Calculation of brickwork for stability. Strength calculation of brickwork Program for brick wall stability analysis

Load on the partition at the level of the bottom of the first floor slab, kN	Values, kN
snow for II snow region	10006,74(23,00,5+0,51+0,25)1,4*0,001=115,7
rolled roof carpet-100N / m 2	1006,74(23,00,5+0,51+0,25)1,1*0,001=9,1
asphalt screed at p \u003d 15000N / m 3 with a thickness of 15 mm	150000,0156,7423,00,51,20,001=20,9
insulation-wood-fiber boards 80 mm thick with a density of p \u003d 3000N / m 3	30000,086,7423,00,51,20,001=22,3
Vapor barrier - 50N / m 2	506,7423,00,51,2*0,001=4,7
prefabricated reinforced concrete slabs covering - 1750N / m 2	17506,7423,00,51,1*0,001=149,2
weight of reinforced concrete farm	69001,10,01=75,9
weight of the cornice on the brickwork of the wall at p \u003d 18000N / m 3	18000((0,38+0,43)0,50,51-0,130,25)* 6,741,1*0,001=23,2
masonry weight above +3.17	18000((18,03-3,17)6,74 - 2,42,13)0,511,1*0,001=857
concentrated from crossbars of floors (conditionally)	1197505,690,530,001=1022
weight of window filling at V n \u003d 500N / m 2	5002,42,131,1*0,001=8,3

The total design load on the wall at elevation level +3.17:

N \u003d 115.7 + 9.1 + 20.9 + 22.3 + 4.7 + 149.2 + 75.9 + 23.2 + 857.1 + 1022 + 8.3 \u003d 2308.4.

It is allowed to consider the wall as divided in height into single-span elements with the location of the supporting hinges at the level of support of the crossbars. In this case, the load from the upper floors is taken to be applied at the center of gravity of the section of the wall of the overlying floor, and all loads P \u003d 119750 * 5.69 * 0.5 * 0.001 \u003d 340.7 kN within this floor are considered to be applied with the actual eccentricity relative to the center of gravity of the section ...

The distance from the point of application of the crossbar support reactions P to the inner edge of the wall in the absence of supports fixing the position of the support pressure is taken to be no more than a third of the crossbar embedment depth and no more than 7 cm.

With a depth of embedment of the crossbar into the wall a 3 \u003d 380 mm, a 3: 3 \u003d 380: 3 \u003d 127 mm\u003e 70 mm, we take the point of application of the reference pressure P \u003d 340.7 kN at a distance of 70 mm from the inner edge of the wall.

Estimated height of the pier on the ground floor

l 0 \u003d 3170 + 50 \u003d 3220 mm.

For the design scheme of the wall of the lower floor of the building, we take a rack with pinching at the level of the foundation cutoff and with a hinge support at the floor level.

Flexibility of a wall made of grade 100 silicate brick on grade 25 mortar, at R \u003d 1.3 MPa with a masonry characteristic α \u003d 1000

λ h \u003d l 0: h \u003d 3220: 510 \u003d 6.31

Buckling coefficient φ \u003d 0.96, in walls with a rigid upper support, buckling in the support sections may not be taken into account (φ \u003d 1) In the middle third of the wall height, the buckling coefficient is equal to the calculated value φ \u003d 0.96. In the supporting thirds of the height, φ varies linearly from φ \u003d 1 to the calculated value φ \u003d 0.96

The values \u200b\u200bof the buckling factor in the design sections of the walls, in the levels of the top and bottom of the window opening:

φ 1 \u003d 0.96 + (1-0.96)

φ 2 \u003d 0.96 + (1-0.96)

The values \u200b\u200bof bending moments at the level of support of the girder and in the design sections of the wall at the level of the top and bottom of the window opening, kNm:

M \u003d Pe \u003d 340.7 * (0.51 * 0.5-0.07) \u003d 63.0

M 1 \u003d 63.0

M 11 \u003d 63.0

The magnitude of the normal forces in the same sections of the wall, kN:

N 1 \u003d 2308.4 + 0.51 * 6.74 * 0.2 * 1800 * 1.1 * 0.01 \u003d 2322.0

N 11 \u003d 2322 + (0.51 * (6.74-2.4) * 2.1 * 1800 * 1.1 + 50 * 2.1 * 2.4 * 1.1) * 0.01 \u003d 2416.8

N 111 \u003d 2416.8 + 0.51 * 0.8 * 6.74 * 1800 * 1.1 * 0.01 \u003d 2471.2.

Eccentricities of longitudinal forces e 0 \u003d M: N:

e 0 \u003d (66.0: 2308.4) * 1000 \u003d 27 mm<0.45y=0.45*255=115мм

e 01 \u003d (56.3: 2322) * 1000 \u003d 24 mm<0.45y=0.45*255=115мм

e 011 \u003d (15.7: 2416.8) * 1000 \u003d 6 mm<0.45y=0.45*255=115мм

e 0111 \u003d 0 mm y \u003d 0.5 * h \u003d 0.5 * 510 \u003d 255 mm.

Load-bearing capacity of an eccentrically compressed rectangular partition

determined by the formula:

N \u003d m g φ 1 RA * (1- ) ω, where ω \u003d 1 + <=1.45,
, where φ is the coefficient of longitudinal bending for the entire section of a rectangular element h c \u003d h-2e 0, m g is a coefficient that takes into account the effect of long-term loading (at h \u003d 510 mm\u003e 300 mm, 1 is taken), A is the sectional area of \u200b\u200bthe wall.

Bearing capacity (strength) of the wall at the level of support of the crossbar at φ \u003d 1.00, e 0 \u003d 27 mm, λ c \u003d l 0: h c \u003d l 0: (h-2e 0) \u003d 3220: (510-2 * 27 ) \u003d 7.1, φ c \u003d 0.936,

φ 1 \u003d 0.5 * (φ + φ s) \u003d 0.5 * (1 + 0.936) \u003d 0.968, ω \u003d 1 +
<1.45

N \u003d 1 * 0.968 * 1.3 * 6740 * 510 * (1-
) 1.053 \u003d 4073 kN\u003e 2308 kN

Bearing capacity (strength) of the wall in section 1-1 at φ \u003d 0.987, e 0 \u003d 24 mm, λ c \u003d l 0: h c \u003d l 0: (h-2e 0) \u003d 3220: (510-2 * 24) \u003d 6.97, φ c \u003d 0.940,

φ 1 \u003d 0.5 * (φ + φ s) \u003d 0.5 * (0.987 + 0.940) \u003d 0.964, ω \u003d 1 +
<1.45

N 1 \u003d 1 * 0.964 * 1.3 * 4340 * 510 * (1-
) 1.047 \u003d 2631 kN\u003e 2322 kN

Bearing capacity (strength) of the wall in section II-II at φ \u003d 0.970, e 0 \u003d 6 mm, λ c \u003d l 0: h c \u003d l 0: (h-2e 0) \u003d 3220: (510-2 * 6) \u003d 6 , 47, φ c \u003d 0.950,

φ 1 \u003d 0.5 * (φ + φ s) \u003d 0.5 * (0.970 + 0.950) \u003d 0.960, ω \u003d 1 +
<1.45

N 11 \u003d 1 * 0.960 * 1.3 * 4340 * 510 * (1- ) 1.012 \u003d 2730 kN\u003e 2416.8 kN

Bearing capacity (strength) of the wall in section III-III at the level of the foundation cutoff with central compression at φ \u003d 1, e 0 \u003d 0 mm,

N 111 \u003d 1 * 1 * 1.3 * 6740 * 510 \u003d 4469 kN\u003e 2471 kN

So the strength of the wall is ensured in all sections of the lower floor of the building.

Working fittings	Design section	Design force M, N mm	Calculated characteristics			Design reinforcement	Accepted fittings
Working fittings	Design section	Design force M, N mm	, mm	, mm	Reinforcement class	Design reinforcement	Accepted fittings
In the lower zone	In extreme spans	123,80*10					, And s \u003d 760mm 2 in two flat frames
In the lower zone	In middle spans	94,83*10					, And s \u003d 628mm 2 in two flat frames
In the upper zone	In the second span	52,80*10					, And s \u003d 308mm 2 in two frames
	In all middle spans	41,73*10					, And s \u003d 226mm 2 in two frames
	On support	108,38*10					, And s \u003d 628mm 2 in one U-shaped mesh
	On a support	94,83*10					, And s \u003d 628mm 2 in one U-shaped mesh

Table 3

Loading scheme

Shear forces, kNm

In extreme spans

In middle spans

Table 7

Arrangement of rods		Sectional reinforcement, mm			R a c e t h a r c te r s t i c
Arrangement of rods		Until rods A break	Cut off	After the break of the rods A		mm x10	A according to table 9
In the lower crossbar area	In the extreme: at support A
	at support B
	Average size: at support B
In the upper crossbar area	At support B: from the extreme span
In the upper crossbar area	from the side of the middle span

Design section	Design force M, kN * m	Section dimensions, mm	Design characteristics	Longitudinal working reinforcement class AIII, mm		Actual bearing capacity, kN * m
Design section	Design force M, kN * m		R b \u003d 7.65 MPa	R s \u003d 355 MPa	Actual adopted	Actual bearing capacity, kN * m
In the lower zone of the outer spans
In the upper zone above the supports B at the column edge
In the lower zone of middle spans
In the upper zone above the supports C at the column edge

Ordinates

I z g and b and y u u h i e m o n t s, to N m

In extreme spans

In middle spans

The ordinates of the main moment diagram when loading according to 1 + 4 schemes

by the amount

M \u003d 145.2 kNm

Plot redistribution ordinates IIa

The ordinates of the main moment diagram when loading according to 1 + 5 schemes

Redistribution of efforts by reducing the reference moment M by the amount

Additional plot ordinates at M \u003d 89.2 kNm

Redistribution ordinates of the plot IIIa

Loading scheme

I z g and b and y u u h i e m o n t s, to N m

Shear forces, kNm

In extreme spans

In middle spans

		Longitudinal reinforcement Breaking fittings	Transverse reinforcement step	Shear force at the point of breaking rods, kN	The length of the launch of the broken rods for the place of the theoretical break, mm	Minimum value ω \u003d 20d, mm	The accepted value ω, mm	Distance from support axis, mm
		Longitudinal reinforcement Breaking fittings	Transverse reinforcement step	Shear force at the point of breaking rods, kN		Minimum value ω \u003d 20d, mm	The accepted value ω, mm	To the place of the theoretical break (to scale according to the material diagram)	To the actual place of the cliff

In the lower crossbar area	In the extreme: at support A
	at support B
	Average size: at support B
In the upper crossbar area	At support B: from the extreme span
In the upper crossbar area	from the side of the middle span

Вр1 with Rs \u003d 360 MPa, АIII with Rs \u003d 355 MPa

In the extreme sections between axes 1-2 and 6-7

In extreme spans

In middle spans

In the middle sections between axles 2-6

In extreme spans

In middle spans

Arrangement of rods		Sectional reinforcement, mm 2			Design characteristics
Arrangement of rods		Until the rods break	cut off	After breaking the rods		b * h 0, mm 2 * 10 -2	М \u003d R b * b * h 0 * A 0, kN * m
In the lower crossbar area	In the extreme span: at support A
	at support B
	Average span: at support B
	at support C
In the upper crossbar area	At support B: from the extreme span
	from the middle span
	At support C from both spans

Location of broken rods		Longitudinal __ fittings__ breakable reinforcement	Transverse reinforcement _quantity_	Shear force at the point of theoretical break of the rods, kN	The length of the launch of the broken rods for the place of the theoretical break, mm	Minimum value w \u003d 20d	The accepted value w, mm	Distance from support axis, mm
Location of broken rods		Longitudinal __ fittings__ breakable reinforcement	Transverse reinforcement _quantity_			Minimum value w \u003d 20d	The accepted value w, mm	To the place of the theoretical break (according to the diagram of materials)	To the actual place of the cliff
In the lower crossbar area	In the extreme span: at support A
	at support B
	Average span: at support B
	at support C
In the upper crossbar area	At support B: from the extreme span
	from the middle span
	At support C from both spans

Let's check the strength of a brick pillar of the bearing wall of a residential building of variable number of storeys in Vologda.

Initial data:

Floor height - Net \u003d 2.8 m;

Number of floors - 8 floors;

The step of the bearing walls is a \u003d 6.3 m;

The dimensions of the window opening are 1.5x1.8 m;

The dimensions of the section of the wall are 1.53x0.68 m;

Internal verst thickness - 0.51 m;

Cross-sectional area of \u200b\u200bthe wall-A \u003d 1.04m 2;

Length of the support platform for floor slabs per masonry

Materials: thickened front silicate brick (250Ch120Ch88) GOST 379-95, grade SUL-125/25, porous silicate stone (250Ch120Ch138) GOST 379-95, grade SRP -150/25 and thickened silicate hollow brick (250x120x88) GOST 379-95 brand SURP-150/25. For masonry of 1-5 floors, cement-sand mortar M75 is used, for 6-8 floors, masonry density \u003d 1800 kg / m 3, multi-layer masonry, insulation - expanded polystyrene brand PSB-S-35 n \u003d 35 kg / m3 (GOST 15588- 86). With multi-layer masonry, the load will be transmitted to the inner verst of the outer wall, therefore, when calculating the thickness of the outer verst and insulation, we do not take into account.

The collection of the load from the pavement and floors is presented in Tables 2.13, 2.14, 2.15. The design wall is shown in Fig. 2.5.

Figure 2.12. Design wall: a - plan; b - vertical section of the wall; c-calculation scheme; d - moment diagram

Table 2.13. Collection of loads on the pavement, kN / m 2

Load name	Standard value kN / m2	Calculated value kN / m2
Constant: 1. Layer of linocrome TKP, t \u003d 3.7 mm, weight of 1m2 material 4.6 kg / m2, \u003d 1100 kg / m3 2. Layer of HPP linocrome, t \u003d 2.7 mm weight of 1m2 material 3.6 kg / m2, \u003d 1100 kg / m3 3. Primer "Bitumen primer"
4. Cement-sand screed, t \u003d 40 mm, \u003d 1800 kg / m3 5. Expanded clay gravel, t \u003d 180 mm, \u003d 600 kg / m3, 6. Insulation - expanded polystyrene PSB-S-35, t \u003d 200 mm, \u003d 35 kg / m3 7. Paroizol 8. Reinforced concrete floor slab

Temporary: S0н \u003d 0.7ЧSqmЧSeЧСt \u003d 0.7Ч2.4 1Ч1Ч1

Table 2.14. Collection of loads on the attic floor, kN / m2

Table 2.15. Collection of loads on the interfloor overlap, kN / m2

Table 2.16. Collection of loads for 1 lm. from the outer wall t \u003d 680 mm, kN / m2

Determine the width of the cargo area using the formula 2.12

where b is the distance between the centerline axes, m;

a - the amount of support of the floor slab, m.

The length of the loading area of \u200b\u200bthe wall is determined by the formula (2.13).

where l is the width of the wall;

l f - width of window openings, m.

The determination of the cargo area (according to Figure 2.6) is carried out according to the formula (2.14)

Figure 2.13. Scheme for determining the cargo area of \u200b\u200bthe wall

The calculation of the effort N on the wall from the higher floors at the level of the bottom of the floors of the first floor is based on the cargo area and the existing loads on the floors, roofs and roofs, and the load from the weight of the outer wall.

Table 2.17. Collection of loads, kN / m

Load name	Calculated value kN / m
1. Cover design
2. Attic floor
3. Interfloor overlap
4. Outer wall t \u003d 680 mm

Calculation of eccentrically compressed unreinforced elements of stone structures should be carried out according to the formula 13

Exterior load-bearing walls should, at a minimum, be dimensioned for strength, stability, localized crushing and heat transfer resistance. To find out how thick should the brick wall be , you need to calculate it. In this article, we will consider the calculation of the bearing capacity of brickwork, and in the following articles, the rest of the calculations. In order not to miss the release of a new article, subscribe to the newsletter and you will find out what the thickness of the wall should be after all the calculations. Since our company is engaged in the construction of cottages, that is, low-rise construction, we will consider all calculations for this category.

Carriers walls are called that perceive the load from floor slabs, coverings, beams, etc. resting on them.

You should also take into account the brand of brick for frost resistance. Since everyone builds a house for themselves, at least for a hundred years, then with a dry and normal humidity conditions of the premises, a brand (M rz) from 25 and above is adopted.

When building a house, a cottage, a garage, outbuildings and other structures with dry and normal humidity conditions, it is recommended to use hollow bricks for external walls, since its thermal conductivity is lower than that of a solid one. Accordingly, with a heat engineering calculation, the thickness of the insulation will turn out to be less, which will save money when buying it. Solid bricks for external walls should be used only if necessary to ensure the strength of the masonry.

Reinforcement of brickwork is allowed only if an increase in the grade of brick and mortar does not allow providing the required bearing capacity.

An example of calculating a brick wall.

The bearing capacity of brickwork depends on many factors - on the brand of brick, the brand of mortar, on the presence of openings and their sizes, on the flexibility of the walls, etc. The calculation of the bearing capacity begins with the definition of the design scheme. When calculating walls for vertical loads, the wall is considered to be supported on hinged fixed supports. When calculating walls for horizontal (wind) loads, the wall is considered rigidly restrained. It is important not to confuse these diagrams, as the moment diagrams will be different.

The choice of the design section.

In deaf walls, the design section is I-I at the level of the bottom of the floor with a longitudinal force N and a maximum bending moment M. Often dangerous section II-II, since the bending moment is slightly less than the maximum and is equal to 2 / 3M, and the coefficients m g and φ are minimal.

In walls with openings, the section is taken at the level of the bottom of the lintels.

Let's take a look at section I-I.

From a previous article Collecting loads on the first floor wall take the obtained value of the total load, which includes the loads from the overlap of the first floor P 1 \u003d 1.8 t and the overlying floors G \u003d G n + p 2 + G 2 = 3.7t:

N \u003d G + P 1 \u003d 3.7t + 1.8t \u003d 5.5t

The floor slab rests on the wall at a distance of a \u003d 150mm. The longitudinal force P 1 from the overlap will be at a distance of a / 3 \u003d 150/3 \u003d 50 mm. Why 1/3? Because the stress diagram under the support section will be in the form of a triangle, and the center of gravity of the triangle is just 1/3 of the support length.

The load from the overlying G floors is considered to be applied centrally.

Since the load from the floor slab (P 1) is applied not in the center of the section, but at a distance from it equal to:

e \u003d h / 2 - a / 3 \u003d 250mm / 2 - 150mm / 3 \u003d 75 mm \u003d 7.5 cm,

then it will create a bending moment (M) in section I-I. Moment is the product of force on the shoulder.

M \u003d P 1 * e \u003d 1.8t * 7.5cm \u003d 13.5t * cm

Then the eccentricity of the longitudinal force N will be:

e 0 \u003d M / N \u003d 13.5 / 5.5 \u003d 2.5 cm

Since the load-bearing wall is 25cm thick, the calculation should take into account the value of the random eccentricity e ν \u003d 2cm, then the total eccentricity is:

e 0 \u003d 2.5 + 2 \u003d 4.5 cm

y \u003d h / 2 \u003d 12.5cm

When e 0 \u003d 4.5 cm< 0,7y=8,75 расчет по раскрытию трещин в швах кладки можно не производить.

The strength of the cage of an eccentrically compressed element is determined by the formula:

N ≤ m g φ 1 R A c ω

Odds m g and φ 1 in the considered section I-I are equal to 1.

III. CALCULATION OF STONE STRUCTURES

Load on the wall (Fig. 30) at the level of the bottom of the first floor slab, kN:

snow for the II snow region

rolled roof carpet - 100 N / m 2

asphalt screed at N / m 3 15 mm thick

insulation - wood-fiber boards 80 mm thick with a density of N / m 3

vapor barrier - 50 N / m 2

prefabricated reinforced concrete roof slabs - 1750 N / m 2

weight of reinforced concrete truss

weight of the cornice on the brickwork of the wall at N / m 3

masonry weight above +3.03

concentrated from the crossbars of the floors (conditionally without taking into account the continuity of the crossbars)

weight of window filling at N / m 2

total design load on the wall at elevation level +3.03

According to clauses 6.7.5 and 8.2.6, it is allowed to consider the wall as divided in height into single-span elements with the location of the supporting hinges at the level of support of the crossbars. In this case, the load from the upper floors is assumed to be applied at the center of gravity of the wall section of the overlying floor, and all kN loads within this floor are considered to be applied with an actual eccentricity relative to the center of gravity of the wall section.

According to clause 6.9, clause 8.2.2, the distance from the point of application of the crossbar support reactions P to the inner edge of the wall in the absence of supports fixing the position of the bearing pressure, no more than one third of the depth of the crossbar embedment and no more than 7 cm is taken (Fig. 31).

When the depth of the bolt is embedded in the wall and h \u003d 380 mm, and s: 3 \u003d 380: 3 \u003d

127 mm\u003e 70 mm assumes the reference pressure point

R \u003d 346.5 kN at a distance of 70 mm from the inner edge of the wall.

Estimated height of the pier on the ground floor

For the design scheme of the wall of the lower floor of the building, we take a rack with pinching at the level of the foundation cutoff and with a hinge support at the floor level.

The flexibility of a wall made of grade 100 silicate brick on grade 25 mortar at R \u003d 1.3 MPa according to table. 2, is determined according to note 1 to table. 15 with the elastic characteristic of the masonry a \u003d 1000;

buckling factor according to table 18 j \u003d 0.96. According to clause 4.14, in walls with a rigid upper support, the longitudinal deflection in the support sections may not be taken into account (j \u003d 1.0). In the middle third of the wall height, the buckling coefficient is equal to the calculated value j \u003d 0.96. In the supporting thirds of the height, j varies linearly from j \u003d 1.0 to the calculated value j \u003d 0.96 (Fig. 32). The values \u200b\u200bof the buckling coefficient in the design sections of the wall, in the levels of the top and bottom of the window opening

Figure: 31

values \u200b\u200bof bending moments at the level of support of the girder and in the design sections of the wall at the level of the top and bottom of the window opening

kNm;

Fig. 32

The magnitude of the normal forces in the same sections of the wall

Eccentricities of longitudinal forces e 0 = M: N:

Mm< 0,45 y \u003d 0.45 × 250 \u003d 115 mm;

Mm< 0,45 y \u003d 115 mm;

The bearing capacity of an eccentrically compressed wall of rectangular section in accordance with clause 4.7 is determined by the formula

where (j is the coefficient of longitudinal deflection for the entire section of a rectangular element; ); m g - coefficient that takes into account the effect of long-term loading h \u003d 510 mm\u003e 300 mm take m g = 1,0); AND - cross-sectional area of \u200b\u200bthe pier.

Brick is a fairly durable building material, especially solid, and when building houses with 2-3 floors, walls made of ordinary ceramic bricks usually do not need additional calculations. Nevertheless, situations are different, for example, a two-story house with a terrace on the second floor is planned. The metal girders, on which the metal beams of the terrace overlap will also be supported, are planned to be supported on brick columns made of facing hollow bricks 3 meters high, there will be more columns 3 meters high on which the roof will rest:

This raises a natural question: what is the minimum column cross-section that will provide the required strength and stability? Of course, the idea of \u200b\u200blaying out columns of clay bricks, and even more so the walls of a house, is far from new, and all possible aspects of calculating brick walls, piers, pillars, which are the essence of the column, are set out in sufficient detail in SNiP II-22-81 (1995) "Stone and reinforced stone structures". It is this regulatory document that should be guided in the calculations. The calculation given below is nothing more than an example of using the specified SNiP.

To determine the strength and stability of the columns, you need to have a lot of initial data, such as: a brick strength grade, the area of \u200b\u200bsupport of the crossbars on the columns, the load on the columns, the cross-sectional area of \u200b\u200bthe column, and if at the design stage none of this is known, then you can do in the following way:

with central compression

Designed by: Terrace measuring 5x8 m. Three columns (one in the middle and two at the edges) of facing hollow bricks with a section of 0.25x0.25 m. The distance between the axes of the columns is 4 m. Brick strength is M75.

With this design scheme, the maximum load will be on the middle bottom column. It is her that should be counted on for strength. The column load depends on many factors, in particular the area of \u200b\u200bconstruction. For example, the snow load on the roof in St. Petersburg is 180 kg / m & sup2, and in Rostov-on-Don - 80 kg / m & sup2. Taking into account the weight of the roof itself 50-75 kg / m & sup2, the load on the column from the roof for Pushkin, Leningrad Region, can be:

N from the roof \u003d (180 1.25 +75) 5 8/4 \u003d 3000 kg or 3 tons

Since the actual loads from the floor material and from people sitting on the terrace, furniture, etc. are not yet known, but the reinforced concrete slab is not exactly planned, but it is assumed that the floor will be wooden, from separately lying edged boards, then for calculating the load from the terrace we can take a uniformly distributed load of 600 kg / m & sup2, then the concentrated force from the terrace acting on the central column will be:

N from the terrace \u003d 600 5 8/4 \u003d 6000 kg or 6 tons

The dead weight of the columns with a length of 3 m will be:

N from the column \u003d 1500 3 0.38 0.38 \u003d 649.8 kg or 0.65 tons

Thus, the total load on the middle lower column in the column section near the foundation will be:

N with rev \u003d 3000 + 6000 + 2 · 650 \u003d 10300 kg or 10.3 tons

However, in this case, it can be taken into account that there is not a very high probability that the temporary load from snow, the maximum in winter, and the temporary load on the floor, the maximum in summer, will be applied simultaneously. Those. the sum of these loads can be multiplied by a probability factor of 0.9, then:

N with rev \u003d (3000 + 6000) 0.9 + 2 650 \u003d 9400 kgor 9.4 tons

The design load on the outermost columns will be almost two times less:

N cr \u003d 1500 + 3000 + 1300 \u003d 5800 kg or 5.8 tons

2. Determination of the strength of brickwork.

Brick grade M75 means that the brick must withstand a load of 75 kgf / cm & sup2, however, the strength of the brick and the strength of the brickwork are different things. The following table will help you understand this:

Table 1... Design compressive strengths for masonry

But that's not all. The same SNiP II-22-81 (1995) clause 3.11 a) recommends that, with the area of \u200b\u200bpillars and piers less than 0.3 m & sup2, multiply the value of the design resistance by the coefficient of working conditions γ c \u003d 0.8... And since the cross-sectional area of \u200b\u200bour column is 0.25x0.25 \u003d 0.0625 m & sup2, we will have to use this recommendation. As you can see, for brick grade M75, even when using M100 masonry mortar, the strength of the masonry will not exceed 15 kgf / cm & sup2. As a result, the design resistance for our column will be 15 0.8 \u003d 12 kg / cm & sup2, then the maximum compressive stress will be:

10300/625 \u003d 16.48 kg / cm & sup2\u003e R \u003d 12 kgf / cm & sup2

Thus, to ensure the required strength of the column, either use a brick of greater strength, for example, M150 (the calculated compressive resistance for the M100 solution grade will be 22 0.8 \u003d 17.6 kg / cm2) or increase the column cross section or use transverse reinforcement of the masonry. For now, let's focus on using a more durable facing brick.

3. Determination of the stability of a brick column.

The strength of the brickwork and the stability of the brick column are also different things and still the same SNiP II-22-81 (1995) recommends determining the stability of a brick column using the following formula:

N ≤ m g φRF (1.1)

m g - coefficient taking into account the effect of long-term load. In this case, relatively speaking, we were lucky, since at a section height h ≤ 30 cm, the value of this coefficient can be taken equal to 1.

φ - coefficient of longitudinal bending, depending on the flexibility of the column λ ... To determine this coefficient, you need to know the estimated length of the column l o, and it does not always coincide with the height of the column. The subtleties of determining the design length of the structure are not set out here, we just note that according to SNiP II-22-81 (1995) clause 4.3: "Design heights of walls and pillars l o when determining the coefficients of buckling φ depending on the conditions of their support on horizontal supports, one should take:

a) with fixed pivot bearings l o \u003d H;

b) with an elastic upper support and rigid pinching in the lower support: for single-span buildings l o \u003d 1.5H, for multi-span buildings l o \u003d 1.25H;

c) for free-standing structures l o \u003d 2H;

d) for structures with partially restrained support sections - taking into account the actual degree of restraint, but not less l o \u003d 0.8Hwhere H - the distance between floors or other horizontal supports, with reinforced concrete horizontal supports, the distance between them in the light. "

At first glance, our design scheme can be considered as satisfying the conditions of item b). i.e. you can take l o \u003d 1.25H \u003d 1.25 3 \u003d 3.75 meters or 375 cm... However, we can confidently use this value only when the lower support is really rigid. If a brick column will be laid out on a waterproofing layer of roofing material laid on a foundation, then such a support should rather be considered as hinged, and not rigidly pinched. And in this case, our structure in a plane parallel to the plane of the wall is geometrically variable, since the structure of the floor (separately lying boards) does not provide sufficient rigidity in the indicated plane. There are 4 ways out of this situation:

1. Apply a fundamentally different design scheme, for example - metal columns, rigidly embedded in the foundation, to which the floor girders will be welded, then, for aesthetic reasons, the metal columns can be overlaid with facing bricks of any brand, since the metal will bear the entire load. In this case, however, it is necessary to calculate the metal columns, but the estimated length can be taken l o \u003d 1.25H.

2. Make another overlap, for example, from sheet materials, which will allow considering both the upper and lower support of the column as articulated, in this case l o \u003d H.

3. Make diaphragm stiffness in a plane parallel to the plane of the wall. For example, lay not columns at the edges, but rather piers. This will also make it possible to consider both the upper and lower support of the column as articulated, but in this case it is necessary to additionally calculate the stiffness diaphragm.

4. Ignore the above options and calculate the columns as free-standing with a rigid bottom support, i.e. l o \u003d 2H... In the end, the ancient Greeks put their columns (albeit not made of bricks) without any knowledge of the resistance of materials, without the use of metal anchors, and there were no such carefully written building codes at that time, nevertheless, some columns stand and to this day.

Now, knowing the calculated column length, you can determine the flexibility factor:

λ h \u003d l o / h (1.2) or

λ i \u003d l o (1.3)

h - the height or width of the column section, and i - radius of gyration.

In principle, it is not difficult to determine the radius of gyration, you need to divide the moment of inertia of the section by the section area, and then extract the square root from the result, but in this case there is no great need for this. Thus λ h \u003d 2,300 / 25 \u003d 24.

Now, knowing the value of the flexibility coefficient, we can finally determine the buckling coefficient from the table:

table 2... Buckling coefficients for stone and reinforced stone structures
(according to SNiP II-22-81 (1995))

In this case, the elastic characteristic of the masonry α determined by the table:

Table 3... The elastic characteristic of the masonry α (according to SNiP II-22-81 (1995))

As a result, the value of the buckling coefficient will be about 0.6 (with the value of the elastic characteristic α \u003d 1200, according to item 6). Then the ultimate load on the central column will be:

N p \u003d m g φγ with RF \u003d 1 0.6 0.8 22 625 \u003d 6600 kg< N с об = 9400 кг

This means that the accepted section of 25x25 cm is not enough to ensure the stability of the lower central centrally compressed column. To increase stability, the most optimal would be to increase the section of the column. For example, if you lay out a column with a void inside one and a half bricks, with dimensions of 0.38x0.38 m, then not only will the sectional area of \u200b\u200bthe column increase to 0.13 m & sup2 or 1300 cm & sup2, but the radius of inertia of the column will also increase to i \u003d 11.45 cm... Then λ i \u003d 600 / 11.45 \u003d 52.4, and the value of the coefficient φ \u003d 0.8... In this case, the ultimate load on the central column will be:

N p \u003d m g φγ with RF \u003d 1 0.8 0.8 22 1300 \u003d 18304 kg\u003e N with rev \u003d 9400 kg

This means that the sections of 38x38 cm are enough to ensure the stability of the lower central centrally compressed column with a margin, and it is even possible to reduce the brick grade. For example, with the originally adopted grade M75, the maximum load will be:

N p \u003d m g φγ with RF \u003d 1 0.8 0.8 12 1300 \u003d 9984 kg\u003e N with rev \u003d 9400 kg

It seems to be all, but it is desirable to take into account one more detail. In this case, it is better to make the foundation tape (single for all three columns), and not columnar (separately for each column), otherwise even small subsidence of the foundation will lead to additional stresses in the column body and this can lead to destruction. Taking into account all of the above, the most optimal section of the columns will be 0.51x0.51 m, and from an aesthetic point of view, this section is optimal. The cross-sectional area of \u200b\u200bsuch columns will be 2601 cm & sup2.

An example of calculating a brick column for stability
with eccentric compression

The extreme columns in the projected house will not be centrally compressed, since the girders will rest on them only on one side. And even if the girders are laid on the entire column, still, due to the deflection of the girders, the load from the floor and the roof will be transferred to the extreme columns not in the center of the column section. Where exactly the resultant of this load will be transmitted depends on the angle of inclination of the crossbars on the supports, the elastic moduli of the crossbars and columns, and a number of other factors. This displacement is called the eccentricity of the load application eo. In this case, we are interested in the most unfavorable combination of factors in which the load from the floor to the columns will be transmitted as close as possible to the edge of the column. This means that in addition to the load itself, the columns will also be affected by a bending moment equal to M \u003d Ne o, and this point must be taken into account in the calculations. In general, stability testing can be performed using the following formula:

N \u003d φRF - MF / W (2.1)

W - moment of resistance of the section. In this case, the load for the lower extreme columns from the roof can be conventionally considered centrally applied, and the eccentricity will only create the load from the floor. With an eccentricity of 20 cm

N p \u003d φRF - MF / W \u003d1 0.8 0.8 12 2601 - 3000 20 2601· 6/51 3 \u003d 19975.68 - 7058.82 \u003d 12916.9 kg\u003eN cr \u003d 5800 kg

Thus, even with a very large eccentricity of the load application, we have more than two times the safety margin.

Note: SNiP II-22-81 (1995) "Stone and reinforced masonry structures" recommends using a different method for calculating the section, taking into account the features of stone structures, but the result will be approximately the same, therefore, the calculation method recommended by SNiP is not given here.

Calculation of brickwork for stability. Strength calculation of brickwork Program for brick wall stability analysis

with central compression

An example of calculating a brick column for stability with eccentric compression

An example of calculating a brick column for stability
with eccentric compression