62
CHAPTER 3. PRINCIPLES OF SIMILITUDE
Example 3.4. Dynamically Similar Breakwater Stability
A model breakwater is constructed of stones that have an average weight of 5
newtons (1.1 lbs). Tests with irregular waves reveal that the model breakwater fails
when significant wave height, Hmo, equals 0.3 m. What weight of stone is necessary
to withstand a significant wave height of 10 m in the prototype? Assume the model
is geometrically similar to the prototype, and that the prototype stone has the same
specific weight as the model stone. (The effect of a salt water prototype and a fresh
water model is neglected in this example, but this effect is examined in Chapter 5.)
The length scale is determined as
Nl =
(f?mo)p
10 m
0.3 m
The weight of the armor stone is the specific weight of the material multiplied by the
volume of the stone, or W, = ys -V3, which can be expressed in scale ratios as
XT
_ (Wj)p _ (7 * )p
‘ " (Ws)m
(7s)m
(Vs)p
(K)m
=
Nv.
Because the armor units are made of the same material in the model and prototype, we know that N^, = 1. We also note that in a geometrically similar model,
the volume scale is simply the length scale cubed, i.e., Nv, = Nf. Making these
substitutions yields
Ww, =
= Nl = (33.3)3 = 37,037
("s )m
Therefore, the model tests indicate that the weight of armor stone needed to
resist significant wave heights of 10 m in the prototype is
(PV,)p = 37,037 (jy,)m = 37,037 (5 W) = 185.2 kN (or 20.8 short tons)
3.3 Hydraulic Similitude
3.3.1 Practical Aspects of Hydraulic Similitude
As mentioned in the previous section, being able to achieve complete similitude where all the force ratios are constant and equal (Eqn. 3.6) is impossible except at prototype scale. However, knowing the requirements for
complete similitude allows us to evaluate the consequences of violating it.
In most problems a combination of experience and common sense will help
in choosing those ratios that need to be in similitude, and an important
CHAPTER 3. PRINCIPLES OF SIMILITUDE
Example 3.4. Dynamically Similar Breakwater Stability
A model breakwater is constructed of stones that have an average weight of 5
newtons (1.1 lbs). Tests with irregular waves reveal that the model breakwater fails
when significant wave height, Hmo, equals 0.3 m. What weight of stone is necessary
to withstand a significant wave height of 10 m in the prototype? Assume the model
is geometrically similar to the prototype, and that the prototype stone has the same
specific weight as the model stone. (The effect of a salt water prototype and a fresh
water model is neglected in this example, but this effect is examined in Chapter 5.)
The length scale is determined as
Nl =
(f?mo)p
10 m
0.3 m
The weight of the armor stone is the specific weight of the material multiplied by the
volume of the stone, or W, = ys -V3, which can be expressed in scale ratios as
XT
_ (Wj)p _ (7 * )p
‘ " (Ws)m
(7s)m
(Vs)p
(K)m
=
Nv.
Because the armor units are made of the same material in the model and prototype, we know that N^, = 1. We also note that in a geometrically similar model,
the volume scale is simply the length scale cubed, i.e., Nv, = Nf. Making these
substitutions yields
Ww, =
= Nl = (33.3)3 = 37,037
("s )m
Therefore, the model tests indicate that the weight of armor stone needed to
resist significant wave heights of 10 m in the prototype is
(PV,)p = 37,037 (jy,)m = 37,037 (5 W) = 185.2 kN (or 20.8 short tons)
3.3 Hydraulic Similitude
3.3.1 Practical Aspects of Hydraulic Similitude
As mentioned in the previous section, being able to achieve complete similitude where all the force ratios are constant and equal (Eqn. 3.6) is impossible except at prototype scale. However, knowing the requirements for
complete similitude allows us to evaluate the consequences of violating it.
In most problems a combination of experience and common sense will help
in choosing those ratios that need to be in similitude, and an important