MAHALAKSHMI ENGINEERING COLLEGE

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1 HLKSH ENGNEENG COLLEGE TUCHPLL -. QUESTON WTH NSWES DEPTENT : CVL SEESTE: V SUB.CODE/ NE: CE / Strngt of atrials UNT DVNCED TOPCS N BENDNG OF BES PT - ( marks). Dfin Unsmmtrical nding T plan of loading (or) tat of nding dos not li in (or) a plan tat contains t principl cntroidal axis of t cross- sction; t nding is calld Unsmmtrical nding.. Stat t two rasons for unsmmtrical nding. (UC a/jun ) (UC p[r /a ) (i) T sction is smmtrical (viz. ctangular, circular, sction) ut t load lin is inclind to ot t principal axs. (ii) T sction is unsmmtrical (viz. ngl sction (or) cannl sction vrtical w) and t load lin is along an cntroidal axs.. Dfin sar cntr. T sar cntr (for an transvrs sction of t am) is t point of intrsction of t nding axis and t plan of t transvrs sction. Sar cntr is also known as cntr of twist. Writ t sar cntr quation for cannl sction. w f = Distanc of t sar cntr (SC ) from t w along t smmtric axis XX w = ra of t w f = ra of t flang. cannl Sction as flangs cm x cm and w cm x cm. Dtrmin t sar cntr of t cannl. Solution: = -. =. cm t = cm, t = cm, = 8 cm UTHUKU.P/sst.Prof./CVL- Pag

2 f = t =. x = cm w = t = 8 x = 8 cm w (.) Writ t sar cntr quation for unsmmtrical sction. t ( xx ) f cm = Distanc of t sar cntr (SC) from t w along t smmtric axis XX t = ticknss of t flang = igt of t w = widt of t flang in rigt portion. = widt of t flang in lft portion. xx =.O. of t sction aout XX axis. 7. Stat t assumptions mad in Winklr s Bac Tor.(UC Nov / Dc ) (UC Nov/Dc ) (UC a/jun ) () Plan sctions (transvrs) rmain plan during nding. () T matrial os Hook s law (limit stat of proportionalit is not xcdd) () adial strain is ngligil. () T firs ar fr to xpand (or) contract witout an constraining ffct from t adjacnt firs. 8. Stat t paralll xs and Principal omnt of inrtia. f t two axs aout wic t product of inrtia is found, ar suc, tat t product of inrtia coms zro, t two axs ar tn calld t principl axs. T momnt of inrtia aout a principal axs is calld t principal momnt of inrtia. 9. Dfin strss concntration.. (UC Nov / Dc ) T trm strss gradint is usd to indicat t rat of incras of strss as a strss raisr is approacd. Ts localizd strsss ar calld strss concntration.. Dfin strss concntration factor. t is dfind as t ratio of t maximum strss to t nominal strss. K t max nom max = maximum strss nom = nominal strss UTHUKU.P/sst.Prof./CVL- Pag

3 . Dfin fatigu strss concntration factor. T fatigu strss concntration factor (K f ) is dfind as t ratio of flang limit of unnotcd spcimn to t fatigu limit of notcd spcimn undr axial (or) nding loads. K q( K ) f t Valu of q rangs from zro to on.. Dfin sar flow. Sar flow is dfind as t ratio of orizontal sar forc H ovr lngt of t am x. Sar flow is acting along t longitudinal surfac locatd at discarg.sar flow is dfind q. H Q z q V x z H = orizontal sar forc. Explain t position of sar cntr in various sctions. (i) n cas of a am aving two axs of smmtr, t sar cntr coincids wit t cntroid. (ii) n cas of sctions aving on axis of smmtr, t sar cntr dos not coincid wit t cntroid ut lis on t axis of smmtr.. Stat t principls involvd in locating t sar cntr. T principl involvd in locating t sar cntr for a cross sction of a am is tat t loads acting on t am must li in a plan wic contains t rsultant sar forc on ac cross-sction of t am as computd from t saring strsss.. Dtrmin t position of sar cntr of t sction of t am sown in fig. Solution: t = cm, = cm, = 8 cm = = cm t ( xx ) x x xx = x() 8 cm x (8 ). 977 (8. Stat t strsss du to unsmmtrical nding. v cos UU u sin VV σ = nding strss in t curvd ar = momnt du to t load applid UU = Principal momnt of inrtia in t principal axs UU VV = Principal momnt of inrtia in t principal axs VV cm UTHUKU.P/sst.Prof./CVL- Pag

4 7. Dfin t trm Fatigu. Fatigu is dfind as t failur of a matrial undr varing loads, wll low t ultimat static load, aftr a finit numr of ccls of loading and unloading. 8. Stat t tps of fatigu strss. (i) Dirct strss (ii) Plan nding (iii) otating nding (iv) Torsion (v) Comind strsss (a) Fluctuating or altrnating strss () vrsd strss. 9. Stat t rasons for strss- concntration. Wn a larg strss gradint occurs in a small, localizd ara of a structur, t ig strss is rfrrd to as a strss concntration. T rasons for strss concntration ar (i) discontinuitis in continuum (ii) contact forcs.. Dfin crp. Crp can dfind as t slow and progrssiv dformation of a matrial wit tim undr a constant strss.. Dfin principal momnt of inrtia. (UC Nov/Dc ) T prpndicular axis aout wic t product of inrtia is zro ar calld principal axs and t momnts of inrtia wit rspct to ts axs ar calld as principal momnts of inrtia. T maximum momnt of inrtia is known as ajor principal momnt of inrtia and t minimum momnt of inrtia is known as inor principal momnt of inrtia. PT B ( KS). Explain t strsss inducd du to unsmmtrical nding. Fig. sows t cross-sction of a am undr t action of a nding momnt acting in plan YY. lso G = cntroid of t sction, XX, YY = Co-ordinat axs passing troug G, UU, VV = Principal axs inclind at an angl θ to XX and YY axs rspctivl T momnt in t plan YY can rsolvd into its componnts in t plans UU and VV as follows: omnt in t plan UU, = sinθ omnt in t plan VV, = cosθ T componnts and av tir axs along VV and UU rspctivl. T rsultant nding strss at t point (u,v) is givn, ' u VV " v UU sin VV cos UU UTHUKU.P/sst.Prof./CVL- Pag

5 VCos UU usin vv t an point t natur of σ will dpnd upon t quadrant in wic it lis. T quation of t nutral axis (N.) can found finding t locus of t points on wic t rsultant strss is zro. Tus t points ling on nutral axis satisf t condition tat σ = VCos usin UU vv VCos usin UU vv v UU Sin u (or) v UU tan u Cos vv vv Tis is an quation of a straigt lin passing troug t cntroid G of t sction and inclind at an angl wit UU wr tan UU vv tan Following points ar wort noting: i. T maximum strss will occur at a point wic is at t gratst distanc form t nutral ii. ll t points of t sction on on sid of nutral axis will carr strsss of t sam natur and on t otr sid of its axis, of opposit natur. iii. n t cas wr tr is dirct strss in addition to t nding strss, t nutral axis will still a straigt lin ut will not pass troug G (cntroid of sction.). Driv t quation of Sar cntr for cannl sction. (UC pril/a ) Fig sows a cannl sction (flangs: x t ; W x t ) wit XX as t orizontal smmtric axis. Lt S = pplid sar forc. (Vrtical downward X) (Tn S is t sar forc in t w in t upward dirction) S = Sar forc in t top flang (tr will qual and opposit sar forc in t ottom flang as sown.) Now, sar strss ( ) in t flang at a distanc of x from t rigt and dg (of t top flang) S t t. x (wr t = t, ticknss of flang) xa St. x.. t xx S x xx UTHUKU.P/sst.Prof./CVL- Pag

6 Sar forc is lmntar ara d t. dx. d t Total sar forc in top flang S (or) dz. t. dx (wr = radt of t flang) S ; t. dx S xx St xx. st xx xdx Lt = Distanc of t sar cntr (sc) from taking momnts of sar forcs aout t cntr O of t w,w gt S S.. St.. S. t xx xx t xx () Now, xx t. t t t. t t t t (nglcting t trm t trms)(or) xx t t Sustitut t valu of xx in quation () w gt,, ing ngligil in comparison to otr Lt t t t t t t t = f (ara of t flang) t = (ara of t w) Tn UTHUKU.P/sst.Prof./CVL- Pag

7 i. w f f w f w f. Driv t quation of Sar cntr for unqual -sction Solution: Fig. sows an unqual sction wic is smmtrical aout XX axis. Sar strss in an lar, S t wr = XX = Sar forc S : d S = t tdx. t. x. d S. x. t XX t xt dx tx = S. x. XX t dx = St XX x St XX Similarl t sar forc (S ) in t otr part of t flang, St S = XX Taking momnts of t sar forcs aout t cntr of t w O, w gt S. = S. + S. (S = S for quilirium) (wr, = distanc of sar cntr from t cntr of t w) or, (S S ) = S. S t( XX ) S. t xx UTHUKU.P/sst.Prof./CVL- Pag 7

8 . Driv t strsss in curvd ars using Winklr Bac Tor. T simpl nding formula, owvr, is not applical for dpl curvd ams wr t nutral and cntroidal axs do not coincid. To dal wit suc cass Winklr Bac Tor is usd. Fig sows a ar BCD initiall; in its unstraind stat. Lt B CD t straind position of t ar. Lt = adius of curvatur of t cntroidal axis HG. Y = Distanc of t fir EF from t cntroidal lar HG. = adius of curvatur of HG = Uniform nding momnt applid to t am (assumd positiv wn tnding to incras t curvatur) = Original angl sutndd t cntroidal axis HG at its cntr of curvatur O and = ngl sutndd HG (aftr nding) a t t cntr of curvatur For finding t strain and strss normal to t sction, considr t fir EF at a distanc from t cntroidal axis. Lt σ t strss in t straind lar EF undr t nding momnt and is strain in t sam lar. EF' EF ( ' ' ) ' ( ) ' ' ' Strain, or. EF ( ) = strain in t cntroidal lar i.. wn = ' ' ' ' '. or () ' ' and + =. Dividing quation () and (), w gt ' ' ' '.. or ' ' ' ccording to assumption (), radial strain is zro i.. =. Strain, ' ' dding and sutracting t trm. /, w gt ().. ' ' UTHUKU.P/sst.Prof./CVL- Pag 8

9 ( )( ) ' () From t fig. t lars aov t cntroidal lar is in tnsion and t lars low t cntroidal lar is in comprssion. ( )( ) Strss, σ = E = E ( ' ) () Total forc on t sction, F =. d Considring a small strip of lmntar ara d, at a distanc of from t cntroidal lar HG, w av ( )( ) F E d E '. d F E. d E ( ) d, F E. E ( ) d (), wr = cross sction of t ar T total rsisting momnt is givn givn ( )( ).. d E. d E ' E. E ( ) d (sinc d ), = E (+ ) d Lt ' Wr = a constant for t cross sction of t ar d d = E (+ ) ' () Now,. d. d d. d Hnc quation () coms = d = d. d (7) UTHUKU.P/sst.Prof./CVL- Pag 9

10 F = E. E (+ ) ' Sinc transvrs plan sctions rmain plan during nding F = = E. E (+ ) ' E. = E (+ ) ' = (+ ) ' (or) (+ ) ' Sustituting t valu of Or = E = E (+ ) ' in t quation () sustituting t valu of in quation () E E * * (or) E * * * E * * (Tnsil) (Comprssiv). T curvd mmr sown in fig. as a solid circular cross sction. m in diamtr. f t maximum tnsil and comprssiv strsss in t mmr ar not to xcd Pa and Pa. Dtrmin t valu of load P tat can safl carrid t mmr. Solution: Givn, d =. m; =. m; G = Pa = N / m (tnsil ) = Pa = N / m (Comprssiv) UTHUKU.P/sst.Prof./CVL- Pag

11 Load P: fr to t fig. ra of cross sction, d. 7.8 m Bnding momnt, m = P (. +.) =. P d.. = 7. x 8 - m. p Dirct strss, d comp Bnding strss at point du to : (tnsil) Total strss at point, d P (tnsil) 7.8 P 7.8.P = -7. P + 8. P x. 7 = P P 88. KN (i) Bnding strss at point du to : (comp) Total strss at point, d P 7.8 P 7.8.P =7. P + 8. P x. =.8 P UTHUKU.P/sst.Prof./CVL- Pag

12 P.8 N P. KN.8 (ii) B comparing (i) & (ii) t saf load P will lssr of two valus Saf load =. KN.. Fig. sows a fram sujctd to a load of. kn. Find (i) T rsultant strsss at a point and ;(ii) Position of nutral axis. (pril/a ) Solution: ra of sction -, = 8 * 8* - = 8. * - m Bnding momnt, = -.* *(+8) * = -. Nm is takn as v caus it tnds to dcras t curvatur. (i) Dirct strss: Hr Dirct strss σ d = P log D.* 8. * D D *.77 N / m = 8 mm =.8 m, D = 8 mm =.8 m.8.8 log (.8).8 (.8).8 (.8) =.8 (log ) =.7 * - m UTHUKU.P/sst.Prof./CVL- Pag

13 (ii) Bnding strss du to at point : ; 8. *. * *..8. * N / m = -9.7 (-.9) = 88.9 N/m (tnsil) (iii) Bnding strss du to at point : 8. *. * *..8 _. * N / m = -. N/m =. N/m (comp) (iv) sultant strss: sultant strss at point, σ = σ d + σ = = 9.7 N/m (tnsil) sultant strss at point, σ = σ d + σ = = 9.8 N/m (comp) (v) Position of t nutral axis:.8.8 *.7 *.7 * = -. m = -. mm Hnc, nutral axis is at a radius of. mm UTHUKU.P/sst.Prof./CVL- Pag

14 7. Fig. sows a ring carring a load of kn. Calculat t strsss at and. Solution: ra of cross-sction = x cm. cm. m Bnding momnt = * * (.* - )Nm = Nm d d = *... 8 Hr d = cm, = 7. + =. cm = * 8. P * Dirct Strss σ d = *. Bnding strss at point du to, = 9.89 cm = 9.89* - m. N / m.. *.*.7. *. = 7.7 N/m (tnsil) 9.89 *.. Bnding strss at point du to,.. *.*.7. *. =. N/m (comp) Hnc σ = σ d + σ = =. N /m (tnsil) and σ = σ d + σ = -.. = 9. N/m (comp) 9.89 *.. UTHUKU.P/sst.Prof./CVL- Pag

15 8. curvd ar is formd of a tu of mm outsid diamtr and 7. mm ticknss. T cntr lin of tis is a circular arc of radius mm. T nding momnt of knm tnding to incras curvatur of t ar is applid. Calculat t maximum tnsil and comprssiv strsss st up in t ar. Solution: Outsid diamtr of t tu, d = mm =. m Ticknss of t tu = 7. mm nsid diamtr of t tu, d = -*7. = mm =.m ra of cross-sction,... m Bnding momnt = knm ra of innr circl,.. 8 ra of outr circl, m.. m For circular sction, d d = *... 8 For innr circl,. = d d = *... 8 For outr circl, * * d d = *...;. = 8 * *. =.*9.* -.8*7.78* - =. m, and / =. /. =.9 aximum strss at, (wr, = mm =. m) *. * * N / m σ = 7. N/m (tnsil) aximum strss at B, UTHUKU.P/sst.Prof./CVL- Pag

16 B B *. * * N / m σ B =.7 N/m (comp) 8. curvd am as a T-sction (sown in fig.). T innr radius is mm. wat is t ccntricit of t sction? Solution: ra of T-sction, = t + t = * + 8* = 8 mm To find c.g of T- sction, taking momnts aout t dg LL, w gt x x x Now x Using t lation: ( *)( ) (8 *)(8 * *) ( *) (8 *) =7. mm = mm; = mm; = 7. mm; = 8 mm (7.) 8 8 *log. log t.log ( ) *log 8 ( ) (7.) UTHUKU.P/sst.Prof./CVL- Pag

17 =.8(.+.) 7. =.8 7. *.8 =. mm ( ) (7.).8 wr = (ccntricit) = distanc of t nutral axis from t cntroidal axis. Ngativ sign indicats tat nutral axis is locats low t cntroidal axis.. Fig. sows a C- fram sujctd to a load of kn. Dtrmin t strsss at and B. (UC Nov/Dc ) Solution: Load (P) = kn ra of cross sction = t + t + t = * + * +8* =. mm To find c.g of t sction aout t dg LL, x x x ( * * ) ( * ) ( * *) ( * ) = = 7 mm =.7 m = mm =. m = + = mm =. m = + = 8 mm =.8 m = + = mm =. m = + = mm =. m log t log (8 * *) (8 * ) log = mm=. m (.8).. log... log... log.. =.8 ( ). =.8 m P * Dirct strss, σ d = * 8.89 N / m ( comp). Bnding momnt, = P* Bnding strss at du to t nding momnt, ( ).8 ( ) P* = 8.89 (+.8) =. N/m (tnsil) Bnding strss at B du to t nding momnt: UTHUKU.P/sst.Prof./CVL- Pag 7

18 ( ) ( ) P* = 8.89 ( - 7.) = -.9 N /m =.9 N/m (comp) Strss at, σ = σ d + (σ ) = =. N/m (tnsil) Strss at B, σ B = σ d + (σ ) B = =.79 N/m (comp). Driv t formula for t dflction of ams du to unsmmtrical nding. Solution: Fig. sows t transvrs sction of t am wit cntroid G. XX and YY ar two rctangular co-ordinat axs and UU and VV ar t principal axs inclind at an angl θ to t XY st of co-ordinats axs. W is t load acting along t lin YY on t sction of t am. T load W can rsolvd into t following two componnts: (i) W sin θ along UG (ii) W cos θ along VG Lt, δ u = Dflction causd t componnt W sin θ along t lin GU for its nding aout VV axis, and Δ v = Dflction causd t componnt W cos θ along t lin GV du to nding aodt UU axis. Tn dpnding upon t nd conditions of t am, t valus of δ u and δ v ar givn u v K W sin E VV K W cos E UU l l wr, K = constant dpnding on t nd conditions of t am and position of t load along t am, and l = lngt of t am T total or rsultant dflction δ can tn found as follows: u v Kl E W sin W cos VV UU Kl E sin VV cos UU UTHUKU.P/sst.Prof./CVL- Pag 8

19 T inclination β of t dflction δ, wit t lin GV is givn : u UU tan v tan VV. 8 mm x 8 mm x mm angl sction sown in fig. is usd as a simpl supportd am ovr a span of. m. t carris a load of kn along t lin YG, wr G is t cntroid of t sction. Calculat (i) Strsss at t points, B and C of t mid sction of t am (ii) Dflction of t am at t mid-sction and its dirction wit t load lin (iii) Position of t nutral axis. Tak E = GN/m (UC pr/a ) Solution: Lt (X,Y) t co-ordinat of cntroid G, wit rspct to t rctangular axs BX and BY. 8 * * 7 * * Now X = Y = 8 * 7 * omnt of inrtia aout XX axis: XX 8 * 8 * *(. ) 8 *7. 7 mm 7 * *(.) = (. + 78) + ( ) = 8898 mm = * mm = YY (sinc it is an qual angl sction) Co-ordinats of G = + (-.), - (.-) = (.,- 8.) Co-ordinats of G = -(.-). + (.) = (-8., +.) (Product of inrtia aout t cntroid axs is zro caus portions and ar rctangular strips) f θ is t inclination of principal axs wit GX, passing troug G tn, UTHUKU.P/sst.Prof./CVL- Pag 9

20 tan XY XY XX tan 9 (sinc xx = ) θ = 9º i.. θ = º and θ = 9º + º = º ar t inclinations of t principal axs GU and GV rspctivl. Principal momnt of inrtia: YY XX UU = ( XX YY ) ( ) ( XY ) 8.89 * * = (8.89 * * ) ( ) ( = ( ) * =.* mm UU + VV = XX + YY VV = XX YY UU = *8.898 x. x =.7 x mm (i) Strsss at t points, B and C: Bnding momnt at t mid-sction, Wl *.*.* Nmm. * ) T componnts of t nding momnts ar; = sin θ =. x sin º =.97 x Nmm = cos θ =. x cos º =.97 x Nmm u,v co-ordinats: Point : x = -., = 8-. =. mm u = x cos θ + sin θ = -. x cos º +. x sin º =. mm v = cosθ + x sin θ =. cos º - (-. x sin º) =. mm Point B: x = -., = -. u = x cos θ + sin θ = -. x cos º + (-. x sin º ) = -. mm v = cosθ + x sin θ = -. cos º - (-. x sin º) = Point C ; x = 8. =., = -. u = x cos θ + sin θ =. cos º -. x sin º =. mm v = cosθ + x sin θ = -. cos º -. sin º) =-. mm ' u " v VV UU UTHUKU.P/sst.Prof./CVL- Pag

21 .97 *.7x (.).97 * (.). x 7.7 N / mm B.97 * (.7x.). x.7 N / mm B.97 *.7x (.).. x (ii) Dflction of t am, δ: T dflction δ is givn :.788 N / mm KWl E sin VV cos UU wr K = /8 for a am wit simpl supportd nds and carring a point load at t cntr. Load, W = N Lngt l =. m E = x N/mm UU =. x mm VV =.7 x mm Sustituting t valus, w gt 8 x(.x E ) sin (.7x ) cos (. x δ =. mm T dflction δ will inclind at an angl β clockwis wit t kin GV, givn tan UU VV tan. x.7 x tan.88 β = 7.º - º =.º clockwis wit t load lin GY. (iii) Position of t nutral axis: T nutral axis will at 9º -.º = 9.7º anti-clockwis wit t load lin, caus t nutral axis is prpndicular to t lin of dflction. ) UTHUKU.P/sst.Prof./CVL- Pag

22 . curd ar of rctangular sction, initiall unstrssd is sujctd to nding momnt of N m tnds to straigtn t ar. t sction is cm wid and cm dp in t plan of nding and man radius of curvatur is cm. Find t position of nutral axis and t strss at t innr and outr fac. (UC pr /a ) UTHUKU.P/sst.Prof./CVL- Pag

23 . cntral orizontal sction of ook is a smmtrical trapzoid mm dp, t innr widt ing mm t outr ing mm. stimat t xtrm intnsitis of strss wn t onk carris a load of kn. t load lin passing mm from t insid dg of t sction and t cntr of curvatur ing in t load lin. also plot t strss distriution across t sction. (UC pr/a ) UTHUKU.P/sst.Prof./CVL- Pag

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