Design of steel beams in houses

NOTE: You can now design Universal Beams on line

There are two values that govern the design of steel beams: (1) the given moment, M, must be less than the beam's moment capacity, PHIM, and (2) limiting the deflection leads to a value for the required moment of inertia, RIX, which must be less than the beam's IX.

We require a database of four values, DESCR, PHIM, IX and KGPM. The database is sorted on self weight SW so that the first satisfactory section found is the lightest. The data is obtained from tables usually provided by the supplier. To set up the database, simply process the next twenty lines. You may, of course, edit the data. If you add or delete a member, please make sure that you add or delete all values with the same index number. Reprocess all lines after editing. However, there is no need to reprocess the lines after the first run if the data is unchanged.

Enter your custom values at lines marked ****, then process all lines. The word "process" means to select (highlight) the relevant text and click the spinning globe in the Engineers' Compendium window. To work with the cue card, you need MATHSERV, which you may download here.

All units must be consistent, ie. kN and m. Therefore, IX should be converted to m4 (IX=IX/10^6) and E should be expressed in kN/m2 (200*10^6). To avoid having to multiply a very small number with a very large number, we leave IX unconverted at 10^6mm4 and express:

When the beam carries a uniformly distributed load, W in kN/m, the relevant formulas are:

M = W*MLF*L^2/8 and DEFL = 5*W*DLF*L^4/(384*E*IX), where MLF and DLF are load factors for moment and deflection respectively, L is the span in m, and the product E*IX is in kNm2.

Often the loads are not uniformly distributed but are concentrated point or distributed loads. In these cases, two equivalent uniformly distributed loads can be calculated, one for moment and one for deflection (request the cue card convert.doc).

Our first task is to find the applied loads W and their load factors. We need to refer to AS1170.1 SAA Loading Code Part 1: Dead and live loads and load combinations. Some relevant values are:

1.Houses
1.1 General        1.5 kPa or 1.8 kN concentrated
1.2 Balconies      3.0 kPa or 1.5 kN/m along the edge
1.3 Stairs and landings    3.0 kPa or 1.8 kN concentrated
1.4 Parking, drives    3.0 kPa or 13.0 kN concentrated

2.Residential and apartment buildings
2.1 Communal, fixed seating  3.0 kPa
2.2 Communal, open space  5.0 kPa
2.3 Balconies      as for houses
2.4 Bedrooms, private rooms  2.0 kPa

Flooring        .25 (timber) to .50 kPa (fibre cement)
Timber floor joists    FLW^2/25 (FLW is floor load width in m)
Ceiling        .22 to .38 kPa
Concrete slabs      2.50 kPa per 100 thick

Superimposed loads (comprise walls, if any, and roof loads on walls):

Metal roof incl. Ceiling  0.40 kPa
Tiled roof incl. Ceiling  0.90 kPa
Studwalls incl. Lining    0.40 kN/m
2.4 m high brick veneer   4.75 kN/m
2.4 m high rendered brick  5.35 kN/m

With all loads collated and the database set up, enter the span and deflection criteria and process the following equations. You will note that the formula for deflection is modified to return a required value for moment of inertia, RIX.

We now loop through the entire database to find the first available section that satisfies both the strength and deflection requirements. Highlight the whole block from For I… to Next I and click the spinning globe: