Load motion on an ice cover in the presence of a liquid layer with velocity shear

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The behavior of an ice cover on the surface of an ideal incompressible fluid of finite depth under the action of a pressure domain that moves rectilinearly at a constant velocity in the presence of a current with velocity shift in the upper layer is studied. It is assumed that the ice deflection is steady in the coordinate system moving with the load. The Fourier transform method is used within the framework of the linear wave theory. The critical velocities, the deflection of ice cover, and the wave forces are studied depending on the current velocity gradient, the shear layer thickness, the direction of motion, and the compression ratio.

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作者简介

L. Tkacheva

Институт гидродинамики им. М.А. Лаврентьева СО РАН

编辑信件的主要联系方式.
Email: tkacheva@hydro.nsc.ru
俄罗斯联邦, Новосибирск

参考

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2. Fig. 1. An example of eigenvectors of the dispersion relation for the system of equations (2.3), (2.4).

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3. Fig. 2. Wave number curves for an ice cover of 50 m thickness, 1–3 –

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4. Fig. 3 Dependence of critical velocities on the thickness of the shear layer at (a, b) and (c, d), y = 0 (a, c) and y = p (b, d): curves 1–5 correspond to the values

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5. Fig. 4. Ice cover deflection at –Q = 0 (a, b) and –Q = 1 (c, d), y = 0 (a, c) and y = p (b, d): 1 – S = 0; 2, 3 – S = 0.3, 0.5 for two layers H1 = 5 m, H2 = 45 m; 4, 5 – S = 0.3, 0.5 for one layer H1 = H = 50 m.

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6. Fig. 5. Dimensionless wave force coefficients A1, A2 depending on the load movement speed at –Q = 0, y = 0 (a), y = p/2 (b) and y = p (c): 1 – S = 0; 2, 3 – S = 0.3, 0.5 for two layers: H1 = 5 m, H2 = 45 m; 4, 5 – S = 0.3, 0.5 for one layer: H = 50 m.

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7. Fig. 6. Dimensionless wave force coefficients A1, A2 depending on the load movement speed at = 1, y = 0 (a), ψ = p/2 (b) and ψ = π (c): 1 – S = 0; 2, 3 – S = 0.3, 0.5 for two layers: H1 = 5 m, H2 = 45 m; 4, 5 – S = 0.3, 0.5 for one layer: H = 50 m.

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8. Fig. 7. Contribution of additional roots to wave forces depending on the speed of load movement at H1 = 5 m, ψ = π/2,  = 0 (a),  = 1 (b): 1–2 – S = 0.3, 0.5.

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