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The 2013 Michell group has made some minor corrections and additions to the original module created by the 2008 PFUG, but this module is still largely the work of the 2008 PFUG.

Notation

We will restrict ourselves to points and vectors in three dimensional Euclidean space, R 3 . A point force is a vector valued measure v δ ( a ) where v is a vector and a is a point in Euclidean space and δ ( · ) is the Kronecker delta mass. If φ is a continuous vector field on Euclidean space we define

( v δ ( a ) , φ ) = φ ( a ) · v

and extend this definition linearly to a sum of point forces. A beam B is a pair of distinct points a and b in Euclidean space with a weight ω R . We can associate with B a mass

Cost ( B ) = | ω | | a - b |

and point forces field

δ B = ω a - b | a - b | ( δ ( a ) - δ ( b ) ) .

δ B corresponds to the reaction force of a cable if ω < 0 and of a bar if ω > 0 with endpoints at a and b . Note that ( δ B , φ ) = 0 if φ is a constant vector field. To see this, note that

( δ B , φ ) = ω a - b | a - b | δ ( a ) - ω a - b | a - b | δ ( b ) , φ = φ ( a ) · ω a - b | a - b | - φ ( b ) · ω a - b | a - b | = 0

because φ ( a ) = φ ( b ) since φ is a constant vector field. We choose the notation δ B to be consistent with the notion of first variation of mass, although in this case it differs by the sign of ω . A truss T is a finite collection of beams and we define Cost ( T ) and δ T by extending the definitions [link] and [link] linearly;

Cost ( T ) = B T Cost ( B )

and

δ T = B T δ B .

If f is a point force field (a sum of point forces) and T is a truss, then T is said to equilibrate f if δ T = f in the sense that

( δ T , φ ) = ( f , φ )

for all continuous vector fields φ . f = f 1 δ ( a 1 ) + + f μ δ ( a μ ) is said to be balanced if

i = 1 μ f i = 0
i = 1 μ f i × a i = 0 .

Existence of equilibrating trusses

One may easily check that δ B is equilibrated if B is a beam and by linearity δ T is equilibrated if T is a truss. The first natural question we deal with is the converse question; is every balanced point force field equal to δ T for some truss T ? In other words, can any balanced point force field be equilibrated by a truss? This section answers this question in the affirmative.The sufficiency will follow from a proof by induction.

Lemma 1 Let f = f δ ( a ) + g δ ( b ) be balanced. Then f is equilibrated by a truss T .

Let T consist of the single beam ( a , b ) with ω = g · ( a - b ) / | a - b | . Then [link] implies f = - g and then [link] implies that f 2 is parallel to a - b , because f × ( a - b ) = 0 . ω = g · ( a - b ) | a - b | = - f · ( a - b ) | a - b | = | - f | | a - b | | a - b | = | f | because f is parallel to a - b . Therefore f = | f | a - b | a - b | because a - b | a - b | is a unit vector parallel to f . Thus f = ω a - b | a - b | . By definition

δ T = ω a - b | a - b | ( δ ( a ) - δ ( b ) ) = f δ ( a ) - f δ ( b ) = f δ ( a ) + g δ ( b ) = f .

Clearly then [link] holds.

Lemma 2 Let f = f δ ( a ) + g δ ( b ) + h δ ( c ) be balanced with a , b and c not lying on the same line. Then f is equilibrated by a truss T .

Without loss of generality, a , b , and c lie in the x y -plane and a = 0 . f , g , and h must lie in the plane, because otherwise it would be impossible to balance the forces. Because a , b , and c do not all lie on the same line, two of the points, say b and c , are linearly independent; dotting [link] with b and then c implies that f · e 3 = g · e 3 = h · e 3 = 0 where e 3 is the unit basis vector parallel with the z -axis. Hence f can be expressed as a linear combination of b and c

f = ω b b | b | + ω c c | c | .

Consider the point force field e = e b δ ( b ) + e c δ ( c ) where e b = g + ω b b | b | and e c = h + ω c | c | . We claim e is balanced;

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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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