Quiz 3

Question 1

Why are convex optimisers better than evolutionary optimisers?

Answer 1

Evolutionary optimisers are very slow and convex optimisers are easier and quicker

Question 2

How does convex optimisation work and what is the standard form?

Answer 2

• Involves minimising (or maximising) convex functions over convex sets
• Standard form:
  minimise f(x)

subject to
gi(x) <= 0, i = 1, …, m
hi(x) = 0, i = 1, …, p
x = set of optimisation variable
gi = inequality constraints
hi = equality constraints

Question 3

What are 5 categories of convex optimisation in order for slowest to quickest?

Answer 3

Cone Problem (CP)
Semidefinite Program (SDP)
Second-Order cone Problem (SOCP)
Quadratic Program (QP)
Linear Program (LP)

Question 4

What does a linear program function comprise of?

Answer 4

• a linear objective function to minimise
  or maximise
• problem variables (continuous)
• a series of linear constraints, either
  equalities or inequalities

Question 5

What makes a a problem solution status unbounded or infeasible?

Answer 5

Unbounded: If there is the objective function can be minimised or maximize without limit

Infeasible: if no solution satisfying all the constraints can be obtained

Question 6

What are integer or mixed integer problems

Answer 6

Integer programming (IP) or Mixed Integer Linear Programming (MILP) can be used if respectively all or some of the variables must take on integer values

Integer problems are not strictly speaking convex and hence are more challenging to solve as a consequence

Question 7

What are the steps to model a given LP problem

Answer 7

1. Identify the problem variables
2. Define the objective function
3. Identify the problem constraints
4. Rearrange in standard form (or in a form
   that can be solved using a standard LP
   tool)

Question 8

What is Maxwell’s Theorem and what does it tell use about the volume of material required to carry tensile and compressive forces?

Answer 8

Maxwell’s theorem states that for any given truss problem, the sum of the products of the tensile forces and the corresponding lengths of the bars minus the sum of the products of the compressive forces and the corresponding lengths of the bars is equal to a constant that is related to the external forces/reactions.

It also assumes that the tensile and compressive strength are equal and if the volume of material required to carry tensile forces 𝑉+ increases the volume of material required to carry compressive forces 𝑉− must also increase.

Question 9

How does the constant c related to external forces/reactions?

Answer 9

The sum of the products of the x coordinate and force in the x direction at each node/support + the sum of the products of the y coordinate and force in the y direction at each node/support equals c.

Question 10

How do we find a minimum volume structure?

Answer 10

When V+ or V- equals 0.

Question 11

What is the more efficient way of connecting tension sand compression elements in a system?

Answer 11

Orthogonally so that there is a 90 degree angle between the two components

Question 12

Why are beams the least efficient structural form?

Answer 12

Lacks of structural depth

Prismatic sections used

Question 13

For a single load case, what is the LP formulation for minimising volume?

How is this adapted more multiple loads?

Answer 13

min V = L^T * a
Where L^T = transpose of the length matrix
a = area matrix

such that B * q = f
B is the angle matrix
q = internal force matrix
f = applied nodal forces

-Stress * a <= q <= Stress *a

The b, q and f matrices are put in a matrix that is n x n where n is the number of nodes

Question 14

What are some key points of numerical layout optimisations?

Answer 14

• Size optimization formulation for ground structures with bars that can have zero cross-sectional area
• Statically determinate solution for single load case problems, minimum volume and stiffest elastic layout
• Assumes truss bars yield plastically for multiple load cases
• Can be applied to both 2D and 3D problems

Question 15

How can numerical layout optimisation be made more practical@

Answer 15

a) First rationalise solution via geometry optimization (moving/merging joints to improve the solution)
* Add nodal positions as optimisation variables
* Resulting problem is non-linear, but relatively small-scale as it only involves a
small subset of the original nodes and bars

b) Then simplify solution either manually or automatically (e.g. reducing numbers of joints or members)
* Minimize number of members or joints subject to given volume increase
* Smooth ‘Heaviside’ representation of 0-1 (off-on) variables
* Advantage: short run time

c) Use enriched formulations to account for e.g. local buckling and/or global instability

Question 16

How can you save material on a floor system?

Answer 16

Avoid long spans between columns
* Reduce span L, reduce M = wL2/8;  = 5wL4/384EI (for simply
supported beam)

• Vary beam sizes across the floorplate
• i.e. avoid standardisation to maximise utilisation of material
• Consider using non-prismatic sections
• Although convenient, prismatic sections (e.g. standard I-beam)
  may only be fully utilised at one location
• Consider changing the layout of beams in the floorplate

Question 17

1. Which of the following is NOT a feature of linear programming (LP)?
   a) It can solve problems involving many thousands of variables and constraints
   b) It can solve many problems more quickly than meta-heuristic methods
   c) It can solve both convex and non-convex problems
   d) It can solve problems involving both equality and inequality constraint

Answer 17

c) It can solve both convex and non-convex problems

Question 18

2. To convert the linear programming constraint 4x - 2y + 7z ≥ 12 into standard form requires:
   a) No action – the constraint is already in standard form
   b) The constraint to be rewritten as a “≤” constraint, with terms on both sides of the expression then multiplied by -1 to convert it
   c) The constraint to be rewritten as an “=” constraint, with terms on both sides of the expression then multiplied by -1 to convert it
   d) The constraint to be rewritten as “≤” and “≥” constraints, with terms on both sides of the “≥” expression then multiplied by -1 to convert it

Answer 18

b) The constraint to be rewritten as a “≤” constraint, with terms on both sides of the expression then multiplied by -1 to convert it

Question 19

How to calculate the volume of a half wheel truss?

Answer 19

Volume of rim = Length * area
Where length equal 2piRadius / 2

Question 20

8. Which of the following statements about the solutions obtained using the linear programming (LP) based truss
   layout formulation is correct?
   a) The generated structures will always be in equilibrium with the applied loads
   b) A unique layout will always be obtained
   c) The generated structures will always be stable
   d) The stresses in the generated structures will only exceed the specified yield stress at the supports

Answer 20

a) The generated structures will always be in equilibrium with the applied loads

The equilibrium constraint (Bq = f) ensures that “a” is true; none of the other statements are correct

Question 21

Consider a three-load case truss layout optimisation problem comprising 10 nodes and 84 bars. How many variables will be present in the plastic linear programming (LP) formulation?

Answer 21

Number of bars = 84
3 loads cases
1 area

Total number of variables = 84 * (3+1) = 336

Question 22

Consider a four-load case truss layout optimisation problem comprising 15 nodes and 200 bars. Assuming that three of these nodes have fixed pin supports, how many equality constraints will be present in the plastic linear programming (LP) formulation? Assume that the problem is two-dimensional

Answer 22

Equality constraints are associated to the equilibrium relation B * q = f

Number of nodes = 15
3 nodes are fixed = -3
x and y direction = 2
4 load cases = 4

Number of equilibrium constraints = (15-3) * 2 * 4 = 96

Question 23

Which of the following statements relating to the basic linear programming (LP) grillage layout optimisation formulation covered in the lectures is correct?
a) It is always assumed that the cross-sectional areas of the beams are fixed between endpoints
b) The problem variables are the internal moments, the shear forces and the cross-sectional areas of the beams
c) The depths of the beams in the grillage can vary according to the magnitude of the moment
d) The cross-sectional areas of most beams in the ground structure will normally be zero

Answer 23

d) The cross-sectional areas of most beams in the ground structure will normally be zero

This is because a layout optimisation problem can be considered as a size optimisation problem, though with many of the beams carrying zero moment and having zero cross-sectional area.

Question 24

12. Consider a grillage layout optimisation problem discretised using 4 nodes. Assuming that each node will be connected to every other node by a potential beam, how many beams (including overlapping beams) will be present in the ‘ground structure’?

Answer 24

6

For a fully connected ground structure containing n nodes, the total number of connections m = n (n – 1) /2 =
4 (4 - 1) / 2 = 6

Question 25

Consider a beam grillage layout optimisation problem comprising 12 nodes and 144 beams. How many equality constraints will be present in the plastic linear programming (LP) formulation, neglecting the impact of supported nodes?

Answer 25

Number of nodes = 12
X and y Plane = 2
out of plane bending = 1

Number of equality constraints = 12 * (2 + 1)

Question 26

Which of the following will not influence the volume of material consumed by the floor-beams in a planned
building?
a) Use non-prismatic beams with varying cross-sections along their length
b) Change of layout of the floor beams
c) Vary the depths of beams across the floorplate
d) All of the above can influence the volume of material consumed

Answer 26

d) All of the above can influence the volume of material consumed

Using non-prismatic beams, changing the layout and varying the depth of beams can all influence, and potentially significantly reduce, the volume of material required

Question 27

How are the following calculated?

a) Plastic moment capacity (Mp)
b) plastic moment of resistance per unit area (mp)
c) Plastic section Modulus (Zp)

Answer 27

a) Mp = Zp * yeild strengt

b) mp = Yield strength * D / 2 (D is the distanced between flange centres)

c) Zp = (a/2) * D