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Lecture 4B.5: Calculation Examples

OBJECTIVE/SCOPE

To make designers familiar with simple methods of calculation of fire resistance time and thickness of insulation for columns and beams (respectively steel and composite steel-concrete).

PREREQUISITES

None.

RELATED LECTURES

Lecture 4B.1: Introduction to Fire Safety

Lecture 4B.2: Background to Thermal Analysis

Lecture 4B.3: Background to Structural (Mechanical Fire) Analysis

Lecture 4B.4: Practical Ways of Achieving Fire Resistance of Steel Structures

SUMMARY

Calculation examples are presented for the following:

• Critical temperatures of tension members, beams and columns (Examples 1, 2, 3).
• Moment resistance of composite beam in fire conditions (Example 5).
• Time equivalent of actual fire (Example 6).
• Fire protection to steel beam (Example 4).

The examples use the principles and design equations presented in the preceding lectures.

For

As the performance of a member in tension is equivalent to the basic performance of the steel, it follows that:

By linear interpolation from the above table, the critical temperature q_cr = 590°C

It is assumed in this example that the beam supports a concrete slab, and hence the upper flange remains cooler than the rest of the section. This benefit is taken into account by use of a load multiplier, or kappa factor, k, such that:

k = 0,7 for a beam supporting a concrete slab

Use the same degree of loading as in Example 1.

In this case,

For = 0,5, it follows that:

y(q) = 0,7 x 0,5 = 0,35

By linear interpolation of the strength reductions in Example 1, the critical temperature of the beam, q_cr = 645°C.

It follows that the critical temperature of a beam supporting a concrete floor slab exceeds that of a member in tension, i.e. uniformly heated, by 55°C for the same degree of loading.

It is assumed in this example that the column is restrained against buckling. The load multiplier, k, for columns is 1,2. This value takes into account the influence of high strains in the column at failure in fire conditions.

For , as in previous examples

y(q) = 1,2 x 0,5 = 0,6

By linear interpolation of the strength reductions in Example 1, the critical temperature of the column, q_cr=560°C.

It follows that the critical temperature of a column is less than that of a member in tension by 30°C for the same degree of loading.

From Example 2, the critical temperature of the beam is 645°C. From Lecture 4B.2, the required thickness of fire protection (in metres) is:

d = 0,0083 l_i (A_m/A) {t/(q_cr -140)}^1,3

Where

A_m/A is the section factor of member (m^-1)

l_i is the thermal conductivity of protection material (W/m°C)

t is the fire resistance period (mins)

q_cr is the critical temperature of beam (°C)

In this Example, use the following parameters:

A_m/A = 200m^-1 (typical of IPE beams)

l_i = 0,15W/m°C (typical of many protection materials)

t = 60 mins

q_cr = 645°C

d = 0,0083 x 0,15 x 200 x {60/(645 -140)}^1,3 ×10^-3

= 15,6 mm (say 16 mm)

This Example follows the use of the principles of plastic analysis to calculate the moment resistance of a composite beam in fire conditions. The following properties are assumed:

Cross-section Temperature Stress blocks

Cross-sectional area of web = area of flange

Temperature of top flange = 2/3 x temperature of web and bottom flange

Effective breadth of slab = 1000 mm

Compressive strength of concrete f_c = 30 N/mm² (Note: for short term load in fire conditions g_mc = 1,0)

Yield strength of steel f_y = 235 N/mm²

Critical temperature of beam q_cr = 600°C (assumed)

Neutral axis depth, x_c, in concrete is obtained by equating tension and compression. Hence:

Moment resistance of composite section is obtained by taking moments about the mid-length of the concrete in compression:

M =

+

M = 0,47 A f_y (0,5 h_a + 1,44 h_c - 0,72 x_c)

For the following data:

h_a = 400 mm

h_c = 120 mm

A = 1000 mm²

x_c = in fire conditions

M = 0,47 x 1000 x 235 x (0,5 x 400 + 1,44 x 120 - 0,72 x 5) x 10^-6

= 40,8 kNm

By comparison under normal conditions, using y = 1,0, moment resistance is 74,2 kNm (note: this value may be calculated using the partial safety factors appropriate for normal conditions, as covered in the lecture on Composite Beams).

Therefore  M/M_u = 40,8/74,2 = 0,58 for q_cr = 600°C

But y(600) = 0,47

If y(600) = k ´ M/M_u, it follows that:

k= 0,47/0,58 = 0,81 for composite beams (compare to 0,7 for non-composite beams)

Refer to Lecture 4B.1. Assume that the fire compartment may be characterized by the following parameters. The time equivalent is:

T_e = c w q_f minutes

c = 0,10 for typical compartment properties

w = 1,5 for typical ventilation conditions

q_f = 450 MJ/m² for office buildings

T_e = 0,1 x 1,5 x 450 = 67,5 minutes

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