Load analysis and maximum deflection permitted

Once the cross-section has been classified, the load 𝑊 must be identified and correctly factorized in the event of dead (𝐷) and live (𝐿) loads taking place at the same time.

There are two different approaches to the factorization of uniformly distributed loads according to Eurocode 0 [2]:

26

Figure 3. Recommended limiting values for vertical deflections from ENV 1993-1-1. [3, Table 7.1]

1. 𝑊 = 1.35 𝑊_𝐷+ 1.50 𝑊_𝐿 (7)

being this the most conservative approach. [4, eq. (6.10)]

2. The higher value from either 𝑊 = 1.35 𝑊_𝐷+ 1.05 𝑊_𝐿 or 𝑊 = 1.25 𝑊_𝐷+ 1.50 𝑊_𝐿, the second one being the most used. [4, eq. (6.10a,b)]

Upon obtaining the value of the load, the maximum deflection permitted due to the total load 𝛿_𝑚𝑎𝑥 is selected through the table shown in Figure 3 while the limit conditions [3, p. 229] have to be considered:

Table 6. Minimum geometrical characteristics of cross-sections.

Uniformly Distributed

27 2.4.3 Resistance verification

2.4.3.1 Shear resistance

Direct analysis of the forces acting on the beam will render the maximum design shear force 𝑉_𝐸𝑑. This design value at each cross-section must never be greater than the design shear resistance 𝑉_{𝑐,𝑅𝑑}, thus 𝑉_𝐸𝑑 ≤ 𝑉_{𝑐,𝑅𝑑} (7) [3, eq. (7.18)]. As to the calculation of the design shear resistance, the approach will vary depending on the method of analysis.

For plastic design, 𝑉_{𝑐,𝑅𝑑} is then regarded as the design plastic shear resistance 𝑉_{𝑝𝑙,𝑅𝑑} and takes the form of:

𝑉_{𝑝𝑙,𝑅𝑑} = 𝐴_𝑣· 𝑓_𝑦

√3 · 𝛾_𝑀0 (9)

where 𝐴_𝑣 is the shear area (see Appendix C for its adequate calculation), 𝑓_𝑦 is the yield strength and 𝛾_𝑀0 is the partial safety factor (see Table B3 in Appendix B). [3, eq. (7.19)]

In the case of elastic design, to verify the design shear resistance in equation 7 the fol-lowing criterion [3, eq. (7.20a,b)]:

𝜏_𝐸𝑑 =𝑉_𝐸𝑑 · 𝑄

𝐼 · 𝑡 ≤ 𝑓_𝑦

√3 · 𝛾_𝑀0 (10)

being 𝜏_𝐸𝑑 the tangential shear stress, 𝑄 the first moment of the area above the examined point on the section, 𝐼 the moment of inertia of the whole cross-section and 𝑡 the thickness at the examined point.

When the shear force is acting on a beam, in the event of the design shear force 𝑉_𝐸𝑑 be-ing less than half of the plastic shear resistance 𝑉_{𝑝𝑙,𝑅𝑑} (as in 𝑉_𝑒𝑑 < 0.5 𝑉_{𝑝𝑙,𝑅𝑑}) then its effect on the moment resistance is to be neglected unless if detected that the shear buck-ling reduces the resistance of the section (see Bending Resistance section). Otherwise, the reduced moment resistance should be based on reduced yield strength, 𝑓_{𝑦,𝑟𝑒𝑑},

28 2.4.3.2 Bending resistance

As with the shear resistance, the design value of the bending moment, 𝑀_𝐸𝑑, must satisfy the condition at each cross-section of 𝑀_𝐸𝑑 ≤ 𝑀_{𝑐,𝑅𝑑} (11) [3, eq. (7.16a)]. The moment capacity 𝑀_{𝑐,𝑅𝑑} can be determined, depending on each class, according to Table 5.

In the event of shear buckling reducing the resistance of the section, the reduced design plastic resistance moment 𝑀_{𝑦,𝑉,𝑅𝑑} can be obtained, alternatively [3, eq. (7.28)], as:

𝑀_{𝑦,𝑉,𝑅𝑑} = (𝑊_{𝑝𝑙,𝑦}− [2 · 𝑉_𝐸𝑑

When a beam is either properly restrained or has a certain type of cross-section, such as square or circular hollow sections, they are not very susceptible to lateral-torsional buckling by default [3, p. 190]. Happen the beam to fail these conditions, verification against the phenomenon is required. Thus, it must be guaranteed that 𝑀_𝐸𝑑 ≤ 𝑀_{𝑏,𝑅𝑑} (12)[3, eq. (7.29)] being 𝑀_𝐸𝑑 the design value of the moment and 𝑀_{𝑏,𝑅𝑑} the design buckling resistance moment, which can be defined as [3, eq. (7.30)]:

𝑀_{𝑏,𝑅𝑑} = 𝜒_𝐿𝑇· 𝑊_𝑦· 𝑓_𝑦

𝛾_𝑀1 (13)

where 𝜒_𝐿𝑇 is the reduction factor for lateral-torsional buckling, 𝑊_𝑦 the class ap-propriate section modulus and 𝛾_𝑀1 the safety coefficient (see Table C2 in Ap-pendix C).

As to the calculation of the reduction factor 𝜒_𝐿𝑇, two procedures can be applied: the general approach, or a more refined approach for doubly symmetrical I- / H-shaped pro-files.

1. General Approach

The reduction factor is given by the following expression [3, eq. (7.31)]:

29 buckling curve [3, p. 191] obtained as detailed in Table 7.

And the relative slenderness for lateral-torsional buckling 𝜆̅_𝐿𝑇 being defined as [3, (eq. 7.33)]:

𝜆̅_𝐿𝑇 = √𝑊_𝑦· 𝑓_𝑦

𝑀_𝑐𝑟 (16)

where 𝑀_𝑐𝑟 is the elastic critical moment for lateral-torsional buckling based on gross cross-sectional properties (see Equation (20)) [3, p. 191].

2. Method for I- / H- Shaped Profiles

The reduction factor is given by the following expression [3, eq. 7.34]:

𝜒_𝐿𝑇 = 1

𝜙_𝐿𝑇+ √𝜙_𝐿𝑇² − 0.75 · 𝜆̅_𝐿𝑇²

being 𝜒_𝐿𝑇 ≤ 1 and 𝜒_𝐿𝑇 ≤ (1 𝜆̅⁄ _𝐿𝑇)² (17)

The term 𝜙_𝐿𝑇 being defined as [3, eq. 7.35]:

𝜙_𝐿𝑇 = 0.5 · [1 + 𝛼_𝐿𝑇(𝜆̅_𝐿𝑇− 0.4) + 0.75 · 𝜆̅_𝐿𝑇² ] (18)

And the relative slenderness defined as Equation (16).

If considering the moment distribution between the lateral restraints of the member, a factor 𝑓 can be applied to the reduction factor as 𝜒_𝐿𝑇/𝑓 [3, p. 193], defining the term 𝑓 according to [2] as:

𝑓 = 1 − 0.5(1 − 𝑘_𝑐) [1 − 2(𝜆̅_𝐿𝑇− 0.8)²] being 𝑓 ≤ 1 (19)

30

Table 7. Recommended values for lateral torsional buckling curves. Adapted from [3, Tables 7.2-7.4].

Stability curve a b c d

𝛼_𝐿𝑇 0.21 0.34 0.49 0.76

GENERAL APPROACH:

Cross-section Limit Stability curve

Rolled I-sections ℎ/𝑏 ≤ 2 a

ℎ/𝑏 > 2 b

Welded I-sections ℎ/𝑏 ≤ 2 c

ℎ/𝑏 > 2 d

Other cross-sections − d

METHOD FOR I-/H-SHAPED PROFILES

Cross-section Limit Stability curve

Rolled I-sections ℎ/𝑏 ≤ 2 b

ℎ/𝑏 > 2 c

Welded I-sections ℎ/𝑏 ≤ 2 c

ℎ/𝑏 > 2 d

where 𝑘_𝑐 is a correction factor depending on the moment distribution as according to Figure 4 [3, Tab. 7.5].

Figure 4. Correction factors 𝒌_𝒄. [3, Table 7.5]

31

The elastic critical moment 𝑀_𝑐𝑟 can be calculated as [2, Annex F]:

𝑀_𝑐𝑟 = 𝐶₁ 𝜋²𝐸𝐼_𝑧 𝑘_𝑧 are the effective length factors that deal with warping end restraint and rotation, 𝑧_𝑔 is the distance between the load application point and the shear centre or neutral axis and 𝑧_𝑗 is an asymmetry parameter [3, Eq. (7.39)].

For further calculation details for either the elastic critical moment or its parameters, refer to [3, pp. 193-199].

2.4.4 Members in compression

For the case of members working predominantly in compression, as is the case of col-umns, it is necessary to assess its column buckling available strength (𝑁_{𝑏,𝑅𝑑}) to deter-mine the safety of the element.

The first step after the correct classification of the cross-section according to its axial loading is to determine the elastic critical buckling load (𝑁_𝑐𝑟) [3, Eq. (6.3)] about both its strong and weak axes as follows:

𝑁_𝑐𝑟 = 𝜋²𝐸𝐼

𝐿_𝑜² (21)

where 𝐿_𝑜 stands for the unbraced member length along the studied axis.

Once the elastic critical buckling stress has been determined, the relative slenderness 𝜆̅

[3, Eq. (6.26)] of each of the member’s axes can be evaluated as:

𝜆̅ = √𝐴 · 𝑓_𝑦

𝑁_𝑐𝑟 (22)

Note: For class 4 sections, 𝐴_𝑒𝑓𝑓 [3, pp. 116,117] is to be used instead of 𝐴.

32

Using Table 8 the stability curve of the section is obtained alongside an imperfection factor 𝛼:

Table 8. Stability curve and imperfection factor in compression. Adapted from [3, Table 6.3a] and [2, Table 6.1].

Stability curve a0 a b c d

Being the stability curve correctly classified, a reduction factor 𝜒 can be calculated as previously formulated in Equations 14 and 15. Thus, with that reduction factor an available column buckling strength 𝑁_{𝑏,𝑅𝑑} value can be obtained as follows:

𝑁_{𝑏,𝑅𝑑} = 𝜒 · 𝐴 𝑓_𝑦

𝑦_𝑀1 (23)

Note: For class 4 sections, 𝐴_𝑒𝑓𝑓 [3, pp. 116,117] is to be used instead of 𝐴.

Safety is then verified when both conditions 𝑁_𝑐𝑟 > 𝑁_𝐸𝑑 (24) and 𝑁_𝑏 > 𝑁_𝐸𝑑(25) are true for the member’s weakest axis.

It is important to note that the buckling effect (thus, the verification of 𝑁_𝑏) can be ne-glected if either 𝜆̅ ≤ 0.2 (26) or 𝑁_𝐸𝑑⁄𝑁_𝑐𝑟 ≤ 0.04 (27) happens to be true.

33

Figure 5. Beam design according to European standards flow chart. (Jordi Mata Garcia, 2021)

Figure 6. Members under compression to European standards flow chart. (Jordi Mata Garcia, 2021)

34

3 METHOD

The method to study the applications of the diverse analytic techniques reviewed in the previous chapter will be as follows:

1. A scenario will be proposed and the possible solutions suggested accordingly.

2. The traditional method of analyses will be conducted to determine the design values affecting the structure such as shear forces, moments, stresses, among many others.

3. Then, regulations established in the European standards on how structural design should be conducted within the European Union will be applied to each structur-al component to determine the ultimate safety of the structure.

4. Each structure will then be designed and analysed using computer software.

Main results will be compared and studied concerning the individual suitability of each method in the specified scenario.

5. An estimation of the most suitable solution will be conducted, and discussion will be made on how to improve each design and on what factors could contrib-ute to the selection of one solution over another.

3.1 Scenario proposal

The following scenario is considered:

• A structure is needed to allow pedestrians to cross over a gap, sized 30 meters long and 10 meters deep. Thus, the required structure is a bridge.

• The safety measures concerning pedestrian bridge loads [5] indicate that a pe-destrian bridge must hold a total live load of 4 𝑘𝑁/𝑚². The dead and resulting total load will be calculated and the profiles readjusted through simulation.

• There are no constraints concerning the number of columns required, their placement or the distance. They will be adapted according to the bridge design specifications.

• A common bridge deck will be used for all bridges, consisting of three platforms sized 10 𝑚 × 4 𝑚. Each platform will be composed of three evenly spaced

lon-35

gitudinal supporting beams, while the girders will be placed according to the support placement requirements.

• Beam cross-sections will be restrained to I-shaped beams for all the beams pre-sent in the structure, while its dimensions will depend entirely on the safety con-ditions established by the Eurocode 3 analyses.

• Structural reinforcement will be ignored by reason of pure beam analysis. Fas-teners/bolts and welded joints will be assumed as perfect with higher resistance to failing than the beam members.

3.2 Proposed solutions and initial analyses

Three different structures are suggested to tackle the given scenario each of them requir-ing a different approach to be analysed: simply supported beam bridge, simple cantile-ver bridge and truss bridge.

3.2.1 Bridge deck design

The bridge deck used by all the different bridge designs consist of a thin 10 𝑚 × 4 𝑚 metal plate to serve as a surface for the live load, three longitudinal beams of 10 𝑚 length that will carry the main weight of the thin plate and two or three 4 𝑚 long beams to act as girders on each end and placed below the beams depending on the situation.

This will create a one-way slab system [1, p. 58] that will distribute the load as shown in Figure 7 being so that the dark grey area will be the load supported by the central beam, represented as CD in the figure, and the rest of it distributed evenly to the side beams AB and EF. It is then observable that the central beam will be the critical component from the bridge deck struc-ture as it will have to support 8 𝑘𝑁/𝑚 while the side beams will support half the amount.

Figure 7. Weight distribution on a one-way slab system. Adapted from [1, fig.2-11(b)]

36 Thus, the central beam can be represented as:

The maximum design shear force 𝑉_𝑒𝑑 and moment 𝑀_𝑒𝑑 can then be calculated as 𝑉_𝑒𝑑 =

8𝑘𝑁 𝑚⁄ ·10 𝑚

2 = 40 𝑘𝑁 and 𝑀_𝑒𝑑 = ^{8𝑘𝑁 𝑚⁄ ·(10 𝑚)2}

8 = 100 𝑘𝑁 · 𝑚.

According to the aforementioned one-way slab system, it is manifest that the shear and moment forces acting on the side beams will be half of those acting on the central beam.

Like so, the girder beams could be represented as:

Which will produce a maximum shear of 𝑉_𝑒𝑑 = 40 𝑘𝑁 and moment of 𝑀_𝑒𝑑 = 40 𝑘𝑁 · 𝑚 on the beam. Despite this direct analysis not being accurate for every bridge model as it will depend on the column placement, it is the worst-case scenario when assuming evenly spaced supports.

Figure 10. Bridge deck 3D design. Dimensions in meters. (Jordi Mata Garcia, 2021)

Figure 8. Deck’s central beam ideal representation. (Jordi Mata Garcia, 2021)

Figure 9. Deck girder-beam ideal representation. (Jordi Mata Garcia, 2021)

37 3.2.2 Simply supported bridge

On an initial investigation and due to the deck spans being 10 𝑚 long, the most logical column placement is at the union of said spans thus resulting in two-column joints as depicted in Figure 11.

At each column joint, there are then different possibilities concerning the number of supporting members. Being the deck light material-wise due to it being a pedestrian bridge it is safe to assume that both the environmental forces and the live loads travers-ing it will create a considerable rocktravers-ing motion effect on the platform. In consequence, a single column system is not recommended as it will be exposed to sudden unexpected bending forces that may cause the beam to suddenly fail, hence a double-column system is the next best option cost-effective wise since it will provide enough stability to com-pensate the bending effort without compromising the stability of the structure. Its placement, nevertheless, will be on the edges of the light and dark grey areas depicted in Figure 7 (at 1 𝑚 from the side edge of the deck) for a more even distribution of the act-ing forces.

The compressive force acting on each of the columns will then be each of the reactions on the girder beam, which due to symmetry are both of 40 𝑘𝑁.

Figure 11. Simply supported bridge representation. (Jordi Mata Garcia, 2021)

38 3.2.3 Simple cantilever bridge

Despite most of the cantilever bridges being built with the help of trusses to adequately distribute and balance the loads, on a span of 30 𝑚 that is very ineffective since the space required by both the support structures and the suspended middle part are very large. Therefore, a simple cantilever structure will be used:

As can be seen from Figure 12, the structure is in fact a double cantilever, each of the side loads balancing each other.

This structure offers the possibility of a single column base at the cost of the column beams taking not only pure com-pressive force as in the simply supported bridge but also bending stress created by the cantilever beam situation pro-duced once the force is divided. Assuming again a double-column system, the 40 𝑘𝑁 received from the girder beam will be divided as shown in Figure 13, rendering that 𝐹_𝑐 = 35.78 𝑘𝑁 and 𝐹_𝑏 = 17.89 𝑘𝑁, the latter creating a bending moment of 200 𝑘𝑁 · 𝑚 on the beam.

Figure 12. Simple cantilever bridge representation. (Jordi Mata Garcia, 2021)

Figure 13. Cantilever column forces. (Jordi Mata Garcia, 2021)

39

Figure 15. Pratt truss bridge design with forces. Red is compression, blue is tension. (Jordi Mata Garcia, 2020)

3.2.4 Truss bridge

Truss bridges are likely one of the preferred options when it comes to short gaps due to their ease of assembly and their weight-carrying capabilities without the need for sup-ports.

Despite the many different truss bridge designs, the Pratt truss was the one chosen for the initial design due to the critical components in the bridge being purely in compres-sion and the possibility of below-deck clearance to keep things in the simplest of ways.

The bridge will then be divided into 6 sections, two of them being triangular sections at each end of the bridge and 5 𝑚 × 5 𝑚 square sections for the middle four sections with diagonal supports as described in Figure 14.

The reason for choosing squared sections is that having 45° angles on the diagonal beams help minimise the forces in both tension and compression members. Since modi-fying that angle decreases the forces in either the tension or compression members while increasing the opposite, the most optimal design to have the structure based on members under compression yet keeping it minimal is the chosen 45° angle.

Breaking down the forces inside the design following the traditional analysis results in:

Figure 14. Pratt truss bridge design. (Jordi Mata Garcia, 2020)

40

As observed in Figure 15, most of the longest members of the structure are kept under tensile load. This is beneficial as beams are prone to buckle under compressive forces, an effect that gets increased dramatically fast with the length of the member. Thus, upon initial inspection, the critical components to be studied will either be the end diagonal members under 169.7 𝑘𝑁 of compressive force or the top members enduring 216 𝑘𝑁 of compressive force.

It is important to note that this calculation is made assuming that the total load of 4 𝑘𝑁/𝑚² is distributed evenly between each side of the truss and then acting solely on the joints. With the help of COMSOL software, it is possible to evaluate the effects of an evenly distributed load of 8 𝑘𝑁/𝑚 on the half truss structure analysed in Figure 15, observing that all the forces detailed in the figure get reduced by about 15%. For the sake of assuming the worst-case scenario, the forces detailed in Figure 13 will be taken as the ones acting in the structure.

3.2.5 Failure analysis applied to the critical components

According to the traditional calculations, the maximum bending stress 𝜎_𝑚𝑎𝑥 [7, Eq. (6-12)] can be calculated according to:

𝜎_𝑚𝑎𝑥 =𝑀_𝐸𝑑 · 𝑐

𝐼 (28) where c is the distance from the neutral axis to the outermost side on the bending axis on the cross-section.

Similarly, the maximum shear stress 𝜏_𝑚𝑎𝑥 caused by the shear force 𝑉_𝐸𝑑 [7, Eq. (7-3)] is obtained as follows:

𝜏_𝑚𝑎𝑥 =𝑉 · 𝑄_𝑚𝑎𝑥

𝐼 · 𝑡 (29) where 𝑄_𝑚𝑎𝑥 is the maximum first moment of inertia of the cross-section.

Note: all the calculations will be made with the use of an Excel sheet where all the re-quired calculations have been automated depending on loads, boundary conditions, cross-section profiles and steel category. Refer to Appendix E for more information.

41 3.2.5.1 Bridge deck – Central Beam

According to the values obtained in section 3.2.1 and using Equations 28 and 29, there is a possibility of evaluating the impact that 𝑀_𝐸𝑑 and 𝑉_𝐸𝑑 may have on the I-beam cross-section according to the IPE standards in terms of bending and shear stress based on a traditional stress analysis as shown in Table 9. It is easily observable that the limiting stress is the bending stress, which is an expected outcome when having a lengthy simply supported beam.

Table 9. Profile - stresses based on the traditional analysis. (Jordi Mata Garcia, 2021)

Profile IPE220 IPE240 IPE270 IPE300 IPE330

𝝈_𝒎𝒂𝒙 [𝑴𝑷𝒂] 396.83 308.33 233.16 179.51 140.19

𝝉_𝒎𝒂𝒙 [𝑴𝑷𝒂] 42.90 37.18 31.15 26.18 22.53

There are many different steel categories depending mostly on its yielding stress, rang-ing from 235 𝑀𝑃𝑎 up to 450 𝑀𝑃𝑎. For the sake of simplicity, the lesser yieldrang-ing stress steel (S 235) will be assumed thus establishing the material’s yielding stress at 235 𝑀𝑃𝑎.

That assumption establishes, then, that IPE270 is the smallest profile that can be used since the maximum stress that it endures is barely below the yielding stress. If a certain safety margin is to be supposed, IPE330 is the next safest option since it is below 75%

that of the yielding stress of the material.

However, when the maximum permitted deflection 𝛿_𝑚𝑎𝑥 obtained from Figure 3 is compared to the limit condition deflection caused by a uniform distributed load on the beam 𝛿_𝐿𝑖𝑚 as in Table 6: The safety condition is failed; thus a different profile is needed. The smallest profile to

satisfy that condition is the IPE400, hence the analysis will proceed with it.

Having it classified as a Class 1 cross-section the plastic analysis can be carried out on its own without any reduction affecting its resistance.

42

Through Equation 9 and Table 5 it is obtained that the design plastic shear resistance is 𝑉_{𝑐,𝑅𝑑𝐼𝑃𝐸400} = 579.21 𝑘𝑁 and the moment capacity 𝑀_{𝑐,𝑅𝑑𝐼𝑃𝐸400} = 307.15 𝑘𝑁 · 𝑚. Both values being bigger than the design values, it is confirmed that the beam will retain its full plastic capacities with ease.

As per the elastic critical moment, knowing that it is a simply supported beam with free ends and with a uniform distributed load applied on the top flange of the beam, it is ob-tained through Equation 20 that 𝑀_{𝑐𝑟𝐼𝑃𝐸400} = 107.85 𝑘𝑁 · 𝑚. That value is greater than the design moment value 𝑀_𝐸𝑑 proving that the beam is indeed resistant to elastic defor-mation.

Finally, for the buckling resistance verification of the beam, there are two alternatives as specified in section “2.3.3.3 – Buckling resistance”: the method for I-shaped profiles or the general approach. If the method for I-shaped profiles is used, Equation 13 renders that 𝑀_{𝑏,𝑅𝑑𝐼𝑃𝐸400} = 100 𝑘𝑁 · 𝑚. Despite that being deemed as safe per the method, when using the general approach it is obtained that 𝑀_{𝑏,𝑅𝑑𝐼𝑃𝐸400} = 86.5 𝑘𝑁 · 𝑚, making it fail.

This does not explicitly mean that the beam is unsafe as it fulfils the analysis dedicated exclusively for its profile shape. Yet, to guarantee maximum safety in the most critical component, the IPE450 profile will be chosen as the definitive profile shape for all the beams present in the bridge deck component.

3.2.5.2 Simply supported bridge – Column

The main threat to a member that endures compressive forces, such as columns and pil-lars, is buckling failure. As per the traditional analysis, the elastic critical buckling stress 𝑁_𝑐𝑟 can be obtained through Equation 21 while applying a certain effective length factor to the length of the member depending on the boundary conditions.

To serve as an example, a cantilevered beam would have a factor of 𝐾 = 2. Using said values and knowing the column would sustain an axial load of 𝑁_𝐸𝑑 = 40 𝑘𝑁, an

To serve as an example, a cantilevered beam would have a factor of 𝐾 = 2. Using said values and knowing the column would sustain an axial load of 𝑁_𝐸𝑑 = 40 𝑘𝑁, an

In document Beam Structure Design Analysis (sivua 25-0)