Indirect (local) reinforcement. Calculation example

Indirect reinforcement is a special transverse reinforcement, which can significantly increase the bearing capacity of compressed elements. Indirect reinforcement of rectangular elements is arranged in the form of often located welded meshes in the amount of not less than 4. Cross-linked nets are used to reinforce concrete under local compression in the zone of application of the greatest load. The essence of the work of indirect reinforcement is as follows: when a load is applied longitudinally, transverse tensile deformations occur in the elements, which lead to the formation of longitudinal cracks in the concrete. Indirect reinforcement works in tension and suppresses the lateral deformations of concrete, thereby increasing its bearing capacity. Indirect reinforced welded wire mesh is made of steel of classes АI, AIII and BpI ¯ 3 ... 10mm. The distance between the rods of the grids is taken from 45 to 100 mm.

The grids should be placed on a section called the reinforcement zone L, which is 10d, where d is the diameter of the longitudinal reinforcement. The distance to the 1st grid Sl is from 10 ... 40mm. The distance between the S2 grids is assigned as

(ranging from 60 to 150mm), so that it is divided by 3.

S2 = (L – S1) / 3

For example:

1) The diameter of the longitudinal reinforcement by calculation turned out to be 28mm.

Then the gain zone is L = 10d; L = 10 Indirect (local) reinforcement. Calculation example 28 = 280mm.

Take S1 equal to 10mm or 40mm,

then S2 = (280-10) / 3 = 90mm or S2 = (280-40) / 3 = 80mm., which ranges from 60 to 150mm.

2) The diameter of the longitudinal reinforcement by calculation turned out to be 12mm.

Then the gain zone is L = 10d; L = 10 Indirect (local) reinforcement. Calculation example 12 = 120mm.

Take S1 equal to 30mm,

then S2 = (120-30) / 3 = 30mm, which is less than the minimum distance of 60mm, therefore we take S2 equal to 60mm.

Type 1 task.

Given:

Ndl = 1000kN

Ncr = 500kN

l0 = 6.4m

b = h = 40cm

concrete B20, Indirect (local) reinforcement. Calculation example bi = 0.85

fittings 8¯18AII

Check the carrying capacity of the section

N £ Nsech

Decision.

Find in the printout:

Rb = 11.5 Indirect (local) reinforcement. Calculation example 0.85 = 9.78mPa

RSc = 280 MPa

A = 40 Indirect (local) reinforcement. Calculation example 40 = 1600cm2

N = 1000 + 500 = 1500kN

Asc = 20.36 cm2 (according to Table 7)

Formula 3 we find a = Rsс • Asc) / Rb • A

a = 280 • 20.36 / 9.78 • 1600 = 0.36

We find Ndl / N = 1000/1500 = 0.7; l = l0 / h = 640/40 = 16

jв = (0.82 + 0.76) /2=0.79 jж = (0.86 + 0.84) /2=0.85

j = jb + 2 (jj-jb) • a £ jj

j = 0.79 + 2 (0.85-0.79) 0.36 = 0.847 <0.85

Check the bearing capacity of the section according to the formula (1):

N £ j • (Rb • A + Rsc • Asc)

1500 £ 0.847 • (9.78 • 1600 + 280 • 20.36) (100) = 1808243Н = 1808kN

1500 < 1808

Conclusion: the carrying capacity of the section provided .

Type 2 task.

Given:

Ndl = 600kN

Ncr = 600kN

l0 = 4.2m

b = h = 30cm

concrete B20, Indirect (local) reinforcement. Calculation example i = 1

fittings AIII

Find Asc.

Decision.

Find in the printout:

Rb = 11.5 Indirect (local) reinforcement. Calculation example 1 = 11.5 mPa

RSc = 365mPa

A = 30 Indirect (local) reinforcement. Calculation example 30 = 900cm2

Set j = 1, define N = Ndl + Ncr

N = 600 + 600 = 1200kN

Find the approximate area of Asc reinforcement

Asc = N / j • Rss-A • Rb / Rsc = 1200 • 103/1 • 365 (100) -900 • 11.5 / 365 = 4.4 cm2

We find Ndl / N = 600/1200 = 0.5; l = l0 / h = 420/30 = 14

Indirect (local) reinforcement. Calculation example a = 4.4 • 365/900 • 11.5 = 0.159

we determine the coefficients jж = 0.88 and jв = 0.86 according to the table and

j = jb + 2 (jj-jb) • a £ jj j = 0.86 + 2 (0.88-0.86) • 0.159 = 0.924 ›0.88, therefore take jzh = jb = 0.88.

Finally we find the area of compressed reinforcement

Asc = 1200 • 103 / 0.88 • 365 (100) -900 • 11.5 / 365 = 8.6 cm2

Taken according to table 7 of Appendix 4¯18 AIII with Asc-fact. = 10.18 cm2

The transverse rods taken from the welding conditions in table 3 of the application

dw = 5ВрI with step Sw = 20d = 20 • 18 = 360mm

Indirect (local) reinforcement. Calculation example

L = 10b = 10 * 25 = 250

Sw = 20b = 20 * 25 = 500

S2 = Indirect (local) reinforcement. Calculation example

4500-20 = 4480-760 = 3720

3720: 500 * 7 = 3500

3720-3500 = 220

MARK

POSITION

NAME

QTY

MASS 1 DETAILS

MASS OF PRODUCT

KR-1

one

¯ 25АIII = 4480

17.24

35.47kg

¯ 8А I = 280

0.11

S-1

one

¯ 8А I = 280

0.11

1,32 kg

0 25 = 3.85-4.48 = 17.24 0 8 = 0.395-0.28 = 0.11

SPECIFICATION OF COLUMN

pos.

designation

name

count in

note

Prefabricated units

one

KPSK-270103-С-31-KZHI-KR-1

Frame flat KP1

KPSK-270103-С-31-КЖИ-С1

Grid C1

four

Line items

¯ 8 AI = 280

0.11

Materials

B15 concrete

0.41

STEEL CONSUMPTION STATEMENT

MARK

ELEMENT

ARMATURE PRODUCTS

TOTAL

Armature class

ASH

GOST 5781 - 82

GOST 5781 -82

¯25

Total

eight

Total

K - 1

68.96

10.78

79.74

¯25 = 17,24-2-2 = 68,96

¯8 = 9 0.11 -2 + 12 -0.11 -4 + 32 -0.11 = 10.78

Check: 35.47 -2 + 1.32 -4 + 32 -0.11 = 79.7

Type 3 task.

Given:

Ndl = 1000kN

Ncr = 1500kN

l0 = 6.4m

concrete B15, Indirect (local) reinforcement. Calculation example bi = 0.85

fittings AIII

Find b = h; Asc.

Decision.

1) find in the printout:

Rb = 8.5 Indirect (local) reinforcement. Calculation example 0.85 = 7.23mPa

Rs = 365 mPa

2) Set by j = 1 and m = 1% = 0.01;

3) determine the N = Ndl + Ncr

N = 1000 + 1500 = 2500 kN

4) Find

A = N / j • (Rb + m • Rsc) = 2500 • 103 / (7.23 + 0.01 • 365) (100) =

1363cm2

b = h = Indirect (local) reinforcement. Calculation example = 37cm; taking into account the unification round to 40cm. A = 40 • 40 = 1600cm2

5) We find Ndl / N = 1000/2500 = 0.4; l = l0 / h = 640/40 = 16

6) According to table 8, interpolating, we find jzh = 0.86, jc = 0.82

7) according to the formula 2 we define j = jв + 2 (jj-jв) • a £ jж

j = jb + 2 (jl-jb) • Rsc • Asc) / Rb • A. Replacing Asc / A = 0.01, we get:

j = 0.82 + 2 (0.86-0.82) • 365 / 7.23 • 0.01 = 0.88 ›0.86, successively, we take j = 0.86

8) According to the formula (4) we find:

Asc = N / j • Rss-A • Rb / Rsc = 2500 • 103 / 0.86 • 365 (100) -1600 • 7.23 / 365 = 15.2 cm2

9) Determine the% reinforcement: m = 15.2 / 1600 • 100 = 1%

Finally, we take a column with a cross section of 40 • 40 cm with a longitudinal working reinforcement 4¯22 AIII with Asc. Fact = 15.2 cm2 (see table 7 of the annex).

The transverse rods taken from the welding conditions in table 3 of the application

dw = 8АI with step Sw = 20d = 20 • 22 = 440mm

Table 1

Calculated resistance concrete for limiting states the first groups Rb and Rbt mPa in dependencies from class concrete by strength on compression

Kind of resistance

Concrete

Concrete compressive strength class

B12.5

B15

IN 20

B 25

VZO

B 35

B40

Compression oseeoe (prismatic strength) Rb

Heavy had a grained

7.5

8.5

11.5

14.5

19.5

Axial stretch Rbt

Heavy

0.6

0.75

0.9

1.05

1.2

1.3

1.4

table 2

Estimated reinforcement resistances for group I of limit states Rs, MPa

Stretched

Compressed rsc

Type and grade of steel;

longitudinal, transverse (clamps and limb) for bending on an inclined section. Rs

transverse (clamps and bends) in the calculation of the transverse force

Rsw

Hot rolled round grade A- I

225

175

225

: '

Hot rolled periodic

class profile:

A-ll

280

225

280

A-III d-8 mm

355

285

355

A-III d- 10. ..40 mm |

365

290

365

Wire reinforcing periodic profile

class bp-i

d, mm:

375

270

375

four

370

265

365

five

360

260

360

table 3

The ratio between the diameters of the welded rods

The diameters of the rods in one direction d1 (mm)

four

five

eight

ten

sixteen

The smallest allowable diameters of rods in a different direction d2 (mm)

four

five

eight

ten

table 4

The modulus of elasticity of reinforcing steel Es

Type and grade of steel

The modulus of elasticity of reinforcement Es.MPa

Core steel AI and A-II

210,000

A-IIIb

200,000

A-IV, A-VI, AT-IIIC

190,000

Reinforcing Wire

B-II, BP-II,

BP-I

200,000

170,000

Reinforcement ropes

K-7, K-19

180,000

table 5

Table: Values Indirect (local) reinforcement. Calculation example at IOAMAH

Reinforcement class

Coefficient

IN 20

VZO

B40

B50

B60

Indirect (local) reinforcement. Calculation example at

0.65

0.59

0.55

AOmax

0.48

0.42

0.4

Indirect (local) reinforcement. Calculation example at

0.62

0.57

0.52

0.47

0.44

AOmax

0.43

0.41

0.38

0.36

0.34

A-III with ¯

6-8 mm

Indirect (local) reinforcement. Calculation example at

0.59

0.54

0.5

0.44

0.41

AOmax

0.42

0.39

0.37

0.34

0.33

table 6

Coefficient values Indirect (local) reinforcement. Calculation example AO,

Indirect (local) reinforcement. Calculation example

0.41

0.795

0.326

0.57

0.715

0.408

0.42

0.79

0.332

0.58

0.71

0.412

0.59

0.705

0.416

0.43

0.785

0.337

0.6

0.7

0.42

0.44

0.78

0.343

0.45

0.775

0.349

0.61

0.695

0.424

0.46

0.77

0.354

0.62

0,69

0.428

0.47

0.765

0.359

0.63

0.685

0.432

0.48

0.76

0.365

0.64

0.68

0.435

0.65

0.675

0.439

0.49

0.755

0.37

0.66

0.672

0.442

0.5

0.75

0.375

0.51

0.745

0.38

0.67

0.665

0.446

0.52

0.74

0.385

0.68

0.66

0.449

0.53

0.735

0.39

0,69

0.655

0.452

0.54

0.73

0.394

0.7

0.65

0.455

0.55

0.725

0.399

0.56

0.72

0.403

0.01

0.995

0.01

0.21

0.895

0.188

0.02

0.99

0.02

0.22

0.89

0.196

0.03

0.985

0.03

0.23

0.885

0.203

0.04

0.98

0.039

0.24

0.88

0.211

0.05

0.975

0.048

0.06

0.97

0.058

0.25

0.875

0.219

0.26

0.87

0,226

0.07

0.965

0.067

0.27

0.865

0.236

0.08

0.96

0.077

0.28

0.86

0.241

0.09

0.955

0.085

0.29

0.855

0.248

0.1

0.95

0.095

0.3

0.85

0.255

0.11

0.945

0,104

0.12

0.94

0.113

0.31

0.845

0,262

0.32

0.84

0.269

0.13

0.935

0.121

0.33

0.835

0.275

0.14

0.93

0.13

0.34

0.83

0.282

0.15

0.925

0.13

0.35

0.825

0,289

0.16

0.92

0.147

0.36

0.82

0,295

0.17

0.915

0.155

0.18

0.91

0.164

0.37

0.815

0.301

0.19

0.905

0.172

0.39

0.805

0.314

0.2

0.9

0.18

0.4

0.8

0.32

table 7

Cross-sectional areas and mass of reinforcement bars

square and cross

sections

(sm.kv.) with the number of rods

weight 1 m

one

four

five

eight

0.071

0.14

0.21

0.28

0.35

0.42

0.49

0.57

0.64

0.055

four

0.126

0.25

0.38

0.50

0.63

0.76

0.88

1.01

1.13

0.098

four

five

0.196

0.39

0.59

0.79

0.98

1.18

1.37

1.57

1.77

0.154

five

0.283

0.57

0.85

1.13

1.42

1.70

1.98

2.26

2.55

0.222

0.385

0.77

1.15

1.54

1.92

2.31

2.69

3.08

3.46

0.302

eight

0.503

1.01

1.51

2.01

2.51

3.02

3.52

4.02

4.53

0.395

eight

0.636

1.27

1.91

2.54

3.18

3.82

4.45

5.09

5.72

0.499

ten

0.785

1.57

2.36

3.14

3.93

4.71

5.50

6.28

7.07

0.617

ten

1,510

2.26

3.39

4.52

5.65

6.79

7.92

9.05

10.18

0.888

1.539

3.08

4.62

6.16

7,69

9.23

10.77

12.31

13.85

1,208

sixteen

2,011

4.02

6.03

8.04

10.05

12.06

14.07

16.08

18.10

1.578

sixteen

2.545

5.09

7.63

10.18

12.72

15.27

17.81

20.36

22.90

1,993

3.142

6.28

9.41

12.56

15.71

18.85

21.99

25.14

28.28

2.466

3,801

7.60

11.40

15.20

19.00

22.81

26.61

30.41

34.21

2,994

4,909

9.82

14.73

19.63

24.54

29.45

34.36

39.27

44.18

3,853

6.158

12.32

18.47

24.63

30.79

36.95

43.10

49.26

55.42

4.83

8,043

16.08

24.13

32.17

40.21

48.25

56.30

64.34

72.38

6.313

10.18

20.36

30,54

40.72

50.9

61.08

71.26

81.44

91.62

7.99

12.56

25.12

37.68

50.24

62,8

75.36

87.92

100.48

113.04

9.87

table 8

The values of the coefficients jж and jв

Ndl

jж and jв at l0 / h

eight

ten

sixteen

jq factor

0.93

0.92

0.91

0.9

0.89

0.86

0.83

0.8

0.5

0.92

0.91

0.9

0.88

0.85

0.8

0.73

0.65

one

0.92

0.91

0.89

0.86

0.81

0.74

0.63

0.55

jj coefficient

0.93

0.92

0.91

0.9

0.89

0.87

0.84

0.81

0.5

0.92

0.91

0.9

0.87

0.84

0.8

0.75

one

0.92

0.91

0.9

0.88

0.86

0.83

0.77

0.7

Note: Ndl - the longitudinal force from the action of constant and prolonged loads;

Ncr - longitudinal force from the action of all loads (constant, long-term, short-term)
Questions for self-control

on the topic: "The calculation of compressed elements with random eccentricities ."

1. What are the three types of compressed elements, depending on the characteristics of their reinforcement?

2. How is the cross-section and characteristics of materials of compressed elements prescribed for random eccentricities?

3. What is random eccentricity and what is it equal to?

4. What are the design features of compressed elements with eccentricities more random?

5. What are the two main cases of destruction of compressed elements with eccentricities more than random?

6. What is the condition that determines each of the two cases of a compressed element?

7. What is the main strength condition of a compressed element and how is the height of a compressed zone determined?

8. How is the influence of deflection in flexible compressed elements taken into account?

9. What is the course of the calculation of compressed elements with eccentricities more random (separately for each of the three types of characteristic practical problems).

10. Name the types of valves according to the nature of work.

11. Name the types of reinforcement by type of surface.

12. How is the diameter and pitch of the transverse rods in the frame of the column.

13. Name the definitions of the coefficients j and l.

14. What is Ndl and Ncr?

15. What is indirect reinforcement mesh; how is their calculation made?

16. What are flexible and rigid reinforcement and structural requirements for the reinforcement of columns.

CALCULATION OF COLUMNS (TYPE I ) Check N < Nce h .

N dl

(kN)

N cr

(kN)

l 0

(m)

b = h

(cm)

Indirect (local) reinforcement. Calculation example vi

Concrete

Fittings

1000

500

6.4

0.85

8¯18 AII

1300

600

5.8

0.9

thirty

8¯20 AII

1500

900

6.2

thirty

0.85

8¯22 AII

1100

400

5.4

0.95

0 ^

8¯20 AIII

900

700

6.0

one

8¯18 All

1700

800

6.4

one

8¯20 AIII

1200

500

5.7

0.9

thirty

4¯25 AIII

1400

600

6.1

0.85

8¯20 AIII

1200

400

5.2

0.9

4¯20 AIII

1000

500

6.4

one

4¯22 All

1200

700

5.8

0.85

thirty

418 AIII

1500

400

6.2

0.9

4¯20 All

1100

600

6.4

one

4¯22 All

1400

500

6.0

one

4¯18 ASH

1300

400

5.2

one

thirty

4¯20 AIII

1100

600

6.2

one

thirty

4¯20 AIII

1000

500

6.2

one

thirty

4¯20 All

1400

400

5.4

0.9

4¯20 All

1200

400

5.8

0.9

8¯22 All

1300

600

5.6

0.85

8¯22 AIII

1200

500

5.2

one

8¯18 All

900

400

6.4

0.85

thirty

8¯20 All

1100

500

6.2

0.9

8¯20 AIII

1500

400

6.0

0.9

8¯20 All

1000

600

5.2

one

thirty

8¯18 AIII

Calculation of columns Type 2.

N dl

(kN)

N cr

(kN)

l 0

(m)

b = h.

( cm )

Indirect (local) reinforcement. Calculation example vi

Concrete

Armat.

one

600

4.2

thirty

one

111

700

600

4.5

thirty

0.85

111

1000

500

5.2

0.8

111

four

900

500

4.8

one

eleven

five

800

1000

5.9

one

111

1100

900

5.4

0.8

thirty

111

1300

1000

4.9

one

eleven

eight

1600

800

6.0

0.9

eleven

900

1200

6.8

0.95

eleven

ten

1000

1500

6.5

one

111

eleven

800

600

4.3

thirty

one

eleven

1400

1300

6.2

0.85

111

500

1100

5.3

0.9

111

As =? -

1000

900

6.4

one

eleven

1400

1200

6.1

0.8

thirty

eleven

sixteen

700

300

4.0

one

111

400

700

5.0

one

111

1000

300

5.8

0.85

eleven

nineteen

700

3.9

0.95

thirty

111

900

1200

8.0

0.9

111

1200

7.7

one

eleven

400

500

6.3

0.8

111

900

1100

7.0

one

eleven

1900

1700

8.7

0.9

thirty

eleven

1700

1000

7.2

one

111

Calculation of columns. Type 3.

N dl

(kN)

N cr

(kN)

l 0

(m)

in i

Concrete

Ar - ra

b =?

h =?

As =?

one

950

1050

7.2

one

1200

800

5.2

one

III

900

1000

one

four

1150

850

5.5

one

III

five

1000

850

6.0

0.8

thirty

1300

1000

4.8

0.8

thirty

III

1400

1100

4.3

0.85

eight

700

400

4.5

0.8

III

900

1200

7.3

0.9

ten

1500

900

6.5

0.8

III

eleven

800

1000

4.7

0.9

1200

750

4.6

0.9

III

800

500

4.3

one

1000

700

5.6

0.9

III

1200

1000

4.2

0.85

sixteen

1300

900

6.1

0.9

III

1100

850

5.7

one

900

500

4.5

0.9

III

nineteen

1200

900

5.1

0.85

900

1350

5.3

III

1000

1500

6.2

0.9

800

450

4.9

one

III

900

1250

5.8

0.85

thirty

1100

750

4.2

0.9

III

700

900

4.0

one

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Indirect (local) reinforcement. Calculation example

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