Question

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3. The fatigue characteristics for a brass alloy are given as follows: Stress Amplitude (MPa) 170 Cycles to Failure 3.7 x 104 1.0 x 105 3.0 x 105 1 x 106 1 x 107 1 x 108 1 x 109 130 114 2 0 (a) Make an S-N plot (stress amplitude vs. logarithm cycles to failure) using these data. (b) Determine the fatigue strength at 4 x 106 cycles. (c) Determine the fatigue life for 120 MPa. 4. Consider 4 materials A, B, C, D that all have edge cracks of width a = plane-strain fracture toughness (i.e, Kic) of A, B, C and D are, respectively, 1 MPa m, 5 MPa m.5, 10 MPa m0.5, and 20 MPa mo.5, If each of these materials A, B, C and D, is subjected to fatigue loading where the stresses vary from a minimum of 0 MPa to a maximum of 100 MPa, and the crack sizes increase due to fatigue: 1 um. The (a) Estimate the crack sizes at which each of these four materials will undergo complete fracture. The stress intensity factor is given by: K = a [Ta].5 where a is the far-field applied stress and a is the crack length. (b) Estimate the fatigue life of each material if the crack growth rate is governed by da/dN expressed, respectively, in the units of m/cycle and MPa mo.5 C (AK)m where C = 2.5 x 10-12 and m 4 (when da/dN and AK are

Answer 1

4 a) For material A Fracture toughness (Kic) = IMPA m KIC= Omax Tdc Crack size at rohich complete fracture 9c= 3.1881X 10m= 31-831 um take place For material B Kic5MPamos Kie= Omar Tac 5= 100 c ac= 7.957X10 m= 7957um For material C Kic= 10 MPa m 10=100 ac ac= 3.183X 10-3 m For material D Kic =20MPa mosS Kic = Omax Tac 20 100 T c ac=0.01273 m For material A 4(b) Initial Crack size Lai) = Lum = 10.0m Final crack size (af)= 3 1831 x10m 40= Oma- Omin= 100-o= 100 MPa 100Ta MPa mos 4K= 40TTa da = c (4K) Governing equation (00/πα) da- 2.5X10-12 =D

- da 2-5X1012x 10 T2a2 This expression will come out same for material A, B,C&D as 40 remains same d expression -P af 3.1831x10S N of AK ill remain same da 2.5X10-4 T2a2 C N=392552343 7 cycles Fatigue life of material A For materialB : Proceeding in similar way ai 10.6m ap= 7.967X10-4m 40 and 4K remains same as per problem af-7.957x1o 4 dN= 2.5X10-4 T2q2 N=404775390-9 cycles Fatigue life of material B For material C ai= 106m af=3.183x 10 m af= 3.183X10-3 da dN 2.5x10 4 T2a2 ai=106 D N= 405157406 7 Cycles Fatigue life oF material c For material D ai= 10-6m af= 0.01273 m O.01273 da 2.5X10-4 T2 a2 106 Fatigu life of material D N=405252897.6 cycles