The bearing capacity of plywood is 20 mm. Characteristics, properties and applications of plywood

So there is a cell with clear dimensions of 50x50 cm, which is planned to be sewn up with plywood with a thickness of h = 1 cm (actually, according to GOST 3916.1-96, the plywood thickness can be 0.9 cm, but to simplify further calculations, we will assume that we have plywood with a thickness of 1 cm), a flat load of 300 kg / m 2 (0.03 kg / cm 2) will act on the plywood sheet. Will stick on plywood ceramic tile, and therefore it is very desirable to know the deflection of the plywood sheet (the calculation of plywood for strength is not considered in this article).

The ratio h/l = 1/50, i.e. such a plate is thin. Since we are technically unable to provide such fastening on supports so that the logs perceive the horizontal component of the support reaction that occurs in the membranes, it makes no sense to consider a plywood sheet as a membrane, even if its deflection is large enough.

As already noted, to determine the deflection of the plate, you can use the appropriate design coefficients. So for a square slab with hinged support along the contour, the calculated coefficient k 1 \u003d 0.0443, and the formula for determining the deflection will be as follows

f = k 1 ql 4 /(Eh 3)

The formula seems to be not complicated and we have almost all the data for the calculation, only the value of the modulus of elasticity of wood is missing. But wood is an anisotropic material and the value of the modulus of elasticity for wood depends on the direction of normal stresses.

Yes, if you believe regulatory documents, in particular SP 64.13330.2011, then the modulus of elasticity of wood along the fibers E = 100000 kgf / cm 2, and across the fibers E 90 = 4000 kg / cm 2, i.e. 25 times less. However, for plywood, the values ​​of the moduli of elasticity are taken not simply, as for wood, but taking into account the direction of the fibers of the outer layers according to the following table:

Table 475.1. Moduli of elasticity, shear and Poisson's ratios for plywood in the sheet plane

It can be assumed that for further calculations it is enough to determine a certain average value of the wood elasticity modulus, especially since the plywood layers have a perpendicular direction. However, this assumption would not be correct.

It is more correct to consider the ratio of elastic moduli as an aspect ratio, for example, for birch plywood b/l = 90000/60000 = 1.5, then the calculated coefficient will be equal to k 1 = 0.0843, and the deflection will be:

f \u003d k 1 ql 4 / (Eh 3) \u003d 0.0843 0.03 50 4 / (0.9 10 5 1 3) \u003d 0.176 cm

If we did not take into account the presence of support along the contour, but calculated the sheet as a simple beam with a width b = 50 cm, a length l = 50 cm and a height h = 1 cm on the action of a uniformly distributed load, then the deflection of such a beam would be (according to the calculated scheme 2.1 of table 1):

f = 5ql 4 / (384EI) = 5 0.03 50 50 4 / (384 0.9 10 5 4.167) = 0.326 cm

where the moment of inertia I \u003d bh 3 /12 \u003d 50 1 3 /12 \u003d 4.167 cm 4, 0.03 50 - bringing the flat load to a linear one, acting across the entire width of the beam.

Thus, support along the contour makes it possible to reduce the deflection by almost 2 times.

For plates that have one or more rigid supports along the contour, the influence of additional supports creating the contour will be less.

For example, if a plywood sheet will be laid on 2 adjacent cells, and we will consider it as a two-span beam with equal spans and three hinged supports, not taking into account the support along the contour, then the maximum deflection of such a beam will be (according to the design scheme 2.1 of Table 2):

f \u003d ql 4 / (185EI) \u003d 0.03 50 50 4 / (185 0.9 10 5 4.167) \u003d 0.135 cm

Thus, laying plywood sheets over at least 2 spans makes it possible to reduce the maximum deflection by almost 2 times even without increasing the thickness of the plywood and without taking into account the support along the contour.

If we take into account the support along the contour, then we have, as it were, a plate with rigid clamping on one side and hinged support on the other three. In this case, the aspect ratio l / b = 0.667 and then the calculated coefficient will be equal to k 1 = 0.046, and the maximum deflection will be:

f \u003d k 1 ql 4 / (Eh 3) \u003d 0.046 0.03 50 4 / (0.9 10 5 1 3) \u003d 0.096 cm

As you can see, the difference is not as significant as with hinged support along the contour, but in any case, an almost twofold decrease in deflection in the presence of a hard jam on one of the sides can be very useful.

Well, now I would like to say a few words about why the elasticity moduli for plywood differ depending on the direction of the fibers, because plywood is such a tricky material in which the directions of the fibers in adjacent layers are perpendicular.

Determination of the modulus of elasticity of a plywood sheet. Theoretical background

If we assume that the modulus of elasticity of each individual layer of plywood depends only on the direction of the fibers and corresponds to the modulus of elasticity of wood, i.e. impregnation, pressing during manufacture and the presence of glue do not affect the value of the elastic modulus, then the moments of inertia for each of the considered sections should first be determined.

In 10 mm plywood, there are usually 7 layers of veneer. Accordingly, each layer of veneer will have a thickness of approximately t = 1.43 mm. In general, the given sections with respect to perpendicular axes will look something like this:

Figure 475.1. The given sections are for a plywood sheet with a thickness of 10 mm.

Then, taking the width b = 1 and b" = 1/24, we get the following results:

I z = t(2(3t) 2 + t(2t 2) + 4 t 3 /12 + 2t(2t 2)/24 + 3t 3 /(24 12) = t 3 (18 + 2 + 1/ 3 + 1/3 + 1/96) = 1985t 3 /96 = 20.67t 3

I x \u003d t (2 (3t) 2 / 24 + t (2t 2) / 24 + 4 t 3 / (12 24) + 2t (2t 2) + 3t 3 / 12 \u003d t 3 (18/24 + 2/24 + 1/72 + 8 + 6/24) = 655t 3 /72 = 9.1t 3

If the moduli of elasticity were the same in all directions, then the moment of inertia about any of the axes would be:

I" x \u003d t (2 (3t) 2 + t (2t 2) + 4 t 3 / 12 + 2t (2t 2) + 3t 3 / 12 \u003d t 3 (18 + 2 + 1/3 + 8 + 1 /4 =43 3 /12 = 28.58t ​​3

Thus, if we do not take into account the presence of glue and other factors listed above, the ratio of the moduli of elasticity would be 20.67/9.1 = 2.27, and when considering the plywood sheet as a beam, the modulus of elasticity along the fibers of the outer layers would be (20.67/28.58)10 5 = 72300 kgf /cm 2 . As you can see, the technologies used in the manufacture of plywood make it possible to increase the calculated value of the elastic moduli, especially when the sheet is deflected across the fibers.

Meanwhile, the ratio of the calculated resistances during bending along and across the fibers of the outer layers (which can also be considered as the ratio of the moments of inertia) is much closer to that determined by us and is approximately 2.3-2.4.

During construction or renovation wooden house using metal, and even more so reinforced concrete floor beams, is somehow not the topic. If the house is wooden, then it is logical to make the floor beams wooden. It’s just that you can’t determine by eye which timber can be used for floor beams and which span to make between the beams. To answer these questions, you need to know exactly the distance between the supporting walls and at least approximately the load on the ceiling.

It is clear that the distances between the walls are different, and the load on the floor can also be very different, it is one thing to calculate the floor if there is an uninhabited attic on top, and it is quite another thing to calculate the floor for the room in which the partitions will be made in the future, stand cast iron bath, bronze toilet and much more. Therefore, take into account all possible options and laying everything out in the form of a simple and understandable table is almost impossible, but I think it will not be very difficult to calculate the cross section of a wooden floor beam and select the thickness of the boards using the example below:

EXAMPLE OF CALCULATION OF A WOODEN FLOOR BEAM

The rooms are different, often not square. It is most rational to fix the floor beams so that the length of the beams is minimal. For example, if the size of the room is 4x6 m, then if you use beams 4 meters long, then the required section for such beams will be less than for beams 6 m long. In this case, the dimensions 4 m and 6 m are conditional, they mean the span length of the beams and not the length the beams themselves. The beams, of course, will be 30-60 cm longer.

Now let's try to determine the load. Typically, floors of residential buildings are calculated for a distributed load of 400 kg/m². It is believed that for most calculations such a load is sufficient, and for the calculation attic floor enough even 200 kg/m². Therefore, further calculation will be carried out for the above load with a distance between the walls of 4 meters.

A wooden floor beam can be considered as a beam on two hinged supports; in this case, the calculation model of the beam will look like this:

1. Option.

If the distance between the beams is 1 meter, then the maximum bending moment:

M max = (q x l²) / 8 = 400x4²/8 = 800 kg m or 80.000 kg cm

Now it is easy to determine the required moment of resistance of a wooden beam

W required \u003d M max / R

where R- design resistance of wood. In this case, the beam on two hinged supports works in bending. The value of the design resistance can be determined from the following table:

Design resistance values ​​for pine, spruce and larch at 12% humidity

And if the material of the beam is not pine, then the calculated value should be multiplied by the conversion factor according to the following table:

Transition factors for other types of wood
according to SNiP II-25-80 (SP 64.13330.2011)

tree species	Coefficient m n for design resistances
	stretching, bending, compression and collapse along the fibers R p , R i , R c , R cm	compression and crushing across the fibers R c90 , R cm90	chipping R sk
Conifers
1. Larch, except for European	1,2	1,2	1,0
2. Siberian cedar, except for the cedar of the Krasnoyarsk Territory	0,9	0,9	0,9
3. Cedar of the Krasnoyarsk Territory	0,65	0,65	0,65
4. Fir	0,8	0,8	0,8
hardwood
5. Oak	1,3	2,0	1,3
6. Ash, maple, hornbeam	1,3	2,0	1,6
7. Acacia	1,5	2,2	1,8
8. Birch, beech	1,1	1,6	1,3
9. Elm, elm	1,0	1,6	1,0
soft deciduous
10. Alder, linden, aspen, poplar	0,8	1,0	0,8
Note: coefficients m n indicated in the table for support structures overhead lines power lines made from larch not impregnated with antiseptics (with a moisture content of ≤25%) are multiplied by a factor of 0.85.

For structures in which the stresses arising from permanent and temporary continuous loads exceed 80% of the total stress from all loads, the design resistance should be additionally multiplied by the factor m d = 0.8. (clause 5.2. in SP 64.13330.2011)

And if you plan the service life of your structure for more than 50 years, then the resulting value of the design resistance should be multiplied by one more coefficient, according to the following table:

Service life factors for wood
according to SNiP II-25-80 (SP 64.13330.2011)

Thus, the design resistance of the beam can be almost halved and, accordingly, the cross section of the beam will increase, but for now we will not use any additional coefficients. If pine wood of the 1st grade is used, then

W required \u003d 80000 / 142.71 \u003d 560.57 cm & sup3

Note: Design resistance 14 MPa = 142.71 kgf/cm². However, to simplify the calculations, you can also use the value of 140, there will not be a big error in this, but there will be a small margin of safety.

Since the cross section of the beam has a simple rectangular shape, then the moment of resistance of the beam is determined by the formula

W required = b x h² / 6

where b- beam width, h- beam height. If the cross section of the floor beam is not rectangular, but, for example, round, oval, etc., i.e. you will use round timber, hewn logs or something else as beams, then you can determine the moment of resistance for such sections using the formulas given separately.

Let's try to determine the required height of the beam with a width of 10 cm. In this case

the height of the beam must be at least 18.34 cm. you can use a beam with a section of 10x20 cm. In this case, you will need 0.56 m³ of wood for 7 floor beams.

For example, if you plan that your structure will stand idle for more than 100 years and at the same time more than 80% of the load will be constant + long-term, then the design resistance for wood of the same class will be 91.33 kgf/cm² and then the required moment of resistance will increase to 876 cm³ and the height of the timber must be at least 22.92 cm.

Option 2.

If the distance between the beams is 75 cm, then the maximum bending moment is:

M max \u003d (q x l & sup2) / 8 \u003d (400 x 0.75 x 4 & sup2) / 8 \u003d 600 kg m or 60000 kg cm

W required \u003d 60000 / 142.71 \u003d 420.43 cm & sup3

but minimally allowable height a beam of 15.88 cm with a beam width of 10 cm, if you use a beam with a section of 10x17.5 cm, then 0.63 m & sup3 of wood will be required for 9 floor beams.

3 Option.

If the distance between the beams is 50 cm, then the maximum bending moment is:

M max = (q x l²) / 8 = (400 x 0.5 x 4²) / 8 = 400 kg m or 40000 kg cm

then the required modulus of the wooden beam

W required \u003d 40000 / 100 \u003d 280.3 cm & sup3

and the minimum allowable beam height is 12.96 cm with a beam width of 10 cm, when using a beam with a section of 10x15 cm, 0.78 m & sup3 of wood will be required for 13 floor beams.

As can be seen from the calculations, the smaller the distance between the beams, the greater the consumption of wood for the beams, but at the same time, the smaller the distance between the beams, the thinner boards or sheet material can be used for flooring. And one more important point- the design resistance of wood depends on the type of wood and the moisture content of the wood. The higher the humidity, the lower the design resistance. Depending on the type of wood, the fluctuations in the design resistance are not very large.

Now let's check the deflection of the beam calculated according to the first option. Most reference books suggest determining the amount of deflection under a distributed load and hinged support of a beam using the following formula:

f=(5q l 4)/(384EI)

- distance between load-bearing walls;
E- elastic modulus. For wood, regardless of species, in accordance with clause 5.3 of SP 64.13330.2011; when calculating for the limit states of the second group, this value is usually taken equal to 10,000 MPa or 10x10 8 kgf/m² (10x10 4 kgf/cm²) along the fibers and E 90 = 400 MPa across the fibers. But in reality, the value of the modulus of elasticity, even for pine, still ranges from 7x10 8 to 11x10 8 kgf / m², depending on the moisture content of the wood and the duration of the load. With a long-term load, according to clause 5.4 of SP 64.13330.201, when calculating the limit states of the first group according to the deformed scheme, it is necessary to use the coefficient m ds = 0.75. We will not determine the deflection for the case when the temporary load on the beam is long, the beams are not treated with deep impregnation before installation, which prevents changes in the moisture content of the wood and the relative humidity of the wood can exceed 20%, in this case, the modulus of elasticity will be about 6x10 8 kgf/m², but remember this value.
I- the moment of inertia, for a rectangular board.

I \u003d (b x h & sup3) / 12 \u003d 10 x 20 & sup3 / 12 \u003d 6666.67 cm 4

f \u003d (5 x 400 x 4 4) / (384 x 10 x 10 8 x 6666.67 x 10 -8) \u003d 0.01999 m or 2.0 cm.

SNiP II-25-80 (SP 64.13330.2011) recommends designing wooden structures so that the deflection for floor beams does not exceed 1/250 of the span length, i.e. allowable maximum deflection 400/250=1.6 cm. We have not fulfilled this condition. Next, you should choose such a section of the beam, the deflection of which suits either you or SNiP.

If you will use glued beams for floor beams LVL(Laminated Veneer Lumber), then the design resistances for such a beam should be determined from the following table:

Design resistance values ​​for bonded laminates
according to SNiP II-25-80 (SP 64.13330.2011)

Calculation for collapse of the supporting sections of the beam is usually not required. But the calculation of strength under the action of shear stresses is not difficult to do here. The maximum shear stresses for the selected design scheme will be in cross sections on the beam supports, where the bending moment is zero. In these sections, the value of the transverse force will be equal to the support reaction and will be:

Q = ql/2 = 400 x 4 / 2 = 800 kg

then the value of the maximum shear stresses will be:

T= 1.5Q/F = 1.5 x 800 / 200 = 6 kg/cm²< R cк = 18 кг/см² ,

where,
F- the cross-sectional area of ​​\u200b\u200bthe beam with a section of 10x20 cm;
R ck- design resistance to shearing along the fibers, is determined from the first table.

As you can see, there is a three-fold margin of safety even for a bar with a maximum section height.

Now let's calculate which boards will withstand the calculated load (the calculation principle is exactly the same).

EXAMPLE OF CALCULATION OF FLOOR COVERING

1 option. Floor covering from floorboards.

With a distance between the beams of 1 m, the maximum bending moment:

M max = (q x l²) / 8 = (400 x 1²) / 8 = 50 kg m or 5000 kg cm

In this case, the design scheme for boards, as for a single-span beam on hinged supports, is adopted very conditionally. It is more correct to consider wall-to-wall floorboards as a multi-span continuous beam. However, in this case, you will have to take into account the number of spans and the method of attaching the boards to the logs. If, however, in some areas boards are laid between two lags, then such boards should really be considered as single-span beams and for such boards the bending moment will be maximum. It is this option that we will consider further. Required modulus of boards

W required \u003d 5000 / 130 \u003d 38.46 cm & sup3

since the load is distributed over the entire design area, the flooring from the boards can be conditionally considered as one board 100 cm wide, then the minimum allowable height of the boards is 1.52 cm, with smaller spans the required board height will be even less. This means that the floor can be laid with standard floor boards 30-35 mm high.

But instead of expensive floorboards, you can use cheaper sheet materials, such as plywood, chipboard, OSB.

Option 2. Plywood flooring.

The design resistance of plywood can be determined from the following table:

Design resistance values ​​for plywood
according to SNiP II-25-80 (SP 64.13330.2011)

Since plywood is made of glued layers of wood, the design resistance of plywood should be close to the design resistance of wood, but since the layers alternate - one layer along the fibers, the second across, the total design resistance can be taken as the arithmetic average. For example, for birch plywood brand FSF

R f \u003d (160 + 65) / 2 \u003d 112.5 kgf / m & sup2

then

W required \u003d 5000 / 112.5 \u003d 44.44 cm & sup3

the minimum allowable plywood thickness is 1.63 cm, i.e. plywood with a thickness of 18 mm or more can be laid on the beams with a distance between the beams of 1 m.

With a distance between the beams of 0.75 m, the value of the bending moment will decrease

M max = (q x l²) / 8 = (400 x 0.75²) / 8 = 28.125 kg m or 2812.5 kg cm

required modulus of plywood

W required \u003d 2812.5 / 112.5 \u003d 25 cm & sup3

the minimum allowable plywood thickness is 1.22 cm, i.e. plywood with a thickness of 14 mm or more can be laid on the beams with a distance between the beams of 0.75 m.

With a distance between the beams of 0.5 m, the bending moment will be

M max = (q x l²) / 8 = (400 x 0.5²) / 8 = 12.5 kg m or 1250 kg cm

required modulus of plywood

W required \u003d 1250 / 112.5 \u003d 11.1 cm & sup3

the minimum allowable plywood thickness is 0.82 cm, i.e. plywood with a thickness of 9.5 mm or more can be laid on the beams with a distance between the beams of 0.5 m. 6.5 mm, which is 3 times the allowable deflection. With a plywood thickness of 14 mm, the deflection will be about 2.3 mm, which practically meets the requirements of SNiP.

General note: in general, when calculating wooden structures a bunch of all sorts of correction factors are used, but we decided not to complicate the above calculation with coefficients, it is enough that we took the maximum possible load, and in addition, there is a good margin when choosing a section.

3 Option. Floor covering made of chipboard or OSB.

Generally use chipboard or OSB as floor covering(even rough) along the floor beams is undesirable, and these sheet materials are not intended for this, they have too many shortcomings. Estimated resistance of pressed sheet materials depends too a large number factors, so what value of the calculated resistance can be used in the calculations, no one will tell you.

Nevertheless, we cannot prohibit the use of chipboard or OSB, we can only add: the thickness of chipboard or OSB should be 1.5-2 times greater than for plywood. Floors with failed chipboard had to be repaired several times, and the neighbor recently leveled the wooden floor OSB boards, also complains about failures, so you can take our word for it.

Note: logs can first be supported on the floor beams, and then boards will be attached to the logs. In this case, it is necessary to additionally calculate the cross section of the lag according to the above principle.

The formwork element of the ceiling, which perceives the pressure of concrete and all other loads, is plywood. The above mentioned types of plywood have, depending on the direction of work different meanings for both modulus of elasticity and ultimate flexural strength:
- in floors with low surface requirements f - in floors with higher surface requirements f Plywood deflection (0 depends on the load (floor thickness), the characteristics of the plywood itself (modulus of elasticity, sheet thickness) and support conditions.
Appendix 1 (Fig. 2.65) shows diagrams for the main types of plywood supplied by PERI - birch plywood (Fin-Ply and PERI Birch) and softwood plywood (PERI-Spruce). The diagrams are based on a sheet thickness of 21 mm. In this case, the dotted line marks the areas where the deflection exceeds 1/500 of the span. All lines end when the tensile strength of the plywood is reached. The main diagrams are made for standard sheets operating as multi-span continuous beams (minimum three spans).
For running sizes sheets, the following options for the pitch of the transverse beams are obtained.
Table 2.7

When evaluating deflections during reaming: for birch plywood, the same values ​​for the modulus of elasticity and tensile strength are taken as for the main sheets, since it is not always known in which direction the additional sheets are laid. For softwood plywood
in which these characteristics change sharply when the sheet is rotated.
From the diagram (Fig. 2.65) for birch plywood with 3 or more spans, we find our floor thickness value (20 cm) along the X axis and determine the values ​​for deflections:


For our sheet length, two options are acceptable - either 50 cm or 62.5 cm. Let's dwell on the second option, since it saves on the number of transverse beams. The maximum deflection in this case is 1.18 mm. We look at the diagram for a single-span system. With this scheme, the line for a span of 60 cm ends just at the value of the overlap thickness of 20 cm (ultimate strength of plywood). The deflection in this case is 1.92 mm.
From this it follows that in order to avoid excessive deformations of the extension, either the span of this extension should be limited to 50 cm, or an additional transverse beam should be placed under this extension (the design scheme of a uniformly loaded 2-span beam has the smallest values ​​for deflections, but it has an increased in relation to reference moment to multi-span schemes).
Determination of the span of the transverse beams (step of the longitudinal beams b)
According to the step of the transverse beams chosen in the previous paragraph, we check according to the table corresponding to our type of beams. 2.11 the maximum allowable span of these beams. As mentioned above, these tables are compiled taking into account all design cases, for transverse beams, first of all, moment and deflection.
When choosing the pitch of the longitudinal beams, it must be taken into account that the extreme longitudinal beam is located at a distance of 15-30 cm from the wall. Increasing this size can lead to the following unpleasant results:
- increase and uneven deflections on the consoles of the transverse beams;
- the possibility of overturning of the transverse beams during reinforcement work.
The reduction complicates the control of the uprights and creates the risk of slipping of the transverse beams from the longitudinal ones.
For the same reason, and also taking into account the normal operation of the end of the beam (especially when using truss beams), a minimum beam overlap of 15 cm is assigned on each side. The actual pitch of the longitudinal beams should in no case exceed the allowable value according to Table. 2.11 and 2.12. Remember that the span in the formula for determining the moment is present in the square, and in the deflection formula even in the fourth power (respectively, formulas 2.1 and 2.2).
Example
For simplicity, we choose a rectangular room. internal dimensions 6.60x9.00 m. Floor thickness 20 cm, PERI Birch plywood 21 mm thick and sheet dimensions 2500x1250 mm.
The permissible value for the span of the transverse beams with their step of 62.5 cm can be found from Table. 2.11 for GT 24 truss beams. In the first column of the table we find the thickness of 20 cm and move to the right to the corresponding step of the transverse beams (62.5 cm). We find the maximum allowable span of 3.27 m.
Here are the calculated values ​​of the moment and deflection for this span:
- maximum moment at the moment of concreting - 5.9 kNm (permissible 7 kNm);
- maximum deflection (single-span beam) - 6.4 mm = 1/511 span.
If we set the longitudinal beams parallel to the length of the side of the room, we get:
6.6 m - 2 (0.15 m) = 6.3 m; 6.3:2 = 3.15 m 3.27 m; 8.7:3 = 2.9 m We get three spans with a beam length of 3.30 m (minimum 2.9 + 0.15 + 0.15 = 3.2 m). The transverse beams are less loaded - most often this is already a sign of material overrun.
In some cases, for example, when it is necessary to install formwork around pre-installed large equipment, it is necessary to calculate the beams. In doing so, the following prerequisites should be taken into account. As a design scheme in MULTIFLEX systems, only a single-span articulated beam without consoles is always considered, since when installing formwork and during concreting, we always have intermediate stages where the beams work exactly according to this scheme. For large spans of beams without additional support, buckling is possible even at low loads. Any floor formwork after concreting must be pulled out from under the finished floor, sometimes from a closed room, therefore it is desirable to limit the length of the beams (a problem of weight and maneuverability).
If there are no values ​​in the table, you can still use it. For example, to increase the span, you want to reduce the step of the beams - as a result, you must check the admissibility of the span. For example, they decided to install the beams in increments of 30 cm, the thickness of the ceiling is 22 cm. The calculated load is 7.6 N / m2 according to the table. We multiply this load by the step of the beams: 7.6-0.3 \u003d 2.28 kN / m. We divide this value by one step of the transverse beams, which are present in the table: 2.28: 0.4 \u003d 5.7 ~ 6.1 (load on floors 16 cm thick); 2.28:0.5 \u003d 4.56 - 5.0 (load on floors 12 cm thick).
In the first case, we find for a ceiling thickness of 16 cm and a beam spacing of 40 cm a span of 4.07 m, in the second case, a thickness of 12 cm and a spacing of 50 cm - 4.12 m.
We can take the smaller of the two values ​​minus the difference between these values ​​(taking into account the change in the live load, which is present only in the calculation for the moment), without wasting time on lengthy calculations. V specific example obtained by exact calculation.
4.6 m, and took 4.02 m.