Chapter 2 Member Design

Chapter 2 : MEMBER DESIGN Summary:

•

Section needs to be be classified classified to avoid local plate buckling

.

A variety of section shapes are available for beams, choice depends on local and span.

• • • • • • • • • • • • • • • • • •

Beams may often be designed designed on basis basis of bending moment resistance. Stiffness under serviceability loads is an important consideration. Beams which are unable to move laterally are termed restrained . Moment resistance is dependent on section classification. Co-existent shear forces below 50% of the plastic shear resistance do not affect moment resistance. Beams bent about the major axis may fail by buckling in a more flexible plane This form of buckling involves both lateral deflection and twisting - lateral-torsional buckling The applied moment at which which a beam buckles by deflecting laterally laterally and twisting reached is the elastic critical moment  A design approach for beams prone to failure by lateral-torsional buckling buckling must account for a large number of factors - including section shape, the degree of lateral restraint , type of loading, residual stress pattern and initial imperfections Stocky beams beams are unaffected by lateral torsional buckling buckling and capacity is governed by the plastic resistance moment of the cross section Slender beams have capacities close to the theoretical elastic critical moment Many practical beams are significantly adversely affected by inelasticity and geometrical imperfections, hence elastic theory provides an upper band solution. A design expression linking the plastic capacity of stocky beams with the elastic behaviour of slender beams is provided by a reduction factor  for  for lateral torsional buckling Structural members subjected to axial compression and bending are known as beam columns. The interaction of normal force and bending may be treated elastically or plastically using equilibrium equilibrium for the classification of cross-section. The behaviour and design of beam-columns are presented within the context of members subjected to uniaxial bending, whose response is such that deformation takes place only in the plane of the applied moments. In the case of beam-columns which are susceptible to lateral-torsional lateral-torsional buckling, the out-of-plane flexural  buckling of the the column has has to be combined combined with the the lateral-torsional lateral-torsional buckling of of the beam using the relevant interaction formulae. • For beam-columns with biaxial bending, the interaction formula is expanded by the addition of an additional term.

• Objectives:

• • • • • 

• • • • 

• •

explain the procedure for section classification explain the procedures used to design restrained beams, design a beam for bending resistance, check a beam for compliance with serviceability criteria, describe how to reduce the bending bending resistance resistance of a beam to allow for high shear loads. describe the difference in behaviour of stocky and slender columns recognise the sources of imperfection in real columns and the need for a probabilistic approach to design compare the ECCS column curves calculate the non-dimensional slenderness of a column calculate the reduction reduction factor for the relevant relevant buckling modes for columns columns of different cross-sectional shapes calculate the in-plane bending and axial compression force for beam-columns calculate the lateral-torsional buckling of beam-columns calculate the biaxial bending and axial compression force for beam-columns

References:

•

Eurocode 3 Design of steel structures Part 1.1 General rules and rules for b uildings

2- 1

• • •

The Behaviour and Design of Steel Structures, N S Trahair and M A Bradord, E & F Span, 1994. Galambos, T.V., Structural Members and Frames, Prentice-Hall, 1968  Narayanan, R., Beams and Beam Beam Columns Columns - Stability Stability and Strength, Applied Science, Science, London, 1983

Contents:

1. 2. 3. 4. 5.

Section classification Compression members Restrained beams Unrestrained beams Members subjected to axial force and moments

2- 1

• • •

The Behaviour and Design of Steel Structures, N S Trahair and M A Bradord, E & F Span, 1994. Galambos, T.V., Structural Members and Frames, Prentice-Hall, 1968  Narayanan, R., Beams and Beam Beam Columns Columns - Stability Stability and Strength, Applied Science, Science, London, 1983

Contents:

1. 2. 3. 4. 5.

Section classification Compression members Restrained beams Unrestrained beams Members subjected to axial force and moments

2- 2

Chapter 2 Member Design 1

Local Buckling and Section Classification

1.1

Introduction.

Local buckling is a phenomena which affects all thin materials when subjected to a compressive force. Its effect is to cause wide plate elements within a member to buckle before they reach the design strength. A typical pattern of local buckling in the outstand flange of a  beam in bending is shown in figure figure 1.1.

Figure 1.1 typical pattern of local buckling in outstand flange:

1.2

Section Classification.

BS 5950 prevents local buckling of the various elements of the cross section by classifying each element according to its b/t or d/t ratio, then designing the cross section accordingly. It is therefore necessary to define the parts of the cross section which are to be considered. Figure 5 of BS 5950- 1 defines the various elements in a number of cross sections. For the  purpose of this lecture two particular shapes shapes will be considered, a universal beam and and a hot finished hollow section as shown in Figure 1.2:

2- 3

 b

T

 b

Outstand

Flange

r  D

Web

Web

t

D

t d

d

B

B

Universal Beam

For a Hot finished hollow section

d=D-2(T + r)

b = B/2

b=B-3t

d=D-3t

Figure 1.2 Section shapes The classification of sections is carried out according to tables 11 and 12 of the code parts of which are shown in Table 1.1 of these notes.

Table 1.1 - Limiting width to thickness ratios for H- or I-section or sections other than CHS and RHS

Compression element

Ratio

Outstand element of Rolled compression flange section Web (Neutral axis at mid depth) Web generally If r 1 is  positive (compression)

Limiting values Class 2 Class 3 Compact Semi-compact 10ε 15ε

 b/T

Class 1 Plastic 9ε

d/t d/t

80ε 80ε 

100ε 100

120ε 120ε 

1 + r 1

1 + 1.5r 1

1 + 2r 2

 but ≥ 40ε 

 but ≥ 40ε 

 but ≥ 40ε 

If the b/t or the d/t for class 3 (semi-compact) semi compact sections is exceeded then the element is class 4 (slender). i.e., it will buckle locally before full axial load is achieved.  Notes to tables: 1. 2. 3.

The term ε=(275/py)1/2 is used to accommodate varying design strengths. F c F c For I and H sections r 1 = but –1
F c

2dtp yw

;

but –1
Factors r 1 and r 2 allow for the applied axial load F c

=

F c  Ag p yw

2- 4

Where Fc is the applied axial load (taken as +ve for compression)  pyw is the design strength of the web Ag is the gross area of the cross section The four classes of cross section given in the code are as follows: Class 1. Plastic cross sections are those in which all elements subject to compression are relatively stocky (small width to thickness ratios) and can sustain high strains without local  buckling. In a class 1 (plastic) cross section plastic hinges can be developed with sufficient rotation capacity to allow redistribution of moments within the structure. Only class 1 (plastic) sections should be used at plastic hinge locations in structures using plastic analysis. Class 2. Compact cross sections contain elements which are less stocky, although the cross section can develop the full plastic moment capacity. However local buckling of the section will prevent development of a plastic hinge with sufficient rotation capacity to permit plastic analysis. Class 2 (compact) sections can be used without restriction except that they may not  be used in plastic design. Class 3. Semi-compact sections are those in which all elements subject to compression can reach the design strength at the extreme fibres but local buckling may prevent the development of full plastic moment. Class 3 (semi-compact) sections are subject to limitations on their moment capacity which are given in clauses 3.5.6, 4.2 and 4.3. Class 4. Slender sections are those which contain slender elements when subject to compression due to moment or axial load. Local buckling will prevent the stress in a slender section from reaching the design strength. Design of class 4 (slender) sections is considered in section 3.6 "Class 4 (slender) cross-sections”.

In general sections will be classified in accordance with the highest classification of any of the elements. For example a beam with a class 1 (plastic), flange and a class 2 (compact) web will  be classified as class 2 (compact). The above classification of cross sections implications as far as the design of the member is concerned. These implications will be dealt with in detail during the lectures on each type of member.

1.3 Class 3 (Semi-compact) and Class 4 (Slender Sections) 1.3.1 Class 3 (Semi-compact sections) For class 3 semi-compact sections subject to bending the elastic section modulus (Z) should  be used for bending calculations. Alternatively an effective section modulus (S eff  )may be used. The code gives formulae for calculating S eff  for various sections in clause 3.5.6. For an I or H section the formulae are given as:

2- 4

S  x,eff 

⎡ ⎛  β   ⎞ 2 ⎤ ⎢ ⎜ 3w ⎟ − 1⎥ ⎢ d / t  ⎥ = Z  x + ( S  x − Z  x ) ⎢ ⎝   ⎠ 2 ⎥    β   ⎞ ⎢ ⎛  ⎜⎜ 3w ⎟⎟ − 1⎥ ⎢⎣ ⎝  β 2 w  ⎠ ⎥⎦

⎡  β 3 f  ⎤ ⎢ − 1⎥ / b T  ⎢ ⎥ but S  x ,eff  ≤ Z  x + ( S  x − Z  x ) ⎢  β 3 f  ⎥ ⎢  β  − 1 ⎥ ⎣ 2 f  ⎦

⎡  β 3 f  ⎤ ⎢ − 1⎥ / b T  ⎥ S  y ,eff  = Z  y + ( S  y − Z  y ) ⎢ ⎢  β 3 f  ⎥ ⎢  β  − 1 ⎥ ⎣ 2 f  ⎦ where:

β2f  β2w β3f  β3w Sx and Sy Zx and Zy

is the limiting value of b/T for a class 2 compact flange is the limiting value of d/t for a class 2 compact web is the limiting value of b/T for a class 3 semi-compact flange is the limiting value of d/t for a class 3 semi cmpact web are the plastic moduli are the elastic moduli

1.3.2 Slender sections BS 5950 deals with class 4 (slender) elements in cross sections by the use of effective section  properties:

•

Effective area - The effective area should be taken as shown in Figure 8 of the code and the examples shown in Figures 1.3 and 1.4 of these notes. 20t

ε

20t

ε

20t 20t

ε

20t

ε

t

Rolled I Section

ε

t 20t

Hot finished RHS

Figure 1.3 Effective cross section subject to pure compression for d etermining A eff

ε

2- 5

20t ε

20t ε 20t ε

20t ε

20tε D t t 20tε

Major axis bending

Minor axis bending

Figure 1.4 Effective cross section subject to pure moment for determining Z eff

• Effective modulus when the web is not slender under pure bending.(i.e. only the

•

flanges are slender).This should be obtained by using Figure 8b in the code and Figure 1.4 of these notes. If the section is not slender when subject to bending about the other axis the full value of the elastic section modulus (Z) should be used when considering  bending about that axis. Effective modulus when the web is slender under pure bending. The effective s ection modulus should be obtained by considering an effective cross section as shown in Figure 9 of the code and Figure 1.5 of these notes. Where: be  ff 

=

120t ε 

⎡  f cw −  f tw ⎤ ⎡  f tw ⎤ ⎢1 + ⎥ ⎢1 + ⎥  p yw ⎥⎦ ⎣  f cw ⎦ ⎢⎣

where: f cw and f tw are the maximum compressive sress and the maximum tensile stress in the web calculated on the gross cross section. 0.4beff 

f cw  Non effective Zone

0.6beff  Elastic neutral axis of gross section

Elastic Neutral axis of effective section

f tw

Figure 1.5 Effective width for slender web under pure bending

2- 6

1.3.3 Section Subject to Axial Force and Moments When dealing with members which are subject to axial load and biaxial bending, the classification of the cross-section should be based upon the combined effects of the axial and  bending actions. The effective cross-sectional area A eff   and the values of effective section modulus Zeff  for bending about the major and minor axes should be determined from separate effective cross-sections as detailed in Clause 3.6. Sections with slender webs under bending should be avoided in beam columns.

1.4. General Guidance when using the Deign Tables in the Appendix (see Handout)

• • •

 None of the universal beam and column sections in grade S275 and S355 are slender under bending only.  None of the universal columns can be slender under compression only, but some universal beams and hollow sections can be slender. Sections that can be slender under axial compression are marked with * in the design tables.  None of the sections listed in the design tables are slender due to the flange being slender. Under combined axial compression and bending, the section would be compact or semi-compact up to given F/P z limits.

1.5. Examples of the use of Tables 11 and 12 from BS5950-1:2000 The use of these tables can be demonstrated as follows: 1.5.1 Consider a S275 steel 457x152x52 universal beam subject to bending about the major axis  b

T

r  t

D

d

B

Universal Beam b = B/2

d=D-2(T + r)

Figure 1.6 Universal Beam

•

2

T<16mm therefore p y=275N/mm  from table 9

2- 7

275

ε  =

•

From section tables the flange b/T = 6.99 and the web d/t = 53.6

•

From table 11 the limit for a class 1 (plastic) rolled flange is 9 ε the flange is therefore class 1 (plastic)

•

From table 1 the limit for a class 1 (plastic) rolled web with the neutral axis at mid depth is 80ε the web is therefore class 1 (plastic)

•

Both the flange and the web are class 1 (plastic). Therefore the section is class 1 (plastic) when subject to bending

 p y

=

275

•

275

=1

Design strength = 275 N/mm2 since T = 10.9mm < 16mm 3 Plastic modulus Sx = 1100 cm For compact section: 3 -6 Bending Moment Capacity = p y x Sx = 275 x 1100x 10  x 10  = 302 kNm Using Design Table Page 197 For a S275 steel 457x152x52 universal beam under pure bending Section classification is plastic. Moment capacity is 301 kNm

1.5.2 Consider the same beam (S275: 457x152x52) when subject to an axial load of 800kN and a bending moment about the major axis.

•

From section tables the flange b/T = 6.99 and the web d/t = 53.6 and ε has been shown to  be 1.

•

From table 11 the limit for a class 1 (plastic) rolled flange is 9 ε  the flange is therefore class 1 (plastic).

•

From table 11 the limit for a class 1 (plastic) rolled web (generally) is 80ε  1 + r 1

 where r 1

=

Fc dtp yw

800 x10 3

•

Therefore r 1 =

• • •

53.6>41.5ε the web is not therefore class 1 (plastic)  by inspection it can be seen that the section is not class 2 (compact)

•

. The limit for a class three web is

407.6 x7.6 x 275

= 0.93  and the d/t limit = 41.5 ε

120 1 + 2r 2

 where r 2

=

Fc  Ag p yw

2- 8

800 x10 3

•

r 2

•

The web is therefore semi -compact

•

The section has therefore got a class 3 (semi-compact) web and a class 1 (plastic) flange and should be treated as a semi compact section.

•

If the axial load were increased to 1500kN it can be shown that the web becomes class 4 (slender)

=

66.6 x10 2 x 275

= 0.44  and the limit = 64 ε

Using Design Table Page 247 For a S275 steel 457x152x52 universal beam under axial load = 800kN and bending about major axis, Section is at least semi-compact when F/P z is less than 0.619 Section is at least compact when F/P z is less than 0.268 Pz = 1830kN Upper limit of F for semi-compact section = 0.619 x 1830kN = 1133kN Upper limit of F for compact section = 0.269 x 1830kN = 492kN Since F = 800kN is between 1133kN and 492 kN, section classification is semi-compact

Note that when axial load is increased to 1500kN, the section becomes slender.

1. 6 Summary of design procedure 1

Select, from experience, a suitable section based on the factored load effects

2

Determine the section classification from Table 11 or 12

3

If necessary calculate effective plastic modulus for Class 3 (semi-compact) sections

4

If necessary calculate effective section properties for class 4(slender sections)

5

Proceed with design procedures suitable for the section classification

2- 9

2. Compression members In order to perform satisfactorily, a compression member must not fail due to: (i) (ii) (iii)

Local buckling Overall yielding Overall buckling

These effects will be examined by reference to BS 5950. 2.1 LOCAL BUCKLING In order to ensure that a member does not fail due to local buckling, the elements of the cross section should be classified by reference to Tables 11 and 12 of the BS code and designed accordingly. If the classification is 1, 2 or 3, the design method is the same. When the section is class 4 then the design differs slightly and these differences will be noted in this lecture. It is therefore important to identify whether or not the section is class 4. b I and H Sections

275 py

=

t

b/t < 15

d

d/t < r2 =

Fc Ag p

T

120 but <40 1+ 2.0 r 2

y

b

Hot finished Rectangular Hollow Sections b/t < 40 t

d

D

d/t <

120 1+ 2.0 r

2

but <40 b=B-3t

B

d=D-3t

Figure 2.1 Classification of class 3 section If the class 3 limit is exceeded for any element, the cross section should be classified as class 4 (slender) and an effective area used throughout the design in accordance w ith clause 3.6. 2.2. OVERALL YIELDING Theoretically, the designer should check that the design stress is not exceeded on the gross area of the section. This check, however, is not necessary in practice because the check for overall buckling will satisfy this condition.

2.3

OVERALL BUCKLING

2.3.1 Slenderness The resistance of the member to overall buckling depends on the slenderness ( λ).

2- 10

For class1, 2 and 3 cross sections the slenderness is taken as λ = LE/r For class 4 (slender) cross sections the slenderness is taken as taken as λ (Aeff /Ag) Where

0.5

LE = the effective length which is a function of the actual length of the member between restraints r

= the radius of gyration, which is a tabulated section property and is different for each axis of buckling.

2.3.2 Compressive Strength and Resistance The designer will select a compressive strength (p c in N/mm2) from Table 24 based on the design strength and the slenderness λ and hence calculate the buckling resistance (P c kN) from clause 4.7.4. i.e.: For class 1 2 or 3 sections the capacity

Pc = Ag pc

For class 4 (slender) cross sections

Pc = Aeff  pc

where  Ag is the gross area of the section  Aeff  is the effective cross sectional area of the section  pc is the compressive strength based on λ and py 0.5  pcs is the compressive strength based on λ(Aeff /Ag)  and py

2.4.

Use of strut table 23 and compressive strength table 24

When referring to compressive strength table 24 it will be noticed that there are four tables one for each strut curve (a) to (d). The four different curves are used for different shaped sections and differences in their buckling behaviour about each principle axis, caused by imperfections such as out of straightness and residual stress. A more detailed explanation of why this occurs is given in appendix A of these notes. It is essential however to use the correct compressive strength table in accordance with the appropriate strut curve as indicated in Table 23 of the code, part of which is reproduced in Table 2.1 of these notes. Table 2.1 Allocation of strut curve Type of Section

Hot finished hollow section Cold formed hollow section I Section (e.g. Universal  beam) H section (e.g.Universal column) Welded I or H section Angles, Channels and T-Sections

Thickness mm

< 40mm >40mm < 40mm >40mm <40mm >40mm

Axis of Buckling x-x y-y (a) (a) (c) (c) (a) (b) (b ) (c) (b) (c) (d) (d) ( b) (c) (b) (d) (c)

2- 11

Fabricated Sections should, in theory, need the use of a further set of strut curves but, for simplification, the use of one of the above tables is used with a reduced value of design stress 2

taken as (py - 20) N/mm  where py is the design stress of the original p late section.

2.5 Effective Length The effective length of a compression member is a function of the actual length between restraints and its value depends on the type of restraint provided, i.e. rotational and/or  positional restraint (see Figure 2.5).

Restraint

Position Direction

Position

Restraint

Position

Position

Practical LE

1.0 L

0.85 L

Figure 2.5

Position Direction

Position Direction 0.7 L

Direction

Position Direction

Position Direction

2.0 L

1.2 L

Effective lengths

In the majority of cases in simple construction, the effective length will be determined from Table 22 of the code (part of which is reproduced below as Table 2.2). Discontinuous angle channel or T section struts and laced and battened members are treated separately, as are members in continuous construction.

Table 2.2 Effective length of members Restraint (in the plane under consideration) by the other parts of the structure Effectively held in position Effectively restrained in direction at both ends at both ends Partially restrained in direction at both ends Effectively restrained in direction at one ends  Not restrained in direction at either end

2.6

LE 0.7L 0.85L 0.85L 1.0L

Members in Lattice Frames and Trusses

Members in lattice frames and trusses using angles channels and T sections are treated in the same way as other compression members, apart from the method of determining the slenderness. For this, reference should be made to Clause 4.7.10 and Table 25. At first glance, this procedure appears very complex. It has, however, been justified on the  basis of test work carried out on large lattice frames and towers and it allows for such things as: (i)

The effective length being influenced by the type of connection.

2- 12

(ii)

The eccentricity caused by using a double angle or a single angle and whether the gusset plate is between or on the back of the angle.

(iii)

The possibility of short members buckling about the stronger axis due to a flexible gusset plate at the end.

For double angles and channels and laced and battened sections, additional rules are also given to ensure that the connection is adequate, eccentricities are allowed for and that proper allowance is made for changes in slenderness about the axis perpendicular to the battens. See Clauses 4.7.8 to 4.7.13. The rules are clearly laid out based on experimental evidence and do not require a detailed explanation in this lecture.

2.7

MEMBERS IN CONTINUOUS CONSTRUCTION

The design of members in continuous construction is dealt with in Section 5 of the code. The  procedures depend on considering the frame as a whole and the stiffness of individual members framing into the column.. The subject is too large to cover as part of this lecture, but it is important to recognize that the simple approach to effective lengths, as given in Table 22, is not applicable to members in continuous frames. Appendix E of the code gives figures to determine the effective length of members in continuous construction

2.8

SUMMARY OF DESIGN PROCEDURE FOR COMPRESSION MEMBERS

1. 2. 3. 4

Select section and determine the value of design strength. Determine if the section is class 4 (slender) For slender class 4 (slender) sections, calculate the effective area Determine the effective length ( LE). For simple members For members in continuous construction For class 1,2 and 3 sections Calculate λ = LE/r For members in lattice frames and trusses determine λ

Table 9 Table 11 or 12 Clause 3.6

For class 4 (slender sections) calculate λ = LE/r (Aeff/Ag) Select appropriate strut curve according to section shape and axis of buckling Obtain the compressive strength from the appropriate strut table and the appropriate value of design strength. Calculate the compressive resistance from the product of the area (effective area for slender sections) and the compressive strength.

Clause 4.7.4

5. 6. 7. 8. 9. 10.

0.5

Table 22 Appendix E Clause 4.7.4 Τable 25

Table23 Table 24

Clause 4.7.4.

2- 13

2.9 EXAMPLES Q1. Check the column shown in S275 steel. It is pin-ended about both x-x and y-y axes and the load shown is factored and includes self weight. The section is classified as not slender. 2500 kN

6m

   C    U    9    2    1   x    0    6    3   x    6    5    3

Solution 356 x 368 x 129 UC, Grade S275 2 T = 17.5 mm, Ag = 164 cm , r x = 15.6 cm, r y = 9.43 cm 2

Flange thickness T  > 16

 py

=

265 N/mm

The section is NOT slender, therefore

Pc

=

 Ag pc 

Table 9 4.7.4

For buckling about the x-x axis, use strut curve (b)

Table 23

For buckling about the y-y axis, use strut curve (c)

Table 23

Slenderness

λx

=

λy

=

 L EX  r  x  L EY  r  y

=

=

6000 15. 6 × 10 6000 9. 43 × 10

 

=

38.5

 

=

63.6

For λx = 38.5 and py = 265

 pcx = 243 N/mm

For λy = 63.6 and py = 265

 pcy = 188 N/mm

 pc is the lesser value of  pcx or  pcy

2

Table 24(b)

2

Table 24(c)

2- 14

Pc

=

2

164 x 10  x 188/10

3

=

3083 kN

Since Fc < Pc i.e 2500 < 3083 Therefore, section OK

Design using Design Table 356 x 368 x 129 UC, Grade S275, Effective length = 6.0m  Note that all UC sections are non slender under pure compression From Page 224 of the design table, Le = 6.0m Pcx = 3990kN Pcy = 3090kN (Control!)

2- 15

3 Restrained Beams 3.1 Introduction In order to perform satisfactorily a restrained beam must be checked for: i) Adequate lateral restraint ii) Local Buckling iii) Shear iv) Bending and combined bending and shear v) Web bearing and buckling vi) Deflection

3.2 Lateral Restraint When a beam is in bending, there is a tendency for the top flange, which is in compression, to  pull the section out of plane and cause buckling. In order t o prevent this and allow the section to achieve its full moment capacity it is important that the compression flange is restrained so that only vertical movement of the beam is allowed Full lateral restraint is defined in clause 4.2.2 as follows: "Full lateral restraint may be assumed to exist if the frictional or positive connection of a floor (or other) construction to the compression flange of the member is capable of resisting a lateral force of not less than 2.5% of the maximum force in the compression flange of the member, [under factored loading]. This lateral force should be considered as distributed uniformly along the flange.........." In practice most floor constructions would be considered adequate to carry this force, but care should be taken with timber floors, where positive fixing should be considered. Where a check is required the maximum force in the flange can be approximated by: Applied Force = Frictional Force =

Maximum Moment / Depth of section Total load on beam x Coefficient of friction / Length of beam

3.3 Local Buckling Local buckling is prevented by the correct classification and design of the cross section. The cross section should be classified as plastic, compact, semi-compact or slender and the section designed accordingly. Reference should be made to Tables 11 and 12 of the code and Figure 3.1 of these notes. Consider first a class 2 (compact) cross section where the breadth to thickness ratio of the flange outstand (b/T) is less than 10 ε and the depth to thickness ratio of the web(d/t) is less than 100ε. In this case the elements of the cross section are relatively stocky and are able to sustain relatively large strains. The stress is able to distribute throughout the section to form a rectangular stress block. The moment capacity Mc = pySx. Most universal beams and columns can be classed as compact cross sections. If the section were a class 1 (plastic) section where the b/T is less than 9 ε and the d/t less than 80ε the moment capacity would still be taken as M c=pySx, although the section has additional

2- 16

rotation capacity. This is not needed for normal design purposes but, if plastic analysis of the structure is used, it would be essential to have a plastic cross section to allow the necessary rotation at the hinge position. b

T

t d

a) Class 1  b/T < 9 d/t < 80

Stress = py

M

Mc = py S

ε ε

 Mc = py Z

Rotation capacity

Rotation

b)

Stress = py

M

Mc = py S

Class 2

 Mc = py Z

ε ε

/T <10 d/t < 100

Rotation Stress = py

M

Mc = py S Mc = py Z

c) Semi - compact  b/T < 1 5 d/t < 120

ε ε

Rotation

M

Stress = py d)

Mc = py S  Mc = py Z

Slender  b/T > 15 d/t >120

ε ε

Mc = py Zeff  Rotation

Figure 3.1 Section classification of rolled sections Consider next a class 3 (semi-compact) cross section in which the extreme fibres are able to withstand a design strength of p  without buckling. In this case the cross section can attain an y

elastic stress distribution although local buckling would occur before the attainment of a rectangular stress block. The moment capacity is then given by M = p  Z  or alternatively by p ySeff  c

y

x

Finally consider a class 4 (slender) cross section in which one or all the elements exceed the limits given for a semi compact section. Such elements will buckle locally before the full capacity of the section is obtained and effective section properties must be used. The effective section properties can be obtained from tables can be obtained from the expressions in section 3.6 of the code. The moment capacity is then given by M = p  Z . c

y

eff 

2- 17

This situation will never occur for beams employing universal beam and column sections. But rectangular hollow sections may need more attention. The classification of the whole section will be governed by the lowest classification of any of the elements (i.e. a section with a class 4 web, but class 3 flanges will be classed as a class 4 section). Having carried out this classification and used the design procedure appropriate to the cross section, then if local buckling is an issue an appropriate allowance will have been made.

3.4 Shear Two modes of shear failure are envisaged by the code, first shear failure due to the shear capacity of the web being exceeded and second shear buckling. The latter will only become a design criterion when the web is relatively thin i.e. when d/t >70 ε for rolled sections and d/t>63ε for welded sections. As such webs only occur in fabricated sections they will not be considered further in this lecture. The shear capacity of the section is defined in clause 4.2.3 as: Pv = 0.6 p y Av where A is the shear area. v

For rolled I and H sections A  is simply the product of the overall depth of the section and its v

thickness.

3.5 Bending and Shear The code deals with this in one of two ways depending on the level of the shear.

3.5.1 Moment capacity with low shear load: If the shear load (F ) is less than 60% of the shear capacity (P ) the effect of shear on the v

v

 bending capacity is so low that it may be ignored. Thus the moment capacity is given in clause 4.2.5.2 as: For Class 1 or Class 2 sections

Mc= py S

For Class 3 sections

Mc = pyZ or alternatively M = p  S

For Class 4 sections

M=p Z

c

c

y

eff 

y eff 

Where S and Z are defined in section tables and S eff and Zeff  are explained in the lecture on local buckling. In order to prevent permanent deformations at working load the moment capacity of simply supported beams and cantilevers employing Class 1 and Class 2 sections should be limited to 1.2 p  Z. y

3.5.2 Moment capacity with high shear load: If the shear load is greater than 60%of the shear capacity the effect of shear should be taken into account when calculating the moment capacity according to clause 4.2.5.3 as follows:

2- 18

For Class 1 or Class 2 sections is given by

M = p  (S-ρS )

For Class 3 sections

Mc = py(Z-ρS /1.5 ) orM = p  (S -ρS )

For Class 4 sections

M = py(Zeff -ρS /1.5 )

c

y

v

v

c

c

y

eff 

v

v

Where

ρ takes account of the level of shear in the section

i.e.

ρ = (2 (Fv/ Pv) - 1)

2

Which tends to zero at Fv=0.5P v although the reduction is trivial until Fv>0.6P v. Sv is, for a section with equal flanges, the plastic modulus of the shear area i.e the plastic modulus of the shear area Dt for a rolled section 2 i.e.S v = 2((D/2 . t) D/4) = D  t / 4.

3.6 Deflection Deflection is a serviceability limit state and in general calculations s hould be based on unfactored imposed load and compared to the suggested limits given in Table 8 of BS 5950 Part 1 Part of which is shown in Table 3.1 of these notes.

Table 3.1 Suggested deflection limits Deflection on beams due to unfactored imposed load Cantilevers Beams carrying plaster or other brittle finish All other beams Crane girders Vertical Horizontal

length/180 span/360 span/200 Span/600 Span/500

These limitations are based on commonly accepted principles, but the clause recognizes that circumstances may arise when greater or lesser values may be more appropriate’. The code also makes it clear that the limitations are given to ensure that finishes are not damaged. For example the traditionally accepted value of span/360 for beams, is based on prevention of damage to plaster ceilings below the beam. In other cases a more relaxed limit of span/200 is allowed. Vertical and horizontal deflection limits are given for crane gantry girders, which appear rather restrictive (span/600 and span/500 respectively). It is recommended that the manufacturer is consulted to ascertain the actual deflections that the crane can tolerate during operation. It should be noted that in this case the total load of the crane as well as the lifted load should be treated as ‘imposed’ load In some cases it may be necessary to calculate deflections due to dead load, to ensure that the structure has an acceptable appearance or that any clearance or tolerance requirements are met. This may be a wise precaution when using long slender composite beams, as high

2- 19

deflections can result due to the weight of the concrete on the non composite beam. This is of  particular significance if there are no ceilings beneath the beams. In some case, such as portal frame rafters and lattice girders the dead load deflection can be 'removed' by carefully  presetting members. In the case of long members dead load deflection can be dealt with by the use of pre-cambering, but cambers less than span/100 are unlikely to be successful. Deflections at the serviceability limit state can be calculated for simply supported beams, from the following standard formulae:

For a udl with total load of W kN

For a central point load of W kN

δ  =

δ  =

For point loads of W kN at 1/3 points δ  =

5 Wl 3 384  EI  1 Wl 3 48  EI  23 Wl 3 648  EI 

3.7 Establishing the Amount of Camber in Beam Beams can be cambered to accommodate part of the dead-load deflection, the full deadload deflection, or dead-load deflection plus part of the live-load deflection, at the discretion of the engineer. This can be influenced by the relative percentages of dead and live load, the  probable frequency and intensity of live load, the performance history of similar members, aesthetics, or other pertinent factors. As previously mentioned, determining the amount of camber is a very inexact process. After the cambering process, performance of the member often is not according to the script. In general, the anticipated amount of beam deflection does not occur. This probably is due to some degree of end fixity of the beam connections.

3.8

Additional checks.

Checks may be required for the buckling and bearing capacity of the web and the effect of holes in the web. As these checks are common to both restrained and unrestrained beams they will be dealt with in a separate lecture.

3.9

Summary of Design Procedure:

1. Select the section and determine the value of p y

Table 9

2. Determine the section classification

Tables11&12

3. For class 4 (slender) sections calculate effective modulus

Clause 3.6

4. Check the shear capacity

Clause 4.2.3

5. Check the moment capacity with low shear with high shear load

Clause 4.2.5.2 Clause 4.2.5.3

2- 20

2- 21

3.9 Example Consider a simply supported beam 914 x 419 x 388 UB, S275 steel subjected to a factored shear force of 2500kN and moment of 4000kNm. Check the shear and bending resistance of the beam if it is fully restrained against lateral-torsional buckling. Section classification The beam section is classified as plastic Check for b/T and d/t

Table 11

Section properties Since T = 36.6 ⇒ ρy = 265N/mm 3 Sx = 17,700cm 3 Zx = 15,600cm

2

Table 9

Check for Shear d t

=

799 21.5

= 37.2 < 70ε

i.e. beam problem 2

Av = Dt = 920.5 x 21.5 = 19,791mm Pv = 0.6ρyAv = 3147kN Fv = 2500kN < P v = 3147kN Shear is OK Check for Moment 0.6Pv = 0.6 x 3147 = 1888.2kN Since Fv = 2500kN > 0.6P v = 1888.2kN Moment capacity needs to be reduced due to high shear Mc = ρy(S - Svρ) 2

≤  1.2ρyZ,

Sv =

tD 2

4

= 4554cm 3

⎛  ⎛ F  ⎞  ⎞ ⎛  2500 ⎞  ⎞ ρ = ⎜⎜ 2⎜⎜ v ⎟⎟ − 1⎟⎟ = ⎜⎜ 2⎛  ⎜ ⎟ − 1⎟⎟ = 0.347 P 3147 ⎝   ⎠ ⎝   ⎠ v ⎝   ⎠ ⎝   ⎠ 2

3

Mc = 265 (17700 – 4554 x 0.347) / 10  = 4191 kNm 1.2ρyZ = 4961 kNm ∴ Mc = 4191 kNm > 4000 kNm (factored moment)

OK!

Using Design Table 914 x 419 x 388 UB, S275 steel under pure bending Page 196: Section is plastic Mcx = 4680kNm Pv = 3130kN  Note that the moment capacity given in the table is for low shear. The above handle calculation procedure needs to be applied for high shear case.

2- 22

4. UNRESTRAINED BEAMS 4.1. Introduction When designing a steel beam it is usual to think first of the need to provide adequate strength and stiffness against vertical bending. This leads naturally to a member in which the stiffness in the vertical plane is much greater than that in the horizontal plane. Sections normally used as beams have the majority of their material concentrated in the flanges which are made relatively narrow so as to prevent local buckling. In addition the need to connect beams to other members suggests the use of open sections i.e. I or H sections. The combination of all these factors results in a section whose torsional stiffness is relatively low, this has a major  bearing on the strength of an unrestrained member.

M

M

L (a) Elevation

(b) Cross Section

Pins prevent end rotation but allow end warping (c) Plan u

z

y

z x

(c) Deformation at centre span

y

Figure 4.1 Lateral Torsional Buckling

It can be seen from considering the behaviour of struts that whenever a structural member is loaded in its stiff plane (axially in the case of a s trut) there is a tendency for it to fail in a more flexible plane (by deflection sideways in the case of a strut). When applying this principle to a  beam in bending it can be seen that by loading the member in its stiffer plane (the plane of the web) the compressive force in the unrestrained flange causes the member to fail in its weakest direction (by deflecting sideways and twisting) (See Figure 4.1)

Many types of construction effectively prevent this type of failure, thereby enabling the member to be designed by considering its performance in the vertical plane only. Normal  beam and slab construction is an example of a situation where the member is restrained to  prevent buckling, but even here it should be recognized that during erection the member may  be unrestrained. Hence although the load may be less than the final design loading, checks on the stability of the member should be carried out at this stage. Situations where lateral torsional buckling has to be taken into account are less common, but typical examples are gantry girders, runway beams and members supporting walls and cladding.

2- 23

4.2. Factors influencing buckling resistance 1. The unbraced span , the distance between points at which lateral deflection is  prevented. 2. The lateral bending stiffness (E I ) y

3. 4.

The torsional stiffness (G J) The shape of the moment diagram i.e. members which are subject to non uniform moments will have a varying force in the compression flange and will therefore be less likely to buckle than members where the compression is uniform. The conditions of restraint provided by the end connection The level of the applied load and whether or not it is free to move as it buckles.

5. 6.

4.3. Behaviour of beams Three distinct stages of failure can be identified depending on the slenderness of the member (see Figure 4.2).

• •

Short stocky members will attain the full plastic moment M p. Slender members will fail at moments approximately equal to the elastic critical moment ME. (This is the theoretical value which takes no account of imperfections and residual stress.) Beams of intermediate slenderness fail through a combination of elastic and plastic  buckling. In addition, imperfections and residual stresses are most significant in this region

•

Mp

Plastic Failure

Resistance Moment Mb

Elasto - Plastic Region

Elastic Critical Moment M E

Elastic Failure

Practical Region 35

150

Slenderness

= L/ry

Fig. 4.2 Behaviour of beams with regard to slenderness

The code deals with this complex situation by expressing the buckling resistance moment (M ) as the lowest root of the Perry type equation i.e.  b

(ME - M b)(M p- M b) = ηLT ME M b ME M

is the elastic critical moment E

=

Μ pπ 2Ε λLT2 p

λLT

y is an expression which reflects the length and slenderness of the beam

M

is the plastic moment ( = p  S )

 p

y

x

2- 24

η

is a coefficient which takes account of imperfections, residual stresses, and strain

LT

hardening Further details are given in appendix A of these notes and Appendix B of the code.

4.4 Code requirements 4.4.1 Moment capacity

The section classification and moment capacity of the section should be found and checked in the same way as for restrained beams. Any reductions for high shear forces should be included in this check, .i.e. M x < Mcx 4.2 Buckling resistance moment The buckling resistance of the section between the either the ends of the section or any intermediate restraints should be checked as follows:

Mx < M b/mLT Where Mx is the applied moment M b is the buckling resistance moment mLT is the equivalent uniform . The buckling resistance moment Mb is calculated as follows: For class 1 and class 2 cross sections

M  = p  S

For class 3 cross sections

M b = p b Zx

For class 4 cross sections

M b = p b Zx,eff 

 b

 b

x

For cross sections with class 1 or class 2 flanges but class 3 webs

M b = p b Sx.eff

where  p is the bending strength  b

S

x

is the plastic modulus

Zx is the elastic modulus. The value of the bending strength s trength p  is obtained from tables 4.5 and 4.6 and depends on the  b

value of the slenderness

λLT. =

u v λ.βw

λLT. Between the points of restraint

1/2

where u is a tabulated section property v is obtained from table 4.7 and depends on λ / x x is a tabulated section property λ is the slenderness = L E/r y LE is the effective length between points of restraint restraint r y is the radius of gyration about the minor axis and is a tabulated section property

2- 25

For class 1 and class 2 cross sections βw = 1.0 For cross sections with class 1 or class 2 flanges but class 3 webs For class 3 cross sections For class 4 cross sections

βw = βw =

βw= Sx.eff / Sx

Zx / S x Zx,eff /Sx

For a quick conservative design for rolled I and H sections with equal flanges: u may be taken as 0.9 v may be taken as 1.0 x may be taken as D/T where D is the depth of the section and T is the thickness of the compression flange.

4.5 Moment Gradient factor ( mLT) The theoretical values of the buckling resistance moment are based on a beam subject to uniform moment . Members which are subject to non uniform moments are less likely to  buckle as the compressive force in the flange flange varies. The factor which takes account of this is mLT which may be obtained from table 18 of BS 5950. Consider the beams shown in Figure 4.3

•

Beam A has a central point load which does not restrain the beam. The unrestrained length is therefore equal to the length of the beam (A-D) The compression flange is subject to a varying compression and the equivalent uniform moment factor from Table 18 is 0.85.

•

Beam B is subject to the same moment but the load is applied as two point loads which do not restrain the beam. Again the unrestrained length is equal to the length of the beam (AD). The central portion of the beam is in uniform compression and the beam is more likely to buckle than Beam A. In this case therefore the equivalent uniform moment factor is 0.925.

•

Beam C is again subject to the same moment but the load is applied as two point loads which do restrain the beam. In this case the unrestrained lengths are between the ends and the intermediate restraint (A-B), between the intermediate restraints (B-C) and between the restraint and the end (C-D). The central portion of the beam is in uniform compression and, providing the three lengths between restraints are equal, it is this length which requires checking with an equivalent uniform moment factor of 1.0.

•

Consider beam D where the lengths are unequal and more than one length may have to be checked. • The equivalent uniform moment factor for A-B would be 0.6 and the unrestrained length would be taken as equal to the length of A-B. • The equivalent uniform moment factor for B-A would depend on the ratio of the end moments β = the smaller end moment divided by the larger, i.e.M 2/M1  should always be less than 1.0 and the unrestrained length would be taken as equal to the length of B-C. • The equivalent uniform moment factor for C-D would be 0.6 and the unrestrained length would be taken as equal to the length of C-D. • In each of these cases the buckling resistance moment is compared to the maximum moment within the unrestrained length. In this case it would be M 1 for A-B and B-C and M2 for C-D. Note that in this particular case C-D would not be critical.

2- 26

a) A

D

A

D

M

b) A

B

C

D

A

D B

M

C

c) A

B

C

D

A B

D

M

D

M

C

c) A

B

C

D

M1

A B

M2 C

Figure 4.3 Effect of Moment Gradient

The equivalent uniform moment factor ( mLT) therefore takes account of the shape of the  bending moment diagram between between restraints and may be obtained from Table 18. The first  part of the table deals with linear moment gradients, i.e.sections with no load between restraints. The second part deals with sections which are subject to transverse loading and the third provides a general formula from m LT may be calculated for more complex cases such as continuous beams and from which the values of m LT in the first two parts of the table may be obtained. 4.6 Effective lengths The theory of lateral torsional buckling is based on the assumption that the ends of the member are effectively pinned in both the vertical and horizontal planes. If connections are  provided which restrict movement movement of the ends of the member or allow allow rotational movement this needs to be considered and this is done by means of an effective length which may be greater or less than the actual length of the member between restraints.

Values of LE are given in table 13 for beams and table 14 for cantilevers. Part of table 13 is shown below as Table 4.1of these notes .

2- 27

Table 4.1 Effective length for beams without intermediate restraint

Conditions of restraint at supports Compression Flange laterally restrained  Nominal restraint against rotation about longitudinal axis

Both flanges fully restrained against rotation on plan (1) Both flanges partially restrained against rotation on plan (2) Compression flange partially restrained against rotation on plan (3) Both flanges free to rotate on plan (4)

Loading condition  Normal destabilizing 0.7Ly  0.85Ly 0.8Ly 

0.9Ly

0.85Ly 

1.0Ly

1.0L y 

1.2Ly

In most cases the effective length will be less than or equal to the actual length. Where the member is torsionally unrestrained at the end, or the load is destabilizing then the effective length may be greater than the actual length. And this is also reflected in the values given in Tables 13 and 14 4.6.1 Torsional Restraints. Torsional restraints may be provided by positive connection of both flanges to another part of the structure or by bearing stiffeners which have a minimum stiffness as specified in clause 4.5.7 (see Figure 4.4).

(a) Torsional restraint using connections

Positive Connection

(b) Torsional restraint using stiffeners

Figure 4.4 Torsional restraint 4.6.2 Destabilising loads Destabilizing loads are loads which are applied to the beam above the shear centre and are free to move with the beam as it deflects laterally and twists (see Figure 4.5). Such loads increase the rotation of the beam and induce additional stresses. Where they are present the

2- 28

effective length should be increased. Theoretically we would expect that the effective length could be decreased if the load was applied below the shear centre but the code makes no allowance for this.

u

Figure 4.5 Destabilizing loads 4.6.3 Intermediate restraints The intermediate restraints must have adequate stiffness and strength. The code defines adequate strength as the restraint being able to resist a force of 2.5% of the maximum factored force in the compression flange divided between the points of restraint in proportion to their spacing. The force in each restraint should not be taken as less than 1% of the total force.

Where several members share a common restraint the force should be taken as the sum of the largest three forces required for each member. If parallel members are taken as sharing the same restraint system the sy stem should be anchored to a robust part of the structure, or a system of triangulated bracing should be  provided in, or close to, the compression flange.

BAY 1 Triangulated

BAY 2 Tied to braced bay

Figure 4.6 Bracing of parallel members

Adequate stiffness is difficult to define but it has been suggested that this can be achieved by making the braced element 25 times stiffer in the lateral direction than the unbraced element. This is a good rule of thumb and is easily achieved with triangulated syste ms. It can however cause problems if the element to be braced is already very stiff in the lateral direction and therefore should be applied with care. Fig. 4.6 suggests the use of triangulated braced bay to  provide effective lateral bracing to the beam. Generally the effective length of members between restraints should be taken as 1.0L y for normal load conditions and 1.2L y for destabilizing load conditions

4.7

Special situations

Lateral torsional buckling need not be checked in the following situations:

2- 29

•

Circular or square hollow sections or solid bars

•

When bending is about the minor axis ( a section bending only about its minor axis cannot fail about its major axis)

λLT is less than

⎛ π 2 E  ⎞ ⎟⎟ = 0.4⎜⎜ ⎝   py  ⎠

1/ 2

•

For I H or channel sections when

•

RHS when LE/r y is less than the values given in table 15 for various values of D/B

λ o

4.8 Design Procedures 4.8.1

Full Design Procedure

1. Select Section -

determine p

Table 6

y

determine section class

Table 7

2. Check M < M cx allowing for shear as for a restrained beam 3. Determine actual unbraced length and effective length

4.3.5

4. Calculate slenderness λ = Le/r y 5. determine v using λ/x

Table 19

4. Calculate λLT = u.v.λ 5. Determine p b depending on λLT

Tables 11 and 12

6. calculate M  from M  = p S  b

 b

 b x

7. Determine equivalent uniform moment factor mLT

Table 18

8. Compare the maximum applied moment with M b/ mLT 9. Check web buckling and bearing as for a restrained beam if necessary

2- 30

4.9 Examples 4.9.1 Simply Supported Restrained Beam The simply-supported beam shown is fully restrained along its length. Check that the shear capacity and the moment capacity are adequate for the factored loading shown, which includes self weight.

225 kN W = 25 kN/m

533 x 210 x 101 UB Grade S355

5m

5m

 A

B

C

The maximum shear occurs at A and C. F vmax

25 × 10

=

2

+

225 2

=

238 kN

The moment at A and C = 0 The maximum moment occurs at B  M max

=

25 × 10 2

The shear at B =

8

+

225 × 10

225 2

4

=

= 875 kNm 113 kN

2- 31

Solution 1.

Check the section classification as Class 1

2.

Check the shear capacity

Pv

=

0.6 py Av

4.2.3

 py

=

345 N/mm

 Av

=

tD

t   D

= =

10.8 mm 536.7 mm

Pv

=

0.6 x 345 x 10.8 x 536.7 x 10  = 1200 kN

4.2.3(a)

2

Section Tables -3

Since  Fvmax <  Pv ,

i.e 238 < 1200

Therefore, 3.

Section adequate in shear.

Check the moment capacity

Check whether the shear is “high”, i.e. F v > 0.6 Pv

4.2.5

or “low”, i.e. F v < 0.6 Pv at the point of maximum moment. The shear at B = 113 kN 0.6 Pv =

0.6 x 1200

=

720 kN

113 < 720 Therefore, the shear is low. For low shear, the moment capacity for a Class 1 section  M c

=

 py S x 

4.2.5.2

For simply supported beam  M c # 1.2 py Z x  p y S x

=

1.2 py Z x=

Therefore, M c = Since M max < M c,

345 x 2612 x 10

4.2.5.1

-3

1.2 x 345 x 2292 x 10

-3

=

901 kNm

=

949 kNm

901 kNm i.e.

875 < 901

Therefore,

Design Using Tables 533x210UB 101 S355 steel From Design Table Page 281 Section is plastic Shar capacity = 1200 kN Moment capacity = 901 kNm

Section adequate in bending.

2- 32

4.9.2 Simply Supported Unrestrained Beam Consider the same beam used in Section 9.1 during construction, when no UDL load or lateral restraint exists except at the point of the point load application B. Check the adequacy of the unrestrained beam between A and B.

The maximum shear occurs at A&C F vmax

225

=

2

+

1. 4 × self  wt  2

112.5 + 0.07

=

113 kN

The maximum moment occurs at B  M max

225 × 10

=

4

The shear at B =

225 2

+  Moment  due to self  wt 

=

575 kNm

 = 112.5 kN

Solution

1.

Check the section classification as in Example 9. 1. The section is Class 1. The design strength py = 345 N/mm2.

2.

Calculate the shear capacity and moment capacity as in Example 9.1. Pv = 193 kN Shear capacity Moment capacity  M c = 901 kNm The shear at B is “low”

3.

Calculate the lateral torsional buckling resistance of the beam between A & B.

a.

Conservatively assume the effective length factor LE = 1.0L = 5 m

 b.

λ

c.

λ/x =

=

 LE/r y

109 33.2

=

=

5000 45. 7 3.3

=

109

Cl 4.3.5.2

2- 33

v  d. = 0.90 For class 1 section βw = 1

e.

λLT  =

f.

for  py = 345

Table 19 4.3.6.9

u v λβw   = 0.5

0.874 x 0.90 x 109 x 1.0 =

λLT = 86

and

86 Table 16

2

 p b = 170 N/mm

g.

For a Class 1 section, the lateral torsional buckling resistance moment  M  b = p b S x 3

4.3.6.4

6

 M  b = 170 x 2610 x 10 /10  = 444 kNm

The bending moment diagram between A & B is as follows: M A= 0 M B= 575

Therefore, β = 0 mLT = 0.6 h.

Table 18

The requirement is that:  M x = M  b/mLT

4.3.6.2

and

 M x < M cx

 M x

=

575 kNm < 444/0.6 = 740 kNm

 M x

=

575 kNm < 901 kNm

from Example 9.1

The section has therefore adequate lateral torsional buckli ng resistance during construction.

Design using Tables 533x210UB 101 S355 steel, effective length = 5m From Design Table Page 281 Section is plastic Moment capacity M b = 446 kNm for compact section  Need to calculate mLT to complete the design

2- 34

4.9.3 Unrestrained Propped Cantilever Beam A propped cantilever spanning 10m is loaded by cross beam spaced at 4m and 7m from its lefthand end. The restraint conditions at both ends of the beam and the load points may be regarded as being fully braced laterally and torsionally. The factored bending moment diagram is shown below. Assume S275 steel, check the adequacy of a beam consisting 457 x 152 x 60 UB against lateral torsional buckling.

Check the section classification as Class 1

Section properties:

T = 13.3 r y = 3.23cm 3 Sx = 1290cm

Table 11

⇒ Design strength = ρy = 275 N/mm2

Table 9

x = 37.5 u = 0.868

The three lengths must be checked separately. (a)

Length AB

(1)

LE = 4m

Beams with effective lateral restraints between the supports Clause 4.3.5.2

r y = 3.24cm λ = LE/r y = 400/3.23 = 124 (2)

x = 37.5, η = 0.5

λ/x = 123.5/37.6 = 3.3 ⇒ v = 0.897

(3)

u = 0.868

(4)

λLT = uvλ β w  = 0.868 x 0.897 x 124 x1.0 ;

Clause 4.3.6.7 or Table 19

β w = 1.0 for class1 sec tion

= 96.5 (5)

β = -150/280 = -0.54 mLT = 0.44

(6)

mLT Mx = 0.44(280) = 123 kNm

λLT= 96.5 , ρy = 275 N/mm2 ⇒ ρ b = 131 M b = ρ bSx

Table 18 

Table 16 

2- 35

-3

-6

-6

= 131 (10 /10 ) x 12980 (10 ) = 169 kNm

mLT Mx = 123 kNm < M b = 169kNm i.e. length AB is OK! (b)

Length BC

(1)

LE = 3m Beams with effective lateral restraints between the supports Clause 4.3.5.2

λ = LE/r y = 300/3.23 = 93 (2

λ/x = 93/37.5 = 2.48 η= 0.5 ⇒ v = 0.935

(3)

u = 0.868

(4)

β = 190/280 = 0.679

Clause 4.3.6.7 or Table 19

mLT = 0.87 mLT Mx = 0.87(280) = 244kNm (5)

Table 18 

λLT = uvλ β w  = 0.868 x 0.935 x 93 x1.0 = 75 ⇒ ρ b = 176N/mm2 M b = ρ bSx -3

-6

Table 16 

-6

= 176(10 /10 ) x 1290(10 ) = 227 kNm

mLT Mx = 244kNm > M b = 227kNm (b)

NG!

Length CD

By inspection length CD is less critical than length BC. Since CD has LE = 3m which is same length as BC, but

mLT (Mmax) = mLT (190) at length CD is less critical. Therefore it is necessary to redesign the member for length BC only. Try 457 x 191 x 67 and check Segment BC again

Section properties: ⇒ Design strength = ρy = 275N/mm2 T = 12.7 r y = 4.12cm x = 37.9 3 Sx = 1470cm u = 0.872

λ/x = 72.8/37.9 = 1.92 η = 0.5 ⇒ v = 0.96 λLT = uvλ β w  = 0.872 x 0.96 x 72.8 x1.0

Table 9

Clause 4.3.6.7 or Table 19

2- 36

= 60.9 ⇒ ρ b = 211.1N/mm2 M b = ρ bSx -3 -6 -6 = 211 (10 /10 ) x 1470 (10 ) = 310kNm > mLT Mx = 244kNm

Design using Tables Using Design Table Page 245 457 x 191 x 67 UB S 275 Steel, Length = 3m Mb = 310kNm for compact section

Table 16 

OK!

2- 37

4.9.4 Example: Design an unrestrained beam, 6m long, simply supported at both ends, carrying the following loads. Check that the beam can resist the design moment and shear.

Dead Load = 8 kN/m Imposed Load = 15kN/m S275 Steel DL = 8 kN/m, IL = 15kN/m

6m Factored design load= 1.4DL + 1.6IL= 35.2kN/m 2 WL = 158.4kNm Design moment = 8 WL = 105.6kN Design shear = 2 Unbraced length = 6m From Table 18, mLT = 0.925 From Design Table Pg. 245 Select 457x191UB67 S275 M b = 159 kNm for L b = 6m Hence M b = 159kNm > mLTMx = 0.925 x 158.4k = 146.5 Nm OK From Page 197 Shear Check: Pvx = 636kN > 105.6kN OK  Note: No need to check high shear since it does not co-exist with maximum moment

2- 38

5 Members Subject To Axial Load and Bending 5.1 Introduction Theoretical approaches to the behavior of members subject to axial load and bending are, for the most part, highly complex, yet do not match the behavior of real members even under test conditions due to imperfections such as out of straightness and residual stresses in the members. When members which form part of a structure framework where the conditions of restraint and the accuracy of construction vary there is even less correlation with the theory and virtually every code of practice including the different parts of BS 5950 and the Eurocodes adopt a slightly different approach.

The intention of this lecture therefore is to consider the approach to members subject to axial load and bending adopted in BS 5950 Part 1, looking at the background theory where it is relevant to the understanding of the approach adopted. The design of such members will be influenced by, the method of frame analysis, the shape of the cross section used and the type of restraint provided. In order to perform satisfactorily a member subject to axial load and bending must not fail due to:

• • •

Local buckling Inadequate local capacity (Tension or compression and / or bending)) Overall buckling major or minor axis buckling due to axial load major axis buckling due to major axis bending and axial load minor axis buckling due to minor axis bending and axial load minor axis buckling due to major axis bending and axial load

Clearly most situations include some combination of these effects and it is necessary to derive equations for design which will cover such combined effects. 5.2 Section Classification According to Clause 4.8.1, the classification of cross-section should be generally based on the combined moment and axial force and the classification should be used in obtaining the resistance terms, Pc, Mc M b for use in the interaction expressions.

In order to ensure that a member does not fail due to local buckling the cross section should be classified according to Table 11 and the design carried out according to the class of cross section. Classification of sections subject to axial load and has already been covered in detail in the lecture on local buckling. The requirements of Table 11 for universal beams and columns are shown in Figure 5.1.

2- 39

b

t d

T

Plastic section b/t < 9 d/t < 80 1+r

=

275 py

Compact section

Semi - compact section

b/t < 10

b/t < 15

when r2 > 0

1

when r2 < 0

d/t < 100 1 + 1.5 r1

d/t <

120 1+ 2.0 r

2

d/t < 100 1 + r1

but > or equal to 40 Where For universal beam and column sections with equal flanges Fc d t py

r = 1

r2 =

Fc Ag p

y

These factors vary for different shapes of section

Figure 5.1 Section Classification for members subject to axial load and bending.

5.3

Local Capacity (Cross section Capacity)

5.3.1 Compression and tension members Failure due to inadequate local capacity can occur in either tension or compression where  buckling is not a possibility and it will be observed that the formula used in BS 5950 Part 1 clause 4.8 are almost identical for these two cases. For tension members the following expression should be satisfied F t 

+

 M  x

Pt   M cx

+

 M  y  M cy

≤1

------------------------- (1)

For compression members the following expression s hould be satisfied F c  Ag p y

+

 M  x  M cx

+

 M  y  M cy

≤ 1 ------------------- (2)

F is the axial load Pt is the tension resistance Ag is the gross area (replaced by A eff  for class 4(slender) sections)

2- 40

 py is the design strength Mx is the applied moment about the major axis Mcx is the moment capacity about the major axis My is the applied moment about the minor axis Mcy is the moment capacity about the minor axis For class 4 (slender) sections, the effective area Aeff  should be used instead of the gross area Ag. The expression is derived from considering that the limit on capacity is reached when yield stress is attained within the cross section (see Figure 5.2). In practice this would suggest that a simple addition of stresses due to axial load and bending should not exceed yield and this is the first and simplest approach used in the code. For class 1 and 2 cross sections a plastic distribution of stress is used and for Class3 and 4 cross sections an elastic distribution of stress is used.

P Mx Mx Yield

+

=

Elastic

Yield +

=

P X

Stress Distribution

X

FIG 5.2 Short column subject to axial load and bending.

Plastic

2- 41

Alternatively where the section is a class 1 or 2 Universal beam or column section the following relationship should be satisfied

 z 2

 z ⎡ M  x ⎤ ⎡ M  y ⎤ ⎢ M  ⎥ + ⎢ M  ⎥ ≤ 1   -----------------(3) ⎣ rx ⎦ ⎢⎣ ry ⎦⎥ 1

where Mrx and Mry are the plastic moment capacity in the presence of axial load. Z1 and Z2 are empirical values varying for the type of section. Typical values given in the code are: Z1 = 2.0 for universal beams columns and hollow circular sections and 5/3 for hollow rectangular sections Z2 = 1.0 for universal beams and columns, 2.0 for circular hollow sections and And 5/3 for hollow rectangular sections There is an advantage in using the second expression as the first is linear and the second is nonlinear. See Figure 5.3.

Linear interaction Non linear interaction F

t

P

t

M

x

M

cx

Figure 5.3 Linear and non linear interaction

5.3.2 Reduced Moment capacity (Mrx and Mry) The terms Mrx and Mry require some explanation. For plastic and compact sections the capacity will approach the full moment capacity but, due to the effects of axial loading, the cross sectional moment capacity will be reduced to the value of Mrx. This is calculated from the design strength multiplied by a reduced plastic modulus .i.e. Mrx = py Srx The reduced plastic modulus of the section is simply the plastic modulus of the area remaining after deduction of the area required to carry the axial load. Consider the reduced plastic modulus about the major axis S rx which may be calculated as follows for sections with equal flanges and low axial load: See Figure 5.4

2- 42

B T

a d a a=

P 2.t.py

t FIG 5.4 Reduction in Plastic Modulus due to axial load

Assuming that the area carrying axial load can be contained in the web: S rx

⎛  d T ⎞ ⎛ d  ⎞ ⎛  ⎛ d  ⎞  ⎞ ⎞ = ⎜⎜ BT⎛  ⎜ + ⎟ + ⎜ − a ⎟ t⎜⎜ a + ⎜ − a ⎟ / 2 ⎟⎟ ⎟⎟2 ⎝  ⎝ 2 2 ⎠ ⎝ 2  ⎠ ⎝  ⎝ 2  ⎠  ⎠ ⎠

Accurate values of S rx for rolled sections are tabulated in section properties tables for varying values of axial load although the above method can give an approximate value when the flanges are equal and the compression area lies within the web. Values of reduced moment capacities are given in the design tables in the appendix for given F/P z values. More accurate approximations for both the major and minor axis are given Appendix J of the code.

5.4

Overall Buckling of Member

5.4.1 Tension Members Generally tension members will not be subject to buckling as the axial tension can assist in  preventing failure due to buckling caused by the bending in the section. Theoretically it should be possible to take account of this beneficial effect but the expressions involved are complex and it would be difficult to ensure that the correct combination of tension and bending is used to ensure the worst effect. Therefore BS5950 simply requires that the  bending effects are checked by themselves and the member treated as a laterally unrestrained  beam. This is to be done even when the tension and bending effects cannot occur independently.

5.4.2 Compression members In general the following relationships should be satisfied (i) Checking buckling about either axis due to axial load and bending.

2- 43

F c Pc

+

m x M  x  p y Z  x

m y M  y

+

 p y Z  y

≤ 1 -----------------(4)

Where: Fc is the applied axial load Pc is the minimum value of A g pcx and Ag pcy Mx is the applied moment about the major axis coincident with the force Fc mx is the equivalent uniform moment factor taking into account the shape of the moment diagram between restraints on the major axis my is the equivalent uniform moment factor taking into account the shape of the moment diagram between restraints on the minor axis

This expression is a combination of the two expressions required to prevent buckling about the major axis due to major axis bending and buckling about the minor axis due to minor axis  bending. See Figure 5.5. Fc

M + mx x Ag p cx  py Zx

≤

1

For major axis bending

my M y

≤

1

For minor axis bendin

Fc

+

Ag p cy

 py Zy P

P My

Mx

My

Mx

P

P y

X

X

y

Figure 5.5 Buckling in the plane of bending (ii) Checking buckling about the minor axis due to major axis bending and axial load This form of buckling is analogous to lateral torsional buckling in beams. The column buckles in a mode involving twisting and minor axis bending. The twisting mode distinguishes it from minor axis buckling and reduces the buckling load. It is significant for I and H sections  buckling at low axial loads. It is generally not relevant for tubular sections which are not liable to suffer from lateral torsional buckling. F c Pcy

+

m LT  M  x  M b

+

m y M  y  p y Z  y

≤ 1 

---------------------(5)

2- 44

In this case the value of Pcy is specifically used as we are considering buckling about the minor axis. See Figure 5.6 P Mx

Mx

P

Figure 5.6 Failure about the minor axis due to x-x bendi ng (lateral torsional buckling)

More exact Approach For I and H sections with equal flanges s the following more exact approach may be used

•

For members with moments about the major axis only F c Pcx

+

m x M  x  M cx

⎡ F c ⎤ ⎢1 + 0.5 P ⎥ ≤ 1  --------------(6) ⎣ cx ⎦

2- 45

The additional term after the moment is an amplification factor which allows for the additional moment created by the axial load at an eccentricity δ. See Figure 5.7. F

Μ

Secondary Moment

δ

Μ

Fδ

Primary Moment

Μ F

Moment diagram Deflected shape

Figure 5.7 Secondary moment induced by axial load acting on member deflection Checking lateral torsional buckling about the minor axis due to axial loads and bending about the major axis F c Pc y

+

m LT  M  x  M b

≤1 

----------------------(7)

In this case the amplification effect is less significant as the axial load is likely to be much less than the major axis strength.

•

For members with moments about the minor axis only F c P y

+

m y M  y  M cy

⎡ F c ⎤ ⎢1 + ⎥ ≤ 1  --------------(8) ⎢⎣ Pcy ⎥⎦

In this case the amplification is more significant than about the major axis and the full value of F c is used. Pcy

Checking buckling about the major axis due to axial loads and bending about the minor axis

2- 46

F c Pc x

+ 0.5

m xy M  y  M cy

≤1 

----------------------(9)

The amplification factor is negligible and the effect of the minor axis moment is small.

•

For members with moments about both axis

For major axis buckling Fc Pcx

+

mxM x M cx

m xy M yx ⎡ Fc ⎤ 1 0.5 0.5 + + ≤ 1 ------------------(10) ⎢ ⎥ P M cx cy ⎣ ⎦

For lateral torsional buckling F c Pc y

+

m LT  M  x  M b

+

⎡ F c ⎤ ⎢1 + ⎥ ≤ 1  ----------------------(11) ⎢⎣ Pcy ⎥⎦

m y M  y  M cy

For interactive buckling m x M  x (1 + 0.5 F c / Pcx  M cx (1 − F c / Pcx )

5.5

m y M  y (1 + F c / Pcy )

+

 M cy (1 − F c / Pcy )

≤ 1  -------------(12)

Design Summary

When a member is subject to axial load and bending a number of checks are required as follows: LOCAL CAPACITY CHECK Check (1) or (2) or (3) Tension

Compression

F t 

+

 M  x

Pt   M cx F c  Ag p y

+

+

 M  y  M cy

 M  x  M cx

+

≤1

 M  y  M cy

------------------------- (1)

≤ 1 ------------------------ (2)

Alternatively where the cross-section is class 1 or class 2  z2

 z ⎡ M  x ⎤ ⎡ M  y ⎤ ⎢ M  ⎥ + ⎢ M  ⎥ ≤ 1   -----------------------(3) ⎣ rx ⎦ ⎢⎣ ry ⎥⎦ 1

2- 47

OVERALL BUCKLING Tension s – check moment capacity and resistance Compression Generally: Check both (4) and (5) F c Pc F c Pcy

+

m x M  x

+

m LT  M  x

 p y Z  x

m y M  y

+

 M b

≤1

 p y Z  y

+

m y M  y  p y Z  y

----------------------------(4)

≤ 1 

------------------------(5)

Alternatively for I and H sections with equal flanges For members with major axis moments only check both (6) and (7) F c Pcx

F c Pc y

•

m x M  x

+

m LT  M  x

 M cx

 M b

≤ 1 

-------------------(7)

For members with minor axis moments only check both (8) and (9)

F c P y

F c Pc x

•

⎡ F c ⎤ ⎢1 + 0.5 P ⎥ ≤ 1  --------------(6) cx ⎦ ⎣

+

+

m y M  y  M cy

+ 0 .5

⎡ F c ⎤ ⎢1 + ⎥ ≤ 1  --------------(8) ⎢⎣ Pcy ⎥⎦

m xy M  y  M cy

≤ 1   ------------------(9)

For members with moments about both axis check (10), (11) and (12)

2- 48

F c Pcx

Fc Pc y

m xy M  y ⎡ F c ⎤ + + ≤ 1 ------------------(10) 1 0 . 5 0 . 5 ⎢ ⎥ P  M  cx ⎦ cy ⎣

+

m x M  x

+

m LT M x

 M cx

Mb

+

m y M yx M cy

m x M  x (1 + 0.5 F c / Pcx  M cx (1 − F c / Pcx )

+

⎡ Fc ⎤ ⎢1 + ⎥ ≤ 1  -------------------------(11) ⎣⎢ Pcy ⎦⎥

m y M  y (1 + F c / Pcy )  M cy (1 − F c / Pcy )

≤ 1  --------------(12)

2- 49

5.6 EXAMPLES

1) What is the axial load capacity of a 203 x 203 UC 60 of 3.1m height assuming that the column is subject to equal end moments of 41.8kN producing a single curvature bending about the minor axis? Steel grade is S275.

Design using the design tables

My = 41.8 kNm

203x203 UC 60, S275 Steel Section is plastic for all axial load

Effective length = 3.1m From design tables in Page 263 L = 3m, Pcy = 1570 kN,

41.8 kNm

L = 3.5m, Pcy = 1430 kN,

→ L = 3.1m,

Pcy = 1542 kN,

 pyZy = 55.3kNm Overall buckling check: Simplified equations F mx M x m yM y + + ≤ 1.0 (←Major axis buckling does not control!) Pc ρx Z x ρyZy Overall buckling check involving minor-axis moment only: F Pcy

+

mxMx M b

+

m yM y

ρy Zy

=

F 1542

+0+

1.0 × 41.8 55.3

=1

gives F = 376kN

Local Capacity Check: use simplified formula

From design tables in Pages 262—263 Mcy = 66.3 kNm, P z = Ag py = 2100 kN F Agρy

+

Mx M cx

+

My M cy

=

F 2100

+0+

F = 756kN i.e. Overall buckling controls!!

41.8 65.3

=1

2- 50

2) Bending about the major axis - lateral torsional buckling What is the axial load capacity of a 203 x 203 UC 60 S275 of 3.1m height assuming that the column is subject to end moments of 168kN producing a double curvature about the major axis?

F? Design using the design tables (Page 263) 203x203 UC 60, S275 Steel

Mx =168 kNm

Section is plastic for all axial load

Effective length = 3.1m L = 3m, Pcy = 1570 kN, M b = 169 kNm L = 3.5m, Pcy = 1430 kN, M b = 161 kNm 168 kNm

L = 3.1m, Pcy = 1542 kN, M b = 167 kNm  pyZy = 55.3kNm

β = -1 (double curvature bending) From Table 18 ⇒ my = 0.44 Overall buckling check using simplified formulas F mxMx myM y + + ≤ 1.0 (By inspection this equation can never Control the design!) Pc ρx Zx ρy Zy The following equation is the most critical case: F mxMx myMy F 0.44 ×168 + + = + +0 Pcy M b 167 ρ y Z y 1542 =

F 1542

+

73.9 167

+ 0  = 1

gives F = 859kN  Note: the use of more exact equations given in BS 5950 clause 4.8.3.3.2 gives less conservative estimate (i.e., higher value of F). Local Capacity Check - Simplified formula Mcx = 180 kNm, P z = Ag py = 2100 kN (from design tables in Pages 262--263) M F M F 168 + x + y = + + 0 =1 A g ρ y M cx M cy 2100 180 F = 140kN i.e. local capacity controls!!

2- 51

Example 3: Biaxial bending Check a 254 x 254 x 167UC of length 5 metre in S275 Steel to resist the following factored axial loads and moments Factored axial load: F = 1000kN, Factored moments: M x = 300kNm M y = 50kNm

Assume that the moments are applied at one end of the column and that the other end is pinned, and the effective length is 5m about both x and y axes. Use the more exact interaction equations to check the cross section capacity and overall buckling. Design Strength Flange thickness T = 31.7

ρ y = 265 N / mm 2

Table 6

Section Classification Section is classified as compact. Local Capacity Check Use Exact Equation 1 ⎛ M x  ⎞2 ⎛ M y  ⎞ . ⎜ ⎟ +⎜ ⎟ ≤ 10 ⎝ M rx ⎠ ⎝ M ry ⎠ F /(A g ρ y ) = 1000 / 5640 = 0.18

From design table (Page 262) F/Pz = 0.1, M rx = 627 and M ry = 237 F/Pz = 0.2, M rx = 580 and M ry = 237 ⇒ Interpolate for F/P z = 0.18 gives M rx = 589 kNm and Mry = 237 kNm Check the more exact interaction equation for plastic section 2

⎛ 300 ⎞ + ⎛  50  ⎞ = 0.47 < 1 ⎜ ⎟ ⎜ ⎟ ⎝ 589 ⎠ ⎝ 237 ⎠ Section is adequate for local capacity Overall Buckling Check F c Pcx F c Pc y

m xy M  y ⎡ F c ⎤ ⎢1 + 0.5 P ⎥ + 0.5  M  ≤ 1 ------------------(10) cx ⎦ cy ⎣

+

m x M  x

+

m LT  M  x

 M cx

 M b

+

m y M  y  M cy

m x M  x (1 + 0.5 F c / Pcx  M cx (1 − F c / Pcx )

+

⎡ F c ⎤ ⎢1 + ⎥ ≤ 1  -------------------------(11) ⎣⎢ Pcy ⎦⎥ m y M  y (1 + F c / Pcy )  M cy (1 − F c / Pcy )

≤ 1  --------------(12)

2- 52

For L Ex

= L Ey = 5.0m , from design table in page 263

= 5090kN Pcy = 3610kN M b = 602kNm Pcx

From design table in page 262 Mcx = 642 kNm Mcy = 237 kNm M

Equivalent Uniform Moment Factor m

Assuming the column is pin-supported at the base for β = 0 Table 26 gives m x

β=0

= m y = m xy = 0.6

Table 18 gives mLT = 0.60 M=0

m xy M y ⎡ Fc ⎤ + + 1 0 . 5 0 . 5 ⎢ ⎥ Pcx M cx ⎣ Pcx ⎦ M cy 1000 0.6 × 300 ⎡ 1000 ⎤ 0.6 × 50 = + + + 1 0 . 5 0 . 5 5090 642 ⎢⎣ 5090 ⎥⎦ 237 = 0.196 + 0.308 + 0.0633 = 0.57 Fc

+

mx Mx

⎡ Fc ⎤ ⎢1 + ⎥ Pc y M b M cy ⎢⎣ Pcy ⎥⎦ 1000 0.6 × 300 0.6 × 50 ⎡ 1000 ⎤ = + + 1+ 3610 602 237 ⎢⎣ 3610 ⎥⎦ = 0.278 + 0.30 + 0.16 = 0.74 Fc

+

m LT M x

+

myMy

m x M x (1 + 0.5Fc / Pcx M cx (1 − Fc / Pcx )

=

OK

+

OK

m y M y (1 + Fc / Pcy ) M cy (1 − Fc / Pcy )

0.6 × 300(1 + 0.5x1000 / 5090) 642(1 − 1000 / 5090)

= 0.383 + 0.224 = 0.607 Section is adequate for buckling

+

0.6 × 50(1 + 1000 / 3610) 2370(1 − 1000 / 3610)

OK

2- 53

Example 4: Biaxial Bending

Check the ability of a S275 steel column of 203 x 203 x 60 and length 3.1m long to carry a compressive load of , assuming that the axial force acts at an eccentricity of 150mm from the column face such as to produce single curvature minor-axis bending and double curvature major-axis bending.

Use the simplified equations for local capacity and overall buckling

checks. Section is class one section 2,

A = 75.8cm r x = 8.98cm, S y = 652cm

3

L/r y = 3100/51.9 = 60 Overall buckling check F c Pc F c Pcy

+

m x M  x

+

m LT  M  x

 p y Z  x

+

 M b

m y M  y  p y Z  y

+

0.15m

≤1

m y M  y  p y Z  y

……………. (1)

340kN 0.15m

≤ 1   …………….

(2)

To find Agpc

λ = L/r y = 3100/52.0 = 60

(Note that minor axis buckling control!)

Select strut curve from Table 23. The column curve “c” is appropriate. The corresponding value of

ρc from Table 24c for L/r y = 60 and

py = 275 N/mm 2 is

ρc = 200N/mm2. Ag pc = 76.4 x 200 x 10 −1  = 1528 kN To find

ρyZx yZy

yZx

&

yZy

= 275x 584 x10 −3 =161 kNm = 275 x 201 x 10 −3 = 55.2 kNm

To find Mb

λLT = uvλ u = 0.846, x = 14.1

λ = L/r y = 60, λ/x = 4.26 v = 0.85

Table 19

λLT = uvλ = 0.846 x 0.85 x 60 = 43.1 From table 16, corresponding value of

ρ b = 255N/mm2

2- 54

-3

M b = 255 x 656 x 10  = 167 kNm

To find mx my, and mLT

βx = -1.0 (double curvature bending) ⇒  mx = 0.40

Table 26 

βy = 1.0 (single curvature bending) ⇒  my = 1.0

Table 26

β = -1 (double curvature bending) ⇒  mLT = 0.44

Table 18

Flexural buckling check: F

+

Pc

=

m x Mx

ρyZx

340

+

+

myMy

My

ρyZy

0.40 x 340 x 0.15

1528 161 = 0.223 + 0.126 + 0.924

+

1.0 x 340 x 0.15

+

1.0 x 340 x 0.15

Mx

55.2

= 1.27 Lateral torsional buckling: F Pcy

=

+

m LT Mx

340

M b

+

+

myMy

ρyZy

0.44 x 340 x 0.15

1528 167 = 0.223 + 0.134 + 0.924

55.2

= 1.28 The check is not satisfactory. A larger column size is required. Alternative you may try to use the more exact equation for buckling checks.

HOMEWORK: Check local capacity

2- 55

Example 5 Check the adequacy of the lattice girder top chord shown below. 1.8 m

150 x 150 x 5 RHS in S275

1.8 m

11.2 kN

11.2 kN

 A

11.2 kN

B

Factored loads

C

Assumptions

The chord member is a 150 x 150 x 5 RHS in S275 steel. The lattice frame is analysed and designed according to the empirical design rules given in Clause 4.7.10 of BS 5950.

4.7.10

The web members connections are taken as nominally pinned. The chord member is continuous and the moment generated by the purlin load = WL/4. Max purlin spacing = 1.8 m. Loading

From analysis, the factored axial load in ABC = 598 kN. The factored bending moment diagram is 5.04

B  A

5.04 C

5.04

You will need to check: The section classification. The cross section capacity. The buckling resistance by the general method 4.8.3.3.1.

4.8.3.2

2- 56

Solution

150 x 150 x 5

RHS S275 steel 2

Area of section

 Ag = 28.7 cm

Depth to thickness of sides Radius of gyration Elastic modulus Plastic modulus Web depth (D – 3t)

d /t  = 27.0 r  = 5.9 cm 3  Z  = 134 cm 3 S  = 156 cm d  = 135 t  = 5

Design Strength

Wall thickness = 5 mm < 16 mm Therefore, design strength  py = 275 N/mm

2

Section Classification ε  =

275

ε  =

 p y

275 275

= 1. 0

Flange b/t = d/t = 27.0

'Flange'

The limitation for a Class 1 section = 28 ε. The flange is therefore Class 1.

'Web'

Web d/t = 27.0 The limitation for a Class 1 section is d/t =

r 1 =

r 1 =

limit

64ε  1 + 0. 6 r 1

F c

2d  t  p y

≤1

598 × 10 3 2 × 135 × 5 × 275

64ε  1 + 0. 6r 1

=

but

= 1.6  

64 × 1 1 + 0. 6 × 1. 0

27.0 < 40

-1.0 < r 1 ≤ 1.0

therefore r 1 = 1.0

= 40 Therefore, web class 1

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Both the flange and the web are Class 1, therefore the section is Class 1. Shear

Maximum possible shear when the purlin is adjacent to a node, say F v = 11.2 kN Shear capacity Pv = 0.6 py Av  Av

=

⎛   AD  ⎞ = 28. 7 × 10 2 × 150 ⎜ ⎟ 150 + 150 ⎝  D +  B ⎠

=

1435 mm

Pv

=

0.6 py Av

=

0.6 x 275 x 1435/10

=

237 kN

F v < Pv

Pv

4.2.3

2

3

11.2 kN < 237 kN

Section adequate in shear Bending

If F v < 0.6 Pv

11.2 < 0.6 x 237 11.2 < 142

Then

=

 M c

 py S x  for class one sections with low shear

4.2.5.2

Also, M c < 1.5 p Z  

4.2.5.1

 M c

=

 py S x

=

 M c

=

1.5 py Z =

275 x 156/10

3

1.5 x 275 x 134/10

3

=

42.9 kNm

=

55.3 kNm

Therefore, M cx = 42.9 kNm  M x < M cx

5.04 < 42.9

Section adequate in bending Combined Axial Load and Bending

4.8.3.2

 Local Capacity Check Generally: F c  Ag p y

+

 M  x  M cx

+

 M  y  M cy

598 × 10 3 28. 7 × 10 2

× 275

+

≤1 5. 04 42. 9

+

0  M cy

Therefore, capacity adequate.

=

0.76 + 0.117 + 0

=

0.88 < 1

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Overall Buckling

Simplified method F  Pc

m x M  x

+

m y M  y

+

 p y z x

 p y z y

4.8.3.3.1

≤1

(No lateral torsional buckling for square hollow section) Slenderness

The effective length in plane taken between the nodes  LEx = 0.85 x 3.6 = 3.06 m

The slenderness in plane  L Ex 3. 06 × 10 3 λ  x = = 51. 9 λ  x = 5. 9 × 10 r  x The effective length out of plane between the purlins  LEy = 1.0 x 1.8 = 1.8 m

The slenderness out of plane λ  y

=

 L Ey

=

r  y

λ  y

=

1. 8 × 10 3 5. 9 × 10

= 30. 5

Pc is the lesser of Pcx and Pcy

For r x = 51.9  py = 275 Pcx

=

 pcx = 249 N/mm

 Ag pcx =

2

2

Table 24a -3

28.7 x 10  x 249 x 10  = 715kN

m x is determined between restraints on the x axis, i.e. A and C

5.04 M1

0 M2

0 M4

5.04 M5

5.04 M3

m x  =

0.2+

0.1 M 2

+ 0.6 M 3 + 0.1 M 4  M max

Therefore, using the simplified method.

= 0.8

Table 26 

2- 59

F c Pc

m x M  x

+

598 715

 p y z x

+

+

4.8.3.1

m y M  y  p y z y

0.8 × 5.04 × 10 6 275 × 134 × 10

3

+

0  p y z

=

0.84 + 0.109 + 0 = 0.95

Also check Lateral Torsional Buckling F c Pcy

+

m LT  M  LT   M b

+

m y M  y  p y z y

≤1

mLT is determined between restraints on the y-y axis, i.e. A and B (or B and C) For RHS, LTB need not be checked unless L/r  y exceeds the limiting value given in Table 15.  M  b  py S x =

M = 5.04

βM = -5.04

For λy = 30.5

Therefore, β = -1

 py = 275 N/mm

2

2

mLT = 0.44

 pcy = 266 N/mm

2

-3

Pcy = Ag pcy = 28.7 x 10  x 266 x 10  = 763kN F c Pcy

+

m LT  M  LT   M b

+

m y M  y  p y z y

=

598 763

+

0.44 × 5.04 42.9

+

0  p y z y

 = 0.84

Both of these factors are below unity. Therefore, the section will not fail due to either in-plane or lateral torsional buckling.

4.8.3.3 4.3.6.1 & Table 15

Table 18  Table 24a

2- 60

Example 6 A cantilever column is subject to factored axial load of 610 kN and horizontal load of15 kN as shown. The top of the column is free to deflect in plane but is braced laterally to  prevent out-of-plane deflection. Check the adequacy of a UB 305 x 127 x 48 Grade 43 member in terms of strength and stability. If the member is found to be inadequate, check whether an additional lateral bracing at point C as shown in the figure is sufficient to enhance the buckling resistance of the column.

F = 610 kN H= 15kN

B

1.5 m C

Add one additional lateral bracing here if necessary

3.5m Figure Forces acting on cantilever column

UB 305 x 127 x 48 Grade S275 A

Fixed

M

F

Given F = 610kN Mx = 15 x 3.5 = 52.5 kNm Lex = 2 x 3.5m = 7m (cantilever) Ley = 0.85 x 3.5m = 2.975 ≈ 3m (bottom end is fixed and the top end is pinned) For UB 305x127x48 Grade S275 steel section Section is plastic (use design table Page 253) Lex = 7m ⇒ Pcx = 1500 kN Ley = 3m ⇒Pcy = 759 kN LLT= 3m ⇒ M b = ρ b Sx = 118kNm  pyZx = 169 kNm Pz =Ag py = 1680 kN Local capacity F/Agρy + Mx/Mcx = 610/1680 + 52.5/195 = 0.36 + 0.27 = 0.63

OK!

In-plane buckling  Determining m x , β=0 → mx= 0.6. However, for cantilever column, m x should not be smaller than 0.85 (Clause 4.8.3.3.4), thus m x=0.85

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F/Pc + mxMx/ pyZx = 610/759 + 0.85x52.5/169 = 0.802 + 0.264 = 1.06 <1 OK! For cantilever without intermediate restraint, m LT = 1.0 (see footnote on Table 18 of BS code) Lateral Torsional Buckling without intermediate bracing F/Pcy + mLTMx/M b = 610/759 + 1.0x52.5/118 =0.802 + 0.445 = 1.25 > 1 NG Provide a lateral bracing at 1.5m from the top Check bottom segment AC Ley = 2m Pcy = 1200kN M b = 156kNm Moment at the bracing point = 15 x 1.5 = 22.5kNm β =22.5/52.5 = 0.43, from table 18 we have m LT = 0.71

Lateral Torsional Buckling with one intermediate bracing F/Pcy + mLTM/M b = 610/1200+ 0.71x52.5/156 = 0.508 + 0.24 = 0.75

OK

Check top segment BC Ley = 1.5m Pcy = 1400kN M b = 178 kNm β = 0, mLT = 0.57 Lateral Torsional Buckling with one intermediate bracing F/Pcy + mLTM/M b = 610/1400+ 0.57x22.5/178 = 0.436 + 0.072 = 0.51 Therefore provide the lateral bracing point at a distance 1.5m from the top.

OK

2- 62

HOMEWORK 1: BEAMS 1 A compound beam consisting of rolled steel beam 610 x 305 x 149 UB and two flange cover plates of 300mm x 30mm is designed for a factored load of 142 kN/m uniformly distributed over a span of 10m. Assuming s275 steel, full lateral support along the length. The beam is supported on a plate with stiff bearing length of 216mm, check the adequacy of the beam in respect of (a) shear and bending, (b) web bearing, and (c) web buckling (1992 supplementary exam.).

2 A simply supported universal beam spanning 12m is loaded by two concentrated factored loads as shown in Fig.2. (i) Assuming the beam is fully restrained against lateral torsional buckling, design the beam in S275 steel for shear and bending. (ii)Assuming the beam is braced laterally and torsionally only at the supporting ends and at the intermediate load  points, check the design for lateral-torsional buckling

Fig. 2

3 The beam shown in Fig. 3 is fully restrained along its length against lateral-torsional buckling. For the loading shown check the adequacy of a beam s ection of 610 x 305 x 179UB (S355 steel) in respect of (a) (b)

maximum shear, maximum moment with co-existent shear, and

(c)

⎛  PL3  ⎞ 5wL4 ⎟⎟ . beam deflection ⎜ δ max = + ⎜ 384 48  EI   EI  ⎝   ⎠

Fig. 3

Unfactored Loads: Dead Load:

Wd =5kN/m;

Pd= 500kN

2- 63

Imposed Load:

4

WI = 10kN/m;

PI = 1050kN

The beam shown in Fig. 3 is laterally restrained at the ends and at the two cross beams which carry uniformly distributed load. For the bending moment diagram shown, design the beam in S275 steel to resist bending only. (Hint: use a trial section UB 457 x 152 x 60kg/m)

Fig. 4

5

An universal beam of 457 x 191 x 74kg/m (S275 steel) is simply supported at both ends and subject to a concentrated factored force of 300kN at the mid-span, as shown in Fig. 3.5. The beam is braced laterally at the supports and at the load point only. Neglecting the self weight, check the adequacy of the beam in  bending and shear. 6 Design an unrestrained beam spanning 6.0m to support an uniform distributed load of the total mass (including mass of beam and casing) of 160kN as shown in Figure 6. Assume the ends of the beam are held against torsion, i.e. a full depth end plate welded against the flanges. The steel is S275 (BCSA, 1991,2nd. Ed.)

Fig. 6

7

A laterally unrestrained beam of 457 x 191 x 67 UB S275 Steel is spanning 12m and carrying secondary beam at third points loaded as shown in figure 7. (1) Check the  bending resistance of beam assuming (a) lateral support at the intermediate point loads, and (b) no lateral support at the point load positions. Also check the web against  bearing and buckling if the stiff bearing length at the load points and supports is 150mm

2- 64

Fig. 7 8 Determine the buckling resistance moment for a 356 x 127 x33 UB in S275 steel for a span of 4.2m, assuming that the applied loading produces moments which vary linearly from a maximum at one end to one quarter of this value at the other, both values being in a clockwise sense (Figure 3.8).

Fig. 8

2- 65

HOMEWORK 2: COMPRESSION MEMBERS 1 A column carrying a floor load is shown in Figure 1. The column can be considered as  pinned at the top and at the base. It is laterally restrained at 3.5m height from the column base to prevent movement in the y direction as shown in Section AA. Design the column using S275 steel for a factored floor load of 2500kN

x

y

Figure 1

2 Determine the compression resistance of a column of UC 254 x 254 x 89 with a clear height of 6m. The top of the column is effectively held in position but not restrained in direction. The bottom of the column is connected to a rigid foundation.

2- 66

HOMEWORK 3: MEMEBRS SUBJECT TO COMPRESSION AND MOMENTS 1 Design a UC column of length 5 metre in S275 steel to resist the following factored loads: Factored axial load F = 1000kN; Factored Moments: M x = 300kNm and M y = 50kNm. Assume that the moments are applied at the top of the column. The bottom of the column is pinconnected. The effective length may be assumed to be 5m about both x- and y-axes. Check local capacity and overall buckling. 2 Check the suitability of a 533 x 210 x 82 UB in S355 steel for used as the column in a portal frames of clear height 5.6m if the factored axial compression force is 160kN, the factored moment at the top of the column is 530kNm and the base is pinned. Both ends of the column are adequately restrained against lateral displacement. If the UB section is not suitable, determine whether an intermediate lateral restraint located at 1.6m below the top of the column is sufficient to provide resistance for column buckling. 160 kN 530 kNm

1.6m

5.6m

Fig. Q2

2- 67

3 A typical column in a multi-storey braced frame is subject to a compression force F = 610 kN

and Mx = 64 kNm in Fig. Q3. The column is laterally braced at the ends and at the midheight. Assume that the ends are simply supported, design a suitable UC section for the column member and check its adequacy for strength and stability. (You may use the simplified method or the more exact approach for member capacity check). F

Mx

Lateral bracing 4m

2m

Mx F

Figure Q3 Forces acting on the column

2- 68

4 Design the beams and columns for the structure steelwork shown below.

Unfactored floor loading: Imposed load = 5kN/m2; Dead load (including estimated self-weight) = 3kN/m 2 Column tops are pinned connected to the be ams. Column bases are fixed. All steels are S275. Use UC and UB sections only. Evaluate the two floor options and determine which option is more efficient in terms of steel usage.

3.5m

6m

9m

Floor plans: Option A 6m

3m

3m

3m

Option B 3m

3m

9m

2- 69

5 Design the canopy system consisting of beam, column and tie for the given loading shown  below:

Lateral restraint

Tie 1.5m

Beam 3m Column

IL = 30kN; DL = 50kN

4m

2- 70

6 A one-storey and two-bay frame consists of 9 floor beams, 6 columns and 3 bracing members

are required to support a floor slab with unfactored dead load of 5 kN/m 2 and imposed load of 2 3.5 kN/m  as shown in Fig. 6. The dead load includes the self-weight of the steel structure. The column bases are assumed to be rigidly connected to the foundations. All other members are assumed to be pin-connected. The column top is braced to prevent side sway in the y-direction  but is free to deflect in the x-direction causing major-axis bending.

a) If the secondary beams are laterally restrained by the slab and the primary beam are laterally unrestrained, select a suitable UB section of S355 steel for both the primary and secondary beams and checks its bending and shear resistance. Assuming that the column is subject to compression and a notional horizontal load (1.5% factored gravity load), select a suitable UC section S275 steel to resist the combined axial force and moment.

Secondary beams

3.5m

2.5m

Primary  beams

x

9m y 5m

9m

Fig. 6(a) Overall view of the steel structure

2.5m

2.5m

x 9m

9m

y Fig 6b Plan View