Practical Design to Eurocode 2 - Concrete Centre · PDF file5 19 Oct Paul Gregory Slabs and Flat Slabs 6 26 Oct Charles Goodchild Deflection and Crack Control ... Practical Design

Practical Design to Eurocode 2 16/11/2016

Week 9: Fire 1

Practical Design to Eurocode 2

The webinar will start at 12.30

Lecture Date Speaker Title

1 21 Sep Charles Goodchild Introduction, Background and Codes

2 28 Sep Charles Goodchild EC2 Background, Materials, Cover and effective spans

3 5 Oct Paul Gregory Bending and Shear in Beams

4 12 Oct Charles Goodchild Analysis

5 19 Oct Paul Gregory Slabs and Flat Slabs

6 26 Oct Charles Goodchild Deflection and Crack Control

7 2 Nov Paul Gregory Detailing

8 9 Nov Jenny Burridge Columns

9 16 Nov Jenny Burridge Fire

10 23 Nov Jenny Burridge Foundations

Course Outline


Week 9: Fire 2

Fire Design

Lecture 8

16th November 2016

Design exercise

From last week

(Lecture 8)


Week 9: Fire 3

Take ‘beam’ width

as, say, half the

bay width

Solution - effective length & slenderness (M2)

41.0

9600

1230030002

8600

1230030002

4200

12500

2 33

4

====××××××××

++++××××××××

========

∑∑∑∑b

b

c

c

L

EI

L

EI

k

From Table (Table 4 of How to…Columns)

F = say 0.74

∴∴∴∴ lo = 0.74 x 4.2 = 3.108 m

Check slenderness:

λ = 3.46 lo/h

= 3.46 x 3.108 /0.5 = 21.5

Using PD 6687 method

Clear span is 4500 – 300 = 4200 mm and k1 = k2

Solution - column moments

MEd = max[M02; M0Ed + M2; M01 + 0.5M2 ; e0NEd]

M02 = M + eiNEd

ei = l0/400 =3108/400 = 7.8 mm

NEd = 7146 kN

= 95.7 + 0.0078 x 7146 = 95.7 + 55.7 = 151.4 kNm

M0Ed = (0.6M02 + 0.4M01) ≥ 0.4M02

= 0.6 × 95.7 + 0.4 × (− 95.7) ≥ 0.4 × 95.7 = 38.3 kNm

M01 = - 95.7 + 0.0078 x 7146 = -95.7 + 55.7 = - 40.0 kNm

e0NEd = 0.02 x 7146 = 142.9 kNm

e0 = Max[h/30,20mm]

= Max[500/30,20mm] = 20 mm

M2 = ?


Week 9: Fire 4

Limiting slenderness:

A = 0.7 (use default value)

B = 1.1 (use default value)

C = 1.7 – rm = 1.7 – M01/M02 = 1.7 – (-40.0/151.4) = 1.96

n = NEd/Acfcd = 7146 x 1000/(5002 x 0.85 x 50/1.5) = 1.01

λlim = 20 ABC/√n

= 20 x 0.7 x 1.1 x 1.96/√1.01

= 30.0

Slenderness:

30.0 > 21.5 ...section is not slender . . . . M2 = 0

Solution – slenderness limit (M2)

Solution- column moments

MEd = max[M02; M0Ed + M2; M01 + 0.5M2 ; e0NEd]

M02 = M + eiNEd

ei = l0/400 =3108/400 = 7.8 mm

NEd = 7146 kN

= 95.7 + 0.0078 x 7146 = 95.7 + 55.7 = 151.4 kNm

M0Ed = (0.6M02 + 0.4M01) ≥ 0.4M02

= 0.6 × 95.7 + 0.4 × (− 95.7) ≥ 0.4 × 95.7 = 38.3 kNm

M01 = - 95.7 + 0.0078 x 7146 = -95.7 + 55.7 = - 40.0 kNm

e0NEd = 0.02 x 7146 = 142.9 kNm

e0 = Max[h/30,20mm]

= Max[500/30,20mm] = 20 mm

M2 = 0

By inspection,

max[M02; M0Ed + M2; M01 + 0.5M2 ; e0NEd] = MEd = 151.4 kNm


Week 9: Fire 5

d2 = cnom + link + ϕ/2 = 35 + 8 + 16 = 59 mm

d2/h = 59/500 = 0.118

MEd/(bh2fck) = 151.4 x 106/(5003 x 50) = 0.024

NEd/(bhfck) = 7146 x 1000/(5002 x 50) = 0.57

Solution – determine As

Interaction Chart

0.09

0.024

0.57


Week 9: Fire 6


Asfyk/bhfck = 0.09

As = 0.09 x 5002 x 50 / 500 = 2250 mm2 Try 8 H20 (2513 mm2)

But > 4 no bars so check d2/h

Average d2 = (3 x 59 + 1 x 250 [say] ) / 4 = 107 mm : d2/h = 0.214

Interpolating between charts for d2/h = 0.20 and d2/h = 0.25

Interaction Chart

Asfyk/bhfck

0.100.57

0.024


Week 9: Fire 7


Asfyk/bhfck = 0.09

As = 0.09 x 5002 x 50 / 500 = 2250 mm2 Try 8 H20 (2513 mm2)

Check d2/h

Average d2 = (3 x 59 + 1 x 250 [say] ) / 4 = 107 mm : d2/h = 0.214

Using chart for d2/h = 0.20

Asfyk/bhfck = 0.10

As = 0.10 x 5002 x 50 / 500 = 2500 mm2

Use 8 H20 (2513 mm2)

Links

Diameter = max (6, 20/4) = 6 say 8mm diam

scl,tmax = min {12 φmin; 0.6b ; 240mm} = 240 mm

Bar centres = 500/2 -59 = 191 mm therefore each bar to be

restrained

With H8 links in 3 legs each way @ 225 cc

Fire


Week 9: Fire 8

a Axis

Distance

Reinforcement cover

Axis distance, a, to centre of bar

a = c + φφφφm/2 + φφφφl

Scope

Part 1-2 Structural fire design gives several methods for fire engineering

Tabulated data for various elements is given in section 5

Structural Fire DesignPart 1-2, Fig 5.2 Figure 4.2

• High strength concrete

• Basis of fire design

• Material properties

• Tabulated data

• Design procedures

– Simplified and advanced calculation methods

– Shear and torsion

– Spalling

– Joints

– Protective layers

• Annexes A, B, C, D and E

• General

Eurocode 2: Part 1.2 Structural Fire Design

100 Pages


Week 9: Fire 9

• Requirements:

– Criteria considered are:

“R” Mechanical resistance (load bearing)

“E” Integrity (compartment separation)

“I” Insulation (where required)

“M” Impact resistance (where required)

• Actions - from BS EN 1991-1-2

– Nominal and Parametric Fire Curves

Chapter 2: Basis of Fire Design

• Verification methods Ed,fi ≤ Rd,fi(t)

• Member Analysis Ed,fi = ηfi Ed

Ed is the design value for normal temperature design

ηfi is the reduction factor for the fire situation

ηfi = (Gk + ψfi Qk.1)/(γGGk + γQ.1Qk.1) ψfi is taken as ψψψψ1 or ψψψψ2 (= ψψψψ1 - NA)

Chapter 2: Basis of Fire Design


Week 9: Fire 10

• Tabulated data (Chapter 5)

• Simplified calculation methods

• Advanced calculation method

Design Procedures

Which method?


Week 9: Fire 11

Provides design solutions for the standard fire exposure up to 4 hours

• The tables have been developed on an empirical basis confirmed by experience and theoretical evaluation of tests

• Values are given for normal weight concrete made with siliceous aggregates

• For calcareous or lightweight aggregates minimum dimension may be reduced by 10%

• No further checks are required for shear, torsion or anchorage

• No further checks are required for spalling up to an axis distance of 70 mm

• For HSC (> C50/60) the minimum cross section dimension should be increased

Section 5. Tabulated DataCl. 5.1 -

Elements

• Approach for Beams and Slabs very similar

– Separate tables for continuous members

– One way, two way spanning and flat slabs

treated separately

• Walls depend on exposure conditions

• Columns depend on load and slenderness


Week 9: Fire 12

Continuous BeamsTable 5.6 Table 4.6

Flat Slabs

Table 4.81992-1-2 Table 5.9


Week 9: Fire 13

Columns Tabular Approach

Columns more Tricky!

• Two approaches

• Only for braced structures

• Unbraced structures – columns

can be considered braced if

there are columns outside the

fire zone

μfi = NEd,fi/ NRd = Gk + ψ1,1 Qk,1/(1.35Gk + 1.5 Qk) NEd/ NRd Conservatively 0.7

where NEd,fi is the design axial load in the fire condition

NRd is the design axial resistance at normal temperature

The minimum

dimensions are

larger than

BS 8110

Columns: Method ATable 5.2a Table 4.4A


Week 9: Fire 14

Limitations to Table 5.2a

Limitations to Table 5.2a for Method A:

• Effective length of the column under fire conditions

l0,fi <= 3m.

• First order eccentricity under fire conditions:

e = M0Ed,fi / N0Ed,fi <= emax = 0.15 h

• Amount of reinforcement:

As < 0.04 Ac

Method A

� = 120(��

��)�.� (Exp 5.7)

where:

�� = 83 1.0 − μ��1 + ω

0.85 α""⁄ + ω

(as αcc = 0.85 in the UK, Rηfi = 83(1.0-μfi))

Ra = 1.6(a-30)

where a is the axis distance


Week 9: Fire 15

Method A

� = 120(��

��)�.� (Exp 5.7)

Rl = 9.6 (5 - l0,fi)

where l0,fi is the effective length in fire.

For an insitu column in a braced structure this can be

taken is 0.5 l for lower storeys and 0.7 l for the top

storey. (2 m ≤ l0,fi ≤ 6m)

Method A

� = 120(��

��)�.� (Exp 5.7)

Rb = 0.09b’

b’ is the width or the diameter of a square or circular

column.

For a rectangular column:

b’ = 2Ac/(b+h)

200mm ≤ b’ ≤ 450mm

h ≤ 1.5 b’


Week 9: Fire 16

Method A

� = 120(��

��)�.� (Exp 5.7)

Rn = 0 where n=4 (corner bars only)

Rn = 12 where n>4

(n is the number of longitudinal bars)

Method A

� = 120(�� + �$ + �% + �& + �'

120)�.�

What is the fire resistance period of a 3.5m long, 300 x 600 column,

NEd = 2950kN, NRd = 3600kN, with 25mm bars, 10mm links, cover

25mm?

μfi = 0.7 x 2950/3600 = 0.57 Rηfi = 83(1-0.57) = 35.4

a = 25+10+25/2 = 47mm Ra = 1.6(47-30) = 27.2

l0,fi = 0.5 x 3.5 = 1.75, so l0,fi = 2m Rl = 9.6(5-2) = 28.8

b’ = 2 x b x 1.5b/(b + 1.5b) = 360mm Rb = 0.09 x 360 = 32.4

n>4 Rn = 12

R = 120((35.4 + 27.2 + 28.8 + 32.4 + 12)/120)1.8 = 150 minutes

(from table


Week 9: Fire 17

Method A

� = 120(�� + �$ + �% + �& + �'

120)�.�

What is the fire resistance period of a 3.5m long, 300 x 600 column,

NEd = 2950kN, NRd = 3600kN, with 25mm bars, 10mm links, cover

25mm?

μfi = 0.7 x 2950/3600 = 0.57 Rηfi = 83(1-0.57) = 35.4

a = 25+10+25/2 = 47mm Ra = 1.6(47-30) = 27.2

l0,fi = 0.5 x 3.5 = 1.75, so l0,fi = 2m Rl = 9.6(5-2) = 28.8

b’ = 2 x b x 1.5b/(b + 1.5b) = 360mm Rb = 0.09 x 360 = 32.4

n>4 Rn = 12

R = 120((35.4 + 27.2 + 28.8 + 32.4 + 12)/120)1.8 = 150 minutes

(from table: 90 minutes)


Week 9: Fire 18

Columns: Method B

ω = 0.1 � 0.4% steel

ω = 0.5 � 2.0% steel

ω = 1.0 � 4.0% steel

(fck = 30MPa, fyk = 500MPa)

Limitations to Table 5.2b

• l/h (or l/b) ≤ 17.3 for rectangular column (λfi ≤ 30)

• First order eccentricity under fire conditions:

e/b = M0Ed,fi /b N0Ed,fi ≤ 0.25 with emax= 100 mm

• Amount of reinforcement, ω = As fyd / Ac fcd ≤ 1

For other values of these parameters see Annex C (e/b ≤ 0.5, emax ≤ 200 mm)


Week 9: Fire 19

• EC2 distinguishes between explosive spalling that can occur

in concrete under compressive conditions, such as in

columns, and the concrete falling off the soffit in the

tension zones of beams and slabs.

• Explosive spalling occurs early on in the fire exposure and is

mainly caused by the expansion of the water/steam particles

trapped in the matrix of the concrete. The denser the

concrete, the greater the explosive force.

− Unlikely if moisture content is less than 3% (NDP) by

weight. The assumption is that in exposure class X0 or

XC1 the moisture class is less than 3%

− Tabular data OK for axis distance up to 70 mm

• Falling off of concrete occurs in the latter stage of fire

exposure

Spalling

Minimum cross section should be increased:

• For walls and slabs exposed on one side only by:

For Class 1: 0.1a for C55/67 to C60/75


• For all other structural members by:



Axis distance, a, increased by factor:

For Class 1: 1.1 for C55/67 to C60/75

For Class 2: 1.3 for C70/85 to C80/95

High Strength Concrete -Tabulated Data


Week 9: Fire 20

• For C 55/67 to C 80/95 the rules for normal strength concrete apply.

provided that the maximum content of silica fume is less than 6% by

weight.

• For C 80/95 to C 90/105 there is a risk of spalling and at least

one of the following should be provided (NA):

Method A: A reinforcement mesh

Method B: A type of concrete which resists spalling

Method C: Protective layers which prevent spalling

Method D: Monofilament polypropylene fibres.

High Strength Concrete -Spalling

Other Methods

• Simplified calculation method for beams, slabs

and columns

• Full Non-linear temperature dependent ……..

• But all of these must have the caveat that they

are unproven for shear and torsion.


Week 9: Fire 21

Rd1,fi

MRd,fi,Span

MRd2,fi

M

M = w l / 8Ed,fi Ed,fi eff

1

1 - Free moment diagram for UDL under fire conditions

Annex E: Simplified Calculation Method for Beams and Slabs

Free bending moment in fire condition MEd,fi:

MEd,fi = wEd,fileff2/8

where

wEd,fi = gk + ψ1qk

Moments of resistance in fire condition:

MRd,fi,Span = (γs /γs,fi ) ks(θ) MEd (As,prov /As,req)

MRd,fi,Support = (γs /γs,fi ) MEd (As,prov /As,req) (d-a)/d

Where a is the required bottom axis distance given in Section 5

As,prov /As,req should not be taken greater than 1.3

Annex E: Simplified Calculation Method for Beams and Slabs


Week 9: Fire 22

500°C Isotherm Method

Ignore concrete > 500°C

Strength of rebar

dependent on steel

temperature

500°C Isotherm Method

Charts in BS EN 1992-1-2


Week 9: Fire 23

Decrease in strength of reinforcement (Figure 4.2)

Zone Method

Divide concrete into zones and work out average

temperature of each zone, to calculate strength

Zones


Week 9: Fire 24

Zone Method

Factors for

concrete

strength

Zone Method

Balance forces with Fst,fi to calculate MRd,fi

Calculate compression forces in zones


Week 9: Fire 25

Worked Example

• NEd= 1824kN

• Myy,Ed= 78.5kNm

• Mzz,Ed= 76.8kNm

• 2 hour fire resistance

required

• External, but no de-icing

salts

• fck = 30MPa

Worked Example

Cover:

cmin,b = diameter of bar (assume 25mm bars with 8mm links)

cmin,dur = (XC3/XC4) 25mm

say Δcdev = 10mm

cnom (to main bars) = max{(25+10),(25+8+10)} = 43mm

Use cnom = 35mm to 8 mm links


Week 9: Fire 26

Worked Example

Check fire resistance of R120 to Method A

eccentricity e < 0.15b

e = MEd/NEd = 78.5x103/1824 = 43mm

0.15 x 350 = 52.5mm ∴ OK

Assume 8 bars ∴ OK

l0,fi = 0.7l = 2.8m < 3m ∴ OK

From Table 5.2a: min dimensions = 350/57

Column is 350mm, axis distance required = 57mm

Check cover – 35mm + 8 (link) + φ/2 = 55.5mm

∴ Increase links to 10 mm diam

or increase nominal cover to 40mm.

Design exercise for next week

Using Equation 5.7, work out the fire resistance of a 250

x 750 column with an axial capacity of 3750kN and an

axial load in cold conditions of 3500kN. The column is on

the ground floor of a three storey building and the length

is 4.5m. The cover is 30mm, main bars are 20mm and

the links are 10mm diameter.


Week 9: Fire 27

Design Exercise

� = 120(�� + �$ + �% + �& + �'

120)�.�

μfi = Rηfi =

a = Ra =

l0,fi = Rl =

b’ = Rb =

n = Rn =

R = 120((Rηfi + Ra + Rl + Rb + Rn)/120)1.8 =

End of Lecture 9

Documents

Practical Design to Eurocode 2 - Concrete Centre · PDF file5 19 Oct Paul Gregory Slabs and Flat Slabs 6 26 Oct Charles Goodchild Deflection and Crack Control ... Practical Design