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Globally stable minimal mass compressive tensegrity structures
Composite Structures 141 (2016) 346–354
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Composite Structures journal homepage: www.elsevier.com/locate/compstruct
Globally stable minimal mass compressive tensegrity structures Robert E. Skelton a, Rosario Montuori b,⇑, Vincenza Pecoraro b a b
University of California, San Diego (UCSD), Jacobs School of Engineering, San Diego, USA University of Salerno, Department of Civil Engineering, Salerno, Italy
a r t i c l e
i n f o
a b s t r a c t
Article history: Available online 4 February 2016
Theory exists for minimal mass tensegrity structures under compressive loads, constrained against local buckling of members. This paper extends that theory to include constraints against global buckling. We design compressive tensegrity structures in the T-Bar configuration defined in Skelton and de Oliveira (2009) by adding constraints against global buckling, not addressed in Skelton and de Oliveira (2009). Formulas are derived for minimal mass design of T-Bar systems of complexity 1 and 2. The result shows explicitly the conditions under which the original design in Skelton and de Oliveira (2009) holds and when it does not. Moreover, the formula for the minimal mass is given when global buckling limits the mass, instead of local buckling. Ó 2016 Elsevier Ltd. All rights reserved.
Keywords: T-Bar system Tensegrity structures Global instability Eulerian instability Minimum mass
1. Introduction Tensegrity structures are composed of bars and cables, with bars typically loaded in compression, and cables in tension [2–12]. To distinguish between various types of systems that fit this general tensegrity definition, the following additional classification can be used: a tensegrity configuration that has no contacts between its rigid bodies is a class 1 tensegrity system, and a tensegrity system with as many as k rigid bodies in contact is a class k tensegrity system. According to this definition the tensegrity system depicted in Fig. 1 is a class 4 tensegrity structure because in the central hinge there are four bars which are in contact. (In this planar example, this contact can by pinned, whereas in three dimensions the contact is represented as a frictionless ball joint). Tensegrity systems are not new concepts, 1950s [13,14], but beyond artistic appeal the analytical tools to design engineering structures from tensegrity concepts are only recently being developed [1]. Tensegrity systems represent one of the more recent structural topologies in the field of civil engineering [15–28]. In this work a particular Tensegrity system is considered, the so called T-Bar depicted in Fig. 1. Such kind of structure can be used in all the cases where there is the need to bear a large amount of axial load [29–32]. If only local member buckling is considered, then Chapter 3 of Ref. [1], shows that, using self-similar iterations (tensegrity ⇑ Corresponding author. E-mail addresses: (R. Montuori).
[email protected]
fractals), to reduce mass of compressive tensegrity structures. Furthermore this chapter shows that there exists an optimal number of self-similar iterations to yield minimal mass. This number is called the optimal complexity of the compressive structure. This paper considers the expansion of the minimal mass compressive structures to include global buckling constraints, but we shall show only the calculations for complexities 1 and 2. As it is well explained in [1] the system can be prestressed to get the same configuration and bar forces f(l1) as when external loads were applied; this provides a rule to choose the magnitude of the prestress in practical designs. This means that the prestress can be chosen in order to obtain a configuration in the factory or lab, when external forces are absent, which is exactly the same that we obtain in the operational environment where the external forces are present. In other words the tensegrity can be designed to have the same configuration in the loaded and unloaded case. This will simplify the mathematic of design, because it sets the rule for choosing the prestress. 2. Evaluation of total mass when only eulerian instability is considered If we consider the string of length s1 subjected to the maximum tensile load t(s) and having a yield strength equal to rs the minimum required radius r can be computed from:
tðsÞ ¼ rs p r2 As a consequence, the required mass is:
(R.E. Skelton), [email protected]
http://dx.doi.org/10.1016/j.compstruct.2016.01.105 0263-8223/Ó 2016 Elsevier Ltd. All rights reserved.
ð1Þ
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R.E. Skelton et al. / Composite Structures 141 (2016) 346–354
ms1 ¼ 4mðs1 Þ ¼ 4qs As s1 ¼ 4cs
l1 f ðl0 Þ 2cs l1 f ðl0 Þ ¼ cos2 a1 cos a1 2 cos a1
ð9Þ
If we consider the ratio between the total mass of the T-Bar and the mass of the solid bars having length l0 we obtain:
l1 ¼ lb1 þ ls1 Fig. 1. T-Bar of order 1: dashed lines are strings (tension), continuous lines are bars (compression).
mðsÞ ¼ qs p r 2 s ¼
qs tðsÞ s ¼ cs s tðsÞ rs
f ðl0 Þ ¼
p2 Eb I 2 l0
¼
p2 Eb p r40
2 l0
4
¼
p3 Eb r40 2
4 l0
ð3Þ
from the above relation the value of r 20 can be obtained as:
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 f ðl0 Þ 4 l0 2 r0 ¼ p3 Eb
ð4Þ
and by substituting the value of r 20 in the calculation of the mass of the considered bar we obtain:
mðl0 Þ ¼ qb p r20 l0 ¼
qb p 2 l20 p
sffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffi f ðl0 Þ 2 ¼ cb l0 f ðl0 Þ p Eb
ð5Þ
pffiffiffiffiffiffiffiffiffiffiffiffi where the coefficient cb ¼ 2 qb = p Eb has been introduced. Now we can consider a T-Bar unit where all eight members of the T-Bar unit are designed to fail at exactly the same value of the external load f ðl0 Þ as for the original bar of length l0 , so that the strength of the two systems is preserved. In the considered T-Bar unit there are four strings having length s1 and four bars, two having length l0 and two having length lv 1 . It is useful to denote the tension in a string by using its label as the argument, so the tension in string s1 is tðs1 ). And in the same way the force in the bar l0 is f ðl0 Þ, the force in the bar l1 is f ðl1 Þ and the force in the bar lv 1 is f ðlv 1 Þ. If we require that f ðl1 Þ ¼ f ðl0 Þ for both the unloaded and loaded cases then we have to consider the strength in strings and bars in both cases in order to find the maximum value to be adopted for the design. It is easy to find that in the unloaded case we obtain:
tðs1 Þ ¼
f ðl0 Þ ; 2 cos a1
f ðl1 Þ ¼ f ðl0 Þ;
f ðlv 1 Þ ¼ f ðl0 Þ tan a1
f ðlv 1 Þ ¼ 0;
tðs1 Þ ¼ 0
2
¼ 2cb
where the following parameter has been introduced:
e¼
cs
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi f ðl0 Þ qs f ðl0 Þ 1 qs f ðl0 ÞpEb ¼ ¼ b ffi rs l0 p2qffiffiffiffiffi 2rs l0 qb c b l0
when l1 is lower than one the use of the T-Bar provides a lower mass than the corresponding solid bar. 3. Evaluation of total mass with constraints against global buckling The results obtained in the previous section did not consider the possibility of the global instability. In fact, the bars l1 can bear the Eulerian critical load f ðl0 Þ only if a sufficient vertical stiffness in the hinge where the bars are connected is provided. If this vertical stiffness is not sufficient then the hinge where the bars are in contact can move vertically, causing the global instability before the local buckling occurs (that is, before the Eulerian critical load f ðl0 Þ is achieved). In order to account for this phenomenon the static scheme reported in Fig. 2 is to be considered, where the restraint of the intermediate node is constituted by an external spring kv 1 which represents the effects of the elongation and of the shortening of the strings constituting the T-Bars when the hinge in C moves from its original position. In order to obtain the vertical stiffness of the spring it is useful to consider the T-Bar in the deformed configuration, in particular, let us consider only the effects due to the springs that lengthen as depicted in Fig. 3. In order to evaluate the vertical stiffness due to the elongation of the strings constituting the T-Bar we can evaluate the vertical displacement x when a vertical force T is applied as depicted in Fig. 3. By considering the vertical components of the tension in the two lateral strings we obtain:
T ¼ 2 cos b½tðs1 Þ þ Dt ðs1 Þ ¼ 2 cos b½t ðs1 Þ þ k1 Ds
The mass of the four strings it is given by:
ð12Þ
ð7Þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ð l v 1 þ xÞ þ l 1 s 1
ð13Þ
The value of cos b is given by:
lv 1 þ x cos b ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ð l v 1 þ xÞ þ l 1
qffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi f ðl0 Þ tan2 a1 tan a1 þ 1
qffiffiffiffiffiffiffiffiffiffi 2 l0 5 f ðl0 Þ tan2 a1 þ 1 2
ð11Þ
p Eb
Ds ¼
qffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ¼ 2mðlv 1 Þ þ 2mðl1 Þ ¼ 2cb lv 1 f ðl0 Þ tan a1 þ 2cb l1 f ðl0 Þ ¼ 2cb l1
ð10Þ
where k1 represents the stiffness of the strings s1 and the value of Ds represents the elongation of the strings s1 which can be obtained from geometrical considerations and it is equal to:
As a consequence the mass of all four bars in the T-Bar unit is given by:
mb1
2cs f ðl0 Þ l0 1 pffiffiffiffiffiffiffiffiffiffi cos2 a1 2 cb l20 f ðl0 Þ 1 5 tan2 a1 þ 1 þ e 1 þ tan2 a1 ¼ 2
ð6Þ
while in the loaded case we have:
f ðl1 Þ ¼ f ðl0 Þ;
qffiffiffiffiffiffiffiffiffiffi 2 tan52 a1 þ 1 l0 f ðl0 Þ pffiffiffiffiffiffiffiffiffiffi 2 2 cb l0 f ðl0 Þ
þ
ð2Þ
where qs is the volumetric mass density and the coefficient cs ¼ qs =rs represents the ratio between the density and the yield strength. If now we consider a solid bar having length l0 , radius r 0 , mass density qb , Young’s modulus of elasticity Eb and mass mðl0 Þ, according to Eulerian instability, the buckling load is given by:
mb1 ms1 ¼ þ ¼ 2cb mðl0 Þ mðl0 Þ
ð8Þ Fig. 2. Static scheme of the T-Bar for the global instability analysis.
ð14Þ
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Fig. 3. T-Bar in a deformed configuration. Fig. 4. Effect on the vertical stiffness due to both shortening and lengthening strings.
By substituting Eqs. (13) and (14) in Eq. (12) the following relation is obtained:
2
3
6 t ðs1 Þ k1 s1 7 T ¼ 2ðlv 1 þ xÞ4qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ k1 5 2 2 ðlv 1 þ xÞ þ l1
ð15Þ
This is a non-linear relation between force and displacement, so that, at the aim of finding a linear relation a Taylor series development can be used. In particular this development provides:
TðxÞ ¼ Tð0Þ þ T 0 ð0Þx
ð16Þ
The value of T for x = 0 is given by:
2
3
6t ðs1 Þ k1 s1 7 T ð0Þ ¼ 2lv 1 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ k1 5 2 2 lv 1 þ l1
ð17Þ
3
6 t ðs1 Þ k1 s1 7 T 0 ðxÞ ¼ 2 4qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ k1 5 2 2 ðlv 1 þ xÞ þ l1 h
i3=2 1 2 2 ðlv 1 þ xÞ þ l1 2ðlv 1 þ xÞ þ 2ðlv 1 þ xÞ ðt ðs1 Þ k1 s1 Þ 2 ð18Þ 2
3
6tðs1 Þ k1 s1 7 T 0 ð0Þ ¼ 24 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ k1 5 2 2 lv 1 þ l1
h i3=2 2 2 þ 2lv 1 ðt ðs1 Þ k1 s1 Þ lv 1 þ l1 ðlv 1 Þ
T ð0Þ ¼ 2lv 1 T 0 ð0Þ ¼ 2
k 1 s1 2 ¼ 2k1 sin a1 s31
when the stiffness kv 1 is known we can determine the critical load Ncrit that causes the global instability of the system when the prestress in the strings is equal to zero. This requires writing the equilibrium conditions of the system for a given displacement X1 (Fig. 5). The first condition can be the rotational equilibrium of the entire structure with respect to point B:
2l1 RA l1 RC ¼ 0 ) 2l1 RA l1 kv 1 X 1 ¼ 0
ð25Þ
The second one can be the rotational equilibrium of the first bar (the one between A and B points) with respect to C point:
l1 RA Ncrit X 1 ¼ 0
ð26Þ
l1 kv 1 T 0 ð0Þl1 k1 s1 sin a1 2 2 ¼ ¼ k1 l1 sin a1 ¼ k1 s1 cos a1 sin a1 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 1 þ tan2 a1 2
ð19Þ
Ncrit ¼
ð27Þ
2
t ðs1 Þ s1
t ð s1 Þ tðs1 Þ k1 s1 2 2lv 1 3 s1 s1
ð20Þ
ð21Þ
If now we consider also the effects due to the springs which shorten as depicted in Fig. 4 the total effect of the springs is given by:
T ¼ T ð0Þ þ T 0 ð0Þx T ð0Þ T 0 ð0Þx ¼ 2T 0 ð0Þx
ð22Þ
Now in order to obtain a solution of minimum mass we can require that the global instability load is equal to the Eulerian instability load. This condition provides the equality between Ncrit and f ðl0 Þ. From this equation it is possible to calculate the value of k1, i.e., the stiffness required to the springs in order to guarantee that the global instability arises simultaneously with Eulerian instability:
k1 ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi f ðl0 Þ 1 þ tan2 a1 s1 sin a1 2
ð23Þ
We note that in the loaded case, the minimum value of the stiffness is obtained for tðs1 Þ ¼ 0, whereas in the unloaded case the tension is not zero. In practice one might choose to have a
ð28Þ
The area required for strings can be evaluated by considering that the value of the string stiffness k1 is also given by:
which means that the vertical stiffness of the external spring showed in in Fig. 3 is equal to:
kv 1 ¼ 2T 0 ð0Þ
ð24Þ
Combining Eqs. (25) and (26), the value of the critical load can be found:
By noting that lv 1 þ l1 ¼ s21 , the expressions of T ð0Þ and T 0 ð0Þ can be written in the following form: 2
kv 1 ¼ T 0 ð0Þ ¼ 2lv 1 2
The first derivative of T(x) and its value for x = 0, can be easily obtained by Eq. (15):
2
prestress greater than the one that yields the same bar force as the loaded condition. This would increase stiffness in the vertical direction in the loaded case. However to develop the theory for minimal mass, we argue that safety margins can always be added later, after minimal mass configurations are defined. So we continue with the prestress (meaning the stress in the unloaded case) assumption tðs1 Þ > 0 for the loaded case, as in (6). In this case the contribution to the stiffness due to the shortening strings is equal to zero, and so the total vertical stiffness kv 1 is given only by the strings which lengthen. Due to this consideration the total vertical stiffness is:
Fig. 5. Global instability configuration of the T-Bar.
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k1 ¼
E s As s1
ð29Þ
that the central support Fig. 5 is not constituted by an elastic support, but by a fixed restraint. In this case it is obvious that global instability cannot arise.
ð30Þ
If we consider only the local buckling assumptions of Section 1, where the ratio between the total mass of the T-Bar and the mass of the solid bars having length l0 is given, then we can note that only the quantity ls1 of Eq. (10) changes, while the quantity lb1 remains unchanged because the design conditions of the bars remain unchanged. Hence, concerning strings mass ratio ls1 we have:
As a consequence the value of the string area is:
As ¼
k 1 s1 f ðl0 Þ ¼ 2 Es sin a1 Es
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ tan2 a1
In the relations reported above no considerations have been made regarding the resistance of the strings. In fact, the only requirement made on the mechanical characteristic of strings is their stiffness. In particular the stiffness obtained in Eq. (28) represents the value which is able to assure that global instability and Eulerian instability appears in the same time. For a stiffness greater than the one expressed by Eq. (28) the T-Bar will collapse due to Eulerian instability, while for a stiffness smaller than the one expressed by Eq. (28) the T-Bar will collapse due to global instability. Now it is clear that also the tensions of the strings have to be checked. For a given yield stress rs of the string material the minimum tension required in the each string is given by tðs1 Þ. As a consequence the following relation has to be satisfied:
As rs tðs1 Þ
ð31Þ
From the above relation we can obtain the minimum value of the string area which is able to assure a sufficient resistance to the string elements.
As ¼
t ðs1 Þ
rs
¼
f ðl0 Þ 2 cos a1 rs
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi f ðl0 Þ 1 þ tan2 a1 ¼ 2rs
ð33Þ
The obtained result can be expressed in the following way:
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
f ðl0 Þ 1 þ tan2 a1 1 2 As ¼ ; 2 max 2 rs sin a1 Es pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 f ðl0 Þ 1 þ tan a1 n o ¼ 2 2 min rs ; sin a1 Es =2
ð35Þ
By means of the ratio between the elastic modulus and the yield strength of the strings material as ¼ Es =rs we have:
min
n
o
n
o
rs ; sin2 a1 Es =2 ¼ rs min 1; as sin2 a1 =2 with as ¼ Es =rs ð36Þ
And new parameter can be introduced:
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð32Þ
By comparing the values of As obtained in Eqs. (30) and (32) we can conclude that the value required for the string area is given by:
( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi) f ðl0 Þ 1 þ tan2 a1 f ðl0 Þ 1 þ tan2 a1 As ¼ max ; 2 2r s sin a1 Es
pffiffiffiffiffiffiffiffiffiffi f ðl0 Þqs 1 þ tan2 a1 n o 2 cb l0 min rs ; sin a1 Es =2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q pEb f ðl0 Þ 1 þ tan2 a1 n o ¼ s 2 2qb l0 min rs ; sin a1 Es =2
ls1 ¼
qs p E f ðl Þ e n b 0 o¼ n o e0 ¼ 2 2 min 1; as sin a1 =2 2qb l0 rs min 1; as sin a1 =2
ð37Þ
where e is the parameter introduced in Eq. (11). Finally the total mass ratio is given by:
l1 ¼ lb1 þ ls1 ¼
1 5 tan2 a1 þ 1 þ e0 1 þ tan2 a1 2
ð38Þ
It is important to emphasize that the results obtained when n o 2 min 1; as sin a1 =2 ¼ 1 represent the case already studied in [1]. In fact, in this case Eqs. (38) and (10) are equivalent. Relations (10) and (38) have been applied for values of
ð34Þ
The meaning of the relation (34) is the following: when
rs sin2 a1 Es =2 the governing instability is represented by the global instability because the area represented by Eq. (32) is smaller than the one represented by Eq. (30). In other words, when
rs sin2 a1 Es =2 the value As of the area requested by the resistance condition is always lower than that one required to avoid global instability, so that, collapse is due to the global instability of the
e ¼ ½103 ; 101 ; 0:05; 1=3 and as ¼ ½5; 10; 100; 200; 500. The obtained results are reported in Figs. 6–9. As expected, the mass ratio is strongly influenced not only from the materials properties but also from the geometrical configuration. In particular for any given material, i.e. for any fixed value of as , there is a value of the angle a1 which minimizes the total mass ratio. In addition it is evident that for as the angle a1 approaches zero, there is no support against vertical displacements at the midpoint, and the total mass approaches infinity because the system guarantees global buckling. This result is also evident from
bar. On the contrary, when rs sin a1 Es =2 the governing instability is represented by the Eulerian one. In fact, in this case the value of As represented by Eq. (30) is always the smaller one; so that the condition representing the global instability is always avoided. To better understand the result it is useful to consider two limit cases: 2
(1) case of rs ! 1: In this case the tension is zero and the required mass of the strings is zero; i.e. every area As provides sufficient tension, while the stiffness required to avoid global instability requires a string mass depending on Es and a1 according to Eq. (30).
(2) case of sin a1 Es =2 ! 1: In this case the required mass for the stiffness of the strings tends to zero; i.e. every area As is able to assure a sufficient stiffness, while the tension required is given by Eq. (31). From a physical point of view the value of the stiffness kv1 tends to infinite, which means 2
Fig. 6. Values of
l1 for e ¼ 103 .
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Fig. 10. T-Bar of order 2 and its schematization for the analysis of global instability.
Fig. 7. Values of
l1 for e ¼ 101 .
In order to analyze the global instability we can consider a generic deformed configuration as reported in Fig. 11 and then we can consider four equilibrium equations. In particular it is useful to define 3 unknown quantities X 1 ; X 2 ; X 3 representing the 3 vertical displacements (Fig. 11). In addition the 3 vertical reactions corresponding to the 3 vertical displacements will be equal to:
R1 ¼ kv 2 X 1 ;
R2 ¼ k v 1 X 2 ;
R3 ¼ k v 2 X 3
ð39Þ
The rotational equilibrium of AC bar with respect to point C is given by:
RA l2 NX 1 ¼ 0 ) RA ¼
NX 1 l2
ð40Þ
The rotational equilibrium of AD part of structure with respect to point D is given by: Fig. 8. Values of
l1 for e ¼ 0:05.
RA 2 l2 R1 l2 N X 2 ¼ 0
ð41Þ
By substituting Eqs. (39) and (40) in Eq. (41) we have:
X 1 ð2N kv 2 l2 Þ NX 2 ¼ 0
ð42Þ
The rotational equilibrium of AE part of structure with respect to point E is given by:
RA 3l2 R1 2l2 R2 l2 NX 3 ¼ 0
ð43Þ
And again, by using Eqs. (39) and (40) we obtain:
X 1 ð3N 2kv 2 l2 Þ kv 1 X 2 l2 NX 3 ¼ 0
ð44Þ
Finally, the equilibrium of the whole structure with respect to point B gives:
RA 4l2 R1 3l2 R2 2l2 R3 l2 ¼ 0
ð45Þ
That by means of Eqs. (39) and (40) becomes: Fig. 9. Values of
l1 for e ¼ 1=3.
a physical point of view, in fact, when the angle a1 decreases, the T-Bar configuration is less able to support vertical loads. Of course when a1 is equal to zero the vertical stiffness Kv1 is equal to zero, and the structure is unstable.
X 1 ð4N 3kv 2 l2 Þ 2kv 1 X 2 l2 kv 2 X 3 l2 ¼ 0
By combining Eqs. (42), (44) and (46) the system able to find the critical load corresponding to the global instability can be found as:
4. The T-Bar of complexity two Now we can analyze the T-Bar of complexity 2, which is obtained from the T-Bar reported in Fig. 1 by substituting the two bars of length l1 with two T-Bars, as depicted in Fig. 10. The same figure represents the structural scheme that is to be analyzed for the study of global instability. It is important to note that in this case in addition to the vertical stiffness kv1 there is another vertical stiffness that we call kv2.
ð46Þ
Fig. 11. Deformed configuration.
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8 > > < > > :
X 1 ð2N kv 2 l2 Þ NX 2 ¼ 0 ð47Þ
X 1 ð3N 2kv 2 l2 Þ kv 1 X 2 l2 NX 3 ¼ 0 X 1 ð4N 3kv 2 l2 Þ 2kv 1 X 2 l2 kv 2 X 3 l2 ¼ 0
As it is well known if we want a solution different from the null one which is given by X 1 ¼ X 2 ¼ X 3 ¼ 0 it is required that the determinant of the coefficients is equal to zero. In this case, it results:
2 3 2 2 2 kv 1 kv 2 l2 þ 4kv 1 kv 2 l2 þ 2kv 2 l2 N ð4kv 1 l2þ 6kv 2 l2 ÞN2 þ 4N3 ¼ 0
ð48Þ The solutions of Eq. (48) are:
Ncrit;1 ¼
Ncrit;3
kv 2 l2 ; 2
kv 2 l2 kv 1 l2 þ þ 2 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 kv 2 l2 þ kv 1 l2
Ncrit;2 ¼
kv 2 l2 kv 1 l2 ¼ þ þ 2 2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 kv 2 l2 þ kv 1 l2 2
2
2
2
; ð49Þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 kv 2 l2 þ kv 1 l2 2
2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 kv 2 l2 þ kv 1 l2 k v 2 l2 k v 1 l2 ¼ Ncrit;2 ¼ þ 2 2 2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 kv 1 l0 kv 2 l0 kv 1 l0 kv 2 l0 ¼ þ þ 8 8 8 8
ð50Þ
ð51Þ
kv 2 ¼ 2k2 sin a2 2
ð52Þ
In order to substitute the values of kv1 and kv2 in Eq. (51) it is useful note that, being s1 ¼ l0 =2 cos a1 and s2 ¼ l0 =4 cos a2 , we have:
i kv 1 l0 2k1 sin a1 2s1 cos a1 1 h 2 ¼ ¼ k1 s1 sin a1 cos a1 2 8 8 2
ð53Þ
kv 2 l0 2k2 sin a2 4s2 cos a2 2 ¼ ¼ k2 s2 sin a2 cos a2 8 8 2
In order to simplify the mathematical details of the following development, it is useful to introduce the following nondimensional parameter:
b¼
k1 s1 Es A1 A1 ¼ ¼ k2 s2 Es A2 A2
ð54Þ
By using the above parameter, the values of kv 1 l0 =8 appearing in Eq. (51) can be written as:
i k v 1 l0 1 h 2 ¼ k2 s2 bsin a1 cos a1 2 8
2
2
2
2
ðk2 s2 Þ ðk2 s2 Þf ðl0 ÞC 1 þ f ðl0 Þ C 2 ¼ 0
ð57Þ
ð58Þ
where the numerical coefficient C1 and C2 are equal to:
bsin a1 cos a1 þ 2sin a2 cos a2 2
C1 ¼ C2 ¼
2
bsin a1 sin a2 cos a1 cos a2 1 2
2
;
bsin a1 sin a2 cos a1 cos a2 2
k2 s2 ¼
2
ð59Þ
ð55Þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi f ðl0 Þ C 1 C 21 4C 2 2
ð60Þ
It is easy to verify that both solutions have positive value, so that, as we are interested in the solution of minimum mass, we adopt the minimum value ðk2 s2 Þ, i.e. the minimum value of string area As2 . In addition we know that k2 s2 ¼ Es As2 and As1 ¼ bAs2 so the value of As1 and As2 can be obtained as:
As2 ¼
By repeating the consideration made for the evaluation of kv1 in the case of T-Bar of first order, the value of kv1 and kv2 can be easily obtained as: 2
By isolating the square root and squaring first and second member we obtain:
The solutions of Eq. (54) are given by:
Given that negative stiffness is impossible here, the minimum among the three values reported in Eq. (49), i.e. the critical load, is (noting that l2 = l0/4):
kv 1 ¼ 2k1 sin a1 ;
ð56Þ
which can be written as:
) kv 2 þ kv 1 < kv 1 ) kv 2 < 0
Ncrit
2
2
2
kv 2 l2 kv 2 l2 kv 1 l2 < þ 2 2 2
bk2 s2 sin a1 cos a1 2 þ k2 s2 sin a2 cos a2 2 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ! u 2 u bk s sin2 a cos a 2 2 2 1 1 2 t þ k2 s2 sin a2 cos a2 ¼ f ðl0 Þ 2
ðk2 s2 Þ b sin a1 sin a2 cos a1 cos a2 h i 2 2 2 ðk2 s2 Þf ðl0 Þ b sin a1 cos a1 þ 2sin a2 cos a2 þf ðl0 Þ ¼ 0
The actual critical load is the minimum among the three reported in relation (49). So that the value Ncrit,3 can be excluded because it is greater than both Ncrit,2. With the aim of determining which is the lower value between Ncrit,2 and Ncrit,1 we can inquiry about the conditions which are to be satisfied in order to assure that Ncrit,1 < Ncrit,2. In particular it results:
Ncrit;1 < Ncrit;2 )
Now, as already accomplished for the T-Bar of complexity 1, we require a solution of minimum mass so that the global instability load expressed by Eq. (51) is equal to the Eulerian instability load. This condition provides the equality between Ncrit and f(l0):
k2 s2 f ðl0 ÞC ¼ ; Es Es
As1 ¼
where
C ¼ Cðb; a1 ; a2 Þ ¼
C1
f ðl0 ÞbC Es
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C 21 4C 2 2
ð61Þ
ð62Þ
As in the T-Bar of complexity 1, in the above relations no consideration have been made regarding the tension in the strings. In fact, the only requirement on the mechanical characteristic of strings is their stiffness. In particular the stiffnesses k2 and k1 (where k1 is obtained from the fixed value b) represent the values that are able to assure that global instability and Eulerian instability occurring at the same time. For a stiffness of the s2 strings greater than the obtained value of k2 or for a stiffness of s1 springs greater than the obtained value of k1 the T-Bar will collapse due to Eulerian instability, while in the opposite case the T-Bar will collapse due to both global and local instability simultaneously. Now it is clear that also the tension of the strings has to be checked. For a given yield strength rs of the string material the minimum tension required depends on the maximum values t(s1) and t(s2). In fact, the following relation has to be satisfied:
As1 rs t ðs1 Þ;
As2 rs t ðs2 Þ
ð63Þ
From the above relation we can obtain the minimum value of the string area which is able to assure a sufficient tension to the string elements:
As1 ¼
t ð s1 Þ
rs
¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi f ðl0 Þ f ðl0 Þ 1 þ tan2 a1 ¼ 2 cos a1 rs 2 rs
ð64Þ
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R.E. Skelton et al. / Composite Structures 141 (2016) 346–354
As2 ¼
t ðs2 Þ
rs
¼
f ð l0 Þ 2 cos a2 rs
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi f ðl0 Þ 1 þ tan2 a2 ¼ 2 rs
By comparing the values of As1 and As2 obtained in Eqs. (61) and (64) we can conclude that the value required for the string areas are given by:
( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi) f ðl0 ÞbC f ðl0 Þ 1 þ tan2 a1 ; ; Es 2rs ( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi) f ðl0 ÞC f ðl0 Þ 1 þ tan2 a2 As2 ¼ max ; Es 2rs As1 ¼ max
ð65Þ
And remembering the parameter as ¼ Es =rs introduced in Eq. (36) the obtained results can be expressed in the following way:
( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi) f ðl0 ÞbC f ðl0 Þ 1 þ tan2 a1 ; 2rs as rs ( ) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 f ðl0 Þ 1 þ tan a1 2bC pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; 1 ¼ max 2rs as 1 þ tan2 a1
As1 ¼ max
ð66Þ
( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi) f ðl0 ÞC f ðl0 Þ 1 þ tan2 a2 ; As2 ¼ max as rs 2rs ( ) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi f ðl0 Þ 1 þ tan2 a2 2C pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; 1 ¼ max 2rs as 1 þ tan2 a2 In the considered T-Bar there are 4 strings having length s1 and 8 strings having length s2 , so that the total mass of all strings can be determined as:
4qs As1 l0 8qs As2 l0 þ 2 cos a1 4 cos a2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 2qs l0 As1 1 þ tan2 a1 þ As2 1 þ tan2 a2
ms2 ¼ 4qs As1 s1 þ 8qs As2 s2 ¼
) 2bC pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; 1 rs as 1 þ tan2 a1 ( )# 2C 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; 1 þ 1 þ tan a2 max as 1 þ tan2 a2 3 2 2 2 7 1 þ tan a1 1 þ tan a2 q l0 f ðl0 Þ 6 6 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
7 ¼ s 5ð67Þ rs 4 as 1þtan2 a1 as 1þtan2 a2 ; 1 ; 1 min min 2bC 2C ¼ q s l0
f ðl0 Þ
"
(
ð1 þ tan2 a1 Þ max
And the ratio between the strings mass and the mass of the solid bar of length l0 is given by:
ls2 ¼
ms2 mðl0 Þ
1 tan5=2 a1 tan5=2 a2 þ 2 4 4 1 þ tan2 a1 1 þ tan2 a2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ e0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ e0 ffi
a 1þtan2 a a 1þtan2 a min s 2bC 1 ; 1 min s 2C 2 ; 1
l2 ¼ lb2 þ ls2 ¼ þ
It is easy to verify that it is equivalent to the value obtained by Skelton and Olivera [1] when both the minimum values are equal to 1, i.e. when global instability is not accounted for. For minimum mass, we need to choose as ; e0 ; a1 and a2 . When these parameters are fixed, then a numerical analysis is to be made in order to find which is the value of b that gives the minimum mass. As an example if we choose a1 ¼ a2 ¼ 10 and e0 ¼ 0:005 then the curves depicted in Fig. 12 are obtained. It is clear that for each given value of as a minimum value of l2 can be found. That is, the value of b is shown that minimizes the mass ratio l2;min of the T-Bar of complexity 2 for the chosen values of as ; e0 ; a1 and a2 . It can be interesting to note that in some cases there are more than one value of b leading to the same minimum value of l2 . In these cases we have more than one possibility to realize a structure of minimum mass. The different solution will differ for the mass of the strings. As an example, in Fig. 12 for a given value of as ¼ 30 the same minimum mass solution is obtained for each value of b such that 0.4 < b < 1.2 . As for T – bar of complexity 1, a comparison with the result already obtained by Skelton [1] can be made. In Figs. 13–16 the results are reported for e ¼ ½103 ; 101 ; 0:05; 1=3 and as ¼ ½5; 10; 100; 200; 500. As in the previous case the mass ratio is strongly influenced not only from the materials properties but also from the geometrical configuration. In particular for any given material, i.e. for any fixed value of as , the total mass ratio depends on the values angles a1 and a2 . In addition it is evident that for little values of these angles the total mass ratio is greater than one: in fact it tends to be infinite, leading to an increase in structural weight with respect to the case of the solid bar. This result is also evident from a physical point of view, in fact, when the angles a1 and a2 decrease, the T-Bar configuration is expected to provide less resistance (stiffness) to vertical loads. As a limiting case, when they are equal to zero the vertical stiffnesses Kv1 and Kv2 are equal to zero, and the structure is unstable. We can conclude that the result is very similar to the one obtained for the T-Bar of complexity 1. Recall that for the T-Bar of order one the relation as sin a1 =2 > 1 is sufficient to assure that the Eulerian instability is the governing phenomenon and so the global instability could be neglected. In a similar way for the T-Bar of complexity pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 the fact that as 1 þ tan2 a1 =2bC > 1 and as 1 þ tan2 a2 =2C > 1 assures that the global instability cannot occur. So the occurrence of global instability depends, also in this case, on the geometrical configuration by means of the angles a1 and a2 , and on the ratio as between the stiffness and the resistance of the string material. 2
3 2 2 7 1 þ tan a1 1 þ tan a2 q s l0 f ð l0 Þ 6 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffi 6 ffi þ ffi 7 4 5 2 2a 2a a 1þtan a 1þtan rs cb l0 f ðl0 Þ min s 1 ; 1 min s 2C 2 ; 1 2bC
2
ð68Þ qb ffi Remembering that cb = p2ffiffiffiffiffiffi it is easy to verify that the common
pEb
factor of Eq. (68) is equal to:
qs f ðl0 Þl0
pffiffiffiffiffiffiffiffiffiffi ¼ rs cb l20 f ðl0 Þ
pffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
qs f ðl0 Þ qs f ðl0 ÞpEb ¼ ¼ e0 rs cb l0 2rs qb l0
ð69Þ
where e0 is the parameter introduced in Eq. (37). So the total mass ratio for the T-Bar of order 2 is given by:
ð70Þ
Fig. 12. Mass ratio versus b for a1 ¼ a2 ¼ 10 and
e0 ¼ 0:005.
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R.E. Skelton et al. / Composite Structures 141 (2016) 346–354
Fig. 13. Values of
l2;min for e ¼ 103 . Fig. 16. Values of
l2;min for e ¼ 1=3.
a1 ¼ a2 ¼ 8 the design is correct, because, in this case, the two curves are coincident for values of angles greater than to 4.91.
5. Conclusions
Fig. 14. Values of
l2;min for e ¼ 101 .
In this work, we have shown how to design tensegrity structures for minimal mass, subject to compressive loads and subject to both local and global buckling constraints. Two fundamental tensegrity structures have been analyzed: the T-Bar of complexity 1 depicted in Fig. 1 and the T-Bar of complexity 2 depicted in Fig. 10. The originality of the work is due to the consideration of global instability that was not accounted for in the previous studies. The obtained results allow one to better understand and design this fundamental structure. In fact, the role played by the geometrical configuration and material properties constituting the strings in avoiding global instability is now explained. The previous calculations minimized mass subject to only local buckling and to prevent global buckling, this study shows that the total mass found in this study is always greater or equal to the one found in the previous study [1]. References
Fig. 15. Values of
l2;min for e ¼ 0:05.
From a physical point of view this result says that if the string material has enough stiffness with respect to the tension then the global instability cannot be a failure mode of the structure. At this point we can better explain the differences between the design rules given in [1] and the ones herein obtained. If we consider a value of ¼ 101 we have to look at Fig. 7 and Fig. 14 of the paper. For a fixed material like steel with yield strength rs equal to 2000 Mpa we have as ¼ 100. From Fig. 7 it is evident that you can find the value of angle alpha that prevents global buckling, it is given by the abscissa where the curve for as ¼ 100 and the Skelton curve intersect. This value is equal approximately to 8. Now with this value of angle a; we can use the self-similar iterations of [1] to design a structure of complexity 2. From Fig. 14, one can find that for the value of as ¼ 100 and for angles
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Composite Structures journal homepage: www.elsevier.com/locate/compstruct
Globally stable minimal mass compressive tensegrity structures Robert E. Skelton a, Rosario Montuori b,⇑, Vincenza Pecoraro b a b
University of California, San Diego (UCSD), Jacobs School of Engineering, San Diego, USA University of Salerno, Department of Civil Engineering, Salerno, Italy
a r t i c l e
i n f o
a b s t r a c t
Article history: Available online 4 February 2016
Theory exists for minimal mass tensegrity structures under compressive loads, constrained against local buckling of members. This paper extends that theory to include constraints against global buckling. We design compressive tensegrity structures in the T-Bar configuration defined in Skelton and de Oliveira (2009) by adding constraints against global buckling, not addressed in Skelton and de Oliveira (2009). Formulas are derived for minimal mass design of T-Bar systems of complexity 1 and 2. The result shows explicitly the conditions under which the original design in Skelton and de Oliveira (2009) holds and when it does not. Moreover, the formula for the minimal mass is given when global buckling limits the mass, instead of local buckling. Ó 2016 Elsevier Ltd. All rights reserved.
Keywords: T-Bar system Tensegrity structures Global instability Eulerian instability Minimum mass
1. Introduction Tensegrity structures are composed of bars and cables, with bars typically loaded in compression, and cables in tension [2–12]. To distinguish between various types of systems that fit this general tensegrity definition, the following additional classification can be used: a tensegrity configuration that has no contacts between its rigid bodies is a class 1 tensegrity system, and a tensegrity system with as many as k rigid bodies in contact is a class k tensegrity system. According to this definition the tensegrity system depicted in Fig. 1 is a class 4 tensegrity structure because in the central hinge there are four bars which are in contact. (In this planar example, this contact can by pinned, whereas in three dimensions the contact is represented as a frictionless ball joint). Tensegrity systems are not new concepts, 1950s [13,14], but beyond artistic appeal the analytical tools to design engineering structures from tensegrity concepts are only recently being developed [1]. Tensegrity systems represent one of the more recent structural topologies in the field of civil engineering [15–28]. In this work a particular Tensegrity system is considered, the so called T-Bar depicted in Fig. 1. Such kind of structure can be used in all the cases where there is the need to bear a large amount of axial load [29–32]. If only local member buckling is considered, then Chapter 3 of Ref. [1], shows that, using self-similar iterations (tensegrity ⇑ Corresponding author. E-mail addresses: (R. Montuori).
[email protected]
fractals), to reduce mass of compressive tensegrity structures. Furthermore this chapter shows that there exists an optimal number of self-similar iterations to yield minimal mass. This number is called the optimal complexity of the compressive structure. This paper considers the expansion of the minimal mass compressive structures to include global buckling constraints, but we shall show only the calculations for complexities 1 and 2. As it is well explained in [1] the system can be prestressed to get the same configuration and bar forces f(l1) as when external loads were applied; this provides a rule to choose the magnitude of the prestress in practical designs. This means that the prestress can be chosen in order to obtain a configuration in the factory or lab, when external forces are absent, which is exactly the same that we obtain in the operational environment where the external forces are present. In other words the tensegrity can be designed to have the same configuration in the loaded and unloaded case. This will simplify the mathematic of design, because it sets the rule for choosing the prestress. 2. Evaluation of total mass when only eulerian instability is considered If we consider the string of length s1 subjected to the maximum tensile load t(s) and having a yield strength equal to rs the minimum required radius r can be computed from:
tðsÞ ¼ rs p r2 As a consequence, the required mass is:
(R.E. Skelton), [email protected]
http://dx.doi.org/10.1016/j.compstruct.2016.01.105 0263-8223/Ó 2016 Elsevier Ltd. All rights reserved.
ð1Þ
347
R.E. Skelton et al. / Composite Structures 141 (2016) 346–354
ms1 ¼ 4mðs1 Þ ¼ 4qs As s1 ¼ 4cs
l1 f ðl0 Þ 2cs l1 f ðl0 Þ ¼ cos2 a1 cos a1 2 cos a1
ð9Þ
If we consider the ratio between the total mass of the T-Bar and the mass of the solid bars having length l0 we obtain:
l1 ¼ lb1 þ ls1 Fig. 1. T-Bar of order 1: dashed lines are strings (tension), continuous lines are bars (compression).
mðsÞ ¼ qs p r 2 s ¼
qs tðsÞ s ¼ cs s tðsÞ rs
f ðl0 Þ ¼
p2 Eb I 2 l0
¼
p2 Eb p r40
2 l0
4
¼
p3 Eb r40 2
4 l0
ð3Þ
from the above relation the value of r 20 can be obtained as:
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 f ðl0 Þ 4 l0 2 r0 ¼ p3 Eb
ð4Þ
and by substituting the value of r 20 in the calculation of the mass of the considered bar we obtain:
mðl0 Þ ¼ qb p r20 l0 ¼
qb p 2 l20 p
sffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffi f ðl0 Þ 2 ¼ cb l0 f ðl0 Þ p Eb
ð5Þ
pffiffiffiffiffiffiffiffiffiffiffiffi where the coefficient cb ¼ 2 qb = p Eb has been introduced. Now we can consider a T-Bar unit where all eight members of the T-Bar unit are designed to fail at exactly the same value of the external load f ðl0 Þ as for the original bar of length l0 , so that the strength of the two systems is preserved. In the considered T-Bar unit there are four strings having length s1 and four bars, two having length l0 and two having length lv 1 . It is useful to denote the tension in a string by using its label as the argument, so the tension in string s1 is tðs1 ). And in the same way the force in the bar l0 is f ðl0 Þ, the force in the bar l1 is f ðl1 Þ and the force in the bar lv 1 is f ðlv 1 Þ. If we require that f ðl1 Þ ¼ f ðl0 Þ for both the unloaded and loaded cases then we have to consider the strength in strings and bars in both cases in order to find the maximum value to be adopted for the design. It is easy to find that in the unloaded case we obtain:
tðs1 Þ ¼
f ðl0 Þ ; 2 cos a1
f ðl1 Þ ¼ f ðl0 Þ;
f ðlv 1 Þ ¼ f ðl0 Þ tan a1
f ðlv 1 Þ ¼ 0;
tðs1 Þ ¼ 0
2
¼ 2cb
where the following parameter has been introduced:
e¼
cs
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi f ðl0 Þ qs f ðl0 Þ 1 qs f ðl0 ÞpEb ¼ ¼ b ffi rs l0 p2qffiffiffiffiffi 2rs l0 qb c b l0
when l1 is lower than one the use of the T-Bar provides a lower mass than the corresponding solid bar. 3. Evaluation of total mass with constraints against global buckling The results obtained in the previous section did not consider the possibility of the global instability. In fact, the bars l1 can bear the Eulerian critical load f ðl0 Þ only if a sufficient vertical stiffness in the hinge where the bars are connected is provided. If this vertical stiffness is not sufficient then the hinge where the bars are in contact can move vertically, causing the global instability before the local buckling occurs (that is, before the Eulerian critical load f ðl0 Þ is achieved). In order to account for this phenomenon the static scheme reported in Fig. 2 is to be considered, where the restraint of the intermediate node is constituted by an external spring kv 1 which represents the effects of the elongation and of the shortening of the strings constituting the T-Bars when the hinge in C moves from its original position. In order to obtain the vertical stiffness of the spring it is useful to consider the T-Bar in the deformed configuration, in particular, let us consider only the effects due to the springs that lengthen as depicted in Fig. 3. In order to evaluate the vertical stiffness due to the elongation of the strings constituting the T-Bar we can evaluate the vertical displacement x when a vertical force T is applied as depicted in Fig. 3. By considering the vertical components of the tension in the two lateral strings we obtain:
T ¼ 2 cos b½tðs1 Þ þ Dt ðs1 Þ ¼ 2 cos b½t ðs1 Þ þ k1 Ds
The mass of the four strings it is given by:
ð12Þ
ð7Þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ð l v 1 þ xÞ þ l 1 s 1
ð13Þ
The value of cos b is given by:
lv 1 þ x cos b ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ð l v 1 þ xÞ þ l 1
qffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi f ðl0 Þ tan2 a1 tan a1 þ 1
qffiffiffiffiffiffiffiffiffiffi 2 l0 5 f ðl0 Þ tan2 a1 þ 1 2
ð11Þ
p Eb
Ds ¼
qffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ¼ 2mðlv 1 Þ þ 2mðl1 Þ ¼ 2cb lv 1 f ðl0 Þ tan a1 þ 2cb l1 f ðl0 Þ ¼ 2cb l1
ð10Þ
where k1 represents the stiffness of the strings s1 and the value of Ds represents the elongation of the strings s1 which can be obtained from geometrical considerations and it is equal to:
As a consequence the mass of all four bars in the T-Bar unit is given by:
mb1
2cs f ðl0 Þ l0 1 pffiffiffiffiffiffiffiffiffiffi cos2 a1 2 cb l20 f ðl0 Þ 1 5 tan2 a1 þ 1 þ e 1 þ tan2 a1 ¼ 2
ð6Þ
while in the loaded case we have:
f ðl1 Þ ¼ f ðl0 Þ;
qffiffiffiffiffiffiffiffiffiffi 2 tan52 a1 þ 1 l0 f ðl0 Þ pffiffiffiffiffiffiffiffiffiffi 2 2 cb l0 f ðl0 Þ
þ
ð2Þ
where qs is the volumetric mass density and the coefficient cs ¼ qs =rs represents the ratio between the density and the yield strength. If now we consider a solid bar having length l0 , radius r 0 , mass density qb , Young’s modulus of elasticity Eb and mass mðl0 Þ, according to Eulerian instability, the buckling load is given by:
mb1 ms1 ¼ þ ¼ 2cb mðl0 Þ mðl0 Þ
ð8Þ Fig. 2. Static scheme of the T-Bar for the global instability analysis.
ð14Þ
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Fig. 3. T-Bar in a deformed configuration. Fig. 4. Effect on the vertical stiffness due to both shortening and lengthening strings.
By substituting Eqs. (13) and (14) in Eq. (12) the following relation is obtained:
2
3
6 t ðs1 Þ k1 s1 7 T ¼ 2ðlv 1 þ xÞ4qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ k1 5 2 2 ðlv 1 þ xÞ þ l1
ð15Þ
This is a non-linear relation between force and displacement, so that, at the aim of finding a linear relation a Taylor series development can be used. In particular this development provides:
TðxÞ ¼ Tð0Þ þ T 0 ð0Þx
ð16Þ
The value of T for x = 0 is given by:
2
3
6t ðs1 Þ k1 s1 7 T ð0Þ ¼ 2lv 1 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ k1 5 2 2 lv 1 þ l1
ð17Þ
3
6 t ðs1 Þ k1 s1 7 T 0 ðxÞ ¼ 2 4qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ k1 5 2 2 ðlv 1 þ xÞ þ l1 h
i3=2 1 2 2 ðlv 1 þ xÞ þ l1 2ðlv 1 þ xÞ þ 2ðlv 1 þ xÞ ðt ðs1 Þ k1 s1 Þ 2 ð18Þ 2
3
6tðs1 Þ k1 s1 7 T 0 ð0Þ ¼ 24 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ k1 5 2 2 lv 1 þ l1
h i3=2 2 2 þ 2lv 1 ðt ðs1 Þ k1 s1 Þ lv 1 þ l1 ðlv 1 Þ
T ð0Þ ¼ 2lv 1 T 0 ð0Þ ¼ 2
k 1 s1 2 ¼ 2k1 sin a1 s31
when the stiffness kv 1 is known we can determine the critical load Ncrit that causes the global instability of the system when the prestress in the strings is equal to zero. This requires writing the equilibrium conditions of the system for a given displacement X1 (Fig. 5). The first condition can be the rotational equilibrium of the entire structure with respect to point B:
2l1 RA l1 RC ¼ 0 ) 2l1 RA l1 kv 1 X 1 ¼ 0
ð25Þ
The second one can be the rotational equilibrium of the first bar (the one between A and B points) with respect to C point:
l1 RA Ncrit X 1 ¼ 0
ð26Þ
l1 kv 1 T 0 ð0Þl1 k1 s1 sin a1 2 2 ¼ ¼ k1 l1 sin a1 ¼ k1 s1 cos a1 sin a1 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 1 þ tan2 a1 2
ð19Þ
Ncrit ¼
ð27Þ
2
t ðs1 Þ s1
t ð s1 Þ tðs1 Þ k1 s1 2 2lv 1 3 s1 s1
ð20Þ
ð21Þ
If now we consider also the effects due to the springs which shorten as depicted in Fig. 4 the total effect of the springs is given by:
T ¼ T ð0Þ þ T 0 ð0Þx T ð0Þ T 0 ð0Þx ¼ 2T 0 ð0Þx
ð22Þ
Now in order to obtain a solution of minimum mass we can require that the global instability load is equal to the Eulerian instability load. This condition provides the equality between Ncrit and f ðl0 Þ. From this equation it is possible to calculate the value of k1, i.e., the stiffness required to the springs in order to guarantee that the global instability arises simultaneously with Eulerian instability:
k1 ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi f ðl0 Þ 1 þ tan2 a1 s1 sin a1 2
ð23Þ
We note that in the loaded case, the minimum value of the stiffness is obtained for tðs1 Þ ¼ 0, whereas in the unloaded case the tension is not zero. In practice one might choose to have a
ð28Þ
The area required for strings can be evaluated by considering that the value of the string stiffness k1 is also given by:
which means that the vertical stiffness of the external spring showed in in Fig. 3 is equal to:
kv 1 ¼ 2T 0 ð0Þ
ð24Þ
Combining Eqs. (25) and (26), the value of the critical load can be found:
By noting that lv 1 þ l1 ¼ s21 , the expressions of T ð0Þ and T 0 ð0Þ can be written in the following form: 2
kv 1 ¼ T 0 ð0Þ ¼ 2lv 1 2
The first derivative of T(x) and its value for x = 0, can be easily obtained by Eq. (15):
2
prestress greater than the one that yields the same bar force as the loaded condition. This would increase stiffness in the vertical direction in the loaded case. However to develop the theory for minimal mass, we argue that safety margins can always be added later, after minimal mass configurations are defined. So we continue with the prestress (meaning the stress in the unloaded case) assumption tðs1 Þ > 0 for the loaded case, as in (6). In this case the contribution to the stiffness due to the shortening strings is equal to zero, and so the total vertical stiffness kv 1 is given only by the strings which lengthen. Due to this consideration the total vertical stiffness is:
Fig. 5. Global instability configuration of the T-Bar.
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k1 ¼
E s As s1
ð29Þ
that the central support Fig. 5 is not constituted by an elastic support, but by a fixed restraint. In this case it is obvious that global instability cannot arise.
ð30Þ
If we consider only the local buckling assumptions of Section 1, where the ratio between the total mass of the T-Bar and the mass of the solid bars having length l0 is given, then we can note that only the quantity ls1 of Eq. (10) changes, while the quantity lb1 remains unchanged because the design conditions of the bars remain unchanged. Hence, concerning strings mass ratio ls1 we have:
As a consequence the value of the string area is:
As ¼
k 1 s1 f ðl0 Þ ¼ 2 Es sin a1 Es
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ tan2 a1
In the relations reported above no considerations have been made regarding the resistance of the strings. In fact, the only requirement made on the mechanical characteristic of strings is their stiffness. In particular the stiffness obtained in Eq. (28) represents the value which is able to assure that global instability and Eulerian instability appears in the same time. For a stiffness greater than the one expressed by Eq. (28) the T-Bar will collapse due to Eulerian instability, while for a stiffness smaller than the one expressed by Eq. (28) the T-Bar will collapse due to global instability. Now it is clear that also the tensions of the strings have to be checked. For a given yield stress rs of the string material the minimum tension required in the each string is given by tðs1 Þ. As a consequence the following relation has to be satisfied:
As rs tðs1 Þ
ð31Þ
From the above relation we can obtain the minimum value of the string area which is able to assure a sufficient resistance to the string elements.
As ¼
t ðs1 Þ
rs
¼
f ðl0 Þ 2 cos a1 rs
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi f ðl0 Þ 1 þ tan2 a1 ¼ 2rs
ð33Þ
The obtained result can be expressed in the following way:
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
f ðl0 Þ 1 þ tan2 a1 1 2 As ¼ ; 2 max 2 rs sin a1 Es pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 f ðl0 Þ 1 þ tan a1 n o ¼ 2 2 min rs ; sin a1 Es =2
ð35Þ
By means of the ratio between the elastic modulus and the yield strength of the strings material as ¼ Es =rs we have:
min
n
o
n
o
rs ; sin2 a1 Es =2 ¼ rs min 1; as sin2 a1 =2 with as ¼ Es =rs ð36Þ
And new parameter can be introduced:
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð32Þ
By comparing the values of As obtained in Eqs. (30) and (32) we can conclude that the value required for the string area is given by:
( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi) f ðl0 Þ 1 þ tan2 a1 f ðl0 Þ 1 þ tan2 a1 As ¼ max ; 2 2r s sin a1 Es
pffiffiffiffiffiffiffiffiffiffi f ðl0 Þqs 1 þ tan2 a1 n o 2 cb l0 min rs ; sin a1 Es =2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q pEb f ðl0 Þ 1 þ tan2 a1 n o ¼ s 2 2qb l0 min rs ; sin a1 Es =2
ls1 ¼
qs p E f ðl Þ e n b 0 o¼ n o e0 ¼ 2 2 min 1; as sin a1 =2 2qb l0 rs min 1; as sin a1 =2
ð37Þ
where e is the parameter introduced in Eq. (11). Finally the total mass ratio is given by:
l1 ¼ lb1 þ ls1 ¼
1 5 tan2 a1 þ 1 þ e0 1 þ tan2 a1 2
ð38Þ
It is important to emphasize that the results obtained when n o 2 min 1; as sin a1 =2 ¼ 1 represent the case already studied in [1]. In fact, in this case Eqs. (38) and (10) are equivalent. Relations (10) and (38) have been applied for values of
ð34Þ
The meaning of the relation (34) is the following: when
rs sin2 a1 Es =2 the governing instability is represented by the global instability because the area represented by Eq. (32) is smaller than the one represented by Eq. (30). In other words, when
rs sin2 a1 Es =2 the value As of the area requested by the resistance condition is always lower than that one required to avoid global instability, so that, collapse is due to the global instability of the
e ¼ ½103 ; 101 ; 0:05; 1=3 and as ¼ ½5; 10; 100; 200; 500. The obtained results are reported in Figs. 6–9. As expected, the mass ratio is strongly influenced not only from the materials properties but also from the geometrical configuration. In particular for any given material, i.e. for any fixed value of as , there is a value of the angle a1 which minimizes the total mass ratio. In addition it is evident that for as the angle a1 approaches zero, there is no support against vertical displacements at the midpoint, and the total mass approaches infinity because the system guarantees global buckling. This result is also evident from
bar. On the contrary, when rs sin a1 Es =2 the governing instability is represented by the Eulerian one. In fact, in this case the value of As represented by Eq. (30) is always the smaller one; so that the condition representing the global instability is always avoided. To better understand the result it is useful to consider two limit cases: 2
(1) case of rs ! 1: In this case the tension is zero and the required mass of the strings is zero; i.e. every area As provides sufficient tension, while the stiffness required to avoid global instability requires a string mass depending on Es and a1 according to Eq. (30).
(2) case of sin a1 Es =2 ! 1: In this case the required mass for the stiffness of the strings tends to zero; i.e. every area As is able to assure a sufficient stiffness, while the tension required is given by Eq. (31). From a physical point of view the value of the stiffness kv1 tends to infinite, which means 2
Fig. 6. Values of
l1 for e ¼ 103 .
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Fig. 10. T-Bar of order 2 and its schematization for the analysis of global instability.
Fig. 7. Values of
l1 for e ¼ 101 .
In order to analyze the global instability we can consider a generic deformed configuration as reported in Fig. 11 and then we can consider four equilibrium equations. In particular it is useful to define 3 unknown quantities X 1 ; X 2 ; X 3 representing the 3 vertical displacements (Fig. 11). In addition the 3 vertical reactions corresponding to the 3 vertical displacements will be equal to:
R1 ¼ kv 2 X 1 ;
R2 ¼ k v 1 X 2 ;
R3 ¼ k v 2 X 3
ð39Þ
The rotational equilibrium of AC bar with respect to point C is given by:
RA l2 NX 1 ¼ 0 ) RA ¼
NX 1 l2
ð40Þ
The rotational equilibrium of AD part of structure with respect to point D is given by: Fig. 8. Values of
l1 for e ¼ 0:05.
RA 2 l2 R1 l2 N X 2 ¼ 0
ð41Þ
By substituting Eqs. (39) and (40) in Eq. (41) we have:
X 1 ð2N kv 2 l2 Þ NX 2 ¼ 0
ð42Þ
The rotational equilibrium of AE part of structure with respect to point E is given by:
RA 3l2 R1 2l2 R2 l2 NX 3 ¼ 0
ð43Þ
And again, by using Eqs. (39) and (40) we obtain:
X 1 ð3N 2kv 2 l2 Þ kv 1 X 2 l2 NX 3 ¼ 0
ð44Þ
Finally, the equilibrium of the whole structure with respect to point B gives:
RA 4l2 R1 3l2 R2 2l2 R3 l2 ¼ 0
ð45Þ
That by means of Eqs. (39) and (40) becomes: Fig. 9. Values of
l1 for e ¼ 1=3.
a physical point of view, in fact, when the angle a1 decreases, the T-Bar configuration is less able to support vertical loads. Of course when a1 is equal to zero the vertical stiffness Kv1 is equal to zero, and the structure is unstable.
X 1 ð4N 3kv 2 l2 Þ 2kv 1 X 2 l2 kv 2 X 3 l2 ¼ 0
By combining Eqs. (42), (44) and (46) the system able to find the critical load corresponding to the global instability can be found as:
4. The T-Bar of complexity two Now we can analyze the T-Bar of complexity 2, which is obtained from the T-Bar reported in Fig. 1 by substituting the two bars of length l1 with two T-Bars, as depicted in Fig. 10. The same figure represents the structural scheme that is to be analyzed for the study of global instability. It is important to note that in this case in addition to the vertical stiffness kv1 there is another vertical stiffness that we call kv2.
ð46Þ
Fig. 11. Deformed configuration.
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8 > > < > > :
X 1 ð2N kv 2 l2 Þ NX 2 ¼ 0 ð47Þ
X 1 ð3N 2kv 2 l2 Þ kv 1 X 2 l2 NX 3 ¼ 0 X 1 ð4N 3kv 2 l2 Þ 2kv 1 X 2 l2 kv 2 X 3 l2 ¼ 0
As it is well known if we want a solution different from the null one which is given by X 1 ¼ X 2 ¼ X 3 ¼ 0 it is required that the determinant of the coefficients is equal to zero. In this case, it results:
2 3 2 2 2 kv 1 kv 2 l2 þ 4kv 1 kv 2 l2 þ 2kv 2 l2 N ð4kv 1 l2þ 6kv 2 l2 ÞN2 þ 4N3 ¼ 0
ð48Þ The solutions of Eq. (48) are:
Ncrit;1 ¼
Ncrit;3
kv 2 l2 ; 2
kv 2 l2 kv 1 l2 þ þ 2 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 kv 2 l2 þ kv 1 l2
Ncrit;2 ¼
kv 2 l2 kv 1 l2 ¼ þ þ 2 2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 kv 2 l2 þ kv 1 l2 2
2
2
2
; ð49Þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 kv 2 l2 þ kv 1 l2 2
2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 kv 2 l2 þ kv 1 l2 k v 2 l2 k v 1 l2 ¼ Ncrit;2 ¼ þ 2 2 2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 kv 1 l0 kv 2 l0 kv 1 l0 kv 2 l0 ¼ þ þ 8 8 8 8
ð50Þ
ð51Þ
kv 2 ¼ 2k2 sin a2 2
ð52Þ
In order to substitute the values of kv1 and kv2 in Eq. (51) it is useful note that, being s1 ¼ l0 =2 cos a1 and s2 ¼ l0 =4 cos a2 , we have:
i kv 1 l0 2k1 sin a1 2s1 cos a1 1 h 2 ¼ ¼ k1 s1 sin a1 cos a1 2 8 8 2
ð53Þ
kv 2 l0 2k2 sin a2 4s2 cos a2 2 ¼ ¼ k2 s2 sin a2 cos a2 8 8 2
In order to simplify the mathematical details of the following development, it is useful to introduce the following nondimensional parameter:
b¼
k1 s1 Es A1 A1 ¼ ¼ k2 s2 Es A2 A2
ð54Þ
By using the above parameter, the values of kv 1 l0 =8 appearing in Eq. (51) can be written as:
i k v 1 l0 1 h 2 ¼ k2 s2 bsin a1 cos a1 2 8
2
2
2
2
ðk2 s2 Þ ðk2 s2 Þf ðl0 ÞC 1 þ f ðl0 Þ C 2 ¼ 0
ð57Þ
ð58Þ
where the numerical coefficient C1 and C2 are equal to:
bsin a1 cos a1 þ 2sin a2 cos a2 2
C1 ¼ C2 ¼
2
bsin a1 sin a2 cos a1 cos a2 1 2
2
;
bsin a1 sin a2 cos a1 cos a2 2
k2 s2 ¼
2
ð59Þ
ð55Þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi f ðl0 Þ C 1 C 21 4C 2 2
ð60Þ
It is easy to verify that both solutions have positive value, so that, as we are interested in the solution of minimum mass, we adopt the minimum value ðk2 s2 Þ, i.e. the minimum value of string area As2 . In addition we know that k2 s2 ¼ Es As2 and As1 ¼ bAs2 so the value of As1 and As2 can be obtained as:
As2 ¼
By repeating the consideration made for the evaluation of kv1 in the case of T-Bar of first order, the value of kv1 and kv2 can be easily obtained as: 2
By isolating the square root and squaring first and second member we obtain:
The solutions of Eq. (54) are given by:
Given that negative stiffness is impossible here, the minimum among the three values reported in Eq. (49), i.e. the critical load, is (noting that l2 = l0/4):
kv 1 ¼ 2k1 sin a1 ;
ð56Þ
which can be written as:
) kv 2 þ kv 1 < kv 1 ) kv 2 < 0
Ncrit
2
2
2
kv 2 l2 kv 2 l2 kv 1 l2 < þ 2 2 2
bk2 s2 sin a1 cos a1 2 þ k2 s2 sin a2 cos a2 2 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ! u 2 u bk s sin2 a cos a 2 2 2 1 1 2 t þ k2 s2 sin a2 cos a2 ¼ f ðl0 Þ 2
ðk2 s2 Þ b sin a1 sin a2 cos a1 cos a2 h i 2 2 2 ðk2 s2 Þf ðl0 Þ b sin a1 cos a1 þ 2sin a2 cos a2 þf ðl0 Þ ¼ 0
The actual critical load is the minimum among the three reported in relation (49). So that the value Ncrit,3 can be excluded because it is greater than both Ncrit,2. With the aim of determining which is the lower value between Ncrit,2 and Ncrit,1 we can inquiry about the conditions which are to be satisfied in order to assure that Ncrit,1 < Ncrit,2. In particular it results:
Ncrit;1 < Ncrit;2 )
Now, as already accomplished for the T-Bar of complexity 1, we require a solution of minimum mass so that the global instability load expressed by Eq. (51) is equal to the Eulerian instability load. This condition provides the equality between Ncrit and f(l0):
k2 s2 f ðl0 ÞC ¼ ; Es Es
As1 ¼
where
C ¼ Cðb; a1 ; a2 Þ ¼
C1
f ðl0 ÞbC Es
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C 21 4C 2 2
ð61Þ
ð62Þ
As in the T-Bar of complexity 1, in the above relations no consideration have been made regarding the tension in the strings. In fact, the only requirement on the mechanical characteristic of strings is their stiffness. In particular the stiffnesses k2 and k1 (where k1 is obtained from the fixed value b) represent the values that are able to assure that global instability and Eulerian instability occurring at the same time. For a stiffness of the s2 strings greater than the obtained value of k2 or for a stiffness of s1 springs greater than the obtained value of k1 the T-Bar will collapse due to Eulerian instability, while in the opposite case the T-Bar will collapse due to both global and local instability simultaneously. Now it is clear that also the tension of the strings has to be checked. For a given yield strength rs of the string material the minimum tension required depends on the maximum values t(s1) and t(s2). In fact, the following relation has to be satisfied:
As1 rs t ðs1 Þ;
As2 rs t ðs2 Þ
ð63Þ
From the above relation we can obtain the minimum value of the string area which is able to assure a sufficient tension to the string elements:
As1 ¼
t ð s1 Þ
rs
¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi f ðl0 Þ f ðl0 Þ 1 þ tan2 a1 ¼ 2 cos a1 rs 2 rs
ð64Þ
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As2 ¼
t ðs2 Þ
rs
¼
f ð l0 Þ 2 cos a2 rs
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi f ðl0 Þ 1 þ tan2 a2 ¼ 2 rs
By comparing the values of As1 and As2 obtained in Eqs. (61) and (64) we can conclude that the value required for the string areas are given by:
( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi) f ðl0 ÞbC f ðl0 Þ 1 þ tan2 a1 ; ; Es 2rs ( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi) f ðl0 ÞC f ðl0 Þ 1 þ tan2 a2 As2 ¼ max ; Es 2rs As1 ¼ max
ð65Þ
And remembering the parameter as ¼ Es =rs introduced in Eq. (36) the obtained results can be expressed in the following way:
( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi) f ðl0 ÞbC f ðl0 Þ 1 þ tan2 a1 ; 2rs as rs ( ) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 f ðl0 Þ 1 þ tan a1 2bC pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; 1 ¼ max 2rs as 1 þ tan2 a1
As1 ¼ max
ð66Þ
( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi) f ðl0 ÞC f ðl0 Þ 1 þ tan2 a2 ; As2 ¼ max as rs 2rs ( ) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi f ðl0 Þ 1 þ tan2 a2 2C pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; 1 ¼ max 2rs as 1 þ tan2 a2 In the considered T-Bar there are 4 strings having length s1 and 8 strings having length s2 , so that the total mass of all strings can be determined as:
4qs As1 l0 8qs As2 l0 þ 2 cos a1 4 cos a2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 2qs l0 As1 1 þ tan2 a1 þ As2 1 þ tan2 a2
ms2 ¼ 4qs As1 s1 þ 8qs As2 s2 ¼
) 2bC pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; 1 rs as 1 þ tan2 a1 ( )# 2C 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; 1 þ 1 þ tan a2 max as 1 þ tan2 a2 3 2 2 2 7 1 þ tan a1 1 þ tan a2 q l0 f ðl0 Þ 6 6 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
7 ¼ s 5ð67Þ rs 4 as 1þtan2 a1 as 1þtan2 a2 ; 1 ; 1 min min 2bC 2C ¼ q s l0
f ðl0 Þ
"
(
ð1 þ tan2 a1 Þ max
And the ratio between the strings mass and the mass of the solid bar of length l0 is given by:
ls2 ¼
ms2 mðl0 Þ
1 tan5=2 a1 tan5=2 a2 þ 2 4 4 1 þ tan2 a1 1 þ tan2 a2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ e0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ e0 ffi
a 1þtan2 a a 1þtan2 a min s 2bC 1 ; 1 min s 2C 2 ; 1
l2 ¼ lb2 þ ls2 ¼ þ
It is easy to verify that it is equivalent to the value obtained by Skelton and Olivera [1] when both the minimum values are equal to 1, i.e. when global instability is not accounted for. For minimum mass, we need to choose as ; e0 ; a1 and a2 . When these parameters are fixed, then a numerical analysis is to be made in order to find which is the value of b that gives the minimum mass. As an example if we choose a1 ¼ a2 ¼ 10 and e0 ¼ 0:005 then the curves depicted in Fig. 12 are obtained. It is clear that for each given value of as a minimum value of l2 can be found. That is, the value of b is shown that minimizes the mass ratio l2;min of the T-Bar of complexity 2 for the chosen values of as ; e0 ; a1 and a2 . It can be interesting to note that in some cases there are more than one value of b leading to the same minimum value of l2 . In these cases we have more than one possibility to realize a structure of minimum mass. The different solution will differ for the mass of the strings. As an example, in Fig. 12 for a given value of as ¼ 30 the same minimum mass solution is obtained for each value of b such that 0.4 < b < 1.2 . As for T – bar of complexity 1, a comparison with the result already obtained by Skelton [1] can be made. In Figs. 13–16 the results are reported for e ¼ ½103 ; 101 ; 0:05; 1=3 and as ¼ ½5; 10; 100; 200; 500. As in the previous case the mass ratio is strongly influenced not only from the materials properties but also from the geometrical configuration. In particular for any given material, i.e. for any fixed value of as , the total mass ratio depends on the values angles a1 and a2 . In addition it is evident that for little values of these angles the total mass ratio is greater than one: in fact it tends to be infinite, leading to an increase in structural weight with respect to the case of the solid bar. This result is also evident from a physical point of view, in fact, when the angles a1 and a2 decrease, the T-Bar configuration is expected to provide less resistance (stiffness) to vertical loads. As a limiting case, when they are equal to zero the vertical stiffnesses Kv1 and Kv2 are equal to zero, and the structure is unstable. We can conclude that the result is very similar to the one obtained for the T-Bar of complexity 1. Recall that for the T-Bar of order one the relation as sin a1 =2 > 1 is sufficient to assure that the Eulerian instability is the governing phenomenon and so the global instability could be neglected. In a similar way for the T-Bar of complexity pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 the fact that as 1 þ tan2 a1 =2bC > 1 and as 1 þ tan2 a2 =2C > 1 assures that the global instability cannot occur. So the occurrence of global instability depends, also in this case, on the geometrical configuration by means of the angles a1 and a2 , and on the ratio as between the stiffness and the resistance of the string material. 2
3 2 2 7 1 þ tan a1 1 þ tan a2 q s l0 f ð l0 Þ 6 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffi 6 ffi þ ffi 7 4 5 2 2a 2a a 1þtan a 1þtan rs cb l0 f ðl0 Þ min s 1 ; 1 min s 2C 2 ; 1 2bC
2
ð68Þ qb ffi Remembering that cb = p2ffiffiffiffiffiffi it is easy to verify that the common
pEb
factor of Eq. (68) is equal to:
qs f ðl0 Þl0
pffiffiffiffiffiffiffiffiffiffi ¼ rs cb l20 f ðl0 Þ
pffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
qs f ðl0 Þ qs f ðl0 ÞpEb ¼ ¼ e0 rs cb l0 2rs qb l0
ð69Þ
where e0 is the parameter introduced in Eq. (37). So the total mass ratio for the T-Bar of order 2 is given by:
ð70Þ
Fig. 12. Mass ratio versus b for a1 ¼ a2 ¼ 10 and
e0 ¼ 0:005.
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Fig. 13. Values of
l2;min for e ¼ 103 . Fig. 16. Values of
l2;min for e ¼ 1=3.
a1 ¼ a2 ¼ 8 the design is correct, because, in this case, the two curves are coincident for values of angles greater than to 4.91.
5. Conclusions
Fig. 14. Values of
l2;min for e ¼ 101 .
In this work, we have shown how to design tensegrity structures for minimal mass, subject to compressive loads and subject to both local and global buckling constraints. Two fundamental tensegrity structures have been analyzed: the T-Bar of complexity 1 depicted in Fig. 1 and the T-Bar of complexity 2 depicted in Fig. 10. The originality of the work is due to the consideration of global instability that was not accounted for in the previous studies. The obtained results allow one to better understand and design this fundamental structure. In fact, the role played by the geometrical configuration and material properties constituting the strings in avoiding global instability is now explained. The previous calculations minimized mass subject to only local buckling and to prevent global buckling, this study shows that the total mass found in this study is always greater or equal to the one found in the previous study [1]. References
Fig. 15. Values of
l2;min for e ¼ 0:05.
From a physical point of view this result says that if the string material has enough stiffness with respect to the tension then the global instability cannot be a failure mode of the structure. At this point we can better explain the differences between the design rules given in [1] and the ones herein obtained. If we consider a value of ¼ 101 we have to look at Fig. 7 and Fig. 14 of the paper. For a fixed material like steel with yield strength rs equal to 2000 Mpa we have as ¼ 100. From Fig. 7 it is evident that you can find the value of angle alpha that prevents global buckling, it is given by the abscissa where the curve for as ¼ 100 and the Skelton curve intersect. This value is equal approximately to 8. Now with this value of angle a; we can use the self-similar iterations of [1] to design a structure of complexity 2. From Fig. 14, one can find that for the value of as ¼ 100 and for angles
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