s1233587_Report

COMPUTING REPORT
Computer Methods in Structural Engineering 3
CIVE09027
The University of Edinburgh
Ralica Bencheva, s1233587

Computer Methods in Structural Engineering 3 - CIVE09027
CONTENTS:
Abstract................................................................................................................... 1
PROBLEM 1 ............................................................................................................ 2
1. Generating a structure on GSA Suite........................................... 2
2. Hand – Calculations of a global stiffness matrix ........................ 3
3. Bending Moment Analysis ............................................................ 5
4. Deflection Analysis and Contraflexure points............................. 9
5. Acting shear and axial forces ....................................................... 11
6. Bending Moment Envelope ........................................................... 12
7. Earthquake Scenario ..................................................................... 14
PROBLEM 2 ............................................................................................................ 16
1. Bridge Design................................................................................. 16
2. Sin Curve Arc ................................................................................. 18
3. Parabola Arc ................................................................................... 20
4. Circular Arc..................................................................................... 22
5. Comparison Analysis .................................................................... 23
REFERENCES..........................................................................................................29
1

ABSTRACT
The main idea of the project is two represent two different structural analysis done on
a computer software – GSA Suite (Oasys Software). The first problem is mainly base
on fully determining the characteristics of a plane frame and its performance depending
on the placed loading. By the usage of different data and diagrams to determine its
stability and security in the worst possible situation. Perform a full structural analysis.
The main points stated in problem 1 would then be used in problem 2, which is focused
on determining efficiency of an arch by comparison between different designs. The
arch structures generated support a bridge deck and differ in their height and the type
of their curve.
PROBLEM 1
1. Generating a structure on GSA Suite
The structure that has to be analysed is a plane frame structure (Fig. 1), consisting of
restrained and free joints and different length members. In order to create a visual analysis of
the performance of the structure – first it has to be recreated node by node on the computer
software GSA Suite as accurate as possible (Figure 2). Looking at Figure 1 all x and y
coordinates of the 9 nodes can be established and placed into the Nodes table. The beam
elements are the connections between the nodes forming the shape of the structure. In order
for the structure to resist the future applied loadings some of the nodes placed have to be
enhanced (fully restrained in x, z direction and rotational movement) or just partly restrained.
Figure 1 Plane Frame structure
2

From the shown figure could be concluded that the frame consists of 9 nodes: 1, 4 and 7 –
enhanced and 3, 6 and 9 restrained in the z direction (vertical movement); 8 beam elements
and three different cross sections. The plane frame has 15 DOFs. A main feature of a frame
structure is to consist of long and thin member structures, which is the result received once the
column cross section is implemented on to the structure. To be able to analyse the structure
all members have to have a cross section and material assigned as those are critical features
of every frame. The beam which will be undertaking the most of the loading of the structure
have a larger T-shape cross section, which could withstand greater loadings. The young’s
modulus given in the data characterizes the material of the structure as it accounts for the
stiffness and elasticity.
Figure 2 3D Model of the frame structure
In this case the plane frame structure, which will be analysed in this report is a part of a
concrete framed structure. The last important part of generating different structure analysis on
a computer software is to place all nodal and beam loading cases, which would be activated
depending on the analysis needed. At this point only the two t-section members (3 and 6)
would have loading cases. The first case of loading is the dead load, which account for weight
load of the frame and any other permanent loads attached to the structure. The second one is
the live load – which is a variable load that could be placed on the beam members. The first
part of the report covers the performance of the structure place under full factored load, which
is simply the two cases acting together.
2. Hand – Calculations of a global stiffness matrix
If the structure on Figure 1 is solved using the direct stiffness method, the global stiffness
matrix K’ and its partitions K’ff, K’rf, K’fr and K’rr will have the following sizes:
• Global Stiffness matrix K’ – for a plane frame structure each joint has 3 degrees of
freedom (2 translational in vertical and horizontal direction and one rotational). On the
structure all the joints are numbers so it could easily be determined that they are 9.
With a total of 9 joints with 3 DOFs at each gives a total of 27 DOFs of the system,
which determines the size of the matrix [27x27].
3

• K’ff – the matrix associated with the free DOFs and caused by the displacements at the
free DOFs. The three enhanced supports are restrained in all three directions (vertical,
horizontal and rotational) and three pin-roller supports only in the vertical direction. This
results in having 27 − (3 × 3 + 3 × 1) = 15 free DOFs. The number of the free DOFs
corresponds to the number of displacements cause from the free DOFs. The size of
the matrix is [15x15].
• K’rf – associated with the support reaction cause by displacements at the free DOFs.
There 3 reactions at the enhanced supports and 1 at each pin-roller so the matrix has
12 support reactions and 15 free DOFs. The K’rf is [12x15] matrix.
• K’fr – associated with the displacements at the free DOFs cause by the support reaction
displacements. Having 15 free DOFs and 12 support reaction give a [15x12] matrix.
• K’rr – associated with support reactions and cause by the support reaction
displacement, which gives a [12x12] matrix.
Local Member stiffness matrix for member 4:
𝐾𝐾4 =
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎡
𝐴𝐴𝐴𝐴
𝐿𝐿
0 0 −
𝐴𝐴𝐴𝐴
𝐿𝐿
0 0
0
12 𝐸𝐸𝐸𝐸
𝐿𝐿3
6 𝐸𝐸𝐸𝐸
𝐿𝐿2
0 −
12 𝐸𝐸𝐼𝐼
𝐿𝐿3
6𝐸𝐸𝐸𝐸
𝐿𝐿2
0
6 𝐸𝐸𝐸𝐸
𝐿𝐿2
4 𝐸𝐸𝐸𝐸
𝐿𝐿
0 −
6 𝐸𝐸𝐸𝐸
𝐿𝐿2
2 𝐸𝐸𝐸𝐸
𝐿𝐿
−
𝐴𝐴𝐴𝐴
𝐿𝐿
0 0
𝐴𝐴𝐴𝐴
𝐿𝐿
0 0
0 −
12 𝐸𝐸𝐸𝐸
𝐿𝐿3
−
6 𝐸𝐸𝐸𝐸
𝐿𝐿2
0
12 𝐸𝐸𝐸𝐸
𝐿𝐿3
−
6 𝐸𝐸𝐸𝐸
𝐿𝐿2
0
6 𝐸𝐸𝐸𝐸
𝐿𝐿2
2 𝐸𝐸𝐸𝐸
𝐿𝐿
0
− 6𝐸𝐸𝐸𝐸
𝐿𝐿2
4 𝐸𝐸𝐸𝐸
𝐿𝐿 ⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎤
The data necessary for solving the matrix:
𝐴𝐴 = 0.3 × 0.3 = 0.09 𝑚𝑚2
𝐿𝐿 = 5 𝑚𝑚
𝐼𝐼 = 0.000675 𝑚𝑚2
𝐸𝐸 = 2.8 × 1010
𝑁𝑁/𝑚𝑚2
𝐾𝐾4
=
⎣
⎢
⎢
⎢
⎢
⎡
504 × 106
0 0 −504 × 106
0 0
0 1814400 4536000 0 −1814400 4536000
0 4536000 151.2 × 105
0 −4536000 75.6 × 105
−504 × 106
0 0 504 × 106
0 0
0 −1814400 −4536000 0 −1814400 −4536000
0 4536000 75.6 × 106
0 −4536000 151.2 × 105⎦
⎥
⎥
⎥
⎥
⎤
Transforming the local member stiffness matrix into global member stiffness matrix [1].
4

Angle of incline: 𝜑𝜑 = 90° so sin(𝜑𝜑) = 1 and cos( φ) = 0. Transforming the matrix from local to
global using geometry:
𝐾𝐾′4
=
⎣
⎢
⎢
⎢
⎢
⎡
1814400 0 −4536000 −1814400 0 −4536000
0 504 × 106
0 0 −504 × 106
0
−4536000 0 151.2 × 105
4536000 0 75.6 × 105
−1814400 0 4536000 181440 0 4536000
0 −504 × 106
0 0 504 × 106
0
−4536000 0 75.6 × 106
4536000 0 151.2 × 105⎦
⎥
⎥
⎥
⎥
⎤
3. Bending Moment Analysis
The main function of the software is to generate solutions with great accuracy for testing the
stability behaviour of structures and analysing them. Once the full factored loading is applied
to the frame the bending diagram can be easily generated (Figure 3) and the software provides
as well a table showing the forces and moments acting at every point of each beam member.
For greater accuracy in solving this system the software was set to calculate data at 19
equidistant points. An assumption is made that separating the beam member into 20 parts
would give viable results. The software output section could provide various information on the
frame structure like displacements, reactions, stresses, strains, etc.
Figure 3 Bending Moment Diagram
K’4 =
69.29
69.29
417.6
389.9
34.9
34.9
-118.5
-314.6
-27.49 23.56
5

Extracting the output information data into an excel table could ease working with the
numerous received data. The software points out most of the important bending moments and
the maximum and minimum values and their position have been clearly listed at the end of the
output table. However, it does not account for all location points where the bending moment is
zero. As it can be noticed on Figure 3 every element has a point where the bending moment
is zero. The method used for calculating the locations of those points is by simply plotting a
graph of the bending moment across the length of each beam element and solving the linear
equation. Element 1 calculations:
Percent
of the
beam Position Fx Fy Fz Mxx Myy Mzz
m [kN] [kN] [kN] [kNm] [kNm] [kNm]
1 0.00 -76.32 0 24.2 0 -27.49 0
5.00% 0.20 -76.32 0 24.2 0 -22.65 0
10.00% 0.40 -76.32 0 24.2 0 -17.81 0
15.00% 0.60 -76.32 0 24.2 0 -12.97 0
20.00% 0.80 -76.32 0 24.2 0 -8.134 0
25.00% 1.00 -76.32 0 24.2 0 -3.295 0
30.00% 1.20 -76.32 0 24.2 0 1.544 0
35.00% 1.40 -76.32 0 24.2 0 6.384 0
40.00% 1.60 -76.32 0 24.2 0 11.22 0
45.00% 1.80 -76.32 0 24.2 0 16.06 0
50.00% 2.00 -76.32 0 24.2 0 20.9 0
55.00% 2.20 -76.32 0 24.2 0 25.74 0
60.00% 2.40 -76.32 0 24.2 0 30.58 0
65.00% 2.60 -76.32 0 24.2 0 35.42 0
70.00% 2.80 -76.32 0 24.2 0 40.26 0
75.00% 3.00 -76.32 0 24.2 0 45.1 0
80.00% 3.20 -76.32 0 24.2 0 49.94 0
85.00% 3.40 -76.32 0 24.2 0 54.78 0
90.00% 3.60 -76.32 0 24.2 0 59.62 0
95.00% 3.80 -76.32 0 24.2 0 64.46 0
2 4.00 -76.32 0 24.2 0 69.29 0
First step is to generate the data for the element in the right case analysis – full factored loading
and plot the length of the beam ( column 2) again the bending moment ( Myy- column 7). The
graph shows a clear linear trend line (Figure 4), which gives a linear equation that could easily
show the x coordinates of the zero moment point.
6

Figure 4 Element 1 Bending Moment
𝑦𝑦 = 24.197𝑥𝑥 − 27.491
𝑥𝑥 =
𝑦𝑦 + 27.491
24.197
To find the zero moment point location y is set to be equal to 0 as this is where the bending
moment line crosses the abscissa.
Setting 𝑦𝑦 = 0 gives us 𝑥𝑥 = 1.136132578 showing the distance from node 1 to node to 2
(beam element 1) where the bending moment is zero. This method has been applied to
elements 1, 2, 4, 5, 7 and 8. However, the plotted graphs for member 3 and 6 follow a parabola
trend line (Fig 5 and 6), which means that each of two member has to points of zero bending
moment. The parabolas can be easily solved like a quadratic equation.
Figure 5 Element 3 Bending Moment Graph
4.00, 69.29y = 24.197x - 27.491
-40
-20
0
20
40
60
80
0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50
BendingMoment
Position of beam length
Element 1
y = 34.061x2 - 228.96x + 69.3
-400
-300
-200
-100
0
100
200
300
400
500
0 1 2 3 4 5 6 7 8 9
BendingMoment
Position of Lenght Beam
Element 3
7

𝑦𝑦 = 34.061𝑥𝑥2
− 228.96𝑥𝑥 + 69.3
𝑥𝑥1/2 =
228.96 ∓ √228.962 − 4 × 34.061 × 69.3
2 × 34.061
𝑥𝑥1 = 6.404370524 and 𝑥𝑥2 = 0.317686961
Figure 6 Element 6 Bending Moment Graph
𝑦𝑦 = 34.058𝑥𝑥2
− 263.27𝑥𝑥 + 387.91
𝑥𝑥1/2 =
263.27 ∓ √263.272 − 4 × 34.058 × 387.91
2 × 34.058
𝑥𝑥1 = 1.981218 and 𝑥𝑥2 = 5.74883
The linear and parabola shape of the plotted graphs is defined by the relationship between
beam loading, shear forces and bending moment. The change in moment between any two
points is equal to the area under the shear diagram, which is defined by the loading intensity
[2].
∆𝑉𝑉 = � 𝑤𝑤(𝑥𝑥)𝑑𝑑𝑑𝑑
∆𝑀𝑀 = � 𝑉𝑉(𝑥𝑥)𝑑𝑑𝑑𝑑
Those two equations define the slope and shape of the shear and bending moment diagrams.
A point load induces a constant shear force along the beam, which results in a linear slope on
the bending diagram. Beam elements 3 and 6 are placed under uniform loading, which
characterises the parabola shape of the bending moment diagram.
Excel was the main tool used for extracting the most valuable information from the generated
output data from the structural analysis made on the frame. Figure 7 shows all main bending
moments’ points – maximum, minimum and zero values for all elements and their position
along the beams. All those are placed on to the bending moment diagram on Figure 8.
y = 34.058x2 - 263.27x + 387.91
-200
-100
0
100
200
300
400
500
0 1 2 3 4 5 6 7
BendingMoment
Position of Bridge Lenght
Element 6
8

Elements
Positionfor
Maximum
Values (m)
Maximum
Values (kNm)
Position
for
Minimum
Values
(m)
Minimum
Values (kNm)
Zero
Moment
Positions
(length of
beam
element)
Second
Position for
Zero
Moment
Element 1 4 69.29 0 -27.4900000 1.136
Element 2 2 0.00001234 0.0 -0.0000123 1.000
Element 3 8 417.6 3.2 -314.6000000 0.318 6.404
Element 4 0 18.93 5.0 -29.6100000 1.950
Element 5 0 0.000003084 2.0 -0.0000031 1.000
Element 6 0 387.9 3.9 -120.8000000 1.981 5.749
Element 7 0 23.56 -4.0 -34.3900000 1.626
Element 8 0 0.000001542 2.0 -0.0000015 1.000
Maximum
Value
417.6
Minimum
Value
-314.6
Bending Moment Analysis
Figure 7 Bending Moment Analysis Table
Figure 8 Bending Moment Analysis Graph and Zero Moment points
4. Deflection Analysis and Contraflexure points
Analysis done on the bending moment diagram (Fig 8) shows that the curvature changes sign
position along the beam length ( from sagging to hogging and vice versa) at the points of
contraflexure. At those points the bending moment produced is zero so they overlap with the
zero moment locations calculated and placed on Figure 8. However, placing them on an
exaggerated deflected shape could be challenging. In order to do position them correctly a
combined graph of deflection and bending moment has been used (Figure 10). The circles on
Figure 10 show the position of contraflexure points on an exaggerated shape. Allocating them
has significance for designing structures as reinforcement could be reduced [3].
69.29
69.29
417.6
387.9
34.4
34.4
-120.8
-314.6
-27.5 23.6
18.9
9

Figure 9 Exaggerated deflected shape
The deflection analysis represents the maximum vertical and horizontal deflections occurring
in each element (Figure 9). The maximum deflection magnitude and its point location are
calculated so a limit state of safety could be determined. The total deflection is a resultant
from the vertical and horizontal deflection acting together at one point of the beam element.
For an analysis of structures the stiffness of the material and the dimensions of the beam
element determine the deflection limits before permanent deformation or collapse. Using the
output from the GSA analysis shown on Figure 10 and extracted into an excel table the
maximum values can be easily allocated:
Elemetns
Maximum Value of
vertical deflection
Maximum Value
of Horizontal
Deflection
Maximum
Magnitude of
Deflection
Position of
Beam Length
(m)
Element 1 -0.1211 1.899 1.903 4.00
Element 2 -0.1211 10.75 10.75 2.00
Element 3 -11.43 1.899 11.59 3.60
Element 4 -0.3284 2.582 2.596 4.00
Element 5 -0.3284 1.88 1.908 0.00
Element 6 -2.108 1.88 2.82 3.60
Element 7 -0.07695 2.254 2.255 2.254
Element 8 -0.07695 1.87 1.871 0.00
Maximum -11.43 10.75 11.59
Deflection Analysis
Figure 10 Maximum Deflection Values Table
The beam section resists the action of the bending moment and the shear forces so it
deforms. Depending on the direction of the acting bending moment one side is put under
tension while the other one is compressed. Figure 11 displays clearly the relationship
between the deflection and the bending moment magnitude – the curvature is proportional to
the bending moment. The location of the maximum bending moment coincides with the
location of maximum deflection [2].
10

Figure 11 Relationship - Bending Moment and Deflection and Contraflexure points
Formulas that define clearly the relationship between the moment produced and the resulting
deflection:
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 = 𝐸𝐸𝐸𝐸
𝑑𝑑2
𝑦𝑦
𝑑𝑑𝑥𝑥2
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 = 𝐸𝐸𝐸𝐸
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 𝐸𝐸𝐸𝐸𝐸𝐸
Where y is the distance that a certain point has moved and EI is the flexural rigidity of the
beam, which is constant. This defines the directly proportional relationship to the bending
moment resulting from the applied loading. [2]
5. Acting shear and axial forces
The shear force slope is defined by the loading intensity applied to the beam element. As
explained in section 2.3. depending on the type of loading the shear diagram follows different
trend line. (Fig 12)
Figure 12 Shear Force Diagram
-3.084E-6
-263.3
-1.542E-6
-145.4
-14.9
-9.708
316.0
-228.812.34E-6
324.20
11

Elements 2, 5 and 8 experience small shear forces as there is no direct loading applied to
them. Almost all beams have a constant shear force distribution except for 3 and 6 - linear
slope determined by the uniform load. Only for the two span beams are given their minimum
and maximum values:
Table 1 Table 2
Element Variation
Element 1 24.2 constant force
Element 2 1.23E-05 constant force
Element 3 -228.9 316
Element 4 -9.708 constant force
Element 5 -3.08E-06 constant force
Element 6 -263.3 145.4
Element 7 -14.49 constant force
Element 8 -1.54E-06 constant force
Shear Force (kN)
Element Axial Force (kN)
Element 1 -76.32
Element 2 1.53E+02
Element 3 -24.2
Element 4 -165.5
Element 5 413.8
Element 6 -14.49
Element 7 -48.48
Element 8 9.70E+01
The axial forces acting on the beams are either compressive or tensile forces, this could be
easily determined by their sign. The top three beams have positive axial forces so they are
under tension while all the rest are place under compression (Table 2). It can be observed that
beams 3 and 6 experience very small values of axial forces, while member 4 and 5 (middle
vertical sections) experience higher forces. The position of the neutral axis can be easily
determined by looking where the force change from positive to negative – which is along
members 3 and 6. All members have constant axial forces acting throughout the beam length
(Fig 13). The Axial force diagram shows clearly the loading of each beam element:
Figure 13 Axial Force Diagram
6. Bending Moment Envelope
A bending moment envelope summarises all possible loading cases, creates a visionary
analysis of the bending moments experienced by the structure. It is a tool identifying clearly
the maximum and minimum moments at each location of the beam members and creates a
closed area between the two bending limits, which should be accounted for. The software has
the option to automatically generate such an envelope for structures. However it could be done
more explicitly, using colours to identify all the cases that the envelope covers.
Figures 14 and 15 represent bought a bending moment envelope analysing four different
loading cases for members 3 and 6. The members of the frame should be designed to
withstand any combination of loading. The cases applied to the structure are:
12

• Dead Loads applied to beams 3 and 6 – 25.80 kN/m ;
• Dead Load of 25.80 kN/m acting on beam 3 and a full factored load of 68.12 kN/m on
beam 6 ( including dead load of 25.80 and live load of 42.32 kN/m);
• Full factored load (68.12 kN/m) acting on beam 3 and only the dead load acting on
beam 6;
• A combination of full factored loading for both beams.
Figure 14 represent the GSA bending moment envelope generated from a compiled task of
all four cases. It can give a rough estimate of the shape the diagram follows, but it does not
indicate clearly which exactly case is the reason for giving that maximum or minimum
bending moment point. Extracting the output data in excel could produce a far more detailed
graph (Figure 16). The table list the three main point for every beam and the point laying on
the lines defining the envelope. Beam 3 is 8 meters long so the significant points will be at
the supports – 0m and 8m and at the midspan – 4m, consequently for element 6 will be the
points at 8, 11 and 14m as it is 6m long.
Figure 14 GSA Bending Moment Envelope
Position [m]
Maximum
Bending Moment
[kNm]
Case
Minimum
Bending Moment
[kNm]
Case2
0 71.04 Full Factored Loading on Member 3 and
Dead loading on Member 6
24.5 Dead Loading on Member 3 and
Full Factored Loading on Member 6
4 -80.5
Dead Loading on Member 3 and Full
Factored Loading on Member 6 -335.2
and Dead loading on Member 6
8.00 417.6 Full Factored Loading on both beams 158.1 Dead Loading Combination
8.00 387.9 Full Factored Loading on both beams 146.9 Dead Loading Combination
11.00 38.27
Full Factored Loading on Member 3 and
Dead loading on Member 6
-169.8
Dead Loading on Member 3 and
14.00 35.49
Dead Loading on Member 3 and Full
Factored Loading on Member 6
11.93
and Dead loading on Member 6
Figure 15 Bending Moment Element Main Points
The values estimated from the excel table are not fully matching the ones from the GSA
Envelope Diagram (Figure 15). This is caused from inaccuracy in the extracted data. The
data output on which all Excel calculations has been done is from dividing each beam into 20
-92.5 kNm
-344.1 kNm
-5.410 kNm
-178.1 kNm
417.6 kNm
387.9 kNm
146.9 kNm158.1 kNm
71.04 kNm
24.5 kNm
35.49 kNm
11.93 kNm
13

part, which would not give as accurate results as the one generated from GSA. However, if
higher accuracy is needed a more sophisticated and detailed table can be generated in
Excel. The colourful bending moment envelope clearly separates the case on the diagram
(Figure 16). It can be observed that the maximum bending moment does not always result
from the full factored loaded structure. Some of the cases which cause high bending
moments in the first beam span, would cause much lower ones in the second. A clear
example is the case of having dead load on element 3 and full factored load on element 6 –
at 4m it causes the highest bending moment value, while at 11m – the lowest.
Figure 16 Bending Moment Element Envelop
7. Earthquake Scenario
To do an earthquake analysis on the frame structure first the new horizontal loads should be
applied to the nodes: to 3, 6 and 9 – 100kN and to node 2 – 200kN. Once they are placed a
new case is created taking 100 % of the dead loading on element 3 and 6 and 50 % of the live
loading, which places 46.96 kN load on the beams acting downwards. Figure 17 shows all the
loads acting on the frame in a case of an earthquake. All forces on Figure 17 are in kN, the
black numbers indicated the nodes of the frame.
14

Figure 17 Earthquake Loading
Figure 18 displays the bending moment diagram from the generated earthquake analysis. Due
to the acting horizontal loads there is a significant change in the values from the vertical
elements. In full factored situation the top three beams experience moments close to 0, while
now the maximum is equal to 200 kNm.
The locations of maximum, minimum and zero moments have changed as well as their values
(Figure 18). From the data generated by the software it can concluded that the maximum
bending moment has increase to 581.6 kNm from 417.6 kNm and is acting at the end support
point of element 6. The new minimum value is equal to -516 kNm, which show a decrease of
201.4 kNm from the previous analysis. The information from the diagram can be used to
assume how the deflections would differ in an earthquake situation. It can be noticed that
element 1 will deform in an opposite direction than in the generated previous analysis. The
number of contra flexure points decreases as well – the bending moment on elements 3 and
6 does not follow a full parabola shape. The differences between the analyses are significant
3.690E-6
Figure 18 Bending Moment Diagram Earthquake Analysis
539.0
200.0
15.26
581.6
145.8E-6
-381.6
429.4
315.2
385.9
200.0
55.02E-6
-292.0
-492.0 -529.1
15

so by relying only on full factored scenario results it cannot be concluded that the structure
would be safe.
Figure 19 Shear Force Diagram Earthquake Analysis
The Shear Force Diagram (Fig 19) shows similar differences as the on explained for bending
moment diagram, which could be easily defined by the relationship between acting shear
forces and bending moment. The main difference is that during an earthquake situation the top
three vertical element would experience greater forces and would deflect, which in previous
analysis was not accounted for.
PROBLEM 2
1. Bridge Design
The objective set in the second problem is by using one main idea of an arch bridge design to
create three different cases for the following arch shapes – sin curve, parabola and circular
and by generating analysis on the GSA software to compare them. The arch is supporting a
bridge deck suspended from cables [4]. The chosen design follows the design of the pedestrian
bridge crossing Humber River in Toronto (Fig.20). It is a double – ribbed arch bridge consisting
of concrete deck connected with cable to the two steel tubes. Its structure would be simplified
in order to generate a structural analysis using a computer software. All calculation would be
done for only one of the ribs of the structure as it is symmetrical. [5]
16

Figure 20 Humber Bay Arch Bridge
Humber Bay Arch Bridge is a pedestrian and bicycle route for crossing the Humber River. It
has 139 meters length and 6.50 meters width. The material of the deck is concrete as it highly
used and well suited for reinforcement of structures and the arch is made out of high-strength
steel. The tubes have a circular cross section with a diameter of 1200 mm and thickness of
100 mm. In order to analyse the performance of the structure - the structure constructed on
the computer software GSA Suite should be as close as possible to the real one. [5]
The weight of the deck is acting as a dead load in the structure. In order for the mass of the
bridge to be reduced the deck is assumed to have a 50 % hollow concrete structure.
𝑉𝑉 = (139𝑚𝑚 × 6.5𝑚𝑚 × 0.5𝑚𝑚) × 50% = 225.875 𝑚𝑚3
The thickness of the concrete plate is set to 0.5 meters as this would give enough stability for
a pedestrian bridge. The density of concrete is𝜌𝜌 = 2400
𝑘𝑘𝑘𝑘
𝑚𝑚3� , which gives the mass of the
deck to be [6]:
𝑀𝑀 = 225.875𝑚𝑚3
× 2400
𝑘𝑘𝑘𝑘
𝑚𝑚3
= 542 100 𝑘𝑘𝑘𝑘
This creates a dead load applied at each node point of the arch equal to 5318 𝑘𝑘𝑘𝑘. Due to
symmetry in the structure it could be further simplified by creating only 69.5 meters of the
length on the software. Only one of the arch steel tubes will be analysed so the load would be
shared between them. Therefore the dead load resulting from the deck is:
𝐹𝐹𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿1 =
5318 𝑘𝑘𝑘𝑘
4
= 1329.5 𝑘𝑘𝑘𝑘
The arch weight load will be added to the dead load once the structure is generated on GSA.
The bridge should be analysed under the worst possible case scenario of loading. Assumption
for the live loading are based on human factors and behavioural aspects research. An average
male body occupies approximately 0.14 m2 area without including personal space preferences.
Minimum desirable occupancy is around 0.5 m2, although in a challenging situation of crowding
due to different events the space deduces down to 0.25 m2. This research data could give us
the capacity of the bridge [7]:
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 = 139 × 6.5 = 903.5𝑚𝑚2
17

𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑜𝑜𝑜𝑜 𝑝𝑝𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠 =
903.5
0.5
= 1807 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑜𝑜𝑜𝑜 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠 =
903.5
0.25
= 3614 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
An average weight per person could be assumed to be 80 kg then:
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 = 3614 × 80 × 9.81 = 2836 𝑘𝑘𝑘𝑘
The live load for one fourth of the bridge:
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝑝𝑝𝑝𝑝𝑝𝑝 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 =
2836 𝑘𝑘𝑘𝑘
4
= 709 𝑘𝑘𝑘𝑘
The factor of safety chosen for the bridge is 2 based on having highly reliable materials and
lack of severe environmental conditions [7] . However, it could be debatable whether a higher
factor of safety in the range of 5 to 8 would not be more suitable for structures placed under
high loading. The results showing the degree of safety of the different arch would show whether
it is necessary to account for higher factor of safety. The three main cases, which would be
generated into the three types of arc– sin curve, parabola and circular, have the following
geometry:
Figure 22 Problem 2 Structure
2. Sin Curve Arc
For constructing the arch bridge and finding the locations of all nodes has been used excel –
as a calculating tool. The sine curve defining the arch shape is simply calculated by
multiplying the height of the bridge by the sine of the angle range from 0° to 180°- from 180°
to 360° are the negative values of the sine curve, which are unnecessary. The x coordinate is
simply the location of the length of the deck corresponding to the angle. The y coordinated is
the product between the sin value of the angle and the height of the bridge, which defines the
different arch shapes in the three cases (Fig. 23).
Length 139
Case 1 ( H = L/2 ) 69.5
Case 2 ( H = L/4) 34.75
Case 3 (H = L/8) 17.375
Length and height (m)
Figure 21 The three heights
18

Figure 23 Sin Curve Cases Graph
Having all coordinates calculated in advance gives an advantage in creating the structure on
GSA as the coordinates of the points can be just exported directly to the software. The structure
on the software is generated by beam elements connecting the nodes placed. The two joint
supports are restrained: node 1 in vertical and horizontal direction and node 13 at horizontal
direction and moment of rotation. The created structure contains 13 points – the number of
points determines the accuracy of the generated analysis results. The cross section applied to
the structure is of a hollow circle creating as similar as possible to the steel tubes on Humber
Bay Arch Bridge. Forming a structure as close as possible of the bridge would show rather real
performance analysis of the bridge. The material assigned to the arch is a steel, with
characteristics mainly applied to structures, it is already set as a given material from the GSA
software.
Once the structure is created and 3-dimentional form and a material is assigned the software
automatically generates the basic data of the structure. It provides a calculated mass of the
arc, which represents the second part of the dead load that should be applied to the structure.
Each of the cases has different mass so the generated arc would undertake different amount
of dead loading. The comparison between the bridges is generated by analysing there
performance under full factored loading, consisting of live loading formed from the worst case
scenario.
Case 1:
𝑀𝑀𝑎𝑎𝑎𝑎𝑎𝑎 = 268.3 𝑡𝑡
𝐹𝐹𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿2 = 268.3 × 9.81 = 2632.023 𝑘𝑘𝑘𝑘
The full factored load, which consists of the dead loads caused by the weight of the deck and
the arc and the estimated live load, is spread through the 13 nodes as point loads:
19

𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑝𝑝𝑝𝑝𝑝𝑝 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 =
𝐹𝐹𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿1 + 𝐹𝐹𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿2 + 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿
𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑜𝑜𝑜𝑜 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛
=
2632.023 + 1329.5 + 709
13
= 359.271 𝑘𝑘𝑘𝑘
Accounting for factor of safety:
359.271 × 2 = 718.542 𝑘𝑘𝑘𝑘
Case 2:
𝑀𝑀𝑎𝑎𝑎𝑎𝑎𝑎 = 207 𝑡𝑡
𝐹𝐹𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 2 = 2030.67 𝑘𝑘𝑘𝑘
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑝𝑝𝑝𝑝𝑝𝑝 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 = �
2030.67 + 1329.5 + 709
13
� × 2 = 626 𝑘𝑘𝑘𝑘
Case 3:
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑝𝑝𝑝𝑝𝑟𝑟 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 = �
1845.261 + 1329.5 + 709
13
� × 2 = 597.5 𝑘𝑘𝑘𝑘
3. Parabola Arc
Similarly to the sin arc curve the parabola arc point’s location have been first calculated on
Excel and then plotted on to GSA Suite. In all constructed arcs the cross-section of the
structure has been kept the same as they all present the Humber Bay Arch Bridge, which
makes them as well easy comparable. For more detailed solution the points forming the full
arc has been increased to 50, which does not interfere with the loading or the dimensions of
the bridge.
The formation of the parabola curve is based on its main characteristic – that it could be defined
by a quadratic equation.
𝑦𝑦 = 𝑎𝑎𝑎𝑎2
+ 𝑏𝑏𝑏𝑏 + 𝑐𝑐
Where y is defined by the height of the bridge due to the three different cases and x is the
length of the bridge. By setting three main boundary conditions all variable in the equation can
be calculated:
• 𝑦𝑦 (0) = 0 the arc starts at (0, 0) location on the graph;
• 𝑦𝑦 �
𝐿𝐿
2
� =
𝐿𝐿
2
this is the mid boundary condition for the first case. The condition for the
mid span point differs for the different cases : case 2 𝑦𝑦 �
𝐿𝐿
2
� =
𝐿𝐿
4
and case 3 𝑦𝑦 �
𝐿𝐿
2
� =
𝐿𝐿
8
;
• 𝑦𝑦 (𝐿𝐿) = 0 the end point of the arc should lay onto the abscissa.
The three cases can now be defined by three different quadratic equations:
• Case 1 𝑦𝑦 = −
2
𝐿𝐿
𝑥𝑥2
+ 2𝑥𝑥
1
𝐿𝐿
𝑥𝑥2
+ 𝑥𝑥
1
2𝐿𝐿
𝑥𝑥2
+
1
2
𝑥𝑥
20

Setting the x coordinates as 50 points between 0 and 139 as the length of the bridge is 139m
and calculating the y coordinates by using the expressed equations results in three different
parabola curves (Fig 24).
Figure 24 Parabola Arc Graph
This analysis follow the same steps as described for the sine curve arc. The full factored
loading has been calculated in the same manner using again a factor of safety equal to 2. The
only difference is in the number of nodes placed for constructing the structure. Calculating the
dead load from the parabola arc weight and dividing it between the nodes gives the following
results:
Case 1: 𝑀𝑀𝑎𝑎𝑎𝑎𝑎𝑎 = 278.8 𝑡𝑡
2735.028 + 1329.5 + 709
26
� × 2 = 367.194 𝑘𝑘𝑘𝑘
Case 2:
2122.884 + 1329.5 + 709
26
� × 2 = 320.106 𝑘𝑘𝑘𝑘
Case 3
21

1923.741 + 1329.5 + 709
26
� × 2 = 304.788 𝑘𝑘𝑘𝑘
4. Circular Arc
The geometry used behind plotting the circular arc is a combination of Pythagoras theorem
and shifting the centre of the defined circle so that the first point of the arch is positioned at (0,
0) point from the coordinate system. First step is expressing the radius using the know variable
– height and length:
𝑟𝑟2
= (𝑟𝑟 − 𝐻𝐻)2
+ �
𝐿𝐿
2
�
2
𝑟𝑟 =
�
𝐿𝐿
2
�
2
+ 𝐻𝐻2
2 × 𝐻𝐻
At each case the arc shold be shiffted down by difference
between the radius and the heigh and moved right with half
the length of the deck. Plotting the calculations for the three
cases into Excel:
Figure 25 Curve Arc Graph
Loading calculations base on the constructed half arc on GSA Suite:
Case 1: 𝑀𝑀𝑎𝑎𝑎𝑎𝑎𝑎 = 296.1 𝑡𝑡
2904.741 + 1329.5 + 709
26
� × 2 = 380.25 𝑘𝑘𝑘𝑘
Case 2: 𝑀𝑀𝑎𝑎𝑎𝑎𝑎𝑎 = 218 𝑡𝑡
r
(r-H)
22

2138.58 + 1329.5 + 709
26
� × 2 = 321.314 𝑘𝑘𝑘𝑘
Case 3
1925.703 + 1329.5 + 709
26
� × 2 = 304.939 𝑘𝑘𝑘𝑘
5. Comparison Analysis
The efficiency of a structures is primarily based on the strength of the structure and its stiffness.
However, the type of the structure, whether it’s a truss or frame, defines the parameters that
be used for comparison. A structural efficiency of a given shape depends strongly on the used
material, which could not be used as a comparison point as all arches are constructed from
structural steel. Another factor that simplifies the comparison between the structures is that
they all have the same cross-section.
Arches are compressive structure, which could be used for reducing the moments in long span
structures [8]. This sets one of the factors for estimating how efficient an arc is. The bending
moment analysis of the cases can easily set a comparison point in estimating which has the
smallest bending moment range. A two – hinged arc has two moments acting on it one at the
supports and one at the crown for the arch.
Figure 26 Curve Arc Graph
To satisfy the criteria for strength of the structure the Von Mises stress, which is the maximum
through thickness stress and torsional stress coexist, should be less than the yield stress limit
of the material [9]. This would estimate how safe the structure is, it provides security against
buckling, which could results in rigid structures like arches. A higher strength efficiency would
have the structure providing a higher degree of security [10]. In case of a structure which would
be placed under higher loads the safety factor should be more than 1.5. The table shows that
the circular curve with 𝐻𝐻 =
𝐿𝐿
2
would not be safe enough. It can be rated as the less efficient
arch structures as strength is the main factor of efficiency. It can be concluded from the results
that all parabola arches give great stability to the bridge and in terms of strength they provide
the highest. The axial forcing on the structure is what determines the von misses stress as it
is a compressive structure. For better evaluation of the strength of the arches the following
axial forcing diagrams represent the difference. As the cross sectional rea for the three cases
shown is the same – the difference in stresses result from the different axial forcing and
bending moments (Figures 27, 28 and 29).
Sine Curve Case 1 163.5 1.53
Case 2 155.6 1.61
Case 3 171 1.46
Prabaola Case1 48.9 5.11
Case 2 53.37 4.68
Case 3 74.21 3.37
Circular Curve Case 1 486.3 0.51
Case 2 138.4 1.81
Case 3 90.06 2.78
Von mises stress [N/mm^2]
Degree of
Safety
23

Figure 27 Parabola Axial Forces Diagram Case 1
Figure 28 Circular Arc Axial Forcing Diagram Case 1
Figure 29 Sin Arch Axial Forcing Diagram Case 1
-9046
-8127
-7542
-7260
-6719
-5567
-4740
-10330
-10000
-9679
-9361
-8735
-8428
-7831
-6985
-6464
-5987
-5216
-4854
-9130
-8222
-7167
-6436
-5721
-5053
-3737
-9602
-9381
-8850
-8547
-7881
-6803
-6077
-5380
-4741
-4189
-3404
-3288 -3159
-9784
-9176
-8573
-7980
-7406
-6856
-6342
-5873
-5464
-5128
-4879
-4727
24

Figure 30 Deflection Table
The sector parameter accounting for the efficiency is the stiffness of the structure. The stiffness
shows the resistance of the structure to deform. Arches are rigid structures so they are
characterised with high stiffness. A deflection analysis is a main point in comparing structure
placed under high loadings. The table shows all deflections – vertical, horizontal and an overall
magnitude. However, as the arch is supporting a horizontal deck only the vertical deflection
would result on the overall system.
All deflections are relatively small in the range of – 360 to 350 mm (Figure 30). It could
concluded from the table that the lowest vertical deflection result in the parabola curve arch,
which makes it more efficient in terms of stiffness. The relationship between the deflection and
the bending moment magnitude – the curvature is proportional to the bending moment. The
cases with lower deflection should show small values of bending moments as well.
Horizontal [mm] Vertical [mm]
Overall
Magnitude [mm]
Case 1
Maximum values 547.2 251.6 655.4
Minimum values -360.8
Case 2
Maximum Values 184.2 167.9 307.3
Minimum Values -246
Case 3
Maximm Values 81.73 111.4 242.4
Minimum Values -228.2
Case 1
Maximum values 0.1432 27.13 70.36
Minimum values -59.36 -70.36
Case 2
Maximum Values 0.4864 14.94 62.03
Minimum Values -18.78 -62.03
Case 3
Maximm Values 1.968 1.482 92.74
Case 1
Maximum values 325.8 2757
Minimum values -1884 -2575
Case 2
Maximum Values 0.2694 146.4 359.5
Case 3
Maximm Values 1.404 30.46 142
Minimum Values -19.55 -142
Parabola Curve Arch
Circular Curve Arch
Sine Curve Arch
Deflection
25

Diagram bending moment analysis and deflection analysis should be generated for the most
efficient arches so that a better evaluation could be done. From the plotted diagram can be
observed that the minimum bending moment is at the crown of the arch while the maximum
value is at around 25 meters of the length of the deck (Figure 31).
Figure 31 Parabola Case 3
Figure 32 Parabola Case 1
Figure 33. Parabola Case 2
253.6
487.5
654.6
802.1
930.1
1038
282486.6631.3
821.7
945.2
1002
1086
1104
1101
-10.22-279.1
-279.1
-1298
-2106
-3088
-185.3E-6
18.85
29.20
41.19
50.92
58.17
62.87
65.03
64.83
62.52
58.40
46.25
37.73
34.53
23.00
167.6
20.40E-6
398.2
533.9
82.21143.2618.1881.2
930.49
930.49
915.1
835.38
0.6846
0.3433
2.287
3.308
3.878
4.325
4.570
10.30
19.21
19.21 31.38
-2527
-1869
-1125
-806.4-309.7
287.4
188.0
-691.8E-6
594.1
948.9
1065
944.6
755.0 495.7
6.836
16.3
23.68
21.65
15.75
12.34
-2588
-1733
-1030-396.9-1006.4
26

Figure 34 Axial, Shear Forces and Bending Moment Comparison Table
For better stability of the arch bridge all shear forcing should be minimised by using the curvature of
the arch. The results from the table clearly outline how large is the difference between the forcing
due to the height and the shape of the bridge (Figure 34). Taller bridge experience less shear force so
it can be concluded that in between the three different cases – the most efficient one in terms of
shear forcing is the case having𝐻𝐻 = 𝐿𝐿/2. However the three shear force diagrams of the sine curve,
parabola and circular curve show the effect of the shape of the arc. The maximum value of shear
forces that the parabola arch bridge experiences is equal to 44.45 kN which is 2797.55 kN less that
what the circular arc bridge forcing.
Axial Forces Position
Bending
Moment [kNm] Position
Shear
Forces Position
Case 1
Maximum values -4727 Beam 12 12770
Node 13
(crown) 530 Beam 8
Minimum values -9784 Beam 1 -12190 Node 4 -697 Beam 1
Case 2
Maximum Values -8136 Beam 12 9921
Node 13
(crown) 556.9 Beam 8
Minimum Values -11020 Beam 1 -11060 Node 4 -910.4 Beam 1
Case 3
Maximm Values -15460 Beam 12 8813
Node 13
(crown) 560.4 Beam 8
Minimum Values -17000 Beam 1 -10800 Node 4 -1040 Beam 1
Case 1
Maximum values -4740 Beam 25 1104 Node 10 41.45 Beam 1
Minimum values -10330 Beam 1 -179.9 Node 25 -169.9 Beam 25
Case 2
Maximum Values -8253 Beam 25 1065 Node 10 66.23 Beam 2
Case 3
Maximm Values -15700 Beam 25 930.4 Node 10 111.7 Beam 21
Case 1
Maximum values -3159 Beam 25 40710 Node 19 2842 Beam 1
Minimum values -9602 Beam 1 -34590 Node 26 -1378 Beam 16
Case 2
Maximum Values -7511 Beam 25 9683 Node 19 934.6 Beam 1
Case 3
Maximm Values -15230 Beam 25 2933 Node 15 308.4 Beam 21
Sine Curve Arch Analisys
Parabola Curve Arch
Circular Curve Arch
27

Figure 35 Parabola Shear Force Diagram Case 1
Figure 36 Circular Arc Shear Force Diagram Case 1
Figure 37 Sin Curve Shear Force Diagram Case 1
530.8
473.5
399.4
-697.0
-377.1
-113.1
101.0
271
486.7
528
362.4
197.9
-1378
-1274
-799.2 -461.7
2842
2279
1747
1253
796.4
379.6
29.28
-313.9
-575.1
-811.1
-1012
-1157
-1269
-1342
-1374
-1346
-1197
-947.3
-636.9 -279.0
-19.82
-13.41
-44.78
-74.13
-8289
-117.0
-126.9
-146.0
41.45
39.50
29.20
26.67
23.94
21.01
8.341
4.555
0.4709
-3.930
-13.82
-50.48
-65.79
-91.99
-136.6
-171.8
-179.9
28

All nine cases were compared between each other following the main criteria set: stability, stiffness
and strength – the three major factors off bridge structures. By the means of using the shear, axial
and bending moment diagrams and the data output from the structural analysis it can be concluded
that the most efficient structure is the parabola arch with height half the length of the deck.
REFERENCES
[1] “Computer Methods in Structural Engineering 3,” IMPORTANT MEMBER STIFFNESS AND
RELATED MATRICES, [Accessed: 10 March 2015].
[2] David Roylance, "Statics of Bending: Shear and Bending Moment Diagrams," November 15,
2000. [Online]. Available: http://www.saylor.org/site/wp-
content/uploads/2012/09/ME1023.1.1.pdf . [Accessed: 7 March 2015].
[3] Course Notes, "BEAMS: STRAIN, STRESS, DEFLECTIONS," 2000. [Online]. Available:
http://courses.washington.edu/me354a/chap3.pdf .[Accessed: 9 March 2015].
[4] Steel Construction, "Tied-arch bridges," November 15, 2000. [Online]. Available:
http://www.steelconstruction.info/Tied-arch_bridges .[Accessed: 9 March 2015].
[5] Kirsty Howie, "Humber Bay Pedestrian Bridge," October 22, 2013. [Online]. Available:
https://firststoryblog.wordpress.com/2013/10/22/humber-bay-pedestrian-bridge/ . [Accessed:
10 March 2015].
[6] The Physics Factbook, "Density of Concrete," October 22, 2013. [Online]. Available:
http://hypertextbook.com/facts/1999/KatrinaJones.shtml . [Accessed: 9 March 2015].
[7] John J. Fruin, "Designs for Pedestrians," 2010. [Online]. Available:
http://ntl.bts.gov/DOCS/11877/Chapter_8.html . [Accessed: 9 March 2015].
[7] Roymech, "Basic Notes on Factor of Safety," 2010. [Online]. Available:
http://roymech.co.uk/Useful_Tables/ARM/Safety_Factors.html . [Accessed: 9 March 2015].
[8] Victor E. Saouma, "STRUCTURAL ENGINEERING," 2013. [Online]. Available:
http://ceae.colorado.edu/~saouma/Lecture-Notes/se.pdf . [Accessed: 10 March 2015].
[9] Tata Steel, "Factors of safety," 2013. [Online]. Available:
http://www.tatasteelconstruction.com/en/reference/teaching-resources/architectural-
teaching-resource/elements/introduction-to-design-code/factors-of-safety . [Accessed: 11
March 2015].
[10] Solidworks help, "Maximum von Mises Stress Criterion," 2013. [Online]. Available:
http://help.solidworks.com/2012/English/SolidWorks/cworks/prot_vonm.htm . [Accessed: 12
March 2015].
29

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