-chremium nickel mm cooling to below that range In stl air at room Mohbdonom anit ton femperatwete oe unlrm stucure Che S00 ofthe metal Hest and corrosion sian Stress aleving -__heatng o, © suberal —chremnevanadum one temperature, about 1300 to 1300*F and foktng _Nitalchronue, 2 tat temperature fers suitable time forthe Shconmarganese wm BO purpose ofreduenginiral residual sveses” — Nesalehoniormaipdenum 2 Tempering - reheating toa temperature below gy the transformation range, folwed by any Commonly Used Metal desired rate of cooling to attain the desired Properties of the metal Case Hardening - process of hardening the surface or case of @ metal to provide a hard, wear resistant surface wile retaining toughnoss inthe core ‘Metal Forming Processes: Rolling - process of forming metal parts by the use of dias after themetalis heated to its plasticrange Forping - process of forming metal parts by the use of powerful pressure from a hammer or press to obtain the desired shape, after the ‘metal has been heated to its plastic range ‘Alsland SAE Designation of sect AISIY 1000 SAE 000 Y isa letter, used in AISI only, to Indicate the method of manufacturing; first number (or fst two numbers) reprosents class of steel; second rhumber indicates the approximate porcentage of the principal alloying element; last two numbers indicate 100 times the approximate percentage of carbon present in the metal, Meta: wroughan season ton forme by tanmetg an rolling operations : Uses Rivets, welded steam and vate pes ‘Meta: Cast ron Description: ran formed by casting Uses: Cylinder blocks, brake drums, gears, machine toc! ways i Metal: Malleable ron Description: Heat treated cast iron which is Strong. ductile and easly machined Uses: Gears ‘Metal: Nodular cas ron Description: Cast iron added with magnesium ardcerim to become stronger and more cute ses Casings crankhats, hubs, ol, toring Metal: cast Stee! Description: Steel formed by casting Uses: Gears, crankshats, cylinder barrels Metal: Wrought Steel Description: Steel formed by hammering, roling ike | 6 rolling Uses: ars, tubes re CNM Metal: Stantess Steet [ecrintion: Steel obtained by adtion of Eitomim Ties: Steam turbine bades, valves Wetal: Brass Description: Alloy of copper and zinc Uses: Propeller shafts, piston rods, screws, ete Metak: Bronze escrintion: Alloy phosphorous ses: Clutch disks, pump rods, shafts, vave of copper, tin and Tabulated Propertios of Materials Tables of efferent materials show the following ukimate strength, yield Important properties: BHN, modulus. of stress, endurance. limit elasticity, elongation, density. Tabies in Fares ‘Tables AT4- ATL, pp. 68-582 (Append). ‘Tables in Valance: Table 2x4 p, 25, Table 2-5, p.27, Table 26, p. 30, Classitieation of Alloy tects: Low-alloy structural steels - were developed for structural uses where light weight is important Low-carbon alloy steels - (0.10-0.25%C] used chiety fer ear burn Mecium-carbon allay stats (0.25 - 050% ©) usually quenched ng: tempered. to hardness between 250 and 400 Snel High-carbon alloy steels. - (0.50 - 0.70% C or more) ordinarily heat treated to hardnessos between 475 and 500 Srinel for use as springs wear resisting parts, ete. High-alay steels - Such as talnles steals Functions of Alloying Elements (chemical symbol in pprenthesis ‘Aluminum {8\) + 1: an effeient deoxidizer, an alloy in nitriding. steels (citralloys), and) it promotes fine grain sie, Boron (8) - in very small amounts (0.001% or less] san economical hardenabilty agent in low ‘or mlm carbon deoxysized stevls Chromium (Cr) - improves. hardenablty ‘economical, resistance to corrosion [with other ‘lloys), strength at high temperatures, and weering properties (high carbon), Cobalt (Co) ~ improves re hardness Columbium (Co) ~ is often used to stablize Stalless steal Copper (Cul - improves steels resistance to atmospheric corasion ‘ead (Pb) - improves machinablity, but affects different alloys differently ‘Manganese (Nic) ~ iryproves strength and inereases hardenability moderately, counteracts Dyitvleness from sulfur Molybdenum (Vo) - increases hardenabiity markedly and economically, tends to counteract temper bitleness, improving creep strength and fed hardness; it improves wear by forming brasion-resetant parties ‘Mickel (Ni) ~ strengthens unquenched. and ‘annealed steels, toughens steel (especial at ow ‘temperatures, and simplifies heat treatment by lessening distortion. It is the most element for reducing the brittleness of steel at very tow temperature Phosphorous (P} - Increases harcenablity, strengthens low carbon steels, improves Imachinablity of fee cutting steels, and improves Fesistance to corrosion, Selenium (Se) - improves machinabilty of Sfalless steel also added to leaded resulfuized tarbon steels forthe same purpose Silicon (S) ~ strengthens low alloy steels and improves resistance to. high temperature ‘oxidation, It Is 2 good general. purpose ‘eoxicizar andl promotes ne gain Tantalum (Fa) - isa stabilizer. Tranlum (7) ~ ' used for deoxition and tor stabllting austentc stainless steels; it increases the haréness and strength of low carbon steel and improves creep strength, Tungsten (W) Inereases_hardenabilty ‘markedly in small amounts and. improves hardness and strength at high temperature Vanadium (v) ~ promotes fine grain structure, Improves the ratio of endurance strength to tUtimate strength of medium carbon steels, Increases hardenablity strongly when clisolved, ‘and results in retention of strength and hardiness {3t high temperature; it Is the most effective ‘clement in etarding softening during temperingUses of Alloy Steels: 1S} 2330: bolts, studs, {tsonal stresses ‘NSI2340:quenched and tempered shattrg, connecting. ods, very highly stressed. bot, forgings ‘lS! 2350: high capacity gears, shafts, heavy duty machine parts ‘Alsi 3130; shafts, bolt, steringknucklos, ‘ANs| 3240: alrcraft and truck engine crakshats, ol wel to! joints, spline shafts, axles, earth moving ‘equiament ‘SI 3150: wear resisting parts In excavating and farm machinery, gears, forings, ‘AISI 3240: shafts, highly stressed pins and keys, ‘ears AISI 3300 seres:_ for heavy pats requitng deep Penetration of the heat treatment and high Fatigue strength per unt weight 1514053: leaf and col springs ‘ANS| 4130, 4149: automotive connecting rods and anes, aircrat parts and tubing, {IS 4340: crankshats, axles, gears, landing gear Parts; perhaps the best general purpose AIS steel ‘AS14640: gears, splined shafts, hand tools, tmiscllaneanus heavy duty machine parts ‘AISI 8530: connecting. rods, boits, shapes; air hardens after welding, Aisi 8640, 3740: knuches, shapes. tubing subjected to 82315, propeler shats, Definitions: ‘Age Hardening ~ Is 2 change in a metal by which its structure recovers from an unstable or ‘metastable condition that has been produced by {quenching or cold working Alley + is a substance with metalic properties, composed of two or more elements of which at leastone isa meta Alfoying elements - in steel are usually considered to be the metalic elements added for the purpose of modifying the propertics, ‘Anisotropy ~ isthe characteristic of exhibiting of exhibiting different properties when tested in diferent directions (as tensile strength “withthe Aral” or “arossthe grain” Snitieness - isthe tendency ta fracture without appreciable deformation Charpy test ~ Is one in which a specimen, supported at both ends as a simple beam, is ‘broken by the impact of falling pendulum, Cold shortness - is britleness of metals ‘ordinary oF lw temperatures, old working - is the process of deforming metal plastically at a temperature. below recrysalizaton temperature and at 9 rate Produce strain hardening. to absorb or damp vibrations, winch a pa of absorbing kinetic energy of vibration owing hysteresis Becarburzation - i 2 oss of carbon fom th Surface of tee, ocauing. during nt olin forang, and heat treating when te surround ‘meclam reacts withthe carbon (as oxygen Carbon combining). Ductity. ~ is that property thet permit permanent deformation before. fracture tension Etsticty - the ability of » material to b deformed and to retur tothe oral shape, Embritiement ~ invelues the loss of duct because of a physicel or chemical change ofthe materia Free carbon - i that part of the carbon content of steel or ion that iin the form of graphite or temper carbon. ‘ard awn ~ is a temper produced ina vite, fod, or tubs by cold drawing. Homogenovs materials - (have homogenety) fave the same structure tal pont. Isotropic ~ materials have the same properties inal direction. nod test ~ is test uieh a specnen,supoorted atone end as a cantlever bea, Is broken by the Impact of fating pend Kile stee = steel that hasbeen deoxened witha strong deoxiiing agent, such 3 slizon or aluminum, in order to eliminate @.reacton between “the carbon and oxygen during solifcation ‘Mochinablty - is a somewhat indefinite Broperty that refers to the relative case with which a materal canbe cut Malleablty - isa materials sucoptbity to ‘xtreme deformation incoing or hammering, Mechonicl properties ~ are those thet have to da ith stress and tran, Percentoge elongation ~ ithe extension inthe vienty of the facture of a tense specinen ceapresied asa percentage of the ciel gage length. Percentage reduction oreo - ithe sales area atthe pont of rupture of 9 tense specimon ‘vided by the eign aro. EERINNG MATERIALS ska! properties exclade_ mechan! Properties and se othr phys! propertes uch 8 deny, condi, cathe of Disty i te aly of met t0 be forme cre who ap sonra «is thera of eter ty Ip iv lot ssin when the deen psi wih ongtcinal tore Precio eat raiment bing: boat the Irecotaton sf a. coettcmt fom = fuvesturatedsokd soon by holding the fovy at an elveed temperature a0 exe sreal agin Proof sve. ~ i tat sree which eases @ Speced permanent deformation ofa material, Vsunly gtr lest Med shortness ~ isa britleness steal vn ised foecton -_ss0ated wih cep, the eceaing ses at a eontat sta, importa fovea nigh temperate serve, reua sts ~ are these not dt pple oa or temptature alent, famed see! = ls incompletely deoddees ‘tea Seltion heat treatment 1 te proce of holding analy ota suaby hi temperature long eneugh to permit one of mare onstuens te patent se soon and then aang ot coo to. hed the contents ao ¢ Siperauted soko, Stifjess «ste sy to est deformation. Strain hr deoing =i icreding the hares and strength by plate. dformaion a temperatures lower than the recrtallaton range. Temper + (s 2 candtion produced in a nor: fercous metal by mechanical or thermal Toughness ~ is the capacity of 8 materiel to witstandashockload tho breaking Tronsverse strength » eters othe rents of 2 tranwvrse bend test, the specimen. bong trountad a 9 simple Beam ato called rupture re ‘wort andeing ea Wout ste! - Is steel that has been hammered le, or drawn i the proces of tmonfactures R may be pli chon of sl9/ see |e the same as strain ‘connosion: Gotrosion is the natural doterioration of 2 metal ‘which metalic atoms leave the metal or form compounds in the presence of water oF gases ‘General corrosion may be minimized by the use of corrosion vesstant materials and the adetion ‘of protective coatings an linoe. Galvanic corrosion occurs when dissimilar metals feist at different lectical potentials in the presence of an electiolte, Galvanic corrosion may be reduced by the careful desien and selection of materials regarding dssiilar metals and the use of serial anodes Localized corrosion can be especially damaging In the presence of ther destructive forces such 26 stress, fatigue, and ather forms of chemical attack Siresscoroson cocking occurs ot srain teundares under ens se. propagetes Stes opens cots ht ae sje carson ctimetey wenkeing the mea oni faire Etectverneans of ediengSct are} proper tesgn 2) redung ses 9] remuing earsve Neer a aang os of te 04 yada on concerto Chloride stress corrosion occurs in austnitic sHoisless steels under tensile stress in. the presence of oxygen, chloride fons, and high fempecature. ff controlled by the removal of ‘oxygen and chloride ons in the environment and ‘the use af low carbon steelsSAMPLE PROBLEMS: 1. ‘carbon steel having Brinell Hardness number 100 should have ultimate tensile strength closer to. (ME 8¢, Apr 2004) 2. 1000 N/mm? © 100N/mm* 8. 350/N/mm! (ans) . 200 N/m? Solution: Tensile strength =BHNxS15 au ae 1,500 pix ACIDS Ni Ta pat 54.98 N/mm? HIN-WALL RESSURE VESSELS inition: thin-well pressure vassl is one in hich the ratio ofthe wal thickness tothe inside eter i less than 0.07, £ < 007 Wall cylinder: Circumferential or Tangential Stress {hoop Ingermal pressure = inside diameter = wall thickness 5, = tangential (tensile strest = inner radius Wen there a seam or joint, the joint efficiency mst be considered Be ee where Joint efficiency. 1. Longitudinal stress i= 2 = [Considering the Joint efficiency =f “Thinwall Sphere ‘Gonsisering the joint efficiency E e- THICK-WALL CYLINDERS Definition: thick-wal evlinder is one in which the ratio of the wall thickness to the iside ‘ameter i more than 0.07 t > oar Lame's Equation ( Vallance p. 451), for internal pressure: all thickness Inside diameter tangential or tense stress internal pressure ‘when subject to internal and extemal pressure Maximum tangential stress a the inside: ete [Maximum tangential stress a the outside y= Het) where internal pressure eternal pessure inside rads ‘uta radiusSAMPLE PROBLEMS: 4, Determine the bursting steam pressure of a steel shell with diameter of 20 inenes and made ‘of 1/4 thick steel plate. Tho joint efficiency is at 70% and the tensile strength is 60 ksl. (ME Bo. 02197), ‘A 4200 ps ans} B. 105ksi © 42841 ©. 8500s, Solution: 5-22 * ta 60.000 = Ta 35K70) P = «2008, 2 Determine the safe wall thickness of a 30 inches steel tank with internal pressure of 7.82 MPa. The veld stress of material is at 275.48 ‘MPa. The factor of safety to use is 2.0. (ME Ba, (oct 97). A 3/4 inch 5 23.6em ©. 21.6 mm (ans) , 5/Binch Solution: oa Base 2 a t * 08516in = 216mm 3A cylindrical tank with 20 inches inside ameter contains oxygen gas at 2500 ps. Calculate the required wall thickness in {nm} under stress of 28,000 ps. (ME Bd. Oct 97), 8. 1134mm(ans)_C 1244mm 3B 1028mm 0. 10.54mm Solution eg 28000 sae t= 0.4a64in © 11.33 mm 4, Determine the thickness ofa steel air recover With 30 inches clameter and pressureload of 120, design stres of 8000 psi, [ME Bd. Oct 95). A W/ain(ans)—. 3/8in 85/8 Dapin Solution: 52 8000 = 29 {= 0.225in: Use 1/49 standard thickness. 5. A steel evlndical air receiver with 5 f lometer and pressure load of 280 pl, des stress of 9500 ps maximum. The pressure Js to be provided with 1.1/2 in diameter vole installed at the bottom of the vessel a safety pressure relief valve installed ether at {top most or at the side with pop-out rating 200 psi. Assume a 100% weld joint eficien The lap welding tensile strength is 65,000 Determine the bursting pressure of this A 1IS4 psi 3 1056 ps © 1354 pst (ans) , 1256 psi Solution: Solving forthe wall thickness ofthe aie recewver The standard plate thickness Is 5/8 in (0.625 ks would be sefe, solving for tne bursting aressut using a wall thiekness of 5/8 in 0 Hons) P= 1354p) ©The work oyinder of a hydraulic system is acted by a hydraulic pressure of 370 psi whi the maximum load ofthe piston is 5500 Ibs, Hh allowable terse sWvess 1s 2000 pst, what is He cequired wal thickness ofthe eying. A 0.402 in © 0.200 H0.408:n(ans) 0. 0.284 in Force = Pressurex aren 500 = 270x (w/8}0" De 435in [Assume thicewal evinger: f= 0.408in W/o = 0.409/4.35 = 0108 ‘Therefore, the evlindr isa thick-wal cylinder. 7. esign a vertical steel elinrial water tank 30 my ioediameter anda m High Culley the prossure variation in the water to fellow 2 HHraight len from the edge of the top to the enter of the bottom and the allowable stress of the steel plate 120 MPa, vuithout einfrcing Angle bacs and. rods, find the thicknass of tho see! pate. (ME Bd Cet 2006) 1 55mm (ANS} B. 75mm © 85mm 6. 25mm Solution: P = prossure = helghtxdensity SxO81 = 441.45 kPa 45130) f= 0.055 m = 58mm 4 azo00 = 4 8. A cylinder tank with 10 Inches inside diameter ontaine onygan gas at 2500 psi. Calculate the Foquired wall thickness ia (run) under stress of 98,000 psi. (ME Ba. Aor 2005) 1. 11.34.eym (ANS) 8. 1024mm © aaamm D. a0samm soliton 5-22 2s) 26,000 = 1 = 0.4868 in = 12.38 mm 9. what isthe longitudinal stress ina wate pe, which is B inches In diameter and 0.25 inches ‘thick if is carrying a fluid at a pressure of 200 psi? (ME Bd, Oct 2003), 1 2600 psi (ans) 8, 800ps) . 2400 psi D, 400 psi te wnstand an interna presse of 200 ps eth a den facto of 4 tned on Su. The ste es the stent eqilent of C1020 anne and the melded joints shoud have love sent (eceney) of 90s, (ME Bd Ot 2003) [AL IN Su for annealed C1020 steel is 57 ksi, ‘compute forst 1, 28,000 ps 8, 42/510 $2,400 951 1D, 14,250 ans) Solution: cee is S700 ps (ee 0 pst B, Determine the suitable plate thickness. A 7i64in © s/a2in 8, 9/82in{ans) 0, 3/641n 200630, 2090). 28 nc use 932 inch thick plate 34250=Compute the stress ona dismetral section, A 2400ps) ©. 6 400psi fans) 8.3240psi —D. L640psi 2086) 5, 5) 1. Compare the longltutinal stress, 8 128,000.si-C. 800,200 psi 8. 821,005). 12,860 ps (ans) 00 Solution ste PD a sex 20036) 240.28) 51= 1285714 ps If Japanese engineer would ask you, how ‘many centimeter isthe minimum plate thickness required? A.D75cm(ans) C017 em 8, 0.27 em B, 05cm =9/32 n= 0.714 cm = use 075 em thick nate 41, Find the thickness of a metal needed is the semispherical end of cylindrical vessel 070 meter in diameter subjected to an internal Pressure of 2.7 Nimm?. The material is mild carbon steel and tensile stress s 63 N/mm? [ME Bd, Oct 2000) A52. 86.02 783 ©. 6.88 fans) Solution PD 5.2 gg 20 1'=0.00684 m= 5.84 mm 12, A oylindsical tank with 10 inches inside ameter contain oxygen gas at 2500 ps Celeuiste the recuired wall thickness in millimeter under stress of 26000 psi, (ME 8d. (oct 2000) A 10se & 1028 B.1134(ans) 0, 12.44 A0375m ——-€ 7/8in oo «8. Xin (ans) ‘D. 0.815 in bs Neat B33.42 pt Suton Dp. 3 +2) 5-22 ro 2) a ae seen «50 ce 4 04 = 200618? + t=075in=% in pe 2c? = 20m? + 10°) nae BD ress vers 420446 in = 11.34mm on 42, oatermine te bursting prose of ‘ohercal tant wih date o 10 nes Make of neh et sel ple The 0 j--26in fenton eae sere ii (Me bd Ox 9) Razoo ¢ 24c0p rg 8 a00Dps| eat ans sour once: aes 285 a A at (13 + 1?)- 2p foes FO) id Joan * 0.23)" (0.70) Derg reste 4200p p= 10,000 psi 2,000 psi goods? +10?) = 200090872 44. Find the minimum thickness in inches steel sphere to contain 2000 ps. The diam of the sphere Is 54 in and the. allowa ‘maximum normal stece Ie 35,000 pei? (nie ct 98) 10? 45, Compute the safe wal thickness ofa 76.2 ameter steel tank. The tank is subjected 7.33 MPa pressure and the steel material Yield stess of 215.4 MPa, The factor of safety ses 3. (ME Bd. Apr 98) A Winches C409. B 3a9emians) 0. 3469em Solution 20 5-22SHAFTS fitons: jute rotating member transmitting power Dxle~ a stationary member earring rotating Pie's, pulleys, te Bpindle - a short shaft or axle on machines Machine shaft 2 shaft which Is an Integra! part tthe machine Transmission shaft = shaft whieh is used to vansmit power between the source sind the machine absorbing the power. Lneshate or mainshate = seven bythe pie mover, transmission shate Countershaf,jackshaft, headshaft shortshaft = ‘vonsmission shaft intermediate betvieen the lineshaft nd the riven machine. Prime Mover {Machine shaft Materials for Transmission Shafts hot-ole, forged carson ste cold role, : 1 4 3 (racans) 5, = 28 or sodcrar sat Sy = i265 (for hollow cicuar shat) riven Machine Commercial Sizes of Shafts, Inches (Fires: p. 269: Vallance: p. 181) oo ere: Sy tral ss see TF quer trial oteng Guanes ram natal anit te outesost fer ras oro cular hat) ete =o forsolé cela shaft 02-04 torhatow caters 0 = damier of sat Clg stngular deformation in length, radlans ‘modilus of iii in shear 1,500,000 sito 12,000,000 ps for steel ‘outside ameter D, inside diameter Stresses in Sold Circular Shaft Subject to Torsion and Bending Sim = SET Sina = 28 (04+ VET) were Scie rab seat ares Svan = maximum tense or compresve sess IM sberding moment T= foionsl moment Strength of Shaft with Assumed Allowable ‘Stresses (PSME CODE p18) For Main Power Transmitting Shatts: or tnesnans Carrying Pulleys For Small, Short Shafts: here: P = Pomer transmitted in HP 1 = Diameter of shaft in inches 1 = Speed in rpm Formals from Machinery Handbook Diameter of ha A fer alowate twist nat exensng 008 de per ‘lena D=021F o p= a6 \f where: = shaft diameter, inchs T = torque, ine He = horsepower N= speed, rom {nS uns (alowable twist 0.26 de per meter lene) b= 2264 or p=1257' wer: = shattcameter, mm = torque, enn = Power, kw N= seed pm . for allowable twist not exceeding 1 deg par 200 length D=01¥T or p=40°[t where: = shaft dlameter, Inches T= torque ind HP = horsepower N= speed, 1pm For shor, solid shaft subjected only to heawy transverse shear \V = maximum transverse sheating loads bs 5, = maximum torsional shearing sees, lin? Lnear Deflection of hating For steel lineshaftng, it Is considered good Practice to limit the linear deflection to 2 ‘maximum of 0.010 inch per foot of length Maximum Distance: ‘A. For shaftng subjected to no bending action ‘except ts own weight L= 895 ,VD7 8. For sheftng subjected to bending action of pulleys, ete b= 52808 here J fnaximum distance betweon bearings ft thameter of shat, inchs Ne Pulloys should be placed as close to the ings 35 possible. In general, shatting up to threo inches in Jnmoter is almost always mado fromeold-oled 0