Plywood is a multilayer construction material, which is made from environmentally friendly raw materials - wood. Namely, it is made of wood veneer. Such veneer is obtained as a result of peeling wood. In this case, the log is first steamed, then sent to a special machine designed for peeling. After that, the resulting veneer is straightened, exposed special treatment and goes to the dryer. The dried veneer is then subjected to a pressing process, after which it is glued together using various adhesives.
Due to the multilayer structure, the quality indicators of the product increase. The thickness and mass of the material in this case is small. For comparison, the strength of a plywood sheet with certain thickness several times stronger than solid wood material. This is due to the fact that the veneer is glued so that the fibers of each layer are perpendicular to each other. Therefore, the strength of plywood products is much higher.
| Nominal plywood thickness, mm | Plywood ply, not less | Sanded plywood | Rough plywood | ||
| Maximum deviation, mm | Variation in thickness | Maximum deviation, mm | Variation in thickness | ||
| Plywood 3 mm | 3 | +0,3/-0,4 | 0,6 | +0,4/-0,3 | 0,6 |
| Plywood 4 mm | 3 | +0,3/-0,5 | +0,8/-0,4 | 1,0 | |
| Plywood 6 mm | 5 | +0,4/-0,5 | +0,9/-0,4 | ||
| Plywood 9 mm | 7 | +0,4/-0,6 | +1,0/-0,5 | ||
| Plywood 12 mm | 9 | +0,5/-0,7 | +1,1/-0,6 | ||
| Plywood 15 mm | 11 | +0,6/-0,8 | +1,2/-0,7 | 1,5 | |
| Plywood 18 mm | 13 | +0,7/-0,9 | +1,3/-0,8 | ||
| Plywood 21 mm | 15 | +0,8/-1,0 | +1,4/-0,9 | ||
| Plywood 24 mm | 17 | +0,9/-1,1 | +1,5/-1,0 | ||
| Plywood 27 mm | 19 | +1,0/-1,2 | 1,0 | +1,6/-1,1 | 2,0 |
| Plywood 30 mm | 21 | +1,1/-1,3 | +1,7/-1,2 | ||
The smallest number of layers is three, that is, one of them is intermediate, covered with two facial ones. If the product has more layers, usually an odd number. Due to several additional layers, the strength increases, and therefore the quality of the material, however, the thickness of the plywood board and its weight increase somewhat.
| Length (width) of plywood sheets | Limit deviation |
| 1200, 1220, 1250 | +/- 3,0 |
| 1500, 1525, 1800, 1830 | +/- 4,0 |
| 2100, 2135, 2440, 2500 | +/- 4,0 |
| 2700, 2745, 3050, 3600, 3660 | +/- 5,0 |
Plywood is classified by grade, material used as raw material and by impregnation, that is, by the glue that is used to glue the material.
Characteristics of plywood varieties
- The third grade is characterized by the presence of black dots, the diameter of which does not exceed 0.5 cm, and the number of such defects should be within 10 when considering one square meter plywood board.
by the most the best option is an elite plywood building material - grade E. There are no flaws on the surface of such a coating, which usually arise due to low-quality raw materials.
First grade plywood may have minor defects or small cracks, but in this case, the length of such sections should be within two centimeters.
The second grade is assigned to a material that has certain adhesive streaks or other inclusions. In this case, the volume of such defects should not exceed two percent of the total area of the material. The length of cracks or drips should be about 18-20 cm.
The fourth grade is characterized by the lowest quality. In this case, wormholes, damage to the edges of the sheet, fallen knots, etc. are allowed. Such building material is used for rough work most often.
plywood material
As a raw material in production plywood material it is possible to use both coniferous trees, such as pine, larch, and deciduous representatives, for example, birch. Valuable species, such as oak or cedar, are used very rarely - to create decorative items. They are of high quality, but have a high cost.
| Name of indicator | Thickness, mm | brand | The value of physical and mechanical indicators | |||
| FSF, FC | ||||||
| Moisture content of plywood, % | 3-30 | FK, FSF | 5-10 | |||
| Ultimate strength in static bending along the fibers of the outer layers, MPA, not less than | 7-30 | 25 | ||||
| Ultimate tensile strength along the fibers, MPa, not less than | 3-6,5 | 30 | ||||
| Hardness, MPa | 9-30 | 20 | ||||
| Sound insulation, dB | 6,5-30 | 23,0 | ||||
| Biological stability, hazard class | 3-30 | 5fDa, St | ||||
Coniferous trees are considered the most popular raw material used for the production of plywood boards. The bulk of such products in the construction markets is made from such raw materials. This type of plywood sheet is mainly used during rough construction work, as well as in rooms where the use of environmentally friendly material is important.
softwood plywood
One of the most important benefits this building material, made on the basis of softwood is low cost. Thanks to this, plywood can be used to assemble furniture and various other structures. Also, plywood products are used for rough work, that is, in work where appearance material does not play a major role.
A big plus of this material is its resistance to moisture. This is due to the fact that coniferous materials have a lot of natural resins in their structure, which provide plywood with high resistance to the negative effects of moisture. And this does not require any additional impregnation. Also, such resins of natural origin have antiseptic properties, that is, mold will not appear on such a surface and plywood will not be destroyed by various insect pests.
Along with the advantages of this material, there are some disadvantages. One of them is low strength. For this reason, plywood made from coniferous wood is not recommended for flooring and other products where the main indicator should be the strength of the material.
The excess resin content in such products is also a disadvantage of this coating. When plywood is heated, the release of these resinous substances may begin, which is unacceptable in principle.
hardwood plywood
In this case, birch veneer is most often used. Birch is the most commonly used type of deciduous tree. Plywood using birch as a raw material is produced much less often, but it is characterized by better quality and strength indicators compared to softwood counterparts.
The advantages of birch plywood include material strength and wear resistance. Thanks to this, such plywood sheets can be used in various construction work and in the creation of any structures. Plywood boards made from hardwood characterized by high wear resistance.
The disadvantage of birch-based material is the high price. For this reason given material is not used as extensively as coniferous counterparts.
Another disadvantage of this material is the lack of natural resins. Birch veneer is not resistant to moisture, therefore, it requires special impregnation, which makes the product environmentally unclean. This can only be avoided if an albumin-casein adhesive is used. But even such processing is not able to increase the moisture resistance of the plywood sheet.
The use of special impregnations and adhesive mixtures is also a kind of disadvantage in the production of products based on hardwood.
To connect the layers of veneer into a solid coating, glue is used, which at the same time is an impregnation. The components of such an impregnation determine what technical indicators the finished product will eventually receive. Depending on the selected adhesive composition plywood is divided into several types.
Classification of plywood by type of glue
In the production of FBA plywood, an albumin-casein adhesive mixture is used, which is based on natural components. Consequently, such plywood will be environmentally friendly, its components will not cause any harm to human health and will not cause allergies. Due to this, such a building material can be used when decorating a children's room.
But this product is also characterized by some disadvantages, such as low strength and moisture absorption. Even impregnation in this case does not give plywood sufficient strength. Since the resistance to wear is more dependent on the type of wood used. Such glue is water-soluble, which means that such a plywood board is highly susceptible to moisture.
FSF plywood is considered the most commonly used type in the construction industry. The basis of such material contains a phenol-formaldehyde adhesive. It is used to impregnate and glue wood fibers. Such glue makes the material more durable and resistant to moisture. Due to this, the scope of phenol-formaldehyde plywood is quite wide, ranging from simple furniture upholstery to its use as a floor covering.
This option is characterized by optimal cost. The disadvantage of such plywood is insufficient environmental friendliness. That is, if such a material is subjected to heating, formaldehyde will begin to be released, which adversely affects human health.
In the construction markets, there is another type of plywood - FB. In this case, the glue is bakelite varnish. Such products have high strength and excellent resistance to moisture. The downside of bakelite plywood is the large weight of one sheet and a rather high toxicity rate.
Page 9 of 30
Features of the calculation of glued elements from plywood with wood
4.23. The calculation of glued elements made of plywood with wood should be carried out according to the reduced cross-section method.
4.24. The strength of the stretched plywood sheathing of slabs (Fig. 3) and panels should be checked using the formula
Where M- calculated bending moment;
R f.r - design resistance tensile plywood;
mφ - coefficient taking into account the reduction in design resistance at the joints of plywood sheathing, taken equal to with a miter joint or with double-sided overlays: m f = 0.6 for plain plywood and m f = 0.8 for bakelized plywood. In the absence of joints m f = 1;
W pr is the moment of section modulus reduced to plywood, which should be determined in accordance with the instructions of clause 4.25.
4.25. The reduced modulus of the cross section of glued elements made of plywood with wood should be determined by the formula
Where y o - distance from the center of gravity of the reduced section to the lower edge of the skin;
I pr - moment of inertia of the section reduced to plywood:
, (40)
Where I f is the moment of inertia of the cross section of plywood sheathing;
I e - moment of inertia of the cross section of the wooden frame ribs;
E d / E f - the ratio of the moduli of elasticity of wood and plywood.
When determining the reduced moments of inertia and the reduced moments of resistance, the calculated width of plywood sheathing should be taken equal to b races = 0.9 b at l³ 6 a And b races = 0.15 b,
at l< 6A (b- full width of the slab section, l- slab span, A- the distance between the longitudinal ribs along the axes).
4.26. The stability of the compressed sheathing of slabs and panels should be checked using the formula
where at ³ 50;
at > 50
(A- distance between the ribs in the light; d - plywood thickness).
The upper skin of the slabs should additionally be checked for local bending from a concentrated load. R= 1 kN (100 kgf) (with overload factor n\u003d 1.2) as a plate embedded in the places of gluing to the ribs.
4.27. Checking for chipping of the ribs of the frame of slabs and panels or sheathing along the seam at its junction with the ribs should be carried out according to the formula
Where Q- estimated shear force;
S pr - static moment of the shifted part of the reduced section relative to the neutral axis;
R cn - design chipping resistance of wood along the fibers or plywood along the fibers of the outer layers;
b ras - the calculated width of the section, which should be taken equal to the total width of the frame ribs.
4.28. The strength calculation of the chords of bent elements of I-section and box sections with plywood walls (Fig. 4) should be carried out according to formula (17), taking W races = W pr, while the stresses in the stretched belt should not exceed R p, and in compressed -j R c (j is the coefficient of buckling from the plane of the bend).
4.29. When checking the wall for a shear along the neutral axis in formula (42), the value R ck is taken equal to R f.sr, and the estimated width b races
b race = åd st, (43)
where åd st is the total wall thickness.
When checking chipping at the seams between the chords and the wall in the formula (42) R sk = R f.sk, and the calculated section width should be taken equal to
b races = nh n, (44)
Where h n - the height of the belts;
n- number of vertical seams.
4.30. The strength of the wall in the dangerous section to the action of the main tensile stresses in the bending elements of the I-section and box sections should be checked according to the formula
, (45)
Where R f.r. a - calculated tensile strength of plywood at angle a determined from the graph in fig. 17 app. 5;
s st - normal stress in the wall from bending at the level of the inner edge of the belts;
t st - shear stresses in the wall at the level of the inner edge of the belts;
a - angle determined from dependence
The stability of a wall with an arrangement of fibers of the outer layers longitudinal with respect to the element axis should be checked for the action of shear and normal stresses under the condition
Where h st - wall height between the inner edges of the shelves;
d is the wall thickness.
The calculation should be made according to the formula
, (48)
Where k and and k t - coefficients determined from the graphs of fig. 18, 19 app. 5;
h race - the estimated height of the wall, which should be taken equal to h st at the distance between the ribs A ³ h st and equal A at a <h Art.
When the outer fibers of the plywood wall are transverse to the element axis, the stability check should be carried out according to formula (48) for the action of only tangential stresses in cases where
B. Calculation of elements of wooden structures according to the limit states of the second group
4.31. Deformations of wooden structures or their individual elements should be determined taking into account the shear and compliance of the joints. The magnitude of the deformations of a pliable connection with the full use of its bearing capacity should be taken from Table. 15, and in case of incomplete - proportional to the force acting on the joint.
Table 15
4.32. Deflections of elements of buildings and structures should not exceed the values given in Table. 16
Table 16
Structural elements | Limit deflections in fractions of a span, no more |
1. Beams of floors | |
2. Beams of attic floors | |
3. Coatings (except valleys): | |
a) runs, rafter legs | |
b) cantilever beams | |
c) trusses, glued beams (except for cantilever beams) | |
e) battens, flooring | |
4. Bearing elements of valleys | |
5. Panels and fachwerk elements |
Notes: 1. In the presence of plaster, the deflection of the 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 can be increased up to 1/200 of the span.
4.33. The deflection of bending elements should be determined from the moment of inertia of the gross cross section. For composite sections, the moment of inertia is multiplied by the coefficient k Well taking into account the shift of pliable joints, given in table. 13.
The greatest deflection of hinged and cantilevered bending elements of constant and variable sections f should be determined by the formula
, (50)
Where f o - deflection of a beam of constant section with a height h without taking into account shear deformations;
h- the greatest height of the section;
l- beam span;
k- coefficient taking into account the influence of the variability of the section height, taken equal to 1 for beams of a constant section;
With- coefficient taking into account the influence of shear deformations from the transverse force.
Coefficient values k And With for the main design schemes of beams are given in table. 3 app. 4.
4.34. The deflection of glued elements made of plywood with wood should be determined, taking the section stiffness equal to 0.7 EI etc. The calculated width of the skins of slabs and panels when determining the deflection is taken in accordance with the instructions in clause 4.25.
4.35. The deflection of compressively-bent hinged-supported symmetrically loaded elements and cantilever elements should be determined by the formula
Where f- deflection determined by formula (50);
x - coefficient determined by formula (30).
| Content |
|---|
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. Ceramic tiles will be glued to the plywood, 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.
So, if you believe the regulatory documents, in particular SP 64.13330.2011, then the modulus of elasticity of wood along the fibers E \u003d 100000 kgf / cm 2, and across the fibers E 90 \u003d 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 effect 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 in the case of hinged support along the contour, but in any case, an almost twofold decrease in the 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.
When building or repairing a wooden house, using metal, and even more so reinforced concrete floor beams, is somehow out of 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 floor.
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 a non-residential 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, there is a cast-iron bath , bronze toilet and much more. Therefore, it is almost impossible to take into account all possible options and lay out everything in the form of a simple and understandable table, 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 even 200 kg / m2 is enough to calculate the attic floor. 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: the coefficients m n indicated in the table for the structures of overhead power transmission towers made of larch not impregnated with antiseptics (at a moisture content ≤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 another factor, 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, 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
and the minimum allowable beam height is 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 m³ 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 calculated 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 sk- 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 in some areas boards are laid between two joists, 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 calculation 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. However, if the plywood deflection is calculated (the calculation is not given in detail), the deflection will be about 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 fact, 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’s enough that we took the maximum possible load, and besides, there is a good margin when choosing a section.
3 Option. Floor covering made of chipboard or OSB.
In fact, it is undesirable to use chipboard or OSB as a floor covering (even a rough one) along the floor beams, and these sheet materials are not intended for this, they have too many disadvantages. The design resistance of pressed sheet materials depends on too many factors, so no one will tell you what value of the design resistance can be used in the calculations.
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 have had to be repaired several times, and a neighbor who recently leveled the wooden floor with 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 values for both the modulus of elasticity and the 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 the running dimensions of the sheets, the following variants of the step 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.
Using 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 deflection values, 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. In no case should the actual pitch of the longitudinal beams exceed the allowable value according to Table. 2.11 and 2.12. Recall 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 with internal dimensions of 6.60x9.00 m. The ceiling thickness is 20 cm, PERI Birch plywood is 21 mm thick and sheet dimensions are 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 put 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 systems of the MULTIFLEX type, only a single-span hinged beam without consoles is always considered, since when installing the formwork and during concreting, we always have intermediate stages where the beams work exactly according to this scheme. For large beam spans 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. In a specific example, it turns out with an exact calculation
4.6 m, and took 4.02 m.
