60min Fire Rating DFT. (mm):. 120min Fire Rating DFT. (mm):. Section. Mass. (kg/m):. A p. /V. (m-1): ... SANS 10400 by limiting the maximum division area allowed in various ..... When O > 0.04, and qt,d < 75, and b < 1160, then Îlim is modified by: ...... Arcelor Mittal - Fire Calculations Download Centre: A number of free ...
ADVANCED DESIGN OF STRUCTURAL STEELWORK:
AN INTRODUCTION TO STRUCTURAL FIRE ENGINEERING By: Mr. RS Walls – March 2016 –
Van Der Sterr Building Fire – Feb 2015 Discussion: 1. What happened? 2. How could the fire have been prevented? 3. How did the structure respond to the fire and what temperatures were experienced? 4. How could the damage caused by the fire have been reduced? 5. How should structural engineers design for such fires? https://www.youtube.com/watch?v=tSQtNR-mPbQ
Introduction to the course Course overview: 1. An introduction to fire engineering 2. Discussions regarding structural fire design and approaches 3. Fire curves and heat transfer equations 4. Characterising the behaviour of steelwork at elevated temperatures 5. Member design at elevated temperatures 6. Advanced design methods
Does this structure need fire protection? Cape Town International Airport
Intro to the course Code basis: Performance-based member design and heat transfer equations: the soon to be released SANS 10162-1 fire design annex will be used, which is based upon the Canadian CSA S16 code. Prescriptive design: British and European design guides. Fire loads, parametric curves and material behaviour: Eurocode (EN) documents. The Eurocodes are the most technically advanced suite of design documents in the world, covering numerous aspects in relation to fire engineering.
Intro to the course - What can I do with what I have been taught? - Examples - Tutorials - Additional reading - Exam
LECTURE 1 Introduction to Fire Engineering
Course Example
Course Example
Steps to Design 1. Classify a building and determine the required fire rating of elements. 2. Do a quick design using prescriptive methods, or 3. Generate a parametric time-temperature fire curve according to the building properties, or use a standard fire curve for a specific period of time (less “scientific” but often a better bench mark). 4. Calculate the heat transfer and maximum temperature of the steelwork. 5. Determine the steel mechanical properties at the elevated temperature. 6. Design the members using simple calculations according to the Canadian steel code. 7. Do performance-based design using more advanced methods for composite floors or entire frames.
What is Structural Fire Engineering
?
What is Structural Fire Engineering "The application of scientific and engineering principles, rules (codes), and expert judgement, based on an understanding of the phenomena and effects of fire and of the reaction and behaviour of people to fire, to protect people, property and the environment from the destructive effects of fire." (IFE, 2014)
What is a Fire and when does it Influence a Building? A fire can be described as the “process in which substances combine chemically with oxygen from the air and typically give out bright light, heat, and smoke” (Oxford, 2014) Smoke normally kills people, not buildings fall over.
What is a Fire and when does it Influence a Building? Interstate Bank Building, Los Angeles (1988) This building burnt for 4 hours causing $50million damage. Four floors were destroyed.
Parque Central East Tower, Caracas (2004) Fire burned for 24 hours across 17 stories. Up to 100 firefighters inside the building. Firefighting stopped after 12 hours due to concerns regarding structural collapse.
The Effects on Society - In South Africa there were 410 deaths due to fires in 2011, which is significantly up from 192 deaths in 2000 and 226 in 2001 (FPASA, 2013). - A total of 37,721 recorded fires in the country caused an estimated damage of R2.1bn during 2011, which does not even include indirect costs such as lost production.
The Role of the Structural Engineer The Commission of the European Communities outlines the general requirements of construction works subjected to fire conditions as: • “the load bearing capacity of the construction can be assumed for a specific period of time, • the generation of and spread of fire and smoke within the works are limited, • the spread of fire to neighbouring construction work is limited, • occupants can leave the works or be rescued by other means, • the safety of rescue teams is taken into consideration.” (CEC, 1988)
How to Protect Steelwork -
Protective boards Spray-on products Intumescent paints Concrete encasement, fire screens, other systems.
The Cost of Protection Fire Protection Costing 60min Fire Rating
120min Fire Rating
Costing
Costing
Intumescent Paint – Nullifire S707-60
Mass
Ap/V
Steel Cost
(kg/m):
(m-1):
(R/m):
UC 152x152x23
23.3
304
R 652.40
R772.79
R418.30 Not possible
UC 203x203x46
46.2
205 R 1,293.60
R503.48
R559.30
R2,075.90
R1,018.51
UC 305x305x137
137
106 R 3,836.00
R472.92
R855.41
R1,779.01
R1,557.74
Section
Vermiculite
Intumescent Paint – Nullifire S707-120
Spray
Vermiculite Spray
R760.81
The Cost of Protection Fire Protection Product Thickness 60min Fire Rating DFT
Section
Mass
Ap/V
(kg/m):
(m-1):
(mm): Intumescent Vermic. Paint – Nullifire Spray S707-60
120min Fire Rating DFT (mm): Intumescent Vermic. Paint – Nullifire Spray S707-120
UC 152x152x23
23.3
304
1.313
25
Not possible
35
UC 203x203x46
46.2
205
0.593
23
4.519
42
UC 305x305x137
137
106
0.329
19
2.486
45
Active Protection Protection Method Public fire brigade Sprinkler
Probability of fire being out of control 10-1 2 x 10-2
High standard fire brigade, combined with alarm system
≤ 10-2 to 10-3
Both sprinkler and high standard residential fire brigade
≤ 10-4
Compartmentation Compartmentation involves the division of fire zones to limit the spread of fire. This is explicitly considered in building codes such as SANS 10400 by limiting the maximum division area allowed in various occupancy categories. Dividing walls must be fire rated and retain their integrity during a fire. Fire walls, fire doors and other methods are commonly used for this. Get your compartmentation right and you have solved most of the problem.
Compartmentation “In tall multi-storey buildings, it can be advisable for each storey to be a separate compartment capable of resisting burn-out. This can protect occupants who might have to exit past the fire storey when a fire is well developed, and can also protect fire fighters who might have to work on storeys immediately above or below a fire when it is well developed.” (BS 9999 - BSI, 2008)
Compartmentation
Structural Fire Engineering Approaches - Prescriptive design - The Yellow Book and Euro-Nomogram - Performance based design
Structural Fire Engineering Approaches
What is Failure?
?
Has this failed?
What is Failure? Various parameters for failure have been identified such as those given in BS 5950 Part 8 for: • Beams: maximum deflection limited to span/20, or for deflections greater than span/30 the rate of deflection must not exceed span2/(9000 x member depth) [mm/min]. • Columns: Failure to support the applied load or a lateral deflection of 120mm. • Insulating materials or floors: objects on the unexposed face must not combust. Temperature on the unexposed side must be limited to 140°C (average) or 180°C (maximum). • Integrity: boundaries required for compartmentation must not allow the passage of smoke or flames from one compartment to another.
Fire Rating Type of occupancy
Stability (min)
Class of occupancy
Single-storey
Double-storey
3 to 10 storey
11 storeys and
Basement in any
building
building
building
more
building
Office
G1
30
30
60
120
120
Hotel
H1
30
60
90
120
120
Dormitory
H2
30
30
60
120
120
Domestic residence
H3
30
30
60
120
120
Detached dwelling house
H4
30
30
60 Not
Not applicable Not
120
Hospitality
H5
30
30
High risk storage
J1
60
90
120
180
240
Moderate risk storage
J2
30
60
90
120
180
Low risk storage
J3
30
30
90
90
120
Parking garage
J4
30
30
30
90
120
applicable applicable
120
Fire Rating EXAMPLE 2: FIRE RESISTANCE RATING Q: Based on SANS 10400 requirements what FRR should be provided for the building shown in Example 1? A: From Table 2.1 the following can be derived: • The building will be used for offices so is Class G1 • The building is between 3 and 10 storeys. • Therefore, a 60 minute fire rating is required for the structure. BUT, 60 MINUTES OF WHAT???
Fire Loading 𝐺𝐺𝑘𝑘 + 𝑄𝑄𝑇𝑇,𝑘𝑘 + 𝛾𝛾𝑄𝑄𝑘𝑘
Gk characteristic permanent load thermal effects due to expansion, contraction or deflection QT,k caused by temperature changes due to the design fire. It can be taken as zero for statically determinate structures or for structures that have sufficient ductility to allow for redistribution of temperature forces before collapse. [Even though these guidelines have generally been sufficient they must be carefully considered in some structures as forces caused in members restrained from expanding can be significant]. 𝛾𝛾 1.0 for storage areas, equipment areas, and service rooms, 0.5 for other occupancies. The Eurocode reduction factors are similar, but vary for some structures. Qk characteristic imposed load
EXAMPLE 3: FIRE LOADING Q: What load should be designed for at the ambient Ultimate Limit State (ULS) and at the Fire Limit State (FLS) for the second floor column on Gridline B3 of the building in Figure 1.1? Assume that the roof may be loaded in the future so the column may carry two full floors above it. A: The loading can be determined as: Permanent / Dead load: 𝐺𝐺𝑘𝑘 = 𝑁𝑁𝑁𝑁. 𝑜𝑜𝑜𝑜 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 × 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑝𝑝𝑝𝑝𝑝𝑝 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 × 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 = 2 × 7.5𝑚𝑚 × 7.5𝑚𝑚 × 3.2𝑘𝑘𝑘𝑘𝑘𝑘 = 360𝑘𝑘𝑘𝑘 Imposed / Live load: 𝑄𝑄𝑘𝑘 = 2 × 7.5𝑚𝑚 × 7.5𝑚𝑚 × 3.4𝑘𝑘𝑘𝑘𝑘𝑘 = 382.5𝑘𝑘𝑘𝑘 ULS Loading = 1.2𝐺𝐺𝑘𝑘 + 1.6 𝑄𝑄𝑘𝑘 = 1044𝑘𝑘𝑘𝑘 FLS Loading = 𝐺𝐺𝑘𝑘 + 𝛼𝛼𝑄𝑄𝑘𝑘 = 360 + 0.5 × 382.5 = 551.3𝑘𝑘𝑘𝑘
LECTURE 2 Fire Curves
What influences the temperature of the braai?
The Fire Triangle
Fire Curves
Gas Temperature (ºC)
1200 1000 800 600 Standard
400
External 200
Hydrocarbon
0 0
50 Time (mins)
100
Fire Curves Standard Fire: θg = 20 + 345 log10 8t + 1 [°C]
Hydrocarbon Fire: 𝜃𝜃𝑔𝑔 = 1080 1 − 0.325𝑒𝑒 −0.167𝑡𝑡 − 0.675𝑒𝑒 −2.5𝑡𝑡 + 20 [°C] External Fire: 𝜃𝜃𝑔𝑔 = 660 1 − 0.687𝑒𝑒 −0.32𝑡𝑡 − 0.313𝑒𝑒 −3.8𝑡𝑡 + 20 [°C] t is the time in minutes
Real Fires
Sprinkler Fire Test Video https://www.youtube.com/watch?v=AH6J-Szo8dw
Curve Comparison Gas Temperature (ºC)
1200 1000 800 600
Standard fires
400 Standard 200
External Hydrocarbon
0 0
50
100 150 Time (mins)
200
Parametric fires
Comments on the Standard Fire It must be understood that the Standard Fire (ISO 834 fire) does not replicate a real fire, does not help in predicting certain failure mechanisms and generally is acknowledged as being technically out-of-date. However, all fire tests, codes and methods are based on it. Hence, it is often just better to use the standard fire as you then at least have a benchmark you are working from, even if the benchmark is arbitrary. Fire officials understand standard fire ratings, not real fire designs. The parametric curves shown below will provide you as the student with an understanding of fire dynamics and fire behaviour, although they are not accepted like the standard fire is.
Eurocode Parametric Fires Fuel controlled Heating phase
Cooling phase tmax
Ventilation controlled
Eurocode Parametric Fires Heating Phase:
𝜃𝜃𝑔𝑔 = 20 + 1325 1 − 0.324𝑒𝑒 −0.2𝑡𝑡∗ − 0.204𝑒𝑒 −1.7𝑡𝑡∗ − 0.472𝑒𝑒 −19𝑡𝑡∗ [°C]
θg = gas temperature in the fire compartment [ºC] t* = t.Γ [hours] (This can be considered as the time period modified by Γ to match the original opening factor of 0.04/1160 utilised in calibration experiments) t = time [h] Γ = (O/b)2/(0.04/1160)2 𝑏𝑏 = 𝜌𝜌𝜌𝜌𝜌𝜌 with 400 ≤ 𝑏𝑏 ≤ 2200 [J/m2s1/2K] – Thermal inertia of the firecell.
Eurocode Parametric Fires Maximum Temperature 𝑡𝑡𝑚𝑚𝑚𝑚𝑚𝑚 = max 0.2 × 10−3 × 𝑞𝑞𝑡𝑡,𝑑𝑑 /𝑂𝑂𝑙𝑙𝑙𝑙𝑙𝑙 ; 𝑡𝑡𝑙𝑙𝑙𝑙𝑙𝑙 [h] t ∗𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑡𝑡𝑚𝑚𝑚𝑚𝑚𝑚 . Γ [h]
Av = area of ventilation openings [m2] h = height of ventilation openings [m] At = total area of enclosure, including openings [m2] 𝑂𝑂 = 𝐴𝐴𝑣𝑣 ℎ𝑒𝑒𝑒𝑒 /𝐴𝐴𝑡𝑡 [m0.5] ℎ𝑒𝑒𝑒𝑒 = ∑𝑖𝑖 𝐴𝐴𝑣𝑣𝑣𝑣 ℎ𝑖𝑖 /𝐴𝐴𝑣𝑣 [m]
Eurocode Parametric Fires Limiting Time 𝑡𝑡 ∗ = 𝑡𝑡. Γ𝑙𝑙𝑙𝑙𝑙𝑙 [h] t ∗𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑡𝑡𝑚𝑚𝑚𝑚𝑚𝑚 . Γ [h]
with Γ𝑙𝑙𝑙𝑙𝑙𝑙 = 𝑂𝑂𝑙𝑙𝑙𝑙𝑙𝑙 /𝑏𝑏 2 / 0.04/1160 where 𝑂𝑂𝑙𝑙𝑙𝑙𝑙𝑙 = 0.1 × 10−3 . 𝑞𝑞𝑡𝑡,𝑑𝑑 /𝑡𝑡𝑙𝑙𝑙𝑙𝑙𝑙
2
Γ𝑙𝑙𝑙𝑙𝑙𝑙 only used with heating the phase, not the cooling phase. 𝑂𝑂𝑙𝑙𝑙𝑙𝑙𝑙 is for the case of large openings where not all the air entering through an opening is used for combustion.
Eurocode Parametric Fires Modifying Γ𝑙𝑙𝑙𝑙𝑙𝑙
When O > 0.04, and qt,d < 75, and b < 1160, then Γlim is modified by: 𝑘𝑘 = 1 +
𝑂𝑂−0.04 0.04
𝑞𝑞𝑡𝑡,𝑑𝑑 −75 75
1160−𝑏𝑏 1160
.
Eurocode Parametric Fires Cooling Phase ∗ ∗ 𝜃𝜃𝑔𝑔 = 𝜃𝜃𝑚𝑚𝑚𝑚𝑚𝑚 − 625 𝑡𝑡 ∗ − 𝑡𝑡𝑚𝑚𝑚𝑚𝑚𝑚 . 𝑥𝑥 if 𝑡𝑡𝑚𝑚𝑚𝑚𝑚𝑚 ≤ 0.5 [h] ∗ ∗ ∗ 𝜃𝜃𝑔𝑔 = 𝜃𝜃𝑚𝑚𝑚𝑚𝑚𝑚 − 250 3 − 𝑡𝑡𝑚𝑚𝑚𝑚𝑚𝑚 𝑡𝑡 ∗ − 𝑡𝑡𝑚𝑚𝑚𝑚𝑚𝑚 . 𝑥𝑥 if 0.5 < 𝑡𝑡𝑚𝑚𝑚𝑚𝑚𝑚 < 2.0 [h] ∗ ∗ 𝜃𝜃𝑔𝑔 = 𝜃𝜃𝑚𝑚𝑚𝑚𝑚𝑚 − 250 𝑡𝑡 ∗ − 𝑡𝑡𝑚𝑚𝑚𝑚𝑚𝑚 . 𝑥𝑥 if 𝑡𝑡𝑚𝑚𝑚𝑚𝑚𝑚 ≥ 2.0 [h]
t* = t.Γ [h] ∗ 𝑡𝑡𝑚𝑚𝑚𝑚𝑚𝑚 = (0.2 × 10−3 × 𝑞𝑞𝑡𝑡,𝑑𝑑 /𝑂𝑂). Γ [h] 𝑥𝑥 = 1.0 if 𝑡𝑡𝑚𝑚𝑚𝑚𝑚𝑚 > 𝑡𝑡lim , or 𝑥𝑥 = 𝑡𝑡𝑙𝑙𝑙𝑙𝑙𝑙 . Γ/t ∗max , if 𝑡𝑡𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑡𝑡lim
Eurocode Parametric Fires Fire Load Densities 𝑞𝑞𝑓𝑓,𝑑𝑑 = 𝑞𝑞𝑓𝑓,𝑘𝑘 . 𝑚𝑚. 𝛾𝛾𝑞𝑞𝑞 . 𝛾𝛾𝑞𝑞𝑞 . 𝛾𝛾𝑛𝑛 [MJ/m2]
m is the combustion factor. For mainly cellulosic materials m = 0.8. 𝛾𝛾𝑞𝑞𝑞 is the partial factor accounting for the risk based on the size of the compartment 𝛾𝛾𝑞𝑞𝑞 is the partial factor accounting for the risk based on the type of occupancy 𝛾𝛾𝑛𝑛 = ∏10 𝑖𝑖=1 𝛾𝛾𝑛𝑛𝑛𝑛 is the differentiation factor taking into account the different active fire-fighting measures available. (It is currently unclear whether this can be applied in South Africa.) qf,k is the characteristic fire load density per unit floor area [MJ/m2]
Fire load densities qf,k [MJ/m2]
Occupancy:
Stand. Deviation:
Average:
Rate of Heat Release
80%
Gumbell
Fire growth
tlim
Fractile:
Alpha:
rate:
[min]:
Dwelling
234
780
948
0.0054782
Medium
20
Hospital (room)
69
230
280
0.018578
Medium
20
Hotel (room)
93
310
377
0.013784
Medium
20
Library
450
1500
1824
0.002849
Fast
15
Office
126
420
511
0.010174
Medium
20
School classroom
85.5
285
347
0.014993
Medium
20
Shopping centre
180
600
730
0.007122
Fast
15
Storage buildings
*
*
*
*
*
*
Theatre (cinema)
90
300
365
0.014243
Fast
15
30
100
122
0.04273
Slow
25
Transport (public square)
Eurocode Parametric Fires Risk of Fire Ignition
25
1.10
Danger of fire activation, 𝛾𝛾𝑞𝑞𝑞
250
1.50
1.00
2500
1.90
1.22
5000
2.00
1.44
10000
2.13
1.66
Compartment Danger of fire floor area Af activation, 𝛾𝛾𝑞𝑞𝑞 [m2]
0.78
Example of occupancies
Artgallery, museum, swimming pool Offices, residence, hotel, paper industry Manufacturer of machinery and engines Chemical laboratory, painting workshop Manufacturer of fireworks or paints
Eurocode Parametric Fires Active Protection
Eurocode Parametric Fires Thermal Inertia λ – Thermal Material:
Conductivity [W/m.K]:
ρ – Density
cp – Specific
[kg/m3]:
Heat [J/kg.k]:
b – Thermal inertia [J/m2s0.5K]:
Brickwork
1.00
2000
1114
1521
CaSi-board
0.069
450
748
151.9
Cerablanket
0.035
128
800
59.9
Gypsum board
0.5
1150
1000
749
Light wt. conc.
1.0
1500
840
1122
Middle wt. conc.
1.0
2000
840
1296
Normal wt. conc.
2.0
2300
900
2034
Structural steel
54.0
7850
425
13422
Wood
0.10
450
1113
223
Eurocode Parametric Fires Thermal Inertia 𝑏𝑏 = 𝑏𝑏 =
𝑐𝑐𝑐𝑐𝑐𝑐 [J/m2s0.5K]
∑ 𝑏𝑏𝑖𝑖 𝐴𝐴𝑖𝑖 ∑ 𝐴𝐴𝑖𝑖
[J/m2s0.5K]
There are ways to consider multiple layers on each boundary.
Eurocode Parametric Fires Comments • Consider multiple “load cases” • Requires feedback from clients and the fire chiefs. • In South Africa these methods are unknown so most people won’t be able to comment on them.
Eurocode Parametric Fires Time Equivalence – Eurocode Formula
?
Eurocode Parametric Fires Time Equivalence – Eurocode Formula • 𝑡𝑡𝑒𝑒 = 𝑘𝑘𝑏𝑏 𝑤𝑤𝑓𝑓 𝑒𝑒𝑓𝑓
[mins]
6.0 0.3 90 0.4−𝛼𝛼𝑣𝑣 4 Ventilation factor: 𝑤𝑤𝑓𝑓 = 0.62 + > 𝐻𝐻𝑟𝑟 1+𝑏𝑏𝑣𝑣 𝛼𝛼ℎ Vertical vent ratio: 𝛼𝛼𝑣𝑣 = 𝐴𝐴𝑣𝑣�𝐴𝐴𝑓𝑓 0.05 ≤ 𝛼𝛼𝑣𝑣 ≤ 0.25 Horizontal vent ratio: 𝛼𝛼ℎ = 𝐴𝐴ℎ�𝐴𝐴𝑓𝑓 𝛼𝛼ℎ ≤ 0.20 Vertical opening factor: 𝑏𝑏𝑣𝑣 = 12.5 1 + 10𝛼𝛼𝑣𝑣 − 𝛼𝛼𝑣𝑣2
Hr is the compartment height [m].
0.5
Eurocode Parametric Fires Time Equivalence – CIB Formula • 𝑡𝑡𝑒𝑒 = 𝑘𝑘𝑐𝑐 𝑤𝑤𝑓𝑓 𝑒𝑒𝑓𝑓
[mins]
Ventilation factor: 𝑤𝑤𝑓𝑓 = Formula:
Term
𝐴𝐴𝑓𝑓
𝐴𝐴𝑣𝑣 𝐴𝐴𝑡𝑡 𝐻𝐻𝑣𝑣 0.5
0.5
b – Thermal inertia (J/m2.s.K)
High
Medium
Low
(> 2500)
(720-2500)
( 𝑡𝑡𝑙𝑙𝑙𝑙𝑙𝑙, 𝑥𝑥 = 1.0 ∗ 0.5 < 𝑡𝑡𝑚𝑚𝑚𝑚𝑚𝑚 ≤ 2.0. ∗ Therefore:𝜃𝜃𝑔𝑔 = 𝜃𝜃𝑚𝑚𝑚𝑚𝑚𝑚 − 250(3 − t ∗max ) 𝑡𝑡 ∗ − 𝑡𝑡𝑚𝑚𝑚𝑚𝑚𝑚 . 𝑥𝑥 x = 1.0 Temperature will return to ambient (20°C) at: t ∗ = 2.048ℎ𝑟𝑟𝑟𝑟, or 𝑡𝑡 = 1.39ℎ𝑟𝑟 = 83𝑚𝑚𝑚𝑚𝑚𝑚 (from substituting 𝜃𝜃𝑔𝑔 = 20°𝐶𝐶 and 𝜃𝜃𝑚𝑚𝑚𝑚𝑚𝑚 in the equation above)
Eurocode Example θg (degC):
Std Fire
1200
Temperature (°C)
1000 800 600 400 200 0 0
20
40
60 Time (min)
80
100
120
Eurocode Example ANSWER 3a: Eurocode Equation 𝑘𝑘𝑏𝑏 = 0.051, by interpolating in Table 3.5 using 𝑏𝑏 = 1849𝐽𝐽/𝑚𝑚2 𝑠𝑠 0.5 𝐾𝐾 Horizontal ventilation: 𝛼𝛼𝑣𝑣 = 𝐴𝐴𝑣𝑣�𝐴𝐴𝑓𝑓 = 31⁄150 = 0.207
Vertical ventilation: 𝛼𝛼ℎ = 𝐴𝐴ℎ�𝐴𝐴𝑓𝑓 = 0⁄150 = 0 Vertical opening factor: 𝑏𝑏𝑣𝑣 = 12.5 1 + 10𝛼𝛼𝑣𝑣 − 𝛼𝛼𝑣𝑣2 = 37.8 Ventilation factor:
6.0 0.3 90 0.4−𝛼𝛼𝑣𝑣 4 𝑤𝑤𝑓𝑓 = 0.62 + > 0.5 𝐻𝐻𝑟𝑟 1+𝑏𝑏𝑣𝑣 𝛼𝛼ℎ 6.0 0.3 90 0.4−0.207 4 = 0.62 + = 0.842𝑚𝑚0.25 4.0 1+37.8×0
Fire load: 𝑒𝑒𝑓𝑓 = 𝑞𝑞𝑓𝑓,𝑑𝑑 = 540.5𝑀𝑀𝑀𝑀/𝑚𝑚2 Thus, the equivalent standard fire time is: 𝒕𝒕𝒆𝒆 = 𝒌𝒌𝒃𝒃 𝒘𝒘𝒇𝒇 𝒆𝒆𝒇𝒇 = 𝟐𝟐𝟐𝟐. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
Eurocode Example ANSWER 3b: CIB Equation 𝑘𝑘𝑐𝑐 = 0.065, but interpolating in Table 3.5 using 𝑏𝑏 = 1849𝐽𝐽/𝑚𝑚2 𝑠𝑠 0.5 𝐾𝐾
𝑤𝑤𝑓𝑓 =
𝐴𝐴𝑓𝑓
𝐴𝐴𝑣𝑣 𝐴𝐴𝑡𝑡 𝐻𝐻𝑣𝑣
0.5 0.5
150 = 31 × 500 × 1.5650.5
0.5
= 1.077𝑚𝑚0.25
𝒕𝒕𝒆𝒆 = 𝒌𝒌𝒄𝒄 𝒘𝒘𝒇𝒇 𝒆𝒆𝒇𝒇 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 × 𝟏𝟏. 𝟎𝟎𝟎𝟎𝟎𝟎 × 𝟓𝟓𝟓𝟓𝟓𝟓. 𝟓𝟓 = 𝟑𝟑𝟑𝟑. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
LECTURE 3 Behaviour of Steelwork at Elevated Temperatures
Thermal Elongation
Δl/l - Steel Elongation (x10-3)
∆𝑙𝑙/𝑙𝑙 = 14 × 10−6 𝜃𝜃𝑎𝑎 . 20 18 16 14 12 10 8 6 4 2 0 0
200
400
600
800
θa - Steel temperature (°C)
1000
1200
Specific Heat ca = 600 J/kgK.
ca - Specific Heat (J/kgK)
3000 2500 2000 1500 1000 500 0 0
200 400 600 800 θa - Steel temperature (°C)
1000
1200
Thermal Conductivity 𝝀𝝀𝒂𝒂 = 𝟒𝟒𝟒𝟒 W/mK λa - Thermal conductivity (W/mK)
60 50 40 30 20 10
0 0
200
400
600
800
θa - Steel temperature (°C)
1000
1200
Reduction factors - Steelwork
Reduction factors - Steelwork Steel temp. θsteel
Reduction factors at temperature θsteel (Tsteel) relative to the value of fy or Ea at 20°C ky,θ – Yield
kE,θ – Young’s
ku,θ – Ultimate
k0.2p,θ – Class 4
Strength
Modulus
Strength
Sections
20 °C
1.000
1.000
1.000
1.250
1.000
100 °C
1.000
1.000
1.000
1.250
1.000
200 °C
1.000
0.900
0.807
1.250
0.890
300 °C
1.000
0.800
0.613
1.250
0.780
400 °C
1.000
0.700
0.420
1.000
0.650
500 °C
0.780
0.600
0.360
0.780
0.530
600 °C
0.470
0.310
0.180
0.470
0.300
700 °C
0.230
0.130
0.075
0.230
0.130
800 °C
0.110
0.090
0.050
0.110
0.070
900 °C
0.060
0.068
0.038
0.060
0.050
1000 °C
0.040
0.045
0.025
0.040
0.030
1100 °C
0.020
0.023
0.013
0.020
0.020
1200 °C
0.000
0.000
0.000
0.000
0.000
(Tsteel)
kp,θ – Prop. limit
NOTE: For intermediate values of the steel temperature linear interpolation may be used
Reduction factors – Bolts & Welds
Section Factor - Ap/V 𝐴𝐴𝑚𝑚 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹: 𝑉𝑉 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑝𝑝𝑝𝑝𝑝𝑝 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 𝑜𝑜𝑜𝑜 𝑡𝑡𝑡𝑡𝑡 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑝𝑝𝑝𝑝𝑝𝑝 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
Section Factor - Ap/V
Heat Transfer – Unprotected Steelwork ∆𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
𝑎𝑎
𝜌𝜌 𝑐𝑐𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝐴𝐴𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑚𝑚 /𝑉𝑉
𝑇𝑇𝐹𝐹 − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ∆𝑡𝑡
Coefficient of heat transfer: 𝑎𝑎 = 𝑎𝑎𝑐𝑐 + 𝑎𝑎𝑟𝑟
Radiative heat transfer coefficient: 𝑎𝑎𝑟𝑟 =
5.67×10−8 𝜖𝜖𝐹𝐹 𝑇𝑇𝐹𝐹 −𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
4 𝑇𝑇𝐹𝐹4 − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
- transfer of electromagnetic energy Convective heat transfer coefficient: ac = 25 W/m°C - transportation of molecular energy
Heat Transfer – Unprotected Steelwork Emissivity factor: εF Type of Assembly
εf
Column, exposed on all sides
0.7
Floor beam: imbedded in the concrete floor slab, with only bottom flange of beam exposed to fire
0.5
Floor beam, with concrete slab resting on top flange of beam: - Flange width : beam depth ratio ≥ 0.5
0.5
- Flange width : beam depth ratio < 0.5
0.7
Box girder and lattice girder
0.7
Heat Transfer – Protected Steelwork 𝜌𝜌𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 When:𝑐𝑐𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝐴𝐴𝑝𝑝 /𝑉𝑉
> 2𝑑𝑑𝑝𝑝 𝜌𝜌𝑝𝑝 𝑐𝑐𝑝𝑝 (the thermal capacity of the
insulation is much less than that of the steel and can be ignored) 𝑘𝑘𝑝𝑝 ∆𝑇𝑇𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 = 𝑇𝑇𝐹𝐹 − 𝑇𝑇𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 ∆𝑡𝑡 𝜌𝜌𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑𝑝𝑝 𝐴𝐴𝑝𝑝 /𝑉𝑉 Otherwise (when the thermal capacity of the insulation must be considered): 𝑘𝑘𝑝𝑝 ∆𝑇𝑇𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 = 𝑑𝑑𝑝𝑝
𝑐𝑐𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑇𝑇𝐹𝐹 − 𝑇𝑇𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 ∆𝑡𝑡 𝑐𝑐 𝜌𝜌 𝑑𝑑 𝜌𝜌𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑝𝑝 𝑝𝑝 𝑝𝑝 + 2 𝐴𝐴𝑝𝑝 /𝑉𝑉
Heat Transfer – Protected Steelwork cp
is the specific heat of the coating (J/kgºC)
ρp
is the coating density (kg/m3)
dp
is the coating thickness (m)
kp
is the thermal conductivity of the coating (W/mºC)
Heat Transfer – Protected Steelwork QUESTION: What temperature does the column on Gridline B3 and primary beam on Gridline B reach for the fire curve generated in EXAMPLE 4. Generate time-temperature curves to show the behaviour. Consider the following for each member: a. The member being bare steel. b. Protection by being boxed out with 12mm gypsum board.
Course Example
Heat Transfer – Protected Steelwork ANSWER – BARE STEEL: Column of GL. B3 - Unprotected For a UC 203x203x46 bare column: Ap / V = 205m-1 (Section 5.2, Euro-Nomogram). 𝑀𝑀 𝐷𝐷
=
𝜌𝜌 𝐴𝐴𝑝𝑝 / 𝑉𝑉
=
7850𝑘𝑘𝑘𝑘/𝑚𝑚3 205𝑚𝑚−1
equation)
= 38.3𝑘𝑘𝑘𝑘/𝑚𝑚2 (useful conversion
εF = 0.7 for a column exposed on all sides.
Heat Transfer – Protected Steelwork ANSWER – BARE STEEL: Time increment to be used: 5 seconds. At the first time increment: TF = 45.1°C, from Example spreadsheet. 5.67 × 10−8 𝜖𝜖𝐹𝐹 4 𝑎𝑎𝑟𝑟 = 𝑇𝑇𝐹𝐹 − 𝑇𝑇𝑠𝑠4 𝑇𝑇𝐹𝐹 − 𝑇𝑇𝑠𝑠 =
∆𝑇𝑇𝑠𝑠 =
5.67×10−8 ×0.7 45.1−20
𝑎𝑎
𝜌𝜌𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑠𝑠 𝐴𝐴𝑚𝑚 /𝑉𝑉
= 0.136°𝐶𝐶
45.14 − 204 = 0.006276 25 + 0.00832 𝑇𝑇𝐹𝐹 − 𝑇𝑇𝑠𝑠 ∆𝑡𝑡 = 45.1 − 20 × 5 600 × 38.3
Heat Transfer – Protected Steelwork ANSWER – CLADDED COLUMN: Perimeter of the boxed out section 𝐴𝐴𝑝𝑝 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 4𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 × (0.203𝑚𝑚) −1 = = = 138.1𝑚𝑚 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 5.88 × 10−3 𝑚𝑚2 𝑉𝑉
For gypsum board:
cp = 1700 J/kgºC. ρp = 800kg/m3. dp = 0.012m. kp = 0.20W/m°C
Heat Transfer – Protected Steelwork ANSWER – CLADDED COLUMN: 𝑐𝑐𝑆𝑆 𝑀𝑀 𝐷𝐷
= 22980 < 2𝑑𝑑𝑝𝑝 𝜌𝜌𝑝𝑝 𝑐𝑐𝑝𝑝 =32640. Thus, equation (4.16) must be used. 𝑘𝑘𝑝𝑝
∆𝑇𝑇𝑆𝑆 = 𝑑𝑑
𝑝𝑝
𝜌𝜌
𝑇𝑇𝐹𝐹 −𝑇𝑇𝑆𝑆
𝑐𝑐𝑝𝑝 𝜌𝜌𝑝𝑝 𝑑𝑑𝑝𝑝 2
𝑐𝑐𝑠𝑠 𝐴𝐴𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑝𝑝 /𝑉𝑉
∆𝑡𝑡 =
0.20 45.1−20 0.012 600× 7850 +1700×800×0.012 138.1
2
5 = 0.0494°𝐶𝐶
The remainder of the calculations have been carried out as shown in the graph below. Maximum temperatures – Unprotected: 837.6°C - Protected: 582.4°C
Heat Transfer – Protected Steelwork 1000
EN 1-1-2 Fire
900
Unprotected
Temperature (°C)
800
Protected
700 600 500 400 300 200 100 0 0
20
40
60 Time (min)
80
100
120
LECTURE 4 - Member Design
Member Design - Prescriptive Design - The Yellow Book - Euro-Nomogram - Performance-based design / Rational Design
The Critical Temperature
The Yellow Book
Member Design to SANS 10162-1 Tensile Resistance The tension resistance of a member is determined in the same way that it is done an ambient temperature, but with the reduced yield strength of members.
Member Design to SANS 10162-1 Compressive Resistance 𝐶𝐶𝑟𝑟 𝑇𝑇 = 1 + 𝜆𝜆 𝑇𝑇
2𝑑𝑑𝑑𝑑 −1/𝑑𝑑𝑑𝑑 𝐴𝐴𝑓𝑓 (𝑇𝑇) 𝑦𝑦
𝐾𝐾𝐾𝐾 𝑓𝑓𝑦𝑦 (𝑇𝑇) = 𝜆𝜆 𝑇𝑇 = 2 𝑟𝑟 𝜋𝜋 𝐸𝐸(𝑇𝑇) d n
𝑓𝑓𝑦𝑦 (𝑇𝑇) 𝑓𝑓𝑒𝑒 (𝑇𝑇)
0.6 1.34 for general steelwork
Member Design to SANS 10162-1 Compressive Resistance Cr(T)/A - Failure Stress (MPa)
350 20°C 300 500°C
250 200
800°C
150 100 50 0 0
50
100
150
KL/r - Slenderness Ratio
200
Member Design to SANS 10162-1 Bending Resistance 𝑀𝑀𝑟𝑟 𝑇𝑇 = 0.12𝑀𝑀𝑝𝑝 (𝑇𝑇) + 0.88𝑀𝑀𝑝𝑝 (𝑇𝑇) 1 −
𝜁𝜁(𝑇𝑇) 0.12𝑀𝑀𝑝𝑝 (𝑇𝑇) 0.5 𝑀𝑀𝑐𝑐𝑐𝑐 (𝑇𝑇)
Mp(T) is the plastic moment at elevated temperature, T Mcr(T) is the elastic critical load at elevated temperature T, given by: 𝑀𝑀𝑢𝑢 𝑇𝑇 =
𝜔𝜔2 𝜋𝜋 𝐿𝐿
𝐸𝐸 𝑇𝑇 𝐼𝐼𝑦𝑦 𝐺𝐺 𝑇𝑇 𝐽𝐽 +
𝑇𝑇 + 800 𝜁𝜁 𝑇𝑇 = ≤ 2.4 500
𝜋𝜋𝜋𝜋(𝑇𝑇) 2 𝐼𝐼𝑦𝑦 𝐶𝐶𝑤𝑤 𝐿𝐿
Member Design to SANS 10162-1 Bending Resistance Mr - Bending resistance (kNm)
450
20°C
400
500°C
350
800°C
300 250 200 150 100 50 0 0.0
2.0
4.0 6.0 8.0 KL - Effective length (m)
10.0
Member Design to SANS 10162-1 Combined Axial Force and Flexure Beam-columns are to be designed in the same manner as done at ambient temperature with the reduced member capacities as defined above.
Member Design to SANS 10162-1 Example 8: QUESTION: Check the design capacity of the UC 203x203x46 column in the fire compartment at the fire limit state. Obtain maximum temperatures from Example 5, considering both the protected and unprotected member.
Member Design to SANS 10162-1 ANSWER: Obtain the reduction factor for fy and E from Table 3.5 by interpolation. UC 203x203x46 Details:
Maximum Temperature:
ky,θ:
kE,θ:
fy((T) (MPa):
E(T) (GPa):
Bare steel
837.6°C
0.091
0.082
32.3MPa
16.4GPa
Protected – 12mm gypsum
582.4°C
0.525
0.361
186.4MPa
72.2GPa
𝜆𝜆 837.6°C =
𝐾𝐾𝐾𝐾 𝑟𝑟
𝑓𝑓𝑦𝑦 (𝑇𝑇)
𝜋𝜋2 𝐸𝐸(𝑇𝑇)
=
1.0×4000 51.2
32.3 𝜋𝜋2 ×16400
= 1.10
Member Design to SANS 10162-1 1 − 𝐶𝐶𝑟𝑟 837.6°C = 1 + 𝜆𝜆 𝑇𝑇 2𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝐴𝐴𝑓𝑓𝑦𝑦 𝑇𝑇 = 1 + 1.10 2×0.6×1.34 −1/(0.6×1.34) × 5880 ×
= 72.4𝑘𝑘𝑘𝑘
32.3
Using the same equations: 𝐶𝐶𝑟𝑟 582.4 = 358𝑘𝑘𝑘𝑘 for the protected member. Thus, the column is insufficient for the 551kN load calculated earlier for the FLS.
Member Design to SANS 10162-1 However, note that for the Eurocode the effective length can be reduced to 0.5 of the original because of the cool columns above and below acting as fixities. For the top floor of a building a factor of 0.7 should be used. If an effective length of 0.5L is used the resistance will increase to around 674kN for the protected member. This would be sufficient for resistance. The Canadian Code advises that effective lengths should be left the same as at ambient temperature.
LECTURE 5 ADVANCED DESIGN
What factors would you need to consider if designing a structure in a real fire situation?
Global structural behaviour considerations and effects 1. Restraint effects. High forces, buckling, localised plastification and other such factors can be results from the restrained expansion of members. 2. P-Δ Effects. The nonlinear behaviour and large deformations that result in a structures in a severe fires causes second-order effects. In simplified analysis methods this is typically neglected. 3. Internal force redistribution. In natural fire conditions force redistribution often occurs as parts of a structure yield. 4. Membrane effects, as discussed below for composite floors.
Global structural behaviour considerations 5. Alternative loads paths are created whereby weakened members transfer loads to adjacent structural elements, or sometime even secondary elements not designed to carry loads (e.g. internal brickwork). 6. Lateral-torsional buckling which might not be an issue at ambient temperatures can become an issue in severe fires. 7. Joint behaviour becomes very complex as large deformations can cause pinned connections to become almost fixed, and local buckling can occur.
ADVANCED MODELLING
JOINT BEHAVIOUR
Full-Scale Testing – The Cardington Tests
SLAB DESIGN IN FIRES - The BRE Cardington full-scale fire tests, amongst others, showed that composite slabs have significant inelastic reserves of strength when subjected to fire conditions.
Has this failed?
SLAB DESIGN IN FIRES
SLAB DESIGN IN FIRES
Outline of Technical Details The Slab Panel Method (SPM) by Clifton or the Tensile Membrane Behaviour model by Bailey function in similar ways when designing composite floors in severe fires. At ambient temperatures the manner that loads are transmitted through a composite building involves: The slab -> Secondary beams -> primary beams -> columns. In a fire this load transfer path becomes: The slab panel -> primary supporting beams -> columns.
Outline of Technical Details In the event of a fully developed fire, the SPM performs as follows: 1. The slab and the unprotected secondary beams may undergo considerable permanent deformation. 2. The primary support beams and columns undergo much less permanent deformation compared to that within the panel. 3. The load-carrying capacity and the integrity of the floor system are preserved. 4. Both local and global collapse is prevented.
Calculation Procedure 1. Determine the total floor load (dead and live) at the fire limit state. 2. Calculate the fire severity based on the fire loads, and building characteristics. Convert this to an equivalent standard fire time. 3. Determine the temperature of the slab concrete, rebar and steel beams. Various equations and tables are provided for this purpose. 4. Evaluate the yieldline load-carrying moment capacity in each direction for hogging and sagging. 5. Calculate the yieldline load-carrying capacity based on the actual support conditions but ignoring tensile membrane action.
Calculation Procedure 6. Determine the deflections that would occur in the slab panel as these have a significant influence on the magnitude of the tensile membrane action. 7. Find the tensile membrane enhancement factor and multiply this by the yieldline load carrying capacity to determine the total capacity of the floor. 8. Check that the total capacity of the floor is greater than the total floor load. 9. Check the shear capacity 10. Ensure that sufficient reinforcement is provided to preserve integrity (no cracks opening up in the floor).
SPM Example
SPM Design By adjusting the rebar in the slab we can obtain a solution that satisfies both the bending and shear requirements of the system, as detailed below. Basically the summary of the calculations is: Design floor load: 4.9kPa Floor capacity: 5.04kPa. OK Design shear load: 18.38kN.m Floor shear capacity: 24.75kN/m. OK Rebar: Mesh ref. 395 (8mm bars at 200 each way) Y12-200 interior support bars in the X-direction Y12-450 trough bars The rebar specific is similar (possibly a bit higher) to that which would be found in a typical composite floor when fire resistance is not considered. Hence, additional capacity has been obtained in fire conditions with significantly less passive protection that typically used.
Software for Fire Design 1. TSLAB: This is a spreadsheet that has been developed in the UK by the Steel Construction Institute (SCI) 2. MACS+: MACS+ is a package freely distributed by Arcelor Mittal and its name stands for Membrane Action of Composite Structures in Case of Fire (Vassart & Zhao, 2012). 3. VULCAN: This is a commercially available software package for the design of structures in fire. It has been produced by Sheffield University over many years through numerous research projects. 4. SAFIR: SAFIR is an advanced finite element structural engineering software package specifically developed for structures in fire. It has been developed by Franssen at the University of Liege.
Software for Fire Design 5. Arcelor Mittal - Fire Calculations Download Centre: A number of free software packages are provided on ArcelorMittal’s website http://amsections.arcelormittal.com. These modules cover aspects such as composite design, column design, beam design etc. 6. ABAQUS or DIANA: ABAQUS and DIANA are general purpose, powerful, finite-element programs that can consider the nonlinear behaviour of structures.
Does this structure need fire protection? Cape Town International Airport
Other Interesting Fire Topics
https://www.youtube.com/watch?v=gmPOIriMiyU
Tutorials to be handed in
Pg. 2-6: Fire resistance ratings Pg. 3-13: Fire curves
THANK-YOU QUESTIONS? 1400 1200 Temperature (ºC)
1000 Steel. O=0.02 Fire. O=0.02 Steel. O=0.10 Fire. O=0.10 Steel. O=0.20
800 600 400 200 0 0
20
40 Time (min)
60
