Figure 1.- Schematic illustration of the new twin pulley fatigue testing machine. Microscope (SEM) was used to determine the cross sectional area at fracture in each case in order to calculate the ultimate tensile strength. Fatigue tests on wires of the Fe78-xCrxSi10B12 alloy series were carried out using both an existing single pulley bend type fatigue testing machine [3] (SP) which produces only tension/tension loading and a modified fatigue machine which uses two pulleys (DP) which can perform tension/compression loading, figure 1. To facilitate comparison between the two testing machines, the single pulley tests were carried out using a constant mean stress of 1550 MPa rather than the constant minimum stress utilised previously [3,5]. The double pulley tests were carried out at constant mean stresses of 200, 900 and 1550 MPa. The HT steel wires were tested in the DP machine at a constant mean stress of 200 MPa. The load frequency for all the fatigue tests was 3.4 Hz and the tests were carried out in ambient humidity (RH~60%). The stress range, Δσa, experienced by the outer surface of the bent wire sample in the SP machine is given by: p w w a d d d E + = ⋅ Δσ (1) where E is the Young’s Modulus (extrapolated from Donald et al [6]) and dw and dp are the wire and pulley diameters, respectively. Thus, for equivalently sized pulleys the stress range in the DP machine is twice that in the SP machine. Figure 2 shows schematically the surface stress profile generated by each machine. In order to determine the fracture toughness for glassy metallic alloys, ribbons of thickness ~45 μm were cast using the chill block melt spinning quenching technique at a roll speed ~30 ms-1 under an atmosphere of Figure 2.- Schematic illustration for the surface bending stress cycles for the single (SP) (a) and (b), and double (DP) (c) and (d) pulley machines; where Δσa, σm, σt, σmin and σmax are the stress range, mean stress, tensioning load, minimum and maximum stresses respectively. a a b b b b c c c d c ΔσD σm BENDING STRESS (+) (-) σmax σ t σmin = Cycle (a) t Cycle (c) ΔσD 2 σmax f σ σ m= σmin (+) (-) a a b b b b c c c c d d d d e e e e B ENDING STRESS E E a b c d e f (d) E a c d b (b)
RkJQdWJsaXNoZXIy MjM0NDE=