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Structural Optimization of 3D Masonry Buildings

(Supplemental Material)

A Matrix Structure

We detail the matrix equation for static equilibrium:

Aeq f + w = 0

wj : 61 vector containing the 3D weight and net torque for blockj. Typically the only non-zero element is the z-component ofweight. For any external loads acting on block j, the force andtorque contributions are added here.

rk: Contains the unknown force vectors fi, for vertices i on inter-

face k. height(rk) is 4vk, where vk is the number of vertices oninterface k and each vertex contributes a 3D force with positive andnegative parts for the axial forces.

Aj,k: Submatrices Aj,k contain coefficients for net force and nettorque contributions from interface k acting on blockj. Each Aj,khas dimension 6height(rk). Rows 1-3 are coefficients for netforce contributions in x,y,zand rows 4-6 are coefficients for nettorque contributions about the x,y,zaxes.

Aj,krk =

Fk Fk . . .

Ti,j,k Ti+1,j,k . . .

fi

fi+1

..

.

Fk = [enk euk evk ] and Ti,j,k = [(enk vi,j) (euk vi,j) (evk vi,j)]. Unit vectors enk , euk and evk are the normal vector andfriction basis vectors for face k (see Figure 1). The subscript xrefers to the x-component of the vector.

The number of submatrices Aj,k in row j of Aeq is equal to the

number of neighbors incident on blockj. There are two submatri-ces in each column k, since rk represents the interaction betweensurfaces of two adjacent blocks.

wj

fi

enk^

vi,j

evk^

euk^

fi+1

interface k

}blockjertex i

Figure 1: Indexing for equations of static equilibrium. Vector enkis the unit normal for interface k, andeuk andevk are the directions

of in-plane friction forces. Vector vi,j is the relative position ofvertex i w.r.t. the centroid of blockj. wj is the 3D weight vector forblockj.

B Partial Derivatives

The partial derivatives of coefficients for net force equilibrium, Fk,on face k are:

Fk

=[en eu ev]k

where is a parameter from the set {ui,k, vi,k, nk, k, k} as de-scribed in 6. The partial derivatives of coefficients for net torque

equilibrium, Tk, on face k are:

Ti,j,k

=[(en vi,j) (eu vi,j) (ev vi,j)]k

(en vi,j)k

= en

vi,j

+en

vi,j

= en

pi,j

cj

+

en

vi,j

The derivative of the centroid position cj for blockj is:

cj

=

vTi,j cTi,jvTi,j

where vTi,j is the volume of tetrahedron i on blockj and cTi,j isthe centroid of tetrahedron i.

vTi,j

=1

6sign(a0 (a1 a2))

(a0 (a1 a2))

cTi,j

=1

4

(a0 + a1 + a2)

where coordinates a0, a1, a2 are three corners of tetrahedron i, off-set such that the fourth coordinate lies at the origin (0, 0, 0).

The derivative of the weight vector w/ for blockj is given by:

wj

= vj

g =

i

vTi,j

g

where is the block density and g is the direction of gravity, and vjis the volume of blockj.

B.1 Constraint Derivatives

In the closed form solution of f, the constraint matrix C is aconcatenation of the matrix Aeq (static equilibrium), the activefriction-cone inequalities of the matrix Afr , and the active lowerbounds on f. The friction constraints and lower bound constraintsare not dependent on block geometry, giving Afr/ = 0 andIlb/ = 0.

C

=

Aeq/

00

,b

=

w/00

where is a parameterization of the structures geometry . Thederivations ofAeq/ and w/ are shown above.

B.2 Energy Derivatives

The expression for the derivative off is then obtained by differen-tiating the closed form expression:

f

= H1CT

E1

b + CTE1b

E

E1

b

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H is the weighting matrix for the objective function and is heldconstant, and E = CH1CT. The derivative of E by application ofthe chain rule is:

E/ = C/ H1CT + C H1CT/

The terms C/ and b/ describe how the constraints changeas the geometry changes according to parameterization . The ex-pression for the gradient ofy() is given by:

y = yuniform + ytorque

where the derivatives of uniform and torque tension energies are:

yuniform

= fTHuniformf

ytorque

= fTHtorque

f

+

1

2

fTHtorque

f

with Htorque/ = (I Hmin)

TDtorque/(I Hmin). Ma-trices Huniform and Hmin are constant since it assumed minimum-tension vertices remain the same for differential movement.

C Cables

For gradient computation, we parametrize cables using the x,y,zcoordinates of their end points. The derivative of the weight vectorw/ for cable is given by:

w

=

L

g

where L is the length of the cable, and is its the mass per unitlength.

L

=

1

2

(p0 p1) P0

p0 p1

where coordinates p0,p1 are two ends of the cable.

The derivative of the cable tension direction is

et

=

(

p0 p1p0 p1

)

=p0 p1 (

p0

) (p0 p1) (

p0p

1

)

p0 p12