Module 3 Design for Strength. Version 2 ME, IIT Kharagpur

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1 Mdule 3 Design fr Strength Versin 2 ME, IIT Kharagpur

2 Lessn 4 Lw and high cycle fatigue Versin 2 ME, IIT Kharagpur

3 Instructinal Objectives At the end f this lessn, the students shuld be able t understand Design f cmpnents subjected t lw cycle fatigue; cncept and necessary frmulatins. Design f cmpnents subjected t high cycle fatigue lading with finite life; cncept and necessary frmulatins. Fatigue strength frmulatins; Gerber, Gdman and Sderberg equatins Lw cycle fatigue This is mainly applicable fr shrt-lived devices where very large verlads may ccur at lw cycles. Typical examples include the elements f cntrl systems in mechanical devices. A fatigue failure mstly begins at a lcal discntinuity and when the stress at the discntinuity exceeds elastic limit there is plastic strain. The cyclic plastic strain is respnsible fr crack prpagatin and fracture. Experiments have been carried ut with reversed lading and the true stressstrain hysteresis lps are shwn in figure Due t cyclic strain the elastic limit increases fr annealed steel and decreases fr cld drawn steel. Lw cycle fatigue is investigated in terms f cyclic strain. Fr this purpse we cnsider a typical plt f strain amplitude versus number f stress reversals t fail fr steel as shwn in figure Versin 2 ME, IIT Kharagpur

4 F- A typical stress-strain plt with a number f stress reversals (Ref.[4]). Here the stress range is Δ. Δε p and Δε e are the plastic and elastic strain ranges, the ttal strain range being Δε. Cnsidering that the ttal strain amplitude can be given as Δε = Δε + Δε A relatinship between strain and a number f stress reversals can be given as ' f Δε = (N) + ε (N) E p e a ' b f where f and ε f are the true stress and strain crrespnding t fracture in ne cycle and a, b are systems cnstants. The equatins have been simplified as fllws: 3.5 ε u p Δε = EN N 0.6 Versin 2 ME, IIT Kharagpur

5 In this frm the equatin can be readily used since u, ε p and E can be measured in a typical tensile test. Hwever, in the presence f ntches and cracks determinatin f ttal strain is difficult. Δε 1 Strain amplitude, ' f E 10-3 c b 1 1 Elastic strain Plastic strain Ttal strain Number f stress reversals fr failure, N F- Plts f strain amplitude vs number f stress reversals fr failure High cycle fatigue with finite life This applies t mst cmmnly used machine parts and this can be analyzed by idealizing the S-N curve fr, say, steel, as shwn in figure The line between 10 3 and 10 6 cycles is taken t represent high cycle fatigue with finite life and this can be given by lgs= blg N+ c where S is the reversed stress and b and c are cnstants. At pint A lg( 0.8 ) 3 u = blg10 + c where u is the ultimate tensile stress 6 and at pint B lg e = blg10 + c where e is the endurance limit. Versin 2 ME, IIT Kharagpur

6 This gives b = lg 3 e u and c = lg ( 0.8 ) e u A S e B N F- A schematic plt f reversed stress against number f cycles t fail Fatigue strength frmulatins Fatigue strength experiments have been carried ut ver a wide range f stress variatins in bth tensin and cmpressin and a typical plt is shwn in figure Based n these results mainly, Gerber prpsed a parablic crrelatin and this is given by 2 m v + = u e 1 Gerber line Gdman apprximated a linear variatin and this is given by m v + = u e 1 Gdman line Sderberg prpsed a linear variatin based n tensile yield strength Y and this is given by Versin 2 ME, IIT Kharagpur

7 m y e v + = 1 Sderberg line Here, m and v represent the mean and fluctuating cmpnents respectively. Variable stress, v e y u Gerber line Gdman line Sderberg line Cmpressive stress Mean stress, m Tensile stress F- A schematic diagram f experimental plts f variable stress against mean stress and Gerber, Gdman and Sderberg lines Prblems with Answers Q.1: A grved shaft shwn in figure is subjected t rtating-bending lad. The dimensins are shwn in the figure and the bending mment is 30 Nm. The shaft has a grund finish and an ultimate tensile strength f 1000 MPa. Determine the life f the shaft. r = 0.4 mm D = 12 mm d = 10 mm F Versin 2 ME, IIT Kharagpur

8 A.1: Mdified endurance limit, e = e C 1 C 2 C 3 C 4 C 5 / K f Here, the diameter lies between 7.6 mm and 50 mm : C 1 = 0.85 The shaft is subjected t reversed bending lad: C 2 = 1 Frm the surface factr vs tensile strength plt in figure Fr UTS = 1000 MPa and grund surface: C 3 = 0.91 Since T 450 C, C 4 = 1 Fr high reliability, C 5 = Frm the ntch sensitivity plts in figure , fr r=0.4 mm and UTS = 1000 MPa, q = 0.78 Frm stress cncentratin plts in figure , fr r/d = 0.04 and D/d = 1.2, K t = 1.9. This gives K f = 1+q (K t -1) = Then, e = e x 0.89x 1x 0.91x 1x 0.702/1.702 = e Fr steel, we may take e = 0.5 UTS = 500 MPa and then we have e = MPa. 32M Bending stress at the utermst fiber, b = 3 πd Fr the smaller diameter, d=0.01 mm, b Since b ' e > life is finite. Fr high cycle fatigue with finite life, lg S = b lg N + C where, b = lg 3 ' c = lg e ( 0.8 ) e u ' 2 0 = = = 305 MPa 1 0.8x1000 lg = ( 0.8x1000) lg = Therefre, finite life N can be given by N=10 -c/b S 1/b if 10 3 N Since the reversed bending stress is 306 MPa, N = 2.98x 10 9 cycles. 2 Versin 2 ME, IIT Kharagpur

9 F F (Ref.[5]) Q.2: A prtin f a cnnecting link made f steel is shwn in figure The tensile axial frce F fluctuates between 15 KN t 60 KN. Find the factr f safety if the ultimate tensile strength and yield strength fr the material are 440 MPa and 370 MPa respectively and the cmpnent has a machine finish. 10 mm F 90 mm 60 mm 15 mm F 6 mm F Versin 2 ME, IIT Kharagpur

10 A.2: T determine the mdified endurance limit at the step, e = e C 1 C 2 C 3 C 4 C 5 / K f where C 1 = 0.75 since d 50 mm C 2 = 0.85 fr axial lading C 3 = 0.78 since u = 440 MPa and the surface is machined. C 4 = 1 since T 450 C C 5 = 0.75 fr high reliability. At the step, r/d = 0.1, D/d = 1.5 and frm figure , K t = 2.1 and frm figure q = 0.8. This gives K f = 1+q (K t -1) = Mdified endurance limit, e = e x 0.75x 0.85x 0.82x 1x 0.75/1.88 = e Take e = 0.5 u. Then e = MPa. The link is subjected t reversed axial lading between 15 KN t 60 KN. This gives max 3 60x10 = = 100 MPa, 0.01x0.06 Therefre, mean = 62.5 MPa and v = 37.5 MPa. min Using Sderberg s equatin we nw have, 3 15x10 = = 25MPa 0.01x = + s that F.S = F.S This is a lw factr f safety. Cnsider nw the endurance limit mdificatin at the hle. The endurance limit mdifying factrs remain the same except that K f is different since K t is different. Frm figure fr d/w= 15/90 = 0.25, K t = 2.46 and q remaining the same as befre i.e 0.8 Therefre, K f = 1+q (K t -1) = This gives e = MPa. Repeating the calculatins fr F.S using Sderberg s equatin, F.S = This indicates that the plate may fail near the hle. Versin 2 ME, IIT Kharagpur

11 Q.3: A.3: A 60 mm diameter cld drawn steel bar is subjected t a cmpletely reversed trque f 100 Nm and an applied bending mment that varies between 400 Nm and -200 Nm. The shaft has a machined finish and has a 6 mm diameter hle drilled transversely thrugh it. If the ultimate tensile stress u and yield stress y f the material are 600 MPa and 420 MPa respectively, find the factr f safety. The mean and fluctuating trsinal shear stresses are τ m = 0 ; τ 16x100 = = 2.36 MPa. πx 0.06 ( ) v 3 and the mean and fluctuating bending stresses are 32x100 = = 4.72 MPa; πx 0.06 m 3 ( ) 32x300 = = MPa. πx 0.06 v 3 ( ) Fr finding the mdifies endurance limit we have, C 1 = 0.75 since d > 50 mm C 2 = 0.78 fr trsinal lad = 1 fr bending lad C 3 = 0.78 since u = 600 MPa and the surface is machined ( figure ). C 4 = 1 since T 450 C C 5 = 0.7 fr high reliability. and K f = 2.25 fr bending with d/d =0.1 (frm figure ) = 2.9 fr trsin n the shaft surface with d/d = 0.1 (frm figure ) This gives fr bending eb = e x 0.75x1x 0.78x 1x 0.7/2.25 = e Fr trsin es = es x 0.75x0.78x 0.78x 1x 0.7/2.9 = 0.11 e And if e = 0.5 u = 300 MPa, eb =54.6 MPa; es = 33 MPa We may nw find the equivalent bending and trsinal shear stresses as: τ τ = τ + τ = MPa ( Taking τ y = 0.5 y = 210 MPa) y eq m v ' es Versin 2 ME, IIT Kharagpur

12 = + = MPa. y eq m v ' eb Equivalent principal stresses may nw be fund as 1eq 2 eq eq 2 τeq = eq 2 eq eq 2 τeq = and using vn-mises criterin 2 2 y eq + 3τeq = 2 F.S 2 which gives F.S = F (Ref.[2]) Versin 2 ME, IIT Kharagpur

13 F (Ref.[2]) Summary f this Lessn The simplified equatins fr designing cmpnents subjected t bth lw cycle and high cycle fatigue with finite life have been explained and methds t determine the cmpnent life have been demnstrated. Based n experimental evidences, a number f fatigue strength frmulatins are available and Gerber, Gdman and Sderberg equatins have been discussed. Methds t determine the factr f safety r the safe design stresses under variable lading have been demnstrated. Versin 2 ME, IIT Kharagpur

14 3.4.6 Reference fr Mdule-3 1) Design f machine elements by M.F.Sptts, Prentice hall f India, ) Machine design-an integrated apprach by Rbert L. Nrtn, Pearsn Educatin Ltd, ) A textbk f machine design by P.C.Sharma and D.K.Agarwal, S.K.Kataria and sns, ) Mechanical engineering design by Jseph E. Shigley, McGraw Hill, ) Fundamentals f machine cmpnent design, 3 rd editin, by Rbert C. Juvinall and Kurt M. Marshek, Jhn Wiley & Sns, Versin 2 ME, IIT Kharagpur