d B a B d A a A a D F D F − − = = 2 , 2 , * 4 4 π π σ (1) where and are the axial force amplitudes (acting on the two cylinders, respectively), which provoke fatigue fracture failure. a A F , a B F , The apparent fatigue strengths for such bodies are equal to : 2 , , 4 A a A a A D F π σ = (2) 2 , , 4 B a B a B D F π σ = (3) Therefore, recalling eqns (1) and (2), equation (3) becomes : d A B a A d A B a A B a B D D D D F D − − = = , 2 , 2 , 4 σ π σ (4a) and in a logarithmic form : ( ) B A a A a B d D D ln ln ln , , − = σ σ (4b) By assuming 1= AD and , where is a generic value of the bar diameter, the last two expressions can be written in a more general form : D D B = D (5a) ( ) d a a D − = ,1 σ σ d D a a ln ln ln ,1 − = σ σ (5b) where the latter equation represents a straight line with slope equal to d − in the diagram shown in Fig.2, ,1 aσ being the fatigue strength for a cylinder with 1 = = A D D . ln σa 1 d ln D ln σa,1 0 Figure 2: Monofractal scaling law for fatigue strength σa 3
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