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Definition 2 A Michell Topology of order q consists of Michell spirals of order less than or equal to q and their conjugate spirals where [link] and [link] hold.

Force equilibrium at a generic node

A force applied to a node can be represented by the vector w i k applied to the node n i k and written in complex form as

w i k = w i k e j φ w i k

where

φ w i k = θ i k + ( i - k ) φ

and θ i k is the angle at which the external force w i k is applied, measured from the radial line to node n k .

At node n i k , the forces along directions m i k , m ¯ k i , m i , k - 1 , m ¯ k , i - 1 , respectively, have magnitudes f i k , t k i , f i , k - 1 , t k , i - 1 . Therefore, the sum of forces at node n i k yields

f i k m i k m i k + t k i m ¯ k i m k i - f i , k - 1 m i , k - 1 m i , k - 1 - t k , i - 1 m ¯ k , i - 1 m k , i - 1 + w i k = 0 .

Multiplying [link] by the vector p i + k e j ( k - i ) φ gives

p i + k [ f i k e j β + t k i e - j β - f i , k - 1 e j ( β + φ ) - t k , i - 1 e - j ( β + φ ) + w i k e j θ i k ] = 0 .

Because both the real and imaginary parts of the above equation must be zero,

cos β cos β - sin β sin β t k i f i k p i + k - cos ( β + φ ) cos ( β + φ ) - sin ( β + φ ) sin ( β + φ ) t k , i - 1 f i , k - 1 p i + k + cos θ i k sin θ i k w i k = 0 0

Linear propogation of forces

Define a vector x α R 2 ( α + 1 ) which contains forces in all members that lie within the radii r α and r α + 1 . That is,

x α = t 0 , α f α , 0 t 1 , α - 1 f α - 1 , 1 t i , α - i f α - 1 , i t α , 0 f 0 , α

Define a vector u α such that

u α = w α + 1 , 0 w α , 1 w α - 1 , 2 w α - 2 , 3 w α - 3 , 4 w 0 , α + 1

Theorem 1 Let a truss be arranged according to the Michell topology of order q , satisfying the conditions in the definitions, having external forces w i k applied at the nodes n i k with i , k 0 and i + k q . Let x α contain the forces(normalized by the member length) in all members within the band of members between radii r α and r α + 1 . Then the forces propagate from one band to the next according to the linear recursive equation x α + 1 = A α x α + B α u α for some matrices A α and B α .

Michell topologies under a single bending load

If t i k f i k < 0 , then one member of the truss is in compression while the other is in tension, so the truss is said to be bending.

Theorem 2 Let a truss be arranged according to the Michell topology of order q satisfying [link] and with a and c given by [link] . Assume the only force applied on the truss is w = w e j θ at node n 00 . If

φ > 0 , 0 < β < | θ | < π - β , φ + 2 β < π ,

then the forces at all nodes satisfy

t i k f i k < 0 for all i q , k i

Furthermore, if

β < θ < π - β

then t i k > 0 for all i q , k i . Otherwise, if

β - π < θ < - β

then t i k < 0 for all i q , k i .

First, take the case where q = 1 . In this case, n 00 is the only node at which external forces are applied:

t 00 f 00 = w 00 sin ( 2 β ) sin ( θ 00 - β ) - sin ( θ 00 + β )

In the case when 0 β π 2 , there are three cases of interest. The first is when | θ 00 | < β < π 2 . Then sin ( 2 β ) 0 and

- π < - 2 β < θ 00 - η < 0 sin ( θ 00 - β ) < 0
0 < θ 00 + β < 2 β < π sin ( θ 00 + β ) > 0

which implies t 00 < 0 and f 00 < 0 , or, in other words, that both members of the truss are in tension.

The second case is when | θ 00 - π | < β < π 2 . A similar analysis shows that t 00 > 0 and f 00 > 0 so both members of the truss are in compression.

Lastly, when β < | θ 00 | < π - β then t 00 f 00 < 0 , indicating that one member of the truss is in tension and the other in compression, so the truss is said to be bending.

Now, consider the case where q > 1 and assume that 0 < β < θ 00 < π - β so that t 00 > 0 and f 00 < 0 (the opposite case is handled similarly). Consider the node n 01 . In the force diagram, w 01 = 0 by assumption, hence t 1 , - 1 = 0 because n 01 is at the lower boundary of the truss. Repeat the analysis from before for the fictitious force w 01 = | f 00 | m 00 . Because the truss has a < 1 , β < θ 01 = φ + β < π , so that t 10 > 0 and f 01 < 0 . The same idea, applied to all nodes on the upper boundary with k q , proves that t k 0 > 0 and f 0 k < 0 for all k q .

The analysis at node n 10 is similar. Repeat the above analysis for a fictitious force w 10 = - | t 00 | m ¯ 00 . The vector has a negative sign because the angle θ 10 is measured clockwise, so the vector needs to be flipped to get the same angle θ as before. Therefore, for all nodes n i 0 with i q , t 0 k > 0 and f i 0 < 0 for all k q .

Continuing on, for the node n 11 , use a fictitious force w 11 = - | t 01 | m ¯ 01 + | f 10 | m 10 . Because a < 1 , one can conclude that β < θ 11 < π - β to get that t 11 > 0 and f 11 < 0 , and continuing in this fashion gives t i i > 0 and f i i < 0 for all i q .

The last type of node to consider is an interior node of the form n i k where i k 0 . At the node n i k create the fictitious force w i k = - | t i - 1 , k | m ¯ 20 + | f i i | m i i . The assumption that a < 1 leads to the conclusion that β < θ i k < π - β , yielding t k i > 0 and f i k < 0 . Therefore, when β < θ 00 < π - β ,

t k i > 0 and f i k < 0 for all i , k q

A similar procedure for - β > θ 00 > β - π yields

t k i < 0 and f i k > 0 for all i , k q

Material volume of optimal structures

Let σ ¯ j , j = 1 , ... , m , denote the yield stress of the material used to construct the j th member, m j , loaded with force density σ i and with constant cross section area A j . To avoid material failure, the force density at any given member should satisfy

σ j ( m j ) σ ¯ j A j m j , j = 1 , ... , m

with equality holding in minimal volume structures. Some algebra yields

v j = m j A j = 1 σ ¯ j σ j ( m j ) m j 2 , j = 1 , ... , m

where v j is the volume of each member. Therefore, the total volume of a structure, V , is

V = j = 1 m v j = j = 1 m 1 σ ¯ j σ j ( m j ) m j 2 .

Let b j and s j be the bar and string vectors for the j th compressive or tensile member in a structure with m b bars and m s strings. Therefore V = V b + V s , where

V b = j = 1 m b 1 λ ¯ j λ j ( b j ) b j 2 and V s = j = 1 m s 1 γ ¯ j γ j ( s j ) s j 2

Assuming that all bars are the same material, λ ¯ j = λ ¯ for all j = 1 , ... , m b . Likewise, if all strings are made of the same material, then γ ¯ j = γ ¯ for all j = 1 , ... , m s . Hence

V b = j = 1 m b 1 λ ¯ λ j ( b j ) b j 2 and V s = j = 1 m s 1 γ ¯ γ j ( s j ) s j 2

so that

V = V s + V b = j = 1 m b 1 λ ¯ j λ j ( b j ) b j 2 + j = 1 m s 1 γ ¯ j γ j ( s j ) s j 2

Now, let J be defined by

J = 2 λ ¯ γ ¯ V + ( γ ¯ - λ ¯ ) = 1 n n i T w i

From [link] and [link] , it is true for a structure in static equilibrium that

= 1 n n i T w i = j = 1 m s γ j ( s j ) s j 2 - j = 1 m b λ j ( b j ) b j 2

Let B = j = 1 m b λ j ( b j ) b j 2 and S = j = 1 m s γ j ( s j ) s j 2 Therefore

J = 2 λ ¯ γ ¯ 1 λ ¯ B + 1 γ ¯ S + ( γ ¯ - λ ¯ ) S - B = 2 γ ¯ B + 2 λ ¯ S + γ ¯ S - γ ¯ B - λ ¯ S + λ ¯ B = γ ¯ B + λ ¯ S + γ ¯ S + λ ¯ B = ( λ ¯ + γ ¯ ) ( B + S )

Thus

J = ( λ ¯ + γ ¯ ) j = 1 m s γ j ( s j ) s j 2 + j = 1 m b λ j ( b j ) b j 2

Assuming that λ ¯ = γ ¯ , we get that

J = 2 λ ¯ 2 V = 2 λ ¯ 2 j = 1 m b 1 λ ¯ λ j ( b j ) b j 2 + j = 1 m s 1 λ ¯ γ j ( s j ) s j 2 = 2 λ ¯ j = 1 m b λ j ( b j ) b j 2 + j = 1 m s γ j ( s j ) s j 2 ,

which is consistent with [link] . This implies that minimization of J is independent of the choice of material properties, and depends only on the variable topology.

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Source:  OpenStax, Michell trusses study, rice u. nsf vigre group, summer 2013. OpenStax CNX. Sep 02, 2013 Download for free at http://cnx.org/content/col11567/1.2
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