Slamming

Question 1

What is slamming?

Answer 1

In rough seas large relative motions between ship and the free water surcafe have to be expected. Due to this large relative motion shocks take place between the vessel hull and the free surface.

Question 2

Why is slamming so important to consider?

Answer 2

These forces are:

• typically the largest local loads in a seaway
• can cause local damage
• can lead to bending vibrations (whipping)
• contributes to material fatigue

Question 3

What ships are particularly affected by slamming?

Answer 3

Fatigue and slamming are problems for light high speed ships but they can also happen for bulk carriers and tankers, which typically operate in the half of their life at ballast draft and with only moderate immersed bow

Question 4

What is the calculation of slamming dependent on?

Answer 4

Contact angle between water and structure. It is a stochastic process

Question 5

What influences the pressure profile?

Answer 5

High slamming pressures are effictive over a short period of time, therefore, the elastic or even plastic behaviour of the structure has great influence on the pressure profile. Also the water elasticity can change considerably with the amount of air bubbles (because of extreme waves). Air entrapment between the ship and water happens by slamming and causes compressible air curents, their speed can exceed the sound speed.

Question 6

What methods can be used to analyse slamming?

Answer 6

Linear methods for calculation of the relative motion between ship and water can be applied to determine of the frequency of occurence of slamming impacts, and based on the relative velocity simplified calculations of the loads can be carried out.
- Calculations of forces and pressures on the frame under water can be done by numerical solutions like Euler or RANS equations for a given speed of immersion. For the ship it is the same ecxept you need the speed of the waves. The numerical effort can increase significantly from method to method

Question 7

How can a slamming investigation be performed in practice?

Answer 7

• High number of linear simulations and more complex calculations for resulting extreme motions can be performed
• Dimensioning wave method

Question 8

Why is slamming so hard to analyze?

Answer 8

• Non-linear and stochastic seaway

Question 9

How can slamming loads be reduced?

Answer 9

• V-shape sections in the bow and deadrise in the forward and after part of the hull floor

Question 10

When does the slamming impact occure and what happens then?

Answer 10

The relative motion between the ship and the water surface can be defined as r(x,t). r is positive, whne the actual water level of the free surface is lower than the average level.

A slamming impact occurs when the ship’s bottom emerges behind a critical x-coordinate Xk. If the water surface hts the ship bottom after this point where the ship bottom is flat and parallel to the water surface, then slamming occurs.

Question 11

What conditions must be satisfied for slamming to occur?

Answer 11

• r(xk,t) = T –> At the ship bottom where T = 0
• r’(xk,t) < 0 –> X goes in the negative direction
• r*(xk,t) < 0 –> Wave goes up because r is defined as positive downwards

Derive r^ with regard to x and t to see if the conditions are fulfilled.

Question 12

Describe the mean frequency of slamming. Explain the picture/graph

Answer 12

x-axis: r and T (relative motion and depth)
r decreases when the wave rises on the hull
y-axis: r* (wave position)
G is the area where the conditions are fulfilled so that slamming occurs
deltaT is the area in which we have the mean depth.

Question 13

Where can slamming occur?

Answer 13

• in heavy seas when the foreship emerges from the water and immerses in the water again
• on the wet deck of catamarans
• on deck or horizontal struts of offshore structures
• in tanks due to high acceleration of free surface of a liquid in a tank agains the tanktop

Question 14

What are the consequences of slamming?

Answer 14

Structure deformation, whipping (which is harmful to people) that can lead to stress concentration and cracks

Question 15

Explain the forces on elliptical bodies (Karman method)

Answer 15

The velocity potential must satisfy the following conditions:

• ø_xx + ø_zz = 0 (continuity equation)
• delta_ø = 0 (boundary condition far below the free surface)
• ø_tt - gø_z + 2delta_ødelta_ø_t + 1/2(delta_ødelta)(delta_ødelta_ø) = 0 (the free surface boundary condition - stoke waves)

=> remaining boundary condition is ø_tt = 0 (for the free water surface)

Question 16

How can we find the force on the ship body? (Karman)

Answer 16

First the potential is found by integrating the speed and the “Ansatzfunktion” amongst others, the the potential in the z direction is found. Thereafter the pressure is found by the use of the bernoulli equation where a lot of the part become zero so we end up with p = - rho*ø_t.

The force is found by integrating the pressure over the surface where we get F = - d/dtV(rhoPi/2c^2). c is the wetted surface and can be found by using phytagoras. The impact force is strongly dependent on the increase of c^2. The maximum impact force occurs at the beginning of the impact.

Karman force: F = -pirhoR*V^2

Question 17

How are the forces found according to Wagner?

Answer 17

Etz (n) increases the rise of the free surface which leads to greater impact forces which again increases the wetted length, c(t). This means that the deformation of the surface is considered. Consequently, the water level increases as strong as the body immerses below the water line. This doubles c^2 in comparison to Kauman. The doubling of c^2 leads to the doubling of determined force based on Wagners theory.

F = - 2pirhoRV^2

Question 18

Explain splashing roots

Answer 18

The splashing root is the point where the water hits the hull and is accelerated in a vertical direction. The flow is stationary in this region and the pressure is constant on the free surface. According to Bernoulli, the velocity is constant. The speed is said to be dc/dt. A stagnation point is located at the vertical free surface on the body controur and the pressure in this point is found by using bernoulli –> P = 1/2rhoV^2(R/Vt). This is also called the acoustic pressure.

Question 19

Acoustic pressure

Answer 19

P = 1/2rhoV^2(R/Vt). infinite pressure takes place at t=0. In reality the compressability of the fluid and the elasticity of the body prevents infinite pressure.

To investigate this problem we look at a vertical elastic liquid column which is pressed at the top of the surface. The load is distributed uniformly on the whole column area. It works basically as a spring.

acoustic pressure: p= rhocsV