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An empirical statistical constitutive relationship for rock joint shearing considering scale effect

Teodor M. Atanackovic

C. R. Mecanique 347 (2019) 626–631Contents lists available at ScienceDirect
Comptes Rendus Mecanique
www.sciencedirect.com
The tallest column problem: New first integrals and estimates
Teodor M. Atanackovic 1
Faculty of Technical Sciences, University of Novi Sad, Serbia
a r t i c l e i n f o a b s t r a c t
Article history:
Received 19 July 2019
Accepted 10 September 2019
Available online 23 September 2019
Keywords:
Tallest column
Pontryagin’s principle
First integrals
We analyze the problem of finding the shape of the tallest column. For the system of 
equations that determine the optimal shape we construct a variational principle and two 
new first integrals. From the first integrals we are able to determine, analytically, the size of 
the cross-sectional area of the optimal column at the bottom, as well as the corresponding 
bending moment and curvature of the elastic line. Our result for critical load is compared 
with the results obtained by other methods.
© 2019 Published by Elsevier Masson SAS on behalf of Académie des sciences.
1. Introduction
In this work, we extend the results for the tallest elastic column in a constant gravity field. The problem of determining 
the shape of the lightest elastic column in a constant gravity field, or the tallest column problem, was treated in many 
publications, for example [1], [2], [3], [4], [8]. In [5], [6], Egorov proved the existence of the optimal design. In earlier 
analysis, the existence of an optimal design was assumed. Our main results in this note are a new variational principle 
for optimally designed column and two new first integrals for the equations, which determine the shape of the optimal 
column. From the first integrals we are able to find the values of the cross-sectional area, of the bending moment, and of 
the curvature of the elastic line of optimal column.
2. Formulation
Consider a column with an inextensible axis, positioned in a constant gravity field with built-in lower, and free upper 
end. Using the dimensionless variables, the system of equations describing the buckled state of the column becomes
(a2ϑ̇)· + λϑ
t∫
0
a(ξ)dξ = 0 (1)
subject to
u(0) = 0; v(0) = 0; lim
t→0a
2(t)ϑ̇(t) = 0; ϑ(1) = 0 (2)
E-mail address: atanackovic@uns.ac.rs.
1 Department of Mechanics, University of Novi Sad.https://doi.org/10.1016/j.crme.2019.09.001
1631-0721/© 2019 Published by Elsevier Masson SAS on behalf of Académie des sciences.
T.M. Atanackovic / C. R. Mecanique 347 (2019) 626–631 627Here a denotes the cross-section, λ is the load parameter, ϑ is the angle between the column axis and the x-axis, and 
d(·)/dt = ·(·), see [7]. The volume of the column is
w =
1∫
0
a(t)dt (3)
The solution ϑ0 = 0 is a trivial one, valid for all λ. For loss of stability, it the necessary that there is a nontrivial solution 
to (1), (2). The tallest column problem is stated as: find a (t) ≥ 0, which satisfies ∫ 10 a (t) dt = 1, and such that the lowest λ in 
(1) has the highest value. For this problem, we have the following result.
Proposition 2.1. Egorov [5], [6]. There exists a unique solution λ to the tallest column problem with a ∈ C ([0,1]) and a ≥ 0, x ∈ (0, 1), 
and ϑ ∈ C1 ([0,1])ϑ ≥ 0, t ∈ (0, 1).
The result of Egorov is the first rigorous proof of the existence and uniqueness of the optimal column. Concerning the 
existence of λ, see also [3]. We state this as, see [7].
Proposition 2.2. [7]. For the lowest eigenvalue λ of (1), (2), the behaviour of a, b, and ϑ near t = 0 is
a (t) ≈ λ
24
t3, b (t) ≈ λ
96
t4, ϑ (t) ≈ 1
t2
(4)
3. Optimal design a
Let b (t) =
t∫
0
a(ξ) dξ and let
ϑ = x1; m = a2ϑ̇ = x2; x3 = b (t) (5)
so that (1), (2) become
ẋ1 = x2
a2
; ẋ2 = −λx1x3; ẋ3 = a (6)
and
x1(1) = 0; x2(0) = 0; x3(0) = 0 (7)
The tallest column problem states as follows: given λ determined as the lowest eigenvalue of the problem (1), (2), let us 
determine the control a∗ (t) that belongs to the admissible set of controls U such that
min
a∈U I = mina∈U
1∫
0
a(t)dt =
1∫
0
a∗(t)dt (8)
if the system is subjected to differential constraints (6), (7). Also, from constraint (3), we conclude that min
a∈U I = 1. For U
we take a set of continuously differentiable nonnegative functions defined on the interval [0, 1]. By using the standard 
procedure of Optimal Control theory [9], the Pontryagin’s function H is defined as
H = a + p1 x2
a2
+ p2 (−λx1x3) + p3a (9)
where
ṗ1 = − ∂H
∂x1
= p2λx3, ṗ2 = − ∂H
∂x2
= − p2
a2
, ṗ3 = − ∂H
∂x3
= λx1 p2 (10)
subject to
p1(0) = 0, p2(1) = 0, p3(1) = 0 (11)
From min
a∈U H, we get a =
(
2p1x2
1+p3
)1/3
. Since p1 = x2, p2 = −x1, we obtain:
a =
(
2x22
1 + p3
)1/3
(12)
628 T.M. Atanackovic / C. R. Mecanique 347 (2019) 626–631Further, ∂
2H
∂a2
= 6 x22
a4
≥ 0; thus, H is minimum. We solve (10), (11) for p3 to obtain:
p3 = λ
1∫
t
x21(ξ)dξ (13)
so that
1 + λ
1∫
t
ϑ2(ξ)dξ = 2aϑ̇2 (14)
Differentiating (14), it follows:
(aϑ̇2)· − λ
2
ϑ2 = 0 (15)
Summing up the above results, we conclude that the shape of the optimal column a may be determined (1) and (15),
(a2 (t) ϑ̇ (t))· + λϑ (t)b (t) = 0, (a (t) ϑ̇2 (t))· + λ
2
ϑ2 (t) = 0 (16)
subject to
lim
t→0a
2(t)ϑ̇(t) = 0; ϑ(1) = 0 (17)
and we determine a and λ. The results (16), (17) agree with the optimality conditions obtained in [6], [8], [2] and [7]. From 
(14), (12), and (7) it follows that
a(1)ϑ̇2(1) = 1
2
; a(0) = 0 (18)
4. First integrals and estimates of the solution to (16) and (17)
Since ḃ(t) = a(t), the system (16) becomes
(ḃ2ϑ̇)· + λϑb = 0; (ḃϑ̇2)· + λ
2
ϑ2 = 0 (19)
subject to
lim
t→0ḃ
2(t)ϑ̇(t) = 0; ϑ(1) = 0; w∗ = b (1) = 1 (20)
At the boundary conditions (20), we took w∗ = 1 to recover the tallest column problem. The system (19), (20) is analyzed 
in [7]. Our main result states as follows.
Theorem 4.1. The solution (b, ϑ) to the system (19), (20) has the following properties. For the functional
J (W ,) = 1
2
1∫
0
(
Ẇ 2̇2 − λ2W
)
dt (21)
defined for k = (W , )
K =
{
k : k = (W ,), W (0) = 0; W (1) = 1; lim
t→0Ẇ
2(t)̇(t) = 0; (1) = 0
}
we have:
i) the functions (b, ϑ) give a stationary value to the functional (21), i.e.
δ J (b,ϑ) = 0 (22)
ii) the value of the functional (21) on the solution to (19), (20) is zero,
J (b,ϑ) = 1
2
1∫ (
ḃ2ϑ̇2 − λϑ2b
)
dt = 0 (23)0
T.M. Atanackovic / C. R. Mecanique 347 (2019) 626–631 629iii) there exist two first integrals of the system (19), (20)
3
2
ḃ2ϑ̇2 + 1
2
λϑ2b = 2; −5ϑ ḃ2ϑ̇ + 8ḃϑ̇2b = 4t (24)
iv) the values of the dependent variables at the bottom of the column are
ḃ(1) = a(1) = 8
3
; ϑ̇(1) =
√
3
4
; a(1)2ϑ̇(1) = m(1) = 16
3
√
3
(25)
Proof. The Lagrangian L of the functional (21), is L = 12
(
Ẇ 2̇2 − λ2W ) so that δ J (b, ϑ) = 0 that is i) holds. Next, we 
multiply (19)1 by ϑ and integrate to obtain 
1∫
0
(
−ḃ2ϑ̇2 + λϑ2b
)
dt = 0, which is ii). Since the Lagrangian L is not explicitly 
dependent on t , we have [9]:
H = ∂L
∂ḃ
ḃ + ∂L
∂ϑ̇
ϑ̇ −L = 3
2
ḃ2ϑ̇2 + λ
2
ϑ2b = const (26)
so that, with ϑ(1) = 0, we have
3
2
ḃ2ϑ̇2 + λ
2
ϑ2z = 3
2
ḃ2(1)ϑ̇2(1) (27)
By multiplying (19)1 by ϑ and (19)2 by −2b and by adding the result, we get
[ϑ(ḃ2ϑ)]· − ḃ2ϑ̇2 − 2[b(ḃϑ̇2)]· + 2ḃ2ϑ̇2 = 0 (28)
Integration of (28) and the use of boundary conditions leads to
2ḃ(1)ϑ̇2(1) =
1∫
0
ḃ2 (t) ϑ̇2 (t) dt
From (18)1 it follows
ḃ(1)ϑ̇2(1) = 1/2
Therefore, (28) becomes
1∫
0
ḃ2 (t) ϑ̇2 (t) dt = 1 (29)
Rewriting (27) as
3ḃ2ϑ̇2 = 3ḃ2(1)ϑ̇2(1) − ϑ2b (30)
integrating and using (23) and (29), we obtain:
ḃ2(1)ϑ̇2(1) = 4
3
(31)
Combining (31) and (27), we get (24)1.
To obtain the second first integral, we multiply (19)1 by ϑ and (19)2 by b and obtain:
(ḃ2ϑ̇ϑ)· = ḃ2ϑ̇2 − λϑ2b; (ḃϑ̇2b)· = ḃ2ϑ̇2 − λ
2
ϑ2b (32)
Using (24)1 to determine ḃ2ϑ̇2 and by substituting the resulting expression in (32), it follows:
(ḃ2ϑ̇ϑ)· = 4
3
− 4
3
λϑ2b, (ḃϑ̇2b)· = 4
3
− 5
6
λϑ2b (33)
In (33), we first eliminate the term λϑ2b. Next, by integration, we have:
−5ḃ2ϑ̇ϑ + 8ḃϑ̇2b = 4t + D (34)
where D is a constant. When (34) is evaluated at t = 0, we get D = 0 and obtain (24)2.
630 T.M. Atanackovic / C. R. Mecanique 347 (2019) 626–631Fig. 1. Cross-sectional area of the optimally shaped column.
We derive now new estimates from the first integrals. From (18)1 and (24)1, we get:
a (1) = 8
3
(35)
so that, with m (1) = a2 (1) ϑ̇ (1), we obtain:
ϑ̇ (1) =
√
3
4
; m (1) = 16
3
√
3
(36)
This completes the proof.
5. Numerical solution
The first integral (24)2 may be written as −5mϑ + 8 m2a3 b = 4t . This expression, together with a3 = 2m
2
1+λp , where we used 
x4 = p, leads to
ϑ = 4 [b (1 + λp) − t]
5m
(37)
From (5) and (37), we obtain:
ṁ = λb 4 [b (1 + λp) − t]
5m
; ḃ =
[
2m2
1 + p
]1/3
ṗ = −λ
{
4 [b (1 + λp) − t]
5m
}2
(38)
with
m(0) = 0; m(1) = 16
3
√
3
; b(0) = 0; b (1) = 1 p(1) = 0 (39)
Also
3
2
m2
ḃ2
+ λ
2
b
{
4 [b (1 + λp) − t]
5m
}2
= 2 (40)
Note that the system (38), (39) is easy to solve since the function ϑ is eliminated. The problem that the variable ϑ intro-
duced into the numerical scheme was that it is unbounded at t = 0, or u = 1, see (4), and yet it is specified at both ends 
of the column. After integration of (38), (39) the optimal cross-sectional area is determined from a =
[
2m2
1+p
]1/3
. We solved 
(38), (39) numerically with λ as a free parameter. The parameter λ = 134.1935084471 was chosen such that b (1) = 0. The 
value m (1) = 0 is satisfied with the error of the order 10−7. The optimal cross-sectional area is shown in Fig. 1.
6. Conclusion
In this work, we treated the tallest column problem as an optimization problem. We derived the known equations, in 
the form given by [7]. Our main results are as follows.
• For the system of differential equations determining the cross-section of the tallest column, we formulated a variational 
principle given by (21) and two new first integrals in the form (24).
T.M. Atanackovic / C. R. Mecanique 347 (2019) 626–631 631• From the first integrals, we determined physically important values of the cross-section. The curvature, and the moment 
at the bottom of the column
a (1) = 8
3
; ϑ̇ (1) =
√
3
4
; m (1) = 16
3
√
3
• We solved the reduced system of equations and determined a new value of the critical load parameter
λ = 134,193 508 447 1
In [8], the value λK−N = 134.19, while in [7] the value λF−N = 134.1944 was determined.
• Our numerical solution agrees well with the asymptotic result obtained in [7].
Acknowledgements
This research was supported by the Serbian Academy of Arts and Sciences, grant F-64.
References
[1] T.M. Atanackovic, Optimal shape of a column with own weight: bi and single modal optimization, Meccanica 41 (2006) 173–196.
[2] S.J. Cox, C.M. McCarthy, The shape of the tallest column, SIAM J. Math. Anal. 29 (1998) 547–554.
[3] S.J. Cox, C.M. McCarthy, The shape of the tallest column: corrected, SIAM J. Math. Anal. 31 (2000) 940–940.
[4] Y.V. Egorov, On the optimization of higher eigenvalues, C. R. Mecanique 332 (2004) 673–678.
[5] Y.V. Egorov, On the tallest column, C. R. Mecanique 338 (2010) 266–270.
[6] Y.V. Egorov, On Euler’s problem, Sb. Math. 204 (2013) 539–562.
[7] J. Farjoun, J. Neu, The tallest column—a dynamical system approach using a symmetry solution, Stud. Appl. Math. 115 (2005) 319–337.
[8] J.B. Keller, F.I. Niordson, The tallest column, J. Math. Mech. 16 (1966) 433–446.
[9] B.D. Vujanovic, T.M. Atanackovic, An Introduction to Modern Variational Techniques in Mechanics and Engineering, Birkhäuser, Boston, MA, USA, 2004.


The tallest column problem: New first integrals and estimates

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