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More detailed and accessible derivations of the equations in section 2 of the paper "Michelle Trusses and Lines of Principal Action" are provided here. See Stress and Measure Theory modules for background material.

Michelle trusses

Unidirectional curve stress

Let C be a simple curve in R d . In other words C is the image of a map r C 1 , 1 ( [ 0 , 1 ] , R d ) such that r ˙ ( t ) 0 for each t [ 0 , 1 ] and r is injective. We will also require without loss of generality that r ( t ) trace out the curve C at a constant speed, making | r ˙ ( t ) | constant on [ 0 , 1 ] . Therefore | r ˙ ( t ) | = 0 1 | r ˙ ( s ) | d s 1 - 0 = H 1 ( C ) for each t [ 0 , 1 ] and the unit tangent vector to C at time t will be

τ ( t ) = r ˙ ( t ) | r ˙ ( t ) | = r ˙ ( t ) H 1 ( C ) .

The curvature κ at time t is then given by

κ ( t ) = d τ / d t | d r / d t | = r ¨ ( t ) / H 1 ( C ) H 1 ( C ) = r ¨ ( t ) ( H 1 ( C ) ) 2 .

We now define a measure σ C that is proportional to the stress along the curve C .

σ C : = ( τ τ ) H 1 C = 1 H 1 ( C ) 2 ( r ˙ r ˙ ) H 1 C

Thus σ C is a rank 1 symmetric matrix measure.

We will now treat σ C as the functional on C c ( R d , R d × d ) that corresponds to the measure σ C . For any ξ C c ( R d , R d × d ) ,

σ C [ ξ ] = ξ ; τ τ d H 1 C = 1 H 1 ( C ) 2 C ξ ( r ) ; r ˙ r ˙ d | r | = 1 H 1 ( C ) 0 1 ξ ( r ( t ) ) ; r ˙ ( t ) r ˙ ( t ) d t .

To find the divergence of the measure σ C , we let an arbitrary vector-valued function u C c 1 ( R d , R d × d ) be given and we evaluate the functional - div σ C [ u ] .

( - div σ C ) [ u ] = σ C [ u ] = R d u d σ C = 1 H 1 ( C ) 0 1 u ( r ( t ) ) ; r ˙ ( t ) r ˙ ( t ) d t

Using the definitions of the inner product and the tensor product, we can convert the matrix product u ( r ( t ) ) ; r ˙ ( t ) r ˙ ( t ) into a vector product:

u ( r ( t ) ) ; r ˙ ( t ) r ˙ ( t ) = u ( r ( t ) ) ; r ˙ ( t ) r ˙ ( t ) T = t r [ u ( r ( t ) ) ( r ˙ ( t ) r ˙ ( t ) T ) T ] = t r [ u ( r ( t ) ) r ˙ ( t ) r ˙ ( t ) T ] = u ( r ( t ) ) r ˙ ( t ) ; r ˙ ( t ) = d d t [ u ( r ( t ) ) ] ; r ˙ ( t ) .

Therefore

( - div σ C ) [ u ] = 1 H 1 ( C ) 0 1 d d t [ u ( r ( t ) ) ] ; r ˙ ( t ) d t = u ( r ( t ) ) ; r ˙ ( t ) H 1 ( C ) t = 0 1 - 1 H 1 ( C ) 0 1 u ( r ( t ) ) ; r ¨ ( t ) d t = u ( r ( 1 ) ) ; r ˙ ( 1 ) H 1 ( C ) - u ( r ( 0 ) ) ; r ˙ ( 0 ) H 1 ( C ) - 1 H 1 ( C ) 2 C u ( r ) ; r ¨ d r = R d u ; τ d δ B - R d u ; τ d δ A - R d u ; κ d H 1 C = τ δ B [ u ] - τ δ A [ u ] - κ H 1 C [ u ] = ( τ δ B - τ δ A - κ H 1 C ) [ u ] ,

where A = r ( 0 ) and B = r ( 1 ) . Since this equation holds for any u C c 1 ( R d , R d × d ) , we say that the measure - div σ C is given by

- div σ C = τ δ B - τ δ A - κ H 1 C .

When the curve C is just equal to the line segment [ A , B ] , τ = B - A | B - A | and κ = 0 . Therefore our equation becomes

- div σ [ A , B ] = ( δ B - δ A ) B - A | B - A | .

Stress in a truss

The stress in a beam is given by 1 2 λ σ [ A , B ] for some λ . A finite collection of such beams is a truss . Therefore the stress in a truss that consists of l nodes at { A 1 , A 2 , , A l } is given by

σ = m , n = 1 l - λ m , n σ [ A m , A n ] ,

where we require that λ m , m = 0 for each m .

Recall that the stress σ of a structure that is in equilibrium must satisfy the restrictions - div σ T = F and σ = σ T . It is easy to see that σ will be symmetric by definition. Therefore, if our truss is in equilibrium, there must be a force distribution F such that

F = - div σ = - div m , n = 1 l - λ m , n σ [ A m , A n ] .

Since the operator - div is linear, we get

F = m , n = 1 l λ m , n div σ [ A m , A n ] = m , n = 1 l λ m , n ( δ A m - δ A n ) A m - A n | A m - A n | ,

which will be our new requirement for a truss to be balanced [link] .

Cost of a truss

If we are given a force distribution F , a set of points A = { A 1 , A 2 , , A l } , and a set of weights Λ = { λ m , n : m , n = 1 , l } such that F = m , n = 1 l λ m , n ( δ A m - δ A n ) A m - A n | A m - A n | , then the cost of the truss structure will be the "total mass" of the structure, which is given by R d | σ | .

R d | σ | = | σ | ( R d ) = | σ | 1 m < n l [ A m , A n ] = 1 m < n l | σ | ( [ A m , A n ] ) = 1 m < n l p , q = 1 l λ p , q σ [ A p , A q ] ( [ A m , A n ] ) = 1 m < n l λ m , n σ [ A m , A n ] + λ n , m σ [ A n , A m ] ( [ A m , A n ] ) = 2 1 m < n l λ m , n A m - A n | A m - A n | A m - A n | A m - A n | H 1 [ A m , A n ] ( [ A m , A n ] ) = m , n = 1 l | λ m , n | | ( A m - A n ) ( A m - A n ) | | A m - A n | H 1 [ A m , A n ] ( [ A m , A n ] ) = m , n = 1 l | λ m , n | | A m - A n |

Cost minimization

Given a truss T , which consists of a set of points A = { A 1 , A 2 , , A l } and a set of weights Λ = { λ m , n : m , n = 1 , l } , and given a force distribution F of finite support, we define a set Σ F T ( Ω ¯ ) of all admissible stress measures for the truss T.

Σ F T ( Ω ¯ ) = σ M ( Ω ¯ , S d × d ) : - div σ = F , and σ = m , n = 1 l - λ m , n σ [ A m , A n ]

The requirement σ = m , n = 1 l - λ m , n σ [ A m , A n ] is equivalent to the restriction that σ be rank 1 and supported on a finite collection of simple curves.

We also define the set

Σ F ( Ω ¯ ) = σ M ( Ω ¯ , S d × d ) : - div σ = F ,

of all stress measures for structures that balance the force distribution F . The stresses in this set do not necessarily correspond to trusses, since we have dropped the requirement that the stress be rank 1 and concentrated on curves in R d . In fact, the stress measures in Σ F ( Ω ¯ ) may be spread out over a region of R d . Also, we may allow the given force F to be diffuse.

Our original optimization problem was to find a truss T 0 such that the cost of T 0 is equal to inf T inf | σ | : σ Σ F T ( Ω ¯ for a given force distribution F of finite support. This is equivalent to finding a stress measure σ 0 T Σ F T ( Ω ¯ ) such that

| σ 0 | = inf | σ | : σ T Σ F T ( Ω ¯ ) .

Such a σ 0 , however, does not always exist [link] .

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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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