PSI - Issue 23

Viacheslav Mokryakov et al. / Procedia Structural Integrity 23 (2019) 143–148 Author name / Structural Integrity Procedia 00 (2019) 000 – 000

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3

After a series of transformations, we obtain the expression for displacements with one coefficient:                                                 2 0 2 0 2 2 0 2 0 2 0 2 0 0 2 0 1 2 2 sin 1 2 cos r z r u U H K J a J a J hr J hr c J ha J ha J r J r z t u U K J a J a J hr H J ha J ha J r z t c                                     where                 2 2 2 2 2 0 2 1 0 2 2 2 1 A C c U c c J a J a c c J ha J ha           Further, the coefficient U is chosen such that the average wave energy density is 1 J/m 3 . Knowing the displacement, we get the stress-strain state using Hooke's law. 3. Stress analysis For the strength calculating , the maximal tension (σ tension ) and shear (σ shear ) at each point are required. The maximum tensile (for all directions) coincides with the first principal stress. Since the considered problem is axisymmetric, one of the principal stresses is known: this is the circumferential stress σ θθ . The other two principal stresses are perpendicular to σ θθ , that is, they are in the plane of the longitudinal section of the rod, and depend on the stresses σ rr , σ zz , and σ rz . We denote these two principal stresses as σ a and σ b , and assume that σ a ≥ σ b . Here 2 2 2 2 , 2 2 2 2 rr zz rr zz rr zz rr zz a rz b rz                             In this problem, the first principal stress σ 1 is the largest of σ θθ and σ a , i.e. max( , ) tension a      . Then, the maximum shear (for all directions) coincides with the von Mises stress. For the axisymmetric problem this value is       2 2 2 2 2 3 shear Mises rr zz zz rr rz                    4. Zero mode study We consider an infinite steel rod of radius a = 1 cm with the following parameters: density ρ = 7.8 · 10 3 kg/m 3 ; Young's modulus E = 2.0 · 10 11 Pa; Poisson's ratio ν = 0.28 (hence c 0 = 5.1 km/s, c 1 = 5.7 km/s, c 2 = 3.2 km/s, c R = = 2.9 km/s). The rod surface is free from loads. Fig. 1 presents the solution of the dispersion relation equation. The dimensionless quantity c / c 1 is marked horizontally, and the dimensionless quantity ω a / c 1 is marked vertically. Consider the zero mode, this is the only one that covers all frequencies (the zero mode in Fig. 1 is highlighted in bold). We calculated σ tension and σ shear on the axis of the rod and on its surface for frequencies up to 3 MHz. The

    tension     tension

    shear     shear

0 max max r r a  

0 max max r r a  

 

 

R

R

 

 

0 max r 

and

are also calculated, where

means the

relations

tension

shear

largest value of the corresponding parameter on the entire axis, and max r a 

on the entire surface. Fig. 2 shows the

distribution of R tension and R shear versus frequency. Let us analyze the obtained dependencies. At low frequencies (up to about 100 kHz), both ratios are close to 1. The phase velocity is close to c 0 . The calculation shows that in this case the stress-strain state is close to the so-called long-wave approximation (i.e., the wavelength is much larger than the rod thickness). Here, the wave is close to