chapter 4 str 4 intr to str steel.doc

Dept. of Civil & Urban Eng., iOTec-HU. Hawassa University

Chapter 4 Introduction to structural steel

4.1 Methods of Analysis

4.1.1 Introduction to frame behaviour

Global frame analysis aims at determining the distribution of the internal forces and the corresponding deformations in a structure subjected to a specified loading.

Achieving this purpose requires the adoption of adequate models which incorporate assumptions about the behaviour of the structure and in particular of its component members and joints. The main purpose of this sub-chapter is to address the question of modelling and analysis of structural behaviour for practical design purposes.

4.1.2 Load displacement relationship of frames

The response of a structure to loading applied to it can be expressed by the relationship between a load parameter and a significant displacement parameter. An example of the behaviour of a typical sway frame under increasing load is shown in Figure 4.1.

The resulting curve for a load parameter and horizontal (lateral) displacement parameter can be considered to characterise the overall structural behaviour. In this case, the slope of the curve is a measure of the lateral stiffness of the frame structure.

Figure 4.1 - Load displacement response of a framed structure

One observes that the response of the structure is quasi-linear up to a certain point (the linear limit). Once the linear limit is reached, the positive slope of the rising part of the curve gradually reduces due to a combination of three kinds of non-linearity: geometrical non-linearity, joint non-linearity and material non-linearity. Joint non-linearity usually __________________________________________________________________________________Structural Engineering IV (CEng 3312) Chapter 4

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manifests itself at relatively low levels of load. Geometrical non-linearity expresses the influence of the actual deformed shape of the structure on the distribution of the internal forces. Typically it becomes evident well before the onset of material yielding, i.e. material non-linearity. Beyond the latter, the response becomes progressively non-linear as the load increases up to a maximum. Once the maximum load is reached, equilibrium would require a decrease in the magnitude of the loads as deformations increase. The slope of the curve (i.e., the stiffness) is zero at the peak load and then it becomes negative indicating that the structure is henceforward unstable. The peak load, often termed the ultimate load, is the point of imminent structural collapse in the absence of the possibility of load shedding.

4.1.3 Modelling of building structures for analysis

Global analysis of frames is conducted on a model based on many assumptions including those for the structural model, the geometric behaviour of the structure and of its members and the behaviour of the sections and of the joints.

Once the analysis is achieved, a number of design checks of the frame and its components (members and joints) must be performed. These checks depend on the type of analysis performed and the type of cross-section verification (i.e. ultimate limit state criteria) used.

4.1.4 Global elastic frame analysis

Linear-elastic analysis implies an indefinite linear response of sections and joints (see Figure 2). Equilibrium is expressed with reference to the non-deformed configuration of the structure in a first-order analysis.

Figure 4.2 - Moment rotation characteristics of member and joint

A priori, no requirements related to the ability of sections and joints to exhibit ductile behaviour (class of member cross-section, ductility class of joint) are imposed. However the class of member cross-section to be finally adopted depends on the type of cross-section verification (ultimate resistance criteria) used.

Frame analysis

Elastic global analysis with linear member and joint behaviour results in a linear load deflection curve (see Figure 3).__________________________________________________________________________________Structural Engineering IV (CEng 3312) Chapter 4

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Figure 4.3 - Load displacement response: First-order elastic analysis

Designers are quite familiar with first-order elastic analysis which is the simplest of all possible types of analysis. Over the years, a variety of methods have been developed aimed at hand calculation such as the slope-deflection method, the moment-distribution method as well as analytical formulae (sometimes presented in non-dimensional graphical form) for rapid analysis. They can be generalised so as to include the joint behaviour. The same applies to procedures based on matrix formulation, which have now almost entirely supplanted the hand methods, as computer use has become common practice in design offices.

4.1.5 Methods of global plastic frame analysis

Plastic methods of analysis are permitted only when certain minimum requirements on steel ductility, member cross-section ductility, joint ductility and lateral support at hinges locations are met. These are needed in order to guarantee that sections and joints, at least at the locations at which the plastic hinges may form, have sufficient rotation capacity to permit all the plastic hinges to develop throughout the structure.

4.1.5.1 Elastic-perfectly plastic analysis (Second-order theory)

In the elastic-perfectly plastic analysis, it is assumed that any section and/or joint remains elastic up to the attainment of the plastic moment resistance, at which point it becomes ideally plastic. Plastic deformations are assumed to be concentrated at the plastic hinge locations which are assumed to have an infinite rotational capacity. That actual rotation capacities are sufficient to meet what is required must usually be checked later.

Figure 4 shows the elastic-perfectly plastic behaviour of a section and a joint. The influence of the normal force and/or the shear force on the plastic moment resistance of the sections may either be accounted for directly or be checked later at the design

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verification stage. (Note: The plastic hinge moment resistance to be adopted is the design value Mpl,Rd.)

The load displacement curve of the frame can be determined. Computation of the plastic rotations at the plastic hinges may also be carried out so as to permit the check that the required rotation capacity is available.

Figure 4.4 - Behaviour of members and joints

4.1.5.2 Elasto-plastic analysis (second-order theory)

Using a second-order elasto-plastic analysis a better estimation of the structural response can be obtained (relative to that provided by a first-order or even a second-order elastic-perfectly plastic analysis for instance).



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Yielding of members and joints is a progressive process and so the transition from elastic behaviour to a plastic one is a gradual phenomenon. Once yielding commences, as the moment in the member cross section continues to increases, the plastic zone extends partially along the member as well as through the depth of the cross-section. This behaviour is considered by the plastic zone theory. Figure 5 shows the moment rotation characteristics of members and joints which are usually adopted in this type of analysis. The beneficial effects of material strain hardening or membrane action in joints have not been included in these models.The ductility requirements for the members and joints, and the procedure for analysis and for checks are the same as those outlined for second-order elastic-perfectly plastic analysis.The elasto-plastic method, because of its complexity, is not used for practical design purposes and is restricted to computer research applications.

4.1.5.3 Rigid-plastic analysis (first-order theory)

Contrary to the elastic-plastic analysis, the elastic deformations (of members, joints and foundations), being small compared to the plastic deformations, are ignored in the rigid-plastic analysis. As for the elastic-perfectly plastic analysis, the plastic deformations are concentrated in sections and joints where plastic hinges are likely to occur. These sections and joints are assumed to have an infinite rotational capacity.Figure 6 shows the idealised rigid-plastic response of the sections and the joints which are adopted for this type of analysis. As a result, the values of the design moment resistance for sections and joints as well as the structural configuration and the loading are the only parameters that affect rigid-plastic analysis.


The ductility requirements for the members and the joints are the same as that for the elastic-perfectly plastic analysis. Rigid-plastic methods are not usually suited for second-order analysis.


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4.2 STRUCTURAL JOINTS

The rotational behaviour of actual joints is well recognised as being often intermediate between the two extreme situations, i.e. rigid or pinned.Consider now the bending moments and the related rotations at a joint (Figure 4.7):

(a) Rigid joint (b) Pinned joint (c) Semi-rigid joint

Figure 4.7 Classification of joints according to stiffness

When all the different parts in the joint are sufficiently stiff (i.e. ideally infinitely stiff), the joint is rigid, and there is no difference between the respective rotations at the ends of the members connected at this joint (Figure 4.77.a). The joint experiences a single global rigid-body rotation which is the nodal rotation in the commonly used analysis methods for framed structures.Should the joint be without any stiffness, then the beam will behave just as simply supported whatever the behaviour of the other connected member(s) (Figure 4.77.b). This is a pinned joint.For intermediate cases (non zero and non infinite stiffness), the transmitted moment will result in there being a difference between the absolute rotations of the two connected members (Figure 4.77.c). The joint is semi-rigid in these cases.The simplest means for representing the concept is a rotational (spiral) spring between the ends of the two connected members. The rotational stiffness S of this spring is the parameter that links the transmitted moment Mj to the relative rotation which is the difference between the absolute rotations of the two connected members.When this rotational stiffness S is zero, or when it is relatively small, the joint falls back into the pinned joint class. In contrast, when the rotational stiffness S is infinite, or when it is relatively high, the joint falls into the rigid joint class. In all the intermediate cases, the joint belongs to the semi-rigid joint class. For semi-rigid joints the loads will result in both a bending moment Mj and a relative rotation between the connected members. The moment and the relative rotation are related through a constitutive law which depends on the joint properties. This is illustrated in Figure 8, where, for the sake of simplicity, the global analysis is assumed to be performed with linear elastic assumptions.


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At the global analysis stage, the effect of having semi-rigid joints instead of rigid or pinned joints is to modify not only the displacements, but also the distribution and magnitude of the internal forces throughout the structure.As an example, the bending moment diagrams in a fixed-base simple portal frame subjected to a uniformly distributed load are given in Figure 9 for two situations, where the beam-to-column joints are respectively either pinned or semi-rigid. The same kind of consideration holds for deflections.

(a) Rigid joint (b) Pinned joint (c) Semi-rigid joint ( = 0) (Mj = 0) (Mj and 0)

Figure 4.8 Modelling of joints (case of elastic global analysis)

(a) Pinned joints (b) Semi-rigid joints

Figure 4.9 Elastic distribution of bending moments in a simple portal frame


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4.3 THE MANUFACTURING PROCESS OF STRUCTURAL STEEL

Hot Rolling Process

Liquid steel is cast into ingots [Fig.(23(d)], which after soaking at 1280-13000 C in the soaking pits [(Fig.23(e)] are rolled in the blooming and billet mill into blooms/billets [(Fig.23(f)] or in slabbing mill into slabs. The basic shapes such as ingots, cast slabs, bloom and billets are shown in Fig.24. The blooms are further heated in the reheating furnaces at 1250-12800 C and rolled into billets or to large structurals[(Fig.23(h)]. The slabs after heating to similar temperature are rolled into plates in the plate mill. Even though the chemical composition of steel dictates the mechanical properties, its final mechanical properties are strongly influenced by rolling practice, finishing temperature, and cooling rate and subsequent heat treatment.

The slabs or blooms or the billets can directly be continuously cast from the liquid state and thereafter are subjected to further rolling after heating in the reheating furnaces.

In the hot rolling operation the material passes through two rolls where the gap between rolls is lower than the thickness of the input material. The material would be repeatedly passed back and forth through the same rolls several times by reducing the gap between them during each pass. Plain rolls (Fig.25) are used for flat products such as plate, strip and sheet, while grooved rolls (Fig. 26) are used in the production of structural sections, rails, rounded and special shapes. The rolling process, in addition to shaping the steel into the required size, improves the mechanical properties by refining the grain size of the material.

Final rolling of structurals, bars/rods and HRC/CRC or sheet product is done in respective mills. In case of cold rolled sheets/coils, the material is annealed and skin passed to provide it the necessary ductility and surface finish


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Ingot Slab Bloom Billet

Basic shapes and their relative proportions

Primary rolls


Cold rolling and cold forming

Cold rolling, as the term implies involves reducing the thickness of unheated material into thin sheets by applying rolling pressure at ambient temperature. The common cold rolled products are coils and sheets. Cold rolling results in smoother surface and improved mechanical properties. Cold rolled sheets could be made as thin as 0.3 mm. Cold forming is a process by which the sheets (hot rolled / cold rolled) are folded in to desired section profile by a series of forming rolls in a continuous train of roller sets. Such thin shapes are impossible to be produced by hot rolling. The main advantage of cold-formed sheets in structural application is that any desired shape can be produced. In other words it can be tailor-made into a particular section for a desired member performance. These cold formed sheet steels are basically low carbon steels (<0.1 % carbon) and after rolling these steel are reheated to about 6500-7230C and at this stage ferrite is recrystalised and also result in finer grain size. Because of the presence of ferrite, the ductility is enhanced.

4.4 DESIGN PRINCIPLE EBCS 3 PROVISIONS

The design of any structure will be assessed by safety economy and appearance Safety is assessed by considering the strength of the structure relative to the loads

which it is expected to carry. Safety assessment is applied to each structural element and overall framework

Economic design result from finding the smallest structural size and weight with the consideration of fabrication and erection process

The appearance of the finished structure is generally of great importance owing to the very size and impact of frames in structures

Partial safety factors Safety factors are used in all designs to allow for variabilities of load, material, workmanship and soon, which can’t be assessed with absolute certainty. They must be sufficient to cover:

i. Load variationsii. Load combinations

iii. Design and detailing procedures iv. Fabrication and erection proceduresv. Material variations

Partial safety factor of steels, 𝛄m=1.1 Resistance of class 1,2or 3 cross-section: 𝛄mo=1.1 Resistance of class 4 cross-section: 𝛄m1=1.1 Resistance of member to buckling: 𝛄m1=1.1 Resistance of net section at bolt holes: 𝛄m2=1.25


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Partial safety factor for loads

Design assumptions

1) Simple Framing i) In simple framing the connection between the members may be assumed not to develop moments. In global analysis, members may be assumed to be effectively pin connected.ii) The connection should satisfy the requirements for nominally pinned connections, as given section 6.1.3 of EBCS-3.2) Continuous Framing Elastic analysis, rigid-plastic analysis and elastic-plastic analysis should be based on the assumption for full continuity, with rigid connections which satisfy the requirements given in section6.1.4 of EBCS-3.3) Semi-continuous Framing i) Elastic analysis should be based on reliably predicated design moment-rotation or force-displacement characteristics for the connections used.ii) Rigid-plastic analysis should be based on the design moment resistance of connections which have been demonstrated to have sufficient rotation capacity iii) Elastic-plastic analysis should be based on the design moment-rotation characteristics of the connections.

Material properties for hot rolled steelThe nominal values of the yield strength fy and the ultimate tensile strength fu for hot rolled steel are given in Tabel3.1 of EBCS-3 for grade Fe 360, Fe 430 and Fe 510


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Loading Load factor

Dead load 1.3

Live ( Imposed) load 1.6


Classification of cross-sectionsDepending on the width to thickness ratio of the elements of a cross-section, under compression, EBCS-3 defines in section 4.3.2, four different classes of cross-sections. These are:

Class1 Plastic cross-sections:- Those which can form a plastic hinge with the rotation capacity required for plastic analysis

Class2 Compact cross-sections:- Those which can develop their plastic moment resistance, but have limited rotation capacity

Class3 Semi-compact cross-sections:- Those in which the calculated stress in the extreme compression fiber of the member can be its yield strength, but local buckling is liable to prevent development of the full plastic moment resistance

Class4 Thin-walled cross-sections:- Those in which it is necessary to make explicit allowances for the effects of local buckling. Yield in extreme fibers can’t be attained because of premature local buckling in elastic range

The classification of compression elements include every element of cross-section which is either totally or partially under compression, due to axial force, bending moment, and


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under combination of loads considered. A cross-section is classified by quoting the least favorable class of its compression element. Alternatively, the classification of a cross-section may be defined by quoting both the flange classification and the web classification. The limiting proportion of class1,2 and 3 compression elements are obtained from Table4.1 of EBCS-3. An element which fails to satisfy the limits of class-3 are taken as class-4.

Thin walled sections Where a thin walled section element is in compression the yield stress fy shall be reduced by the factor given in tabel4.2 page-43. Alternatively the effective cross-sectional properties for class-4 may be determined based on effective area of compression element as described in section 4.3.4 (page-45)A reduction factor ρ is to be determined for finding effective width of element.


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Effective cross-section for class 4 in compression and bending


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ExamplesExample1

Determine the classification of 305X102X28UB of grade Fe430if the sections is used as;(a)column under axial compression load, and (b) Flexural member bending about its major axis Solution Relevant cross-section properties h=308.9mm tf=8.9mm b=101.9mm tw=6.1mm r=7.6mm

For Fe430steel grade fy=2754N/mm2, thus ε=

a) Section used as column under axial compression loadi) Outstand element of compression flange (Table4.1 of EBCS-3) c/tf=(b/2)/tf=(101.9/2)/8.9=5.73<8.5ε=8.5*0.92=7.82Hence, based on flange classification the cross-section is classified as class-1ii) Web where whole section is subject to compression (Table4.1 of EBCS-3) d=h-2tf-2r=308.9-2(8.9)-2(7.6)=275.9mm d/tw=275.9/6.1=42.23 > 44ε=44*0.92=40.48 for class-2 < 51ε=51*0.92=46.92 for class-3Hence, based on web classification, the cross-section is classified as class-3 The whole cross-section has to be classified as class-3b) Section used as flexural member bending about major axisi) Outstand element of compression flange As the flange of this cross section is under compression it will same as in part(a) above, thus the section is class-1.ii) Web with neutral axis at mid-height d/tw=275.9/6.1=42.23 < 79ε=79*0.92=72.68Hence the cross-section is classified as class-1 The whole cross section has to be classified as class-1 Example 2Determine the effective area of 406X140X39UB of Fe430 grade if the section is used as a centrally loaded column.Solution Relevant cross-section properties h= 397.3mm r=10.2mm b=141.8mm A=4940mm2

tw=6.3mm tf=8.6mm

For Fe430steel grade fy=2754N/mm2, thus ε=

i) Outstand element of compression flangec/tf=(b/2)/tf=(141.8/2)/8.6=8.2 >8.5ε=8.5*0.92=7.82 < 9.5ε=9.5*0.92=8.74Hence, the flange is class-2 and is fully effective ii) Web, where whole section is subject to compression d/tw=359.7/6.3=57.1 > 51ε=51*0.92=46.92Hence, the web is considered as class-4Stress distribution: since the column is axially loaded, the stress distribution is uniform (i.e. σ1=σ2). Table

4.3 of EBCS-3 is used to calculate the effective width. For .0, the corresponding value of Kσ=4

According to section 4.3.4 of EBCS-3


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Substituting,

(359.7/6.3) /(28.4X0.92* )=1.09 >0.673

ρ= (λρ—0.22)/ λρ2= (1.09—0.22)/1.092=0732

beff=ρb=0.732*359.7=263.4mmTherefore, the area that should be ignored at center of the web is ΔA=(b-beff)tw=(359.7—263.3)*6.3=607mm2

The effective area becomes Aeff=A—ΔA=4940—607=4333mm2

Example 3Classify the sections: ISA (Indian Standard Angle) 200X200X12 with steel grade of Fe360

fy for Fe360=235MPa ε=

8.5ε=8.5*1=8.5

9.5ε=9.5*1=9.5

15ε=15 and 23ε=23

The element fails to satisfy the limits of class-3; therefore the section is classified as class-4(Thin walled section)Example 4 classify the section ISA 75X50X8→Grade 430h=75mm fy=275MPa

b=80mm ε= t=8mm

b/t=50/8=6.25 8.5ε=8.5*0.92=7.82h/t=75/8=9.375 9.5ε=9.5*0.92=8.74(h+b)/8=15.625 15ε=15*0.92=13.8 and 23ε=23*0.92=21.16Short leg: - b/t=6.25 < 7.82(i.e. 8.5ε) →It is clas-1(plastic section) Long leg:- h/t=9.375> 9.5ε<15ε and (h+b)/t=15.625<23ε →It is class-3( Semi compact section)The section is said to be a semi compact sections4.2 Deign of tension and compression members 1) Tension members

Tension members are efficient carries of axial load and are used in many types of structures


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The form of tension member is governed to a large extent by the type of structure, which is it part and by the method of jointing it to connecting portions of structure.

The simplest tension members are made of wire rope or cable, round and square bars, and rectangular bars or plate. Single shapes, such as angle, the plate, the W and S shapes, and the T may be used as tension members. However two or more shapes are often combined to form a ‘Built-up’ memberI) Axially loaded tension members

Where Npl,Rd and Nu,Rd =Design plastic &Design ultimate resistance respectively A and Aeff = Gross and Effective areas respectively γM1 and γM2 =safety factors at solid and hole sections respectively fy and fu = Yield and Ultimate tensile strengths of steel.Failure of tension member is considered to occur either when the gross section has reached the yield strength, or when the net cross-section (where holes occur at connections) reaches the ultimate strength.

Example 1 Determine the design strength of an angle, ISA 90X90X10, is grade of Fe360 used as welded bracing members under tension SolutionThe gross area of the section is A= 17.03cm2=1703mm2(it is taken from ISA table for that section)The yield strength for Fe360 steel grade is, fy=235MPaThe partial safety factor for the section (section 4.1.1(2)) is γm1=1.1The design plastic resistance is: Npl,Rd=Afy/ γm1=(1703mm2*235N/mm2)/1.1=363822.73N=363.82KN


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Example 2 Determine the design strength of an angle, ISA 80X50X6 is grade of Fe430 used as a welded bracing members, connected by its smaller leg. The member is under tension.SolutionAs per specification in page129, of EBCS-3For an unequal angle, connected by its smaller leg, effective area is equal to the cross section area of equal angel with leg length equal to the smaller leg. Therefore area of cross section = Area of cross section of ISA 50X50X6 =5.68cm2=568mm2

The yield strength for Fe430 steel grade is, fy=275MPaThe partial safety factor for the section is γm1=1.1The design plastic resistance is: Npl,Rd=Afy/ γm1=(568mm2*275N/mm2)/1.1=142000N=142KN

Example 3Determine the design strength of an angle 100X100X10 in grade Fe430 used as a bolted bracing member with single row of 16.5mm holes at each leg of the angle.SolutionThe gross area of the section is A= 19.2cm2=1920mm2 The net area of the section is Aeff=1920-2X16.5X10=1590mm2

The ultimate strength for grade Fe430 is, fu=430N/mm2

The partial safety factor for the net section (section4.1.1(2)) is γm2=1.25The design ultimate resistance of the net section is: Nu, Rd=0.9Aefffu/ γm2=0.9*1590*430/1.25=492KNThe yield strength for Fe430steel grade is, fy=275MPaThe partial safety factor for the section (section 4.1.1(2)) is γm1=1.1The design plastic resistance is: Npl,Rd=Afy/ γm1=(1920mm2*275N/mm2)/1.1=480000N=480KN

Nu, Rd=492KN > Npl,Rd=480KN →the design strength of the bolted bracing members is controlled by the yield strength of full section.

II) Tension members with moment Members subjected to axial tensile force and bending moments do not occur frequently in reality, while members subjected to axial compression force and bending moment are quite common and are dealt with great detail in next chapter. To design members subjected to axial tensile force and moment the following steps should be followed:1. Determine the tensile axial force and bending moment applied to the member2. Select a trial section 3. Check the section for local buckling under moment alone4. Determine the design plastic resistance of the gross section or the design ultimate resistance of the net section5. Determine the resistance moment about one or both axes as required __________________________________________________________________________________Structural Engineering IV (CEng 3312) Chapter 4

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6. Check the interaction expression given by Eq.4.25 of EBCS-3 to determine whether the cross section is adequate for the applied tensile axial force and the moments

Eq. 4.25 of EBCS-3

Example 1 A tension members have to resist a load 600KN. Select suitable angle sections if the connection is made by welding to gusset plate and the steel grade is Fe360.Solution: The design plastic resistance of the gross c/s Npl,Rd=Afy/ γm1=(A*235N/mm2)/1.1=600000N

A=600000/235=2808.5mm2

If it is a single angle, try 130X130X12mm → Area=2918mm2

If it is double angle, try 80X80X10mm-(2numbers) →A=2*15.05mm2=30.1mm2

Example 2 A 406 X178 X67 UB grade Fe430 beam is laterally restrained against lateral buckling subjected to a factored bending moment of 120KNm about the major axis and 20KNm about the minor axis. Determine the maximum axial force the beam can carry Solution: Relevant cross-section properties h= 409.4mm r=10.2mm b=141.8mm A=8550mm2

tw=8.8mm d=360.5mm tf=14.3mmPlastic section modulus Elastic section modulusWpl,y=1350cm3 Wel,y=1190cm3

Wpl,z=237cm3 Wel,z=153cm3

Step1 and Step 2: the loads and the section is already given no need the deal these steps

Step 3: check for local buckling

For Fe430steel grade fy=275 N/mm2, thus ε=

i) Outstand element of compression flange (Table4.1 of EBCS-3) c/tf=(b/2)/tf=(178.8/2)/14.3=6.25<8.5ε=8.5*0.92=7.82Hence, based on flange classification the cross-section is classified as class-1 :- No local buckling will take placeii) Web with neutral axis at mid height d/tw=360.5/8.8=40.97< 79ε=79*0.92=72.68 for class-1This element also is class-1; no local buckling takes place in whole section Step 4: calculation the plastic resistance of the gross section Npl,Rd=Afy/ γm1=(1350 X 103 mm2*275N/mm2)/1.1=2351000N/1.1=2137.27KN

Step 5: calculation the resistance moment about both axes (a) Major axis bending for class 1 section


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My,Rd=Wpl,y fy/γm1=(1350 X 10 3 X 275 X10 -6)/1.1=337.5KNm (b) Minor axis bending for class 1 section Mz Rd= Wpl,zfy/γm1=(237 X 10 3 X 275 X 10-6 )/1.1= 59.3KNmStep 6: check for combined effect:According to equation 4.25 of EBCS-3

Nsd=656.5KN

Example 3 Find the adequacy of the section 457 X 152 X 82 UB of grade to act as a member subjected to an axial load of 550KN and a biaxial bending about its major axis of 150KNm and minor axis of 25KNmSolution: Relevant cross-section properties h= 409.4mm b=153.5mm A=10400mm2

tw=10.7mm d=406.9mm tf=18.7mmPlastic section modulus Elastic section modulusWpl,y=1800cm3 Wel,y=1560cm3

Wpl,z=235cm3 Wel,z=149cm3

Check for local buckingFor Fe430, fy=275MPa and ε=0.92(i) Outstand element of compression flange (Table4.1 of EBCS-3) c/tf=(b/2)/tf=(153.5/2)/18.9=4.06<8.5ε=8.5*0.92=7.82Hence, based on flange classification the cross-section is classified as class-1 :- No local buckling will take placeii) Web with neutral axis at mid height d/tw=406.9/10.7=38.02< 79ε=79*0.92=72.68 for class-1This element also is class-1; no local buckling takes place in whole section

Plastic resistance of gross section Npl,Rd=Afy/ γm1=(10400 mm2*275N/mm2)/1.1=2600KNCalculation for resistance moment about major and minor axes Major axis: My,Rd=Wpl,y fy/γm1=(1800X 10 3 X 275 X10 -6)/1.1=450KNmManor axis: Mz Rd= Wpl,zfy/γm1=(235 X 10 3 X 275 X 10-6 )/1.1= 58.75KNmCheck for combined effect


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0.2115 + 0.3333 +0.4255 <1.0 0.97< 1 hence ok

2) Design of compression members Compression members are usually given names given names which identify them as particular members in structure

In building frames are called columns Compression members in truss are known according to their position as chord

members or web members The principal compression member in crane is called a boom Some types of compression member are called struts They are sometimes called posts in bridge structure

Test on axially loaded, pin-ended struts show that their behavior can be represented by a number of curves which relate to the type of section and the axis of buckling. These curves are dependent on material strength and initial imperfection, which affect the inelastic behavior and the inelastic buckling load.

Axially loaded compression members should be designed for the following limit states: (1) yield strength, (2) overall column buckling (flexure buckling, torsional buckling, or torsional-flexural bucking) and (3) local buckling of individual elements (class 4 cross-section). The governing failure mode depends on the configuration of cross-section, thickness of material, unbraced length, and end restraint.


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Design steps for buckling resistance of axially loaded compression members

1. Determine the axial load in the compression member 2. Determine the buckling length Leff. In EBCS-3 section 4.5.2.1(page 48-51), it is

given the definition for buckling length Leff is given as

Also is given that in the absence of better information the theoretical buckling length for elastic critical buckling may conservatively be adapted

3. Select a trial section 4. Determine the class of the cross section

5. Determine the non-dimensional slenderness ratio from section 4.5.4.3 of EBCS-3

Where

βA=defined already

6. Using table 4.11 of EBCS-3, determine the appropriate buckling curve 7. Using table 4.9 find the value of χ. interpolation shall be used for more exact

values.8. Calculate the design buckling resistance Nb,RD of the member

9. Check the computed buckling resistance against the applied load. If the calculated value is inadequate or is too high, select another section and back to step-4.


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chapter 4 str 4 intr to str steel.doc

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