SP 52-101-2003, Reference Materials ENG

REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures without reinforcement pretensioning based on sp 52-101-2003

Association «Zhelezobeton» Central Scientific-Research and Design-Experimental Institute of Industrial Buildings and Structures (CNIIPromzdanii)

Scientific-Research Drawing and Design Institute of Concrete and Reinforced Concrete (NIIZhB)

REFERENCE MANUAL FOR DESIGNING OF CONCRETE AND REINFORCED CONCRETE STRUCTURES MADE OF HEAVY CONCRETE WITHOUT REINFORCEMENT PRETENSIONING (BASED ON SP 52-101-2003)

MOSCOW 2005

REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures without REINFORCEMENT PRETENSIONING based on sp 52-101-2003

UDC 624.012.4.04 Manual for concrete and reinforced concrete structures without reinforcement pretensioning (to SP 52-101-2003). CNIIPromzdanii, NIIZhB.M.: OJSC “CNIIPromzdanii, 2005. – p. 214. It contains instructions of SP 52 -101-2003 for designing of concrete and reinforced concrete structures made of heavy concrete without reinforcement pretensioning as well as recommendations necessary for designing. The reference manual is meant for design engineers as well as for construction institutes. Table 26 , Figure 74.

© OJSC CNIIPromzdanii, 2005


PREFACE The present reference manual has been developed on the basis and in elaboration of the set of rules SP 52-101-2003 “Concrete and reinforced concrete structures made without reinforcement pretensioning” ALL THE INSTRUCTIONS FOR DESIGNING, PROVISIONS, SPECIFICATIONS, INSTRUCTIONS, CALCULATION EXAMPLES FOR ELEMENTS AS WELL AS RECOMMENDATIONS FOR DESIGNING WHICH ARE A PART OF SP 52-101-2003 ARE GIVEN IN THE PRESENT REFERENCE MANUAL. DATA ON DESIGNING UNUSUAL NON-TYPICAL STRUCTURES WITH UNTENSIONED HIGH-STRENGTH REINFORCEMENT (А600 CLASS AND HIGHER) ARE NOT COVERED BY THE PRESENT REFERENCE MANUAL, HOWEVER THEY ARE PRESENTED IN THE REFERENCE MANUAL FOR “DESIGNING OF PRETENSIONED REINFORCED CONCRETE STRUCTURES MADE OF HEAVY CONCRETE”. SPECIAL DESIGN FEATURES FOR TYPES OF BUILDINGS AND STRUCTURES FOR WHICH INTERNAL FORCES ARE ACCOUNTED FOR ARE NOT GIVEN IN THIS REFERENCE MANUAL. THESE ISSUES ARE COVERED BY THE CORRESPONDING SETS OF RULES AND REFERENCE MANUALS. THE FOLLOWING MEASUREMENT UNITS ARE USED IN THE REFERENCE MANUAL: FORCES ARE EXPRESSED IN NEWTONS (N) OR KILONEWTONS (KN); LINEAR DIMENSIONS ARE EXPRESSED IN MM (FOR CROSS-SECTIONS) AND IN M (FOR ELEMENTS AND THEIR PARTS); STRESS, STRENGTH, MODULUS OF ELASTICITY ARE GIVEN IN MEGAPASCALS (MPA); DISTRIBUTED LOADS AND FORCES ARE EXPRESSED IN KN/M AND N/MM. SINCE 1 MPA = 1 N/MM2, WHEN VALUES IN MPA (STRESS, STRENGTH, ETC.) ARE USED IN EXAMPLES OF FORMULA CALCULATION, THE REST OF THE VALUES ARE GIVEN ONLY IN N AND MM (MM2). VALUES OF CHARACTERISTIC AND DESIGN STRENGTH AND MODULUS OF ELASTICITY ARE PRESENTED IN THE TABLES IN MPA AND KGF /CM2.


THE REFERENCE MANUAL HAS BEEN DEVELOPED BY “CNIIPROMZDANII” (ENGINEER I.К. NIKITIN, DOCTORS OF ENGINEERING E.N. KODISH AND N.N. TRIOKIN) WITH THE PARTICIPATION OF «NIIZHB» (DOCTORS OF ENGINEERING А.S. ZALESOV, Е.А. CHISTIAKOV, A.I. ZVEZDOV, T.A.MUHAMEDIEV). PLEASE FORWARD YOUR COMMENTS AND OBSERVATIONS TO THE FOLLOWING ADDRESSES: 127238, MOSCOW, DMITROVSKOE SHOSSE, 46/2, OJSC «CNIIPROMZDANII»; 109384, MOSCOW, 2-YA INSTITUTSKAYA STREET, 6, SUE «NIIZHB».


1.

GENERAL RECOMMENDATIONS BASIC PROVISIONS

1.1. Recommendations of the reference manual cover designing of concrete and reinforced concrete buildings and structures made of heavy concrete which belongs to compression strength class from B10 to B60 without reinforcement pretensioning and are operated under conditions of systematic thermal exposure in the range not higher than + 50оС and no lower than - 40оС in non-corrosive environment with static load impact. Recommendations of the reference manual do not cover designing of concrete and reinforced concrete hydraulic structures, bridges, tunnels, pipes under embankments, highway and aerodrome surface and some other special structures. Note. The term «heavy concrete» is used in accordance with GOST 25192.

1.2. When concrete and reinforced concrete structures are designed not only design and construction requirements of the present aid shall be made, but also process requirements to manufacture and erection of a structure. Conditions of proper service and maintaining of structures shall be taking into consideration environmental requirements in accordance with the corresponding regulatory documents. 1.3. For assemblies it is necessary to pay specific attention to strength and long service life of connections. 1.4. Concrete elements are used: а) mainly for structures which are in compression with normal force within the limits of the element cross section with normal force along the element cross-section; б) in specific cases for structures which are in compression with normal force beyond the limits of the element cross section as well as in bending structures when their failure does not directly endanger people’s life and equipment safety (for example, elements located on solid base). Structures are considered as concrete in case their strength is provided for by concrete only. 1.5. Design winter temperature of outdoor air is taken as average temperature of the coldest five-day period depending on construction zone in accordance with SNIP 23-01-99. Design process temperatures are specified in the design task.


BASIC DESIGN REQUIREMENTS 1.6. Calculation of concrete and reinforced concrete structures shall be performed for limit states including: limit states of the first group (total unserviceability due to loss of sustaining capacity); - limit states of the second group (unsuitability to normal service due to cracks formation or excessive crack opening, occurrence of unallowable deformations, etc.). Calculation of limit states of the first group containing in this reference manual include strength calculation with taking into consideration structure deformation state failure. Calculation of limit states of the second group containing in this reference manual include evaluation of crack opening and deformation. Calculation of limit states of the second group for concrete structures containing in this reference manual is not performed. Limit state calculation for a structure in general as well as its separate elements shall be performed for all stages: manufacturing, transportation, erection and service, herewith calculation models shall be in agreement with taken structural Schematics. 1.7 Calculation of forces and deformation due to different impact on structures and systems of buildings shall be carried out taking into consideration potential cracks formation and non-elastic deformation in concrete and reinforcement (material nonlinearity) as well as structure deformation state before its failure (geometric nonlinearity). Calculation method has not been developed for statically indeterminate structures taking into consideration material non linearity and it is permissible to determine forces on the assumption of material linear elasticity. 1.8 Standard values of loads and impacts, combination coefficient, partial safety factor for loads, intended use reliability factor as well as classification of loads for constant and temporary (long-term and short-term ones) are set in accordance with SNIP 2.01.07-85*. When force effects brought about by lifting, transportation and mounting are calculated, load of element weight shall be taken with service factor equal to: 1.60 for transportation, 1.40 for lifting and mounting. In this case partial safety factors of loads shall be taken into account as well.


It is permissible to assume lower values of service factor justified in the established procedure but no lower than 1.25

MATERIALS FOR CONCRETE AND REIFORCED CONCRETE STRUCTURES 2.

CONCRETE CONCRETE QUALITY CHARACTERISTICS AND THEIR DESIGN APPLICATION 2.1. For concrete and reinforced concrete structures it is necessary to provide the following concrete classes and grades: a) Class of compression strength: В10; В15; В20; В25; В30; В35; В40; В45; В50; В55; В60; b) Class of axial tensile strength: Bt0,8; Bt1,2; Вt1,6; Bt2,0; Вt2,4; Bt2,8; Bt3,2; c) Frost resistance grade: F50; F75; F100; F150; F200; F300; F400; F500; г) Watertightness grade: W2; W4; W6; W8; W10; W12. 2.2. Age of concrete which corresponds to its class of compression strength and axial tensile strength (design age) is assigned in the design relying on potential actual terms of loading the structure with design loads. In the absence of these data, the concrete class is assigned at 28 days. Value of concrete handling strength for assembly elements is assigned in accordance with GOST 13015.0 and the corresponding standards for structures of certain types. 2.3 Class of concrete compression strength is assigned in all cases. Class of concrete axial tensile strength is assigned in case if this characteristic is dominating and it is monitored during manufacture (for example for concrete flexural elements). Frost resistance grade is assigned for structures which during their service life are alternately subject to freezing and thaw (aboveground structures, subject to weather impact, located in one cant ground, under water, etc.).


Water tightness grade is assigned for structures which have waterproof restrictions (water houses, supporting walls, etc.). 2.4. For reinforced structures it is recommended to assign class of concrete compression strength which is no lower than В15; herewith for heavy-loaded compressed axial elements it is recommended to assign concrete class no lower than В25. For concrete compressed elements it is not recommended to assign concrete class higher than В30. 2.5. For aboveground structures which are subject to weather impact at design winter temperature from - 5оС to - 40оС, frost resistance concrete grade shall be no lower than F75; herewith, in case if these structures are protected against atmospheric fallout, frost resistance grade might be applied no lower then F50. Concrete frost resistance grade is not specified for above described structures in case if design winter temperature is above - 5оС. Note. Design winter temperature of outdoor air is assigned in accordance with paragraph 1.5.


SPECIFIED AND DESIGN CONCRETE CHARACTERISTICS 2.6. Specified characteristic of concrete axial compression strength (prism strength) Rb,n and axial tension strength (when compression strength class is assigned) Rbt,n is taken depending on concrete of class В in accordance with Table 2.1. Table 2.1. Specified and design values of concrete strength Rb,п and Rbt,n for limit states of the second group Rb,ser and Rbt,ser,, MPa (kgf/cm2) with quality Strength class of concrete compression strength type В10 В15 В20 В25 В30 В35 В40 В45 В50 В55 В60 Axial 7,5 11,0 15,0 18,5 22,0 25,5 29,0 32,0 36,0 39,5 43,0 compression (76,5) (112) (153) (188) (224) (260) (296) (326) (367) (403) (438) Rb,п,Rb,ser Tension 0,85 1,10 1,35 1,55 1,75 1,95 2,10 2,25 2,45 2,60 2,75 (8,7) (11,2) (13,8) (15,8) (17,8) (19,9) (21,4) (22,9) (25,0) (26,5) (28,0) Rbt,п,Rbt,ser

When concrete class is assigned in accordance with axial tensile strength Bt, specified concrete resistance to axial tension Rbt,n in MPa is taken to be equal to numerical characteristics. 2.7. Specified concrete resistance to axial compression Rb and axial tension Rbt for limit states of the first group is calculated using formula: Rb =

Rb , ï

γb

; Rbt =

Rbt , n

γ bt

,

(2.1)

where γb is a safety factor for concrete compression strength which is taken equal to 1.3; γbt - safety factor for concrete compression strength which is taken equal to: 1.5 – when concrete class is assigned regarding compression strength; 1.3 – when concrete class is assigned regarding tensile strength. Concrete design strength Rb and Rbt (with approximation) depending on concrete quality class with respect to compression strength and axial tensile are presented respectively in Tables 2.2 and 2.3 Design values of concrete axial tensile strength Rb,ser and axial tensile Rbt,ser for limit states of the second group


Table 2.2 Concrete design strength for limit states of the first group Rb and Rbt, MPa (kgf/cm2) with concrete quality class regarding compression strength Strength type В2 В10 В15 В25 В30 В35 В40 В45 В50 В55 В60 0 11. Axial 8.5 6.0 5 14.5 17.0 19.5 22.0 25.0 27.5 30.0 33.0 compression, (86.6 (61.2) (117 (148) (173) (199) (224) (255) (280) (306) (336) ) Rb ) 0.9 0.56 0.75 0 1.05 1.15 1.30 1.40 1.50 1.60 1.70 Axial tension, 1.80 (5.7 (7.6 (9. (10.7) (11.7) (13.3) 14.3) (15.3) (16.3) (17.3) (18.3) Rbt ) ) 2) Таble 2.3 Concrete design strength with respect to axial tensile for limit states of the first group Rbt, MPa (kgf/cm2) with concrete quality class regarding axial tensile strength Вt0.8 Вt1.2 Вt1.6 Вt2.0 Вt2.4 Вt2.8 Вt3.2 0.62 0.93 1.25 1.55 1.85 2.15 2.45 (6.3) (9.5) (12.7) (15.8) (18.9) (21.9) (25.0)

are taken equal to the corresponding specified strength, i.e. they are introduced into calculation along with partial safety factor for concrete strength γb = γbt = 1.0. Values Rb,ser and Rbt,ser are given in Table 2.1. 2.8. Concrete design strength might be multiplied by the following service factors γbi if required: а) γb1 = 0.9 used for concrete and reinforced concrete structures with impact of only constant and long-term loads introduced to design values of Rb and Rbt; б) γb2 = 0.9 used for concrete structures introduced to design value Rb; в) γb3 = 0.9 used for concrete and reinforced concrete structures encased in concrete vertically is introduced to design value Rb. 2.9. Value of initial elasticity modulus with compression and tension Еb is taken depending on concrete quality class regarding compression strength В in accordance with Table 2.4 2.10. It is permissible to assume value of Poisson’s ratio as νb,P = 0.2. Shearing modulus of elasticity G is taken equal to 0.4 of the corresponding value Еb, specified in Table 2.4.


2.11. Values of linear thermal deformation coefficient for concrete with temperature gradient in the range from -40 up to +50оС are taken as αbt =1.10-5 °C-1. Table 2.4 Values of initial tangent modulus of concrete elasticity with compression and tension Еb.10-3, MPa (kgf/cm2), with concrete compression strength class В10 В15 В20 В25 В30 В35 В40 В45 В50 В55 В60 19,0 24,0 27,5 30,0 32,5 34,5 36,0 37,0 38,0 39,0 39,5 (194) (245) (280) (306) (331) (352) (367) (377) (387) (398) (403)

2.12. In order to calculate mass of reinforced concrete or concrete structure concrete density is taken to be equal to 2400 kg/m3. Reinforced concrete density with percentage of reinforcement 3% and less is taken to be equal to 2500 kg/m3; with percentage of reinforcement more than 3% density is calculated as sum of concrete and reinforcement mass per volume unit of a reinforced concrete structure. Herewith mass of 1 m of reinforcement steel is taken in accordance with Appendix 1 and mass of sheet steel and shaped bars are set in accordance with state standards. When gravity weight of a structure is calculated it is permissible to assume specific gravity to be equal to 0.01 of density in kg/m 3. 2.13. Values of concrete relative deformations which characterize state diagram of compressed concrete (εb0, εb1,red, εb2) and tensile concrete (εbt0, εbt1red and εbt2) as well as coefficient of concrete creep ϕb,cr are given in paragraphs 4.27 и 4.23.

REINFORCEMENT REINFORCEMENT QUALITY CHARACTERISTICS 2.14. For reinforced concrete structures designed in accordance with requirements of the present aid it is necessary to provide the following reinforcement types: - hot-rolled plain rods of reinforcement of class А240 (А-I); - hot-rolled and thermo-mechanical hardened Isteg reinforcement А300 (А-II), А400 (А-III, A400С), A500 (A500С); - cold-deformed Isteg reinforcement of class В500 (Вр-I, В500С).


As reinforcement used in structures in accordance with calculation it is recommended to apply mainly: Isteg reinforcement of classes А500 and А400 ; Isteg reinforcement of class В500 in fabricated frames and nets. Reinforcement gauge is given in Appendix 1. 2.15. For structures which are in outdoor service or in unheated buildings in zones with design winter temperature lower than - 30оС it is permissible to apply reinforcement class А300 of steel grade St5ps 18 - 40 mm in diameter as well as class А240 of steel grade St3kp. These types of reinforcement might be applied in structures of heated buildings located in the specified zones if in construction phase load carrying capacity of a structure is provided relying on design reinforcement strength with reduction factor 0.7 and design load with load safety factor γf = 1.0. Other types and classes of reinforcement might be applied without restrictions. 2.16. Hot-rolled reinforcement of class А240 of steel grade St3sp and St3ps as well as of class А300 of steel grade 10GТ shall be used for mounting (limiting) eyes of elements of concrete and reinforced concrete assemblies. SPECIFIED AND DESIGN REINFORCEMENT CHARACTERISTICS 2.17. Basic strength characteristic of reinforcement is a specified value of tensile strength Rs,п, taken depending on reinforcement class given in Table 2.5 2.18. Design values of reinforcement tensile strength Rs for limit states of the first group are taken using formula Rs =

Rs , n

γs

,

(2.2)

where γs is partial safety factor of reinforcement strength, which is taken equal to: 1.1 – for reinforcement class А240, А300 and А400; 1.15 – for reinforcement class А500; 1.2 – for reinforcement class В500.


Design values Rs are given (with approximation) in Table 2.6. Herewith, value Rs,is taken equal to the smallest monitored value in accordance with the corresponding GOST. Design values of reinforcement tensile strength Rs,ser


Reinforcement class

Specified diameter of reinforcement, mm

А240 А300 А400 А500 В500

6 - 40 10 - 70 6 - 40 6 - 40 3 - 12

Table 2.5 Specified values of tensile strength Rs,n and design values of tensile strength for limit states of the second group Rs,ser,MPa (kgf/cm2) 240 (2450) 300 (3060) 400 (4080) 500 (5100) 500 (5100)

for limit states of the second group are taken equal to the corresponding specified values of strength Rs,n (see Table 2.5). Design values of reinforcement compression strength Rsc is taken equal to reinforcement tensile strength Rs with exception for reinforcement class А500 for which Rsc = 400 MPa and reinforcement class В500 for which Rsc = 360 MPa (see Table 2.6). When structure is analyzed regarding impact of constant and long-term loads, it is permissible to assume values Rsc to be equal to Rs for reinforcement classes А500 and В500.

Reinforcement class

А240 А300 А400 А500 В500

Table 2.6. Design values of reinforcement strength for limit states of the first group, MPa (kgf/cm2) Compression transverse (stirrups Tension, Rsc longitudinal, Rs and diagonal bars), Rsw 215 (2190) 170 (1730) 215 (2190) 270 (2750) 215 (2190) 270 (2750) 355 (3620) 285 (2900) 355 (3620) 435 (4430) 300 (3060) 400 (4080) 415 (4230) 300 (3060) 360 (3670)

2.19. Design values of crosswise reinforcement strength (stirrups and diagonal bars) Rsw are reduced in comparison with Rs by means of multiplying by service factor γs1 = 0.8, however, taken no more 300 MPa. Design values Rsw are given (with approximation) in Table 2.6. 2.20. Value of modulus of reinforcement elasticity Еs is taken to be the same for both compression and tension and equal to Es = 2.0.105 MPa = 2.0.106 kgf/cm2.


3.

CALCULATION LIMIT STATES OF THE FIRST GROUP FOR CONCRETE AND REINFORCED CONCRETE ELEMENTS STRENGTH ANALYSIS OF CONCRETE ELEMENTS GENERAL PROVISIONS

3.1. Strength analysis of concrete elements is performed regarding impact of longitudinal compressed forces, moments of flections as well as local compression. 3.2. Concrete elements are calculated with or without regard to resistance of tensile zone of concrete depending on their service conditions and specified requirements. Calculation of eccentrically compressed elements specified in paragraph 1.4,a is performed without regard to resistance of tensile zone concrete assuming that reaching of limit state is characterized by failure of compressed concrete. Calculation of elements specified in paragraph 1.4,b as well as elements for which cracking is not permissible in accordance with operational requirements (elements subject to pressure of water, drop aprons, barrier walls, etc.) is performed with allowance for resistance of tensile zone concrete. Herewith, it is taken that limit state is characterized by reaching of limit state in tensile zone concrete. 3.3. If forces (moment, transverse or normal force) F1 of constant and long-term loads exceed 0.9 of forces of all loads, including short-term ones, calculation regarding impact of forces F1 shall be performed, assuming concrete design strength Rb and Rbt with allowance for coefficient γb1 = 0.9. 3.4. Strength of concrete elements regarding impact of local compression is performed in accordance with instructions of paragraphs 3.81 and 3.82. 3.5. Constructional reinforcement shall be provided in concrete elements under conditions specified in paragraph 5.12.


CALCULATION FOR ECCENTRICALLY COMPRESSED ELEMENTS 3.6. Calculation for eccentrically compressed elements shall be carried out with allowance for accidental eccentricity еа taken no less than: 1/600 of element length or distance between its cross-sections fixed against movement; 1/30 of section depth; 10 mm. For statically indeterminate structures (for example, fixed-end poles) amount of eccentricity of normal force with respect to cross-section gravity center е0 is taken equal to eccentricity value derived from static calculation, however, no less than еа. For statically determinate structures eccentricity е0 is taken equal to a sum of eccentricities from static structure calculation and statically distributed. 3.7. With element flexibility l0/i > 14 (for a rectangular cross-section with l0/h > 4) it is necessary to take into account impact of deflection on their load carrying capacity by means of multiplying value е0 by coefficient η specified in accordance with paragraph 3.10. 3.8. Calculation for concrete eccentrically compressed elements with normal force within the limits of cross-section is performed without regard to concrete resistance of tensile zone in the following way. For elements of rectangular cross-sections, T- and I-sections, when force is applied in the plane of mirror symmetry, calculation is carried out using condition (3.1), N ≤ Rb Ab, where Ab is square of concrete tensile zone determined using the condition that its gravity center coincides with a point of application of normal force N (with allowance for deflection) (Drawing. 3.1.). For elements of a rectangular cross-section ⎛ 2å η ⎞ Ab = bh⎜1 − 0 ⎟, h ⎠ ⎝

(3.2)

where for η refer to paragraph 3.10. Symmetrical trapezoidal and V-shaped sections might be calculated using condition (3.1) provided that maximum compression is at the bigger cross-section side.


Ab

Rb

eoη

N

h

1

2

b DRAWING.3.1. SCHEMATIC REPRESENTATION OF FORCES AND STRESS LINE FOR A CROSS-SECTION NORMAL TO LONGITUDINAL AXIS OF ECCENTRICALLY COMPRESSED CONCRETE ELEMENT STRENGTH OF WHICH IS CALCULATED WITHOUT REGARD TO RESISTANCE OF TENSILE ZONE CONCRETE 1-GRAVITY CENTER OF COMPRESSED ZONE AREA AB, 2 - SAME, AREA OF THE WHOLE CROSS-SECTION

In other cases calculation is performed on the basis of non-linear deformational model in accordance with paragraphs 3.72 - 3.76 assuming that in specified relationship steel area is equal to zero. With oblique eccentrical compression calculation for a rectangular cross-section is performed on the basis of condition (3.1) when Аb is determined using the formula ⎛ 2e η ⎞⎛ 2e0 yη y ⎞⎟ , Ab = bh⎜1 − 0 x x ⎟⎜⎜1 − h ⎠⎝ b ⎟⎠ ⎝

(3.3)

where e0x and e0y are eccentricities of force N in direction with respect to cross-section size h and b. ηx and ηy are coefficients η specified in accordance with paragraph 3.10 separately for each direction. 3.9. Eccentrically compressed concrete elements with normal force beyond the limits of the element cross-section and also elements for which cracks formation is not permissible regаrdless calculation using condition (3.1) shall be checked with allowance for resistance of concrete of tensile zone using the condition N≤

Rbt A

A e0ηyt − 1 I

,

(3.4)


where уt is distance from gravity center of element cross-section to the most tensile fiber; for η refer to paragraph 3.10. For elements of a rectangular cross-section condition (3.4) can be written in the form N≤

Rbt bh . 6e0η −1 h

(3.5)

It is permissible to analyze concrete elements with allowance for concrete tensile zone on the basis of non-linear deformation model in accordance with paragraphs 3.72-3.76 assuming that steel area is equal to zero. 3.10. Value of coefficient η with allowance for deflection effect on amount of eccentricity of normal force е0 is calculated using formula η=

1

N 1− N cr

,

(3.6)

where Ncr is nominal critical force determined using formula N cr =

π 2D l02

,

(3.7)

where D is element stiffness in strength limit state calculated using the formula D = Eb I

0,15 ; ϕ l (0,3 + δ e )

l0 is determined using Table 3.1. Type of wall and pole bearing 1. With supports upwards and downwards: а) with flap hinges on either side regardless of bearing displacement value б) with one end restraint and possible bearing displacement of: single-aisle building multiple-aisle building в) with partial restraint of fixed supports 2. Free standing buildings

(3.8) Table 3.1. Design length l0 of eccentrically compressed concrete elements Н 1.2 Н 1.5 Н 0.8 Н 2Н


Note. Н is either distance between floor structures and other horizontal supports (for floor structures which are monolithically connected to the wall (pole) with the deduction of floor structure width) or height of a free-standing building.

For elements of a rectangular cross-section formula (3.8) can be written in the form Eb bh 3 D= . 80ϕ l (0.3 + δ е )

(3.8а)

In formulae (3.8) and (3.8а): ϕl is a coefficient taking into consideration impact of long-term load on vertical deflection in limit state which is equal to ϕl = 1 +

M 1l , M1

(3.9)

however, no more than 2; М! is moment of relatively tensile or the least cross-section edge due to impacts of constant, long-term and short-term loads; M1l is the same for constant and long-term loads; δе is a coefficient taken to be equal to е)/h, however no less than 0.15. For walls and poles with elastically fixed supports the specified value η is taken for calculation of cross-sections in the middle 1/3 of height Н. When calculation is carried out for support cross-sections, it is taken that η = 1.0, for all other cross-sections using linear interpolation. If a lower support is stiffly restrained, then with an elastic upper support value η determined using the formula (3.6) is taken for sections of the lower section with height of 2/3Н.. 3.11. Calculation with allowance for deflection of eccentrically compressed concrete elements of a rectangular cross-section of class not higher than В20 with l0 ≤ 20h is permissible to carry out using condition N ≤ αnubby, (3.10) Where the value is determined using the drawing (3.2) depending on values E0/h and λ = lo/h. 3.12. When impacts of normal forces are sufficient, the following condition shall be met σ mt Rbt

+

σ mc Rb

≤ 1,0 ,

(3.11)

where σtm и σmc are main tensile and compressed stress calculated using the formula


σ mt

mc

2

σx

⎛σ ⎞ =m + ⎜ x ⎟ +τ 2 , 2 ⎝ 2 ⎠

(3.12)

σх and τ - normal and shearing stress in the considered section fiber

calculated the same way as for elastic section fiber.

αN

1.000 0.900

λ=0

0.800

λ=5

0.700 0.600

λ=10

0.500 0.400

λ=15

0.300 0.200

λ=20

0.100 0.000 0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0.50 E0/H

Drawing.3.2. Diagram of load-carrying capacity of eccentrically compressed concrete elements Graphic symbols: with M 1l M 1 = 1,0 ; with M 1l M 1 = 0,5 ;

For a rectangular cross-section condition check (3.11) is performed for fiber at the level of gravity center of the section and for T- and Isections at the level of contact of compression flanges and section wall. CALCULATION OF FLEXURAL ELEMENTS


3.13. Calculation of concrete flexural elements shall be performed using the following condition M ≤ RbtW, (3.13) where W is Z-modulus for outermost tensile fiber; for a rectangular crossbh 2 . section W = 6

In addition for elements of T- and I-sections the following condition shall be met: τ ≤ Rbt, (3.14) where τ - shearing stress calculated the same way as for elastic material at the level of gravity center. CALCULATION EXAMPLES Example 1. Given: a separation concrete panel with thickness h = 150 mm, height Н = 2.7 m, manufactured upright (in a cassette); concrete of class В15 (Еb= 24000 MPa, Rb = 8.5 MPa); total load per 1 m of the wall is N = 700 kN including constant and long-term load Nl = 650 kN. It is required to check panel durability. C a l c u l a t i o n is carried out in accordance with paragraph 3.8. regarding impact of normal force applied with accidental eccentricity еа, determined in accordance with paragraph 3.6. Since h = 150 = 5 mm < 10 mm and Í = 2700 = 4,5 mm < 10 mm, is taken еа= е0 30

30

600

600

= 10 mm. It is taken that there is hinge support upwards and downwards. Consequently, design length l0 is equal to l0 = Н = 2.7 m in accordance with Table 3.1. Taking into consideration that l0/h = 2.7/0.15 = 18 > 4, calculation is carried out with allowance for deflection influence in accordance with paragraph 3.10. Using formula (3.9) one can determine coefficient ϕl taking into consideration M1l/M1 = Nl/N =650/700 = 0.93, ϕl = 1 + Mll/M1 = 1+ 0.93 = 1.93. Since е0/h = 10/150 = 0.067 < 0.15, assuming δе = 0.15. Stiffness D is calculated using formula (3.8,а) assuming cross-section width b = 1 m = 1000 mm


Ebbh3 24000 ⋅ 1000 ⋅ 1503 . 2 = D= = 1,166 ⋅1012 Н mm . 80ϕ l (0,3 + δ å ) 80 ⋅ 1,93(0,3 + 0,15)

Then N cr =

η=

π 2D l

2 0

=

π 2 ⋅ 1,166 ⋅ 1012 2700

2

= 1578,3 ⋅ 10 3 N = 1578,3 kN;

1 1 = = 1,797. N 700 11− N cr 1578

Concrete design strength Rb is taken in accordance with paragraph 2.8 with allowance for coefficients γb2 = 0.9 and γb3 = 0.9. Taking into consideration occurrence of short-term loads one can assume that γb1 = 1.0. Then Rb = 8.5 . 0.9. 0,9 = 6,89 MPa. Let us check condition (3.1) using formula (3.2) ⎛ 2å η ⎞ Rb Ab = Rb bh⎜1 − 0 ⎟ = 6.89 ⋅ 1000 ⋅ 150(1 − 2 ⋅ 0,067 ⋅ 1.797) = 784635 N = 784.6 kN h ⎠ ⎝

> N =700

kN, which means panel strength regarding total load impact is ensured. Since Nl/N = 0.93 > 0.9 in accordance with paragraph 3.3, let us check panel strength regarding only constant and long-term loads, that means with N = 650 kN. In this case ϕl = 2, and then N cr = 1578.3

1,93 1 = 1523,4 and η = = 1.745. 2 1 − 650 / 1523

Design strength Rb is taken with allowance for γb1 = 0.9: Rb = 6.89. 0.9 = 6.2 N. ⎛ 2 ⋅ 10 ⋅ 1,745 ⎞ Rb Ab = 6.2 ⋅ 1000 ⋅ 150⎜1 − ⎟ = 713620 N = 713.6 kN > N = 650 kN , ⎝

150

⎠

i.e. panel strength is ensured with any combination of loads.

CALCULATION OF DURABILITY FOR REINFORCED CONCRETE ELEMENTS 3.14. Durability of reinforced concrete elements is calculated regarding moments of flection, transverse forces, normal forces, torque moments and local load impact (local compression, pushing, cleavage). FLEXURAL ELEMENTS CALCULATION OF DURABILITY FOR REINFORCED CONCRETE ELEMENTS UNDERTHE IMPACT OF FLECTION MOMENTS


General provisions 3.15. Calculation of durability of reinforced concrete elements regarding impact of flection moments shall be performed for cross-sections which are normal in relation to axle. Calculation of normal cross-sections of flexural elements shall be performed on the basis of non-linear deformation model in accordance with paragraphs 3.72-3.76 assuming that N = 0. Calculation for a rectangular, T- and I-sections with reinforcement located at element edges which are perpendicular to bending plane with moment effect in the cross section plane of mirror symmetry might be performed with respect to critical forces in accordance with paragraphs 3.17 – 3.27. Calculation for elements with such cross-sections regarding impact of biaxial bending in restrained terms might be performed with respect to critical forces in accordance with paragraphs 3.28 and 3.29. 3.16. For reinforced concrete elements with ultimate bending moment with respect to durability less than moment of crack formation (paragraphs 4.5-4.8), area of longitudinal tensile reinforcement shall be increased in comparison with specified design value by no less than 15% or shall satisfy durability analysis regarding moment of crack formation. 3.17. Durability of normal cross-sections shall be performed depending on correlation between value of relative height of concrete compressed zone

ξ=

x , h0

determined using respective equilibrium

conditions, and value of boundary relative height of compressed zone ξR, with which limit state is reached simultaneously with reaching tensile reinforcement stress which is equal to design strength Rs. Value ξR is calculated using formula ξR =

0,8 , Rs 1+ 700

(3.15)

or Table 3.2. Reinforcement class Value ξR Value αR

А240

А300

А400

А500

Table 3.2 В500

0,612 0,425

0,577 0,411

0,531 0,390

0,493 0,372

0,502 0,376


Rectangular cross-sections 3.18. Calculation for rectangular cross-sections (Drawing.3.3) is carried out in the following way depending on compressed zone height Rs As − Rsc As' x= : Rbb

а) with ξ =

(3.16)

x ≤ ξ R using the condition h0

M < Rbbх(h0 – 0,5x) + Rsc As' (h0 – a’); (3.17) б) with ξ > ξR using the condition (3.18) M < αRRbbh 02 + Rsc As' (h0 - a'), where αR =ξR(1 – 0.5ξR) or see Table 3.2. The right part of condition (3.18) might be increased to a certain extent if required replacing value αR (0.7αR + 0.3αm), where αm = ξ(1 – 0.5ξ) and assuming that ξ is no more than 1. If х ≤ 0, durability is checked using the condition M ≤ Rs As (h0 – a'). (3.19) Rsc A 's

x

a'

Rb

A's

Rb Ab

M h ho

Ab

As

a

Rs As

b Drawing.3.3. Schematic representation of forces and stress diagram in a rectangular cross section of a flexural reinforced concrete element

If compressed zone height х calculated without regard to compressed reinforcement ( As' = 0.0) is less than 2а', condition (3.19) might be checked with the use of replacing а' by х/2.


3.19. It is recommended to design flexural elements in such a way to provide fulfillment of the condition ξ ≤ ξR. A failure of the above condition is permissible only in case if area of tensile reinforcement is calculated regarding limit states of the second group or taken due to design considerations. 3.20. Durability check of rectangular cross-sections with a single reinforcement is performed: with x < ξRh0 using the condition M ≤ RsAs (h0 – 0,5x), (3.20) where х is height of compressed zone which is equal to x =

Rs As ; Rbb

ξR – see

paragraph 3.17; with х ≥ ξRh0 using the condition (3.21) M ≤ αRRb h02 , where for αR refer to Table 3.2; herewith, load carrying capacity shall be increased to a certain extent using the recommendation from paragraph 3.18,b. 3.21. Longitudinal reinforcement is selected in the following way. Value α m =

M Rb bh02

is calculated.

(3.22)

If αm < αR (see Table 3.2), compressed reinforcement is not required. When there is no compressed reinforcement, area of tensile reinforcement is calculated using the formula (3.23) As = Rbbh0 (1 − 1 − 2α m ) / Rs. In case if αт > αR, it is required to increase cross-section or to enhance concrete quality class, or otherwise to install compressed reinforcement in accordance with paragraph 3.22. 3.22. Areas of tensile As and compressed As' reinforcements corresponding to minimum value of their sum in case if compressed reinforcement is required (see paragraph 3.21) are calculated using the formulae: As' =

M − α R Rbbh02 ; Rs (h0 − à ' )

As = ξRRbbh0/Rs + As' , Where for ξR and αR refer to Table 3.2.

(3.24) (3.25)


If value of the taken area of compressed reinforcement AS' sufficiently exceeds the value calculated using formula (3.24), area of tensile reinforcement might be a little decreased in comparison with the value calculated using formula (3.25) by means of the formula As = Rbbh0 (1 − 1 − 2α m ) / Rs + As' , (3.26) αm =

where

M − Rsc As' (h0 − a ' ) Rbbh02

≥ 0.

Herewith, condition αm < αR shall be fulfilled (see Table 3.2). T-sections and I-sections 3.23. Calculation for cross-sections with a flange in the compressed zone (T-sections, I-section, etc.) is performed depending on boundary of compressed zone: a) if boundary is in the flange (Drawing. 3.4,а), the following condition is met RsAs ≤ Rb b f ' h 'f + Rsc As' , (3.27) Calculation is performed in accordance with paragraphs 3.18 and 3.20 in the same way as for rectangular cross-section with width b 'f ; b) if boundary is in the jack rib (Drawing. 3.4,b), i.e. condition (3.27) is not met, calculation is performed using the condition: a'

)

bf

As

As

b

h0 a

a

h

h0

x

hf

hf

a)

x

a'

bf

As

As

b

Drawing.3.4. Boundary of compressed zone in a T-section of a flexural reinforced concrete element a – in the flange; b –in the jack rib

REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures without REINFORCEMENT PRETENSIONING based on sp 52-101-2003 M ≤ Rb bx(h0 − 0.5 x) + Rb Aov (h0 − 0,5h 'f ) + Rsc As' (h0 − a ' ) ,

(3.28)

where Aov – area of flange overhangs, which is equal to (b 'f − b)h 'f , herewith, compressed zone height is calculated using the formula Rs As − Rsc As' − Rb Aov , x= Rbb

(3.29)

and taken no more than ξRh0 (see Table 3.2). If x >ξR h0, condition (3.28) might be written in the form M ≤ α R Rb bh02 + Rb Aov (h0 − 0,5h 'f ) + Rsc As' (h0 − à ' ) , Where for αR refer to Table 3.2.

(3.30)

Note: 1. When overhang height varies, it is permissible to assume value h 'f equal to average overhang height. 2. Compressed flange width b 'f , introduced into the calculation shall not exceed values specified in paragraph 3.26.

3.24. Required area of compressed reinforcement is determined using the formula As'

=

M − α R Rbbh02 − Rb Aov (h0 − 0,5h 'f ) Rsc (h0 − à ' )

,

(3.31)

where for αR refer to Table 3.2; Aov = (b 'f − b)h 'f . Herewith, condition h 'f ≤ ξ R h0 shall be met. In case if h 'f > ξ R h0 , area of compressed reinforcement is determined in the same way as for a rectangular cross section with width b = b 'f using formula (3.24). 3.25. Required area of tensile reinforcement is determined in the following way: a) If boundary is in the flange, i.e. the following condition is met: M ≤ Rb b 'f h 'f (h0 − 0,5h 'f ) + Rsc As' (h0 − a ' ), (3.32) Area of tensile reinforcement is determined in the way as for a rectangular cross-section with width b 'f in accordance with paragraphs 3.21 and 3.22; б) if boundary is in the jack rib, i.e. condition (3.32) is not met, area of tensile reinforcement is calculated using the formula Rbbh0 (1 − 1 − 2α m ) + Rb Aov + Rsc As' As = , Rs

(3.33)


where

αm =

M − Rb Aov (h0 − 0.5h 'f ) − Rsc As' (h0 − a ' ) Rb bh02

.

(3.34)

Herewith, condition αm ≤ αR shall be met (see Table 3.2). 3.26. Value b 'f introduced into the calculation is taken on the basis of condition that width of flange overhang on either side is no more than 1/6 of the element bay: а) when there are transverse jack ribs or with h 'f ≥ 0.1h is 1/2 of clearance between longitudinal jack ribs; б) when there are transverse jack ribs (or distance between them is more than distance between longitudinal jack ribs) and with h 'f < 0.1h - 6h 'f ; в) with cantilevers of the flange with h 'f ≥ 0.1h - 6h 'f ; with 0.05h ≤ h 'f < 0.1h − 3h 'f ; with h 'f ' < 0.05h overhangs are not taken into account. Calculation examples Rectangular cross-sections Example 2. Given: cross-section size b = 300 mm, h = 600 mm; а= =40 mm; moment of flection with allowance for short-term loads М = =200 kN.m; concrete quality class В15 (Rb = 8.5 MPa); reinforcement class А300 (Rs = 270 MPa). It is required to determine area of longitudinal reinforcement. C a l c u l a t i o n. h0 = 600 – 40 = 560 mm. Longitudinal reinforcement is selected in accordance with paragraph 3.21. One can calculate value αт using formula (3.22): αm =

M 200 ⋅ 10 6 = = 0.25. Rb bh02 8.5 ⋅ 300 ⋅ 560 2

One can use Table 3.2. to find αR = 0.41. Since αт = 0.25 < αR, compressed reinforcement is not required. Required area of tensile reinforcement is calculated using formula (3.23) 2 As = Rb bh0 (1 − 1 − 2α m ) / Rs = 8.5 ⋅ 300 ⋅ 560(1 − 1 − 2 ⋅ 0.25 ) / 270 = 1550 mm . One can assume 2∅28 + 1∅25 (Аs = 1598 mm2).


Example 3. Given: a cross-section with dimensions b = 300 mm, h = 800 mm; а = 70 mm; tensile reinforcement А400 (Rs = 355 MPa); area As = 2945 mm2 (6∅25); concrete of class В25 (Rb = 14.5 MPa); moment of flection М = 550 kN.m. It is required to check cross-section durability. C a l c u l a t i o n h0 = 800 – 70 = 730. Durability is checked in accordance with paragraph 3.20: Value х is calculated: x=

Rs As 355 ⋅ 2945 = = 240 mm. Rbb 14,5 ⋅ 300

One can find ξR = 0,531 in Table 3.2. Since ξ =

x 240 = = 0,329 < ξ R , h0 730

condition (3.20) is checked: Rs As(h0 – 0,5x) = 355 .2945 (730 – 0.5 . 240) = 636.8 . 106 Н. mm =636.8 kN.m > М = 550 kN.m, i.e. cross-section durability is ensured. Example 4. Given: a cross-section with dimensions b= 300 mm, h = 800 mm; a = =50 mm; reinforcement class А400 (Rs = Rsc = 355 MPa); moment of flection M = 780 kNm; concrete of class В15 ( Rb = 8.5 MPa). It is required to determine area of longitudinal reinforcement. C a l c u l a t i o n. h0 = h – a = 800 – 50 =750 mm. Required area of longitudinal reinforcement is determined in accordance with paragraph 3.21. One can find value αт: using formula (3.22). αm =

M 780 ⋅ 10 6 = = 0.544. Rb bh02 8.5 ⋅ 300 ⋅ 750 2

Since αm = 0.544 > αR = 0.39 (see Table 3.2) section dimensions and concrete class, compressed required. Assuming а′ = 30 mm and ξR = 0.531 (see Table of compressed and tensile reinforcement might be formulae (3.24) and (3.25):

with given crossreinforcement is 3.2) required area calculated using

M − α R Rb bh02 780 ⋅ 10 6 − 0.39 ⋅ 8,5 ⋅ 300 ⋅ 750 2 2 = = 863 mm ; Rsc (h0 − a' ) 355(750 − 30) ξ bh R 0.531 ⋅ 300 ⋅ 750 ⋅ 8.5 2 + 863 = 3724 mm . As = R 0 b + As' = Rs 355 As' =

One can assume As' = 942 mm2 (3∅20); As = 4021 mm2 (5∅32).


Example 5. Given: a cross-section with dimensions b = 300 mm, h = 700 mm; a = 50 mm; a’ = 30 mm; concrete of class В30 (Rb = 17 MPa); reinforcement А400 (Rs= Rsc = 355 MPa); area of compressed reinforcement As' = 942 mm2 (3∅20); moment of flection М = 580 kN.m. It is required to determine area of tensile reinforcement. C a l c u l a t i o n h0 = 700 – 50 = 650 mm. Calculation is performed taking into consideration the presence of compressed reinforcement in accordance with paragraph 3.22. Value αт:is calculated M − Rsc As' (h0 − a ' ) 580 ⋅ 106 − 355 ⋅ 942(650 − 30) αm = = = 0,173 . Rb bh02 17 ⋅ 300 ⋅ 650 2

Since αm = 0,173 < αR = 0,39 (see Table 3.2), required area of tensile reinforcement is calculated using formula (3.26) As = Rb bh0 (1 − 1 − 2α m ) / Rs + As' = 17 ⋅ 300 ⋅ 650(1 − 1 − 2 ⋅ 0.173 ) / 355 + 942 = = 2727 mm 2 .

One can take 3∅36 (As = 3054 mm2). Example 6. Given: a cross-section with dimensions b = 300 mm, h = 700 mm; a = 70 mm; a’ = 30 mm; concrete of class В20 (Rb = 11.5 MPa); reinforcement class А400 (Rs = Rsc= 355 MPa); area of tensile reinforcement is As = 4826 mm2 (6∅32), area of compressed reinforcement is A 's = 339 mm2 (3∅12); moment of flection М = 630 kN.m. It is required to check cross-section durability. C a l c u l a t i o n h0 = 700 – 70 = 630 mm. Cross-section durability is checked in accordance with paragraph 3.18. Height of compressed zone х is calculated using formula (3.16): x=

ξ=

x h0

Rs As − Rsc As' 355(4826 − 339) = = 461.7 Rb b 11.5 ⋅ 300

mm.

One can find ξR = 0.531 and αR = 0.39 in Table 3.2. Since 461.7 = = 0,733 > ξ R = 0.531, cross-section durability is checked using 630

condition (3.18): α R Rb bh02 + Rsc As' (h0 − a' ) = 0.39 ⋅ 11.5 ⋅ 300 ⋅ 630 2 + 355 ⋅ 339 ⋅ (630 − 30) = = 606.2 ⋅ 10 6 H ⋅ мм = 606.2кН ⋅ м < M = 630кН ⋅ м,

i.е. durability in accordance with this condition is not ensured. One can specify the right part of the condition (3.18) replacing value αR by (0.7αR + 0.3αm), where αm = ξ(1 - 0.5ξ) = 0.733(1- 0.5 . 0.733) = 0.464:


(0.7 . 0.39 + 0.3 . 0.464)11.5 . 300 . 6302 + 355 . 339 . 600 = 636.6 106 N.mm = 636.6 kN. m > М = 630 kN.m, i.e. durability is ensured.

.

T-sections and I-sections Example 7. Given: a cross section with dimensions b 'f = 1500 mm, h 'f = 50 mm, b = 200 mm, h = 400 mm; а = 80 mm; concrete of class В25

(Rb = 14.5 MPa), reinforcement class А400 (Rs = 355 MPa); moment of flection М = 260 kN.m. It is required to determine area of longitudinal reinforcement. C a l c u l a t i o n h0 = 400 – 80 = 320 mm. Calculation is performed in accordance with paragraph 3.25 on the assumption that compressed reinforcement is not required. One can check condition (3.32) assuming that As' = 0: Rbb 'f h 'f (h0 – 0.5 h 'f ) = 14.5 . 1500 . 50(320 – 0.5 . 50) = 320.8 . 106 N.mm = = 320.8 kN.m > М = 260 kN.m, i.е. compressed reinforcement zone is in the flange and calculation is performed in the same way as for a rectangular cross-section with width b = b 'f = 1500 mm in accordance with paragraph 3.21. The value is calculated: αm =

As

M 260 ⋅ 10 6 = = 0,117 < Rb bh02 14.5 ⋅ 1500 ⋅ 320 2

αR = 0.39 (see Table 3.2),

i.е. compressed reinforcement is in fact not required. Area of tensile reinforcement is calculated using formula (3.22) 2 = Rb bh0 (1 − 1 − 2α m ) / Rs = 14.5 ⋅ 1500 ⋅ 320(1 − 1 − 2 ⋅ 0.117 ) / 355 = 2446 mm . One can take 4∅28(As = 2463 mm2). Example 8. Given: a cross section with dimensions b 'f = 400 mm,

h 'f = 120 mm, b = 200 mm, h = 600 mm; а = 65 mm; concrete of class В15

(Rb = 8.5 MPa); reinforcement class А400 (Rs = 355 MPa); moment of flection М = 270 kN. m. It is required to determine area of tensile reinforcement. C a l c u l a t i o n.h0 = 600 – 65 = 535 mm. Calculation is performed in accordance with paragraph 3.25 on the assumption that compressed reinforcement is not required. Since


Rbb 'f h 'f (h0 – 0.5 h 'f ) = 8.5 . 400 . 120(535 – 0.5 . 120) = 193.8 . 106 N.mm = = 193.8 kN.m < M = 270 kN. m, boundary of compressed zone is in the jack rib and area of tensile reinforcement is determined using formula (3.33), assuming area of overhangs equal to Aov = (b 'f − b)h 'f = (400 − 200) ⋅120 = 24000 mm2. Value αm is calculated with As' = 0 αm =

M − Rb Aov (h0 − 0,5h 'f ) Rb bh02

=

270 ⋅ 10 6 − 8.5 ⋅ 24000(535 − 0.5 ⋅ 120) = 8.5 ⋅ 200 ⋅ 535 2

= 0.356 < α R = 0.39 (see Table3.2),

consequently, compressed reinforcement is not required. As =

Rb bh0 (1 − 1 − 2α m ) + Rb Aov Rs

=

8.5 ⋅ 200 ⋅ 535(1 − 1 − 2 ⋅ 0.356 ) + 8.5 ⋅ 24000 = 355

= 1949.5мм 2

One can take 4∅25(As = 1964 mm2). Example 9. Given: a cross-section with dimensions b 'f = 400 mm,

h 'f = 100 mm, b= 200 mm, h = 600 mm; a = 70 mm, concrete of class В25

(Rb = 14.5 MPa); tensile reinforcement class А400 (Rs = 355 MPa); area As = 1964 mm2 (4∅25); As' = 0.0; moment of flection M = 300 kN.m. It is required to check cross-section durability. C a l c u l a t i o n. h0 = 600 – 70 = 530 mm. Durability is checked in accordance with paragraph 3.23, assuming As' = 0.0. Since RsAs = 355 х.1964 = 697220 H >Rb b 'f h 'f =14.5 . 400 . 100 = 580000 N, boundary of compressed zone is in the jack rib and cross-section durability is checked using condition (3.28). For this purpose height of compressed zone is calculated using formula (3.29) assuming area of overhangs equal to 2 Aov = (b 'f − b)h 'f = (400 − 200) ⋅ 100 = 20000 mm : x=

Rs As − Rb Aov 355 ⋅ 1964 − 14.5 ⋅ 20000 = = 140 Rb b 14.5 ⋅ 200

mm < ξRh0 = 0.531 х

530 = 281 mm (ξR is found in Table 3.2). Rbbx (h0 – 0,5x) + RbAov(h0 – 0.5h 'f ) = 14.5 . 200 . 140 . (530 – 0.5 . 140) + + 14.5 . 20000(530 – 0.5 . 100) = 326 . 106 N.mm = 326 kN.m>М = 300 kN.m, i.е. cross-section durability is ensured.

.


Elements in biaxial bending 3.27. Calculation of rectangular cross sections, T-sections, I-sections and L-shaped cross sections in biaxial bending is permissible to perform assuming that the form of compressed zone is the same like in Drawing. 3.5; herewith, the following condition shall be met (3.35) Mx ≤ Rb[Awеb(h0 – x1/3) + Sov,x] + RscSsx, where Мх is a component of flection moment in the plane of axis х (two mutual perpendicular axes crossing gravity center of tensile reinforcement in parallel with cross-section sides are taken as axes х and у; for a cross section with a flange axis х is taken in parallel with the jack rib plane); Awеb = Ab – Aov; (3.36) Ab – area of concrete compressed zone, which is equal to Rs As − Rsc As' Ab = ; Rb 1

x b'f

x

A'S

b)

A'S

1

b

y

2

b0i b0 b

AS x

y

x1 b0i

y AS

h oi

b'ov

θ

Ab β

ho

bov A b β

h oi

ho

x1

h'f

a)

(3.37)

y

b0 x

2

DRAWING.3.5 FORM OF COMPRESSED ZONE IN A CROSS-SECTION OF A REINFORCED CONCRETE ELEMENT IN BIAXIAL BENDING a – T-section; b- rectangular cross-section; 1-plane of flection moment effect; 2- gravity center of tensile reinforcement cross-section

Aov – square of the most compressed flange overhang; х1 – size of concrete compressed zone along the most compressed side face of the cross-section calculated using the formula (3.38) x1 = −t + t 2 + 2 Aweb ctgβ , where

⎛ S ov , y ctgβ − S ov , x ⎞ t = 1.5⎜⎜ + b0 ctgβ − h0 ⎟⎟; Aweb ⎝ ⎠


Sov,y,,Sov,x – static moments of area Aov in relation to axes х and y; β − bending angle of plane of flection moment to axis х, i.е. ctgβ = Mx/My (My – a component of flection moment in the plane of axis у); b0 – distance from gravity center of tensile reinforcement crosssection to the most compressed side edge of the jack rib (side). When rectangular cross-sections are calculated, values Aov, Sov,x, Sov,y are taken to be equal to zero. If Ab < Aov or x1 < 0,2h 'f , calculation is performed in the same way as for a rectangular section with width b = b 'f . If the condition 1.5 Aweb , is met x1 < b + bov

(3.39)

(where bov is width of the least compressed flange overhang), calculation is performed without taking biaxial bending into account, i.e. using formulae from paragraphs 3.18 and 3.23 regarding moment effect М = Мх, herewith, condition (3.40) shall be checked, assuming х1 is the same like with biaxial bending. When value Ab is calculated using formula (3.37), stress in tension bar closest to the boundary of compressed zone shall not be less than Rs, which is ensured when the following condition is met ξ=

' bov tgθ + x1 ' (b0i + bov ) tgθ + h0i

≤ ξR ,

(3.40)

where for ξR refer to Table 3.2; b0i and h0i are distances from the considered bar respectively to the most compressed edge (side) normal to axis (see Drawing. 3.5); ' bov is width of the most compressed overhang; θ is inclination of line which limits compressed zone to axis y; value tgθ is calculated using the formula x12 tgθ = . 2 Awåb

If condition (3.40) is not met, calculation for a cross-section is performed using stepwise approximation by means of replacing value Rs in formula (3.37) for each tension bar by stress values equal to


⎞ − 1⎟⎟ (MPa), however no more than Rs. ⎠ ⎝ ξi ⎛ 0,8

σ si = 700⎜⎜

When structures are designed it is not permissible to use value ξi exceeding ξR by more than 20%, herewith only one structural recalculation might be performed with replacement in formula (3.37) values Rs for tension rods for which ξi > ξR regarding stress equal to σ s1 =

700(0,8 / ξ i − 1) + 2 Rs . (3.41) 3

When formula (3.37) is used, it is recommended to assume reinforcement located close to tensile edge as tensile reinforcement with area As in parallel with axis y; reinforcement located close to compressed zone in parallel with axis у, however, at one the most compressed side of axis х as compressed reinforcement with area As' (see Drawing. 3.5). It can be used in case if a condition is fulfilled: for rectangular cross-sections, T- and L-shaped sections with a flange in compressed zone x1 < h; for rectangular cross-sections, T- and L-shaped sections with a flange in tensile zone x1 < h – hf - bov,ttgθ, where hf and bov,t are height and width of the least tensile flange overhang (Drawing. 3.6). Otherwise, calculation is performed on the basis of non-linear deformation model in accordance with paragraphs 3.72-3.76 assuming that N = 0. 3.28. It is recommended to use graphs of Drawing 3.7 in order to determine required quantity of tensile reinforcement with biaxial bending for elements of rectangular, T- and L-shaped sections with a flange in compressed zone. Location of gravity center of tensile reinforcement is taken for this purpose and value αs is determined using the graph depending on:

MANUAL to CP 52-101-2003 concrete and REinforced concrete structures without prior reinforcement stress

Drawing. 3.6. T-section with compressed zone spreading into the least tensile flange overhang

α mx = α my =

M x − Rb S ov, x − Rsc S sx Rb b0 h02 M y − Rb S ov, y − Rsc S sy Rb b02 h0

; ,

where Ssx and Ssy are first moments of area As' in relation to axis y and axis x. All the other designations see in paragraph 3.27. If αmx < 0, calculation is performed in the same way as for a rectangular cross-section assuming b = b'f . If value αs in the graph is at the left side of the curve corresponding to the parameter

bov + b , reinforcement is selected b0


Drawing.3.7. Graph of load carrying capacity of a rectangular cross-section, Tand L-shaped sections for elements in biaxial bending

α mx =

M x − Rb S ov, x − Rsc S sx Rb b0 h02

; α my =

M y − Rb S ov, y − Rsc S sy Rb b02 h0

;


αs =

Rs As − Rb Aov − Rsc A's Rbb0 h0

is performed without regard to biaxial bending, i.е. in accordance with paragraphs 3.22 and 3.26 regarding moment effect М = Мх. Required square of tensile reinforcement in case if it is in service with total design strength is calculated using the formula As = (α s b0 h0 + Aov )

Rb + As' , Rs

(3.42)

where for Aov refer to formula (3.36). Gravity center of a taken tensile reinforcement shall be apart from tensile edge not further than design gravity center. Otherwise, recalculation is required assuming new gravity center of the tensile reinforcement crosssection. Tensile reinforcement can be used with total strength only if condition (3.40) is fulfilled. With reinforcement of class А400 and less condition (3.40) is always fulfilled if value αs in graph 3.7 is inside the zone limited by coordinate ' / b0 . axes and a curve corresponding to the parameter bov If condition (3.40) is not fulfilled, it is necessary to install (to increase) compressed reinforcement or to enhance concrete of class, otherwise to increase dimensions of the cross-section (especially, the most compressed flange overhang). Values αs in the graph shall not be between the axis αmy and the curve corresponding to parameter h0 / h. Otherwise, х1 is becoming more than h and calculation shall be performed in accordance with paragraphs 3.72-3.76. Calculation examples Example10. Given: reinforced concrete wall plate with sloping 1:4 (ctgβ=4); reinforcement cross-section and location are determined using Drawing. 3.8; concrete of class В25 (Rb=14,5MPa); tensile reinforcement of class А400 (Rs=355MPa); As = 763 mm2 (3∅18); Às' = 0.0; flection moment in the vertical plane is М = 82,6 kN.m. It is required to check cross-section durability. C a l c u la t i o n. Using Drawing. 3.8:

MANUAL to CP 52-101-2003 concrete and REinforced concrete structures without prior reinforcement stress 2 ⋅ 120 + 1 ⋅ 30 1 ⋅ 30 = 90 mm; = 360 mm; b0 = 3 3 300 − 150 20 bov' = bov = = 75 mm; h f ' = 80 + = 90 mm. 2 2 h0 = 400 − 30 −

1 80

x

20

30 0

θ

h 0i

Ab h0

y :4 i=1

400

b0

2

b 0i

30

30

30

y

30

90 150 x

Drawing 3.8 For calculation example 10 1- plane of flection moment; 2-gravity center of tensile reinforcement area

Using formula (3.37) one can determine area of concrete compressed zone Ab: Ab =

Rs As 355 ⋅ 763 2 = = 18680 mm . 14,5 Rb

Area of the most compressed flange overhang and first moments of the area in relation to axes х and у are respectively equal to: . 2 ' ' h f = 75 90 =6750 mm ; Аov= bov Sov,y = Aov(b0 + b 'ov /2)=6750(90 + 75/2) = 86.06 . 104 mm3; Sov,x= Aov(h0 - h 'f /2) = 6750(360 – 90/2) = 212,6 . 104 mm3. Since Ab > Aov, one can continue calculation in the same way as for a T-section. Aone canb = Ab – A ov = 18680 – 6750 = 11930 mm2. Compressed zone size х1 is determined using formula (3.38). For this purpose one can calculate

MANUAL to CP 52-101-2003 concrete and REinforced concrete structures without prior reinforcement stress ⎛ S ov , y ctgβ − S ov , x ⎞ ⎛ 860600 ⋅ 4 − 2126000 ⎞ + b0 ctgβ − h0 ⎟⎟ = 1.5⎜ + 90 ⋅ 4 − 360 ⎟ = t = 1.5⎜⎜ Aweb 11930 ⎝ ⎠ ⎝ ⎠ = 165.5 мм. x1 = −t + t 2 + 2 Aweb сtgβ = −165.5 + 165.5 2 + 2 ⋅ 11930 ⋅ 4 = 185 мм.

Condition (3.39) is checked: 1.5 Aweb 1.5 ⋅ 11930 = = 79.5 mm < x1 = 185 mm, b + bov

150 + 75

consequently, calculation is continued using formulae of biaxial bending. One can check condition (3.40) for the least tension bar. From Drawing. 3.8 one can have b0i= 30 mm, h0i = 400 – 30 = 370 mm; x12 185 2 = = 1.434; 2 Aweb 2 ⋅ 11930 b' ov tgθ + x1 75 ⋅ 1.434 + 185 ξi = = = 0.562 > ξ R = 0.531 (b0i + b' ov ) tgθ + h0i (30 + 75)1.434 + 370 tgθ =

(see Table 3.2).

Condition (3.40) is not met. Recalculation is performed with replacement in formula (3.37) value Rs for the least tension bar with stress σs, determined using formula (3.41) and correction of values h0 and b0. σs =

700(0.8 / ξ i − 1) + 2 Rs 700(0.8 / 0.562 − 1) + 2 ⋅ 355 = = 335.5 MPa 3 3

=

=0.945 Rs. Since all the rods are of the same diameter new values Ab,b0 and h0 are equal to: 2 + 0.945 2 ⋅ 120 + 0.945 ⋅ 30 = 91.1 mm; = 18338 mm 2 ; b 0 = 3 2 + 0.945 1 ⋅ 30 = 359.8 mm. h0 = 400 − 30 − 2 + 0.945 Ab = 18680

Similarly one can determine values Sov,y, Sov,x, Aone canb and x1: Sov,y = 6750(91,1 + 75/2 = 86,8 . 104 mm3; Sov,x = 6750(359,8 – 90/2) = 212,5 . 104 mm3; Aone canb = 18338 – 6750 = 11588 mm2; ⎛ 868000 ⋅ 4 − 2125000 ⎞ t = 1.5⎜ + 91.1 ⋅ 4 − 359,8 ⎟ = 181.3 mm; 11588 ⎝ ⎠ x1 = −181.3 + 181.3 2 + 2 ⋅ 11588 ⋅ 4 = 173.1 mm.

Let us check cross-section durability using condition (3.35) assuming Ssx =0 and M x = M ⋅ сosβ = M ctgβ 2 = 82.6 4 2 = 80.1 kN.m: 1 + ctg β

=

1+ 4

Rb[Aone canb(h0 – x1/3) +Sov,x] = 14.5[11588(359.8 – 173,1/3) + 212.5 . 104] =81.57 . 106 N.mm> Mx = 80,1 .106 N.mm,


i.е. cross-section durability is ensured. Example 11. It is required to find tensile reinforcement area with moment in the vertical plane М = 64 kN.m using the data from Example 10. C a l c u l a t i o n. Components of flection moment in the plane of axes у and х are: M 64 = = 15.52 kNm; M y = Msinβ = 2 2 1 + ctg β

1+ 4

Mx = My сtgβ = 15.52 4 = 62.1 kN.m. One can determine required quantity of reinforcement in accordance with paragraph 3.28. Assuming values Rb, h0, Sov,x and Sov,y from Example 10 with Ssy = Ssx= 0 one can calculate values αmx and αmy: .

α mx = α my =

M x − Rb S ov , x Rb b0 h02

M y − Rb S ov , y Rb b02 h0

=

62.1 ⋅ 10 6 − 14.5 ⋅ 212.6 ⋅ 10 4 = 0.185; 14.5 ⋅ 90 ⋅ 360 2

=

15.52 ⋅ 10 6 − 14,5 ⋅ 86.06 ⋅ 10 4 = 0.072. 14.5 ⋅ 90 2 ⋅ 360

Since αmx> 0, calculation is continued for a T-section. Since a point with coordinates αmx = 0.185 and αmy = 0.072 in the graph of Drawing. 3.7 is located at the right side from the curve corresponding to the parameter b + bov = 150 + 75 = 2.5 , and at the left side of the b0

90

curve corresponding to the parameter bov' / b0 = 75 / 90 = 0.83 , calculation is continued with allowance for biaxial bending and total design strength of reinforcement, i.е. condition (3.40) is fulfilled. In the graph value αs = 0.20 corresponds to coordinates αmx = 0.185 and αmy = 0.072. Then in accordance with formula (3.42) area of tensile reinforcement will be equal to As = (αsb0h0 + Aov)Rb/Rs = (0.2 . 90 . 360 + 6750)14.5/355 = 540.4 mm2. Rods are taken as 3∅16 (As = 603 mm2) and located in the way specified in the Drawing 3.8. CALCULATION OF REINFORCED CONCRETE ELEMENTS UNDER THE IMPACT OF TRANSVERSE FORCES

3.29. THE CALCULATION OF ELEMENTS UNDER THE IMPACT OF TRANSVERSE FORCES SHALL ENSURE THE DURABILITY: - ALONG THE STRIP BETWEEN THE INCLINED SECTIONS ACCORDING TO PARA 3.30;


- TO THE ACTION OF TRANSVERSE FORCE ALONG THE INCLINED SECTION ACCORDING TO PARAS 3.31-3.42; - TO THE IMPACT OF MOMENT ALONG THE INCLINED SECTION ACCORDING TO PARAS 3.43-3.48.


CALCULATION OF REINFORCED CONCRETE ELEMENTS ALONG THE STRIP BETWEEN THE INCLINED SECTIONS 3.30. THE FLEXURAL ELEMENTS ALONG THE CONCRETE STRIP BETWEEN THE INCLINED SECTIONS ARE CALCULATED TAKING INTO ACCOUNT THE CONDITION Q ≤ 0,3RBBH0,, (3.43) WHERE Q IS THE TRANSVERSE FORCE APPLIED TO THE PERPENDICULAR SECTION AT A DISTANCE FROM THE SUPPORT OF AT LEAST H0. CALCULATION OF REINFORCED CONCRETE ELEMENTS ALONG THE INCLINED SECTIONS FOR THE IMPACT OF TRANSVERSE FORCES CONSTANT HEIGHT ELEMENT, REINFORCED BY STIRRUPS PERPENDICULAR TO THE ELEMENT AXIS 3.31. THE FLEXURAL ELEMENTS ALONG THE INCLINED SECTION (DRAWING 3.9) ARE CALCULATED TAKING INTO ACCOUNT THE CONDITION Q ≤ QB + QSW, (3.44) WHERE Q IS THE TRANSVERSE FORCE APPLIED TO THE INCLINED SECTION WITH A PROJECTION LENGTH C FROM EXTERNAL FORCES LOCATED ON ONE SIDE FROM THE CONSIDERED INCLINED SECTION; UNDER A VERTICAL LOAD APPLIED TO THE TOP SIDE OF THE ELEMENT VALUE Q IS TAKEN FOR THE PERPENDICULAR SECTION, WHICH PASSES AT A DISTANCE C FROM THE SUPPORT; THEREBY THE POSSIBILITY OF THE ABSENCE OF TEMPORARY LOAD ON THE SUPPORT-ADJACENT SECTION WITH A LENGTH C SHOULD BE CONSIDERED; QB IS THE TRANSVERSE FORCE APPLIED TO THE CONCRETE IN INCLINED SECTION; QSW IS THE TRANSVERSE FORCE APPLIED TO THE STIRRUPS IN INCLINED SECTION. THE TRANSVERSE FORCE QB IS DETERMINED USING THE FOLLOWING FORMULA


Qb =

Mb , c

(3.45)

WHERE MB = 1.5 RBTBH20. (3.46) VALUE QB IS TAKEN AS NO MORE THAN 2.5 RBTBH0 AND NO LESS THAN 0.5 RBTBH0. VALUE IS DETERMINED ACCORDING TO PARA 3.32. THE FORCE QSW IS DETERMINED USING THE FORMULA QSW = 0.75QSWC0, (3.47) WHERE QSW IS THE FORCE IN THE STIRRUPS PER UNIT OF LENGTH OF THE ELEMENT, EQUAL TO q

h'f

Sw

b'f

Sw

Qb RswA sw

h0

Sw

F

Rsw A sw RswA sw

Asw b

Co

C

Q max

PROJECTIO Эпюра Q NQ

Q=Q max -qC-F

Q max

DRAWING 3.9. DIAGRAM OF FORCES IN THE INCLINED SECTION OF ELEMENTS WITH STIRRUPS FOR ITS CALCULATION WITH RESPECT TO THE ACTION OF A TRANSVERSE FORCE

qsw =

Rsw Asw , sw

(3.48)

C0 IS THE LENGTH OF THE PROJECTION OF THE OBLIQUE CRACK TAKEN EQUAL TO C, BUT NO MORE THAN 2H0. THE STIRRUPS ARE TAKEN INTO CONSIDERATION, IF THE FOLLOWING CONDITION IS MET QSW ≥ 0.25RBTB. (3.49) THIS CONDITION MAY BE DISREGARDED, IF SUCH A REDUCED VALUE OF RBTB IS CONSIDERED IN FORMULA (3.46), FOR WHICH CONDITION (3.49) IS TRANSFORMED INTO AN 2

EQUALITY, I.E., ASSUME MB = 6H 0 QSW.


3.32. DURING VERIFICATION IN THE GENERAL CASE THE CONDITIONS (3.44) ARE PROVIDED BY A RANGE OF INCLINED SECTIONS WITH DIFFERENT VALUES OF C THAT DO NOT EXCEED THE DISTANCE FROM THE SUPPORT TO THE SECTION WITH THE MAXIMUM FLECTION MOMENT AND NOT EXCEEDING 3H0. WHEN CONCENTRATED FORCES ARE APPLIED TO THE ELEMENT VALUES OF C ARE TAKEN AS EQUAL TO THE DISTANCES FROM THE SUPPORT TO APPLICATION POINTS OF THESE FORCES (DRAWING 3.10) AND ALSO EQUAL TO c=

Mb BUT NO LESS THAN H0, IF THIS VALUE IS LESS THAN 0,75qsw

THE DISTANCE FROM THE SUPPORT TO THE 1ST LOAD. WHEN CALCULATING AN ELEMENT FOR THE IMPACT OF UNIFORMLY DISTRIBUTED LOAD Q THE LEAST FAVORABLE VALUE

OF

C

F1

IS

TAKEN

EQUAL

TO

Mb , q1

AND

F2

2 1

C1 C2

Q1

Q=Q - F1 2 1

DRAWING 3.10. ARRANGEMENT OF CALCULATED INCLINED PROJECTIO SECTIONS FOR CONCENTRATED Эпюра Q NQ FORCES 1 – INCLINED SECTION CHECKED FOR THE IMPACT OF TRANSVERSE FORCE Q1; 2 – THE SAME, FORCE Q2 Mb Mb 2h0 q < or sw > 2 , c = SHOULD BE q1 1 − 0.5 qsw Rbt b 0.75qsw + q1

IF THEREBY

Rbt b

TAKEN, WHERE THE Q1 VALUE IS DETERMINED AS FOLLOWS: A) IF A CONTINUOUS UNIFORMLY DISTRIBUTED LOAD Q, Q1 = Q IS APPLIED;


B) IF THE LOAD Q INCLUDES A TEMPORARY LOAD, WHICH IS BROUGHT TO A UNIFORMLY DISTRIBUTED LOAD QV WITH AN EQUIVALENT MOMENT (I.E. WHEN MOMENT PROJECTION M OF THE LOAD QV TAKEN FOR THE CALCULATION ALWAYS GOES AROUND THE M PROJECTION FROM ANY ACTUAL TEMPORARY LOAD) Q1 = Q – 0.5 QV. THEREBY IN CONDITION (3.44) VALUE Q IS TAKEN AS EQUAL TO QMAX - Q1C, WHERE QMAX IS THE TRANSVERSE FORCE IN THE BEARING SECTION. 3.33. THE REQUIRED INTENSITY OF STIRRUPS, EXPRESSED AS QSW (SEE PARA 3.31), IS DETERMINED AS FOLLOWS: A) WHEN THE ELEMENT IS EXPOSED TO CONCENTRATED FORCES, LOCATED AT THE DISTANCE OF CI FROM THE SUPPORT, FOR EACH ITH INCLINED SECTION WITH A LENGTH OF PROJECTION CI NOT EXCEEDING THE DISTANCE TO THE SECTION WITH THE MAXIMUM FLECTION MOMENT, VALUE QSW(I) IS DETERMINED AS FOLLOWS DEPENDING ON THE COEFFICIENT αι = CI/H0, TAKEN AS NOT EXCEEDING 3: ε грi Qi 1.5 ≤ ε грi = + 0.1875α 0i , qsw(i ) = 0.25 Rbt b ; IF ε i = (3.50) εi Rbt bh0 αi ε i − 1.5 / α i q = R b ( ) sw i bt IF εI > εГРI, (3.51) 0.75α 0i WHERE α0I IS THE SMALLEST OF VALUES αI AND 2; QI IS THE TRANSVERSE FORCE IN THE ITH PERPENDICULAR SECTION, SITUATED AT A DISTANCE CI FROM THE SUPPORT; FINALLY THE LARGEST VALUE OF QSW IS TAKEN; B) WHEN THE ELEMENT IS EXPOSED ONLY TO A UNIFORMLY DISTRIBUTED LOAD Q THE REQUIRED INTENSITY OF THE STIRRUPS QSW IS DETERMINED AS FOLLOWS, DEPENDING ON Qb1 = 2 M b q1 : IF Qb1 ≥ 2M b / h0 − Qmax , q sw

2 − Qb21 Qmax ; = 3M b

(3.52)

IF QB1<2MB/H0 – QMAX, q sw =

Qmax − Qb1 ; 1,5h0

(3.53)


THEREBY, IF QB1 < RBTBH0, qsw =

Qmax − 0,5 Rbt bh0 − 3h0 q1 , 1,5h0

(3.54)

WHERE MB – SEE PARA 3.31; Q1 – SEE PARA 3.32. IF THE OBTAINED VALUE OF QSW DOES NOT SATISFY CONDITION (3.49), IT SHOULD BE CALCULATED ACCORDING TO THE FORMULA 2

Q / h + 8q1 ⎛ Q / h + 8q1 ⎞ ⎛ Qmax ⎞ ⎟⎟ qsw = max 0 − ⎜ max 0 ⎟ − ⎜⎜ 1.5 1.5 ⎠ ⎝ 1.5h0 ⎠ ⎝

2

(3.55)

Qmax / h0 − 3q1 AND TAKEN NO LESS THAN . 3.5

3.34. WITH THE INTENSITY OF THE STIRRUPS DECREASING FROM THE SUPPORT TO THE SPAN FROM QSW1 TO QSW2 (FOR EXAMPLE, DUE TO AN INCREASED SPACING OF THE STIRRUPS) CONDITION (3.44) SHOULD BE CHECKED WITH VALUES OF C EXCEEDING L1, THE LENGTH OF THE SECTION WITH STIRRUPS INTENSITY QSW1 (DRAWING 3.11). THEREBY VALUE QSW IS TAKEN AS EQUAL TO: IF C < 2H0 + L1, QSW = 0.75QSW1C0 – (QSW1 –QSW2) (C – L1); (3.56) IF C > 2H0 + L1, QSW = 1.5QSW2H0, (3.57) C0 – SEE PARA 3.31. WHEN THE ELEMENT IS EXPOSED TO A UNIFORMLY DISTRIBUTED LOAD, THE LENGTH OF SECTION WITH STIRRUPS INTENSITY QSW1 IS TAKEN NO LESS THAN VALUE L1, DETERMINED DEPENDING ON ΔQSW = 0.75 (QSW1 – QSW2) AS FOLLOWS: - IF ΔQSW < Q1, l1 = c −

M b / c + 0,75qsw1c0 − Qmax + q1c , Δqsw

(3.58)

MANUAL to CP 52-101-2003 concrete and REinforced concrete structures without prior reinforcement stress F

s w1

s w1

s w2

s w2

l1 C0 C

DRAWING 3.11. FOR CALCULATION OF INCLINED SECTIONS AFTER A CHANGE OF STIRRUPS INTENSITY

WHERE c =

Mb , BUT NO MORE THAN 3H0, q1 − Δqsw

Ml Ml 2h0 < , c= WHEREBY, IF q1 − Δqsw q q1 + 0.75qsw 2 ; 1 − 0.5 sw Rbt b - IF ΔQSW ≥ Q1,

l1 =

Qmax − (Qb. min + 1.5qsw 2 h0 ) − 2h0 ; q1

(3.59)

HERE MB, C0 – SEE PARA 3.31; Q1 – SEE PARA 3.32; QB,MIN = 0.5 RBTBH0. IF CONDITION (3.49) IS NOT SATISFIED FOR VALUE QSW2, THE LENGTH L1 IS CALCULATED WITH VALUES MB = 6H02QSW2 AND QB,MIN = 2H0QSW2 CORRECTED ACCORDING TO PARA 3.31; THEREBY THE TOTAL (QB,MIN + 1.5QSWH0) IN FORMULA (3.59) IS TAKEN NO LESS THAN THE UNCORRECTED VALUE OF QB,MIN. 3.35. THE SPACING OF STIRRUPS CONSIDERED IN THE CALCULATION SHALL NOT EXCEED VALUE: s w. max

Rbt bh02 = Q

(3.60)

FURTHERMORE, THE STIRRUPS SHALL SATISFY THE DESIGN SPECIFICATIONS STIPULATED IN PARAS 5.20 AND 5.21. ELEMENTS WITH VARIABLE HEIGHT AND CROSSWISE REINFORCEMENT


3.36. THE ELEMENTS WITH INCLINED COMPRESSED OR TENSIONED FACES IN THE SECTIONS NEXT TO THE SUPPORT ARE CALCULATED ACCORDING TO PARA 3.31, TAKING THE LARGEST VALUE OF H0 WITHIN THE LIMITS OF THE CONSIDERED INCLINED SECTION AS THE EFFECTIVE DEPTH OF SECTION (DRAWING 3.12). q

q b)

01

β

h

0

h

h

h

01

01

β

a)

C

C

DRAWING 3.12 BEAMS WITH VARIABLE DEPTH OF SECTION AND AN OBLIQUE FACE

3.37. FOR BEAMS WITHOUT OFFSET BENDS WITH A HEIGHT INCREASING EVENLY FROM THE SUPPORT TO THE SPAN CALCULATED FOR THE IMPACT OF A UNIFORMLY DISTRIBUTED LOAD Q, THE INCLINED SECTION IS VERIFIED USING CONDITION (3.44) WITH THE LEAST FAVORABLE VALUE OF C, EQUAL TO c = h01

1 .5 , q1 /( Rbt b) + 1,5tg 2 β

WHEREBY, IF THIS VALUE IS LESS THAN cгр = 2h01

(3.61) 1 − 2 tgβ (1 − 2 tgβ ) 2 − 0.5

qsw Rbt b

OR, IF QSW/(RBTB) > 2(1-2TGβ)2, THEN LEAST FAVORABLE VALUE OF C IS EQUAL TO c = h01

1.5 . (0,75qsw + q1 ) /( Rbt b) + 1,5tg 2 β

(3.62)

THE TAKEN VALUE OF C SHALL NOT EXCEED 3H01/(13TGβ), OR THE LENGTH OF THE BEAM SECTION WITH A CONSTANT VALUE β. HERE: H01 IS THE EFFECTIVE DEPTH OF THE BEARING BEAM SECTION;


Q1 - SEE PARA 3.32; β IS THE ANGLE BETWEEN THE COMPRESSED AND TENSIONED BEAM FACES. THE EFFECTIVE DEPTH IS TAKEN EQUAL TO H0 = H01+ C · TG β. WHEN STIRRUPS INTENSITY DECREASES FROM QSW1 AT THE SUPPORT TO QSW2 AT THE SPAN, CONDITION (3.44) SHOULD BE CHECKED WITH VALUES OF C EXCEEDING L1, THE LENGTH OF ELEMENT SECTION WITH STIRRUPS INTENSITY QSW1; THEREBY VALUE QSW IS DETERMINED USING FORMULA (3.56) OR FORMULA (3.57) PARA 3.34 DEPENDING ON THE FULFILLMENT OR NON FULFILLMENT OF CONDITION c <

2h01 + l1 . 1 − 2 tgβ

WHEN THE BEAM IS EXPOSED TO CONCENTRATED FORCES, VALUE C IS TAKEN AS EQUAL TO THE DISTANCE FROM THE SUPPORT TO THE APPLICATION POINTS OF THESE FORCES AND ALSO DETERMINED USING FORMULA (3.62) WITH Q1 = 0, IF THIS VALUE OF C IS LESS THAN THE DISTANCE FROM THE SUPPORT TO THE 1ST LOAD. 3.38. FOR CANTILEVERS WITHOUT OFFSET BENDS WITH HEIGHT EVENLY INCREASING FROM THE FREE END TO THE SUPPORT (DRAWING 3.13) IN GENERAL CONDITION (3.44) IS CHECKED ASSIGNING INCLINED SECTIONS WITH VALUES OF C DETERMINED USING FORMULA (3.62), WITH Q1 = 0 AND TAKEN AS NOT EXCEEDING THE DISTANCE FROM THE BEGINNING OF THE INCLINED SECTION IN THE TENSION AREA TO THE SUPPORT. THEREBY H01 AND Q ARE TAKEN TO MEAN THE EFFECTIVE DEPTH AND THE TRANSVERSE FORCE AT THE BEGINNING OF THE INCLINED SECTION IN THE TENSION AREA RESPECTIVELY. FURTHERMORE, IF C > 2H01/(1-2TGβ), THE INCLINED SECTIONS LEADING TO THE SUPPORT ARE CHECKED.


F

F

h0

h01

F

β

c

Q

PROJECTION Эпюра Q Q

DRAWING 3.13. CANTILEVER WITH HEIGHT DECREASING FROM THE SUPPORT TO THE FREE END

WHEN THE CANTILEVER IS EXPOSED TO CONCENTRATED FORCES THE BEGINNING OF THE INCLINED SECTION IS LOCATED IN THE TENSION AREA OF NORMAL SECTIONS PASSING THROUGH THE APPLICATION POINTS OF THESE FORCES (SEE DRAWING 3.13). UNDER THE IMPACT OF A UNIFORMLY DISTRIBUTED LOAD OR A LOAD LINEARLY INCREASING TOWARDS THE SUPPORT THE CANTILEVER IS CALCULATED AS A CONSTANT DEPTH OF SECTION ELEMENT ACCORDING TO PARAS 3.31 AND 3.32, ASSUMING THE EFFECTIVE DEPTH H0 AT THE BEARING SECTION. ELEMENTS REINFORCED WITH OFFSET BENDS 3.39. THE DURABILITY OF AN INCLINED SECTION TO THE IMPACT OF A TRANSVERSE FORCE FOR AN ELEMENT WITH OFFSET BENDS IS CHECKED USING CONDITION (3.44) WITH THE ADDITION OF FOLLOWING VALUE TO ITS RIGHT PART: QS,INC=0.75RSWAS,INCSINθ, (3.63) WHERE AS,INC IS THE SECTIONAL AREA OF THE OFFSET BENDS INTERSECTING THE OBLIQUE CRACK SITUATED AT THE END OF THE INCLINED SECTION WITH


PROJECTION LENGTH EQUAL TO C, BUT NO MORE THAN 2H0 (DRAWING 3.14); θ IS THE ANGLE OF OFFSET BENDS SLOPE TO THE LONGITUDINAL ELEMENT AXIS. As,inc

θ

F

2h0 C

DRAWING 3.14. FOR THE DETERMINATION OF THE MOST DANGEROUS OBLIQUE CRACK FOR ELEMENTS WITH OFFSET BENDS DURING THE CALCULATION OF THE IMPACT OF A TRANSVERSE FORCE

VALUES OF C ARE TAKEN EQUAL TO DISTANCES FROM THE SUPPORT TO THE ENDS OF OFFSET BENDS AND TO THE APPLICATION POINTS OF CONCENTRATED FORCES; FURTHERMORE, INCLINED SECTIONS ENDING AT A DISTANCE OF 2H0 FROM THE START OF THE NEXT-TO-LAST AND LAST OFFSET BENDS PLANE SHOULD BE CHECKED (DRAWING 3.15). 3.40. THE DISTANCES BETWEEN THE SUPPORT AND THE END OF THE OFFSET BEND NEAREST TO SUPPORT S1, AS WELL AS BETWEEN THE END OF PREVIOUS AND THE BEGINNING OF THE NEXT OFFSET BEND S2 (DRAWING 3.16) SHALL NOT EXCEED RBTBH 02 /Q.

MANUAL to CP 52-101-2003 concrete and REinforced concrete structures without prior reinforcement stress As,inc2

As,inc1

θ

θ

4 3 θ

2

1

2h0

C1 Qmax

As,inc3

2h0

C2 C3 C4

DRAWING 3.15. FOR DETERMINATION OF INCLINED SECTIONS IN AN ELEMENT WITH OFFSET BENDS 1- 4 ARE THE CALCULATED INCLINED SECTIONS

S1

S2

S

DRAWING 3.16. DISTANCES BETWEEN STIRRUPS, SUPPORT AND OFFSET BENDS

FURTHERMORE, THE OFFSET BENDS SHALL SATISFY THE DESIGN SPECIFICATIONS STIPULATED IN PARA 5.22. ELEMENTS WITHOUT CROSSWISE REINFORCEMENT 3.41. THE CALCULATION OF ELEMENTS WITHOUT CROSSWISE REINFORCEMENT FOR THE IMPACT OF A TRANSVERSE FORCE IS BASED ON THE CONDITIONS A) QMAX < 2.5RBTBH0; (3.64) WHERE QMAX IS THE MAXIMUM TRANSVERSE FORCE AT THE SUPPORT FACE;


1.5Rbt bh02 Q≤ B) , (3.65) c WHERE Q IS THE TRANSVERSE FORCE AT THE END OF THE INCLINED SECTION STARTING AT THE SUPPORT; VALUE C IS TAKEN AT NO MORE THAN CMAX = 3 H0. FOR CONTINUOUS FLAT SLABS WITH THE RESTRICTED EDGES (CONNECTED WITH OTHER ELEMENTS OR WITH SUPPORTS) AND A WIDTH OF B> 5H IT IS ALLOWED TO TAKE CMAX = 2.4H0. UNDER THE IMPACT OF CONCENTRATED FORCES ON THE ELEMENT VALUES OF C ARE TAKEN EQUAL TO THE DISTANCES FROM THE SUPPORT TO THE APPLICATION POINTS OF THESE FORCES (DRAWING 3.17), BUT NO MORE THAN CMAX. WHEN CALCULATING THE ELEMENT FOR THE IMPACT OF DISTRIBUTED LOADS, IF THE FOLLOWING CONDITION IS SATISFIED q1 ≤

Rbt b , 6

(3.66)

CONDITION (3.65) TAKES THE FORM QMAX< 0.5RBTBH0 + 3H0Q1 (WHICH CORRESPONDS TO C = 3H0),

(3.67)

F2

F1 2

1

C1

PROJECTION Эпюра Q

Q1

Q

Q=Q 1 1- F 2

C2

DRAWING 3.17. SCHEMATIC REPRESENTATION OF THE LEAST FAVORABLE INCLINED SECTIONS IN ELEMENTS WITHOUT CROSSWISE REINFORCEMENT 1- INCLINED SECTION CHECKED FOR THE IMPACT OF A TRANSVERSE FORCE Q1; 2 - SAME, FORCE Q2

Benefit to CP 52-101-2003 concrete and ferroconcrete constructions without the preliminary stress of the reinforcement

AND WHEN CONDITION (3.66) IS NOT FULFILLED Qmax ≤ 6 Rbt bh02 q1

(WHICH CORRESPONDS TO

c = h0

1.5 Rbt b q1

).

FOR THE AFOREMENTIONED FLAT SLABS WITH RESTRICTED SIDE EDGES THE RIGHT SIDE OF CONDITION (3.66) IS DIVIDED BY 0.64, AND CONDITION (3.67) TAKES THE FORM QMAX ≤ 0.625RBTBH0 + 2.4H0Q1. (3.67A) HERE Q1 IS TAKEN IN ACCORDANCE WITH PARA 3.32 WHEN UNDER THE IMPACT OF A UNIFORMLY DISTRIBUTED LOAD, AND UNDER THE IMPACT OF A CONTINUOUS LOAD WITH LINEARLY CHANGING INTENSITY IT IS TAKEN EQUAL TO THE AVERAGE INTENSITY IN THE SECTION NEXT TO THE SUPPORT WITH A LENGTH EQUAL TO A FOURTH OF THE BEAM (SLAB) SPAN OR HALF OF THE CANTILEVER PROTRUSION, BUT NO MORE THAN CMAX. 3.42. FOR ELEMENTS WITH VARIABLE DEPTH OF SECTION WHEN CHECKING CONDITION (3.64) VALUE H0 IS TAKEN FOR THE BEARING SECTION, AND WHEN CHECKING CONDITION (3.65) IT IS TAKEN AS THE AVERAGE VALUE OF H0 WITHIN THE LIMITS OF THE INCLINED SECTION. FOR ELEMENTS WITH DEPTH OF SECTION THAT INCREASES WITH AN INCREASE IN THE TRANSVERSE FORCE 3h01 , WHILE FOR 1 + 1,5tgβ 2.4h01 FLAT SLABS, MENTIONED IN PARA 3.41, cmax = , 1 + 1,2tgβ

VALUE CMAX IS TAKEN EQUAL TO cmax =

WHERE H01 IS THE EFFECTIVE DEPTH AT THE BEARING SECTION; β IS THE ANGLE BETWEEN THE TENSIONED AND COMPRESSED FACES. WHEN SUCH AN ELEMENT IS EXPOSED TO A DISTRIBUTED LOAD, VALUE C IN CONDITION (3.65) IS TAKEN EQUAL TO c=

h01 tg 2 β / 4 + q1 /(1.5 Rbt b)

,

(3.68)

BUT NO MORE THAN CMAX, WHERE Q1 – SEE PARA 3.32.


CALCULATION OF REINFORCED CONCRETE ELEMENTS ACCORDING FOR THE IMPACT OF MOMENTS ALONG INCLINED SECTIONS

Sw

Nb

Zs ho

Sw

x

3.43. THE CALCULATION OF REINFORCED CONCRETE ELEMENTS FOR THE IMPACT OF MOMENT ALONG INCLINED SECTIONS (DRAWING 3.18) IS CARRIED OUT BASED ON THE CONDITION M ≤ MS + MSW, (3.69) WHERE M IS THE MOMENT AT THE INCLINED SECTION WITH PROJECTION LENGTH C ONTO THE LONGITUDINAL AXIS OF THE ELEMENT DETERMINED FOR ALL THE EXTERNAL FORCES LOCATED TO ONE SIDE FROM THE CONSIDERED INCLINED SECTION WITH RESPECT TO THE END OF THE INCLINED SECTION (POINT 0), OPPOSITE TO THE END WHERE THE CHECKED LONGITUDINAL REINFORCEMENT IS LOCATED THAT EXPERIENCES TENSION FROM THE MOMENT AT THE INCLINED SECTION (DRAWING 3.19)

ls

R sw A sw

Rs As

Rsw A sw

R swA sw Q

C

DRAWING 3.18. DIAGRAM OF FORCES AT THE INCLINED SECTION FOR ITS FLECTION MOMENT CALCULATION


q

a)

Fi

ai

б) 0

q

Fi Ns

Nb

Ns

0

Nb C

l1

Q C y PROJECTION Эпюра M M =Qy- qy2- Fi ai M 2

PROJECTION Эпюра M M

M =ql1 (C+l21 ) +Fi C

DRAWING 3.19. DETERMINATION OF THE CALCULATED VALUE OF MOMENT WHEN CALCULATING AN INCLINED SECTION A - FOR A FREE BEAM; B - FOR A CANTILEVER

MS IS THE MOMENT EXPERIENCED BY THE LONGITUDINAL REINFORCEMENT INTERSECTING THE INCLINED SECTION WITH RESPECT TO THE OPPOSITE END OF THE INCLINED SECTION; MSW IS THE MOMENT EXPERIENCED BY THE CROSSWISE REINFORCEMENT INTERSECTING THE INCLINED SECTION WITH RESPECT TO THE OPPOSITE END OF THE INCLINED SECTION (POINT 0). MOMENT MS IS DETERMINED USING THE FORMULA MS = NSZS, (3.70) WHERE NS IS THE FORCE IN THE TENSIONED LONGITUDINAL REINFORCEMENT TAKEN EQUAL TO RSAS, AND DETERMINED ACCORDING TO PARA 3.45 IN THE ANCHORING ZONE;


ZS IS THE ARM OF THE INTERNAL COUPLE, DETERMINED USING THE FORMULA z s = h0 −

Ns (WHERE B IS THE WIDTH OF THE COMPRESSED 2 Rbb

FACE); WHICH IS TAKEN AT NO LESS THAN H0 – A’ FOR COMPRESSED REINFORCEMENT; IT IS ALSO PERMITTED TO TAKE ZS = 0.9H0. MOMENT MSW FOR TRANSVERSE FITTINGS IN THE FORM OF STIRRUPS, PERPENDICULAR TO THE LONGITUDINAL AXIS OF THE ELEMENT, IS DETERMINED USING THE FORMULA MSW = 0.5QSW C2, (3.71) WHERE QSW IS DETERMINED USING FORMULA (3.48) PARA 3.31, AND C IS TAKEN NOT GREATER THAN 2H0. IF THE STIRRUPS CHANGE THEIR INTENSITY OVER THE LENGTH C FROM QSW1 AT THE BEGINNING OF THE INCLINED SECTION TO QSW2, MOMENT MSW IS DETERMINED USING THE FORMULA: MSW = 0.5QSW1C2 – 0.5(QSW1 – QSW2) (C – L1)2 (3.72) WHERE L1 IS THE LENGTH OF THE SECTION WITH STIRRUPS INTENSITY QSW1. VALUE C IS DETERMINED ACCORDING TO PARA 3.46. 3.44. THE MOMENT IMPACT CALCULATION IS CARRIED OUT FOR INCLINED SECTIONS LOCATED AT LONGITUDINAL REINFORCEMENT BREAK POINTS AND AT THE FACE OF THE OUTER FREE BEAM SUPPORT AND AT THE FREE END OF CANTILEVERS IN THE ABSENCE OF SPECIAL ANCHORS FOR LONGITUDINAL FITTINGS. FURTHERMORE, INCLINED SECTIONS AT POINTS OF ABRUPT ELEMENT HEIGHT CHANGE (FOR EXAMPLE, AT CUTOFFS) ARE CALCULATED. 3.45. WHEN AN INCLINED SECTION WITH LONGITUDINAL TENSIONED REINFORCEMENT THAT DOES NOT HAVE ANCHORS WITHIN THE LIMITS OF THE ANCHORING ZONE IS INTERSECTED, THE FORCE NS IS DETERMINED USING THE FORMULA: N s = Rs As

ls , laï

(3.73)


WHERE

LS IS THE DISTANCE FROM THE END OF REINFORCEMENT TO ITS POINT OF INTERSECTION WITH THE INCLINED SECTION; LAN IS THE LENGTH OF THE ANCHORING ZONE EQUAL TO LAN = λANDS,

WHERE

λan =

Rs α 4 Rbond

(3.74)

RBOND IS THE CALCULATED COHESION RESISTANCE OF THE REINFORCEMENT WITH THE CONCRETE, EQUAL TO RBOND = η1η2RBT, η1 IS THE COEFFICIENT THAT CONSIDERS THE IMPACT OF THE REINFORCEMENT SURFACE TYPE AND IS TAKEN EQUAL TO: 2.5 FOR A300, A400, A500 REINFORCEMENT CLASSES; 2.0 - FOR V500 REINFORCEMENT CLASS; 1.5 - FOR A240 REINFORCEMENT CLASS; η2 IS THE COEFFICIENT THAT CONSIDERS THE IMPACT OF THE REINFORCEMENT DIAMETER AND IS TAKEN EQUAL TO: 1.0 FOR A DIAMETER DS ≤ 32 MM, 0.9 FOR DIAMETERS 36 AND 40 MM; α IS THE COEFFICIENT THAT CONSIDERS THE IMPACT OF THE TRANSVERSE CONCRETE REDUCTION AND CROSSWISE REINFORCEMENT AND IS TAKEN EQUAL TO: A) FOR THE FREE OUTER SUPPORTS, IF 0.25 ≤ σB/RB ≤ 0.75 - 0.75; IF σB/RB < 0.25 OR σB/RB > 0.75 - 1.0, HERE σB = FSUP/ASUP; FSUP, ASUP ARE THE BEARING PRESSURE AND THE AREA OF BEAM SUPPORT; THEREBY, IF THERE IS A TRANSVERSE A REINFORCEMENT ENCOMPASSING THE LONGITUDINAL REINFORCEMENT WITHOUT WELDING, THE α COEFFICIENT IS DIVIDED BY VALUE 1+

6 Asw (WHERE ASW AND S ARE THE SECTIONAL AREA OF THE as

ENVELOPING STIRRUP AND ITS SPACING) AND IS TAKEN AT NO LESS THAN 0.7; B) FOR THE FREE CANTILEVER ENDS – 1.0.


IN ANY CASE THE λAN COEFFICIENT IS TAKEN AT NO LESS THAN 15 AND THE LENGTH OF THE ANCHORING ZONE LAN IS TAKEN AT NO LESS THAN 200 MM. FOR RODS WITH A DIAMETER LESS THAN 36 MM THE λAN VALUE MAY BE TAKEN ACCORDING TO TABLE 3.3. WHEN TRANSVERSE OR DISTRIBUTIVE REINFORCEMENT IS WELDED TO LONGITUDINAL TENSIONED RODS THE FORCE NS IS INCREASES BY VALUE NW = 0.7 NWϕW d w2 RBT, (3.75) TAKEN AT NO MORE THAN 0.8 RS d w2 NW. HERE: NW IS THE NUMBER OF WELDED-ON RODS FOR THE LENGTH LS; ϕW IS THE COEFFICIENT TAKEN FROM TABLE 3.4; DW IS THE DIAMETER OF THE WELDED RODS. THEREBY VALUE NS IS TAKEN AT NO MORE THAN THE VALUE CALCULATED USING FORMULA (3.73) WITH THE USE OF THE α = 0.7 COEFFICIENT FOR THE DETERMINATION OF LAN. WHEN SPECIAL ANCHORS IN THE FORM OF PLATES, WASHERS, NUTS, CORNERS, CLOSING HEADS, ETC, ARE SET UP AT THE ENDS OF RODS MEETING THE REQUIREMENTS OF PARA 5.35 AND WHEN THE ENDS OF RODS ARE WELDED TO RELIABLY ANCHORED INSET COMPONENTS, THE FORCE NS IS TAKEN EQUAL TO RSAS. 3.46. FOR FREE BEAMS THE LEAST FAVORABLE INCLINED SECTION BEGINS FROM THE SUPPORT FACE AND A PROJECTION C TAKEN AT NO MORE THAN 2H0 AND DETERMINED AS FOLLOWS:


TABLE 3.3 REIN- α RELATIVE LENGTH OF REINFORCEMENT ANCHORING λAN = FORCEM COEFLAN/DS FOR CONCRETE CLASSES -ENT FICIE V10 V15 V20 V25 V30 V35 V40 V45 V50 V55 V60 CLASS NT A240 0.7 45 33 28 24 22 19 18 17 16 15 15 0.75 48 36 36 26 23 21 19 18 17 16 15 1.0 64 48 40 34 31 28 26 24 22 21 20 A300 0.7 34 25 21 18 16 15 15 15 15 15 15 0.75 36 27 23 19 18 16 15 15 15 15 15 1.0 48 36 30 26 23 21 19 18 17 16 15 A400 0.7 44 33 28 24 22 19 18 17 16 15 15 0.75 48 36 30 25 23 20 19 18 17 16 15 1.0 63 47 39 34 31 27 25 24 22 21 20 A500 0.7 54 41 34 29 26 23 22 20 19 18 17 0.75 58 44 36 31 28 25 23 22 20 19 18 1.0 78 58 48 41 38 33 31 29 27 26 24 V500 0.7 65 48 40 35 32 28 26 24 23 21 20 0.75 69 52 43 37 34 30 28 26 24 23 22 1.0 93 69 58 49 45 40 37 35 32 31 29 NOTE. WHEN CALCULATING TAKING INTO ACCOUNT ONLY CONSTANT AND LONG-TERM LOADS, VALUES OF λAN SHOULD BE DIVIDED BY γB1 = 0.9. TABLE 3.4. DW ϕW

6 200

8 150

10 120

12 100

14 80

A) IF THE ELEMENT IS EXPOSED TO CONCENTRATED FORCES, VALUES OF C ARE TAKEN EQUAL TO THE DISTANCES FROM THE SUPPORT TO THE APPLICATION POINTS OF THESE FORCES AND EQUAL TO QMAX/QSW, IF THIS VALUE IS LESS THAN THE DISTANCE TO THE 1ST LOAD; B) IF THE ELEMENT IS EXPOSED TO A UNIFORMLY DISTRIBUTED LOAD Q, VALUE C IS DETERMINED USING THE FORMULA: c=

Qmax , qsw + q

HERE QSW IS - SEE FORMULA (3.48).

(3.76)


IF OVER THE LENGTH THE STIRRUPS CHANGE THEIR INTENSITY FROM QSW1 AT THE BEGINNING OF THE INCLINED SECTION TO QSW2, VALUE C IS DETERMINED USING FORMULA (3.76) WITH A DECREASE OF THE NUMERATOR BY ΔQSWL1 AND OF THE DENOMINATOR BY ΔQSW, (WHERE L1 IS THE LENGTH OF THE SECTION WITH INTENSITY QSW1, ΔQSW1=QSW1 - QSW2). FOR BEAMS WITH AN INCLINED COMPRESSED FACE UNDER THE IMPACT OF A UNIFORMLY DISTRIBUTED LOAD THE INCLINED SECTIONS WITH C VALUES EQUAL TO c=

Qmax − N s tgβ , qsw + q

c=

Qmax − N s tgβ − 4qsw h0 tgβ q + 4qsw tg 2 β

,

(3.77)

ARE CHECKED, WHERE H0 IS THE EFFECTIVE DEPTH AT THE BEARING SECTION; β IS THE ANGLE OF SLOPE OF THE COMPRESSED FACE TO THE HORIZONTAL. WHEN THE TENSIONED FACE IS SLOPED AT AN ANGLE β TOWARD THE HORIZONTAL, VALUE TGβ IN THESE FORMULAS IS REPLACED BY SINβ. FOR CANTILEVERS LOADED BY CONCENTRATED FORCES (DRAWING 3.19, B) THOSE INCLINED SECTIONS ARE CHECKED THAT START AT THE APPLICATION POINTS OF CONCENTRATED FORCES NEAR THE FREE END WITH VALUES OF c =

Q1 , (WHERE qsw

Q1 IS THE TRANSVERSE FORCE AT THE START OF THE INCLINED SECTION), BUT NO MORE THAN L1, THE DISTANCE FROM THE START OF THE INCLINED SECTION TO THE SUPPORT. THEREBY, IF

Q1 > 2h0 , C = L1 SHOULD BE TAKEN. IF THESE q sw

CANTILEVERS HAVE AN INCLINED COMPRESSED FACE, VALUE Q1/QSW IS REPLACED BY (Q1- NSTGβ)/QSW. FOR CANTILEVERS LOADED ONLY WITH A UNIFORMLY DISTRIBUTED LOAD Q, THE LEAST FAVORABLE SECTION ENDS AT THE BEARING SECTION AND HAS A PROJECTION LENGTH c=

BUT NO MORE THAN 2H0.

Rs As z s , laï (qsw + q )

(3.78)


IF C < L–LAN, IT IS POSSIBLE NOT TO CALCULATE THE INCLINED SECTION. HERE: AS IS THE SECTIONAL AREA OF REINFORCEMENT BROUGHT TO THE FREE END; ZS IS – SEE PARA 3.43; LAN IS – SEE PARA 3.45. IN THE ABSENCE OF CROSSWISE REINFORCEMENT VALUE C IS TAKEN EQUAL TO 2H0, WHERE H0 IS THE EFFECTIVE DEPTH AT THE END OF THE INCLINED SECTION. 3.47. IN ORDER TO ENSURE THE DURABILITY OF INCLINED SECTIONS FOR THE IMPACT OF MOMENT IN CONSTANT HEIGHT ELEMENTS WITH STIRRUPS, THE LONGITUDINAL TENSIONED RODS BROKEN WITHIN THE SPAN SHALL BE BROUGHT BEYOND THE THEORETICAL BREAK-OFF POINT (I.E. PAST THE NORMAL SECTION WHERE THE EXTERNAL MOMENT BECOMES EQUAL TO THE LIMIT MOMENT MULT WITHOUT TAKING INTO ACCOUNT THE BROKEN OFF REINFORCEMENT, DRAWING 3.20) FOR A LENGTH NO LESS THAN VALUE W DETERMINED USING THE FORMULA

w Mult

1

2

DRAWING 3.20. TENSIONED RODS BREAKING-OFF WITHIN THE SPAN 1 - THEORETICAL BREAK-OFF POINT; 2 - PROJECTION M

w=

Q + 5d s ; 2q sw

(3.79)


WHEREBY, IF

⎛ q h ⎞ Q > h0 , w = 2h0 ⎜⎜1 − sw 0 ⎟⎟ + 5d s , 2qsw Q ⎠ ⎝

(3.80)

WHERE Q IS THE TRANSVERSE FORCE IN THE PERPENDICULAR SECTION PASSING THROUGH THE THEORETICAL BREAK-OFF POINT; QSW IS – SEE PARA 3.31; DS IS THE DIAMETER OF THE BROKEN-OFF ROD. FOR A BEAM WITH AN INCLINED COMPRESSED FACE WITH TGβ ≤ 0.2 VALUE W IS TAKEN EQUAL TO W = αH0 + 5DS, (3.81) THEREBY, IF α >1, W = H0(2.2 – 1.2/α) + 5DS, (3.82) WHERE α =

Q − N s tgβ , 2q sw h0

β IS THE ANGLE OF THE FACE SLOPE TO THE HORIZONTAL. FOR A BEAM WITH AN OBLIQUE TENSIONED FACE W IS DETERMINED SIMILARLY WITH THE REPLACEMENT OF TGβ BY SINβ. FOR ELEMENTS WITHOUT CROSSWISE REINFORCEMENT VALUE W IS TAKEN EQUAL TO 2H0. FURTHERMORE, THE REQUIREMENTS OF PARAS 5.32 AND 5.33 SHALL BE TAKEN INTO ACCOUNT. 3.48. IN ORDER TO ENSURE THE DURABILITY OF INCLINED SECTIONS TO THE IMPACT OF MOMENT THE BEGINNING OF THE OFFSET BEND IN THE TENSION AREA SHALL BE AT LEAST 0.5H0 DISTANT FROM THE PERPENDICULAR SECTION, WHERE THE UNBENT ROD IS FULLY USED WITH THE MOMENT, AND THE END OF THE OFFSET BEND SHALL BE LOCATED NOT CLOSER THAN THE PERPENDICULAR SECTION WHERE A OFFSET BEND IS NOT REQUIRED ACCORDING TO THE CALCULATION (DRAWING 3.21).


h0

≥ nc s, i

Z

ZS

Zs

Rs As

Zs

,in

c<

Z

s

a

Nb

X ≥ 0,5h0 PROJECTION Эпюра M M

M

DRAWING 3.21. FOR THE DETERMINATION OF THE POSITION OF THE LONGITUDINAL TENSIONED REINFORCEMENT OFFSET BEND


EXAMPLES OF CALCULATION

CALCULATION OF INCLINED SECTIONS OF REINFORCED CONCRETE ELEMENTS FOR THE IMPACT OF TRANSVERSE FORCES EXAMPLE 12. GIVEN: THE EDGE OF A TT-SHAPED OVERLAP SLAB WITH SECTION DIMENSIONS: H = 350 MM, B = 85 MM; A = 35 MM; V15 CLASS CONCRETE (RB = 8.5 MPA, RBT = 0.75 MPA); THE EDGE IS REINFORCED BY A FLAT FRAME WITH TRANSVERSE RODS FROM CLASS A400 REINFORCEMENT (RSW = 285 MPA) WITH A DIAMETER OF 8 MM (ASW = 50.3 MM2) WITH A SPACING SW = 100 MM; FULLY UNIFORMLY DISTRIBUTED LOAD ACTING ON THE EDGE, Q = 21.9 KN/M; TEMPORARY EQUIVALENT LOAD QV = 18 KN/M; TRANSVERSE FORCE AT THE SUPPORT QMAX = 62 KN. IT IS REQUIRED TO CHECK THE DURABILITY OF INCLINED SECTIONS AND THE CONCRETE STRIP BETWEEN THEM. C A L C U L A T I O N . H0 = H – A = 350 – 35 = 315 MM. LET US CHECK THE DURABILITY OF THE CONCRETE STRIP USING CONDITION (3.43): 0.3RBBH0 = 0.3 . 8.5 . 85 . 315 = 68276 N > QMAX = 62 KN, I.E. STRIP DURABILITY IS ENSURED. LET US CHECK THE DURABILITY OF THE INCLINED SECTION TO THE TRANSVERSE FORCE ACCORDING TO PARA 3.31. LET US DETERMINE THE STIRRUPS INTENSITY ACCORDING TO FORMULA (3.48) Rsw Asw 285 ⋅ 50.3 = = 143.3 N/MM. sw 100 q sw 143.3 = = 2.25 > 0.25 , I.E., CONDITION (3.49) IS Rbt b 0.75 ⋅ 85 qsw =

SINCE

SATISFIED, ONE CAN CONSIDER THE STIRRUPS FULLY AND VALUE MB IS DETERMINED USING FORMULA (3.46) MB= 1.5RBTBH02 = 1.5 . 0.75 . 85 . 3152 = 9.488 . 106 N.MM.


LET US DETERMINE THE LENGTH OF THE LEAST FAVORABLE INCLINED SECTION C PROJECTION ACCORDING TO PARA 3.32. Q1 = Q – QV/2 = 21.9 – 18/2 = 12.9 KN/M (N/MM). SINCE

qsw = 2.25 > 2.0 , ONE CAN DETERMINE VALUE C Rbt b

USING THE FORMULA Mb 9.488 ⋅10 6 c= = = 280.7 mm < 2h0 0.75q sw + q1 0.75 ⋅143.3 + 12.9

ONE CAN ASSUME C0 = C = 280.7 MM. THEN QSW = 0.75QSWC0 = 0.75 . 143.3 . 280.7 = 30168 N = 30.17 KN. M b 9.488 ⋅ 10 6 Qb = = = 33801 N = 33.8 KN. c 280.7

Q = QMAX – Q1C = 62 – 12.9 . 0.28 = 58.4 KN. ONE CAN VERIFY CONDITION (3.44) QB + QSW = 33.8 + 30.17 = 63.97N > Q = 58.4 KN, I.E. INCLINED SECTIONS DURABILITY IS ENSURED. LET US CHECK THE REQUIREMENT PARA 3.35: s w , max

Rbt bh02 0.75 ⋅ 85 ⋅ 315 2 = = = 102 mm > s w = 100 MM, Q 62000

I.E. THE REQUIREMENT IS MET. PARA 5.21 CONDITIONS SW

EXAMPLE 13. GIVEN: A FREE BEAM WITH SECTION DIMENSIONS: B = 200 MM, H = 400 MM; H0 = 370 MM; V25 CLASS CONCRETE (RBT = 1.05 MPA); TWO-BRANCH STIRRUPS WITH A DIAMETER OF 8 MM (ASW = =101MM2) WITH A SPACING OF SW = 150 MM; A240 CLASS REINFORCEMENT (RSW = 170 MPA); TEMPORARY MOMENT EQUIVALENT LOAD QV = 36 KN/M, CONSTANT LOAD QG = 14 KN/M; TRANSVERSE FORCE AT THE SUPPORT QMAX = 137.5 KN. IT IS REQUIRED TO CHECK THE INCLINED SECTIONS DURABILITY. C A L C U L A T I O N . INCLINED SECTIONS DURABILITY IS CHECKED ACCORDING TO PARA 3.31. LET US DETERMINE STIRRUPS INTENSITY ACCORDING TO FORMULA (3.48) q sw =

R sw Asw 170 ⋅ 101 = 114.5 N/MM. = sw 150

SINCE

qsw 114.5 = = 0.545 > 0.25 , I.E., CONDITION (3.49) IS Rbt b 1.05 ⋅ 200

SATISFIED, THE STIRRUPS ARE CONSIDERED FULLY AND VALUE MB IS DETERMINED FROM FORMULA (3.46) MB =1.5RBTBH02 = 1.5 . 1.05 . 200 . 3702 = 4.312 . 107 N.MM. ONE CAN DETERMINE THE LEAST FAVORABLE INCLINED SECTION PROJECTION LENGTH ACCORDING TO PARA 3.32: Q1 = QG + 0.5QV = 14 + 0.5 . 36 = 32 KN/M (N/MM). SINCE Mb 4.312 ⋅10 7 = = 1161 mm > q1 32

2h0 1 − 0.5

q sw Rbt b

=

2 ⋅ 370 = 1017 MM, 1 − 0.5 ⋅ 0.545

VALUE C IS TAKEN EQUAL TO 1161 MM > 2H0 = 740 MM. THEN C0 = 2H0 = 740 MM QSW = 0.75QSWC0 = 0.75 . 114.5 . 740 = 63548 N = 63.55 KN; M b 4.312 ⋅ 10 7 Qb = = = 37140 N = 37.14 KN; c 1161 Q = QMAX - Q1C = 137.5 – 32 . 1.161 = 100.35 KN.

LET US CHECK CONDITION (3.44) QB + QSW = 37.14 + 63.55 = 100.69 KN > Q = 100.35 KN, I.E. INCLINED SECTIONS DURABILITY IS ENSURED. EXAMPLE 14. GIVEN: A FREE BEAM WITH A SPAN OF L = 5.5 M; FULLY UNIFORMLY DISTRIBUTED LOAD ON THE


BEAM Q = 50 KN/M; TEMPORARY LOAD EQUIVALENT QV = 36 KN/M; CROSS SECTION DIMENSIONS B = 200 MM, H = 400 MM; H0 = 370 MM; V15 CLASS CONCRETE (RBT = 0.75 MPA); A240 CLASS REINFORCEMENT STIRRUPS (RSW = 170 MPA). IT IS REQUIRED TO DETERMINE THE DIAMETER AND SPACING OF STIRRUPS AT THE SUPPORT AND TO FIND OUT AT WHAT DISTANCE AND HOW CAN THE STIRRUPS SPACING BE INCREASED. C A L C U L A T I O N . THE GREATEST TRANSVERSE FORCE IN THE BEARING SECTION IS EQUAL TO ql 50 ⋅ 5.5 = 137.5 KN. QMAX = = 2 2

LET US DETERMINE REQUIRED STIRRUPS INTENSITY OF THE SECTION NEXT TO THE SUPPORT ACCORDING TO PARA 3.33, B. ONE CAN DETERMINE MB ACCORDING TO FORMULA (3.46) MB = 1.5RBTBH02 = 1.5 . 0.75 . 200 . 3702 = 30.8 . 106 N.MM. ACCORDING TO PARA 3.32 Q1 =Q - 0.5QV = 50 – 0.5 . 36 = 32 KN/M (N/MM). Qb1 = 2 M b q1 = 2 30.8 ⋅ 10 6 ⋅ 32 = 62790 N.

SINCE 2MB/H0 –QMAX = 2 . 30.8 . 106/370 – 137500 = 28986 N
ACCORDING TO PARA 5.21 THE STIRRUPS SPACING SW AT THE SUPPORT SHALL NOT EXCEED H0/2 = 185 AND 300 MM AND WITHIN THE SPAN – 0.75H0 = 278 AND 500 MM. THE MAXIMUM PERMISSIBLE SPACING AT THE SUPPORT ACCORDING TO PARA 3.35 IS EQUAL TO sw,max

Rbt bh02 0.75 ⋅ 200 ⋅ 370 2 = = = 149.3 ≈ 150 MM. Q 137500

LET US TAKE THE STIRRUPS SPACING AT THE SUPPORT SW1= 150 MM AND 250 MM WITHIN THE SPAN. HENCE Asw =

qsw sw1 161.9 ⋅ 150 = = 142.9 MM2. 170 Rsw

WITHIN THE CROSS SECTION ONE CAN TAKE TWO


STIRRUPS OF 10 MM (ASW = 157MM2). THUS, THE STIRRUPS INTENSITY TAKEN AT THE SUPPORT AND WITHIN THE SPAN IS EQUAL TO, RESPECTIVELY: R A 170 ⋅ 157 qsw1 = sw sw = = 177.9 N/MM; sw1 150

170 ⋅ 157 = 106.7 N/MM. QSW2 = 250 LET US CHECK CONDITION (3.49): 0.25 . RBTB = 0.25 . 0.75 . 200 = 37.5 N/MM < QSW1 AND 37.5 < QSW2. CONSEQUENTLY, VALUES OF QSW1 AND QSW2 ARE NOT CORRECTED. LET US DETERMINE THE LENGTH OF SECTION L1 WITH STIRRUPS INTENSITY QSW1 ACCORDING TO PARA 3.34. SINCE ΔQSW = 0.75 (QSW1 - QSW2) = 0.75(177.9 – 106.7) = =53.4N/MM >Q1 = 32 N/MM, LET US CALCULATE VALUE L1 ACCORDING TO FORMULA (3.59), TAKING QB,MIN = 0.5RBTBH0 = 0.5 . 55500 = 27750 N l1 = =

Qmax − (Qb ,min + 1.5qsw 2 h0 ) q1

− 2h0 =

137500 − (27750 + 1.5 ⋅ 106.7 ⋅ 370) − 2 ⋅ 370 = 839 mm 32

LET US TAKE THE SECTION LENGTH WITH A STIRRUPS SPACING OF SW1 = 150 MM EQUAL TO 0.9 M. EXAMPLE 15. GIVEN: A COVER BEAM LOADED BY CONCENTRATED FORCES AS SHOWN IN DRAWING 3.22,A; SECTION DIMENSIONS ACCORDING TO DRAWING 3.22,B; V15 CLASS CONCRETE (RBT = 0.75 MPA); A240 CLASS REINFORCEMENT STIRRUPS (RSW = 170 MPA). IT IS REQUIRED TO DETERMINE THE DIAMETER AND SPACING OF THE STIRRUPS AND TO FIND OUT AT WHAT DISTANCE FROM THE SUPPORT AND HOW CAN THE STIRRUPS SPACING BE INCREASED.

F=42кН 1

F=34кН 1

б)

220

80 1500

PROJECTION Q, KN

24,1

58,1

63,2

Эпюра Q, кН

110

105,2

1350

80

890

60

a

100


DRAWING 3.22. FOR CALCULATION EXAMPLE 15

C A L C U L A T I O N . H0 = 890 – 80 = 810 MM. LET US DETERMINE THE REQUIRED STIRRUPS INTENSITY QSW ACCORDING TO PARA 3.33, AND, ASSUMING SECTION PROJECTION LENGTH C EQUAL TO THE DISTANCE FROM THE SUPPORT TO THE FIRST LOAD C1 = 1350 MM. THEN α1 = C1/H0 = 1350/810 = 1.667 < 2, AND, CONSEQUENTLY, α01 = α1 = 1.667. ONE CAN DETERMINE ε гр1 =

1.5

α1

+ 0.1875α 01 =

1.5 + 0.1875 ⋅1.667 = 1.212 . 1.667

ACCORDING TO DRAWING 3.22 THE TRANSVERSE FORCE AT A DISTANCE OF C1 FROM THE SUPPORT IS EQUAL TO Q1 = 105.2 KN. THEN ε 1 =

Q1 105200 = = 2.165 > ε гр1 , Rbt bh0 0.75 ⋅ 80 ⋅ 810

AND, CONSEQUENTLY, FORMULA (3.51): q sw1 = Rbt b

QSW

IS

DETERMINED

USING

ε 1 − 1 .5 / α 1 2.165 − 1.5 / 1.667 = 0.75 ⋅ 80 = 60,7 N/MM. 0.75α 01 0.75 ⋅ 1.667

LET US DETERMINE QSW FOR A VALUE OF C EQUAL TO THE DISTANCE FROM THE SUPPORT TO THE SECOND LOAD C2 = 2850 MM: α 2 = C2/H0 = 2850/810 = 3,52 > E; ONE CAN ASSUME α 2 = 3,0. SINCE α 2 > 2, ONE CAN TAKE α02 = 2.0.


ε гр 2 =

1.5

α2

+ 0.1875α 02 =

1.5 + 0.1875 ⋅ 2 = 0.875. 3

THE CORRESPONDING TRANSVERSE FORCE IS EQUAL TO Q2 = 58.1 KN. THEN ε2 =

Q2 58100 = = 1.195 > .ξ гр 2 = 0.815 , Rbt bh0 0.75 ⋅ 80 ⋅ 810

AND, CONSEQUENTLY, ε − 1.5 / α 2 1.195 − 1.5 / 3.0 qsw 2 = Rbt b 2 = 0.75 ⋅ 80 == 27.8 N/MM. 0.75 ⋅ 2.0 0.75α 02 ONE CAN TAKE THE MAXIMUM VALUE OF QSW = QSW1 =60.7. THE WELDING CONDITION GIVES US THE STIRRUPS DIAMETER 8 MM (ASW = 50.3 MM2). THEN THE MAXIMUM PERMISSIBLE STIRRUPS SPACING IN THE SECTION NEXT TO THE SUPPORT IS EQUAL TO Rsw Asw 170 ⋅ 50.3 = = 140.9 MM. SW1 = qsw1 60.7

ONE CAN ASSUME SW1 = 100 MM. ONE CAN DESIGNATE THE STIRRUPS SPACING WITHIN THE SPAN TO BE EQUAL TO SW2 = 300 MM. THEN THE STIRRUPS INTENSITY OF THE SECTION NEXT TO THE SUPPORT IS qsw1 =

Rsw Asw 170 ⋅ 50.3 = = 85.5 N/MM, 100 sw1

AND WITHIN THE SPAN SECTION: R A 170 ⋅ 50.3 qsw2 = sw sw = = 28.5 N/MM sw 2 300 LET US ASSIGN THE LENGTH OF THE SECTION WITH A STIRRUPS SPACING OF SW1 EQUAL TO THE DISTANCE FROM THE SUPPORT TO THE FIRST LOAD L1 = 1350 MM, AND LET US CHECK CONDITION (3.44) WITH A VALUE OF C EQUAL TO THE DISTANCE FROM THE SUPPORT TO THE SECOND LOAD C = 2850 MM. HOWEVER, SINCE 3H0 = 3 . 810 = 2430 MM < C, ONE CAN TAKE C = =2430MM. VALUE QSW IS DETERMINED ACCORDING TO PARA 3.34. SINCE 2H0 + L1 = 2 . 810 + 1350 = 2970 MM > C, VALUE QSW IS DETERMINED USING FORMULA (3.56). THEREBY, SINCE C > 2H0, C0 = 2H0 = 1620 MM. QSW = 0.75[QSW1С0 – (QSW1 - QSW2)(С – L1)] = 0.75[85.5 . 1620


б)

150

– (85.5 – 28.5)(2430 –1350)] = 57712 N = 57.7 KN. WITH C=3H0 QB = QB MIN = 0.5RBTBH0 = 0.5 . 0.75 . 80 . 810 = 24300 N = 24.3 KN. THE TRANSVERSE FORCE AT A DISTANCE OF C = 2430 MM FROM THE SUPPORT (DRAWING 3.22) IS EQUAL TO 2.43 − 1.35 (63.2 − 58.1) = 59.5 KN. Q = 63.2 − 1.5 ONE CAN CHECK CONDITION (3.44) QB + QSW = 24.3 + 57.7 = 82.0 KN > Q = 59.5 KN, I.E. THE DURABILITY OF THIS INCLINED SECTION IS ENSURED. ONE CAN DO NOT CONSIDER A GREATER VALUE OF C, SINCE THAT GREATLY REDUCES THE TRANSVERSE FORCE. THUS, THE LENGTH OF SECTION WITH THE STIRRUPS SPACING SW1 = 100 MM IS TAKEN EQUAL TO 1.35 M. EXAMPLE 16. GIVEN: A DOUBLE SLOPE BEAM WITH A 8.8 M SPAN (DRAWING 3.23,A); CONTINUOUS UNIFORMLY DISTRIBUTED LOAD ON THE BEAM Q = 46 KN/M; BEARING SECTION DIMENSIONS ACCORDING TO DRAWING 3.23,B; V20 CLASS CONCRETE (RBT = 0.9 MPA); A400 CLASS REINFORCEMENT STIRRUPS (RSW = 285 MPA) WITH A DIAMETER OF 10 MM (ASW = 78.5 MM2) AND A SPACING OF SW = 100 MM. IT IS REQUIRED TO CHECK THE INCLINED SECTION DURABILITY TO A TRANSVERSE FORCE. C A L C U L A T I O N . THE EFFECTIVE DEPTH OF THE BEARING SECTION IS EQUAL TO H0 = 600 – 40 = 560 MM (SEE DRAWING 3.23,B). LET US DETERMINE STIRRUPS INTENSITY ACCORDING TO FORMULA (3.48) 300

8800

40

c

100

i=1:12

100


600

a)


Rsw Asw 285 ⋅ 78.5 = = 223.7 N/MM. 100 sw LET US DETERMINE THE PROJECTION OF THE LEAST FAVORABLE INCLINED SECTION C ACCORDING TO PARA 3.37. FROM DRAWING 3.23,A ONE CAN HAVE TGβ = 1/12, B = 100 MM, RBTB = 0.9 . 100 = 90 N/MM; 1 –2TGβ = 1 – 2/12 = 0.833. SINCE QSW/(RBTB) = 223.7/90 = 2.485 > 2 (1 – 2TGβ)2 = 2 . 0.8332 = 1.389, VALUE C IS CALCULATED ACCORDING TO FORMULA (3.62). 1,5 1,5 = 560 = c = h01 (0.75qw + q) /( Rb b) + 1.5tg 2 β (0.75 ⋅ 223.7 + 46) / 90 + 1.5 / 12 2 q sw =

= 444mm.

THE EFFECTIVE DEPTH OF THE CROSS SECTION H0 AT A DISTANCE C = 444 MM FROM THE SUPPORT IS EQUAL TO H0 = H01 + С . TGβ = 560 + 444/12 = 597 MM. SINCE C = 444 MM < 2H0, CO = C = 444 MM; 1.5Rbt bh02 1.5 ⋅ 90 ⋅ 597 2 Qb = = = 108370 Н = 108.4 KN; c 444 QSW = 0.75QSWС0 = 0.75 . 223.7 . 444 = 74492 N = 74.5 KN. LET US CHECK CONDITION (3.44) TAKING THE TRANSVERSE FORCE AT THE END OF THE INCLINED SECTION EQUAL TO 46 ⋅ 8.8 − 46 ⋅ 0.444 = 182.0 KN: Q = QMAX – QC = 2 QB + QSW = 108.4 + 74.5 = 182.9 KN > Q = 182 KN, I.E. THE DURABILITY OF THE INCLINED SECTIONS TO A TRANSVERSE FORCE IS ENSURED. EXAMPLE 17. GIVEN: A CANTILEVER WITH DIMENSIONS ACCORDING TO DRAWING 3.24, A CONCENTRATED FORCE APPLIED TO THE CANTILEVER F = 130 KN LOCATED AT A DISTANCE L1 = 0.8 M FROM THE SUPPORT; V15 CLASS CONCRETE (RBT = 0.75 MPA); TWO-BRANCH STIRRUPS WITH A DIAMETER OF 8 MM (ASW = 101 MM2) MADE OF A240 CLASS REINFORCEMENT (RSW = 170 MPA) WITH A SPACING OF SW = 200 MM.


1

F=130kH

1-1

650 1

800 950

50

300

C

400


IT IS REQUIRED TO CHECK THE DURABILITY OF INCLINED SECTIONS TO A TRANSVERSE FORCE. C A L C U L A T I O N . ONE CAN CHECK THE LEAST FAVORABLE INCLINED SECTION USING CONDITION (3.44) ACCORDING TO PARA 3.38, WHICH STARTS AT THE APPLICATION POINT OF THE CONCENTRATED FORCE, WITH A VALUE OF C DETERMINED ACCORDING TO FORMULA (3.62) 650 − 300 = 0.369. WITH Q1 = 0 AND TGβ = 950 THE EFFECTIVE DEPTH AT THE APPLICATION POINT OF THE CONCENTRATED FORCE IS EQUAL TO H01 = 650 – (650 – 300) 800 − 50 = 305 MM (SEE DRAWING 3.24); RBTB = 0,75. 200 = 150 N/MM. 950

VALUE QSW IS EQUAL TO qsw = c = h01

Rsw Asw 170 ⋅ 101 = = 85.8 N /MM; sw 200

1,5 1 .5 = = 305 0.75 ⋅ 85.8 / 150 + 1.5 ⋅ 0.369 2 0.75q sw /( Rbt b) + 1.5tg 2 β

= 469.4 mm < l1 = 800 mm.

2h01 2 ⋅ 305 = SINCE 1 − 2tgβ 1 − 2 ⋅ 0.369 = 1164 mm > c , ONE CAN LEAVE C = 469.4 MM. LET US DETERMINE THE EFFECTIVE DEPTH H0 AT THE END


OF THE INCLINED SECTION H0 = H01 + С TGβ = 305 + 469 . 0.369 = 478 MM. SINCE C = 469.4 > 2H0, C0 = C = 469 MM. 1.5 Rbt bh02 1.5 ⋅ 150 ⋅ 4782 Qb = = = 109614 N = 109.6 KN; c 469

QSW = 0.75 QSWС0 = 0.75 . 85.8 . 469 = 30180 N = 30.18 KN; QB + QSW = 109.6 + 30.2 = 139.8 KN > F = 130 KN. I.E. THE DURABILITY OF INCLINED SECTIONS TO A TRANSVERSE FORCE IS ENSURED. Example 18. Given: a solid slab of the tank bottom without crosswise reinforcement with dimensions 3х6 m and thickness h = 160 mm, monolithically connected throughout the perimeter with beams; complete uniform load 50 kN/m2; concrete of class В15 (Rbt = 0.75 MPa). It is required to check slab durability regarding transverse force. C a l c u l a t i o n. h0 = 160 – 20 = 140 mm. Calculation is carried out for a strip with width b=1.0 м = 1000 mm, span l = 3 м. Thereafter, q = 50·1.0 = 50 kN/m and transverse force on the support is equal to Qmax =

ql 50 ⋅ 3 = = 75 kN. 2 2

Let us check condition (3.64) 2.5Rbtbh0 = 2.5 . 0.75 . 1000 . 140 = 262500 N > Qmax = 75 kN. Let us check condition (3.66) assuming q1 = q = 50 kN/m (N/mm). Since side edges of the slab are monolithically connected with beams, condition (3.66) is written in the form Rbt b 0.75 ⋅ 1000 = = 195 N / mm > q1 = 50 N/mm, 6 ⋅ 0.64

6 ⋅ 0.64

Consequently, slab durability is checked using condition (3.67a) 0.625Rbtbh0 + 2.4h0q1 = 0.625 . 0.75 . 1000 . 140 + 2.4 . 140 . 50 = 82425 H = 82.4 kN > Qmax = 75 kN, i.е. durability of the slab regarding transverse force is ensured. Example 19. Given: a tank wall panel of cantilevered type with graduated thickness from 262 (in embedment) up to 120 mm (at the loose end) with overhanging length 4.25 m; lateral earth pressure taking into account traffic load on the ground surface linearly decreases from q0 = 55 kN/m2 in embedment to q = 6 KN/m2 at the loose end; а=22 mm; concrete of class В15 (Rbt = 0.75 MPa) . It is required to check panel durability regarding transverse force. C a l c u l a t i o n. Effective depth of panel cross section in


embedment is equal to h01 = 262 – 22 = 240 mm. Let us determine tgβ (β - angle between compressed and tensile faces): tgβ =

262 − 120 = 0.0334. 4250

Let us check the conditions from para 3.41. Transverse force in embedment is equal to 55 + 6 Qmax = 4.25 = 129.6 kN. 2

Calculation is carried out for a strip with width b = 1,0 м = 1000 mm. Let us check condition (3.64) assuming h0 = h01 = 240 mm. 2.5Rbtbh0 = 2.5 . 0.75 . 1000 . 240 = 450000 H = 450 kN > Qmax, i.е. the condition is fulfilled. Since panels are interconnected and tank wall width is admittedly more than 5h, value сmax is calculated using formula 2.4h01 2.4 ⋅ 240 c max = = 554 mm. = 1 + 1.2 tgβ

1 + 1.2 ⋅ 0.0334

Medium intensity of the load directed to the area near the support with length сmax =554 mm is equal to 554 q1 = 55 − (55 − 6) = 51.8 N/mm. 4250 ⋅ 2

Since c = h0

1 1 = 240 = 2 tg β / 4 + q1 /(1,5 Rbt b) 0.0334 / 4 + 51.8 /(1.5 ⋅ 0.75 ⋅ 1000) 2

= 1115 ìì > c max = 554 mm,

one can assume с = сmax = 554 mm. Let us determine effective depth of the cross-section at the distance of с/2 from the support (i.е. average value h0 within length с): h0 = h01 - c tgβ = 240 − 554 ⋅ 0.334 = 231 mm. 2

2

Transverse force at the distance of с = 554 mm from the support is equal to: Q = Qmax – q1с = 129.6 – 51.8 . 0.554 = 100.9 kN. Let us condition (3.65): 1,5 Rbt bh02 1.5 ⋅ 0.75 ⋅ 1000 ⋅ 2312 = = 108360 N = 108.4 kN > Q = 100.9 c 554

kN,

i.е. durability of the plane regarding transverse force is ensured. Calculation of inclined sections for the impact of the moment of flection


Example 20. Given: a freely supported beam with span l = 5.5 m, uniform load q = 29 kN/m; near support area structure of the beam is assumed in accordance with Diagram 3.25; concrete of class В15 (Rb = 8.5 MPa); longitudinal reinforcement without foundation bolts of class А400 (Rs = 355 MPa) with cross-section area Аs = 982 mm2 (2∅25); stirrups made of reinforcement of class А240 (Rsw = 170 MPa) 8 mm in diameter with spacing sw = 150 mm are welded to longitudinal rods. It is required to check durability of inclined sections regarding the impact of moment of flection. C a l c u l a t i o n. h0 = h - a = 400 – 40 = 360 mm. Since tensile reinforcement does not have foundation bolts, calculation of the flection moment impact for sloping cross-sections is necessary. Let us determine the force in tensile reinforcement using formula (3.73). One can assume beginning of the slopping section at the support face. Thereafter, ls = lsup - 10 mm = 280 – 10 = 270 mm (see Drawing 3.25). Beam reaction is equal to Fsup =

ql 29 ⋅ 5,5 = = 80 kN, 2 2

and beam bearing area is Asup = blsup = 200 . 280 = 56000 mm2, from which

σb =

Fsup

Asup

=

σ 80000 1.43 = 1.43 MPa, b = = 0.168 < 0.25, Rb 56000 8.5

consequently,α = 1.0. From Table 3.3 with concrete of class В15, reinforcement class А400 and α = 1/0 one can find λаn= 47. Thereafter, anchorage length is equal to lап = λаnds = 47 . 25 = 1175 mm. N s = Rs As

ls 270 = 355 ⋅ 982 = 80106 H. 1175 lan

MANUAL to CP 52-101-2003 concrete and REinforced concrete structures without prior reinforcement stress 1

400

35

1-1 2 12

40

10

lsup=280 93 Fsup

1

C=557,5 l=5500

2 25 200

Drawing 3.25. For calculation example 20

Since 4 vertical and 2 horizontal transverse rods are welded to tensile rods within the length (see Drawing 3.25), one can increase force Ns by value Nw. Assuming dw = 8 mm, nw = 6, ϕw = 150 (see table. 3.4), one can obtain Nw = 0.7 nw ϕw dw2 Rbt = 0.7 . 6 . 150 . 2 . 0.75 = 3024 . 103 N. Therefore Ns = 80106 + 30240 = 110346 N. Let us determine maximum permissible value Ns. From Table 3.3 with α =0.7 one can find λаn = 33; thereafter, N s , max = Rs As

ls

λan d

= 355 ⋅ 982

270 = = 114040 H > N s , i.e. one can keep Ns = 33 ⋅ 25

110346 N. Let us determine lever arm z s = h0 −

Ns 110346 = 360 − = 327.5 mm > h0 − à ' = 360 − 35 = 325 2 ⋅ 8.5 ⋅ 200 2 Rb b

mm.

Thereafter, moment applied to longitudinal reinforcement is equal to Ms = Nszs = 110346 . 327.5 = 36.1 . 106 N.mm. One can calculate qsw using formula (3.48) qsw =

Rsw Asw 170 ⋅ 101 . = = 114,5 N mm. 150 sw

One can determine projected contact length of the least favorable inclined section using formula (3.76), assuming value Qmax equal to beam reaction, i.e. Qmax = Fsup = 80 kN. c=

Qmax 80 ⋅ 10 3 = = 557.5 q sw + q 114.5 + 29

mm <2h0.

Thereafter, moment applied to crosswise reinforcement is equal to Msw =0.5qswс2 = 0.5 . 114.5 . 557.52 = 17.8 . 106 N. mm. Moment in a inclined section is determined as moment in a normal


cross-section located at the end of the inclined section, i.е. at the distance from point of application of beam reaction equal to х = lsup/3 + с = 280/3 + 557.5 = 650.8 mm M = Qx −

qx 2 29 ⋅ 650.8 2 . = 80 ⋅ 10 3 ⋅ 650.8 − = 45.9 ⋅ 10 6 N mm 2 2

= 45.9 kN. m.

Let us check condition (3.69) Мs + Msw = 36.1 + 17.8 = 53.9 kN.m > M = 45.9 kN.m, i.е. durability of inclined sections regarding moment of flection is ensured. Example 21. Given: a cross-beam of the multistoried frame with diagrams of moments and transverse forces from uniform load q = 228 kN/m in accordance with Drawing 3.26; concrete of class В25; transverse and longitudinal reinforcement of class А400 (Rs = 355 MPa, Rsw = 285 MPa); transverse cross-section of the near support area is in accordance with Drawing 3.26; three-branch stirrups 10 mm in diameter (Asw = 236 mm2) with spacing sw equal to 150 mm. It is required to determine distance from the left support to the spot of breaking off of the first rod of the upper reinforcement. C a l c u l a t i o n. From Drawing 3.26 one can have: h0 = h - a = 800 – 60 = 740 mm; a’= 50 mm; area of the upper tensile reinforcement without taking into account one broken-off rod ∅32 As = 1609 mm2 (2∅32); As' = 2413 mm2 (3∅32). One can determine critical moment corresponding to this reinforcement using formula (3.19). Since As < A 's , i.е. х < 0: Mult = RsAs(h0 - а’) = 355 . 1609(740 – 50) = 394.1 . 106 N. mm = 394.1 kN.m. In accordance with diagram of moments one can determine distance from the support to the spot of possible breaking off using equation M = M sup −

from which Z=

' M sup − M sup

l

x = Z − Z2 −2

x−

M sup − M ult q

ql q x + x 2 = M ult , 2 2

, where Z =

' l M sup − M sup + ; ql 2

4.9 600 − 300 600 − 394.1 + = 2.719 m; x = 2.719 − 2.719 2 − 2 = 0.355 2 228 ⋅ 4.9 228

Transverse force in the spot of possible breaking off is equal to Q = Qmax – qx = 620 – 228 . 0.355 = 539 kN. Let us determine value qsw R A 285 ⋅ 236 q sw = sw sw = = 448.4 N/mm. sw

150

m.

MANUAL to CP 52-101-2003 concrete and REinforced concrete structures without prior reinforcement stress 1 1116 w=761

1 l=4,9 M

Эпюра M,kH M 394

M sup =600

M'sup=300

x=0,355 M 539

Эпюра Q ,kH

620 488 1-1

800

60

3 32A400

50

3 32A400

300


Since

Q 539 = = 0.601 ì < h0 = 0,74 2q sw 2 ⋅ 448.4

m, length w, for which a broken-

off rod is carried under behind the point of the possible breaking off is determined using formula (3.79) w=

Q + 5d s = 601 + 5 ⋅ 32 = 761mm. 2qsw

Consequently, distance from the support to the spot of rod breaking off might be assumed equal to x + w = 355 + 761 = 1116 mm. Let us determine necessary length lan from the spot of the rod breaking off to the support cross-section assuming complete use of this rod in the


support cross-section. For this purpose using Table 3.3 with α = 1.0 concrete of class В25, reinforcement class А400 one can find λ = 34. Thereafter lan = λand = 34 . 32 = 1088 mm < 1116 mm. Consequently, the rod is broken off at the distance 1116 mm from the support. ECCENTRICALLY COMPRESSED ELEMENTS GENERAL PROVISIONS

3.49. When calculation is carried out for concrete reinforcement elements regarding effect of compressed normal force, accidental eccentricity еа shall be taken into account assumed no less than: 1/600 of element length or distance between its cross-sections fixed against displacement; 1/30 of section depth; 10 mm. For elements of statically indeterminate structures (including columns of framed buildings) value of normal force eccentricity in respect with gravity center of transformed section е0 is assumed equal to eccentricity value obtained using static analysis, however, no less than еа. For elements of statically determinate structures (for example, framed structures, power line posts, etc.) eccentricity е0 is assumed equal to sum of accidental eccentricity and eccentricity of construction static analysis . 3.50. Calculation for normal cross-sections of eccentrically compressed elements is carried out in the plane of normal force (in the plane of flection) eccentricity and separately in the plane normal to it with eccentricity е0 equal to accidental еа (see para 3.49). Calculation in the plane of flection is not strictly required in case if element flexibility l0/i (for rectangular cross-sections – l0/h) in the plane of flection exceed flexibility in plane normal to the plane of flection. Calculation of element with allowance for eccentricities in planes of both main axes (biaxial eccentrical compression) shall be carried out in case if both eccentricities exceed accidental еа. In all cases eccentricities е0 are determined with allowance for element deflection influence (see paras 3.53-3.55). 3.51. Calculation for normal cross-sections of eccentrically compressed elements in general case is carried out on the basis of nonlinear deformation model in accordance with paras 3.72-3.76.


It is permissible to carry out calculation for elements of rectangular, T- and I-sections, with reinforcement located at element edges perpendicular to the plane of flection and with eccentricity directed in plane of cross section symmetry, in accordance with paras 3.56-3.61. In addition, regarding critical forces calculation might be carried out: - for element of annular and circular cross-section with reinforcement uniformly distributed at the circumference with a number of rods more than 6 in accordance with paras 3.62-3.65; - for elements of rectangular cross-section with reinforcement in the form of 4 similar corner rods regarding biaxial eccentrical compression in accordance with para 3.66. ANALYSIS OF TRANSVERSE FORCE ACTION

3.52. Calculation for eccentrically compressed elements regarding action of transverse forces is carried out similarly to calculation of bending elements in accordance with paras 3.29-3.35 and the following instructions: а) with N/Nb > 0.5 the right part of condition (3.43) is multiplied by coefficient ϕn1 = 2(1-N/Nb), (3.83) where Nb = 1.3RbA, however, no less than N; b) value of transverse force accepted by concrete in inclined section Qb as well as the right part of condition (3.49) are multiplied by coefficient 2

ϕ n2

⎛ N ⎞ N ⎟⎟ ; =1+ 3 − 4⎜⎜ Nb N ⎝ b⎠

(3.84)

Value Mb connected with Qb is also multiplied by coefficient ϕп2. CONSIDERING EFFECTS OF ELEMENT DEFLECTION

3.53. Effect of element deflection on the moment of normal force (or its eccentricity е0) is taken into account as a rule by means of structure calculation using deformational model and considering inelastic deformations of concrete and reinforcement as well as presence of cracks. It is permissible to carry out calculation using non-deformational model and to take into account element flection by means of moment multiplying by coefficients ηv and ηh in accordance with formula (3.85) М = Мvηv + Мhηh + Мt, where Мv is moment of vertical loads which do not cause noticeable horizontal end movements; ηv is a coefficient assumed equal to:


1.0 for cross-sections at element ends with pliable embedment; value determined using formula (3.86) with stiff embedment; value determined using formula (3.86) for cross-sections in middle 1/3 of element length; value determined using linear interpolation (3.86) for all other cross-sections; Мh is moment of loads causing horizontal end movement (wind loading, etc.) ηh is a coefficient determined using formula (3.86); Мt is moment of forced horizontal end movements (i.е. movements which do not depend on element stiffness, for example, thermal deformation of floor structure, etc.). It is permissible to determine moments used in this para with respect to gravity center of concrete cross-section. Note. In case if vertical loads cause noticeable horizontal movements (for example with non-symmetric frames), moments Мv are determined with fictitious horizontal fixed supports, and moments of horizontal forces which are equal to reactions in these supports shall be referred to moments Мh, i.е. added together with moments of horizontal forces.

3.54. Value of coefficient ηv(h) with structural analysis in accordance with non-deformation model is determined using formula ηv ( h ) =

1 N 1− N cr

,

(3.86)

where Nсr is conditional critical force determined using formula N ñr =

π 2D l02

;

(3.87)

l0 – design element length determined for coefficients ηv и ηh in accordance with para 3.55,a and 3.55,b; D – stiffness of concrete reinforcement element in limit state is determined using formulae: for elements of any form of section 0.15 Eb I + 0 .7 E s I s ; D= (3.88) ϕ l (0.3 + δ e )

for elements of rectangular cross section with reinforcement located at the most compressed or tensile (less compressed) element edge

MANUAL to CP 52-101-2003 concrete and REinforced concrete structures without prior reinforcement stress 2 ⎡ 0.0125 ⎛ h0 − а' ⎞ ⎤ D = Eb bh ⎢ + 0,175μα ⎜ ⎟ ⎥. ⎝ h ⎠ ⎥⎦ ⎢⎣ϕ l (0.3 + δ e ) 3

(3.89)

In formulae (3.88) and (3.89): I and Is are moments of inertia of concrete cross-section and cross-section of the entire reinforcement with respect to gravity center of concrete cross-section, respectively; ϕl is a coefficient taking into account long-term load effect on element deflection and equal to ϕ=1+Mll / М1 (3.90) however, no more than 2; M1 and M1l are external moments with respect to an axis normal to the plane of flection and crossing the center of the most tensile or the least compressed rod (with completely compressed cross-section) of effect of all loads and constant and permanently-acting loads, respectively; for elements calculated in accordance with paras 3.56-3.61 it is permissible to determine M1 and M1l with respect to an axis crossing the gravity center of the entire reinforcement S; δе is a coefficient assumed to be equal to е0/h, however, no less than 0.15 (for annular and circular cross-sections value h is substituted with Dcir); As + As' E s μα = ⋅ . bh Eb

Stiffness D when coefficients ηv and ηh are determined taking into account all loads. In case if it is required coefficient ηv might be reduced calculating stiffness D without regard to loads causing end movement. With element flexibility l0/i < 14 (with l0/h < 4 for rectangular crosssections) one can might assume ηv(h) = 1.0. When N > Ncr, it is required to increase cross-section dimensions. 3.55. Design length l0 is assumed equal to the following values: a) when coefficient ηv is calculated and also calculation is performed regarding action of normal force with accidental eccentricity for elements: with hinge support at both ends l0=1.0 l ; with hinge support at one end and at other end: with stiff embedment l0=0.7l with pliable embedment l0=0.9l; with embedment at both ends: with stiff embedment l0=;0.5l with stiff embedment l0=0.8l; with pliable embedment at one end and stiff embedment at the other


l0=0.7l; b) when coefficient ηh is calculated for elements: with hinge support at one end and at the other end: with stiff embedment l0= 1,5 l; with pliable embedment l0= 2,0 l; with embedment at both ends: with stiff embedment l0= 0,8 l; with pliable embedment – l0=1,2 l; with pliable embedment at one end and stiff embedment at the other l0= l; with stiff embedment at one end and the other loose end (cantilever) l0= 2 l. l is distance between element ends. Some other values of l0 might be assumed for specific structures and constructers. CALCULATION FOR NORMAL CROSS-SECTIONS REGARDING CRITICAL FORCES

Rectangular cross-sections with symmetrical reinforcement 3.56 Durability of rectangular cross-sections with symmetrical reinforcement (when Rs As = Rsc A 's ) is checked using condition M ≤ Rbbx(h0 – 0,5x) + (RscA 's – N/2)(h0 – à' ), (3.91) where М is moment with respect to gravity center of cross-section determined taking into account element deflection in accordance with paras 3.53-3.55; х is height of compressed area assumed to be equal to а) x = αnh0 when α ï =

N ≤ ξ R (Drawing 3.27) −; Rb bh0

б) x = ξ h0 when αn > ξR , where ξ is determined using formula


Drawing 3.27. Schematic of forces in transverse rectangular cross-sections of an eccentrically compressed element

ξ=

where αs =

α n (1 − ξ R ) + 2α sξ R ; 1 − ξ R + 2α s

(3.92)

Rs As ; Rbbh0

for ξR see Table 3.2. 3.57. Required amount of symmetrical reinforcement is determined in the following way depending on relative value of normal force α n = a) with αn ≤ ξR As = As' =

b) with αn > ξR

N : Rbbh0

Rb bh0 α m1 − α n (1 − α n / 2) ⋅ , 1−δ Rs

(3.93)

Rb bh0 α m1 − ξ (1 − ξ / 2) ⋅ , 1−δ Rs

(3.94)

α m1 − ξ1 (1 − ξ1 / 2) 1−δ

(3.95)

As = As' =

where ξ is relative height of compressed area determined using formula (3.92) where it is permissible to assume value αs to be equal to αs =

with ξ1 = (αn + ξR)/2, however, no more than 1.0. In formulae (3.93 – 3.95):


α m1 =

M + N (h0 − à' ) / 2 R b bh02

,

δ=

a' ; h0

for М see paras 3.53 – 3.55. If value а’ does not exceed 0.15h0, required amount of reinforcement might be determined by means of the graph of Drawing 3.28 using formula As = As' = α s

Rbbh0 , Rs

(3.96)

where αs is determined using the graph of Drawing 3.28 depending on values αm =

M Rb bh02

αn =

N . Rb bho

Drawing 3.28. Graphs of load carrying capacity of eccentrically compressed elements of rectangular cross-sections with symmetrical reinforcement M N Rs As αn = ; αm = ; α = s Rbbh0 Rb bh0 Rbbh02

3.58. Calculation of compressed elements made of concrete class В15В35 regarding action of normal force applied with eccentricity assumed in accordance with para 3.49 which is equal to accidental eccentricity е0 = h/30, with l0 ≤ 20h might be carried out using condition N ≤ ϕ(RbA + RscAs,tоt), (3.97) Where ϕ is a coefficient determined using formula


ϕ=ϕb+2(ϕsb - ϕb)αs, (3.98) however, assumed no more than ϕsb. ϕb and ϕsb are coefficients assumed in accordance with Tables 3.5 and

3.6.

Nl N

6

8

0 0,5 1,0

0.93 0.92 0.92

0.92 0.91 0.91

Nl N

6

8

Coefficient ϕb with l0/h 10 12 14 16 0.91 0.90 0.89

0.90 0.89 0.87

0.89 0.86 0.83

0.88 0.82 0.76

Coefficient ϕsb with l0/h 10 12 14 16

Table 3.5 18

20

0.86 0.77 0.68

0.84 0.71 0.60 Table 3.6.

18

20

1

a

>

h-2a 4

1

A. When а = а'< 0.15h and there are no intermediate rods (see Sketch) or sectional area of these rods is less than As,tоt./3 0 0.93 0.92 0.91 0.90 0.89 0.88 0.86 0.83 0.5 0.92 0.91 0.91 0.90 0.88 0.87 0.83 0.79 1.0 0.92 0.91 0.90 0.90 0.88 0.85 0.80 0.74 B. When 0.25h > а = а' ≥ 0.15h or area of intermediate rods is equal or more than As,tоt/3 regardless а(see Sketch) 0 0,92 0,92 0,91 0,89 0,87 0,85 0,82 0,79 0,5 0,92 0,91 0,90 0,88 0,85 0,81 0,76 0,71 1,0 0,92 0,91 0,89 0,87 0,83 0,77 0,70 0,62 Designations used in Tables 3.5 and 3.6: рассматриваемая плоскость Nl – normal force due to constant and long term loads. N – normal force due to all loads.

1

a

h-2a 1 > 4

h-2a

1

промежуточные стержни

αs =

Rs As ,tot Rb A

;

(3.99)

As,tot is area of the entire reinforcement in cross-section; When αs > 0.5, ϕ =ϕsb might be assumed without using formula (3.98).


Rectangular cross-sections with non-symmetrical reinforcement 3.59. Durability of rectangular cross-sections with non-symmetrical reinforcement is checked using conditions (3.91) from para 3.56 defining height of compressed area using formula x=

hereby, if

N + Rs As − Rsc As' ; Rb b

(3.100)

x > ξ R (see Table 3.2), height of compressed area is h0

corrected by means of calculation using the formula x=

1 + ξR − Rsc As' 1− ξR . 2 Rs As Rb b + h0 (1 − ξ R )

N + Rs As

(3.101)

3.60. Areas of compressed and tensile reinforcement corresponding to minimum of their sum are determined using formulae: Ne − α R Rb bh02 = ; Rsc (h0 − a ' ) ξ R bh − N As = R b 0 + As' , Rs As'

(3.102) (3.103)

whereαR and ξR are determined using Table 3.2 and assumed to be no more than 0.4 and 0.55, respectively; е = М/N + (h0 – а ' )/2. With negative value As calculated using formula (3.103) area of reinforcement S is taken equal to the minimum in accordance with structural requirements, however, no less than value As, min =

N (h0 − a'−e) − Rb bh(h / 2 − a ' ) , Rsc (h0 − a ' )

(3.104)

and area of reinforcement S’ is determined: with negative value As,min using formula As' = [( N − Rb ba' ) − ( N − Rb ba' ) 2 − N ( N − 2 Rb bh0 + 2 Rb be) ] / Rsc ; (3.105)

with positive value As,min using formula As' =

N − Rbbh − As , min . Rsc

(3.106)

If assumed area of compressed reinforcement A’s,fact, sufficiently exceeds its value calculated using formula (3.102) (for example, with its negative value), sectional area of tensile reinforcement might be reduced


using formula ξRb bh0 − N + Rsc As' , fact

As =

Rs

,

(3.107)

ãäå ξ = 1 − 1 − 2α m ,

αm =

Ne − Rsc As' , fact (h0 − à ' ) Rb bh02

.

If there is no compressed reinforcement or it is not taken into account in calculation, sectional area of tensile reinforcement is always determined only using formula (3.107), hereby, condition αm < αR shall be fulfilled.. I-sections with symmetric reinforcement 3.60. Durability of I-sections with symmetric reinforcement concentrated in flanges (Drawing 3.29) is checked in the following way. If condition N ≤ Rb b 'f h 'f (3.108) is met (i.е. boundary of compressed area is in a flange), calculation is carried out in the same way as for rectangular cross-sections with width b 'f in accordance with para 3.56. b' f Rsc A's

A's x

a'

h'f

b

h

h0

h0 -a'

e

e0

R

b As hf

a

R s As

bf

Drawing 3.29. Schematic of forces in transverse I-section of an eccentrically compressed element

If condition (3.108) is not fulfilled (i.е. boundary of compressed area is in the jack-rib), cross-section durability is checked using condition


M ≤ Rbbx ( h0 − x / 2) + Rb Aov ( h0 − h 'f / 2) + ( Rsc As' − N / 2)( h0 − à ' ), (3.109)

where height of compressed area х is assumed to be equal to: a) with ξ =

N − Rb Aov ≤ ξ R x = ξh0 ; Rb bh0

b) with ξ > ξR

(α n − α ov )(1 − ξ R ) + 2α sξ R , 1 − ξ R + 2α s R A A N ; α ov = ov ; where α s = s s ; α n = Rb bh0 Rbbh0 bh0

(3.110)

x = h0

(

)

Aov is area of compressed flange overhangs equal to Aov = b 'f - b h 'f ; for ξR see Table 3.2.

Note. With varying height of flange overhangs value h 'f is assumed to be equal to average overhang height.

3.61. Required amount of reinforcement of I-sections is determined in the following way. When condition (3.108) is met, reinforcement is selected in the same way as for a rectangular cross-section with width b 'f in accordance with para 3.57. If condition (3.108) is not met, reinforcement is selected depending on relative height of compressed area which is equal to ξ=αn - αov: (3.111) a) with ξ ≤ ξR As = As' =

Rb bh0 α m1 − ξ (1 − ξ / 2) − α m, ov ; 1−δ Rs

(3.112)

As = As' =

Rbbh0 α m1 − ξ1 (1 − ξ1 / 2) − α m, ov , 1−δ Rs

(3.113)

b) ξ > ξR

where relative height of compressed area ξ1 = x/h0 is determined using formula (3.110) calculating αs from formula αs =

α m1 − ξ (1 − ξ / 2) − α m, ov , 1−δ

(3.114)

hereby,ξ is assumed no more than 1.0. In formulae (3.111) – (3.114): for αп,,αov see para 3.60; α m1

⎛ h 'f M + N (h0 − a' ) / 2 a' ⎜ = ; δ = ; α m ,ov = α ov 1 − 0.5 ⎜ h0 h0 R b bh02 ⎝

⎞ ⎟. ⎟ ⎠


Annular cross-sections 3.62. Durability of annular cross-sections (Drawing 3.30) with ratio of inner and outer radiuses r1/r2 ≥ 0.5 and reinforcement uniformly distributed at the circumference (with longitudinal rods no less than 7) in the following way depending on relative area of concrete compressed area N + Rs As ,tot

ξ cir =

Rb A + (Rsc + 1,7 Rs )As,tot

(3.115)

;

a) with 0.15 < ξcir < 0.6 using condition M ≤ ( Rb Arm + Rsc As ,tot rs )

sin πξ cir

π

+ Rs As,tot rs (1 − 1,7ξ cir )(0,2 + 1,3ξ cir ) ; (3.116)

b) with ξcir ≤ 0.15 using condition M ≤ ( Rb Arm + Rsc As ,tot rs )

sin πξ cir1

π

(3.117)

+ 0.295 Rs As ,tot rs

Drawing 3.30. Schematic taken for calculation for annular cross section of compressed element

where

ξ сir1 =

N + 0.75 Rs As ,tot Rb A + Rsc As ,tot

c) with ξcir ≥ 0.6 using condition M ≤ ( Rb Arm + Rsc As ,tot rs )

ξ cir 2 =

where

(3.118)

;

sin πξ cir 2

N . Rb A + Rsc As ,tot

π

;

(3.119) (3.120)

in formulae (3.115) – (3.120): As,tоt is sectional area of the entire longitudinal reinforcement; rm =

r1 + r2 ; 2


rs is circle radius crossing gravity centers of the rods of longitudinal reinforcement; Moment М is determined with allowance for element flection in accordance with paras 3.53 – 3.55. 3.63. It is permissible to check durability as well as to determine amount of longitudinal reinforcement for annular cross-sections specified in para 3.62 with rs ≈ rm and reinforcement class not higher than А400 by means of graphs of Drawing 3.31 using formulae: (3.121) M≤αmRbrmA; As , tot = α s

Rb A , Rs

(3.122)

where values αт and αs are determined using a graph depending on values αs =

Rs As ,tot Rb A

и αm =

M Rb Arm

,respectively, also α n =

N . Hereby, moment М is Rb A

determined with allowance for element flection in accordance with paras 3.53-3.55.

Drawing 3.31. Graphs of load carrying capacity of eccentrically compressed


elements of annular cross-section Rs As ,tot Ne0 N αn = ; αm = ; αs = Rb A Rb A Rb Arm


Circular cross-sections 3.64. Durability of circular cross-sections (Drawing 3.32) with reinforcement uniformly distributed at the circumference (with a number of longitudinal rods no less than 7), with reinforcement class no higher than А400 is checked using condition sin 3 πξ cir 2 ⎛ sin πξ cir ⎞ + Rs As , tot ⎜ + ϕ ⎟rs , M ≤ Rb Ar 3 π ⎝ π ⎠

(3.123)

where r is radius of cross-section; ξciris relative area of concrete compressed area determined in the following way:

Drw.3.32. Schematic assumed for calculation for circular cross-sections of eccentrically compressed elements

with fulfillment of the condition N ≤ 0.77RbA + 0.645RsAs,tot from calculation of the equation ξ cir =

sin 2πξ cir 2π ; Rb A + 2,55 Rs As , tot

N + Rs As , tot + Rb A

(3.124)

(3.125)

in case of failure to meet condition (3.124) from calculation of the equation

ξ ñir =

sin 2πξ cir 2π ; Rb A + Rs As ,tot

N + Rb A

(3.126)


ϕ is a coefficient taking into account tensile reinforcement work and assumed to be equal to: ϕ = 1.6(1 – 1.55ξcir)ξcir, however, no more than 1.0 when condition (3.124) is fulfilled; ϕ = 0 when there is a failure of condition (3.124); As,tot is sectional area of the entire longitudinal reinforcement; rs is circle radius going through gravity center of the rods of longitudinal reinforcement. Moment М is determined with allowance element deflection in accordance with paras 3.53-3.55. 3.65. It is permissible to check durability as well as necessary to determine amount of longitudinal reinforcement for annular cross-sections specified in para 3.64 by means of graphs in Drawing 3.33, using formulae (3.127) M ≤ αmRrod;

Graphical symbols:


with а/Dcir=0,05 with а/Dcir=0,10 Drawing 3.33. Graphs of load carrying capacity of eccentrically compressed elements of an annular cross-section Rs As ,tot Ne0 N αn = ; αm = ; αs = Rb A Rb A Rb Ar R A (3.128) As , tot = α s b , Rs

where values αs and αm are determined using graphs depending on values α s =

Rs As ,tot M è αm = , Rb A Rb Ar

respectively, and also on value α n =

N . Rb A

Hereby, moment М is determined with allowance for element deflection in accordance with paras 3.53-3.55. Calculation of elements regarding biaxial eccentrical compression 3.66. For elements of a rectangular cross-section with symmetrical reinforcement in the form of four corner rods calculation regarding biaxial eccentrical compression might be carried out using condition k

k ⎛ M x ⎞ ⎛⎜ M y ⎞⎟ ⎟ + ⎜ ≤ 1, ⎜M0 ⎟ ⎜M0 ⎟ ⎝ x⎠ ⎝ y⎠

(3.129)

where Mx and Mу are external moments with respect to gravity center of the cross-section in planes of mirror symmetry х and y; 0 0 M x è M y are critical moments in planes of mirror х and y with respect to gravity center of cross-section equal to the right part of condition (3.91) from para 3.56. Values M x0 è M 0y might be also determined by means of the graph in Drawing 3.28 using formula (3.130) M x0 ( M 0y ) = α m Rb bh02 , where αm is determined using the graph in Drawing 3.28 depending on α s =

Rs As N è αn = ; Rbbh0 Rbbh0

b and h0 – width and effective depth of the cross-section as applied to direction of the considered moment; hereby, condition a′≤ 0,15h0 shall be performed for the considered direction. Index of power k in condition (3.129) is determined using formulae:


if αп ≤ 0.4, if αп > 0.4,

k=

3.44 − 0.023α s (0,4 − α n ) 2 + k 0 ; 0.254 + α s

⎡ (1.7 − α s ) 2 ⎤ + 0.1775⎥ (α n2 − 0.16) + k 0 , k=⎢ 4 ⎣ ⎦

(3.131) (3.132)

however, no more than 1.6, where

αn =

Rs As ,tot 0.275 + α s N ; αs = ; k0 = Rb bh Rb bh 0.16 + α s

.

Calculation examples Rectangular cross-sections with symmetrical reinforcement Example 22. Given: a column of the middle floor of a framework with cross-section dimensions b = 400 mm, h = 500 mm; а = а′ = 40 mm; concrete of class В25 (Еb = 300000 MPa, Rb = 14.5 MPa); reinforcement class А400 (Rs = Rsc = 355 MPa); its area Аs = A 's = 1232 mm2 (2∅28); normal force and moments of flection in a support cross-section due to vertical loads: due to all loads Nv = 650 kN, Mv = 140 kN.m, constant and permanently-acting Nl = 620 kN, Ml = 130 kN.m.; due to wind loading Nh = 50 kN, Mh = 73 kN.m; storey height is l = 6 m. It is required to check durability of the column supporting crosssection. C a l c u l a t i o n. h0 = 500 – 40 = 460 mm. Calculation is carried out taking into account deflection effect in accordance with para 3.53. Since the considered cross section is supporting and the column of the support has pliable embedment, one can assume ηv = 1.0. In order to calculate coefficient ηh one can assume design length of the column to be equal to l0 = 1.2 . 6 = 7.2 m in accordance with para 3.55,b. Hereby, l0/ h =7,2/0,5 = 14,4 > 4, i.е. deflection shall be taken into account. Forces from all loads are equal to М = Mv + Mh = 140 + 73 = 213 kN. m, N =Nv + Nh= 650 + 50 = 700 kN. Hereby e0 =

M 213 = = 0,304 м > ea = N 700

h/30, i.е. in accordance with para 3.49 value of moment М is not corrected. One can determine moment М1 and М1l with respect to tensile reinforcement from all loads and from constant and long-term loads, respectively. h − à' 0.46 − 0.04 . M1 = M + N 0 = 213 + 700 = 360 kN m; 2

2


M 1l = M l + N l

h0 − à' 0.46 − 0.04 = 130 + 620 = 260.2 2 2

kN.m.

Thereafter ϕl = 1 + M1l/M1 = 1 + 260.2/360 = 1.72. Since e0 = 304 = 0.608 > 0.15, one can assume δе = 0.608. h

500

As + As' E s 2 ⋅ 1232 ⋅ 2 ⋅ 10 5 μα = ⋅ = = 0,0821. bh Eb 400 ⋅ 500 ⋅ 3 ⋅ 10 4

One can determine stiffness D (3.89) 2 ⎡ 0.0125 ⎛ h0 − à' ⎞ ⎤ D = Eb bh ⎢ + 0.175μα ⎜ ⎟ ⎥= ⎝ h ⎠ ⎥⎦ ⎢⎣ ϕ l (0,3 + δ å ) 3

2 ⎡ 0,0125 ⎛ 460 − 40 ⎞ ⎤ = 3 ⋅ 10 ⋅ 400 ⋅ 500 ⎢ + 0.175 ⋅ 0.0821⎜ ⎟ ⎥= ⎝ 500 ⎠ ⎦⎥ ⎣⎢1.72(0.3 + 0.608) 4

3

= 2.721 ⋅ 1013 N ⋅ mm 2 .

Therefore, ηh =

1 N 1− N cr

N cr =

=

π 2D l 02

=

3.14 2 ⋅ 2.721 ⋅ 1013 = 5180830 7200 2

N = 5181 kN.

1 = 1.156 . 700 1− 5181

Design moment with allowance for deflection is calculated using formula (3.85) assuming Mt = 0.0. M = Mvηv + Mhηh = 140 + 73 . 1.156 = 224.4 kN.m. Let us check cross-section durability in accordance with para 3.56. N 700000 αn = = = 0.262 < ξ R = 0.531 (see Table 3.2). Rb bh0

14.5 ⋅ 400 ⋅ 460

Consequently, x = αnh0 = 0.262 . 460 = 120.5 mm. Rbbx(h0 – x/2) + (RscA 's - N/2)(h0 -а ' ) = 14.5 . 400 . 120.5 (460 – - 120.5 /2) +(355 . 1232 – 700000/2)(460 – 40) = 316.07 . 106 N.mm = =316.1 kN.m > M = 224.4 kN.m, i.е. cross-section durability is ensured. Example 23. Given: column cross-section of the middle floor of the framework with dimensions b = 400 mm, h = 400 mm; а = а′ = 50 mm; concrete of class В25 (Rb = 14,5 MPa, Eb = 3 . 105 MPa); symmetrical reinforcement of class А400 (Rs = Rsc = 355 MPa); normal force and moments of deflection in the support cross-section due to vertical loads: due to all loads - Nv = 900 kN, Mv = 160 kN.m; due to constant and long-term loads - Nl = 800 kN, Ml = 150 kN.m; due to wind load Nh = 100 kN.m, Mh= 110 kN.m; storey height is 4.8 m. It is required to determine area of reinforcement cross-section.


C a l c u l a t i o n. h0 = 400 – 50 = 350 mm. In accordance with para 3.53 one can assume ηv = 1.0, and in accordance with para 3.55,b column design length is assumed equal to l0 = 1,2 . 4.8 = 5.76 m. Hereby, l0/h = 5.76/0.4 = 14.4 > 4, i.е. column deflection is taken into account. Forces from all loads are equal to M = Mv + Mh = 160 + 110 = 270 kN.m; N =Nv + Nh = 900 + 100 = 1000 kN. Hereby, e0 =

M 270 = = 0,27 ì N 1000

> e0 = h / 30 , i.е. value М is not corrected.

Coefficient ηh is determined in accordance with para 3.54. h0 − à' 0,35 − 0,05 . = 270 + 1000 = 420 kN m; 2 2 h − à' 0,35 − 0,05 . M 1l = M l + N l 0 = 150 + 800 = 270 kN m; 2 2 M1 = M + N

ϕl = 1 + M1l/M1 = 1 + 270 /420 = 1,64. Since

e0 270 e = = 0,675 > 0,15, one can assume δ e = 0 = 0,675. h 400 h

In first approximation one can assume that μ = 0,01, μα = 0,01

20 ⋅ 10 4 3 ⋅ 10 4

= 0,067 .

One can determine stiffness D using formula (3.89): 2 ⎡ 0,0125 ⎛ h0 − à ' ⎞ ⎤ D = Ebbh ⎢ + 0,175μα ⎜ ⎟ ⎥= + h ( 0 , 3 ) ϕ δ ⎝ ⎠ ⎦⎥ å ⎣⎢ l 3

2 ⎡ 0,0125 ⎛ 0,3 ⎞ ⎤ = 3 ⋅ 10 ⋅ 400 ⎢ + 0,175 ⋅ 0,067⎜ ⎟ ⎥ = 1,107 ⋅ 1013 Í ⋅ ìì ⎝ 0,4 ⎠ ⎥⎦ ⎢⎣1,64(0,3 + 0,675) 4

4

Therefore, N cr =

π 2D

2

.

3,14 2 ⋅ 1,107 ⋅ 1013

= = 3293000 Í = 3293 êÍ; l02 5760 2 1 1 = = 1,436; ηh = N 1000 1− 1− N cr 3293

M = M v ηv + M hη h = 160 ⋅1,0 + 110 ⋅1,436 = 318 êÍ ⋅ ì.

Required area of reinforcement cross-section is determined in accordance with para 3.57. For this purpose one can calculate the value:


αn = α m1 = δ=

N 1000000 = = 0.493 Rb bh0 14.5 ⋅ 400 ⋅ 350 M + N (h0 − a ' ) / 2 318 ⋅ 10 6 + 1000000(350 − 50) / 2 = = 0.659; Rb bh02 14.5 ⋅ 400 ⋅ 350 2

50 a' = = 0.143. h0 350

One can find ξR = 0.531 from Table 3.2. Since αп < ξR,, As = A 's one can determine using formula (3.93) As = As` =

Rb bh0 α m1 − α n (1 − α n / 2) 14.5 ⋅ 400 ⋅ 350 0.659 − 0.493(1 − 0.493 / 2) = = 1−δ 355 1 − 0.143 Rs

= 1918mm 2

Thereafter

μ=

As + As' 2 ⋅ 1918 = = 0.024 > 0.01. 400 ⋅ 400 bh

Since obtained additional reinforcement exceeds reinforcement which has been assumed for calculation of D, and moment Mh = 110 kN.m which is a high percentage of total moment М = 270 kN.m, value As = 1918 mm2 is determined with certain margin, which might be reduced by recalculation taking value μ = 0,024 in formula (3.89) : 20 ⎛ 0.0125 ⎞ + 0.175 ⋅ 0,024 0.75 2 ⎟ = 1.81 ⋅ 1013 N ⋅ mm 2 ; D = 3 ⋅ 10 4 ⋅ 400 4 ⎜ 3 ⎝ 1,64 ⋅ 0,975 ⎠ 2 13 3.14 ⋅ 1.81 ⋅ 10 N cr = = 5379000 N = 5379 kN; 5760 2 1 1 = = 1.228; ηh = 1 − N / N cr 1 − 1000 / 5379 M = 160 + 110 ⋅ 1,228 = 295 kN ⋅ m; 295 ⋅ 10 6 + 1000000 ⋅ 150 = 0.6263; α m1 = 14.5 ⋅ 400 ⋅ 350 2 14.5 ⋅ 400 ⋅ 350 0.6263 − 0.493(1 − 0.493 / 2) As = As' = = 1700 mm 2 . 355 0.857

Taking value As = As' =1847 mm2 (3∅28) which is close to value As used for calculation of D. Example 24. Given: a column of the lower floor of the multistoried framework with cross-section dimensions b= 400 mm, h = 500 mm; а = а’ = 50 mm; concrete of class В25 (Eb = 3 .104 MPa, Rb = 14,5 MPa); reinforcement class А400 (Rs = Rsc = 355 MPa) with area As = A 's = 1847 mm2 (3∅28); normal forces and moments of flection in the lower support cross-section: from vertical loads: all loads Nv = 2200 kN, Mv = 250 kN. m, from constant and long-term loads Nl = 2100 kN, Ml = =230 kN.m; from wind loading Nh = 0.0, Mh = 53 kNm; storey height is 6 m.


It is required to check durability of the column support cross-section. C a l c u l a t i o n. h0 = h - а = 500 – 50 = 450 mm. Calculation is carried out taking into account column deflection in accordance with para 3.53. Since the column is fixed with stiff embedment at the considered cross-section, coefficient ηv is calculated using formula (3.86) assuming that column design length is equal to l0 = 0.7 . 6 = 4.2 m in accordance with para 3.55а. When either coefficient ηv or coefficient ηh are determined, stiffness D is calculated using formula (3.89) with allowance for all loads. Forces from all loads are equal to М = Мv + Mh = 250 + 53 = 303 kN, N = Nv = 2200 kN. Hereby, e0 = M = 303 = 0.137 m > eà = h / 30. N 2200 h0 − à ' 0.45 − 0.05 = 743 kN ⋅ m; = 303 + 2200 M1 = M + N 2 2 h −α' 0.45 − 0.05 = 230 + 2100 650 kN ⋅ m; M 1l = M l + N l 0 2 2 ϕ1l = 1 + M 1l / M 1 = 1 + 650 / 743 = 1,875. e0 137 e = = 0.274 > 0.15, assuming δ e = 0 = 0.274. h 500 h 5 2A E 2 ⋅ 1847 2 ⋅ 10 μα = s s = ⋅ = 0.1231. bh Eb 400 ⋅ 500 3 ⋅ 10 4

Since

2 ⎡ 0.0125 ⎛ h − a' ⎞ ⎤ + 0.175μα ⎜ 0 D = Eb bh 3 ⎢ ⎟ ⎥= ϕ δ + ( 0 , 3 ) h ⎝ ⎠ ⎦⎥ e ⎣⎢ l 2 ⎡ 0,0125 ⎛ 450 − 50 ⎞ ⎤ = 3 ⋅ 10 ⋅ 400 ⋅ 500 ⎢ + 0.175 ⋅ 0.1231⎜ ⎟ ⎥= ⎝ 500 ⎠ ⎥⎦ ⎢⎣1.875(0.3 + 0.274) = 3.81 ⋅ 1013 N ⋅ mm 2 . 2 2 13 Thereafter, N cr = π 2D = 3.14 ⋅ 3.812 ⋅ 10 = 21318000 N = 21318 kN ; l0 4200 1 1 = = 1.115. ηv = 2200 N 1− 1− 21318 N cr 4

3

Coefficient ηh is calculated similarly assuming that design length is equal to l0 = 1.0 . 6 = 6 m in accordance with para 3.55,b. Therefore π 2D

3.14 2 ⋅ 3.81 ⋅ 1013 = 10446 ⋅ 10 3 H = 10446 l 02 6000 2 1 1 = = 1.267. ηh = 2200 N 1− 1− 10446 N cr N cr =

=

kN.

Design moment with allowance for deflection is equal to M=Mvηv + Mhηh = 250 . 1.115 + 53 . 1.267 = 345.9 kN.m.


One can check cross-section durability in accordance with para 3.56. N 2200000 αn = = = 0.843 > ξ R = 0.531 (see Table 3.2). Rb bh0

14.5 ⋅ 400 ⋅ 450

Consequently, compressed area height х is calculated using formula (3.92). For this purpose one can calculate αs =

Rs As 355 ⋅ 1847 = = 0.251; Rb bh 0 14.5 ⋅ 400 ⋅ 450

α n (1 − ξ R ) + 2α sξ R 0.843 ⋅ 0.469 + 2 ⋅ 0.251 ⋅ 0.531 = = 0.682; 1 − ξ R + 2α s 0.469 + 2 ⋅ 0.251 x = h0ξ = 450 ⋅ 0,682 = 306.7 cm.

ξ=

Rb bx(h0 − x / 2) + ( Rsc As' − N / 2)(h0 − а' ) = 14.5 ⋅ 400 ⋅ 306.7(450 − 306.7 2) + + (355 ⋅ 1847 − 2200000 / 2)(450 − 50) = 3.5 ⋅ 10 8 N ⋅ mm = = 350 kN ⋅ m > M = 345.9 kN ⋅ m,

i.е. cross-section durability is ensured.


Example 25. Given: a lower storey column of supported framework with cross-sectional dimensions 400х400 mm; а = а’ = 50 mm; concrete of class В40 (Eb= 36 . 103 MPa, Rb = 22 MPa); longitudinal reinforcement of class А500 (Rs = 435 MPa, Rsc = 400 MPa); normal forces and moments of flection in lower support section from vertical loads Nv = 6000 kN, Mv = 120 kN.m, from constant and long-term loads Nl = 5800 kN, Ml= 100 kN.m; forces from wind load are neglected; storey height l = 3,6 m. It is required to determine the cross-sectional area of longitudinal reinforcement. C a l c u l a t i o n. h0 = 400 – 50 = 350 mm. Calculation is performed with consideration of the column deflection according to para 3.53. As the column of the section in view is fully fixed into the footing, coefficient ηv is determined from formula (3.85), taking the design length according to para 3.55,a, being equal to l0 = 0,7 . 3,6 = 2,52 m. At that l0/h = 2,52/0,4 = 6,3 > 4, i.е. consideration of deflection is necessary. Stiffness D is determined from formula (3.89), considering all loads, i.е. M = Mv = 120 kN.m and N = Nv= 6000 kN. Eccentricity е0 = M 120 = = 0,02 m = 20mm > e0 = h/30 = 400/30 = 13,3 mm, therefore, the N 6000

moment does not have to be adjusted.

h0 − à ' 0,35 − 0,05 . = 120 + 6000 = 1020 kN m; 2 2 h − à' . M 1l = M l + N l 0 = 100 + 5800 ⋅ 0,15 = 970 kN m; 2 ϕ l = 1 + M 1l / M 1 = 1 + 970 / 1020 = 1,951 e 20 = 0,05 < 0,15 , taking δе = 0,15. As 0 = h 400 M1 = M + N

In = 0,02

the

20 ⋅ 10 4 3,6 ⋅ 10

4

first = 0,1111 .

approximation,

taking

μ = 0,02, then

μα =

MANUAL to CP 52-101-2003 concrete and REinforced concrete structures without prior reinforcement stress 2 ⎡ 0,0125 ⎛ h0 − a ' ⎞ ⎤ D = Eb bh ⎢ + 0,175μα ⎜ ⎟ ⎥= ⎝ h ⎠ ⎥⎦ ⎢⎣ϕ l (0,3 + δ e ) 3

2 ⎡ 0,0125 ⎛ 300 ⎞ ⎤ 13 = 36000 ⋅ 400 ⎢ + 0,175 ⋅ 0,1111⎜ ⎟ ⎥ = 2,32 ⋅ 10 Í ⋅ ìì ⎝ 400 ⎠ ⎥⎦ ⎢⎣1,951(0,3 + 0,15) 4

N cr =

Îòñþäà

ηv =

1 1−

=

N N cr

π 2D l02

=

3,14 2 ⋅ 2,32 ⋅ 1013 2520

2

2

.

= 36057 ⋅ 103 H = 36057 êÍ;

1 = 1,2; 6000 1− 36057

M= Mvηv = 120 . 1,2 = 144 kN. m. The required cross-sectional area of the reinforcement can be determined according to para 3.57. To do that, one can determine the values: N 6000 ⋅ 103 αn = = = 1,948; Rb bh0 22 ⋅ 400 ⋅ 350

α m1 = δ1 =

M + N (h0 − à ' ) / 2 Rb bh02

=

144 ⋅ 10 6 + 6000 ⋅ 103 ⋅ 300 / 2 22 ⋅ 400 ⋅ 350 2

= 0,9685;

à 50 = = 0,143. h0 350

From Table 3.2 one can find ξR = 0,493. As αn > ξR, value As= =A 's can be determined from formula (3.94). At that, as compressed reinforcement determines the durability here, one takes Rs = Rsc = 400 MPa. Value ξ is determined from formula (3.92), calculating αs according to formula (3.95) at αn + ξR

1,948 + 0,493 = 1,22 > 1, i.е. at ξ1 = 1,0, 2 2 α − ξ (1 − ξ1 / 2) 0,9685 − 1,0(1 − 1,0 / 2) = 0,547; α s = m1 1 = 1−δ 1 − 0,143 α (1 − ξ R ) + 2α sξ R 1,948 ⋅ 0,507 + 2 ⋅ 0,547 ⋅ 0,493 = = 0,954; ξ= n 1 − ξ R + 2α s 0,507 + 2 ⋅ 0,547

ξ1 =

As = As' = =

=

Rbbh0 α m1 − ξ (1 − ξ / 2) = 1−δ Rs

22 ⋅ 400 ⋅ 350 0,9685 − 0,954(1 − 0,954 / 2) = 4219,3 ìì 400 0,857

2

.


Taking As = A 's = 4539 mm2 (2∅40 + 2∅36). Example 26. Given: a middle storey column of supported framework with cross-sectional dimensions 400х400 mm; concrete of class В25 (Rb = 14,5 MPa), longitudinal reinforcement of class А400 (Rs = Rsc = 355 MPa): normal forces moments of flection from vertical loads in support section: from all loads Nv = 2200 kN, Mv =20 kN.m, from constant and long-term loads Nl = 1980 kN, Ml = 0,0; storey height Н = 6 m. It is required to determine cross-sectional area of longitudinal reinforcement. C a l c u l a t i o n. As the column is fixed at both ends with hinged cross-beams, the design length of the column is taken according to para 3.59,а being equal to l0 = H = 6 m. Then l0/h = 6/0,4 = 15>4, i.е. consideration of the column deflection is necessary. Eccentricity of the normal force from all loads equals e0 =

Mv 20 = = 0,009 ì = 9 mm. As h/30 = 400/30 = 13,3 mm > >l0/600 N v 2200

=6000/600 = 10 mm, according to para 3.49 accidental eccentricity is taken equal to еα= 13,3 mm > е0 . Therefore, calculation of the column is carried out for the impact of normal force with eccentricity е0 = еα according to para 3.58. From Tables 3.5 and 3.6 with Nl/N = 1980/2200 = 0,9, suggesting absence of intermediate rods at а = а’ < 0,15 h one can find ϕb = 0,804 and ϕsb = 0,867. Taking in the first approximation ϕ = ϕsb = 0,867, from condition (3.97) one can find Rs As , tot

2200 ⋅ 103 = − Rb A = − 14,5 ⋅ 400 ⋅ 400 = 2537,5 ⋅ 103 − 232 ⋅ 10 4 = . 0,867 ϕ N

= 217,5 ⋅ 103 H.

Hence α s =

Rs As, tot Rb A

=

217,5 ⋅ 103 232 ⋅ 10

4

= 0,094 .

Since αs < 0,5, one can revise value ϕ by calculating it from formula (3.98): ϕ = ϕb + 2(ϕsb - ϕb)αs = 0,804 + 2(0,867 – 0,804)0,094 = 0,816. Similarly one can determine R s As, tot

2200 ⋅ 103 = − 232 ⋅ 10 4 = 376,1 ⋅ 103 H. 0,816

The obtained value RsAs,tot significantly exceeds the one taken in the


first approximation, so one should revise ϕ again: αs =

376,1 ⋅ 103 232 ⋅ 10

4

= 0,162 ;

ϕ = 0,804 + 2(0,867 – 0,804)0,162 = 0,824; Rs As , tot

2200 ⋅ 103 = − 232 ⋅ 10 4 = 348,5 ⋅ 103 H. 0,824

As the obtained value RsAs,tot is close to the one taken in the second approximation, the total cross-sectional area of the reinforcement is taken equal to As , tot

348,5 ⋅ 103 2 = = 982 mm . 355

Finally, taking As,tot = 1018 mm2 (4∅18). Rectangular sections with asymmetrical reinforcement Example 27. Given: a column with semirigid attachments at both ends of the section with dimensions b = 400 mm, h = 500 mm; а = а’ = 40 mm; concrete of class В25 (Rb = 14,5 MPa), reinforcement of class А400 (Rs = Rsc = 355 MPa); forces in support section from vertical loads: normal force N = 800 kN.m; moment M = 400 kN.m; forces from wind loads are not applied. It is required to determine the cross-sectional area of the reinforcement S and S’. C a l c u l a t i o n. h0 = 500 – 40 = 460 mm. At the moment from wind load is not applied and according to para 3.53 ηv = 1,0, there is no impact of the element deflection to the moment. Then e = =

400 ⋅ 10 6 800 ⋅ 103

+

M h0 − à ' + = N 2

460 − 40 = 710 mm. 2

The required cross-sectional area of reinforcement S’ and S can be determined from formulae (3.102) and (3.103), taking from Table 3.2 αR = 0,39, ξR = 0,531: As'

Nå − à R Rbbh02 800 ⋅ 103 ⋅ 710 − 0,39 ⋅ 14,5 ⋅ 400 ⋅ 460 2 = = = 599,3 ìì Rsc (h0 − à ' ) 355(460 − 40)

As =

ξ R Rbbh0 − N Rs

+

As'

2

> 0,

0,531 ⋅ 14,5 ⋅ 400 ⋅ 460 − 800 ⋅103 = + 599,3 = 2336,5 ìì 355

As both values are greater than zero, they do not have to be revised.

2

> 0.


Taking As' = 628 mm2 (2∅20), As= 2413 mm2 (3∅32). I-sections Example 28. Given: a column of a single-storey industrial building: section dimensions and reinforcement location – as per Drawing 3.34; concrete of class В30 (Eb = 32500 MPa, Rb = 17,0 MPa); reinforcement of class А400 (Rs = Rsc = 355 MPa), cross-sectional area As=A 's =5630 mm2 (7∅32); normal forces and moments of flection in lower support section from vertical loads: all Nv = 6000 kN, Mv = 1000 kN.m, from constant and long-term loads Nl = 5000 kN, Ml = 750 kN.m; from wind loads Nh = 0,0, Mh = 2000 kN.m; column height H = 15 m. It is required to check the section durability.

Drawing 3.34. For calculation examples 28 and 29 C a l c u l a t i o n i n t h e p l a n e o f f l e c t i o n. Calculation is performed with consideration of the column deflection according to para 3.53. AS the column of the section is fully fixed to the footing, coefficient ηv is determined from formula (3.86), taking the design length of the column according to para 3.55,а being equal to l0 = 0,7Н = 0,7 . 15=10,5 m. Now one can define stiffness D from formula (3.88), taking all loads into consideration. Taking the design thickness of the flange equal to average height of overhangs h 'f = h f = 200 + 30/2 = 215 mm. The area and the moment of inertia of concrete section can be calculated: А = 200 . 1500 + 2 . 400 . 215 = 472 . 103 mm2;


400 ⋅ 2153 200 ⋅ 15003 4 ⎛ 1500 215 ⎞ 8 I= +2 + 2 ⋅ 400 ⋅ 215⎜ − ⎟ = 1279 ⋅ 10 mm . 12 12 2 ⎠ ⎝ 2

Radius of inertia for the section i =

I 1279 ⋅ 108 = = 520 mm. A 472 ⋅ 103

As l0/i = 10500/520 = 20,2 > 14, it is necessary to consider the column deflection. Forces from all loads: M = Mv + Mh = 1000 + 2000 = 3000 kN.m; N =N v = 6000 êÍ;

å0 =

M 3000 = = 0,5 ì = 500 ìì . N 6000

Now one can determine the moment of inertia for the section of all reinforcement. The reinforcement gravity center As and As' is located at the distance of à = à' =

5 ⋅ 50 + 2 ⋅ 150 = 79 mm from the nearest face, therefore h0 5+2

= h - а = 1500 – 79 = 1421 mm. 0,5h - а = 750 – 79 = 671 mm. Is = 2As(0,5h - а)2 = 2 . 5630 . 6712 = 5,07 . 109 mm4. Determining coefficient ϕl:

h0 − a ' 1,421 − 0,079 = 3000 + 6000 = 7026 êÍ ⋅ ì; 2 2 h − a' 1,421 − 0,079 M 1l = M l + N l 0 = 750 + 5000 = 4105 êÍ ⋅ ì; 2 2 M 4105 ϕ l = 1 + 1l = 1 + = 1,584. M1 7026 e e 500 = 0,333 > 0,15, taking δ e = 0 = 0,333 . As 0 = h 1500 h 3 0,15 Eb I 0,15 ⋅ 32,5 ⋅ 10 ⋅ 1279 ⋅ 108 + 0,7 E s I s = + 0,7 ⋅ 2 ⋅ 105 ⋅ 5,07 ⋅ 109 = D= 1,584(0,3 + 0,333) ϕ l (0,3 + δ e ) M1 = M + N

= 1,331 ⋅ 1015 Í ⋅ ìì

Hence, N cr = ηv =

π 2D l02

1 N 1− N cr

=

2

.

3,14 2 ⋅ 1,331 ⋅ 1015

= 1−

10500 1

2

6,0 ⋅ 10 6

= 1,192 ⋅ 108 N;

= 1,053.

119,2 ⋅ 10 6

Similarly, one can determine coefficient ηh, taking, according to para 3.55,б, the design length being equal to l0 = 1,5H = 1,5 . 15 = 22,5 m:


N cr =

3,14 2 ⋅ 1,331 ⋅ 1015 22500

2

1

= 2,595 ⋅ 10 7 H, η h = 1−

6 ⋅ 10

6

= 1,30.

25,95 ⋅ 10 6

Design moment with consideration of deflection equals M=Mvηv + Mhηh= 1000 . 1,05 + 2000 . 1,3 = 3653 kN.m. Verifying condition (3.108): Rb b 'f h 'f = 17 ⋅ 600 ⋅ 215 = 2193 ⋅ 103 H = 2193 kN < N = 6000 kN, i.е. calculation shall be carried out as for an I-section. The area of compressed overhangs equals: Aоv = (b 'f - b) h 'f = (600 – 200)215 = 86000 mm2. Calculating the height of compressed zone х. As ξ =

N − Rb Aîv 6000 ⋅ 103 − 17 ⋅ 86000 = = 0,939 > ξ R = 0,531 (see Table Rb bh0 17 ⋅ 200 ⋅ 1421

3.2), value х is determined from formula (3.110). To do that, calculating Rbbh0 = 17 . 200 . 1421 = 4831400 N;

Rs As 355 ⋅ 5630 N 6 ⋅ 10 6 αs = = = 0,414; α n = = = 1,242; Rb bh0 4831400 Rb bh0 4831400

α ov =

Aov 86000 = = 0,303; bh0 200 ⋅ 1421

x = h0

(α n − α ov )(1 − ξ R ) + 2α sξ R (1,242 − 0,303)(1 − 0,531) + 2 ⋅ 0,414 ⋅ 0,531 = 1421 = 1 − ξ R + 2α s 1 − 0,531 + 2 ⋅ 0,414

= 964 ìì.

The durability is checked based on condition (3.109): Rbbx(h0 - x/2)+RbAov(h0 - h’f/2)+(RscA’s – N/2)(h0 - a’) = 17 . 200 × 964 . (1421 - 964/2) + 17 . 86000(1421 – 215/2)+(355 . 5630 – 6× ×106/2)(1421 – 79) = 3,654 . 109 N.mm = 3654 kN. m > M = 3653 kN.m, i.е. durability of the section in the plane of flection is ensured. C a l c u l a t i o n o u t o f t h e p l a n e o f f l e c t i o n. First one shall determine the radius of inertia for the section out of the plane of flection: I= i=

2 ⋅ 215 ⋅ 6003 1070 ⋅ 2003 4 + = 845 ⋅ 10 7 mm ; 12 12 I 845 ⋅ 10 7 = = 134 mm. A 472 ⋅ 103

As the flexibility out of the plane of flection l0/i = 10500/134 =78,4


noticeably exceeds the flexibility in the plane of flection l0/i = 20,2, according to para 3.50 one should check the section durability out of the plane of flection taking the eccentricity е0 being equal to accidental eccentricity еа. The depth of section at that equals h = 600 mm. Determining value еа according to para 3.49.

l h 600 10500 h = = 20 ìì > 0 = = 17,5 mm and > 10 mm, taking 30 30 600 600 30 l h 10500 ea = , which at 0 = = 17,5 < 20 allows to carry out the calculation 30 h 600

Asу

according to para 3.58; at that, coefficient ϕ is determined as for the one for rectangular section not considering “in reserve” the rib section, i.е. at b = 2 . 215 = 430 mm. As the number of intermediate rods ∅32 located along both flanges being equal 6 exceeds 1/3 of the number of all rods ∅32 14/3 = 4,67, Table 3.6 (section Б) is used in the calculation. From this table at Nl/N = 5000/6000= 0,833 and l0/h =17,5 one can find ϕsb = 0,736. As,tot = 11260 mm2 (14∅32). value α s =

Rs As, tot Rb bh

=

355 ⋅ 11260 = 17 ⋅ 430 ⋅ 600

= 0,911 >0,5.

Therefore, ϕ = ϕsb = 0,736. Verifying condition (3.97): ϕ(RbA + RscAs,tot) = 0,736(17 . 472 . 103 + 355 . 11260) = 8848 . 103 N > N = 6000 kN, i.е. durability out of the plane of flection is ensured. Example 29. Given: a column with semirigid attachments at both ends; section and reinforcement location – as per Drawing 3.34; concrete of class В30 (Rb = 17,0 MPa); symmetrical reinforcement of class А400 (Rs = Rsc = 355 MPa); normal force and moment if support section from vertical loads N = 6000 kN, m = 3000 kN.m, forces from wind loads are not applied (Mh = 0,0, Nh = 0). It is required to determine the cross-sectional area of the reinforcement for support section of the column. C a l c u l a t i o n i n t h e p l a n e o f f l e c t i o n. According to para 3.53 coefficient ηv =1,0, and as Mh = 0, coefficient ηh is not calculated. Therefore, the element deflection in the plane of flection will not be considered. From example 28 one has: h 'f = 215 mm, h0 = 1421 mm, а’ = 79 mm. Verifying condition (3.108):


Rbb 'f hf' = 17 . 600 . 215 = 2193 . 103 H = 2193 kN < N = 6000 kN, i.е. calculation is parried out as for an I-section according to para 3.61. The area of compressed overhangs of the flange equals: 2 Aov = (b 'f − b)h 'f = (600 − 200) ⋅ 215 = 86000 mm . Determining values αn, αm1, αov, αm,ov, δ. Rbbh0 = 17 . 200 . 1421 = 4831400 N. 6000 ⋅ 103 N = = 1,242; αn = Rb bh0 4831400

α m1 =

M + N (h0 − a ' ) / 2 Rb bh02

3000 ⋅ 10 6 + 6000 ⋅ 103 (1421 − 79) / 2 = = 1,023; 4831400 ⋅ 1421

' ⎞ ⎛ h Aov 86000 215 ⎞ f ⎟ ⎛ = = 0,302; α m, ov = α ov ⎜1 − α ov = = 0,302⎜1 − ⎟= ⎜ 2h0 ⎟ bh0 200 ⋅ 1421 2 ⋅ 1421 ⎠ ⎝ ⎝ ⎠ = 0,279;

δ=

à' 79 = = 0,055. h0 1421

From Table 3.2 one can find ξR = 0,531. As ξ = αn - αov = 1,242 – 0,302 = 0,94 > ξR = 0,531, the crosssectional area of the reinforcement is determined from formula (3.113). To do that, using formulae (3.114) and (3.110) one can calculate values αs and ξ1 = x/h0. α m1 − ξ (1 − ξ / 2) − α m, ov 1,023 − 0,94(1 − 0,94 / 2) − 0,279 = = 0,26; 1− δ 1 − 0,055 (α − α ov )(1 − ξ R ) + 2α sξ R 0,94(1 − 0,531) + 2 ⋅ 0,26 ⋅ 0,531 = = 0,725. ξ1 = n 1 − ξ R + 2α s 1 − 0,531 + 2 ⋅ 0,26

αs =

Hence As = As' = =

Rb bh0 α m1 − ξ1 (1 − ξ1 / 2) − α m, ov = 1−δ Rs

4831400 1,023 - 0,725(1 - 0,725/2) - 0,279 == 4058,6 ìì 355 1 − 0,055

2

.

Taking As = A 's = 4310 mm2 (7∅28). Calculation out of the ;plane of flection is carried out using procedure similar to the one in example 28. Annular sections


EXAMPLE 30. GIVEN: CANTILEVER COLUMN WITH HEIGHT H = 6 M, SECTION WITH INTERNAL RADIUS R1 = 150 MM, EXTERNAL RADIUS – R2 = 250 MM; CONCRETE OF CLASS B25 (EB = 3·104 MPA, RB= 14,5 MPA); LONGITUDINAL REINFORCEMENT OF CLASS A400 (RS= RSC = 355 MPA) IS LOCATED IN THE MIDDLE OF THE ANNULAR WALL, ITS CROSS-SECTIONAL AREA IS AS,TOT = 1470 MM2(13∅12); NORMAL FORCE AND MOMENT IN THE FOOTING: FROM VERTICAL LOADS: NV = 120 KN, MV =40 KN.M; FROM WIND LOADS: NH = 0, MH = 70 KN.M. IT IS REQUIRED TO CHECK THE SECTION DURABILITY. C A L C U L A T I O N. INTERNAL AND EXTERNAL DIAMETERS EQUAL D1=2R1=300MM, D2=DCIR=2R2=500MM. AS FOR A CANTILEVER COLUMN AN ECCENTRICALLY APPLIED VERTICAL FORCE CAUSES DISPLACEMENT OF THE UPPER PART, ACCORDING TO PARA 3.53 ONE TAKES ML = 0 AND MH = 40 + 70 =110KN.M. COEFFICIENT ηH IS DETERMINED FROM FORMULA (3.85), TAKING UP ACCORDING TO PARA 3.55,Б THE DESIGN LENGTH OF THE COLUMN BEING EQUAL TO L0 = 2H = 3·6 = 12 M. LOADS FROM ALL FORCES EQUAL: N = 120 KN, M = MH = 110 KN.M; M 110 = = 0,917 m = 917 mm. N 120

e0 =

DETERMINING STIFFNESS D FROM FORMULA (3.88); RS = RM=

r1 + r2 150 + 250 = = 200 MM; 2 2

M1 = M + N ⋅ rs = 110 + 120 ⋅ 0,2 = 134 KN.M; M1L = MV + NV·RS = 40 + 120 ⋅ 0,2 = 64 KN.M;

ϕL = 1 + M1L/M1 = 1+ 64/134 = 1,478. e e0 917 = = 1,834 > 0,15, TAKING δE = 0 = 1,834. Dcir 500 Dcir

AS

Moments of inertia for the concrete section and all reinforcement respectively equal I=

(

π D24 − D14

IS =

64 As ,tot rs2 2

) = 3,14(5004 − 3004 ) = 2,67 ⋅ 109 MM ; 4

64

1470 ⋅ 200 2 = = 2,94 ⋅ 107 MM4. 2


D =

0,15Eb I 0,15 ⋅ 3 ⋅ 10 4 ⋅ 2,67 ⋅ 109 + 0,7ESIS= + 0,7 ⋅ 2 ⋅105 ⋅ 2,94 ⋅107 = ϕ l (0,3 + δ e ) 1,478(0,3 + 1,834 ) 2

7,925 ⋅1012 N.MM .

HENCE NCR = 1

ηH = 1−

N N cr

=

π 2D l02

=

3,14 2 ⋅ 7,925 ⋅1012 12000

2

= 543,2 ⋅ 103 Н = 543,2 KN.

1 = 1,284. 120 1− 543,2

THE MOMENT WITH CONSIDERATION OF DEFLECTION EQUALS M = 110 ⋅ 1,284 = 141,2 KN.M, THE CROSS-SECTIONAL AREA EQUALS A = π r22 − r12 = 3,14(2502 – 1502) = 125600 MM2. Calculating relative area of the compressed zone of the concrete from formula (3.115):

(

ξ cir =

)

N + Rs As,tot Rb A + 2,7 Rs As,tot

=

120 ⋅ 10 3 + 355 ⋅ 1470 = 0,274. 14,5 ⋅ 125600 + 2,7 ⋅ 355 ⋅ 1470

AS 0,15 < ξ cir < 0,6, THE SECTION DURABILITY IS CHECKED BASED ON CONDITION (3.116): (RB ARM + RSC AS,TOT RS)

sin πξ cir

π

+ RSAS,TOTRS(1 – 1,7 ξ cir )(0,2 +

(

)

sin 180 o ⋅ 0,274 + 1,3 ξ cir ) = =(14,5 ⋅ 125600 ⋅ 200 + 355 ⋅ 1470 ⋅ 200 ) 3,14

+ 355 ⋅ 1470 ⋅ 200(1 − 1,7 ⋅ 0,274 ) X (0,2 + 1,3 ⋅ 0,274 ) = 144,1 ⋅ 10 6 N.MM > M = =141,2KN.M, i.е. durability of the section is ensured. Circular sections EXAMPLE 31. GIVEN: A LOWER STOREY COLUMN OF SUPPORTED FRAMEWORK 4,8 M LONG; SECTION WITH DIAMETER DCIR = 400 MM; A = 35 MM; CONCRETE OF CLASS B25 (EB = 3 ⋅ 10 4 MPA, RB = 14,5 MPA); LONGITUDINAL REINFORCEMENT OF CLASS A400 (RS= RSC= 355MPA); ITS CROSS-SECTIONAL AREA AS,TOT = 3140MM2 (10∅20); NORMAL FORCES AND MOMENTS IN UPPER SUPPORT SECTION: FROM VERTICAL LOADS NV = 1700 KN; MV = 60 KN.M; FROM WIND


LOADS NH = 100 KN, MH = 45 KN.M, SHORT-TERM VERTICAL LOADS ARE NOT APPLIED. IT IS REQUIRED TO CHECK THE DURABILITY OF UPPER SUPPORT SECTION. C A L C U L A T I O N. AS THE SECTION IN VIEW IS LOCATED AT THE SEMIRIGID ATTACHMENT, ACCORDING TO PARA 5.53 ηV = 1,0. DETERMINING COEFFICIENT ηH ACCORDING TO PARA 5.54. AT THAT, THE DESIGN LENGTH IS TAKEN ACCORDING TO PARA 5.55,Б BEING EQUAL TO L0 = H = 4,8 M. FORCES FROM ALL LOADS EQUAL: M = MV + MH = 60 + 45 = 105 KN.M, N = NV + NH = =1700 + 100 = 1800 KN; E0 =

105 M = = 0,0583 M= 58,3 MM. N 1800

DETERMINING STIFFNESS D FROM FORMULA (3.88). FOR THAT PURPOSE, CALCULATING: R = DCIR /2 = 400/2= 200 MM, RS = R – A = 200 – 35 = 165 MM; M1 = M + NRS = 105 + 1800 ⋅ 0,165 = 402 KN.M; DUE TO ABSENCE OF SHORT-TERM VERTICAL LOADS ML = MV = 80 KN.M N = NV = 1700 KN; THEN M1l = Ml + Nlrs = 80 + 1700 ⋅ 0,165 = 360,5 kN.M;

ϕl = 1 + AS e 0

D cir

M 1l 360,5 =1+ = 1,897. M1 402 58,3 = = 0,146 < 0,15, TAKING δe = 0,15. 400

MOMENT OF INERTIA OF THE concrete section AND ALL reinforcement RESPECTIVELY EQUAL: 4 πDcir

3,14 ⋅ 400 4 = = 1256 ⋅ 10 6 mm4. I= 64 64 2 A r 3140 ⋅ 165 2 = 42,74 ⋅ 10 6 mm4. Is = s,tot s = 2 2

THEN

0,15 Eb I 0,15 ⋅ 3 ⋅ 10 4 ⋅ 1256 ⋅ 10 6 + 0,7Es Is = + 0,7 ⋅ 2 ⋅ 10 5 ⋅ 42,74 ⋅ 10 6 = D= ϕ l (0,3 + δ e ) 1,897(0,3 + 0,15)

=1,26 ⋅ 1013 n.mm2. Ncr =

π 2D l 02

=

3,14 2 ⋅ 1,26 ⋅ 1013 4800 2

= 5397 ⋅ 10 3 n = 5397 kN;


1

ηh = 1−

N N cr

=

1 = 1,5; 1800 1− 5397

DESIGN MOMENT WITH CONSIDERATION OF DFLECTION EQUALS M = Mv + Mhηh = 60 + 45 ⋅ 1,5 = 127,5 kN.м. The section durability is checked based on condition (3.127) using the graph shown in Drawing 3.33. Determining the area of concrete section 2 πD 3,14 ⋅ 400 2 A = cir = = 125600 mm2. 4 4 N 1800 ⋅ 10 3 = = 0,988, using values α n = Rb A 14,5 ⋅ 125600 Rs As,tot a 355 ⋅ 3140 35

αs =

Rb A

=

14,5 ⋅ 125600

= 0,612 and δ =

Dcir

=

400

= 0,0875 in the

graph one can find α m = 0,375. . . α m RbAr = 0,375 ⋅ 14,5 ⋅ 125600 ⋅ 200 = 136,6 ⋅ 10 6 N mm > M = 127,5 kN м, I.е.durability of the section isensured. Elements in oblique eccentrical compression Example 33. Given: RECTANGULAR SECTION OF A column WITH DIMENSIONS b = 400 mm, h = 500 mm; concrete OF class B25 (Rb = 14,5 MPa); longitudinal reinforcement OF class A400 (Rs = Rsc = 355 MPa) IS LOCATED IN THE SECTION AS PER DRAWING 3.35; THE section IS EXPOSED TO SIMULTANEOUS IMPACTS OF FORCE N = 2600 kN AND MOMENTS OF FLECTION IN THE PLANE PARALLEL TO DIMENSION h, Mx = 150 kN.м; IN THE PLANE PARALLEL TO DIMENSION b, My = 100 kN.м; MOMENTS Mx AND My ARE GIVEN WITH CONSIDERATION OF THE COLUMN DEFLECTION.


50 x

50

50

2

y

h=500

y

32

50

2

32

b=400 x Drawing 3.35. For calculation example 33

It is required to check the section durability. C a l c u l a t i o n. as the reinforcement is presented in the form of 4 corner rods, The section durability is checked according to para 3.66. Axes of symmetry parallel to dimensions h and b shall be denoted x and y. Now one can determine limit moments M x0 and M 0y . when the moment acts in the plane of axis x one can take b = 400 mm, h0 = 500 – 50 = 450 mm. As = As' = 1609 mm2(2∅32). as a = 50mm < 0,15h0 = 0,15 ⋅ 450 = 67,5mm, calculation can be performed using the graph shown in drawing 3.28. for that purpose determining α s =

Rs As Rbbh0

=

2600 ⋅ 10 3 N 355 ⋅ 1609 = = 0,219 and α n = = = 0,996. 14,5 ⋅ 400 ⋅ 450 Rb bh0 14,5 ⋅ 400 ⋅ 450

in the graph, α m = 0,24 corresponds to these values. Therefore, . . M x0 = α m Rb b h02 = 0,24 ⋅ 14,5 ⋅ 400 ⋅ 450 2 = 281,9 ⋅ 10 6 n mm = 281,9 kN м. when the moment acts in the plane of axis y one can take b = 500 mm, h0 = 400 – 50 = 350 mm. as a = 50 mm < 0,15h0 = 0,15 ⋅ 350 = 52,5 mm, moment M 0y can also be determined using the graph shown in drawing 3.28. αm = 0,23. corresponds to values in the graph, R A 355 ⋅ 1609 αs = s s = =0,225 Rb bh0 14,5 ⋅ 500 ⋅ 350

and

N αn = Rb bh0

2600 ⋅ 10 3 = = 14,5 ⋅ 500 ⋅ 350

=1.

therefore, M 0y = α m Rb bh02 = 0,23 ⋅ 14,5 ⋅ 500 ⋅ 350 2 = 204,3 ⋅ 10 6 Н.mm = 204,3 kN.м.


now one can determine index of power k. as α n =

2600 ⋅ 10 3 N = = Rb bh 14,5 ⋅ 400 ⋅ 500

0,8965 > 0,4, one can use formula (3.132), calculating values α s = =

R s As,tot Rb bh

=

0,275 + α s 355 ⋅ 2 ⋅ 1609 0,275 + 0,394 = 0,394 and k0 = = = 0,16 + α s 14,5 ⋅ 400 ⋅ 500 0,16 + 0,394

1,21.

(

⎡ (1,7 − α s )2

)

⎤ ⎡ (1,7 − 0,394 )2 ⎤ + 0,1775⎥ α n2 − 0,16 + k0 = ⎢ + 0,1775⎥ × 4 4 ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ · 0,8965 2 − 0,16 + 1,21 = 1,484.

k= ⎢

(

)

verifying condition (3.129): k

k 1, 484 1, 484 ⎛M ⎞ ⎛Mx ⎞ ⎛ 100 ⎞ ⎜ ⎟ + ⎜ y ⎟ = ⎛⎜ 150 ⎞⎟ = 0,738 < 1, +⎜ ⎟ ⎜M0 ⎟ ⎜M0 ⎟ 281 , 9 204 , 3 ⎝ ⎠ ⎝ ⎠ ⎝ x⎠ ⎝ y⎠

т.е. the section durability is ensured. Calculation of inclined sections Example 34. Given: a column of multi-storeyed supported framework with section dimensions b = 400 mm, h = 600 mm; а = а′ = 50 mm; concrete of class B25 (Rb = 14,5 MPa, Rbt = 1,05 MPa); stirrups located at column faces, made of reinforcement of class A240 (Rsw= 170 MPa) with diameter 12 mm (Asw = 226 mm2) and spacing sw = 400 mm; moments of flection in upper and lower support sections equal Msup = 350 kN.м, Minf = 250 kN.м and tension the left and the right column faces respectively; normal force N = 572 kN; column length (distance between the support sections) l = 3,3м . It is required to check the column durability to the impact of transverse force. C a l c u l a t i o n. h0 = h – a = 600 – 50 = 550 mm. Calculation is carried out according to para 3.30-3.32 with consideration of recommendations of para 3.52. Transverse force in the column equals Q=

M sup + M inf l

=

350 + 250 = 181,8 kN. 3,3

as the transverse force is constant along the coumn length, theprojection length of the inclined section is takenas the maximum possible one, i.е. equal to сmax = 3h0 = 3 ⋅ 550 = 1650 mm < l = 2800 mm.


from formula (3.84) one can determine coefficient ϕn2, taking Nb = 1,3Rb bh =1,3 ⋅ 14,5 ⋅ 400 ⋅ 600 = 4524 ⋅ 10 3 N = 4524 kN > N = 572 kN, 2

572 ⎛ 572 ⎞ ϕn2 = 1 + 3 − 4⎜ ⎟ = 1,0625. 4524 ⎝ 4524 ⎠

as с = сmax, Qb = Qb,min = 0,5Rbtbh0 = 0,5 ⋅ 1,05 ⋅ 400 ⋅ 550 = =115500 N, and after multiplying by ϕn2 Qb = 111,5 ⋅ 1,0625 = 122,7 kN. value qsw is determined from formula (3.48) qsw =

Rsw Asw 170 ⋅ 226 = = 96 N/mm. 400 sw

now one can determine the force in stirrups Qsw, taking с0 = 2h0 = 2 ⋅ 550 = 1100 mm, Qsw = 0,75 qsw с0 = 0,75 ⋅ 96 ⋅1100 = 79200 N = 79,2 kN. verifying condition (3.49) by multiplying its right part by ϕn2: 0,25Rbt b ⋅ ϕ n 2 = 0,25 ⋅ 1,05 ⋅ 400 ⋅1,0625 = 111,6 n/м > qsw = 96 n/mm. as condition (3.49) is not satisfied, one can take Rbtbϕn2 = 4qsw = 4 ⋅ 96 = 384 n/mm, аnd therefore, Qb = 0,5h0 ⋅ Rbt bϕ n 2 = 0,5 ⋅ 550 ⋅ 384 = 105600 n = 105,6 kN verifying condition (3.44): Qb + Qsw = 105,6 + 79,2 = 184,8 > Q = 181,8 kN, i.е. durability of the inclined section to the impact of transverse force is ensured. CENTRALLY AND ECCENTRICALLY TENSIONED ELEMENTS CENTRALLY TENSIONED ELEMENTS

3.67. Calculation OF CENTRALLY TENSIONED ELEMENTS DURABILITY SHOULD BE CARRIED OUT BASED ON CONDITION N ≤ Rs As, (3.133) where As – cross-sectional area of all longitudinal reinforcement. ECCENTRICALLY TENSIONED ELEMENTS

3.68. Calculation OF NORMAL SECTIONS OF ECCENTRICALLY TENSIONED ELEMENTS IN GENERAL CASE IS PERFORMED BASED ON NON-LINEAR DEFORMATIONAL MODEL ACCORDING TO PARAS 3.72-3.76.


Calculation OF RECTANGULAR sectionS OF ECCENTRICALLY TENSIONED ELEMENTS WITH reinforcement BEING LOCATED AT ELEMENT FACES PERNEDICULAR TO THE PLANE OF FLECTION, AT ECCENTRICITY DIRECTION IN THE SECTION PLANE OF SYMMETRY IS PERMITTED TO CARRY OUT BY LIMIT FORCES ACCORDING TO PARAS 3.69 AND 3.70. 3.69. DURABILITY CHECKUP OF RECTANGULAR SECTIONS OF ECCENTRICALLY TENSIONED ELEMENTS SHOLD BE PERFORMED DEPENDING ON LOCATION OF NORMAL FORCE N: а) IF NORMAL FORCE N IS APPLIED BETWEEN THE RESULTANTS OF FORCES IN reinforcement S AND S ' (DRAWING 3.36,а), I.е. AT e ' ≤ h0 – a ' , − BASED ON CONDITIONS (3.134) N e ' ≤ Rs As(h0 – a ' ); (3.135) N e≤ Rs A ' s(h0 – a′); )

Rs A's

A's

Rsc A's RA

x

a'

A's

a'

a)

h h

0

e'

h h0

Ab R s As

As

a

a

e

R s As

e

e'

b b

b As

Drawing3.36. Schematic representation of forces and stress diagram in section normal to longitudinal axis of eccentrically tensioned reinforced concrete element, at its durability calculation б) if normal force N is applied beyond the distance between the resultants if firces in reinforcement S and S ' (drawing 3.36,б), i.е. at e ' > h0 – a ' − based on condition (3.136) N e ≤ Rbbx(h0 – 0,5x) + RscA ' s(h0 – a’) at that, the compressed zone height x is determined from formula Rs As − Rsc As' − N . x= Rbb

(3.137)

if the obtained from calculation from formula (3.137) value x > ξRh0, then x =ξRh0 is plugged into condition (3.136) where ξR is found form table 3.2. at х < 0 section durav=bility is checked based on condition (3.134).


at symmetrical reinforcement the durability ireespective of value e‘ is checled based on condition (3.134). note. if at e ' > h0

- a ' height of compressed zone determined without Rs As − N is less than 2а ' , the design Rb b

consideration of compressed reinforcement, x =

carrying capacity can be increased to some extent by performing calulation using formulae (3.136) and (3.137) without consideration of compressed reinforcement.

3.70. required amount of longitudinal reinforcement elements is determined in the following way: а) at e ' ≤ h0 - а ' one determines the cross-sectional area of reinforcement and S and S ' respectively from formulae: Ne ′ ; Rs (h0 − a ′) Ne As ′ = ; Rs (h0 − a ′)

(3.138)

As =

(3.139)

б) at e ' > h0 - a ' one determines the cross-sectional area of reinforcement Аs from formula: As =

ξbh0 Rb + N Rs

+ As′

Rsc ; Rs

where ξ is found using formula ξ =1 − 1 − 2α m , here

αm =

Ne − Rsc As ′ (h0 − a ′) Rb bh02

(3.140) (3.141)

.

(3.142)

At that, condition αm ≤ αR (see table 3.2) shall be satisfied. otherwise the cross-section of compressed reinforcements As' shall be increased, a higher class concrete shall be used of the section dimensions increased. if αm<0, the cross-sectional area of tensioned reinforcement As is found from formula (3.138). the area of symmetrical reinforcement irrespective of value e' is selected using formula (3.138). notе. at e ' > h0 – a ' the required amount of reinforcement elements determined from formula (3.138) can be decreased if value ξ, determined from formula (3.141) at Ne ' , turns less than 2 a /h0 .in this case the cross-sectional area of tensioned 2 Rbbh0 reinforcements As is determined from formula (3.140), using the said value ξ at AS′ =

αm =

0.


3.71. Calculation of inclined sections of tensioned elements at the impact of transverse forces is carried out in similar way to calculation of flexural elements in accordance with paras 3.30 - 3.35. at that value of transverse force supported by concrete in the inclined section Qb as well as the right part of condition (3.49) is divided by coefficient ϕ nt = 1 +

N . 1,5 Rbt A

value Mb related to Qb is divided by the same coefficient ϕnt.

(3.143)


Calculation examples Example 35. Given: tensioned stanchion of a two-stanchioned column with cross-section with dimensions b = 500 mm, h=200 mm; a = a'= 40 mm; longitudinal reinforcement of class А400 (Rs= Rsс=355 MPa); its cross-sectional area As=A's=982 mm2 (2∅25); concrete of class В25 (Rb = 14,5 MPa); longitudinal tensile force N = 44 kN; maximum moment of flection m = 43 kN⋅м. It is required to check the durability of normal section C a l c u l a t i o n. h0 = 200 – 40 = 160 mm. M 43 ⋅ 10 6 = e0 = = 977 mm; N 44 ⋅ 10 3 200 h e′ = e0 + − a ′ = 977 + − 40 = 1037 mm; 2 2 200 h e = e0 − + a = 977 − + 40 = 917 mm. 2 2

as the reinforcement is symmetrical, the durability will be checked using condition (3.134): RsAs(h0 – a') = 355 ⋅ 982(160 - 40) = 41,8 ⋅ 106 H ⋅ ìì h0-a'=120 mm, and the hieght of compressed zone х determined without consideration of compressed reinforcement i.е. equal to Rs As − N 355 ⋅ 982 − 44 ⋅ 103 x= = = = 42 ìì , is less than 2а' = 2 ⋅ 40 = 80 mm, Rbb 14.5 ⋅ 500

according to the note to para 3.69 one should check the durability based on condition (3.136), taking х = 42 mm and As′ =0: Rb bx(h0 − 0,5 x) = 14,5 ⋅ 500 ⋅ 42(160 − 0,5 ⋅ 42) = 42,3 ⋅ 10 6 n.mm > Ne= = 44 ⋅ 10 3 ⋅ 917 = 40,4 ⋅ 10 6 n.mm, i.е. the durability is ensured.

Example 36. Given: rectangular section with dimensions b = 1000 mm, h = 200 mm; a = a' = 35 mm; concrete of class В15 (Rb = 8,5 MPa); longitudinal reinforcement of class А400 (Rs= Rsc = 355 MPa); crosssectional area of reinforcement As′ = 1005 mm2 (5∅16); tensile force N = =160 kN; moment of flection m = 116 kN.м. It is required to determine the cross-setional area of reinforcement S. C a l c u l a t i o n. h0 = 200 – 35 = 165 mm;


M 116 ⋅106 = = 725 ìì ; N 160 ⋅103 h 200 e = e0 − + a = 725 − + 35 = 660 mm; 2 2 200 h e′ = e0 + − a ′ = 725 + − 35 = 790 mm. 2 2 e0 =

as e' = 790 mm > h0 - a' =165 - 35 = 130 mm, one can determine the required cross-sectional area of tensioned reinforcement according to para 3.70,б. Calculating value

αm =

Ne − Rsc As ′ (h0 − a ′) Rb bh02

=

160 ⋅ 10 3 ⋅ 660 − 355 ⋅ 1005 ⋅ 130 8,5 ⋅ 1000 ⋅ 165 2

= 0,256 .

as 0 < αm < αR = 0,39 (see table 3.2), value As is determined from formula (3.140). for that purpose, one shoual calculate ξ = 1 1 − 2α m = 1 − 1 − 2 ⋅ 0,256 = = 0,301 . As =

ξbh0 Rb + N Rs

Rsc 0,301 ⋅1000 ⋅ 165 ⋅ 8,5 + 160 ⋅ 103 2 ′ + As = + 1005 = 2645 mm . Rs 355

taking As = 3079 mm2 (5∅28). Example 37. Given: tensioned stanchion of a two-stachined column with sectional dimensions b = 500 mm, h = 200 mm; a =a' = 40 mm; concrete of class В25 (Rbt = 1,05 MPa); stirrups located at ee\lement faces, made of reinforcement of class А400 (Rsw = 285 MPa); normla tensile force N = 44 kN; transverse force Q = 130 kN; clearance between the bulkheads of two-stanchioned column l = 600 mm. It is required to determine diameter and spacing ot stirrups. C a l c u l a t i o n. h0 = 200 - 40 =160 mm. Calculation is carried out according to para 3.33,а with cosideration of recommendations of para 3.71. using formula (3.143) one can determine coefficient ϕnt , taking А = = bh = 500 ⋅ 200 =100000 mm2: 44000 N = 1+ = 1,279. 1,5 Rbt A 1,5 ⋅ 1,05 ⋅ 100000 As within the bulkheads the transverse force is constant, the projection length on the inclined section is taken equal to the maximum possible one, i.е. с = сmax = 3h0 = 3 ⋅ 160 = 480 mm < l = 600 mm.

ϕnt = 1 +

at α = с/h0 = 3 and α0 = 2 < 3 calculating

εгр =

1,5 1,5 130000 Q + 0,1875α 0 = + 0,1875 ⋅ 2 = 0,875 < ε = = = 1,548 . 3 a Rbt bh0 1,05 ⋅ 500 ⋅ 160


therefore, the required intensity of stirrups can be determined from formula (3.48),at that the value 1,5 characterising value Qb is divided by ϕnt = 1,279: ε− q sw = Rbt b

1.5

αϕ nt = 1,05 ⋅ 500 0,75α 0

1,5 3 ⋅ 1,279 = 405 n/mm. 0,75 ⋅ 2

1,548 −

maximum allowable spacing, according to para 3.35 equals s w,max

Rbt bh02 1,05 ⋅ 500 ⋅ 160 2 = = = 103,4 mm. Q 130000

taking he stirrups spacing sw = 100 mm < sw,max, and then Asw =

q sw s w 405 ⋅ 100 2 = = 142,1mm . Rsw 285

taking two stirrups with diameter 10 mm (Asw=157 mm2). CALCULATION OF NORMAL SECTIONS BASED ON NON-LINEAR DEFORMATIONAL MODEL 3.72. At carrying out durability calculation, forces and deformations in normal section are determined based on non-linear deformational model which uses equations of equilibrium of external forces and internal stresses in element section as well as the following stipulations: - distribution of relative deformations of concrete and reinforcement along the element section is taken linear (Bernoulli hypothesis, see Drawing 3.39); - relation between axial compressing stresses of concrete σb and its relative deformations εb is taken in the form of a bilinear diagram (Drawing 3.37), according to which stresses σb are determined in the following way: at 0 ≤ εb ≤ εb1,red σb = Eb,red εb; at εb1,red < εb ≤ εb2 σb = Rb; where Eb,red – transformed modulus of deformation of the concrete equal to Eb,red = Rb /εb1,red; εb1,red = 0,0015; εb2 = 0,0035; Rb – see Table 2.2; - concrete resistance in the tensioned zone is not considered (i.е. one takes σb = 0) except for calculation of concrete elements specified in para 1.4,б as well as those concrete elements in which no cracks are permitted;


in these elements, relation between axial tensile stresses of concrete σbt and its relative deformations are also taken in the form of bilinear diagram with σb Rb

arctg Eb,red

εb ε b2

ε b1,red

Drawing 3.37. Bilinear diagram of the state of compressed concrete

εb1,red being substituted with εbt1,red = 0,0008; εb2 with εbt2 = 0,00015; Eb,red with Ebt,red = Rbt /εbt1,red, where Rbt – see Table 2.2; - relation between reinforcement stresses σs and its relative deformations εs is taken in the form of a bilinear diagram (Drawing 3.38) according to which stresses σs are taken equal to: at 0 < εs < εs0 σs = εs Es; at εs0 ≤ εs < εs2 σs = Rs, where εs0 = Rs/Es; Rs – see Table 2.6; Es = 2 . 105 MPa; εs2 = 0,025. σs

Rs

arctg Es ε s0

εs ε s2

Drawing 3.38. State diagram of tensioned reinforcement


3.73. Transition from stress diagram in the concrete to generalized internal stresses is recommended to be carried out using the procedure of numerical integration for normal section. For this purpose, in the direction of the plane of flection (normal neutral axis), a normal section is divided into segments of small width, and stresses in them are taken to be uniformly distributed and corresponding to deformations at the level of the middle of the segment. In general case, the position of neutral axis and maximum deformations (Drawing 3.39) are determined from equations of equilibrium of internal and external forces: Mx = ΣσbiAbiZbxi + ΣσsjAsjZsxj; (3.144) My = ΣσbiAbiZbyi + ΣσsjAsjZsyj; (3.145) N = ΣσbiAbi + ΣσsjAsj ; (3.146) where Mx and My – moments of external forces in relation to the chosen axes of reference, acting in the planes of axes х and y respectively; x Z byi b /2 i

a) Zbxi

/2 bi

Abi y

y

strain diagram

i

y

y

concrete stress Эпюра напряжений бетона

h

ε2

σs

yb

i

x

j

Эпюра напряжений concrete stress бетона

ε bi

diagram

Rs

σ bi

Эпюра strainдеформаций diagram

d ,re

Rb

ε b1

ус

ε1

conventional neutral axis

diagram

ε sj

Z bxi

Abi

Rb

Уs

j

Уb

x

ax x a m

s

Asj

Zbyix

d /R s ε bi b1,re Es ε

б)

εm ε

x

Эпюра деформаций

reinforcement Эпюра напряжений арматуры stress diagram

вн ло ая не йт л ра ьн ая ь ос


Drawing 3.39. Strain and stress diagrams of a section normal to longitudinal axis of a reinforced concrete element in general case of durability calculation а) – two-valued strain diagram б) –one-valued strain diagram

Abi, Zbхi, Zbyi, σbi – area, coordinates of the gravity center of the ith segment of concrete and stress at the level of its gravity center. Asj,, Zsx,j, Zsyj, σsj – area, coordinates of the gravity center of the ith reinforcement rod and stress in it. Stresses σbi and σsj are determined in accordance with diagrams shown in Drawings 3.37 and 3.38. Tensile stresses of reinforcement σsj and of concrete σbi as well as longitudinal tensile force N are recommended to consider in equations (3.144) - (3.146) with the “minus” sign. Axes of reference х and y are recommended to be drawn through the gravity center of the concrete section. 3.74. Durability calculation of normal sections of reinforced concrete elements is carried out based on the following conditions εb,max ≤ εb,ult ; (3.147) |εs,max |≤ εs,ult , (3.148) where εb,max and εs,max – relate deformations of the most compressed concrete fiber and the most tensioned rod of reinforcement respectively from the impact of external loads determined form solutions of equations (3.144) – (3.146); εb,ult and εs,ult – limit values of relative deformations of compressed concrete and tensioned reinforcement respectively, taken according to para 3.75. For flexural and eccentrically compressed concrete elements in which no cracks are permitted, the calculation is performed with consideration of tensioned concrete behavior in the element cross-section based on condition (3.149) |εbt,max| ≤ εbt,ult where εbt,max – relative deformation of the most tensioned concrete fiber determined from the solution of equations (3.144) – (3.146); εbt,ult – limit value of relative deformation of tensioned concrete taken according to para 3.75.


3.75. The limit values of relative deformations of concrete εb,ult (εbt,ult) are taken according to the two-valued strain diagram (compression and tension) in the element cross-section being equal to εb2 (εbt2) (see para 3.72). At eccentrical compression or tension and distribution concrete deformation of the same sign in the element cross-section, the limit values of relative deformations of concrete εb,ult (εbt,ult) are determined depending on the relation of relative deformations of concrete on the opposite sides of the section

⎛ ε ⎞ ε1 < 1 from formulae ε b,ult = ⎜⎜ 35 − 15 1 ⎟⎟10 −4 ; ε2 ε2 ⎠ ⎝ ⎛ ε ⎞ ε bl ,ult = ⎜⎜15 − 5 1 ⎟⎟10 −5 . ε ⎝

2

⎠

(3.150) (3.151)

The limit value of relative deformation of tensioned reinforcement εs,ult is taken equal to 0,025. 3.76. Calculation based on non-linear deformational model is carried out in a computer-assisted way. If two moments mх and mу act along both axes of reference х and y in normal section and there is a compressing normal force, the computer software shall be developed on the basis of the following algorithm: 1. the direction of neutral axis is set: in the first approximation, this direction is determined as for elastic material, i.е. the inclination angle of the neutral axis to reference axis y is taken equal to θ = arctg

Ix M y . Iy Mx

2. the character of strain diagram is determined by comparing external normal force N and internal stress N, determined from formula (3.146) at values εb in extreme points being equal to εb2 and 0. At N > Nc – the diagram is one-valued, at N < Nc – the diagram is two-valued. 3. At two-valued strain diagram, one should use step-by-step approximations to choose the height of compressed zone x so that congruence (3.146) is fulfilled; at that, in the extreme compressed point εb= εb2, is taken, deformations of compressed concrete of each ith segment are taken equal to εbi = εb2ybi /x, and deformations of each jth reinforcement rod εsi = εb2ysj /x, where ybi and ysj - distances from the neutral axis to the gravity center of the ith concrete segment and the jth reinforcement rod, respectively. In case εs,max> 0,025, one takes εs,max=0,025, and then εbi= εs,max ybi /(h0 – x), εs,j=εs,max ysi /(h0 – x), where h0 – distance between the most tensioned reinforcement rod and the most compressed point of concrete in the


direction normal to the neutral axis. Deformations of tensioned reinforcement are taken with the “minus” sign. 4. At one-valued strain diagram, one should use step-by-step approximations to choose the ratio of deformations in extreme points α =ε1/ε2 < 1 so that congruence (3.146) is fulfilled; at that, in the extreme compressed point, deformation εb,ult is taken as determined from formula (3.150), compressed concrete deformations of every ith segment are taken equal to ε bi = ε b,ult [α + ε sj = ε b,ult [α +

yi th (1 − α )] , and deformation of every j rod h

y si (1 − α )] , where yi and ysi – distances from the least h

compressed point to the gravity centers of the ith concrete segment and the jth reinforcement rod in direction normal to the neutral axis, h – see Drawing 3.39,б. 5. Formulae (3.144) and (3.145) are used to determine internal forces Mx.ult and My,ult. If these two moments turn to be greater or smaller than the corresponding external moments mх and mу in relation to the same axes, then the section durability is deemed to be ensured or not ensured. If one of the moments (for instance, My,ult) is less than the corresponding external moment (i.е. My,ult < mу) and the other one is greater (i.е. Mx,ult > Mx), another inclination angle of the neutral axis θ is set (greater than the one taken previously), and a similar calculation is carried out again. AT the impact of tensile force or its absence, calculation can be carried out in a similar way. At calculation of concrete elements with consideration of tensioned concrete behavior, values εb2 are replaced with εbt2, а εb,ult with εbt,ult (see paras 3.72 and 3.75). ELEMENTS EXPOSED TO BENDING WITH TORSION RECTANGULAR SECTION ELEMENTS

Calculation OF COMBINED IMPACT OF TORqUE MOMENT AND MOMENT OF FLECTION 3.77. Calculation OF THE ELEMENT durability between spatial sections for the impact of torque moment T is carried out based on condition


T ≤ 0,1Rbb2h, (3.152) where b and h – smaller and greater dimensions of the element crosssection, respectively 3.78. at combined action of torque and bending moments, one analyzes a spatial section with a compressed side at the element face perpendicular ti the plane of moments of flection (drawing 3.40). Calculation of this section is performed based ont condition 2

T ≤ T0

C

⎛ M ⎞ ⎟⎟ , 1 − ⎜⎜ ⎝ M0 ⎠

(3.153)

C/2

C/2

q

SW1

C

SW

C 2 SW

C = δC 1

C 2 SW

h

R

S

A

S1

SW

A1 S

A

SW1

b

drawing 3.40. schematic representation of forces in spatial sections at calculation of the impact of torque and bending moments; tensioned reinforcement at he lower boundary of the element

where M0 T0

-the limit moment of flection supported by the normal section and determined according to para 3.14; - the limit torque moment supported by the spatial section and determined from formula T0 = Tsw1 + Ts1 , (3.154)


Tsw1 - torque moment supported by crosswise reinforcement located at the tensioned face within the spatial section, equal to Tsw1= 0,9qsw1δ1ch, (3.155) Ts1 - torque moment supported by longitudinal tensioned reinforcement of the spatial section, equal to b c

Ts1 = 0,9RsAs1 h.

(3.156)

in formulae (3.155) and (3.156): c – projection length of the compressed side of the spatial section to the longitudinal axis of the element;

δ1 =

b ; 2h + b

(3.157)

b и h - width of the face tensioned from the bending and the face perpendicular to it, respectively (see drawing 3.40); qsw1 = Rsw

Asw1 ; sw

(3.158)

Asw1 and sw – cross-sectional area if one crosswise rod at the tensioned from the bending face with width b and spacing of these rods; As1 – cross-sectional area of longitudinal reinforcement AT the face tensioned from bending, with width b. value RsAs1 in formula (3.156) is takenno greater than value 2qsw1b, and value qsw1 in formula (3.155) is takenno greater than value 1,5RsAs1/b. torque moment T and moment of flection M in condition(3.153) are takenin cross-section located in the middle of projection length c along the longitudinal axis of the element. in general case, Calculation is performed for spatial sections with various values c, takenno greater than 2h + b and no greater than b

2

δ1

.

spatial sections are recommended to locate in the following way: а) for whole beams or beams built into supports (cross beams) as well as for cantilevers, the spatil section is located at the support; б) for any elements loaded with spot loads and torque moments, the spatial sections are located at the force application points from the side of the area with large torque moments (drawing 3.41). For these cases, at calculation of the limit torque moment T0, in formulae (3.155) and (3.156) it is recommended to use the projection length value c=c0 corresponding t the minimum value T0, but at that, moments T and M are determined based on the projection length being equal to c=(1,2


-0,4Mmax/M0)c0 but no greater than c0, where Mmax –maximum moment of flection in the beginning of the spation section. value c0 is determined from formula Rs As1 (2h + b ) , q sw1

c0 =

(3.159)

with fulfillment of the said constraints on values consideration RsAs1 and qsw1. F1

C

C

1

2 T

T

M1

F2

C/2

M2

C/2 Эпюра M

T1 T2

Эпюра T

Q2

Эпюра Q

Q1

Drawing 3.41. location of design spatial sections in a beam loaded with spot loads 1, 2 – dsign spatial sectionя; M1,T1, Q1 – design forces for spatial section 1; M2, T2, Q2 – same for spatial section 2

in particular: - at RsAs1 ≥ 2qsw1b formula (3.159) takes the form c0=b

2

δ1

, and then

T0 =1,8qsw1bh 2δ 1 ; (3.160) - at RsAs1<2qsw1b<3RsAs value T0 after substituting c = c0 equals T0 =1,8bh

Rs As1q sw1 ; 2h + b

(3.161)

в) for elements loaded with uniformly distributed load, if there si a torque moment in a span section with the largest moment of flection, the middle of projection c is located in the said cross-


section; in this case, the most unfavorable value c is takenequal to c0. Calculation of combined impact of moment of flection and transverse force

3.79. durability Calculation of an element between spatioal section to the impact of a torque moment T and a transverse force Q is carried out based on condition ⎛

T ≤ T01 ⎜⎜1 − ⎝

Q ⎞ ⎟, Q01 ⎟⎠

(3.162)

where T01 – limit torque moment supported bu the element between spatial sections and takenequal to the right part of condition (3.152); Q01- limit transverse force supported by the concrete between inclined sections and takenequal to the right part of condition (3.43) para 3.30. forces T and Q are takenin normal section at distance 2b + h from the support where b and h – see drawing 3.40. 3.80. at combined action of a torque moment and a transverse force one shall analyze a spatial section with a compressed side at the element face which is parallel to the plane of bending moment (i.е. with a width h, drawing 3.42) Calculation of such section is carried out based on condition ⎛

Q ⎞

⎝

0

T ≤ T0 ⎜⎜1 − ⎟⎟ , Q ⎠

(3.163)

where Q0 – limit transverse force supported by inclined section and takenequal to the right part of condition (3.44) para 3.31; T0 - limit torque moment supported by spatial section and determnied from formula T0 = Tsw2 + Ts2; (3.164) Tsw2 – torque moment supported by crosswise tensioned reinforcement located at one of the faces with width h within the spatial section, equal to (3.165) Tsw2 = 0,9qsw2δ2cb; Ts2 – torque moment supported by longitudinal reinforcementоlocated at one of the faces with width h, equal h c

Ts2 = 0,9 Rs As 2 b . in formulae (3.165) and (3.166):

(3.166)


c – projection length of a compressed side of the psatial section to the longitudinal axis of the element h ; 2b + h R A qsw2 = sw sw2 ; sw

δ2 =

(3.167) (3.168)

Asw2 and sw – cross-sectional area of one crosswise rod at the face

C C/2 C/2

h

SW

q2c

SW

Rs A s2

CSW / 2

A

CSW / 2

C = Cδ SW

2

S2

A

SW2

b

Drawing 3.42.Shcematic representation of forces in spatial sections at calculation of the impact of torque moment and transverse force; tensioned reinforcement and lateral face of the element with width h , and spacing of these rods; As2 - cross-sectional area of longitudinal tensioned reinforcement located at one of the faces with width h. value RsAs2 in formula (3.166) is takenno greater than value 2qsw2h, and value qsw2 in formula (3.165) is takenno greater than value 1,5RsAs2/h. torque moment T and transverse force Q in condiition (3.163) is determined in normal section located in the middle of projection length c along the longitudinal axis of the element. In general case, Calculation is performed for spatial sections with various values of c takenno greater than 2b + h and no greater than h

2

δ2

.


value c in formulae (3.165) and (3.166) is permitted to be calculated from formula c=

Rs As 2 (2b + h ) q sw2

(3.169)

with fulfillment of the said constraints on consideration of values RsAs2 and qsw2; at that, values Q and Q0 are determined according to para 3.31 with account for values c takenaccording to para 3.32, and force T is takenas maximum one at the isegment in view. At using value c сorresponding to formula (3.169), value T0 can be calculated from formulae: T0 = 1,8qsw2 bh 2δ 2 ; at RsAs2 ≥ 2qsw2h at RsAs2 < 2qsw2h ≤ 3RsAs2 at 2qsw2>3RsAs2

R s As 2 q sw2 ; 2b + h T0=1,8Rs As2b 1,5δ 2 .

T0 = 1,8bh

Calculation examples Example 38. Given: an end frame flooring joist of a multistoreyed industrial building loaded with uniformly distributed load q = 154,4 kN/м and uniformly distributed torque moments t = 34,28 kN.м/м; cross-section of the fllooring joist at the support – see drawing 3.43,a; torque moments diagram from vertical loads – see drawing 3.43,б; bending moments and transverse forces diagrams from the most unfavorable combination of vertical loads and wind load for the support section – see drawing 3.43,в and г; bending moments diagram from the most unfavorable combination of vertical loads for the span section – see drawing 3.43,д; concrete of class B25 (Rb=14,5 MPа, Rbt=1,05 MPа), longitudinal and crosswise reinforcement of class A400 (Rs=355 MPa, Rsw=285 MPa).

MANUAL to CP 52-101-2003 concrete and REinforced concrete structures without prior reinforcement stress Эпюра T ,kH м T

б)

a)

20 300 20 50 50

84

∅14 100 мм

84

2,71

60

2,45 3 32

в)

Эпюра M,kH м 490

100

800

155

2,45

250

2,98

68 50

50 300

300

45 50

2 20

Mmax

г) Эпюра Q ,kH

460

2 22 д) Эпюра M ,kH м 245

49 2,71

Mmax = 321

Drawing 3.43. To example 38

It is required to check the element durability to the impact of torque and bending moments as well as to combined impact of torque moments and transverse forces. C a l c u l a t i o n. the section is viewed as a rectangular one, not considering the flooring joist flane “in reserve”. dimensions of this section are taken equal to b= 300 mm, h= 800 mm. Calculations are performed according to paras 3.77-3.80. verifying condition (3.152) for the impact of maximum torque moment T = 84 kN.м. 0,1Rbb2h = 0,1 ⋅ 14,5 ⋅ 300 2 ⋅ 800 = 104,4 ⋅ 10 6 Н.mm = 104,4 kN.м > T = 84 kN.м, i.е. condition is met. now one should check the spatial section with a compressed side at the lower face located ta the support section to the cobimned impact of torque and bending moments based on condition (3.153). determining according to para 3.19 the limit bending moment. from drawing 3.43,а one can find: As1 = 2413 mm2(3∅32), A' s1 = 1388 mm2 (2∅20+2∅22), a' = 68 mm; h0 = 800 – 60 = 740 mm. from formula (3.16) one has x=

Rs As1 − Rsc A' s1 355(2413 − 1388) = =83,6mm < ξRh0 = 0,531 ⋅ 720 = Rb b 14,5 ⋅ 300

382,3mm.


Then M0 = Rbbx(h0 - 0,5x) + Rsc As' (h0 − a') = 14,5 ⋅ 300 ⋅ 83,6(740 − 0,5 ⋅ 83,6) + + 355 ⋅ 1388(740 − 68) = 585 ⋅ 10 6 Н.mm. Now one can determine the limit torque moment T0. horizontal crosswise rods according to drawing 3.43,a ∅14 and spaced at sw = 100 mm. then Rsw Asw1 285 ⋅ 154 = = 439 H/mm. sw 100 b 300 δ1 = = = 0,158. 2h + b 2 ⋅ 800 + 300 as RsAs1 = 355 ⋅ 2413 = 856620 n > 2qsw1b = 2 ⋅ 439 ⋅ 300 = =263400 n,

qsw1 =

value T0 is determined from formula (3.160) T0 = 1,8qsw1bh 2δ 1 = 1,8 ⋅ 439 ⋅ 300 ⋅ 800 2 ⋅ 0,158 = 106,6 ⋅ 10 6 n.mm = =106,6 kN.м, ⎛

M

⎞

аnd moments M and T are determined at с = ⎜⎜1,2 − 0,4 max ⎟⎟c0 = M0 ⎠ ⎝ ⎛

⎞ 2 490 ⎞ 2 ⎟⎟b = ⎛⎜1,2 − 0,4 = 923 mm = 0,923 m, ⎟300 585 0 , 158 δ ⎝ ⎠ 1 ⎝ ⎠ 154,4 ⋅ 0,923 2 0,923 qc 2 = 490 - 460 + = 294,2 i.e. m = Mоп – Qопс/2 + 8 8 2

= ⎜⎜1,2 − 0,4

M îï Ì 0

kN.м, T = Tоп - t

0,923 c = 84 – 34,3 = 68,2 kN.м. 2 2

verifying condition (3.153): ⎛ M ⎞ ⎟⎟ T0 1 − ⎜⎜ M ⎝ 0⎠

2

294,2 ⎞ = 106,6 1 − ⎛⎜ ⎟ ⎝ 585 ⎠

2

= 92,1 kN.м > T = 68,2 kN.м,

i.e. durability with regard to the combined impact of bending and torque moments at the support is ensured. now one can check the durability of spatial section with a compressed side at a lateral face, locating this section at the support. first one can verify condiiton (3.162), taking, according to the aforesaid, T01 = 104,4 kN.м and, having calculated from condition (3.43) Q01 = 0,3Rbbh0 = 0,3 ⋅ 14,5 ⋅ 300 ⋅ 740 = 965700 Н = 965,7 kN. values T and Q оare determined in the section at distance a = 2b + h = 2 ⋅ 300 + 800 = 1400 mm = 1,4 m from the support, i.e T = Tоп – ta = 84 – 34,3 ⋅ 1,4 = 36 kN.м;


Q = Qоп – qa = 460 – 154,4 ⋅ 1,4 = 243,8 kN.м. Тhen ⎛

T01 ⎜⎜1 − ⎝

Q ⎞ ⎛ 243,8 ⎞ ⎟⎟ = 104,4 ⎜1 − ⎟ = 78 kN.м > T = 36,0 kN.м, Q01 ⎠ ⎝ 965,7 ⎠

i.e condition (3.162) is met. from drawing 3.43,а one can find As2 = 804 + 314 +380 = 1498 mm2 (∅32 + ∅20 + ∅22). spacing and diameter of vertical stirrups are the same as for horizontal rods, so qsw2 = qsw1 = 439 n/mm. as RsAs2 = 355 ⋅ 1498 = 531790 n < 2qsw2h = 2 ⋅ 439 ⋅ 800 = 702400 n, value T0 equals T0 = 1,8bh

Rs As 2 q sw2 531790 ⋅ 439 = 1,8 ⋅ 300 ⋅ 800 = 176,4 ⋅ 10 6 n.mm 2 ⋅ 300 + 800 2b + h

= 176,4 kN.м. determining acording to para 3.31 value Q and value Q0 as the right part of condition (3.44). Mb = 1,5Rbtbh02 = 1,5 ⋅1,05 ⋅ 300 ⋅ 7402 = 258,7 ⋅106 n.mm. with two-part stirrups qsw = 2 qsw2 = 2 ⋅ 439 = 878 n/mm. one can determine the most unfavorable value c according to para 3.32, taking q1 = 100 kN/м. as с=

Mb = 0,75qsw + q1

878 qsw = = 2,787 > 2, value с equals Rbt b 1,05 ⋅ 300

258,7 ⋅106 = 584 mm. 0,75 ⋅ 878 + 100

taking с0 = с = 584 mm < 2h0, one has Mb 258,7 ⋅106 + 0,75qswс0 = + 0,75 ⋅ 878 ⋅ 584 = 827550 n; Q0 = c 584 Q = Qоп – q1с/2 = 460 - 100 ⋅ 0,584 / 2 = 430,8 kN; T = Tоп – tс/2 = 84 − 34,3 ⋅ 0,584 / 2 = 74,0 kN.м.

verifying condition (3.163) ⎛

T0 ⎜⎜1 − ⎝

Q⎞ ⎛ 430,8 ⎞ . . ⎟⎟ = 176,4 ⎜1 − ⎟ = 90,8 kN м > T = 74 kN м, Q0 ⎠ ⎝ 827,5 ⎠

i.e. durability at combined impact of bending moments and transverse forces is ensured. As one can see from drawings 3.43,б and д, in normal section with the largest span bending moment there is a torque moment, so one should check the spatial section which has the middle of its projection located in


this normal section, to the impact of moments M = 321 kN.м and T = =84

2,71 − 2,45 = 8,9 kN.м 2,45

At that, the tensioned side of the spatial section is located at the lower face. now one can determine the limit bending moment M0. for this part of the flooring joist the intermediate upper rod ∅ 32 is broken off, therefore according to drawing 3.43,a one has As′1 = 1609 mm2(2∅32); a′ = 60 mm; As1 = 1388 mm2 (2∅20+2∅22); a = 68 mm; h0 = 800 – 68 =732 mm. height of the compressed zone equals 355(1388 − 1609 ) Rs As1 − Rsc As' 1 = < 0, x= Rb b 14,5 ⋅ 300

therefore, value m0 is determined from formula (3.19): М0 = Rs As1 (h0 − a') = 355 ⋅1388(732 − 60) = 331,1⋅106 n.mm = 331,1 kN.м horizontal crosswise rods ∅14 in this part of the flooring joist have the spacing \ sw = 200 mm; hence 285 ⋅154 Rsw1 Asw = = 219,5 n/mm. sw 200 as Rs As1= 355⋅1388 = 492740 n > 2qsw1b = 2 ⋅ 219,5 ⋅ 300 =

qsw1 =

=131700 n, value T0 is determined from formula (3.160): T0 = 1,8qsw1bh 2δ1 = 1,8 ⋅ 219,5 ⋅ 300 ⋅ 800 2 ⋅ 0,158 = 53,3 ⋅106 n.mm = =53,3 kN.м. verifying condition (3.153) 2

2

⎛M ⎞ ⎛ 321 ⎞ . . ⎟⎟ = 53,3 1 − ⎜ T0 1 − ⎜⎜ ⎟ = 13,06 kN м > T = 8,9 kN м, ⎝ 331,1 ⎠ ⎝ M0 ⎠

i.e. durability of this section is ensured.

CALCULATION OF REINFORCED CONCRETE ELEMENTS FOR THE LOCAL IMPACT OF LOADS CALCULATION OF LOCAL COMPRESSION 3.81. Calculation of local compression (bearing) for elements in the absence of lateral reinforcement is carried out based on condition (3.170) N ≤ψRb,locAb,loc,


where N – local compressing force from an external load; Ab,loc – area of application of compressing force (bearing area); ψ − coefficient taken equal to: at unifromaly distributed local load over the bearing area –1,0; at non-uniformly distributed local load ver the bearing area (under the ends of beams, girders, joists etc,) – 0,75; Rb,loc – design compression strength of concrete at the local load impact determined from formula (3.171) Rb,loc =ϕbRb where

ϕb = 0,8

Ab, max , Ab,loc

(3.172)

but no less than 2,5 and no less than 1,0; Ab,max – maximum design area set according to the following rules: - gravity centers of areas Ab,loc and Ab,max coincide; - boundaries of design area Ab,max are removed from each oh the area Ab,loc sides to the distance equal to the dimension of these sides (drawing 3.44);


- if there are several loads, design areas are limited with lines passing through the center 2

4 б)

a2

a1 2

a2 2

e)

4

4

5

3 a1

a2

5

a2

3

a2

3

д)

1

3

a2

a1 a2

4

a1

a2

2

2 4

c

a1 г)

1

a2

a2

3

a2

в) 2 4

a1

a2

a2

ж)

2

a2

a2 l1 /2

l2 /2

a2

4 3

2 a2

2

a1

c
a)

3

a1

1

a1

l
a1

l
a1

drawing 3.44. schematic for calculation of local compresseion for elements, at location of the local load а – far from the element edges; б – over the whole width of the element; в –at the edge (abutment) of the element over the whole of its width; г –at the corner of the element; д –at one edge of the element; e – close to one edge of the element; ж –if there are several loads 1 – element exposed to the local load; 2 –bearing area Ab,loc; 3 –maximum designarea Ab,max; 4 – gravity center of areas Ab,loc and Ab,max; 5 –minimum zone of reinforcement with meshes, with lateral reinforcement being considered in calculation

- of distances between the application points of two adjacent loads (drawing 3.44, ж). note. at local loads from beams, girders and other elements exposed to bending, the support depth taken into account during the calculation at determining Ab,loc and Ab,max is taken no greater than 20 см.


3.82. Calculation of local compression for elements with lateral reinforcement in the form of welded meshes is carried out based on condition (3.173) N≤ψ Rbs,loc Ab,loc, where Rbs,loc – transformed with consideration of lateral reinforcement in the local compression zone design compression strength of the concrete determined from formula (3.174) Rbs,loc = Rb,loc + 2ϕs,xyRs,xyμs,xy. here:

ϕs,xy =

Ab.loc,ef Ab,loc

;

(3.175)

Ab,loc,ef – area enclosed by the contour of the meshes of lateral reinforcement counting by their extreme rods and taken in formula (3.175) no greater than Ab,max; Rs,xy - design tensile strength of lateral reinforcement; μs,xy – reinforcement factor determined from formula

μs,xy =

n x Asx l x + n y Asy l y Ab,loc,ef s

;

(3.176)

nx , Asx , lx – number of rods, cross-sectional area and rod length, counting in the axes of extreme rods in direction x; ny, Asy, ly – same in direction y: s – spacing of meshes of lateral reinforcement. values Rb,loc, Ab,loc,ψ and N are taken accoridng to para 3.81. values of local compressing force supported by the element with lateral reinforcement (right part of condition 3.173) is taken no greater than double value of local compressing force supported by the element wihtout lateral reinforcement (right part of condition 3.170). meshes of lateral reinforcement are located within design area Ab,max. at that, for schematics of drawings 3.44 в and г meshes of lateral reinforcement are located over the area with dimensions in each direction no less than the total of two mutually perpendicular sides. If the loading area is located at the edge of the element (see drawing 3.44,б-д,ж), at determining values Ab,loc and Ab,loc,ef the area occupied by the protective layer of concrete for the extreme rods of meshes is not taken into consideration. The meshes are located: - at the element thickness being greater than the double greater dimension of the leadng area – within the double dimension of the loading area;


- at the element thickness being less than the double greater dimension of the loading area – within the element thickness. - meshes of lateral reinforcement shall meet structual requirements specified in para 5.27. Calculation examples Example 39. Given: steel stanchion supported by the footing and centrally loadedwith force N=1000 kN (drawing 3.45); footing made of concrete of class B10 (Rb = 6,0 MPa). N =1000 kH

1-1

C1

200

ly=100×7=700

200 300 200

c=200

b1

b =800

b2

1

lx=100×6=600

100×6=600

1

c =200

C1

Drawing 3.45. To calculation example 39

It is required to check the concrete durability under the stanchion for the local compression. C a l c u l a t i o n is carried out in accordance with paras 3.81 and 3.82. design area Ab,max оis determined in accordance with drawing 3.44, e. According to drawing 3.45, one has с = 200mm < a1 = 300 mm; b1 = 200 ⋅ 2 + 200 = 600 mm; b2 = 200 ⋅ 2 + 300 = 700mm; Ab,max = b1b2 = 600 ⋅ 700 = =420000mm2: bearing surface equals Ab,loc = 300 ⋅ 200 = 60000 mm2. Coefficient ϕb equals

ϕb = 0,8

Ab,max Ab,loc

= 0,8

420000 = 2,12. 60000


then Rb,loc = ϕbRb = 2,12 ⋅ 6,0 = 12,72 MPa. verifying condition (3.170), taking ψ = 1,0 as with uniform distribution of local load: ψ Rb,loc Ab,loc =1 ⋅12,72 ⋅ 60000 = 763200 N = 763,2 kN < N =1000 kN, i.е. local compression strength of the concrete is not provided for, so it is necessary to apply lateral reinforcement in the form of meshes made of reinforcement of class B500 with diameter 4 mm, cells 100 × 100 mm and height spacing s = 100 mm (Rs,xy = 415 MPa). checking the durability according to para 3.82. determining coefficient of lateral reinforcement from formula (3.176). from drawing 3.45 one has: nx=8, lx = 600 mm; ny = 7; ly = 700 mm; Asx = Asy = 12,6 mm2(∅4); Ab,loc,ef = 600 × 700 = 420000 mm2; then

μs,xy =

n x Asx l x + n y Asy l y Ab,loc,ef s

=

8 ⋅ 12,6 ⋅ 600 + 7 ⋅ 12,6 ⋅ 700 = 0,00291. 420000 ⋅ 100

Coefficient ϕs,xy equals

ϕs,xy =

Ab,loc,ef Ab,loc

=

420000 = 2,65. 60000

transformed design strength of concrete Rbs,loc is determined from formulaе (3.174) Rbs,loc = Rb,loc +2ϕs,xyRs,xyμs,xy = 12,72 + 2 ⋅ 2,65 ⋅ 415 ⋅ 0,00291 = 19,12 MPa. verifying condiiton ( 3.173) ψRbs,locAb,loc = 1,0 ⋅19,12 ⋅ 60000 = 1147200 n = 1147,2 kN > N = 1000 kN, i.е. durability of the concrete is ensured. the meshes are installed into the depth of 2 ⋅ 300 = 600mm. CALCULATION OF THE ELEMENTS FOR PUSHING General provisions 3.83. Calculation OF PUSHING OF THE ELEMENTS IS CARRIED OUT FOR FLAT REINFORCED CONCRETE ELEMENTS (SLABS) AT THE IMPACT (NORMAL TO THE ELEMENT PLANE) OF LOCAL FOCUS-APPLIED FORCES – A CONCENTRATED FORCE AND A MOMENT OF FLECTION.


AT PERFORMING calculation OF PUSHING ONE ANALYZES A DESIGN CROSS-SECTION LOCATED AROUND THE AREA OF FORCE TRANSMISSION TO THE ELEMENT AT DISTANCE h0 /2 NORMAL TO ITS longitudinal AXIS, WITH TANGENT SHEARING FORCES FROM THE CONCENTRATED FORCE AND THE MOMENT OF FLECTION ACTING ON ITS SURFACE. THE ACTING TANGENT SHEARING FORCES OVER THE AREA OF THE DESIGN CROSS-SECTION SHALL BE SUPPORTED BY THE CONCRETE WITH TENSILE STRENGTH OF CONCRETE Rbt AND THE CROSSWISE REINFORCEMENT LOCATED AT BOTH SIDES FROM THE DESIGN CROSS-SECTION AT DISTANCE h0 /2 WITH TENSILE STRENGTH OF REINFORCEMENT Rsw. Design contour of the cross-section is taken: at location of the load transmission zone inside the flat element – a closed one and located around the load transmission area (Drawing 3.46,а), at location of the load transmission zone at a free edge or a corner of the flat element in the form of two options: a closed one and located around the load transmission area and a non-closed one, going from an edge of the flat element (Drawing 3.46,б,в), in this case one should take into account the least carrying capacity out of two location options of the design contour if the crosssection. Y

Y1

h 0 /2 x0

5 в)

Lx

Y

2'

3 e0x

4 3 h 0 /2

X(X1 )

1 Lx

Y1

1 4

X(X1)

Y (Y1 )

h 0 /2 h 0 /2

3

Ly

2'

Ly

h 0 /2

2

e0y

1 4

X X1

y0

а)

Ly

5

h 0 /2

б)

x0

Lx

e0

Drawing 3.46. Schematic representation of design contours of the cross-section at pushing:


а –load application zone inside the flat element; б, в -same,at the edge of a flat element; 1 – load application area; 2 –design contour of the cross-section; 2'-second option of design contour location; 3 –gravity center of design contour ( intersection point of axes X1 and Y1); 4 – gravity center of the load application zone (intersection point of axes X and Y); 5 -boundary (edge) of the flat element.

With moment mloc in the application zone of the concentrated load, a half of this moment is taken into account at calculation of pushing, and the other one is taken into account while calculating of normal sections with width including the width of the load transmission zone and depth of section of the flat element to the both sides of the load transmission zone. While concentrated moments and force act under strength conditions, the ratio between the acting concentrated moments m considered at pushing and the limit ones Mult is taken no greater than the ratio between the acting concentrated force F and the limit ones Fult. Calculation of pushing for elements without crosswise reinforcement 3.84. Calculation of pushing for elements without crosswise reinforcement exposed to a concentrated force is carried out based on condition (3.177) F ≤ Rbtuh0 , where F – concentrated force from external load; u − perimeter of the design cross-section contour located at distance 0,5h0 from the boundary of the concentrated force F bearing surface(Drawing 3.47); h0 - effective depth of the element equal to normal value of effective depths for longitudinal reinforcement in directions of axes x and y. At dimensions of rectangular bearing surface a×b u=2(a+b+2h0). If the bearing surface is located close to a free end of the slab, in addition to the said calculation (if the cross-sectional contour does not fall beyond the free end of the slab) one shall check durability of the non-closed design cross-section (see Drawing 3.46, в) to the impact of eccentrically applied concentrated force in relation to the gravity center of the crosssectional contour based on condition


F≤

Rbt h0 , 1 e0 y + u I

(3.178)

u – length of the non-closed design section contour, equal to u = 2 Lx + L y , (3.179) I – moment of inertia of the design section, equal to L3 x 2(Lx + L y ) + Lx L y I= ; 3 u2 2

(3.180)

y – distance from the gravity center of the design section contour to the fiber in view, equal to y=

(

Lx Lx + L y u

) - for a fiber at a free edge of the slab;

L2x y= - for a fiber at the opposite edge of the slab; u

e0 - eccentricity of the concentrated force as related to the gravity center of the design section contour, equal to e0 =

Lx ( Lx + L y ) u

(3.181)

− x0 ;

х0 – distance of the concentrated force application point from a free edge of the slab; Lx и Ly – design cross-section contour dimensions, Ly – dimension parallel to a free edge of the slab. Mloc h 0 /2

F

h 0 /2

h 0 /2

h0

1

b

X 3

h 0 /2

2 a Y

Drawing 3.47. Schematic for calculation of pushing for reinforced concrete elements without crosswise reinforcement 1 – design cross-section; 2 – contour of the design cross-section; 3 – contour of the load application zone.


Concentrated force F is taken less loads applied to the opposite edge of the slab within the area with dimensions exceeding those of bearing surface by h0 in all directions. 3.85. Calculation of pushing for elements without crosswise reinforcement at combined impact of concentrated forces and a moment of flection is performed based on condition F M + ≤ Rbt h0 ; u Wb

(3.182)

where ratio M/Wb is taken no greater than F/u; Wb – stiffness moment of contour of the design cross-section; F, u, h0 – see para 3.84. Concentrated moment m taken into account in condition (3.182) equals to a half of the concentrated moment from external load mloc. In a reinforced concrete building frame with flat flooring slabs the moment mloc equals to the total moment of flection in sections of upper and lower columns abutting the slab in the node in view, and force F is directed upward. If the bearing surface is located close to a free edge of the slab, when the concentrated force is applied eccentrically in relation to the contour of a non-closed design cross-section, a moment from eccentrical force application (with its own sign) shall be added to the moment m in condition (3.182), being equal to F⋅e0, where e0 – see formula (3.181). With a rectangular bearing surface and closed contour of design cross-section, value Wb is determined from formula ⎛ a + h0 ⎞ Wb = (a + h0 )⎜ + b + h0 ⎟; ⎝ 3 ⎠

(3.183)

where а and b – dimensions of bearing surface in the direction of the moment impact and the direction normal to the moment impact, respectively. With a non-closed contour of the design cross-section (see Drawing 3.46,в) value Wb is taken equal to Wb = I/y, where I and y – see para 3.84. At additional moment My in the direction normal to the direction of moment M, the left part of condition (3.182) increases by

My

Wb, y

where Wb,y –

stiffness moment of the contour of design section in the direction of moment My; at that, the total

My M is also taken no greater than F/u. + Wb Wb, y


Calculation of pushing for elements with crosswise reinforcement 3.86. Calculation of pushing for elements with crosswise reinforcement at the impact of a concentrated force (Drawing 3.48) is performed based on condition F ≤ Fb,ult + Fsw,ult , (3.184) where Fb,ult – right part of condition (3.177); Fsw,ult – limit force supported by crosswise reinforcement at pushing and equal to Fsw,ult = 0,8q sw u , (3.185) but taken no greater than Fb,ult, , where qsw – force in crosswise reinforcement per unit of length of the contour of design cross-section, at uniform distribution of crosswise reinforcement equal to q sw =

Rsw Asw , sw

(3.186)

Asw – section area of crosswise reinforcement with spacing sw located within distance 0,5h0 at both sides from the contour of the design cross-section (see Drawing 3.48) sw – spacing of crosswise rods in the direction of the contour of crosssection With uniform distribution of crosswise reinforcement along the contour of the design cross-section, value u is taken as for concrete design cross-section according to para 3.84. With crosswise reinforcement being concentrated at the axes of bearing surface (cruciform location of crosswise reinforcement, Drawing 3.49), the contour perimeter u for crosswise reinforcement is taken according to actual lengths of the crosswise reinforcement location area Lsw,x and Lsw,y on design pushing contour [i.е. u=2(Lsw,x+Lsw,y)]. Crosswise reinforcement is taken into account in calculation with Fsw,ult being no less than 0,25Fb,ult. Beyond the location of crosswise reinforcement, the calculation of pushing is carried out according to para 3.48, viewing the contour of design cross-section at distance 0,5h0 from the boundary of crosswise reinforcement location. At concentrated location of crosswise reinforcement at the axes of bearing surface, in addition to this, the design contour of the concrete cross-


section is taken along diagonal lines going from the edge of crosswise reinforcement location (see Drawing 3.49). Crosswise reinforcement shall meet structural requirements specified in para 5.26. 3.87. Calculation of pushing for elements with crosswise reinforcement at combined impact of concentrated force and moment of flection (see Drawing 3.48) is carried out based on condition Fbt ,ult

F M + ≤ 1, + Fsw,ult M b,ult + M sw,ult

(3.187)

where ratio M/(Mb,ult + Msw,ult) is taken no greater than F/(Fbt,ult + Fsw,ult);


M F h 0 /2

h

0

h 0 /2

1 b

h 0 /2

h 0 /2

S w < b/4

2

S w < a/4

>h0 /3
4 Asw

>h0 /3
X

a

a1

5

h 0 /2 h 0 /2

>1,5h0

b1

3

>h0 /3

>h0 /3

Asw

>1,5h0

h 0 /2 3

h 0 /2

Y

Drawing 3.48 Schematic for calculation of pushing for reinforced concrete slabs with vertical uniformly distributed crosswise reinforcement 1 – design cross-section; 2 – contour of the design cross-section


3 – boundaries of the zone within which crosswise reinforcement is taken into account during calculation; 4 – contour of the design cross-section without consideration of crosswise reinforcement in the calculation; 5 – contour of the load application zone. L = 2s SW,X

s

W

s

W

W

3

2

h0 /2 Asw

s

W

Asw

h 0/2

s

W

Lsw,y=2sw

1

h0 /2

h 0/2

Drawing 3.49. Schematic of the design contour of the cross-section during pushing and with cruciform location of crosswise reinforcement 1 – load application area; 2 – contour of the design cross-section at consideration of crosswise reinforcement Asw; 3 – contour of the design cross-section without consideration of crosswise reinforcement

Fbt,ult and Fsw,ult – see para 3.86; Mb,ult – limit concentrated moment supported by the concrete in the design cross-section, equal to M b,ult = Rbt Wb h0

Wb – see para 3.85; M – see para 3.85; Msw,ult – limit concentrated moment supported by crosswise reinforcement in the design cross-section, equal to Msw,ult = 0,8qswWsw; but taken no greater than Mb,ult; qsw – see para 3.86 ; Wsw – stiffness moment of the contour of crosswise reinforcement.


At uniform location of crosswise reinforcement along the contour of design cross-section, value Wsw is taken equal to Wb. At concentrated location of crosswise reinforcement at the axes of bearing surface (Drawing 3.49), the stiffness moment Wsw is determined according to the same rules as for the stiffness moment Wb, taking actual lengths of crosswise reinforcement location areas Lsw,x and Lsw,y at the design contour of pushing. At uniform location of crosswise reinforcement around the bearing surface, instead of condition (3.187) one can make use of condition (3.182) with increasing the right part by value 0,8qsw taken no greater than Rbth0. With additional moment My in the direction normal to direction of moment M, the left part of condition (3.187) is increased by my =

My M by ,ult + M sw, y ,ult

, where Mby,ult and Msw,y,ult – limit concentrated

moments supported by the concrete and the crosswise reinforcement respectively in the design cross-section and in the direction of moment My; at that, the total

M b, ult

M F + m y is also taken no greater , Fbt ,ult + Fsw, y ,ult + M sw, ult

and Msw,y,ult no greater than Mby,ult. Calculation examples Example 40. Given: a flat monolithic flooring slab 220 mm thick; columns abutting the flooring at the top and at the bottom, with section 500 × 800 mm; a load transferred from the flooring to the column N = 800 kN; moments in column sections at the upper and the lower faces of the slab are equal: in the direction of the column dimension 500 mm – Mx,sup = 70 kN.m, Mx,inf = 60 kN.m, in the direction of the column dimension 800 mm – My,sup = 30 kN.m, My,inf = 27 kN.m; concrete of class B30 (Rbt = 1,15 MPa). It is required to check the flooring slab for pushing. C a l c u l a t i o n. Averaged effective depth of the slab is taken equal to h0 = 190 mm. The load of the flooring is taken as the concentrated pushing force F = N =800 kN; column section a × b = 500 × 800 mm is taken as the bearing area of this force. Now one can determine geometrical characteristics of the contour of design cross-section according to paras 3.84 and 3.85:


perimeter u = 2(a + b + 2h0 ) = 2(500 + 800 + 2 ⋅ 190 ) = 3360 mm; stiffness moment in the direction of moment Mx (i.е. at a = 500 mm, b = 800 mm) ⎛ 500 + 190 ⎞ ⎞ ⎛ a + h0 Wb,x = (a + h0 )⎜ + 800 + 190 ⎟ + b + h0 ⎟ = (500 + 190 )⎜ 3 ⎠ ⎝ 3 ⎝ ⎠ 2 = =841800 mm ; stiffness moment in the direction of moment My (i.е. at a = 800 mm, b= 500 mm) ⎛ 800 + 190 ⎞ Wb,y = (800 + 190 )⎜ + 500 + 190 ⎟ = 1009800 mm2. 3 ⎝ ⎠ For the design concentrated moment in each direction, one can take a half of the total of moments in the section at upper and lower faces of the slab, i.е. Mx = (M x ,sup + M x ,inf )/ 2 = (70 + 60 ) / 2 = 65 kN.m; My = (M y ,sup + M y ,inf )/ 2 = (30 + 27 ) / 2 = 28,5 kN.m. Verifying condition (3.182), taking M = Mx = 65 kN.m, Wb=Wb,x= = My 28,5 ⋅106 2 841800 mm and adding to the left part = = 28,2 N/mm. Wb, y 1009800 At

that

M y 65 ⋅ 105 Mx F 800 ⋅ 103 + = + 28,2 = 105,4 Í ìì < = = Wb, x Wb, y 841800 u 3360

= 238,1 Í ìì , therefore, the moment shall not be adjusted.

F Mx My + + = 238,1+105,4=343,5 N/mm > Rbth0 = 1,15 ⋅ 190 = u Wb, x Wb, y 218,5N/mm, i.е. condition (3.182) is not met and it is necessary to install crosswise reinforcement into the slab. According to requirements of para 5.26, the spacing of crosswise rods is taken sw = 60 mm < h0/3 = 63,3 mm, 1st row of the rods shall be placed at the distance of 75 mm from the column, as 75 mm < h0/2 and 75 mm> h0/3 (Drawing 3.50). Then within the limits at the distance of 0,5h0 = 95 mm to both sides from the design cross-section contour, 2 rods can be placed within one section. Rods made of reinforcement of class A240(Rsw = 170 MPa) of minimum diameter 6 mm are taken. Then Asw = 57 mm2 and 170 ⋅ 57 R A = 129,2 N/mm < Rbth0=218,5N/mm. 0,8qsw = 0,8 sw sw = 0,8 sw 60


At that, according to para 3.86, the limit force supported by the crosswise reinforcement and equal to 0,8qswu = 129,2u shall be no less than



0,25Fb,ult = 0,25Rbth0u = 0,25 ⋅ 218,5u = 54,6u. As one can see, this requirement is met. Verifying condition (3.182), adding values 0,8qs to the right part. My F Mx + + = 343,5 N/mm < Rbth0 + 0,8qsw = 218,5 + 129,2 = u Wb, x Wb, y =347,7 N/mm, i.е. durability of the design section with consideration of installed crosswise reinforcement is ensured. Now one can check durability of the design section with contour at distance 0,5h0 beyond the boundary of crosswise reinforcement location. According to requirements specified in 5.26, the last row of crosswise rods is placed at a distance from the load zone (i.е. from the column) equal to 75 + 4 ⋅ 60 = 315 mm >1,5h0 = 1,5 ⋅ 190 = 285 mm. Then the contour of a new design section has the dimensions: a = 500 + 2 ⋅ 315 + 190 = 1330 mm; b = 800 + 2 ⋅ 315 + 190 = 1620 mm. Its geometrical characteristics: u = 2(1320 + 1620 + 2 ⋅ 190) = 6640 mm; 1320 + 190 ⎞ 2 + 1620 + 190 ⎟ = 3,49 ⋅ 10 6 mm ; 3 ⎝ ⎠ 1620 + 190 ⎞ 2 + 1320 + 190 ⎟ = 3,825 ⋅ 10 6 mm . Wb,y = (1620 + 190)⎛⎜ 3 ⎝ ⎠

Wb,x = (1320 + 190)⎛⎜

Verifying condition (3.182) with consideration of moment My. At that, the decrease of the pushing force F is neglected “in reserve” due to the load located at an segment with dimensions (a + h0 ) × (b + h0 ) around the column. 800 ⋅ 10 3 65 ⋅ 10 6 28,5 ⋅ 10 6 + + = 146,5 N/mm < Rbth0 = 218,5 N/mm, 6 6 6640 3,49 ⋅ 10 3,825 ⋅ 10

i.е. durability of this cross-section is ensured. Example 41. Given: a flat monolithic flooring slab 230 mm thick; columns abutting the flooring at the top and at the bottom, with section 400 × 500 mm; a load transferred from the flooring onto the column N = 150 kN; moments in column sections at the upper and the lower slab faces in the direction of the column dimension 500 mm – Msup = 80 kN.m, Minf = 90 kN.m; the column section center is located at distance x0 = 500 mm from a free edge of the slab (Drawing 3.51); concrete of class B25 (Rbt = 1,05 MPa).


400

M sup 1

h0 =200

1-1

1

500

1 M int

3

500

2

F

Drawing 3.51. For calculation example 41 1 – force F application point; 2 –gravity center of a non-closed contour; 3 – non-closed contour of a design section

It is required to check the flooring slab for pushing. C a l c u l a t i o n. Averaged effective depth of the slab is taken equal to h0 = 200 mm. The load of the flooring is taken as the concentrated pushing force F = N =150 kN; column section a × b = 4 mm is taken as the bearing area of this force. Now one can check durability of the design sections of the non-closed contour. Dimensions of this contour are: Lx = x0 + (a+h0)/2 = 500+(500+200) / 2 = 850 mm; Ly = b+h0 = 400+200=600 mm. Perimeter and the moment of inertia of the contour are equal to u = 2Lx + Ly = 2 · 850+600 = 2300 mm; L3x 2(Lx + L y ) + Lx L y 8503 2·14502 + 850·600 I= = = 1,825·106 ìì 2 2 3 3 u 2300 2

3

.

Eccentricity of force F e0 =

(

Lx Lx + L y u

)− x

0

=

850·1450 − 500 = 35,9 ìì . 2300

With the taken directions of moments Msup and Mint (see Drawing 3.51), the most stressed fiber of the design section is located at the section edge farthest removed from the free end of the slab. This fiber is situated at the distance from the gravity center equal to 2 2 y = Lx u = 850 2300 = 314,1 ìì . Then the stiffness moment is equal to : Wb = I y = 1,825·106 314,1 = 581025 ìì 2 . Design moment from the column equals


M = Mloc /2 = (Msup + Minf)/2=(80+90)/2=85 kN.m. The moment of eccentrical application of force F equals Fе0 = =150·0,0359=5,4 kN.m. This moment is of the opposite sign to the moment Mloc, therefore М = 85-5,4=79,6 kN.m. Checking the durability based on condition (3.182) 150000 79,6·106 F M + = + = 202,2 N/mm < Rbth0 = 1,05·200=210 2300 581025 u Wb N/mm, i.е. durability of the section with a non-closed contour is ensured. Now one can check the section durability of a closed contour. Determining its geometrical characteristics: Perimeter u=2(a+b+2h0)=2(500+400+2·200)=2600mm; ⎛ a + h0 ⎞ + b + h0 ⎟ = Stiffness moment Wb = (a + h0 )⎜ ⎝ 3 ⎠ ⎛ 700 ⎞ = 700⎜ + 600 ⎟ = 58333 ìì 2 ; ⎝ 3 ⎠ The moment equals M=Mloc / 2 = 85 kN.m. 150000 85·106 F M = + = 203,4 N/mm < Rbth0 = 210 N/mm, i.е. + 2600 583333 u Wb the slab durability against pushing is ensured in all sections.

4. CALCULATION OF LIMIT STATES OF THE SECOND GROUP FOR ELEMENTS OF REINFORCED CONCRETE STRUCTURES CALCULATION OF CRACKS OPENING IN REINFORCED CONCRETE ELEMENTS GENERAL PROVISIONS 4.1. Calculation of short-term and long-term cracks opening can be carried out for reinforced concrete elements.

Short-term opening of cracks is determined as resulting from combined impact of constant and temporary (long-term and short-term) loads; long-term opening – only from constant and temporary long-term loads.


4.2. Calculation of cracks opening is carried out using condition (4.1) acre ≤ occult, where acre – width of cracks opening due to the impact of external load determined according to paras 4.10-4.14; acrс,ult – maximum allowable width of cracks opening. Values acrс,ult are taken equal to: a) from condition of reinforcement integrity (for any structures) 0,3 mm – at long-term opening of cracks; 0,4 mm – at short-term opening of cracks; б) from condition of limiting the permeability of structures exposed to direct pressure of liquids, gases, granular material) 0,2 mm – at long-term opening of cracks; 0,3 mm – at short-term opening of cracks. 4.3. Calculation of cracks opening is not carried out if the following condition is met М < Mcrс , (4.2) where М – moment from external load in relation to the axis normal to the plane of the moment and passing through the gravity center of transformed cross-section of the element; at that, all loads (constant and temporary ones) are considered with safety factor γf = 1; Mcrс – moment supported by normal section of the element during cracks formation and determined according to paras 4.4-4.8. For centrally tensioned elements condition (4.2) is transformed into condition N < Ncrс, (4.3) where Ncrc – longitudinal tensile force supported by the element during cracks formation and determined according to para 4.9 DETERMINING THE CRACKS FORMATION MOMENT 4.4. Moment of flection Mcrс during cracks formation is determined on the basis of deformational model with consideration of inelastic deformation of tensioned concrete according to para 4.7 and 4.8. It is permissible to determine the moment Mcrс without considering inelastic deformations of concrete according to paras 4.5 and 4.6. If at that condition (4.1) is not met, then the moment of cracks formation should be determined with consideration of inelastic deformations of concrete.


4.5. Moment of cracks formation without consideration of inelastic deformations of concrete is determined as for a solid elastic body from formula (4.4) Mcrc = Rbt , ser W ± Ne ÿ , where W – stiffness moment of transformed section for an extreme tensioned fiber of concrete: eя – distance form the gravity center of the transformed section of the element to the core point farthest removed from the tensioned area being checked for cracks formation. Values W and eя are determined according to para4.6. In formula (4.4) the “plus” sing is taken at compressing normal force N , the “minus” sign – at tensile force. 4.6. Stiffness moment W and distance eя are determined form formulae: I red , yt W , eя = Ared

W=

(4.5) (4.6)

where Ired – moment of inertia of transformed section in relation to its gravity center and determined from formula (4.7) Ired=I+Isα+Is′α ; I, Is, I s' – moments of inertia of the section of concrete and , tensioned and compressed reinforcement respectively; Ared – area of transformed section equal to (4.8) Ared = A + Asα + As′ α , α =

Es – modular ratio of reinforcement to concrete; Eb

yt – distance from the most tensioned fiber of concrete to the gravity center of transformed section of the element. At μ =

As < 0,005 values W and eя are permitted to be determined A

without consideration of reinforcement. 4.7. Determining of crack opening moment using nonlinear deformational model is performed using conditions specified in paras 3.723.75 (Diagram 4.1). At that one considers behavior of concrete in tensioned zone determined from a bilinear diagram according to para 3.72. Transformed moduluses of deformations of compressed and tensioned concrete in the bilinear diagram


a)

σb

εb εs

x

σs

Mcrc

yt

h

1

εs εb1,red

σs Rbt,ser

εbt2

yc

εs

Mcrc

σs 1

εb1,red εs εbt2

yt

yp

x

N

σb

εb

б)

σs Rbt,ser

Diagram 4.1. Schematic of tensile-deformational state of an element section for calculation of cracks formation at the impact of bending moment (а), bending moment and normal force (б) 1 – level of the gravity center of transformed section

are taken equal to, respectively Eb,red =

Rb, ser

ε b1, red

and Ebt,red =

Rbt , ser

ε bt1, red

,

where εb1,red = 15 ⋅ 10 − 4 and εbt1,red = 8 ⋅ 10 − 5 . Value Mcrc is determined by solving the system of equations (3.144) -(3.146), taking relative deformation of concrete εbt,max at the tensioned face being equal to: at a two-valued strain diagram in the cross-section εbt2 = 15 ⋅ 10− 5 ; at a single-value strain diagram εbt,ult = (15 - 5

ε1 )10−5 , where ε2

ε1/ε2 < 1 – ratio of concrete deformations at opposite sides of the section.

4.8. For rectangular, T- and I-sections at the impact of the moment in the plane of axis of symmetry, the cracks formation moment with consideration of inelastic deformations of tensioned concrete is permissible


to be determined from formula (4.4), replacing value W with Wpl = Wγ, where γ - see Table 4.1 Таble 4.1 Section

Factor γ

1. Rectangular

1,30

2.Т-section with a flange located in compressed zone

1,30

3. Т-section with a flange (widening) located in tensioned zone: а) at bf / b ≤ 2 irrespective of ratio hf /h б) at bf /b > 2 and hf /h ≥ 0,2 в) at bf /b > 2 and hf /h < 0,2

4.Symmetrical I-section (box-section): а) at bf' /b = bf /b ≤ 2 irrespective of ratio hf‘/h = hf /h б) at 2 < bf‘/b = bf /b ≤ 6 irrespective of ratio hf‘/h = hf /h в) at bf‘/b = bf /b > 6 and hf‘/h = hf /h ≥ 0,2 г) at 6< bf‘/b = bf /b ≤ 15 and hf‘/h = hf /h < 0,2 д) at bf‘/b = bf /b ≥ 15 and hf‘/h = hf /h < 0,2 5.Asymmetrical I-section meeting condition bf‘/b ≤ 3: а) at bf /b ≤ 2 irrespective of ratio hf /h б) at 2< bf /b ≤ 6 irrespective of ratio hf /h в) at bf /b > 6 and hf /h > 0,1

1,25 1,25 1,20

1,30 1,25 1,25 1,20 1,15

1,30 1,25 1,25

Shape of crosssection


Factorγ

Section 6.Asymmetrical I-section meeting condition 3 < bf‘/b < 8: а) at bf /b ≤ 4 irrespective of ratio hf /h б) at bf /b > 4 and hf /h ≥ 0,2 в) at bf /b > 4 and hf /h < 0,2

Shape of crosssection

1,25 1,25 1,20

7. Asymmetrical I-section meeting condition bf‘/b ≥ 8: 1,35 а) at hf /h > 0,3 1,30 б) at hf /h ≤ 0,3

4.9. Force Ncrc during formation of cracks in centrally tensioned elements is determined from formula Ncrc = Rbt,serA + 20 As, (4.9) where 20 (MPa) – stress in all reinforcement before cracks are formed in the concrete. DETERMINING THE OPENING WIDTH OF CRACKS NORMAL TO LONGITUDINAL AXIS OF THE ELEMENT 4.10. Width of normal cracks opening is determined from formula acrc = ϕ1ϕ2ϕ3ψs

σs Es

ls ,

(4.10)

where σs – stress in longitudinal tensioned reinforcement in a section with a crack from a corresponding external load, determined according to para 4.11; ls – basic (without considering the impact of the reinforcement surface type) distance between adjoining normal cracks, determined according to para 4.12; ψs – factor taking into account the nonuniform distribution of relative deformations of tensioned reinforcement between the cracks; it is permitted to take ψs = 1; if at that condition (4.1) is not met, value ψs should be determined according to para 4.13;


ϕ1 – factor taking into account the long-term impact of the load and taken equal to: 1,0 – at short-term impact of the load; 1,4 – at long-term impact of the load; ϕ2 – factor taking into account the profile of longitudinal reinforcement and taken equal to: 0,5 – for reinforcement of periodic profile (classes A300, A400, A500, B500); 0,8 – for plain reinforcement (class A240); ϕ3 – factor taking into account the loading condition and taken equal to: 1,0 – for flexural and eccentrically compressed elements; 1,2 – for tensioned elements. 4.11. Value of stress σs in tensioned reinforcement of flexural elements (Diagram 4.2,a) is determined from formula M (h0 − x) α s1 I red

σs = a)

εs

εs

N

σs = Esεs

σb = ε b Εb,red

εb As

εs h0

e

Nc= σs As + Nb

σs = Esεs

zs

h0

1

As

σs = Es εs

x

1

As

σs = Es εs

εs

σb = ε b Εb,red

εs

σs = Es εs

1

e

h0

As

x

εb

a'

в)

x

M

б)

σb = ε b Εb,red

εb As

(4.11)

e

As N

εs

σs = Es εs

Diagram 4.2. Schematics of stressed deformational state of elements with cracks at the impact of: moment of flection (а), compressing normal force (б), tensile normal force (в)


1 –level of the gravity center of transformed section

where Ired and x – moment of inertia and height of compressed zone of transformed cross-section including the area of cross-section only of compressed zone of concrete and cross-sectional area of tensioned and compressed reinforcement multiplied Es

by modular ratio of reinforcement to concrete αs1 = Factor

αs1 =

where Eb,red - see para 4.7. αs1 can also be determined

from

E b,red

,

formula

300 . Rb, ser

Height of compressed zone is determined from solution of equation

Sb = αs1(Ss – Ss‘), (4.12) ‘ where Sb, Ss, Ss – static moments of compressed zone of concrete, areas of tensioned and compressed reinforcement respectively in relation to the neutral axis. For rectangular, T- and I-sections, it is permissible to determine stress σs from formula

σs =

M , z s As

(4.13)

where zs – lever arm equal to zs = ζh0, and factor ζ is determined from the graph in Diagram 4.3. Graphical symbols at δ ≤ 0,2 at δ ≥ 0,3

Diagram 4.3. Graph of factor ζ=zs /h0 for determining the lever arm at calculation of cracks opening in flexural elements b 'f − b h 'f + α s1 As' As E s ;γ = μ sα s1 = bh0 bh0 Eb, red

(

)

δ = h 'f h0 , for sections without a compressed flange δ = 2a ' h0


Stress value σs for eccentrically compressed element as well as for eccentrically tensioned element at application of force N beyond the distance between reinforcement S and S‘ (Diagram 4.2,б, в) is determined from formula σs = N (h0 − x)α s1 , (4.14) S red

where Sred – static moment in relation to the neutral axis; value Sred is calculated from formula (4.15) Sred = Sb + αs1( S s' - Ss), and the height of compressed zone x is found from solutiоn of equation I red = e – (h0 – x), S red

(4.16)

where Ired – moment of inertia of transformed section in relation to the neutral axis. For eccentrically tensioned elements eccentricity e in formula (4.16) is taken with the “minus” sign. Value of stress σs for eccentrically tensioned elements at application of force N between the gravity centers of reinforcement S and S' (i.е. at e' < h0 – a) is determined from formula Ne ' σs = . As (h0 − a ′) For centrally tensioned elements

σs =

(4.17)

N . As

(4.18)

For eccentrically compressed element of rectangular section, stress σs is permitted to be determined from formula

σs =

Ne ϕ crñ , As h0

(4.19)

where ϕсrc – factor determined from Table4.2. For eccentrically tensioned elements of rectangular section, stress σs is permitted to be determined from formulae: N e (1,2 + 1); As h0 Ne′

a) at e' > h0 – a' and at As' = 0

σs =

б) at As' ≥ As irrespective of e'

σs =

As (h0 − a ' )

.

(4.20) (4.21)


At 0 < As' < As, value σs is determined by means of linear interpolation between values σs calculated from formulae (4.20) and (4.21). In all cases, value σs shall not exceed Rs,ser. Таble 4.2 Factors ϕcrс ' e/h0 at As ≥ As and values μαs1, equal to

≤0,8 1,0 1,2 1,5 2,0 3,0 4,0

at As' = 0 and values μα1, equal to

0,01

0,05

0,10

0,20

≥ 0,40 0,01

0,05

0,10

0,20

≥ 0,40

0,01 0,13 0,25 0,42 0,56 0,73 0,80

0,06 0,20 0,33 0,48 0,63 0,79 0,86

0,07 0,23 0,37 0,52 0,66 0,82 0,90

0,08 0,25 0,39 0,54 0,68 0,84 0,93

0,08 0,26 0,40 0,55 0,69 0,85 0,93

0,06 0,20 0,33 0,48 0,63 0,79 0,86

0,10 0,26 0,38 0,53 0,67 0,82 0,91

0,20 0,31 0,43 0,58 0,72 0,88 0,96

0,18 0,36 0,49 0,64 0,78 0,93 1,01

μαs1 =

0,01 0,13 0,25 0,42 0,56 0,73 0,80

As 300 bh0 Rb , ser

Note. At 0 < As' < As coefficients ϕcrс are determined by means of linear interpolation.

4.12. Value of basic distance between the cracks ls is determined from formula ls = 0,5

Abt ds As

(4.22)

and taken no less than 10ds and 100 mm and no greater than 40ds and 400 mm (for elements with effective depth of cross-section no greater than 1 m). Here Abt – cross-sectional area of tensioned concrete determined in general case according to provisions of para.4.7. At that the height of tensioned zone is taken no less than 2a and no greater than 0,5h. For rectangular, T- and I-sections it is permissible to determine the height of tensioned zone of the concrete from formula (4.23) with consideration of the said constraints: y = ytk, (4.23) where yt – height of tensioned zone of the concrete determined as for elastic material at modular ratio of reinforcement to concrete α = Es/Eb;


k – correcting coefficient equal to: for rectangular sections and T-sections with a flange in compressed zone - 0,90; for I-sections (box-shaped) and T-sections with a flange in tensioned zone – 0,95. Value yt is taken equal to: for flexural elements yt =

S red , Ared

for eccentrically loaded elements yt =

Ared

S red , ± N / Rbt , ser

where Sred – static moment of complete transformed section in relation to the tensioned face; Ared – see formula (4.8); the “plus” sign is taken at compressing normal force N, the “minus” sign – at tensile force N. At various diameters of tensioned reinforcement rods value ds is taken equal to 2 n1d s21 + ... + n k d sk , ds = n1d s1 + ... + n k d sk

(4.24)

where ds1,…dsk – diameters of tensioned reinforcement rods; n1,…nк – number of rods with diameters ds1,…dsk respectively. 4.13. Value of coefficient ψs is determined from formula

ψs = 1 – 0,8

σ s ,crc σs

,

(4.25)

where σs,crс – stress in longitudinal tensioned reinforcement in the section immediately after normal cracks formation determined according to provisions of para 4.11, taking in corresponding formulae M = Mcrc, e =

M crc M ± y p and e' = crc + yc , where yр N N

and yс – distances from the gravity center of tensioned and compressed reinforcement respectively to the axis passing through the gravity center of the transformed section taken at determining Mcrc; at that, the “plus” sign is taken for eccentrical compression, the “minus” sign – for eccentrical tension; Mcrc – see paras 4.4-4.8; N – normal force at the impact of the load in view;


σs – stress in longitudinal tensioned reinforcement at the impact of the load in view. If σs,crc > σs,, ψs is taken equal to 0,2. For flexural elements value of coefficient ψs is permitted to be determined from formula ψs = 1 – 0,8 M crc , (4.26) M

and be taken no less than 0,2. 4.14. The width of cracks opening is taken equal to: at long-term opening acrc = acrc1; (4.27) at short-term opening acrc = acrс,1 + acrс,2 – acrс,3, (4.28) where acrc,1 – width of cracks opening determined according to para 4.10 at ϕ1 = 1,4 and at the impact of constant and long-term loads; acrc2 – same, at ϕ1 = 1,0 and the impact of all loads (i.e. including short-term ones); acrc3 – same, at ϕ1 = 1,0 at the impact of constant and long-term loads. The width of short-term cracks opening can also be determined from formula ⎛

acrc = acrc2 ⎜⎜1 + 0,4 ⎝

σ sl − 0,8σ s,crc ⎞ ⎟, σ s − 0,8σ s,crc ⎟⎠

(4.28a)

where values σsl and σs are determined according to para 4.11 at the impact of the total of constant and long-term loads, and all loads, respectively; σs,crc – see para 4.13. In the absence of requirement to the structure in terms of limitation of permeability and at meeting the following condition σ sl − 0,8σ s,crc > 0,68, σ s − 0,8σ s,crc

(4.29)

One can check only long-term opening of cracks, and if condition (4.29) is not met – only short-term opening. For flexural element in formulae (4.28a) and (4.29) values σs,crc, σs and σsl can be replaced with Mcrc, M and Ml respectively – moment from the impact of constant and long-term loads. CALCULATION EXAMPLES


Example 42. Given: reinforced concrete flooring slab with crosssectional dimensions (for half of the slab section) as per Diagram 4.4; concrete of class B25 (Rbt,ser = 1,55 MPa, Rb,ser = 18,5 MPa, Eb = 30000 MPa); cross-sectional area of tensioned reinforcement of class A400 As = 760 mm2 (2∅22); total moment in the middle of the span M = 69 kN.m; all loads are constant and long-term ones.

Diagram 4.4.For calculation example 42

It is required to carry out calculation of opening of normal cracks. C a l c u l a t i o n. From Diagram 4.4 one has: b = 85 mm, h = 400 mm, a = 58 mm, bf' = 725 mm; hf' = 50 mm. Let us determine moment of cracks formation Mcrс according to para 4.5. To do that, one should determine geometrical characteristics of the Es 2 ⋅ 10 5 = = 6,67 and As' = 0: transformed section at α = 4 Eb 3 ⋅ 10

Ared = A + αAs = bh + (bf' – b)hf' + αAs = 85 ⋅ 400 + (725 − 85)50 + + 6,67 ⋅ 760 = 34000 + 32000 + 5069 = 71069 mm2; yt = Sred/Ared = [ 34000 ⋅ 400 / 2 + 32000(400 − 50 / 2) + 5069 ⋅ 58 ]/71069 = 268,7 mm; bh3 Ired = + bh( yt − h / 2) 2 + (b'f − b)h′f (h. − h'f / 2 − yt ) 2 + 12 85 ⋅ 4003 ' '3 2 + (b f − b)h f / 12 + + αAs ( yt − a) = + 34000(268,7 − 200) 2 + 12 640 ⋅ 503 + + 32000(375 − 268,7) 2 + 5069(268,7 − 58) 2 = 1,207 ⋅109 mm4’; 12 I red 1,207 ⋅109 W= = = 4,49 ⋅106 mm3. yt 268,7


To consider inelastic deformations of tensioned concrete by multiplying W by coefficient γ equal to, according to Table 4.1 1,30, i.е. W= 4,49 ⋅106 ⋅1,3 = 5,84 ⋅10 6 mm3. Тhen Mcrc = Rbt,serW = 1,55 ⋅ 5,84 ⋅106 = 9,052 ⋅106 N.mm = 9,05 kN.m < M = =69 kN.m. i.е. cracks are formed and the calculation of cracks opening width is necessary. One should determine stress in reinforcement σs from formula (4.13). Effective depth of section h0 = h – a = 400 – 58 = 342 mm; modulus ratio 760 ⋅ 16,2 Aα 300 300 αs1 = = 16,2. Then at μsαs1 = s s1 = = = 0,424 = Rb, ser 18,5 bh0 85 ⋅ 342

( b 'f − b )h 'f and γ =

32000 = 1,1 > 0,80 from the graph in Diagram 4.2 bh0 85 ⋅ 342 one can find coefficient ζ = 0,9 and the lever arm equals zs = ζh0 = 0,9 ⋅ 342 = 308 mm. 69 ⋅106 M σs = = = 294,8 MPa. zs As 308 ⋅ 760 Now one can determine distance between cracks ls from formula (4.22). As the height of tensioned concrete equal to y = ytk = 268,7 ⋅ 0,9 = = 247,8 mm > h/2 = 200 mm, the cross-sectional area of tensioned concrete is taken equal to Abt = b ⋅ 0,5h = 85 ⋅ 200 = 17000 mm2. Тhen 17000 A ls = 0,5 bt d s = 0,5 ⋅ 22 = 246 mm, As 760 which is less than 40ds = 880 mm and less than 400 mm, therefore ls is kept equal to 246 mm. Value ψs should be determined from formula (4.26)

=

ψs = 1 – 0,8

M crc 9,05 = 1 – 0,8 = 0,895. 69 M

Let us use formula (4.10) to determine the width of long-term opening of cracks, taking ϕ1 = 1,4, ϕ2 = 0,5 and ϕ3 = 1,0, 294,8 σ acrc = ϕ1ϕ2ϕ3ψs s ls = 1,4 ⋅ 0,5 ⋅ 0,895 ⋅ 246 = 0,227 mm, 5 Es 2 ⋅10


which is less than maximum allowable width of long-term opening of cracks equal to, according to para 4.2, acrc,ult = 0,3 mm. Example 43. Given: reinforced concrete foundation slab with crosssectional dimensions h = 300 mm, b = 1150 mm; a = 42 mm; concrete of class B15 (Rbt,ser = 1,1 MPa, Rb,ser = 11 MPa); principal reinforcement of class A400 with cross-sectional area As = 923 mm2 (6∅14); moment in design section from constant and long-term loads Ml = 50kN.м, from shortterm loads Msh=10kN.m; foundation is operated in non-aggressive environment (above the upper level of groundwater). It is required to carry out calculation of opening of normal cracks. C a l c u l a t i o n. Let us determine moment of cracks formation Mcrc 923 A according to paras 4.5 – 4.8. As μ = s = = 0,0027 < 0,005, bh 1150 ⋅ 300 elastic section modulus W can be determined without consideration of reinforcement , i.е. bh 2 1150 ⋅ 300 2 W= = = 1,725 ⋅ 10 7 mm3. 6 6 Inelastic deformations can be taken into account by multiplying W by coefficient γ equal to, according to Table 4.1, 1,30, i.e. W = 1,3 ⋅ 1,725 ⋅ 107 = 2,24 ⋅107 mm3. Тhen Mcrc = Rbt,serW = 1,1 ⋅ 2,24 ⋅107 = 24,67 ⋅106 N.mm = 24,67 kN.m < M = Ml + Msh = 50 + 10 = 60 kN.m. i.e. under the impact of full load cracks are formed and a calculation of cracks opening is necessary. One can check condition (4.29) with replacement of stresses σs with corresponding moments M l − 0,8M crc 50 − 0,8 ⋅ 24,67 = = 0,752 > 0,68, 60 − 0,8 ⋅ 24,67 M − 0,8M crc therefore, only long-term opening of cracks should be checked. Stress in reinforcement σs is determined from formula (4.13), taking M = Ml. Effective depth of section h0 = h – a = 300 – 42 = 258 mm; modulus ratio 300 300 Aα 923 ⋅ 27,27 αs1 = = = 27,27. Тhen at μsαs1 = s s1 = = 0,085 1150 ⋅ 258 bh Rb, ser 11 0 and γ = 0,0 from the graph in Diagram 4.3 one can find ζ = 0,89. Lever arm equals zs = ζh0 = 0,89 ⋅ 258 = 229,6mm. 50 ⋅106 M σs = = = 235,9 MPa. z s As 229,6 ⋅ 923


For a rectangular section, the height of tensioned zone with consideration of inelastic deformations equals y = 0,5hk = 0,5 ⋅ 300 ⋅ 0,9 = 135 mm > 2a = 2 ⋅ 42 = 84 mm and, besides, y = 135 mm < 0,5h = 150 mm therefore y is kept equal to 135 mm and then Abt = by = 1150 ⋅135 = 155250 mm2. Distance between cracks can be determined from formula (4.22) 155250 A ls = 0,5 bt d s = 0,5 14 = 1177 mm, which is greater than 40ds = As 923 40 ⋅14 = 560 mm and greater than 400 mm, so one can take ls = 400 mm. Value ψs is determined from formula (4.26), taking M = Ml = =50 . kN m. 24,67 M ψs = 1 – 0,8 crc = 1 – 0,8 = 0,605. M 50,0 The width of long-term opening of cracks can be determined from formula (4.10), taking ϕ1 = 1,4, ϕ2 = 0,5 and ϕ3 = 1,0: 235,9 σ acrc = ϕ1ϕ2ϕ3ψs s ls = 1,4 ⋅ 0,5 ⋅ 0,605 ⋅ 400 = 0,2 mm, Es 2 ⋅ 105 which is less than the maximum allowable width of long-term cracks opening being equal to acrc,ult = 0,3 mm. Example 44. Given: reinforced concrete column of an industrial building, with cross-sectional dimensions h = 500 mm, b = 400 mm; a = a' = 50 mm; concrete of class B15 (Eb = 24000 MPa, Rb,ser = 11 MPa, Rbt,ser = 1,1 MPa); principal reinforcement of class A400 with crosssectional area As = As' = 1232 mm2 (2∅28); forces from constant and longterm loads Nl = 500 kN, Ml = 150 kN.m; forces from short-term (wind) load: Nsh = 0,0; Msh = 90 kN.m. It is required to calculate opening of cracks for the column. C a l c u l a t i o n. Moment оf cracks formation Mcrc can be determined according to paras 4.5 – 4.8. As 1232 = = 0,0062 > 0,005, value W and eя can be bh 400 ⋅ 500 E determined with consideration of reinforcement at modulus ratio α = s = Eb

As μ =

2 ⋅105 = 8,333. For a rectangular section with symmetrical reinforcement 2,4 ⋅10 4 yt = h/2 = 250 mm, and moment of inertia Ired equals


400 ⋅ 5003 bh 3 2 Ired= + 2 Asα (0,5h − a ) = + 2 ⋅ 1232 ⋅ 8,333(250 − 50) 2 = 12 12 4 9 = 4,988 ⋅10 mm . I red 4,988 ⋅ 10 9 Тhen W = = = 19,95 ⋅106 mm3. 250 yt

Area of transformed section equals Ared = bh + 2Asα = 400 ⋅ 500 + 2 ⋅ 1232 ⋅ 8,333 = 220533 mm3. W 19,95 ⋅ 10 6 = = 90,5 mm. Then eя = 220533 Ared

Inelastic deformations of tensioned concrete can be considered by multiplying W by coefficient γ = 1,3 (see Table 4.1), i.e. W = =19,95 ⋅106 ⋅1,3 = 25,94 ⋅106 mm, Moment Mcrc can be determined from formula (4.4), taking N = Nl = 500 kN, Mcrc = Rbt,serW + Neя = 1,1 ⋅ 25,94 ⋅106 + 500000 ⋅ 90,5 = 73,76 ⋅106 N.mm = 73,76 kN.m < M = Ml + Msh = 150 + 90 = 240 kN.m, i.e. under the impact of all loads cracks are formed and a calculation of opening of cracks is necessary. Stress in tensioned reinforcement under the impact of all loads can be determined from formula (4.19). h0 = h – a = 500 – 50 = 450 mm = 0,45 m. e=

At

240 0,45 − 0,05 M h0 − a ′ + = + = 0,68 m. 2 500 2 N

0,68 e = = 1,51 and h0 0,45

μαs1 =

As 300 1232 300 ⋅ = ⋅ = bh0 Rb, ser 400 ⋅ 450 11

0,187 from Table .4.2 one can find ϕсrc = 0,54. Then σs = 500 ⋅ 10 3 ⋅ 680 ⋅ 0,54 = =331,2 MPa. 1232 ⋅ 450

Ne ϕ crc = As h0

Similarly, one can determine stress σs under the impact of constant and long-term loads, i.e. taking M = Ml = 150 kN.m and N = Nl = 500 kN. e=

M h0 − a ′ 150 + = + 0,2 = 0,5 m. 2 500 N


e 0,5 = = 1,111 and h0 0,45

At

μαs1 = 0,187 from Table 4.2 one can find

ϕсrc=0,32.

500 ⋅ 10 3 ⋅ 500 σs = σsl = ⋅ 0,32 = 144,3 MPa. 1232 ⋅ 450

One can also determine stress σs under the impact of moment M = Mcrc = 73,76 kN.m and force N = 500 kN. e=

å 73,76 0,3475 + 0,2 = 0,3475 m; from = = 0,772 and h0 500 0,45

0,187 one can find ϕсrc = 0,08; σs = σs,crc =

μαs1 =

500 ⋅ 10 3 ⋅ 347,5 ⋅ 0,08 = 25 1232 ⋅ 450

MPa. To check condition (4.29)

σ sl − 0,8σ s,crc 144,3 − 0,8 ⋅ 25 = = 0,399 < 0,68, 331,2 − 0,8 ⋅ 25 σ s − 0,8σ s,crc

i.e. condition (4.29) is not met, therefore, one should check only shortterm opening of cracks, determining acrс from formula (4.28a). To do that, one should determine acrс,2 from formula (4.10) at ϕ1 = 1,0 and σs = 331,2 MPa. From formula (4.25) one has ψs = 1 – 0,8

σ s,crc 25 = 1 – 0,8 = σs 331,2

0,94. Distance between cracks ls can be determined according to para 4.12. For this purpose, one should calculate the height of tensioned zone of concrete from formula (4.23), taking k = 0,90, аnd yt =

Ared

S red = + N / Rbt , ser

bh 2 / 2 + 2 Asαh / 2 400 ⋅ 500 2 / 2 + 2 ⋅ 1232 ⋅ 8,33 ⋅ 250 = = 81,7mm, Ared + N / Rbt , ser 220533 + 500 ⋅ 103 / 1,1

y= ytk = 81,7 ⋅ 0,9 = 73,5 mm < 2a = 2 ⋅ 50 = 100 mm. One takes y= 100 mm and then the cross-sectional area of tensioned concrete equals Abt = yb = 100 ⋅ 400 = 40000 mm2

and

ls = 0,5

Abt 40000 d s = 0,5 ⋅ 28 = 454,5 mm > 400 mm. As 1232

One takes ls equal to 400 m. acrc,2 = ϕ1ϕ2ϕ3ψs

σs Es

ls = 0,5 ⋅ 0,94

331,2 2 ⋅ 10 5

400 = 0,31 mm ;


⎛

acrc = acrc,2 ⎜⎜1 + 0,4 ⎝

σ sl − 0,8σ s ,crc ⎞ ⎟ = 0,31(1 + 0,4 ⋅ 0,399) = 0,36 mm, σ s − 0,8σ s ,crc ⎟⎠

which is less than the maximum allowable width of cracks opening, equal to 0,4 mm.

CALCULATION OF DEFORMATIONS FOR REINFORCED CONCRETE STRUCTURES GENERAL PROVISIONS 4.15. Calculation of deformations for elements of reinforced concrete structures is carried out with consideration of operational requirements to be met by the structure. Calculation of deformations should be performed for the impact of constant, temporary, long-term and short-term loads if there are limitations for deformations imposed by technological or structural requirements; constant and temporary long-term loads if there are limitations for deformation imposed by esthetic and psychological requirements. 4.16. Values of maximum allowable deformations of elements are taken according to SNiP 2.01.07-85* and regulatory documents for certain types of structures. CALCULATION OF DEFLECTION FOR REINFORCED CONCRETE ELEMENTS 4.17. Calculation of deflection for reinforced concrete elements is carried out using condition f ≤ fult, (4.30) where f – deflection of reinforced concrete element from the impact of external load; fult – value of maximum allowable deflection of a reinforced concrete element. Deflections of reinforced concrete structures are determined according to general rules and constructional mechanics depending on flexural, shearing and axial deformational characteristics of reinforced concrete elements in sections along its length (skewness, shearing angles, relative longitudinal deformations).


In cases when deflections of reinforced concrete elements mainly depend on flexural deformations, the value of deflections is determined from skewness of the element according to paras 4.18 and 4.19. 4.18. Deflection of reinforced concrete element resulting from flexural deformation is determined from formula l − 1 f= ∫ M x ⎛⎜ ⎞⎟ dx , 0

−

⎝ r ⎠x

(4.31)

where M x - moment of flection in section x from the impact of single force applied to the section for which deflection is calculated, in the direction of this deflection; ⎛1⎞ ⎜ ⎟ - total skewness of the element in section from external load at ⎝ r ⎠x

which the deflection is calculated. In general case, formula (4.31) can be used by dividing an element into a number of segments, determining skewness at the boundaries of these segments (with consideration of presence or absence of cracks and the sign − 1 of skewness) and multiplying the areas of moments M x and skewness ⎛⎜ ⎞⎟

⎝ r ⎠x

along the element length, taking linear distribution of skewness within each segment. In this case, at determining deflection in the middle of the span, formula (4.31) takes the form n / 2−1 ⎡ 1 l 2 ⎧⎪⎛ 1 ⎞ ⎛1⎞ ⎛ ⎞ ⎛1⎞ ⎤ ⎛ 1 ⎞ ⎫⎪ f = +⎜ ⎟ + 6 ∑ i ⎢⎜ ⎟ + ⎜ ⎟ ⎥ + (3n − 2 )⎜ ⎟ ⎬ , (4.32) ⎨⎜ ⎟ ⎝ r ⎠ c ⎪⎭ 12n 2 ⎪⎩⎝ r ⎠ sup,l ⎝ r ⎠ sup,r i =1 ⎣⎝ r ⎠ il ⎝ r ⎠ ir ⎦

1 where ⎛⎜ ⎞⎟

⎝ r ⎠ sup,l

1 , ⎛⎜ ⎞⎟

⎝ r ⎠ sup,r

- skewness of the element on the left and the right supports respectively;

⎛1⎞ ⎛1⎞ ⎜ ⎟ , ⎜ ⎟ - skewness of the element in symmetrically located ⎝ r ⎠ il ⎝ r ⎠ ir

sections i and i' (at i = i') to the left and to the right from the axis of symmetry (of the middle of the span, Diagram 4.5), respectively; ⎛1⎞ ⎜ ⎟ - skewness of the element in the middle of the span; ⎝ r ⎠c

n – even number of equal segments to which the span is divided, no less than 6; l – element span.


In formulae (4.31) and (4.32) skewnesses

1 are determined according r

to provisions of paras 4.21 – 4.27. At that, the sign of skewness is taken in accordance with diagram of skewness distribution.

0

1

i

0

1

i

l/n

1 r sup,l

,,

c

n/2-1

,,

1 r il

c

n/2-1

1 r il

,,

i'

(n/2-1)'

,,

1'

i'

(n/2-1)'

,,

1'

,,

1 r c

0'

l/n

1 r ir

1 r sup,r 1 (r)

l/2

0'

1r

l/2

Diagram 4.5. Diagram of skewness distribution in a reinforced concrete element in a general case of deflection calculation

4.19. For flexural elements of uniform section having cracks in each segment where the sign of the moment of flection remains unchanged, it is permitted to calculate akewness for the most stressed section and take skewnesses for other sections of this segment to be changing in proportion to values of the moment of flection (Diagram 4.6).


Diagram 4.6. Distribution diagram of moments of flection and skewness in a reinforced concrete element of uniform section а – schematic of load location; б – distribution of moments; в – distribution of skewness

In this case for freely supported and cantilever elements the maximum deflection is determined from formula 1 f=Sl2 ⎛⎜ ⎞⎟

⎝ r ⎠ max

1 where ⎛⎜ ⎞⎟

⎝ r ⎠ max

,

(4.33)

- total skewness in the section with the largest moment of

flection from the impact of the load at which the deflection is calculated; S – coefficient taken from Table 4.3. If deflection determined from formula (4.33) exceeds the allowable one, then for under0reinforced elements (μs < 0,5%) its value is recommended to be revised at the account of increased stiffness at segments without cracks and consideration of variable stiffness at segments with cracks; for freely supported beams loaded with uniformly distributed load, it corresponds to formula f =

⎤ l 2 ⎡⎛ 1 ⎞ (5 + 3m + 3mk + k )(1 − m ) + 4m 2 ⎛⎜ 1 ⎞⎟ ⎥ , ⎢⎜ ⎟ 48 ⎣⎝ r ⎠ max ⎝ r ⎠ el ⎦

where m = Mcrc/ Mmax; 1 k = ⎛⎜ ⎞⎟

⎝ r ⎠ crc

⎛1⎞ /⎜ ⎟ ; ⎝ r ⎠ max

⎛1⎞ ⎜ ⎟ - skewness under the impact of moment Mcrc with ⎝ r ⎠ crc

consideration of the cracks; ⎛1⎞ ⎜ ⎟ - same, without consideration of cracks; ⎝ r ⎠ el

(4.34)

MANUAL to CP 52-101-2003 concrete and REinforced concrete structures without prior reinforcement stress Таble 4.3

Схема загружения Schematic of loading of a свободно опертой freely supported beam балки

КоэффиCoefficient S

циент

S

Схема загружения Schematic of cantilever loading консоли

КоэффиCoefficient S

циент

1

5 l

l

48

4

1 l/2

a
1

12

l/2

S

3

l

1

a2

8

6l2

a a l

6l

(3 -

a ) l

Note. At loading anзагружении element simultaneously according to several schematics 1,i Примечание. При элемента сразу по нескольким схемам SS=ΣS =∑S1iM M1i/ΣM / ∑M, where S and M are coefficient S and moment M respectively in the middle of a beam span 1 - соответственно 1 где Si и M коэффициент S и момент M в середине пролета балки i i or a cantilever embedment for each loading schematic. In this skewness (l/r)max(is1 ) или в заделке консоли для каждой схемы загружения. В case этомthe случае кривизна r max determined at value M equal to ΣM1 . определяется при значении M равном ∑Mi

Mcrc – moment of cracks formation determined according to paras 4.4-4.8. For flexural elements with restrained ends, deflection in the middle of the span can be determined from formula ⎡⎛ 1 ⎞ ⎤ ⎛1 ⎛1⎞ ⎞⎫⎪ 2 S − 0,5⎢⎜ ⎟ S (4.35) +⎜ ⎟ × − ⎥ ⎜ ⎟⎬l , 8 r r r ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎪⎭ ⎪⎩ ⎢⎣ max sup,l sup,r ⎥⎦ 1 1 1 where ⎛⎜ ⎞⎟ , ⎛⎜ ⎞⎟ , ⎛⎜ ⎞⎟ - skewness in the middle of the span at the ⎝ r ⎠ max ⎝ r ⎠ sup,l ⎝ r ⎠ sup,r ⎧⎪ 1 f= ⎨⎛⎜ ⎞⎟

left and the right supports respectively; S – coefficient determined from Table 4.3 as for a freely supported beam. In all cases, deflection is taken no less than the deflection determined from skewnesses without consideration of cracks. 4.20. For flexural elements at l/h < 10 it is necessary to consider the impact of transverse forces to their deflection. In this case total deflection


equals to the total of deflections resulting from the flexural deformation (see paras 4.18 and 4.19) and shearing deformation fq. Deflection fq resulting from shearing deformation is determined from formula l −

f = ∫ Q x γ x dx ,

(4.36)

0

−

where Q x – transverse force in section x from the impact of a single force applied in the section for which deflection is calculated, in the direction of this deflection; γx – shearing angle of the element in the section under the impact of external load at which the deflection is calculated. Value γx is determined according to provisions of para 4.28. DETERMINING OF SKEWNESS FOR REINFORCED CONCRETE ELEMENTS GENERAL PROVISIONS

4.21.Skewness of reinforced concrete elements for calculation of their deflections is determined: а) for elements or element parts where cracks normal to longitudinal axis are not formed in tensioned zone -- according to para 4.23; б) for elements or element parts where there are cracks in tensioned zone -- according to paras 4.24-4.26. Elements or parts of elements are viewed without cracks, if the cracks are not formed (i.е. condition 4.2 is met) at the impact of all loads (i.е. including short-term ones, too) with a safety factor γf = 1 Skewness of reinforced concrete elements with cracks and without cracks can be determined based on deformational model according to para 4.27. 4.22. Total skewness of flexural, eccentrically compressed and eccentrically tensioned elements is determined from formula: - for segments without cracks in tensioned zone 1 1 1 = ⎛⎜ ⎞⎟ + ⎛⎜ ⎞⎟ , r ⎝ r ⎠1 ⎝ r ⎠ 2

(4.37)


1 1 where ⎛⎜ ⎞⎟ and ⎛⎜ ⎞⎟ - skewnesses from short-term impact of shot-term ⎝ r ⎠1

⎝ r ⎠2

loads and from long-term impact of constant and longterm loads; - for segments with cracks in tensioned zone 1 1 ⎛1⎞ ⎛1⎞ = ⎜ ⎟ − ⎜ ⎟ + ⎛⎜ ⎞⎟ , r ⎝ r ⎠1 ⎝ r ⎠ 2 ⎝ r ⎠ 3

(4.38)

1 where ⎛⎜ ⎞⎟ - skewness from short-term impact of all loads included into the ⎝ r ⎠1

calculation of deformation; ⎛1⎞ ⎜ ⎟ - skewness from short-term impact of constant and long-term ⎝ r ⎠2

loads; ⎛1⎞ ⎜ ⎟ - skewness from long-term impact of constant and long-term loads. ⎝ r ⎠2 Note. At using formula (4.37), a shot-term load which includes a reduced value according to SNiP 2.01.07-85*, should be taken decreased by this value, considered in this formula as a long-term load. SKEWNESS OF REINFORCED CONCRETE ELEMENTS IN SEGMENTS WITHOUT CRACKS IN TENSIONED ZONE

4.23. Skewness of a reinforced concrete element in a segment without cracks is determined from formula M 1 = , r Eb1I red

(4.39)

where M – moment of flection from external load (including the moment from normal force N in relation to the axis passing through the gravity center of transformed section); Ired – moment of inertia of transformed section in relation to its gravity center determined as for a solid body according to general rules of elastic materials resistance with consideration of all cross-sectional area of concrete and cross-sectional areas of reinforcement with modulus ratio of reinforcement to concrete being equal to α = E s / Eb1 ; Eb1 – modulus of deformation of compressed concrete, taken equal to: at short-term impact of a load Eb1= 0,85Eb ; (4.40)


at long-term impact of a load Eb1=

Eb

1 + ϕ b,cr

,

(4.41)

where ϕb,cr – creep coefficient of concrete taken depending on relative air humidity and class of concrete from Table 4.4. Таble 4.4 Relative ambient air humidity ,%

Values of creep coefficient ϕb,cr at compression class of concrete B10 B15 B20 B25 B30 B35 B40 B45 B50 B55 B60

over 75 (increased) 40-75 (normal)

2,8

2,4

2,0

1,8

1,6

1,5

1,4

1,3

1,2

1,1

1,0

3,9

3,4

2,8

2,5

2,3

2,1

1,9

1,8

1,6

1,5

1,4

below 40 5,6 4,8 4,0 3,6 3,2 3,0 2,8 2,6 2,4 2,2 2,0 (reduced) Note. Relative air humidity is taken according to SNiP23-01-99 as a monthly average relative humidity of the warmest month in the construction area. SKEWNESS OF REINFORCED CONCRETE ELEMENTS IN SEGMENTS WITH CRACKS IN TENSIONED ZONE 4.24. Skewness of a flexural reinforced concrete element in segments with cracks in tensioned zone is determined from formula M 1 = , (4.42) r E b,red I red

where Ired – moment of inertia of transformed section in relation to its gravity center determined according to general rules of elastic materials resistance with consideration of cross-sectional area of concrete only in compressed area, cross-sectional areas of compressed reinforcement with modulus ratio αs1 and tensioned reinforcement with modulus ratio αs2 (Diagram 4.7); Eb,red – transformed modulus of deformation of compressed concrete taken equal to Eb,red =

Rb, ser

ε b1,red

, where value εb1,red equals:

at short-term impact of a load − 15 ⋅ 10 −4 ; at long-term impact of a load depending on relative ambient air humidity w%:


б)

a'

A's α s1 ε's xm

Ab

M

h0

σb= εb Eb,red

εb

σ'= s ε's Es

Xm - Ycm

a)

1 εsm σs = εsmEs /ψs A s αs2

Diagram 4.7. Transformed cross section (а) and schematic of stressed deformational state of a flexural element with cracks at calculation of its deformations (б) 1 – level of the gravity center of transformed section

at w > 75% at 75% ≥ w ≥ 40% at w < 40% Relative ambient air

− 24 ⋅ 10 −4 ; 28 ⋅ 10 −4 ; − − 34 ⋅ 10 − 4 . humidity is taken according to the note to Table

4.4. Values of modulus ratios of reinforcement to concrete are taken equal to: for compressed reinforcement - αs1 = for tensioned reinforcement -αs2 =

Es E b,red

;

Es ψ s E b,red

where ψs – see para 4.13. Coefficient αs1 can also be determined from formulae: at short-term impact of a load - αs1 =

300 ; Rb, ser

at long-term impact of a load and normal ambient air humidity (w = 40…75%) - αs1 =

560 ; Rb, ser

аnd coefficient αs2 - from formula αs2 =

α s1 . ψs

MANUAL to CP 52-101-2003 concrete and REinforced concrete structures without prior reinforcement stress The height of compressed zone is determined by solving the following equation

Sb =αs2Ss - αs1S's (4.43) where Sb, Ss and S's – static moments of compressed zone of concrete, areas of tensioned and compressed reinforcement respectively in relation to neutral axis. For rectangular, T- and I-sections, the height of compressed zone is determined from formula ' ⎞ ⎡ ⎤ ⎛ a' 2 ' ' hf ⎟ ⎜ ⎢ − z⎥ , x = h0 z + 2⎜ μ sα s 2 + μ sα s1 + μ f ⎟ ⎢ ⎥ 2h0 h0 ⎝ ⎠ ⎣ ⎦

where z = μsαs2 + μ'sαs1 + μf';

(

(4.44)

)

b 'f − b h 'f As As' μs = ; μs' = ; μf' = . bh0 bh0 bh0

4.25. For flexural elements of rectangular, T- and I-sections operated at relative ambient air humidity exceeding 40%, skewness in segments with cracks is permitted to be determined from formula M − ϕ 2 bh 2 Rbt , ser 1 = r ϕ1 E s As h02

(4.45)

N 1 , =± r S red Eb,red

(4.46)

where ϕ1 – see Table 4.5; ϕ2 - see Table 4.6; 4.26. Skewness of eccentrically compressed elements as well as eccentrically tensioned element at application of force N beyond the distance of reinforcements S and S ' in segments with cracks in tensioned zone is determined from formula

μf' 0,0 0,2 0,4 0,6 0,8 1,0

Coefficient ϕ1 at values μαs1, equal to ≤ 0,07 0,60 0,69 0,73 0,75 0,76 0,77

0,10 0,15 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,00 0,55 0,65 0,69 0,72 0,74 0,75

0,49 0,59 0,65 0,68 0,71 0,72

0,45 0,55 0,61 0,65 0,69 0,70

0,38 0,48 0,55 0,59 0,62 0,65

0,34 0,43 0,50 0,55 0,58 0,61

0,30 0,39 0,46 0,51 0,54 0,57

0,27 0,36 0,42 0,47 0,51 0,54

0,25 0,33 0,40 0,45 0,48 0,52

0,23 0,31 0,37 0,42 0,46 0,49

0,22 0,29 0,35 0,40 0,44 0,47

0,20 0,27 0,33 0,38 0,42 0,45


(

)

b'f − b h 'f As As' μαs1 = α s1 ; μ'f = + α s1 bh0 bh0 bh0

at long-term impact of loads αs1 = 560/Rb,ser, at short-term impact of loads αs1 = 300/ Rb,ser.

Таble 4.5 DETERMINING SKEWNESS OF REINFORCED CONCRETE ELEMENTS BASED ON NONLINEAR DEFORMATIONAL MODEL 4.27. Value of skewness is taken equal to:

at two-valued deformational distribution diagram for the section ε b,max x

1 = r

;

at single-valued deformational distribution diagram for compressed concrete for the section -

ε −ε 1 = b,max b,min , r h

where εb,max and εb,min – maximum and minimum deformations of concrete determined based on provisions specified in paras 3.72-3.75; x – height of compressed zone in the direction normal to the neutral axis; h – depth of section in the direction normal to the provisional neutral axis (Diagram 3.39,б); at single-valued deformational distribution diagram for tensioned reinforcement for the section -

ε −ε 1 = s, max s, min , r hs

where εs,max and εs,min – absolute values of deformations of reinforcement rods tensioned to the maximum and minimum extent and determined according to paras 3.72-3.75; hs – distance between these rods in the direction normal to the provisional neutral axis. At that, for an element with cracks in tensioned zone of stress in reinforcement crossing the crack is determined from formula σsi=Es(εsi+0,8εs,crc), (4.50)


where εsi – averaged relative deformation of tensioned reinforcement at the current stage of calculation, corresponding to the linear law of strain distribution over the section; εs,crc – relative deformation of tensioned reinforcement in the section with a crack immediately after the crack formation (i.е. at the impact of moment Mcrc) equal to εs,crc = σs,crc /Es, where σs,crc – see para 4.13. If there are cracks in the element, the stressed deformational state of compressed concrete is determined using a bilinear diagram σb-εb using a transformed modulus of deformation for compressed concrete Eb,red, determined according to para 4.24, and values εb0 and εb2, taken according to Table 4.7. In the absence of cracks, the stressed deformational state of compressed concrete is determined using a three-line diagram (Diagram 4.8), where εb1 = 0,6

Rb, ser Eb1

, εb0 and εb2 – see Table 4.7; Eb1 is taken equal

to: at short-term impact of a load – Eb, at long-term impact of a load – see formula (4.41). Stressed deformational state of tensioned concrete can also be determined using a three-linear diagram (see Diagram 4.8) with replacement of Rb,ser with Rbt,ser, εb0 with εbt0, εb2 with εbt2,, where values εbt0 and εbt2 – see Table 4.7 Таble 4.7 Relative deformations of concrete at compression

Type of the load ε b 0 ⋅ 10

3

ε b 2 ⋅ 10

3

ε b1, red ⋅ 10

at tension 3

ε bt 0 ⋅ 10

3

ε bt 2 ⋅ 10

3

ε bt1, red ⋅ 103

short-term

2,0

3,5

1,5

0,10

0,15

0,08

long-term at relative ambient air humidity ,% over 75 40…..75 below 40

3,0 3,4 4,0

4,2 4,8 5,6

2,4 2,8 3,4

0,21 0,24 0,28

0,27 0,31 0,36

0,19 0,22 0,26

MANUAL to CP 52-101-2003 concrete and REinforced concrete structures without prior reinforcement stress Rb,ser

σb

σb1=0,6R b,ser arctgE b εb1

εb0

εb2

εb

Diagram 4.8.Three-line diagram of the state of compressed concrete.

Skewness can also be determined in a computer-assisted way using a non-linear deformational model. At calculation of statically indeterminate structures with consideration of physical nonlinearity for certain segments of the element, stiffnesses 1 are used where М – maximum moment in relation to r 1 geometrical axis of the element in the segment in view, - corresponding r

equal to D=M:

skewness determined according to para 4.27.


DETERMINING SHEARING ANGLES OF REINFORCED CONCRETE ELEMENTS

4.28. Shearing deformation angle is determined from formula

γx=

1,2Q x ϕ bϕ crc , Gbh

(4.51)

where Qx – transverse force in the section from an impact of external load; ϕb – coefficient considering the concrete creep effect and taken equal to: at long-term impact of loads ϕb = 1+ϕb,cr where ϕb,cr – see Table 4.4; at short-term impact of loads ϕb = 1,0; ϕcrc – coefficient considering the influence of cracks to shearing deformation and taken equal to: - in segments along the element length having no normal cracks or cracks inclined to the normal axis of the element, - ϕcrc = 1,0; - in segments having only inclined cracks - ϕcrc = 4,0; - in segments having only normal or normal and oblique cracks, coefficient ϕcrc is determined from formula

ϕcrc=

3Eb I red ⎛ 1 ⎞ ⎜ ⎟ M x ⎝ r ⎠x

(4.52)

1 where Мx and ⎛⎜ ⎞⎟ - moment and skewness from an external load at its

⎝ r ⎠x

short-term impact, respectively; Ired – moment of inertia of total transformed section at modulus ratio of reinforcement to concrete α = Es/Eb. Inclined cracks formation corresponds to meeting condition Q > 0,5Rbt,serbh0 . (4.53) CALCULATION EXAMPLES

Example 45. Given: reinforced concrete flooring slab of a civil building of rectangular section with dimensions h = 200 mm, b = 1000 mm; h0 = 173 mm; span l = 5,6 m; concrete of class B15 (Eb = 24000 MPa; Rb,ser = 11 MPa, Rbt,ser = 1,1 MPa); tensioned reinforcement of class A400 (Es = 2 ⋅ 10 6 MPa) with cross-sectional area As = 769 mm2 (5∅14); total uniformly distributed load q = 7,5 kN/m, including its part resulting from constant and long-term loads ql = 6,5 kN/m; deflection is limited by esthetic requirements.


It is required to calculate deformations of the slab. C a l c u l a t i o n. Let us determine skewness in the middle of the span resulting from constant and long-term loads as the deflection is limited by esthetic requirements. Moment in the middle of the span equals ql l 2 6,5 ⋅ 5,6 2 Mmax = Ml = = = 25,5 kN.m = 25,5 ⋅ 10 6 N.mm. 8 8

One can assume without calculation that the slab has cracks in tensioned zone, so the skewness shall be determined from formula (4.45). Reinforcement factor equals

μ=

As 769 = = 0,0045. 1000 ⋅ 173 bh0

At long-term impact of the load, the modulus ratio of the reinforcement equals αs1 =

560 560 = = 50,9. From Table 4.5 at 11 Rb, ser

μαs1 =

0,0045 ⋅ 50,9 = 0,226 and μf' = 0 one can find ϕ1 = 0,43, and from Table 4.6 300 = 0,121 and μ'f = μf = 0 one can find the coefficient at μαs1 = 0,0045 ⋅ 11

corresponding to long-term impact of the load ϕ2 = 0,13. M − ϕ 2bh 2 Rbt , ser ⎛1⎞ = Then ⎜ ⎟ ⎝ r ⎠ max ϕ1Es As h02 1 . = 9,99 ⋅ 10 − 6 ìì

=

25,5 ⋅ 10 6 − 0,13 ⋅ 1000 ⋅ 200 2 ⋅ 1,1 0,43 ⋅ 2 ⋅ 10 5 ⋅ 769 ⋅ 173 2

=

Deflection can be determined from formula (4.33) taking according to

Table 4.3 S = 1 f = Sl2 ⎛⎜ ⎞⎟

⎝ r ⎠ max

5 : 48 5 = ⋅ 5600 2 ⋅ 9,99 ⋅ 10 −6 = 32,6 mm. 48

According to SNiP 2.01.07-85* Table 19, item one can find the deflection maximum allowable according to esthetic requirements for a span of 5,6 m by means of linear interpolation fult = 20 + (30 – 20)

5,6 − 3 = 28,7 mm < f = 32,6 mm, i.е. condition 6−3

(4.30) is not met. Let us revise the value of the slab deflection by means of considering the variable stiffness in a segment with cracks by determining it from


formula (4.34). To do that, one should determine the moment of cracks formation Mcrc according to paras 4.5 and 4.8. Geometrical characteristics of transformed section can be calculated at modulus ratio α =

Es 200 = = 8,33: Eb 24

Ared = bh + Asα = 1000 ⋅ 200 + 769 ⋅ 8,33 = 2 ⋅ 10 5 + 6408 = 2,064 ⋅ 10 5 mm2; yt =

(

)

S red = 2 ⋅ 10 5 ⋅ 100 + 6408 ⋅ 27 / 2,064 ⋅ 10 5 = 97,7 mm; Ared

1000 ⋅ 2003 4 + 2 ⋅ 105 (100 − yt ) 2 + 6408( yt − 27 )2 = 6,998 ⋅ 108 mm ; 12 6,998 ⋅ 108 I red W= = = 7,16 ⋅ 10 6 mm3. yt 97,7

Ired =

Replacing value W in formula (4.4) with value Wpl = Wγ, where according to Table 4.1 γ = 1,3 one can determine value Mcrc Mcrc = Rbt,serWγ = 1,1 ⋅ 7,16 ⋅ 10 6 ⋅ 1,3 = 10,24 ⋅ 10 6 N.mm. Similarly, from formula (4.45) one can determine skewness at M = Mcrc

M crc − ϕ 2 bh 2 Rbt ,ser 1 10,24 ⋅ 10 6 − 0,13 ⋅ 1000 ⋅ 200 2 ⋅ 1,1 ⎛1⎞ 2,28 ⋅ 10 −6 = = . ⎜ ⎟ = ìì ⎝ r ⎠ crc ϕ1 E s As h02 0,43 ⋅ 2 ⋅ 10 5 ⋅ 769 ⋅ 173 2 1 Let us determine skewness ⎛⎜ ⎞⎟

⎝ r ⎠ el

at M = Mcrc without consideration of

cracks under long-term impact of the load and, taking from Table 4.4 for concrete of class B15 ϕb,cr = 3,4 and therefore Eb1=

Eb

1 + ϕ b,cr

=

24000 = 1 + 3,4

5455MPa. As the influence of value 1r to the element deflection is not el significant, one can determine the skewness from formula (4.38) without recalculation of value Ired:

( )

M crc 1 10,24 ⋅ 10 6 ⎛1⎞ 2,68 ⋅ 10 −6 = = . ⎜ ⎟ = E b1 I red ìì ⎝ r ⎠ el 5455 ⋅ 6,998 ⋅ 108

Coefficients m and k equal: 1 m = Mсrс/Mmax = 10,24/25,5 = 0,4019; k = ⎛⎜ ⎞⎟

⎝ r ⎠ crc

Then

⎛1⎞ /⎜ ⎟ =2,28/9,99=0,228. r ⎝ ⎠ max


f=

[

⎤ l 2 ⎡⎛ 1 ⎞ (5 + 3m + 3mk + k )(1 − m) + 4m 2 ⎛⎜ 1 ⎞⎟ ⎥ = ⎢⎜ ⎟ 48 ⎣⎝ r ⎠ max ⎝ r ⎠ el ⎦

]

5600 2 9,99(5 + 3 ⋅ 0,4019 + 3 ⋅ 0,4019 ⋅ 0,228 + 0,228)(1 − 0,4019 ) + 4 ⋅ 0,4019 2 ⋅ 2,68 × 48 × 10 − 6 = 27,6 mm < fult = 28,7 mm,

15 ∅ 22Α400 65

31

100

300

635

Ось симметрии

45

50

30

i.е. revised deflection value does not exceed the allowable one. Example 46. Given: reinforced concrete roofing slab with design span 5,7 m; sectional dimensions (for a half of the slab section) as per Diagram 4.9; concrete of class B25 (Eb = 30000 MPa, Rb,ser = 18,5 MPa, Rbt,ser = 1,55 MPa); principal reinforcement of class A400 with crosssectional area As=380mm2(1∅22); constant and long-term uniformly distributed load ql= 11 kN/m; slab deflection is limited by esthetic requirements; ambient air humidity is reduced (w < 40%). It is required to calculate deformations of the slab. C a l c u l a t i o n . As approximated formula for skewness (4.45) does not apply to structures operated at air humidity less than 40%, the skewness shall be determined from general formula (4.42) as for elements with cracks in tensioned zone. Moment in the middle of the slab resulting from constant and long-term loads for a half of the slab section equals:

Diagram 4.9. For calculation example 46 ql l 2 11 ⋅ 5,7 2 Mmax = Ml = = = 22,34 kN.m. 2 ⋅8 2⋅8

First, the moment of cracks formation Mcrc should be determined according to paras 4.5 and 4.8. Let us determine geometrical characteristics Es 200 ⋅ 10 3 == = 6,67: of transformed section at modulus ratio α = Eb 30 ⋅ 10 3

Ared= 300 ⋅ 50 + 100 ⋅15 + 300 ⋅ 45 / 2 + 30 ⋅ 635 + 380 ⋅ 6,67 =


= 15000 + 1500 + 6750 + 19050 + 2533,5 = 44833 mm ; yt = (15000 ⋅ 150 + 1500 ⋅ 50 + 6750 ⋅ 200 + 19050 ⋅ 285 + 2533,5 ⋅ 31) / 44833 =

=204,84 mm; 3 3003 ⋅ 50 2 100 ⋅ 15 + 15000( yt − 150 ) + + 1500( yt − 50 )2 + Ired = 12 12 3 3003 ⋅ 45 2 30 ⋅ 635 + 6750( yt − 200 ) + + 19050(285 − yt )2 + 2533,5( yt − 31)2 36 12 4 8 = 4,291 ⋅ 10 mm .

=

I red 4,291 ⋅ 108 Elastic section modulus W = = = 2,095 ⋅ 10 6 mm3. yt 204,8

Replacing value W in formula (4.9) with Wpl = W ⋅ γ , where γ = 1,3 (see Table 4.1), one can determine value Mcrc: Mcrc = Rbt,serWpl = 1,55 ⋅ 2,095 ⋅ 10 6 ⋅ 1,3 = 4,22 ⋅ 10 6 N.mm = 4,22 kN.m. From formula (4.26) one can determine coefficient ψs

ψs = 1 –0,8

M crc 4,22 = 1 – 0,8 = 0,849. 22,34 M

Transformed modulus of deformation under long-term impact of the load at w < 40% equals Eb,red =

Rb, ser

ε b1,red

and then αs2 =

=

18,5 34 ⋅ 10 −4

Es

Eb,red ⋅ ψ s

= 5441,2 MPa, 2 ⋅ 10 5 = = 43,29. 5441,2 ⋅ 0,849

Height of compressed zone can be determined from formula (4.44), taking the averaged rib width equal to 85 mm and the area of compressed overhangs being equal to Acв = (bf'-b)hf' = 635 ⋅ 30 = 19050 mm2, and effective depth h0 = 300 – 31 = 269 mm: As 380 α s2 = 43,29 = 0,719; 85 ⋅ 269 bh0 A 19050 μf' = ñâ = = 0,833; μ's = 0,0; bh0 85 ⋅ 269

μsαs2 =

z = μsαs2 + μf' = 0,719 + 0,833 = 1,552;

' ⎞ ⎡ ⎤ ⎛ h f ⎟ 2 ' ⎜ ⎢ x = h0 z + 2⎜ μ sα s 2 + μ f − z⎥ = ⎟ ⎢ ⎥ 2h0 ⎝ ⎠ ⎣ ⎦


⎡

=269 ⎢ 1,552 2 + 2⎛⎜ 0,719 + 0,833 ⎝

⎣

⎤ 15 ⎞ ⎟ − 1,552 ⎥ = 116,4 mm. 269 ⎠ ⎦

2

⎛ h 'f ⎞⎟ bx 3 ⎜ Ired = + Acâ x − + Asα s 2 (h0 − x )2 = ⎜ 3 2 ⎟ ⎝ ⎠ 85 ⋅ 116,4 3 4 = + 19050(116,4 − 15)2 + 380 ⋅ 43,29(269 − 116,4 )2 = 6,236 ⋅ 10 8 mm . 3

From formula (4.42) one has M max 1 22,34 ⋅ 10 6 ⎛1⎞ = = = . 6,58 ⋅ 10 −6 ⎜ ⎟ 8 ìì r E I ⎝ ⎠ max 5441,2 ⋅ 6,236 ⋅ 10 b, red red

Deflection can be determined from formula (4.33), taking according to 5 : 48 1 5 f = Sl2 ⎛⎜ ⎞⎟ = 5700 2 ⋅ 6,58 ⋅ 10 −6 = 22,3 mm. 48 ⎝ r ⎠ max

Table 4.3 S =

According to SNiP 2.01.07-85* Table 19, line 3, the maximum allowable deflection according to esthetical requirements for a span of 5,7 m equals fult = 29 mm > f = 22,3 mm, i.е. condition (4.30) is met.

5. STRUCTURAL REQUIREMENTS GENERAL PROVISIONS 5.1. To provide for carrying capacity, normal serviceability and longevity of concrete and reinforced concrete structures, apart from requirements determined by calculation, the following structural requirements shall be met: - to geometrical dimensions of structure elements; - to reinforcement (content and location of reinforcement, thickness of protective layer of concrete, anchoring and reinforcement joints);

- to protection of structures against negative environmental impact. GEOMETRICAL DIMENSIONS OF STRUCTURES 5.2. Minimum geometrical dimensions of structure sections shall be set so as to provide for:


- possibility of appropriate location of reinforcement (distances between the rods, protective layer of concrete etc.), its anchoring and combined behavior with concrete; - sufficient stiffness of structures; - required fireproofness, watertightness of structures, heat and sound insulation, corrosion resistance, protection against radiation etc.;

- possibility of quality manufacturing at structures concreting. 5.3. Sectional dimensions of eccentrically compressed elements to provide for their stiffness are recommended to set at such values that their flexibility

l0 in any direction does not exceed: i

- for reinforced concrete elements - 200 (for rectangular sections at l0/h ≤ 60); - for column serving as building elements - 120 (at l0/h≤35); - for concrete elements - 90 (at l0/h ≤ 2b). 5.4. Thickness1 of monolithic ribbed floorings thickness shall be taken no less than, mm: for roofs …………………………………40; for interstorey floorings of living and public buildings…………………………………………………50; for interstorey floorings of industrial buildings …………………………………………………60. 5.5. In structures of buildings and facilities one should envisage their slitting with permanent and temporary contraction joints with distances between them set according to climatic conditions, structural features of the facility, sequence of works execution etc. If nonuniform settlement of foundation is possible, one should envisage structures slitting with settlement joints. REINFORCEMENT PROTECTIVE LAYER OF CONCRETE

5.6. Reinforcement located inside the section of a structure 1

Here and further in the text values of section dimensions, thickness of protective layer of concrete etc. specified in the present Training Aid pertain to nominal values used at design stage and specified in Diagrams. In kind, there can be deviations form these nominal values, not exceeding values specified in corresponding state standards, technical specifications etc.


shall have a protective layer of concrete (distance from reinforcement surface to surface of a corresponding face of a structure) to provide for: - composite action of reinforcement and concrete; - anchoring of reinforcement in the concrete and possibility of jointing of reinforcement elements; - protection of reinforcement from environmental impact (including that of aggressive environment); - fire resistance and fire endurance. 5.7. Thickness of protective layer of concrete is set based on requirements of para 5.6 with account of the structure type, intended purpose of reinforcement in the structures (longitudinal principal, crosswise, distributional, structural reinforcement), environmental conditions and diameter of reinforcement. Minimum thickness values of protective layer of concrete for principal reinforcement should be taken according to Table 5.1.

№

Operational conditions of the building structures

1. In closed rooms at normal and reduced humidity In closed rooms at increased humidity (if no additional 2. protective measures are envisaged) In the open air (if no additional protective measures are 3. envisaged) In the ground (if no additional protective measures are 4. envisaged), in foundations, if concrete blinding is available In monolithic foundations if no concrete blinding is 5. available

Таble 5.1 Thickness of protective layer of concrete, mm, no less than 20 25 30 40 70


For precast elements, minimum thickness values of protective layer of concrete for principal reinforcement specified in Table 5.1 shall be decreased by 5 mm. For reinforced concrete slabs made of concrete of class B20 and higher manufactured at plants in metallic moulds and protected in a facility with concrete blinding or screed, the thickness of protective layer for upper reinforcement is permitted to be set at 5 mm. For structural reinforcement, minimum thickness values of protective layer of concrete are set by 5 mm lower than the required ones for principal reinforcement. In all cases, thickness of protective layer of concrete should also be taken no less than a diameter of a reinforcement rod. 5.8. In flexural, tensioned and eccentrically compressed (with Ml/Nl > 0,3 h) elements, except for foundations, thickness of protective layer for tensioned principal reinforcement should not, as a rule, exceed 50 mm. In a protective layer over 50 mm thick, structural reinforcement in the form of meshes should be installed. At that, cross-sectional area of longitudinal reinforcement of meshes shall be no less than 0,05 As, spacing of crosswise reinforcement shall not exceed section depth and meet requirements specified in para 5.18. MINIMUM DISTANCES BETWEEN REINFORCEMENT RODS

5.9. Minimum clearances between reinforcement rods should be taken so as to provide combined behavior of reinforcement and concrete and quality manufacturing of structures associated with placement and compaction of concrete mixture but no less than the largest diameter of a rod, as well as no less than: 25 mm – at horizontal or inclined position of rods during concreting – for lower reinforcement located in one or two rows; 30 mm - same, for upper reinforcement; 50 mm - same, at lower reinforcement location in greater than two rows (except for rods of two lower rows) as well as at vertical position of rods during concreting. In elements or nodes with highly concentrated reinforcement or embedded items manufactured without usage of a platform vibrator or a falsework-fixed vibrator, in certain places clearance of 60 mm shall be provided for in order to allow the heads of spud vibrators compacting the


concrete mixture to pass between the reinforcement rods. Distances between these places shall be no greater than 500 mm. Under constrained conditions it is permitted to locate rods in groups – bunches (without clearances between them). At that, clearances between the bunches shall also be no less than a transformed diameter of a rod equivalent to cross-sectional area of a reinforcement bunch taken equal to n

ds,red = ∑ d si2 where dsi – diameter of one rod in the bunch, п – number of i

rods in the bunch. 5.10. Clearances between the rods of ribbed profile specified in para 5.9 are determined based on nominal diameter without consideration of projecting parts and ribs. At determining reinforcement location in a section under constrained conditions with consideration of other adjoining reinforcement elements and embedded items, one should take into account diameters of rods with consideration of projecting parts and ribs (Appendix 1) as well as allowable deviations from nominal dimensions of reinforcement rods, welded meshes and frames, embedded items, shape and location of reinforcement and embedded items in the section. LONGITUDINAL REINFORCEMENT

5.11. In reinforced concrete elements, cross-sectional area of longitudinal tensioned reinforcement as well as compressed one if it is required according to calculation, as percentage of cross-sectional area of concrete, equal to the product of a rectangular section width or a rib width

of a T- or I-section by the effective depth of section, μs =

As ⋅ 100% should bh0

be taken no less than the one specified in Table 5.2 In elements with longitudinal reinforcement located uniformly over the section contour as well as in centrally tensioned elements, the minimum cross-sectional area of all longitudinal reinforcement should be taken twice as large as that specified in Table 5.2 and include them into the total area of cross-sectional area of concrete. Elements not meeting the requirements of minimum reinforcement are regarded as concrete elements. Requirements of the present paragraph are not taken into account at determining a cross-sectional area of reinforcement installed along the slab


or panel contour based on calculation of deflection in the plane of slab (panel). 5.12. In concrete structures one should envisage structural reinforcement: - in places of sharp changes of element section dimensions; - in concrete walls under and above apertures; - in eccentrically compressed elements calculated for durability without considering behavior of tensioned concrete, at faces where tensile stresses appear; at that, factor of reinforcement μs is taken no less than 0,025%. Таble 5.2. Operational conditions of reinforcement

μs,min ,%

0,10 1.Reinforcement S in flexural and eccentrically tensioned elements with normal force being applied beyond the effective depth of the section 0,10 2.Reinforcement S and S’ in eccentrically tensioned elements with normal force being applied between the reinforcement S and S’ 3.Reinforcement in eccentrically compressed elements at: 0,10 а) l0 /i < 17 (for rectangular sections - at l0/h ≤ 5) 0,15 б) 17 < l0 /i ≤ 35 (5 < l0 /h ≤ 10) 0,20 в) 35 < l0 /i < 83 (10 < l0 /h <25) 0,25 г) l0 /i ≥ 83 (l0 /h > 25) Note: For eccentrically compressed elements at 17 < l0 /i <83 value μs,min ,(%) can also be determined by means of linear interpolation between values 0,10 and 0,25

5.13. In reinforced concrete linear structures and slabs the ;largest distances between axes of longitudinal reinforcement rods providing for efficient engagement of concrete, uniform distribution of stresses and deformations as well as limitation of cracks opening width between reinforcement rods shall be no greater than: - in reinforced concrete beams and slabs: 200 mm – at cross-section depth h ≤ 150 mm; 1,5 h and 400 mm – at cross-section depth h > 150 mm; - in reinforced concrete columns: 400 mm – in the direction perpendicular to the plane of flection;


500 mm - in the direction of flection. In reinforced concrete walls distances between rods of vertical reinforcement are taken no greater than 2t and 400 mm (t- wall thickness), and those of horizontal reinforcement – no greater than 400 mm. In hollow-core floorings the distances between axes of principal rods are permitted to be increased in accordance with hollow parts of section but no greater than 2h. At reinforcing continuous slabs with welded rolled meshes, close to intermediate supports it is permitted to transfer all lower rods into upper zone. Continuous slabs no greater than 80 mm thick are permitted to be reinforced with single flat meshes without bendings. 5.14. In beams and ribs over 150 mm thick, the number of longitudinal principal tensioned rods in the cross-section shall be no less than two. At the element thickness of 150 mm and less it is permitted to install one longitudinal rod in the cross-section. 5.15. In beams, rods of longitudinal principal reinforcement with cross-sectional area no less than 1/2 of the section area of the rods in the span and no less than two rods should be brought to the support In slabs, rods of longitudinal principal reinforcement over 1 m of the slab width with cross-sectional area no less than 1/3 of the section area of rods over 1 m of the slab thickness in the span and no less than two rods should be brought to the support. 5.16. In flexural elements at section depth over 700 mm, structural longitudinal rods shall be placed at lateral faces, with distances between rods being no greater than 400 mm and cross-sectional area no less than 0,1 % of the cross-sectional area of concrete having a dimension equal by the element height to the distance between these rods, by the width – to a half of the element rib width, but no greater than 200 mm (Diagram 5.1).

Diagram 5.1. Installation of structural longitudinal reinforcement by the depth of beam section


5.17. Diameter of longitudinal rods of eccentrically compressed linear elements of monolithic structures shall be no less than 12 mm. In columns with a dimension less than the section side of 250 mm and over, diameter of longitudinal rods is recommended to be set at no less than 16 mm. In reinforced concrete walls, diameter of longitudinal rods is recommended to be set at no less than 8 mm. CROSSWISE REINFORCEMENT 5.18. Crosswise reinforcement should be installed based upon calculation of force bearing as well as in order to limit cracks formation, retention of longitudinal rods in design position and prevention of their lateral deflection in any direction. Crosswise reinforcements installed at all surfaces of reinforced concrete elements with longitudinal reinforcement installed close to them. At that, distances between crosswise rods at each surface of the element shall be no greater than 600 mm and no greater than doubled width of the element face. Crosswise reinforcement is permitted not to be installed at faces of thin ribs with width of 150 mm and less, having only one longitudinal rod installed in the width. 5.19. In eccentrically compressed elements with their carrying capacity at the given eccentricity of normal force if used at less than 50 %,

as well as in elements with flexibility

l0 < 17 (for instance, windowsills), i

where according to calculation no compressed reinforcement is needed and the number of tensioned reinforcement does not exceed 0,3 % it is permitted not to install crosswise reinforcement required according to para 5.18, along faces parallel to the plane of flection. At that, reinforcement along faces perpendicular to the plane o flection is carried out with welded frames and meshes with protective layer of concrete less than 50 mm thick and no less than two diameters of longitudinal reinforcement. 5.20. Diameter of crosswise reinforcement (stirrups) in tied frames of eccentrically compressed elements is taken no less than 0,25 of the largest diameter of longitudinal reinforcement and no less than 6 mm. Diameter of crosswise reinforcement in tied frames of flexural elements is taken no less than 6 mm. 5.21. In reinforced concrete elements in which according to calculation a transverse force cannot be supported by concrete alone, one


should envisage installation of crosswise reinforcement with spacing no greater than 0,5h0 and no greater than 300 mm. In solid slabs as well as in hollow-core and multiribbed plates less than 300 mm high and in beams (ribs) less than 150 mm high at the element segment where according to calculation a transverse force is supported by concrete only, it is permitted not to install any crosswise reinforcement. In beams and ribs with height of 150 mm and over as well as in multiribbed plates with height of 300 mm and over where according to calculation a transverse force is supported by concrete only, one should envisage installation of crosswise reinforcement with spacing no greater than 0,75 h0 and no greater than 500 mm. 5.22. Bent reinforcement rods shall be envisaged in flexural elements when reinforcement is provided by means of tied frames. Bendings shall be carried out in an arc with radius no less than 10d (Diagram 5.2). In flexural elements at the ends of bent rods straight parts shall be provided with length of no less than 0,8lап, taken according to provisions of para 5.32 but no less than 20d in tensioned zone and 10d — in compressed zone. ≥0,8lап; ≥10d r≥10d

≤50 mm ≥0,8lап ; ≥20d Diagram 5.2. Structure of reinforcement bendings

Straight parts of bent plain rods shall end with hooks. Distance from the face of free support to the upper end of the first bending (counting from the support) shall be no greater than 50 mm. The inclination angle of the bendings to the longitudinal axis of the element shall be taken within 30 - 60°, it is recommended to take the angle equal to 45°. 5.23. in eccentrically compressed linear elements as well as in flexural elements, if proven necessary by calculation compressed longitudinal reinforcement is available, in order to prevent bending of longitudinal reinforcement, one should envisage installation of crosswise reinforcement with spacing no greater than 15d and no greater than 500 mm (d – diameter of compressed longitudinal reinforcement).


If concentration of compressed longitudinal reinforcement installed at one of the element faces exceeds 1,5%, crosswise reinforcement shall be installed with spacing no greater than 10d and no greater than 300 mm. Distances between stirrups of eccentrically compressed elements in places of principal reinforcement jointing by overlapping without welding shall be no greater than 10d. 5.24. Structure of stirrups (crosswise rods) in eccentrically compressed linear elements shall be chosen the way that longitudinal rods (at least every second rod) would be located in places of bending, and these bendings – at a distance of no greater than 400 mm over the face width. At the face width of no greater than 400 mm and the number of longitudinal rods at this face no greater than four, it is permitted to use a single stirrup to wrap all longitudinal rods (Diagram 5.3).

Diagram 5.3. Structure of spatial reinforcement frames in compressed elements

5.25. In reinforced concrete walls crosswise rods normal to the plane of the wall are located at distances no greater than 20d vertically and no greater than 600 mm horizontally. At that, if, according to calculation, the required longitudinal reinforcement has lower concentration than the minimum percentage of reinforcement (see Table 5.2), crosswise rods can be located at distances no greater than 600 mm vertically and no greater than 1000 mm horizontally. At concentration of longitudinal reinforcement in reinforced concrete walls of over 2%, crosswise rods shall be located at distances no greater than 15d vertically and no greater than 500 mm, and horizontally – no greater than 400 mm and no greater than 2 spaces of vertical rods. In this paragraph, d – diameter of vertical rods. 5.26. Crosswise reinforcement in slabs in the zone of pushing in the direction perpendicular to the sides of design contour is installed with spacing no greater than h0/3 and no greater than 300 mm. The closest to the contour of load area rods are installed no closer than h0/3 and no farther than h0/2 from this contour. At that, the width of the crosswise


reinforcement installation area (from the contour of the load area) shall be no less than 1,5h0. Distances between the rods of crosswise reinforcement in the direction parallel to the sides of the design contour are taken no greater than 1/4 of the length of the corresponding side of the design contour. 5.27. Crosswise reinforcement in the form of welded meshes of lateral reinforcement at exposure to local compression (bearing) shall meet the following requirements: а) the mesh rods area per unit of length in one and the other directions shall not differ greater than 1,5 fold; б) spacing of meshes (distance between the meshes in the axes of rods of the same directions) should be taken no less than 60 and no greater than 150 mm; в) mesh cell clearance size shall be no less than 45 and no greater than 100 mm; г) the first mesh is located at the distance of 15-20 mm from the loaded surface of the element. 5.28. Crosswise reinforcement intended to support transverse forces and torque moments shall have a closed contour with reliable anchorage at the ends by means of welding or wrapping with longitudinal reinforcement providing for full-strength connections in crosswise reinforcement. REINFORCEMENT ANCHORING

5.29. reinforcement is anchored using of the following methods or a combination thereof: - as a straight rod end (straight anchoring); - with a bend at the rod end in the form of a hook, fold (pawl) or a loop; - with welding or installation of transverse rods; - using special anchoring devices at the rod end. 5.30. Straight anchoring and anchoring with pawls is allowed for use only with periodic profile reinforcement. hooks, loops, welded transverse rods or special anchoring devices should be provided For extended plain rods. Pawls, hooks and loops are not recommended for use with compressed reinforcement anchoring, except for smooth reinforcement, which can undergo tension under certain load combinations.


5.31. WHEN calculating the length of reinforcement anchoring the following should be considered: anchoring method, reinforcement class and profile, reinforcement diameter, concrete durability and its stressed state in the anchoring zone, design of the element in the anchoring zone (transverse reinforcement, position of rods in the element section, etc.). 5.32. The basic (principal) anchoring length necessary for the transmission of effort in the reinforcement with the total computed value of resistance Rs for concrete is determined using the formula

l0, an=

R s As , Rbond u s

(5.1)

where As and us are the cross-sectional area of the anchored reinforcement rod and the perimeter of its sections determined along the nominal rod diameter respectively; Rbond is the calculated resistance of reinforcement cohesion with concrete taken to evenly distributed along the leng of anchoring and determined using the formula Rbond = η1η 2 Rbt , (5.2) here η1 is the coefficient taking into account the influence of the reinforcement surface type taken to equal: 1.5 for smooth reinforcement (a240 class); 2.0 cold-deformable periodic profile reinforcement (v500 class) 2.5 for hot-rolled and thermomechanically reinforced periodic profile reinforcement (a300, A400 and A500 classes); η2 is the coefficient taking into account the influence of the reinforcement diameter size taken to equal: 1.0 for reinforcement diameter ds ≤ 32 mm; 0.9 for reinforcement diameters 36 and 40 mm. 5.33. The required calculated reinforcement anchoring length taking into account element design in the anchoring zone is determined using the formula lan = αl 0,an

As,cal As ,ef

,

(5.3)

where l0, an is the basic anchoring length determined using formula (5.1); As,cal, As,ef are the cross-sectional areas of reinforcement required according to the calculation with the full design impedance and actually established, respectively;


α is the coefficient taking into account the influence of the concrete and reinforcement stressed state and element design in the anchoring zone on the anchoring length. when anchoring periodic profile rods with the straight ends (straight anchoring) or smooth reinforcement with hooks or loops without additional anchoring devices for extended rods α is taken equal to α = 1.0, and for compressed rods α = 0.75. It is allowed to decrease the anchoring length depending on the quantity and diameter of the transverse reinforcement and the value of transverse concrete compression in the anchoring zone (for example, from bearing pressure), in accordance with the indications in p.3.45. The values of the relative anchoring length λan = lan/ds for rods that work with the full design impedance with diameters less than 36 mm, are provided in table.3.3 p.3.45. in any case the actual anchoring length is taken at no less than 0.3 ⋅ l0,an , and no less than 15ds and 200 mm. 5.34. The effort experienced to the anchored reinforcement rod Ns, is determined using the formula Ns=RsAs

ls ≤ rsAs l an

(5.4)

where lan is determined according to p.5.33 with As,cal/As,ef = 1.0; ls is the distance from the anchored rod end to the cross section of the considered element. 5.35. At the outer free supports of elements the length of overreach of the tie bars over the internal face of the free support, if the Q ≤ 0.5Rbtbh0 condition is met, must be at least 5ds. If the indicated condition is not met, the length of reinforcement overreach over the support face is checked by calculation according to paragraphs.3.43 – 3.46. 5.36. Special measures for the anchoring of longitudinal rods must be adopted when it is impossibile to meet the requirements of p. 5.33: A) a device at the ends of special anchors in the form of plates, washers, nuts, corners, closing heads, etc. (Drawing 5.4). thereby the anchor-to-concrete contact area must meet the concrete crumpling condition (see Section 3.81), and the thickness of the anchoring plate must be at least 1/5 of the total width (diameter) and meet the welding conditions; the length of rod fixation must be calculated for riving and taken at no less than 10d; b) the 90° fold of the anchored rod over the circular arc with a radius in the clear no less than 10d(1– l1/lan) [where l1 IS THE length of THE


straight section at the START of the fixation (Drawing 5.5.)], and no less than the values stipulated in p.5.41; additional stirrups against rods unbending are placed in the bent section; c) welding over the fixation length ls of the transverse anchoring rods; in this case the length of anchoring lap, determined according to p. 5.32, decreases by the length ΔL = lan

Nw [where for Nw see formula (3.75) in Rs As

p.3.45], but more than for 0.43ls; ΔL ≥ 150 mm, plain rods may be supplied without hooks, whereby the value of lan does not decrease.

DRAWING 5.4. reinforcement Anchoring using devices at the ends of special anchors, as a – a welded-on plate; b – pressed plate; C – closing head; d – closing head with washer; e – rod welded-on to the corner; f – nut with an outside washer; g – internal nut


DRAWING 5.5. ANCHORING OF REINFORCEMENT BY FOLDING REINFORCEMENT CONNECTIONS

5.37. one of the following joints types is used to connect reinforcement: A) overlapped joints without welding: - with straight periodic profile rod ends; - with straight rod ends with welding or installation of transverse rods along the length of the overlap; - with bends at the ends (hooks, pawls, loops); thereby only hooks and loops are used for plain rods. b) welded and mechanical butting joints: - reinforcement welding; - using special mechanical devices (joints with the pressed sockets, screwed sockets, etc.). 5.38. the provisions of p.5.30 apply To overlapping reinforcement connections. The joints of extended or compressed reinforcement must have a bypass (overlap) length no less than the value of ll determined using the formula

ll= α ⋅ l0, an

As , cal As, ef

,

(5.5)


where l0,an is the reference anchoring length determined using formula (5.1); As,cal, As,ef - see p.5.33; α is the coefficient taking into account the influence of the stressed reinforcement state, element design in the rods connection zone, amount of butted reinforcement within one section with respect to the total amount of reinforcement in this section, distance between butting rods. when connecting periodic profile reinforcement with straight ends and plain rods with hooks or loops without additional anchoring devices, the α coefficient for extended reinforcement is taken equal to 1.2, and 0.9 for compressed reinforcement. trhereby the following conditions must be met: - the relative amount of working extended periodic profile reinforcement butted in one design level element cross-section must not exceed 50%, for smooth reinforcement (with hooks or loops) this value should not be more than 25%; - the effort experienced by the entire transverse reinforcement placed within the joint limits must be no less than half the effort experienced by the working extended reinforcement butting in one design level element cross-section; - the distance between the butted working reinforcement rods must not exceed 4ds (DRAWING 5.6, A); - the distance between adjacent overlapped joints (along the width of the reinforeced concrete element) must be no less than 2ds and no less than 30 mm (Drawing 5.6, b).

ds


ll

≥2ds; >30mm

ll

ds

0 ≤ e ≤ 4ds

≥2ds; >30mm ≥1.3ll Drawing 5.6 the arrangement of overlapping butted rods and joints a – the arrangement of rods in the joint; b – the arrangement of joints

an element section with a length of 1.3ll along the butted reinforcement is adopted As one design element cross-section examined to determine the relative amount of butted reinforcement within one section. It is considered that the reinforcement joints are located in one design crosssection, if the centers of these joints are within the limits of the length of this section (Drawing 5.6,b). It is allowed to increase the relative amount of working extended reinforcement butted in one design element cross-section to 100%, taking the α coefficient value equal to 2.0. when the relative amount of periodic profile reinforcement butted in a design cross-section is more than 50% or more than 25% for smooth reinforcement, the α coefficient value is determined by linear interpolation. The length of butting rods overlap may can reduced when there are additional anchoring devices at the ends of butting rods (welding of transverse reinforcement, bending of periodic profile butting rod ends, etc.), but not more than by 30%. in any case the actual overlap length must be no less than 0.4α ⋅ l0, an , no less than 20 ds and no less than 250 mm.


5.39. When connecting reinforcement using welding the welded joint types and welding methods are selected taking into account the facility operating conditions, steel weldability and production technology requirements in accordance with the valid regulatory documents (gost 14098-91). 5.40. When using socket-type mechanical devices to connect reinforcement elements (threaded sockets, pressed sockets, etc.) the bearing capacity of the socket connection must be the same as of the butting rods (during extension or compression respectively). The ends of butting rods should be inserted into the socket for the required length determined by calculation or experimentally. When using threaded sockets they must be tightened to eliminate gaps in the thread must. BENT RODS

5.41. when using bent reinforcement (folds, rod ends bends), the minimum diameter of the bend of a separate rod must be such as to avoid the destruction or splitting of concrete inside the reinforcment rod bend and its destruction at the bending point (see paragraphs 5.22, 5.36). The minimum diameter of the dop mount for reinforcement is adopted depending on the diameter of rod ds but no less than: - for the plain rods dop = 2.5 ds for ds < 20 mm; doP = 4 ds for ds ≥ 20 mm; - for periodic profile rods doP = 5ds for ds < 20 mm; doP = 8 ds for ds ≥ 20 mm.

REQUIREMENTS TO CONCRETE AND REINFORCED CONCRETE STRUCTURES 5.42. The sizes of composite concrete and reinforced concrete elements should be assigned taking into account load capacity and dimensional limitations of the technological, transport and installation equipment at the manufacturing plants and at the construction sites. when necessary the possibility of lifting the reinforced concrete item together with the form should be considered.


5.43. To avoid damages from local stress concentrations during an abrupt change in the directions of the element faces (for example, within the internal corners) the outline should be softened with, if possible, small drafts, bevels or roundings (up to 50 mm) so that local reinforcement would not be required (DRAWING 5.7, A, B, C). sharp external corners should be equiped with slants or roundings to avoid concrete splitting off (Drawing 5.7, e). A)

B) C)

D)

Node a

E)

Drawing 5.7. Roundings and bevels a – roundings in a ribbed slab; b – bevel between a flange and the wall in a T-beam; c – A combination of bevel and rounding at a truss knot; d – softening a sharp cross bar corner; e – rounding the opening for communications, slinging, etc.

5.44. Openings in reinforced concrete elements intended for communication lines, slinging etc. should be taken as small as possible and located within the reinforcement grids cells and frames so that it would not be necessary to cut the reinforcement and reinforce on the spot. It is desirable to make the openings corners smooth (Drawing 5.7, d) 5.45. During the design of concrete and reinforced concrete structures their outline should be adopted taking into account the design and method of exploitation of the forms (moulding). when applying forms with the flaps the outline of the article must not prevent the edge from turning (Drawing 5.8,A) during the removal of moulding. when applying permanent forms the possibility of removing the article from them must be ensured using technological inclines of at least 1:10 (Drawing 5.8, b, c). when applying permanent forms using pressing-off the incline must be no less than 1:15 (Drawing 5.8,D). during the immediate removal of the molds while ensuring a fixed (to avoid the damage of concrete) vertical displacement of the molding rigging element (Drawing 5.8, d, e) the incline must not be less than 1:50.

i≥1:50


Drawing 5.8. Technological inclines a – in a form with flaps; b and c – in a permanently attached form; d – the same with the application of the press-off; e and f – for immediate molds removal; g – in a form with a solid edge; h – the same, with the press-off 1 – article; 2 – form; 3 – the flap; 4 – press off; 5 insert

when using forms with one fixed and one flap side the shift from a larger to a smaller article width [for example, from the lower flange to the wall (Drawing 5.8, G)] should be adopted as smooth with an of AT least 45° to ensure the possibility of vertical lift during the removal of moulding. This requirement may be disregarded, if the form is equiped with a pressing device (Drawing 5.8.3). The use of pressing off and the immediate removal of molds must be coordinated with the item’s manufacturer. 5.46. when designing composite reinforced concrete articles a convenient methods of their acquisition using load-gripping devices during the removal from the form (removal of the molds) and during loadingunloading and installation should be provided.


the Methods and points of acquisition should be designated taking into account the items’s production and installation technology and its design features. the item must be checked by calculation for the operating conditions under the adopted method and arrangement of acquisiotion points. 5.47. slinging devices should be provided In concrete and reinforced concrete items: sling openings (including for inventory loops), grooves, ledges, etc. or stationary steel sling loops, which must be made from hotrolled steel according to p. 2.16. items should be acquired as far as possible without applying devices that require the expenditure of steel through the creation of depressions, grooves, openings, ledges, etc. (Drawing 5.9). 5.48. standard loops should be used for the design of items with sling loops. in the absence of standard loops with the required characteristics it si recommended to design loops of the types indicated in Drawing 5.10. The minimum parameters for loops with straight and bent branches of p1,1 and P2,1 types (see Drawing 5.10) are indicated in table 5.3. 5.49. The diameter of loop d rod should be adopted according to table 5.4 depending on the mass of the item applied to the loop. The mass of the item article is determined according to the provisions in p. 2.12. when lifting flat items using four loops, the mass of the items is considered to be distributed over three loops.

daig.5.9. Examples of sling devices without loops a – during block slinging; b – sling openings in a column; c – a combination of two different sling devices in one item 1 – cargo slings; 2 – acquisition cut; 3 – acquisition opening; 4 – acquisition loop for extraction from form


DRAWING 5.10. TYPES OF SLING LOOPS a – freely placed in items made of steel classes A240 and A300; b – placed in constricted conditions made of steel class A240; c – the same, made of A300 steel

Table 5.3 Loops

Designations of sizes

Sizes

d

6-12

14; 16

18-22

25

R

30

30

40

60

R

20

30

40

60

a1

3d

a2

6d


Table 5.4 Diameter of loop rod, mm

Mass of item m, kg, assigned during lifting to one loop made of steel of class:

Diameter of loop rod, mm

Mass of item m, kg, assigned during lifting to one loop made of steel of class:

A 240 A 300 A 240 A 300 ' 150 18 2500 3100 − 8 300 20 3100 3900 − 10 700 22 3800 4700 900 12 1100 25 4900 6100 1500 14 1500 28 6100 7600 1900 16 2000 32 8000 9900 2500 Notes: 1. the Values of t correspond to the angle between the slings and the horizon equal to 45° and more; smaller slope angles are not allowed. If items slinging using vertical slings is provided, during the selection of loop diameter it is allowed to decrease the item mass applied to the loop 1.4 times. 2. for a loop rod diameter from 8 to 22 mm (incl.) it is allowed to increase the provided values of t by 25% with a special substantiation.

When lifting using three loops and more located at the same side of the item (for example, on the wall panel), item mass is tken as distributed to only two loops; therefore, in this case it is not recommended to install more than two loops. when using devices (self-balancing traverses) ensuring the selfbalancing of the efforts between the slings, it is allowed to distribute the item mass between the loops in accordance with the device design. 5.50. The gudgeon height he (Drawing 5.10) that corresponds to the sizes of the cargo slings hoisting hooks should be taken equal to (in mm): 60… for loop rod diameter from 6 to 16 mm; 80… „ „ „ „ „ „ from 18 and 22 mm; 150. „ „ „ „ „ „ from 25 to 32 mm. Length ls and penetration depth hb of the ends of loop branches into the item’s concrete (see Drawing 5.10) should be taken according to table 5.5 when arranging sling loops in standard deepenings (Drawing 5.11,A) the value of hb may be counted off from the upper surface of the concrete element.


Table 5.5 standard cubic durability of concrete at the moment of first item lifting, MPa From 3 to 7

Length of penetration into concrete ls

Depth of penetration into concrete hb

45d (50d) 35d (40d) 30d (35d) 25d (30d) 20d (25d)

35d (40d) 25d (30d) 20d (25d) 15d (20d) 15d (20d)

SV 5 to 10 ,, 10 ,, 20 ,, 20 ,, 30 SV 30 Note. The values indicated in brackets refer to the cases of vertical lifting single-layer thin-walled elements (such as heavy-aggregate concrete wall panels) with a thickness not exceeding 220 mm.

Drawing 5.11. dimensions of holes for penetrating sling loops openings a – closed deepenings; b – extended deepenings (at the item edges) for a loop rod diameter 6-16mm: R1=125mm, a=30mm, b1=50mm, l1=25mm, l2=30mm; for a loop diameter of 18-22mm: R1=150mm, a=40mm, b1=65mm, l1=30mm, l2=30mm

In all cases the value of ls should be taken no less than 200 mm. For loops made of reinforced steel ∅ 25A240 and ∅28a300 and more the values of ls and hb should be increased by 20%. loop branches made of class a240 steel, as well as straight (without folds) loop branches made of class a300 steel must end in hooks.


where necessary, it is allowed to arrange the branches at an angle to each other not exceeding 45°. The distance between the side surface of the loop hook tail section and the item surface, measured in the hook plane, should be adopted no less than 4d (Drawing 5.10, A). if it is not possible to penetrate the loop ends for the necessary length, the loop must be anchored using other methods, for example, by welding to inset components, by passing behind operating longitudinal reinforcement, etc. the reliability of the adopted loop anchoring should be confirmed by calculation or checks. 5.51. It is allowed to okace sling loops in deepenings so that their openings were located below the concrete or reinforced concrete item surface. This arrangement is especially recommended for mechanized concrete surface finishing when protruding loops would impede such finishing. loop Deepenings may be closed (see Drawing 5.11, A) or open (Drawing 5.11, b). In the latter case the accumulation of water that may freeze in them is prevented, and the conditions of the loop fixation are also improved. due to the condition of passing a sling hoisting hook into the opening the hole should be displaced with respect to the middle of the item in the opening’s plane. REINFORCEMENT FIXATION

5.52. THE CORRESPONDENCE OF REINFORCEMENT ARRANGEMENT TO ITS DESIGN POSITION MUST BE ENSURED BY APPLYING FIXATION MEANS. reinforcement should be fixated using: A) SINGLE USE DEVICES THAT REMAIN IN THE CONCRETE; b) inventory devices extracted from the concrete before or after its hardening: c) special components fastened to the working surface of form or moulding that do not impede the extraction of the reinforced concrete element from the form or the removal of the moulding from it. 5.53. the following single use lock pins should be used: A) to ensure the required protective concrete layer thickness – according to Drawing 5.12;


Drawing 5.12. single use lock pins ensuring the required protective concrete layer thickness S a – c – with a large form contact surface, manufactured from cement mortar; d – with a small form contact surface, made from cement-sand solution; e – the same, made asbestos cement; f-h – the same, made of plastic (perforated); i – the same, made of a perforated aluminum strip; j-k – the same, made of reinforcement steel; 1 – form working surface; 2 – lockng pin; 3 – fixed reinforcement; 4 – twist from binding wire; 5 – binding wire sealed in the locking pin; 6 – possible elastic ring; 7 – supports welded to the reinforcement

b) in order to ensure the required distance between the separate reinforcement items or the rods - according to Drawing 5.13; c) in order to meet the requirements indicated in subparagraphs “a” and “b” - according to Drawing 5.14. The type of locking pin to ensure the thickness of the protective concrete layer on the top faces of elements should be selected according to the requirements in table 5.6. It is not allowed to use cut sections of reinforcement rods, plates, etc as locking pins.


Drawing 5.13. single use locking pins ensuring the required distance a-c – between separate reinforcement items; d – between rods 1 – reinforcement steel separator installed between grid rows; 2 – locking pin - lining to ensure the protective concrete layer; 3 – elongated transverse form rods bent around the grid rods; 4 – locking pin to connect crossing rods (spring wire helix); 5 – truss point

IT IS NOT ALLOWED TO SET PLASTIC LINING UNDER WORKING REINFORCEMENT RODS OR CLOSE TO THEM, UNDER DISTRIBUTIVE REINFORCEMENT RODS IN THE CONCRETE TENSION AREA OF ELEMENTS OPERATED UNDER AGGRESSIVE CONDITIONS. IN SUCH ITEMS DENSE CEMENTSAND SOLUTION, CONCRETE OR ASBESTOS CEMENT LINING SHOULD BE USED PREDOMINANTLY. 5.54. when using single use locking pins in accordance with the requirements of table 5.6 the working drawings should indicate, which of these locking pins are allowed in this element. The thickness of the protective concrete layer at the locking pin lining installation spot should be adopted in multiples of 5 mm.


Drawing 5.14. single use locking pins simultaneously ensuring the required protective concrete layer thickness and the distance between separate reinforcing elements a – in flat plates; b and C – in rectangular cross section girders; d – in circular cross section elements 1 – inverted U-type form locking pin; 2 – reinforcement grids; 3 – the working surface of the form; 4 – rack-body type locking pin; 5 – flat reinforcing form; 6 – locking pin rod welded additionally to the forms; 7 – applied clamp type locking pin made of reinforcement wire; 8 – concentrical forms; 9 – truss point

drawings should be made For single use locking pins made of reinforcement steel. the working drawings of reinforcement items and, if necessary, the general drawings of reinforced concrete elements reinforcement should show the location of these locking pins or supporting rods, and the specifications should stipulate the expenditure of steel for their manufacture. The arrangement and quantity of nonmetallic locking pins linings may not be provided in the working drawings.


Table 5.6 TYPE OF LOCKING PIN element Operating conditions open air

rooms with normal humidity

type of element top face

Solution, concrete, asbestos cement rM rB

Plastic (polyethylene)

Steel

PM

PB

SZ

CH

Clean concrete for painting; coated by ceramic slab, etc. during concreting

+

-

+

-

+

-

Processed by the mechanical method

+

-

-

-

-

-

Clean concrete

+

-

+

-

+

-

+

×

+

×

+

×

Concrete for painting with the aqueous compounds

Concrete for painting with oil, enamel and + + + + + + synthetic paints; concrete for revetment Concrete for wall+ + + + + papering Notes: 1.legend: R - solution, concrete, asbestos cement locking pins; P – plastic, polyethylene locking pins; C – steel locking pins; M - small locking pin contact surface with the form (moulding); B – large lokcing pin contact surface with the form (moulding); z - locking pins protected from corrosion; N - locking pins not protected from corrosion. 2. sicn “+”allowed; sign “-” not allowed; sign “×” Allowed but not recommended.

APPENDIX 1 STANDARD REINFORCEMENT GAUGE ROD NOMIN AL DIAME TER OF, mm

EFFECTIVE AREA OF CROSSWISE RODS, MM2, WITH CERTAIN NUMBER OF RODS

Теоретическая масса 1м длины арматуры, кг

Диаметр арматуры классов A240 A400 A500

Максимальный размер сечения стержня A300 B500 периоди -ческого профиля

1

2

3

4

5

6

7

8

9

3

7.1

14.1

21.2

28.3

35.3

42.4

49.5

56.5

63.6

0,052

-

-

+

-

4

12.6

25.1

37.7

50.2

62.8

75.4

87.9

100.5

113

0,092

-

-

+

-

5

19.6

39.3

58.9

78.5

98.2

117.8

137.5

157.1

176.7

0,144

-

-

+

-

6

28.3

57

85

113

141

170

198

226

254

0,222

+

-

+

6,75

8

50.3

101

151

201

251

302

352

402

453

0,395

+

-

+

9,0

10

78.5

157

236

314

393

471

550

628

707

0,617

+

+

+

11,3

12

113.1

226

339

452

565

679

792

905

1018

0,888

+

+

+

13,5

14

153.9

308

462

616

769

923

1077

1231

1385

1,208

+

+

-

15,5

16

201.1

402

603

804

1005

1206

1407

1608

1810

1,578

+

+

-

18

ROD NOMIN AL DIAME TER OF, mm

EFFECTIVE AREA OF CROSSWISE RODS, MM2, WITH CERTAIN NUMBER OF RODS

Теоретическая масса 1м длины арматуры, кг

Диаметр арматуры классов A240 A400 A500

Максимальный размер сечения стержня A300 B500 периоди -ческого профиля

1

2

3

4

5

6

7

8

9

18

254.5

509

763

1018

1272

1527

1781

2036

2290

1,998

+

+

-

20

20

314.2

628

942

1256

1571

1885

2199

2513

2828

2,466

+

+

-

22

22

380.1

760

1140

1520

1900

2281

2661

3041

3421

2,984

+

+

-

24

25

490.9

982

1473

1963

2454

2945

3436

3927

4418

3,84

+

+

-

27

28

615.8

1232

1847

2463

3079

3685

4310

4926

5542

4,83

+

+

-

30,5

32

804.3

1609

2413

3217

4021

4826

5630

6434

7238

6,31

+

+

-

34,5

36

1017.9 2036

3054

4072

5089

6107

7125

8143

9161

7,99

+

+

-

39,5

40

1256.6 2513

3770

5027

6283

7540

8796

10053

11310

9,865

+

+

-

43,5

45

1590.4 3181

4771

6362

7952

9542

11133

12723

14313

12,49

-

+

-

49

50

1963.5 3927

5891

7854

9818

11781

13745

15708

17672

15,41

-

+

-

54

55

2376

4752

7128

9504

11880

14256

16632

19008

21384

18,65

-

+

-

59

60

2827

5654

8481

11308

14135

16962

19789

22616

25443

22,19

-

+

-

64

70

3848

7696 11544

15392

19240

23088

26936 30784

34632

30,46

-

+

-

74,0

NOTE: 1. Nominal diameter of RODS for isteg reinforcement steel corresponds to

nominal diameter of equareal RODS. actual dimensions of high bond RODS ARE SPECIFIED IN ACCORDANCE WITH GOST 5781 – 82. 2. symbol “+” denotes availability of the diameter in mix of this class

207

REFERENCE MANUAL FOR CONCRETE AND REINFORCED CONCRETE STRUCTURES WITHOUT REINFORCEMENT PRETENSIONING BASED ON SP 52-101-2003

Appendix 2 BASIC LETTER DESIGNATIONS FORCES FROM EXTERNAL LOADS AND ACTIONS IN ELEMENT CROSS SECTION

М – MOMENT OF FLECTION; N – NORMAL FORCE; Q – TRANSVERSE FORCE; Т – TORQUE MOMENT. MATERIAL CHARACTERISTICS

Rb,n, - characteristic axial compressive strength of concrete; Rb, Rb,ser - design axial compressiON strength of concrete for limit states of the first and second groups, respectively; Rbt,n - characteristic axial tensile strength; Rbt,Rbt,ser - design axial tensile strength of concrete for limit states of the first and second groups, respectively; Rs, Rs,ser- design axial tensile strength of reinforcement for limit states of the first and second groups, respectively; Rsw - design tensile strength of transverse reinforcement; Rsc - design compressiON strength of reinforcement for limit states of the first group; Eb - initial modulus of concrete elasticity for compression and tension; Es - initial modulus of reinforcement elasticity. CHARACTERISTICS OF LONGITUDINAL REINFORCEMENT LOCATION

S – designation of longitudinal reinforcement: a) located in tensile area when compressed and tensile areas are present due to external loads; b) located at the least compressive cross section edge when crosssection is completely compressive due to external load;


c) located at the most tensile cross-section edge when cross section is completely tensile due to external load; completely in transverse cross-section for centric tensile elements; S ' – designation of longitudinal reinforcement: а) located in compressed area when compressed and tensile areas are present due to external loads; б) located at the most compressive cross section edge when crosssection is completely compressive due to external load, в) located at the least cross-section tensile edge when crosssection of eccentric tensile elements is completely tensile. GEOMETRIC DATA

- WIDTH OF RECTANGULAR CROSS SECTION; rib width OF T- AND I-SECTIONS; bf , b'f - WIDTH OF FLANGE OF T- AND I-SECTIONS IN TENSILE AND COMPRESSIVE AREAS RESPECTIVELY; h - HEIGHT OF RECTANGULAR, T- AND I-SECTIONS; hf, h'f - HEIGHT OF FLANGE OF T- AND ISECTIONS IN TENSILE AND COMPRESSed AREAS RESPECTIVELY; a, a' - DISTANCE FROM FORCE RESULTANT IN REINFORCEMENT S AND S', RESPECTIVELY, TO THE NEAREST CROSS SECTION EDGE; h0 - OPERATING HEIGHT OF CROSS SECTION EQUAL TO h - а; x - HEIGHT OF CONCRETE COMPRESSIVE AREA; ξ - RELATIVE HEIGHT OF CONCRETE COMPRESSed b

AREA EQUAL TO sw е0

x ; h0

- DISTANCE BETWEEN STIRRUPS MEASURED ALONG THE ELEMENT LENGTH; ECCENTRICITY OF NORMAL FORCE N WITH RESPECT TO CENTER OF GRAVITY OF THE TRANSFORMED CROSS SECTION DEterminED WITH ALLOWANCE FOR INSTRUCTIONS OF PARAGRAPH 3.6;


- DISTANCE FROM POINT OF APPLICATION OF NORMAL FORCE N TO FORCE RESULTANT IN REINFORCEMENT S AND S′, RESPECTIVELY. l - ELEMENT BAY; l0 - DESIGN LENGTH OF THE ELEMENT SUBJECT TO ACTION OF COMPRESSive NORMAL FORCE; i - radius of inertia of element cross-section in relation to the gravity center of its section; ds, dsw - NORMINAL DIAMETERS OF RODS OF LONGITUDINAL AND TRANSVERSE REINFORCEMENT, RESPECTIVELY; As, A's - SECTIONAL AREA OF REINFORCEMENT S AND S', RESPECTIVELY; Asw - SECTIONAL AREA OF STIRRUPS LOCATED IN THE SINGLE PLANE NORMAL TO LONGITUDINAL ELEMENT AXIS CROSSING A SLOPING CROSS SECTION; μs - PERCENTAGE OF REINFORCEMENT DEFINED AS THE RATIO OF REINFORCEMENT SECTIONAL AREA S TO ELEMENT CROSS SECTIONAL AREA bh0 WITHOUT REGARD TO OVERHANGS OF COMPRESSed AND TENSILE FLANGES; A - SQUARE OF THE ENTIRE CONCRETE IN NORMAL CROSS SECTION; Ab - SECTIONAL AREA OF CONCRETE OF COMPRESSed AREA: I - MOMENT OF INERTIA OF A SECTION OF THE ENTIRE CONCRETE WITH RESPECT TO gravity center OF CROSS SECTION; Dcir - DIAMETER OF ANNULAR AND circular CROSS SECTIONS. е, е'


TABLE OF CONTENTS Preface ……………………………………………….………………….…3 1. General recommendations………………………………………………..4 Basic provisions ……..…………………………………………………...4 Basic design requirements………..……………………………………....5 2. Materials for concrete and reinforced concrete structures…….….…...…6 Concrete..…………………………………………………………………6 Concrete quality characteristics and their design application…..……..6 Specified and design concrete characteristics………..………………..8 Reinforcement..………………………………………………………...10 Reinforcement quality characteristics………………………………..10 Specified and design reinforcement characteristics…….…………....11 3. Calculation of limit states of the first group for concrete and reinforced concrete elements……………………… …………………...………...13 Calculation of durability for concrete elements…….…………………..13 General provisions…………………………………………………...13 Calculation for eccentrically compressed elements..…………………14 Calculation of flexural elements……………………………………...18 Calculation examples…………………………………………………19 Calculation of durability for reinforced concrete elements……………..20 Flexural elements……..………………………………………………20 Calculation of durability of reinforced concrete elements under the impact of flection moments…….…………………………………20 General provisions………………………………………………20 Rectangular sections……………………………………………..22 Т- and I-sections………………….……………………….……..24 Calculation examples………………………….…………………26 Rectangular sections………………………..…………………26 Т- and I-sections……….……………………………………...28 Elements in biaxial bending…………….……………………….30 Calculation examples……………………..…..………………35 Calculation of reinforced concrete elements under the impact of transverse forces…………..……………………………………...38 Calculation of reinforced concrete elements along the strip between the inclined sections……………………….……….....39 Calculation of reinforced concrete elements along the inclined sections for the impact of transverse forces………………….…39


Constant height element, reinforced by stirrups perpendicular to the element axis…………………..……………….……………39 Elements with variable height and crosswise reinforcement..…………………………………………….….43 Elements reinforced with offset bends…………………….…..45 Elements without crosswise reinforcement..……….……...…..47 Calculation of inclined sections of reinforced concrete elements for the impact of transverse forces……………………..……..49 Calculation examples…………………………………....…....56 Eccentrically compressed elements..……..……………………….…70 General provisions…………………………………………….…..70 Calculation for the impact of transverse forces….………………...71 Consideration of elements deflection..………………………….….71 Calculation of limit forces for normal sections…………. …….…..74 Rectangular sections with symmetrical reinforcement ...…….….74 Rectangular sections with asymmetrical reinforcement …….......78 I-sections with symmetrical reinforcement ………….…….….…79 Ring sections……………………………..………………….…...81 Circular sections..……………………….…………………….…83 Elements in biaxial eccentrical compression..……………..….…85 Calculation examples…………..………...………………………86 Rectangular sections with symmetrical reinforcement ………..86 Rectangular sections with asymmetrical reinforcement ….…...94 I-sections………………….…………………………………...95 Annular sections……..………………………………………...99 Circular sections......………………………………………….101 Elements in oblique eccentrical compression…………..……102 Calculation of inclined sections..… ……………..….….……104 Centrally and eccentrically tensioned elements……...……..……….105 Centrally tensioned elements……… …………………………..105 Eccentrically tensioned elements…….…………………..…….105 Calculation examples …………………………...………..….108 Calculation of normal sections based on nonlinear deformational model 110 Elements exposed to bending with torsion…………..……….……..115 Rectangular section elements……….………..………………...115 Calculation of combined impact of torque moment and moment of flection………………..…………………………….…....115


Calculation of combined impact of torque moment and transverse force……………………………………………...118 Calculation examples………………...………………..……..121 Calculation of reinforced concrete elements for the local impact of loads…..……………………………………………………………125 Calculation of local compression..………………………..……...125 Calculation examples…………………………………………..128 Сalculation of elements for pushing.........................…………..…129 General provisions……………………………………………..129 Calculation of pushing for elements without crosswise reinforcement.………………………………………………...131 Calculation of pushing for elements with crosswise reinforcement…………………………..……………………..133 Calculation examples………...…………………………………133 4. Calculation of limit states of the second group for reinforced concrete elements…………………………………………………………….….142 Calculation of cracks opening in reinforced concrete elements....……142 General provisions.…….……………………………………………142 Determining the cracks formation moment….…………..………….143 Determining the opening width of cracks normal to longitudinal axis of the element ……………………….………………………………..147 Calculation examples…….………………………………………..153 Calculation of deformations for reinforced concrete structures..……...159 General provisions.……...…………………………………………..159 Calculation of deflection for reinforced concrete elements…………160 Determining of skewness in reinforced concrete elements………….164 General provisions..……………………………………….………164 SKEWNESS OF REINFORCED CONCRETE ELEMENTS IN SEGMENTS WITHOUT CRACKS IN TENSIONED ZONE……………………………………..……………………..……………….…….165 SKEWNESS OF REINFORCED CONCRETE ELEMENTS IN SEGMENTS WITH CRACKS IN TENSIONED ZONE……………………………….…………………………………..……….….….166 DETERMINING SKEWNESS OF REINFORCED CONCRETE ELEMENTS BASED ON NONLINEAR DEFORMATIONAL MODEL ……………………………..……………………...…170 DETERMINING SHEARING ANGLES OF REINFORCED CONCRETE ELEMENTS.........………173

Calculation examples………...……………………………………173 5.Structural requirements……..…………………………………………178 General provisions.…..………………………………………………..178 Geometrical dimensions of structures….………………………...…...178


Reinforcement...……………..………………………………………...179 Protective layer of concrete..……………………………………..179 Minimum distances between reinforcement rods………….…...…181 Longitudinal reinforcement……...………………………………..182 Crosswise reinforcement …………...…………...………………..184 Reinforcement anchoring..………….……..……………………...188 Reinforcement connections……..……………………………….191 Bent rods……….………………………...………………………..194 Requirements to concrete and reinforced concrete structures……...194 Reinforcement fixation.……………..……………………………...201 Appendix 1…………..……………………………………………….….206 Appendix 2………………………………..…………………….……….208

SP 52-101-2003, Reference Materials ENG

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