1
Homework 2
Q1 Calculate the section factor of a steel H-section column of dimension 300 x 300mm.
The column is exposed to fire on all four sides. Make calculations for a) box type
protection b) spray-on protection.
Answer:
Geometry of the H-section:
Height: h=300mm
Width: b=300mm
Thickness of flange: T=20mm
Thickness of web: t=8mm
(a) Box-type Protection
Area of the cross-section:
A = 2b × T + ( h − 2T ) × t = 0.01408m 2
Perimeter of the section:
H p = 2(b + h ) = 1.2m
Section factor:
H p A = 1.2 0.01408 = 85.2m −1
(b) Spray-on Protection
Perimeter of the section:
H p = 2(b + h + (b − t )) = 1.78m
Section factor:
H p A = 1.78 0.01408 = 126m −1
2
Q2 Use the step-by-step procedure to calculate the steel temperature of an unprotected
beam exposed to ISO fire. The beam section factor is 200 m-1. Use a conservative heat
transfer coefficient hc = 25 W/m2K and emissivity 0.6. The density of steel is 7850
kg/m3 and the specific heat is 600 J/kgK. Use a time step of 0.5 minutes. Plot the
temperature-time curve of ISO fire and steel temperature.
Answer:
Part of the result is listed below. The whole spreadsheet in .xls format is attached.
Step by step method to solve the steel member temperature
Parameters:
hc = 25 W/m2K
emissivity=0.6
density=7850kg/m3
Specific heat=600J/kgK
section factor=200m(-1)
Change in
Time at ISO fire at Steel Difference in
Time hr h steel
half step half step temperature Temperature
temperature
0 0.25 184.61 20.00 7.54 32.54 164.61 6.82
0.5 0.75 311.56 26.82 12.99 37.99 284.74 13.78
1 1.25 379.28 40.60 17.21 42.21 338.68 18.21
1.5 1.75 425.75 58.81 20.98 45.98 366.94 21.49
2 2.25 461.17 80.31 24.56 49.56 380.86 24.04
2.5 2.75 489.80 104.35 28.09 53.09 385.45 26.07
3 3.25 513.82 130.42 31.66 56.66 383.40 27.67
3.5 3.75 534.52 158.09 35.31 60.31 376.43 28.92
4 4.25 552.70 187.01 39.08 64.08 365.69 29.85
4.5 4.75 568.92 216.86 42.99 67.99 352.06 30.49
5 5.25 583.55 247.35 47.05 72.05 336.19 30.86
5.5 5.75 596.87 278.21 51.27 76.27 318.66 30.96
6 6.25 609.11 309.17 55.65 80.65 299.94 30.81
6.5 6.75 620.43 339.98 60.16 85.16 280.44 30.42
7 7.25 630.94 370.41 64.81 89.81 260.54 29.81
7.5 7.75 640.77 400.21 69.55 94.55 240.56 28.97
8 8.25 650.00 429.19 74.36 99.36 220.81 27.95
8.5 8.75 658.68 457.14 79.21 104.21 201.55 26.76
9 9.25 666.90 483.89 84.06 109.06 183.00 25.43
9.5 9.75 674.68 509.32 88.88 113.88 165.36 23.99
10 10.25 682.08 533.31 93.62 118.62 148.77 22.48
The time-temperature curves of ISO fire and the steel member for the first 30 minutes are
plotted below.
At 30 minutes fire, maximum temperature in steel is almost equal to that of fire and
exceeding 800 oC
3
900.00
800.00
700.00
Temperature (oC)
600.00
500.00
400.00
300.00
200.00 ISO fire
Steel temperature
100.00
0.00
e
.5
.5
.5
.5
12
17
22
27
2
7
5
5
m
4.
9.
14
19
24
29
Ti
Time (min)
Fig. 1 Time-Temperature curve of ISO fire and steel member
Q3 Use the step-by-step procedure to calculate the steel temperature of a protected beam
exposed to ISO fire. The beam is same as Q 2. The beam is protected with 50mm
lightweight insulating material which has thermal conductivity of 0.2 W/mK, and specific
heat 1100 J/kgK and density 300 kg/m3. Plot the temperature-time curve of ISO fire and
steel temperature.
Answer: The time-temperature curve is shown below. The detailed calculation procedure
is attached in Microsoft Excel Worksheet. The maximum temperature in the protected
steel reduces to slightly less than 600oC at 60 minutes ISO fire.
1000.00
900.00
800.00
700.00
Temperature
600.00
500.00
400.00
300.00 ISO fire
200.00
100.00
Steel
temperature
0.00
0.25
4.75
9.25
13.8
18.3
22.8
27.3
31.8
36.3
40.8
45.3
49.8
54.3
58.8
Time(min)
Fig. 2 Fig. 1 Time-Temperature curve of ISO fire and protected steel member
4
Q4 Use the step-by-step procedure to calculate the steel temperature of an unprotected
beam exposed to a parametric fire. The beam section factor is 200 m-1. The fire
compartment is made from lightweight concrete with density 2000 kg/m3, thermal
conductivity of 0.8 W/mK, and specific heat 840 J/kgK . The room is 5 m square and 3 m
high with one window 2.4m wide and 1.5m high. The fuel load is 900MJ/m2 floor area.
Plot the temperature-time curve of the design fire and steel temperature.
Answer:
The parametric fire is defined as below:
The temperature in the heating phase is given by
(
Tg = 1325 1 − 0.324e −0.2 t − 0.204e −1.7 t − 0.427e −19 t + 20
* * *
)
(v b) 2
where t * = t = 8.41 × 10 8 t (v b ) (for ventilation controlled fire)
2
2
(0.04 / 1160)
t = time, in hours,
b = ρc p k e
where ρ = density of boundary of enclosure, in kg/m3,
cp = specific heat of boundary of enclosure, in J/kgK,
ke = thermal conductivity of boundary of enclosure, in W/mK.
Or
( f b) 2
t* = t k = 8.41 × 10 8 t ( f b ) k (for fuel controlled fire)
2
2
(0.04 / 1160)
qt ,d
f = 0.0001 ×
t lim
⎛ v − 0.04 ⎞⎛ q t ,d − 75 ⎞⎛ 1160 − b ⎞
k = 1+ ⎜ ⎟⎜⎜ ⎟⎟⎜ ⎟ for v>0.04, qt,d<75, b<1160
⎝ 0.04 ⎠⎝ 75 ⎠⎝ 1160 ⎠
k = 1, otherwise
A maximum temperature, Tg,max, is reached at tmax, in hours, where
Fuel controlled fire
⎛ qt ,d ⎞
t max = greater ⎜⎜ 0.0002 × ; t lim ⎟⎟
⎝ v ⎠
and
Ventilation controlled
qt,d =Fire Load per enclosure area, valid only if 50 ≤ qt ,d ≤ 1000 (MJ/m2)
5
Aw h
v = opening factor, v = (in m0.5)
At
tlim = Table 1 below.
Table 1: Growth rate for buildings
Type of compartment Fire growth rate tlim
(min.)
Residential Medium 20
Offices Medium 20
Schools Medium 20
Hospitals Medium 20
Hotels Medium 20
Library Fast 15
Shopping centre Fast 15
Theatre (cinema) Fast 15
Transport (public space) Slow 25
The fire curve for the cooling phase is a linear one, given by
(
Tg = Tg ,max − 625 t * − t max
*
)x for *
t max ≤ 0.5
Tg = Tg ,max − 250(3 − t )(t − t
*
max
* *
max x) for 0.5 < t max
*
<2
Tg = Tg ,max − 250(t − t x )
* *
max for *
t max ≥2
where
( v b) 2
= t max = 8.41 × 10 8 t max (v b )
* 2
t max 2
(16)
(0.04 / 1160)
qt ,d
t max = 0.0002 ×
v
x = 1.0 if tmax > tlim
t lim
8.41 × 10 8 (v b )
2
= *
if tmax = tlim
t max
For this problem,
A f = 5 × 5 = 25m 2
At = 5 × 5 × 2 + 4 × 5 × 3 = 110m 2
qt ,d = 900 × 25 110 = 204.545MJ / m 2 , satisfying 50 ≤ qt ,d ≤ 1000 (MJ/m2)
Aw = 2.4 × 1.5 = 3.6m 2
6
Aw h
v= = 0.04m 0.5
At
Assume ventilation controlled combustion.
t max = 1.023h =61.38min
b = ρCk = 2000 × 0.8 × 840 = 1159.3
t* = t
(v b)2 = 1.001t
(0.04 1160)2
χ = 1.0
when t = t max , t * max = 1.001 × 1.023 = 1.024
Tg ,max = 1325 × (1 − 0.324e −0.2×1.024 − 0.204e −1.7×1.024 − 0.427e −19×1.024 ) + 20 = 947.8
The final expression for the Parametric Fire is
Tg = ⎨
(
⎧1325 1 − 0.324e −0.2 t − 0.204e −1.7 t − 0.427e −19 t + 20 t * < 1.024
* * *
)
⎩ 947.8 − 250(3 − 1.024 )(t * − 1.024 ) t * ≥ 1.024
The Parametric fire curve and the corresponding steel member temperature is calculated
in the attached Excel worksheet.
The time-temperature relationship for the Parametric Fire and the steel member is plotted
below:
1000.00
900.00
Parametric fire
800.00
Steel temperature
700.00
Temperature
600.00
500.00
400.00
300.00
200.00
100.00
0.00
0
9.5
19
28.5
38
47.5
57
66.5
76
85.5
95
105
114
124
133
143
152
162
171
Time (min)
Fig. 3 Time-temperature curves for the parametric fire and the steel member
7
Q5 Repeat Q4 with the beam protected with 25mm and 50mm lightweight insulating
material which has thermal conductivity of 0.2 W/mK, and specific heat 1100 J/kgK and
density 300 kg/m3. Plot the temperature-time curve of the design fire and steel
temperature with different thickness of fire protection.
Answer:
The parametric fire model is the same as Q.4. The only difference comes from the fire
protection. The calculation procedure is in the attached Excel worksheet for 25mm
protection and 50mm protection respectively.
The time-temperature curve for the fire model, the steel member with 25mm protection
and the steel member with 50mm protection are shown below in Fig. 4.
1000.00
900.00
800.00
700.00
Temperature
600.00
500.00
400.00
300.00 Parametric fire
200.00 Steel temperature (50mm protection)
100.00 Steel temperature (25mm protection)
0.00
1 17 33 49 65 81 97 113 129 145 161 177 193 209 225 241 257 273 289 305 321 337
Time (min)
Fig. 4 The time-temperature curves of the Parametric fire and the protected steel members
