PSI - Issue 42

Rita Dantas et al. / Procedia Structural Integrity 42 (2022) 1676–1683 Rita Dantas / Structural Integrity Procedia 00 (2019) 000–000

1680

5

Meanwhile, Eqs. 1, 2 and 3 can be reduced to a single equation applied to a fatigue specimen of variable cross section:

∂ 2 u ∂ t 2

∂ f ∂ x

ρ S ( x )

(7)

=

where S ( x ) is the cross-section area and f is the force acting on the section, which is defined as:

∂ u ∂ x

(8)

f = ES ( x )

Thus, the di ff erential equation can be rewritten: ∂ 2 u ( x , t ) ∂ t 2 = c 2 p ( x ) ∂ u ( x , t ) ∂ x + ∂ 2 u ( x , t ) ∂ x 2

(9)

where:

S ( x ) S ( x )

p ( x ) =

(10)

For the cross-section area determination, the specimen’s profile function is defined as piecewise: a hyperbolic cosine for the variable section and a constant function for the constant section: y ( x ) = R 2 , L 2 < | x | ≤ L (11) y ( x ) = R 1 cosh ( α x ) , | x | ≤ L 2 (12) where: L = L 1 + L 2 (13) For this geometry, p ( x ) is defined as: p ( x ) = 0 , L 2 ≤ | x | ≤ L (14) p ( x ) = 2 α tanh ( α x ) , | x | ≤ L 2 (15) where: In order to fulfil the boundary conditions of the specimen under analysis, the solution of Eq. 9 can be given by the following function: u ( x , t ) = U ( x ) sin ( ω t ) (17) where U ( x ) is the amplitude of vibration and ω is the frequency of vibration. Thus, from Eq. 9 can be defined the di ff erential equation for the amplitude of vibration at each point throughout the specimen: U ( x ) + p ( x ) U ( x ) + k 2 U ( x ) = 0 (18) where: k = ω c (19) Eq. 18 can be simplified for each specimen section as follows: U ( x ) + k 2 U ( x ) = 0 , L 2 ≤ | x | ≤ L (20) α = 1 L 2 arcosh R 2 R 1 (16)