MAHALAKSHMI ENGINEERING COLLEGE

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This intrnal rsistanc offrd b th bod pr unit ara is calld th strss inducd in th bod.. Dfin principal plans and principal strss.

1 MAHALAKSHMI ENGINEERING COLLEGE TIRUCHIRAPALLI - 6. QUESTION WITH ANSWERS DEPARTMENT : CIVIL SEMESTER: V SUB.CODE/ NAME: CE 5 / Strngth of Matrials UNIT 4 STATE OF STRESS IN THREE DIMESIONS PART - A ( arks). Dfin strss Whn a crtain sst of xtrnal forcs acts on a bod thn th bod offrs rsistanc to ths forcs. This intrnal rsistanc offrd b th bod pr unit ara is calld th strss inducd in th bod.. Dfin principal plans and principal strss.(auc Nov/Dc ) (AUC Apr/Ma ) (AUC Nov/Dc ) (AUC Apr/Ma ) Th plan in which th shar strss is zro is calld principal plans. Th plan which is indpndnt of shar strss is known as principal plan. Th noral strss acting on prinicipal plans is calld principal strss. Dfin sphrical tnsor. ii ij It is also known as hdrostatic strss tnsor x z is th an strss. 4. Dfin Dviator strss tnsor ij x x xz l l x z xz z z MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag

2 5. Dfin volutric strain. (AUC Nov/Dc ) It is dfind as th ratio btwn chang in volu and original volu of th bod and is dnotd b v v = Chang in volu Original volu v v 6. Stat th principal thoris of failur.. Maxiu principal strss thor. Maxiu shar strss (or) strss diffrnc thor. Strain nrg thor 4. Shar strain nrg thor 5. Maxiu principal strain thor 6. Mohr s Thor 7. Stat th Liitations of Maxiu principal strss thor. On a ild stl spcin whn spil tnsion tst is carrid out sliding occurs approxiatl 45 o to th axis of th spcin; this shows that th failur in this cas is du to axiu shar strss rathr than th dirct tnsil strss.. It has bn found that a atrial which is vn though wak in sipl coprssion t can sustain hdrostatic prssur for in xcss of th lastic liit in sipl coprssion. 8. Explain axiu principal strss thor. (AUC Nov/Dc ) According to this thor failur will occur whn th axiu principl tnsil strss ( ) in th coplx sst rachs th valu of th axiu strss at th lastic liit ( t ) in th sipl tnsion. 9. Dfin axiu shar strss thor This thor iplis that failur will occur whn th axiu shar strss axiu in th coplx sst rachs th valu of th axiu shar strss in sipl tnsion at lastic liit (i.) t l ax (or) t. Stat th liitations of axiu shar strss thor. i. Th thor dos not giv accurat rsults for th stat of strss of pur shar in which th axiu aount of shar is dvlopd (i.) Torsion tst. ii. Th thor dos not giv us clos rsults as found b xprints on ductil atrials. Howvr, it givs saf rsults.. Explain shar strain Enrg thor. This thor is also calld Distortion nrg Thor or Von Miss - Hnk Thor. MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag

3 According to this thor th lastic failur occurs whr th shar strain nrg pr unit volu in th strssd atrial rachs a valu qual to th shar strain nrg pr unit volu at th lastic liit point in th sipl tnsion tst.. Stat th liitations of Distortion nrg thor.. Th thor dos to agr th xprint rsults for th atrial for which at is quit diffrnt tc.. This thor is rgardd as on to which confor ost of th ductil atrial undr th action of various tps of loading.. Explain Maxiu principal strain thor Th thor stats that th failur of a atrial occurs whn th principal tnsil strain in th atrial rachs th strain at th lastic liit in sipl tnsion (or) whn th in iniu principal strain (i ) axiu principal coprssiv strain rachs th lastic liit in sipl coprssion. 4. Stat th Liitations in axiu principal strain thor i. Th thor ovrstiats th bhaviour of ductil atrials. ii. Th thor dos no fit wll with th xprintal rsults xcpt for brittl atrials for biaxial tnsion. 5. Stat th strss tnsor in Cartsian coponnts ' ij x x. x xz z xz z z 6. Explain th thr strss invariants. (AUC Nov/Dc ) (AUC Ma/Jun ) Th principal strsss ar th roots of th cubic quation, whr I I I I x z I z x z x z xz x xz z x I x Z x x z xz 7. Stat th two tps of strain nrg i. Strain nrg of distortion (shar strain nrg) ii. Strain nrg of dilatation. MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag

4 8. Explain Mohr s Thor Lt f Th nvloping curv f ust rprsnt in this abscissa and ordinats, th noral and sharing strsss in th plan of slip. Lt P p l 9. Stat th total strain nrg thor. Th total strain nrg of dforation is givn b U E v and strain nrg in sipl tnsion is U E. Stat th shar strain nrg pr unit volu s C whr C E MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 4

5 Explain th concpt of strss? Whn crtain sst of xtrnal forcs act on a bod thn th bod offrs rsistanc to ths forcs. This intrnal rsistanc offrd b th bod pr unit ara is calld th strss inducd in th bod. Th strss a b rsolvd into two coponnts. Th first on is th noral strss n, which is th prpndicular to th sction undr xaination and th scond on is th shar strss, which is oprating in th plan of th sction.. Stat th Thoris of failur. Th principal thoris ar:. Maxiu principal strss thor. Maxiu shar strss (or) strss diffrnc thor. Strain nrg thor 4. Shar strain nrg thor 5. Maxiu principal strain thor 6. Mohr s Thor. Dfin factor of saft. (AUC Nov/Dc ) Th ratio of ultiat strss to th working strss is known as factor of saft. Howvr cas of lastic atrial, it is takn as th ratio of ild strss or.% proof strss to working strss. PART - B (6 arks). For th stat of strss shown in fig. find th principal plan and principal strss. MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 5

6 . For th stat strss shown in fig. Find th principal plan and principal strss and axiu shar strss. (AUC Nov/Dc ) MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 6

7 MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 7

8 . Th rctangular strss coponnts of a point in thr dinsional strss sst ar dfind as a σ x =Mpa, σ =- 4Mpa, σ z = 8Mpa, x =4Mpa,, z= -6Mpa,, xz=mpa. Dtrin th principal strsss and principal plans. (AUC Apr/ Ma ) MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 8

9 MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 9

10 4. A shaft is subjctd to a axiu torqu of KN- and a axiu bnding ont of 8KN- at prpndicular sction. if th allowabl quivalnt strss in sipl is 6MN/, find th diatr of th shaft accoding to th axiu shar strss thor (AUC Nov/ Dc ) MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag

11 5. In a stl br, at a point th ajor principal strss is MN/ and th inor principl strss is coprssiv. if th tnsil ild point of th stl is 5MN/. Find th valu of th inor principal strss at which ilding will conc, according to ach of th following crtria of failur.. Maxiu shar strss. Maxiu total strain nrg. Maxiu shar strain nrg. tak passion ratio.6 (AUC Nov/ Dc ) MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag

12 6. A circular shaft has to tak a bnding ont of 9N/ and torqu 675 N. th strss at lastic liit of th atrial is 7 x 6 N / both in tnsion and coprssion. E = 7 x 6 KPa and µ =.5. Dtrin th diatr of th shaft using octahdral shar strss thor an th axiu shar strss thor. Factor of saft =. (AUC Apr/ Ma ) MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag

13 7. In a atrial, th principal strsss ar 5 KN/, 4 KN/, and - KN/,. calculat th total nrg, volutric strain nrg, shar strain nrg and factor saft on th total strain nrg critrion if th atrial ilds at N/. (AUC Nov/ Dc ) MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag

14 MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 4

15 8. A stl shaft is subjctd to an nd thrust producing a strss of 9Mpa and th axiu sharing strss on th surfac arising fro th torsion is 6Mpa. th ild point of th atrial in sipl in sipl tnsion was found to b Mpa. calculat th factor of saft of th shaft according to. Maxiu shar strss thor. Maxiu distortion nrg thor ( AUC Apr / Ma ) MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 5

16 9. At a point, th ajor principal strsss is N/ (tnsil), and th inor th principal strss is coprssiv. if th ild strss of stl is 5N/. Find th valu of inor principal strss at which ilding tak plac according to ach of th following thorizs of failur.. Maxiu shar strss thor. Maxiu principal strss thor ( AUC Apr / Ma ) MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 6

17 . Th strss coponnts at a point ar givn b th following arra Mpa 6 6 Calculat th principal strss and principal plans. Solution: Th principal strsss ar th roots of th cubic quation I I I () MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 7

18 whr, I x z I x z z x x z x z x z x z xz z x I x z xz ar thr strss invariants Th strss tnsor x. x xz ij x z zx z z B coparing strss tnsor and th givn awa, I x = =4 z I x z z x x z = ( x 8) + (8 x 6) + (6 x ) - (5) () (6) = =7 x z x z xz z x I = x 8 x 6 - () -8 (6 ) - 6 (5) + (5) () (6) = =-58 xz Substitut ths valus in () quation () W know that x z xz Cos 4Cos Cos Fro this 4 Cos Cos Cos MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 8

19 put, Cos Cos Cos () 4 4 r Cos rcos rcos rcos I 4 8 Equation () bcos 5 4r Cos 9 rcos 4 r Cos 64 rcos 7 (r cos + 8) + 58 = r Cos r Cos r Cos - 4 r Cos r Cos + 7 r Cos = r Cos - 65 r Cos - 45 = Dividd b r Cos Cos (4) r r Coparing quation () and (4),w gt, 8 and 65 r 4 r = r Cos Cos Cos =.5568 = 8.84 o = + = 8.84 o = + = o = r Cos + 8 MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 9

20 = 4.84 Cos (8.84 o ) + 8 =.4 MPa = 4.84 Cos o + 8 = -.7 MPa Rsult: = r cos + 8 = 4.84 Cos o + 8 = 5. MPa = 8.84 o =.4 MPa = 8.84 o = -.7 MPa = o = 5. MPa. Obtain th principal strsss and th rlatd dirction cosins for th following stat of strss. (April / Ma ) Solution: MPa Th principal strsss ar th roots of th cubic quation. I I I I I () x z = + + = 6 x z x x z xz = ( x ) + ( x ) + ( x ) - (4) - (5) - (6) = I = -66 I x z x z xz z x x =( x x ) - (5) - (6) - (4) + (4 x 6 x 5) z xz = I = 8 Substitut ths valus in quation () MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag

21 () W know that Cos 4Cos Cos 4 Cos Cos Cos Cos Cos Cos 4 4 Cos Cos Cos 4 4 () I Put rcos rcos Equation () bcos rcos 6 rcos 66 rcos 8 r Cos 8 r Cos rcos 4 6r Cos r Cos 4r cos 4 66r cos 7rCos 66rCos 79 8 r Cos Dividd b r 9rCos Cos Cos r r (4) B coparing () and (4) 9 r 4 r = 56 r =.48 and 79 r Cos 4 MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag

22 76 = Cos x (.48 ) 76 Cos Cos =.6857 = =.79 o = + = 4.79 = + = 6.79 r cos =.48 Cos (.79) +.56MPa rcos =.48 Cos (4.79) MPa rcos =.48 Cos (6.79) + =.468 MPa Rsult: =. 79 o =.56 MPa = o = MPa = o =.468 MPa MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag

23 . Th stat of strss at a point is givn b MPa Dtrin th principal strsss and principal dirction. Solution: Th cubic quation I I I () I x z = = 7 I x z z x x z zx =( x ) + ( x 7) + (7) x + (6) + (64) + () = I =6 I x z x z xz z x x z zx =( x x 7) - (64) - () - 7 (6) + (6) (8) () = =8 Substitut ths valus in quation () W know that () Cos 4Cos Cos 4Cos Cos Cos Cos Cos 4 Cos 4 Cos Cos 4 Cos 4 () Put rcos I MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag

24 rcos. Equation () bcos rcos. 7 rcos. 6 rcos. 8 r Cos r Cos r Cos rCos 7 r Cos 6 r Cos = 6.99r Cos 6 r Cos = rCos 7r Cos 4.66rCos 9.r Cos r r Cos Cos Cos (4) r r B coparing () & (4) 4 95 r r = 8 r = 4.5 and Cos 4 Cos r = 6.69 =. o = + = 4. o MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 4

25 = 6. o rcos. = 4.5 Cos (4. o ) MPa rcos. = 4.5 Cos (4. o ) MPa rcos. = 4.5 Cos (6. o ) Rsult: =. o = 6. o = MPa = 4. o = 44.5 MPa = 6.56 MPa. Explain th Enrg of Distortion (shar strain nrg) and Dilatation. (AUC Nov/Dc ) Th strain nrg can b split up on th following two strain nrgis. i. Strain nrg of distortion (shar strain nrg) ii. Strain nrg of Dilatation (Strain nrg of unifor coprssion (or)) tnsion (or) volutric strain nrg ) Lt an d b th principal strain in th dirctions of principal strsss, and. Thn E E E Adding th abov quation w gt, E MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 5

26 E But + + = v (Volutric strain) v E If, v. This ans that if su of th thr principal strss is zro thr is no volutric chang, but onl th distortion occurs. Fro th abov discussion,. Whn th su of thr principal strsss is zro, thr is no volutric chang but onl th distortion occurs.. Whn th thr principal strsss ar qual to on anothr thr is no distortion but onl volutric chang occurs. Not: In th abov six thoris, t, c = Tnsil strss at th lastic liit in sipl tnsion and coprssion; load.),, = Principal strsss in an coplx sst (such that > > ) It a b assud that th loading is gradual (or) static (and thr is no cclic (or) ipact 4. Explain th Maxiu Principal strss Thor: ( Rankin s Thor) This is th siplst and th oldst thor of failur According to this thor failur will occur whn th axiu principl tnsil strss ( ) in th coplx sst rachs th valu of th axiu strss at th lastic liit ( t ) in th sipl tnsion (or) th iniu principal strss (that is, th axiu principal coprssiv strss), rachs th lastic liit strss ( ) in sipl coprssion. (i.) = t (in sipl tnsion) ac (In sipl coprssion) Mans nurical valu of If th axiu principal strss is th dsign critrion, th axiu principal strss ust not xcd th working for th atrial. Hnc, MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 6

27 This thor disrgards th ffct of othr principal strsss and of th sharing strsss on othr plan through th lnt. For brittl atrials which do not fail b ilding but fail b brittl fractur, th axiu principal strss thor is considrd to b rasonabl satisfactor. This thor appars to b approxiatl corrct for ordinar cast irons and brittl tals. Th axiu principal strss thor is contradictd in th following cass:. On a ild stl spcin whn sipl tnsion tst is carrid out sliding occurs approxiatl 45 o to th axis of th spcin; this shows that th failur in th cas is du to axiu shar strss rathr than th dirct tnsil strss.. It has bn found that a atrial which is vn though wak in sipl coprssion t can sustain hdrostatic prssur for in xcss of th lastic liit in sipl coprssion. 5. Explain th Maxiu shar strss (or) Strss Diffrnc thor. (AUC Nov/Dc ) This thor is also calld Gusti s (or) Trsca s thor. This thor iplis that failur will occur whn th axiu shar strss axiu in th coplx sst rachs th valu of th axiu shar strss in sipl tnsion at th lastic liit i.. ax t in sipl tnsion. (or) t In actual dsign t in th abov quation is rplacd b th saf strss. This thor givs good corrlation with rsults of xprints on ductil atrials. In th cas of two dinsional tnsil strss and thn th axiu strss diffrnc calculatd to quat it to t. Liitations of this thor: i. Th thor dos not giv accurat rsults for th stat of strss of pur shar in which th axiu aount of shar is dvlopd (i) Torsion tst. ii. Th thor is not applicabl in th cas whr th stat of strss consists of triaxial tnsil strsss of narl qual agnitud rducing, th sharing strss to a sall agnitud, so that failur would b b brittl factur rathr than b ilding. iii. Th thor dos not giv as clos rsults as found b xprints on ductil atrials. Howvr, it givs saf rsults. MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 7

28 6. Explain th Shar strain Enrg Thor (April / Ma ) This thor is also calld Distortion Enrg Thor : (or) Von Miss Hnk Thor According to this thor th lastic failur occurs whr th shar strain nrg pr unit volu in th strssd atrial rachs a valu qual to th shar strain nrg pr unit volu at th lastic liit point in th sipl tnsion tst. Shar strain nrg du to th principal strsss,, and pr unit volu of th strss atrial. U S C But for th sipl tnsion tst at th lastic liit point whr thr is onl on principal strss (i) t w hav th shar strain nrg pr unit volu which is givn b U s C t at Equating th two nrgis, w gt t t Th abov thor has bn found to giv bst rsults for ductil atrial for which approxiatl. t c Liitations of Distortion nrg thor:. T thor dos to agr with th xprintal rsults for th atrial for which t is quit diffrnt fro c.. Th thor givs t for hdrostatic prssur (or) tnsion, which ans that th atrial will nvr fail undr an hdrostatic prssur (or) tnsion. Whn thr qual tnsions ar applid in thr principal dirctions, brittl factur occurs and as such axiu principal strss will giv rliabl rsults in this cas.. This thor is rgardd as on to which confor ost of th ductil atrial undr th action of various tps of loading. MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 8

29 7. Explain th Maxiu principal strain Thor? This thor associatd with St Vnnt Th thor stats that th failur of a atrial occurs whn th principal tnsil strain in th atrial rachs th strain at th lastic liit in sipl tnsion (or) whn th iniu principal strain (i) axiu principal coprssiv strain rachs th lastic liit in sipl coprssion. Principal strain in th dirction of principal strss, E Principal strain in th dirction of th principal strss, E Th conditions to caus failur according to h axiu principal strain thor ar: and t ( ust b +V) E E E c ( ust b -V) E To prvnt failur: t E t E t c t c At th point of lastic failur: and t c MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 9

30 For dsign purposs, t c (whr, t and c ar th saf strsss) Liitations: i. Th thor ovrstiats th bhavior of ductil atrials. ii. T thor dos not fit wll with th xprintal rsults xcpt for brittl atrials for biaxial tnsion. 8. Explain th Strain nrg thor? Th total stain nrg of dforation is givn b U v E and th strain nrg undr sipl tnsion is U E Hnc for th atrial to ild, v Th total lastic nrg stord in a atrial bfor it rachs th plastic stat can hav no significanc as a liiting condition, sinc undr high hdrostatic prssur, larg aount of strain nrg a b stord without causing ithr fractur (or) prannt dforation. 9. Explain Mohr s Thor? A atrial a fail ithr through plastic slip (or) b fractur whn ithr th sharing strss in th plans of slip has incrasd. Lt f Th nvloping curv f ust rprsnt in thir abscissa and ordinats, th noral and sharing strsss in th plan of slip. Now Lt P MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag

31 thn p This quation rprsnts th fail of ajor principal strss circls in paratr for. Th quation of this nvlop is obtaind b partiall diffrntiating with rspct to P p P d. dp strss in paratr for. d p P. This is to quation of Mohr s nvlop of th ajor principal dp. In a stl br, at a point th ajor principal strss is 8 MN/ and th inor principal strsss is coprssiv. If th tnsil ild point of th stl is 5 MN/, find th valu of th inor principal strss at which ilding will conc, according to ach of th following critria of failur. i. Maxiu sharing strss ii. Maxiu total strain nrg iii. Maxiu shar strain nrg Tak Poisson s ratio =.6 Solution: Major principal strss, Yild point strss MN / MN / To calculat inor principal strss ( ) (i) Maxiu sharing strss critrion = 8-5 = - 45 MN/ = 45 MN/ (cop) MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag

32 ii. Maxiu total strain nrg critrion: = (8) + - x.6 x 8 = (5) = = MN / (Onl V sign is takn as is coprssiv) = 96.8 MN/ (coprssiv) iii. Maxiu shar strain nrg critrion: putting = (8) + ( ) = (5) ( ) = MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag

33 MN / = 7.5 MN/ (Coprssiv). In a atrial th principal strsss ar 6 MN/, 48 MN/ and - 6 MN/. Calculat i. Total strain nrg ii. Volutric strain nrg iii. iv. Shar strain nrg Factor of saft on th total strain nrg critria if th atrial ilds at MN/. Tak E = GN/ + and / =. Solution: Givn Data: Principal strsss: = + 6 MN/ = + 48 MN/ = - 6 MN/ Yild strss, = MN / E = GN/, / =. i. Total strain nrg pr unit volu: U E U U U = 9.5 KN/ MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag

34 ii. Volutric strain nrg pr unit volu: / v E v v =.78 KN/ iii. shar strain nrg pr unit volu s c Whr, C E. 76.9GN / s s s KN / 96 iv. Factor of saft (F.O.S) Strain nrg pr unit volu undr uniaxial loading is E 6 9 6KN / 6 F.O.S MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 4

35 . In a atrial th principal strsss ar 5 N/, 4 N/ and - N/, calculat: i. Total strain nrg ii. Volutric strain nrg iii. Shar strain nrg and iv. Factor of saft on th total strain nrg critrion if th atrial ild at N/. Tak E = x N/ and poission ratio =.8 Solution: Givn, Principal strsss: N / 5 N / 4 N / Yild strss, N / i. Total strain nrg pr unit volu: U E U =.55 KN/ ii)volutric strain nrg pr unit volu: / v E MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 5

36 v v 6. v =. K N / iii. Shar strain nrg s C whr C s E / N / s s.5knn / iv. Factor of saft (F.O.S) Strain nrg pr unit volu undr uniaxial loading is E 54 KN / F. O. S MUTHUKUMAR.P/Asst.Prof./CIVIL-III Pag 6

ME311 Machine Design

ME311 Machine Design ME311 Machin Dsign Lctur 4: Strss Concntrations; Static Failur W Dornfld 8Sp017 Fairfild Univrsit School of Enginring Strss Concntration W saw that in a curvd bam, th strss was distortd from th uniform