1 Introduction 2015

CE3166:  R Liew

CE3166 Part I Structural Steel Design and System J Y Richard Liew Professor PhD, FSEng, PE, CEng, ACPE, StEr

National University of Singapore Department of Civil & Environmental Engineering E-MAIL: [email protected] TEL: 65162154 Room: E1A-05-13 1

CE3166:  R Liew

Introduction to Eurocodes for Steel Design

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Objectives

This module introduces the limit-state design of structural steel frame and components based on Eurocode 3, EN 1993-1-1. At the end of the module, you should be able to DESIGN a structure like this: beams trusses connections

columns Frames

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10/8/2016

Intended Learning Outcomes

CE3166:  R Liew

Upon successful completion of the course, students should be able to (a) understand response of various structural steel systems under typical loads. (b) identify and appreciate the behaviour and purpose of different elements/components in a steel structure. (c) perform safe, economical and efficient designs of structural steel systems and various structural elements/components to suit their intended functions and according to current design codes. (d) be aware of typical constraints in engineering design and come up with different solutions/options to achieve the desired outcomes. (e) communicate relevant thoughts and ideas effectively to others in verbal and written forms. (f) understand the impact of engineering solutions in a societal context and to be able to respond effectively to the needs for sustainable development. (g) understand professional, ethical and moral responsibility as engineers. (h) engage in continuous and life-long learning and seek relevant information to solve engineering problems beyond materials covered in this module. 4

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Course Outline (1) First part of CE3166 is covered by Prof. Richard Liew

• • • • • • •

Introduction to limit state design Local buckling & Section classification Compression members Tension members Restrained Beams Laterally unrestrained beams Quiz 1.

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Course Outline (2) Second part of CE3166 is covered by Prof. Pang Sze Dai • beam-columns • Multi-Storey Frames • Plate Girders • Connections – welded and bolted

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References 1. 2. 3. 4. 5. 6. 7. 8. 9.

BS EN 1993-1-1: 2005 Eurocode 3: Design of steel structures – Part 1-1: General rules and rules for buildings. (IVLE) BS EN 1993-1-5:2006 Eurocode 3: Design of steel structures – Part 1-5: Plated structural elements. (IVLE) BS EN 1993-1-8:2005 Eurocode 3: Design of steel structures – Part 1-8: Design of joints. (IVLE) BS EN 1994-1-1:2004 Eurocode 4: Design of composite steel and concrete structures – Part 1-1: General rules and rules for buildings. (IVLE) BS EN 1991-1-1: 2002 Eurocode 1: Actions on structures – Part 1-1: General actions – Densities, self-weight, imposed loads for buildings. (IVLE) BS EN 1990:2002 Eurocode – Basis of structural design. (IVLE) Buick Davison & Graham Owens (Editors), Steel designers’ manual, WileyBlackwell, 2012. Gardner, L. and Nethercot, D. A. (2005). Designers’ guide to Eurocode 3: Design of steel structures. Thomas Telford Limited. TA684 Gar 2005 (RBR). Trahair, N.S., Bradford, M. A., Nethercot, D. A. and Gardner, L. (2008). The behavior and design of steel structures to EC3. Fourth Edition. Taylor & Francis. TA 684 Tra 2008 7

CE3166:  R Liew

Final Examination • Items allowed in the exam

mandatory

– – – – – –

Lecture notes EN 1993-1-1 EN 1993-1-5 EN 1993-1-8 Section tables Tutorials and Homeworks

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Overview on Structural Eurocodes 10 Structural Eurocodes  EN1990: Basis of structural design  EN1991: Actions on structures  EN1992: Design of concrete structures  EN1993: Design of steel structures  EN1994: Design of composite steel and concrete structures  EN1995: Design of timber structures  EN1996: Design of masonry structures  EN1997: Geotechnical design  EN1998: Design of structures for earthquake resistance  EN1999: Design of aluminium structures

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Standards relating to EC3 and EC4

(4Parts)

(3 Parts)

(21 Parts)

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Overview on Eurocode 3 (EN1993) Part 1 (General)  Part 1-1: General rules and rules for buildings  Part 1-2: Structural fire design  Part 1-3: Supplementary rules for cold formed members and sheeting  Part 1-4: Supplementary rules for stainless steel  Part 1-5: Plated structural elements  Part 1-6: Strength and stability of shell structures  Part 1-7: Plated structures subject to out of plane loading  Part 1-8: Design of joints  Part 1-9: Fatigue  Part 1-10: Material toughness and through thickness properties  Part 1-11: Design of structures with tension components  Part 1-12: Additional rules for the extension of EN1993 up to steel grades S700 11

CE3166:  R Liew

Advantages of Structural Steel • High strength: reduced weight • Uniformity: isotropic properties do not change appreciably with time • Elasticity : Hook’s law applies up to fairly high stresses • Ductility : offer additional safety • Toughness: ability to absorb large amount of energy; ease of fabrication and erection • Speed of Construction: no formwork, speed of erection, low self-weight, modifications (strengthening, extensions), good dimensional control.

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CE3166:  R Liew

Conventions Property

Symbol

Subscript Definition

area

A

k

characteristic

section modulus

W

d

design

radius of gyration

i

E

effect

second moment of area

I

Rd

design resistance

el

elastic

pl

plastic

Loads

Symbol

Permanent action

G

Variable action

Q

Accidental action

A

z y

Member axes y

z

z–z y–y x–x

Minor axis Major axis Longitudinal axis

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Material Properties & Notation (EC3) EN 1993-1-1 Clause 3.2.6

Modulus of elasticity:

E  210 GPa

Poisson’s ratio:

  0.3

Coefficient of thermal expansion:

Shear Modulus

G = 81 GPa

  12  10 6 / o C b

z r

tw y

y

h

d

x-x axis: along member axis z

tf 14

CE3166:  R Liew

Mechanical Properties of Steel Stress f

E  210 GPa

fu

fy

1

• fu = ultimate tensile strength • fy = yield strength • E = Young’s modulus • u = ultimate strain • y = yield strain • Elongation measured in percentage

Est

E 1

Elastic Plastic

y

Strain hardening

 sh

Necking and failure

u

Strain 

Elongation at failure, f 15

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Nominal values of yield strength fy & ultimate strength fu for non-alloy structural steel EN 10025-2:2004

Nominal Thickness [mm]

S235

S275

S355

S450

fy [MPa]

fu [MPa]

fy [MPa]

fu [MPa]

fy [MPa]

fu [MPa]

fy [MPa]

fu [MPa]

t ≤ 16

235

360

275

410

355

470

450

550

16 < t ≤ 40

225

360

265

410

345

470

430

550

40 < t ≤ 63

215

360

255

410

335

470

410

550

63 < t < 80

215

360

245

410

325

470

390

550

80 < t < 100

215

360

235

410

315

470

380

550

100 < t < 150

195

350

225

400

295

450

380

530

150 < t < 200

185

340

215

380

285

450

-

-

200 < t < 250

175

340

205

380

275

450

-

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An Example (*) S275 UB 457×191×98

19.6 mm

11.4 mm

fy = 275 MPa for web fy = 265 MPa for flange

S275 Thickness range (mm)*

fy (MPa)

16

275

40

265

63

255

80

245

100

235

What is the strength of the entire section?

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Design Philosophies • Conceptual development – Recognize the main structural system – Trace the “load paths” through elements to the foundation Gravity load: self weight + Imposed load

wind

Deflected shape

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Industrial Buildings Portal frames

Truss types

Warren truss

Fink truss

Pratt truss

Vierendeel truss (fixed joint, large opening)

Bowstring truss 19

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Bridges Cable stayed bridge

Sutong Bridge (China) Longest cable-stayed bridge (longest span = 1108m, total length = 8206 m)

Suspension bridge Main span L ~ L/11 Truss bridge

Golden gate bridge,1937 (longest span = 1280 m, total length = 2737 m)

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Multi-Storey Buildings Rigid

Moment frame

Moment connection

Pinned

Braced frame

Shear wall frame

Pin connection 21

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Structural Members • Tension member – subjected to tensile axial loads only • Column (or compression) member – subjected to compressive axial loads only • Beam member – subjected to flexural loads, i.e., shear force and bending moment only. The axial force in a beam member is negligible • Beam-column member – subjected to combined axial force and flexural loads. • Connections • Plate girders 22

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Common Steel Sections Hot-rolled sections

Build up by plates

UB

UC

RHS

CHS

A Connections A

Section A-A 23

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Ultimate Limit state design Design Resistance

m Characteristic resistance ÷ Partial factor for resistance (Decrease characteristic resistance)

Design Effect x  G 

Characteristic action × Partial factor for action (Increase characteristic action)

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Design Approach of Eurocodes The approach in Eurocode is based on Limit State Design and the following are the three main types of limit states: Ultimate Limit States states associated with collapse or with other similar forms of structural failure  yielding  buckling  overturning Serviceability Limit States states that correspond to conditions beyond which specified service requirements for a structure or structural member are no longer met.  excessive deflection  excessive vibration

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CE3166:  R Liew

Combination of actions Fundamental combination of actions can be determined from Eqs. 6.10, 6.10a or 6.10b of EN1990.

 j 1

G, j

Gk , j   Q ,1Qk ,1    Q ,i 0,i Qk ,i

Permanent actions

(6.10)

i 1

Leading variable action

Accompanying variable actions

ψ : combinations factors Details for γ and ψ in EN 1990: 2002.

Examples Dead Load or Self weight = permanent action. Wind load, imposed load = variable action

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CE3166:  R Liew

Actions 3 types: permanent (G), variable (Q) and accidental (A). Partial factors for actions γ :

Actions

ULS

Unfavourable conditions: Permanent action (G) / Dead load (DL) Variable action (Q) / Imposed load (IL)

1.35 1.5

Favourable conditions: Permanent action (G) / Dead load (DL) Variable action (Q) / Imposed load (IL)

1.0 0

Favourable: action results in lower load resultant/effect. Unfavourable: action results in higher load resultant/effect. Load resultant/effect: bending moment, shear, tension, compression, overturning, etc.

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Load Combinations for ULS EN 1990 Clause 6.4.3.2

The fundamental combination for ULS is given as follows:

 j 1

G, j

Gk , j   Q ,1Qk ,1    Q ,i 0,i Qk ,i

(6.10)

i 1

NON‐LEADING variable action

Example for  Unfavourable Condition

 1.35G j 1

k, j

LEADING variable action 

 1.5Qk ,1   1.5 0,i Qk ,i

Examples

i 1

Values of 0,i are found in NA to SS EN 1990: 2008+A1: 2010): Action Imposed loads Category A: domestic, residential areas Category B: office areas Category C: congregation areas Category D: shopping areas Category E: storage areas Category H: roofs Wind loads on buildings *

ψ0 0.7 0.7 0.7 0.7 1.0 0.7

For permanent + imposed action,

1.35Gk  1.5Qk For permanent + imposed action + other variable action,

1.35Gk  1.5Qk  1.5  0Qk , i

0.5 28

CE3166:  R Liew

Typical ULS combinations: For beam and floor slab design 1.35DL + 1.5IL (unfavourable DL and IL) For frame design 1.35DL + 1.5IL + 0.75WL+EHF (unfavourable DL, IL and WL; IL dominant) 1.35DL + 1.5WL + 1.05IL + EHF (unfavourable DL, IL and WL; WL dominant) EHF = Equivalent horizontal forces (to be covered in Lecture 9: frames)

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Serviceability Limit States (SLS) Deflection: • should not affect the appearance of the structure • should not cause discomfort to the users • should not affect the function of the structure (including functioning of machines or services) • should not cause damage to finishes or non-structural members Vibration and oscillation • should not cause discomfort to people • should not limit the functional effectiveness of the structure Other damages: • should not adversely affect appearance • should not adversely affect durability • should not adversely affect the functioning of the structure 30

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Load Combinations for SLS EN 1990 Clause 6.5.3

Leading variable action

Qk ,1 



0 , i Qk , i

i 1

If the leading variable action is the imposed load, Qk, we can simplfy it as follows:

Qk  0.5Wk

0 = 0.5 based on SS NA

If the leading variable action is the wind load, Wk, we can simplify it as follows:

Wk  0.7Qk SS National Annex ignores the permanent action in evaluating serviceability Note: Equivalent Horizontal Force (EHF) needs to be considered for SLS design 31

CE3166:  R Liew

Load Combinations for SLS Partial Factor for Serviceability Limit State  (SLS) Action

Partial Factor

Permanent Action (G)

G=0.00

Variable Action (Q)

Q=1.00

The combination of variable actions for serviceability limit states is given as follows:

Qk ,1 

 i 1

Beam Design Frame design

Action

ψ0

Imposed loads Category A: domestic, residential areas Category B: office areas Category C: congregation areas Category D: shopping areas Category E: storage areas Category H: roofs

0.7 0.7 0.7 0.7 1.0 0.7

Wind loads on buildings

0.5

0 , i Qk , i

Typical SLS combinations:

1.0IL (Imposed load is the only variable) 1.0IL + 0.5WL (Imposed load is the leading variable 1.0WL+0.7IL (wind load is leading variable) 32

Deflection Check

CE3166:  R Liew

(a) Vertical deflection due to imposed load Cantilevers

Length/180

Internal beams carrying plaster or other brittle finish

Span/360 or 40mm

Other beams (except purlins and sheeting rails)

Span/200 or 40mm

Edge beam

Span/300 to span/500 or 20mm

(b) Horizontal deflection of columns due to imposed load and wind load Tops of columns in single-storey buildings, except portal frames In each storey of a building with more than one storey

Height/300 Height of that storey/300

(c) Crane girders Vertical deflection due to static vertical wheel loads from overhead traveling cranes

Span/600

Horizontal deflection (calculated on the top flange properties alone) due to horizontal crane loads

Span/500 33

CE3166:  R Liew

Classification of Actions Actions (loads) shall be classified by their variation in time as follows: - Permanent actions (G), e.g., self-weight of structures, fixed equipment and road surfacing, prestressing force, indirect actions (e.g., settlement of supports). - Variable actions (Q), e.g., imposed loads on building floors, beams and roofs, wind action and snow actions, indirect actions (e.g., temperature effects).

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Permanent Actions, Gk, (EN 1991-1-1) Materials

Also refer to as dead load Density  (kN/m3)

Light weight concrete

9.0 to 20.0

Normal weight concrete

24.0 to 25.0

Cement mortar

19.0 to 23.0

Gypsum mortar

12.0 to 18.0

Wood

3.5 to 10.8

Plywood

4.5 to 7.0

Particle boards

7.0 to 12.0

Fibre building board

4.0 to 10.0

Steel

77.0 to 78.5

Glass

22.0 to 25.0

Acrylic sheet

12.0

Hot rolled asphalt

23.0

(refer to BS EN 1991- 1 - 1 : 2002 Annex A for full details) 35

CE3166:  R Liew

Imposed Loads (EN 1991-1-1) Uniformly distributed load

Point load

qk (kN/m2)

Qk (kN)

Residential - Floors - Stairs - Balconies

1.5 to 2.0 2.0 to 4.0 2.5 to 4.0

2.0 to 3.0 2.0 to 4.0 2.0 to 3.0

Office

2.0 to 3.0

1.5 to 4.5

Cafe, restaurant

2.0 to 3.0

2.0 to 4.0

Theatres

3.0 to 4.0

2.5 to 7.0 (4.0)

Shopping mall

4.0 to 5.0

3.5 to 7.0

Usage

Recommended values are underlined!

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Some values of imposed loads on floors, balconies and stairs in buildings qk (kN/m2)

Qk (kN)

Bedrooms and dormitories except those in hotels and motels

1.5

2.0

Bedrooms in hotels and motels; hospital wards; toilet areas

2.0

2.0

Office area (At or below ground floor level)

3.0

2.7

Office area (Above ground floor level)

2.5

2.7

Public, institutional and communal dining rooms and lounges, cafes and restaurants

2.0

3.0

Reading rooms with no book storage

2.5

4.0

Classrooms

3.0

3.0

Assembly areas with fixed seating

4.0

3.6

Places of worship

3.0

2.7

Corridors, hallways, aisles in institutional type buildings (not subjected to crowding)

3.0

4.5

Stairs, landings in institutional type buildings not subjected to crowding

3.0

4.0

Corridors, hallways, aisles in all buildings (subjected to crowding)

4.0

4.5

Stairs, landings in all buildings (subjected to crowding)

4.0

4.0

Walkways – Light duty

3.0

2.0

Walkways – General duty

5.0

3.6

Walkways – Heavy duty

7.5

4.5

Museum floors and art galleries for exhibition purposes

4.0

4.5

Dance halls and studios, gymnasia, stages

5.0

3.6

Assembly areas without fixed seating, concert halls, bars and places of worship

5.0

3.6

Balconies in hotels and motels

7.5

4.5

Areas in general retail shop, department stores

4.0

3.6

Specific Use

(refer to NA to SS for full details)

CE3166:  R Liew

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CE3166:  R Liew

Some values of imposed loads due to storage qk (kN/m2)

Qk (kN)

General areas for static equipment not specified elsewhere (institutional and public buildings)

2.0

1.8

Reading rooms with book storage, e.g. libraries

4.0

4.5

General storage other than those specified

2.4 per metre of storage height

7.0

File rooms, filing and storage space (offices)

5.0

4.5

Stack rooms (books)

2.4 per metre of storage height but with a minimum of 6.5

7.0

Paper storage for printing plants and stationery stores

4.0 per metre of storage height

9.0

Dense mobile stacking (books) on mobile trolleys, in public and institutional buildings

4.8 per metre of storage height but with a minimum of 9.6

7.0

Dense mobile stacking (books) on mobile trucks, in warehouse

4.8 per metre of storage height but with a minimum of 15.0

7.0

Cold storage

5.0 per metre of storage height but with a minimum of 15.0

9.0

Specific Use

(refer to NA to SS for full details) 38

CE3166:  R Liew

Example 1 A beam of span 9 m is simply supported at its ends. It is loaded by two concentrated loads at its third-points. Calculate the moment and shear forces required for beam design. The dead and imposed loads are given as follows: DL

Distributed load Concentrated load

3 kN/m 40 kN

IL

Concentrated load

60 kN

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CE3166:  R Liew

Design loads 54 kN + 90 kN

54 kN + 90 kN 4.05 kN/m

162 kN

3m

3m

3m

Design loads: DL Distributed load Concentrated load

3 × 1.35 = 4.05 kN/m 40 × 1.35 = 54 kN

IL

60 × 1.5 = 90 kN

Concentrated load

162 kN

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Load resultants/effects 54 kN + 90 kN

54 kN + 90 kN

4.05 kN/m

162 kN

3m

3m

3m

162 kN

Maximum bending moment occurs at mid-span: MEd = 162×4.5 – 4.05×4.5×4.5/2 – (54+90)×1.5 = 472 kNm. Maximum shear force occurs at the supports: VEd = 162 kN.

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Example 2: Design of primary beam with one-way spanning slabs (office building) Determine the design loads on Beam 1 supporting one-way spanning slabs. The uniformly distributed dead and imposed load are 5kN/m2 and 3kN/m2 respectively. 7m

Design permanent loads 1.35Gk = 1.35*5*4 = 27kN/m Beam 1

Design imposed loads 1.5Qk = 1.5*3*4 = 18kN/m

4m

4m

Design loads 1.35Gk + 1.5Qk = 45kN/m 45kN/m MEd = (45x72 )/8 kNm Beam 1

158kN

7m

158kN

VEd = 158kN

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CE3166:  R Liew

One-Way Spanning versus Two-Way Spanning Slab L D

C w

kN/m2

wL/2 kN/m A

0 C

C

D Beam CD

Beam AC & Beam BD B A One-way spanning slab L C

H

D w kN/m2

wL/2 kN/m A

C Beam AC

A B Two-way spanning slab

wH/2 C

D Beam CD

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Example 3

One-way spanning

One-way spanning F

One-way spanning B

A 4m

Balcony Extension

5.0 m

D

One-way spanning E

2.0 m

The architect decided to extend beams AD, BE and CF to support a balcony of 2 m width under the same distributed loading (a dead load of 4 kN/m2 and an imposed of 5 kN/m2) and an imposed load of 160 kN at point H. The architect requires no columns to be located below G, H and I. Evaluate the design load on beam BEH. I H G

C 4m

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Example 3 Design load for BE: Permanent: 1.35 × 4 × 4= 21.6 kN/m; Imposed: 1.5 × 5 × 4= 30 kN/m Point load? ? kN

21.6 + 30 = 51.6 kN/m

? kN

E E ? kN

H 5.0 m

B

2.0 m

H

B 4m

4m 45

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Example 3

4m H

• Due to one-way spanning, there is no distributed load on HE • Point load at H = the load intensity × area of upper rectangle • Point load at E = the load intensity × area of lower rectangle

1m 1m E

Design load for EH: Permanent at H or E: 1.35 × 4 × 4 × 1 = 21.6 kN; Imposed at H: 1.5 × 5 × 4 × 1 + 1.5 × 160 = 270 kN; Imposed at E: 1.5 × 5 × 4 × 1 = 30 kN

B 12.4 kN

E 588.8 kN (max)

2.0 m

E 291.6 kN

5.0 m

51.6 kN 21.6 + 30 = 51.6 kN/m

H

H B 4m

4m

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Example 3 Minimize the contribution of loads on span BE, which are counteracting the overturning moment caused by the force at point H: (γf = 1.0 for DL on span BE)

76.6 kN (Uplift)

291.6 kN

E 499.8 kN

Favorable loading condition

f

Permanent action Variable action

1.0 0.0

H

H

2.0 m

B

51.6 kN

E

5.0 m

16 + 0 = 16 kN/m

B 4m

4m

Proper design is required to resist the uplifting force at B, i.e. to anchor the beam at B. 47

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Wind Loads (EN 1991-1-4) Wind loads on external surfaces of the building:

We  q p  ze   ceq

q p  ze  peak velocity pressure

ze ceq

reference height for the external pressure pressure coefficient for the external pressure

qp  v

2 m

48

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Load combinations – Example 4 A gantry structure experiences the following loads. Evaluate the load combinations that need to be considered in the ultimate limit state design of the legs. G = 3, , Q = 3.5

Permanent action G Self-weight of beam Self-weight of each column

= 3 kN = 2 kN

Imposed action on beam Q

= 3.5 kN

Wind load W

= 5 kN

W =5

7m

2

2

4m

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Maximum compression at support A G, Q 1.35×3

1.5 x 3.5

Unfavorable – increases RA. Favorable – reduces RA. G and Q are unfavorable while W = 0 is favorable.

W=0

7m

1.35×2

Use equilibrium of moments about right support to calculate RA: RA×4 + 0×5×7 = 1.35×2×4 + 1.35×3×2 + 1.5×3.5×2  RA = 7.35 kN.

4m RA

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Maximum tension at support A • •

Unfavorable – increases RA. Favorable – reduces RA.

G = 1.0×3 Q =0 x 3.5

W= 1.5x5

7m

•

W is unfavorable while G and Q are favorable.

• •

Use equilibrium of moments about right support to calculate RA:

1.0×2

4m

• • •

RA×4 + 1.0×2×4 + 1.0×3×2 + 0×3.5×2 = 1.5×5×7  RA = 9.63 kN.

RA must be designed for both compression and tension.

RA

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Maximum compression at support B 1.35×3 1.5 x 3.5

• •

Unfavorable – increases RB. Favorable – reduces RB.

W= 1.5x0.5x5

7m

•

G, Q and W are unfavorable.

• •

Use equilibrium of moments about left support to calculate RB:

1.35×2

4m

• • •

Imposed load as leading variable action: RB×4 = 1.35×2×4 + 1.35×3×2 + 1.5×3.5×2 + (1.5×0.5)×5×7

•

 RB = 13.9 kN.

RB

Reduction factor for wind load

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CE3166:  R Liew

Maximum compression at support B

G = 1.35×3 Q =1.5x 0.7x 3.5

W= 1.5x5 Unfavorable – increases RB. Favorable – reduces RB. 7m

G, Q and W are unfavorable.

1.35×2

Use equilibrium of moments about left support to calculate RB: 4m

Wind load as leading variable action: RB×4 = 1.35×2×4 + 1.35×3×2 + (1.5×0.7)×3.5×2 + 1.5×5×7  RB = 19.7 kN.

Wind load as leading variable action is the critical case for maximum RB in compression.

RB

54

CE3166:  R Liew

Maximum tension at support B

G = 1.0×3 Q = 0 x 3.5

W= 0x5

• •

Unfavorable – increases RB. Favorable – reduces RB.

•

G, Q and W are favorable.

• •

Use equilibrium of moments about left support to calculate RB:

7m

1.0×2

4m

•

RB×4 + 1.0×2×4 + 1.0×3×2 + 0×3.5×2 + 0×5×7

•

 RB = -3.50 kN.

RB only needs to be designed for compression.

RB

55

CE3166:  R Liew

Summary • • • •

Reaction at support A: Support A must be designed for both compression and tension. Maximum compression = 7.35 kN. Maximum tension = 9.63 kN.

• • •

Reaction at support B: Support B only needs to be designed for compression. Maximum compression = 19.7 kN.

Summary: Since wind loads can act in reverse direction, the design forces for the supports are: Compression = 19.7 kN Tension: 9.63 kN

CE3166:  R Liew

Additional Problem The roof structure is subject to the following characteristic loads: Dead Load, Live Load (or imposed load), Wind Load, Rain Load. Determine the design loads (factored loads). They are three variable loads. Need to find suitable load combination depending whether they are favaourable or unfavourable. Dead load Live load Rain load Wind load downward)

Wind load ( upward)

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