Fire Resistance of Materials & Structures - Analysing the Steel Structure

Fire Resistance of Materials & Structures
3rd Homework - Steel Structure
Date of Submission
2016
Submitted by
Seyed Mohammad Sadegh Mousavi
836 154
Submitted to
Prof. R. Felicetti
Prof. P. G. Gambarova
Dr. P. Bamonte
Structural Assessment & Residual Bearing
Capacity, Fire & Blast Safety
Civil Engineering for Risk Mitigation
Politecnico di Milano
[ 3 r d H o m e w o r k - M o d e l l i n g o f f i r e s c e n a r i o ]

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Politecnico di Milano – Lecco Campus
Prof. R. Felicetti & Prof. P. G. Gambarova & Dr. P. Bamonte
Seyed Mohammad Sadegh Mousavi (836154)
Fire Resistance of Materials and Structures
Prof. R. Felicetti, Prof. P.G. Gambarova and Dr. P. Bamonte
3rd Homework - Steel structure
The figures below show the plan and the section of a library room, whose structural steel members are to be checked
in fire conditions (in terms of bearing capacity, R criterion).
The required rate is 120 minutes under the fire scenario assessed in Homework 2 (Parametric Fire)

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The structural scheme of the beam can be assumed as follows:
(balanced load, no interaction between the beam and the column)
The two spans are L1 = L2 = 4 (1 + L/100)
The upper floors are subjected to the following loads:
· permanent load: Gk = 5.00 kN/m2
· variable load: Qk = 2.00 kN/m2 (office not open to public)
· snow load: Qsk = 1.20 kN/m2
1. Design of the beam and the column at room temperature
a) design the beam capacity at the ULS and the check the deflection at the SLS (d ≤ L1/250 in the rare combination)
b) design the column for its buckling resistance
2. Design the beam fire protection (boards) for the required fire resistance under the quasi-permanent load
combination and assuming a three-side exposure (concrete deck on top)
suggested steps: design load under fire
ultimate load of the beam at time = 0
ductility class
global failure or just a critical section?
increased capacity of the critical sections by the adaptation factors
degree of utilization of the structure (or the critical section)
critical temperature
protection design & final check
3. Design the column fire protection
for the required fire resistance under the quasi- permanent
load combination (optional: accounting for the effect of the
thermal elongation of the beam).
suggested steps: design load under fire
thermal elongation of the beam
assessment of the equiv. uniform moment
critical temperature (spreadsheet file)
protection design & final check
If needed, the member cross-sections designed at room temperature may be adjusted in order to meet the required
fire resistance (parts 2 and 3)
L= 21 (U)
𝐿1 = 𝐿2 = 4 × (1 +
𝐿
100
) = 4 × (1 +
21
100
) = 4.84 𝑚

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Figure 1 – Span
i. Design of the beam and the column at room temperature
a. Design of the beam at room temperature
Design the beam capacity at the ULS and the check the deflection at SLS
(d ≤
𝐿1
250
= 0.01936 in the rare combination)
Unit load on the floor:
𝑞 𝑈𝐿𝑆 = 1.35 × 5 (
𝐾𝑁
𝑚2
) + 1.5 × 2 (
𝐾𝑁
𝑚2
) = 9.75 (
𝐾𝑁
𝑚2
)
𝑞 𝑆𝐿𝑆 = 5 (
𝐾𝑁
𝑚2
) + 2 (
𝐾𝑁
𝑚2
) = 7.00 (
𝐾𝑁
𝑚2
)
The unit load on the beam depends on the width of the floor pertaining to it. If we assume bending continuity
between the two bays of the slab, this width is:
2 ×
5
8
× 6 = 7.5 𝑚
Figure 2 – Distance of 2 bays of the slab
Figure 2

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Hence, the distributed load on the beam is:
𝑃 𝑈𝐿𝑆 = 9.75 (
𝐾𝑁
𝑚2
) × 7.5 (𝑚) + 1.35 × 0.5 (
𝐾𝑁
𝑚
) = 73.8 (
𝐾𝑁
𝑚
)
𝑃𝑆𝐿𝑆 = 7.00 (
𝐾𝑁
𝑚2
) × 7.5 (𝑚) + 0.5 (
𝐾𝑁
𝑚
) = 53.00 (
𝐾𝑁
𝑚
)
Where the dead weight of the steel profile has been assessed as 0.5 (
𝐾𝑁
𝑚
) (on the safe side up to IPE 300)
Figure 3 – Permanent & Variable loads on the beam
Flexure Design (ULS):
The bending moment & Shear force diagrams are obtained by the SAP2000.
Figure 4 - Bending Moment Diagram
𝑀 𝐸𝑑 = 216.10 𝐾𝑁. 𝑚
𝑉𝐸𝑑 = 223.25 𝐾𝑁
Check
Applied Design Moment: MEd= puls*L2
/8 = 216.10 kN.m
Resisting Design Moment: MRd= fyd . Wpl

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Steel grade is S355 with its mechanical properties:
E=210 Gpa=210000000 (
𝐾𝑁
𝑚2)
𝑓𝑦,𝑘 = 355000 (
𝐾𝑁
𝑚2
)
𝑓𝑦,0 =
𝑓𝑦,𝑘
𝛾 𝑀0
= 355000 (
𝐾𝑁
𝑚2
)
Where
𝛾 𝑀0= Material Safety factor, applied to charactrsitic value of strength of section of any class, recommended value
is 1.0.
𝜀 = √
235
𝑓𝑦,𝑘
= √
235
355
= 0.814
For design at ULS, requirement that bending moment resistance of the steel section has to be equal or
greater that design bending moment value has to be satisfied:
𝑀 𝐸𝑑 ≤ 𝑀 𝑅𝑑
For the following, Steel profile section should be chosen and check it with the ULS requirements. On one hand, for
the beam section, best choice is IPE profile that it’s characterized with small width of flange versus height of the
section. Also thickness of web and flange are small and so their area as well as weight.
On the other hand, IPE is basically used for the beams in order to high uniaxial bending moment capacity. They
have large moment of inertia, therefore they are very effective under bending moment and in limiting vertical
deflections, which is very important for horizontal elements such as beams.
𝑊𝑝𝑙 =
216.10 × 100
35.5
= 608.74 𝑐𝑚3
Steel section profile chosen is IPE 300 with the following properties:
IPE 300
𝑊𝑝𝑙 628.4 cm3
h 300 mm
b 150 mm
A 53.81 cm2
r 15 mm
𝑡𝑓 10.7 mm
𝑡 𝑤 7.1 mm
𝐼 𝑥 8356 cm3
Figure 5 - IPE 300 Properies
Classification is to identify the extent to which resistance and rotation capacity of the section is limited by its local
buckling resistance.
Resisting Design Moment:
MR,d= fyd . Wpl = 35.5 * 628.4/100 =223.011 kNm. > MEd = 216.10 kN.m
Table below gives conditions for web and flange. (The ductility class of the chosen profile is checked with respect
to the standard EN-1991)

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Figure 6 - Web & Flange Table (IPE)
Web check:
𝐶 𝑤 = ℎ − 2. 𝑡𝑓 − 2. 𝑟 = 300 − 2 × 10.7 − 2 × 15 = 248.6 𝑚𝑚
𝐶 𝑤
𝑡 𝑤
=
248.6
7.1
= 35.01 < 72 × 𝜀 = 58.58 (𝐶𝑙𝑎𝑠𝑠 1)
Flange Check:
𝐶𝑓 =
𝑏 − 𝑡 𝑤
𝑤
− 𝑟 =
150 − 7.1
2
− 15 = 56.45 𝑚𝑚
𝐶𝑓
𝑡𝑓
=
56.46
10.7
= 5.28 < 9 × 𝜀 = 7.29 (𝐶𝑙𝑎𝑠𝑠 1)
Therefore, The section is classified to class 1. For class 1 cross sections, the total plasticization of the critical cross
section is assumed and the rotation capacity of the plastic hinges allows the formation of a failure mechanism (it is
a benefit for redundant structures).
Check the Bending moment resistance for ULS:
𝑀 𝑅𝑑 = 𝑀 𝑝𝑙,𝑅𝑑 =
𝑊𝑝𝑙. 𝑓𝑦
𝛾 𝑀0
=
628.4 × 35.5
1
= 223.08 𝐾𝑁. 𝑚 > 𝑀 𝐸𝑑 = 216.10 𝐾𝑁. 𝑚
Check Shear force effect:
According to the presence of shear the theoretical plastic resistance moment of the cross section will be reduced.
For small values of shear force, this reduction is so small that it is counter-balanced by strain hardening and may
be neglected. Provided that the design value of the shear force 𝑉𝑠𝑑 doesn’t exceed 50% of the design plastic shear
resistance 𝑉𝑃𝑙,𝑅𝑑 no reduction need to be made in the resistance moments. On the other hand, when 𝑉𝑠𝑑 exceeds
50% of 𝑉𝑃𝑙,𝑅𝑑, the design resistance moment of the cross section should be reduced to 𝑀 𝑉,𝑅𝑑, allowing for the shear
force.

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Design value for shear force 𝑉𝑠𝑑 at each cross section should satisfy the following condition:
𝑉𝐸𝑑 = 5/8 ∗ 4. 84 ∗ 73.8 = 223.245 𝑘𝑁 ≤ 𝑉𝑃𝑙,𝑅𝑑 = 526.33 𝑘𝑁 (𝑆𝑎𝑓𝑒)
Where 𝑉𝑃𝑙,𝑅𝑑is the design plastic shear resistance given by the expression:
𝑉𝑃𝑙,𝑅𝑑 =
𝐴 𝑣 × 𝑓𝑦
√3 × 𝛾 𝑀0
𝐴 𝑣 = 𝐴 − 2 ∙ 𝑏 ∙ 𝑡𝑓 + (𝑡 𝑤 + 2𝑟) × 𝑡𝑓 = 53.81 × 102
− 2 × 150 × 10.7 + (7.1 + 2 × 15) × 10.7
𝐴 𝑣 = 2567.97 𝑚𝑚2
(Rolled I-beam)
𝑉𝑃𝑙,𝑅𝑑 =
2567.97 × 35.5
√3 × 1
= 526.33 𝐾𝑁
𝐼𝑓
𝑉 𝑠𝑑≤0.5𝑉 𝑃𝑙,𝑅𝑑
⇒
𝑉𝐸𝑑
𝑉𝑃𝑙,𝑅𝑑
=
206.6
526.33
= 0.395 < 0.5
So, we can neglect the effect of shear force on our beam.
Check deflection (SLS)
𝛿𝑙𝑖𝑚 ≤
𝐿1
250
≤
484
250
= 1.936 𝑐𝑚
Maximum deflection of beam:
𝛿 𝑚𝑎𝑥 =
3 × 𝑃𝑆𝐿𝑆 × 𝐿4
384 × 𝐸𝐼
=
3 × 53 × 4.844
384 × 𝐸𝐼
= 1.295 𝑐𝑚 < 𝛿𝑙𝑖𝑚 = 1.936 𝑐𝑚 (𝑆𝑎𝑓𝑒)
Maximum Deflection is satisfied.
b. Design of the Column for Buckling resistance
Floor area sustained by the column at eache story:
𝐴𝑖𝑛𝑓𝑙𝑢𝑒𝑛𝑐𝑒 = 7.5 × 2 × (
5
8
) × 4.84 = 45.375 𝑚2
Loads on the columns
All columns should be checked for buckling resistance. Procedure is the same for all the columns, taking into
account different values of actions for the columns on different storeys. Service load is assumed as dominant, thus
snow load is reduced by the coefficient 𝜓0 = 0.5 (Following Fig.4)

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Figure 7 - 𝝍 Factor
Design values of permanent and variable load have been used. Actions on columns on all four storeys
have been computed and listed below:
 permanent load: Gk = 5.00 kN/m2
 variable load: Qk = 2.00 kN/m2 (office not open to public)
 snow load: Qsk = 1.20 kN/m2
(The factor 1.01 is considered as self-weight of the cross section)
Last Floor
𝑁𝑒𝑑,4(𝐾𝑁) = (𝐺 𝑑 + 𝑄𝑠,𝑑) × 𝐴𝑖𝑛𝑓𝑙𝑢𝑒𝑛𝑐𝑒 = 1.01 × (1.35 × 5 + 1.5 × 0.5 × 1.2) × 45.375 𝑚2
3rd
Floor
𝑁𝑒𝑑,3 (𝐾𝑁) = (𝐺 𝑑 + 𝑄 𝑑) × 𝐴𝑖𝑛𝑓𝑙𝑢𝑒𝑛𝑐𝑒 + 𝑁𝑒𝑑,4
2nd
Floor
𝑁𝑒𝑑,2(𝐾𝑁) = (𝐺 𝑑 + 𝑄 𝑑) × 𝐴𝑖𝑛𝑓𝑙𝑢𝑒𝑛𝑐𝑒 + 𝑁𝑒𝑑,3
1st
Floor
𝑁𝑒𝑑,1(𝐾𝑁) = (𝐺 𝑑 + 𝑄 𝑑) × 𝐴𝑖𝑛𝑓𝑙𝑢𝑒𝑛𝑐𝑒 + 𝑁𝑒𝑑,2
Last Floor Ned,4 (KN) 347.12
3rd Floor Ned,3 (KN) 789.53
2nd Floor Ned,2 (KN) 1231.931
1st Floor Ned,1 (KN) 1674.338
Figure 8 - Loads on each storey (Obtained by Excel)
For designing the column is better to use the HE cross sections instead of IPE, due to it has large thickness, width
and height of the flanges of the HE sections is almost the same and also it has the biaxial bending moment resistance
in case of the direction of for e.g. earthquake and etc is not known and larger area, so great axial load capacity. For
the assumption, different profiles assumed for different columns,
𝐶 𝑤
𝑡 𝑤
and
𝐶 𝑓
𝑡 𝑓
ratios are checked in order to determine
class of profile.

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Compression members should be verified against buckling resistance as follows:
𝑁𝑒𝑑
𝑁𝑏,𝑅𝑑
≤ 1
𝑁𝑏,𝑅𝑑 = 𝜒 ∙ 𝐴 ∙
𝑓𝑦
𝛾 𝑀𝐼
(The column Resistance force)
Where:
𝜒 = reduction factor for the relevant buckling mode.
𝛾 𝑀𝐼 = material safety factor, related to buckling strength of column, recommended value is 𝛾 𝑀𝐼= 1.1.
For constant axial compression in members of constant cross section, value of 𝜒 for appropriate non dimensional
slenderness λ may be determined from:
𝜒 =
1
∅ + √∅2 − 𝜆2
≤ 1
∅ = 0.5 × [1 + 𝛼(𝜆 − 0.2) + 𝜆2
]
Where:
𝛼 = Imperfection factor corresponding to the appropriate buckling curve that depends on the geometry of the
section.
Buckling Curve 𝑎0 a b c d
Imperfection factor (𝛼) 0.13 0.21 0.34 0.49 0.76
λ = the non dimensional slenderness for the relevant buckilng mode and can be determined from the formula 
λ = √
𝐴×𝑓𝑦
𝑁 𝑐𝑟
In the following Fig.6, table shows how to assign the buckling curve and imperfection factor.

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Figure 9 – Selection of Buckling curve for a cross section
𝑁𝑐𝑟 =
𝜋2
× 𝐸𝐼
𝑙 𝑐𝑟
2
Where:
𝑁𝑐𝑟: Elestic critical force for the relevant buckling mode.
𝐼𝑐𝑟: Buckling Length (Critical length) & can be equal to height of the column=3.5 m
Calculations should be done twice for each buckling directions. Direction of maximum slenderness, i.e.
minimum 𝜒 is then considered.
E = 210000 N/mm^2
l cr = 3500 mm
h = 220 mm
b = 206 mm
h/b (<1.2) = 1.068
tf (<100 mm) = 25 mm
I y = 106400000 mm^4
I z = 36510000 mm^4
f y = 355 N/mm^2
A = 13130 mm^2
γ M1 = 1.1
y-y z-z
Buckling Curve = b c
α = 0.34 0.49
EI = 2.234E+13 7.6671E+12
N cr = 18002158.43 6177244.4
λ = 0.5088 0.8687
∅ = 0.6820 1.0411
χ = 0.8803 0.6192
N b,Rd (KN)=
N ed (KN)=
N ed/N b,rd =
Safe
2623.8149
1674.34
0.6381
Buckling
HE 200 M
1st Floor Column
E = 210000 N/mm^2
l cr = 3500 mm
h = 200 mm
b = 200 mm
h/b (<1.2) = 1.000
tf (<100 mm) = 15 mm
I y = 56960000 mm^4
I z = 20030000 mm^4
f y = 355 N/mm^2
A = 7810 mm^2
γ M1 = 1.1
y-y z-z
α = 0.34 0.49
EI = 1.196E+13 4.2063E+12
N cr = 9637245.71 3388940.16
λ = 0.5364 0.9045
∅ = 0.7010 1.0817
χ = 0.8677 0.5971
N b,Rd (KN)=
N ed (KN)=
N ed/N b,rd =
Safe
1504.9116
1231.93
0.8186
2nd Floor Column
HE 200 B
Buckling

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Figure 10 – HE Properties cross sections for each floors
In the previous tables (Figure.7), the imperfection factors (h/b) and tf are computed for each cross sections and their
requirements was checked.
Column on the fourth floor is definitely underutilized, but in case of simplicity of implemnetation it was chosen the
same profile section as for the third floor column.
Check the members’ class (Ductility class):
Figure 11 – Check the members’ Class
E = 210000 N/mm^2
l cr = 3500 mm
h = 180 mm
b = 180 mm
h/b (<1.2) = 1.000
tf (<100 mm) = 14 mm
I y = 38310000 mm^4
I z = 13630000 mm^4
f y = 355 N/mm^2
A = 6530 mm^2
γ M1 = 1.1
y-y z-z
α = 0.34 0.49
EI = 8.045E+12 2.8623E+12
N cr = 6481792.19 2306103.57
λ = 0.5980 1.0026
∅ = 0.7465 1.1993
χ = 0.8380 0.5384
N b,Rd (KN)=
N ed (KN)=
N ed/N b,rd =
Safe
1134.6857
789.53
0.6958
3rd Floor Column
HE 180 B
Buckling
E = 210000 N/mm^2
l cr = 3500 mm
h = 180 mm
b = 180 mm
h/b (<1.2) = 1.000
tf (<100 mm) = 14 mm
I y = 38310000 mm^4
I z = 13630000 mm^4
f y = 355 N/mm^2
A = 6530 mm^2
γ M1 = 1.1
y-y z-z
α = 0.34 0.49
EI = 8.045E+12 2.8623E+12
N cr = 6481792.19 2306103.57
λ = 0.5980 1.0026
∅ = 0.7465 1.1993
χ = 0.8380 0.5384
N b,Rd (KN)=
N ed (KN)=
N ed/N b,rd =
Safe
1134.6857
347.12
0.3059
4th Floor Column
HE 180 B
Buckling
Cw (mm)= 134
Cw/tw= 8.93 < 26.85
Cf (mm)= 77.5
Cf/tf= 3.10 < 7.32
Flange & Web Check (HE 200 M)
Class 1
Cw (mm)= 134
Cw/tw= 14.89 < 26.85
Cf (mm)= 77.5
Cf/tf= 5.17 < 7.32
Flange & Web Check (HE 200 B)
Class 1
Cw (mm)= 122
Cw/tw= 14.35 < 58.32
Cf (mm)= 70.75
Cf/tf= 5.05 < 7.29
Flange & Web Check (HE 180 B)
Class 1

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ii. Design the beam fire protection (boards) for the required fire resistance under the quasi-
permanent load combination and assuming a three-side exposure (concrete deck on top)
Design action in fire condition: quasi permanent load combination:
𝐹𝑓𝑖,𝑑 = 𝛾 𝐺𝐴 ∙ 𝐺 𝑘 + ∑ 𝜓2𝑖 ∙ 𝑄 𝑘𝑖
𝑖≥1
Where:
𝛾 𝐺𝐴 = Partial coefficient for permanent actions in accidental conditions. (Sugessted value = 1)
𝜓2𝑖 = Coefficient for quasi-permanent combination of loads. (equal to 0.3)
Figure 12 – Coefficient for Quasi-permanent combination of loads
Therefore, in this case the design load under fire is: 𝑞 𝑓𝑖,𝑑 = 1 × 5 + 0.3 × 2 = 5.6 (
𝐾𝑁
𝑚2)
The uniformly distributed load on the beam is: 𝑃𝑓𝑖,𝑑 = 5.6 × 7.5 = 42 (
𝐾𝑁
𝑚2)
For the IPE 300 (1st
Class) we have checked the ductility class and so it can be concluded that it will be the same
class in fire conditions and checking again the slenderness of the web and flange is not necessary.
We considering Global failure, If n is degree of redundancy of a structure, then n+1 plastic hinges would have to
form in the critical sections of the beam in order for beam to failure. Our beam is one times redundant, so two
plastic hinges to be formed where the moment reaches the plastic resisting moment. The following scheme is
considered for the failure mechanism:
Figure 13 – Location of x
𝑥
𝐿
=
1
[((
𝑀 𝑅𝑑
−
𝑀 𝑅𝑑
+ ) + 1)
0.5
+ 1]

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𝑃𝐶𝑜𝑙 = (
2𝑀 𝑅𝑑
+
𝐿2 ) × [((
𝑀 𝑅𝑑
−
𝑀 𝑅𝑑
+ ) + 1)
0.5
+ 1]
2
Where:
Pcol = Collapse Load (Depending on the coorection factor (κ1, κ2), temperature for the whole beam is assumed as a
single temp and so there is the value of ky,θ for the yield strength reduction for the entire beam.
κ1 = The correction factore accounting for the position of the section, in our case, it’s equal to 0.85 for protected
beam exposed on 3-sides.
κ2 = The correction factore accounting for the position of the section, in our case, it’s equal to 0.85 at the support
and 1 at the mid-span.
Calculation of the resistant bending moments for the 2 sections:
𝑀 𝑅𝑑,𝜃
−
=
𝑘 𝑦,𝜃 . 𝑓𝑦,𝑘 . 𝑊𝑝𝑙
κ1κ2
= 1.384 𝑘 𝑦,𝜃 . 𝑓𝑦,𝑘 . 𝑊𝑝𝑙 (𝑘𝑁. 𝑚)
𝑀 𝑅𝑑,𝜃
+
=
𝑘 𝑦,𝜃 . 𝑓𝑦,𝑘 . 𝑊𝑝𝑙
κ1κ2
= 1.177 𝑘 𝑦,𝜃 . 𝑓𝑦,𝑘 . 𝑊𝑝𝑙 (𝑘𝑁. 𝑚)
𝑀 𝑅𝑑
−
𝑀 𝑅𝑑
+ = 1.176
𝑃𝐶𝑜𝑙 = (
2𝑀 𝑅𝑑
+
𝐿2 ) × [((
𝑀 𝑅𝑑
−
𝑀 𝑅𝑑
+ ) + 1)
0.5
+ 1]
2
= (
2 × 1.177 𝑘 𝑦,𝜃 . 𝑓𝑦,𝑘 . 𝑊𝑝𝑙
4.842 ) × [(1.176 + 1)0.5
+ 1]2
IPE 300
Wpl (cm^3) = 628.4
fy,k (KN/m^2)= 355000
𝑃𝐶𝑜𝑙,𝜃 = 137.22 𝑘 𝑦,𝜃 (
𝑘𝑁
𝑚
)
As a result, when Pfi, Ed = Pcol, the failure of the beam is occuring so that means when 𝑘 𝑦,𝜃 = 0.31
At fire time t=0, the collapse load is equal to Pcol,0=137.22 kN/m, and the applied load is Pfi, Ed=42 kN/m
Utilization factor :
𝜇0 =
𝐴𝑝𝑝𝑙𝑖𝑒𝑑 𝑙𝑜𝑎𝑑
𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑙𝑜𝑎𝑑 𝑎𝑡 𝑡𝑖𝑚𝑒 𝑡 = 0
=
42
137.22
= 0.31
So, the value of the utilization factor is the same as the strength reduction factor 𝑘 𝑦,𝜃.
According to the utilization factor and using it, the critical temperature at which the collapse load equals to the
design fire load can be calculated in the following:
𝜃 𝑎,𝑐𝑟 = 482 + 39.29 × ln (
1
0.9674 × 𝜇0
3.833 − 1) = 𝟔𝟔𝟏. 𝟏𝟗 ℃
i. Thermal analysis of the Unprotected members
∆𝜃 𝑎,𝑡 = 𝑘 𝑠ℎ
𝐴 𝑚
𝑉⁄
𝑐 𝑎 𝜌 𝑎
ℎ̇ 𝑛𝑒𝑡∆𝑡

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Where:
𝑘 𝑠ℎ = Shadow effect correction factor (EN 1993-1-2)
𝐴 𝑚 = Lateral surface area of the member (𝑚2
)
ℎ̇ 𝑛𝑒𝑡 = Net heat flux per unit area (
𝑊
𝑚2)
V = Volume of the member (𝑚3)
𝑐 𝑎 = Specific heat of steel (
𝐽
𝐾𝑔 ℃
)
𝜌 𝑎 = Steel Density (
𝑘𝑔
𝑚3)
Comparing 𝒉 𝒏𝒆𝒕
ℎ 𝑛𝑒𝑡 = 𝛼 𝑐(𝜃𝑐 − 𝜃 𝑚) + 𝜎𝜀 𝑟[(𝜃𝑟 + 273)4
− (𝜃 𝑚 + 273)4]
Where:
𝛼 𝑐= Convective heat transfer coefficient (
𝑊
𝑚2 ℃
)  25 (
𝑾
𝒎 𝟐 ℃
) with no distinction
𝜃𝑐= Gas temperature next to the member (℃)
𝜃 𝑚= Surface temperature of the member (℃)
𝜀 𝑟= Resultant emissivity between the combusted gas and member surface
𝜃𝑟= Radiant temperature of the fire compartment (℃)
𝜎= Boltzman constant (5.77 × 10−8 𝑊
𝑚2 𝐾4)
𝛼 𝑟= Radiative heat transfer coefficient (
𝑊
𝑚2 ℃
)
In general it is assumed 𝜃𝑐 = 𝜃𝑟 = 𝜃𝑔 where 𝜃𝑔is the gas temperature.
𝜀 𝑟 =
1
(
1
𝜀 𝑟
+
1
𝜀 𝑚
− 1)
≅ 𝜀 𝑟 × 𝜀 𝑚 = 1 × 𝜀 𝑚 = 0.5 𝑓𝑜𝑟 𝑐𝑎𝑟𝑏𝑜𝑛 𝑠𝑡𝑒𝑒𝑙, 0.4 𝑓𝑜𝑟 𝑠𝑡𝑎𝑖𝑛𝑙𝑒𝑠𝑠 𝑠𝑡𝑒𝑒𝑙.
𝑁𝑜 𝑎𝑙𝑙𝑜𝑤𝑎𝑛𝑐𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑠ℎ𝑎𝑑𝑜𝑤 𝑒𝑓𝑓𝑒𝑐𝑡.
Section Factor – Calculation
Beam (IPE 300) Exposed on three sides
𝐴 𝑚
𝑉
=
𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑟𝑒𝑐𝑒𝑖𝑣𝑖𝑛𝑔 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑓𝑙𝑢𝑥
𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑠𝑡𝑜𝑟𝑖𝑛𝑔 𝑡ℎ𝑒 ℎ𝑒𝑎𝑡 (𝐴𝑟𝑒𝑎)
𝐴 𝑚 = 2ℎ + 3𝑏 − 2𝑡 𝑤 − 4 (2 −
𝜋
2
) 𝑟 = 1010 𝑚𝑚
𝐴 𝑚
𝑉
=
1.001
0.005381
= 187.8 𝑚−1
The assumption of uniform temperature inside the steel profile for
𝐴 𝑚
𝑉
>30, and for
𝐴 𝑚
𝑉
>300 profiles are very thin,
so the profile temperature is virtually the same as the gas temperature. With the exception of very massive members

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(small
𝐴 𝑚
𝑉
), after 30min 𝜃 𝑎>700°C, and after one hour, the temperature is very high to neglect all the mechanical
strength.
Figure 14 – Shadow Effect
Correction factor for shadow effect for IPE Cross Sections:
𝑘 𝑠ℎ = 0.9 ×
[
𝐴 𝑚
𝑉⁄ ]
𝑏𝑜𝑥
𝐴 𝑚
𝑉⁄
= 0.9 ×
𝐴 𝑚𝑏
𝐴 𝑚
= 0.9 ×
𝑏 + 2ℎ
1010
= 0.9 ×
750
1010
= 0.668
In order to find out the steel temperature evolution in time “Steel Temperature” Excel sheet is used.
Figure 15 – Steel temperature evolution in time
According the 2nd
homework, the value of gas temperature is taken, for the 2 hours fire duration and time dependent
openings.
Specific heat of carbon steel is temperature dependent (phase change at 735°C):
0
25 W/m2
°C
0.70
187.8 1/m
0.668
5 s
steel type (1 = stainless)
coefficient of heat transfer by convection ac =
surface emissivity of the member = resultant emissivity er =
section factor Am / V =
correction factor for the shadow effect ksh =
time interval Dt =
0
1000
2000
3000
4000
5000
6000
0 200 400 600 800 1000 1200
specificheat(J/kg°C)
temperature (°C)
carbon
stainless

of 21
Figure 16 – Specific heat of Carbon & Stainless Steel
Temperature in the steel profile after 2 hours is 𝜃 𝑎 = 1227.9 ℃ that is higher than critical temperature computed
in the previous section. As a result, protection for the beam should be design.
ii. Thermal analysis of the Protected members
Several materials available for using as a fire protection of a member such as: Concrete, gypsum, Plaster, Fiber
gypsum and fire calcium-silicate boards, lightweight plasters, mineral fibre plasters, thin coating films and so on.
In the equation, due to assumption of surface protection temperature is the same as hot gas in the fire compartment
(insignificant thermal resistance compared to the insulation), the factors of thermal exchange at the surface
(convection & radiation) does not exist.

of 21
Figure 17 – Protected steel graph
Assumption: Three sides of the section is exposed to the fire (Concrete slab on the top)
- Section Factor (
𝐴 𝑝
𝑉
)
𝐴 𝑝
𝑉
=
𝐸𝑥𝑝𝑜𝑠𝑒𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 (𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟)
𝑉𝑙𝑜𝑢𝑚𝑒 (𝐴𝑟𝑒𝑎)𝑜𝑓 𝑠𝑡𝑒𝑒𝑙 𝑝𝑟𝑜𝑓𝑖𝑙𝑒
Figure 18 – Section factor
For Hollow encasement reduces the exposed surface:
𝐴 𝑝
𝑉
=
2ℎ + 𝑏
𝑉𝑙𝑜𝑢𝑚𝑒 (𝐴𝑟𝑒𝑎)𝑜𝑓 𝑠𝑡𝑒𝑒𝑙 𝑝𝑟𝑜𝑓𝑖𝑙𝑒
=
2 × 300 + 150
0.005381
= 139.38 𝑚−1
According to design parameters for protected steel members:
 Thermal ductility λp=0.135 W/m°C
 Density ρ=550 kg/m3
 Specific heat of the protection material cp= 1200 J/kg°C
 Moisture content of the protection material
 Protection thickness
 Section Factor
And using thickness of the protection dp= 3 cm. The thermal analysis for IPE300 – protected beam exposed to the

of 21
parametric fire evaluated in Homework 2 has done and excelsheet presented based on a time interval Δt= 15 sec.
The excelsheet is using the following equation that balance the heat flux input (through the conduction in the
protective material) to the stored thermal energy in the beam and the protection layer, assuming that the
temperature of the outer layer of the protection equals the gas temperature.
The parameters extracted from excelsheet (Steel+Temperature) for protected section:
steel type (1 = stainless) 0
conductivity of protection system p = 0.135 W/m°C
specific heat of protection system cp = 1200 J/kg°C
density of protection system p = 550 kg/m3
thickness of protection system dp = 0.030 m
moisture content (by mass) = 0%
section factor for protected members Ap / V = 139.38 1/m
time interval Dt = 15 s
Outputs:
Check:
 Time Domain: θa = θa,cr = 661.9 °C, at time tcr=150 min > treq = 120 min.
 The temperature domain: at time treq =120 min, θa =572.7 °C < θa,cr = 661.9 °C.
 The load domain: at time treq = 120 min, ky,θ =0.627, Pcol,θ= 86.04 kN/m > Pfi,Ed =42 kN/m
iii. Design the column fire protection
For Quasi-permanent load combination, snow load was not taken into account.
𝑞 𝑓𝑖 = 𝐺 𝑘 + 0.3 𝑄 𝑘 = 5.6 𝑘𝑁/𝑚2
Design load for the 2nd
floor columns under fire (Neglecting the snow load)
𝑁𝑓𝑖,𝐸𝑑 = Σ(𝐺 + 𝜓2𝑄) × 𝐴𝑖𝑛𝑓𝑙𝑢𝑒𝑛𝑐𝑒 = [(5 + 0 × 1.2) + 2 × (5 + 0.3 × 2)] × 45.375 = 𝟕𝟑𝟓. 𝟎𝟕𝟓 𝑘𝑁
step # time time steel temp. gas temp.
t t qa qg Dqg ca f Dqa
(s) (min) (°C) (°C) (°C) (J/kg°C) (1) (°C)
0 0 0.0 20.0 20.0 108.8 440 0.799 -9.06
1 15 0.3 20.0 128.8 91.6 440 0.799 -7.39
2 30 0.5 20.0 220.4 77.2 440 0.799 -5.99
3 45 0.8 20.0 297.6 65.1 440 0.799 -4.82
4 60 1.0 20.0 362.7 55.1 440 0.799 -3.85
5 75 1.3 20.0 417.8 46.7 440 0.799 -3.03
6 90 1.5 20.0 464.5 39.7 440 0.799 -2.35
7 105 1.8 20.0 504.2 33.8 440 0.799 -1.77
8 120 2.0 20.0 538.0 28.9 440 0.799 -1.29
9 135 2.3 20.0 567.0 24.9 440 0.799 -0.89
10 150 2.5 20.0 591.8 21.4 440 0.799 -0.55
11 165 2.8 20.0 613.3 18.6 440 0.799 -0.27
12 180 3.0 20.0 631.8 16.2 440 0.799 -0.03
478 7170 119.5 571.0 1172.4 0.3 729 0.482 0.84
479 7185 119.8 571.8 1172.7 0.3 730 0.482 0.83
480 7200 120.0 572.7 1173.0 0.3 730 0.481 0.83

of 21
Last Floor 𝑁𝑒𝑑,4(𝐾𝑁) = 𝐺 𝑘 × 𝐴𝑖𝑛𝑓𝑙𝑢𝑒𝑛𝑐𝑒 = 5 × 45.375 𝑚2
3rd
Floor 𝑁𝑒𝑑,3 (𝐾𝑁) = 𝐺 𝑘 + 0.3𝑄 𝑘 × 𝐴𝑖𝑛𝑓𝑙𝑢𝑒𝑛𝑐𝑒 + 𝑁𝑒𝑑,4
2nd
Floor 𝑁𝑒𝑑,2(𝐾𝑁) = 𝐺 𝑘 + 0.3𝑄 𝑘 × 𝐴𝑖𝑛𝑓𝑙𝑢𝑒𝑛𝑐𝑒 + 𝑁𝑒𝑑,3
Figure 19 – Quasi-Permanent load combination
According to the Hot Buckling spreadsheet file the value of Critical temperature can be found. The process of
calculating buckling strength of the cross sections is same as the steps done for room temperature, with difference
buckling length equal to 0.5 L = 1.75 and decay of mechanical properties (yield strength & modolus of elasticity)
at elevated temperature that should be takein into consideration at design in fire conditions. Critical temperature
for the 2nd
floor: 𝜃 𝑎 = 642 ℃
Figure 20 – HE200B (Hot Buckling Spreadsheet)
i) Thermal analysis of the Unprotected member
Section Factor – Assumption: Exposed on 4 sides
Last Floor Ned,4 (KN) 226.88
3rd Floor Ned,3 (KN) 480.98
2nd Floor Ned,2 (KN) 735.075
Loads on the Columns without snow
profile name = HE 200 B
bending axis (y: strong, z:weak) = z
cross-secton area - A = 7810 mm2
radius of gyration - i = 50.7 mm
buckling length - lf l = 1.75 m
yielding stress at 20°C - fy = 355 N/mm2
material factor - gM,f i = 1.0
imperfection factor - a = 0.52885073
non-dimensional slenderness a t = 0 -  = 0.456
steel temperature - qa = 642.0 °C
ky ,q = 0.369
kE,q = 0.234
hot non-dimensional slenderness - q = 0.572
factor jq = 0.815
reduction factor cf i = 0.717
buckling strength - Nb,f i,q,Rd = 733 kN
design load Nf i,Ed = 735.075 kN
safety margin - Nb,f i,q,Rd / Nf i,Ed = 100%
profile name = HE 200 B
bending axis (y: strong, z:weak) = y
cross-secton area - A = 7810 mm2
radius of gyration - i = 85.4 mm
buckling length - lf l = 1.75 m
yielding stress at 20°C - fy = 355 N/mm2
material factor - gM,f i = 1.0
imperfection factor - a = 0.52885073
non-dimensional slenderness a t = 0 -  = 0.271
steel temperature - qa = 663.0 °C
ky ,q = 0.319
kE,q = 0.197
hot non-dimensional slenderness - q = 0.345
factor jq = 0.651
reduction factor cf i = 0.832
buckling strength - Nb,f i,q,Rd = 735 kN
design load Nf i,Ed = 735.075 kN
safety margin - Nb,f i,q,Rd / Nf i,Ed = 100%

of 21
Figure 21 – HE200B Properties
𝐴 𝑚
𝑉
=
2ℎ + 4𝑏 − 2𝑡 𝑤 − 4 (2 −
𝜋
2) 𝑟
𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑠𝑡𝑜𝑟𝑖𝑛𝑔 𝑡ℎ𝑒 ℎ𝑒𝑎𝑡 (𝐴𝑟𝑒𝑎)
=
1.151
0.00781
= 147.38 𝑚−1
Correction factor for shadow effect for I profiles:
𝑘 𝑠ℎ = 0.9 ×
2𝑏 + 2ℎ
𝐴 𝑚
= 0.9 ×
800
1151
= 0.625
Figure 22 – Steel Temperature spreadsheet
Calculated temp in the steel column after 2 hours is 𝜃 𝑎 = 1227.9 ℃ that is higher than the critical temp, so the
fire protection should be design for the column.
ii) Thermal analysis of the Protected member
(Hollow encasement reduces the exposed surface)
The Section Factor
𝐴 𝑝
𝑉
=
200 × 4 × 10−3
0.00781
= 102.43 𝑚−1
According to design parameters for protected steel members:
 Thermal ductility λp = 0.135 W/m°C
 Density ρ = 550 kg/m3
 Specific heat of the protection material cp = 1200 J/kg°C
 Moisture content of the protection material
 Protection thickness
 Section Factor
And using thickness of the protection dp= 3 cm. The thermal analysis for HE 200 B – protected beam exposed to
the parametric fire evaluated in Homework 2 has done and excelsheet presented based on a time interval Δt= 15
sec.
Wpl (cm^3) = 642.5
h (mm)= 200
b (mm)= 200
A (cm^2)= 78.1
r (mm)= 18
tf (mm)= 15
tw (mm)= 9
HE 200 B
0
25 W/m2
°C
0.70
147.38 1/m
0.625
5 s
coefficient of heat transfer by convection ac =
surface emissivity of the member = resultant emissivity er =
section factor Am / V =
correction factor for the shadow effect ksh =
time interval Dt =

of 21
Figure 23 – Steel Temperature spreadsheet – Protected Member
Oputputs
Checks:
The time domain: θa = θa,cr = 642 °C, at time tcr=175 min > treq = 120 min.
The temperature domain: at time treq =120 min, θa =492.7 °C < θa,cr = 642 °C.
The load domain: at time treq = 120 min, θa =492.7 °C,
 The safety margins are: y-axis: Nb,fi,Rd/Nfi,Ed = 255%
z-axis: Nb,fi,Rd/Nfi,Ed = 223%
0
0.135 W/m°C
1200 J/kg°C
550 kg/m3
0.030 m
0%
102.43 1/m
15 s
section factor for protected members Ap / V =
time interval Dt =
conductivity of protection system p =
specific heat of protection system cp =
density of protection system p =
thickness of protection system dp =
moisture content (by mass) =
step # time time steel temp. gas temp.
t t qa qg Dqg ca f Dqa
(s) (min) (°C) (°C) (°C) (J/kg°C) (1) (°C)
0 0 0.0 20.0 20.0 108.8 440 0.587 -6.58
1 15 0.3 20.0 128.8 91.6 440 0.587 -5.36
2 30 0.5 20.0 220.4 77.2 440 0.587 -4.33
3 45 0.8 20.0 297.6 65.1 440 0.587 -3.48
4 60 1.0 20.0 362.7 55.1 440 0.587 -2.76
5 75 1.3 20.0 417.8 46.7 440 0.587 -2.16
6 90 1.5 20.0 464.5 39.7 440 0.587 -1.66
7 105 1.8 20.0 504.2 33.8 440 0.587 -1.24
8 120 2.0 20.0 538.0 28.9 440 0.587 -0.88
9 135 2.3 20.0 567.0 24.9 440 0.587 -0.59
10 150 2.5 20.0 591.8 21.4 440 0.587 -0.34
11 165 2.8 20.0 613.3 18.6 440 0.587 -0.13
477 7155 119.3 490.4 1172.1 0.3 659 0.392 0.79
478 7170 119.5 491.2 1172.4 0.3 660 0.391 0.79
479 7185 119.8 491.9 1172.7 0.3 661 0.391 0.79
480 7200 120.0 492.7 1173.0 0.3 661 0.391 0.79

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Fire Resistance of Materials & Structures - Analysing the Steel Structure