1) The document presents Von Karman 3D procedures as a generalization of Westergaard 2D formulae for estimating hydrodynamic forces on dams during earthquakes. 2) Westergaard's 2D formulae have limitations and can produce singularities, while Von Karman's approach is non-periodic and non-singular with easier generalization to 3D. 3) Numerical examples show that accounting for 3D effects through Von Karman procedures can increase estimated overturning moments on dams by up to 7% compared to 2D Westergaard formulae.

1. Submittal 120611 to Hydrovision International - for July 18, 2012 presentation in Louisville KY, USA
Page 1 of 14
Luc-Marie A. Jeudy de Sauceray, X67, Ph.D., P.E.
LucMarie.JeudydeSauceray@aecom.com
AECOM Canada Ltd - Suite 970 - 789 West Pender
Vancouver, BC - V6C 1H2 - Canada - Ph: (604) 828-0503
VON KARMAN 3D PROCEDURES AS A GENERALIZATION OF WESTERGAARD 2D
FORMULAE WITH IMPLICATIONS FOR IMPROVED SEISMIC UPGRADE ESTIMATES
Abstract
Westergaard two dimensional (2D) pressure formulae have been used for a number of years to
evaluate hydrodynamic forces acting on water retaining structures during earthquakes. These
forces have a sizable impact on the cost of structural refurbishment and their correct estimation is
critical to optimize dollar amounts versus risk minimization.
We develop the Von Karman 3D procedures which, in many cases (within the assumption of
constant & uniform dam acceleration), provide more accurate seismic forces and overturning
moments estimates. Examples are worked out numerically and presented in a table for easy
comparison.
_______________________
Introduction:
Hydrodynamic forces generated by earthquakes on water retaining structures have been analyzed
by numerous authors, as can be found in reference [1]. In this context, our topic is:
1. to review H.M. Westergaard (HMW) 2D closed formulae,
2. to summarize their various limitations and compare them to the 2D Theodor Von Karman
(TVK) solutions, addressing various issues, then
3. to develop the TVK 3D procedures and formulae, providing a few numerical examples,
showing revisions (up to 7% increase) in overturning (OT) moment estimates (cf (30)I
).
I. - HMW 2D Formulae:
Following professor1
H.M. Westergaard (HMW), in his paper [2], having defined:
- g the earth gravitational acceleration (9.81m/s2
),
- α the seismic ratio (expressing the average seismic acceleration as a fraction of g,
- x axis, horizontal and perpendicular to the upstream face of the dam, oriented upstream
- y axis, along the upstream face of the dam and directed vertically downwards (cf fig. 1)
- t being the time “coordinate”,
the general solution for a liquid element displacements of volume dV = dxdydz, in the x
and y directions respectively (ξ,η), with its pressure (p) is written as follows:
ξ(x,y,t) = - (αgT2
/π3
) cos(2πt/T) Σn(1/n) exp(-qn) sin(nπy/2h) (1)
1
At the time (1933) professor with University of Illinois, Urbana, Ill.

2. Submittal 120611 to Hydrovision International - for July 18, 2012 presentation in Louisville KY, USA
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η(x,y,t) = (αgT2
/π3
) cos(2πt/T) Σn(1/ncn) exp(-qn) cos(nπy/2h) (2)
p(x,y,t) = (8αρgh/π2
) cos(2πt/T) Σn[1/(n2
cn)] exp(-qn) sin(nπy/2h) (3)
where T is the period of the assumed HMW sinusoidal seismic excitation function2
in the
x direction, h the water depth at the upstream face of the dam, with summation Σn for all
odd integers (n = 1, 3, 5…), with qn defined as:
qn = nπcnx/2h
and cn as:
cn = {1 - (16h2
/n2
λ
2
)}1/2
(4)
where λ = veT is the wavelength of sound vibration in water and,
ve = (k/ρ)1/2
= √(2200E6/E3) = 1483m/s the propagation velocity;
λ2
= kT2
/ρ where k is the water bulk modulus,
k = 2200E6Pa (Pa = N/m2
),
ρ = 1000kg/m3
in kilograms per cubic meters the water density
Alternately: an expression for cn using the period T can be derived from (4):
cn = {1 - (16ρh2
/n2
kT2
)}1/2
(5)
Note#1: Numerically, HMW, at the time, to avoid singularities, assumed the period T to
be large enough and the dam height (h) small enough, so that cn could be considered very
close to unity and assume cn = 1 for all n.
Numerically,
we ended up with: cn = {1 - (16x1000h2
/2200E6n2
T2
)}1/2
(5)I
and finally: cn = {1 – (7.2727E-6)h2
/n2
T2
}1/2
(5)II
In particular, for the first coefficient c1,
which can be a problem in some cases (undefined due to square root of a negative value),
we have:
c1 = {1 – (7.2727E-6)(h/T)2
}1/2
(6)
h in meters and T in seconds
Comment#1: Even with h = 40m (120’) and T=0.2s we have c1 = 0.842 (close to 1)
Note#2: Water incompressibility: As it is noted elsewhere in [2] p.452, Bratz3
&
Heilbron4
(B&H) mention the assumption of water incompressibility leading to cn = 1
2
There is no loss of generality since a seismic spectrum can be expanded into a Fourier series
where each term would be similar to the above mentioned HMW excitation function
3
California Institute of Technology, Pasadena
4
Pasadena Water Department

3. Submittal 120611 to Hydrovision International - for July 18, 2012 presentation in Louisville KY, USA
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for all n, as being due to the fact that k becoming infinite (∞) in formula (5), for any
given h and any given T, gives us:
cn = {1 - (16ρh2
/n2
kT2
)}1/2
1 as k ∞, for all n, h and T
B&H concluded that compressibility would become an issue only for high dams.
Note#3: Modified equation for incompressibility:
This would translate into a modified elasticity equation with zero volume change, which
in two dimensions is written as:
0 = ∂ξ/∂x + ∂η/∂y, (7)
where “∂” stands for the partial derivative symbol,
The two other equations (as shown in [2] by HMW) being derived from the Cauchy
equations (cf [3], p.59) are the following:
∂σ/∂x = ρ ∂2
ξ/∂t2
∂σ/∂y = ρ ∂2
η/∂t2
,
for a total of 3 equations, with boundary conditions to be specified, and to be solved for
the 3 unknown functions ξ , η and σ .
Remark: Now, solving for the cn coefficients, using the above modified equation (7),
produces all the “cn“ equals to one as found in Note#2 earlier.
_cn = 1, for all n in the case of incompressibility_
Note#4: Damping:
As can be seen in notes 2 and 3 above, the incompressibility assumption becomes a way
of removing singularities (since the cn coefficients are constant, and equals to unity).
If one wants to keep the compressibility and avoid the aforementioned singularities,
following A.K. Chopra, damping can be introduced (cf [1]).
Remark#1: Example#1:
As we now know, earthquake frequencies can generate high levels of energy in the sub-
second periodic range;
typically we would consider T = 0.2s, so that (for example)
with h = 70 m, we obtain using (5)I
:
cn = {1 – 0.8909/n2
}1/2
,
obtaining the following first coefficients:
c1 = 0.3303, c3 = 0.949, c5 = 0.982, ... (7)I

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Figure 1: HMW 2D reference frame Oxy: It shows the origin O and the pressure
diagram at the face of the dam depending on downward ordinate y:
HMW 2D Setup
Legend: pb is the seismically generated pressure at the bottom of the reservoir,
w the constant width of the reservoir along the z coordinate (perpendicular to the plane of
the paper) and m = w/h the width to water depth ratio (to be used in equations further
below).
Calculating the maximum pressure at the bottom of the reservoir, using (3), we have
(cosine and exponential functions becoming 1):
pb = p(x = 0, y = h, t = 0, T,…)
pb = (8αρgh/π2
)Σn[1/(n2
cn)](-1)(n-1)/2
for all odd valued n
pb = (8/π2
)[Σn[1/(n2
cn)](-1)(n-1)/2
]αρgh
pb = (8/π2
)[1/c1-1/32
c3+1/52
c5+ ]αρgh (8)
Example#1 (cont.): Continuing with (7)I
, we have:
pb = 0.8106x(1/0.3303-1/(9x0.949)+1/(25x0.982)…)αρgh
= 0.8106x(3.03-0.118+0.041…)αρgh
= 2.39αρgh, (8)I
Comment#2: comparing (8)I
with formula (9) below, we have 2.39/0.743=3.2
that is a 320% increase.
Obviously, we are getting close to a singularity (due to the c1 coefficient becoming too
small). As mentioned in Note#4 earlier, introducing damping would probably help.
We do not pursue this avenue for the moment.
Instead, we consider the Von Karman (TVK) approach which is essentially non-periodic
and non-singular with easier generalization to the 3D case.

5. Submittal 120611 to Hydrovision International - for July 18, 2012 presentation in Louisville KY, USA
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Note#5: In fact HMW never went to the point of considering such high coefficients, and
was making an assumption equivalent to the water incompressibility leading to all cn
coefficients equal to unity.
Remark#2: HMW 2D Bottom Pressure Formula:
When all the cn coefficients are taken as unity, we have:
pb = (8/π2
)[1 - 1/32
+ 1/52
- 1/72
+ 1/92
- 1/112
...]αρgh
pb = 0.8106x0.9166xαρgh
pbhmw = 0.743αρgh, (9)
where index “bhmw” stands for Bottom H.M. Westergaard
and as it is published in [2] p.436.
Remark#3: HMW 3D Equations: Introducing the third coordinate z would probably
lead to significantly modified pressures in some cases, as can be presumed from reference
[2], when HMW was discussing the 1D (one dimensional) case in relation to his more
general 2D solution. The HMW equations would have to be modified, introducing a new
unknown function ζ(x,y,z,t) describing the liquid element displacement in the z direction.
Assuming incompressibility, equation (7) would become:
0 = ∂ξ/∂x + ∂η/∂y + ∂ζ/∂z, (10)
with an additional equation, namely:
∂σ/∂z = ρ ∂2
ζ/∂t2
(11)
for a total of 4 equations with 4 unknowns: ξ(x,y,z,t), η(x,y,z,t), ζ(x,y,z,t), p(x,y,z,t),
with corresponding boundary conditions modified accordingly. Continuing along those
lines is beyond the scope of this short presentation.
Remark#4: HMW 2D Force Formula: Denoting by (F) the force acting against the
dam, “ƒ” being the integration symbol from y=0 to y=h, and w the width of the dam (cf
legend of Fig. 1), we have, assuming cn = 1 for all n:
F = ƒp(x,y,t)wdy = (8αρgh/π2
)Σn{[1/(n2
)]ƒsin(nπy/2h)w}dy =
F = (8αρgh/π2
)w Σn (2h/n3
π) = (16αρgh2
/π3
)wΣn[1/(n3
)] (n odd)
Fwi = 0.543αρgh2
per unit length (12)
as published in [2] p.436, and where “wi” index stands for Westergaard Incompressible
and where w has been removed, Fwi being a force per unit length.
The total force Ftwi would be, multiplying (12) by mh:

6. Submittal 120611 to Hydrovision International - for July 18, 2012 presentation in Louisville KY, USA
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Total Force: Ftwi = α(ρgh3
)x0.543xm (12)I
Remark#4I
: HMW 2D Moment Formula: (w not shown)
M = ƒ(h-y)p(x,y,t)dy
= hF - (8αρgh/π2
)Σnƒysin(nπy/2h)dy/(n2
)
= hF - (32αρgh3
/π4
)Σn odd (-1)(n-1)/2
/n4
= (0.543 – 0.3248) αρgh3
Mwi = 0.218αρgh3
per unit length (12)II
Total moment Mtwi is obtained by multiplying (12)II
by mh:
Total Moment: Mtwi = α(ρgh4
)x0.218xm (12)III
Remark#4II
: HMW 2D Resultant Location: Dividing the moment by the force,
we have the distance:
Mhmw / Fwi = 0.218αρgh3
/ 0.543αρgh2
= 0.401h (13)
II. - - TVK 2D Formulae:
Following professor5
Theodor Von Karman (TVK) in [2], we have:
1. During the time interval ∆t (during which the earthquake is occurring), the dam
displacement is (1/2)ax∆t2
where ax is the average dam acceleration in the x direction
perpendicular to the upstream face of the dam and oriented in the upstream direction,
the volume of water ABCD (cf Fig. 2) is modified by the volume amount (w not
shown for clarity):
∆V = y’(1/2)ax∆t2
(14)
which corresponds to the volume variation due to the dam displacement (1/2)ax∆t2
multiplied by the section height, which is AB = y’ and multiplied by w (not shown).
2. Since the rest of the water beyond the CD boundary in an upstream direction is
assumed to be at rest, this volume amount ∆V has to go upwards through boundary
BC (cf Figure 2) and/or sideways in the z direction which is not considered here (cf
3D derivation in III further below where it is considered). It equals b(1/2)ay’∆t2
,
where ay’ is the average water acceleration of the body of water ABCD in the upward
y’ direction, so that we have the volume conservation equation (assuming
incompressible fluid):
5
At the time (1933) Professor & Director at Caltech, Pasadena, California

7. Submittal 120611 to Hydrovision International - for July 18, 2012 presentation in Louisville KY, USA
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∆V = y’(1/2)ax∆t2
= b(1/2)ay’∆t2
or after simplification: (15)
y’ax = bay’ (16)
Figure 2: Water volume ABCD being analyzed during the time interval ∆t.
TVK 2D Setup
Legend: The dam width is denoted by w as in Fig. 1,
(in the direction perpendicular to the plane of this figure).
3. Next step: We express the force equilibrium in the horizontal direction by equating
the horizontal pressure force on the dam with the water mass times the acceleration
for the differential water element of thickness dy’ as it can be seen, colored in blue,
in Figure 2:
pdy’=ρbdy’ax (w not shown) becoming after simplification:
p = ρbax (17)
4. Following step: Express the force equilibrium in the vertical direction. We are
equating the top and bottom forces on the water element with its mass times its
vertical acceleration ay’, we have:
[d(pb)/dy’]dy’ = - ρbay’dy’ (w not shown)
where d(pb)/dy’ is the vertical force rate of change between the upper face and the
lower face, becoming:
d(pb)/dy’ = - ρbay’ (18)
combining with (17) and (16), we obtain:
d(ρb2
ax)/dy’ = - ρy’ax

8. Submittal 120611 to Hydrovision International - for July 18, 2012 presentation in Louisville KY, USA
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and since ax does not depend on y’
(because it is the uniform dam seismic acceleration)
and simplifying by ρ and ax , we have:
- y’ = d(b2
)/dy’ or - d(b2
) = d(y’2
)/2
- b2
= y’2
/2 + cste (18)I
for y’ = h, b = 0 at the surface, so that cste = - h2
/2 (18)II
5. TVK 2D Bottom Pressure Formula:
The expression for the volume element length b(y’), expressed as a function of y’
(measured from the bottom of the reservoir, cf Fig. 2), using (18)I
and (18)II
,
becomes:
b(y’)2
= [h2
– y’2
]/2 (19)
The pressure at the bottom is given by (17): pbvk = ρbax , so that with y’ = 0 in (19),
b = h/√2 and ax = αg we have:
pbvk = αρgh/√2
pbvk = 0.707αρgh (20)
where “bvk” index stands for “bottom von karman”, and further,
comparing with (9) we see a difference of about (0.743/0.707) -1 = 5%
We have good substantial agreement with formula (9).
6. TVK 2D Force:
Ftvk = ƒp(y)dy
Ftvk = 0.707(π/4)αρgh2
Ftvk = 0.555αρgh2
per unit length (20)I
(about 2% difference with (12))
Total Force (multiplying (20)I
by mh = reservoir width):
Fttvk = α(ρgh3
)x0.555xm (20)II
7. TVK 2D Moment Formula:
Using (19) (with y instead of y’), where “ƒ” is the integration sign from y=0 to y=h,
we have:
Mtvk = (αρg/√2)ƒy√[h2
– y2
]dy
Mtvk = αρgh3
/3√2 (20)III
Finally: Mtvk = 0.236αρgh3
per unit length (20)IV

9. Submittal 120611 to Hydrovision International - for July 18, 2012 presentation in Louisville KY, USA
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Note: Comparing with (12)II
, this gives .236/.218 = 1.08, so that we have an 8%
difference between the two approaches (HMW 2D vs TVK 2D),
TVK 2D being more conservative.
The total Moment (multiplying (20)IV
by mh) is:
Mttvk = α(ρgh4
)x0.236xm (20)V
8. TVK 2D Resultant Location:
Mtvk / Ftvk = 0.236αρgh3
/ 0.555αρgh2
= 0.425h (20)VI
Remark#5: In a way similar to the HMW 2D solution discussed earlier, no allowance
was made for a third dimension in the TVK 2D general derivation shown above.
III. - - - TVK 3D Formulae:
Following TVK and an example in [2], we have the following:
Example#2: An example below (cf [2] p.471) is shown, where the z coordinate is
introduced in a direction perpendicular to the upstream direction. Let us assume a cross
section with bottom width h/2 and top surface width 3h/2 with sloping sides at 63.4
degree angle from the horizontal (cf Fig. 3).
Figure 3: Example#2 reservoir cross section, with B(y) width function:
TVK 3D:
B(y) = h/2+y, B(0)=h/2 at the bottom, B(h)=3h/2 on top
Legend: The section plane is perpendicular to the upstream direction.

10. Submittal 120611 to Hydrovision International - for July 18, 2012 presentation in Louisville KY, USA
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The fluid cross section area of height (y) is denoted by A(y), and B(y) denotes the width
of the water reservoir at elevation y (y being zero at the bottom) so that:
A(y) = ƒB(u)du where u is a dummy integration variable (20)VII
integrated from 0 to the current value y.
TVK equation (15), for the volume expression, becomes:
∆V = A(1/2)ax∆t2
= B(1/2)bay ∆t2
so that: Aax = Bbay (21)
(17) is unchanged: p(y) = ρb(y)ax (22)
and (18) is modified as follows:
d(pBb)/dy = - ρBbay (23)
Introducing (21) and (22) in equation (23), we have:
d(ρBb2
ax)/dy = - ρAax (24)
simplifying by ρax , it leads to:
d(Bb2
)/dy = - A (25)
where the fundamental TVK 3D expression reads: _
B(y)b(y)2
= - ƒAdu + cste (26)
where u is a dummy integration variable running from 0 to y.
Note#6:
B(y) is given as a function of y, (describing the shape of the reservoir side walls),
A is the integral of B(y) (cf (20)VII
) with boundary condition A(0) = 0, so that b is a
function of y, which can be entirely calculated explicitly, as shown in (26). The cste in
(26) is determined by the boundary condition b(h) = 0, because b = 0 at the water surface
(y=h), the y direction being oriented upwards with zero origin at the bottom.
Example#2 (cont.): using B = h/2 + y (cf Fig. 3), we find:
A = hy/2 + y2
/2 and:
Bb2
= - ƒAdu = 5h3
/12 – hy2
/4 – y3
/6
b(0) = h√(10/12) = 0.913h (27)

11. Submittal 120611 to Hydrovision International - for July 18, 2012 presentation in Louisville KY, USA
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Finally pbz2 = 0.913αρgh (28)
index “bz2” stands for bottom z case example#2 (“z” for 3D).
Note#7: As a comparison, had we used, ... cf Example#3 below:
Example#3: B = h + y/2, we would have found:
A = hy + y2
/4 and
Bb2
= - y3
/12 – hy2
/2 + 7h3
/12 , so that:
b(0) = h√(7/12) = 0.764h < 0.913h (29)
Finally pbz3 = 0.764αρgh
Continuing with our TVK 3D general derivation: _
From formula (26),
the total force Ftvk3D is given by:
Fttvk3D = ƒBp(y)dy = αρgƒBbdy where Bb = [- BƒAdu]1/2
leading to the TVK 3D expressions for the total force and overturning moment: _
Fttvk3D = αρgƒBbdy (29)I
Total 3D Force: Fttvk3D = αρgƒ{[- BƒAdu]1/2
}dy (30)
Mttvk3D = αρgƒBbydy
Total 3D Moment: Mttvk3D = αρgƒ{[- BƒAdu]1/2
}ydy (30)I
where small “ƒ” is the integration symbol from 0 to current y and “ƒ” is the integration
symbol from 0 to h.
Example#2 (cont.): we find the TVK 3D force:
Bb = [(h/2 + y)(5h3
/12–hy2
/4–y3
/6)]1/2
= h2
[(1/2+y/h)(5/12–y2
/4h2
–y3
/6h3
)]1/2
so that, using (29)I
, we end up integrating a non-dimensional
function of v=y/h with dy=hdv from 0 to 1, giving:
3D TVK force: F3Dtvk2 = αρgƒBbdy = 0.498αρgh3
(cf [2] for details);
Note: To make a comparison, it is possible to use the TVK 2D formulae
by integrating along the z variable (the 3rd
coordinate) where the water depth h(z)
becomes a function of z. In the case of example#2, we have h(z)=2z along the slopes,
so that with integrals from 0 to h/2, plus the first term for the flat area, we have:
F2Dint = 0.555αρgh2
xh/2 + 2x0.555αρgƒ(2z)2
dz = (5/6)0.555αρgh3
= 0.463αρgh3

12. Submittal 120611 to Hydrovision International - for July 18, 2012 presentation in Louisville KY, USA
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slightly less than the 3D value shown above (i.e. 0.498αρgh3
).
3D TVK OT moment can be found using:
M = αρgƒBbydy = αρgh2
ƒ[(1/2+y/h)(5/12–y2
/4h2
–y3
/6h3
)]1/2
ydy
M = αρgh4
√(5/24)ƒ[(1+2v)(1–3v2
/5–2v3
/5)]1/2
vdv (int. 0 to 1)
where y = vh and dy = hdv, and ƒ integration symbol from v=0 to
v=1, having √(5/24) = 0.456, M becomes:
M = αρgh4
x 0.456 x 0.5136
finally,
3D TVK moment M = 0.234αρgh4
(31)
Resultant location: 0.234h/0.498 = 0.47h (32)
Note#8: Comparing with 2D results using M = 0.236αρgh3
; and a width of h/2 plus side
integrals along the slopes (y=2z), we arrive7
at 0.223αρgh4
vs 0.234αρgh4
We see, in this example, that the 2D formulae leads to an overturning moment
(0.223αρgh4
) slightly less than the more accurate value provided by the 3D TVK formula
(0.234αρgh4
).
Table 1: Summary of 2D versus 3D approaches.
Symbol
& coef.
b
pb times
αρgh
F times
αρgh2
M times
αρgh3
Pt of
act.
B(y) cf
M adapt.
2D
3D
add.
HMW 0.743h 0.743 0.543 0.218 0.4h Cste
2D
TVK 0.707h 0.707 0.555 0.236 0.4h Cste
Ex. 2 0.913h 0.913 0.50 (1) 0.23 (2) 0.5h .5h+y (3) .22 5%
Ex. 3 0.764h 0.764 0.67 (1) 0.29 (2) 0.4h h+.5y (4) .27 7%
TVK
3D
Ex. 4 1.080h 1.080 0.72 (1) 0.35 (2) 0.5h .5h+2y (5) .33 6%
Legend: Values rounded as shown.
(1) TVK 3D - use coefficient αρgh3
(2) TVK 3D - use coefficient αρgh4
(3) Ex 2: a narrow base (h/2) and sloping sides (cf. Fig. 3) is used.
(4) Ex 3: a wider base (h) and steeper slopes are used.
(5) Ex 4: a narrow base (h/2) and wider surface with milder slopes are used.
Example#3 (cont.): we find:
Bb = [(h + y/2)(7h3
/12–hy2
/2–y3
/12)]1/2
= h2
[(1+y/2h)(7/12–y2
/2h2
–y3
/12h3
)]1/2
so that we end up integrating a non-dimensional function
of u=y/h from 0 to 1.
3D TVK example 3 force: _
F3Dtvk3 = αρgƒBbdy
6
Obtained from numerical integration
7
M2Dex2 = 0.118αρgh4
+ 2x0.555αρgh4
ƒ[(0.425+1.15z/h)(1-4z/h+4(z/h)2
)]dz/h = 0.223αρgh4
(with integration ƒ from z=0 to z=h/2)

13. Submittal 120611 to Hydrovision International - for July 18, 2012 presentation in Louisville KY, USA
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= αρgh2
ƒ[(1+y/2h)(7/12–y2
/2h2
–y3
/12h3
)]1/2
dy
= αρgh3
ƒ[(1+u/2)(7/12–u2
/2–u3
/12)]1/2
du
3D TVK force ex3: F3Dtvk3 = 0.6678
αρgh3
Note#9: Comparing with an integral of the TVK 2D formula (cf explanations in
example#2 above), it gives 0.648αρgh3
(slightly less)
The overturning moment ex3 can be found using: _
M3Dtvk3 = αρgƒBbydy
= αρgh4
ƒ[(1+u/2)(7/12–u2
/2–u3
/12)]1/2
udu
where y = uh and dy = hdu, and ƒ integration symbol from 0 to 1
M3Dtvk3 = αρgh4
x 0.2949
finally,
3D TVK moment: M3Dtvk3 = 0.29αρgh4
(33)
Resultant location: .29h/.64 = 0.4h (34)
Note#10: The reader can easily verify that the TVK 3D formulae are reducing into the
TVK 2D formulae in the case of rectangular channel cross section (i.e. the width function
B(y) being considered a constant and permitting easy mathematical integration).
Example 4: We use:
B(y) = h/2+2y corresponding to a reservoir with a bottom width identical to the case of
example 2 and a much larger surface width (5h/2 instead of 3h/2).
We obtain A(y) = hy/2 + y2
providing a total cross section equals 3h2
/2.
Following the TVK 3D procedure, we derive:
B(y)b(y)2
= 7h3
/12 – hy2
/4 – y3
/3
and b(0)2
= 7h2
/6 so that: b(0) = 1.08h.
The overturning moment is given by (y=hu):
M3Dtvk4 = αρgƒBbydy = αρgh4
ƒ[(1/2+2u)(7/12–u2
/4–u3
/3)]1/2
udu
= αρgh4
√(7/24)ƒ[(1+4u)(1–3u2
/7–4u3
/7)]1/2
udu
= αρgh4
0.540 x 0.64610
= .349αρgh4
0.35αρgh4
to be compared to the 2D estimate:
M2Dex4 = 0.118αρgh4
+ 2x0.555αρgh4
ƒ[(0.4+0.6z/h)(1-2z/h+(z/h)2
)]dz/h =
= 0.118αρgh4
+ 2x0.555αρgh4
x 0.192 = 0.33αρgh4
(with integration ƒ from z=0 to z=h)
8
By numerical integration
9
Obtained from numerical integration
10
By numerical integration

14. Submittal 120611 to Hydrovision International - for July 18, 2012 presentation in Louisville KY, USA
Page 14 of 14
and the 3D force being given by:
F3Dtvk4 = αρgƒBbdy = αρgh3
ƒ[(1/2+2u)(7/12–u2
/4–u3
/3)]1/2
du
= αρgh3
√(7/24)ƒ[(1+4u)(1–3u2
/7–4u3
/7)]1/2
du
= αρgh3
0.540 x 1.32611
= 0.72αρgh3
Conclusion:
The comparison between the H.M. Westergaard and the T. Von Karman (TVK) (both 2D)
approaches indicates a profound similitude between the two methods, with the TVK
methodology easier to adapt for the three dimensional (3D) cases.
Originally the HMW and TVK methods are 2D approaches only valid for very wide and
rectangular section shaped reservoir (cf Note#10).
We presented example#3 & 4 and elaborated further example#2, originally from reference [2],
where HMW shows how to model a reservoir in 3D using the TVK method.
Our finding is that the 2D adaptation of the original HMW and TVK methods in the case of
irregularly shaped reservoir (i.e. reservoir having sides at an angle other than 90 degree),
leads to seismic overturning moments slightly underestimated by about 6% (cf table 1) when
compared to those obtained using the 3D TVK method.
It would be interesting to compare the results obtained through the TVK 3D formulae shown in
this paper with those obtained from numerical integration of the fluid mechanical equations
which could be derived in the 3D case as explained earlier in this paper. Interested readers are
very welcome to send a copy to the author for comparative study.
Acknowledgements:
Numerous comments from colleagues, particularly the very useful written comments by
Normand Thibeault Technical Director at AECOM, are gratefully acknowledged.
References:
[1] USACE (2003) “Time-History Dynamic Analysis of Concrete Hydraulic Structures”
- EM 1110-2-6051 (22 Dec 03) – Washington, DC - 401 pages
[2] Westergaard, H.M. (1933) “Water pressures on dams during earthquakes” with discussions,
(ASCE) American Society of Civil Engineers, 98, 418 – 472.
formerly at 345 East 4th
St., New York, NY
[3] Mandel, J. (1966) ”Cours de mécanique des milieux continus” Ecole Polytechnique, Paris,
published by Gauthier-Villars, Paris, France – 848 pages – reprinted by Éditions Jacques Gabay
11
By numerical integration