nrc concrete fire design

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Winter 2003 Concrete Structures 7-1

CONCRETE STRUCTURES


OVERVIEW

• This section will describe:– methods of designing reinforced concrete structures to

resist fires– information on thermal and mechanical properties of

concrete– briefly composite steel-concrete structures


BEHAVIOUR OF CONCRETE STRUCTURES IN FIRE

• Concrete structures behave well in fires• Concrete is non-combustible and has a low

thermal conductivity• Usually concrete remains in place during fires

and protects the reinforcing steel• Calculation of behaviour of concrete structures in

fire depends on many factors:– applied loads on the structure– elevated temperatures in the concrete and reinforcing – mechanical properties of steel/concrete at high temp.


BEHAVIOUR OF CONCRETE STRUCTURES IN FIRE

• Under fire, temp. of steel and concrete increase in reinforced concrete structures, leading to high deformation and possible failure

• Different concrete types include lightweight concrete and high-strength concrete

• Observations have shown that failure of concrete buildings in real fires is usually due to the inability of other parts of the structure to absorb the large imposed thermal deformations causing shear or buckling failures of columns or walls


High-strength Concrete

• Recently, interest in high-strength concrete has increased

• High-strength concrete contains additives which result in compressive strength of 50 - 120 MPa

• High-strength concrete shows a higher rate of strength loss than normal concrete at temp. up to 400°C and explosive spalling

• Kodur (1997) reported fire tests on high-strength columns

• Tomasson (1998) gave design recommendations


Lightweight Concrete

• Lightweight concrete is usually made with normal cement and some form of lightweight aggregate

• Lightweight concrete has good fire resistance due to low thermal conductivity compared with normal weight concrete

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Fibre Reinforced Concrete

• Steel-fibre reinforced concrete uses steel fibresadded to the concrete mix (improved strength)

• Fibres are typically 25-40 mm long and 0.5 mm diameter with hooked ends to improve bonding

• Lie and Kodur (1996) described mechanical and thermal properties of steel-fibre reinforced concrete at elevated temp.

• The steel fibres presence increases the ultimate strain and improves the ductility of the concrete


Spalling

• In concrete structures, design methods are often based on concrete remaining intact in fires

• This is not always valid due to spalling in fires• Tests and real fires have shown that most normal

concrete members do not show serious spalling• The spalling phenomenon is not well understood

because it is a function of many factors• Spalling often occurs when water vapour is driven

off from the cement paste during heating


Spalling

• Tests have shown that spalling results from high moisture content concrete

• High-strength concrete is more susceptible tospalling than normal concrete

• A method of preventing spalling is to add fine polypropylene fibres to concrete mix (0.15-0.3%)

• The fibres melt during fire exposure, increasing the porosity in the concrete for vapour escape


Masonry

• Concrete masonry consists of hollow concrete blocks mortared together (often used in walls)

• Concrete masonry blocks are often manufactured from lightweight concrete

• Unfilled unreinforced masonry has demonstrated excellent fire resistance

• Brick masonry also behaves well in fires• Thermal bowing of tall unreinforced cantilever

masonry walls can lead to collapse during a fire


Prestressed Concrete

• Prestressed concrete means concrete structures stressed prior to applying external loads

• Two types: pre-tensioned and post-tensioned• Pre-tensioned prestressed concrete is often used

for precast components for flooring (flat panels)• Post-tensioning is used within large components

(beams or slabs)• Full-scale fire tests have shown that bond failures

of pre-tensioned tendons failed prematurely


FIRE-RESISTANCE RATINGS

• Verification Methods• Generic Ratings

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Verification Methods• The design for fire resistance requires that:

provided fire resistance > design fire severity• Verification may be in:

– time domain, temperature domain or strength domain• In time domain, compare fire resistance ratings

with code-specified fire resistance (often used)• In strength domain, compare load-bearing

capacity with expected loads at time of fire• In temp. domain, compare critical temp. with max.

reached temp. in fires (not usually used)


Generic Ratings• Very few proprietary ratings for concrete structures• Generic ratings are often used• Many codes list generic ratings for concrete

members (min. sizes and concrete cover to steel)• Some ratings distinguish between concrete types• Information applies to standard fire exposure only• Table below shows generic fire resistance ratings

for reinforced concrete members (BSI, 1985)• Min. thickness of floors/walls are based on the

insulation criterion


Generic Ratings

Min. width (mm) and min. cover (mm) for generic fire resistance rating of reinforced concrete members

Beams Columns Slabs Walls0.5 hours Width

Cover8020

15020

7515

7515

1.0 hours WidthCover

12030

20025

9520

7515


15040

25030

11025

10025

2 hours WidthCover

20050

30035

12535

10025

3 hours WidthCover

24070

40035

15045

15025

4 hours WidthCover

28080

45035

17055

18025


Generic Ratings

• Appendix D gives generic approvals from 10 different sources from around the world

• Concrete structures show good fire resistance, thus many will meet the generic approvals with no increase in cover from normal temp. design

• Calculations may be useful for members which are thin or slender and members with little concrete cover to the reinforcing


CONCRETE AND REINFORCING TEMPERATURES

• Fire Exposure• Calculation Methods• Thermal Properties• Published Temperatures


Fire Exposure

• To design concrete structures under fire, temp. of concrete and reinforcing steel must be known

• The fire exposure may be standard or real curve• Design charts are available giving thermal

gradients in beams, columns and slabs exposed to the standard fire

• It is better to use computer programs to calculate temperature gradients in concrete members exposed to realistic fires

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Calculation Methods

• Thermal calculations in reinforced or prestressedconcrete members usually assume that:– heat transfer is a function of thermal properties of

concrete alone– temperature of the reinforcing is the same as the

temperature of the surrounding concrete• Different thermal conductivity is not a factor since

most steel is parallel to the fire-exposed surfaces• Only accurate way to calculate temperatures is to

use a 2-D finite-element computer program


Calculation Methods

• For simple members of normal-weight concrete, empirical hand calculation methods are available

• Wickstrom proposed a method of calculating the temp., Tw, in a normal weight concrete slab exposed to a standard fire on its surface as:

Tw = ηw Tfηw = 1 - 0.0616 th-0.88

• Tf is the fire temperature (°C) • th is the time (hours)


Calculation Methods

• Temp. Tc (°C) at any depth x (m) into the slab is:Tc = ηx ηw Tf

ηx = 0.18 ln(th/x2) - 0.81• The method can be used for corners of beams

where heat is conducted in 2 directions, using ηycalculated in the same way as ηx so that Tc is:

Tc = [ηw (ηx + ηy – 2 ηx ηy) + ηx ηy] Tf

• Wickstrom stated that these equations can be modified for other types of concrete


Thermal Properties

• To calculate temp. in structural assemblies, thermal properties of materials must be known

• Concrete density depends on aggregate and mix design (typical concrete density of 2300 kg/m3)

• When heated to 100°C, density of concrete reduces by up to 100 kg/m3 (water evaporation)

• Limestone aggregate concrete decomposes over 800°C with a corresponding decrease in density

• Thermal conductivity of concrete varies with temperature in a broad range (see figure below)


Thermal PropertiesThermal conductivity of concrete


Thermal Properties• Values for design purposes in Eurocode:

– 1.6 W /m-K for siliceous concrete– 1.3 W /m-K for limestone aggregate concrete– 0.8 W /m-K for lightweight concrete

• Specific heat of concrete varies in a broad range with temperatures (see Figure below)

• Peak between 100 and 200°C allows for water being driven off during the heating process

• Values for design purposes in Eurocode:– 1000 J/kg-K for siliceous/limestone aggregate concrete – 840 J/kg-K for lightweight concrete

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Thermal PropertiesSpecific heat of concrete


Published Temperatures

• There is good published information available on temperatures within concrete members exposed to the standard fire

• The availability of this information makes it much easier to design for standard fire exposure especially for the simple hand calculated design methods


Published Temperatures - Slabs

• Figure below shows typical temp. in concrete slabs exposed to standard fire (Wade, 1991)

• For exposure to real fires, limited information is available (see Figure below, Wade 1994)

• Figure below shows progression of temp. vs. time at various depths within the slab (for a real fire)

• The greater the cover, the greater delay in reaching the peak temperature


Published Temperatures - SlabsTemp. in concrete slabs exposed to the standard fire


Published Temperatures - Slabs

Peak temperatures in concrete slabs exposed to design fires


Published Temperatures - SlabsTemperature-time curves inside concrete slabs

exposed to design fires

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Published Temperatures - Beams• Temperature

contours in concrete beams exposed to standard fire


MECHANICAL PROPERTIES OF CONCRETE AT ELEVATED

TEMPERATURES• Components of Strain


Components of Strain

• The deformation of concrete is described by the total strain ε as:

ε = εth(T) + εσ(σ,T) + εcr(σ,T,t) + εtr(σ,T)• εth(T) is the thermal strain • εσ(σ,T) is the stress related strain• εcr(σ,T,t) is the creep strain• εtr(σ,T) is the transient strain


Components of Strain• Under heat and load, strains described above

produce deformations as shown in Figure below• For simple structures such as S.S. beams, only

the stress-related strain needs to be considered• For more complex systems, all strains must be

considered (using structural analysis programs)


Thermal Strain

• Approximate expressions for thermal elongation ∆L/L of concrete from Eurocode are given by: – ∆L/L = 18 x 10-6 Tc for siliceous aggregate concrete – ∆L/L = 12 x 10-6 Tc for calcareous aggregate concrete– ∆L/L = 8 x 10-6 Tc for lightweight concrete

• where Tc is the concrete temperature


Creep Strain and Transient Strain

• Creep strain and transient strain are linked• Figure below shows measured creep strains

under constant temperature and stress• During heating, there are changes in moisture

content and composition of concrete (transient)

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Creep Strain and Transient StrainCreep in concrete one day after loading at 10% of

the initial strength


Stress-related Strain

• Stress-related strains (elastic and plastic) result from applied loads

• Figures below show:– typical stress-strain curves for concrete at elevated

temperatures– reduction in ultimate compressive strength with temp.

for typical structural concrete• Ultimate compressive strength drops and strain at

peak stress increases with increasing temp.


Stress-related StrainStress-strain relationships for concrete at elevated

temperatures


Stress-related StrainReduction in compressive strength with temp.


Stress-related Strain -Design values

• The tensile strength of concrete is assumed to be zero at elevated temperatures

• Ultimate strength varies with temperature • Simple expression is necessary for design

purposes• Figure below shows an example of compressive

strength with temperature



Design values for reduction of compressive strength with temperature

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• The equations in the Figure are given by:– for normal weight concrete

kc,T = 1.0 for T < 350°Ckc,T = (910 - T)/560 for T > 350°C

– for lightweight concrete

kc,T = 1.0 for T < 500°Ckc,T = (1000 - T)/500 for T > 500°C


Stress-related Strain -Modulus of elasticity

• The modulus of elasticity of concrete drops with increasing temperature

• Figure below shows an example of reduction in modulus of elasticity with equations given by:

kE,T = 1.0 for T < 150°CkE,T = (700 - T)/550 for T > 150°C


Stress-related Strain -Modulus of elasticity

Design values for reduction of modulus of elasticity with temperature


DESIGN OF CONCRETE MEMBERS EXPOSED TO FIRE

• The overall strategy for structural limit states design of fire-exposed concrete structures is:(1) for S.S slabs or tee-beams exposed to fire from

below, concrete in the compressive zone remains at normal temp., so the structural design needs only consider the effect of elevated temp. on yield strength of the reinforcing steel (simple hand calculations)

(2) for continuous slabs or beams, some of fire-exposed surfaces are in compression, so simple calculation methods must consider effects of elevated temperature on the compressive strength of the concrete



(3) similar methods can be applied to fire-exposed concrete walls and columns, but these methods are less accurate because of deformations caused by non-uniform heating and the possibility of instability failures

(4) For moment-resisting frames, or structural members affected by axial restraint or non-uniform heating, it is recommended to use computer programs for structural analysis under fire conditions



• Eurocode gives three methods of design:– generic tabulated method– simplified calculation method – general calculation methods, which provide a realistic

analysis of concrete structures exposed to fire• Complex structures should be designed using

general calculation methods (computer program)• Simplified method is useful for single members,

using hand calculations (used for normal temp.)• Temp. profile within the members is essential

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• For standard fire exposure, use design charts, Wickstrom's method or a computer calculation

• For S.S. slabs or beams, consider the reduction of steel strength only (ignore concrete in tension)

• For concrete in compression (e.g., columns or continuous beams or slabs), the effect of temp. on concrete strength is important

• No consensus on limiting temp. to use for design• Most simple approach is to assign full strength to

concrete < 500°C and zero strength > 500°CWinter 2003 Concrete Structures 7-50


• This simple method is:– excellent for large members– inaccurate for thin concrete members (temp. > 500°C)

• Another approach is to ignore concrete over 750°C and assign a single strength to remaining concrete core based on its average temp.

• These simplified methods give similar results• The first method is recommended (500°C limit)


Member Design• Verification in the strength domain requires that:

U*fire ≤ Rfire

• U*fire is the design force from applied loads (code)

and Rfire is the load-bearing capacity• U*

fire may be axial force N*fire, bending moment

M*fire or shear force V*

fire

• Rfire may be calculated as axial force Nf, bending moment Mf or shear force Vf

• Calculations of Rfire are based on the mechanical properties of concrete/steel at elevated temp.


Member Design• Below is a description of design of members using

the simple method with zero strength for concrete above 500°C

• The design method uses normal assumptions for reinforced concrete design, assuming:– concrete has no tensile strength– parabolic compressive stress block in the concrete can

be approximated by an equivalent rectangle• The stress block is calculated assuming that the

characteristic strength is 85% x crushing strength• For beams/slabs, ignore compression reinforcing


Simply Supported Slabs/Beams

• Simplest reinforced concrete members to design are S.S. slabs and tee beam (see Figures below)

• Design equation for members with moment M*fire:

M*fire ≤ Mf

• The flexural capacity under fire conditions Mf is:Mf = As fy,T (d - af/2)

• As is the area of the reinforcing steel• fy,T is the yield stress of reinforcing steel reduced

for temperature (fy,T = ky,T fy)



• d is the effective depth of the cross section• af is the depth of the rectangular stress block:

af = (As fy,T) / (0.85 f’c b)• f’c is the characteristic compressive strength of

the concrete and b is the width of the slab/beam• The calculations assume that temp. of concrete in

the compressive zone is not high enough to cause any reduction in strength

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Simply Supported Slabs/BeamsSimply supported slab exposed to fire


Simply Supported Slabs/BeamsSimply supported Tee-beam exposed to fire



• A S.S. beam with a non-composite slab (see Figure below) is slightly more affected by fire

• Depth of the rectangular stress block is:af = (As fy,T) / (0.85 f’c bf)

• bf is the fire-reduced effective width of the beam


Simply Supported Slabs/BeamsS.S. non-composite beam exposed to fire


WORKED EXAMPLE 1

SIMPLY SUPPORTED REINFORCED CONCRETE SLAB (Refer to Figure page 7-55)

For a simply supported reinforced concrete slab with known span, load, geometry and reinforcing, check the flexural capacity after 60 minutes exposure to the standard fire. Use Wickstrom'sformula to calculate the reinforcing temperature.

Given information: • Dead load (excluding self weight) G1 = 0.5 kN/m• Live load Q = 2.5 kN/m • Slab span L = 7.0 m


WORKED EXAMPLE 1• Slab thickness h = 200 mm • Concrete density ρ = 24 kN/m3

• Concrete strength f’c = 30 MPa • Yield stress fy = 300 MPa • Bars:

– diameter Db = 16 mm– spacing s = 125 mm – bottom cover cv = 15 mm

• Design a 1 m wide strip - b = 1000 mm • Self weight G2 = ρ h b = 4.8 kN/m• Total dead load G = G1+G2 = 0.5+4.8 = 5.3 kN/m • Steel area As = n π r2 b/s = 1608 mm2

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WORKED EXAMPLE 1• Effective depth d = h - cv - Db/2 = 177 mm • Effective cover ce = cv + Db/2 = 23 mm = 0.023 m

COLD CALCULATIONS (for a 1 m wide strip) • Strength reduction factor Φ = 0.85 • Stress block depth a = Asfy/0.85f’c b

a = 1608x300/0.85x30x1000 = 18.9 mm • Internal lever arm jd =d-a/2=177-18.9/2=168 mm • Design load (cold) wc = 1.2G+1.6Q= 10.4 kN/m • Bending moment

M*cold = wcL2/8 = 10.4x7.02/8 = 63 kNm


WORKED EXAMPLE 1• Bending strength

Mn = Asfy jd = 1608x300x168/106 = 81 kNm Φ Mn = 69 kNm Φ Mn > M*

cold so design is OK.

FIRE CALCULATIONS • Revised strength reduction factor Φ = 1.0• Design load (fire) wf = G1+G2+0.4Q = 6.3 kN/m • Bending moment

M*fire = wf L2/8 = 6.3x7.02/8 = 38.6 kNm


WORKED EXAMPLE 1

• After 60 minutes of standard fire exposure t = 60 min (th = 1.0 hour)

• Fire temperature Tf = 20+345 log(8t+1) = 945°C• Surface temperature Tw = [1-0.0616 th-0.88] Tf

Tw = [1-0.0616x1.0-0.88] x 945 = 887°C• Concrete temperature Tc =[0.18 In(th/ce

2)-0.81] TwTc = [0.18 In(1.0/0.0232)-0.81] x 887 = 486°C

• Steel temperature Ts = Tc = 486°C • Reduced yield stress • fy,T = fy(720-Ts)/470=300(720-486)/470= 149 MPa


WORKED EXAMPLE 1

• Stress block deptha = As fy,T / 0.85 f’c b a = 1608x149/0.85x30x1000 = 9.4 mm

• Internal level arm jd = d-a/2 = 177-9.4/2 = 172 mm

• Bending strengthMnf = As fy,T jd = 1608x149x172/106 = 41.2 kNmΦ Mnf = 1.0x41.2 = 41.2 kN.mΦ Mnf > M*

fire so design is OK.


Shear Strength

• Shear is not usually a problem in fire-exposed concrete structures

• For shear design, Eurocode recommends using normal temp. design methods with mechanical properties reduced with temp. and the cross section reduced to the 500°C contour


Continuous Slabs and Beams

• Slabs or beams which are built into one or more supports usually have enhanced fire resistance because of moment redistribution

• If calculations show that the slab or beam can resist the fire as a simply supported member, no calculations for continuity are necessary

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Continuous Slabs and Beams -Negative flexural capacity

• To allow for flexural continuity effects, need to calculate negative moment capacity at supports

• For a slab of uniform thickness, negative flexural capacity M-

f at supports is (see Figure below)M-

f = As fy,T (df - af/2)• df is the effective depth of the slab• af is the depth of the rectangular stress block:

af = As fy,T / 0.85 f’c b• b = bf for a beam with its compression edge

exposed to fire (see Figure below) Winter 2003 Concrete Structures 7-68


Support region of continuous slab exposed to fire



Support region of continuous beam exposed to fire



• When compression region of a slab/beam is exposed to fire, it is important to ensure that the compression capacity is not reduced so low as to cause a sudden compression failure

• This can be ensured by checking that:(As fy,T) / (b df f’c,T) < 0.30

• This check is not necessary if there is significant longitudinal reinforcing in the compression zone


Axial Restraint

• Axial restraint can have a significant influence on fire resistance of reinforced concrete members

• Hot concrete members tend to undergo larger horizontal displacements with lower vertical deflections (large horizontal restraint forces)

• Axial restraint is beneficial for reinforced orprestressed concrete slabs or beams

• The line of thrust must be below the compressive stress block to utilize the benefits of axial restraint


Axial Restraint -PCA method for calculating restraint

• The PCA method is the only method of assessing restraint without a computer analysis package

• The step-by-step guide to the PCA procedure is: (1) Calculate the bending moment at mid-span under

fire-reduced loads M*fire assuming S.S. behaviour

(2) Calculate the flexural capacity at mid-span during fire M+

f. If M+f > M*

fire no continuity or restraint is necessary(3) Calculate the flexural capacity at the supports during

fire M-f. If M+

f + M-f > M*

fire continuity is sufficient and no restraint is necessary. If member is not symmetrical, flexural capacity M-

f should be average of the 2 ends

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Axial Restraint -PCA method for calculating restraint(4) Estimate the mid-span deflection ∆ using

∆ = (L2 ∆0) / (89000 yb)∆0 is the mid-span deflection of the reference specimen,

see Figure below, L is the heated length of member (mm), and yb is the distance from the neutral axis of the member to the extreme bottom fibre (mm)

(5) Estimate the distance of the line of thrust from the top of the member dT (mm) at the supports

• For built-in construction, assume that line of thrust is 0.1 h above the bottom of member where h is member overall depth

• For other support conditions, an independent estimate may be necessary



Mid-span deflection of reference specimens


Axial Restraint -PCA method for calculating restraint(6) Calculate the magnitude of the required axial thrust T

(kN) to prevent collapse as: T = 1000 (M*

fire – (M+f + M-

f)) / (dT – af/2 - ∆)af (mm) is the height of the rectangular compression

stress block in the member, approximated byaf ≈ a M*

fire / M+f where a is af = (As fy,T) / (0.85 f’c bf)

(7) Calculate af more accurately, usingaf = (T + A+

s f+y,T) / (0.85 f’c,T bf) A+

s f+y,T is the tensile strength of the bottom steel at mid-span. Repeat steps (6) and (7) if necessary


Axial Restraint -PCA method for calculating restraint(8) Calculate the non-dimensional thrust parameter T/AE

and the effective thickness z = A/s (mm) where A is the member cross-section area (mm2), E is the modulus of elasticity of concrete (usually about 25 GPa), and s is the heated perimeter of the member (mm)

(9) Determine the strain parameter ∆L/L, from Figure below, using T/AE and z. (Enter Figure with T/AE, pass through diagonal line at calculated value of z, and read-off ∆L/L. Example shown by dashed line)



Nomogram for thrust in concrete members


Axial Restraint -PCA method for calculating restraint(10) Calculate the maximum permitted displacement

∆L/L (mm) by multiplying the strain parameter ∆L/L by the heated length L (mm).

(11) Determine independently whether the surrounding structure can withstand the thrust T with a displacement no greater than ∆L. If so, the structure can withstand the fire.

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WORKED EXAMPLE 2

AXIAL RESTRAINT Consider a reinforced concrete floor constructed

from precast concrete tee-beams as shown in the Figure below. The slabs are simply supported over a span of 6.0 m, carrying a live load of 3.0 kPa. The dead load is 4.8 kPa (including the self weight). Calculate the restraint condition necessary to give a fire-resistance rating of 120 minutes.


WORKED EXAMPLE 2

Beam for Worked Example 2


WORKED EXAMPLE 2

Given information: • Slab span L = 6.0 m • Concrete strength f’c = 25 MPa • Overall depth h = 300 mm• Steel strength fy = 350 MPa • Web width bw = 200 mm • Concrete MOE E = 25 GPa • Overall width bf = 1200 mm • Cross-sectional area A = 185000 mm2

• Heated perimeter s = 1550 mm


WORKED EXAMPLE 2

Load combinations: • Dead load per metre G = 1.2x4.8 = 5.76 kN/m • Live load per metre Q = 1.2x3.0 = 3.6 kN/m • Load combination for cold conditions:

wc = 1.2G+1.6Q = 12.7 kN/m Load combination for fire conditions: • wf = G + 0.4Q = 7.20 kN/mReinforcing: • Number of bars n = 4


WORKED EXAMPLE 2• Bar diameter Db = 16 mm • Single bar area As1 = π(Db/2)2 = 201 mm2

• Total bar area As = n As1 = 804 mm2

• Cover cv = 20 mm • Effective depth d = h - cv - Db/2 = 272 mm

• Effective cover ce = cv + Db/2 = 28 mm

COLD CALCULATIONS• Strength reduction factor Φ = 0.85 • Mid-span bending moment

M*cold = wcL2/8 = 12.7x6.02/8 = 57.2 kNm


WORKED EXAMPLE 2• Stress block depth a = As fy/0.85 f’c bf

a = 804x350/0.85x25x1200 = 11.0 mm • Internal lever arm jd = d-a/2 = 272-11/2 = 266 mm• Flexural strength Mc =As fy jd

Mc =804x350x266/106 = 75.0 kNm Φ Mc = 0.85x75.0 = 63.8 kNmΦ Mc > M*

cold so cold design is OK. FIRE CALCULATIONS• Strength reduction factor Φ = 1.0 • Mid span bending moment

M*fire = wfL2/8 = 7.2x6.02/8 = 32.4 kNm

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WORKED EXAMPLE 2• Steel bar temperature from the isotherms on

Figure 7-31:• Bar group (1) (corner bars) Ts1 = 830°C• Bar group (2) (corner bars) Ts2 = 640°C • Reduced yield strength

– fy,T1 = 350 x (720-830)/470 = 0 MPa– fy,T2 =350 x (720-640)/470 = 59.5 MPa

• As fy,T = (2 As1 fy,T1+ 2 As2 fy,T2) = (2x201x0+2x201x59.5)/1000 = 23.9 kN

• Stress block depth af = As fy,T/0.85f'c bfaf = 23.9x1000/(0.85x25x1200) = 0.93 mm


WORKED EXAMPLE 2• Internal lever arm

jdf = d-af/2 = 272-0.93/2 =271.5 mm• Flexural strength

Φ Mf = Mf = Asfy,T jdf = 24x271.5/106 = 6.4 kNmΦ Mf < M*

fire so slab will fail unless restraint or continuity is provided.

Provide axial restraint (numbers in brackets are steps from notes)

(4) Estimate the mid-span deflection: • Mid-span deflection of the reference specimen

∆0 = 65 mm (from Figure 7-74)


WORKED EXAMPLE 2• Heated length L = 6000 mm • Distance from neutral axis to extreme bottom

fibre yb = 290 mm (assume that neutral axis is 10 mm from top of slab)

• Mid-span deflection ∆ = L2 ∆0/89000 yb∆ = 60002 x 65/(89000 x 290) = 90.7 mm

• Distance to line of thrust from top of beamdT = 0.9h = 280 mm

(assume that the slab is built-in to the surrounding construction, thrust 0.1h from bottom.)


WORKED EXAMPLE 2(6) Calculate the required thrust to prevent collapse

T = 1000 (M*fire-Mf)/(dT-af/2-∆)

T = 1000 (32.4-6.4)/(270-0.9/2-90.7) = 145 kN (7) Recalculate af

af = (T+ As fy,T)/0.85 f’c bfaf =(14500+ 23700)/(0.85x25x1200) = 6.6 mm

• Recalculate T T = 1000 (32.4-16.1)/(270-6.6/2-90.7) = 148 kN

(8) Non-dimensional thrust parameterT/AE = 147700/(185000x25000) = 32 x 10-6


WORKED EXAMPLE 2

• Shape parameter z = A/S = 185000/1550 = 119 mm

• Strain parameter∆L/L = 0.0065 (from Figure 7-77)

• Maximum permitted displacement∆L = 0.0065x6000 = 39 mm

So, this slab will have a fire resistance of 120 min if the surrounding structure at each end is capable of resisting an axial thrust of 148 kN with an axial elongation of less than 39 mm.


Columns

• Columns are more difficult to design than flexural members due to possible instability problems

• As a simple method, minimum dimensions and cover, as shown in Tables below, can be used

• The recommended conservative design approach is to use the simplified method assuming zero strength for all concrete above 500°C

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ColumnsMinimum width (mm) and minimum cover (mm) for

generic fire-resistance rating of reinforced concrete members

Beams Columns Slabs Walls0.5 hours Width

Cover8020

15020

7515

7515


12030

20025

9520

7515


15040

25030

11025

10025

2 hours WidthCover

20050

30035

12535

10025

3 hours WidthCover

24070

40035

15045

15025

4 hours WidthCover

28080

45035

17055

18025


Columns

Generic fire-resistance ratings for concrete columns

• Width is the minimum dimension of the column and cover is the location of the centre line of the reinforcing relative to the outside of the column. Cover required for durability will control in some cases

Column exposed on more than one side Exposed on one sideLoad ratio 0.2 0.5 0.7 0.7

Width(mm)

Cover(mm)

Width(mm)

Cover(mm)

Width(mm)

Cover(mm)

Width(mm)

Cover(mm)

30 minutes 150 10 150 10 150 10 100 1060 minutes 150 10 180 10 200 10 120 1090 minutes 180 10 210 10 240 35 140 10120 minutes 200 40 250 40 280 40 160 45180 minutes 240 50 320 50 360 50 200 60240 minutes 300 50 400 50 450 50 300 60


Walls• Non-load-bearing concrete walls or partitions do

not require structural design (provide thickness to meet insulation criterion)

• Previous Table shows some ratings• Load-bearing walls are usually axial load-carrying

members that should be designed as slender columns (using columns methods)

• The main difference is fire exposure, columns are often designed for exposure on all sides, but most walls are exposed to fire on one side only


COMPOSITE STEEL-CONCRETE CONSTRUCTION EXPOSED TO

FIRE• Composite slabs• Composite beams• Composite columns


Composite slabs

• Composite steel-concrete slabs are popular • Fire behaviour of composite slabs is discussed

under the three categories:– integrity– insulation– stability


Composite slabs - Integrity

• Composite steel-concrete slabs generally have excellent integrity because there is no passage of flames or hot gases through the floor system

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Composite slabs - Insulation

• To meet the insulation criterion it is simply necessary to provide sufficient thickness of slab

• Generic and proprietary listings are available• Calculation of temperatures has been discussed

previously


Composite slabs - Stability

• The strength of composite steel-concrete slabs is influenced by fire because steel sheeting acting as external reinforcing loses strength rapidly

• Composite slabs behave well in fire because of axial restraint and moment re-distribution


Composite beams

• Hot-rolled steel beams• Light steel joists


Composite beams -Hot-rolled steel beams

• Composite steel-concrete beams often consist of reinforced concrete slabs connected to a supporting hot-rolled structural steel beams

• The most common system is for composite steel-deck slabs to run over the top of the steel beams

• Composite beams act as tee-beams• Structural calculation for fire is essentially the

same procedure as in normal temp. conditions


Composite beams -Light steel joists

• A common system of composite construction uses open web or light I-gauge steel joists combined with concrete

• The fire resistance of this type of composite construction is very poor without joist protection

• Additional fire resistance can be provided with a fire resisting ceiling membrane or fire protection material sprayed onto the steel joists


Composite columns

• The concrete filling has three beneficial effects:– it acts as a heat sink to slow the rise in temp. of steel– it provides lateral stability to prevent local buckling of

the column wall – it carries some of axial load as steel strength reduces

• NRC has carried out research on concrete filling in hollow steel columns exposed to standard fires using the column furnace

• Lie and Kodur developed a design formula that is used in National Building Code of Canada

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Composite columns• The empirical equation gives fire resistance tr

(min) of circular/square steel columns completely filled with concrete as:

tr = (f (f’c + 20) d2.5) / ((KL – 1000) √N)• f is a factor from Table below• f’c is the strength of the filling concrete (MPa)• d is the outside diameter/width of column (mm)• L is the unsupported length of the column (mm)• KL is the effective length of the column (mm)• N is the applied load on the column (kN)


Composite columns

Value of factor f for fire resistance of concrete-filled steel columns

Note that tabulated values of f may be increased cumulatively as follows Tabulated values are for siliceous aggregate concreteFor carbonate aggregate concrete, add 0.01. Bar-reinforced concrete values are for cover < 25 mmFor cover ≥ 25 mm, add 0.005. Bar-reinforced concrete values are for reinforcing < 3%For reinforcing ≥ 3%, add 0.005

Filling concrete Square columns Circular columnsPlain concrete 0.06 0.07Bar-reinforced concrete 0.065 0.075Fibre-reinforced concrete 0.065 0.075


Composite columns

• This formula is valid for:– fire-resistance times up to 2 hours for plain concrete

and 3 hours for reinforced concrete– for column sizes from about 140 to 410 mm, except

that bar reinforcing cannot be used in columns smaller than about 200 mm

• Square columns with fibre-reinforced concrete can be as small as 100 mm


WORKED EXAMPLE 3Calculate the fire resistance of a circular steel

column filled with concrete.

Given information:• Column length L = 4800 mm • Effective length factor

K = 0.67 (fixed-fixed end conditions) • Axial load N = 1344 kN• Concrete strength

f’c = 40 MPa (carbonate aggregate concrete)


WORKED EXAMPLE 3

• Column size HSS 324x6.4• Column diameter d = 324 mm • Bar reinforcing 3% • Cover 25 mm

• From Table 7-104, f = 0.095.• From Eq. (7-103) the fire resistance is given by• tf = f(f’c + 20) d2.5/(KL-1000) √N • tf = 0.095 (40+20) 3242.5(0.67x4800-1000)√1344

= 133 minutes