Methods for calculating slabs on an elastic foundation. Description of the stove program. Study of soil characteristics
The fact is that today there is no ideal model of an elastic foundation. One of the most common is the Fuss-Winkler model, according to which the support reaction of an elastic foundation, in other words, a distributed load q acting on the beam is not evenly distributed, but proportional to the deflection of the beam f at the point in question:
q = - kf (393.1)
k = k about b (393.2)
k about- the bed coefficient, constant for the considered base and characterizing its rigidity, is measured in kgf / cm 3.
b- beam width.
Figure 393.1 a) a model of a beam on a solid elastic foundation, b) the reaction of the foundation q to an acting concentrated load.
From this, at least two conclusions can be drawn that are disappointing for a person who is going to quickly calculate the foundation of a small house, besides, even the foundations of theoretical mechanics and the theory of resistance of materials have been comprehended with difficulty:
1. Calculation of a beam on an elastic foundation is a statically indefinable task, since the static equations only allow determining the total value of the load q (foundation reaction). The distribution of the load along the length of the beam will be described by a rather complicated equation:
q / EI = d 4 f / dx 4 + kf / EI (393.3)
which we will not solve here.
2. Among other things, when calculating such beams, it is necessary to know not only the bedding coefficient of the base, but also the rigidity of the beam EI, i.e. all the parameters of the beam - material, width and height of the section, must be known in advance, meanwhile, when calculating ordinary beams, determining the parameters is the main task.
And in this case, what should an ordinary person do, who is not burdened with deep knowledge of the strength of materials, theories of elasticity and other sciences?
The answer is simple: order engineering and geological surveys and a foundation project in the appropriate organizations. Yes, I understand that the cost of the house may increase by several thousand $, but still this is the optimal solution in this case.
If, in spite of everything, you want to save on geological exploration and calculation, i.e. perform the calculation yourself, then be prepared for the fact that you will have to spend more money on the foundation. For such a case, I can offer the following calculation prerequisites:
1. As a rule, a solid foundation slab is adopted as a foundation in cases where the bearing capacity of the foundation is very low. In other words, the soil is sand or clay, not rocks. For sand, clay and even gravel, the bed coefficient, determined empirically, depending on various factors (moisture, grain size, etc.) k o = 0.5-5 kgf / cm 3. For rocks k o = 100-1500 kg / cm 3. For concrete and reinforced concrete k o = 800-1500 kgf / cm 3. As can be seen from formula 393.1, the lower the value of the bed coefficient, the greater the deflection of the beam at the same load and beam parameters. Thus, to simplify further calculations, we can assume that weak soils do not affect the deflection of the beam; more precisely, this insignificant effect can be neglected. In other words, bending moments, shear forces, angles of rotation of cross-sections and deflections will be the same as for a beam loaded with a distributed load. The result of this assumption will be an increased margin of safety, and the greater the strength characteristics of soils, the greater the margin of safety.
2. If the concentrated loads on the beam are symmetric, then, to simplify the calculations, the reaction of the elastic foundation can be taken as uniformly distributed. The following factors serve as the basis for this assumption:
2.1. As a rule, the foundation, considered as a beam on an elastic foundation, in low-rise construction has a relatively short length - 10-12 m. In this case, the load from the walls, considered as concentrated, in fact, is evenly distributed over an area equal to the width of the walls. In addition, the beam has a certain height, which is not taken into account at the first stage of the calculation, and yet even a concentrated load applied to the top of the beam will be distributed in the body of the beam, and the higher the beam height, the larger the distribution area. So for example for foundation slab 0.3 m high and 12 m long, considered as a beam on which three walls rest - two outer and one inner, all 0.4 m thick, it is more correct to consider the loads from the walls not as concentrated, but as evenly distributed over 3 sections with a length of 0.4 + 0.3 2 = 1 m. the load from the walls will be distributed over 25% of the length of the beam, which is not a small amount.
2.2. If a beam lying on a solid elastic foundation has a relatively short length and several concentrated loads are applied to it, then the base reaction will not change from 0 at the beginning of the beam length to a certain maximum value in the middle of the beam and again to 0 at the end of the beam length (for the variant shown in fig. 393.1), and from some minimum value to maximum. And the more concentrated loads are applied to a beam of relatively short length, the less the difference between the minimum and maximum values of the support reaction of the elastic foundation will be.
The result of this assumption will again be a certain margin of safety. However, in this case, the possible margin of safety will not exceed a few percent. For example, even for a single-span beam, on which a distributed load acts, uniformly varying from 1.5q at the beginning of the beam to 0.5q in the middle of the beam and again to 1.5q at the end of the beam (see the article "Reducing a distributed load to an equivalent uniformly distributed load"), the total the load will be ql, as for a beam on which a uniformly distributed load acts. Meanwhile, the maximum bending moment for such a beam will be
М = ql 2 / (8 2) + ql 2/24 = 10ql 2/96 = ql 2 /9.6
This is 20% less than for a beam that is subject to an evenly distributed load. For a beam, the change in the support reaction of which is described by a rather complex equation, especially if there are many concentrated loads, the difference will be even smaller. Well, don't forget about clause 2.1.
As a result, when using these assumptions, the task of calculating a beam on a solid elastic foundation is simplified as much as possible, especially when the applied loads are symmetrical, asymmetric loads will lead to a roll of the foundation and this should be avoided in any case. Moreover, the calculation is practically not affected by the number of applied concentrated loads. If for a beam on hinged supports, regardless of their number, the condition of zero deflection on all supports must be met, which increases the static uncertainty of the beam by the number of intermediate supports, then when calculating a beam on an elastic foundation, it is sufficient to consider the deflection as zero at the points of application of extreme concentrated loads - external walls. In this case, the deflection under concentrated loads - the inner walls is determined according to the general equations. Well, to determine the settlement of the foundation at the points where the deflection is assumed to be zero, you can use the existing regulatory documents on the calculation of bases and foundations.
And you can simply choose the length of the beam consoles in such a way that the deflection under the inner walls is also zero. An example of how you can take advantage of these design assumptions is described
Program Plate is based on the finite element method, however, the user sees this only in mesh images on the slab field, the breakdown into elements occurs without his participation. The user defines the geometry of the slab, loads, supports, places the piles, as it is done on a sheet of paper or in AUTOCAD, using the mouse cursor and clicking on the button. The procedure for specifying the initial data in the program brings pleasure in its simplicity, does not require special computer skills, even experience in calculating structures. However, an experienced calculator should work. Program Plate only a convenient tool, the design scheme is always only mathematical model, which can be changed to achieve the desired result.As a result of the calculation on the program Plate the colored fields of displacements, stresses and reinforcement of the slab are displayed with palettes by color values. The fields of the longitudinal and transverse reinforcement are drawn, the calculation is made for punching by a point load and a support (column, pile). Calculation of draft and roll is performed. User Slabs in one calculation receives a full range of results required for the design of the slab.
Program Plate allows you to calculate flat reinforced concrete slabs of arbitrary geometry in the plan, with stiffeners, thickenings and holes, any type of load, based on oblique soil layers, piles of programmed stiffness, columns or supports of arbitrary configuration. It is possible to take into account karst phenomena in the form of funnels, which should simply be drawn, the bed coefficient is automatically calculated using 5 different methods, the user is only prompted to choose a method. There are many small conveniences that can be appreciated just by starting to work with the program.
Features of the program:
The book discusses approximate methods for calculating beams and slabs located on an elastic foundation, beyond the elastic limit. The basic principles of the theory of limiting equilibrium are briefly stated, the problem of determining the limiting bearing capacity of a beam on an elastic foundation at various loads is considered. Determination of the ultimate load for frames and grillages is shown, taking into account the influence of an elastic foundation. The solution of problems for a prestressed beam is given. The influence of a two-layer base is considered. Problems related to slabs located on an elastic foundation, with a concentrated load in the center, at the edge and in the corner of the slab, have been solved. The calculation of the pre-stressed and three-layer slab has been made. At the end of the work, experimental data related to beams and slabs are presented, as well as a comparison with theoretical results. The book is intended for design engineers and may be useful for senior students of construction universities.
Preface to the first edition
Preface to the second edition
Introduction
Chapter 1. General principles calculation
1.1. Conditions for the transition of beams on an elastic foundation beyond the elastic limit
1.2. Limiting equilibrium for bending members
1.3. General case
1.4. Formation of plastic areas at the base
1.5. Conditions for creating the lowest weight foundations
Chapter 2. Beam on an elastic half-space
2.1. The greatest load in the elastic stage
2.2. Distribution of reactions beyond the elastic limit
2.3. Maximum load value
2.4. Two concentrated forces
2.5. Three concentrated forces
2.6. Uniformly distributed load
2.7. Tapered beam
2.8. Grillage from two cross beams
2.9. Three-layer beam
2.10. Concentrated force applied asymmetrically
2.11. Concentrated force at the edge of the beam
2.12. Prestressed beam
2.13. Prestressed annular beam
2.14. Infinitely long beam
2.15. Simple frame
2.16. Complex frame
Chapter 3. Beam on a two-layer base
3.1. The greatest load in the elastic stage
3.2. Determination of ultimate load
3.3. Applying group plots
3.4. Prestressed beam at finite thickness
3.5. Grillages on an elastic layer
Chapter 4. Beam on a layer of variable stiffness
4.1. Drawing up differential equations
4.2. Taking into account the influence of own weight
4.3. Selection of the design model of the limiting state
4.4. An example of determining the limiting force
4.5. Calculation of a truss of a layered floor
4.6. Calculation of a laminated frame
4.7. Beams on a non-linear foundation
4.8. An example of calculating a beam on a non-linear foundation
4.9. Regulation of base reactions
4.10. Determining the Optimal Stiffness for a Beam
Chapter 5. Calculation of slabs
5.1. Approximate solution for an infinite slab
5.2. Infinitely Rigid Square Slab
5.3. Load at the corner of the slab
5.4. Square slab on a two-layer base
5.5. Prestressed plate
5.6. Influence of local and general plate deformations beyond the elastic limit
5.7. Three-layer slab
5.8. Load at the edge of the slab
5.9. Prefabricated slabs
Chapter 6. The use of computers to determine the limiting state of the foundation
6.1. Finite element method
6.2. Maximum load of a high foundation beam
6.3. Definition of plastic areas in the base
6.4. High foundation beam on elastoplastic base
6.5. The ultimate load of the beam, determined from the condition for the formation of plastic regions in the base
6.6. Using Beam Finite Elements
6.7. Calculation of limit displacements and loads
Chapter 7. Maximum precipitation of multi-storey frame buildings
7.1. Basic design provisions
7.2. Method for solving the problem and drawing up general equations
7.3. Features of the calculation, depending on the structure of the foundation (solid slabs, strip foundations, individual pillars)
7.4. Calculation examples
Chapter 8. Test results
8.1. Frames, grillages and slabs
8.2. Comparison of theoretical and experimental data
8.3. Base deformation modulus
Bibliography
Introduction
Beams and slabs on resilient foundations are mainly used as design models for foundations, which are the main elements that ensure the overall strength and reliability of a structure.
For the calculation of the foundation, as a rule, increased requirements are imposed with respect to its condition during the operation of structures. Small deviations from the established values in the area of deformations or stresses, which are often found in other structural elements, are completely unacceptable for the foundation.
This essentially correct position sometimes leads to the fact that foundations are designed with an excessive margin of safety and turn out to be uneconomical.
To assess the value of the bearing capacity of the foundation, it is necessary to study the distribution of forces in such structures beyond the elastic limit, only then it will be possible to establish correctly those most rational dimensions at which the required reliability of the structure is ensured at its minimum cost.
The difficulty of the problem of calculating beams on an elastic foundation beyond the elastic limit is that it is impossible directly, without special techniques, to apply the general method for calculating structures by limiting equilibrium.
The method of limiting equilibrium, created as a result of the work of our domestic scientists, professors V.M. Keldysh, N.S. Streletsky, A.A. Gvozdev, V.V. Sokolovsky, N.I. Bezukhova, A.A. Chirasa, A.R. Rzhanitsyn, A.M. Ovechkin and many others, has received universal recognition and is widely used in practice. In foreign literature, this method is also used and highlighted in the works of B.G. Nile, F.G. Hodge, R. Hill, M. R. Horn, F. Bleich, V. Prager, I. Guyon and others; some of these works have been translated into Russian.
Example 9 deals with the static analysis and design of a reinforced concrete slab. The objectives of the example are as follows:
demonstrate the procedure for constructing the design model of the slab;
show the technique for setting loads and compiling DCF;
show the procedure for the selection of fittings.
A reinforced concrete slab with a size of 3x6m and a thickness of 150mm is calculated. The short side of the slab is supported along its entire length, the opposite side is supported by its ends on the columns. The long sides of the slab are free. It is required to perform a static calculation, draw up a DCS table and select the slab reinforcement.
Loads are set:
load 1 - dead weight;
load 2 - concentrated loads P = 1mc, applied according to the diagram in Fig. 1.13, zag 2;
load 3 - concentrated loads P = 1mc, applied according to the diagram in Fig. 1.13, zag 3.
The calculation is made for a 6 x 12 grid.
Rice. 1.13. Calculation scheme of the slab
"LIRA" EXAMPLES |
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9.1. Creation |
dialogue |
"Sign |
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set the task name: "Example9" and the attribute |
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schemes: "3". |
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9.2 Defining Geometry |
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In the Create Plane |
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fragments and networks "activate |
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the "Slab Generation" tab, then |
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set the FE step along the first and second |
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9.2.1 Generation |
Step along the first axis: |
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Step along the second axis: |
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Then click on the button |
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Apply. |
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9.3 Setting the Boundary Conditions |
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Display the node numbers. |
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Select the support nodes No. 1, 7, 85 - 91. |
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9.3.3 Purpose |
activate |
bookmark |
"Assign |
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boundary conditions |
communications "and mark the directions by |
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in selected nodes |
prohibited |
displacement |
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(Z) and click the Apply button. |
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9.4 Setting the stiffness parameters of the slab elements |
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9.4.1 Formation |
dialogue |
"Rigidity |
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elements "form a list of types |
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types of stiffness |
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rigidity. |
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9.4.1.1 selection |
Click the Add button and select |
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tab for the numerical description of stiffness, |
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"Plates" |
activate the "Plates" section. |
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In the Set Stiffness dialog box |
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9.4.1.2. Task |
for plates ”set the section parameters: |
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Elastic modulus - E = 3e6 t / m2; |
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section parameters |
Coef. Poisson - V = 0.2; |
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"Plates" |
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Plate thickness - H = 15 cm; |
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The specific gravity of the material is Ro = 2.75 t / m2. |
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9.4.2 Assigning stiffnesses |
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9.4.2.1. Purpose |
Highlight |
rigidity |
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the current |
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list and click the Install button |
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rigidity |
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as the current type. |
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"1.Plate Н 15" |
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Select all the elements of the diagram. |
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Assign the selected elements the current stiffness type. |
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comments |
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9.5 Setting loads |
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9.5.1 assignment |
Execute |
Loads |
The elements |
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load |
automatically |
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the elements |
Add your own weight. |
loaded with load |
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own weight |
own weight. |
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9.5.2 Change |
dialogue |
"Active |
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the current |
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load ”set the load number 2. |
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loading |
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Select nodes 18, 46, 74. |
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activate the tab “Loads in |
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nodes ". Then use the radio buttons to indicate |
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coordinates |
"Global", |
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9.5.4 assignment |
direction - along the "Z" axis. By clicking on |
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loads in |
button focused |
force call |
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dedicated nodes |
Load Parameters dialog box. |
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In this window, enter the value P = 1 tf and |
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confirm your entry. After that in |
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the dialog box "Specify loads" |
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click the Apply button. |
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9.5.5 Shift |
dialogue |
"Active |
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the current |
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load ”set the load number 3. |
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loading |
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Display the numbers of the elements of the calculation scheme. |
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In the "Specify Loads" dialog box |
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activate the tab “Loads on |
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plates ". |
radio buttons |
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coordinates |
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"Global", direction - along the axis |
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9.5.7 assignment |
"Z". Clicking the button centered |
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call |
interactive |
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load |
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"Parameters |
load ". V |
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dedicated |
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enter the parameters in the window: |
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elements |
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P = 1 tf; |
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A = 0.25 m; |
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B = 0.25 m and confirm the entry. After |
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this in the dialog box "Job |
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loads " |
click |
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Apply. |
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In the Calculated Combinations dialog box |
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9.6. Generation |
efforts "set the types of loadings: |
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The first is Constant (0); |
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DCS tables |
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The second is the Temporary duration. (one); |
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Third - Temporary duration. (one). |
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The launch of the calculation task and the transition to the visualization mode of the calculation results is carried out in the same way as in the previous examples.
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9.7. Output on display |
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isofields |
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displacement |
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direction Z |
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9.8. Output on display |
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stresses Мх |
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9.9. Running |
Run Windows commands: Start h |
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Programs h Lira 9.0 h LirArm. |
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In the dialog box of the LIR-ARM system |
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9.10. Import |
"Open" |
highlight |
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design scheme |
Example9 # 00.example9 "and click on |
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the Open button. |
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9.11 Assignment and selection of material |
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In the Materials dialog box, check |
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radio button Type and click on the button |
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9.11.1 assignment |
Add. |
output |
Rest |
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dialog window " General characteristics |
the General dialog box |
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reinforcement ", in which set the module |
specifications |
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characteristics |
reinforcement - |
stove and |
click |
reinforcement "remain |
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reinforcement |
the Apply button. |
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dialogue |
"Materials" |
accepted by default. |
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click |
Assign |
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9.11.2 assignment |
In the Materials dialog box |
operation |
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activate |
radio button |
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characteristics |
click |
Add |
default |
accepted |
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concrete of class B25. |
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Default and Set Current. |
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9.11.3 Assignment |
In the same window, activate the radio |
operation |
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the Armature button and click on the buttons |
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characteristics |
Add |
default |
Assign |
default |
accepted |
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fittings |
fittings of class A-III. |
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9.12 Purpose of material |
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9.12.1 Selection |
Select all the elements of the diagram. |
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frame elements |
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9.12.2 Purpose |
You can also assign |
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dialogue |
"Materials" |
material |
using |
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material |
click the Assign button. |
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frame elements |
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toolbar). |
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9.13. Payment |
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reinforcement |
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9.14. View |
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bottom reinforcement in |
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plates |
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9.16. View |
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results |
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reinforcement |
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2.14. View |
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results |
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reinforcement in the form |
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HTML tables |
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1.11. Investigation of the stress-strain state of structures working in conjunction with the base
All finite elements in SP LIRA perceive an elastic foundation in accordance with the Pasternak model. However, the Winkler base model is most often used.
The mechanical properties of the Winkler model are characterized by the stiffness (bed) coefficient C1. According to the physical meaning, the stiffness coefficient is the force that must be applied to 1 m2 of the base surface in order for the latter to settle by 1 m. Dimension C1 - tf / m3 (kN / m3).
To implement the Winkler model, FE No. 51 are used.
For a nonlinear problem of a system with one-way connections, the software package uses FE No. 261. This element models one-way discrete connections of the Winkler base and allows one to take into account the effects of separation of the structure from the base.
Stages and operations
Your actions
comments
Save
under the new
"Example10".
10.2 Removing Imposed Boundary Conditions
Select the nodes of the calculation scheme.
In the Node Links dialog box
10.2.2 Deletion
activate the "Delete links" tab
and mark the directions in which
boundary conditions
remove the anchors (Z) and click on
the Apply button.
10.3. Exercise
dialogue
"Rigidity
elements "
click
characteristics
Change and in the new window "Task
elastic foundation
rigidity
for plates "
enter the coefficient.
C1 = 1000 tf / m3.
Run the task for calculation, go to
calculation results visualization mode
and display displacements and
stresses in the plates.
1.11.2. Slab on a resilient base with finite stiffness ties. Example 11
The main purpose of this example is to demonstrate the technique of using finite element No. 51, which models the Winkler base with finite stiffness constraints.
Here we use the initial data from Example 9 (see Fig. 1.13).
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Save the task with a new name: |
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"Example11". |
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superimposed connections |
similarly |
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Example 10. |
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11.3. Setting constraints of finite stiffness
11.3.1. Select all nodes of the circuit
11.4. Setting stiffness parameters for FE No. 51
V in the Stiffness dialog box
11.4.1 selection |
elements " |
click |
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section "FE |
Add and by selecting the numerical tab |
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numerical " | ||||||||||
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Modern houses erected on different foundations. The choice directly depends on the loads, the topography of the selected area, the structure and composition of the soil itself and, of course, climatic conditions. This article reveals complete information about the slab foundation, intelligibly answers the question of how to correctly make a complete calculation that will help build the required foundation.
Peculiarities
The tiled type of foundation consists of the base of the building, which is a flat or reinforced concrete slab with stiffeners. The structure of this foundation is of several types: prefabricated or monolithic.
Prefabricated foundations are prefabricated slabs made at the factory. Plates are laid with construction equipment on a previously prepared, that is, leveled and compacted, base. Airfield slabs (PAG) or road slabs (PDN, PD) can be used here. This technology has a major drawback. It is associated with the lack of integrity, and, as a result, with the corresponding impossibility of resisting even the smallest movements of the ground. It is for this reason that the prefabricated type slab foundation mainly used only on surfaces made of rocky soil or on non-porous coarse-grained soils for the construction of small structures made of wood in areas where the minimum freezing depth is.
But a monolithic slab foundation is one rigid whole reinforced concrete structure, which is being built under the area of the building itself.
Geometrically, this type of foundation is of several types.
- Simple. When the underside of the foundation tile is flat and level.
- Reinforced. When the underside has stiffeners that are arranged in a specially calculated order.
- UWB. This is the name of the insulated type of Swedish slabs, which belong to the type of reinforced foundation slabs. During construction, a unique technology is used: the concrete mixture is poured into a separately developed factory type permanent formwork, which allows in the future to form on an elastic foundation, or rather, in its lower part and on the surface, a mesh of reinforced and small-sized stiffeners. Also, the USHP has a heating system.
This article tells about the simplest monolithic slab foundation.
Advantages and disadvantages, selection criteria
The first advantage is almost perfect versatility. Sometimes on the Internet you can come across articles that say that foundation tiles can be built everywhere.
Even if construction work is carried out on a swampy area, nothing terrible will happen to the tiles: during the period of severe cold weather, it will rise, and during the hot period, on the contrary, it will sink, so to speak, to swim.
It turns out a kind of "concrete ship", which has a superstructure on top of the whole house.
And yet, the following remark will be fair here: the only foundation that allows for a fairly reliable erection on planting and highly heaving soils, including swampy soil type, is pile foundation... This type of foundation is used when the piles have enough own length to be anchored in the lowest bearing soil layers.
Frosty type of heaving, including subsidence, during thawing or subsidence of the foundation due to moistening of the soil surface (for example, during lifting groundwater) cannot occur equally under the surface of the entire tile. In any case, only one of the sides will move more. A simple example spring thawing of the soil surface can become. The thawing process will proceed much faster and with greater intensity on the south side of the house than on the north. In the meantime, the tile will be subject to enormous loads, which, by the way, it does not always withstand. All this will affect the structure: the house can simply tilt. It won't be so scary if this building is made of wood. And if it was built of bricks or blocks, cracks may appear on the walls.
The slab foundation allows you to build houses even on the most difficult soils, which include the medium-porous type of soil, which has the smallest bearing capacity than, for example, tape soil. Just overestimate this opportunity no need.
Are slab foundations used when building large structures? Some argue that only the lightest and at the same time insufficiently durable structures can be built on a monolithic slab. This statement is not entirely true, since when choosing favorable conditions and a correctly designed foundation with competent conduct construction work, the slab foundation is able to withstand even the capital's TSUM. By the way, this building was built on a slab.
The price is too high. For some reason, this opinion is widespread. Almost everyone is sure that the slab type of foundation is very expensive, more expensive existing species grounds. Also, for some reason, most believe that the cost will be about half of the available costs for all subsequent construction work.
At the same time, no one and never any comparative analysis did not. Also, for some reason, many do not take into account that during the construction of a house, for example, there is no need to make floors. Of course, this refers to a rough floor surface.
The complexity of the work itself. The following statement is often heard: "For the construction of a slab-type foundation, the experience of qualified workers is required." And yet, if you think about it, it becomes clear that such "masters" greatly overestimate the prices for their work. In fact, only ignorance of the technology usually leads to mistakes, and you can twist it with any other foundation.
So what kind of difficulties can you face when working with a slab foundation? When leveling the site? No, everything here is also no more complicated than when leveling a buried strip foundation foundation. Maybe the difficulty with waterproofing or insulation? Here, rather, it is better to perform these operations on a flat horizontal surface than on vertical planes.
Maybe it's the knitting of the reinforcement cage? Again, you need to compare and understand which is easier, for example, you can take the fittings laid out on a flat platform, or climb into yourself strip foundation with its formwork. Maybe it's the filling itself concrete mix? In this option, everything depends not on the chosen foundation, but rather on the characteristics of a separate site, on whether the mixer can drive up to the construction site or whether the concrete will have to be mixed manually.
In fact, erecting foundation slabs is a physically challenging task. Due to the rather large construction area, this work can be called tedious, but it does not say that the help of qualified builders will be required. Therefore, ordinary "handy" men will be able to cope with such a case. In addition, if you correctly follow the construction technology and SNiP of a columnar, slab and other foundation, everything will definitely work out.
Calculations
Each zero cycle will require a calculation, which consists, first of all, in determining the thickness of the slab itself. This choice cannot be made approximately, since such an unprofessional solution to the issue will lead to a weak base that can crack in cold weather. They do not make too massive a deep foundation so as not to waste unnecessarily extra money.
For self-construction of houses, you can use the calculation below. And even if these calculations do not compare with engineering ones, which are carried out in design organizations, it is these calculations that will help in the implementation of high-quality foundation laying.
Examine the ground
The soil on the selected building site should be examined.
For further calculations, you will need to select a certain thickness for the foundation slab with the appropriate mass. This will help to obtain the best specific pressure for the type of soil available. When the loads are exceeded, the structure usually begins to "sink", with minimal loads, a slight frosty swelling of the soil surface will tilt the foundation. All this will cause corresponding not too pleasant consequences.
The optimal specific pressure for the soil surface on which construction is usually started:
- fine sand or dusty type of high density sand - 0.35 kg / cm³;
- fine sand with an average density of 0.25 kg / cm³;
- sandy loam in solid and plastic form - 0.5 kg / cm³;
- plastic and hard loams - 0.35 kg / cm³;
- plastic grade of clay - 0.25 kg / cm³;
- hard clay - 0.5 kg / cm³.
Total weight / weight of the house
Based on the developed project of the future structure, it is possible to determine what the total mass / weight of the house will be.
The approximate value of the specific gravity of each structural element:
- brick wall with a thickness of 120 mm, that is, half a brick - up to 250 kg / m²;
- aerated concrete wall or 300 mm D600 aerated concrete blocks - 180 kg / m²;
- wall of logs (diameter 240 mm) - 135 kg / m²;
- 150 mm timber wall - 120 kg / m²;
- 150 mm frame wall (insulation is required) - 50 kg / m²;
- attic made of wooden beams with mandatory insulation, with a density reaching 200 kg / m³, - 150 kg / m²;
- hollow concrete slab - 350 kg / m²;
- interfloor or basement made of wooden beams, insulated, density reaches 200 kg / m³ - 100 kg / m²;
- monolithic reinforced concrete floor - 500 kg / m²;
- operating load for overlapping interfloor and basement - 210 kg / m²;
- with a roof made of sheet steel, corrugated board or metal tiles - 30 kg / m²;
- operating load for overlapping the attic - 105 kg / m²;
- with a two-layer roofing material made of roofing material - 40 kg / m²;
- with a ceramic tile roof - 80 kg / m²;
- with slate - 50 kg / m²;
- snow load type applied to the middle lane Russian territory, - 100 kg / m²;
- snow type of load for the northern regions - 190 kg / m²;
- snow type of load for the southern part - 50 kg / m².