Zanaboni Theorem Is Not True
Jian-zhong Zhao
Geophysics Department, College of Resource, Environment
and Earth Sciences, Yunnan University，Kunming 650091，China
ABSTRACT
Zanaboni Theorem is not true and its proof is wrong.
Key Words
Zanaboni Theorem，Saint-Venant’s Principle， proof
AMS Subject Classifications: 74G50
Introduction
Saint-Venant’s Principle in elasticity (see [1,2])
has its over 100 year’s history.
Boussinesq [3] and Love [4] announced general statements of Saint-Venant’s Principe.
The early and important researches contributed to the principle are the articles [3-9].
Zanaboni “proved” a theorem [7]，trying to concern Saint-Venant’s Principle, but in
the present paper we will prove that Zanaboni’s theorem is false.
1. Zanaboni Theorem and Its Proof
In 1937, Zanaboni published a theorem dealing with energy decay of bodies of
general shape [7]. The result played a remarkable role in the history of research on
Saint-Venant’s Principle, restoring confidence in formulating the principle (see [10] ).
The theorem is described as follows (see [7,11] ):
Let an elastic body of general shape be loaded in a small sphere B
by P , an
arbitrary system of self-equilibrated forces, otherwise the body is free. Let S ' and
S ' ' be two arbitrary nonintersecting cross sections outside of B and S ' ' be farther
away from
B than
S ' . Suppose that the body is cut into two parts at S ' . The
system of surface tractions acting on the surface S ' is R ' , and the total strain
energy that would be induced in the two parts,
denoted by U R' . Similarly, we use
if they were loaded by R ' alone, is
R ' ' and U R '' for the case of the surface S " which
would also imaginarily cut the body into two pieces (see Fig.1 ). Then, according to
Zanaboni,
0  U R''  U R' .
(1.1)
The proof of Zanaboni Theorem is (see [7] and pp 300-303 of [11]) :
Assume that the stresses in the enlarged body C1  C2 are constructed by the
following stages (see Fig.2 ). First， C1 is loaded by
P . Second， each of the
separate surfaces S1 and S 2 is loaded by a system of surface tractions R .
Suppose that R is distributed in such a way that the deformed surfaces S1 and S 2
fit each other precisely, so that displacements and stresses are continuous across the
joint of S1 and S 2 . Then C1 and C2 are brought together and joined with S as
an interface. The effect is the same if C1 and C2
were linked in the unloaded state
and then the combined body C1  C2 is loaded by P . Thus
U12  U1  U R1  U R 2  U PR
where U 1 2 is the strain energy stored in
the first stage,
(1.2)
C1  C2 ,
U 1 is the work done by P in
U R 2 is the work done by R on C2 in the second stage, U R1 is the
work done by R on C1 if
C1 were free , U PR is the work done by P on C1
due to the displacements caused by
R in the second stage.
Now the minimum complementary energy theorem is used. All the actual forces
R are considered as varied by the ratio 1 : (1   ) , then the work
U R1 and
U R2
will be varied to (1   ) 2 U R1 and (1   ) 2 U R 2 respectively because the load and the
deformation will be varied by a factor
(1   )U PR because the load
factor (1   ) . Hence, U 1 2
(1   ) respectively.
P is not varied and the deformation is varied by a
will be changed to
U '1 2  U1  (1   ) 2 (U R1  U R 2 )  (1   )U PR .
The virtual increment of
U PR will be varied to
(1.3)
U 1 2 is
U 1 2   (2U R1  2U R 2  U PR )   2 (U R1  U R 2 )
For U 1 2 to be a minimum, it is required from (1.4) that
(1.4)
2U R1  2U R 2  U PR  0
(1.5)
Substituting (1.5) into (1.2), he obtains
U12  U1  (U R1  U R 2 )
(1.6)
By repeated use of (1.6) for U 1 ( 23) and U (1 2) 3 (see Fig.1), then
U 1 ( 23)  U 1  (U R '1  U R '( 23) )
，
（1.7）
U (1 2) 3  U 1 2  (U R ''(1 2)  U R ''3 )
 U1  (U R1  U R 2 )  (U R ''(1 2 )  U R ''3 )
（1.8）
Equating（1.7） with（1.8），he obtains
U R '1  U R '( 23)  U R1  U R 2  U R ''(1 2)  U R ''3 .
（1.9）
It is from（1.9）that
U R '1  U R '( 23)  U R ''(1 2)  U R ''3
because U R1 and
writing U R' for
（1.10）
U R 2 are essentially positive quantities . Eq.(1.10) is Eq.(1.1), on
U R '1  U R '( 23) , etc. And Eq.(1.1) is proved.
However, Theorem (1.1) is wrong. Our argument
is as follows:
2. Zanaboni Theorem Is Not True
2.1 In fact,
U R''  U R'  0
(2.1)
considering the continuity of displacements on S ' and S ' ' . The distributions of the
tractions of R ' on the two sides of S ' are oppositely directed, but the displacements
on the two sides are identical because of continuity, and so the work done by tractions
on one side will be
negative to the work done by tractions on the other side, and
the algebraic sum of the two works will
vanish.
2.2 If (1.1) were correct and (2.1) were wrong, one could accumulate strain energy
simply by increasing the “imaginary” cuts sectioning the elastic body. That would
violate the law of conservation of energy because energy would be created from
nothing only by imagination according to Zanaboni’s Theorem.
3. The Proof of Zanaboni Theorem is Wrong
3.1 The false proof is responsible for the invalidity of Zanaboni’s Theorem The main
mistake of the proof of Zanaboni’s Theorem is misuse of the minimum
complementary energy theorem ([12]) ,
for which only virtual
variations of
forces or stresses are allowed and virtual variations of deformation or displacements
are forbidden. Zanaboni includes in his proof the variations of deformation, and so
what he used is not the minimum complementary energy theorem, or he misused
the
minimum complementary energy theorem.
To deal with Zanaboni’s problem by the minimum complementary energy theorem
([12]), the revised proof in the present paper should be :
All the forces
U R1 and
R are considered as varied by the ratio 1 : (1   ) , then the work
U R 2 will be varied to (1   ) U R1 and (1   ) U R 2
because only the loads
will be varied by a factor
respectively
(1   ) respectively and the
displacements on S1 and S 2 remain unchanged. U 1 and U PR will not be varied
because the load
Hence,
P is fixed
and also the variations of deformation are forbidden.
from the minimum complementary energy theorem
 (U12  U1  U R1  U R 2  U PR )  0
comes
（3.1）
U12  U R1  U R 2  0 .
(3.2)
According to （3.2），if and only if
U R1  U R 2  0
（3.3）
is satisfied, then
U12  0 . .
（3.4）
Putting（3.3）into（1.2），we have the stationary value of the energy
U12  U1  U PR .
（3.5）
Equation（3.5）is obviously different from（1.6）. In addition to that, we notice that
Eq.(2.1) is exactly the deduction of Eq.(3.3).
3.2
The second error of Zanaboni’s proof is the confusion between energy and
work. In fact, Eq. (1.2) should be
U12  W1  WR1  WR 2  WPR
(3.6)
U12  W1  (WR1  WR 2 ) .
(3.7)
and Eq. (1.6) should be
And then the use of (3.7) should result in (see Fig.1)
U 1 ( 23)  W1  (WR1  WR ( 23) ) ,
U (1 2) 3  W1 2  (WR (1 2)  WR 3 ) .
(3.8)
Then Zanaboni’s theorem, which would be
0  WR''  WR'
(3.9)
(see Eq.(1.1)), is not deducible from (3.7) and the two equations of (3.8) because of
W12  U12 .
4. Discussion
(3.10)
In Zanaboni’s proof U R1 is supposed to be the work done by R on C1 if
C1
were free . If so, the stored energy in the combined body C1  C2 should be
U12  U1  U R1  U R 2  U PR  U RP
（4.1）
where U RP should be the work done by R on C1 due to the displacements caused
by P .
Using the minimum complementary energy theorem for（4.1）and supposing R
to be varied by the ratio 1 : (1   ) , we obtain
U12  U R1  U R 2  U RP  0 .
And the condition of stationary value of
(4.2)
U 1 2 is
U R1  U R 2  U RP  0
（4.3）
according to (4.2). Putting（4.3） into （4.1）, the energy in the combined body
C1  C2 is
U12  U1  U PR
（4.4）
which is identical with（3.5）.
From the analysis above we realize that U R1 in (1.2) is actually identical to
U R1  U RP in (4.1). Zanaboni’s supposition that U R1 would be the work done by
R on C1 if C1 were free is unreasonable. U R1 in (1.2) should be the work done by
R on C1
loaded by P where R and
make contribution
the displacements caused by
R and P
together to the work, then to the energy stored in C1 . Without
the judgement here the combination of C1 and C2 to construct the body C1  C2
and the continuity of stress and displacement across the joint of S1 and S 2 could
not be justified.
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