There are only two types of foundations that are suitable for the construction of almost any building: pile and slab. They allow the construction of buildings on soils with poor characteristics at minimal cost. It is worth choosing a monolithic slab as a foundation for many reasons, but in order for it to be strong and reliable, it must be properly calculated.

The advantages of the design include:

• construction on soils with poor characteristics;
• the possibility of constructing large objects;
• possibility of self-filling;
• high load-bearing capacity;
• prevention of local deformations;
• resistance to frost heaving forces.

The weaknesses of this type of foundation include:

• inappropriate for use on sloped areas;
• high consumption of concrete and reinforcement;
• Compared to ready-made foundation elements, the installation of a monolithic slab requires additional time for concrete to gain strength;
• complex calculation.

Study of soil characteristics

Before starting to calculate any type of foundation, the characteristics of the foundation under it are determined. The main and most important points include:

• water saturation;
• bearing capacity.

When constructing large facilities, before starting the development of project documentation, full-fledged geological surveys are carried out, which include:

• drilling of the wells;
• laboratory research;
• development of a report on the characteristics of the foundation.

The report provides all the values ​​obtained during the first two steps. A full range of geological surveys is expensive. When designing a private house, it is most often not necessary. Soil studies are carried out using two methods:

• pits;
• wells.

The cutting of pits is carried out manually. To do this, dig a hole with a shovel, 50 cm deep below the expected level of the base of the foundation. The soil is studied by section, approximately the type of bearing layer and the presence of water in it are determined. If the soil is too saturated with water, it is recommended to use pile supports for the building.

The second option for studying the characteristics of the foundation for a house is carried out using a hand drill. The analysis is carried out on pieces of soil on the blades.

Important! When conducting events, it is necessary to select several points to study. They should be located under the building site. This will allow you to study the soil type most thoroughly.

Having decided on the base, the optimal specific pressure on the ground is determined for it. The value will be required in further calculations, an example of which is presented below. The value is taken according to the table.

*With this type of foundation soil, the strip option may be more economical, so you need to calculate the estimate for two types of foundation and choose the one that will cost less.

Calculation of slab thickness

For different loads, the coefficient is different and ranges from 1.05-1.4. The exact values ​​are also given in the table. For a concrete foundation using monolithic technology, a coefficient of 1.3 is taken.

Important! If the roof slope is more than 60 degrees, the snow load is not taken into account in the calculation, since with such a steep slope, snow does not accumulate on it.

The total area of ​​all structures is multiplied by the mass given in the table and the coefficient, after which, adding, the total weight of the house is obtained without taking into account the foundations.

The basic formula for calculations is as follows:

where P1 is the specific load on the ground without taking into account the foundation, M1 is the total load from the house obtained when collecting loads, S is the area of ​​the concrete slab.

where P is the tabulated value of the bearing capacity of the soil.

where M2 is the required mass of the foundation (it is impossible to build a foundation greater than this mass), S is the area of ​​the concrete slab.

The following formula:

t = (M2/2500)/S,

where t is the thickness of concrete pouring, and 2500 kg/m 3 is the density of one cubic meter of reinforced concrete foundation.

Next, the thickness is rounded to the nearest larger and smaller multiple of 5 cm. Afterwards, a check is performed in which the difference between the calculated and optimal pressure on the ground should not exceed 25% in any direction.

Advice! If the calculation turns out that the thickness of the concrete layer exceeds 350 mm, it is recommended to consider such types of construction as a strip foundation, columnar or slab with stiffeners.

In addition to the thickness, you will need to select the appropriate reinforcement diameter, as well as calculate the amount of reinforcement for concrete.

Important! If, as a result of the calculation, you get a slab thickness of more than 35 cm, this indicates that the slab foundation is redundant in the given conditions, you need to calculate strip and pile foundations, perhaps they will be cheaper. If the thickness is less than 15 cm, then the building is too heavy for the given soil and accurate calculations and geological studies are needed.

Calculation example

The example provides the following input data:

• a one-story house with an attic measuring 8 m by 10 m in plan;
• the walls are made of sand-lime brick 380 mm thick, the total area of ​​the walls (4 external walls 4.5 m high) is 162 m²;
• the area of ​​internal plasterboard partitions is 100 m²;
• metal roof (hatched, slope 30ᵒ), area equals 8 m * 10 m/cosα (roof inclination angle) = 8 m * 10 m/0.87 = 91 m² (also needed when calculating the snow load);
• soil type - loam, bearing capacity = 0.32 kg/cm² (obtained from geological surveys);
• wooden floors with a total area of ​​160 m2 (also needed when calculating the payload).

Collection of foundation loads is carried out in tabular form:

The area of ​​the slab for the building is taken into account that the width of the slab is 10 cm greater than the width of the house. S = 810 cm * 1010 cm = 818100 cm² = 81.81 m2.

Specific load on the ground from the house = 210696 kg/818100 cm2 = 0.26 kg/cm2.

Δ = 0.32 - 0.26 = 0.06 kg/cm2.

M = Δ*S = 0.06 kg/cm 2 * 818100 cm 2 = 49086 kg.

t = (49086 kg/2500 m3)/81.81 m2 = 0.24 m = 24 cm.

The thickness of the slab can be 20 cm or 25 cm.

We check for 20 cm:

1. 0.2 m * 81.81 m 2 = 16.36 m 3 - volume of the slab;
2. 16.36 m 3 * 2500 kg/m 3 = 40905 kg - mass of the slab;
3. 251601 kg/ 818100 cm2 = 0.31 kg/cm² - actual ground pressure is less than optimal by no more than 25%;
4. (0,32-0,31)*100%/0,32 = 3% < 25%(максимальная разница).

There is no point in checking a foundation of greater thickness, since the size requiring less consumption of concrete and reinforcement satisfied the requirements. This completes the thickness calculation example. We accept a slab with a thickness of 20 cm. The next step will be the calculation of reinforcement and the amount of reinforcement.

The reinforcement for the slab structure is selected depending on the thickness. If the slab with concrete thickness is 150 cm or less, one reinforcement mesh is laid. If the concrete thickness is more than 150 mm, it is necessary to lay reinforcement in two layers (bottom and top). The diameter of the working rods is 12-16 mm (the most common is 14 mm). Reinforcement bars with cross-sectional dimensions of 8-10 mm are installed as vertical clamps.

For good reason, the slab should also be calculated for bending loads, but these calculations are complex and are performed by professionals using special software. To understand exactly what diameter of reinforcement and its spacing is needed in your case, you need to carry out precise calculations, or lay reinforcement with a large margin of safety and a minimum spacing, respectively, greatly overpaying.

Reinforcement calculation

Calculation of the amount of reinforcement for the slab calculated above:

1. slab 20 cm thick - two working grids;
2. rod diameter - 12 mm, pitch - 150 mm;
3. the rods are laid so as to provide a protective layer of concrete on each side of 0.02-0.03 m. The length of the rods in the example = 8.1 m - 0.02 * 2 = 8.06 m and 10.06 m;
4. number of rods in one direction = (8.1 m (side length)/0.15 m (step) + 1) *2 (two layers) = 110 pcs;
5. number of rods in other direction = (10.1 m (side length)/0.15 m (step) + 1) * 2 (two layers) = 136 pcs;
6. total length of rods = 110*8.06 + 136*10.06 = 886.6 m + 1368.16 = 2254.76 m;
7. total weight of reinforcement 2254.76 m * 0.888 kg/m = 2002.2 kg.

When purchasing, you must provide a reserve of 3-5% to avoid the need to purchase additional material. You will also need to calculate the volume of concrete. In the case under consideration, it is equal to: 8.1m*10.1m*0.2m = 16.36 m³. This value will be required when ordering concrete mix.

A simplified calculation of the thickness of the foundation slab and the amount of materials for it is a simple task that does not require a lot of time. But completing this stage will ensure reliability without wasting materials, which will save the nerves and money of the future home owner.

Important! This article is for informational purposes only. To accurately calculate the foundation, a geological study is necessary. Trust the calculations only to professionals.

Advice! If you need contractors, there is a very convenient service for selecting them. Just send in the form below a detailed description of the work that needs to be performed and you will receive proposals with prices from construction teams and companies by email. You can see reviews about each of them and photographs with examples of work. It's FREE and there's no obligation.

Example 9 is devoted to the static calculation and design of a reinforced concrete slab. The objectives of the example are as follows:

demonstrate the procedure for constructing a design diagram for a slab;

show the technique of specifying loads and drawing up DCS;

show the procedure for selecting fittings.

A reinforced concrete slab measuring 3x6m and thickness 150mm is calculated. The short side of the slab is supported along its entire length, the opposite side is supported with 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 specified:

load 1 – own weight;

load case 2 – concentrated loads P = 1ts, applied according to the diagram in Fig. 1.13, chapter 2;

load case 3 – concentrated loads P = 1ts, applied according to the diagram in Fig. 1.13, chapter 3.

The calculation is made for a 6 x 12 grid.

Rice. 1.13. Design diagram of the slab

"LIRA" EXAMPLES											http://www.lira.com.ua
Stages and operations					Your actions
											comments
9.1. Creation				dialogue			"Sign
			set the task name: “Example9” and the characteristic
			schemes: "3".
9.2.Specifying geometry
			In the Create Planar Dialog Box
			fragments and networks" activate
			tab “Slab generation”, then
			set the FE step along the first and second
9.2.1.Generation			Step along the first axis:
			Step along the second axis:
			After this, click on the button
			Apply.
9.3. Setting boundary conditions
		Display the node numbers.
		Select support nodes No. 1, 7, 85 – 91.
9.3.3. Purpose			activate			bookmark		"Assign
boundary conditions			communications" and mark the directions according to
in dedicated nodes					prohibited		movements
			(Z) and click the Apply button.
9.4. Setting the rigidity parameters of the slab elements
9.4.1.Formation				dialogue				"Rigidity
			elements" create a list of types
types of stiffness
			rigidity.
9.4.1.1.Selection			Click the Add button and select
			tab for numerical description of stiffness,
"Plates"			activate the “Plates” section.
			In the Set Stiffness dialog box
9.4.1.2.Task			for plates" set the section parameters:
			Modulus of elasticity – E = 3е6 t/m2;
section parameters			Coef. Poisson – V = 0.2;
"Plates"
			Slab thickness – H = 15 cm;
			Specific gravity of the material – Ro = 2.75 t/m2.
9.4.2. Purpose of rigidities
9.4.2.1. Purpose			Highlight				rigidity
current
			list and click the Install button
rigidity
			as current type.
"1. Plate N 15"
		Select all elements of the diagram.
		Assign the current stiffness type to the selected elements.

http://www.lira.com.ua										"LIRA" EXAMPLES
Stages and operations					Your actions
										comments
9.5.Specifying loads
9.5.1.Task			Execute				Loads			Elements
loads										automatically
elements			Add your own weight.							loaded with load
own weight										own weight.
9.5.2.Shift				dialogue				"Active
current
			load case" set load case number 2.
loading
		Select nodes No. 18, 46, 74.
			activate the “Loads in” tab
			nodes." Then use the radio buttons to specify
					coordinates		"Global"
9.5.4.Task			direction – along the “Z” axis. Click on
loads in			centered button				call for strength
dedicated nodes			Load Parameters dialog box.
			In this window, enter the value P = 1 tf and
			Confirm your entry. After that in
			"Specify Loads" dialog box
			Click the Apply button.
9.5.5.Shift				dialogue				"Active
current
			load case" set load case number 3.
loading
		Display the numbers of elements of the calculation scheme.
			In the Define Loads dialog box
			activate the “Loads on” tab
			plates."				radio buttons
								coordinates
			“Global”, direction – along the axis
9.5.7.Task			"Z". By clicking the focused button
				call		interactive
loads
			"Options			loads." IN
highlighted
			window enter the parameters:
elements
			P = 1 tf;
			A = 0.25 m;
			B = 0.25 m and confirm the entry. After
			this in the "Task" dialog box
			loads"		click
			Apply.
			In the Result Combinations dialog box
9.6. Generation			effort" specify the types of loads:
			The first is Constant (0);
DCS tables
			Second – Temporary duration. (1);
			Third – Temporary duration. (1).

Launching a calculation task and switching to the visualization mode of calculation results is carried out similarly to the previous examples.

http://www.lira.com.ua

Stages and operations

Your actions

comments

9.7. Output on display

isofields

movements

Z direction

9.8. Output on display

voltage Mx

9.9. Launch

Run Windows commands: Start h

Programs h Lira 9.0 h LirArm.

In the dialog box of the LIR-ARM system

9.10. Import

"Open"

highlight

design scheme

“example9#00.example9” and click on

Open button.

9.11.Specification and selection of material

In the Materials dialog box, check

radio button Type and click the button

9.11.1.Task

Add.

is displayed

Rest

General Features Dialog Box

General dialog box

reinforcement", in which specify the module

characteristics

characteristics

reinforcement –

stove and

click

reinforcement" remain

reinforcement

button Apply.

dialogue

"Materials"

accepted by default.

click

Assign

9.11.2.Task

In the Materials dialog box

operation

activate

radio button

characteristics

click

Add

default

accepted

concrete class B25.

Default and Set as current.

9.11.3.Task

In the same window, activate the radio

operation

Reinforcement button and click on the buttons

characteristics

Add

default

Assign

default

accepted

fittings

Class A-III fittings.

9.12. Purpose of the material

9.12.1.Selection

Select all elements of the diagram.

frame elements

9.12.2. Purpose

You can also assign

dialogue

"Materials"

material

using

material

Click the Assign button.

frame elements

toolbar).

9.13. Calculation

reinforcement

9.14. View

lower reinforcement in

plates

X axis direction

Stages and operations

Your actions

comments

9.16. View

results

reinforcement

2.14. View

results

reinforcement in the form

HTML tables

1.11.Study of the stress-strain state of structures working together with the base

All finite elements in PC LIRA perceive an elastic base 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 coefficient (bed) C1. In physical terms, the stiffness coefficient is the force that must be applied to 1 m2 of the base surface in order for the latter to settle 1 m. Dimension C1 - tf/m3 (kN/m3).

To implement the Winkler model, FE No. 51 is used.

For the 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 new

"example10".

10.2.Removing imposed boundary conditions

Select the nodes of the design diagram.

In the Node Links dialog box

10.2.2.Removal

activate the “Delete connections” tab

and mark the directions in which

boundary conditions

remove anchors (Z) and click on

button Apply.

10.3. Exercise

dialogue

"Rigidity

elements"

click

characteristics

Edit and in the new window “Task”

elastic base

rigidity

for plates"

enter odds

C1 = 1000 tf/m3.

Run the calculation task, go to

calculation results visualization mode

and display the movements and

stress in the plates.

1.11.2. A slab on an elastic base with bonds of finite rigidity. Example 11

The main purpose of this example is to demonstrate the technique of using finite element No. 51, which models the Winkler foundation with finite stiffness links.

Here the initial data of example 9 is used (see Fig. 1.13).

Stages and operations			Your actions
					comments
		Save the task under a new name:
		"example11".
			superimposed connections	similarly
		example 10.

11.3. Specifying connections of finite stiffness

11.3.1. Select all circuit nodes

11.4. Setting stiffness parameters for FE No. 51

IN Hardness dialog box

11.4.1.Selection			elements"	click
sections "FE			Add and by selecting the numerical tab
numerical"
Stages and operations						Your actions

The goal is to familiarize yourself with the methodology for creating design diagrams of flat structures in the SCAD software package by generating a diagram using parametric prototypes of slabs on an elastic foundation.

2. Theoretical background

When calculating structures on an elastic foundation, problems arise taking into account the distribution properties of the foundation, which are ignored in the simplest case of a Winkler foundation (keyboard model). Most real soils have distributive capacity when, in contrast to the Winkler design scheme, not only the directly loaded parts of the foundation are involved in the work. Consequently, to take into account the distributive capacity of the foundation, it is necessary, firstly, to use foundations different from the Winkler model and, secondly, to introduce into the calculation scheme those parts of the foundation that are located outside the foundation structure.

Accounting for the part of the base located behind the area W occupied by the structure itself in SCAD can be done using “infinite” finite elements such as a wedge or strip. These elements make it possible to model the entire environment of the area W, if it is convex and polygonal (Figure 6.1).

The polygonality of the area is almost always ensured with varying degrees of accuracy. If the region W is non-convex or not simply connected, then it must be supplemented to a convex region with finite elements of limited sizes. In this case, in the complemented parts, the thickness of the slab is assumed to be zero.

Figure 6.1 – Location of contour finite elements such as wedge and strip: 1 – slab; 2 – addition of the region W to a convex one; 3 – strip element; 4 – wedge element

The SCAD computing system provides users with procedures for calculating buildings and structures in contact with foundations. These procedures consist of calculating the generalized characteristics of natural or artificial foundations. Typically, designers experience certain difficulties when assigning these characteristics, especially for heterogeneous layered bases, because Obtaining appropriate experimental data requires special full-scale tests, and the accumulated tabular data is not always adequate to real design conditions.

3. Equipment and materials

Computer class for 25 seats. SCAD software package. Regulatory and technical documentation in construction.

4. Safety instructions

Only students who have undergone safety instructions are allowed to perform laboratory work.

The distance from the workplace to the monitor must be at least 1 m. Do not touch the monitor screen with your hands or move the system unit in working condition.

5. Methodology and order of work

Create New project.

Choose Schema type.

Form Scheme - a rectangular grid with variable (Figures 6.3 – 6.4) or constant pitch (Figure 6.5), located in the XoY or XoZ plane. Grid parameters are assigned in the dialog box shown in Figure 6.2.

Figure 6.2 – Dialog box

The type of diagram and its position in space are assigned using the buttons located at the top of the window. If you select the correct circuit type, the end elements will automatically be assigned a type and you will not have to change it while working with the circuit. Slabs are assigned type 11 by default.

Figure 6.3 – Scheme of a slab with different grid spacing along the X and Y axes

Figure 6.4 – Scheme of a slab with a variable grid pitch along the X and Y axes

Figure 6.5 – Rectangular slab with a constant finite element mesh pitch

When assigning different mesh spacing, it should be remembered that the highest quality solution will be obtained when the aspect ratio of the four-node finite elements is close to 1. It is not recommended to assign a ratio of more than 1/5. The ideal in this sense is a square.

Enter loads.

Setting the type, direction and value of loads is carried out in a dialog box (Figure 6.6), which opens after clicking the button Plate loads in the toolbar Downloads. In the window, you should set the coordinate system in which the load is specified (general or local), the type of load (concentrated, distributed, trapezoidal), enter the value of the load and its binding (for distributed and trapezoidal loads, the binding is not specified). The dialog box displays an icon showing the positive direction of the load.

Figure 6.6 – Dialog box Specifying loads on plate elements

After pressing the button OK in the dialog box you can begin assigning loads to circuit elements. Before entering loads, it is advisable to enable the appropriate display filter.

When you enter concentrated loads, the program controls the binding of loads within the boundaries of the element. If the load does not fall on an element, a message is displayed and elements in which a binding error was made are marked on the diagram.

The load on plate elements can be specified and distributed along a line connecting two user-specified nodes of the element. To set this load you need to:

– in the dialog box, assign the type of load (uniformly distributed or trapezoidal) and activate the corresponding button Along the line;

– set the direction and enter the load value;

– press the button OK in the dialog box;

– select elements on the diagram to whose nodes the load is attached;

– press the button OK In chapter Downloads;

– in the dialog box (Figure 6.7) assign nodes to which the load is attached (the nodes are circled in the diagram with green and yellow rings for the first and second anchor nodes, respectively);

– press the button or .

Figure 6.7 – Dialog box Assignment of load binding nodes along the line

When using the button Assign to selected item only the load will be assigned to one element (its number is indicated in the window). After assignment, the selection marker for this element will be extinguished, and control will pass to the next element in order.

If the button was pressed Repeat for all selected items, the toneload will be automatically assigned to all selected elements. Naturally, in this case it is necessary to be sure that the position of the nodes between which the load is specified in all selected elements corresponds to the loading intent.

Perform calculation.

Get various forms of presenting calculation results.

Print the results.

Report structure:

– methodology and procedure for performing work;

- results;

- conclusions.

The results are presented in the form of tables and graphic material, in accordance with the data obtained.

7. Test questions and defense of work

What is the peculiarity of calculating structures on an elastic foundation?

How to generate a rectangular mesh with variable spacing for a plate element in PC SCAD?

How to generate a rectangular mesh with a constant pitch for a plate element in PC SCAD?

What is special about entering loads for a plate element in PC SCAD?

Specifying loads distributed along a line on plate elements.

How to take into account the part of the base located behind the area occupied by the structure itself?

What type of slab on an elastic base is it?

Lab 7

Please tell me on what basis are stiffnesses assigned for 51 FE?

Why bother so much - you need to fill out the table in the cross-section once, set the approximate dimensions of the site, the well and save the cross-section file, and when you create the calculation diagram in scsd, select the site you created.
And step number 2 raises doubts - initially the coefficients of the elastic foundation can be assigned “from the bulldozer” and all elements of the slab are the same, which is why CROSS is needed to calculate them through several iterations

I cannot give a qualified answer to the question about rigidity. This is taken from many people's calculations as the best solution. Options such as firmly pinching it at two or three points or leaving the slab without any support at all also have a right to life. In the first case, we may get reinforcement peaks at pinch points; in the second case, we will get large settlements or errors in calculations. All these options are comparable to each other.

An anonymous reply to an anonymous comment. I described the same thing in general terms. Yes, I suffered until I understood the subtleties, so I shared my experience. Why is step 2 questionable? If because “originally. the coefficient can be assigned from the bulldozer. “, then let me note that there are many methods for applying the load to the foundation slab. The method of distributed load on a slab that I described in the second step was popular before the advent of CAD and still has fans. Therefore, it is always useful to analyze the calculation results using it. Often its results do not differ from the results of endless iterations, also described in the second step.

for 51 elements, the stiffness is assigned from the coe of the element bed 0.7C1 x A^2
C1 bed coefficient
And the area of ​​the element

Thanks for the information.

On the issue of rigidities of 51 FE, see “Calculation models of structures and the possibility of their analysis” by A.V. Perelmuter V.I. Slivker 2011 pp. 449-450

Calculation of the foundation slab in SCAD. Calculation of the foundation slab. Calculation in CROSS. Calculation in SCAD

6.5.7. Calculation of structures on an elastic foundation using tables (Part 1)

The full calculation of beams and slabs on an elastic foundation according to the hypothesis of an elastic half-space or a compressible layer according to tables of ready-made calculated values ​​is given in the book. Here only basic information is given on the classification of beams and slabs for selecting the necessary tables, as well as tables for the most important calculation cases.

Calculation of beams (strips) in a plane problem. The tables give reactive pressures, transverse forces and bending moments for strips taken as absolutely rigid, for strips of finite length and rigidity, infinite and semi-infinite. Cases of uniform load and load in the form of a concentrated force or moment applied in any section are provided.

A strip is considered absolutely rigid if its flexibility is t(dimensionless quantity) satisfies the inequality

Where E and ν - deformation modulus and Poisson's ratio of the soil, E and ν - elastic modulus and Poisson's ratio of the strip material, I- moment of inertia of the strip section, l- half-length of the strip, h- height, b‘ - width equal to 1 m.

Second approximation for t in formula (6.131) refers to strips of rectangular cross-section. Table 6.8 serves to calculate rigid strips for the most important case of loading with a concentrated force applied in any section of the strip.

The table has two inputs: by α, reduced to the half-length of the strip l- abscissa of load application points, and according to ξ, reduced to l- abscissas of sections for which the calculated value is established. The reference point is the middle of the strip, and it is assumed that for sections located to the right of the middle of the strip, the values ​​of ξ are positive, and those to the left are negative. Values ​​α and ξ are rounded to the first decimal place.

The table shows the ordinates of dimensionless quantities that allow you to determine the true values ​​of reactive pressures R, shear forces Q and bending moments M using equalities:

(implying that the force R is given in kN, and the half-length is in m).

In the tables, values ​​to the left of the force are marked with an asterisk. R. On the right are the values. If a force is applied in the left half of the band in the table for, all values ​​change sign.

Strips are considered to have finite length and stiffness if their flexibility index satisfies the inequality

(detailed tables for this case are given in the book).

Finally, long stripes when t> 10, when calculating, they are approximately taken as either infinitely long or semi-infinite. The strip is considered infinite when the force R applied at a distance a l, from the left end of the strip and at a distance a r from the right end satisfying the inequalities:

Where L- elastic characteristic of the beam, m:

If inequality (6.134) is valid only for or only for a r, the strip is called semi-infinite. In table Table 6.9 shows the values ​​of dimensionless quantities for an infinite strip, and table. 6.10 - for semi-infinite. The rules for using these tables are the same as the table. 6.8, with the only difference that in formulas (6.132) the quantity l must be replaced by the value L .

If the strip is loaded with a number of concentrated forces, then the diagrams for each force are determined separately, and then they are summed up.

The book also contains tables for the case of bending moment loading m .

Calculation of beams in the conditions of a spatial problem. In this case, the calculation method also depends on the beam flexibility index

Where A And b- half-length and half-width of the beam.

The beam is accepted as rigid if the flexibility indicator t≤ 0.5. A beam is considered long if

Where L is determined by equality (6.135),

and the conditions are satisfied:

» 0.15 ≤ β ≤ 0.3 λ > 2

The remaining beams are calculated as short, i.e. having a finite length and rigidity.

Rigid beams are calculated by replacing the actual load on the beam with an equivalent one in the form of a total vertical load R and moment m, applied in the middle of the beam.

Calculation of a slab on an elastic foundation
6.5.7. Calculation of structures on an elastic foundation using tables (Part 1) The full calculation of beams and slabs on an elastic foundation according to the hypothesis of an elastic half-space or compressible layer using tables of ready-made calculated values ​​is given in the book. Here only basic information is given on the classification of beams and slabs for selecting the necessary tables, as well as tables for the most important calculation cases.

Calculation of beams and slabs on an elastic foundation beyond the elastic limit (a manual for designers). Sinitsyn A.P. 1974

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 limit equilibrium are briefly outlined, and the problem of determining the maximum load-bearing capacity of a beam on an elastic foundation under various loads is considered. The determination of the maximum load for frames and grillages taking into account the influence of the elastic base is shown. A solution to problems for a prestressed beam is given. The influence of a two-layer base is considered. Problems related to slabs located on an elastic base with concentrated load in the center, at the edge and in the corner of the slab have been solved. A calculation was made for a prestressed and three-layer slab. At the end of the work, experimental data related to beams and slabs are presented, and a comparison with theoretical results is made. The book is intended for design engineers and may be useful to senior students of construction universities.

Preface to the first edition
Preface to the second edition
Introduction

Chapter 1. General principles of calculation
1.1. Conditions for the transition of beams on an elastic foundation beyond the elastic limit
1.2. Limit equilibrium for bending elements
1.3. General case
1.4. Formation of plastic areas at the base
1.5. Conditions for creating foundations of the least weight

Chapter 2. Beam on an elastic half-space
2.1. The greatest load is 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. Evenly distributed load
2.7. Beam of variable section
2.8. Grillage made of 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 ring 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 is in the elastic stage
3.2. Determination of ultimate load
3.3. Application of group diagrams
3.4. Pre-stressed beam on a layer of 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 your own weight
4.3. Selection of limit state design scheme
4.4. Example of definition of ultimate force
4.5. Calculation of a layered floor truss
4.6. Layered frame calculation
4.7. Beams on a nonlinear foundation
4.8. An example of calculating a beam on a nonlinear 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 plate
5.3. Load in the corner of the slab
5.4. Square slab on a two-layer base
5.5. Prestressed slab
5.6. The influence of local and general deformations of the slab beyond the elastic limit
5.7. Three-layer board
5.8. Load at the edge of the slab
5.9. Prefabricated slabs

Chapter 6. Application of computers to determine the limit state of the foundation
6.1. Finite element method
6.2. Ultimate load of high foundation beam
6.3. Defining plastic areas at the base
6.4. High foundation beam on an elastic-plastic base
6.5. Ultimate load of the beam, determined from the condition of the formation of plastic regions in the base
6.6. Using Beam Finite Elements
6.7. Calculation of limit displacements and loads

Chapter 7. Limit settlements of frame multi-storey buildings
7.1. Basic design provisions
7.2. Method for solving the problem and drawing up general equations
7.3. Calculation features depending on the foundation design (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. Modulus of deformation of the base
Bibliography

Beams and slabs on an elastic foundation are used mainly as design models for foundations, which are the main elements that ensure the overall strength and reliability of the structure.

As a rule, the calculation of the foundation is subject to increased requirements regarding its condition during the operation of the structures. Small deviations from the established values ​​in the area of ​​deformations or stresses, which are often present 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 will it be possible to correctly establish those most rational dimensions that ensure the necessary reliability of the structure 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 to directly, without special techniques, apply the general method of calculating structures using limit equilibrium.

The limit equilibrium method, created as a result of the work of our domestic scientists professors V.M. Keldysh, N.S. Streletsky, A.A. Gvozdeva, 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. Nila, F.G. Hoxha, R. Hill, M.R. Horn, F. Bleich, V. Prager, I. Guyon and others, some of these works have been translated into Russian.

Library: books on architecture and construction
Architectural and construction library Totalarch. Book: Calculation of beams and slabs on an elastic foundation beyond the elastic limit (a manual for designers). Sinitsyn A.P. Stroyizdat. Moscow. 1974. 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 limit equilibrium are briefly outlined,

5.11.1 To calculate slab foundations on an elastic foundation, it is recommended to use the following calculation models:

a) method of local elastic deformations,

b) method of linearly deformable half-space,

c) the method of an elastic layer on an incompressible base or with a variable modulus of soil deformation along its depth.

Method a), as a rule, should be used for weak, low-strength foundations, b) and c) - for low- and medium-compressible foundations when calculating flexible structures: beams, strips (including cross ones) and massive slabs.

5.11.2 Foundations on an elastic foundation should be calculated taking into account their flexibility. Beams
and tapes, with the ratio of their length and width l/b 1, are considered absolutely rigid in the transverse direction, and at 7 £ l/b£20 and t£ 1 - in the longitudinal direction. The flexibility index of beams (strips), taking into account the rigidity of the beam and base, is determined by formula (5.69), for slabs in the shape of a circle - by formula (5.70), polygon, with l/b

Where E and n are the deformation modulus, MPa, and Poisson’s ratio of the soil, respectively,

E 1, n 1 - elastic modulus, MPa, and Poisson’s ratio of the foundation material,

I- moment of inertia of the cross section of the foundation, m 4,

l And h- length and height of the foundation, m,

R- radius of the slab, m.

5.11.3 The calculation of foundations on an elastic foundation is carried out depending on the foundation model according to 5.11.1 and its operating conditions using numerical methods using appropriate programs, using a PC, or practical calculation methods using the appropriate tables.

Calculation of slab foundations loaded with various loads (Figure 5.13), using tables, is carried out according to the flexibility indicator a:

where n is the coefficient of transverse soil deformations,

E- soil deformation modulus, MPa,

L And b- length and width of the beam, m,

IN- beam rigidity, MPa∙m 4.

When a beam is loaded with several forces, the total forces are found by adding their ordinates of the same name. The calculation of a slab foundation on an elastic foundation is given in example D.7 of Appendix D.

Figure 5.13 - Schemes for loading beams with various loads:

a) evenly distributed,

b) concentrated,

Principles for calculating slab foundations on an elastic foundation
Principles for calculating slab foundations on an elastic foundation 5.11.1 To calculate slab foundations on an elastic foundation, it is recommended to use the following calculation models: a) local method