Ice Actions on Sloping-Sided Structures Sveinung Løset1,2,3 1Professor, Norwegian University of Science and Technology, Trondheim 2Adjunct professor, University Centre in Svalbard, Longyearbyen 3Professor honoris causa, St. Petersburg State Polytechnical University, St. Petersburg ICE ACTIONS Ice features Ice properties Design philosophy Interaction geometry Failure modes Level Crystallography Limit stress Single Creep Rafted Temperature Limit momentum Multileg Crushing Ridge Salinity Limit force Out- of-plane shape Bending Rubble Porosity Splitting Water depth Buckling Iceberg Surface tension Waterline shape Splitting Dimensions Concentration Velocity Strength Friction Adhesion Material Roughness Compressive Flexure Tensile Shearing Spalling Sloping structures Face of structure: plane, cone or facet - slope angle α The slope changes the failure mode -> the ice loads are less than on vertical ones (σf < σc) Influence of ice strength, ice thickness, slope and friction on the ice load The slope affects the characteristic breaking frequencies and thus reduces potential resonance problems The advantage of sloping structures may be reduced by: – – rubble accumulation at the structure high velocity of the advancing ice sheet Downward breaking cone Windmill foundation, Denmark Upward breaking cone Confederation Bridge, Canada Different stages Upward/downward breaking Same treatment? (Weight Æ Buoyancy) Effects on vertical load and overturning moment Load prediction models Ice loads induced by horizontal and vertical components Limited by: – bending strength, shear stress capacity and thickness of ice – friction and sloping of the structure Models: – – – – Croasdale (1980), 2D beam theory Ralston (1977), 3D plate theory FEM simulations (Määttänen et al.) + several other models, see Chao (1992) Forces on structure μNcos α N x μN Ncos α Nsin α μNsin α y α H = ∑ Fx = N sin α + μ N cos α V = ∑ Fy = N cos α − μ N sin α Simple 2D theory Croasdale (1980) 2D beam on elastic foundation Note: Only valid for wide structures H = N sin α + μ N cos α V = N cos α − μ N sin α 8 ⎛ sin α + μ cos α ⎞ H =V ⎜ = V ⋅ξ ⎟ ⎝ cos α − μ sin α ⎠ Beam M y I M M σ u = hu , σ o = − ho I I σ= I = I z = ∫ y 2dA A h/2 bh 3 Rectangular cross section : I z = ∫ y bdy = 12 −h / 2 2 σo x I = second moment of area b = beam width, h = ice thickness ho hu y σu Simple 2D theory continue Vertical load V limited by the bending strength of ice Ice sheet assumed as a beam on elastic foundation Strength limited by the bending moment as: σ f ,max M 6M h/2 = 2 = 3 bh /12 bh Simple 2D theory continue The maximum bending moment capacity for a semi-infinite beam on elastic foundation (Hetenyi, 1946): M= V sin(π / 4) β exp(π / 4) where 1/β is characteristic length defined by 1/ 4 K ⎞ β = ⎛⎜ ⎟ ⎝ 4 EI ⎠ K = ρ w gb foundation constant ρ w = density of water g = acceleration due to gravity E = Young ' s modulus I = second moment of area of the cross section (bh 3 /12) Simple 2D theory continue By combining the previous equations, the limits of the vertical and horizontal loads read: 1/ 4 ⎛ ρ w gh ⎞ V = 0.68σ f W ⎜ ⎟ E ⎝ ⎠ hence 5 1/ 4 ⎛ ρ w gh ⎞ H = 0.68σ f W ⎜ ⎟ E ⎝ ⎠ 5 sin α + μ cos α cos α − μ sin α where W = b is beam width (breath along the water line on the sloping face) Simple 2D theory continue Force needed to push ice blocks up the slope: Z hW ρ i g (sin α + μ cos α ) sin α where P= ρ i = density of ice Z = heigth reached by the ice on the slope ⎛ sin α + μ cos α ⎞ H = (V + P sin α ) ⎜ ⎟ + P cos α cos α μ sin α − ⎝ ⎠ substituting for V and P 1/ 4 ⎛ ρ w gh 5 ⎞ H = W ⋅ 0.68σ f ⎜ ⎟ ⎝ E ⎠ ⎛ ( sin α + μ cos α ) 2 ( sin α + μ cos α ) ⎞ ⎛ sin α + μ cos α ⎞ ⎟ ⎜ cos α − μ sin α ⎟ + W ⋅ Zhρ i g ⎜⎜ cos α − μ sin α + ⎟ α tan ⎝ ⎠ ⎝ ⎠ simplified 1/ 4 ⎛ ρ w gh 5 ⎞ H = W ⋅σ f ⎜ ⎟ ⎝ E ⎠ ⋅ C1 + W ⋅ Zhρ i g ⋅ C2 Simple 2D theory continue Constants C1 and C2 predicted vs α and μ Simple 2D theory continue Breaking force 1/ 4 ⎛ ρ w gh ⎞ H /W = σ f ⎜ ⎟ E ⎝ ⎠ 5 Ride-up force ⋅ C1 + Zhρ i g ⋅ C2 (1) Example 1: Effects of thickness Eq. (1), conical structure Parameters: α = 45D , μ = 0.3 σ f = 0.7 MPa, Z = 5 m h = 1 m → H / W = 144 kN / m h = 3 m → H / W = 462 kN / m Ride-up force dominates for h > 1 m Force ∝ h Average failure pressure p almost independent of h p ≈ 0.15MPa for h = 1-3 m, Example 2: Effects of α and μ Eq. (1), conical structure Parameters: σ f = 0.7 MPa, E = 5 GPa, h = 1 m α = 45D , μ = 0.1, 0.5 : H / W = 95 kN / m, 235 kN / m α = 55D , μ = 0.1, 0.5 : H / W = 125 kN / m, 430 kN / m Friction effects significant for slopes steeper than 45˚ Steeper angles → more crushing → higher loads Important to maintain smooth surfaces for sloping structures to minimize ice loads Effects of ice strength Affects the breaking part Wide structures: – ride-up part > breaking part (2D situation) Narrow structures: – 3D effects, ice strength important Effects of ice thickness Breaking: Fbreaking ( h 1.25 ) Ride-up: Fride up ( h ) Ice thickness is the most important parameter for load estimation on all sloping structures Effects of velocity – upward cone Influence of velocity only if V > 0.5 m/s (F0.5 is the load at 0.5 m/s) η = FV / F0.5 ⎧1 if V < 0.5 m / s η=⎨ ⎩1 + 0.5(V − 0.5) if V > 0.5 m / s 2D vs 3D model Wide structures: – 2D assumption valid – simple 2D beam on elastic foundation may be assumed Narrow structures: – 3D effects will dominate – failure zone wider than structure – plate theory more valid than beam theory 3D model Ralston (1977) 3D plate model based on plastic limit analysis (ice as a ductile material) H = A4 ⎡⎣ A1σ f h 2 + A2 ρ w ghD 2 + A3ρ w gh ( D 2 − DT ) ⎤⎦ 2 2 V = B1H + B2 ρ w gh( D − DT ) DT - top diameter D – waterline diameter A, B coefficients Adfreeze on sloping structures Croasdale (1980) Fadfreeze Fadfr. h q W π hqW = tan α - horizontal ice load due to adfreezing (MN) - ice thickness (m) - adfreeze bond strength (0.3-1 MPa) - width of struture (m) Kulluk Full-scale - Kulluk Beaufort Sea Resistance of a ship Resistance Inertia Force Hydrodynamic Force Open Water Resistance Breaking Friction, Buoyancy Speed Ice barriers – Caspian Sea Ice barriers – Caspian Sea
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