Optimizing a Suspension Bridge
Robert Powell
APMA 1210
12/12/09
Objective
The cables chosen to be used to construct the suspension bridge could be made from a
variety of materials. My project is to determine which material to use for the vertical cables of
a suspension bridge in order to minimize cost. Furthermore, the spacing and diameter of the
cables will be determined.
Background
Suspension bridges consist of a deck, vertical suspension cables, main cables, and towers
(shown in the figure below). The weight of the deck and automobiles transfers into the vertical
cables which then distribute the force into the main cables. The main cables distribute the
force into the ground through the towers. Both the towers and main cables are anchored into
the ground. Suspension bridges of this type date from the early 19th century. There are several
advantages that suspension bridges have over other types of bridges. They can achieve longer
main spans. They also require less material which leads to a reduced construction cost.
Force
The load (P) that a cable can support is
P = σA
(1)
where σ is the yield stress and A is the cross sectional area of cable. The yield stress represents
the strength of the material. A material with a large yield stress will be able to withstand a
large force. A force that creates a stress beyond this yield stress leads to permanent
deformation. The total cross sectional area of the cables has to be large enough to be able to
support the weight of the bridge and traffic traveling on the deck.
Deformation
When a force is applied on an object, deformation will occur. In this case, with a tensile
force being applied, the material will become elongated. If the deformation is elastic, the
object will return to its original length. The amount of deformation (δ) is determined by
δ = (PL)/(AE)
(2)
where L is the original length of cable and the coefficient E is the modulus of elasticity or
Young’s modulus. The modulus of elasticity is different for every material. It represents the
stiffness of a material. A stiff material will have a large modulus of elasticity and only a small
deformation will occur. The bridge should be designed to only allow a small deformation.
Cost
Want to minimize the cost of the vertical cables of the suspension bridge. The cost of
the material depends on the amount of material. The amount of material (V) is determined by
the cross sectional area.
V = AL
(3)
Prices are represented by the unit price per pound of material. To determine the weight (W),
the density (ρ) of the material must be known
W = Vρ
(4)
Therefore, the cost (C) is calculated by the following equation
C = ALρPr
(5)
where Pr is the price per pound. If A represents the total cross sectional area (area of all the
cables), then C will be the total cost for the suspension cables.
Problem
Want to choose a material that will minimize the cost of the suspension cables (need a
material that is cheap but strong enough that the cross sectional area is not too large). There
are four materials to choose from: Steel, Titanium, Aluminum, or Brass. These are all metal
alloys. Steel is cheap and is relatively strong. Titanium is the strongest out of the four but a bit
expensive. Aluminum is the least expensive but also the weakest. Strength can be measured
several different ways. In this case, I am judging the strength based on the yield stress and
modulus of elasticity. For this model L will be the average length of the cables, which is 100 ft.
Objective Function
Need to minimize the total cost which is the sum of the cost (equation 5) for all 4
materials. However, only 1 material will be chosen.
C = A1LSρSPrSy1 + A2LTρTPrT y2 + A3LAρAPrAy3 + A4LBρBPrBy4
(6)
The subscripts represent the four metal alloys. y is a binary variable (It is either 1 or 0). It
represents the choice of metal alloy. If material j minimizes cost then yj will equal 1 and the
rest of the y’s will be zero. The variable that will be adjusted (optimized) is the cross sectional
area (A). A represents the total cross sectional area of all cables.
Constraints
The cables must be able to withstand a total force of P = 100,000 tons = 2*10^8 lbs. This is double
the total weight of the bridge and automobiles. Using this weight for the force will give a factor of
safety of 2. The first constraint is that the force of the cables (equation 1) must be able to
withstand P.
σsA1y1 + σTA2y2 + σAA3y3 + σBA4y4 > P
(7)
The deformation (δ) should be kept small. For this model it should be less than 6 inches. An
elongation of more than 6 inches is not desirable. The second constraint (using equation 2) is:
(PL)/(A1Es)*y1 + (PL)/(A2ET)*y2 + (PL)/(A3EA)*y3 + (PL)/(A4EB)*y4 < δ
(8)
Only one of the binary variables (y) can be equal to one. The rest should be zero. This
means that that only one material should be chosen. Constraint 3 is:
y1 + y2 + y3 + y4 = 1
(9)
As stated earlier. y is a binary variable. Constraint 4:
y1, y2, y3, and y4 are binary
(10)
The cross sectional area must be positive. Constraint 5:
A1, A2, A3, and A4 > 0
(11)
Model
Minimize:
C = A1LSρSPrSy1 + A2LTρTPrT y2 + A3LAρAPrAy3 + A4LBρBPrBy4
(6)
Subject to:
σsA1y1 + σTA2y2 + σAA3y3 + σBA4y4 > P
(7)
(PL)/(A1Es)*y1 + (PL)/(A2ET)*y2 + (PL)/(A3EA)*y3 + (PL)/(A4EB)*y4 < δ
(8)
y1 + y2 + y3 + y4 = 1
(9)
y1, y2, y3, and y4 are binary
(10)
A1, A2, A3, and A4 > 0
(11)
Again, the only variables to be adjusted are y and A. The rest of the terms are constants. The
values for yield stress (σ), Young’s modulus (E), density (ρ), and price per pound (Pr) are shown in the
following table.
This model can be solved in excel. The excel spreadsheet is shown below.
To solve this problem, excel solver was used:
And the optimal solution is shown below.
Since y1 equals 1, steel is the optimal material to choose for the suspension cables. Steel will be
the cheapest metal alloy to use for the construction of the bridge with a price of $1.4 million.
Part 2
Now that we know steel will be the cheapest material to use for the suspension cables, we
want to determine the best way to space the cables. The cables can be spaced far apart with
each cable having a large cross sectional area or close together with a smaller area. To optimize
the bridge, the total amount (volume, V) of steel should be minimized. This will minimize the
cost of steel.
V = L*A*#cables
(12)
where L is the average length of cable (100 ft), A is the cross sectional area of each cable, and
#cables is the total number of cables. #cables must be an integer. The area and number of
cables will vary to minimize the volume. The strength and deformation constraints are the
same as before.
Model
Minimize:
V = L*A*#cables
(12)
Subject to:
σ*A*(#cables) > P
(13)
(PL)/(AE)*(#cables) < δ
(14)
#cables is an integer
(15)
A > 0
(16)
This model was optimized using excel solver.
And the optimal solution is:
The optimal solution is to have 43 steel suspension cables each with a diameter of 7.7 inches.
This diameter seems to be a bit large compared to the diameter actual used in suspension
bridges. The difference could be due to the numbers chosen for the force of the bridge or the
yield strength of steel.