Lecture 2 Intro Continued_2072

Design of steel structures*Lecture 2February 8, 2012Extra class5.30 pm 7.00 pmCross section PropertiesClassificationMoment rotation behaviourCross section properties used in design

Design of steel structures

Design of steel structures*Cross section PropertiesElastic section propertiesPlastic section properties


Design of steel structures*Elastic Section PropertiesThe exact section dimensionsThe location of the centroid if the section is asymmetric about one or both axesArea of cross sectionMoments of inertia about various axesRadii of gyration about various axesModuli of section for various axes


Design of steel structures*Plastic section propertiesPlastic moduli of section


Design of steel structures*Other important properties of universal beams, joists and channels used for determining the buckling resistance momentsBuckling parameter, uTorsional index, xWarping constant, HTorsional constant , JThese properties are given in standard tables or can be calculated using formulae given in BS 5950 part 1


Design of steel structures*Elastic section properties Area = 2BT+dt

Symmetrical I sectionModulus of section is the elastic modulus of section


Design of steel structures*ProblemDetermine the properties of a plated UB section 610 UB 125 strengthened by welding a 300 mm x 20 mm plate to each flange. Determine the section properties Ix and ZxSolution:The properties of the UB are available in tables and are shown above.Because of the symmetry of the section the centroid of the plated UB Is at the web centre.


Design of steel structures*The properties Ix and Zx are elastic properties i.e. the whole section is effectiveMoment of inertia of plate is small compared to other values , so omitted.


Design of steel structures*Plastic moment of a sectionM= X ZxxMe=Py x ZxxMp=Py x SxxDerivation:Plastic Moment = Py x area in compression x d/2 = Py x area in tension x d/2 = Py ( area in compression x d/4 +area in tension x d/4) = Py x algebraic sum of first moments of area about equal area axis = Py x Sxx where Sxx= plastic section modulusNeutral axis


Design of steel structures*Plastic modulus and shape factor

Shape factor of a section is defined as , whereProblem: Determine the elastic moment, plastic moment, elastic section modulus, plastic modulus and shape factor for the rectangular section 10 mm500mm


Design of steel structures*Elastic properties Plastic propertiesEqual area axis coincides with the centroid of section


Design of steel structures*Determine the shape factor for 610x229x125 UB S275 From the table of properties of Universal Beams, the properties can be obtained:Zxx = 3220 cm3Sxx=3680 cm3

The value of shape factor for most I-sections is about 1.15


Design of steel structures*Types of elementsTwo distinct types of elementsOutstand element elements attached to adjacent element at one edge only , the other end being freeInternal elements elements attached to other elements at both longitudinal edgesInternal elementsOutstand elements


Design of steel structures*More exampleInternalelementsOutstand elements


Design of steel structures*Local BucklingThe outstand element (flange) of an I beam will buckle prematurely) if it is too thinWebs also will buckle under compressive stresses from bending and shearTo prevent local buckling from occurring it is necessary to limit their outstand/thickness ratios for flanges and depth /thickness ratios for websd/t ratio for web

b/t ratio for flanges


Design of steel structures*Effective width, effective area; effective moment of inertia and effective section modulusWhen the plate buckles, the whole width or area is not effective The effective width limitation under different conditions are given in the code.The effective area is calculated based on the effective widths multiplied by thicknessesEffective moment of inertia is moment of inertia calculated using the effective areasEffective section modulus is determined from Effective moment of inertia divided by extreme fibre distance from Neutral axis


Design of steel structures*Section classificationClause 5.5.2 BS EN 1993-1-1:2005 classifies the compression elements of structural members into 4 categories depending upon their resistance to local buckling effects, which may affect their load carrying capacity.The compression may be due to direct axial force , bending moment or a combination of bothCross sections are classified as Class 1 cross sectionsClass 2 cross sectionsClass 3 cross sectionsClass 4 cross sections


Design of steel structures*The role of cross section classification is to identify the extent to which the resistance and rotation capacity of cross sections is limited by its local buckling resistanceThe class (refers to Class 1,2,3,4) of cross section is determined based on the c/t values of the compression parts compared to the limits given in table 5.2 of Code.Classification


Design of steel structures*


Design of steel structures*The limiting aspect ratio depends on the design strength fy as shown below where fy = 235, 275, 355, 420 or 460


Design of steel structures*Moment rotation behaviourThe following classification should be applied.

Class 1 cross sections: Cross-sections which can form a plastic hinge with rotation capacity required for plastic analysis. Elements subject to compression that meet the limits for class 1 given in Table 5.2 should be classified as class 1 plastic.

Class 2 compact: Cross-sections which can develop their plastic moment resistance, but have limited rotation capacity because of local buckling. Elements subject to compression that meet the limits for class 2 given in Table 5.2 should be classified as class 2 compact.

Class 3 semi-compact: Cross-sections in which the stress at the extreme compression fibre can reach the yield strength, but the plastic moment capacity cannot be developed due to local buckling. Elements subject to compression that meet the limits for class 3 given in Table 5.2 should be classified as class 3 semi-compact.

Class 4 slender: Cross-sections in which local buckling will occur before attainment of yield stress in one or more parts of the cross section. It is necessary to make explicit allowance for the effects of local buckling. Elements subject to compression that do not meet the limits for class 3 semi-compact given in Table 5.2 should be classified as class 4 slender.


Design of steel structures*Moment rotation behaviourMpMeMrotationClass 1Class 2Class 3Class 4Moment


Design of steel structures*Moment rotation behaviourM= X ZxxMe=fy x ZxxMp=fy x SxxNeutral axisClass 4 SlenderFigure aClass 3 Semi compactFigure bClass 2 CompactClass 1 PlasticFigure cZxx = elastic section modulusSxx = plastic section modulus = elastic stressPy = design strength


Design of steel structures*DescriptionsSlender section: When aspect ratio is relatively high, then local buckling may prevent any part of the cross section from reaching the design strength (Py). Such sections are called slender sections or class 4; their reduced capacity is based on effective cross section properties.Semi compact section or class 3: Semi compact sections may be prevented from reaching their full plastic moment capacity by local buckling of one or more elements of the cross section. The aspect ratio may be such that only the extreme fibre stress can attain the design strength before local buckling occurs. Such sections are classified as Class 3 and their capacity is therefore based on the limiting elastic moment as shown in figure b above.Compact section or class 2: Sections which have full plastic moment capacity but not sufficient hinge rotation capacityPlastic section or class 1: sections which have full plastic moment and hinge rotation capacity


Design of steel structures*DETERMINING CLASSIFICATIONS FOR SECTIONSExample 1Determine the classification for section 356 x 127 UB 33 Grade 275 subject bendingSx = 539.8 cm3, Ixx = 8200cm4, B = 125.4 mm, D = 348.5 mm t = 5.9 mm T = 8.5mm, r = 10.2 mm (figure next slide)Table 5.2 is to be used.


Design of steel structures*Cf/tf=(125.4-5.9-2x10.2)/(2x8.5)=5.833d/t = 311.1/5.9 =52.7

Design of steel structures*Since all elements are plastic, the section is also plastic

Compression FlangeActual Cf/tf= 5.83Limitsc/t99x0.924=8.32.element is plasticWebsince it is beam, the NA is at mid depth..d/t = 52.7

Design of steel structures*rMinor axis or z axisMajor axis or y axisDFlangeWEBC for web = d = D-2T-2rC for flange = (B-tw-2r)/2So even if d and c are not given; they can be calculated from D and B


Design of steel structures*Cross section properties of sectionscl.6.2.2 of EC3The cross section properties are dependent on the classification seen earlierThe capacity of class 4 : slender sections is calculated using effective section properties like effective cross section area and effective section modulusEffective plastic section modulus is used for class 3 : semi-compact sectionsNet cross sectional allowing for reduced area due to bolt holesGross cross section properties


Design of steel structures*Gross cross section 6.2.2.1Gross cross section properties are determined from the specified shape and nominal dimensions of the member or elementHoles for bolts should not be deducted but due allowance must be made for larger openingsMaterial used solely in splices or as battens should not be included


Design of steel structures*Net cross sectional area, anThe reduction in the cross sectional area due to bolt holes The clear hole size is deducted.20 mm dia bolt requires 22 dia holes- so deduction is area corresponding to 22 mm diameter.The holes can have different arrangements the strength depends on the hole arrangementCase 1: Holes are not staggeredCase 2: Holes are staggered


Design of steel structures*Holes are not staggeredThe net area of a cross-section = gross area minus appropriate areas for all holes and other openings. If the fastener holes are not staggered, then total area to be deducted shall be the maximum sum of the sectional areas of the holes in any cross-section perpendicular to the member axis (perpendicular to the direction of the applied stress in the member.btDiameter of bolt =d


Design of steel structures*Plate with staggered holes

When the holes are staggered it is necessary to use special formulae to calculate the deduction It is necessary to consider all possible failure paths extending progressively through the memberThe net width now must account for the change in direction of the line of fracture.


Design of steel structures*g1g1g1s1s1s1S is staggered pitchG is the gaugeT is the thickness


Design of steel structures*Minimum amount of staggerThe minimum amount of stagger Spm for which a hole no longer reduces the area of the member depends on the diameter d of the hole and the inclination Sg/Sp of the failure path, where sg is the gauge distance between the rows of holes.An approximate expression for this minimum stagger is

When the actual stagger sp is less than spm, deduction is reduced by sp2 x t/4sg


Design of steel structures*First, consider the different ways a tension member with staggered holes can fracture. These pictures depict the different lines of failure. When analyzing a member like this, it is important to find all the lines of failure and then determine which line of failure is the weakest cross section. That cross section will be taken as the net width


Design of steel structures*Area to be deducted = 2x dttddtddsgCase 1Case 2


Design of steel structures*tddgWhere n= number of slant failure linessFind the maximum area to be deducted from above three cases. This will produce the minimum area and therefore the critical sectionWhere s and g are different , different expressions have to be written

Net area, an = Gross area deductions for holes


Design of steel structures*Problem 1: Determine the gross cross sectional area, net cross sectional area40 mm50 mm40 mm80 mm55 mm80 mm55mm180mm wide x 10 mm thick plate


Design of steel structures*CalculationGross area, Ag = 180 x 10 = 1800 mm2Net cross sectional area, An Three failure paths can be considerded.Case 1: area to be deducted


Design of steel structures*40 mm50 mm40 mm80 mm55 mm80 mm55mmCase 1Case 1 failure sectionNote that this section is similar to one drawn through 1,2,4, and 5 since in all these cases deduction is the same1245


Design of steel structures*tddsgCase 2Note here that there are many failure paths similar to this as shown in next figure


Design of steel structures*Case 2Case 2


Design of steel structures*Case 3: S is same for all slant failure lines.G is 40 for outer lines and 50 for inner line.

Case one is the most severe. Since highest deuction occurs for this Case one 360 mm2; case two 351 mm2 and case 3 190.62 mm2

NET AREA = GROSS AREA DEDUCTION = 180 X 10 360 =1440 MM2.


Design of steel structures*Ag = 180x10 = 1800 mm2An =1440mm2


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Lecture 2 Intro Continued_2072

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