MITOITUS html.stx

Transkriptio

1 ELEMENT DESIGN MITOITUS html.stx 131 Kuva: TRY ry:n / Risto Liljan Oppimisympäristön mitoitusperusteet - Yleistä - Runkojärjestelmät - Palkit - Profiilipellit - Katto- ja seinäorret - Ohutuumapalkit - Ristikot - Pilarit - Jäykistys - Liitokset - Palosuojaus - Liittorakenteet - Tuotemallinnus ja suunnitelmat

2 13 Kuva: Määräykset ja ohjeet TRY ry:n / Risto Liljan Oppimisympäristö Kuva: Rajatilamitoitus TRY ry:n / Risto Liljan Oppimisympäristö

3 133 Kuva: Kuormat TRY ry:n / Risto Liljan Oppimisympäristö Kuva: Kuormat TRY ry:n / Risto Liljan Oppimisympäristö

4 134 Kuva: Kuormat TRY ry:n / Risto Liljan Oppimisympäristö Kuva: Materiaaliominaisuudet TRY ry:n / Risto Liljan Oppimisympäristö

5 135 Kuva: Poikkileikkausluokat TRY ry:n / Risto Liljan Oppimisympäristö Kuva: Mitoitusta TRY ry:n / Risto Liljan Oppimisympäristö

6 136 Kuva: Mitoitusta TRY ry:n / Risto Liljan Oppimisympäristö Kuva: Mitoitusta TRY ry:n / Risto Liljan Oppimisympäristö

7 137 Kuva: Mitoitusta TRY ry:n / Risto Liljan Oppimisympäristö Kuva: Palkin mitoitusta TRY ry:n / Risto Liljan Oppimisympäristö

8 138 Opiskelija perehtyy teräsrakenteiden mitoitukseen. ks ESDEP oppimisympäristö: suomenkielinen versio DESIGN OF ELEMENTS Course Contents ELEMENTS Lecture 7.1 : Methods of Analysis of Steel Structures Lecture 7. : Cross-Section Classification Lecture 7.3 : Local Buckling Lecture : Tension Members I Lecture 7.4. : Tension Members II Lecture : Columns I Lecture 7.5. : Columns II Lecture 7.6 : Built-up Columns Lecture 7.7 : Buckling Lengths Lecture : Restrained Beams I Lecture 7.8. : Restrained Beams II Lecture : Unrestrained Beams I Lecture 7.9. : Unrestrained Beams II Lecture : Beam Columns I Lecture : Beam Columns II

9 139 RAKENNUSTEKNIIKKA Lecture : Beam Columns III Lecture 7.11 : Frames Lecture 7.1 : Trusses and Lattice Girders

10 140 RAKENNUSTEKNIIKKA Workgroup Contents Lecture 7. : Cross-Section Classification Top 1. INTRODUCTION. REQUIREMENTS FOR CROSS-SECTION CLASSIFICATION 3. CRITERIA FOR CROSS-SECTION CLASSIFICATION 4. CONCLUDING SUMMARY 5. REFERENCES 6. ADDITIONAL READING

11 Previous Next Contents ESDEP WG 7 ELEMENTS Lecture 7.: Cross-Section Classification OBJECTIVE To describe the classification of cross-sections and explain how this controls the application of the methods of analysis given in Eurocode 3 [1]. PREREQUISITES Lecture 7.1: Methods of Analysis of Steel Structures RELATED LECTURES Lecture 7.3: Local Buckling Lectures & Lecture 7.5.: Columns Lectures 7.8: Restrained Beams Lectures 7.9: Unrestrained Beams Lectures 7.10: Beam Columns Lecture 7.11: Frames Lecture 14.10: Simple Braced Non-Sway Multi-Storey Buildings RELATED WORKED EXAMPLES Worked Example 7.1: Cross-Section Classification SUMMARY The analysis methods used are primarily dependent upon the geometry of the crosssection and especially on the width-thickness ratio of the elements which make it up. The lecture describes how sections are classified as plastic, compact or semi-compact and gives the limiting proportions of the elements by which these classifications are made. 141

12 1. INTRODUCTION When designing a structure and its components, the designer must decide on an appropriate structural model. The choice of model effects: the analysis of the structure, which is aimed at the determination of the stress resultants (internal forces and moments), and the calculation of the cross-section resistance. Thus a model implies the use of a method of analysis combined with a method of cross-section resistance calculation. There are several possible combinations of methods of analysis and methods of cross-section calculation, for the ultimate limit state, involving either an elastic or plastic design approach; the possible combinations are listed in Table 1. Table 1 Ultimate Limit State Design - Definition of Design Models 14 Model I II III IV Method of Global Analysis (Calculation of internal forces and moments) Plastic Elastic Elastic Elastic Calculation of Member Cross- Section Resistance Plastic Plastic Elastic Elastic Plate Buckling Model I is related to plastic design of structures. Full plasticity may be developed within cross-sections, i.e. the stress distribution corresponds to a fully rectangular block, so that plastic hinges can form. These have suitable moment rotation characteristics giving sufficient rotation capacity for the formation of a plastic mechanism, as the result of moment redistribution in the structure. For a structure composed of sections which can achieve their plastic resistance, but have not sufficient rotation capacity to allow for a plastic mechanism in the structure, the ultimate limit state must refer to the onset of the first plastic hinge. Thus, in Model II, the internal forces are determined using an elastic analysis and are compared to the plastic capacities of the corresponding cross-sections. For statically determinate systems, the onset of the first plastic hinge produces a plastic mechanism; both methods I and II should thus give the same result. For statically indeterminate structures, Model II, in contrast to Model I, does not allow moment redistribution. When the cross-sections of a structure cannot achieve their plastic capacity, both analysis and verification of cross-sections must be conducted elastically. The ultimate

13 limit state, according to Model III, is achieved when yielding occurs at the most stressed fibre. Sometimes yielding in the extreme fibre cannot even be attained because of premature plate buckling of one component of the cross-section; in such cases, the above ultimate limit state should apply only to effective cross-sections (Model IV). It is obviously not possible to have a model where a plastic method of analysis is combined with an elastic cross-section verification. Indeed, the moment redistribution which is required by the plastic analysis cannot take place without some cross-sections being fully yielded.. REQUIREMENTS FOR CROSS-SECTION CLASSIFICATION In the previous section, the models are defined in terms of structural design criteria; these are actually governed by conditions related to stability problems. Plastic redistribution between cross-sections and/or within cross-sections can take place provided that no premature local buckling occurs, as this would cause a drop-off in load carrying capacity. It must be guaranteed that no local instability can occur before either the elastic (Model III), or the plastic (Model II), bending resistance of the cross-section, or the formation of a complete plastic mechanism (Model I), is achieved. Such a mechanism, as envisaged by Model I, can occur provided that the plastic hinge, once formed, has the rotational capacity required for the formation of a plastic mechanism. To ensure sufficient rotation capacity, the extreme fibres must be able to sustain very large strains without any drop-off in resistance. In tension, the usual steel grades have sufficient ductility to allow for the desired amount of tensile strains; in addition, no drop-off is to be feared before the ultimate tensile strength is reached. With compressive stresses, however, it is not so much a question of material ductility, as of ability to sustain these stresses without instability occurring. Table gives a summary of the requirements for cross-sections in terms of behaviour, moment capacity and rotational capacity. As can be seen from this table, the limits are referred to cross-section classes, according to Eurocode 3 [1], each corresponding to a different performance requirement: Class 1 Plastic cross-sections: those which can develop a plastic hinge with sufficient rotation capacity to allow redistribution of bending moments in the structure. Class Compact cross-sections: those which can develop the plastic moment resistance of the section but where local buckling prevents rotation at constant moment in the structure. Class 3 Semi-compact cross-sections: those in which the stress in the extreme fibres should be limited to yield because local buckling would prevent development of the plastic moment resistance of the section. 143

14 Class 4 Slender cross-sections: those in which yield in the extreme fibres cannot be attained because of premature local buckling. Table Cross-section requirements and classification 144 Model of Behaviour Moment Resistance Rotation Capacity Class M Plastic moment on full section M Mpl Sufficient fy Mpl Local buckling 1 1 φ 1 φrot φpl φ φpl M Plastic moment on full section M Mpl Limited fy Mpl Local buckling 1 φ 1 φ φpl M Elastic moment on full section M Mpl None fy Mpl Mel Local buckling 1 Mal Mpl 3 φ 1 φ φpl M Elastic moment on effective section M Mpl None fy Mpl Mel Local buckling 1 4 φ 1 φ φpl The moment resistances for the four classes defined above are: for Classes 1 and : the plastic moment (M pl = W pl. f y ) for Class 3: the elastic moment (M el = W el. f y ) for Class 4: the local buckling moment (M o < M el ). The response of the different classes of cross-sections, when subject to bending, is usefully represented by dimensionless moment-rotation curves. The four classes given above are recognised for beam sections in bending. For struts loaded in axial compression, Classes 1, and 3 become one, and, in the absence of overall buckling are referred to as "compact"; in this case Class 4 is referred to as "slender".

15 3. CRITERIA FOR CROSS-SECTION CLASSIFICATION The classification of a specific cross-section depends on the width-to-thickness ratio, b/t, of each of its compression elements. Compression elements include any component plate which is either totally or partially in compression, due to axial force and/or bending moment resulting from the load combination considered; the class to which a specified cross-section belongs, therefore, partly depends on the type of loading this section is experiencing. a. Components of cross-section A cross-section is composed of different plate elements, such as web and flanges; most of these elements, if in compression, can be separated into two categories: internal or stiffened elements: these elements are considered to be simply supported along two edges parallel to the direction of compressive stress. outstand or unstiffened elements; these elements are considered to be simply supported along one edge and free on the other edge parallel to the direction of compressive stress. These cases correspond respectively to the webs of I-sections (or the webs and flanges of box sections) and to flange outstands (Figure 1). 145 b. Behaviour of plate elements in compression

16 For a plate element with an aspect ratio, α = a/b (length-to-width), greater than about 0,8, the elastic critical buckling stress (Euler buckling stress) is given by: 146 σ cr = k σ (1) where k σ is the plate buckling factor (see below), υ Poisson's coefficient, E Young's modulus. The critical buckling stress is proportional to (t/b) and, therefore, is inversely proportional to (b/t). The plate slenderness, or width-to-thickness ratio (b/t), thus plays a similar role to the slenderness ratio (L/i) for column buckling. In accordance with the definition of Class 3 sections, the proportions of the plate element, represented by the b/t ratio, must be such that σ cr would exceed the material yield strength f y so that yielding occurs before the plate element buckles. The ideal elastic-plastic behaviour of a perfect plate element subject to uniform compression may be represented by a normalised load-slenderness diagram, where the normalised ultimate load: = σ u /f y and the normalised plate slenderness: p = are plotted as ordinates and abscissae respectively (Figure ).

17 147 For p < 1, = 1 which means that the plate element can develop its squash load σ u = f y. For p > 1, decreases as the plate slenderness increases, σ u being equal to σ cr. Substituting the Equation (1) value for σ cr into the above and taking υ = 0,3 gives: p = () This expression is quite general as loading, boundary conditions and aspect ratio all influence the value of the buckling factor k σ. The factor k σ is a dimensional elastic buckling coefficient, depending on edge support conditions, on type of stress and on the ratio of length to width (a/b), aspect ratio, of the plated element. In general, the plated elements of a section have an aspect ratio much larger than unity and most of them are submitted to uniform compression. For such cases, Table 3 gives buckling factors for plated elements having various long edge conditions. Table 3 Elastic buckling factor k s Support conditions at long edges Clamped + clamped Clamped + simply supported Simply supported + simply supported Clamped + free Simply supported + free Free + free Buckling factor kσ 6,97 5,40 4,00 1,8 0,43 (b/a) a b Various support conditions a/b >> 1

18 When plated elements of sections are submitted to any kind of direct stress, other than uniform compression (e.g. webs of a girder in bending), the buckling factor k σ has to be modified to take account of the stress gradient, given by the stress ratio, ψ. Table 4 gives the buckling factors for different stress ratios ψ, for internal or outstand elements. In the latter case a distinction is made for elements with tip in compression or in tension. Table 4 Buckling factors and stress distribution 148 I II III σ1 σ σ1 σ σ σ1 σ1 = maximum compressive stress and s1 is positive ψ = σ σ1 ψ = σ / σ ψ 0 0 > ψ > > > 0 Case I internal element 4,0 8,4 1,1 + ψ 7,64 7,64-6,6 ψ + 10ψ 3,9 Buckling factor kσ Case II outstand element 0,43 0,578 ψ + 0,34 1,70 1,7-5 ψ + 17,1ψ 3,8 Case III outstand element 0,43 0,57-0,1 ψ + 0,07ψ 0,57 0,57-0,1 ψ + 0,07ψ 0,85 c. Limit plate element slendernesses The actual behaviour is somewhat different from the ideal elastic-plastic behaviour represented in Figure because of: i. initial geometrical and material imperfections, ii. strain-hardening of the material, iii. the postbuckling behaviour. Initial imperfections result in premature plate buckling, which occurs for p < 1. The corresponding limit plate slenderness p3, for Class 3 sections, may differ substantially from country to country because of statistical variations in imperfections and in material properties which are not sufficiently well known to be quantified accurately; a review of the main national codes shows that it varies from

19 0,5 to 0,9 approximately. Eurocode 3 [1,] has adopted p3 = 0,74 as the limit plate slenderness of Class 3 compression elements and p3= 0,9 for elements in bending where the yield strength may be reached in the extreme fibre of the cross-section. For plate elements for which p < p3, no plate buckling can occur before the maximum compressive strength reaches the yield strength. A Class 1 section must develop a resistance moment equal to the plastic capacity of the section and must maintain this resistance through relatively large inelastic deformations. In order to fulfil these conditions without buckling, the entire plate element must be yielded and the material must be strained in the strain-hardening region (see Table ); this is only possible for elements with low reference slendernesses ( p < p1), see Figure. 149 On the basis of certain theoretical approaches [3, 4, 5] values of p1 between 0,46 and 0,6, are proposed in various standards. The difference can be explained in the choice of the amount of necessary rotation capacity. A value of p1= 0,6 corresponds to a limited rotation capacity which is estimated to be sufficient for usual plastic design (continuous beams, non-sway frames, etc.). In Eurocode 3 [1], the proposed value is: p1 = 0,5 A Class (or compact section) is one which can just reach its plastic moment resistance but has a rapid drop-off in resistances at that point (Table ). The plate element is yielded and the material strained in the plastic range; it occurs for elements with medium reference slendernesses p where: p1 < p < p3 In Eurocode 3 [1], the proposed value is p1 = 0,6. Using formula (), and the appropriate values of p and k σ, the limiting b/t ratios can be calculated. Table 5 gives some limiting value of b/t for the elements of the cross-section of a rolled I-profile in compression or bending.

20 150 Table 5 Maximum slenderness ratios for the elements of a rolled section in compression or in bending Element Class 1 Class Class 3 cross-section crosssection crosssection Formula k σ b*/t or d/t w Flange (1) (b*/t) 9ε 10ε 1ε 0,43 14ε (1) Web in compression d/t w 33ε 38ε 1 1,0 4ε Web in pure bending d/t w 7ε 83ε 5,4ε 3,9 14ε f y ε = ε 1,0 0,9 0,81 (1) In practice, b, the half-width of the flange is considered instead of b*. For this reason, the values given in the "Essentials of Eurocode 3" is b = 15 ε > b*. The most important limiting proportions of the elements of a cross-section, which enable the appropriate classifications to be made, are specified in Eurocode 3 [1]. Appendix 1 gives the limiting proportions for compression elements of Class 1 to 3. The limiting values of the width-to-thickness ratio (b/t) of the plate elements of sections apply to members in steel of a specific yield strength. In order to cover all grades of steel, Eurocode 3 presents local buckling data non-dimensionally, in terms of a reduction factor ε =, where 35 represents the yield stress of mild steel and f y that of the steel considered.

21 The various compression elements in a cross-section (such as a web or a flange) can, in general, be in different classes and a cross-section is normally classified by quoting the least favourable (highest) class of its compression elements. It is important, particularly in plastic design, that the sections selected for various members should be, in all cases, appropriate for the assumed mode of behaviour. When any of the compression elements of a cross-section fail to satisfy the limits given in Table 5 for Class 3, the section is classified as "slender" and local buckling shall be taken into account in the design. This may be done by means of the effective cross-section method which is discussed in detail in Lecture CONCLUDING SUMMARY The methods of analysis used are influenced by the geometry of the cross-sections and, more particularly, by the width to thickness ratios of the plate elements in compression. It must be guaranteed that no local instability can occur before a complete mechanism is achieved or before the plastic or elastic moment can be reached. Four cross-sectional classes are identified, each corresponding to a different performance requirement: plastic, compact, semi-compact and slender cross-sections. Limiting proportions for the elements of a cross-section, which enable the appropriate classifications to be made, are given in the lecture. When any of the compression elements of a cross-section fail to satisfy the limiting proportions for Class 3 (semi-compact), local buckling shall be taken into account in the design. 151

22 15 3. DESIGN OF MEMBERS 3.1 Columns in Compression As shown by the effective cross-sections 1 and, in Figure 4, the neutral axes of doubly-symmetrical cross-sections will not change with the formation of effective holes. Hence the compression load N Sd that is central to the gross cross-sections will also be central to the effective cross-section. The column buckling design check, therefore, is based on the non-dimensional slenderness =, where N cr is calculated on the basis of the gross crosssection and N pl is calculated using the effective cross-sectional area A eff (N pl = A eff. f y ). The design buckling resistance is given by: N brd = χ. N pl /γ M where χ is the reduction factor for the relevant buckling curve. For singly symmetrical cross-sections - type 3 in Figure 4 - or unsymmetrical cross-sections, the formation of effective holes may lead to a shift, e N, in the neutral axes position. The compression load, N Sd, that is central to the gross cross-section will, therefore, be eccentric to the effective cross-section and hence will cause an additional bending moment M=N Sd.e N. The member is now a beam-column and must be checked in accordance with Section Beams in Bending Beams must be checked using the section modulus determined for the effective cross-sections, as given in Figure 5. In general the attainment of the yield strength at the compression face will limit the design bending resistance of effective crosssections: M o,rd = W eff. f y /γ M1 For cross-sections similar to type 3 of Figure 5, W eff will be determined for the actual edge (e): W eff = and not the edge of the effective hole.

23 153 RAKENNUSTEKNIIKKA The lateral-torsional buckling check for beams is analogous to that for columns. The non-dimensional slenderness LT =, is calculated with M u =W eff.f y for the effective cross-section, and with M cr calculated for the gross cross-sectional values. The design lateral-torsional buckling resistance is then given by: M b,rd = χ LT. M u /γ M1 where χ LT is the reduction factor for the relevant buckling curve. 3.3 Beam-Columns In the case of members that are subject to compression and monoaxial or biaxial bending (e.g. in the case of columns with monosymmetrical or unsymmetrical crosssections) the design check is carried out using an interaction formula in which the checks for a centrally compressed column, for a beam with bending about the y-axis only, and for a beam with bending about the z-axis only, are combined. If lateral-torsional bucking is prevented the interaction formula is as follows: If lateral-torsional bucking can occur: where e Ny or e Nz are the eccentricities due to the shift of the neutral axis for compression only. The resistances N brd and N bz.rd are related to the case of central compression; M oy.rd and M by.rd are related to bending about the y-axis only; and M oz.rd is related to bending about the z-axis (weak axis) only. 4. CONCLUDING SUMMARY The design of members with Class 4 sections is carried out as for members with Class 3 sections (elastic analysis, elastic cross-sectional resistance limited by yielding in the extreme fibres) except that an effective cross-section (derived from gross sections with "effective holes", where buckles may occur) is used. The buckling checks for columns and the lateral-torsional buckling check for beams, requires the critical values, N cr and M cr, to be calculated using the gross cross-sectional data without considering "effective holes".

24 In the case of columns with non-doubly symmetric cross-sections, the formation of "effective holes" may cause a shift in the neutral axis position resulting in eccentric compression and hence a beam-column problem. Beam-columns (compression and biaxial bending) are verified by using an interaction formula where the checks for the column in compression only, for the beam with bending about the y-axis only, and for the beam with bending about the z-axis only, are combined. 5. REFERENCES [1] "Eurocode 3: "Design of steel structures" ENV : Part 1.1, General rules and rules for buildings, CEN, ADDITIONAL READING 1. "Eurocode 3: Part 1.3: "Cold formed thin gauge members and sheeting", CEN, (in preparation).. Eurocode 3: Background Document 5.5. (Justification of the design resistances, for buckling verifications) 154

25 3. 5. Basis of the ECCS Buckling Curves From 1960 onwards, an international experimental programme was carried out by the ECCS to study the behaviour of standard columns []. More than 1000 buckling tests, on various types of members (I, H, T, U, circular and square hollow sections), with different values of slenderness (between 55 and 160) were studied. A probabilistic approach, using the experimental strength, associated with a theoretical analysis, showed that it was possible to draw some curves describing column strength as a function of the reference slenderness. The imperfections which have been taken into account are: a half sine-wave geometric imperfection of magnitude equal to 1/1000 of the length of the column; and the effect of residual stresses relative to each kind of cross-section. The European buckling curves (a, b, c or d) are shown in Figure 14. These give the value for the reduction factor χ of the resistance of the column as a function of the reference slenderness for different kinds of cross-sections (referred to different values of the imperfection factor α). 155 The mathematical expression for χ is: χ = 1/ {φ + [φ ] 1/ } 1 (10) where: φ = 0,5 [1 + α ( - 0,) + ] (11) Table 1 gives values of the reduction factor χ as a function of the reference slenderness. The imperfection factor α depends on the shape of the column cross-section considered, the direction in which buckling can occur (y axis or z axis) and the fabrication process used on the compression member (hot-rolled, welded or coldformed); values for α, which increase with the imperfections, are given in Table.

26 Curve a represents quasi perfect shapes: hot-rolled I-sections (h/b > 1,) with thin flanges (t f 40mm) if buckling is perpendicular to the major axis; it also represents hot-rolled hollow sections. Curve b represents shapes with medium imperfections: it defines the behaviour of most welded box-sections; of hot-rolled I-sections buckling about the minor axis; of welded I-sections with thin flanges (t f 40mm) and of the rolled I-sections with medium flanges (40 < t f 100mm) if buckling is about the major axis; it also concerns cold-formed hollow sections where the average strength of the member after forming is used. Curve c represents shapes with a lot of imperfections: U, L, and T shaped sections are in this category as are thick welded box-sections; cold-formed hollow sections designed to the yield strength of the original sheet; hot-rolled H-sections (h/b 1, and t f 100mm) buckling about the minor axis; and some welded I-sections (t f 40mm buckling about the minor axis and t f > 40mm buckling about the major axis). Curve d represents shapes with maximum imperfections: it is to be used for hotrolled I-sections with very thick flanges (t f > 100mm) and thick welded I-sections (t f > 40mm), if buckling occurs in the minor axis. Table 4 helps the selection of the appropriate buckling curve as a function of the type of cross-section, of its dimensional limits and of the axis about which buckling can occur. For cold-formed hollow sections, f yb is the tensile yield strength and f ya is the average yield strength. If the cross-section in question is not one of those described, it must be classified analogously. It is important to note that the buckling curves are established for a pin-ended, end loaded member; it is necessary carefully to evaluate the buckling lengths if the boundary conditions are different, see Lecture Equivalent Initial Bow Imperfection To study a column using second order theory, it is necessary to choose geometrical imperfections (initial out-of-straightness and eccentricities of loading) and mechanical imperfections (residual stresses and variations of the yield stress). Eurocode 3 proposes values for a bow imperfection, e o, whose effect is equivalent to a combination of the two previous kinds of imperfections [1]. If the column is designed using elastic analysis, e o is as follows: 156 e o = α ( - 0,) W pl /A for plastic design of cross-sections or, e o = α ( - 0,) W el /A for elastic design of cross-sections (1)

27 If it is designed with an elastic-plastic analysis (elasto-plastic or elastic-perfectly plastic), the values of e o are functions of the buckling length L, and are given in Table DESIGN OF BEAMS FOR SIMPLE BENDING For a doubly-symmetrical section or a singly-symmetrical section bent about the axis of symmetry, the basic theory of bending, assuming elastic behaviour, gives the distribution of bending stress shown in Figure Since the maximum stress σ max is given by: σ max = (1) where M is the moment at cross-section under consideration. d is the overall depth of section. I is the second moment of area about the neutral axis (line of zero strain).

28 it follows that limiting this to a fraction of the material yield stress gives a design condition of the form: W M/f d () 158 where W = f d is the limiting normal bending stress. When M is taken as the moment produced by the working loads, this approach to design is termed 'elastic' or 'permissible stress' design and is the method traditionally used in many existing codes of practice. In more modern Limit States codes f d is taken as the material strength f y possibly divided by a suitable material factor γ M and M is taken as the moment due to the factored loads i.e. the working loads suitably increased so as to provide a margin of safety in the design. Equation (), in this case, represents the condition of first yield. Values of W for the standard range of sections are available in tables of section properties. Selection of a suitable beam therefore comes down to: 1. determination of the maximum moment in the beam,. extraction of the appropriate value of f d from a suitable code, 3. selection of a section with an adequate value of W subject to considerations of minimum weight, depth of section, rationalisation of sizes throughout the structure, etc... Clearly, sections for which the majority of the material is located as far away as possible from the neutral axis will tend to be the most efficient in elastic bending. Figure gives some quantitative idea of this for some of the more common structural shapes. I-sections are most often chosen for beams because of their structural efficiency; being open sections they can also be connected to adjacent parts of the structure without undue difficulty. Figure 3 gives some typical examples of beam-to-column connections.

29 159 RAKENNUSTEKNIIKKA

30 Utilisation of the plastic part of the stress-strain curve for steel enables moments in excess of those which just cause yield to be carried. At full plasticity the distribution of bending stress in a doubly-symmetrical section will be as illustrated in Figure 4, with half the section yielding in compression and half in tension. 160 The corresponding moment is termed the fully plastic moment M pl. It may be calculated by taking moments of the stress diagram about the neutral axis to give: M pl = f y W pl (3) where f y is the material yield stress (assumed identical in tension and compression). W pl is the plastic section modulus. Basing design on Equation (3) means that the full strength of the cross-section in bending is now being used, with the design condition being given by: W pl M/f yd (4) where M is the moment at cross-section under consideration. f yd is the design strength (material yield strength divided by a suitable material factor). When M is due to the factored loads Equation (4) represents the design condition of ultimate bending strength used in Eurocode 3 for beams whose cross-sections meet at least the Class limits [1]. It is usual in codes such as Eurocode 3 for the value of f yd to be taken as the material yield strength, reduced slightly so as to cover possible variations from the expected value. For continuous (statically indeterminate) structures, attainment of M pl at the point of maximum moment will not normally imply collapse. Providing nothing triggers unloading at this point, e.g. local buckling does not occur, then the local rotational stiffness will virtually disappear, i.e. the cross-section will behave as if it were a hinge, and the pattern of moments within the structure will alter from the original elastic distribution as successive plastic hinges form. This redistribution of moments

31 will enable the structure to withstand loads beyond that load which produces the first plastic hinge, until eventually collapse will occur when sufficient hinges have formed to convert the structure into a mechanism, as shown in Figure 5. Utilisation of this property of redistribution is termed "plastic design". It can only be used for continuous structures and then only when certain restrictions on cross-sectional geometry, member slenderness, etc.., are observed. The topic is covered in detail in Lecture 7.8. in the context of beams and in Lecture 7.11 in the context of frames. 161

32 4. DESIGN OF BEAMS FOR SHEAR Although bending will govern the design of most steel beams, situations will arise, e.g. short beams carrying heavy concentrated loads, in which shear forces are sufficiently high for them to be the controlling factor. Figure 6 illustrates the pattern of shear stress found in a rectangular section and in an I-section assuming elastic behaviour. In both cases shear stress varies parabolically with depth, with the maximum value occurring at the neutral axis. However, for the I-section, the difference between maximum and minimum values for the web, which carries virtually the whole of the vertical shear force, is sufficiently small for design to be simplified by working with the average shear stress, i.e. total shear force/web area. Since the shear yield stress of steel is approximately 1/ 3 of its tensile yield stress, a suitable value for "permissible" shear stress when using elastic design is 1/ 3 of the permissible tensile stress. 16 The ultimate shear resistance (based on plastic principles) used in Eurocode 3 [1] is f yd / 3; this is used in conjunction with a shear area A v, examples of which are: rolled I-section, load parallel to web A v = A-bt f + (t w +r)t f plates and solid bars A v = A circular hollow sections A v = A/π In cases where high shear and high moment coexist, e.g. the internal support of a continuous beam, it may sometimes be necessary to allow for interaction effects. However, since the full shear capacity may be developed in the presence of quite large moments and vice versa (Eurocode 3 only requires reductions in moment resistance when the applied shear exceeds 50% of the ultimate shear resistance [1]) this will not often be required.

33 5. DEFLECTIONS Excessive deflections, whilst not normally leading to a structural failure, may, nonetheless, impair the serviceability of the structure. Deflection might, for example, cause: cracking of plaster ceilings. misalignment of crane rails. difficulty in opening large doors. Since these affect the performance of the structure in its working conditions, it is usual to conduct this type of check at service load levels. Eurocode 3 suggests a limit of span/300 for the maximum deflection of beams supporting floors under service live load [1]. Such limits should, however, be regarded as advisory only, and smaller or larger values can be more appropriate in any given situation. Wherever possible, specialist advice should be sought; crane manufacturers, for example, should be able to give detailed guidelines regarding permissable deflection for gantry beams. 7.3 Sway Frames Sway frames shall be analysed under those arrangements of the variable loads which are critical for failure in a sway mode. In addition, sway frames shall also be analysed for the non-sway mode. The initial sway imperfections, and member imperfections where necessary, shall be included in the global analysis of all frames. The allowance for imperfections in the analysis of sway frames is intended to cover effects such as lack of verticality, lack of straightness, residual stresses, etc. It is expressed in Eurocode 3 by means of a set of equivalent geometrical imperfections [1]. These imperfections are not actual construction tolerances but, because they are intended to represent the effect of a number of factors, are likely to be larger than such tolerances. The form specified in Eurocode 3 is: The effects of imperfections shall be allowed for in frame analysis by means of an equivalent geometric imperfection in the form of an initial sway imperfection φ determined from: 163 φ = k c k s φ o with φ o = 1/00 k c = [0,5 + 1/n c ] 0,5 but k c 1,0 and k s = [0, + 1/n s ] 0,5 but k s 1,0 where n c is the number of columns per plane.

34 n s is the number of storeys. Columns which carry a vertical load N Sd of less than 50% of the mean value of the vertical load per column in the plane considered, shall not be included in n c. Columns which do not extend through all the storeys included in n s shall not be included in n c. Those floor levels and roof levels which are not connected to all the columns included in n c shall not be included when determining n s. These initial sway imperfections apply in all horizontal directions, but need only be considered in one direction at a time. The possible torsional effects on the structure of anti-symmetric sways, on two opposite faces, shall also be considered. If more convenient, the initial sway imperfection may be replaced by a closed system of equivalent horizontal forces, see Figure 7. In beam-and-column building frames, these equivalent horizontal forces should be applied at each floor and roof level and should be proportionate to the vertical loads applied to the structure at that level, see Figure 8. The horizontal reactions at each support should be determined using the initial sway imperfection and not the equivalent horizontal forces. In the absence of actual horizontal loads, the net horizontal reaction is zero. 164

35 First-order or second-order analysis may be used. If the analysis is first-order, second-order effects may be allowed for in an appropriate way when designing the columns by using the results of a first-order analysis and either: - using amplified sway moments, or - using the sway-mode buckling lengths. When second-order elastic global analysis is used, in-plane buckling lengths for the non-sway mode may be used for member design. In the amplified sway moments method, the sway moments found by a first-order elastic analysis should be increased by multiplying them by the ratio: 165 where V Sd is the design value of the total vertical load. V cr is its elastic critical value for failure in a sway mode. The amplified sway moments method should not be used when the elastic critical load ratio V Sd /V cr is more than 0,5. Sway moments are those associated with the horizontal translation of the top of a storey relative to the bottom of that storey. They arise from horizontal loading and may also arise from vertical loading if either the structure or the loading is asymmetrical. As an alternative to determining V Sd /V cr directly, the following approximation may be used in beam-and-column type frames: where δ, h, H and V are as defined previously. When the amplified sway moments method is used, in-plane buckling lengths for the non-sway mode may be used for member design. When first-order elastic analysis with sway-mode in-plane buckling lengths is used for column design, the sway moments in the beams and the beamto-column connections should be amplified by at least 1, unless a smaller value is shown by analysis to be adequate. Rules for the application of plastic analysis procedures to sway frames are given in Clause of Eurocode 3 [1].

36 166 Lecture 7.1: Trusses and Lattice Girders OBJECTIVE/SCOPE To introduce two-dimensional trusses: types, uses and principal design considerations. PREREQUISITES None. RELATED LECTURES None. SUMMARY This lecture presents the types and uses of trusses and lattice girders and indicates the members that are most often used in their construction. A discussion of overall truss design considers primary analysis, secondary stresses, rigorous elastic analysis, cross- braced trusses and truss deflections. The practical design of truss members is discussed.

37 INTRODUCTION The truss or lattice girder is a triangulated framework of members where loads in the plane of the truss or girder are resisted by axial forces in the individual members. The terms are generally applied to the planar truss. A 'space frame' is formed when the members lie in three dimensions. The main uses are: in buildings, to support roofs and floors, to span large distances and carry relatively light loads, see Figure 1. in road and rail bridges, for short and intermediate spans and in footbridges, as shown in Figure. as bracing in buildings and bridges, to provide stability where the bracing members form a truss with other structural members such as the columns in a building. Examples are shown in Figure 3.

38 168 The principle of a truss is simple. The structure is composed of top and bottom chords triangulated with diagonals in the webs so that each member carries purely axial load. Additional effects do exist but in a well designed truss these will be of a secondary nature. A global moment on a truss is carried as compression and tension in the chords. A global shear is carried as tension or compression in the diagonal members. In the simplified case, where joints are considered as pinned, and the loads are applied at the panel points, the loading creates no bending moment, shear, or torsion in any

39 single member. Loads applied in such a way as to cause bending, shear, or torsion will usually result in inefficient use of material. Trusses and lattice girders are classified in accordance with the overall form and internal member arrangement. Pitched trusses are used for roofs. Parallel chord lattice girders are used to support roofs and floors and for bridges, although in continuous bridges, additional depth is often required at the piers. In the past, proper names were given to the various types of trusses such as the Fink truss, Warren girder, etc.. The most commonly used truss is single span, simply supported and statically determinate with joints assumed to act as pins. The Vierendeel girder should also be mentioned. It consists of rigid jointed rectangular panels as shown in Figure 1d. This truss is statically indeterminate and will not be further considered in this lecture, although it has a pleasing appearance and is often used in foot-bridges. The saving over a plate girder is clear when the webs are considered. In a truss the webs are mainly fresh air - hence less weight and less wind force. A truss can be assembled from small easily handled and transported pieces, and the site connections can all be bolted. Trusses can have a particular advantage for bridges in countries where access to the site is difficult or supply of skilled labour is limited.. TYPICAL MEMBERS Truss, lattice girder and bracing members for buildings are selected from: open sections, primarily angles, channels, tees and joists. compound sections, i.e. double angle and channels. closed sections, in practice structural hollow sections. For bridges, members are selected from: rolled sections. compound sections. built-up H, top hat and box sections. Typical sections are shown in Figure

40 170 The selection of members depends on the location, use, span, type of connection and the appearance required. Hollow sections are more expensive than open sections but are cheaper to maintain and have a better appearance. However, in exposed trusses corrosion can occur at the crevices which are formed at gusset positions. Angles are the sections traditionally used for small span truss construction.

41 LOADS ON TRUSSES AND LATTICE GIRDERS The main types of loads on buildings are shown in Figure 5, namely: 1. Dead loads. These are caused by self-weight, sheeting, decking, floor or roof slabs, purlins, beams, insulation, ceilings, services and finishes. Dead loads for the construction to be used in any particular case must be

42 carefully estimated from material weights given in handbooks and manufacturers' literature.. Imposed loads. These are given in Eurocode 1 [1] for floors in various types of building and for roofs with or without access. The imposed load may cover the whole or part of the member and should be applied in such a way as to cause the most severe effect. 3. Wind loads. These are given in Eurocode 1 [1] and can be estimated from the location of the building, its dimensions and the sizes of openings on its faces. Wind generally causes uplift on roofs and this can lead to reversal of load in truss members in light construction. In multi-storey buildings, wind gives rise to horizontal loads that must be resisted by the bracing. In special cases, trusses resist dynamic, seismic and wave loads. A careful watch should be kept for unusual loads applied during erection. Failures may occur at this stage when the final lateral support system is not fully installed. For bridges, in addition to the dead loads and the vertical effects of live loads due to highway or railway loading, horizontal effects of live load have to be considered. These include braking and traction effects, centrifugal loads and accidental skidding loads. Temperature effects are significant in some bridges. 4. ANALYSIS OF TRUSSES 4.1 General Trusses may be single span, statically determinate or indeterminate, or may be continuous over two or more spans, as shown in Figure 6. Only single span, statically determinate, trusses are considered in this section. 17 A truss is usually statically determinate when: m = j - 3, where m is the number of members in a truss

43 j is the number of joints. However compliance with this formula for the truss as a whole does not preclude the possibility of a local mechanism in part of the truss. Manual methods of analysis for trusses, where the loads are applied at the nodes, are joint resolution, the method of sections and the force diagram. Joint resolution is the quickest method for analysing parallel chord lattice girders when all the forces are required. The method of sections is useful where the values of the forces in only a few critical members are required. The force diagram is the best general manual method. Computer programs are also available for truss analysis. 4. Secondary Stresses in Trusses In many cases in the design of trusses and lattice girders, it is not necessary to consider secondary stresses. These stresses should, however, be calculated for heavy trusses used in industrial buildings and bridges. Secondary stresses are caused by: Eccentricity at connections Loads applied between the truss nodes Moments resulting from rigid joints and truss deflection. They are discussed in detail below: 1. Eccentricity at connections 173 Trusses should be detailed so that either the centroidal axes of the members or the bolt gauge lines meet at a point at the nodes. Otherwise, members and connections should be designed to resist the moments due to eccentricity. These moments should be divided between members meeting at joints in proportion to their rotational stiffnesses. Stresses due to small eccentricities are often neglected.. Loads applied between the truss nodes Moments due to these loads must be calculated and the stresses arising combined with those due to primary axial loads; that is the members concerned must be designed as beam-columns. This situation often occurs in roof trusses where the loads are applied to the top chord through purlins which may not be located at the nodes, as shown in Figure 7. The manual method of calculation is first to analyse the truss for the loads applied at the nodes which gives the axial forces in the members. Then a separate analysis is made for bending in the top chord which is considered as a continuous beam. The ridge joint E is fixed because of symmetry, but the eaves joint A should be taken as pinned; otherwise, moment will be transferred into the bottom chord if the joint between the truss and column is assumed to be

44 pinned. The top chord is designed for axial load and bending. Computer analysis is mentioned below Moments resulting from rigid joints and truss deflection Stresses resulting from secondary moments are important in trusses with short thick members. Approximate rules specify when such an analysis should be made. Secondary stresses will be insignificant if the slenderness of the chord members in the plane of the truss is greater than 50 and that of most of the web members is greater than 100. In trusses in buildings, the loads are predominantly static and it is not necessary to calculate these stresses. The maximum stresses from secondary moments occur at the ends of members and are not likely to cause collapse. However, where fatigue effects are significant, these secondary stresses have to be considered. The method of analysis for secondary moments is set out below. 4.3 Rigorous Elastic Analysis Rigid jointed, redundant or continuous trusses or trusses with loads applied between the nodes can be analysed using a plane frame program based on the matrix stiffness method of frame analysis. The truss can also be modelled taking account of joint eccentricity. Member sizes must be determined in advance using a manual analysis. All information required for design is output including joint deflections. It is important that a consistent approach is adopted for analysis and design. This means that if secondary moments are to be ignored then the primary axial forces to be used in design must be obtained from the simple analysis of the truss as a pin jointed frame. The axial forces obtained from a rigid frame computer analysis may be modified considerably by the joint moments.

45 5.4 Design Steps for Compression Members To design a simple compression member it is first necessary to evaluate its two effective lengths, in relation to the two principal axes, bearing in mind the expected connections at its ends. Secondly, the required second moment of area to resist the Euler critical loads should be calculated to give an idea of the minimum cross-section necessary. The verification procedure should then proceed as follows: the geometric characteristics of the shape, and its yield strength give the reference slenderness (Equation (9)). χ is calculated, taking into account the forming process and the shape thickness, using one of the buckling curves and (Equations 10 and 11). Buckling resistance of a compression member is then taken as: N b.rd = χ A f y /γ M1 (13) the plastic resistance as: N pl.rd = A f y /γ M0 (14) and the local buckling resistance as: N o.rd = A eff f y /γ M1 (15) If these are higher than the design axial load, the column is acceptable; if not, another larger cross-section must be chosen and checked. In addition, torsional or flexural-torsional buckling of the column must be prevented. 6. CONCLUDING SUMMARY o Many different kinds of cross-sections are used as compression members; these include simple members, built-up, tapered and stepped columns. o A stub column (with 0,) can achieve the full plastic resistance of the cross-section and buckling does not need to be checked. 175 o If > 0,, reduction of the load resistance must be considered because of buckling. Columns with medium slenderness fail by inelastic buckling and slender columns by elastic buckling. o European buckling curves give the reduction factor for the relevant buckling mode depending on the shape of the cross-section, the forming-process, the reference slenderness and the axis about which buckling can occur. They take into account experimental and theoretical approaches and give reliable results.

46 176 LASKUESIMERKKEJÄ: Saarinen, Eero. Teräsrakenteiden suunnittelu, Eurocode. Sähköinen versio. Teräsrakenneyhdistys r.y. 00. Julkaisematon Saarinen, Eero. Teräsrakenteiden suunnittelu, Eurocode. Sähköinen versio. Teräsrakenneyhdistys r.y. 00. Julkaisematon Esimerkki 3.1 Määritä kuvan 3.1 vedetyn sauvan kestävyys, kun teräs on S355 ja levyn paksuus t = 10 mm. On tutkittava leikkaukset I - I ja II - II: A I = 10 mm (10-18) mm = 100 mm A II = 10 mm ( /(4 60)) mm = 944 mm Tässä tapauksessa A net = A II. γ M0 = 1,1 (poikkileikkausluokka 1) γ M = 1,5 A = 10 mm 10 mm = 100 mm N pl.rd = 100 mm 0,355 kn/mm /1,1 = 387 kn N u.rd = 0,9 944 mm 0,510 kn/mm /1,5 = 347 kn Vetokestävyys N t.rd = 347 Kn

47 177 Kuva 3.8. Vapaasti tuettu vesikaton pääkannattaja Esimerkki 3. Mitoitettava vapaasti tuettu vesikaton pääkannattaja (kuva 3.8), jonka jänneväli L = 10 m. Kehäväli on 6,0 m ja orsiväli L 1 =,0 m. Teräs S35JG3.. Palkin kuormitus: kate + eriste 0,30 kn/m orret 0,10 pysyvä kuorma yhteensä 0,40 kn/m lumikuorma 1,80 kn/m Ominaislumikuorma orrelta palkille F qk = 6,0 m,0 m 1,8 KN/m = 1,6 kn max V qk = 1,6 kn = 43, kn max M qk = (43, kn 5,0 m - 1,6 kn (3,0 + 1,0)m) = 130 knm Arvioidaan palkin omaksipainoksi g k 0,7 kn/m 1 max M k = 0, knm = 139 knm 8 Laskentakuormat: kn g d = 1, 0,7 kn/m = 0,84 kn/m F d = 1, 6,0 m,0 m 0,4 kn/m + 1,5 1,6 kn = 38,16 max V Sd = 1 0,84 kn/m 10 m + 38,16 kn = 80,5 kn

48 max M Sd = (80, , ,16 4) knm = 39 knm S35 f y = 35 N/mm (t < 16 mm) Poikkileikkausluokka 1 γ M0 = 1,1 6 11, W pl mm 3 = mm 3 35 Taipuman rajatila on L/50 (vrt. taulukko.4), sillä kokonaistaipuman rajatila antaa pienemmän tuloksen jäyhyysmomentille. k = 50 δ f = 5 48 I mm 4 = mm 4 (kaava 3.4) 48, 1 10 Vaaditun taivutusvastuksen ja jäyhyysmomentin perusteella valitaan IPE 400. Poikkileikkausarvot: I y = mm 4 W pl.y = mm 3 (W el = mm 3 ) i z = 39,5 mm t w = 8,6 mm h s = (400-13,5) mm = 387 mm r = 1 mm (pyöristyssäde) g k = 0,663 kn/m < 0,7 kn/m (oletus) 180mm 13, 5mm + 8, mm 13, 5mm = 473 mm A v = 8450mm - ( ) Leikkauskestävyys V pl.rd = 473mm λ LT = , / 39, / 13, 5 0, 35kN / mm 11, 3 = 47,3 (kaava 3.30) 178 = 57 kn > V Sd (kaava 3.7) λ 1 = 93,9 λ _ LT = 47, 3 = 0,504 ( kaava 3.6 kun β w = 1) 93, 9 χ LT = 0,99 (taulukosta 3.4 interpoloimalla) 3 3 0, 35kN / mm M b.rd = 0, mm = 58 knm > MSd 11, Yhteisvaikutusta ei tarvitse tarkistaa, koska normaalivoimaa ei esiinny ja taivutusmomentin sekä leikkausvoiman suurimmat arvot vaikuttavat eri poikkileikkauksissa. IPE 400 on siten sopiva profiili. Jos teräkseksi valitaan S355JG3, voidaan profiiliksi valita IPE 360.

49 179 Esimerkki : Vapaasti tuettu vesikaton pääkannattaja Mitoitettava vapaasti tuettu vesikaton pääkannattaja, jonka jänneväli L = 14 m. Kehäväli on 6,0 m ja orsiväli L1 =,0 m. Teräs S35JG3 kn := 1000N Kuva: Vapaasti tuettu vesikaton pääkannattaja. knm := 1000N m Palkin kuormitus: kate + eriste 0,30 kn/m orret 0,10 pysyvä kuorma yhteensä 0,40 kn/m lumikuorma 1,80 kn/m gk1 := qk1 := 0.40 kn m 1.8 kn m palkin omapaino 0,70 kn/m gk := kn 0.70 m Palkin geometria: jänneväli L = 10 m orsiväli L1 = m kehäväli kk = 6 m L := L1 := kk := 14 m m 6 m Voimasuureet: KÄYTTÖRAJATILA: Orren pistekuorma rakenteista: Fgk1 := kk L1 gk1 Fgk1 = 4.8kN Orren pistekuorma lumesta: Fqk1 := kk L1 qk1 Fqk1 = 1.6kN Palkin op:n metrikuorma: pgk := kn 0.70 m

50 Muuttuvan kuorman= lumikuorman aiheuttama voimasuureet käyttötilassa: 180 maxvqk1 := Fqk1 3 maxmq1:= 3 Fqk1 L ( Fqk1) ( 5 m + 3 m + 1 m) maxvqk1 = maxmq1 = 64.8kN 59.kNm Muuttuvien ja pysyvien kuormien aiheuttamat voimasuureet käyttötilassa: maxvk := 3( Fgk1 + Fqk1) + L pgk maxvk = 84.1kN maxmk := 3 ( Fgk1 + Fqk1) L MURTORAJATILA [( Fgk1 + Fqk1) ( 5 m + 3 m + 1 m) ] + maxmk = pgk L kNm γg := 1. γq := 1.5 Orren pistekuorma rakenteista: Orren pistekuorma lumesta: Palkin op:n metrikuorma: Fgd1 := Fqd1 := pgd := γg kk γq kk γg pgk L1 gk1 L1 qk1 Fgd1 = Fqd1 = 5.76kN 3.4kN pgd = kn 0.84 m Muuttuvien ja pysyvien kuormien aiheuttamat voimasuureet käyttötilassa: maxvsd:= 3( Fgd1 + Fqd1) + L pgd maxvsd = 10.36kN maxmsd:= 3 ( Fgd1 + Fqd1) L [( Fgd1 + Fqd1) ( 5 m + 3 m + 1 m) ] + maxmsd = pgd L kNm

51 MITOITUS Lujuudet S35 Taivutus PL1 Taipuma fy := 35 E := N mm N mm Msd := maxmsd γm := 1.1 fy fyd := γm Msd Wpltarv:= fyd fyd = N mm Wpltarv = cm 3 Taipuman rajatila on L/50 (vrt. taulukko.4), sillä kokonaistaipuman rajatila antaa pienemmän tuloksen jäyhyysmomentille. k = δf := 5 48 k := 50 Mk := maxmk Itarv := k δf Mk E L Itarv = cm 4 Vaaditun taivutusvastuksen ja jäyhyysmomentin perusteella valitaan IPE 550: Poikkileikkausarvot: A = 8450 mm Iy = mm4 Wpl.y= mm3 (Wel= mm3) iz = 43.1 mm tw = 11,1 mm tf = 17, mm hs = (400-13,5) mm = 387 mm r = 4 mm (pyöristyssäde) gk = 0,663 kn/m < 0,7 kn/m (oletus) h =550 mm b = 10 mm A := 8450 mm Wply := mm 3 iz := tw := tf := r := h := b := 43.1 mm 11.1 mm 17. mm 4 mm 550 mm 10 mm L1 = m

52 Leikkaus 18 Av := A b tf + ( tw + r) tf Av = 4.5mm Av := 1.04 h tw Av = 6349.mm VplRd:= Av fy γm 3 VplRd = > VSd kN C1 := 1 λlt := λl := π E fy ( λltmuunn ) C1 := ( ) 1 L1 iz + λlt λl 1 0 βw ( ) λlt = λl = λltmuunn= 0.47 φlt := αlt λltmuunn 0. + λltmuunn L1 iz h tf βw := 1 αlt := 0.1 φlt = χlt := 1 φlt+ φlt λltmuunn χlt = MbRd := fy χlt βw Wply γm MbRd = kNm >MSd Yhteisvaikutusta ei tarvitse tarkistaa, koska normaalivoimaa ei esiinny ja taivutusmomentin sekä leikkausvoiman suurimmat arvot vaikuttavat eri poikkileikkauksissa. IPE 550 on siten sopiva profiili.

53 Esimerkki 3.3 Kuvan 3.9 jatkuva palkki saa välipohjalta pysyvää kuormaa g 1k = 15,0 kn/m ja oleskelukuormaa q k = 7, kn/m. Mitoita palkki teräksestä S355JG Saarinen, Eero. Teräsrakenteiden suunnittelu, Eurocode. Sähköinen versio. Teräsrakenneyhdistys r.y. 00. Julkaisematon Kuva 3.3. Esimerkin 3.3 jatkuva palkki. Arvioidaan omaksipainoksi g 0k 0,4 kn/m g k = 15,4 kn/m Laskentakuormat: g d = 1, 15,4 kn/m = 18,5 kn/m q d = 1,5 7, kn/m = 10,8 kn/m Pysyvä kuorma g d kuormittavaa koko palkkia. Liikkuvan kuorman q d vaarallisimmat kuorma-asennot eri voimasuureille on esitetty kuvassa a) Lasketaan voimasuureet kimmoteorialla. Murtotilassa: max M 0-1 = [0,080 18,5 + 0,101 10,8] 6,0 knm = 9,5 knm(kuormitustap. 1) vast. M 1 = - [0,100 18,5 + 0,050 10,8] 6,0 knm = - 86,0 knm min M 1 = - [0,100 18,5 + 0,117 10,8] 6,0 knm = - 11 knm(kuormitustap. 3) vast. M = - [0,100 18,5 + 0,033 10,8] 6,0 knm = - 79,4 knm max M 1- = [0,05 18,5 + 0,075 10,8] 6,0 knm = 45,8 knm(kuormitustap. ) , 0, +,, kn kn = 73, 6kN 6, = ,,, kn kn 107kN 6, , 4, +,, kn + = 93, 3kN 6, 0 max V 0.1 = ( ) min V 1.0 = ( ) max V 1. = ( )

54 Käyttötilassa (kuormitustapaus 1): max M 0-1 = [0,080 15,4 + 0,101 7,] 6,0 knm = 70,5 knm vast. M 1 = - [0,100 15,4 + 0,050 7,] 6,0 knm = - 68,4 knm 5 δ f = = , 4 0, , 5 Poikkileikkauksen valinta: 6 11, W pl mm 3 = mm , 5 10 I 50 0, mm 4 = mm 4, 1 10 Valitaan IPE 70 (painoltaan pienin valssattu I-profiili, joka täyttää vaatimukset). Tarkistetaan leikkauskestävyys: A v = [ , + (6,6 + 15) 10,] mm = 13 mm V pl.rd = 13 mm 0 355, kn / mm = 41 kn > V Sd 11, 3 Kiepahduskestävyyttä kentässä ei tartvitse tarkistaa, koska välipohjalaatta tukee ylälaipan sivusuunnassa, mutta välitukien kohdalla tarkistus on tarpeen. Kokeillaan aluksi onko kiepahduskestävyys riittävä, jos alalaippa on tukematon. Tällöin kuorman vaikuttaessa ylälaipan tasossa saadaan kiepahduskestävyys laskettua seuraavasti: i z = 30, mm t f = 10, mm h s = (70-10,) mm = 59,8 mm W pl = mm 3 z g = 70 mm = 135 mm k = kw = 1,0 C 1 1,879 C = 0 λ LT = λ 1 = π , / 30, 1, / 10, = 74, = 76,4 λ _ 74, LT = 76, 4 1 = 0,971 χ LT = 0,685 M b.rd = 0, mm , kn / mm = 107 knm 11, Tämän likimääräisen tarkastelun perusteella välituilla tarvittaisiin jäykisteet, mutta tarkemmin laskettaessa ne todennäköisesti olisivat tarpeettomia. 184

55 185 RAKENNUSTEKNIIKKA Yhdistetyt rasitukset välituella: b) Voimasuureet plastisuusteorialla Mekanismi murtotilassa on kuvan 3.11 mukainen. Työyhtälöksi saadaan ( g d + qd ) L = M p + + ξ L ( 1 ξ) L ( 1 ξ) L josta ratkaisemalla seuraa täysplastiselle momentille M p lauseke 1 ξ( 1 ξ) M p = ( g d + qd ) L (3.33) 1+ ξ Parametrin ξ arvo, jolla M p saavuttaa suurimman arvonsa, voidaan ratkaista derivoimalla yhtälö 3.43 ja asettamalla derivaatta yhtäsuureksi kuin 0. M p 1 1 ξ ξ = ( gd + q d ) L = 0 ξ 1+ ξ ( ) ξ = 1 0, 414 Vaatimukseksi taivutuskestävyydelle saadaan siten 1 0, 414 ( 1 0, 414) M c.rd M p = ( 18, , 8) 6, 0 knm = 90,5 knm 1, , 5 10 W p mm 3 = mm IPE 0 riittäisi taivutuskestävyyden perusteella, mutta taipumarajatila on tässä tapauksessa määräävä. Sen vuoksi on valittava profiili IPE 70. Plastisuusteorian soveltamisesta ei siten tällä kertaa ollut saavutettavissa hyötyä. Palkin teräksen lujuusluokaksi riittäisi S35, mutta silloin alalaippa olisi jäykistettävä välitukien molemmin puolin.

56 Esimerkki 3.4 Mitoitetaan kaksitukinen vapaasti tuettu suora kattopalkki, jonka jänneväli L = 5,0 m. Kehäväli on 8,0 m ja orsiväli L 1 =,5 m. Teräs S355JG3. Katon kuormitus: kate + eriste 0,7 kn/m orret 0,1 pysyvä kuorma yhteensä 0,39 kn/m lumikuorma 1,80 kn/m Arvioidaan omaksi painoksi 1,7 kn/m Kuormat palkille: g k = 8,0 m 0,39 kn/m + 1,7 kn/m = 4,8 kn/m q k = 8,0 m 1,8 kn/m = 14,4 kn/m Käyttötilan voimasuureet , +, kn / m, m = kn max V k = ( ) max M k = 1 19, kn / m 5, 0 m = 150 knm Murtorajatilan voimasuureet 1 max V Sd = ( ) 5 0,, +,, kn / m, m = 34 kn 1 max M Sd = ( 1, 4, , 14, 4) kn / m 5, 0 m = 139 knm 8 Poikkileikkauksen valinta Koska rakennekorkeutta ei ole rajoitettu, valitaan optimikorkeuden perusteella (ks. Hitsatut profiilit käsikirja) mahdollisimman kevyt poikkileikkaus. 6 11, h opt = 7, 0 mm = 1315 mm 355 Valitaan palkki WI x300 Valitun palkin poikkileikkaus on esitetty kuvassa 3.19 ja sen poikkileikkaussuureilla on seuraavat arvot: h f = h s = ( ) mm = 185 mm h w = d = ( ) mm = 170 mm (hoikkuutta määritettäessä ollaan varmalla puolella kun hitsejä ei oteta huomioon) A = ( ) mm = 1700 mm I y = ( / / ,5 )mm 4 = m 4

57 187 Kuva Esimerkin 3.4 hitsatun palkin poikkileikkaus. Saarinen, Eero. Teräsrakenteiden suunnittelu, Eurocode. Sähköinen versio. Teräsrakenneyhdistys r.y. 00. Julkaisematon Tehollinen poikkileikkaus Kyseessä on poikkileikkausluokka 4, sillä uuma ei täytä poikkileikkausluokka 3:n ehtoa: d/t w = 17 > 14 ε = 100 Pristetulle laipalle k σ = 4,0 ja ε = 0,81 (S355) λ _ 150 / 15 p = 8, 4 0, 81 4 = 0,17 < 0,673 ρ = 1 b eff = b f = 300 mm Uumalle ψ = -1 k σ = 3,9 λ _ 170 / 10 p = = 1,19 > 0,673 8, 4 0, 81 3, 9 ( 119, 0, ) ρ = = 0,713 d 119, eff = 0, mm = 453 mm d e1 = 0,4 453 mm = 181 mm d e = ( ) mm = 7 mm tehoton alue a = ( )mm = 18 mm (ks. kuva 3.0) Tehollinen pinta-ala: A eff = 1700 mm - 10 mm 18 mm = mm 1 e t = ( ) 19880, +, + +, mm = 64 mm e c = ( )mm = 676 mm Neutraaliakseli laskee matkan e = (650-64)mm = 6 mm I eff = ( 676 7, 5) ( , 5) W eff = W ec = ( ) ( ) ] +, mm 4 = mm mm 3 = mm 3

58 188 Kuva 3.0. Tehollinen poikkileikkaus. Tarkistetaan laipan taipumisesta johtuva lommahdus. Hoikkuussuhteen raja-arvo lasketaan kaavasta c 150 = = 10 < 14 0, 81 = 11, 34 t f 15 puristettu laippa kuuluu poikkileikkausluokkaan 3 k = 0,55 d , = 547 t w d = 170 = 17 < 441 t w 10 Taivutuskestävyys Taivutuskestävyys M c.rd = mm 3 3 0, 355kN / mm 11, = 469 knm > M Sd Kiepahduskestävyys I z = mm 4 = mm I w = mm mm = 7, mm , , 5 e pl = mm = 45,3 mm W pl = 45,3 mm 1700 mm = mm 3 1 I t = ( ) mm 4 = 10, mm 4 3

59 π, , , , M cr = 1, , 61 10, , 1 10 Nmm π M cr = knm 3 λ = LT π , = 1, λ = π = 76, λ _ 1, 476 LT = 76, 409 χ LT = 1, = 0,0171 < 0, tai λ_ LT = = 0,0171 < 0, Leikkauskestävyys Leikkauslommahduskestävyys on mitoittava. Sen määrittämiseksi käytetään yksinkertaista ylikriittiseen tilaan perustuvaa menetelmää. S355 ε = 0,81 λ _ 170 / 10 w = = 1,814 (uuman jäykisteet vain tuilla) 37, 4 0, 81 5, 34 0, τ ba = N/mm = 101,7 N/mm 1814, 3 0, 1017kN / mm V ba.rd = 170 mm 10 mm = 1174 kn 11, V Sd = 0,5 7,38 kn/m 5,0m = 34 kn < V ba.rd Taipumarajatila max δ = Nmm 48, 1 10 N / mm mm 15mm mm 6 = 90,1 mm< L 00 =

60 190 Esimerkki 3.5 Molemmista päistään vapaasti tuettu kotelopoikkileikkauksen omaava välipohjapalkki (kuva 3.1 a) on mitoitettava hyötykuormalle q k =,0 kn/m, kun jänneväli L 5,75 m (pilariväli 6,0 m). Ontelolaattojen pituus palkin vasemmalla puolella 6,0 m ja oikealla puolella 8,4 m. Teräs S355. Välipohjan (ontelolaatat + pintabetoni) pysyvä kuorma g k = 5,0 kn/m. Valitaan pikavalintataulukon perusteella (ontelolaattojen paksuus 65 mm) WQ 65x5-15x40-1x450 (vrt. kuva 3.1 b). g 0k = 0,915 kn/m. Kuva 3.1. Esimerkki 3.5:n palkki. [Kuva 3.1 (alkuper. 3.37) korjattava alalaipan paksuus 15 1 kuvassa b)] Poikkileikkaussuureet Osa A i (mm ) A i a i (mm 3 ) e - a i A i (e - a i ) (mm) , x , x , x 10 3 Σ x 10 3

61 e = mm = 118,9 mm e 1 = 158,1 mm 3 40 I y = mm 4 = mm W c = mm 3 = mm 3 158, W t = mm 3 = mm 3 118, 9 Uuma: 118, 9 1 ψ = = 0, 747 ( 158, 1 15) k σ = 7,81 + 6,9 0, ,78 0,747 = 17,97 ( ) λ _ / 5 p = 8, 4 0, 81 17, 97 = 0,513 ρ = d/t w = = 50 < 7 ε = 58,3 uuma kuuluu poikkileikkausluokkaan 5 1 Puristettu laippa: b/t f = 40/15 = 16 < 33 ε = 6,7 poikkileikkausluokka 1 Mitoitus taivutukselle Kuorma palkille: 1 g k = 0,915 kn/m + ( ) 1 q k = ( ),, m 5,0 kn/m = 36,9 kn/m,, m,0 kn/m = 14,4 kn/m 191 p k = 51,3 kn/m p d = (1, 36,9 + 1,5 14,4) kn/m = 65,9 kn/m max M Sd = , 9 kn/m 5,75 m = 7 knm α d = W pl = mm mm =,5 mm (d - αd) = (65 -,5) mm = 4,5 mm ,,, +, ( 4, 5+ 6)

62 19 RAKENNUSTEKNIIKKA W pl = mm , 355kN / mm M c.rd = mm = 417 knm > 11, M Sd Mitoitus leikkaukselle d/t w = 50 < 69 ε = 55,9 ei leikkauslommahdusvaaraa A v = 5 mm (65-15) mm = 500 mm V pl.rd = 500 mm 0, 355kN / mm 11, 3 = 466 kn max V Sd = 1 65, 9 kn/m 5,75 m = 189 kn < V pl.rd Mitoitus väännölle Kuormitustapaus 1 (hyötykuorma oikealla): Ontelolaatan tukireaktion suurin mahdollinen etäisyys palkin painopisteestä (ks. kuva 3.1c) = 0,5 mm. p vas = 1 6, 0 m 5,0 kn/m = 15,0 kn/m p d.vas = 1, 15,0 kn/m = 18,0 kn/m p d.oik = 1 8, 4 m (1, 5,0 + 1,5,0) kn/m = 37,8 kn/m 1 max T Sd = ( ),, kn/m 0,05 m 5,75 m = 11,5 knm A 0 = 63,5 mm 45 mm = mm 1 V Sd = ( ), +, +,, kn/m 5,75 m = 164 kn 6 115, 10 Nmm τ V = = 17, 8 N/mm 64558mm 5mm VSd / Av τv / , 8 βv = + = + f / 3 / γ f / 3 / γ 355 / 11, / 11, 3 ( ) ( ) ( ) ( ) y M 0 y M 0 = 0,448 < 0,5 leikkauskestävyys on riittävä, eikä taivutuskestävyyttä tarvitse pienentää. Kuormitustapaus (täysi kuorma): p vas = 1 6, 0 m (1, 5 + 1,5 ) kn/m = 7,0 kn/m 1 T Sd = ( ) ,, kn / m, m, m =, knm 6 6, 9 10 τ v = N/mm = 9,74 N/mm

63 / 500 9, 74 β V = + = 0,458 < 0,5 355 / 11, / 11, 3 Taipumarajatila ( ) ( ) 193 max M qk = 1 14, 4 5, 75 knm = 59,5 knm 8 δ q = , , mm , mm = 5,81 mm < δ adm = mm = 16,4 max M k = 1 51, 3 5, 75 knm = 1 knm 8 max δ = , mm , mm = 0,7 mm < δ adm = 5750 mm = 3,0 50 WQ-palkin mitoittamiseksi on käytettävissä tietokoneohjelma, jonka avulla voidaan myös alustavasti tarkistaa ontelolaattojen leikkauskestävyys.

64 Esimerkki 3.6. Keskisesti kuormitettu pilari. Tarkista kuvan 3.40 mukaisen pilarin normaalivoimakestävyys, kun pilari pääsee nurjahtamaan poikkileikkauksen heikompaan suuntaan. Pysyvän kuorman osuus normaalivoimasta N gk = 00 kn Lähtöarvot Teräs S35 f y = 35 N/mm Profiili HE00A A = 5383 mm i z = 49,8 mm poikkileikkausluokka 1, koska uuman d/t w = ( )/6,5 = 0,6 < 33ε = 33 laipan c/t f = 00 = 10 = 10ε 10 Nurjahduspituus L eff = γ L = 0, mm = 300 mm Koska profiilin h/b <1, nurjahduskäyrä c α = 0,49 _ 300 1, 0 35 λ = 49, 8 π = 0,684 φ = 0,5 [1 + 0,49 (0,684-0,) + 0,684 ] = 0,857 1 χ = = 0,7344 0, , 857 0, 684 0, 35kN N b.rd = 0, , mm = 844 kn 11, mm N Sd = (1, ,5 380) kn = 810 kn < N b.rd profiili on sopiva.

65 Esimerkki 3.7. Keskisesti kuormitettu putkipilari. Määritettävä kuvan 3.41 putkipilarin nurjahduskuormat Lähtöarvot Teräs S355JH EN1019 f y = 355 N/mm Profiili : suorakaideputki 300 x 00 x 6 A = 5763 mm I y = mm 4 i y = 113,1 mm I z = mm 4 i z = 8,9 mm Toimivat leveydet _ = Lyhyempi sivu: b b 3 t = 00 mm mm = 18 mm λ _ 18 / 6 p = = 0,534 < 0,673 ρ = 1 8, 4 1, 0 4 lyhyt sivu toimii täysin tehollisena Pidempi sivu: b _ = ( ) mm = 8 mm λ _ 8 / 6 0, 875 0, p = = 0,875 > 0,673 ρ = = 0,887 8, 4 4 0, 875 b eff = 0,887 8 mm = 50mm b = (8-50) mm = 3 mm Tehollinen poikkileikkaus (ks. kuva 3.4): A eff = A - b t = ( ) mm A eff = 5379 mm

66 196. Nurjahduskuormat.1 Nurjahdus y - y akselin suhteen (suuntaan z - z) nurjahduskäyrä b α = 0,34 L eff = 1,0 L = 8000 mm β A = = 0,9334 λ _ , y = = 0, , π φ y = 0,5 [1 + 0,34 (0,8944-0,) + 0,8944 ] = 1,018 1 χ y = = 0,6648 1, , 018 0, , 355kN N b.rd = 0, , mm 11, mm = 1154 kn. Nurjahdus z - z akselin suhteen (suuntaan y - y) _ , λ z = = 1,0 8, 9 π φ z = 0,5 [1 + 0,34 (1,0-0,) + 1,0 ] = 1,418 1 χ z = = 0,4673 1, , 418 1, 0 0, 355kN N b.rd = 0, , mm = 811 kn 11, mm Pilarin kestävyys on siten = 811 kn.

67 Esimerkki 3.8. Mitoita kuvan 3.43 mukaisen kehärakenteen puristettu ja taivutettu pilari, kun pilarit on vaakaorsilla tuettu pienemmän jäykkyyden suunnassa. 197 Kuva Esimerkki 3.8:n rakennemalli. 1. Lähtöarvot Teräs S355 g k = 4,0 kn/m q 3k = 14,0 kn/m (lumikuorma) q 1k =,6 kn/m q k = 1,1 kn/m F wk = 3,0 kn. Voimasuureiden ominaisarvot Pilarin arvioitu omapaino 7,0 kn (sisältää seinäorsilta tulevan kuorman) N gk = 7,0 kn + 10m 4,0 kn/m = 47,0 kn N qk = 10 m 14,0 kn/m = 140 kn Tuulikuorman aiheuttama 1.kertaluvun taivutusmomentti pisteessä A: M Ak = 1 F h q1 k H + H 8 3 F h = ( q1 k + q k ) H + Fwk 8 3 F h = ( ) 8, 4 + 1, 7 kn/m 7,0 m + 3,0 kn = 13,8 kn M Ak = , 8, kn m, m + kn 7, 0m = 63, 0 knm 3. Kuormitusyhdistelmien aiheuttamat rasitukset Koska pilari on tuettu heikommassa suunnassa seinäorsilla siten,että λ _ z _ < λ ja y λ _ LT < 0, 4, pilari nurjahtaa jäykempään suuntaan ei kiepahdusta tapahdu. a) voimakas tuuli + samanaik. lumi N Sd = 1, 47,0 kn + 1,5 0,7 140 kn = 03 kn

68 1 k c = 0, 5 + = 0, k s = , = 1,10 φ = 0, , = lisävaakavoima pilarien mahdollisesta vinoudesta = N d /44 M Ad = 1,5 63,0 knm + 03 kn 7, 0m = 100 knm 44 Valitaan alustavasti IPE300 A = 5381 mm i y = 14,6 mm W pl.y = mm 3 L eff =,1 7,0 m = 14,7 m _ λ = = 1544, y π 14, nurjahduskäyrä a (nurjahdus suuremman jäykkyyden suunnassa, kun h/b > 1,) α = 0,1 φ = 0,5 [1 + 0,1 (1,544-0,) + 1,544 ] = 1,833 1 χ y = = 0,3545 1, , 1544, Tarkistetaan poikkileikkausluokka: N f d = 355 = 3,7 N/mm 11, mm Uuma: d = (300-7,1-15) mm = 48,6 mm 1 03 α = + = 0,856 48, 6 7, 1 0, 37 d 48, , 81 = = 35, 0 < t w = 36, 5 poikkileikkausluokka,, c 150 laippa: = = 7, 01 < 10ε = 8, 1 poikkileikkausluokka 1 t f 10, 7 M Q = 0,15 1,5,6 7,0 knm = 3,9 knm M = 100 knm ,, 6 7 knm = 113,4 knm β My = 1, 3, ( ) ,, 3 1, 8 = 1,69 W el.y = mm µ y = 1, 544 ( 1, 69 4) + = - 0,

69 199 RAKENNUSTEKNIIKKA 0, k y = 1+ = 1,5 0, , 355 Yhteisvaikutusehto: 03 0, , 37 1, , 37 = 0,947 < 1 (kaava 3.56) b) runsas lumi + samanaik. tuuli N Sd = 1, 47,0 kn + 1,5 140 kn = 66 kn M Ad = 1,5 0,5 63,0 knm + 66 kn 7, 0m = 54, 9 knm , 5 54, 9 + = 0,771 < 1 0, , , 37 IPE300 on täten painoltaan pienin I-profiili, joka soveltuu pilariksi.

70 Esimerkki 3.9. Teollisuushallin päädyn tuulipilarin mitoitus (ks. kuva 3.44). Lähtöarvot: Koko kohdetta käsittelevästä rakenneselostuksesta saadaan: teräs S35 rakennemalli kuvassa 3.44b HEA-pilari on sivusuunnassa tukematon. N gk = 8,0 kn N qk = 30,0 kn q w.ref = 0,331 kn/m 00 Kuva Esimerkin 3.9 tuulipilari. Voimasuureet: w k = w e + w i = 0,331 kn/m 1,7 (0,6 + 0,5) = 0,619 kn/m M 1k = 1 8 5, 0 m 0, 619 kn / m 6, 0 m = 13,9 knm 9 M fk = 5, 0 0, 619 6, 0 knm = 7,83 knm 18 Mitoituksessa määräävä kuormitustapaus on omapaino + tuuli + samanaikainen lumi. N Sd = (1, 8,0 + 1,5 0,7 30,0) kn = 41,1 kn M 1d = 1,5 13,9 knm = 0,9 knm M fd = 1,5 7,83 knm = 11,7 knm 1. Alustava mitoitus N A = 108 mm 40N / mm 6 0, 9 10 Nmm W el.y 140N / mm = mm 3 Kokeillaan profiilia HE 140A : A = 3140 mm W el.y = mm 3 h = 133 mm W pl.y = mm 3 b = 140 mm i y = 57,3 mm