Course Bases, foundations, staircase march.

Contents

Contents

Introduction

Architectural and structural solution of the building

Load collection per 1 m2 of slab, covering

Calculation of stair flight of 1LM grade 27.12.14-

Determination of strength characteristics of materials

Determining the Loads That Affect the March

Calculation of stair platform of grade 2LP 25.15-

Determination of strength characteristics of materials

Calculation of stairway slab

Calculation of the frontal edge of the site

Calculation of the tape foundation for the external wall of the building

Determining the Foundation Load

Defining the Width of the Foundation Floor

Calculation of foundation slab for strength during pressing

Calculation of plate strength by normal section

Conclusion

Literature

Introduction

Reinforced concrete consists of concrete and steel reinforcement, working together, due to the adhesion that occurs between them. Concrete is an artificial stone material, resists compression well, and works 15... 20 times weaker for stretching. To increase the strength of concrete, reinforcement is used, which has a significantly higher tensile strength and allows to significantly increase the bearing capacity of building structures.

The main advantages of concrete are high strength, durability, fire resistance, weather resistance, the possibility of using local building materials, ease of forming, low operating costs.

Architectural - structural solution of the building.

In accordance with the task, a rectangular building with dimensions along the extreme axes of 13.32 x 17.4m. The height of the floor is 2.8 m. Under the building there is a basement 2.2m high.

The building is designed with transverse bearing walls. Spatial rigidity and stability of the building is provided by dressing of vertical seams between bricks, reinforcement of angles and places of adjoining internal walls to external and anchoring slabs with a wall (L-shaped anchor) and to each other (linear anchor).

In terms of durability, the building belongs to the II degree, since its structural elements are designed for a service life of at least 50 years.

In terms of fire resistance in accordance with SNB 2.02.0198, the building belongs to the III degree.

Building liability class as per SNiP 2.01.0785 - III.

The building is located on a site with a calm relief. In the base lies soil - clay semi-solid. Groundwater level - 1.3 m from the day surface.

Note: Mass of 1m ² ceramic tile roof - 50 kg. Lattice of 50x50mm bars with pitch of 350mm. Rafter legs with a section of 50x150mm with a pitch of 1100mm. Mauerlat section 100kh100mm, loading on 1pog. meter Mnmauer = 0,1∙0,1∙10∙500/10³=0.05 kN/m, Mmauer = Mnmauer ∙γf = makes 0,05∙1,1=0,055 kN/m. A standard snow load for the II B of the snow region of Sn = S0 μ=1.2∙1=1.2 (f. 5 [1]), where [mu] = 1, since [alpha] = 25 [deg.] 3[1]). Coefficient αf = 1.6, since gn/S0 = = 0,677/1,2 = 0,564 < 0.8 (cl. 5.7[1]).

Calculation of stair flight of 1LM grade 27.12.144.

Source data. Calculate and design a reinforced concrete ribbed march of a two-march staircase of a residential building of the brand 1LM27.12.144. Width of march B = 1200mm, length L = 2720mm, height H = 1400mm. Floor height Nat = 2.8 m, stage size 14 × 30 cm. The mass of the flight of stairs is equal to 1.52 t, the volume of concrete is 0, 607m3.

Length of horizontal projection of march L1 = mm

March slope: tgα = 1400/2332 = 0.600

Angle of the march α ≈ 300, cos α = 0.857

Temporary standard load on the stairwell of a residential building pn = 3.0 kPa (Table 3 SNiP 2.01.0785), load reliability factor αf = 1.2 (item 3.7 SNiP 2.01.0785).

The stairway is made of concrete of class C20/25, longitudinal working reinforcement of the frontal rib of class S400, working reinforcement of the slab of class S500.

Determination of strength characteristics of materials.

For class C 20/25 concrete:

- normative resistance of concrete to axial compression fck = 20 MPa and axial tension fctk = 1.5 MPa (Table 6.1 SNB 5.03.01 02);

- safety factor for concrete αс = 1.5 (for reinforced concrete structures);

- design strength of concrete to axial compression fcd = fck/αс = 20/1.5 = 13.33 MPa, axial tension fctd = fctk/αс = 1.5/1.5 = 1 MPa;

- module of elasticity of concrete at the brand on placeability of Zh2 of Ecm = 39⋅103 MPas (a tab. 6.2 SNB 5.03.01 02).

Longitudinal working reinforcement S400:

- design resistance fyd = 365 MPa (Table 6.5 SNB 5.03.01 02).

Transverse reinforcement of S500 class:

- design resistance fyd = 450 MPa (Table 6.5 SNB 5.03.01 02).

- design resistance fywd = 324 MPa (Table 6.5 SNB 5.03.01 02).

Elastic modulus of reinforcement of all classes Es = 200⋅103 MPa.

Defines the loads that affect the march.

2.1. Calculation of Kosor strength by normal section.

We replace the actual section of the march with the calculated T-ray with the shelf in the compressed zone

Checking the condition

M = 8.26 kN⋅m ≤kN⋅m

The condition is met, the neutral axis passes in the shelf, the section is calculated as rectangular with a width of m

We determine the boundary value of the relative height of the compressed zone of concrete

,

where ω = 0.85 - 0.008fcd = 0,85 - 0,008⋅13,33 = 0,743

We define αlim = ξ lim (1 - 0.5αlim) = 0.601 (1 - 0,5⋅0,601) = 0.42

We determine the coefficient

Check the condition α0 ≤ αlim

α0 = 0,048 < αlim = 0,42

The condition is fulfilled, by the value α0 determines the coefficient? = 0.975

Determine the area of stretched reinforcement

According to the grade we accept 2 rods ∅10 mm with As1 = 1.57 cm2

We check the thickness of the protective layer of concrete. Thickness of protective layer of concrete is accepted

not less than the diameter of the working rod and not less than 15 mm in beams up to 250 mm high;

cm < s = 25 cm

The condition is met.

2.2. Calculation of Kosor strength by inclined section.

Longitudinal edges 90 mm, in the middle of flight of s2 = 200 mm are reinforced by two frameworks of KP1 with cross fittings of ∅4 mm of class S500 with a step on prioporny sites of s1 =. Section width b = 22 cm. Design transverse force Q = 15.04 kN. Rib length l = 2.720 m.

Let us determine the coefficient, which is to be 1 - 0.01fcd = 1 - 0,01⋅13,33 = 0.867

Define αe = Es/Ecm = 200/39 = 5.13

We determine the reinforcement factor

Let's determine the coefficient, which is ¼ w1 = 1 + 5αe⋅ρsw = 1 + 5 ⋅ 5.13 ⋅ 0.00127 = 1,033≤ 1.3

Checking the condition

Q = 15,04 кН < 0,3ηw1⋅ηc1⋅ fcd⋅ b⋅ d = 0,3⋅1,033⋅0,867⋅13,33⋅103⋅0,22⋅0,162 = 127,65кН

The condition is met.

Checking the condition

Q = 15.04 kN < 0.6 fctd ⋅ b⋅ d = 0,6⋅1⋅103⋅0,22⋅0,162 = 21.38 kN

The condition is fulfilled, inclined cracks are not formed, transverse reinforcement is installed structurally.

4. Constructs a stair flight.

Longitudinal ribs are reinforced by flat frames KK1. Longitudinal stretched reinforcement is accepted according to the calculation of ∅14 mm class S400, longitudinal compressed reinforcement is accepted structurally ∅6 mm class S400. Transverse extreme rods are accepted as ∅4 mm of class S500, transverse reinforcement is accepted according to calculation of ∅4 mm of class S500 in support sections with pitch s1 = 90 mm, in the middle of span s2 = 200 mm.

Cross edges are reinforced structurally by frameworks of KP2, KP3. Longitudinal rods in the stretched and compressed zone are accepted ∅5 mm class S500. Transverse rods are adopted ∅4 mm class S500 with spacing of 200 mm.

The staircase plate is reinforced with a C1 grid of ∅4 mm class S500 rods with a pitch of 200 mm in the longitudinal direction and ∅5 mm class S500 rods with a pitch of 150 mm in the transverse direction.

The remaining reinforcement frames and embedded parts are accepted in accordance with the working drawings of the collection "Typical structures, products and assemblies of buildings and structures," staircase reinforced concrete routes (series 1.152.1-8 issue 1).

4. Calculation of stair platform of grade 2LP 25.15-4-K

Source data. Calculate and design the ribbed slab of the stair platform of the 2LP grade double-staircase 25.15 - 4 - K. Plate width B = 1600 mm, plate height H = 320 mm, length L = 2500 mm, plate thickness h = 90 mm. The width of the staircase in the light is 2.5 m. The height of the floor Nat = 2.8 m. The mass of the staircase of the 1LM27.12.14 - 4 brand is 1520 kg, the own weight of the site will be 1345 kg, the volume of heavy concrete is 0.462 m3, decorative concrete is 0.076 m3. Temporary standard load on the stairwell of a residential building pn = 3.0 kPa (Table 3 SNiP 2.01.0785), load reliability factor αf = 1.2 (item 3.7 SNiP 2.01.0785).

Ladder platform is made of concrete of class C25/30, longitudinal working reinforcement of frontal rib of class S400, working reinforcement of slab of class S500.

When calculating the platform plate, a shelf elastically embedded in the ribs, a frontal rib on which the marches rest, and a wall rib that receives a load from the half span of the platform plate are considered separately.

Determination of strength characteristics of materials

Calculation of stairway slab

Required section area of reinforcement in span:

We accept the frame from wire reinforcement of S500 class with pitch in longitudinal direction S1 = 150 mm, in transverse direction S2 = 200 mm.

We take in longitudinal direction 6∅4 S500 with As = 0.754 cm2 per 1 linear meter of the plate, and in transverse napravlenii5∅4 S500 with As = 0.628 cm2 per 1 linear meter of the plate.

(1m/S1 = 1000/150 = 6 rods, 1m/S2 = 1000/200 = 5 rods).

Calculation of the frontal edge of the site

Estimated rib span is accepted as per Appendix VI.

l0 = 2.66 m

The platform plate as a shelf located in the compressed zone is involved in the operation of the rib. The calculated section has the following geometric characteristics:

h = 30 cm (excluding the mosaic layer), b = (b1 + b2 )/2 = (16 + 10 )/2 = 13 cm, bf = 18 cm,

If the calculated width of the compressed shelf is less than two values:

mass of march mm = 1.52 t, width of march bm = 1.2 m, design span of march (detailed drawings of march of 1LM27.12.14 - 4 grade).

Checking the condition

α0 = 0,0248 < αlim = 0,408

By tab. We define 4 applications ξ = 0.025, η = 0.988

The value of the compressed zone of concrete xeff = μ ⋅ d = 0.025 ⋅ 0.27 = 0.0068 m <, that is, the compressed zone of concrete is within the shelf, the section is calculated as rectangular in width.

Section area of stretched reinforcement

Accept 2∅12 mm of S400 class with As1 = 2.26 cm2

Reinforcement percentage

μ = 0,13% > μmin = 0,05%

Condition met.

3.2. Calculation of strength of inclined sections

The front edge is reinforced by two frameworks of KP1 with cross fittings of ∅4 mm of class S500 with s step = 150 mm. Section width b = 13 cm. Design transverse force Q = 22.01 kN. Rib length l = 2.78 m.

Let us determine the coefficient, which is to be 1 - 0.01fcd = 1 - 0,01⋅16,67 = 0.833

Define αe = Es/Ecm = 200/38 = 5.263

We determine the reinforcement factor

Let's determine the coefficient, which is, "w1 = 1 + 5αe⋅ρsw = 1+5⋅5,263⋅0,00129 = 1,034≤ 1.3

Checking the condition

Q = 22,01 кН < 0,3ηw1⋅ηc1⋅ fcd⋅ b⋅ d = 0,3⋅1,034⋅0,833⋅16,67⋅103⋅0,13⋅0,27 = 151,19 кН

The condition is met.

Checking the condition

Q = 22.01 kN < 0.6 fctd ⋅ b⋅ d = 0,6⋅1,2⋅103⋅0,13⋅0,27 = 25.27 kN

The condition is fulfilled, therefore, inclined cracks are not formed in the section, transverse reinforcement is installed structurally.

Designing a Stair Landing

The frontal edge is reinforced by two flat frames KR1. Longitudinal stretched reinforcement is accepted according to the calculation of ∅12 mm class S400, longitudinal compressed reinforcement is accepted structurally ∅6 mm class S400. Transverse extreme rods are accepted ∅6 mm of class S400, transverse internal rods are accepted according to the calculation of ∅4 mm of class S500 with a spacing of 150 mm.

Transverse ribs are reinforced structurally by KR3 frame. Longitudinal rods in the stretched and compressed zone are accepted ∅5 mm class S500. Transverse rods are accepted ∅4 mm of class S500 with a pitch of 150 mm.

Longitudinal wall rib is structurally reinforced by flat frame KK2. Longitudinal stretched reinforcement is adopted ∅10 mm class S400, longitudinal compressed ∅5 mm class S500. Transverse extreme rods are adopted ∅5 mm of class S500, transverse internal ∅4 mm of class S500 with a pitch of 150 mm.

The staircase plate is reinforced by a flat frame KK4 of rods of ∅4 mm class S500 with a pitch of 200 mm in the longitudinal direction and a pitch of 150 mm in the transverse direction.

The remaining reinforcement frames and sling loops are adopted in accordance with the working drawings of the collection "Typical structures, products and assemblies of buildings and structures," staircase reinforced concrete platforms (series 1.152.1-8 issue 1).

Source Data:

You want to calculate the tape foundation for the exterior wall along axis 1

5-storey residential building.

The roof is made of ceramic tiles, the angle of inclination of the roof is α = 25 °, the attic is cold, through, the floor is plank. Construction district - Bykhov, Mogilev region. UHF at elevation 2.310. Foundation depth d = 2.33 m. Soil - semi-solid clay with the following characteristics:

- porosity coefficient e = 0.6;

- specific adhesion of soil cn = 15 kPa;

- the angle of internal friction,, is,

- specific gravity of the soil lying above the base of the basement αII = 17.2kN/m ³

- specific gravity of the soil lying below the base of the foundation αII = 17.4 kN/m ³

Payload on the floor of a residential building pn = 1.5 kPa (tab.3 SNiP 2.01.0785)

Note: Mass of 1m ² ceramic tile roof - 50 kg. Lattice of 50x50mm bars with pitch of 350mm. Rafter legs with a section of 50x150mm with a pitch of 1100mm. Mauerlat with cross section 100x100mm ,

load on 1og.meter is Mnfiber = 0,1∙0,1∙10∙500/10³=0,05 kN/m,

Firewall = Mnmauer∙γf= 0,05∙1,1=0,055 kN/m.

Standard snow load for II B snow area Sn=S0∙μ=1,2∙1=1,2 (f. 5 [1]), where [mu] = 1, since [alpha] = 25 [deg.] 3[1]). Coefficient αf = 1.6, since gn/S0 = = 0,677/1,2 = 0,564 < 0.8 (cl. 5.7[1]).

5.2. Defines the width of the foundation floor.

For preliminary determination of foundation slab width we use table values of design soil resistance (Table 2, Appendix 3 [2]).

For semi-solid clay, R0 = 450 kPa.

Define the width of the foundation slab

Mean pressure on foundation bottom

Checking the condition

The condition is met.

If there is an active soil pressure on the basement wall in the basement building.

We determine the equal of the active soil pressure on the 1m wall

T =

where: q = 10kN/m - pressure by 1m2 along the upper cut of the soil wall

Moment of equilibrium with respect to the center of gravity of the foundation sole surface:

Find eccentricity e = Mn/Nnf = 29.76/269.9 = 0, 11m

We determine the maximum pressure on the bottom of the foundation

The condition is not met, we accept the width of the foundation slab b = 1.6m (mass of the unit 2.15t; length of 2380mm unit).

We determine the design soil resistance at the width of the foundation b = 1.6m

Mean pressure on foundation bottom

Checking the condition

The condition is met.

Find eccentricity e = Mn/Nnf = 29.76/279.08 = 0, 106m

We determine the maximum pressure on the bottom of the foundation

The condition is fulfilled, we finally accept the foundation slab of FL grade 16.24 (weight 2.15t; block length L = 2380mm; concrete volume 0.86m3).

5.3 Calculation of foundation slab for strength during pressing.

Design soil pressure under the foundation base Rgr = N/b =

= 299.37/1.6 = 187.11 kPa

Conventionally, the second side of the foundation a = 1m.

Transverse force in slab section at foundation block face:

Required height of plate section on the condition of strength in calculation for transverse force Q:

Design forcing force:

We check the condition:

Forcing strength is ensured.

We check the strength of concrete on the action of compressive forces:

Q = <

where (table 6.1 [3]).

Concrete strength is ensured.

5.4 Calculation of plate strength by normal section.

Moment occurring in section 1-1 at wall face (Fig. 5)

We determine the coefficient:

By tab. 4 Annexes we define 0.083; 0,961

Boundary value of relative height of compressed zone of concrete is determined by formula 7.5 [3].

where w = 0,850,0080,850,008∙10,67=0,765

(tab. 6.5 [3] for valves of class S400).

; the condition is met.

Reinforcement section area by 1 m of plate length:

According to the grade, we accept by 1 m the length of the unit 6 Ø12 S400 with a spacing of 200 mm and.

Transverse reinforcement is accepted by Ø5 S500 with spacing of 250 mm.