sci student worked examples 2nd draft

Student worked examples book Draft Introductory text 14 May 2007 Rev B

The Steel Construction Institute P:\EDN\EDN399\Student worked examples\Second draft-book\22746^DRAFT_SWE book Introductory text_02B.doc

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STUDENT WORKED EXAMPLES BOOK – INTRODUCTORY TEXT

1. SCOPE This book gives a brief introduction to the design of steel structures to the Eurocode system. Worked examples for steel elements within a building are given to demonstrate basic design processes.

2. EUROCODE SYSTEM The structural Eurocodes consist of 9 main codes which except for EN 1990 are sub divided in to parts. Each Eurocode and its parts consist of: • Main Eurocode text • Informative and normative annexes • National Annex The guidance given in the main Eurocode text and the informative and normative annexes are the same across Europe. Within those sections national choice is allowed for some design methods and factors, which are collectively known as Nationally Determined Parameters (NDP). The National Annex, where allowed in the Eurocode will: • Specify which design method to use • Specify what value to use for a factor • State whether an informative annex may be used In addition the National Annex may give references to publications that contain non conflicting complimentary information that will assist the designer. The guidance given in a Nation Annex applies to structures that will be constructed within that country. National annexes may differ between countries within Europe. Therefore the guidance given in the National Annex for the country where the structure will be constructed should be used and not that given in the National Annex of the country where the structure is designed. EN 1990 presents the basic principles that apply to structural design to the Eurocodes. It includes the combinations of actions and associated factors that should be used to determine design loads for ultimate and serviceability limit states. Most parts of the structural Eurocodes present design guidance in principle and application rules. Principle rules are denoted by the use of a letter P after the clause number e.g. 1.2(3)P, whereas application rules do not contain a letter P e.g. 1.2(3). Guidance given in principle rules must be used when designing to the Eurocodes. However, different guidance to that given in an application rule may be used if it can be shown that it gives the same reliability as that of the Eurocode rule. Structures designed using different guidance to that given in an application rule can not be said to be designed to the Eurocodes, but may be designed to the principles of the Eurocodes.



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The structural Eurocodes do not present design guidance in more than one part, e.g. guidance given in EN 1990 on combination of actions is not repeated in the material Eurocodes. Also guidance given in one part of a Eurocode is not repeated in other parts of the same Eurocode. Therefore when design a steel structure more than just part 1-1 of EN 1993 will be required. The structural Eurocodes that may be required for the design of a steel and concrete composite building are: EN 1990 Basis of structural design EN 1991 Actions on structures EN 1992 Design of concrete structures EN 1993 Design of steel structures EN 1994 Design of composite steel and concrete structures EN 1997 Geotechnical design EN 1998 Design of structures for earthquake resistance (depending on the location) In addition to references between structural Eurocodes and parts of a Eurocode references to other codes may be given e.g. product standards. EN 1993-1 comprises twelve parts (EN 1993-1-1 to EN 1993-1-12), when designing a building the following parts of EN 1993-1 will be required. EN 1993-1-1 General rules and rules for buildings EN 1993-1-2 Structural fire design EN 1993-1-5 Plated structural elements EN 1993-1-8 Design of joints EN 1993-1-10 Material toughness and through-thickness properties

3. STEEL GRADES Steel is graded based on its material properties e.g. yield and ultimate strength. EN 10025 specifies the requirements for different steel grades and the tests that should be carried out to determine the material properties. The design guidance given in EN 1993-1-1 covers steel grades S235, S275, S355 and S450. EN 1993-1-12 gives guidance on the applicability of the rules in EN 1993-1-1 to high strength steels. Some of the yield and ultimate strengths for S275 and S355 given in EN 1993-1-1 Table 3.1 are reproduced below. Standard and Nominal thickness t < 40 mm Nominal thickness t > 40 mm steel grade Yield strength

(fy) (N/mm2) Ultimate strength (fu) (N/mm2)

Yield strength (fy) (N/mm2)

Ultimate strength (fu) (N/mm2)

EN 10025-2 S275 275 430 255 410 S355 355 510 335 470 EN 10025-3 S275 N/Nl 275 390 255 370 S355 N/Nl 355 490 335 470



3

The National Annex may specify that the product standard EN 10025 should be used instead of the values given in Table 3.1 of EN 1993-1-1. Where the product standard values are used they should be considered as the minimum characteristic values. In addition to the main grade, steel is divided into sub-grades. The guidance given in EN 1993-1-10 may be used to determine the required steel sub-grade for a member. EN 1993-1-10 gives guidance for the selection of steel grade for fracture toughness and gives a maximum steel thickness that may be used. The steel material properties, the structural member characteristics and the design situations are used to determine the maximum thickness for a steel sub-grade.

4. LIMIT STATE DESIGN EN 1990 defines a limit state as a ‘state beyond which the structure no longer fulfils the relevant design criteria’. Limit state design provides a consistent reliability against the failure of structures by ensuring that limits are not exceeded when design values of actions, material and product properties, and geotechnical data are considered. Design values are obtained by applying partial factors to representative values of actions and properties. Limit state design considers the resistance, serviceability and durability of a structure. All relevant design situations should be considered for the structure. The design situations considered by the Eurocodes are:

• Persistent –the normal use of the structure. • Transient – temporary situations e.g. execution*. • Accidental – exceptional events e.g. fire, impact or explosion. • Seismic – seismic events that may act on the structure.

Two limit states are considered during the design process, ultimate and serviceability. 4.1 Ultimate Limit States

Ultimate limit states are those that relate to the failure of a structural member or a whole structure. These limit states usually consider the maximum resistance of a member or a structure. Design checks that relate to the safety of the people in and around the structure are ultimate limit state design checks. Limit states that should be considered where relevant are:

• Loss of equilibrium of the structure or a structural member. • Failure of the structure or a structural member caused by, excessive deformation

causing a mechanism, rupture, loss of stability, fatigue or other time dependent effects.

• Failure of the supports or foundations, including excessive deformation of the supporting ground.

* Execution is the term used in the structural Eurocodes for the construction process.



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4.2 Serviceability Limit States

Serviceability limit states concern the functioning of the structure under normal use, the comfort of the people using the structure and the appearance of the structure. Serviceability limit states may be irreversible or reversible. Irreversible limit states occur where some of the consequences remain after the actions that exceed the limit have been removed e.g. permanent deformation of a beam. Reversible limit states occur when none of the consequences remain after the actions that exceed the limit have been removed. Criteria that are considered during serviceability limit state design checks are:

• Deflections that affect the appearance of the structure, the comfort of its users and its functionality.

• Vibrations that may cause discomfort to users of the structure and restrict the functionality of the structure.

• Damage that may affect the appearance or durability of the structure. The Eurocodes do not specify any limits for serviceability criteria. The limits should be defined for each project based on the use of the element and the requirements of the Client. EN 1990 specifies that the serviceability limit state deflection should be checked for actions that occur after the execution of the structure. However, when members supporting only brickwork or cladding will be visible in the finished state the deflection due to the self weight of the brickwork or cladding should be determined to ensure that the deflection does not result in an aesthetically unpleasing deflected element.

5. COMBINATION OF ACTIONS EN 1990 requires the structure or member to be designed for the critical load cases that are determined by combining actions that can occur simultaneously. This implies that all variable actions that occur concurrently should be considered in a single combination. However, for buildings, note 1 of clause A1.2.1(1) of EN 1990 allows the critical combination to be determined from not more than two variable actions. Therefore engineering judgement may be used to determine the two variable actions that may occur together to produce the critical combination of actions for the whole building or the particular structural member under consideration within the building. 5.1 Ultimate Limit State

Two methods for determining the combination of actions to be used for the persistent or transient ultimate limit state (ULS) are presented in EN 1990. The methods are to use equation (6.10) on its own or to determine the least favourable combination from equations (6.10a) and (6.10b). The National Annex for the country in which the building is to be constructed must be consulted for guidance on which method to use. Where multiple independent variable actions occur simultaneously the Eurocodes consider one to be a leading variable action (Qk,1) and the other(s) to be accompanying variable


actions (Qk,i). A leading variable action is one that has the most onerous effect on the structure or member. The equations for the combinations of actions given in EN 1990 for ultimate limit state design are shown below. Persistent or transient design situation

ii

ij

jj QQPG k,0,

1iQ,k,1Q,1Pk,

1G, ψγγγγ ∑∑

≥≥+++

BS EN 1990 Eq. (6.10)

iii

ijj

j QQPG k,0,1

Q,k,10,1Q,1Pk,1

G, ψγψγγγ ∑∑≥≥

+++ Eq. (6.10a)

iii

ijjj

j QQPG k,0,1

Q,k,1Q,1Pk,G,1

∑γξ γ


5

ψγγ∑≥≥

+ ++ Eq. (6.10b)

i

ii

jj QQAPG k,

12,k,12,11,1d

1k, )or( ∑∑

≥≥+ + + ψ + ψAccidental design situation ψ Eq. (6.11b)

i

ii

jj QAPG k,

12,Ed

1k, ∑ψ Eq. (6.12b) ∑

≥≥+ + +Seismic design situation

Where: Gk,j unfavourable permanent action P prestressing action Qk,1 leading variable action Qk,I accompanying variable action Ad is an accidental action AEd is a seismic action γ, ψ and ξ are partial, combination and reduction factors given in EN 1990 Persistent or transient design situation The combination of actions given for the persistent or transient design situation is used for static equilibrium, structural resistance and geotechnical design checks. It should be noted that for structural checks involving geotechnical actions and ground resistance checks additional guidance on the approach to determining the combination of actions is given. Annex A of EN 1990 presents three different approaches and allows the National Annex to specify which approach to use. Guidance contained in EN 1997 should also be used when considering geotechnical actions. Accidental design situation The combination of actions for the accidental design situation can be used to determine a design value that either:

• Contains an accidental action (e.g. impact, fire); or • Applies to a situation after an accidental action has occurred (e.g. after a fire).

In the later case Ad = 0. Seismic design situation This combination of actions and guidance given in EN 1998 should be used when seismic actions are being considered.


5.2 Serviceability Limit State

The equations for the combinations of actions given in EN 1990 for serviceability limit state design are shown below. Characteristic combination

ii

ij

j QQPG k,1

0,k,11

k, ∑∑≥≥

+++ ψ BS EN 1990 Eq. (6.14b)

Frequent combination i

ii

jj QQPG k,

12,k,11,1

1k, ∑∑

≥≥+++ ψψ Eq. (6.15b)

Quasi-permanent combination

ii

ij

j QPG k,1

2,1

k, ∑∑≥≥

++ ψ Eq. (6.16b)

Characteristic combination This combination of actions should be used when considering an irreversible serviceability limit state. An irreversible limit state is one where after the removal of the action(s) that exceed the specified limit some of the consequences remain, e.g. permanent deformation of a beam or cracking of partition wall. The characteristic combination should be used when considering the functioning of the structure, damage to finishes or non-structural elements. Frequent combination Reversible serviceability limit states are covered by the frequent combination of actions. . A reversible limit state is one where after the removal of the action(s) that exceed the specified limit none of the consequences remain i.e. the element stays within its elastic region. This combination could be used when checking the non permanent vertical displacement of a floor that supports a machine that is sensitive to vertical alignment. Quasi-permanent combination The quasi-permanent combination of actions should be used when considering a reversible limit states or long term effects. When considering the appearance of a structure the quasi-permanent combination should be used.

6. DESIGN PROCESS The procedures that should be followed when designing a structure are:

• Concepts of structural frame o Layout of structural members o Type of connections, i.e. simple, semi-rigid or moment resisting o Stability of structure at all stages (during construction, use and demolition)

• Loading on structure and members • Design of individual members and connections • Choice of steel sub-grade • Protection of steel, e.g. fire and corrosion • Robustness checks • Frame stability calculations


6



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7. ROBUSTNESS Connections between building elements should be designed so that they prevent the building from failing in a disproportionate manner to the event that has caused the structural damage. EN 1991-1-7 gives the design requirements for making structures robust against accidental actions. The Eurocodes separate buildings into 4 classes, with different design requirements for each classes of structure. In addition to the requirements given in the Eurocodes any national requirements should also be satisfied. In England and Wales the requirements for the control of disproportionate collapse are given in Approved Document A of the Building Regulations. In Scotland the requirements are given in The Scottish Building Standards Technical Handbook Domestic and for Northern Ireland they are given in The Building Regulations (Northern Ireland) Technical booklet D.

8. FIRE DESIGN / PROTECTION Structural steelwork must be either protected or designed in such a way as to avoid premature failure of the structure when exposed to fire. Fire protection may be given to structural steelwork members by the use of

• Intumescent paints • Mineral boards • Concrete encasement

Design guidance for the accidental design situation for fire exposure is given in EN 1993-1-2 for structural steelwork and EN 1994-1-2 for steel and concrete composite structures. The guidance given identifies differences or supplements that given for ambient temperature design in EN 1993-1-1 and EN 1994-1-1.

9. CORROSION PROTECTION The main points to be considered during the design process when deciding on the type of corrosion protection to be applied to the structural steelwork are:

• Application of coating – need to ensure that the chosen coating can be efficiently applied.

• Contact with other materials. • Entrapment of moisture and dirt around the steelwork. • Other factors e.g. provide suitable access for maintenance and inspection during the

life of the structure. Types of corrosion protection for structural steelwork elements include painted coatings, hot-dip galvanising and thermal (metal) spraying. Guidance on corrosion protection can be found in the Corrosion Protection Guides produced by Corus.



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10. SOURCES OF FURTHER INFORMATION • Add SCI Eurocode design guides • Steel Designers’ Manual 6th Edition, Blackwell Publishing 2003, ISBN0-632-04925-1 • Architectural Teaching Resource Studio Guide, Second Edition • Corrosion protection guides – various titles available from Corus web site:

www.corusconstruction.com • Access steel web site: www.access-steel.com • Designers’ Guide to EN 1993-1-1 Eurocode 3: Design of Steel Structures - General

Rules and Rules for Buildings, L Gardner and D Nethercot, Thomas Telford 2005, ISBN 9780727731630

Drafting Note: Need to add further references.

http://www.corusconstruction.com/

http://www.access-steel.com/


Worked examples The worked examples contained in the next section have been formatted in to a common format. However, checks on consistency (e.g. symbols) between the examples have yet to be undertaken. Below is a table showing the worked examples that are included in this second draft. Example 2nd draft

included Simply supported restrained beam (including classification of cross section)

Yes

Simply supported unrestrained beam Yes

Simply supported composite beam Yes

Edge beam Yes

Plate girder No

Column in simple construction Yes

Roof truss Yes

Choosing a steel sub-grade Yes

Slab design No

Bracing and bracing connections Yes

Beam-to-column flexible end plate connection Yes

Column base connection Yes

Frame stability Yes

The two examples not included will be circulated separately once they gave been received and formatted.

Blank page

Structural arrangement and loading Sheet 1 of 3 Rev

Made by MEB Date Sept 06 Checked by Date

Reference Output

P:\EDN\EDN399\Student worked examples\Second draft-book\00_Structural_arrangement_&_loading_meb.doc 1

Unless stated otherwise all references are to BS EN 1991-1-1:2002

Characteristic loads – Floors above ground level

Permanent actions

Self weight of floor 3.5 kN/m2 Self weight of ceiling, raised floor and services 0.2 kN/m2

Total permanent action is

gk = 3.5 + 0.2 = 3.7 kN/m2

Variable actions

Table 6.2

6.3.1.2(8)

Imposed floor load for offices (category B) 3.0 kN/m2

Imposed floor load for moveable partitions of less than 2 kN/m run 0.8 kN/m2

Floors above ground level

Permanent action is

gk = 3.7 kN/m2

Total variable action is

qk = 3.0 + 0.8 = 3.8 kN/m2

Variable action is

qk = 3.8 kN/m2

Characteristic loads – Roof

Permanent actions

Self weight of roof construction 0.75 kN/m2 Self weight of ceiling and services 0.15 kN/m2

Total permanent action is

gk = 075 + 0.15 = 0.9 kN/m2

Roof Permanent action is

gk = 0.9 kN/m2

Variable actions

Table 6.9

Table 6.10

The roof is only accessible for routine maintenance (category H)

Imposed roof load 0.4 kN/m2

The imposed roof load due to snow obtained from EN1991-1-3 is less than 0.4 kN/m2, therefore the characteristic imposed roof load is taken from BSEN1991-1-1.

Roof Variable action is

qk = 0.4 kN/m2

Characteristic loads – Wind1)

BS EN 1991-1-4:2005

The total wind force acting on the length of the building is

Fw = 925 kN

Wind force acting on the length of the building is: Fw = 925 kN



6 m

6 m

6 m

6 m

6 m

6 m

6 m

6 m

8 m 6 m

14 m

48 m

1 2 3

A

B

C

Stairwell

D

E

F

G

H

J

Liftshaft

Winddirection

Plan at level 1

Precastfloorunits

6 m

6 m

6 m

6 m

6 m

6 m

6 m

6 m

8 m 6 m

14 m

48 m

1 2 3

A

B

C

Stairwell

D

E

F

G

H

J

Liftshaft

AA

Winddirection

Typical plan

Seco

ndar

ybe

amSe

cond

ary

beam

Seco

ndar

ybe

am

Seco

ndar

ybe

am

Seco

ndar

ybe

am

Seco

ndar

ybe

am

Seco

ndar

ybe

am

Seco

ndar

ybe

am

Seco

ndar

ybe

am

Seco

ndar

ybe

am

Seco

ndar

ybe

am

Seco

ndar

ybe

am

Seco

ndar

ybe

am

Seco

ndar

ybe

am

Insitucompositefloor

4.5 m

4.5 m

4.5 m

5 m

6 m

Typical braced bay

1

Ground

Roof

3

2

4.5 m

4.5 m

4.5 m

5 m

8 m 6 m

Section along grid line E

Plate girder

1

Ground

Roof

3

2

4.5 m

4.5 m

4.5 m

5 m

8 m 6 m

A - ATypical section



Commentry and notes

The highlighted values are ones that may be altered by the National Annex. The value recommended by the Eurocode is used here. The National Annex for the country where the building will be constructed must be consulted.

1) The wind load considered here is only for one direction. Other directions must be considered during the design process.

Example 01 Sheet 1 of 6 Rev Simply supported fully restrained beam

Made by DL Date Nov-06 Checked by JTM Date Dec-06

Reference Output

P:\EDN\EDN399\Student worked examples\Second draft-book\01_Simply supported restrained beam_meb.doc 1


Simply supported fully restrained beam

This example gives the details for the design of a fully restrained non-composite beam under uniform loading. The steel beam is horizontal and because it is supporting the concrete slabs the compression (top) flange is fully restrained by the floor. It can be taken that the resistance of the web to transverse forces at the ends of beam is acceptable. 1)

Consider floor beam at Level 1 – Gridline G1-2

Beam span, L = 8.0 m

Bay width, w = 6.0 m

Actions

See structural arrangement and loading

Permanent action gk = 3.7 kN/m2

Variable action qk = 3.8 kN/m2

gk = 3.7 kN/m2

qk = 3.8 kN/m2

Partial factors for actions

BS EN 1990 Table A.1(2)B

For the design of structural members not involving geotechnical actions the partial factors for actions to be used for ultimate limit state design should be obtained from Table A1.2(B).

Partial factor for permanent actions Gγ = 1.35 Partial factor for variable actions Qγ = 1.5

Reduction factor ξ = 0.85

BS EN 1990 6.4.3.2

Note for this example the combination coefficient (ψ0) is not required as the variable actions are not independent of each other.

Combination of actions for Ultimate Limit State

ξ kqg QkGγ + γBS EN 1990 Eq. (6.10b)

95.9)8.95.395.1()7.335.185.0( +× × =× kN/m2

UDL per metre length of beam accounting for bay width of 6 m,

7.590.695.9Ed =×=F kN/m

ULS design load

FEd = 59.7 kN/m

Design moment and shear force

Maximum design moment, MEd, occurs at mid-span, and for bending about the major (y-y) axis is determined as:

4788

0.87.598

22ED

Edy, =×

==LF

M kNm

Maximum bending moment at mid-span is My,Ed = 478kNm


Reference Output


Maximum design shear force, VEd, occurs at the end supports, and for transverse loading is

2392

87.592

EDEd =

×==

LFV kN

There are no other forces and moments acting on the beam member.

Maximum vertical shear force at supports is VEd = 239kN

Partial factors for resistance

6.1(1) γM0 = 1.0

Trial section

Table 3.1 Grade of steel is S275 and because a standard Universal Beam (UB) is to be use the nominal thickness (t) of the flange and web is less than 40 mm. Therefore the yield strength is:

=yf 275 N/mm2

Yield strength is fy = 275 N/mm2

The required section needs to have a plastic modulus about the major-axis (y-y) that is greater than:

Wpl,y = 275

0.110478 3

y

M0Edy, ××=

f

M γ= 1737 cm3.

Dimensions and properties from Structural Sections to BS4: Part 1: 1993 and BS EN10056: 1999, Corus

From the tables of section properties try section 457 × 191 × 82 UB, S275, which has the Wpl,y = 1831 cm3

b

htr

t

w

f

d

Section 457 × 191 × 82 UB has the following dimensions and properties

Depth of cross-section h = 460.0 mm Web depth hw = 428.0 mm (hw = h – 2tf) Width of cross-section b = 191.3 mm Depth between fillets d =407.6 mm Web thickness tw = 9.9 mm Flange thickness tf = 16.0 mm Radius of root fillet r = 10.2 mm Cross-section area A = 104 cm2 Second moment of area(y-y) Iy = 37050 cm4 Second moment of area/(z-z) Iz = 1871 cm4 Elastic section modulus (y-y) Wel,y = 1610 cm3 Plastic section modulus (y-y) Wpl,y = 1831 cm3


Reference Output


3.2.6(1) Modulus of elasticity E = 210000 N/mm2

5.5 & Table 5.2

Classification of cross-section

For section classification the coefficient depending of fy is:

92.0275235235

y

===f

ε

Outstand flange: flange under uniform compression

C = ( )

2 2 -- w rtb

= ( )

22102993191 ... ×−−

= 80.5 mm

ftC

= 016580

.

. = 5.03

The limiting value for Class 1 is 28.892.099 =×=≤ εtC

5.03 < 8.28 Therefore, the flange outstand in compression is Class 1.

Internal compression part: web under pure bending

C = d = 407.6 mm

wtC

= 99

6407.

. = 41.17

The limiting value for Class 1 is 24.6692.07272 =×=≤ εtC

41.17 < 66.24

Therefore, the web in pure bending is Class 1.

Therefore the section is Class 1 under pure bending.

Section is Class 1

Member resistance checks

6.2.6 Shear resistance

6.2.6(2)

The basic design requirement is:

0.1Rdc,

Ed ≤VV

Vc,Rd =Vpl,Rd = ( )

M0

yv 3/

γfA

(for Class 1 sections)

6.2.6(3) For a rolled I-section with shear parallel to the web the shear area is

( )rttbtAA 22 wffv ++−= but not < ηhw tw

104 × 102 – (2 × 191.3 × 6.0)+(9.9+10.2)×16.0 = 4600 mm2

#.# η = 1.0 (conservative)


Reference Output


ηhw tw = 1.0 × 428.0 × 9.9 = 4237 mm2

4600 mm2 > 4237 mm2

Therefore, Av = 4600 mm2

6.2.6(2) The design shear resistance is therefore

Vc,Rd =Vpl,Rd = ( ) 310

0132754600 −×

×.

/ = 730 kN

Design shear resistance is:

Vc,Rd = 730 kN

730239

c.Rd

Ed =VV

= 0.33 < 1.0

Therefore, the shear resistance of the section is adequate.

Shear resistance is adequate

Shear buckling

6.2.6(6) Shear buckling of the unstiffened web need not be considered provided

ηε

th

72w

w ≤

4399

0428th

w

w ==.

. 66

0.1924.0

7272 =⎟⎠⎞

⎜⎝⎛×=

ηε

43 < 66

Therefore the shear buckling check is not required.

Moment Resistance

6.2.5 The basic design requirement is:

01Rdc,

Ed .MM

≤

6.2.5(#) M0

yypl,Rdpl,Rdc, γ

fWMM

× (For Class 1 sections)

==

6.2.8(2) At the point of maximum bending moment the shear force is zero. Therefore the design moment resistance does not need to be reduced due to the presence of sheer. 2)

6.2.5(#) 310

0.12751831 −×

× = 504 kNm

Design shear resistance is:

Mc,Rd = 504 kNm

== Rdpl,Rdc, MM

01950

504478

MM

Rdc,

Edy, .. <==

Therefore, the design moment resistance of the section is adequate.

Moment resistance is adequate

Therefore, the section satisfied all the ULS checks.

Serviceability Limit State


Reference Output



BS EN 1990 A1.4.1(1)

Partial factor for permanent actions Gγ = 1.0

Partial factor for variable actions Qγ = 1.0

Vertical deflection of beam

Here the check is concerned with the performance of the structure and its finishes. Therefore the irreversible serviceability limit state should be checked using the characteristic combination of actions.

BS EN1990 A1.4.3(3)

As the permanent actions all occur during the construction phase, only the variable actions need to be considered. Therefore, the vertical deflection at the mid-span is determined as:

y

kQ4

384

5

EI

QγLw =

8.220.68.3k =×=Q kN/m

6.15

10370502100003848.220.180005

4

4

=×××

×××=w mm

Vertical mid-span deflection w = 15.6 mm

Vertical deflection limit fo this example is taken as: 3)

2.223608000

360Span

yy

== mm

15.6 mm < 22.2 mm

Therefore, the vertical deflection of the section is satisfactory.

Vertical deflection is acceptable

Therefore, the use of a 457 × 191 × 82 UB in S275 steel is acceptable.

Adopt 457×191×82 UB in S275 steel



Commentary and Notes

The highlighted values are ones that may be altered by the National Annex. The value recommended by the Eurocode is used here. The National Annex for the country where the building will be constructed must be consulted to determine if a different value should be used.

1) For the detailed procedure for the design of a non-composite beam under uniform loading see Access Steel flow chart SF001a-EN-EU (www.access-steel.com).

2) Providing the shear force for the rolled section is less than half of Vpl.Rd where the maximum bending moment is its effect on the moment resistance may be neglected. However, at mid-span where the bending moment is maximum there is no shear force present. The maximum shear force occurs at the end supports where for the uniformly distributed load the bending moment is zero. Therefore there is no reduction to the section’s design strength, which is fy.

3) The Eurocodes do not give limits for serviceability checks. The National Annex for the country where the structure is to be constructed should be consulted for guidance on limits.


Example 02 Simply Supported Laterally Unrestrained Beam Sheet 1 of 7 Rev

Made by YGD Date 30/11/2006 Checked by NS Date

Reference Output

P:\EDN\EDN399\Student worked examples\Second draft-book\02_SCI-Example 1-Unrestrained Beam-457X191X98UB-2 (Final)-jbw.doc 1

Simply Supported Laterally Unrestrained Beam

Description


This example shows the details of the design of a simply supported and laterally unrestrained beam under an action of a uniformly distributed load (udl) on its top flange. The beam is on gridlines G2-3 at level 1 with a span, L = 6.0m, and is used to support a pre-cast concrete floor with a bay width, B = 6.0m. Both ends of the beam are fully laterally restrained by connection to the steel columns, but the connection with the pre-cast floor is, in this case, assumed not capable of providing suitable lateral restraint. This beam will therefore be designed as laterally unrestrained.

Scope of Example

Taking a rolled I-shape (UB) as the cross section of the beam, this example covers:

• Design for beam lateral buckling

• Design for beam shear resistance

Design Value of Combined Actions for ULS of the Beam

Using the method described in Example 1 the design value of actions for ultimate limit state design is determined as:

FEd = 60.8 kN/m

Design value of udl FEd = 60.8 kN/m

Design Values of Bending Moment and Shear Force

FEd = 60.8 kN/m

L = 6.0 m

VEd=182.4 kN

MEd=273.6 kNm

The span of the simply supported beam L = 6.0 m

Maximum bending moment at the midspan

MEd = FEdL2/8 = 60.8 × 6.02 / 8 = 273.6 kNm

Design Moment

MEd = 273.6 kNm


Reference Output


Maximum shear force nearby beam support

VEd = FEdL/2 = 60.8 × 6.0 / 2 = 182.4 kN

Design Shear Force VEd = 182.4 kN

6.1(1)


γM0 = 1.0 γM1 = 1.0

Try Section 457 × 191 × 98UB

Section Dimensions and Properties of 457 × 191 × 98UB, S275

z

z

y y

b

d h

t fr

t w

Universal Beam Section Table

Depth h = 467.2 Width b = 192.8 mm Web depth between fillets d = 407,6 mm Web thickness tw = 11.4 mm Flange thickness tf = 19.6 mm Root radius r = 10,2 mm Section area A = 125 cm2

Second moment, y-y Iy = 45730 cm4

Second moment, z-z Iz = 2347 cm4

Warping constant Iw = 1180000 cm6

Torsion constant It = 121 cm4

Elastic modulus, y-y Wel,y = 1957 cm3

Plastic modulus, y-y Wpl,y = 2232 cm3

Section Dimensions and Properties

Yield Strength, fy of Steelwork

Table 3.1 Steel grade = S275,

Flange thickness of the section tf =19.6 mm ≤ 40.0 mm Hence, yield strength of the steelwork fy = 275 N/mm2

Yield Strength

fy = 275 N/mm2

Section Classification

92.0

275235235

y

===f

ε


Reference Output


Table 5.2 (sheet 1 of 3)

Outstand of uniform compression flange

C = (b – tw – 2 × r)/2 = (192.8-11.4-2 ×10.2)/2 = 80.5 mm

Flange Ratio: C / tf = 80.5/19.6 = 4.11 < 9.0ε = 8.28, Class 1

Table 5.2 (sheet 1 of 3)

Internal compression part: web under pure bending

C = (h- 2 × tf – 2 × r) = (467.2 - 2 × 19.6 - 2 × 10.2) = 407.6 mm

C/tw = 407.6 / 11.4 =35.8 < 72 ε = 66.24 Class 1

This section is Class 1

Moment Resistance of Beam Section (If Fully Restrained)

l.6.2.5 Eq.6.13

The design resistance for bending about major axis y-y

M0

yRdpl,Rdpl,Rdc,

γ

fWMM ==

=2232 × 1000 × 275 × 10-6/1.0 =613.8 kNm

Moment Resistance

Mc,Rd =613.8 kNm

l.6.2.5 Eq.6.12 00.1446.0

8.6136.273

Rdc,

Ed <==M

M OK

Non-dimensional slenderness ratio of unrestrained beam

A less conservative value of non-dimensional slenderness ratio may be obtained by taking account of bending moment diagram, section geometry and lower yield strength, as follows below.

Concise Guide to EC3 Table 5.5

for simply supported beam uniform distributed load, 94,01

1

=C

Radius of gyration about Z-Z axis 3.43

1250023470000z

z ===A

Ii

Span to radius ratio 6.138

3.436000

zz ===

iLλ

6.3.1.3

Limiting slenderness ratio 275

210000

y1 ππλ ==

fE

=86.8

596.18.86

13.43

60001

1zz =×==

λλ

iL

For Class 1 and 2 sections, 0.1

ypl,

ypl,

ypl,

yw ===

W

W

W

Wβ

Concise Guide to EC3 Expn (6.55)

Hence, non-dimensional slenderness ratio

35.10.1596.190.094.09.01

wz

1

LT =×××=×= βλλC

slenderness ratio 35.1LT =λ

Reduction factor for lateral torsional buckling


Reference Output


6.3.2.3

For rolled I or H section, the reduction factor for torsional buckling

2LT

2LTLT

LT1

λβΦΦχ

−+= but

2LT

LT

LT

100,1

λχ

χ

≤

≤

Where, the value for reduction factor

( ) ⎥⎦⎤

⎢⎣⎡ +−+=

2LTLT,0LTLTLT 10.5 λλαΦ βλ

6.3.2.3 The recommended maximum value of ⎯λ LT,0 = 0.4 The recommended maximum value of β = 0.75

Table 6.5 Table 6.3

For rolled Section with h/b = 467.2/192.8 = 2.42 > 2.0,

the buckling curve should be c, and imperfection factor αLT = 0.49

Hence, the value for reduction factor

[ ]( ) 415.135.175.040.035.149.010.5 2LT =×+−×+=Φ

Value 415.1LT =Φ

Eq.6.57 Reduction factor

452.035.175.0415.1415.1

122

LT =×−+

=χ

00.1452.0LT = <χ Check:

548.035.11 22

LT ==λ= 452.0LT <χ 1

So, reduction factor, χ LT = 0.452

Reduction factor, χ LT = 0.452

Correction factor for moment distribution 94.0c =k 6.3.2.3 Table 6.6

Modification factor for LTχ

])8.0(0.21)[1(5.01 2LTc −−−−= λkf

988.0])8.035.1(0.21)[94.01(5.01 2 =−×−−×−=

but ≤ 1.0

6.3.2.3 Eq.6.58

Modified reduction factor

457.0988.0452.0LT

modLT, ===χ

χf

Modified Reduction factor

457.0mod,LT =χ

Design buckling resistance of the unrestrained beam

6.3.2.1 Eq.6.55

M1

yplyLTRdb, γ

fWχM = = 610

00.12752232000

457.0 −××

×

= 280.5 kNm

Buckling Resistance Mb,Rd = 280.5 kNm

6.3.2.1 Eq.6.54 0.1975.0

5.2805.273

Rdb,

Ed <==M

M OK

Bending resistance adequate


Reference Output


Shear Resistance

6.2.6.3

In the absence of torsion, the shear resistance the of beam depends on its shear area

=++−= fwfv )2(2 trtbtAA

2mm 55666.19)2.1024.11(6.198.192212500 =××++××−=

6.2.6.2 Eq.6.18

Hence, the shear resistance

( )88310

00.1)3/275(55663/ 3

M0

yvRdpl,Rdc, =×

×=== −

γfA

VV kN

Shear Resistance is Vc,Rd = 883 kNm

l.6.2.6 Eq.6.17 0.1207.0

8834.182

Rdc,

Ed <==V

V OK


Adopt 457 × 191 × 98 UB in S275




The highlighted values are ones that may be altered by the National Annex. The value recommended by the Eurocode is used here. The National Annex for the country where the building will be constructed must be consulted to determine if a different value should be used. 1. Reference: 1a. BS EN 1993-1-1:2005 Eurocode 3: Design of Steel Structures-Part 1-1: General rules and rules

for buildings 1b. SCI and Corus, Concise Guide to Eurocode 3

1c. SN002a-EN-EU: NCCI: Determination of non-dimensional slenderness ratio of I and H sections 1d. SN003a-EN-EU: NCCI: Elastic critical moment for lateral torsional buckling

LTλ2. The other three different approaches to determine the non-dimensional slenderness ratio

Reference Calculation Output 2a. The Simplified Method From the SN002a-EN-EU, a conservative value of non-dimensional

slenderness ratio of unrestrained beam can be calculated as follows,

Radius of gyration about Z-Z axis

3.4312500

23470000zz ===

A

Ii

SN002a-ENEU Table 1.1

a conservative value of slenderness ratio

Span to radius ratio 6.1383.43

6000

z

==iL

Non-dimensional slenderness ratio 444.196

6.13896

/ zLT ===

iLλ 444.1LT =λ

2b. The Economic Method A less conservative value of non-dimensional slenderness ratio may be

obtained by taking account of bending moment diagram, section geometry and lower yield strength, as follows below.


for simply supported beam uniform distributed load, 94.01

1

=C

881.0

11800002347

457302347

11252232

1w

z

y

zypl,

=

×−×=×−×=I

I

I

I

A

WU

Radius of gyration about Z-Z axis 3.43

1250023470000z

z ===AI

i

Effective length parameter k = 1.0

Span to radius ratio 3.43

60000.1 ×==

zz i

kLλ = 138.6



Slenderness parameter

781.0

6.19/2.4676.138

201

1

1

/201

1

1

4

2

4

2

f

z

=

⎥⎦⎤

⎢⎣⎡+

=

⎥⎦

⎤⎢⎣

⎡+

=

th

Vλ

Limiting slenderness ratio

275210000

y1 ππλ ==

fE

= 86.8

for Class 1 and 2 sections ypl,

ypl,

ypl,

yw W

W

W

W==β = 1.0

SN002a-ENEU Eq. (1)

Hence, a less conservative value of non-dimensional slenderness ratio a less conservative value of slenderness ⎯λLT = 1.032

032.10.18.866.138

781.0881.094.01

w1

z

1

LT =××××== βλλ

λ UVC

2b. Using Elastic Critical Moment A more economical section may be checked using the more complex

approach given in equation (1) as follows: -

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−++⎥⎦

⎤⎢⎣

⎡= g2

2g2

z2

t2

z

w2

w2z

2

1cr )()(

)(zCzC

EI

GIkL

I

I

kk

kL

EICM

ππ

Where:

Young Modulus E = 2100000 N/mm2

Shear Modulus G = 80770 N/mm2

Distance between lateral supports L = 6000 mm No device to prevent beam end from warping kw = 1


Compression flange free to rotate about z-z k = 1 For uniformly distributed load on beam, C1 = 1.127, C2 = 0.454

Distance from load point to shear centre 2

2.4672g ==h

z = 233.6 mm

kN 1351

)60000,1(

1000/23470000210000

)( 2

2

2

2

zcr, =×××

==ππ

kL

EIN z

22

z

w2

w

cm 7.5022347

11800000.10.1

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛I

I

kk

29t kNm 73.9710/121000080770 =×=GI

m 10605.02336.0454.0g2 =×=zC

kNm 396

10605.0)10605.0(1351

73.9705027.01351127.1 2

cr

=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−++×=M

6.3.2.2 Eq.6.56

Hence, Non-dimensional slenderness ratio

245.110396

27522320006

cr

yypl,LT =

××

==M

fWλ

Slenderness ratio 245.1LT =λ

Example 03 Sheet 1 of 8 RevSimply supported composite beam

Made by BK Date Checked by WT Date

Reference Output

P:\EDN\EDN399\Student worked examples\Second draft-book\03_Simply supported composite beam_meb.doc 1


Simply supported composite beam

This example shows the design of a 6m long composite beam subject to UDL at 3m centres. The composite slab is 130mm deep with ComFlor 51 (Corus, 2002) perpendicular to the steel beam. The design checks include the moment resistance of the composite beam, the number of shear connectors, vertical shear and transverse reinforcement.

See “Structural arrangement and loading”

Consider the secondary composite beam between CD and 3 on the typical floor.

51

40 112.5 152.5 144.5

Cover width 610 mm

Dimensions of ComFlor 51 (Corus, 2002)

Design data

Beam span L = 6.0 m Beam spacing s = 3.0 m Slab depth hc = 130 mm Depth above profile ht = 79 mm Deck profile height hp = 51 mm Width of the bottom trough bbottom = 122.5 mm Width of the top trough btop = 112.5 mm

Shear connectors

Diameter d = 19 mm Overall height 100 mm Height after welding hsc = 95mm

Materials

BS EN1993-1-1

Table 3.1

Structural Steel:

For gradeS275 and maximum thickness (t) less than 40 mm

Yield strength fy = 275 N/mm2 Ultimate strength fu = 430 N/mm2

Steel reinforcement:

Yield strength fy = 460 N/mm2

BS EN1992-1-1 Table 3.1

Concrete:

Normal weight concrete strength class C25/30

Density 2400 kg/m3 (23.5 kN/m3) Cylinder strength fck = 25 N/mm2 Ecm = 31 kN/mm2


Reference Output


Actions

Permanent actions

Self weight of the concrete slab

( ) ( )[ ] 80.2105.23513071000130 6 =××××−× − kN/m2

Construction stage kN/m2

Concrete slab 2.80 Steel deck (allow) 0.15 Reinforcement (allow) 0.04 Steel beam 0.20Total 3.19

Composite stage kN/m2

Concrete slab 2.80 Steel deck (allow) 0.15 Reinforcement (allow) 0.04 Steel beam 0.20 Ceiling and services 0.15 Total 3.34

Permanent

Construction stage: gk = 3.19 kN/m2

Composite stage: gk = 3.34 kN/m2

Variable actions Variable

Construction stage kN/m2

Construction loading 0.50

Composite stage kN/m2

Floor load 3.80 (See structural arrangement and loading)

Construction stage: qk = 0.50 kN/m2

Composite stage: qk = 3.80 kN/m2

Combination of actions for Ultimate Limit State

Using equation 6.10b) the design loads are found as: 1)

Construction stage: Distributed load 4.41 kN/m2

40.790.30.641.4Ed = × × =F kN Total load

Composite stage: Distributed load 9.53 kN/m2

50.1710.30.653.9Ed = × × =Total load F kN

Construction stage FEd = 79.40 kN Composite stage FEd = 171.50 kN

Design moments and shear forces for ULS

Construction stage

Maximum design moment (at mid span)

kNm

Construction stage:

My,Ed = 60kNm 608

0.64.798Ed

Edy, =×

==LF

M

Composite stage

Maximum design moment (at mid span)

1298

0.65.1718Ed

Edy, =×

==LF

M kNm

Maximum design shear force (at supports)

862

5.1712Ed

Ed ===F

V kN

Composite stage:

My,Ed = 129kNm

VEd = 86 kN


Reference Output


BS EN1993-1-1 6.1(1)

BS EN1992-1-1 Table 2.1N 2.4.1.2


Structural steel γM0 = 1.0 Concrete γC = 1.5 Reinforcement γS = 1.15 Shear connectors γV = 1.25 Longitudinal shear γVS = 1.25

Trial section

The plastic modulus that is required to resist the construction stage maximum design bending moment is determined as:

Wpl,y = 275

0.11060 3

y

M0Edy, ××=

f

M γ= 218 cm3

From the tables of section properties try section 254 × 102 × 22 UB, S275, which has the Wpl,y = 259 cm3

Depth of cross-section h = 254.0 mm Width of cross-section b = 101.6 mm Depth between fillets d =225.2 mm Web thickness tw = 5.7 mm Flange thickness tf = 6.8 mm Radius of root fillet r = 7.6 mm Cross-section area Aa = 28 cm2

Second moment of area(y-y) Iy = ? cm4

Second moment of area/(z-z) Iz = ? cm4

Elastic section modulus (y-y) Wel,y = ? cm3

Plastic section modulus (y-y) Wpl,y = 259 cm3

BS EN1993-1-1 3.2.6(1)

Modulus of elasticity E = 210000 N/mm2

Section classification

The section is Class 1 under bending. 2)

Section is Class 1

Composite stage member resistance checks

Compression resistance of concrete slab

5.4.1.2 At mid-span the effective width of the compression flange of the composite beam is determined from:

ei0eff bbb ∑+=

75.086

88e

ei ====LL

b m (for simply supported beams)

Assume single shear studs, therefore, 00 =b m

( ) 50.175.020eff =×+=b m < 3 m (beam spacing)

Effective width

beff = 1.50 m


Reference Output


Compression resistance of concrete slab is determined from:

ceffC

ckc,f

85.0hb

fN

γ=

1679107915005.1

2585.0 3c,f =×××

×= −N kN

Compressive resistance of slab Nc,f = 1679 kN

Tensile resistance of steel section

M0

ayada γ

AfAfN ==

770100.1

1028275 32

a =×××

= −N kN

Tensile resistance of steel section Na = 770 kN

Location of neutral axis

Since Na < Nc,f the plastic neutral axis lies in the concrete flange.

Moment resistance with full shear connection

6.2.1 As the plastic neutral axis lies in the concrete flange, the moment resistance of the composite beam with full shear connection is:

⎥⎦

⎤⎢⎣

⎡×−++=

22 c

c,f

apcsRdpl,

hNN

hhh

RM

184 10279

1679770

1302

254770 3

Rdpl, =×⎥⎦⎤

⎢⎣⎡ ×−+= −M kNm

Moment resistance of composite beam Mpl,Rd = 184kNm

Bending moment at mid span My,Ed = 129 kNm

7.0

184129

Rdpl,

Edy, ==M

M < 1.0

Therefore, the design moment resistance of the composite beam is adequate.

Design moment resistance is adequate

Shear connector resistance

6.6.3.1 The design shear resistance of a single shear connector is the smaller of:

E d 29.0

v

cmck2

Rd γα f

=P

or

/4)d ( 8.0

v

2u

Rd γπf

=P

5

1995sc ==

dh

As 0.4sc >dh

0.1=α


Reference Output


73.7 10

25.1103125 19 1.0 29.0 3

32

Rd =×××××

= −P kN

or

81.7 10 25.1

/4)19 ( 450 8.0 3-2

Rd =××××

=π

P kN

As 73.7 kN < 81.7 kN 73.7 Rd =P kN

Shear resistance of a single shear stud PRd = 73.7 kN

Influence of deck shape

6.6.4.2 Deck crosses the beam. (i.e. ribs transverse to the beam)

One stud per trough, nr = 1.0

Reduction factor

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛= 1

7.0

p

sc

p

0

rt h

hhb

nk ≤ 1.0

1.33 15195

51112.5

1

7.0t =⎟

⎠⎞

⎜⎝⎛ −×⎟

⎠⎞

⎜⎝⎛×⎟

⎠⎞

⎜⎝⎛=k > 1.0

As kt > 1.0 no reduction in shear connector resistance. therefore, PRd = 73.7 kN

Number of shear studs in half span

Use one shear connector per trough, therefore,

Stud spacing along beam = 152.5 mm

Considering the primary beam width or the column width (assume 254mm).

stud shear connectors per half span

Provide a stud per trough, total 36 stud shear connectors for the whole span.

185.152

)2/254(3000=

−

Degree of shear connection

13277.731818 Rdq ×== =R P kN

7.17701327

a

q ==N

R > 1.0

Therefore, full shear connection is provided and no reduction in moment resistance calculated earlier is required.

Full shear connection is provided


Reference Output


6.2.2.2

BS EN1993-1-1 6.2.6

Resistance to vertical shear

Shear resistance of the composite beam is:

3d

vRdpl,

fAV =

where wv htA =

230 103

2757.5 3-Rdpl, =×

××=

254V kN

Vertical shear resistance

Vpl,Rd = 230 kN

4.0

23086

Rdpl,

Ed ==VV

< 1.0

Therefore the design resistance for vertical shear is adequate.

Design resistance for vertical shear is adequate

6.2.2.4 As there is no shear force at the point of maximum bending moment (mid soan) no reduction (due to shear) in moment resistance is required.

Design of the transverse reinforcement

For simplicity neglect the contribution of the decking and check the resistance of the concrete flange to splitting.

The area of reinforcement (Asf) can be determined using the following equation:

f

ydsf

s

fA >

6.6.6 fEdhV therefore,

θcot sfA > θcotyd

ffEd

fshV

Where

4001.15460

s

yyd ===

γf

f

4427.736EdL, =×

N/mm2

For compression flanges 26.5º ≤ θ ≤ 45º

6.6.6.1 The longitudinal shear force can be determined from the resistance of shear connectors. Since there are 6 shear connectors per metre, the longitudinal shear force per unit length is:

=V kN/m

Due to symmetry, the critical shear plane a-a is to resist half of this force. For minimum area of transverse reinforcement assume θ = 26.5º

sfA > 276

6.52cot10400221

cot 3yd

ffEd =××

= −θfshV

mm2/m

Therefore, provide A314 mesh reinforcement in the slab. 3)

Use A314 mesh reinforcement


Reference Output


Serviceability limit states

Serviceability limit states checks should be performed. However, they are not included here as the example is not country specific and the Eurocodes do not give any recommended limits for serviceability. The National Annex for the country where the building is to be constructed should be consulted for guidance.

Example XX Sheet 8 of 8 RevSimply supported composite beam




1) See example 1 for further details of loading combination equation 6.10b).

2) See example 1 for classification method.

3) If the contribution of decking is included, the transverse reinforcement provided can be reduced.

Example 04 Edge beam Sheet 1 of 8 Rev

Made by MXT Date Dec 2006 Checked by AB Date April 2007

Reference Output

P:\EDN\EDN399\Student worked examples\Second draft-book\04_Edge beam with torsion_meb02.doc 1


Edge beam with torsion

The 6m span beam shown below is unrestrained along its length. It carries permanent loads only. The brickwork load is applied with an eccentricity of 172mm to the centroidal axis and induces torsion to the beam. Design the beam using an RHS 1) in S355 grade steel. Only the ultimate limit state need be considered.5)

Block Brick

RHS

172

End detail

Actions

Permanent actions

Uniformly Distributed Load (brickwork) G1 = 4.8 kN/m Uniformly Distributed Load (blockwork) G2 = 3.0 kN/m Uniformly Distributed Load (assumed self weight) G3 = 0.47 kN/m


BS EN1990 A1.3.1(4)

For the design of structural members not involving geotechnical actions the partial factors for actions to be used for ultimate limit state design should be obtained from Table A1.2(B).

BS EN1990 Table A1.2(B)

Partial factor for permanent actions γG = 1.35 Reduction factor ζ = 0.85

Design values of actions for Ultimate Limit State

BS EN1990 Table A1.2(B) & Eq. (6.10a)

UDL (total permanent)

16.11)47.00.38.4(35.1)( 321GEd,1 =++×=++= GGGF γ kN/m

UDL (permanent/inducing torsion)

48.68.435.11GEd,2 =×== GF γ kN/m

Design moments and shear forces


Reference Output


Span of beam L = 6000 mm Eccentricity of brickwork e = 172 mm

Maximum design moment occurs at the mid-span

2.508

616.118

22Ed,1

Ed =×

==LF

M kNm

Design moment MEd = 50.2 kNm

Maximum design shear occurs at the supports

5.332

616.112

Ed,1Ed =

×==

LFV kN

Design shear force VEd = 33.5 kN

Maximum design torsional moment occurs at the supports

0.42

6172.048.62

Ed,2Ed =

××=

××=

LeFT kNm

Design torsional moment TEd = 4.0 kNm

The design bending moment, torsional moment and shear force diagrams are shown below.

Trial section

Try 160 × 80 × 8.0 RHS in S355 steel. The RHS is class 1 under the given loading.

Bending moment

Shear force

Torsional moment

50.2 kNm

33.5 kN

33.5 kN

4 kNm

4 kNm


Reference Output


Hot finished 160 × 80 × 8.0 RHS in grade S355 steel

Depth of section h = 160.0 mm Width of section b = 80.0 mm Wall thickness t = 8.0 mm Second moment of area about the y-axis Iy = 1090 cm4 Second moment of area about the z-axis Iz = 356 cm4 Torsional constant IT = 883 cm4 Torsional modulus CT = 151 cm3 Plastic modulus about the y-axis Wpl,y = 175 cm3 Cross-sectional area A = 35.2 cm2

Table 3.1

For steel grade S355 and t < 40 mm

Yield strength fy = 355 N/mm2 Ultimate tensile strength fu = 510 N/mm2


6.1(1) γM0 = 1.0 Member resistance checks

Shear buckling check

6.2.6(6)

The shear buckling resistance for webs should be checked according to section 5 of EN1993-1-5 if:

Eq. 6.22

ηε72

w

w >th

Table 5.2

81.0355235235

y

===f

ε

η = 1.0 (conservative) 6.2.6(6)

1440.821602w =×−=−= thh mm

180.8

144

w

w ==th

580.1

81.07272=

×=

ηε

18 < 58


Reference Output


Therefore the shear buckling resistance of the web does not need to be checked.

Shear buckling check is not required.

Shear capacity


6.2.6(1)

Eq. 6.17 0.1

Rdc,

Ed ≤VV

Vc,Rd is equal to the design shear resistance for plastic design (Vpl,Rd).

6.2.6(2) Eq. 6.18

M0

yvRdpl,Rdc,

)3/(

γfA

VV ==

Av is the shear area and is determined as follows for RHS sections with the load applied to the depth.

7.2346

160801603520

v =+×

=+

=hb

AhA mm2

Eq. 6.18

48110

0.1)3/355(7.2346)3/( 3

M0

yvRdc, =×

×== −

γfA

V

5.33Ed

kN

Shear capacity:

Vc,Rd = 481 kN

=V kN Sheet 2

6.2.6(1)

Maximum design shear

0.1071.0481

5.33

Rdc,

Ed <==VV


Torsional moment capacity


6.2.7(1)

Eq. 6.23 0.1

Rd

Ed ≤TT

6.2.7(7) As a simplification, in the case of closed hollow-sections the warping effects can be neglected. Only Saint-Venant torsion needs be considered.

4Edt,Ed == TT kNm

TEd = 4 kNm

6.2.7(8)

BS EN 1993-1-1 does not provide a formula for calculating the torsional resistance. By considering the shear flow in a closed hollow section, the torsional resistance is (see Figure 4.3) given by:

M,0

yinscrRd

3

2

γ×××

=fA

T .


Reference Output


Shearcentre

Shearcentre

Centre-line

Inscribed area bycentre-line A insc.

Shear flow

Figure 4.3

The term equals the torsional modulus provided by cross-sectional tables. Hence

inscrA×2

3113

10355151

3

3

M,0

yRd =

×××

=×

×=

−

γfC

T kNm

TRd = 26 kNm

113.0

314

Rd

Ed <==TT

Torsional resistance is adequate

6.2.7(9) For combined shear and torsional moment the plastic shear resistance should be reduced.

In this case, 071.0

Rdc,

Ed =VV

and 13.0Rd

Ed =TT

By inspection the effects of combined shear and torsion are negligible so may be neglected.

Bending moment capacity


6.2.5(1) Eq (6.12) 0.1

Rdc,

Ed ≤MM

6.2.8(1) At the point of maximum moment (mid-span) check if the shear force will reduce the moment capacity of the section. From Figure 4.2 it can be seen that at the point of maximum bending moment no shear force is present. Hence no moment-shear interaction check is required.

No moment-shear interaction check is required

6.2.5(2) The design resistance to bending moment for Class 1 and 2 cross-sections is:

Eq. 6.13 6210

0.135510175 6

3

M0

yypl,Rdpl,Rdc, =×

××=== −

γfW

MM kNm

MRd = 50 kNm


Reference Output


6.2.5(1) Eq. 6.12

81.062

2.50

Rdc,

Ed ==MM

< 1.0

Therefore, the moment capacity is adequate.

Moment capacity is adequate

Member buckling check

6.3.2.1 The beam considered, is laterally unrestrained and should therefore be verified against lateral-torsional buckling. 4)

6.3.2.2(4) For slendernesses 0,LTLT λλ < lateral torsional buckling effects may be ignored.

6.3.2.3 0,LTλ = 0.4

6.3.2.2(1) The slenderness LTλ is given by

cr

yyLT

M

fW ×=λ

Access-steel document SN003a-EN-EU

No equation for the calculation of Mcr is given in BS EN 1993-1-1. The NCCI document SN003a-EN-EU proposes the following equation for doubly symmetric cross-sections:

⎪⎭

⎪⎬⎫

−⎪⎩

⎪⎨⎧

++⎟⎟⎠

⎞⎜⎜⎝

⎛= g2

2g2

z2

T2

z

w

2

w2

2

1cr )()(

)(zczc

EIGIkL

II

kk

kLEI

cMπ

π

Access-steel document SN003a-EN-EU 3.3 & Table 3.2

Where

E is the Young modulus (E = 210000 N/mm2)

G is the shear modulus (G = 80770 N/mm2)

Iz is the second moment of area about the minor axis

IT is the torsion constant

Iw is the warping constant 2)

L is the beam length between points of lateral restraint

zg is the distance between the shear centre and the point of application of the transverse loading 3)

k & kw are effective length factors. kw = 1.0 (unless provision for warping fixity is made) k = 1.0 (for simply supported beams) (equivalent to the effective length factor for minor axis buckling).

C1 accounts for actual moment distribution C1 = 1.127 (for simply supported beam with a UDL and k = 1.0)

C2 accounts for the effect of loads applied above (destabilizing effect) or below (stabilizing effect) the shear centre. It equals (C2 = 0.454 for a simply supported beam with a UDL and k = 1.0)


Reference Output


Here, the effect of destabilizing loads is very small and will be neglected in this example.

The term z

w

2

w II

kk

⋅⎥⎦

⎤⎢⎣

⎡is negligible compared to

z2

T2)(EIGIkL

π

(0.22%) of the later) and may therefore be omitted for hollow sections.

Hence, Mcr=431 kNm Mcr = 431 kNm

6.3.2.2(1)

38.0

43110355142 3

cr

yyLT =

××=

×=

−

M

fWλ < 0.4 4)

38.0LT =λ

Therefore, lateral torsional buckling effects may be neglected.

Lateral-torsional buckling effects may be neglected

The RHS 160 × 80 × 8.0 in S355 steel passes all the ULS checks.5) All ULS checks passed. 5)





1) When dealing with torsion, closed hollow sections are the best choice, due to their high torsional stiffness.

2) No value for Iw was given in the RHS section tables. However the influence of warping may be neglected for closed hollow sections.

3) The stabilizing or destabilizing effects of loading may most times be neglected for closed hollow sections.

4) CHS and SHS are not susceptible to lateral-torsional buckling (both axes have the same moment of area). RHS are almost always torsionally stiff enough to be expelled from LTB checks.

5) Serviceability limit state checks should also be carried out in addition to the ultimate limit state checks. The serviceability limit state limits may be given in the National Annex for the country where the structure will be built.

SCI publication P057 suggests that an acceptable serviceability limit for the maximum angle of torsional twist is 2°.

When members supporting brickwork or cladding will be visible in the finished structure the deflection limits should be chosen carefully to avoid aesthetically unpleasing ‘sagging’ beams due to the self weight of the brickwork or cladding.

Example 05 – Plate Girder Awaiting draft of this example 14/05/07

Example 06 Column in Simple Construction Sheet 1 of 5 Rev

Made by LG Date Checked by PS Date

Reference Output

P:\EDN\EDN399\Student worked examples\Second draft-book\06_Column calculation - LG - Rev B_jbw.doc 1

Column in Simple Construction Unless stated otherwise all references are to BS EN 1993-1-1:2005

Description

This example gives the details for the design of an internal column in simple construction.

Internal column at ground level – Gridline G2

Column height = 5.0 m

See structural arrangement and loading

Actions

Reactions at each of the three floor levels from 8 m span beams:

Permanent = 88.8 kN Variable = 91.2 kN

Reactions at each of the three floor levels from 6 m span beams:

Permanent = 66.6 kN Variable = 68.4 kN

The total axial load acting on the column between the ground floor and level 1 is given by:

Permanent Gk = 3 × (88.8 + 66.6) = 466.2 kN Variable Qk = 3 × (91.2 + 68.4) = 478.8 kN

Gk = 466.2 kN Qk = 478.8 kN

BS EN 1990: 2002 Table A1(2)B

Partial factors

For permanent actions γG = 1.35 For variable actions γQ = 1.5

BS EN 1990: 2002 Table A1(2)B

Reduction factor

ξ = 0.85

BS EN 1990: 2002 6.4.3.2

Combination of actions

= ξγG Gk + γQ Qk

= 0.85 × 1.35 × 466.2 + 1.5 × 478.8 = 1253 kN

The ULS load combination is NEd = 1253 kN

6.1(1)


γM0 = 1.0 γM1 = 1.0

Section properties

Try 254 × 254 × 73 UC, S275


Reference Output


Depth h = 254.1 mm Flange thickness tf = 14.2 mm Web thickness tw = 8.6 mm Radius of gyration iz = 6.48 cm Section area A = 93.1 cm2

Plastic modulus, y-y Wpl, y = 992 cm3

Buckling length about y-y axis Lcr,y = 5.0 m Buckling length about z-z axis Lcr,z = 5.0 m

For columns in simple construction the beam reactions are assumed to act at 100 mm from the face of the column.

Factored reactions at level 1:

From 8 m beam: R = (0.85 × 1.35 × 88.8) + (1.5 × 91.2) = 238.7 kN

From 6 m beam: R = (0.85 × 1.35 × 66.6) + (1.5 × 68.4) = 179.0 kN

Moments at level 1

Level 1

100 mm 100 mm h

239 kN 179 kN

M1,y,Ed = ((h/2) + 100) × (238.7 – 179.0) = 13.6 kNm

The moment is distributed between the column lengths above and below level 1 in proportion to their bending stiffness (EI/L). The moment acting on the column length between the ground floor and level 1 is therefore:

My,Ed = 13.6 × 4.5/9.5 = 6.4 kNm My,Ed = 6.4 kNm

Mz,Ed = 0 because the beam reactions are equal on either side of the column in this direction since the spans are equal (6 m).

Mz,Ed = 0


Reference Output


Table 3.1

Yield Strength, fy

Steel grade = S275

Nominal thickness of the element, t ≤ 40 mm

Yield strength is fy = 275 N/mm2

6.3.1.3

Flexural buckling resistance

λ1 = 93.9ε = 93.9 × (235/275)0.5 = 86.8

zλ = (Lcr,z/iz)/λ1 = (5000/64.8)/86.8 = 0.89

Table 6.2

Figure 6.4

h/b < 1.2 and tf < 100 mm, so use buckling curve ‘c’ for the z-axis

From graph, χ z = 0.61

Eq. (6.47) Nb,z,Rd = χ zAfy/γ M1 = 0.61 × 9310 × 275 × 10-3/1.0 = 1553 kN Nb,z,Rd = 1553 kN

6.3.2 Lateral torsional buckling resistance

Take zLT 9.0 λλ = = 0.9 × 0.89 = 0.80

6.3.2.3 Utilising Clause 6.3.2.3 for rolled and equivalent welded sections

2LT

2LTLT

LT1

λβφφχ

−+=

where ( )[ ]2LTLT,0LTLTLT 15.0 λβλλαφ +−+=

81.0]80.075.0)40.080.0(34.01[5.0 2LT =×+−+=φ

Table 6.5 Rolled bi-symmetric I-section, h/b ≤ 2, so use buckling curve ‘b’.

82.080.075.081.081.0

122

LT =×−+

=χ

0.1LT ≤

6.3.2.3 But the following restrictions apply:

χ

56.180.0

1122

LTLT ==≤

λχ

82.0LT =

∴ χ

6.3.2.1(3) Mb,Rd = χ LTWyfy/γ M1 = 0.82 × 992 × 275 × 10-3/1.0 = 223 kNm Mb,Rd = 223 kNm

Combined bending and axial compression buckling check (simplified)

With reference to ‘Verification of columns in simple construction – a simplified interaction criterion’, UK localised NCCI, The Steel Construction Institute, www.access-steel.com.

The simplified expression given below will be used:

0,15,1Rdz,

Edz,

Rdb,

Edy,

Rdz,b,

Ed ≤++M

M

M

M

N

N

0.184.00223

4.615531253

≤=++

UK localised NCCI



Reference Output


Therefore a 254 × 254 × 73 UC is adequate. Section used is 254×254×73 UC, S275





Cross-section is assumed to be Class 1 or 2.

Simplified formula used for axial + bending interaction.

Example 07 Roof Truss Sheet 1 of 9 Rev

Made by LYL Date Checked by IB Date

Reference Output

P:\EDN\EDN399\Student worked examples\Second draft-book\07_Truss_Li_IWB_Final_jbw.doc 1

Roof Truss


The truss to be designed is to support a roof which is only accessible for routine maintenance (category H). The truss is 14 m long with 15° pitch. The dimensions of the truss are shown in the figure below. The imposed roof load due to snow obtained from EN1991-1-3 is less than 0.4 kN/m2, therefore the characteristic imposed roof load is taken from BSEN1991-1-1. The truss uses compound (back-to-back) angles for its tension chord and rafters, and single angles for its internal members. The top and bottom chords of the truss are to be welded at joints A, H, and D, while the internal diagonal members are to be bolted to the chords using two M18 bolts through gusset plates at joints B, C, D, E and G. Truss analysis is carried out by placing concentrated loads at the joints of the truss. All of the joints are assumed to be pinned in the analysis and therefore only axial forces are transferred by members. No moments or shear forces are assumed at joints.

B

C

G

D

E H

F F

F

Ed

Ed Ed

Ed

Ed

A

15°

F /2 F /2

30°

35 353535

64983751 3751

BS EN 1991-1-1 Tables 6.9 and 6.10

Characteristic Loads

Permanent actions

Self weight of roof construction 0.75 kN/m2

Self weight of services 0.15 kN/m2

Total permanent loads 0.9 kN/m2

Variable actions

Imposed roof load 0.4 kN/m2

Total imposed loads 0.4 kN/m2

BS EN 1991-1-1: 2002 Tables 6.9 and 6.10

Load Factors

Permanent load factor 0.85 × 1.35 = 1.15 Imposed load factor 1.5

Design Loads on Roof

Unfactored load 0.9 + 0.4 = 1.3 kN/m2

Factored load 1.15 × 0.9 + 1.5 × 0.4 = 1.64 kN/m2


Reference Output


Design Loads on Purlins Supported By Truss

For the distance of 3.5 m between purlins centre to centre

Unfactored load 1.3 × 3.5/cos15o = 4.71 kN/m Factored load 1.64 × 3.5/ cos15o = 5.94 kN/m

W = 28.26 kN for unfactored load W = 35.64 kN for factored load

Design Loads at Joints of Top Chords (Rafters)

For a purlin span of 6 m

Unfactored load 4.71 × 6 = 28.26 kN Factored load 5.94 × 6 = 35.64 kN

Truss analysis (due to factored loads FEd =35.64 kN)

Reaction force at support A RA=2 × FEd = 71.28 kN

At joint A FAB × sin15o + (RA-W/2) = 0 FAB × cos15o + FAC = 0

FAB = -206.55 kN FAC = 199.52 kN

At joint B FBC + W × cos15o = 0 FBD - FAB - W × sin15o = 0

FBC = -34.43 kN FBD = -197.33 kN

At joint C FBC × sin75o + FCD × sin30o = 0 FCE – FAC - FBC × cos75o + FCD × cos30o= 0

FCD = 66.51 kN FCE = 133.01 kN

6.1(1)


γM0 = 1.0 γM1 = 1.0 γM2 = 1.25

Design of Top Chords (members AB, BD, DG, GH)

Maximum force (member AB and GH) = 206.55 kN (in compression)

Try 125 × 75 × 10 unequal angles with long legs back to back with 8 mm back to back spacing to allow for gusset plates connecting the internal members, steel grade EN 10025-4, S275 M/ML.

NEd =206.55 kN

Table 3.1

Material properties:

Young’s modulus E = 210000 N/mm2 for steel grade S275 and thickness ≤ 40 mm Yield strength fy = 275 N/mm2

Ultimate strength fu = 370 N/mm2

92.0275235235

y

===f

ε

Steel Designers’ Manual

Cross-section dimensions:

Depth of section h = 125 mm Width of section b = 75 mm Thickness t = 10 mm


Reference Output


y y

z

z

8 mm

Table 5.2 Classification of the cross-section:

58.105.111010212575

2=<=

×+

=+ εthb

8.13155.1210125

=<== εth

The section is Class 3


Compound cross-section properties:

Cross-section area A = 38.3 cm2

Second moment of area about y-axis Iy = 606 cm4

Radius of gyration about y-axis iy = 3.98 cm Second moment of area about z-axis Iz = 345 cm4

Radius of gyration about y-axis iz = 3.00 cm

Individual cross-section properties:


Second moment of area about u-axis Iu = 336 cm4

Radius of gyration about u-axis iu = 4.19 cm Second moment of area about v-axis Iv = 50.2 cm4

Radius of gyration about v-axis iv = 1.62 cm

Eq.(6.10) for Class 3 sections

Calculate the compression resistance of the cross-section:

10500.1

102753830 3

M0

yRdc, =

××==

−

γAf

N kN

Nc,Rd > NEd

Eq.(6.50) for Class 1,2 and 3 cross-sections

Calculate the resistance to flexural buckling:

Determine the non-dimensional slenderness for flexural buckling(1):

39.11

1z

cr

cr

y ===λ

λi

L

N

Af

where 3620

15cos

35000.1 ABcr ==×=

oLL mm

mm 30z == ii

81.86y

1 ==fEπλ


Reference Output


Eq.(6.49) and Tables 6.1 and 6.2 for T cross-section

Determine the reduction factor due to buckling

353.039.176.176.1

112222=

−+=

−+=

λΦΦχ

where [ ] 76.1)2.0(15.0 2 =+−+= λλαΦ

49.0=α = the imperfection factor tee section

3710.1

102753830353.0 3

M1

yRdb, =

×××==

−

γχAf

N kN

Nb,Rd > NEd


Calculate the resistance to flexural buckling of individual member:


711.01

1v

cr

cr

y ===λ

λi

L

N

Af

where mm (between two welded spacer plates)1000cr =L (2)

mm 2.16v == ii

81.86y

1 ==fEπλ

Determine the reduction factor due to buckling Eq.(6.49) and Tables 6.1 and 6.2 for L cross-sections

777.0711.084.084.0

112222=

−+=

−+=

λΦΦχ

[ ] 840.0)2.0(15.0 2 =+−+= λλαΦ

34.0=

where

α = the imperfection factor tee section

4100.1

102751920777.0 3

M1

yRdl, =

×××==

−

γχAf

N kN

2Nl,Rd > NEd

Therefore, the top chord members are satisfactory.

Design of Bottom Chords (members AC, CE, EH)

Maximum force (member AC and EH) = 199.52 kN (in tension)

For economic reasons, together with ease of fabrication of the truss, the bottom chord will also be 125 × 75 × 0 compound unequal angles with 8 mm back to back spacing.

NEd =199.52 kN

Eq.(6.7)

Calculate the tension resistance of the cross-section:

For the design plastic resistance of the gross cross-section:

25.10530.1

102753830 3

M0

yRdpl, =

××==

−

γAf

N kN

Npl,Rd > NEd


Reference Output


Eq.(6.7)

For the design ultimate resistance of the net cross-section at holes (20 mm diameter holes) for fasteners(3):

91425.1

1037034309.09.0 3

M2

unetRdu, =

×××==

−

γfA

N kN

where 3430)102(203830)2(onet =××−=×−= tdAA mm2

Nu,Rd > NEd

60

40

Therefore, the bottom chord members are satisfactory.

Design of Internal Members (members BC, EG, CD, DE)

Maximum compression force (BC and EG) = 34.43 kN Maximum tension force (CD and DE) = 66.51 kN Try a 60 × 60 × 6 angle

NEd > 34.43 kN

Table 3.1

Material properties:

Young’s modulus E = 210000 N/mm2 for steel grade S275 and thickness ≤ 40 mm Yield strength fy = 275 N/mm2

Ultimate strength fu = 370 N/mm2

92.0275235235

y

===f

ε


Cross-section dimensions:

Depth of section h = 60 mmWidth of section b = 60 mmThickness t = 6 mm

Table 5.2

Classification of the cross-section

58.105.1110660

2=<==

+= ε

thb

th

The section is Class 3


Cross-section properties:


Second moment of area about u-axis Iu = 36.8 cm4

Radius of gyration about u-axis iu = 2.3 cm Second moment of area about v-axis Iv = 9.64 cm4

Radius of gyration about v-axis iv = 1.18 cm


Reference Output


Positioning of two holes for bolts:

Diameter of holes do = 20 mm Diameter of bolts d = 18 mm End distance e1 = 40 mm (e1>1.2do) Spacing between two holes p1 = 60 mm (14t>p1>2.2do)

Eq.(6.10)

Calculate the compression resistance of the cross-section:

kN 13.1910.1

10275695 3

M0

yRdc, =

××==

−

γAf

N

Nc,Rd > NEd


Calculate the resistance to flexural buckling:


95.01

1

cr

cr

y ===λ

λi

L

N

Af

where mm 83.970sin15 3751 oBCcr === LL

mm 8.11v == ii

81.86y

1 ==fEπλ

Determine the reduction factor due to buckling

63.095.008.108.1

112222=

−+=

−+=

λΦΦχ

[ ] 08.1)2.0(15.0 2 =+−+= λλαΦ

34.0=

where Eq.(6.49) and Tables 6.1 and 6.2 for L cross-sections

α = the imperfection factor angle section

kN 41.1200.1

1027569563.0 3

M1

yRdb, =

×××==

−

γχAf

N

Therefore, the internal members BC and EG are satisfactory.

Nb.Rd > NEd

Calculate the tension resistance of the cross-section:

For the design plastic resistance of the gross cross-section:

Eq.(6.47)

kN 13.1910.1

10275695 3

M0

yRdpl, =

××==

−

γAf

N

NEd =66.51 kN

Npl.Rd > NEd

For angles connected by one leg with 2 bolts (20 mm diameter holes):

kN 7.8925.1

1037057552.0 3

M2

unet2Rdu, =

×××==

−

γβ fA

N

where 52.02 =β for pitch 601 =p

2onet mm 575620695 =×−=×−= tdAA

Nu.Rd > NEdEN 1993-1-8: 2005 Eq (3.12) and Table 3.8


Reference Output


For bearing resistance:

kN 28.5325.1

10618370667.05.2 3

M2

ub1Rdb, =

×××××==

−

γα dtfk

F

2Fb.Rd > NEd

where 5.21 =k

667.0203

403 o

1b =

×==

d

eα

Therefore, the internal members CD and DE are satisfactory.

Secondary Moments(4):

The forces obtained from an elastic analysis of the truss are given in Table 1, from which it is noticed that the axial forces do not differ significantly from those calculated based on the pin joint assumption. Both the moments and shear forces are low.

Deflection(5):

The maximum allowable deflection is assumed to be span/300;

Span/300 = 14000/300 = 46.67 mm.

The maximum deflection of the truss is obtained for unfactored loads (i.e. W = 28.26 kN). The deflection at the apex was found to be 10.58 mm when all of joints are assumed to be rigid (Model C), or 10.60 mm when the bolted joints are all assumed to be pinned (Model B), and 10.80 mm when all of the joints are assumed to be pinned (Model A).

It is obvious that the difference in the deflection between the three models is negligible. The deformed shapes of the truss under the full dead and imposed loads for three different analysis models are plotted in the figures below. It should be pointed out that the maximum deflection in the bottom chord is smaller in model A than in other two models, because the central member CE is not bent when the member is jointed by pins.

Model A: Deformed and undeformed shapes with all pinned joints.

Model B: Deformed and undeformed shapes of the truss with continuous chords but pinned internal members.


Reference Output


Model C: Deformed and undeformed shapes of the truss with all rigid joints.





(1) Only the buckling occurring in the weakest plane is considered, as this is the worst case.

(2) The two angles are welded together by plates at intervals of 1 m.

(3) Gusset plates 8 mm thick are used to connect the internal members to chords.

(4) Model A assumes all joints are pinned (i.e. all members are bar elements). Model B assumes chords are continuous but the internal members are pinned (i.e. chords are beam elements and internal members are bar elements). Model C assumes all joints are rigid (i.e. all members are beam elements).

Table 1. Axial forces, shear forces and bending moments in members (where N is the axial force, S is the shear force and M is the bending moment.)

Model A Model B Model C Members N, kN N, kN S, kN M, kNm N, kN S, kN M, kNm

AB A B

-206.55 -206.55

-202.39 -202.39

1.02 1.02

0.58 -3.15

-202.30 -202.30

-1.00 -1.00

0.53 -3.08

BD B D

-197.33 -197.33

-193.17 -193.17

-1.48 -1.48

-3.15 2.22

-193.01 -193.01

1.49 1.49

-3.18 2.23

AC A C

199.52 199.52

195.23 195.23

0.12 0.12

0.58 1.02

195.16 195.16

0.18 0.18

0.53 1.17

CE C E

133.01 133.01

133.74 133.74

0 0

1.02 1.02

133.82 133.82

0 0

0.98 0.98

BC B C

-34.43 -34.43

-31.93 -31.93

0 0

0 0

-31.95 -31.95

0.07 0.07

-0.09 -0.02

CD C D

66.51 66.51

61.47 61.47

0 0

0 0

61.24 61.24

-0.07 -0.07

0.17 -0.10

(5) The deflections are calculated from elastic analysis of the truss, and are obtained using the unfactored loads. There is no specified allowable deflection in EN 1993-1-1:2005.

Example 08 Choosing a steel sub-grade Sheet 1 of 4 Rev

Made by LPN Date May 2007 Checked by Date

Reference Output

P:\EDN\EDN399\Student worked examples\Worked Examples\Second Drafts\Formatted\08_Choose steel grade_jbw01.doc 1


Choosing a steel sub-grade

Introduction

Determine the steel sub-grade that may be used for the simply supported restrained beam (UB 457 × 191 × 82 steel grade S275).

Floor beam at Level 1 – Gridline G1-2

Beam span, L = 8,0m Bay width, w = 6,0m

Actions

2Permanent action : gk = 3.7 kN/m 2Variable action : qk = 3.8 kN/m

Section Properties

From example 01: = 9.9 mm Web thickness tw = 16.0 mm Flange thickness tf

= 1610.9 cm3 Elastic modulus, y-y Wel, y

= 275 N/mm2Yield strength fBS EN

1993-1-1 Table 3.1

y

is the leading action) Load combination (T 2.2.4 (i)

Ed

] "+" ∑G BS EN 1990 A.1.2.2 (1)

Ed = E { A[TEd K "+" ψ 1 QK1 "+" ∑ ψ Q2,i Ki } not relevant for this example Where: ψ = 0,5 1

Calculation of reference temperature T Ed

εcfεσ TTTTTTT Δ Δ + Δ + Δ++= &RrmdEd + Δ 2.2

Where:

= − 5°C (lowest air temperature) mdT

rTΔ = − 5°C (maximal radiation loss)

σTΔ = 0°C (adjustment for stress and yield strength)

RTΔ = 0°C (safety allowance to reflect different reliability levels for different applications)

ε&TΔ = 0°C (assumed strain rate equal to reference

strain rate 0ε& )

εcfTΔ = 0°C (no cold forming for this member)

= −10 °C EdT


Reference Output


Calculation of relevant loads

QK + ψ 1 GK1 = 3.8 × 3.6 + 0.5 × 3.7 × 3.6= 20.3 kN/m Moment diagram

162.4 kNm

Maximum moment at mid span :

= 20.3 × 8² / 8 = 162.4 kNm My,Ed

Calculation of maximum bending stress:

9.1610

10004.162

yel,

Edy,Ed

×==

W

Mσ = 100.8 N/mm²

Stress level as a proportion of nominal yield strength 2.3.2 = 100.8 N/mm² σEd

0

nomy,y 25.0t

tf(t)f ×−=

where: t = 16 mm (flange thickness) t = 1 mm 0

fy(t) = 1

1625.0 ×−275 = 271N/mm²

Note: fy(t) may also be taken as ReH-value from the product standard EN10025

proportion of the nominal yield strength

(t)f(t)f yyEd 37.0271

8.100=×=σ

Choice of steel sub-grade

Two different methods can be used to select an appropriate steel sub-grade. The first one is conservative (without interpolation). The second method uses linear interpolation and possibility leads to more economic values. Both methods are presented here.

Conservative method

Input values: >(t)f yEd 50.0=σ (t)f yEd 37.0=σProportion of yield strength:

10Ed −=T °C = Temperature:

Element thickness: t = 16 mm

Table 2.1 Therefore the steel sub-grade obtained without interpolation is:

S275JR provides a limiting thickness of 55 mm > t = 16 mm f


Reference Output


Exact Determination

Linear interpolation can be used in applying Table 2.1 of EN 1993-1-10 for the actual value of the proportion of the nominal yield strength.

Proportion of yield strength: (t)f yEd 50.0=σ a)

(t)f yEd 25.0=σ b)

Temperature: 10Ed −=T °C

Table 2.1

Max permissible element thickness for S355JR

55)50.0( yEd == (t)ft σ mm

95)25.0( yEd == (t)ft σ mm

with linear interpolation: > 8.75)37.0( yEd == (t)ft σ mm 16f =t mm

Steel sub-grade S275JR is adequate.

S275JR provides a limiting thickness of 75.8 mm > t = 16mm f

Example 09 Composite slab Sheet 1 of 8 Rev

Made by LPN Date 04/04/07 Checked by Date

Reference Output

P:\EDN\EDN399\Student worked examples\Worked Examples\Second Drafts\Formatted\09_Composite slab.doc 1

Unless stated otherwise all references are to BSEN 1994-1-1

EN1991-1-1

Table 6.2

Composite slab

3.00 3.00

A

A

6.00

130 51

CF51 profile

Plan view

Section A - A

152.5

Slab depth: 130 mm

Deck properties (CF51)

Thickness: 0.9 mm

Yield strength: 350 N/mm2

Sagging moment resistance: MRd=5.69 kNm/m

Effective steel area: Ap = 1579 mm2/m

Inertia: Ip =55.7 cm4/m

Resistance to horizontal shear: =τ Rd,u 0.13 N/mm2

The slab is designed for both the construction stage and the composite stage. In the construction stage, the steel sheeting acts as shuttering and has to carry its own weight, the wet concrete and the construction loads. In the composite stage the slab has to carry its own weight, floor finishes and the live load. The following loads are considered in this example:

Permanent actions:

Self weight of floor: 3.5 kN/m2

Self weight of ceiling, raised floor and services: 0.2 kN/m2


Reference Output


§9.4.2 (5) Variable actions:

Imposed floor loads for offices (cat. B): 3.0 kN/m2

Imposed floor loads for partitions: 0.8 kN/m2

Verification of the sheeting as shuttering: The sheeting resistance needs to be verified in the construction stage at both ultimate and serviceability limit state according to EN 1993-1-3. Construction stage checks are not carried out in this example.

Verification of the composite slab: Ultimate limit state: The continuous slab will be designed as a series of simply supported spans. This approach is conservative, because it does not take into account the positive effect of the continuity (bending moment diagram less important). However, hogging moments through reinforcement at the support are normally used at the fire limit state.

Slabs designed

4.00 4.00 3.00 3.00

( )

8Lqg

M2

qgEd

⋅⋅γ+⋅γ=

( )

838.35.17.335.1M

2

Ed⋅⋅+⋅

=

m/kNm12MEd =

Design bending resistance: The sagging bending moment resistance should be calculated from the stress distribution in the figure below, if the neutral axis is above the sheeting.

Figure 9.5

+

-

Centriodal axis of theprofiled steel sheeting

z x

0.85 f cd

pl

fyp,d

M

M

M

p

c,f

plRd

http://www.access-steel.com/discovery/linklookup.aspx?id=EC056



Reference Output


cd

d,ypppl fb85.0

fAx

⋅⋅

⋅=

2

0M

k,ypd,yp mm/N350

0.1350f

f ==γ

=

2

c

ckcd mm/N7.16

5.125ff ==

γ=

mm9.387.16100085.0

3501579xpl =⋅⋅

⋅=

For full shear connection: ( )2/xdfAM plpydpRd,pl −⋅⋅=

( )2/9.383.1133501579M Rd,pl −⋅⋅=

m/kNm9.51M Rd,pl =

Mpl,Rd > MEd (=12 kNm/m) OK

§9.7.3 (8) Longitudinal shear by partial connection method: Shear span required for full shear connection cfxRd,uc NLbN ≤⋅⋅τ=

The distance to the nearest support, Lx, required for full shear

connection can be determined by

Rd,u

ydp

Rd,u

cfx b

fAb

NLτ⋅

⋅=

τ⋅=

2

VS

Rk,uRd,u mm/N104.0

25.113.0

==γτ

=τ

mm5314104.01000

3501579Lx =⋅⋅

=

Hence, the slab cannot develop a full shear connection as the span is equal to 3000 mm.

Design check using the simplified partial interaction diagram: For any cross section along the span it has to be shown that the corresponding design bending moment, MEd, does not exceed the design bending resistance, MRd. In the following, x is the distance from the support.



Reference Output


Dead loads: 3.7 kN/m2

Live loads: 3.8 kN/m2

Factored: 2m/kN7.108.35.17.335.1 =⋅+⋅x05.16x35.5)x(M 2

Ed +−=

69.5x7.8)x(MRd +=

)x(MEd )x(MRdx

0 0 5.69 0.25 3.68 7.87 0.50 6.69 10.04 0.75 9.03 12.22 1.00 10.70 14.39 1.25 11.70 16.57 1.50 12.04 18.74

for all cross sections RdEd MM < Vertical shear:

( )

2Lqg

V qgEd

⋅⋅γ+⋅γ=

( )

238.35.17.335.1VEd

⋅=

+ ⋅ ⋅

m/kN16VEd =

Design vertical shear resistance: EN 1992-1-1 §6.2.2

[ ]( ) pwcp13/1

ck1c,RdRd,v dbkf100kCV ⋅⋅σ⋅+⋅ρ⋅⋅⋅=

with a minimum of

( ) pwcp1minmin,Rd,v dbkvV + ⋅σ ⋅ ⋅=

See Note in EN 1992-1-1 §6.2.2

12.05.118.018.0C

cc,Rd ==

γ=

3.23.113

2001d2001k

p

=+=+=

02.0db

A

pw

1s1 ≤

⋅=ρ

Asl is the area of the tension reinforcement in [mm], and is usually taken as the area of the deck. However, in our case, it could be unconservative to consider the deck area, due to the end conditions of the profile. We should also consider the equivalent tensile reinforcement generated by the deck bond, and the equivalent tensile reinforcement from anchorage force generated by studs, and takes the minimum of these two values.




Reference Output


Bond: yeqRd,u fA

2L

⋅≤⋅τ

y

Rd,udeck,eq f2

LA

⋅⋅τ

=

m/mm7.445103502

3000104.0A 23deck,eq =⋅

⋅⋅

=

Bearing of studs: d,yp0dRd,pb ftdkP ⋅⋅⋅= ϕ

0.6da1k

0d

≤+=ϕ

mm9.20191.1d1.1d 00d =⋅=⋅=

a should be not less than 1.5 =31.35 say 40 mm 0dd

91.29.20

401k =+=ϕ

stud/kN15.193509.09.2091.2P Rd,pb =⋅⋅⋅=

m/kN6.1251525.0

15.19P Rd,pb ==

m/mm9.35810350

6.125A 23stud,eq =⋅=

02.0db

A

pw

1s1 ≤

⋅=ρ

m/mm9.358A)A;Amin(A 2stud,eqstud,eqdeck,eq1s ===

And

bw = 737 mm/m, i.e. the smallest width in [mm] of the section in the tension area. See Note in

EN 1992-1-1 §6.2.2 02.0103.4

3.1137379.358 3

1 <⋅=⋅

=ρ −

0AN

c

Edcp ==σ , since NEd = 0, i.e. no axial forces or

prestress. k1 = 0.15

( )[ ] 3.113737015.025103.41003.212.0V 3/13Rd,v ⋅⋅⋅+⋅⋅⋅⋅⋅= −

m/kN9.50V Rd,v =

Minimum value

61.0253.2035.0fk035.0v 2/12/32/1ck

2/3min =⋅⋅=⋅⋅=

( ) m/kN9.503.113737015.061.0V min,Rd,v =⋅⋅⋅+=

EdRd,v VV > (=12 kN/m) OK

All design checks of the composite slab in the ultimate limit state are OK.



Reference Output


Serviceability limit state: §9.8.1 (2) Cracking of concrete:

As the slab is designed as simply supported, only anti-crack reinforcement is needed. The cross-sectional area of the reinforcement above the ribs should be not less than 0.4% of the cross-sectional area of the concrete above the ribs.

min m/mm316791000004.0hb004.0A 2cs =⋅⋅=⋅⋅=

A393 mesh (393 mm2/m) will be enough for this purpose.

§9.8.2 (5)

Deflection: For the calculations of the deflections of the slab, the slab is considered to be continuous. The following approximations apply:

• the second moment of area may be taken as the average of the values for the cracked and un-cracked section;

• for concrete, an average value of the modular ratio, n, for both long-term and short-term effects may be used.

10

000,3132

000,210

3EE

21

EEE

ncm

cm

p

cm'p ≈

⋅=

⎟⎠⎞

⎜⎝⎛ +

==

Second moment of area for the cracked section

( ) p2

cpp

3c

bc IxdAn3

xbI +−⋅+⋅⋅

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

⋅

⋅⋅+⋅

⋅=

⋅=∑∑ 1

Andb2

1bAn

AzA

xp

pp

i

iic

mm1.461157910

3.1131000211000

157910xc =⎟⎟⎠

⎞⎜⎜⎝

⎛−

⋅⋅⋅

+⋅⋅

=

( )

m/mm1069.7

107.551.463.1131579103

1.461000I

46

423

bc

⋅=

⋅+−⋅+⋅⋅

=

h =51 p

h =79 c

152.5b '= 132.5

d = 113.3pxu

x

o




Reference Output


Second moment of area for the un-cracked section

( ) p2

upp

2p

utp0

3p0

2c

uc

3c

bu

IxdA2

hxh

nhb

n12hb

2hx

nhb

n12hbI

+−⋅+⎟⎟⎠

⎞⎜⎜⎝

⎛−−⋅

⋅

+⋅

⋅+⎟

⎠⎞

⎜⎝⎛ −⋅

⋅+

⋅⋅

=

pp0c

ppp

tp0

2c

u Anhbhb

dAn2

hhhb

2hb

x⋅+⋅+⋅

⋅⋅+⎟⎟⎠

⎞⎜⎜⎝

⎛−⋅⋅+⋅

=

With mm9.8685.1325.152

1000'bb 00 =×==∑

mm6.68157910519.868791000

3.113157910251130519.868

2791000

x

2

u

=⋅+⋅+⋅

⋅⋅+⎟⎠⎞

⎜⎝⎛ −⋅⋅+⋅

=

( ) 422

323

bu

107.556.683.11315792516.68130

10519.868

1012519.868

2796.68

10791000

1012791000I

⋅+−⋅+⎟⎠⎞

⎜⎝⎛ −−⋅

⋅

+⋅⋅

+⎟⎠⎞

⎜⎝⎛ −⋅

⋅+

⋅⋅

=

m/mm1018.21I 46bu ⋅=

Average Ib of the cracked and un-cracked section

( ) m/mm1044.14

21018.2169.7

2III 46

6bubc

b ⋅=⋅+

=+

=

EN1992-1-1 §7.4.1(4)

Deflections The total deflection under the worst load case should not exceed L/250.

Live load, worst case:

q q

Considered span

See Access-Steel SX009a

mm7.01044.1410210

30008.37.00099.0IE

Lq0099.063

4

b

41

q,c =⋅⋅⋅⋅⋅⋅

=⋅

⋅⋅ψ⋅=δ

Deflection due to finishes negligible, as load only 0.2 kN/m2.

mm12250

3000250Lmm7.0q,c ==<=δ OK



Example 10 Sheet 1 of 12 RevBracing and bracing connections B3

Made by JPR Date Nov-06 Checked by AGK Date Dec-06

Reference Output

P:\EDN\EDN399\Student worked examples\Second draft-book\10_Bracing bracing connections_meb01 REV B3.doc 1


Bracing and bracing connections

Assumptions:

This calculation assumes that:

(a) The wind loading1) at each floor is transferred to two vertically braced end bays on grid lines ‘A’ and ‘J’ by the floors acting as diaphragms 2).

(b) The braced bays, acting as vertical pin-jointed frames, transfer the horizontal wind load to the ground.

(c) The beams and columns that make up the bracing system have already been designed for gravity loads3). Therefore, only the diagonal members have to be designed and only the forces in these members have to be calculated.

(d) All the diagonal members are of the same section; thus, only the most heavily loaded member has to be designed.

Forces in the bracing elements of the Warren truss 4) bracing system

BSEN 1991-1-4 Total overall unfactored wind load 5), Fw = 925 kN

Vertical bracing located at each end of the building, therefore, total unfactored load to be resisted by each braced bay is 463 kN

Design wind load at ultimate limit state

Using equation 6.10b) 6) gives the design wind load per braced bay as:

695Ed =F kN

Apply this horizontal load as a proportional share of the cladding area:

Roof level 856955.18

25.2=× kN

3rd & 2nd floor levels 1696955.185.4

=× kN

1st floor level 1786955.18

75.4=× kN

See Figure B1 Ground at column base level 94695

5.185.2

=× kN*

*Assume that this load is taken out in shear through the ground slab and is therefore not carried by the frame.

Horizontal reaction, H = (85 + 169 +169 + 178) = 601 kN


Reference Output


Vertical reactions,

( ) ( ) ( ) ( )6

0.51785.91690.141695.1885 ×+×+×+×±=F

1072±=F kN

A computer analysis of the bracing system can be performed to obtain the member forces. Alternatively, hand calculations can be carried out to find the member forces. The use of the ‘method of sections’ is straightforward in this case.

Thus, make a cut at y-y shown in figure B1 to find the forces in the members and in particular, the bracing member ‘bk’ (worst case).

y

Figure B1

85 kN

169 kN

169 kN

178 kN

H = - 601

F = 1072 kN F = - 1072 k

4.5 m

4.5 m

4.5 m

5.0 m

Roof - (18.5 m)

3rd - (14.0 m)

2nd - (9.5 m)

1st - (5.0 m

Grd - (0.0 m)

6.0 m

b

c

d

e f

g

h

j

y

a k

θ

5.0 m

6.0 m

a

b j

k s

Fbk Fjk Fab

601 kN

1072 kN 1072 kN


Reference Output


θ = o81.39

65

tan 1 =⎟⎠⎞

⎜⎝⎛−

s = m ( ) 841.381.39sin6.0 =× o

Take moments at ‘a’ ( M∑ = 0):

( ) 00.610720.6841.3 jkbk =×−+ FF

Take moments at ‘b’ ( M∑ = 0):

( ) ( ) 00.56010.610720.6 jk =×+×−F

Take moments at ‘k’ ( M∑ = 0):

( ) 00.610720.6 ab =×−F

Therefore,

Fab = 1072 kN Tension

Fjk = 571 kN Compression Fbk = 783 kN Compression in bracing member ‘bk’


6.1(1)

BSEN 1993-1-8 Table 2.1

γM0 = 1.0

γM1 = 1.0

γM2 = 1.25 (for shear)

Trial section


Check diagonal bracing member, ‘bk’ (worst case).

Try: 219.1 × 10.0 mm thick Circular Hollow Section (CHS), grade S275 steel

Section Properties

Area A = 65.7 cm²

Second moment of area I = 3598 cm4 Radius of gyration i = 7.40 cm Thickness t = 10.0 mm

Ratio for local Buckling d / t = 21.9

Material properties

Table 3.1 As t ≤ 40 mm, for S275 steel

Yield strength fy = 275 N/mm²

3.2.6 (1) Young’s modules E = 210 kN/mm²


Reference Output


5.5

Table 5.2

Section classification

Class 1 limit for section in compression, d/t ≤ 50 ε2

ε = (235/fy)0.5, fy = 275 N/mm², ε = 0.92, ε2 = 0.85

d/t ≤ 50(0.85) = 42.5 Therefore, 21.9 < 42.5

For axial compression the section is Class 1.

Design of axially loaded compression members

Cross sectional resistance to axial compression

6.2.4(1) Eq. 6.9 Basic requirement 0.1

Rdc,

Ed ≤NN

NEd is the design value of the applied axial force NEd = Fbk = 783 kN

Nc,Rd is the design resistance of the cross-section for uniform compression

M0

yRdc, γ

fAN

×= (For Class 1, 2 and 3 cross-sections)

6.2.4(2) Eq. 6.10

kN

Compressive resistance Nc,Rd = 1807 kN

1807100.12756570 3

Rdc, =××

= −N

43.0

1807783

Rdc,

Ed ==NN

< 1.0

Therefore, the compressive resistance of the cross section is adequate.


6.3.1.1(1) Eq. 6.46

For a uniform member under axial compression the basic requirement is:

0.1Rdb,

Ed ≤NN

Nb,Rd is the design buckling resistance and is determined from:

6.3.1.1(3) Eq. 6.47

M1

yRdb, γ

χ fAN = (For Class 1, 2 and cross-sections)

6.3.1.2(1) χ is the reduction factor for the buckling curve and may be determined from Figure 6.4.

Table 6.2 For hot finished CHS in grade S275 steel use buckling curve a Use buckling curve a

For flexural buckling the slenderness is determined from:

6.3.1.3(1) Eq. 6.50 ⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛==

1

cr

cr

y 1λ

λi

LNAf

(For Class 1, 2 and 3 cross-sections)


Reference Output


Where:

Lcr is the buckling length

As the bracing member is pinned at both ends:

Lcr = L = 781060005000 22 =+ mm

Lcr = 7810 mm

Table 5.2

ελ 9.931 =

924.0275235235

y

===f

ε

8.86924.09.931 =×=λ 8.861 =λ

6.3.1.3(1) Eq. 6.50 22.1

8.861

747810

=⎟⎠⎞

⎜⎝⎛×⎟

⎠⎞

⎜⎝⎛=λ 22.1=λ

Figure 6.4 For 22.1=λ and buckling curve a

51.0=χ

51.0=χ

6.3.1.1(3) Eq. 6.47

Therefore,

kN


921Rdb, =N kN 921100.1

275107.6551.0 32

Rdb, =××××

= −N

6.3.1.1(1) Eq. 6.46 85.0

921783

Rdb,

Ed ==NN

< 1.0

Therefore, the flexural buckling resistance of the section is adequate.

6.2.3 Design of axially loaded tension member and connection

When the wind is applied in the opposite direction, the bracing member considered above will be loaded in tension. Therefore, check section for the same magnitude of loading.

Cross sectional resistance of the CHS to axial tension

6.2.3(1) Basic requirement 0.1Rdt,

Ed ≤NN

NEd is the design value of the applied axial tension NEd = Fbk = 783 kN

Nt,Rd is the design tensile resistance of the cross-section

For the CHS

M0

yRdpl,Rdt, γ

AfNN ==

180710

0.12756570 3

Rdt, =××

= −N kN Tensile resistance of CHS Nt,Rd = 1807 kN


Reference Output

P:\EDN\EDN399\Student wo ed examples\Second draft-book\10_Bracing bracing connections_meb01 REV B3.doc 6

rk


Table 3.1

43.01807783

Rdt,

Ed ==NN

< 1.0

Therefore the tensile resistance of the CHS cross-section is adequate.

Resistance of connection

Assume the CHS are connected to the frame via gusset plates. Flat end plates fit into a slotted hole in the CHS section and fillet welded together. Bolts in clearance holes transfer the load between the CHS end plate and gusset plates.

Check connection to transfer 783 kN tensile force.

Try: 6No non-preloaded Class 8.8 M24 diameter bolts in 26 mm diameter clearance holes

Assume shear plane passes through the threaded part of the bolt

Single bolt thread area, A = As = 353 mm²

Clearance hole diameter, do = 26 mm

For Class 8.8 non-preloaded bolts:

Yield strength fyb = 640 N/mm2 Ultimate tensile strength fub = 800 N/mm2

Positioning of holes for bolts:

(Minimum) End distance (e1) 1.2 do = 31.2 mm< e1 = 40 mm

(Minimum) Edge distance (e2) 1.2 do = 31.2 mm < e2 = 60 mm (Minimum) Spacing (p1) 2.2 do = 57.2 mm < p1 = 80 mm (Minimum) Spacing (p2) 2.4 do = 62.4 mm < p2 = 130 mm

(Maximum) e1 & e2, larger of 8t = 120 mm or 125 mm > 40 mm & 60 mm (Maximum) p1 & p2

Smaller of 14t = 210 mm or 200 mm > 80 mm & 130 mm

Therefore, bolt spacings are acceptable.


Reference Output


EN 1993-1-8 Figure 3.1 See sheet 6

219.

1 D

ia. C

HS

e1

e2

p1

p2

e2

p1

Positioning of holes for bolts

CHS sealing plate

CHS end plate details

250

15

Section X – X

6 mm fillet weld

6 mm fillet weld

Slotted 219.1 Dia. CHS

All dimensions shown in mm (Not to Scale)

500 21

9.1

Dia

. C

HS

40 80 80 50 250

250

60

130

60

Grade S275 end plate 500 x 250 x 15 mm thick to fit into a slotted hole in the CHS

X

Centre line

X

Bracing setting out and connection detail – Figure B2

Universal beam (UB)

Uni

vers

al c

olum

n

Gusset plate

CHS end plate

219.1 Dia. CHS

Centre line

Extended UB end plate

Centre line


Reference Output


Shear resistance of bolts

Table 3.4 The resistance of a single bolt in shear is determined from:

1351025.1

3538006.0 3

M2

ubvRdv, =×

××== −

γα Af

F kN

Resistance of a single bolt to shear:

135Rdv, =F kN

Minimum No of bolts required is 8.5

135783

Rdv,

Ed ==FN

bolts

Therefore, provide 6 bolts in single shear.

Resistance of all six bolts in single shear is:

81013566 Rdv, =×=F kN

Resistance of 6 bolts in shear 810 kN

97.0

810783

= < 1.0

Therefore shear resistance of 6 bolts is adequate.

Bearing resistance of bolts

Assume gusset plate has a thickness greater than the end plate. Bearing failure will tend to take place in the end plate.

Table 3.1

End plate is a grade S275 with t < 40 mm, therefore:≤ 40 mm, for S275 steel

Yield strength fy = 275 N/mm²

Ultimate tensile strength fu = 430 N/mm²

Table 3.4 The bearing resistance of a single bolt is determined from:

M2

ub1Rdb, γ

α dtfkF =

αb is the least value of αd,

pu,

ub

ff

and 1.0

For end bolts 51.0

26340

3 o

1d =

×==

deα

For inner bolts 78.041

26380

41

3 o

1d =⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛

×=−=

dpα

86.1430800

u,fp

ub ==ff

Therefore:

For end bolts 51.0b =α

For inner bolts 78.0b =α

Conservatively consider 51.0b =α for each bolt.

51.0b =α


Reference Output


For edge bolts k1 is the smaller of 7.18.2o

2 −de or 2.5

8.47.12660

8.2 =−⎟⎠⎞

⎜⎝⎛ ×

For inner bolts k1 is the smaller of 7.14.1o

2 −dp or 2.5

3.57.126130

4.1 =−⎟⎠⎞

⎜⎝⎛ ×

Therefore:

For both end and inner bolts 5.21 =k

5.21 =k

The bearing resistance of a single bolt may be conservatively taken as:

1581025.1

152443051.05.2 3Rdb, =×

××××= −F kN

158Rdb, =F kN

Resistance of all six bolts in bearing may be conservatively taken as: 94815866 Rdb, =×=F kN

Resistance of 6 bolts in bearing 948 kN

83.0

948783

= < 1.0

Therefore the bearing resistance of 6 bolts is adequate.

Design of fillet weld

BSEN 1993-1-8 4.5

Assume 6 mm fillet weld is used on both sides, top and bottom, of the fitted end plate.

BSEN 1993-1-8 4.5.3.3(3) Design shear strength,

M2

u,

3/γβw

dvwf

f =

4.26.00.7lengthleg7.0

BSEN 1993-1-8 Table 4.1

Correlation factor, βw = 0.85

Throat thickness of weld

× ==×=a mm

Therefore,

6.23325.185.03/430

, =×

=dvwf N/mm2

BSEN 1993-1-8 4.5.3.3(2)

Design resistance of weld per unit length is:

1.9812.46.233,, =×== afF dvwdvw N/mm


Reference Output


See sheet 8 Hence, for a weld with an effective length of:

( ) 2380.62250 =×−=effl mm

the shear resistance of the weld is

934102381.98144 3Rdw, =×××= −

efflF kN

Shear resistance of 4 by 238 mm long 6mm fillet welds is:

934 kN

84.0

934783

= < 1.0

Therefore the shear resistance of the fillet welds is adequate.

Tensile resistance of end plate

Two modes of failure to be considered: i) cross-sectional failure and ii) block tearing failure.

BSEN 1993-1-8 3.10.2

219.

1 D

ia. C

HS

NedNed

i) Cross-sectional failure

Ned21

9.1

Dia

. C

HS

Ned

ii) Block tearing failure

6.2.3(1)

i) Cross-sectional failure

Basic requirement: 0.1Rdt,

Ed ≤NN

For a cross-section with holes, the design tensile resistance is taken as the smaller of Npl,Rd and Nu,Rd:

6.2.3(2)

Eq. 6.6 a)

M0

yRdpl, γ

fAN

×=

A is the gross cross-sectional area:

375015250 =×=A mm2

1031100.12753750 3

Rdpl, =××

= −N kN Npl,Rd = 1031 kN

b) M2

unetRdu,

9.0

γfA

N××

= Eq. 6.7

( ) 2970152623750net =××−=A mm2 6.2.2.2


Reference Output


9191025.1

43029709.0 3Rdu, =×

××= −N kN

Nu,Rd = 919 kN

Nu.Rd < Npl.Rd

Therefore, the tensile resistance of the end plate is:

919Rdu,Rdt, == NN kN

Nt,RD = 919 kN

6.2.3 (1) 85.0919783

Rdu,

Ed ==NN

< 1.0

Therefore, the tensile resistance of the cross-section of the end plate is adequate.

ii) Block tearing failure

EN 1993-1-8 3.10.2 (2)

For a symmetric bolt group subject to concentric loading the design block tearing resistance ( ) is determined from: RdEff,1,V

( )M0

nvy

M2

ntuRdEff,1, 3/1

γγAfAf

V +=

Where:

ntA is the net area subject to tension

nvA is the net area subject to shear

See Sheet 8 ( ) ( ) 15601526130w02nt − × ==−= tdpA

4050151352)5.22(2 011nv

mm2

−+= wtdepA = × × = mm2

( ) 1179100.140502753/1

1025.11560430

33RdEff,1, =×+×

×=V × V

× kN Eff,1,Rd =

1179 kN

66.0

1179783

RdEff,1,

Ed ==V

N < 1.0

Therefore, the resistance of the end plate to block tearing failure is adequate.

As both checks are passed, the tensile resistance of the end plate is adequate.

Similar design checks would be required to be carried out for the other bracing member where the loads and overall lengths reduce.

7) The gusset plates would also require checking for shear, bearing

and welds, together with full design check for the extended beam end plates.





1) Any notional horizontal forces must also be considered and applied to determine if they give more onerous conditions than the applied wind loading.

2) Horizontal bracing is required for the temporary condition during frame erection. Design check required for the horizontal bracing arrangement and connections.

3) It should be checked that these members can carry the loads imposed by the wind.

4) Alternative member arrangements can be used for the bracing system to those considered here. For example, the bracing system could consist of cross-braces so that all of the braces only act in tension and could therefore consist of flat bars, or similar, and be of much smaller sections.

5) The wind load considered is only for the direction shown on structural arrangement and loading sheet 2 GA. Other directions must also be considered.

6) See example 1 for more information on equation 6.10b)

7) The gusset plate must be checked for yielding across an effective dispersion width of the plate. When the bracing member is in compression, buckling of the gusset plate must be prevented and therefore a full design check must be carried out.

Example 11 Beam-to-column flexible end plate connection Sheet 1 of 11 Rev

Made by MS Date Nov-06 Checked by PA Date Dec-06

Reference Output

P:\EDN\EDN399\Student worked examples\Second draft-book\11_Flexible end plate_meb.doc 1


Beam-to-column flexible end plate connection

Design the beam-to-column connection at level 1 between gridlines ?? and ??.

Initial sizing of the components of the connection

Column 254 × 254 × 73 UC in S275 steel Beam 457 × 191 × 82 UB in S275 steel From example 1 the design shear force at ULS is:

VEd = 239 kN

Based on the above data the initial sizing of the connection components is shown below.

VEd = 239 kN

p3 100gv e2,c 7750

a a = 5 mm

hp

e 551p 851 p1e1

mp e2

280 8555

50

6 No. M20 8.8 bolts

(a) parameters definition (b) Profile adopted based on initial sizing

Connection parameters definition and sizes

End plate thickness 1)

Choose a plate thickness tp = 10 mm

End plate thick-ness tp = 10 mm

Choice of bolt type, number of bolts, and positioning of holes 2)

The bolts are fully threaded, non-preloaded, size M20 with a length of 60 mm and class 8.8 (class 8.8 bolts are the ones generally used in the UK).

Total number of bolts n = 6 Tensile stress area of bolt As = 245 mm2

Diameter of the shank d = 20 mm Diameter of the holes d0 = 22 mm Diameter of the washer dw = 37 mm Yield strength fyb = 640 N/mm2

Ultimate tensile strength fub = 800 N/mm2

Use 3 rows of two bolts each (i.e. n = 6 bolts)


Reference Output


Check limits for locations and spacings of bolts

End distance e1 = 55 mm 3.5, Table 3.3 Limits 1.2 404 p1o +≤≤ ted

26.4221.21.2 0 =×=d mm < 55 mm

( ) 8040104404 p =+×=+t mm > 55 mm

Therefore, adopt e1 = 55 mm

e1 = 55 mm

Edge distance e2 = 50 mm Limits are the same as those for end distance, therefore adopt this edge distance.

e2 = 50 mm

Spacing p1 = 85 mm ≤ ≤Limits 2.2 10 pd

min(14tp;200 mm) 4.48222.22.2 0 =×=d mm < 85 mm

140101414 p =×=t mm > 85 mm

Therefore, adopt p1 = 85 mm

p1 = 85 mm

Spacing p3 = 100 mm ≤ ≤Limits 2.4 30 pd

8.52224.22.4 0 =×

min(14tp;200 mm) =d mm < 100 mm

140101414 p =×=t mm > 100 mm

Therefore, adopt p3 = 100 mm

p3 = 100 mm

Joint details

Column 254 × 254 × 73 UC in S275 steel

Depth hc = 254 mm Width bc = 254 mm Thickness of web tw,c = 8.6 mm Thickness of flange tf,c = 14.2 mm Fillet radius rc = 12.7 mm Area Ac = 92.9 cm2

Second moment of area Iy, c = 11370 cm4

Depth between fillets dc = 200.2 mm

BS EN 1993-1-1

Table 3.1


Yield strength fy,c = 275 N/mm2 Ultimate strength fu,c = 430 N/mm2


Reference Output


Supported beam 457 × 191 × 82 UB in S275 steel

Depth hb = 460.2 mm Width bb = 191.3 mm Thickness of web tw,b = 9.9 mm Thickness of flange tf,b = 16.0 mm Filled radius rb = 10.2 mm Area Ab = 105 cm2

Second moment of area Iy,b = 37090 cm4

Depth between fillets db = 407.8 mm

BS EN 1993-1-1:2005

Table 3.1


Yield strength fy,b = 275 N/mm2 Ultimate strength fu,b = 430 N/mm2

End plate 280 × 200 × 10 in S275 steel 3)

Vertical gap gv = 50 mm Plate depth 4) hp = 280 mm Plate width bp = 200 mm Plate thickness tp = 10 mm

BS EN 1993-1-1:2005

Table 3.1


Yield strength fy,p = 275 N/mm2 Ultimate strength fu,p = 430 N/mm2

In the direction of load transfer Number of horizontal bolt rows n1 = 3 End distance e1 = 55 mm Pitch between bolt rows p1 = 85 mm In the perpendicular direction to load transfer Number of vertical lines of bolts n2 = 2 Edge distance e2 = 50 mm Column edge to bolt line e2,c = 77 mm Gauge (i.e. distance between cross centres) p3 = 100 mm

4.7.3 Welds

End plate is welded to the beam web through a pair of partial penetration butt welds. If the throat thickness of the weld is a, then:

2a ≥ tw,b, therefore a 9.429.9

2bw, ==

t mm

Rounding off to the nearest whole number, a = 5 mm

Weld throat thickness, a = 5 mm


Reference Output


BS EN1993-1-1:2005 6.1(1)

Table 2.1


γM0 = 1.0

γM2 = 1.25 (for shear)

BS EN 1993-1-1:2005 3.2.2 & Table 3.1.

Ductility requirements 5)

Steel grade used is S275 with fy = 275 N/mm2 and fu = 430 N/mm2 to EN 10025-2, therefore, according to BS EN 1993-1-1:2005 cl.3.2.2(2), the ductility requirement of the material is satisfied.

Ductility is adequate

Weld design 6)5.3.3

Beam is steel grade S275

Fw.Rd = fvw.d a

With fvw.d = M2w

bu, 3/

γβf

βw = 0.85 for S275 grade steel

Therefore,

Fw.Rd = 1168.31.250.85

3430/=×⎟⎟

⎠

⎞

⎝

⎛

×5

65410280 3 =××× −1168.32

⎜⎜ N/mm

Over 280 mm and for weld on two sides:

Fw.Rd = kN > VEd = 239 kN

Therefore, the weld is adequate.

Weld is adequate

Joint shear resistance

The following table gives the complete list of design checks that need to be performed for the joint shear resistance. Only the critical checks are shown in this example. The critical checks are denoted with an * in the table.

Mode of failure Bolts in shear* VRd,1

End plate in bearing* VRd,2

Supporting member (column) in bearing VRd,3

End plate in shear (gross section) VRd,4

End plate in shear (net section) VRd,5

End plate in block shear VRd,6

End plate in bending VRd,7

Beam web in shear* VRd,8


Reference Output


Bolts in shear 8)

Assuming the shear plane passes through the threaded portion of the bolt, the shear resistance Fv,Rd of a single bolt is given by:

M2

ubvRdv, γ

AfαF =

For bolt class 8.8, αv = 0.6, therefore

941.25

5x100.6x800x24 3

Rdv, ==−

F kN

For 6 bolts, VRd,1= 6 × 94 = 564 kN

For use of values in Table 3.4 of EN 1993-1-8, see cl.3.6.1(3)

VRd,1 = 564 kN

3.6.1 & Table 3.4

End plate in bearing

3.6.1 & Table 3.4

The bearing resistance of a single bolt, Fb,Rd is given by:

M2

ppu,b1Rdb, γ

α dtfkF =

Where,

⎟⎟⎠

⎞⎜⎜⎝

⎛−= ;1.0;

41

3;

3min

pu,

ub

o

1

o

1b f

fdp

deα

⎟⎠⎞

⎜⎝⎛ ==−

×=

×= 1.74;1.0

430800

1.04;41

22385

0.83;223

55minbα

83.0b =α

Therefore,

⎟⎠

⎞⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛×=⎟⎟

⎠

⎞⎜⎜⎝

⎛= 4.66;2.51.7-

2250

8.2min1.7;2.5-8.2mino

21 d

ek

kN

Therefore, k1 = 0.83

Therefore,

142.8101.25

10204300.832.5 3Rdb, =×

× ×××= −F

For 6 bolts, 8578.1426Rd,2 =×=V kN VRd,2 = 857 kN


Reference Output


Beam web in shear 8)

Shear resistance is checked only for the area of the beam web connected to the end plate.

The design plastic shear resistance is given by:

M0

by,vRd,8Rdpl,

)3/(

γfA

VV ==

3101.0

)3(275/9.9)(280 −×××

= = 440 kN

VRd,8 = 440 kN

BS EN 1993-1-1:2005 6.2.6(2)

Tying resistance of end plate 9)

The following table gives the complete list of design checks that need to be performed for the tying resistance of the end plate. Only critical checks are shown. The critical checks are denoted with an * in the table. The check for bolts in tension is also carried out because the tension capacity of the bolt group is also needed for the end plate in bending check.

Mode of failure

Bolts in tension NRd,u,1

End plate in bending* NRd,u,2

Supporting member in bending NRd,u,3

Beam web in tension NRd,u,4

Bolts in tension

3.6.1 & Table 3.4

The tension resistance for a single bolt is given by:

M2

sub2Rdt, γ

AfkF =

k2 = 0.9

160.4101.1

2458000.9 3Rdt,u,1Rd, =×

× ×== −FN

962160.46u,1Rd,

kN

For 6 bolts, = × =N kN

NRd,u,1 = 962 kN

End plate in bending 10)

6.2.4 & Table 6.2

NRd,u,2 = min(FRd,u,ep1; FRd,u,ep2)

For mode 1: )(n2

)2(8

ppwpp

Rdpl,1,wpRdT,1,u,ep1Rd, nmem

MenFF

+−

−==

For mode 2: pp

Rdt,pRdpl,2,RdT,2,u,ep1Rd,

2

nm

FnMFF

+

+== ∑


Reference Output


Where,

np = min(e2; e2,c; 1.25mp)

mp =2

)20.82( bw,3 atp ×−−

ew =4wd

= 25.9437

= mm

(dw is the diameter of washer or width across points of bolt head or nut)

P3

a0.8a 2

mp

tw,b

Here,

( )[ ]4.39

2250.829.9100

p =×××−−

=m

3.494.3925.125.1 p

mm

=×=m mm

np = min(50; 77; 49.3) = 49.3 mm

M0

py,2peff,1

Rdpl,1, 41

γftL

M ∑=

M0

py,2peff,2

Rdpl,2, 41

γftL

M ∑=

∑ 1,effL Calculate the effective length of the end plate for mode 1

( ) and mode 2 (∑ 2,effL ).

6.2.5.6 & Table 6.6

For bolt row considered as part of a group of bolt rows: If non circular patterns are assumed

for inner bolt row: Leff = p1 =85 mm

for end bolt rows: Leff = 2(2mp+0.625e2+0.5p1)

( ) ( ) ( )[ ] 305585.050625.04.3922eff =+×+××=L mm

∑ eff,1L = = 85+305 = 390 mm > h∑ eff,2L p = 280 mm

Therefore, = ∑ eff,1L ∑ eff,2L = hp = 280 mm


Reference Output


M0

py,2pp

uRd,pl,1, 41

γfth

M =

1.93101.0

2751028041 6

2

uRd,pl,1, =×⎟⎟⎠

⎞⎜⎜⎝

⎛ ×××= −M kNm

Mode 1:

( )( ) 723

49.3)(39.49.2549.3)39.4(2101.939.25)2(49.3)(8 3

RdT,1, =+×−×××××−×

=F kN

Mode 2:

In this case uRd,pl,1,uRd,pl,2, MM =

( ) ( )578

49.339.496249.3101.932 3

RdT,1, =+

×+××=F kN

237578)min(237;RdT,1, ==F kN

Therefore, kN 237N u,2Rd, = NRd,u,2 = 237 kN

Summary of the results

The following tables give the complete list of design checks that need to be performed for the tying resistance of the end plate. Only critical checks are shown in this example The critical checks are denoted with an * in the tables.

Joint shear resistance

Mode of failure Resistance

Bolts in shear* VRd,1 564 kN End plate in bearing* VRd,2 857 kN Supporting member (column) in bearing VRd,3 End plate in shear (gross section) VRd,4 End plate in shear (net section) VRd,5 End plate in block shear VRd,6 End plate in bending VRd,7 Beam web in shear* VRd,8 440 kN

The governing value (i.e. the minimum value) is shown in bold type, and VRd = 440 kN

Tying resistance of end plate

Mode of failure Resistance

Bolts in tension NRd,u,,1 962 kN End plate in bending* NRd,u,2 237 kN Supporting member in bending NRd,u,3 N/A Beam web in tension NRd,u,4

The governing value (i.e. the minimum value) is shown in bold type, and NRd,u = 237 kN


Reference Output


Since the ultimate design shear force is VEd = 239 kN, the connection is safe as far as the shear resistance is concerned (VEd < VRd) but is not safe for tying resistance (VEd > NRd,u), as shown from the summary of all results in the tables above.

To make the connection safe for tying resistance, one solution would be to increase the end plate thickness to, say 12 mm, and recheck.

Example XX Sheet 10 of 11 RevBeam-to-column flexible end plate connection




1) The thickness of the end plate should be thinner than that of the column flange to which it is bolted, otherwise the column flange must be checked for bending. Since the column flange thickness is tf,c = 14.2 mm, and in order to prevent column flange bending, a thickness of the end plate tp = 10 mm is chosen.

2) The minimum number of bolts, n, may be determined using document SN013a-EN-EU using the equation: n = VED/75. In this example VED = 239 kN and number of bolts will be n = 239/75 = 3.2. Rounded off to the nearest whole number, the minimum number of bolts is n = 4. (n = 6 has been used in the example).

3) The end plate should be welded near the top flange of the beam and be of sufficient depth to provide torsional restraint.

4) There is no specific guidance in EC3 on the choice of length of the end plate. Guidance from document SN013a (Table 5.1) may be used to determine the depth of the end plate as follows:

⎟⎟⎠

⎞⎜⎜⎝

⎛≥ p

bw,by,

Edp ;0.6

t1900

max hf

Vh and mm07.84bp ≤ =dh

where db is the depth between fillets of the beam, in this case db = 407.8 mm

⎟⎠⎞

⎜⎝⎛ ×

××

≥ 460.2;0.69.92752391900

maxph = 276 mm

Therefore, hp may be taken as 280 mm (to the nearest 10 mm).

5) For the UK, and based on SN014, the ductility requirement is satisfied if the supporting element (column flange in this case), complies with the following conditions:

py,

ubp 8.2 f

fdt ≤

y,c

ubf,c 8.2 f

fdt ≤ or

12275800

8.220

8.2 py,

ub =×⎟⎠⎞

⎜⎝⎛=

ffd

b1w,39.0

mm

Since tp = 10 mm < 12 mm, ductility is ensured.

6) Document SN014a-EN-EU gives the following condition for the design of the weld: ta ≥ for S275 supported beam.

7) In document SN014a-EN-EU, the shear resistance of the bolt is multiplied by a reduction factor equal to 0.8 to allow for the presence of tension in the bolts.

According to EC3, in addition to the shear resistance check, the bolts should be checked for tension resistance as well as combined effects of tension and shear (see Table 3.4 of BS EN 1993-1-8:2005).

8) In the old version of Eurocode 3, BS ENV 1993-1-1:1993, cl.5.4.6 there is a reference to the design plastic shear resistance for a plate. This has been omitted from the new version, BS EN 1993-1-1:2005, cl.6.2.6

In document SN014a-EN-EU, the area of the beam connected to the end plate is multiplied by a factor of 0.9. This is because the plastic shear resistance for a plate is not covered in EC3

Example XX Sheet 11 of 11 RevBeam-to-column flexible end plate connection


9) The tying force to be resisted should be determined following the guidance in BS EN 1991-1-7 or any other National Regulations i.e. Building Regulations.

10) Document SN10a-EN-EU uses the actual depth of the end plate for the calculation of moments Mpl,1,Rd and Mpl,2,Rd instead of the effective depth of the end plate as is required by BS EN 1993-1-8: 2005, cl.6.2.4, Table 6.2.

Example 12 Column base connection Sheet 1 of 5 Rev

Made by MC Date Apr 2007 Checked by KS Date May 2007

Reference Output

P:\EDN\EDN399\Student worked examples\Second draft-book\12 Column base connection_meb.doc 1


Column base connection

The interaction between the column, the base plate and the supporting concrete foundation is complex but the simplified approach used for sometime is applied here.

Design conditions for column G2

From previous calculations the following design conditions should be considered in design the column base plate.

• The column is assumed to be pin-ended. However it is crucial the column is stable during the erection phase therefore 4 bolts should be used as foundation bolts.

Characteristic Permanent action Gk = 466kN

Characteristic Variable action Qk = 479kN


BS EN 1991.1.1 Tables A1.1 and A1.2

Variable ξγGj = 0.85×1.35 = 1.1475

Permanent γQ1 = 1.5

Design axial force

( ) ( ) 12534795.1466147.1Ed × = +×=N

From previous design section classified as Class 1.

kN

Axial force NEd = 1253 kN

Column details

Column G2 is a typical internal column that has previously designed

Details from CORUS “AdvanceTM” section properties

Serial size UKC254x254x73kg/m (S275)

Height of section h = 254.1mm Breadth of section b = 254.6mm Thickness of flange tf = 14.2mm Thickness of web tw = 8.6mm Cross sectional Area A = 93.1cm2


Reference Output


Table 3.1 As t < 40 mm and steel grade S275

Design yield strength fy = 275 N/mm2

artial factors for resistance

.1(1)

P

6 =M2γ 1.25

ase plate details

S EN

f the column base plate are as follows:

ted on

2. oncrete to be 37/30 2

able 3.1

3. from UKC assume 50mm.

than 40mm 2

5. aken as follows

BS EN

.7

B

B

Design assumptions o

1. The axial load in the column will be uniformly distributo the concrete base.

Strength of foundation c

1992-1-1Table 3.1

(i.e. fck =30 N/mm )

Outstand of plate edge

T 4. Plate material as per column S275 and assume lessthick, therefore fy=275 N/mm

Maximum pressure on concrete t

1992-1-1Clause 3.1

c

ckccα ff cd γ

=

αcc to be taken between 0.8 & 1.0. Assume the lesser, therefore

8.0cc =α

ax allowable

Mpressure on concrete

165.1308.0

cd ==f N/mm2 /mm2

Check minimum thic plate (tp)

× fcd = 16 N

kness of base

21

27550

y

cdp =⎟

⎠⎜⎝

×=⎟⎟⎠

⎜⎜⎝

=f

ct mm

Adopt 25mm thick base plate in S275 material = 25 mm

tion

16335.05.0

⎞⎛ ×⎞⎛ f

tp

Check bearing pressure on concrete founda

Acceptable bearing pressure, allow fcd = 16 N/mm2

Assume ‘c’ to be 50 mm

h+2c

b+2c

h-2c b/2


Reference Output


Effective area:

( ) ( )[ ] ( )[ ]chbcbchA 222eff −×−+×+=

( ) ( )[ ]

( )[ ]5021.2546.2545026.2545021.254 ×−×−×+××+=

86330= mm2 = 863.3 cm2

Effective area Aeff = 563.3 cm2

Therefore, the bearing pressure on the concrete foundation is:

5.1486330

101253 3

ff

Ed =×

==eA

N N/mm2 < fcd = 16 N/mm2

on concrete is acceptable.

Therefore, adopt 25th 310×310 S275 base

Bearing pressure on concrete is

ceptable

Connection of base plate to column

Connection of the base plate is to be a welded connection and it is assumed s (FW). The loading on the column

BS EN 1993-1-8

Clause 4.5.3.2

Clause 4.5.3.3

To determine the minimum length of 8mm fillet weld two options are available:

• Directional Method, and

this example.

Bearing pressure

ac

to be 8mm fillet weld

should transfer through the direct contact with the plate.

• Simplified Method, which we will use in

Shear strength of weld, M2w

3/f

uvw,d γβf =

Table 3.1

As both the column and the plate are S275 steel and the thicknesses are less then 40 mm

fu = 430 N/mm2

β

BS EN 1993-1-8 Table 4.1

w = 0.85 Shear strength of weld

234vw,d ==f N/mm2

fvw,d =234 N/mm

25.185.0 ×3/430

S EN 993-1-8 4.5.2

Shear on the weld is taken through the throat thickness (a).

nce value for an 8 mm fillet weld per mm is:

,d

B1

BS EN 1993-1-84.5.3.3

Throat thickness is normally taken as 0.7 × the leg length a = 5.6 mm

Design weld resistafF vwRdvw, = a

Minimum effective length =

95631.1

1253= mm


Reference Output


Length of section profile is: ( ) ( ) 4.10176.2541.25422 =+×=+ bh mm > 956 mm

However for practical reasons of ensuring the secure fixing of the base plate and the reduced risk of corrosion a full profile weld will be used.

(Design capacity = 1332 kN)

Adopt an 8mm fillet weld (full rofile)

p





1)

Example 13 Frame Stability Sheet 1 of 4 Rev

Made by KP Date Checked by BD Date

Reference Output

P:\EDN\EDN399\Student worked examples\Second draft-book\13_Frame Stability Rev B_jbw.doc 1

Frame Stability

Introduction

Unless stated otherwise all references are to BS EN 1993-1-1:2005 This example tests the building for susceptibility to sway instability.

Beam-and-column type plane frames in buildings may be checked for sway mode failure with first order analysis using the approximate formula

5.2.1(4) 10EdH,Ed

Edcr ≥

∂=

h

V

Hα

where the definition of each parameter is given later in this example and the load and reactions in the system are caused by a combination of horizontal wind loading and notional horizontal force applied at each floor level.

The diagram shows the structural layout of the braced bays which are present in each end gable of the building. Similar internal unbraced bays occur at 6m spacing along the 48m length of the building (ie 8 bays in total). The braced bay therefore attracts one half of the total wind loading on the windward face of the building which is assumed to be transferred to the bracing via a wind girder in the roof and diaphragm action in the floor slabs at each floor level. The notional horizontal load is calculated as 0.5% of the total factored dead and imposed load acting on each frame at roof and floor level. This force is also distributed to the end bracing via wind girder and floor diaphragm action such that each braced bay receives the equivalent of one half of the notional horizontal forces calculated for the whole building.

Factored Wind Loading

Total wind load on windward face of building = 1.5 × 925 =1388 kN

Total wind load resisted by braced bay = 0.5 × 1388 = 694 kN

Distribution : At roof level = 694 / 8 = 86.8 kN At floor levels = 694 / 4 = 173.5 kN

Wind loading on braced bay

Factored Vertical Loading

Roof loading on one frame = 14 × 6 [1.35 × 0.9 + 1.5 × 0.4] = 152.5 kN

Total roof loading = 8 × 152.5 = 1220 kN

Notional horizontal loading at roof level in end bracing = 0.5 × 0.5% × 1220 = 3.1 kN

Notional loading at roof level


Reference Output


Floor loading on one frame = 14 × 6 [1.35 × 3.7 + 1.5 × 3.8] = 898.4 kN

Total floor loading = 8 × 898.4 = 7187 kN

Notional horizontal loading at each floor level in end bracing = 0.5 × 0.5% × 7187 = 18.0 kN

Notional loading at each floor level

Braced Bay Layout

14 m

8 m

5.0 m

4.5 m

4.5 m

4.5 m

6 m

wind load

notional horizontal forces

86.8 + 3.1 = 89.9 kN

173.5 + 18.0 = 191.5 kN

173.5 + 18.0 = 191.5 kN

173.5 + 18.0 = 191.5 kN

Section at Braced Bay

Section at braced bay

Assumed Section Sizes

Columns 203 UC 53 Beams 254 UB 31 Bracing 193 CHS 10

Sway Analysis

The sway analysis is carried out for horizontal loading only as shown

6.0 m

5.0 m

4.5 m

4.5 m

89.9 kN

4.5 m

191.5 kN

191.5 kN

191.5 kN

42.3 mm

Storey sway δ H,Ed = 7.8 mm

34.5 mm


22.4 mm

12.5 mm

0 mm

Deflections from computer analysis (QSE)



Sway in each storey height



Reference Output


Assumptions

Column bases both pinned Columns continuous over full height Bracing and beams effectively pinned to columns

Critical Check for Stability

Check as follows :-

10EdH,Ed

Edcr ≥

∂=

h

V

Hα for non-sway case

Where HEd = horizontal reaction at bottom of storey

VEd = total vertical load at bottom of storey

h = storey height

δH,Ed = storey sway

Storey Checks

Fourth Storey :

HEd = 89.9 kN

VEd = 152.5 kN

10340

8.7

4500

5.152

9.89cr ≥==α Not sway critical

Third Storey :

HEd = 89.9 + 191.5 = 281.4 kN

VEd = 152.5 + 898.4 = 1050.9 kN

10100

1.12

4500

9.1050


Second Storey :

HEd = 281.4 + 191.5 = 472.9 kN

VEd = 1050.9 + 898.4 = 1949.3 kN

10110

9.9

4500

3.1949


First Storey :

HEd = 472.9 + 191.5 = 664.4 kN

VEd = 1949.3 + 898.4 = 2847.7 kN

1093

5.12

5000

7.2847


Frame is classified as non-sway and P- δ can be ignored

FRAME NOT CRITICAL IN SWAY MODE





1)

sci student worked examples 2nd draft

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