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ME5554

Advanced Solid Body Mechanics

AME Mechanical Engineering

School of Engineering and Design

Brunel University

Uxbridge

UK

Structural Design and FEA, Mechanical Engineering, March 2011 1 of 8

W.R. Shipway

Structural Design of a Bridge

W.R. Shipway, BEng Hons. 2008

1. Introduction

Three types of truss design are tested with varying x-sections and varying materials. A parametric

program is created in MATLAB to test different designs, and the economy of each design is calculated.

The sub-routines are explained and certain sub-routines are expanded un-successfully. Their use is

stated. Comparisons of the final design are made with ANSYS for verification, and justified according

to economic parameters. Practical issues are considered and highlighted.

1.1 - Assumptions

2. The design will use a truss structure to connect two positions 150m apart.

3. The height of the bridge base is 50m above ground. The height of the bridge is arbitrary.

4. Weight and self collapse are ignored. However, structural performance is paramount.

5. Modelling is 2-dimensional and point loading is applied at the nodal positions. The node

connections are pinned, hence bar elements are used, with moments ignored.

6. Other bridge design factors are ignored, e.g. aerodynamic, torsional, and buckling stability, and

the transients that give rise to such problems, like vibrations and environmental forces (winds,

earthquakes, thermals like extreme heat or cold exposure).



W.R. Shipway

2. Modelling

2.1 - Bridge Design

A truss structure is considered as the viable method for the bridge design, i.e. triangular sections. The

forces and reactions are considered at the nodes only, which are pinned. The design is constrained by

the following parameters: -

1. The loads are applied along the base, at each node (assuming the equidistant node distribution).

If a design dictates non-equidistant node locations then the load applied will be dependent on

the design, although it is likely the former statement will still hold true.

2. The total load is 1MN.

3. Boundary conditions are typical of bridge modelling, being pinned at one end (allowing

rotational freedom) and rollers at the other (see Fig. 1a).

4. Economy of the structure and safety factors are also considered.

Referring to statement 1 of Modelling – Bridge Design, it is not considered representative to model the

applied loads at the boundaries. Although the loading would be applied here practically, reactions

would be attributed to the ground, rather than the bridge. Fig. 1b highlights the modelling issues raised

by applying loads at the boundaries and is credited to modelling with nodes rather than UDL’s across

the element. In Fig. 1b four elements and four nodes are used, but all lie along the boundary plane w.r.t

the force vector. The vertical (and horizontal) displacement would thus be zero, and is not a

representative model. Hence no loads are applied at the boundaries.

(a) boundary conditions for the model (b) reactions from the ground rather than bridge nodes

Figure 1. Basic conditions of the bridge design.

Rb

150m

Ra



W.R. Shipway

2.2 - Modelling - Justification

The bridge design model is evaluated on a number of factors. The main factors are,

1. It does not deflect by more than 30mm.

2. Safety Factor. The safety factor is considered after the maximum deflection is verified, and

is a subroutine algorithm computed after the FEA model has been processed. The safety

factor of a bridge or component is usually taken as a load/stress multiplier implicitly;

however it is not here, rather the deflection itself is the main design consideration.

3. Material. Typical materials for bridge structures are used [3][4][5]:

a. Structural Steel. Relevant properties E = 200 GPa, £577 per tonne, ρ = 7800kg/m3

b. Reinforced Pre-stressed Concrete. E = 26 GPa , £210 per tonne, ρ = 2400kg/m3

c. Timber (oak). E = 11 GPa, £50 per tonne, ρ = 750kg/m3

Note: other factors pertaining to the material used, like corrosion (steel), weather

resistance (timber), or erosion (concrete) is ignored. Usually if one of these materials

were used appropriate protection would be applied to the material, and may change the

overall price non-linearly. Equally the anisotropic properties of concrete are not taken

into account.

4. Economy. The quantity of bar elements represent increasing complexity and time for

assembly. Thus the amount of bars are taken into account when considering the economy of

the structure. The material itself is also considered, with the prices given above. Whilst this

may not be exact it gives an accurate ratio of the cost vs. structural integrity.

5. The MATLAB algorithm is verified with ANSYS.



W.R. Shipway

2.3 - Modelling – FEA algorithm [2]

The full algorithm is shown in the Appendix, and is set-up as a group of sub-routines as follows,

%% Bridge Structure

clear all

clc

format short

%% Material properties in m,N,£ and kg

E=200e9; A=0.2*0.2; EA=E*A;

cost_per_tonne = 577; density = 7800;

%% Generation of elements and nodes

elements = [1 2;2 3;1 3;2 4;4 5;2 5;3 5;4 6;5 6;6 7;5 7;6 8;7 8];

nodes = [0 0;37.5 0;37.5 50;75 0;75 50;112.5 0;112.5 50;150 0];

element_num = size(elements,1);

node_num = size(nodes,1);

xx = nodes(:,1);

yy = nodes(:,2);

GDof = 2*node_num;

U = zeros(GDof,1);

%% Applied loads, in N

UDL = 1e6;

F = zeros(GDof,1);

F(4)= -UDL/3;

F(8)= -UDL/3;

F(12)= -UDL/3;

Clear workspace

Set material properties

Create spatial coordinates

of the elements and nodes

in 2D arrays (x and y)

Create coordinate system arrays as

components of the horizontal (xx) and

vertical (yy) directions. Initialise (pre-

allocate) the degrees of freedom matrix

(U) of the nodes in the x- and y-

directions.

Apply the forces distributed along xx and

yy as matrix force.



W.R. Shipway

%% Stiffness matrix

stiffness = zeros(GDof);

element_length_total = zeros(size(elements,1),1);

% computation of the system stiffness matrix

for e = 1:element_num

indice = elements(e,:);

elementDof = [indice(1)*2-1 indice(1)*2 …

indice(2)*2-1 indice(2)*2];

xa = xx(indice(2))-xx(indice(1));

ya = yy(indice(2))-yy(indice(1));

element_length = sqrt(xa*xa+ya*ya);

element_length_total(e) = element_length;

C = xa/element_length;

S = ya/element_length;

% stiffness matrix(element - local)

ke = EA/element_length*[C*C C*S -C*C -C*S;…

C*S S*S -C*S -S*S; -C*C -C*S C*C C*S;-C*S -S*S C*S S*S];

stiffness(elementDof,elementDof) = …

stiffness(elementDof,elementDof)+ke;

end

%% Boundary conditions at the ends

prescribedDof = [1; 2; 16];

%% Solution

activeDof = setdiff(1:GDof,prescribedDof);

U = stiffness(activeDof,activeDof)\F(activeDof);

displacements = zeros(GDof,1);

displacements(activeDof) = U;

us = 1:2:2*node_num-1;

vs = 2:2:2*node_num;

Compute the element dimensions and

create the local stiffness matrix

F = KU

Where Ke,1 =

[

] locally.

Converting into principle coordinates

with the transformation matrix

T=[

] then Ke = TT

K T.

Compute the global stiffness matrix by

adding Ke element by element

Adding BC’s by fixing the displacements

as per Figure 1a.

To solve the initial problem

F = KU U = K-1

F

Programming U = F.inv(K) = K\U using

the matrix division operator “\” for

quicker execution time.

Creating displaced nodes.



W.R. Shipway

%% Plot drawing displacements

L = xx(2)-xx(1);

XX = displacements(us);

YY = displacements(vs);

dispNorm = max(sqrt(XX.^2+YY.^2));

clf

hold on

%% Plot controls

scaleFact = 1e2;

ux = displacements(1:2:2*size(nodes,1)-1);

uy = displacements(2:2:2*size(nodes,1));

deformedcoordinates = [nodes(:,1)+scaleFact*...

displacements(1:2:2*size(nodes,1)-1) nodes(:,2)+...

scaleFact*displacements(2:2:2*size(nodes,1))];

for e = 1:element_num;


% undeformed plot

M = [nodes(elements(e,1),1) nodes(elements(e,1),2);

nodes(elements(e,2),1) nodes(elements(e,2),2)];

plot(M(:,1),M(:,2),'green');

% deformed plot

N = [deformedcoordinates(elements(e,1),1) deformedcoordinates(elements(e,1),2);

deformedcoordinates(elements(e,2),1) deformedcoordinates(elements(e,2),2)];

plot(N(:,1),N(:,2),'red');

end

max_deflection = min(uy);

Plotting both the initial coordinates and

displaced coordinates computed above.

Compute the maximum vertical

deflection.



W.R. Shipway

%% Graphics controls

xlabel('Length, m'); ylabel('Height, m');

title('Bridge Design','FontSize',16);

text = text(80,-(max(abs(uy))*scaleFact),...

['\leftarrow \Delta_m_a_x=',num2str(max_deflection), 'm']);

set(text,'Rotation',30,'FontSize',14);

legend('Undeformed','Deformed\newlineScale Factor = 100');

axis([-5 155 -5 52]);

%% Verification and Safety Factor

maximum_permitted_deflection = 0.03; % m, vertical

if abs(max_deflection) < maximum_permitted_deflection

safety_factor = maximum_permitted_deflection/abs(max_deflection);

margin_of_safety = safety_factor - 1;

fprintf('The bridge can sustain another %3.1f times the current load\n',margin_of_safety')

else fprintf('BRIDGE INADEQUATE, deflection exceeded\n')

end

%% Material Costs

vol_element = A*element_length_total; % m^3

mass = sum(density*vol_element)/1000; % array, tonne

material_cost = mass*cost_per_tonne;

fprintf('The mass of the bridge is%4.0f tonne. The material cost of the bridge is £%6.0f \n',...

mass,material_cost)

The margin of safety is only calculated once it is within the absolute design requirement, so memory is

not pointlessly used. The cost of material is attributed to the total length only, and ignores the cutting

cost and amount of sections. The base elements consist of a single bar length, by which the nodes are

created on the (physical) construction. However, the amount of individual sections will affect the

assembly costs, and a sub-routine was attempted to calculate this, but was unsuccessful (see Section

3.2).

Annotating the plot and

highlighting the maximum vertical

displacement.

Computing economy factor and

safety factor.

Printing the margin of safety.

Computing physical properties of

the structure and the total cost.



W.R. Shipway

3. Results and Discussions

3.1 - Modelling – Design Testing

Three types of truss design are tested, the Warren Truss, Pratt Truss and Howe Truss. The Warren

Truss requires the triangles be isosceles and hence determine the height of the bridge. By computing the

costs and the margin of safety different designs can be immediately compared. Also, by setting the

material properties at the top of the program different materials can be immediately tested and stated

whether they will fail or not. For example, the results of the structure shown in Fig. 3a for structural

steel perform as follows: The bridge can sustain another 0.9 times the current load. The mass of

the bridge is 215 tonne. The material cost of the bridge is £124208. But, if the height of the truss

were 26m then results show (Fig. 3b): The bridge can sustain another 0.0 times the current load.

The mass of the bridge is 162 tonne. The material cost of the bridge is £93468. Obviously the

design is still within the absolute limits (being 29mm) and is 33% cheaper, but has a factor of safety of

1. Testing the design of Fig. 3b for concrete shows that the bridge is inadequate as the deflection is

exceeded. Increasing the x-sectional area A = 0.55m2 for an equivalence SF of 1 gives: The mass of the

bridge is 377 tonne. The material cost of the bridge is £79157. Clearly it is cheaper to create it with

concrete, but it is over twice the weight, which is an important consideration.

(a) height 50m (b) Height 26m

Figure 3. A typical Truss (not Warren design). Material: steel.



W.R. Shipway

The results of structures shown in Fig. 4 are stated in Table 1, with a safety factor of 2 in each case.

Fig. 4 represents the actual designs used in MATLAB, and are standard designs. [6] Fig.’s 5 and 6

highlight the comparisons by weight and cost respectively.

Table 1. Results for multiple variable designs, all with SF = 2.

Type Height, m Material x-section (m2) £ cost mass (t)

Warren

Steel 0.24 x 0.24 145863 253

32.5 Concrete 0.67 x 0.66 125401 597

Timber 1.05 x 1 22155 443

Steel 0.27 x 0.27 170935 296

Howe 32.5 Concrete 0.75 x 0.75 147702 703

Timber 1.2 x 1.1 25789 516

Steel 0.26 x 0.25 152411 264

Pratt 32.5 Concrete 0.7 x 0.7 128665 613

Timber 1.05 x 1 22566 451

Double

Warren*

Steel 0.2 x 0.17 141542 245

43.3 Concrete 0.52 x 0.5 121210 577

Timber 0.78 x 0.78 21104 422

Double

Howe

Steel 0.22 x 0.22 174749 303

43.3 Concrete 0.61 x 0.61 150449 716

Timber 1 x 0.9 27075 542

Double

Pratt

Steel 0.22 x 0.21 166806 289

43.3 Concrete 0.6 x 0.61 147983 705

Timber 0.95 x 0.9 25722 514

*Note: the double Warren Truss exceeds the height limit of 50m, hence to maintain the evenly

distributed point loads 50.tan30o = 29m, with the closest even horizontal width of 25m.

(a) Warren Truss (b) Howe Truss (c) Pratt Truss

(d) Double Warren (e) Double Howe (f) Double Pratt

Figure 4. Typical bridge designs.



W.R. Shipway

Figure 5. Mass of each bridge design.

Figure 6. The cost of each design.

0

100

200

300

400

500

600

700

800

Mas

s (t

onne)

Bridge Design

Mass of Different Bridge Designs

Steel

Concrete

Timber

0

20000

40000

60000

80000

100000

120000

140000

160000

180000

200000

Cost

(£)

Bridge Design

Cost of Different Bridge Designs

Steel

Concrete

Timber

Warren Howe Pratt Double Double Double

Warren Howe Pratt

Warren Howe Pratt Double Double Double

Warren Howe Pratt



W.R. Shipway

3.2 - Modelling – Incomplete Algorithms

A couple of sub-routines were attempted but not successful: -

1. The applied forces were as a number of point loads at regular intervals, each equally

distributed (divided by the number of points). Each force was separately defined as F(a)= -

load/b where a = the node point, load = total load (1MN) and b = number of nodes the

forces are applied to. The more complex the design becomes the more nodes are distributed

along the bottom surface, hence b increases. Instead of repeatedly inputting b a sub-routine

was created as distribution = size(find(F),1) which counts the number of non-zero (find)

elements in the force array. But, this is a kind of self perpetuating circle, as the array

distribution itself increases the number of non-zero elements in F.

2. In computing an approximate cost evaluation for justifying the design the cost of materials

is calculated. A sub-routine was also created to evaluate the number of bar elements needed

to create the structure. This differs from the total elements as the nodes do not separate the

in-plane elements if such a structure were created. For example, in Fig. 1a the horizontal

elements are separated by seven nodes, but can be constructed by just two bars. In

assembling the stiffness matrix K the angles sine and cosine are computed. If these angles

are the same, C = S, then the bars lie along the same plane. This is stored as countCS within

the global stiffness loop: -

countCS = zeros(element_num,1); i = 0

for ....

ke = EA/element_length*[C*C C*S -C*C -C*S; C*S S*S -C*S -S*S;...

-C*C -C*S C*C C*S;-C*S -S*S C*S S*S];

if C/S == 1

countCS(e) = i + 1

end

....end

The array countCS may be set to i and will store the amount of times an element lies along

the same plane. However, this will eventually store all elements. If countCS(e) = C it will

store all horizontal element lengths, which may be used in conjunction with elements. The

sub-routine would keep all elements of elements when countCS is non-zero, and then

compare the array values. If the second value of elements is the same as the first value in a

later row then the total is added. For example in Fig. 7 the element-node (3,5) needs to be



W.R. Shipway

omitted, and as such is not related to any of the other values. However, this would only

work depending on how elements is initially sequenced by the user.

if,

elements [

]

and,

countCS [ ]

then

horizontalBars [

]

Figure 7. Nodal positions

The matrices stated in the above are the results of the sub-routine stated in this paragraph.

The sub-routine only computes an output for the connecting horizontal nodes, and all non-

zero terms can be added together and subtracted from element_length_total to give an

assembly cost approximation. But, this is not a parametric solution, and as previously stated

is dependent on the input geometry statements elements and nodes.

1

2 4

6

5 3



W.R. Shipway

3.3 – Model verification

It is clear that although timber possesses inferior mechanical properties, but from Fig. 6 it stated as

having a much lower cost for the given structure, and Fig. 5 shows it has a comparably average total

mass. More specifically, from Table 1 the Double Warren Design has a total mass of only 4.7% more

than the minimum (Warren design) of the same material, but reduced the x-sectional area by 42%. The

Double Warren design model is therefore chosen to be verified using ANSYS. Note however, that

almost half the designs have x-sections approaching (or over) 1m2, which in reality is not viable. Indeed

it may be difficult to even procure oak of such thickness.

The ANSYS model is shown in Fig. 8 below. Results of the verification from the model are shown in

Fig. 9. The ANSYS results are directly comparable, both showing a vertical displacement of 15.131mm

(i.e. margin of safety = 1).

(a) Element design (b) Application of forces and constraints

Figure 8. Pre-process plot using ANSYS



W.R. Shipway

(c) FEA results plot highlighting the maximum displacement and scale factor.

Figure 9. Results of deformed and undeformed structure using ANSYS and MATLAB.

(a) Displacement nodal solution (b) Deformed and un-deformed shape

in the vertical direction



W.R. Shipway

3.4 - General Discussions

The x-sectional area stated in Table 1 is not considered applicable, as in reality varied sections would

be used (I- or H- sections among others), and varied thicknesses would be maximised. But it is obvious

that the material variables like the density and cost have a dramatic effect, and the cost will fluctuate

according to the economy.

For the criteria explored in Table 1 certain design optimisation methods could have been used, like

fminbnd (Golden Section Search method) to obtain the optimum x-section and the optimum material.

The accuracy of the results is also significant, where %3.1f i.e. 1d.p. is not necessarily accurate enough,

due to the high quantity and costs involved.

4. Conclusions

The ANSYS verification proved a close approximation to that of the FEA program. The program

showed versatility in finding solutions to new designs, and can be expanded to accommodate new

criteria. Certain sub-routines were attempted but un-successful, but other methods may have provided

solutions to problems like cost of assembly and varying x-sectional elements.

The final design used a Double Warren Truss structure, using timber. The width of the structure (150m)

limited the height according to the isosceles configuration. The design had a total mass of only 4.7%

more than the minimum (Warren design) but a reduced x-sectional area of 42%, and a total cost of 88%

less than the most expensive design (Double Howe - steel). Although steel proved to consistently have

the lowest x-section (owing to its high Young’s Modulus), it had the highest cost, and this was

disproportionate in the design.

In reality other factors affect the design. With the use of timber the compressive strength is about half

that of its tensile strength [7]. For example the Howe bridge design will put the diagonal members in

compression, thus the stiffness matrix would change accordingly. This variation in strength is also true

for concrete, as they are both anisotropic, and the pre-stressing of concrete members will have a

dramatic effect on its performance.



W.R. Shipway

The x-section of each member is rather redundant when deliberating the best design as in reality more

complex x-sections would be used, and would likely resolve to show steel a more viable option.

However, due to its high cost, along with other pertaining options like aesthetics, alternatives are viable

as potential materials.

The margin of safety was arbitrarily chosen as 1, but in reality this would depend on the nature of the

bridge, for example whether it were for public use, and the commissioners and legalities pertaining to

the application of the design. However, the algorithms for calculating the structural efficiency are

parametric, and will allow changes according to better material specifications (E, £, etc.).

5. References

[1] Logan, D., 2007, A first Course in the Finite Element Method, 4rd Ed., Thomson

[2] Ferreira, A., 2009, MATLAB Codes for Finite Element Analysis - Solids and Structures,

Springer

[3] Management Engineering & Production Services, 2011,

http://www.meps.co.uk/world-price.htm, MEPS International Ltd. Accessed 08.04.2011

[4] http://www.wolframalpha.com, accessed 08.04.2011

[5] Caltrans, 2010, Price Index for Selected Highway Construction Items, California Department of

Transportation

[6] Jackson, D., 1995, Great American Bridges and Dams, Wiley & Sons

[7] Record, S., 1914, The Mechanical Properties Of Wood, Wiley & Sons

http://www.meps.co.uk/world-price.htm



W.R. Shipway

6. Appendix

The final code is shown: -

%% Bridge Structure Design - W. Shipway, April 2012

clear all

clc

format short

%% Material properties in m,N,£ and kg

E=200e9; A=0.22*0.21; EA=E*A; cost_per_tonne = 577; density = 7800; % steel

%E=26e9; A=0.6*0.61; EA=E*A; cost_per_tonne = 210; density = 2400; % concrete

%E=11e9; A=0.9*0.95; EA=E*A; cost_per_tonne = 50; density = 750; % timber

%% Generation of coordinates and connectivities

elements = [1 2;1 3;2 3;2 4;2 5;3 5;4 5;3 6;5 6;4 7;5 7;5 8;6 8;7 8;6 9;8 9;8 10;7 10;8 11;10 11;9

11;9 12;11 12;10 13;11 13;11 14;13 14;12 14;13 16;12 15;14 15;14 16;14 17;16 17;15 18;15 17;17

18];

nodes = [0 0;12.5 21.65;25 0;25 43.3;37.5 21.65;50 0;50 43.3;62.5 21.65;75 0;75 43.3;87.5

21.65;100 0;100 43.3;112.5 21.65;125 0;125 43.3;137.5 21.65;150 0];

element_num = size(elements,1);

node_num = size(nodes,1);

xx = nodes(:,1);

yy = nodes(:,2);

GDof = 2*node_num;

U = zeros(GDof,1);

%% Applied loads in N

load = 1e6;

F = zeros(GDof,1);

F(6)= -load/5;

F(12)= -load/5;

F(18)= -load/5;

F(24)= -load/5;

F(30)= -load/5;

%% Stiffness matrix

stiffness = zeros(GDof);

element_length_total = zeros(size(elements,1),1);

% computation of the system stiffness matrix

for e = 1:element_num


elementDof = [indice(1)*2-1 indice(1)*2 indice(2)*2-1 indice(2)*2];

xa = xx(indice(2))-xx(indice(1));

ya = yy(indice(2))-yy(indice(1));

element_length = sqrt(xa*xa+ya*ya);

element_length_total(e) = element_length;



W.R. Shipway

C = xa/element_length;

S = ya/element_length;

% stiffness matrix(element - local)

ke = EA/element_length*[C*C C*S -C*C -C*S; C*S S*S -C*S -S*S;...

-C*C -C*S C*C C*S;-C*S -S*S C*S S*S];

% stiffness matrix (global)

stiffness(elementDof,elementDof) = stiffness(elementDof,elementDof)+ke;

end

%% Boundary conditions at the ends (node 1 fixed in x,y; other side fixed in x)

prescribedDof = [1 2 36]';

%% Solution

activeDof = setdiff(1:GDof,prescribedDof);

U = stiffness(activeDof,activeDof)\F(activeDof);

displacements = zeros(GDof,1);

displacements(activeDof) = U;

us = 1:2:2*node_num-1;

vs = 2:2:2*node_num;

%% Plot drawing displacements

L = xx(2)-xx(1);

XX = displacements(us);

YY = displacements(vs);

dispNorm = max(sqrt(XX.^2+YY.^2));

clf

hold on

%% Plot controls

scaleFact = 1e2;

ux = displacements(1:2:2*size(nodes,1)-1);

uy = displacements(2:2:2*size(nodes,1));

deformedcoordinates = [nodes(:,1)+scaleFact*...

displacements(1:2:2*size(nodes,1)-1) nodes(:,2)+...

scaleFact*displacements(2:2:2*size(nodes,1))];

for e = 1:element_num;


% undeformed plot

M = [nodes(elements(e,1),1) nodes(elements(e,1),2);

nodes(elements(e,2),1) nodes(elements(e,2),2)];

plot(M(:,1),M(:,2),'green');

% deformed plot

N = [deformedcoordinates(elements(e,1),1) deformedcoordinates(elements(e,1),2);

deformedcoordinates(elements(e,2),1) deformedcoordinates(elements(e,2),2)];

plot(N(:,1),N(:,2),'red');

end

max_deflection = min(uy);



W.R. Shipway

%% Graphics controls

xlabel('Length, m'); ylabel('Height, m');

title('Bridge Design','FontSize',16);

text = text(76,-(max(abs(uy))*scaleFact),['\leftarrow \Delta_m_a_x=',...

num2str(max_deflection),'m']);

set(text,'Rotation',30,'FontSize',14);

legend('Undeformed','Deformed\newlineScale Factor = 100');

axis([-5 155 -5 52]);

%% Verification and Safety Factor

maximum_permitted_deflection = 0.03; % m, vertical

if abs(max_deflection) < maximum_permitted_deflection

safety_factor = maximum_permitted_deflection/abs(max_deflection);

margin_of_safety = safety_factor - 1;

fprintf('The bridge can sustain another %3.1f times the current load\n',...

margin_of_safety')

else fprintf('BRIDGE INADEQUATE, deflection exceeded\n')

end

%% Material costs

vol_element = A*element_length_total; % m^3

mass = sum(density*vol_element)/1000; % tonne

material_cost = mass*cost_per_tonne;

fprintf('The mass of the bridge is%4.0f tonne.\nThe material cost of the bridge is £%6.0f \n',...

mass,material_cost)

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Structural Design of a Bridge - Tumblrstatic.tumblr.com/9qomcnd/yYXlspl9z/structural... · Structural Design of a Bridge ... Three types of truss design are tested with varying x