Quiz 2

Question 1

What is the process of form finding and how does it differ from traditional design analysis?

Answer 1

Form finding is when you are given a problem with boundary conditions and you give the connectivity first. You prescribe how you want the internal forces to be and then you find the unknown geometry afterwards.

This is the direct inverse of design analysis.

Question 2

What were some disadvantages of physical form-finding?

Answer 2

• They required meticulous care in both the set-up and in taking measurements so that the data can be used in real buildings
• The types of structures that could be feasibly modelled and their complexity was very restricted.

Question 3

How does the Force Density Method work?

Answer 3

It seeks the equilibrium of a pin-jointed network with fixed boundary nodes.

Question 4

If there is a 3 bar systems with 4 nodes (1 free node and 3 fixed nodes, What does m, n, ni, nf equal?

Answer 4

m = bars = 3
n = nodes = 4
ni = free nodes = 1
nf = fixed nodes = 3

Question 5

In this equation to find xi:
Dixi + Dfxf + px = 0
What do all the values mean?

Answer 5

Di = CiT * Q * Ci
CiT = transpose of the sub-matrix of the connectivity matrix for the free nodes
Q = the diagonal matrix of q which is the force density of a bar which = Fi / Li

Xi = unknown internal coordinates

Df = CiT * Q * Cf
Cf = the sub-matrix of the connectivity matrix for the fixed nodes

Xf = Known fixed boundary points

Px = External load

Question 6

How can you fix the problem with the FDM in hypar form-finding when the geometry pulls in too much?

Answer 6

When all the force densities are constant the form pulls itself in too much. To fix this increase the size of (tighten) the force densities at the edges.

Question 7

How can we improve FDM even more in compression-tension-only structures?

Answer 7

We can be more intelligent in how we distribute the force densities. For example, structural weight-optimised force density or even better member length-optimised solutions which use the minimum amount of material or have more uniform member lengths.

Question 8

How can we used the data produced in FDM to determine the material required?

Answer 8

The FDM provides us with the length and forces within the elements. We can use this to find the best section and we can decide what material we use. For an elastic material the area of the material is:

Ai = Fi / fy where fy is the yield stress for the material which should be factored in practice.

The lengths provided by FDM are the stressed lengths. To find the unstressed lengths:

L0,i = Li / ( Fi/(E*Ai) + 1 )

Question 9

What is Dynamic Relaxation (DR) and what is it used for?

Answer 9

DR is a pseudo-dynamic analysis where we find the rest state after oscillations have damped. This is used for form-finding and analysis of membrane structures, Ties/nets.

Question 10

Using the DR method how is the next displacement (x(t+Δt) ) found?

Answer 10

x(t+Δt) = xt + Δt * vx(t+Δt)
x = displacement
v = velocity

vx(t+Δt) = vt + Δt * rx/m
rx = residule force = px - k*x
px = external force
K = stiffness
m = nodal mass

Question 11

How does kinetic damping for in DR?

Answer 11

It is effective to dampen a dynamic system at peak kinetic energy. To do this the system’s kinetic energy is tracked and nodal velocities are set to 0 at a local kinetic energy peak. Kinetic energy declines with time as energy are extracted form the system. It approaches the final rest state with no kinetic energy and no residual forces.

Question 12

How to calculate residual forces for all nodes after a system has finished oscillating

Answer 12

rx = px - CT * Q * C*x
Px = external force
CT = transpose of the connectivity matrix
Q = Diagonal matrix of element force densities
C = Connectivity matrix
x = coordinates of nodes

Question 13

Why are mass-spring models used in engineering?

Answer 13

To simplify member stiffness is static/dynamic analysis

Question 14

For particle springs what does velocity and acceleration equal?

Answer 14

V = dx/dt
a = dv/dt = 1 / m * (p - cv - Σki(Li - Li,0))
m = partical mass
p = applied load
c = drag coefficient
v = velocity
ki = stiffness
Li = current length
Li,0 = Rest length

Question 15

How is Euclidean norm calculated in 3D?

Answer 15

L = II b-a II = √( (xb - xa)^2 + (yb - ya)^2 + (zb - za)^2)

Question 16

How is the dot product calculated in 3D and how is this used to find then angle between two vectors?

Answer 16

ab = xaxb + yayb + zazb

ab = IIaII * IIbII * cosθ
So,
θ = arccos( (ab) / (IIaII * IIbII) )

Question 17

How is the cross-product found in 3D?

Answer 17

a x b = I i,j,k ; xa,ya,za ; xb,yb,zb I
Which equals

(yazb - zayb)i - (xazb - zaxb)j + (xayb - yaxb)k

Question 18

What are the uses of vector length, Dot Product and Cross-product?

Answer 18

Vector length find the length of a vector

The dot product finds the angle between to vectors

Cross-product Returns a vector that is perpendicular to the original vector. It can be used as a IIaxbII to find the area of a parallelogram.

Question 19

How does translation work?

Answer 19

You add a vector to a point/vertex and this will translate the point by the translation vector.

Question 20

How does scaling work and what are the different ways you can do it?

Answer 20

Scaling is when you multiply a point by a scaler
(c). There is uniform scaling about the origin or a defined point such as an object’s centroid. You can also non uniformly scale a point or objects by using cx, cy and cz

Question 21

To scale point (x,y,z) by scale factors of cx, cy and cz what would the matrix equation look like?

Answer 21

[x’ ; y’ ; z’ ; 1] = [ cx,0,0,0 ; 0,cy,0,0 ; 0,0,cz,0 ; 0,0,0,1] [x ; y ; z ; 1]

Question 22

How would you rotate point (x,y) through an angle θ to get (x’,y’)

Answer 22

[ x’ ; y’ ; 1] = [ cosθ,-sinθ,0 ; sinθ,cosθ,0 ; 0,0,1] * [ x ; y ;1]

Question 23

How do you rotate in 3D and what are Rx, Ry, Rz?

Answer 23

[x’ ; y’ ; z’] = Ri * [x ; y ; z]
Rx = [ 1,0,0 ; 0,cosα,-sinα ; 0,sinα,cosα]
Ry = [ cosβ,0,sinβ ; 0,1,0 ; -sinβ,0,cosβ]
Rz = [ cosγ,-sinγ,0 ; sinγ,cosγ,0 ; 0,0,1]

Question 24

What are the types of arrays and how do they work?

Answer 24

1) Rectangular arrays repeat geometry in the x and y direction by some factor and count

2) Box arrays repeat geometry in the x, y and z direction by some factor and count

3) Polar arrays provide a geometry, a plane for rotation, and the count and angle to rotate through.

Question 25

How would you find the normal to a triangular mesh face with vertices of v0,v1,v2?

Answer 25

find two vectors in the plane of the triangular face y-x and z-x. Using the cross product find the normal vector:

N = (v1-v0) x (v2-v0)

As we are not interested in the length we divide it by its length:

N^ = N / IINII

Question 26

Define what is meant by a convex and a concave polygon. Which type can always be triangulated easily?

What is a convex Hull?

Answer 26

Convex = draw a line between any two vertices without crossing the boundary. These can always be triangulated easily

Concave = interior angles exceed 180º and one or more lines will cross a boundary.

Convex Hull = is a convex polygon that contains all the points.

Question 27

How are distance matrices (S) created and describe there form?

What are they useful for?

Answer 27

They are created by setting up a matrix nxn where n is the number of nodes and inputting the distance between each node in the corresponding location. The rows and columns dictate the node number and there is a diagonal of 0s.

They are useful for point clouds and algorithms on points in space with neighbours. They can also be used to find the smallest distance between nodes.

Question 28

How does optimisation work?

Answer 28

it finds model input parameter values that lead to improved/best output values

Question 29

How would you optimise a system by maximise the object function g(x)?

Answer 29

Optimization is usually minimization problem so to optimize a maximisation problem you have to define an object function g(x). Then you minimise the negative of the object function.

Question 30

How would you minimise a function between two bounds?

Answer 30

min f(x)
x€R
s.t. XL ≤ x ≤ Xu

s.t. subject to
Xl = lower bound
Xu = upper bound

Question 31

Outline the steps to create a brute force for finding the minimum volume of elements ina. truss.

Answer 31

1) Set up parameters
2) Set up Dummy/starting optimum - start with a large default optimum
3) Loop over the range of x
4) Input the geometry of the problem
5) Resolve the forces or whatever is specific to the problem
6) Calculate the volume
7) Check is volume is better. If it is start again. only stop when the volume is worse.
8) Print results

Question 32

What are the advantages and disadvantages of the brute force method?

Answer 32

Advantage
- Simple procedure that will always find the global optimum

Disadvantage
- Iterations grow rapidly with finer step sizes

Question 33

What are Heuristic-evolution methods?

Answer 33

Candidate solutions of a population evolve or improve through a number of generations. It guarantees the population improves as the candidates converge to a LOCAL optimum. They require a well define object function

Question 34

What are some examples of Heuristic-evolution methods and what are the 4 most common phases?

Answer 34

Genetic algorithms, Differential evolution, swarm algorithms

The 4 stages are:
1) Initialise the population of candidate solutions (usually randomly)
2) Get the candidates to interact and share information
3) generate new candidate solutions
4) assemble a new population with fitter candidates and the other less fit candidates are ‘killed’.

Question 35

What are the advantages and disadvantages of Heuristic-evolution methods?

Answer 35

Advantages
• Much faster than brute force and can give good results.
• Very robust and resistant to getting trapped in local minima.

Disadvantages
• Random processes causes different answers each time.
• No guarantee a global optimum will be found.

Question 36

Outline the general procedure for gradient methods of optimisation.

Answer 36

1) start from an initial guess point (x0 , fo)
2) find the direction vector Po to improve upon fo
3) Follow Po for some length αo to arrive at x1
4) Repeat iteratively for each fk of iteration k

Question 37

What does the next point x(k+1) depend on for each iteration K?

Answer 37

1) the current location xk
2) the search direction Pk
3) Some movement along direction Pk

x(k+1) = xk + αk*Pk

Question 38

What are key things to think about when doing a line search using the gradient method?

Answer 38

• If f(x) is convex the minimum will be the global but if concave it may be a local minimum
• if αk is too large/small it can lead to poor convergence or instability

Question 39

What is the simplest way to decide on which direction to use for px?

Answer 39

Take the negative of the gradient at the current point:

Pk = - grad( f(xk) )

Question 40

Define the forward and central differences.

Answer 40

Forward difference = grad( f(xk)i ) = ( f(xk + εv) - f(xk) ) / ε
Central difference = grad( f(xk)i ) = ( f(xk+εv) - f(xk-εv)) / ε

Question 41

What is vector Projection and how is it done to project vector a onto vector b?

Answer 41

You project a vector to find its component onto another vector

IIaII * cos(0) = (a.b)/IIbII