The use of higher steel grades in building elements

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The use of higher steel grades in building

elements

Hampus Berggren

Civil Engineering, master's level 2021

Luleå University of Technology

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Abstract

Structural steel is one of the main building materials used commercial. Its strength is its formality and high yield strength which allows less material overall. Steel is often divided into steel grades such as S275, S355, S420 and so forth. The purpose of this thesis is to look at the strength and weaknesses of the different steel grades and how the area gets effected by higher steel grades. This is done by looking at a real life project where only S355 is used and optimizing for S355, S420 and S460. This is done to shred some light on how much of an improvement the upgrade of steel grade may allow.

For each steel grade an optimized cross section is created with the goal of having the lowest area possible. The difference in area for the optimized beams will act as the benchmarking for projected reductions. This optimization is done with the help of Python and follows Eurocode.

When looking at the case study, the projected weight reduction is 4, 2% for S420

and 7, 2% for S460 compared to S355. This reduction is higher on some beams

and lower on others. The main driving force for the higher reduced beams is the possibility to utilize more than one type of load condition, example both moment and shear force. Out of the 12 beams, beam 10 have the highest reduction between S355 and S420 which land around 10%. Between S355 and S460 Beam 12 comes in at a area reduction of 14%. As the higher steel grades have a higher equivalent CO2 emission the reduction of equivalent CO2 is lower than the area reduction. The

projected reduction of equivalent CO2 for a switch from S355 to S420 is 3, 3% and

the reduction between S355 and S460 are 5, 0%

Keywords: Area reduction , Change of steel grade , Emission reduction , High strength steel , Higher steel grade , Steel

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Acknowledgements

I would like to offer my appreciation to Structor and Johan Jansson who have given me a project to study and guidance. Sincerely thank you.

I would also like to thank Professor Ove Lagerqvist at Lule˚a Tekniska Universitet for being my supervisor and helping me in times of trouble.

A thanks goes out to family and friends as well who have been showing support, interest and acted as a sounding board throughout the thesis.

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4 Calculations 34 4.1 Imposing loads . . . 34 4.2 Beam 1 & 2 . . . 35 4.2.1 Beam 1 . . . 36 4.2.2 Beam 2 . . . 38 4.3 Beam 3 & 4 . . . 39 4.4 Beam 5 . . . 40 4.5 Beam 6 . . . 42 4.6 Beam 7 & 8 . . . 43 4.6.1 Beam 7 . . . 45 4.6.2 Beam 8 . . . 46 4.7 Beam 9 . . . 47 4.8 Beam 10 . . . 49 4.9 Beam 11 . . . 51 4.10 Beam 12 . . . 52 5 Analysis 54 5.0.1 Transverse force . . . 57

5.1 Lateral torsional buckling . . . 58

5.2 Savings for the project . . . 59

5.2.1 Difference in environmental impact . . . 61

5.3 Trendlines for dimensioning . . . 62

6 Discussion 66 7 Further work 67 A Optimized beams 71 A.1 Beam 1 . . . 71 A.2 Beam 2 . . . 74 A.3 Beam 3 4 . . . 76 A.4 Beam 5 . . . 79 A.5 Beam 6 . . . 82 A.6 Beam 7 . . . 86 A.7 Beam 8 . . . 89 A.8 Beam 9 . . . 92 A.9 Beam 10 . . . 94 A.10 Beam 11 . . . 97 A.11 Beam 12 . . . 100

A.12 Original beams . . . 102

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List of Figures

1 Steel grades taken from EN 1993-1-1 . . . 1

2 A illustration of the the building ”˚Akaren 1” . . . 4

3 The steel frame for ˚Akaren 1 . . . 5

4 The effect on different cooling speeds (Smirnov et al. 2012). The steel shown on the left have been cooled at 0.9◦C / Sec, and the steel shown on the right have been cooled at 18◦C / Sec. They are otherwise identical 7 5 The stress / strain curve . . . 9

6 Yield strength and Ultimate strength . . . 9

7 The linearity of steel before yield . . . 10

8 Elongation and yield strength of various steels (Shaw et al. 2002). HSS stands for high strength steel and AHSS stands for advanced high strength steel and have a multi phase micro structure. . . 11

9 Typical DBTT (Ductile-Brittle Transition Temperature) graph for steel from 2:Ductile to Brittle Transition. . . 12

10 Graville diagram (Weglowski 2012) . . . 14

11 S355, S420 and s460’s weldability according to Gravilles diagram . . . 14

12 Measurement for an I beam . . . 21

13 Centre of mass for each element . . . 22

14 Class division for width-to-thickness ratios of internal compressed parts (Eurocode 3: Design of steel structures : part 1-1 : General rules and rules for buildings 2005) . . . 23

15 Class division for width-to-thickness ratios of external compressed parts (Eurocode 3: Design of steel structures : part 1-1 : General rules and rules for buildings 2005) . . . 25

16 Table 6.1 from Eurocode 1993-1-1 (Eurocode 3: Design of steel struc-tures : part 1-1 : General rules and rules for buildings 2005) . . . 26

17 From table 6.2 Eurocode 1993-1-1 (Eurocode 3: Design of steel struc-tures : part 1-1 : General rules and rules for buildings 2005) . . . 26

18 Bucklingcases, K-factor (Fig. 2: Column effective length factors for Euler’s critical load. In practical design, it is recommended to increase the factors as shown above) . . . 27

21 Illustration of the HVAC system resting on beam 1, beam 2 and a third support . . . 35

22 Supports for the HVAC system . . . 35

23 Illustration of beam 1 . . . 36

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26 Illustration of supports and acting loads on beam 2 . . . 38

27 Illustration of beam 3 & 4 . . . 39

29 Top down view of the structure showing all the load bearing elements. The roof is assumed to be simply supported in the left-right direction in this picture with a uniformly distributed load. . . 40

30 The load configuration on beam 5 . . . 41

31 Reaction forces acting on beam 5s supports . . . 41

32 Moment distribution in beam 5 . . . 41

33 Shear forces in beam 5 . . . 41

34 Axial forces in beam 5 . . . 41

36 The load configuration on beam 6 . . . 43

37 Reaction forces present in supports at beam 6 . . . 43

38 Moment in beam 6 . . . 43

39 Shear in beam 6 . . . 43

40 Axial loads in beam 6 . . . 43

41 Illustration of the HVAC system resting on beam 7 and beam 8 . . . 43

43 Load configuration on beam 7 . . . 45

44 Reaction forces present in supports at beam 7 . . . 46

45 Moment in beam 7 . . . 46

46 Shear in beam 7 . . . 46

49 Supports and loads on beam 9 . . . 48

52 Supports and loads for beam 11 . . . 51

54 Areas for all optimized cross sections . . . 54

55 Area vs Moment plotted with trendlines . . . 55

56 The reduction of area in % based on the area of S355 . . . 55

57 Efficiency for the cross section measured in kN m/mm2 _{. . . .} ₅₆

58 Correlation between height and Moment . . . 63

59 Correlation between width and Moment . . . 63

60 Slenderness of the web for the twelve beams sorted by Moment . . . . 64

61 Slenderness of the flange for the twelve beams sorted by Moment . . . 64

62 Flange classes . . . 83

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1 Introduction

1.1 Background

Steel is one of the three main building materials at this time. Its high tensile and compressive strength gives it an unique place in the building industry. In recent times the environmental aspect is more in focus and unfortunately steel is not one to take a spot on the podium.

The classification of steel is divided into three parts, counting the steel parameter. The first one declaring what material it is, as it is structural steel that is looked at each grade will have a ’S’. Thereafter is the characteristic yield strength of the steel. Lastly there might be symbols that explains part of the testing or manufacturing process.

Figure 1: Steel grades taken from EN 1993-1-1

As seen in figure 1, the steel grade S460N/N L refers to a structural grade steel with a characteristic yield strength of 460M P a which are normalized or normalized rolled weldable fine grain structural steels. EN 10025-3 refers to the European stan-dard that dictates these steel classes, where EN 10025-2 is the stanstan-dard for non-alloy structural steel such as S275.

The steel grade S355 is often used in buildings in Sweden, but the use of higher grade steels are on the rise (May 2015). The use of a higher steel grade will have the benefit of a higher yield strength for each kilo of steel used. This increase will not only reduce the need to transport as much steel, but will also allow for different designs of elements.

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The Swedish St˚albyggnadsinstitutet (hereafter they will be referred to as SBI) is an independent organisation which strives for an competitive, durable and long-term profitability in the industry. In 2002 a report was written by SBI with the focus on what makes different steel qualities and their properties (Sehl˚a 2002). They stated that with coal as the only alloy the steel grade will only reach up to around S250. This allows the steel to be soft, but as the coal percentage reaches too high rates the weldability goes down as well. Other alloys are often added such as manganese. The manganese gives the steel an additional increase to yield strength. The alloys gives the steel different properties which might be beneficial in some cases and drawbacks in others. Thus the steel used is different situation may differ.

By looking at a project being built south of V¨aster˚as, Sweden, by Structor, a case study takes place. The goal of the study is to find out how a change in steel grade will effect the design of the structural elements by optimising them in three steel grades. The building is part of a service station that caters to people on the road and the main goal of the building is to be a diner. The outside is made to look like shipping containers stacked on top of each other. Large windows and a variety of coloured containers are adding to a modern look. The buildings load-bearing elements is made out of steel trusses and frameworks. A total of 14 unique cross sections are used and the steel grades used is S355 with qualities of S355J2 and S355J2H.

To increase the variate in the optimizing of beams, the buildings geographical location is moved from snow zone qk,snowzone = 2kN/m2 to Sweden’s highest snow

zone, qk,snowzone = 5, 5kN/m2. What this results in is the characteristic snow load

increasing from qk,snow = 1, 6kN/m2 to qk,snow = 4, 4kN/m2 when factors of roof

slope, heat flow trough the roof, and exposure are all factored in. This will allow the beams to have a wider range of possible dimensions. This will help with determining impactful dimensions as well as singling out week areas.

1.2 Contributing parties

This thesis is written with help from Structor. They have kindly offered their help by allowing insight into the project and a mentor, Johan Jansson, too help with the calculations and navigate through the project.

From Lule˚a University of Technology, Ove Lagerqvist acts as both a mentor and examiner. He have help out with the structure of the report and issues surrounding the thesis.

1.3 Purpose

This study is conducted to find how an exchange of steel quality will affect the final design. Thus conclude what element would benefit from using a higher steel grade and which elements that are better off with a lower steel grade.

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take place by looking at how the steel grade effects the final design of the beams in the building. This will in return reveal more insight when higher steel grades should be used.

1.4 Aim and Objective

The goal for this study is to find when is it suitable for an increase in steel grade between S355, S420 and S460. This will be based on different factors such as what loads are applied, what the highest degree of usage is and where the loads are coming from.

The key components of analysing this is: How large of an reduction can be estab-lished by changing steel grade? How the height and width change with the change of steel grade? And How will the emissions change with a change in steel grade?

1.5 Method

For the literature study previous works around higher steel grades are looked at which gives an understanding about the difference in steel grades and their strength and weaknesses outside of Eurocodes dimensioning.

The load bearing beams from the case study are looked at and optimized. Con-clusions are drawn from the optimization of area reduction for each beam and pooled together and analyzed for results.

The optimization of beams is done by writing a program that follows the rules of Eurocode and brute forces the best dimensions. Some parts are simplified for reducing the calculation time and reducing the pre processing.

1.6 Limits

Due to the time constraints, calculations for the whole design will not be done. The two calculation steps that will be disregarded are, calculations for fatigue of the web and flanges, and the entirety of welds. The focus is solely on the steels load capacity. This also includes the deformation as the deformation is in serviceability limit state. As the project is quite large, an limitation on the number of members looked at is introduced. Out of the 84 elements, 12 beams are selected and looked at. These 12 beams are the major load bearing beams.

The Eurocode chapter EN 1993-1-12 is an extension of EN 1993 which allows the use of steel grades up to S700. But even as steel grades up too S700 could be calculated using EN 1993-1-12, this thesis will be limited to the non extended EN 1993 thus only up to the steel grade S460 are used.

The cross sections will be designs that could be premade but not limited to stan-dardised beams and columns. This limit is due to the fact that in site solutions is

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both cost and time ineffective, which could result in conclusions that are impractical (Isaksson, M˚artensson, and Thelandersson 2010).

1.7 Guiding documents

Eurocodes is the European unions standards for structural design, but is widely used by countries outside EU as well. It contains a set of rules and approaches for design. Sweden have their own national annex. A Swedish version of the eurocode which includes small changes that is tailored for design in Sweden with respect too Sweden’s prerequisites such as the weather and wind loads.

1.8 The project

In 2015, the decision that a new truck stop was needed in Väster˚as was taken (“Truck-stop byggs i Väster˚as” 2015). Three main goals are the cause, reducing the number of lorries parked on the cities streets, improve the working environment for drivers, and improve the perception of Väster˚as as a logistics city. A total of 28 000m2 _of

mark is dedicated for the project and sold to an external partner. The truck stop will house services as parking, hygienic and food.

Figure 2: A illustration of the the building ”˚Akaren 1” ˚

Akaren 1 is the scope of this thesis. The buildings load-bearing elements is made out of steel trusses and frameworks. There are in total 84 elements with 14 unique cross sections being used, and the steel grades is S355 with qualities of both S355J2 and S355J2H. The frame includes three steel trusses which are premade and from a external source.

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Figure 3: The steel frame for ˚Akaren 1

The steel frame have a total weight of 8951 kg with details included but trusses discarded.

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2 Literature study

2.1 What is steel

Steel is an iron-carbon alloy where the iron content greatly out ways the carbon one (Green, Batchelor, and Yuen). It involves two steps in the process. The first one produces molten iron, it includes different steps to ensure a clean iron product. Thereafter the creation of steel begins. Even as steel is classified as an iron-carbon alloy, more elements are often present in the alloy which create different characteristics from the pure iron-carbon alloy.

This would be refereed to as an iron ore-based steel production as the components come from each element (Steel production). The iron ore-based steel production account to around 70% of the total steel production.

Scrap-based steel production makes up the remaining 30% of all steel production (Steel production). It have been present for more than 150 years (National Minerals Information Center ). Waste from the production of ore-based production is reused for the scrap-based production (Green, Batchelor, and Yuen). But other sources are also being used such as recycled steel from scrapyards. In the year 2050 it is estimated that the scrap-based production will be equal to the ore-based production (Steel production). It is also good to note that steel is one of the most recycled materials in the world.

2.2 Difference between steel

Each steel grade have different additives which is resulting in different characteristics. The difference in steel grades are due to the manufacturing process and the material used.

In table 1, The steel grades looked at are normalized or normalized-rolled and are assumed to have a thickness of 20 mm. All the values are taken from Structural steels — Part 3: Technical delivery conditions for fine-grain structural steels 2012 and indicate the maximum values in % for all except aluminium which are the minimum values. There is other ways to divide the steel in categories (Krauss 2015). There is non alloy steel which is without any coal content. Low carbon steels which have below 0, 25% coal, all three steels from table 1 are under this category. There is also medium and high carbon steels which are high strength and high fatigue resistant materials. When the carbon content extends 2%, it is classified as cast iron. Cast iron is often portrayed as a brittle material, but in recent times additives are allowing the brittle nature of cast iron to subside.

Even as the amount of alloys added are quite small as seen in table 1, the strength proprieties are greatly increased from non alloyed steel (Kuoppa 2012). Too achieving even higher strength hardening phases are needed. By allowing the steel to rest at certain temperature intervals, microstructures are created (Ravi et al. 2020). One

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Table 1: Chemical compounds in construction steel S355 S420 S460 Carbon 0,20 0,20 0,20 Silicon 0,50 0,60 0,60 Manganese 0,60-1,65 1,00-1,70 1,00-1,70 Phosphorus 0,030 0,030 0,030 Sulfur 0,025 0,025 0,025 Niobium 0,05 0,05 0,05 Vanadium 0,12 0,20 0,20 Aluminum 0,02 0,02 0,02 Nickel 0,50 0,80 0,80 Molybdenum 0,10 0,10 0,10 Copper 0,55 0,55 0,55 Nitrogen 0,015 0,025 0,025

disadvantage with high strength steel is that the higher the strength of the steel, the tougher it is to work with (Krauss 2015).

The microstructure that are the most studied are bainite (Ravi et al. 2020). It can be created at temperature as low as 250 ◦C and are a function of time (Ravi, Sietsma, and Santofimia 2020). The bainites allow the strength of the steel to reach above 700 MPa, and even 800 MPa in some cases (Kuoppa 2012). What creates a difference in strength are both the cooling process but also what the steel contains. As the toughness of the bainites are correlated with how much carbon are present in the steel. The more carbon that are in the steel, the tougher it allows the bainites to become. But thanks too new metallurgy practices low carbon steels are also on the rise in yield strength (Krauss 2015).

Figure 4: The effect on different cooling speeds (Smirnov et al. 2012). The steel shown on the left have been cooled at 0.9 ◦C / Sec, and the steel shown on the right have been cooled at 18◦C / Sec. They are otherwise identical

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The toughened steel sheets in the range of 700 – 1300 MPa hardened with rapid cooling have increasing slenderness correlating with the increase in yield strength (Weglowski 2012). This is due too a limitation in the production where thicker slabs of steel are not available too be cooled as quickly as thinner sheets, which is critical for the tempering process.

Other hardening processes are also available (Kuoppa 2012). This allows very precise steel to be created for a precise occasion as there is pros and cons with each of them. The research is rapidly going forward and within years, problems that are a hinder now will be solved (Weglowski 2012).

2.2.1 Hooke’s law and its effect on steel cross-sections

As the strength of the material increases, the material needed decreases. This de-crease of material will effect the stiffness of the elements as all current steels have roughly the same module of elasticity (Mahendran 1996).

In chapter 3.2.6 ”Design values of material coefficients” from Eurocode 1993-1-1, the value used for calculations are 210 GP a (Eurocode 3: Design of steel structures : part 1-1 : General rules and rules for buildings 2005).

Hooke’s law is the model used to determine the elongation of an element by looking at the forces that are applied (“Hooke’s law” 2020).

F = kx (1)

F is the force, k is a non dimensional factor and x is the elongation. For equation 1 to hold true the following must be true:

k = F

x (2)

Thus:

k = F orce

Length (3)

As k is just a factor this holds true for linear elastic systems. This is often demon-strated by a spring. By looking at the Stress-Strain relation Hooke’s law can also be applied too steel elements.

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Figure 5: The stress / strain curve

Figure 5 shows the relation between stress and strain in steel. This is a typical curve for hot rolled steel grades (Isaksson, M˚artensson, and Thelandersson 2010). As seen in the figure a top and a plateau is present.

Figure 6: Yield strength and Ultimate strength

Until fy in figure 6, the curve is estimated to be linear. Thus the material is seen

as linear-elastic up until that point. Because it’s seen as linear-elastic Hooke’s law can be applied but needs to be modified.

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Figure 7: The linearity of steel before yield From figure 7 the following expression can be extracted:

∆σ

∆ = E (4)

As it’s seen as linear and the line crosses the origin the following is true:

∆ = − 0 (5)

0 = 0 (6)

With equation 6 in equation 5

∆ = (7)

Also

∆σ = σ − σ0 (8)

σ0 = 0 (9)

With equation 9 in equation 8

∆σ = σ (10)

With equation 7 and 10 in 4 the following will subside: σ

= E (11)

or:

σ

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As the Young’s modulus is roughly the same for all steel grades, the elongation or compression for a specific steel grade on a member will be the same for a different steel grade on the same member. With the premise that the stress will be lower than the yield strength and that the same load is applied. That means if the elongation or compression is whats the criteria of dimensioning, a steel grade change will not solve the problem on it’s own.

2.2.2 Brittle failure

Brittle failure can be described as ”a sudden fracture under a load which is not sufficient to result in general yielding across the whole of the fractured section” (Boyd 2016). Its failure will lead to a total failure of the element which is affected too the brittle failure (Eriksson 2006).

One hardening method which is often used are cold forming (Isaksson, M˚artensson, and Thelandersson 2010). It involves plasticization of the steel which is done to a degree which often removes the yields strength but hardens the material. This is beneficial if the steel in the construction is not allowed too yield or strain too much. For steels without a yield strength fy can not be used and a different measurement

is used, f0,2. It represent the strength for the steel when the strain is 0,2%.

Figure 8: Elongation and yield strength of various steels (Shaw et al. 2002). HSS stands for high strength steel and AHSS stands for advanced high strength steel and have a multi phase micro structure.

Aging of the steel will occur and is also something that needs to be taken into account (Isaksson, M˚artensson, and Thelandersson 2010). With time the steels yield strength increases, and at the same time the extensibility decreases. What this does is make the steel more brittle. This is especially true for cold hardened steel elements.

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For testing of steel specimens the normal rate of stress-increase is below 100M P a (Isaksson, M˚artensson, and Thelandersson 2010). This represent a elongation rate (δ_δt) of 0, 00051_s. At higher rates of δ_δt a critical intersection point will occur between fy and fu. While the elongation rate is above this intersection a failure will be brittle.

This may happen in the case of sudden impact or explosion.

Figure 9: Typical DBTT (Ductile-Brittle Transition Temperature) graph for steel from 2:Ductile to Brittle Transition.

But brittle failure is not limited too failure due to brittle material properties (Boyd 2016). Figure 9 represent the transition of regular structural steel between brittle and ductile. Steel is one of the materials where temperature have an effect on if the failure will be brittle or ductile (Eriksson 2006). Too the left, on the cooler side, there is a higher risk of a brittle failure occurring. On the right, on the warmer side, there is a higher chance of the failure being ductile. What dictates if the failure will be brittle is associated with the chance that a critical particle to initiate a fissure dominated failure. The transition is often in a narrow and rapid span and in the transition temperature either a brittle failure or a ductile failure may occur.

A brittle steel have a few downsides outside the fact that the failure is very sudden (Isaksson, M˚artensson, and Thelandersson 2010). One of them is that local stresses can not be evened out over the cross section. This can make local stress peaks devastating. What might happen is that the stress peaks will extend the strength of the steel and create a crack. With the inability too dispense the stress the crack will extend and a failure without much of a warning signal will occur.

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2.2.3 Welding

Often are higher steel grades associated with poor weldability (Kuoppa 2012). One of the main factors for determining the weldalility is the calculating of carbon equivalent (Odebiyi et al. 2019). But for moderns steel grades the amount of carbon used in higher steel grades are reduced (Kuoppa 2012). This decrease in carbon allow the steel to be more easily welded. But as stated in the chapter Difference between steel, too reach the hardest bainites a higher concentration of carbon are needed. As the concentration is rising, some methods might be needed to improve the weldability and too not form decreased quality in the form of cold cracks and other problems (Odebiyi et al. 2019).

The equation for calculating the carbon equivalent is as following (Lundin et al. 1989): CE = C + M n + Si 6 + Cr + M o + V 5 + N i + Cu 15 (13)

The carbon equivalent also points at if any pre, mid or post processes is required (Odebiyi et al. 2019).

Table 2: Breakpoints for processing (Odebiyi et al. 2019)

Carbon equivalent Recommendation

CE < 0,45 No pre heating required 0,45 < CE < 0,52 Pre heating is required

0,52 < CE Can’t be efficiently welded

A Graville diagram may also be used for for determining the weldability (We-glowski 2012). It uses both the carbon equivalent and the carbon content for the reading.

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Figure 10: Graville diagram (Weglowski 2012)

As seen in figure 10. Even as the carbon equivalent is high, the steel is easily welded if the carbon content is low. By adding the steel grades maximum values from table 1 into equation 13 and plotting on figure 10. Aluminium and manganese are exceptions which have their minimum taken.

Figure 11: S355, S420 and s460’s weldability according to Gravilles diagram As shown in figure 11 S355, S420 and S460 are all in Zone II. This classifies them as weldable steels according to Gravilles diagram. But as seen, S420 and S460 are on the edge between Zone II and Zone III. If the manganese are increased above its minimum value, S420 and S460 will shift into Zone III. This is for the maximum values, the real values used in commercial steels may sway away from the values used for this diagram.

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If these are needed preheating is recommended to at least 10 ◦C above the tem-perature where creating Marentiste starts (Odebiyi et al. 2019). As the temtem-perature required is high in some steel alloys, the producer sometimes accommodates the steel with a preheating temperature. This will prevent cold cracks and lack of fusion, which may occur if is not done correctly. Controlled cooldown is also used to slow down the cooling process which in turn allows the hydrogen to defuse from the weld pool before the welds fully settles.

What affects the quality of welds is not only the welded plates, but also the filler material (Weglowski 2012). To successively identify the filler that is best accustomed for the weld, these following should be considered:

• The chemical composition for the steel being welded together – This includes how the micro structure in the steel are formed and the mechanical properties such as the ultimate strength and the toughness.

• Design requirements for the welds – Regarding minimum ultimate strength and toughness of the welds.

• Chemical properties for the filler material – Including restrictions on the carbon content and temperatures when the welding takes place.

• Mechanical properties for the filler material – The ultimate strength and tough-ness of the filler material. This is one of the most crucial aspects to take into account.

If the right precautions are applied beforehand, meanwhile and after the welding process, higher steel grades can be advantageous in medium load cycles when looking at fatigue (Pedersen et al. 2010).

2.3 Environmental impact of steel

The steel industry is a large part of the CO2 emissions both when comparing to

industrial and to all emissions (Pasquale 2016). Worldwide, a total of 30% of all CO2

emissions are from the steel industry. Looking at all emissions this figure is down to 6.7%. This number is a little bit bleaker if counting only Europe. If so, 21% of the industrial CO2 emission is from the steel industry.

A large part of the pure CO2 emission come from the blast furnace (Pasquale

2016). It is roughly calculated that the blast furnace produces 71% of all the plants emissions released while contributing too 88% of the CO2 emission. It is a little

kinder in terms of electrical usage with with 82% of the whole plants electrical usage. By doing a Life Cycle Assessment (LCA) of the steel traveling trough the steel mill, a rough idea of what emissions exists on its journey will be reveal (Renzulli et al. 2016). There are 4 facility’s inside the steel mill which are used before the steel slab are set to cool.

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Table 3: The air emissions that are produced while making 1 million tones steel (Renzulli et al. 2016)

Air emissions C.O S.P B.F B.O.F Total

CH4 1 t 1 t Benzene 32 t 32 t CO2 310 133 t 245 724 t 551 845 t 50 873 t 1 158 575 t CO 20 t 344 t 32 t 80 t 476 t Dust 121 t 371 t 313 t 263 t 1 068 t NO2 634 t 1 040 t 599 t 74 t 2 347 t SO2 818 t 1 617 t 1,32 t 65 t 2 501 t HCN 1 t 1 t Diffused emissions 86 t 62 t 205 t 62 t 415 t PAH 3 t 1 t 4 t PCDD & PCDF 0,028 g 5,65 g 0,0057 g 6 g Benzo(a)pyrene 37 kg 37 kg Pb 7 t 1 t 8 t Zn 2 t 2 t HCI 1 t 1 t HF 54 t 1 t 55 t VOC 1 t 1 t PCB 7,1 kg 7 kg

Table 4: The water emissions that are produced while making 1 million tones steel (Renzulli et al. 2016)

Water emissions C.O S.P B.F B.O.F Total

COD 159 t 17 t 176 t N 52 t 52 t NH3 54 t 54 t Phenols 2 t 2 t N-kjeldhal 30 t 30 t Suspended solids 4 t 20 t 24 t

The emissions varies greatly dependent on the quality of products (Pasquale 2016). For each percent of purer iron-ore, theoretically the fuel ratio will decrease 1.5%.

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Table 5: The solid emissions that are produced while making 1 million tones steel (Renzulli et al. 2016)

Solid emissions C.O S.P B.F B.O.F Total

Waste 137 t 3 070 t 445 t 3 652 t

Sludge and Dust 34 925 t 34 925 t

Oil 2 t 2 t

Slag 146 672 t 146 672 t

Water usage is not taken into account in table 3, 4 or 5. A total 180-200 m3 of water is estimated for each ton of finished steel (Pasquale 2016). A large portion, nearly half of it, is used for cooling. Half of what is used for cooling is used for cleaning the waste gas and whats left after that is split between metal processing and hydraulic transport with a little cut in other services.

By converting the emission into carbon equivalent, an estimate of how much difference it is between steel grades can be looked at (Sperle 2012). Based on data from SSAB the following equation is established:

kgCO2,equivalent = 4, 424 ∗ 10−4∗ fy + 2, 1826 (14)

The equation is based on cradle to gate so it does not include the CO2 emitted during

shipping and installation (Sperle 2012). The emission produced from transport and shipping differs a lot dependent on which mode of transportation is used (McKinnon and Piecyk 2010). The highest is for Airfreight and the lowest is for transportation in pipelines and deep-sea tankers, with a difference above ten times in between.

Table 6: The equivalent CO2 produced for freight (McKinnon and Piecyk 2010)

Mode of transportation CO2equivalent[g/tonne ∗ km]

Road transport 62 Rail transport 22 Barge transport 31 Short sea 16 Intermodal road/rail 26 Intermodal road/barge 34

Intermodal road/short sea 21

Pipelines 5

Deep-sea container 8

Deep-sea tanker 5

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2.4 Steel grades in Eurocode

Table 3.1 in Eurocode 3: Design of steel structures : part 1-1 : General rules and rules for buildings 2005 allows the use of steel grades from S235 to S460. Two years after the release of en 1993-1-1, Eurocode 3 - Design of steel structures - Part 1-12: Additional rules for the extension of EN 1993 up to steel grades S 700 2007 was released. This new release expands the possibility to use steel grades up too S700, it includes rules and exceptions too build on the original steel eurcodes. It directs which paragraphs still apply and which are not applicable for steel grades greater than S460.

2.5 Conclusions

Welding in higher steels grades may cause problems. To gain increased strength of the steel, additives are added. This will in turn inflate the carbon equivalent of the steel and thus making the product less weldable. And the less weldable product may require at least one of pre or post treatment. Some of which requires heating of the whole elements which are impractical for larger elements. But as the processes develop and evolve, problems such as the one mentioned may be eliminated by either other welding techniques or improved pre and post welding.

As a larger percentage of steel are projected to be scrap-based in the future, what will result of that is a larger percent of the steel produced will have residue from pre-viously made steel within. Today’s steel are a product of the chemical compounds, and processes which are limited too certain geometrical frames and mechanical prop-erties. As mechanical properties are hard to alter, it is the chemical compounds which may be. With residue from previous steels, the creation of highest quality and highest steel grade steels that rely on a perfect combination of chemical compound may be bottlenecked to the ore-based steel production.

The pure CO2 emission from the steel accounts for less than half of the total

equivalent CO2 emission.

As Eurocode allows the use of steels grades up too S700 and the expansion in methods used for handeling weak areas areas, higher steel strength are prone to have a spot in one way or another in the future of structural engineering.

2.6 Previous works

A study called ”ENVIRONMENTAL ADVANTAGES OF USING ADVANCED HIGH STRENGTH STEEL IN STEEL CONSTRUCTIONS” published in September of 2012 by Jan-Olof Sperle explores the use of high strength steel in two case studies. One of the two case studies are on Swedbank Arena with the aim too recalculate the building in S355 too look at how much of a environmental saving have been achieved. The roof is shown in the study and the existing roof consisting of S355, S460, S690 and S900. 32% of the building consists of steel that have a higher grade than S355.

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What they conclude from their study is that a weight reduction of 20% have been achieved for the upgraded elements. Or a 13% reduction by looking at all elements and steel classes. The reduction amounts to a saving of above 1 million kg CO2equ

for the steel production over the buildings entire life cycle. However are the recycling taking a hit and the higher steel grade building are less recyclable. This results in the total saving being below the 1 million kg CO2equ over the buildings entire life cycle.

Ricky Andersson wrote a master thesis called ”Concept study for cost and weight reduction of a barge container sized module” where he studied how a increase in steel grade will reduce weight and cost. His findings were that with cleaver design and increase from S250 to S700 the sandwich panels weight is reduced with 58%. For the beam system this figure is 47%. This is a high weight reduction but also a high yield strength jump.

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3 Design process

3.1 Design criterias

The design criteras looked at are that for each category the design values have to be larger than the applied load. This is not in line with the whole of Eurecode as interactions between different load types such as shear and moment is also checked. What this results in will be that designed beams that is going to have greater than one fully utilized load is going to be slightly under dimensioned. This will keep all the designs more in line with each other. Thus if:

MEd≤ MRd

NEd≤ NRd

VEd≤ VRd

FEd≤ FRd

FEd,supp ≤ FRd,supp

The design for the beam will be deemed sufficient.

3.2 Moment of inertia

The moment of inertia may be called the second moment of area. It is a measurement of the resistance for the cross section when it is on bending. The complete formula for the moment of inertia is:

Ixx =

Z

y2dA (15)

As for a rectangle this will result in the following formula: Ixx =

b ∗ h3

12 (16)

The total moment of inertia may then be built up using the moment of inertia of each rectangle in the cross section and Parallel axis theorem, sometimes called Steiner’s theorem.

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Figure 12: Measurement for an I beam

Thus are the formula for an I-beam without any buckling with regard to section 3.2.1 and 3.2.2: II−beam= wlf ∗ t3lf 12 + wlf ∗ tlf ∗ (cmlf − cm) 2₊tw ∗ h3w 12 + + tw∗ hw ∗ (cmw− cm)2+ wuf ∗ t3uf 12 + wuf ∗ tuf ∗ (cmuf − cm) 2 (17)

If the cross section is uniformed, wlf = wuf = wf and tlf = tuf = tf, the following

may be used:

I = wf ∗ (2 ∗ tf + hw)3− (wf − tw) ∗ h3w (18)

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3.2.1 Centre of mass

Figure 13: Centre of mass for each element

By using figure 12 and figure 13 the following formulas may be constructed:

Centre of mass for the lower flange will be equal to half the height of the lower flange.

cmlf =

tlf

2 (19)

Centre of mass for the web will be equal to: cmweb = tlf + hweb 2 (20) cmuf = tlf + hweb+ tuf 2 (21)

cm = cmlf ∗ Arealf + cmweb∗ Areaweb+ cmuf ∗ Areauf Arealf + Areaweb+ Areauf

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3.2.2 Parallel axis theorem

Parallel axis theorem or Steiner’s theorem is a theorem regarding the moment of inertia of an object that is not rotating on its axis. It states that the moment of inertia of an object not rotating on its axis is equal too:

I = Itp+ A ∗ d2 (23)

Where A is the area, d is the distance between the centre of mass and the centre of rotation. Itp is the moment of inertia for the element.

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3.3 Local buckling

Buckling is an instability problem which occurs for slender compressed elements and results in failure. Two types of buckling may occur, global and local.

The global buckling mode is for the whole element, it causes a deflection calculated with Euler’s buckling cases. It is a theory by the famous mathematician Leonhard Euler whom lived in the 18th century.

Local buckling are present in internal or external elements of a cross section. The cross sections parts are divided into classes which will determine if a risk of buckling is present.

Figure 14: Class division for width-to-thickness ratios of internal compressed parts (Eurocode 3: Design of steel structures : part 1-1 : General rules and rules for buildings 2005)

The ratios in figure 14 are referring to classes, 1 too 4. Each class will determine how the cross section can be utilized. If the part is in class 4, buckling may occur and a reduction of the element is necessary. If there is a risk of buckling the calculations

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used are according too Eurocode 3: design of steel structures : part 1-5 : plated structural elements 2010. First of, the stress ratio is calculated by comparing the stress in each side of the element. For an internal buckling of an I-Beam, this would relate too the top of the web and the bottom of the web.

ψ = σ2 σ1 (24) ¯ λp = s ¯_b t28, 4√kσ (25) ¯

λpis the slenderness parameter which will determine the severity of the instability.

From ¯λp, ρ will be calculated which is directly correlated with the effective area of

the cross section.

3.3.1 Internal element

Table 7: Buckling factor from internal compression elements (Eurocode 3: design of steel structures : part 1-5 : plated structural elements 2010)

ψ Buckling factor kσ 1 4,0 1 > ψ > 0 _1,05+ψ8,2 0 7,81 0 > ψ > −1 7,81-6,29ψ+9,78ψ2 -1 23,9 −1 > ψ > −3 5, 98(1 − ψ)2

If the factor ¯λp is less than 0,673, ρ is equal to 1.

Otherwise when ¯λp is above 0,673, and ψ ≥ −3, the following formula is used:

ρ = λ¯p− 0, 055(3 + ψ) ¯

λp

2 ≤ 1 (26)

3.3.2 External element

Outstand compression elements are calculated in the same fashion as internal ele-ments. The limits of ratios are one thing that differs from internal eleele-ments. The limits and how they are calculated can be seen in figure 15.

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Figure 15: Class division for width-to-thickness ratios of external compressed parts (Eurocode 3: Design of steel structures : part 1-1 : General rules and rules for buildings 2005)

If the outstanding element is in class 4, the calculating approach is nearly identical as internal elements but some limits have been shifted.

If the factor ¯λp is less than 0,748, ρ is equal to 1.

Otherwise when ¯λp is above 0,748 the following formula is used:

ρ = λ¯p− 0, 188 ¯ λp

2 ≤ 1 (27)

And if the outstanding element is in class 1 - 3, it will not be subjected to buckling.

3.4 Global buckling

The calculations for global buckling is according too chapter 6.3 in 1993-1-1. Too satisfy that global buckling failure will not occur the following must apply:

NEd

Nb,Rd

≤ 1 (28)

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The buckling resistance is equal too: Nb,Rd =

χAfy

γM 1

(29) With the area A being Aef f for cross sections in class 4.

The factor χ is the reduction in strength according too buckling. It is a factor relating too the non dimensional slenderness factor ¯λ and the imperfection factor α.

χ = 1

Φ +pΦ2_{− ¯}_λ2 ≤ 1 (30)

Where Φ is:

Φ = 0.5[1 + α(¯λ − 0.2) + ¯λ2] (31)

Values for α comes from EN-1993-1-1 table 6.1 and table 6.2:

Figure 16: Table 6.1 from Eurocode 1993-1-1 (Eurocode 3: Design of steel structures : part 1-1 : General rules and rules for buildings 2005)

The imperfection factor α is dependent on which buckling curve the cross section belongs too. It is dependent on the form of the cross section and the assembly method.

Figure 17: From table 6.2 Eurocode 1993-1-1 (Eurocode 3: Design of steel structures : part 1-1 : General rules and rules for buildings 2005)

The slenderness factor ¯λ are calculated as: ¯

λ =r Afy Ncr

(32) Once again, the area A being Aef f for cross sections in class 4.

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Ncr =

π2_EI

(KL)2 (33)

Values for K is dictated by the restraints acting upon the element.

Figure 18: Bucklingcases, K-factor (Fig. 2: Column effective length factors for Eu-ler’s critical load. In practical design, it is recommended to increase the factors as shown above)

If the slenderness is below 0.2 or if the imposing load is below 4% of the crit-ical buckling load Ncr, the global buckling effect may be ignored. But a check for

compression of the cross section is still in effect.

3.5 Lateral torsional buckling

Lateral torsional buckling is present for an element which are subjected too both compression and bending may experience lateral torsional buckling if the element is not restrained laterally.

The criteria to be followed is:

Mb,Rd = χLTWy

fy

γM 1

(34) The bending resistance is dependent on the cross section class.

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Table 8: Bending resistance

Cross section class Class 1 Class 2 Class 3 Class 4

Wy = Wpl,y Wpl,y Wel,y Wef f,y

χLT is calculated in the same fashion as χ:

χLT = 1 ΦLT + p Φ2 LT − ¯λ2LT ≤ 1 (35) Where ΦLT is: ΦLT = 0.5[1 + αLT(¯λLT − 0.2) + ¯λ2LT] (36)

The imperfection factor α comes from Eurocode 1993-1-1 table 6.3 and 6.4:

The factor α is dependent on which buckling curve the cross section belongs too. It is dependent on the form of the cross section and the assembly method.

The slenderness for lateral torsional buckling are calculated with the critical mo-ment instead of the critical load.

¯ λLT =

r Wyfy

Mcr

(37) Where the critical moment Mcr are often calculated by programs. There is also a

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Mcr = C1 π2_EI z L2 s Iw Iz + L 2_GI t π2_EI z (38) It involves a lot of variables which are influenced by how the moment are dis-tributed, section properties and buckling lengths.

A different method for calculating ¯λLT is:

¯ λLT = 1 √ C1 U V ¯λz p βw (39)

It may look equality as frightening in the beginning but a lot of the components may be chosen on the safe side. This will also reduce the number of steps taken and Mcr is skipped completely.

C1 is a factor which gets it value from how the moment diagram is distributed.

To keep on the safe side, 1 may be chosen.

U is a section property and may be taken as 0,9.

V are dependant on the slenderness of the beam. It may also be conservatively taken as 1.

βw is also conservatively taken as 1.

¯ λz = λz λ1 (40) Where: λz = kL iz (41) And λ1 is calculated according too Eurocode 3: Design of steel structures : part

1-1 : General rules and rules for buildings 2005 Chapter 6.3.1.3:

λ1 = 93, 9 (42) = s 235 fy (43) With all these conservative assumptions, the formula for ¯λLT will be the following:

¯ λLT = 3 313 L iz q 235 fy (44) Where: iz = r Iz A (45)

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3.6 Transverse force

For the calculations of resistance to transverse force, EN-1993-1-5 chapter 6 is the guiding document. With the assumption that the compressed flange is restrained in the lateral direction, the following line of calculations may be used:

FRd =

fywLef ftw

γM 1

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Lef f = χF`y (47)

χF is the factor that takes into account the risk of buckling for the web.

χF = 0, 5 ¯ λF (48) Where: ¯ λF = r `ytwfyw Fcr (49) The effective load length `y is always shorten than the distance between adjacent

transverse stiffeners. If the beam is supported on both sides and is being compressed vertically from either one or two directions, the following formulas apply:

`y = ss+ 2tf(1 + √ m1+ m2) (50) Where: m1 = fyfbf fywtw (51) And: m2 = 0.02(hw tf ) 2_{, if ¯}_λ F > 0, 5. 0, if ¯λF ≤ 0, 5. (52) On the other hand. If the beam is a cantilever and is being vertically with a single load, the effective length is the smallest of:

`y = è+ tf s m1 2 + ( è tf )2_{+ m} 2 (53) And: `y = è+ tf √ m1+ m2 (54) Where:

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`y =

kFEt2w

2fywhw

≤ ss+ c (55)

If it is designed properly, the following should apply:

FEd≥ FRd (56)

3.7 Shear

In case there is no torsion. the plastic shear resistance is calculated as following: Vpl,Rd =

Avfy

γM 0

√

3 (57)

Where Av is the shear area and is dependent on the type of cross section and

manufacturing process.

Table 9: Shear area

Cross section Formula for Av

Rolled I or H section loaded parallel with web A - 2 b tf + (tw+ 2r)tf > ηhwtw

Rolled I or H section loaded parallel with flanges A - 2 b tf + (tw+ r)tf

Welded I, H or box section loaded parallel with web ηΣ(hwtw)

Welded I, H or box section loaded parallel with flanges A - Σ(hwtw)

η in table 9 may be taken as 1.

A check for if buckling may be present is done for shear as well. If the web is unstiffened, the following equation is used:

hw

tw

≥ 72

η (58)

In the case that the web is stiffened, the following equation is used: hw tw ≥ 31 η p kτ (59)

Where epsilon is calculated the same way for both cases:

= s

235 fy

(60) If the equation for respective webs holds true, the web should be checked for resistance too shear buckling and be provided with transverse stiffeners at supports.

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The calculations for the elastic shear stress, Vc,Rd, may be calculated the following

way given that buckling may not occur: τEd √ 3γM 0 fy ≤ 1, 0 (61) Where: τEd= VEdS It (62)

Where VEd is the shear force acting on the cross section. I is the moment of

inertia. t is the width at the examined point. As the maximum shear will always be at the centre of mass for these beams, the examined point will be in the web. Therefor t will be the thickness of the web. S is the first moment of area.

Or if the cross section is a I-section with the web area being 3/5 or lager than the flange area:

τEd=

VEd

Aw

(63) These expressions can be re written as:

Vel,Rd=

Itfy

S√3γM 0

(64) Or if the cross section is a I-section with the web area being 3/5 or lager than the flange area: Vel,Rd= fyAw √ 3γM 0 (65)

3.8 Not included in the thesis

3.8.1 Self weight

The difference in self weight from the original beam to the optimized beams will be minimal in relation to the overall weight. Thus are the self weight always calculated as the self weight for the original beam.

3.8.2 Welds

The change of steel grade have an impact on how the welds will be designed. But as the welds are not taken into account when calculating the area, the size, length and filler used does not have an effect on this study.

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3.8.3 Resonance

As self resonance is something that must be taken into account for large buildings and structures that are effected by wind and similar natural effects. However, as ˚Akaren 1 is not a large structure, resonance calculations are not being taken into account.

3.8.4 Fire

As the fire protecting of steel is often done as either fire protective paint or built in with fire protective boards. Therefor as the fire not something that will be affecting the final design, it will be excluded.

3.8.5 Deformation

Deformation is a failure which results in that the deformation of the element is large enough that functions of the element is either lost or impaired (Downling 2017). Two classes of defamation may occur. Either it is a time independent failure which involves elastic or plastic deformation. Elastic and plastic deformation is the result of loads imposing on the structure and are displayed in real time. Creep on the other hand is a slow time dependent process. It is also a result of the loads acting on the structure but the deformation will not happen instant, instead it is spread out over a longer time frame.

No universal formula for the calculation of deformation is available. It is depen-dent of the cross section, material, load configuration and support. The restrictions of how large the deformation may be is also not specifically described in the Eu-rocode or any of the national compendiums and is often dependant on each building. The regular restrictions are often described in deformation/length, such as L/250 or L/400. This is because a deformation may be a lot more noticeable on a smaller span width than on a larger one.

For this project, Structor have not taken into account any specific criteria for the deformation and thus such a criteria will not be present. It will not be calculated because of its complexity for a setup for each beam.

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4 Calculations

4.1 Imposing loads

The loads are taken from Structors foundation calculations. These involve the char-acteristic loads and which load cases is used where. The loads and load cases are being used but the foundation calculations lack the load path which is calculated in this chapter. As the beams and pillars are connected, the load case used for the top beam is then continuously used trough out the structure.

Table 10: Loads Load Values qk,snow 4,4 kN/m2 qk,roof 0,4 kN/m2 qk,wall 0,3 kN/m2 qk,HV AC DL 2,3 kN/m2 qk,HV AC N L 0,5 kN/m2

Table 11: Load combination factors

Factor Value

Ψ0 0,7

Ψ1 0,7

Ψ2 0,6

Table 12: Gamma factor

Factor Value

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4.2 Beam 1 & 2

Figure 21: Illustration of the HVAC system resting on beam 1, beam 2 and a third support

The forces acting on the beam is based on the snow load and the load from the HVAC system.

STR-A:

qd,hvac = γd∗ 1, 35 ∗ qk,HV ACDL+ γd∗ 1, 5 ∗ Ψ0∗ (qk,HV ACN L+ qk,snow) (66)

With values:

qd,hvac = 1 ∗ 1, 35 ∗ 2, 3 + 1 ∗ 1, 5 ∗ 0, 7 ∗ (0, 5 + 4, 4) = 8, 25kN/m2

STR-B:

qd,hvac = γd∗ 1, 2 ∗ qk,HV ACDL+ γd∗ 1, 5 ∗ (qk,HV ACN L+ qk,snow) (67)

With values:

qd,hvac = 1 ∗ 1, 2 ∗ 2, 3 + 1 ∗ 1, 5 ∗ 0, 7 ∗ (0, 5 + 4, 4) = 10, 11kN/m2

STR-B is the dominant load combination.

The HVAC shipping container is assumed to be stiff enough to carry the loads according to beam theory. As this is a statically indeterminate beam, Linpro is used to calulcate the reaction forces for each support.

Figure 22: Supports for the HVAC system

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LRa = 0, 94m (68)

LRb = 3, 71m (69)

LRc = 1, 25m (70)

Thus are the reaction forces:

Ra= LRa∗ qd,hvac (71)

Rb = LRb∗ qd,hvac (72)

Rc= LRc∗ qd,hvac (73)

Moment distribution, shear force calculations and deflections are not taken into account for the shipping container.

4.2.1 Beam 1

Beam 1 have the cross section of a HEA 160 with a span of 2450mm. It is assumed too be simply supported as there is no major fasteners clamping it in place. The external line load acting on the beam is assumed to be 9, 50kN/m which is calculated from equation 67, 68 & 71. The self weight of a HEA 160 is assumed to be 0, 31kN/m.

Figure 23: Illustration of beam 1

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Figure 24: Illustration of supports and acting loads on beam 1

By setting the beam up as a simply supported beam with symmetry in the middle, the following hold true:

Vlef t = Vright (75)

As the beam is at rest, according to Newton’s second law the sum of forces must be zero:

ΣF = 0 (76)

Thus, with the forces acting on the beam:

q ∗ L − Vlef t− Vright = 0 (77)

Because of the symmetry the following holds true:

qL = 2V (78)

Or:

V = qL

2 (79)

With values the result becomes: Vb1 =

9, 50 ∗ 2, 450

2 = 11, 64kN

The moment will be at a maximum in the middle:

Mmax =

Vb4∗ Lb4

4 (80)

With values the maximum moment will be:

Mb1,max =

11, 64 ∗ 2, 450

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4.2.2 Beam 2

As beam 2 have the same geometrical boundary conditions as beam 1 and the same cross section, the calculations are the same for beam two as for beam 1.

The line load comes from the distributed load from the container, same as beam 1:

qd,b2= RB+ 1, 2 ∗ qk,HEA160 = 37, 88kN/m (81)

Figure 26: Illustration of supports and acting loads on beam 2 Thus the vertical force in the supports are equal too:

Vb2=

37, 88 ∗ 2, 450

2 = 46, 40kN

With calculations for the maximum moment:

Mmax =

46, 18 ∗ 2, 450

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4.3 Beam 3 & 4

Beam 1 and 2 are held up by two beams and have a span of 5250 mm. These two beams are identically loaded with the forces from beam 1 and 2 and their self weight. It is also assumed that the beams are simply supported.

Figure 27: Illustration of beam 3 & 4

The force from beam 1 (P1) is equal to 11,64 kN and the force from beam 2 (P2) is 46,40 kN. The profile for beams 3 and 4 is a HEA 220 which have a self weight of 0,50 kN/m. Thus:

qd,b3−4 = γd∗ 1, 2 ∗ qk,HEA220 (82)

qd,b3−4 = 1 ∗ 1, 2 ∗ 0, 50 = 0, 60kN/m (83)

The reaction force for the support are calculated by two relations, first the vertical force equilibrium:

ΣF = qd∗ L + P1+ P2− RA− RB = 0 (84)

Secondly, moment around A gives:

y A : P1∗ 2450 + P2∗ 5000 + qd∗ L2 2 − RB∗ 5250 = 0 (85) With values: RB = 11, 64 ∗ 2450 + 46, 40 ∗ 5000 + 0,0006∗5250₂ 2 5250 = 51, 20kN (86)

With equation 86 in equation 84 with a rewrite, RA may also be solved:

RA= 9, 99kN (87)

As qd∗ LP1 < RA and qd∗ LP1+ P1 > RA. the maximum moment are at L = LP1.

Mmax= RA∗ 2, 450 −

qd,b3−4∗ 2, 4502

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4.4 Beam 5

Beam 5 is a continuous beam that have 5 supports and a length of 13600mm with the longest span ranging 4150mm. The profile used is a IPE240 with a self weight of 0, 3kN/m and the roof are assumed to carry all its load perpendicular to the beam. This will result in an evenly distributed load. The snowload is qk,snow = 4, 4kN/m2

and the roofing have a weight of qk,roof ing = 0, 4kN/m2. A point load P coming from

beam 3 and 4 are P = 51, 20kN .

Figure 29: Top down view of the structure showing all the load bearing elements. The roof is assumed to be simply supported in the left-right direction in this picture with a uniformly distributed load.

Due to its complex structure, a program called Linpro is used to determine the reaction forces and moment.

The distance from beam 5 to the next truss is 5550mm, thus are the loaded length affecting beam 5 2775mm.

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The area between beam 4 and beam 5 is not effected by the distributed snow load as it is assumed to be laying on the HVAC system. Thus are the load between beam 4 and 5 reduced with this taken into account.

qq,reduced = 2, 775 ∗ 1, 2 ∗ qk,roof + 1, 2 ∗ qk,IP E240= 1, 69kN/m (90)

Figure 30: The load configuration on beam 5 This configuration gives the following:

Figure 31: Reaction forces acting on beam 5s supports

Figure 32: Moment distribution in beam 5 The maximum moment is 36, 68kN m in beam 5.

Figure 33: Shear forces in beam 5

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4.5 Beam 6

On the other end of the building, beam 6 is present. As its counterpart, beam 5, beam 6 is an IPE 240. This will result in a line load of 0, 3kN/m. It is assumed that the beam rests on three supports spaced, from the left: 7700mm and 6600 mm apart.

No external point loads are present on the beam. This leaves only three different loads acting on the beam, the beams self weight, the roofs self weight and the snow load acting on the roof. The distance between beam 6 and the second roof tile support are 5690 mm. Thus are the area beam 6 are taking roof and snow load from 2745mm. This will result in a line load of 12, 52kN/m for the characteristic snow load and 1, 14kN/m for the characteristic roof load.

STR-A gives:

qd,str−A= γd∗ 1, 35 ∗ (qk,roof + qk,beam6) + γd∗ 1, 5 ∗ Ψ0∗ qk,snow (91)

With values:

qd,str−A = 1, 35 ∗ (1, 138 + 0, 3) + 1, 5 ∗ 0, 7 ∗ 12, 52 = 15, 09kN/m

STR-B gives:

qd,str−B = γd∗ 1, 2 ∗ (qk,roof + qk,beam6) + γd∗ 1, 5 ∗ qk,snow (92)

With values:

qd,str−B = 1, 2 ∗ (1, 138 + 0, 3) + 1, 5 ∗ 12, 52 = 20, 51kN/m

Thus are STR-B the determining load case. The line load of 20,51 kN/m are placed on the beam and because of the fact that the beam is statically indeterminate, Linpro is once again used.

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Figure 36: The load configuration on beam 6

Figure 37: Reaction forces present in supports at beam 6

Figure 38: Moment in beam 6

Figure 39: Shear in beam 6

Figure 40: Axial loads in beam 6

4.6 Beam 7 & 8

The second HVAC system rests on two supports, beam 7 and beam 8

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The imposing loads are from the HVAC and the snow load on the HVAC system. The hvac system is a total of 10100 mm long and have a width of 2450 mm and the distance between the two supports are 8100 mm.

STR-A gives:

qd,str−A = γd∗ 1, 35 ∗ qk,HV ACDL+ γd∗ 1, 5 ∗ Ψ0∗ (qk,snow+ qk,HV ACN L) (93)

With values:

qd,str−A = 1 ∗ 1, 35 ∗ 2, 3 + 1 ∗ 1, 5 ∗ 0, 7 ∗ (4, 4 + 0, 5) = 8, 25kN/m2

STR-B:

qd,str−B = γd∗ 1, 20 ∗ qk,HV ACDL+ γd∗ 1, 5 ∗ (qk,snow+ qk,HV ACN L) (94)

With values:

qd,str−B = 1 ∗ 1, 20 ∗ 2, 3 + 1 ∗ 1, 5 ∗ (4, 4 + 0, 5) = 10, 11kN/m2

Thus are STR-B dominant and will be the one taken into account. Force equilibrium in horizontal direction gives:

RA+ RB = L ∗ qd (95) Moment around A: y A : L ∗ L 2 ∗ qd− RB∗ LAB = 0 (96) Rewritten as: RB = L2 2LAB ∗ qd (97)

Equation 97 in equation 95 gives:

RA = L ∗ qd∗ (1 −

L 2 ∗ LAB

) (98)

With values from equation 94 in equation 98 and equation 97: RA= 10, 1 ∗ 10, 11 ∗ (1 − 10, 1 2 ∗ 8, 1) = 38, 45kN/m And: RB = 10, 11 ∗ 10, 12 2 ∗ 8, 1 = 63, 66kN/m

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4.6.1 Beam 7

Beam 7 is a continuous beam with three supports. It is a HEA 160 with a self weight of 0,31 kN/m. Between the first two supports, a line load from the HVAC system of 38,45 kN/m is present. Between the second and third support the beams only load is its self weight.

Figure 42: Illustration of beam 7 The load effecting the beam under the HVAC system.

qd,beam7 = γd∗ 1, 2 ∗ qk,beam7+ RA (99)

With values:

qd,beam7= 1 ∗ 1, 2 ∗ 0, 31 + 38, 45 = 38, 82kN/m

The part of the beam that is only loaded with its self weight have its load calcu-lated the following way:

qd,beam7reduced= γd∗ 1, 2 ∗ qk,beam7 (100)

With values:

qd,beam7reduced = 1 ∗ 1, 2 ∗ 0, 31 = 0, 372kN/m

From Linpro the following calculations are derived:

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Figure 44: Reaction forces present in supports at beam 7

Figure 45: Moment in beam 7

Figure 46: Shear in beam 7

4.6.2 Beam 8

Beam 8 is a simply supported HEA 160 with the same length as the first span of Beam 7, 2450mm. Its only load is the HVAC system and its self weight.

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qd,beam8= γd∗ 1, 2 ∗ qk,beam7+ RB (101)

With values:

qd,beam7= 1 ∗ 1, 2 ∗ 0, 31 + 63, 66 = 64, 03kN/m

As it is simply supported with a evenly distributed load, the calculations from beam 1 apply: From equation 79: VM ax = 64, 03 ∗ 2, 45 2 = 78, 44kN From equation 4.2.2: MM ax = 78, 44 ∗ 2, 45 4 = 48, 04kN m

4.7 Beam 9

Beam 9 is simply supported with a length of 4950mm. It is a HEA 180 which makes its self weight 0,35 kN/m. Additional to its self weight, a point load is also present. The point load comes from one of beams 7s support. The load is located 1000mm from the end of beam 9 and have a magnitude of 40,26 kN.

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qd,beam9= γd∗ 1, 2 ∗ qk,beam9 (102)

qd,beam9= 1 ∗ 1, 2 ∗ 0, 35 = 0, 42kN/m

Figure 49: Supports and loads on beam 9 Force equilibrium in the vertical direction gives:

ΣF = RA+ RB− qd,beam9∗ L − P = 0 (103) Moment around A: y A : qd,beam9∗ L 2 2 + P ∗ (L − LP −B) − RB∗ L = 0 (104) Or: RB = qd,beam9∗ L 2 + P (1 − LP −B L ) (105)

Equation 105 in equation 103 gives: RA+ qd,beam9∗ L 2 + P (1 − LP −B L ) − qd,beam9∗ L − P = 0 (106) Or:

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RA=

qd,beam9∗ L

2 + P

LP −B

L (107)

The maximum moment will occur when the shear force is 0: qd,beam9∗ L 2 + P LP −B L − qd,beam9∗ Lmax (108) Or: Lmax= L 2 + LP B L P qd,beam9 (109) when Lmax ≤ L − LP B = LAP.

Mmax = RA∗ Lmax− qd,beam9∗

L2_max

2 (110)

With values in equation 105:

RB = 0, 472 ∗ 4, 95 2 + 40, 26(1 − 1, 00 4, 95) = 33, 29kN Values in equation 107: RA = 0, 42 ∗ 4, 95 2 + 40, 26 1, 00 4, 95 = 9, 17kN Length where the shear force is 0:

Lmax = 4, 95 2 + 1, 00 4, 95 40, 26 0, 42 = 21, 84m

As Lmax > L − LP B and Lmax > L, the maximum moment occurs at the point

load. Thus:

Mmax = 9, 17 ∗ 3, 95 − 0, 42 ∗

3, 952

2 = 32, 94kN m

4.8 Beam 10

Beam 10 is identical to beam 9 but with a different point load. Its distrubuted load consists of the self weight of a HEA 180, it is 4950 mm long and have a point load P at 1000mm from support B.

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The point load P is from beam 8 and have a magnitude of 78,44 kN and is places the same as on beam 9. This will result in the same equations being used.

RA = qd,beam10∗ L 2 + P LP −B L (112) RB = qd,beam10∗ L 2 + P (1 − LP −B L ) (113) Lmax = L 2 + LP B L P qd,beam10 (114) Where Lmax <= LAP. This will result in:

Mmax = RA∗ Lmax− qd,beam10∗

L2_max 2 (115) With values: RA = 0, 42 ∗ 4, 95 2 + 78, 44 ∗ 1, 00 4, 95 = 16, 89kN RB = 0, 42 ∗ 4, 95 2 + 78, 44(1 − 1, 00 4, 95) = 63, 63kN Lmax = 4, 95 2 + 1, 00 4, 95 78, 44 0, 42 = 40, 20m Thus are Lmax = LAP:

Mmax = 16, 89 ∗ 3, 95 − 0, 42 ∗

3, 952

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4.9 Beam 11

Beam 11 is a HEA 220 with a span of 5300 mm. An HEA 220 have a line load of

0,5 kN/m. There is also two point loads acting on beam 11. P1 = 66, 66kN and

P2 = −11, 25kN .

Figure 51: Illustration of beam 11 The design line load is:

qd,beam11 = γd∗ 1, 2 ∗ qk,beam11 (116)

With values:

qd,beam11 = 1 ∗ 1, 2 ∗ 0, 5 = 0, 6kN/m

Figure 52: Supports and loads for beam 11 Force equilibrium in vertical direction gives:

ΣF = RA+ RB+ P2 − P1− qd∗ L = 0 (117) Moment around A: y A : P1∗ LAP1 − P2∗ LAP2 + qd L2 2 − RB∗ L = 0 (118) Or:

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RB = P1 LAP1 L − P2 LAP2 L + qd L 2 (119)

Equation 119 in equation 117 gives: RA= qd L 2 + P1(1 − LAP1 L ) − P2(1 − LAP2 L ) (120) With values: RB = 66, 66 1, 35 5, 30 − 11, 25 2, 90 5, 30 + 0, 6 5, 30 2 = 12, 41kN And: RA= 0, 6 5, 30 2 + 66, 66(1 − 1, 35 5, 30) − 11, 25(1 − 2, 90 5, 30) = 46, 18kN If: RA− qd,beam11∗ LAP1 ≤ 0 (121)

The maximum moment is in between 0 and L = LAP1:

46, 18 − 0, 6 ∗ 1, 35 = 45, 37

As P1 > 46, 37, the maximum moment will occur at L = LAP1.

Mmax = RA∗ LAP1 − qd,beam11∗ L2 AP1 2 (122) With values: Mmax = 46, 18 ∗ 1, 35 − 0, 6 ∗ 1, 352 2 = 61, 80kN m

4.10 Beam 12

Beam 12 is identical to beam 11 with different point loads. P1 = 78, 44 and P2 = 0.

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qd,beam12= qd,beam11 = 0, 6kN/m (123)

From equation and equation : RA= 0, 6 5, 30 2 + 78, 44(1 − 1, 35 5, 30) = 60, 05kN And: RB = 78, 44 1, 35 5, 30 + 0, 6 5, 30 2 = 21, 57kN Mmax = 60, 05 ∗ 1, 35 − 0, 6 ∗ 1, 352 2 = 80, 52kN m

The use of higher steel grades in building elements