Interfloor overlap on wooden beams: calculation for prefabricated loads and permissible deflection

Recommendation points

• Requirements for interfloor floors
• Types of floor slabs
• Types of fasteners and connections of wooden beams
• About timber floor beams
• Calculation of load-bearing beams
• Calculation of loads
• Calculation of the allowable deflection

If you are planning the construction of a two-story or one-story house, but with a basement or an attic, it is necessary to correctly calculate and erect interfloor floors. Consider the stages and nuances of flooring on wooden beams and calculate the cross-sections of beams that provide sufficient strength.

The device of interfloor ceilings needs special attention, because made “by eye”, they may not withstand the loads falling on them and collapse, or require unnecessary, not motivated costs. Therefore, you need to comprehensively consider and calculate one or more possible options. The final decision can be made by comparing the cost or availability of materials.

Requirements for interfloor floors

Interfloor floors must withstand constant and variable loads, that is, in addition to their own weight, they must withstand the weight of furniture and people. They must be sufficiently rigid and not allow exceeding the maximum deflection, provide sufficient noise and heat insulation.

Before starting work, we advise you to familiarize yourself with the materials set out in SNiP II-25-80 (SP 64.13330.2011), there is a lot of useful information.

Specific loads from furniture and people for living quarters are taken in accordance with the standards. However, if you plan to install something massive, for example, a 1000 l aquarium or a natural stone fireplace, this must be taken into account..

The stiffness of the beams is determined by calculation and is expressed in the allowable bending per span. The permissible bending depends on the type of floor and coating material. The main limiting deflections, determined by SNiP, are given in table 1.

Table 1

Structural elements	Limit deflections in parts of span, no more
1. Beams of intermediate floors	1/250
2. Beams of attic floors	1/200
3. Covers (except for valleys):
a) purlins, rafter legs	1/200
b) cantilever beams	1/150
c) trusses, glued beams (except for cantilevers)	1/300
d) plates	1/250
e) lathing, flooring	1/150
4. Bearing elements of valleys	1/400
5. Panels and half-timbered elements	1/250
Notes: 1. In the presence of plaster, the deflection of floor elements only from a long-term temporary load should not exceed 1/350 of the span. 2. In the presence of a building lift, the maximum deflection of glued beams may be increased to 1/200 of the span.

Please note that ceramic tile or concrete screed flooring that is prone to cracking can further increase the allowable deflection requirements, especially with sufficiently long spans..

To reduce the loads on the beams, if possible, they should be placed parallel to short walls, with the same pitch. Maximum span length when overlapping with wooden beams – 6 m.

Types of floor slabs

By purpose, overlappings are divided into:

• interfloor;
• attic;
• basement (basement).

Features of their design are in the permissible loads and the device of steam and heat insulation. If the attic is not intended for living or storing massive objects, variable loads when calculating the deflection can be reduced to 50-100 kg / m².

Thermal insulation between two residential floors may seem excessive, but sound insulation is a desirable parameter for most, and this is achieved, as a rule, with the same materials. It should be taken into account that attic and basement floors need a thicker layer of thermal insulation material. The film material for vapor barrier in the attic floor should be located under the layer of insulation, and in the basement – above it. To prevent the occurrence of dampness and damage to structures by fungus, all rooms should be equipped with ventilation.

Overlap options: 1 – board board; 2 – vapor barrier; 3 – thermal insulation; 4 – sparse flooring; 5 – boards; 6 – floor covering

The design of the floors can also be different:

• with open and hidden beams;
• with various types of supporting beams;
• with different filling materials and floor sheathing.

The hidden beams are sewn on both sides and are not visible. Open – protrude from the ceiling and serve as decorative elements.

The figure below shows what the structure of the attic floor overlap with a panel board and with a filing of boards can be.

a – with board roll; b – with a filing of boards; 1 – plank floor; 2 – plastic film; 3 – insulation; 4 – vapor barrier; 5 – wooden beams; 6 – cranial bars; 7 – board roll; 8 – finishing; 9 – boarding from boards

Types of fasteners and connections of wooden beams

Depending on the design and material of the bearing walls, wooden beams are attached:

• in the nests provided for in brick or block masonry, deepening a bar or log at least 150 mm, and a board at least 100 mm;
• on the shelves (ledges) provided in brick or block masonry. It is used if the wall thickness of the second floor is less than that of the first;
• into the cut grooves in the log walls to a depth of at least 70 mm;
• to the bar of the upper strapping of the frame house;
• to metal supports-brackets fixed to the walls.

1 – support on a brick wall; 2 – solution; 3 – anchor; 4 – insulation with tar paper; 5 – wooden beam; 6 – support on a wooden wall; 7 – bolt

If the length of the beam is not enough, you can lengthen it by connecting (splicing) along the length using one of the well-known methods using wooden pins and wood glue. When choosing the type of connection, be guided by the direction of the load application. It is desirable to strengthen the spliced ​​beams with metal overlays.

a – compression; b – stretching; c – bend

About timber floor beams

In construction, beams are used with rectangular, circular or partially circular sections. The most reliable are rectangular lumber, and the rest are used in the absence of a bar or for reasons of economy, if such materials are available on the farm. Glued wood materials are even more durable. Glued laminated timber or I-beams can be installed on spans up to 12 m.

The most inexpensive and popular type of wood is pine, but other coniferous species are also used – larch, spruce. Spruce is used to make ceilings in summer cottages, small houses. Larch is good for the construction of premises with high humidity (sauna, pool in the house).

Materials also differ in grade, which affects the bearing capacity of the beams. Grades 1, 2 and 3 (see GOST 8486–86) are suitable for floor beams, but grade 1 for such a structure can be unnecessarily expensive, and grade 3 is better for small spans.

Calculation of load-bearing beams

To determine the cross-section and spacing of the beams, it is necessary to calculate the load on the floor. The collection of loads is performed according to the methodology and taking into account the coefficients set forth in SNiP 2.01.07–85 (SP 20.13330.2011).

Calculation of loads

The total load is calculated by summing the constant and variable loads, determined taking into account the standard coefficients. In practical calculations, they are first set by a certain structure, including a preliminary layout of beams of a certain section, and then corrected based on the results obtained. So in the first step, sketch out all the layers of the “pie” overlap.

1. Own specific gravity of the floor

The specific gravity of the floor is made up of its constituent materials and is divided by the horizontal total length of the floor beams. To calculate the mass of each element, you need to calculate the volume and multiply by the density of the material. To do this, use table 2.

table 2

Material name	Density or bulk density, kg / m3
Asbestos-cement sheet	750
Basalt wool (mineral)	50-200 (from the degree of compaction)
Birch	620-650
Concrete	2400
Bitumen	1400
Drywall	500-800
Clay	1500
Chipboard	1000
Oak	655-810
Spruce	420-450
Reinforced concrete	2500
Expanded clay	200-1000 (from the coefficient of foaming)
Expanded clay concrete	1800
Solid brick	1800
Linoleum	1600
Sawdust	70-270 (from fraction, wood species and moisture content)
Parquet, 17 mm, oak	22 kg / m2
Parquet, 20 mm, panel board	14 kg / m2
Foam concrete	300-1000
Styrofoam	60
Ceramic tiles	18 kg / m2
Roofing material	600
Wire mesh	1.9-2.35 kg / m2
Pine	480-520
Carbon steel	7850
Glass	2500
Glass wool	350-400
Plywood, glued	600
Cinder block	400-600
Plaster	350-800 (from the composition)

For wood materials and waste, the density depends on the moisture content. The higher the humidity, the heavier the material.

Partitions (walls) also belong to permanent loads, the specific weight of which is approximately 50 kg / m².

2. Variable load

The furnishings of the room, people, animals – all these are variable loads on the floor. According to the table. 8.3 SP 20.13330.2011, for residential premises the normative distributed load is 150 kg / m².

3. Total load

The total load is not determined by simple addition, it is necessary to take the reliability coefficient, which, according to the same SNiP (clause 8.2.2), is:

• 1.2 – with a specific gravity less than 200 kg / m²;
• 1.3 – with a specific gravity of more than 200 kg / m².

4. Calculation example

As an example, let’s take a room 5 m long and 3 m wide. Every 600 mm of length we put beams (9 pcs.) Of pine with a section of 150×100 mm. We will overlap the beams with a 40 mm thick board and lay 5 mm thick linoleum. From the side of the first floor, we will sew the beams with 10 mm plywood, and inside the ceiling we will lay a layer of mineral wool 120 mm thick. No partitions.

1 – beam; 2 – board; 3 – insulated linoleum 5 mm

Calculation of the constant specific load on the area of ​​the room (5 x 3 = 15 m²) is given in Table 3.

Table 3

Material	Volume, m3	Density, kg / m3	Weight, kg	Specific load, kg / m2
Beam (pine)	9 x 0.15 x 0.1 x 3.3 = 0.4455	500	222.75	14.85
Board (pine)	15 x 0.04 = 0.6	500	300	20.0
Plywood	15 x 0.01 = 0.15	600	90	6.0
Linoleum	15 x 0.005 = 0.075	1600	120	8.0
Minvata	15 x 0.12-0.405 = 1.395	one hundred	139.5	9.3
Total:				58.15
Considering k = 1.2				70

Variable load – 150 x 1.2 = 180 kg / m².

Total load – 70 + 180 = 250 kg / m².

Design load on the beam (qr) – 250 x 0.6 m = 150 kg / m (1.5 kg / cm).

Calculation of the allowable deflection

We accept the permissible deflection of the interfloor floor – L / 250, i.e. for a three-meter span, the maximum deflection should not exceed 330/250 = 1.32 cm.

Since the beam with both ends lies on the support, the calculation of the maximum deflection is carried out according to the formula:

• h = (5 x qp x L4) / (384 x E x J)

Where:

• qр – design load on the beam, qр = 1.5 kg / cm;
• L – beam length, L = 330 cm;
• E – modulus of elasticity, E = 100,000 kg / cm² (for wood along the grain according to SNiP);
• J – moment of inertia, for rectangular bar J = 10 x 153/12 = 2812.5 cm⁴.

For our example:

• h = (5 x 1.5 x 3304) / (384 x 100000 x 2812.5) = 0.82 cm

The result obtained has a margin of 60% compared to the permissible deflection, which seems excessive. Consequently, the distance between the beams can be increased by reducing their number and repeat the calculation..

In conclusion, we suggest watching a video on calculating the floor over wooden beams using a special program: