Usually, a beam is placed on two support points in order to be able to absorb a certain load and the load is transferred to the support points. Steel is a strong material, but the loads can also be high. Especially when it comes to many floors, a lot of weight must be transferred to the supports in a safe manner. How can you calculate a steel beam yourself and what are the starting points for obtaining a structurally safe beam?
Why does a beam usually have an H or L shape?
The strength and stiffness of a beam are achieved by the material present in the flange cross-section. Flanges are on the ends, held together with an intermediate web. In principle, the body is also less heavy, because the forces are not actually absorbed by that material.
A beam is normally subjected to a moment due to the load placed on it, which can cause the beam to deform. To prevent the beam from collapsing or bending too much, sufficient steel must be present to absorb the load.
Properties of steel
To test whether steel can withstand a certain load, two properties must be checked. For this purpose, the following two properties are considered:
This forms the ductility of the steel and amounts to 2.1*10^5 N/mm2. When steel is subjected to compression or tension, it can absorb a certain amount of deformation before it collapses. This factor is therefore applicable to check the deformation or deflection. In doing so, the I value or the moment of inertia of the beam is checked.
This is for plain steel FE360 235 N/mm2. This is the limit at which the steel will actually fail. This factor is therefore applicable to testing the strength of the beam. The W-value or the resistance moment of the beam must be checked.
Reading Suggestions: How To Make A Conservatory At Your House Yourself?
Many types of load can be placed on a beam from the floors and walls. This of course depends on what kind of floor it is and what it is used for. For homes you can simply assume the following representative values before taxes:
|Wand 100 mm||2,0 kN/m2||–|
|Wooden floor||1,2 kN/m2||1,75 kN/m2|
|Concrete floor 200 mm||4,8 kN/m2||1,75 kN/m2|
|Concrete floor 225 mm||5,4 kN/m2||1,75 kN/m2|
|Concrete floor 250 mm||6,0 kN/m2||1,75 kN/m2|
Overvoltage and load
Suppose you make a large opening of 2.5 m, where you have to catch a piece of wall and floor from above. In principle, you also have to take some space with you for support in order to have the correct length for the beam from a calculation point of view. Imagine that you have a floor of 225 mm concrete, where you have to absorb 3 meters per linear meter of the beam.
In addition, there is also a wall on the girder of 0.5 meters of old masonry, which must be absorbed after the breakthrough. This means that the following tax will be reduced:
- take 0.4 kN/m for the beam’s own weight as standard, or view a profile and take the actual weight;
- the masonry weighs 0.5 * 20 * 0.1 = 1 kN/m;
- the floor weighs 0.225 * 24 * 3 = 16.2 kN/m;
- the variable load from the floor is 1.75 * 3 = 5.25 kN/m.
In principle, the steel girder can be checked with this data.
Two moments lines apply. The moment is usually greatest in the center of the beam if the load is evenly distributed. How do you calculate that then? Suppose the evenly distributed load Q is large over spring L. Then 0.5*Q*L goes to both supports.
If you want to know how high the moment is in the middle, you look at the half beam, holding the middle, as it were. The support point is 0.5*Q*L large at a distance of 0.5*L and in addition, the actual Q load acts over half the beam at a distance of 0.25*Q*L, i.e.:
- M = moment = 0,5*Q*L*0,5*L – 0,5*Q*L*0,25*L = 0,25*Q*L^2 – 0,125*Q*L^2 = 0,125*Q*L^2.
- Q; representative = 0.4+1+16.2+5.25 = 22.85 kN/m;
- M; representative = 0.125*22.85*2.5^2 = 17.85 km
- Q; calculate = (0.4+1+16.2)*1.2 + 5.25*1.5 = 29.00 kN/m;
- M; reken = 0,125*29,00*2,5^2 = 22,66 kNm
Strength of a steel beam
The resistance moment is used to check the strength of the beam. This is the W-value of a beam which you usually find in a book of tables. Because this concerns strength, some things must be determined with calculation values. The required moment of resistance is calculated as follows:
- W = M;design/yield stress steel = 22.66*10^6 / 235 = 96.4*10^3 mm^3;
- with a HEA100 the W = 72.8*10^3 mm^3 is not sufficient and with a HEA120 the W = 106*10^3 mm^3. In other words, the HEA120 has a UC (Unity Check) value of 96.4 / 106 = 90%.
Deflection of the beam
The moment of inertia is used to check the beam for deflection. Because there is a wall on the floor, the following requirement applies:
- deflection may be a maximum of 0.002*L for beams with walls on top;
- deflection may be a maximum of 0.003*L for beams without crack-sensitive walls.
The formula for deflection is structured as follows:
- deflection = ( 5 * M;representative * L^2 ) / (48 * E * I) = 0.002 * L = 0.002 * 2500 = 5 mm;
- I = 5 * M;representative * L / (48 * E * 0.002) = 5 * 17.85*10^6 * 2500 / (48 * 2.1*10^5 * 0.002) = 1.107*10^4 mm ^4;
- with a HEA160 the I = 1.673*10^4 mm^4;
- So actual deflection is: 5*17.85*10^6*2500^2 / (48 * 2.1*10^5 * 1.637*10^4) = 3.38 mm. In other words, for the deflection, the beam has a UC value of 3.38/5 = 67%.
Due to the deflection requirements, a heavier profile must be used.
Calculate beam choice for steel beam
The beam choice is based on the calculated moment of resistance and moment of inertia. You can find these values in table booklets, the Binas, or profile books. A number of HEA profiles are listed below:
|Profile name||I *10^4 mm^4||W * 10^3 mm^3|
In case of major drastic actions at home, it is always advisable to call in a construction company and constructor to prevent unforeseen problems.