The load on the beam from the wall. Leaning a wooden beam on a brick wall: doing it right. From floor slabs

This article discusses the schemes of classical constructive solutions support nodes of load-bearing metal beams of floors (coverings) on the brick walls of buildings. Using these schemes in design beam floors will save the designer from a lot of routine calculations related to the layout of the support nodes of the beams, the selection of sections of individual elements (ensuring the operability of the nodes) and the calculation of their field connections.

The decision to choose one of the options proposed below for the design of the nodes supporting the beams on the walls is based on the magnitude of the support reaction (support pressure at the end of the beam).

According to the requirements of current regulations, steel beams must be supported by load-bearing stone walls through steel or reinforced concrete distribution pads, the main function of which is to equalize the pressure under the ends of the beams and prevent local collapse of the masonry (local destruction of the masonry under the supporting sections of the beams from collapse).

Nodes No. 1, 2, 3, 4 provide articulated supporting beams directly on the brickwork of the walls through the layer cement-sand mortar 15 mm thick. The support pressure under the end of the beam embedded in the wall is transmitted to the masonry through the support metal plates 20 mm thick, the dimensions of which are assigned in such a way that the average pressure under the plate (within the compression area) does not exceed the minimum admissible the value of the design resistance of the masonry, provided that the masonry is made of solid ceramic brick normal strength on hard cement mortar.

If the value of the support pressure exceeds 100 kN (≈10 tons), then, in accordance with the requirements of SNiP ll-22-81 *, it is necessary to install a reinforced concrete distribution pad with a thickness of at least 100 mm, reinforced with two grids according to the calculation (supporting the carrier steel floor beams directly on the brickwork of the walls in this case is not allowed). In this case, the supporting nodes of the beams are performed tough- see Knots No. 4, 5.

Node No. 1 (articulated)
Brick wall thickness b=380 mm. The limiting value of the support reaction R=0.6 t.

Knot No. 2 (articulated)
Brick wall thickness b>380 mm. The limiting value of the support reaction R=0.7 - 3.0 t.

Knot No. 3 (articulated)
Brick wall thickness b>380 mm. The limiting value of the support reaction R=3.1 - 5.0 tons.

Knot No. 4 (articulated)
Brick wall thickness b>380 mm. The limiting value of the support reaction R=5.1 - 7.0 tons.

Node #5 (hard)
Brick wall thickness b>380 mm. The limiting value of the support reaction R=10.1 - 18.0 t.

Knot #6 (hard)
Brick wall thickness b>380 mm. The limiting value of the support reaction R=18.1 - 20.0 t.

Notes (important!!!):

• All friction connections of elements (in all nodes) are performed on anchor bolts accuracy class B, strength classes 5.8 and 8.8. High-strength bolts may also be used.
• The legs of all fillet welds (in all nodes) should be taken according to the smallest thickness of the elements being welded, but not less than the values ​​\u200b\u200bspecified in Table 38 of SNiP II-23-81 *.
• If the mode of operation of the building is characterized by the presence of dynamic loads, all elements and details of the nodes must be checked by the endurance calculation.
• The steel grade of all metal elements and parts of assemblies is accepted according to table 50x of SNiP II-23-81 *, as for structures of the 2nd group (in the absence of dynamic, vibration and moving loads).

Table of contents:

• Cover material and design
• Beam termination methods
• Brick wall thick and leaning beams on it
• Beam support when the wall thickness is reduced
• Installation and fastening of beams
• Floor installation
• Supporting a metal beam brick wall
• Summarizing

The most important element in the construction of any house is the ceiling. The floor structure can be based on the use of beams and slabs, which, in turn, can be wooden, metal, or concrete. Of particular interest is the specificity of the installation of floors on a brick wall, since the construction is precisely brick houses very common. Supporting a beam on a brick wall or, accordingly, supporting a slab on a brick wall is the most important factor in the reliability and safety of the entire floor.

The choice of support design depends on the material, the depth of embedment, fastening (anchoring) in the wall.

Main hallmark supporting the structure on a brick wall is the possibility of fairly free deformation of the ends of the beam during its deflection. The safety and reliability of the structure can only be achieved by ensuring the correct connection of the beam with the wall, which excludes dangerous stresses in the material even when exposed to extreme temperature conditions. When choosing a support design, the material, embedment depth, fastening (anchoring) in the wall are fully taken into account.

Cover material and design

Table for calculating the cross section of floor beams.

In general, the floor is the carrier building construction, subdivided by purpose: interfloor, attic, attic. Structurally, the ceiling can be divided into two types: prefabricated (longitudinal beam and transverse flooring) and monolithic (slab).

In the construction of private houses, prefabricated floors using wooden beams are most used. Such material is made from durable hardwood and softwood. The size of a standard instance, depending on the purpose of the floor and loads, varies within:

• height 150-300 mm,
• width 100-250 mm.

To increase durability, the timber is impregnated with an antiseptic and oiled.

Reinforced bearing structures sometimes performed using metal beams. Standard steel beams are offered for this purpose. Safety standards establish that in the case of the use of such beams, their ends must rest on the brickwork through distribution pads.

Monolithic floors are made from reinforced concrete slabs. Factory plates are used, consisting of reinforcement and concrete mass co standard sizes. To reduce the weight of the slab, as a rule, they are made hollow.

Beam termination methods

Scheme for sealing the ends of wooden beams in attic floor into a wall 2 bricks thick.

The reliability and safety of the floor is largely determined by the correct installation of the beam into the wall. The termination determines the nature of the bearing on the brick wall, and this stage of construction is the most important.

A wooden beam is installed in a niche made in brickwork, up to 150 mm deep. The butt ends undergo a certain treatment: the butt is crimped at an angle of about 60º, impregnated with an antiseptic and resin, wrapped with roofing felt or roofing material. The wrapped ends are laid in a brick wall with a gap of 30-50 mm from the back wall of the niche. The gap is filled with thermal insulation ( mineral wool, felt, etc.). The laid ends, as a rule, are coated (sealed) with a solution of concrete, bitumen, or covered with a layer of roofing.

Thick brick wall and beams supported on it

In the case where the thickness of the brick wall exceeds 600 mm (2.5 bricks), several great way terminations. The nest in the brickwork is made in such a way that between the end of the beam and back wall niches remained a distance of at least 100 mm. The total depth of the niche is selected taking into account the fact that the beam must rest on the wall for a length of at least 150 mm. The gap left allows you to put in it thermal insulation material and provide an air gap.

The lower part of the nest is reinforced with concrete mortar, bituminous layer and two layers of roofing felt or roofing material. This creates a cushion for laying, which at the same time levels the surface of the masonry. The niche in its upper and lateral parts is covered with roofing felt.

Beam support when the wall thickness is reduced

The scheme of embedding the ends of the beam into a wall with a thickness of 0.64 m or more.

When performing overlapping on brick walls with a thickness of about 500 mm (2 bricks), the termination method should be changed. In a niche up to 250 mm deep, left in the brickwork, is installed wooden box(box) with 2-3 walls. Tarred felt is placed between the back wall of the niche and the box. The walls of the box are treated with an antiseptic and impregnated with resin.

The lower part of the niche is leveled with two layers of roofing felt or roofing felt. Side walls nests are insulated with felt. The box is installed in a niche so that it presses the felt. The floor beam rests on the lower part of the box for a length of at least 150 mm.

With a reduced thickness of a brick wall, the thickness of the wall remaining after the formation of a niche should be controlled. With a wall thickness of less than 50 mm, there is a danger of cold penetration and, therefore, it is necessary to provide additional insulation in the area where the beam rests on a brick wall.

Installation and fastening of beams

The process of mounting beams in the manufacture of floors depends on the purpose of the floor, its area and loads. Usually wooden beam distributed along load-bearing brick walls at a distance of 600 to 1500 cm from each other. The termination of the beams begins with the extreme ones and is evenly distributed along the length of the wall. It is recommended to provide a gap between the end beam and the edge of the wall of at least 5 cm.

Scheme of laying floors and subsequent fixation.

An important element installation of the floor is to check the horizontal position of the fastening of the beams and the equal level of the location of all beams relative to the floor. Horizontal deviation or level unevenness will cause additional load in the bearing area on the brick wall, especially after further laying of the transverse floor boards.

It is possible to increase the reliability and rigidity of bearing on a brick wall by using additional fasteners. Steel anchors have found the greatest application. The anchor is strengthened so that between outer surface wall and its end there was a distance of at least 15 mm. The anchor and the floor beam are fastened with nails and a metal plate with a size of at least 6x50 mm.

Floor installation

After the installation is completed and the beams are sealed, the transverse flooring of the floor is installed. For the manufacture of flooring, boards with a thickness of 25-45 mm, thick plywood are used. The flooring is installed on top of the layers of thermal insulation. In the manufacture of interfloor ceilings, a soundproofing layer is also laid. The installation of the flooring is carried out on top of the bars (lags), which are attached across the load-bearing beams.

In the manufacture of the floor, you must use a standard tool. The following set of tools is recommended.

For processing and fastening wooden elements:

• hacksaw,
• axe,
• a hammer,
• Bulgarian,
• drill,
• perforator (for working with bricks).

To carry out measurements and measurements:

• roulette,
• ruler,
• level.

Before calculating the support of jumpers on a brick wall, you should find out a few important points. What is a jumper? This is the part of the wall that covers the door and window openings and holds the masonry over the openings. When the gravity of the ceiling falls directly on the wall above the window and doorways, use prefabricated carriers reinforced concrete elements. If there is no load on the walls, and the width of the openings does not exceed 2 m, use non-bearing reinforced concrete or ordinary brickwork using solutions increased strength laying reinforcing bars to support the bottom row of bricks. It happens that instead of ordinary jumpers, wedge-shaped ones are made, which also serve as a decoration for the facade. For the same purpose, arched lintels are erected in 4 meter spans. This type of masonry is used to organize floors in buildings with vaults. In this case, all transverse and longitudinal seams must be filled during the laying of the lintels.

Jumpers are designed to cover door and window openings in the brick walls of buildings for various purposes.

If the vertical seams are not sufficiently filled, then the loads will first shift the individual bricks, and then the destruction of the masonry. Be sure to observe the horizontal rows of ordinary jumpers, the rules for dressing masonry from a whole brick. In masonry, a solution of at least 25 grades is used. The height of an ordinary jumper should be about 5 rows of masonry, and the length should exceed the width of the opening itself by 50 cm. The laying of lintels is done using formwork from boards 40-50 mm thick. Spread the mortar over the formwork about 2 cm thick. Reinforcement bars are then embedded in the solution. The layer goes under the first row of bricks of ordinary lintels. The lintels are reinforced with rods 6 mm in diameter made of round steel. The calculation of the number of the rod is simple, it is placed one for every half-brick, but for the entire lintel at least 3. In the lintel, the reinforcement works in tension, it perceives tensile forces from the masonry. The ends of the rods are released 25 cm beyond the edges of the opening, ending with a hook.

Brick lintel

The formwork is supported on bricks released from the masonry. After removing the formwork, the bricks are cut down. If the width of the opening exceeds 1.5 m, the formwork rests on a circle of boards, which are placed on the edge. It happens that, in addition to plank formwork, they use tubular inventory support-circles designed by Ogarkov. Such a support structure is very simple to manufacture. To do this, 2 cut pipes 48 mm in diameter are made and inserted into a pipe cut with a diameter of 60 mm. During masonry, they circled the pipes apart in such a way that the ends of smaller diameter went inside the grooves that were left in the masonry. 2 circles are placed on each opening and are used if window and door blocks. When using other types of circles, window and door blocks are inserted after the circles are removed. Wedge-shaped and arched lintels with wedge-shaped seams are laid out of ordinary clay bricks, the thickness of which is 25 mm at the top of the lintel, and about 5 mm at the bottom.

Masonry of ordinary lintels: a - facade of the lintel, b - section of the lintel, c - laying of the lintel on the plank formwork, d - laying of the lintel on inventory tubular circles; 1 - reinforcing bars, 2 - boards, 3 - wooden circles, 4 - Ogarkov's tubular circles.

Of course, at first the walls are erected to the level of the lintels, while the supporting heel is laid out of brick, hewn before the laying of the lintels is arranged. The direction of the reference plane is determined using a template, that is, the angle of deviation from the plane along the vertical. According to the prepared formwork, which is held by the circles, masonry is carried out in transverse rows. The calculation of masonry rows on the formwork is marked in such a way that their number, taking into account the thickness of the seam, is odd. In this case, the rows of masonry are considered horizontally. The odd row of bricks in the center is called the castle, and it is clearly located in vertical position at the center of the jumper. Laying arched and wedge-shaped lintels on both sides evenly from the heel to the lock so that it is wedged in the lock with an odd central brick. Using a lace, check the correct direction of the seams. At the point of mating lines of intersection of the supporting parts, a lace is attached. Wedge-shaped lintels cannot be laid if the span exceeds 2 m.

Arched lintels, vaults

Scheme of bricklaying of arches and vaults (arched lintels).

Arched lintels, vaults and arches are laid in the same sequence as wedge-shaped ones. The seams formed by the masonry between the rows should be perpendicular to the masonry of the outer and lower surfaces of the arch. The seams form a wedge shape with expansion upwards and narrowing below. In vaults and arches, the force from the load to the curve of the arch acts tangentially. Beds of rows lie perpendicular to the direction of pressure. With this arrangement of rows, this is the first rule for cutting brickwork. The masonry in the seams is tightly filled with mortar. In the process of doing work from above, the surface of the vaults is rubbed with a mortar 1/4 brick thick. The correctness of the seams and laying of the rows is checked with a cord, which is fixed in the center of the arch. The position of each row is checked with a square template and cord. The design of the formwork for laying arches and vaults must ensure its uniform lowering when stripping. They put wedges under the circles, if they are loosened, the formwork drops. Keeping arched and wedge-shaped lintels in the formwork in time depends on the brand of mortar and air temperature.

Reinforced concrete lintels

prefabricated reinforced concrete products(Reinforced concrete products) used in construction are produced at specialized factories and mounted directly at construction sites.

During the construction of houses, prefabricated reinforced concrete lintels are used to cover door and window openings.

Prefabricated reinforced concrete lintels are distinguished by types: bar, beam with a quarter for support (PG), slabs with a width of more than 250 mm (PP) and facade (PF).

Reinforced concrete products are made for openings using reinforcing embedded products 0.4-0.6 cm and heavy concrete mix M 250. Structurally, jumpers are distinguished as bearing and non-bearing. Bearings are considered to be those that carry the load of the floor in addition to the mass of the masonry above it. Non-bearing ones include those that carry the load of their own weight and those masonry sections that are located above them.

By type, they distinguish: bar up to 250 mm wide (PB), beams with a quarter for support (PG), slabs with a width of more than 250 mm (PP) and facade (PF), which are designed to cover openings with quarters with the thickness and width of the part protruding in the opening masonry over 250 mm. Using a level, check the supports and installation sites before installation and spread the solution.

Leaning on a brick wall should be at least 250 mm deep, on partitions - at least 200 mm. Non-load-bearing elements with a two-meter span can also be laid by hand, while heavy load-bearing elements are slinged by mounting loops and installed by crane. Laying is checked by level. They are assembled from several elements to cover the entire width of the masonry opening, the side parts of the lintels enter the plane of the brick wall without protrusions. During installation, jumpers should only be laid in a certain position. The bearing capacity may vary depending on the location of the reinforcement and the quantity.

Metal jumper device

The advantage of prefabricated jumpers is the speed of installation, ease of selection and reliability.

When it is not possible to install prefabricated jumpers, you can mount metal ones. Brickwork perfectly holds its own weight after the mortar has gained strength. Of course, this is the case if there is no overlap load and a moderate window width. But until the mortar has hardened and gained strength, the brickwork above the opening needs support. The advantage of prefabricated jumpers is the speed of installation, ease of selection and reliability. There is no need to do a calculation. But they, as we have already said, are produced only in the factory and are very heavy. For monolithic reinforced concrete, it is necessary to make a calculation, selection of height and reinforcement. It is made more difficult, using formwork. But installation can be done directly at the construction site. Well, you can install jumpers from metal rolled profiles, such as corners, channels or I-beams.

when selecting metal elements, it is necessary to make a calculation so that there is no more deflection of the jumper than the permissible one, and to determine the required strength of the selected metal elements. In this case, the calculation is made according to the following conditions:

• strength is determined by the formula,

Mp \u003d 1.12 * W * R,

where Mp - depends on the load and the length of the jumper, as well as the reliability factor;

W - resistance metal element, which is taken from reference books. If the jumpers are made up of 2 corners or 2 channels, then the moment of resistance of the composite element is equal to the sum of each of the elements of the moments of resistance;

R is the resistance of steel.

• deflection is determined by the formula:

Mn * L / (10EI) \u003d 1/200,

where Mn is the standard moment, depending on the load and the length of the jumper;

L is the estimated length, which is equal to the sum of the clear width on the third of each side of the web of the support length;

I is the moment of inertia;

E is the modulus of elasticity of steel;

1/200 is the maximum allowable deflection.

Calculation for door and window openings

In order to fit a metal jumper, the load from the brickwork is calculated per 1 rm of the jumper.

Consider, for example, the selection of a jumper for a door, and then. Suppose that the opening of the proposed door in the wall has a thickness of 0.25 mm. The opening will not be supported by the ceiling. Above the lintel, the height of the masonry is 0.9 m, and the width of the opening is 1 m. We select a metal lintel. To do this, it is necessary to calculate the load from brickwork per 1 pm of the jumper, if specific gravity bricks 1.8 t/cu.m.

q \u003d 0.25 * 0.9 * 1.8 * 1 \u003d 0.41 t / m.

Now, according to the formula, we determine the moment М = qL2/8,

L is the estimated length;

200 - depth of support of the jumper. Mn = 0.065 t*m;

L \u003d 1000 + 2 * 200 / 3 \u003d 1130 mm;

Мр = 73 kN*cm.

Required moment of resistance according to the strength condition:

W \u003d 65 / (1.12 * 21) \u003d 2.76 cu. cm.

Moment of inertia:

I \u003d 200Mn * L / (10E) \u003d 7.85 cm4.

You need to use a jumper, which consists of 2 corners 50 * 50 * 5

W = 7.88 cu. cm > 0.5 * 2.76 cc cm, I \u003d 11.2 cm4\u003e 0.5 * 7.85 cm4.

Simplified calculation of a metal jumper

It will not be difficult for a person who has encountered resistance to materials to deal with such a calculation, but for the rest, these concepts can be complex and incomprehensible. The calculation of the cross section of metal jumpers for . The calculation consists of determining the load acting on the jumper; determining the maximum bending moment acting on the cross section of the jumper; selection of the cross section of the jumper.

We determine the load per 1 pm jumper by the formula:

q 1 \u003d p * b * h,

It is necessary to calculate the cross section of a metal lintel for a brick partition.

where p (kg / m3) is the density of the partition material, taking into account masonry mortar and plaster mortar. Density cement mortar- up to 2200, which must be taken into account when laying from hollow bricks, you can multiply the density of the material by 1.1. The density of a solid brick is 1600 - 1900; the density of a hollow brick is 1000 - 1450.

b (m) - wall thickness. For example, a half-brick brick partition will be 15 cm.

h - the height above the lintel of the brick wall, taking into account the bricks that will be laid on the corner in the case of a lintel from the corners.

For a meter-wide opening for a brick partition half a brick thick, the load will be q 1 = 142.5 kg / m.

In this case, we carried out the calculation for the partition. For load-bearing walls, it is also necessary to take into account the load from the ceiling.

Required modulus and design load

Let's use the following formula:

where n is the number of metal profiles;

The density of the cement mortar is up to 2200, you can multiply the density of the material by 1.1.

P - own weight per 1 pm of the profile, determined by the assortment. As a rule, for metal lintels, the weight does not exceed 1-2% of the total weight of the wall above the lintel, so it can be calculated as a correction factor of 1.1.

Thus, for a meter-long opening of a brick partition half a brick thick, the total design load is equal to: q = 157 kg / m.

Now the selection of the required section. For a beam resting on 2 supports, the maximum bending moment will be at the middle of the beam:

M max \u003d (q * 1 sq.m) / 8 \u003d 19.6 kg / m

For a one-meter opening half a brick wide, the required moment of resistance will be:

W required \u003d M max / R y \u003d 0.933 cu. cm,

where R y is the design resistance of steel equal to 2100 kgf / sq. cm

The resulting value is divided by the number of profiles that we will use when arranging the jumper. It is more rational to use at least 2 profiles for brick partitions. Wreq = 0.47 cc Next, in the assortment, select the type of profile and find a value greater than in the calculation. For a meter wide opening for a brick partition half a brick thick, 2 equal-shelf corners 28 * 28 * 3 mm are enough. The support of metal lintels on the walls must be at least 250 mm.

Calculation of jumpers for load-bearing brick walls

The calculation of the lintel for load-bearing walls is almost the same as the previous calculation, only you need to determine the load on the lintel and select correct scheme calculation. If the lintel is a load-bearing beam above the opening, it can be calculated as a beam on hinged supports.

b - in this case it will be 2 bricks, that is, 0.51 - 0.55 m.

Calculation of the cross section of a metal lintel for a brick load-bearing wall.

h - masonry above the jumper, which will carry the load, can be defined as h \u003d L / 2. So, for a one and a half meter opening with a length and width of 2 bricks, the load will be 755.3 kg / m. Floor slabs weigh a lot. You can take their weight in the range of 800-1000 kg / sq.m. Hollow core slabs weigh approximately 320 kg/sq.m. In addition, 100 kg/sq.m. will be provided with insulation and screed. So, with 6 m hollow core slabs, the load will be 2400 kg/m. The linear design load will be 3167 kg/m. The maximum bending moment for a web, which is affected by a distributed and concentrated load, is calculated by the formula:

M max \u003d (q * l 2) / 8 + (Q * l) / 4 \u003d 1133.7 kg / m

Required drag moment: Wreq = (1133.7 * 100) / (2100 * 2) = 27.0 cc

You can make a jumper from steel hot-rolled corners, unequal or equal-shelf, profile pipes. For an opening 1.5 m long and 2 bricks wide, 2 unequal corners 110 * 70 * 8 mm are enough. Instead of 2 required corners, you can use 4 90 * 56 * 5.5. The support of such a jumper on the walls must be at least 250 mm. Bending:

f = (5 * q * L 4) / (384 * E * I z), where

E - the modulus of elasticity is 2 * 10 10 kg / sq.m - for steel.

I z - moment of inertia, according to the assortment, according to the selected profile.

For a jumper of 2 corners f = (5 * 3167 * 1.5 4) / (384 * 2 * 10 10 * 2 * 171.54 * 10 -8) = 0.003045 m. According to the requirements of "Load and Impact" SNiP 2.01.07-85, the maximum deflection for lintels must not exceed 1/200 of the span. According to our calculation, 150/200 = 0.75 cm. The condition is met.

Exterior load-bearing walls should, at a minimum, be designed for strength, stability, local collapse and resistance to heat transfer. To find out how thick should a brick wall be , you need to calculate it. In this article, we will consider the calculation bearing capacity 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, coatings, beams, etc. resting on them.

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

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

Reinforcement of masonry allowed only in the case when the increase in the grade of brick and mortar does not allow to provide the required bearing capacity.

An example of the calculation of a brick wall.

The bearing capacity of brickwork depends on many factors - on the brand of brick, 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 by hinged-fixed supports. When calculating walls for horizontal loads (wind), the wall is considered to be rigidly clamped. It is important not to confuse these diagrams, since the moment diagrams will be different.

Choice of design section.

In blank walls, the section I-I at the level of the bottom of the floor with the longitudinal force N and the maximum bending moment M is taken as the calculated one. It is 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 look at the section I-I.

From a previous article Collection of loads on the wall of the first floor we take the obtained value of the total load, which includes the loads from the floor of the first floor P 1 \u003d 1.8t and the overlying floors G \u003d G P + 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 a=150mm. The longitudinal force P 1 from the overlap will be at a distance a / 3 = 150 / 3 = 50 mm. Why 1/3? Because the stress diagram under reference area will be in the form of a triangle, and the center of gravity of the triangle is just 1/3 of the length of support.

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

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

e = h / 2 - a / 3 = 250mm / 2 - 150mm / 3 = 75 mm = 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 = P 1 * e = 1.8t * 7.5cm = 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

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

e 0 \u003d 2.5 + 2 \u003d 4.5 cm

y=h/2=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 masonry 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 section under consideration, I-I are equal to 1.

It is required to collect loads on a monolithic floor beam of a residential building (a beam along the “2” axis in the “B-C” axes in Fig. 1). Dimensions of the beam section: h = 0.5 m, b = 0.4 m. Take the floor design according to the drawing in.

Decision

This type of building belongs to II class of responsibility. Reliability factor for responsibility γn = 1.0.

The composition of the floor and the values ​​of permanent loads will be taken from example 1.1.

The loads acting on the beam are assumed to be linearly distributed (kN/m). To do this, uniformly distributed loads on the floor are multiplied by the width of the cargo section, which is equal to the frame pitch for the middle beams. In our example, see fig. 1 the width of the cargo section is B = 6.6 m. It remains to multiply the constant load calculated in example 1.1 by this value and write it down in table 1:

q1 \u003d 5.89 * B \u003d 5.89 * 6.6 \u003d 38.87 kN / m;

q1p \u003d 6.63 * B \u003d 6.63 * 6.6 \u003d 43.76 kN / m.

Table 1

Collection of loads on the floor beam

short-term ν1

long p1

Calculate the load from the own weight of the beam.

The volumetric weight of reinforced concrete is 2500 kg/m3 (25 kN/m3). With a beam height h = 0.5 m and its width b = 0.4 m, the standard value of the load from its own weight is

q2 \u003d 25 * h * b * γn \u003d 25 * 0.5 * 0.4 * 1.0 \u003d 5.0 kN / m.

Load safety factor γt = 1.1, then the calculated value will be:

q2р = q2*γt =5*1.1 =5.5 kN/m.

The total standard constant load is

q = q1 + q2 = 38.87 + 5.0 = 43.87 kN/m;

calculated:

qр = q1р + q2р = 43.76 + 5.5 = 49.26 kN/m.

Reducing factors φ1, φ2, φ3 or φ4, when calculating beams, the standard values ​​of loads can be reduced depending on the cargo area A, m2, of the calculated element by multiplying by the combination factor φ. With a cargo area A = 6.6 * 7.2 = 47.52 m2 and with A = 47.52 m2 > A1 = 9.0 m2 for premises, the combination coefficient φ1 is determined by the formula:

φ1 \u003d 0.4 + 0.6 / √ (A / A1) \u003d 0.4 + 0.6 / √ (47.52 / 9.0) \u003d 0.66.

The full (short-term) standard value of the load from people and furniture for apartments in residential buildings is 1.5 kPa (1.5 kN/m2). Taking into account the reliability factor for the responsibility of the building γn = 1.0 and the combination factor φ1 = 0.66, the final normative short-term payload is:

ν1 \u003d 1.5 * V * γn * φ1 \u003d 1.5 * 6.6 * 1.0 * 0.66 \u003d 6.53 kN / m.

When the standard value of the live load is less than 2.0 kPa, the load safety factor γt is taken equal to γt = 1.3. Then the calculated value is:

ν1р = ν1*γt = 6.53*1.3 = 8.49 kN/m.

The continuous payload is obtained by multiplying it full value by a factor of 0.35 i.e.:

p1 = 0.35*ν1 = 0.35*6.53 = 2.29 kN/m;

р1р = р1*γt = 2.29*1.3 = 2.98 kN/m.

The normative value of a uniformly distributed load from partitions is at least 0.5 kN/m2. We bring it to a linearly distributed load on the beam by multiplying it by the width of the cargo area B = 6.6 m:

р2 = 0.5*В*γн = 0.5*6.6*1.0 = 3.3 kN/m.

The calculated value of the load is then:

р2р = р2*γt = 3.3*1.3 = 4.29 kN/m.

I combination: constant load (self-weight of the floor and beams) + useful (short-term).

When taking into account the main combinations, including permanent loads and one live load (long-term or short-term), the coefficient Ψl, Ψt should not be entered.

q1 = q + ν1 = 43.87 + 6.53 = 50.4 kN/m;

q1р = qр + ν1р = 49.26 + 8.49 = 57.75 kN/m.

II combination: constant load (self-weight of the floor and beams) + useful (short-term) + load from partitions (long-term).

For the main combinations, the coefficient of combinations of long-term loads Ψ1 is taken: for the first (according to the degree of influence) long-term load - 1.0, for the rest - 0.95. The coefficient Ψ2 for short-term loads is taken: for the first (according to the degree of influence) short-term load - 1.0, for the second - 0.9, for the rest - 0.7.

Since in combination II there is one short-term and one long-term load, the coefficient Ψl and Ψt = 1.0.

qII = q + ν1 + р2 = 43.87 + 6.53 + 3.3 = 53.7 kN/m;

qIIр = qр+ ν1р + р2р = 49.26 + 8.49 + 4.29 = 62.04 kN/m.