report into brigde stability and design

Computer Methods in Structural Engineering 3

Daniel Rose

S1203681

16/03/2015

Problem 1

A frame structure was created in GSA software suit by mapping out nodes connected with

beam elements. The dimensions of the structure can be seen in figure 1 alongside an

isometric view of the frame modelled in GSA. The beams are colour coded in figure one to

highlight the different geometries of the beam sections. The first section of this report will

display the analysis results and comment on how they can be used in the design process.

Figure 1. Dimensions of the frame (left), pictorial view of GSA model (right)

1a.

Table 1 shows the global stiffness matrix for the whole structure, K, and its partitions Kff,

Krf, Kfr and Krr. By inspection of the frame there are 15 degrees of freedom and 12 reaction

forces.

Table 1. Matrices shown with associated size

Matrix size

K 27 x 27 Kff 15 x 15 Krf 12 x 15 Kfr 15 x 12 Krr 12 x 12

K is an nxn matrix where n equals the total possible reaction forces on the structure

if the structure was fully constrained.

Kff is an nxn matrix where n represents the total number of DOFs

Krf is the matrix used to find the reaction forces from the DOFs

Kfr is the matrix used to find the displacements from the reaction forces

Krr is an nxn matrix where n represents the total number of reaction forces

The local stiffness matrix for member four was calculated and can be seen in table 2.

Table 2. Local stiffness matrix for member 4

P4 V4 M4 P5 V5 M5 P44

0 0

0 0

V44 0 12

3

6

2

0

12

3

6

2

M44 0

6

2

4

0 6

2

2

P45

0 0

0 0

V45 0

12

3

6

2

0 12

3

6

2

M45 0 6

2

2

0

6

2

4

1b.

The points of 0 moment were found on the bending moment diagram by inspection. It was

simply where the moment graph crossed the beams local x axis. These points were

highlighted with black circles and shown in figure 2 for visualisation. A more accurate

location of the points of zero moment is shown in table 3. This was achieved using GSAs

moment output function. Halfway between the points where the bending moment changed

sign were approximated as the points of zero moment. The percentage distance along the

beam in the output was converted to meters.

Table 3. Points of zero moment across the frame

Figure 2. Full bending moment diagram for full factored load to highlight zero moment locations

Member X coordinates (m) Y coordinates (m) % of individual member 1 0 2.128 28.2

3 (i) 0.36 5 4.4

3 (ii) 6.376 5 79.7

4 8 1.955 60.9

6 (i) 1.99 5 33.2

6 (ii) 5.73 5 95.5

7 34.30 2.635 58.9

Represents points of zero moment

Figure 3. Full bending moment diagram for full factored load with maximum values shown in kNm

To obtain the locations of the maximum moments along the span, elements 3 and 6 were

analysed. This was achieved through GSAs output function. The desired output was set to

Beam and Spring Forces and Moments to show the results on beams 3 and 6 for the full

factored load case. The locations of diplacement were initially given as a percentage

distance along a member, these were converted to a displacement using the beam lengths

which were known. Table 4 sumarises the results. Because the programe could reference a

maximum of 100 points the error in the displacement value was 0.5%. This was considered

reasonable since a 4cm can easily be analysed for defects if required.

Table 4. Positive and negative max moments along beams 3 and 6.

1c.

Figure 4. Deflected shape diagram showing the maximum displacements for beams 3 and 6.

Element Max Moments Displacement from node

3 +ve 417.6 8.000m from 2

3 ve -315.4 3.328m from 2 6 +ve 387.5 0.000m from 5

6 -ve -120.9 3.864m from 5

Beam 3 Beam 6

Figure 4 illustrates the approximate locations for maximum deflections on an exaggerated deflected shape graph. GSAs output function for beam displacement was used to more accurately show the locations of maximum displacement. The results are shown in table 5.

Table 5. Max displacement of beam 3 an 6 with corresponding locations

Element Max displacement Location

3 0.002796 3.684m from node 5

6 0.011520 3.640m from node 2

The exaggerated deflected shape show in figure 5 has been annotated with black circles to

show the frames points of contraflexure. A point of contraflexure is a point in a beam where

no bending occurs. This will be when the rate of change of gradient transitions from positive

to negative. These points are at the same locations as the points of zero moment on the

bending moment diagram so the locations are also shown in table 3. To summarise, when

the moment diagram is positive the rate of change of the displacement diagram curve will be

negative. When the moment diagram is negative the rate of change of the displacement

curve will be positive.

Figure 5. Deflected shape diagram showing points of contraflexure

Beam 3 Beam 6

-9.714

3.100E-6

-14.46

15.50E-6

Values shown are the shear maxima for each member (N)

1d.

Figure 6. Full shear force for the full factored load case to show maximum member values.

Figure 6 shows that the sign of the shear force diagram is directly related to the sign of the

bending moment diagrams gradient. The shear force diagram for beams 1, 4 and 7 are

straight lines and do not change in sign or magnitude. Looking at the bending moment

diagram in figure 5 the bending moment gradient is constant. Beams 3 and 6 have uniformly

distributed loads so the bending moments gradient changes in sign and magnitude along

the length. We can see that the shear force diagram has a constant gradient but its

magnitude changes along the length with a sign change corresponding to a peak on the

moment diagram.

Figure 7. Full axial force diagram for the full factored load case

The axial force along the length of all beams is constant since when a static system is at

equilibrium the applied forces and reaction forces do not change. The applied forces are

transferred as a constant down to the supports.

Values shown are the axial force maxima for each member

(N)

1e.

To analyse the beam for the worst possible loading scenarios three cases additional to the

full factored load were created. These are summarised in figure 9. The dead load and full

factored load was given in the problem description, the active load was simply the full minus

the dead load.

Dead load DL= 25.8kN/m

Active load AL= 42.32kN/m

Figure 8. Beam and node location across the span with associated numbering

Full Factored Partial Left

Partial Right Dead Load

Figure 9. Force diagrams for the 4 load cases analysed

Each of the cases were analysed using GSA. The output function was utilised to extract

results from the analysis. Results for the maximum midspan moments, support moments

and support shear stresses were noted then tabulated in tables 6, 7 and 8. The numbers in

bold in these tables represent the maximum value for each study.

2 5 8 Beam 3 Beam 6

Table 6. Showing Max moment at midspan of 3 and 6 for full factored load

i. Load Case Beam 3 (kNm) Beam 6 (kNm)

Max. moment at midspan

Dead -114.1 -36.06

Full Factored -301.4 -95.21

Partial Left -335.1 -38.39

Partial Right -80.41 -169.7

Table 7 Showing max moments from the supporting nodes 2 and 5

Table 8 Showing max moments from the supporting nodes 2 and 5

ii. Load case Node 2 (kNm) Node 5 (kNm)

Max. moments at supports (Left)

Full Factored -69.21 29.60

Partial Left -70.98 51.60


Load case Node 5 (kNm) Node 8 (kNm)

Max. moments at supports (Right)


Partial Left -70.98 51.60


iii. Load case Node 2 (kN) Node 5 (kN)

Max shear at supports (Right)


Partial left -237.8 307.2

Partial Right -77.81 128.6

Load case Node 5 (kN) Node 8 (kN)

Max shear at supports (Left)

Full Factored -263.4 -145.3

Partial left -124.9 29.85

Partial Right -238.2 170.5

A bending moment envelope can be easily created in GSA for two or more loading cases.

The envelope gives a quick visualisation of the maximum and minimum bending moments at

every point along the beam. The envelope diagram shown in figure 10 gives this

representation and is annotated with maximum ve/+ve moments and moments at the end

points. The maximum values from this diagram are the maximum from all cases considered

in analysis so for the purposes of bridge design it is a very useful tool. That is if the bridge is

designed to cope with these maximum bending moments it should not fail.

Figure 10. Annotated bending moment diagram

1f.

A fifth scenario of loading during an earthquake was considered for the final part of the

analysis. This loading case is shown in figure 11. There were UDL loads across beams 3

and 6 and unlike the other cases there were Point loads acting along the x axis. These point

loads will cause greater horizontal reaction forces at the base and a different distribution and

magnitude of shear forces and bending moments across the frame. These will be discussed

in the following section.

Figure 11. Loading diagram to show earthquake load case, values in kN

-323.7

70.98

24.45 -5.4 -92.46

-343.5

-177.7

238.3

418.0

11.87

35.44 227.5

388.4

The bending diagram in figure 2 shows the different magnitude and distribution when

compared to figure 3 in section 1b. We can see that the maximum bending moments along

the span and at the end points have increased. The moments along the vertical components

have also increase dramatically. Particularly in beams 2, 5, and 8 where they were

approximately 0Nm before. This changes the strength requirements of the top beams since

for the vertical loads they did not need to be considered as load bearing elements. For the

bottom vertical beams 1, 4 and 7 the maximum bending moment is increased by as much as

1600%. This means a cross section with larger outer dimensions would be required for this

case than for previous loading if the bridge was being designed efficiently.

Figure 12. Full bending moment diagram for earthquake load case

The shear force diagram shown in figure 12 shows that the shear force in the vertical section

also increases, by as much as 1400%. The greater shear force means that there would be a

greater cross sectional area in the vertical beams since the shear force varies inversely with

area, equation 1.

, =

1.

Figure 13. Full shear force diagram for earthquake load case

581.6

70.98 200.0

315.1 385.8

200.0

15.43 200.0

-381.6

539.2

429.4

-468.9 -526.3

-100.0

-138.9

-162.3

-137.9

-100.0

406.1

-199.8

-100.0

164.9 10.1

Problem 2

To approach the problem a set of geometric data had to be calculated for the bridge

structures. The basic design of the bridge is shown in figure 14. The calculation procedure

can be found in the Appendix.

Figure 14. Diagram to show basic dimensions of bridge

There were three different curve forms to compute and for each, three different basic

parameters. That gives nine cases in total and these are highlighted below in table 9.

Table 9. Highlighting the different geometric curves

L = 50m, constant H1 H2 H3 Sin curve L/2 L/4 L/8

Parabola L/2 L/4 L/8

Circle arc L/2 L/4 L/8

The bridge was designed to be a pedestrian/cycle bridge going over a road or river - many

bridges similar to this are currently in operation, for example a bridge in Berkley [1]. The

Bridge deck length, L, is 50m and width is 3m in accordance with the Highways Agency [2].

The deck would be 0.3m thick of concrete. Concrete is widely used as a deck material due to

its wide availability, low cost and ease of manufacture.

The load used in designing the bridge was considered to be worst case. Firstly the dead load

was calculated using the volume multiplied by the density of standard density concrete,

2400kg/m2 [3].

The maximum load case was considered to be a crowd of people bunched together with

very little space for walking, a common scene after a football game in a large city. The

dimensions of the pedestrian were taken as the 95th percentile man to allow some room. The

weight was taken as a 50th percentile man as an average. From Anthropometric tables each

person would occupy 0.1484m2 and weigh 90kg [4]. The bridge deck area is known so the

live load can be calculated as the number of people that would fit on the deck multiplied by

the weight of each person.

There was assumed to be two arches supporting the deck using 17 loading points on each.

This meant the total load from the dead load and live load could be halved. This was

calculated as 53.72kN, calculations can be seen in the appendix.

The material for the cross section was chosen in line with BS EN 10025 which is a British

standard code for practice in the design of bridges and use of materials [5]. S275 steel

(275MPa yield strength [6]) sheet welded into a hollow rectangular section was chosen in

accordance with the standard. To match this design decision a user designed material was

created in GSA, see appendix for a screenshot.

The cross section dimensions were chosen by a method of trial and error. The case of the

L/2 was found by testing to give the largest displacement and stress for constant load, cross

section, materials. This was used as a test case to find an appropriate cross section, a

dimensioning reference is shown in figure 15. The cross section was initially taken as 0.5mW

x 0.5mD x 5mmt x 5mmT. The W and D values where take as an estimation from looking at

similar bridges and the 5mm thickness is the minimum required by the standards for

highways.

Figure 15. Dimensioning parameters for the beam cross section from GSA

The analysis was started with an aim of getting a safety factor of 1.5. This is recommended

where structures are not under extreme loading conditions and the materials are reliable.

The loading for a footbridge is predictable and I have chosen a worst case scenario for this.

The materials used are in line with British standards so by definition will be reliable if

responsibly sourced. The safety factor was calculated using equation 2.

=

2.

After several trials for the L/2 circular arc case a cross section of 0.8mW x 0.6mD x 12mmt x

12mmT was chosen. This gave a stress safety factor of 1.55 which was suitable for the

required 1.5.

The above cross section gave a deflection of L/312.5 where L is the length of the bridge.

The width and depth dimensioning lends itself to minimising deflection since when the

moment of area is calculated there is a cubed term which relates directly to the width or

depth, depending on the inertial axis.

I have kept the cross section constant for all three curve geometry cases (sin, parabola and

circle) since the analysis on each case for the three bridge heights is purely relative. That is

that every case for curve geometry does not have to be optimised in order to give an

accurate comparison. The final outcome being which curve geometry is the most efficient for

each of the cases of H.

Efficiency Criteria

For judging the efficiency of the bridge types a standard set of criteria were adopted. The

criteria aimed to analyse three different desirable features of the bridge; the structural

efficiency, strength, and its stiffness. These would then be combined together to give an

overall Efficiency rating. This section will take each of these in turn and explain how and why

they were analysed.

Strength

The main criteria for strength was how much load the arch could take without breaking. This

value was computed to compare the three geometric curve cases for each bridge height.

The cross section a material of the arch was held constant. A safety factor calculation

method was adopted using the yield strength of S275 steel, (275MPa) and the max stress

from the case being analysed. The Von Mises Stress is often taken as the maximum stress

that will occur in a cross section under load. This is true providing the material has not

become brittle. For this study it was assumed that if the Yield strength is larger than the Von

Mises stress the material would not fail [7]. The output tool in GSA was used to obtain values

for the max Von Mises stress for each case. Equation 3 could then be used to calculate a

safety factor.

=

3.

Stiffness

The beam stiffness is defined as a beams resistance to bending. The equation used to

calculate this is stiffness (k) = force/displacement. Since the force on the structures were all

the same a different method was adopted using the maximum allowable displacement of

160mm - as stated in section 2a. This analysis used the difference between allowable

displacement and the maximum displacement as a percentage of the allowable

displacement for each case under loading. This gave an easily workable factor which had a

significant influence when multiplied with the other coefficients to achieve the overall

efficiency rating. This was important since the deflection would be the second most

important factor after strength. Once these requirements are satisfied the design can then be

optimised for structural efficiency.

=

4.

The Structural Efficiency

When analysing the structural efficiency the main focus was on weight. After some brief

research and thought it was hypothesised that a structurally efficient bridge is one that can

withstand a high load with a low self-weight. This suggests the geometry of the bridge is

transferring the load through the material to the supports effectively and use of material can

be minimised through optimisation. The equation used to calculate a Structural Efficiency

factor was the weight of the load bearing arch divided by the weight of the live load plus the

deck load equation 5. This gave a high factor output for a bridge that was lower weight and

carried a greater load.

= +

5.

Results

It was considered that the results should be presented in table form to keep them concise

and easily comparable. Table 10 and 11 show a complete set of data found using GSA and

calculated efficiency data using the methods described in the previous section. An example

of the bending moment, shear force and axial force diagram is shown in figure 16. For

completeness the diagrams for all cases are shown in the subsequent section.

Table 10. Results for the bridge cases analysis

Design Parameters and Measurements Efficiency Ratings

H Curve Weight (kN)

Max Deflection

(mm)

Max Stress (MPa)

Safety factor

Stiffness Factor

Structural Factor

L/2

Sin 644.35 34.130 53.82 5.11 0.7867 18.957

Parabola 647.32 2.962 18.83 14.60 0.9815 19.155

Circle 664.14 167.600 186.50 1.47 -0.0475 20.276

L/4

Sin 581.54 18.790 -58.26 4.72 0.8826 14.769

Parabola 583.07 4.724 -23.01 11.95 0.9705 14.872

Circle 585.23 25.66 -59.25 4.64 0.8396 15.015

L/8

Sin 561.70 19.80 -73.51 3.74 0.8763 13.447

Parabola 562.22 13.57 -36.76 7.48 0.9152 13.481

Circle 562.40 17.32 -43.48 6.32 0.8918 13.493

Table 11. Summary of main results table showing combined efficiency ratings

H Curve Combined Efficiency Rating

L/2

Sin 11.39

Parabola 40.20

Circle -0.19

L/4

Sin 15.15

Parabola 41.90 Circle 13.94

L/8

Sin 13.10 Parabola 27.28

Circle 22.46

Figure 16. GSA diagrams for the L/4 Parabolic curve which gave the highest efficiency

Bending Moment Shear Force Axial Force

Max = -8.532kNm Max = -0.4668kN Max = -751.7kN

Additional diagrams for all Cases

Circle L/2

Bending moment on deflected shape

Shear force diagram Axial force diagram

Circle L/4



Circle L/8



Parabola L/2



Parabola L/4



Parabola L/8



Sin L/2



Sin L/4



Sin L/8



Result Analysis and Conclusions

Looking at the parabola deflected shapes we can see that they are much more uniform. The

deflection is only downward compared with the circle and sin arches who buckle and contort

under the load. The shear force diagrams reflect this and for the parabola are unidirectional

forces but are not for the circle and sin arches which tend to alternate in sign.

By looking at the combined efficiency from each case simple conclusions can be drawn from

the results.

For the L/2 case it was clear that parabolic curve was much more efficient than the circle and

the sinusoidal curve. This was due to it performing better in safety factor, stiffness factor and

structural factor than the other curves. Another important conclusion from the L/2 case was

that the circular arch performed very badly in the stiffness rating meaning that it deflected far

beyond what was acceptable.

For the L/ 4 case the parabolic curve had the highest efficiency rating again with it

performing well in all categories. This was the largest efficiency rating recorded at 41.90.

This time the circular arch had a much higher efficiency due to it not deflecting as much, it

performed similarly to the L/2 case in the safety factor and structural ratings however.

In the L/8 case the parabolic curve again had the highest efficiency rating of the three curve

types but was lower than its result for the L/2 and L/4 cases. The circle increased in

efficiency again.

The sign curve was consistent for all three cases with its safety factor decreasing but the

stiffness rating increasing as the height of the bridge decreased. This meant the overall

efficiency rating did not change by more than 5 units, compare that with a 20 unit change for

the circle and a 13 unit change for the parabolic curve.

The circle arch first appeared very poor but as the height of the arch decrease it

performed better. It is possible that for smaller bridge heights this trend would

continue and it may be the best case for these, this could be a topic of further

research.

The sinusoidal curve was the most consistent in its efficiency rating but was never

particularly high.

In conclusion the parabolic curve was found to be much more suitable for bridge arch

geometry design with it having performed the best for all three cases of height. It also

has a more predictable deflection shape and shear force distribution. That is that it

only bends one way and the shear force is unidirectional. This may make designing

of the cross section easier since one side would always be in compression and one

in tension.

References List

1. M. Yashinsky. Arch bridges. Pedestrian Overcrossing 13 February 2009. [Online]

Available from http://www.bphod.com/2009/02/i-80-pedestrian-overcrossing.html

[Accessed 10 March 2015]

2. British Government. Standards for highways. Volume 2. August 2004. [Online]

Available from http://www.standardsforhighways.co.uk/dmrb/vol2/section2/bd2904.pdf


3. G. Elert. Density of concrete. 2001 [Online]

Available from http://hypertextbook.com/facts/1999/KatrinaJones.shtml


4. R. Beardmore. Anthropometric notes. 10 January 2013 [Online]

Available from http://www.roymech.co.uk/Useful_Tables/Human/Human_sizes.html


5. BSI euro codes. Standard for steel, concrete and composite bridges October 2006 [Online]

Available from http://www.bsigroup.com/en-GB/about-bsi/media-centre/press-

releases/2006/10/Standard-for-steel-concrete-and-composite-bridges/#.VQXIT46sWuJ


6. The Steel Construction institute. Bridge design to the euro codes. February 2011

Available from http://discus.steel-sci.org/Content/documents/RT1156-simplified-bridge-

eurocodesv02.pdf


7. Learn Engineering. What is Von Mises Stress? 2011 [Online]

Available from http://www.learnengineering.org/2012/12/what-is-von-mises-stress.html


Appendix

Problem 1:

Output results for max bending moment

3 A2 5 -24.20 0.0 316.0 0.0 417.6 0.0

3 A2 41.6% -24.20 0.0 -2.331 0.0 -315.4 0.0 6 A2 5 -14.49 0.0 -263.3 0.0 387.9 0.0

6 64.4% -14.49 0.0 -0.2477 0.0 -120.9 0.0

Output results for max displacement

6 A2 61.4% 0.001875 0.0 -0.002075 0.002796 0.0 -0.001243 0.0 0.001243

3 A2 45.5% 0.001892 0.0 -0.01136 0.01152 0.0 0.001762 0.0 0.001762

Max deflection for 0.8 x 0.6 x 10mm

Circle L/2

Maxima

15 A1 90.1% 0.09185 0.0 0.02285 0.09465 0.0 75.95E-6

0.0 75.95E-6

1 A1 1 0.0 0.0 0.0 0.0 0.0 -0.01083

0.0 0.01083

15 A1 92.1% 0.09185 0.0 0.02286 0.09465 0.0 223.5E-6

0.0 223.5E-6

8 A1 9 0.0 0.0 -0.1343 0.1343 0.0 0.0

0.0 0.0

1 A1 1 0.0 0.0 0.0 0.0 0.0 -0.01083

0.0 0.01083

4 A1 5 -0.02881 0.0 -0.04910 0.05693 0.0 0.01108

0.0 0.01108

1 A1 1 0.0 0.0 0.0 0.0 0.0 -0.01083

0.0 0.01083

4 A1 5 -0.02881 0.0 -0.04910 0.05693 0.0 0.01108

0.0 0.01108

Minima

2 A1 9.9% -0.09185 0.0 0.02285 0.09465 0.0 -75.95E-6

0.0 75.95E-6

1 A1 1 0.0 0.0 0.0 0.0 0.0 -0.01083

0.0 0.01083

8 A1 9 0.0 0.0 -0.1343 0.1343 0.0 0.0

0.0 0.0

1 A1 1 0.0 0.0 0.0 0.0 0.0 -0.01083

0.0 0.01083

1 A1 1 0.0 0.0 0.0 0.0 0.0 -0.01083

0.0 0.01083

12 A1 13 0.02881 0.0 -0.04910 0.05693 0.0 -0.01108

0.0 0.01108

1 A1 1 0.0 0.0 0.0 0.0 0.0 -0.01083

0.0 0.01083

8 A1 9 0.0 0.0 -0.1343 0.1343 0.0 0.0

0.0 0.0

Max deflection for 0.8 x 0.6 x 10mm

Circle L/8

Maxima

5 A1 6 753.4E-6 0.0 -0.01062 0.01064 0.0 860.8E-6

0.0 860.8E-6

1 A1 1 0.0 0.0 0.0 0.0 0.0 184.4E-6

0.0 184.4E-6

1 A1 1 0.0 0.0 0.0 0.0 0.0 184.4E-6

0.0 184.4E-6

8 A1 9 0.0 0.0 -0.01540 0.01540 0.0 0.0

0.0 0.0

1 A1 1 0.0 0.0 0.0 0.0 0.0 184.4E-6

0.0 184.4E-6

4 A1 5 591.4E-6 0.0 -0.007693 0.007715 0.0 910.6E-6

0.0 910.6E-6

1 A1 1 0.0 0.0 0.0 0.0 0.0 184.4E-6

0.0 184.4E-6

4 A1 5 591.4E-6 0.0 -0.007693 0.007715 0.0 910.6E-6

0.0 910.6E-6

Minima

11 A1 12 -753.4E-6 0.0 -0.01062 0.01064 0.0 -860.8E-6

0.0 860.8E-6

1 A1 1 0.0 0.0 0.0 0.0 0.0 184.4E-6

0.0 184.4E-6

8 A1 9 0.0 0.0 -0.01540 0.01540 0.0 0.0

0.0 0.0

1 A1 1 0.0 0.0 0.0 0.0 0.0 184.4E-6

0.0 184.4E-6

1 A1 1 0.0 0.0 0.0 0.0 0.0 184.4E-6

0.0 184.4E-6

12 A1 13 -591.4E-6 0.0 -0.007693 0.007715 0.0 -910.6E-6

0.0 910.6E-6

1 A1 1 0.0 0.0 0.0 0.0 0.0 184.4E-6

0.0 184.4E-6

9 A1 9 0.0 0.0 -0.01540 0.01540 0.0 0.0

0.0 0.0

Problem 2:

Geometric calculations

The calculation of the bridge geometry was set up by finding the equation of the curve for

each case. These equations were manipulated to a form where y was being calculated as a

function of x, L and H where x is the horizontal coordinate along the beam, L is the total

length of the bridge and H is the height of the bridge. The x coordinates corresponded to the

locations of the 17 nodes which were distributed evenly along the length. Taking node 1 as

the origin coordinate (0, 0) the rest of the nodal coordinates could be calculated. From this

an excel spreadsheet was set up so that the values of H, L and x could be changed as an

input and y displacement was given as an output. Below details the geometric relationships

used for the 3 curve types.

Sinusoidal curve equation

= (2

) where = 2 and =

= (sin (

))

X coordinates of the nodes where plugged into this formula to get the y coordinates.

Parabolic equation

General equation: y = ax2 + bx + c

Initial conditions:

1. y = 0 x =L

2

2. y = 0 x = [0, L]

3. x =L

2 y = H

The constants were found by substituting the initial conditions into the equation. Three conditions and

three unknowns meant the equation was solvable.

a = 4H

L2 b =

4H

L c = 0

The equation used in the spreadsheet was therefore y = 4H

L2x2 +

4H

Lx

Circle equation

General equation:

2 + 2 = 2

R = d + h =h

2+

L2

8h

Specific equation:

y = h

2+

L2

8h x2

yrelative = y d

L

R

h

d

Design Calculations for Loading Case

Pedestrian/cycle Bridge

Length of bridge = 50m

Number of loading nodes for deck = 17

Width = 3m - 1.5m, 1.5m split by line

The bridge will be designed to BS 5400 BS 5400 is a British Standard code of practice for

the design and construction of steel, concrete and composite bridges. It is applicable to

highway, railway and pedestrian bridges.

Cross section Steel plate (S275) 5mm welded in accordance to BS EN 10025. Better

machinability and weld ability than S355 which also conforms to the standard.

Max Loading

Took average weight of man to be = 90kg

Dimensions of a person = 530mm, 280mm = 0.1484m2

Bridge surface area = 50 x 3 = 150m2

Weight of deck = Volume of deck x density of material (concrete)

3m x 50m x 0.2m x 2400kg/m2 = 72000kg = 72tonnes

Load of deck per node = 72t / 15 = 4800kg = 48kN

Number of people = 1010 (rounded down to nearest whole number)

Weight of people = 1010 x 90kg x 9.81ms-2 = 90900kg x 9.81ms-2 = 891.729kN

Per node P = 891.729kN / 15 = 59.449kN

Self-weight of circle L/2 arch

Arch length 78.21m

Cross section - 0.8D x 0.6W x 0.01T

Steel density 7850kg/m2

Weight = 16.946

Total load per node = 48kN + 59.449kN = 107.449kN

BUT two arches therefore total node load = 53.72kN + 16.946kN = 70.667kN

User defined steel material

Max Bending Moments

Max Moment

L/2

Sin -291.000

Parabola -2.532

Circle 132.100

L/4

Sin -288.800

Parabola -8.532

Circle 302.700

L/8

Sin -297.400

Parabola -31.660

Circle -90.720

report into brigde stability and design

Documents