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Hydraulics 3 Waves: Loading – 1 Dr David Apsley
4. WAVE LOADING ON STRUCTURES AUTUMN 2022
4.1 Pressure Distribution
The pressure force (pressure  area) on a structure such as a breakwater is
∫ 𝑝 d𝐴
or, per unit width,
∫ 𝑝 d𝑧
where the integral is over the submerged depth of the structure (which may be surface-piercing
or fully submerged).
For a progressive wave the pressure distribution has hydrostatic and wave components:
𝑝 = −𝜌𝑔𝑧
⏟
hydrostatic
−𝜌
𝜕𝜙
𝜕𝑡
⏟
wave
= −𝜌𝑔𝑧 + 𝜌𝑔𝐴
cosh 𝑘(ℎ + 𝑧)
cosh 𝑘ℎ
cos(𝑘𝑥 − 𝜔𝑡)
but here we are only interested in the dynamic component, as the hydrostatic force is a static
contribution equal (at least below the SWL) to that in still water.
Assuming total reflection (the worst case) the dynamic pressure component for an incident
regular wave 𝜂 = 𝐴 cos(𝑘𝑥 − 𝜔𝑡) is
𝑝 = 𝜌𝑔𝐴
cosh 𝑘(ℎ + 𝑧)
cosh 𝑘ℎ
{cos(𝑘𝑥 − 𝜔𝑡) + cos(−𝑘𝑥 − 𝜔𝑡)}
= 2𝜌𝑔𝐴
cosh 𝑘(ℎ + 𝑧)
cosh 𝑘ℎ
cos 𝑘𝑥 cos 𝜔𝑡
and hence the maximum wave pressure over a cycle at 𝑥 = 0 is
𝑝 = 𝜌𝑔𝐻
cosh 𝑘(ℎ + 𝑧)
cosh 𝑘ℎ
where 𝐻 (= 2𝐴) is the height of the incident wave. (Since the water surface elevation due to
incident plus reflected waves has amplitude 𝜂max = 2𝐴 = 𝐻, then 𝐻 is also the maximum crest
height above SWL.)
Hydraulics 3 Waves: Loading – 2 Dr David Apsley
4.2 Surface-Piercing Structure
The maximum wave pressure force per unit width is (with the linear-wave-theory
approximation 𝜂 ≪ ℎ, so that the upper limit of integration can be taken as 0 rather than η):
𝐹 = ∫ 𝑝 d𝑧 =
𝜌𝑔𝐻
cosh 𝑘ℎ
∫ cosh 𝑘(ℎ + 𝑧) d𝑧
0
𝑧=−ℎ
=
𝜌𝑔𝐻
cosh 𝑘ℎ
[
sinh 𝑘(ℎ + 𝑧)
𝑘
]
−ℎ
0
=
𝜌𝑔𝐻 sinh 𝑘ℎ
𝑘 cosh 𝑘ℎ
Hence,
𝐹 =
𝜌𝑔𝐻 tanh 𝑘ℎ
𝑘
In the shallow-water limit (tanh 𝑘ℎ ~𝑘ℎ):
𝐹 = 𝜌𝑔𝐻ℎ
This is essentially hydrostatic; (constant excess wave pressure 𝜌𝑔𝐻 over depth ℎ).
In the deep-water limit (tanh 𝑘ℎ → 1):
𝐹 =
𝜌𝑔𝐻
𝑘
This is independent of depth (because the wave disturbance does not extend the whole way to
the bed).
4.3 Fully-Submerged Structure
The only change is that the upper limit of integration is −(ℎ − 𝐵), where 𝐵 is the height of the
structure. The maximum wave pressure force per unit width is
𝐹 = ∫ 𝑝 d𝑧 =
𝜌𝑔𝐻
cosh 𝑘ℎ
∫ cosh 𝑘(ℎ + 𝑧) d𝑧
−(ℎ−𝐵)
𝑧=−ℎ
=
𝜌𝑔𝐻 sinh 𝑘𝐵
𝑘 cosh 𝑘ℎ
This is always less than the surface-piercing case (since sinh 𝑘𝐵 < sinh 𝑘ℎ).
Hydraulics 3 Waves: Loading – 3 Dr David Apsley
4.4 Loads on a Vertical (Caisson-Type) Breakwater
As we can see from the plots above, a conservative approach
(i.e., one which tends to overestimate pressure on the
structure) is to assume that the excess pressure over that
which would exist in the absence of waves is (see diagram):
• hydrostatic in the wave crest above the SWL;
• varies linearly with depth between the wave
pressures at the SWL and the bed.
Additionally, if water can get underneath the breakwater
(porous foundation) there may be upthrust due to wave-
induced pressure, which contributes to the overturning
moment about the back heel of the structure. For this,
assume a linear distribution between the bed pressure at the
front and zero at the back (see diagrams).
For a surface-piercing structure, the force distribution can be
broken down into:
1: triangular pressure distribution above SWL;
2, 3: uniform + triangular pressure distributions bed to SWL;
and, if water can get underneath the breakwater:
4: triangular distribution pressure distribution below.
The relevant pressures are:
𝑝1 = 0
𝑝2 = 𝜌𝑔𝐻
𝑝3 =
𝜌𝑔𝐻
cosh 𝑘ℎ
The forces and moment arms are, per unit length of breakwater:
p1
3
p
3
p
1
2
3
4
2
p
heel
F
x
My
F
z
bed
SWL
crest
p1
2
p
3
p
3
p
Hydraulics 3 Waves: Loading – 4 Dr David Apsley
Force moment arm
𝐹1𝑥 =
1
2
𝑝2𝐻 𝑧1 = ℎ +
1
3
𝐻
𝐹2𝑥 = 𝑝3ℎ 𝑧2 =
1
2
ℎ
𝐹3𝑥 =
1
2
(𝑝2 − 𝑝3)ℎ 𝑧3 =
2
3
ℎ
𝐹4𝑧 =
1
2
𝑝3𝑏 𝑥4 =
2
3
𝑏
The net horizontal force (per unit length of breakwater) is
𝐹1𝑥 + 𝐹2𝑥 + 𝐹3𝑥
The net overturning moment (per unit length of breakwater) about the heel is
𝐹1𝑥𝑧1 + 𝐹2𝑥𝑧2 + 𝐹3𝑥𝑧3 (+𝐹4𝑧𝑥4)
with the last term only applying for porous foundation.
For a fully-submerged structure we can perform a similar
breakdown of forces with a similar piecewise-linear
pressure distribution, but in this instance we only need the
following pressures:
𝑝SWL = 𝜌𝑔𝐻
𝑝bed =
𝜌𝑔𝐻
cosh 𝑘ℎ
and, by interpolation,
𝑝top = 𝑝bed +
𝐵
ℎ
(𝑝SWL − 𝑝bed)
Example.
A regular wave of period 8 s and height 1.2 m is normally incident to a caisson-type breakwater
of 4 m breadth and located in water depth of 5 m. Using linear theory, determine the following:
(a) Caisson height required for 1 m freeboard above the peak wave elevation at the
breakwater.
(b) Maximum wave-induced horizontal force per metre of wave crest.
(c) Maximum wave-induced overturning moment per metre of wave crest.
bed
SWL
crest
B
top

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WavesLoading.pdf

  • 1. Hydraulics 3 Waves: Loading – 1 Dr David Apsley 4. WAVE LOADING ON STRUCTURES AUTUMN 2022 4.1 Pressure Distribution The pressure force (pressure  area) on a structure such as a breakwater is ∫ 𝑝 d𝐴 or, per unit width, ∫ 𝑝 d𝑧 where the integral is over the submerged depth of the structure (which may be surface-piercing or fully submerged). For a progressive wave the pressure distribution has hydrostatic and wave components: 𝑝 = −𝜌𝑔𝑧 ⏟ hydrostatic −𝜌 𝜕𝜙 𝜕𝑡 ⏟ wave = −𝜌𝑔𝑧 + 𝜌𝑔𝐴 cosh 𝑘(ℎ + 𝑧) cosh 𝑘ℎ cos(𝑘𝑥 − 𝜔𝑡) but here we are only interested in the dynamic component, as the hydrostatic force is a static contribution equal (at least below the SWL) to that in still water. Assuming total reflection (the worst case) the dynamic pressure component for an incident regular wave 𝜂 = 𝐴 cos(𝑘𝑥 − 𝜔𝑡) is 𝑝 = 𝜌𝑔𝐴 cosh 𝑘(ℎ + 𝑧) cosh 𝑘ℎ {cos(𝑘𝑥 − 𝜔𝑡) + cos(−𝑘𝑥 − 𝜔𝑡)} = 2𝜌𝑔𝐴 cosh 𝑘(ℎ + 𝑧) cosh 𝑘ℎ cos 𝑘𝑥 cos 𝜔𝑡 and hence the maximum wave pressure over a cycle at 𝑥 = 0 is 𝑝 = 𝜌𝑔𝐻 cosh 𝑘(ℎ + 𝑧) cosh 𝑘ℎ where 𝐻 (= 2𝐴) is the height of the incident wave. (Since the water surface elevation due to incident plus reflected waves has amplitude 𝜂max = 2𝐴 = 𝐻, then 𝐻 is also the maximum crest height above SWL.)
  • 2. Hydraulics 3 Waves: Loading – 2 Dr David Apsley 4.2 Surface-Piercing Structure The maximum wave pressure force per unit width is (with the linear-wave-theory approximation 𝜂 ≪ ℎ, so that the upper limit of integration can be taken as 0 rather than η): 𝐹 = ∫ 𝑝 d𝑧 = 𝜌𝑔𝐻 cosh 𝑘ℎ ∫ cosh 𝑘(ℎ + 𝑧) d𝑧 0 𝑧=−ℎ = 𝜌𝑔𝐻 cosh 𝑘ℎ [ sinh 𝑘(ℎ + 𝑧) 𝑘 ] −ℎ 0 = 𝜌𝑔𝐻 sinh 𝑘ℎ 𝑘 cosh 𝑘ℎ Hence, 𝐹 = 𝜌𝑔𝐻 tanh 𝑘ℎ 𝑘 In the shallow-water limit (tanh 𝑘ℎ ~𝑘ℎ): 𝐹 = 𝜌𝑔𝐻ℎ This is essentially hydrostatic; (constant excess wave pressure 𝜌𝑔𝐻 over depth ℎ). In the deep-water limit (tanh 𝑘ℎ → 1): 𝐹 = 𝜌𝑔𝐻 𝑘 This is independent of depth (because the wave disturbance does not extend the whole way to the bed). 4.3 Fully-Submerged Structure The only change is that the upper limit of integration is −(ℎ − 𝐵), where 𝐵 is the height of the structure. The maximum wave pressure force per unit width is 𝐹 = ∫ 𝑝 d𝑧 = 𝜌𝑔𝐻 cosh 𝑘ℎ ∫ cosh 𝑘(ℎ + 𝑧) d𝑧 −(ℎ−𝐵) 𝑧=−ℎ = 𝜌𝑔𝐻 sinh 𝑘𝐵 𝑘 cosh 𝑘ℎ This is always less than the surface-piercing case (since sinh 𝑘𝐵 < sinh 𝑘ℎ).
  • 3. Hydraulics 3 Waves: Loading – 3 Dr David Apsley 4.4 Loads on a Vertical (Caisson-Type) Breakwater As we can see from the plots above, a conservative approach (i.e., one which tends to overestimate pressure on the structure) is to assume that the excess pressure over that which would exist in the absence of waves is (see diagram): • hydrostatic in the wave crest above the SWL; • varies linearly with depth between the wave pressures at the SWL and the bed. Additionally, if water can get underneath the breakwater (porous foundation) there may be upthrust due to wave- induced pressure, which contributes to the overturning moment about the back heel of the structure. For this, assume a linear distribution between the bed pressure at the front and zero at the back (see diagrams). For a surface-piercing structure, the force distribution can be broken down into: 1: triangular pressure distribution above SWL; 2, 3: uniform + triangular pressure distributions bed to SWL; and, if water can get underneath the breakwater: 4: triangular distribution pressure distribution below. The relevant pressures are: 𝑝1 = 0 𝑝2 = 𝜌𝑔𝐻 𝑝3 = 𝜌𝑔𝐻 cosh 𝑘ℎ The forces and moment arms are, per unit length of breakwater: p1 3 p 3 p 1 2 3 4 2 p heel F x My F z bed SWL crest p1 2 p 3 p 3 p
  • 4. Hydraulics 3 Waves: Loading – 4 Dr David Apsley Force moment arm 𝐹1𝑥 = 1 2 𝑝2𝐻 𝑧1 = ℎ + 1 3 𝐻 𝐹2𝑥 = 𝑝3ℎ 𝑧2 = 1 2 ℎ 𝐹3𝑥 = 1 2 (𝑝2 − 𝑝3)ℎ 𝑧3 = 2 3 ℎ 𝐹4𝑧 = 1 2 𝑝3𝑏 𝑥4 = 2 3 𝑏 The net horizontal force (per unit length of breakwater) is 𝐹1𝑥 + 𝐹2𝑥 + 𝐹3𝑥 The net overturning moment (per unit length of breakwater) about the heel is 𝐹1𝑥𝑧1 + 𝐹2𝑥𝑧2 + 𝐹3𝑥𝑧3 (+𝐹4𝑧𝑥4) with the last term only applying for porous foundation. For a fully-submerged structure we can perform a similar breakdown of forces with a similar piecewise-linear pressure distribution, but in this instance we only need the following pressures: 𝑝SWL = 𝜌𝑔𝐻 𝑝bed = 𝜌𝑔𝐻 cosh 𝑘ℎ and, by interpolation, 𝑝top = 𝑝bed + 𝐵 ℎ (𝑝SWL − 𝑝bed) Example. A regular wave of period 8 s and height 1.2 m is normally incident to a caisson-type breakwater of 4 m breadth and located in water depth of 5 m. Using linear theory, determine the following: (a) Caisson height required for 1 m freeboard above the peak wave elevation at the breakwater. (b) Maximum wave-induced horizontal force per metre of wave crest. (c) Maximum wave-induced overturning moment per metre of wave crest. bed SWL crest B top