-
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
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
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 (600 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.
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
THE REFERENCE MANUAL HAS BEEN DEVELOPED BY
CNIIPROMZDANII (ENGINEER I.. NIKITIN, DOCTORS OF ENGINEERING
E.N. KODISH AND N.N. TRIOKIN) WITH THE PARTICIPATION OF NIIZHB
(DOCTORS OF ENGINEERING .S. ZALESOV, .. 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.
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
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 + 50 and no lower than - 40 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: ) 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;
) 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 peoples 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.
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
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.
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
It is permissible to assume lower values of service factor
justified in
the established procedure but no lower than 1.25
2. MATERIALS FOR CONCRETE AND REIFORCED CONCRETE STRUCTURES
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: 10; 15; 20; 25; 30; 35; 40; 45; 50; 55;
60;
b) Class of axial tensile strength: Bt0,8; Bt1,2; t1,6; Bt2,0;
t2,4; Bt2,8; Bt3,2;
c) Frost resistance grade: F50; F75; F100; F150; F200; F300;
F400; F500;
) 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.).
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
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 15;
herewith for heavy-loaded compressed axial elements it is
recommended to assign concrete class no lower than 25.
For concrete compressed elements it is not recommended to assign
concrete class higher than 30.
2.5. For aboveground structures which are subject to weather
impact at design winter temperature from - 5 to - 40, 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
- 5.
Note. Design winter temperature of outdoor air is assigned in
accordance with paragraph 1.5.
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
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 in accordance with Table 2.1.
Table 2.1.
Strength type
Specified and design values of concrete strength Rb, and Rbt,n
for limit states of the second group Rb,ser and Rbt,ser,, MPa
(kgf/cm2) with quality
class of concrete compression strength 10 15 20 25 30 35 40 45
50 55 60
Axial compression Rb,,Rb,ser
7,5 (76,5)
11,0 (112)
15,0 (153)
18,5 (188)
22,0(224)
25,5 (260)
29,0 (296)
32,0 (326)
36,0 (367)
39,5 (403)
43,0 (438)
Tension Rbt,,Rbt,ser
0,85 (8,7)
1,10 (11,2)
1,35 (13,8)
1,55 (15,8)
1,75 (17,8)
1,95 (19,9)
2,10 (21,4)
2,25 (22,9)
2,45 (25,0)
2,60 (26,5)
2,75(28,0)
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:
, ; ,,
bt
nbtbt
b
bb
RR
RR == (2.1)
where b is a safety factor for concrete compression strength
which is taken equal to 1.3;
bt - 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
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
Table 2.2
Strength type
Concrete design strength for limit states of the first group Rb
and Rbt, MPa (kgf/cm2) with concrete quality class regarding
compression strength
10 15 20 25 30 35 40 45 50 55 60
Axial compression,
Rb
6.0 (61.2)
8.5 (86.6
)
11.5
(117)
14.5 (148)
17.0 (173)
19.5 (199)
22.0 (224)
25.0 (255)
27.5 (280)
30.0 (306)
33.0 (336)
Axial tension, Rbt
0.56(5.7
)
0.75 (7.6
)
0.90
(9.2)
1.05 (10.7)
1.15 (11.7)
1.30 (13.3)
1.40 14.3)
1.50 (15.3)
1.60 (16.3)
1.70 (17.3)
1.80 (18.3)
ble 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
t0.8 t1.2 t1.6 t2.0 t2.4 t2.8 t3.2 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 b = bt = 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 bi if required:
) b1 = 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;
) b2 = 0.9 used for concrete structures introduced to design
value Rb;
) b3 = 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 b is taken depending on concrete quality class regarding
compression strength in accordance with Table 2.4
2.10. It is permissible to assume value of Poissons ratio as b,P
= 0.2. Shearing modulus of elasticity G is taken equal to 0.4 of
the
corresponding value b, specified in Table 2.4.
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
2.11. Values of linear thermal deformation coefficient for
concrete
with temperature gradient in the range from -40 up to +50 are
taken as bt =1.10-5 C-1.
Table 2.4
Values of initial tangent modulus of concrete elasticity with
compression and tension b
.10-3, MPa (kgf/cm2), with concrete compression strength class
10 15 20 25 30 35 40 45 50 55 60 19,0 (194)
24,0 (245)
27,5 (280)
30,0 (306)
32,5 (331)
34,5 (352)
36,0 (367)
37,0 (377)
38,0 (387)
39,0 (398)
39,5 (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 (b0, b1,red, b2)
and tensile concrete (bt0, bt1red and bt2) as well as coefficient
of concrete creep b,cr are given in paragraphs 4.27 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 240 (-I); -
hot-rolled and thermo-mechanical hardened Isteg reinforcement
300 (-II), 400 (-III, A400), A500 (A500); - cold-deformed Isteg
reinforcement of class 500 (-I, 500).
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
As reinforcement used in structures in accordance with
calculation it
is recommended to apply mainly: Isteg reinforcement of classes
500 and 400 ; Isteg reinforcement of class 500 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 -
30 it is permissible to apply reinforcement class 300 of steel
grade St5ps 18 - 40 mm in diameter as well as class 240 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 f = 1.0.
Other types and classes of reinforcement might be applied
without restrictions.
2.16. Hot-rolled reinforcement of class 240 of steel grade St3sp
and St3ps as well as of class 300 of steel grade 10G 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,, 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
s
nss
RR
,= , (2.2) where s is partial safety factor of reinforcement
strength, which is taken equal to:
1.1 for reinforcement class 240, 300 and 400; 1.15 for
reinforcement class 500; 1.2 for reinforcement class 500.
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
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
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
Table 2.5
Reinforcement class Specified diameter of
reinforcement, mm
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)
240 300 400 500 500
6 - 40 10 - 70 6 - 40 6 - 40 3 - 12
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 500 for which Rsc = 400 MPa and reinforcement
class 500 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 500 and 500.
Table 2.6.
Reinforcement class
Design values of reinforcement strength for limit states of the
first group, MPa (kgf/cm2)
Compression
Tension, Rsc longitudinal, Rs transverse (stirrups and diagonal
bars),
Rsw240 215 (2190) 170 (1730) 215 (2190) 300 270 (2750) 215
(2190) 270 (2750) 400 355 (3620) 285 (2900) 355 (3620) 500 435
(4430) 300 (3060) 400 (4080) 500 415 (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 s1 = 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 s is taken to
be the same for both compression and tension and equal to Es =
2.0.105 MPa = 2.0.106 kgf/cm2.
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
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 b1 = 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.
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
CALCULATION FOR ECCENTRICALLY COMPRESSED
ELEMENTS
3.6. Calculation for eccentrically compressed elements shall be
carried out with allowance for accidental eccentricity 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 0 is taken equal to eccentricity value
derived from static calculation, however, no less than .
For statically determinate structures eccentricity 0 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 0 by coefficient 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
N Rb Ab, (3.1), 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
,21 0
=hbhAb
(3.2) where for 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.
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
h
bR
eo
N
b
bA
2
1
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 b is determined using the formula
=
be
hebhA yyxxb
00 2121 , (3.3) where e0x and e0y are eccentricities of force N
in direction with respect to
cross-section size h and b. x and y are coefficients 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 regrdless calculation
using condition (3.1) shall be checked with allowance for
resistance of concrete of tensile zone using the condition
10
t
bt
yeIA
ARN
, (3.4)
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
where t is distance from gravity center of element cross-section
to the most
tensile fiber; for refer to paragraph 3.10.
For elements of a rectangular cross-section condition (3.4) can
be written in the form
16 0
he
bhRN bt . (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 with allowance for deflection effect
on amount of eccentricity of normal force 0 is calculated using
formula
crNN
=1
1 , (3.6)
where Ncr is nominal critical force determined using formula
20
2
lDNcr
= , (3.7) where D is element stiffness in strength limit state
calculated using
the formula ;
)3,0(15,0
elb IED += (3.8)
l0 is determined using Table 3.1. Table 3.1.
Type of wall and pole bearing
Design length l0 of eccentrically
compressed concrete elements
1. With supports upwards and downwards: ) with flap hinges on
either side regardless of bearing
displacement value ) with one end restraint and possible
bearing
displacement of: single-aisle building
multiple-aisle building ) with partial restraint of fixed
supports
2. Free standing buildings
1.2 1.5 0.8
2
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
Note. 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
)3.0(80
3
l
bbhED += . (3.8) In formulae (3.8) and (3.8): l is a coefficient
taking into consideration impact of long-term load
on vertical deflection in limit state which is equal to
1
11MM l
l += , (3.9) however, no more than 2;
! 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; is a
coefficient taken to be equal to )/h, however no less than 0.15.
For walls and poles with elastically fixed supports the specified
value
is taken for calculation of cross-sections in the middle 1/3 of
height . When calculation is carried out for support
cross-sections, it is taken that = 1.0, for all other
cross-sections using linear interpolation.
If a lower support is stiffly restrained, then with an elastic
upper support value determined using the formula (3.6) is taken for
sections of the lower section with height of 2/3..
3.11. Calculation with allowance for deflection of eccentrically
compressed concrete elements of a rectangular cross-section of
class not higher than 20 with l0 20h is permissible to carry out
using condition
N nubby, (3.10) Where the value is determined using the drawing
(3.2) depending on values
E0/h and = lo/h. 3.12. When impacts of normal forces are
sufficient, the following
condition shall be met 0,1+
b
mc
bt
mtRR
, (3.11) where tm mc are main tensile and compressed stress
calculated using the
formula
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
22
22 +
+= xx
mcmt
m , (3.12)
and - normal and shearing stress in the considered section fiber
calculated the same way as for elastic section fiber.
0.000
0.100
0.200
0.300
0.400
0.500
0.600
0.700
0.800
0.900
1.000
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50
Drawing.3.2. Diagram of load-carrying capacity of eccentrically
compressed concrete elements
Graphic symbols: with 0,111 =MM l ; with 5,011 =MM l ;
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 I-sections at the level of contact of compression flanges
and section wall.
CALCULATION OF FLEXURAL ELEMENTS
E0/H
N
=5
=0
=10
=15
=20
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
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 cross-
section 6
2bhW = . In addition for elements of T- and I-sections the
following condition
shall be met: Rbt, (3.14)
where - 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 = 2.7 m, manufactured upright (in a cassette);
concrete of class 15 (b= 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 , determined in accordance with paragraph 3.6.
Since mm, 10 mm 5,46002700
600 and mm 10 mm 5
30150
30
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
1233
10166,1)15,03,0(93,180
150100024000)3,0(80
=+=+= l
bbhED . mm2.
Then
.797,1
15787001
1
-1
1
kN; 3,1578103,15782700
10166,1 32
122
20
2
=
==
====
cr
cr
NN
Nl
DN
Concrete design strength Rb is taken in accordance with
paragraph 2.8 with allowance for coefficients b2 = 0.9 and b3 =
0.9. Taking into consideration occurrence of short-term loads one
can assume that b1 = 1.0. Then Rb = 8.5 . 0.9. 0,9 = 6,89 MPa.
Let us check condition (3.1) using formula (3.2) 6.784N
784635)797.1067,021(150100089.6
21 0 ===
=
hbhRAR bbb
kN > 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 l = 2, and then
.745.11523/6501
1 and 4,1523293,13.1578 ==== crN
Design strength Rb is taken with allowance for b1 = 0.9: Rb =
6.89. 0.9 = 6.2 N.
kN 650 kN 6.713N713620150
745,1102115010002.6 =>==
= NAR bb , 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
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
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
0hx= , determined using respective equilibrium
conditions, and value of boundary relative height of compressed
zone R, with which limit state is reached simultaneously with
reaching tensile reinforcement stress which is equal to design
strength Rs.
Value R is calculated using formula
7001
8,0s
R R+= , (3.15)
or Table 3.2. Table 3.2
Reinforcement class
240 300 400 500 500
Value R 0,612 0,577 0,531 0,493 0,502 Value R 0,425 0,411 0,390
0,372 0,376
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
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
:'
bRARARx
b
sscss = (3.16)
) with Rhx =0
using the condition
M < Rbb(h0 0,5x) + 'ssc AR (h0 a); (3.17) ) with > R using
the condition M < RRbbh 20 + 'ssc AR (h0 - a'), (3.18)
where R =R(1 0.5R) or see Table 3.2. The right part of condition
(3.18) might be increased to a certain
extent if required replacing value R (0.7R + 0.3m), where m = (1
0.5) and assuming that is no more than 1.
If 0, durability is checked using the condition M Rs As (h0 a').
(3.19)
A
hh
a
sR
M
a'
s
b
sA
xA '
bR Ab
Rsc
A'
o
Rb
A
s
s
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 calculated without regard to
compressed reinforcement ( 'sA = 0.0) is less than 2', condition
(3.19) might be checked with the use of replacing ' by /2.
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
3.19. It is recommended to design flexural elements in such a
way to
provide fulfillment of the condition R. 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 < Rh0 using the condition M RsAs (h0 0,5x), (3.20)
where is height of compressed zone which is equal to bRARx
b
ss= ; R see paragraph 3.17;
with Rh0 using the condition M RRb 20h , (3.21)
where for R 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
2
0bhRM
bm = is calculated. (3.22)
If m < R (see Table 3.2), compressed reinforcement is not
required. When there is no compressed reinforcement, area of
tensile
reinforcement is calculated using the formula .0 /)211( smbs
RbhRA = (3.23) In case if > R, 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 'sA reinforcements
corresponding to minimum value of their sum in case if compressed
reinforcement is required (see paragraph 3.21) are calculated using
the formulae:
;)'( 0
20'
hRbhRMA
s
bRs
= (3.24) As = RRbbh0/Rs + 'sA , (3.25)
Where for R and R refer to Table 3.2.
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
If value of the taken area of compressed reinforcement 'SA
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
'0 /)211( ssmbs ARbhRA += , (3.26) where 0)'(2
0
0'
=bhR
ahARM
b
sscm .
Herewith, condition m < R 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,), the following
condition is met
RsAs Rb ''' sscff ARhb + , (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
'fb ; 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:
b
b f
h
sA
a) a'
fh
ba'
b
ha
sA
x0
As)
fh
ha
s
x0
f
A
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
)'()5,0()5.0( 0''00 ahARhhARxhbxRM sscfovbb ++ , (3.28) where
Aov area of flange overhangs, which is equal to '' )( ff hbb ,
herewith, compressed zone height is calculated using the
formula
bR
ARARARxb
ovbsscss ='
, (3.29)
and taken no more than Rh0 (see Table 3.2). If x >R h0,
condition (3.28) might be written in the form )'()5,0( 0''020
hARhhARbhRM sscfovbbR ++ , (3.30)
Where for R refer to Table 3.2. Note: 1. When overhang height
varies, it is permissible to assume value 'fh
equal to average overhang height. 2. Compressed flange width 'fb
, 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
)'(
)5,0(
0
'0
20'
hRhhARbhRM
Asc
fovbbRs
= , (3.31) where for R refer to Table 3.2; Aov = '' )( ff hbb
.
Herewith, condition 0' hh Rf shall be met. In case if 0' hh Rf
> , area of compressed reinforcement is determined in the same
way as for a rectangular cross section with width 'fbb = 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: ),()5,0( '0''0'' ahARhhhbRM sscfffb + (3.32) Area of tensile
reinforcement is determined in the way as for a
rectangular cross-section with width 'fb in accordance with
paragraphs 3.21 and 3.22;
) if boundary is in the jack rib, i.e. condition (3.32) is not
met, area of tensile reinforcement is calculated using the
formula
s
sscovbmbs R
ARARbhRA
'0 )211( ++= , (3.33)
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
where 20
0''
0 )'()5.0(bhR
ahARhhARM
b
sscfovbm
= . (3.34) Herewith, condition m R 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:
) when there are transverse jack ribs or with hh f 1.0' is 1/2
of clearance between longitudinal jack ribs;
) when there are transverse jack ribs (or distance between them
is more than distance between longitudinal jack ribs) and with ;6 -
1.0 '' ff hhh <
) with cantilevers of the flange with hh f 1.0' - ;6 'fh with ;3
1.005.0 '' ff hhhh
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
Example 3. Given: a cross-section with dimensions b = 300 mm, h
=
800 mm; = 70 mm; tensile reinforcement 400 (Rs = 355 MPa); area
As = 2945 mm2 (625); concrete of class 25 (Rb = 14.5 MPa); moment
of flection = 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 is calculated:
2403005,14
2945355 ===
bRARx
b
ss mm.
One can find R = 0,531 in Table 3.2. Since Rhx = 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 400 (Rs = Rsc = 355
MPa); moment of flection M = 780 kNm; concrete of class 15 ( 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 : using formula (3.22).
.544.07503005.8
107802
6
20
===
bhRM
bm
Since m = 0.544 > R = 0.39 (see Table 3.2) with given
cross-section dimensions and concrete class, compressed
reinforcement is required.
Assuming = 30 mm and R = 0.531 (see Table 3.2) required area of
compressed and tensile reinforcement might be calculated using
formulae (3.24) and (3.25):
863)30750(355
7503005,839.010780)'(
26
0
20' =
==
ahRbhRMA
sc
bRs
mm2;
3724863355
5.8750300531.0'0 =+=+= ss
bRs AR
RbhA
mm2.
One can assume 'sA = 942 mm2 (320); As = 4021 mm2 (532).
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
Example 5. Given: a cross-section with dimensions b = 300 mm, h
=
700 mm; a = 50 mm; a = 30 mm; concrete of class 30 (Rb = 17
MPa); reinforcement 400 (Rs= Rsc = 355 MPa); area of compressed
reinforcement 'sA = 942 mm
2 (320); moment of flection = 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 :is calculated 173,0
65030017)30650(94235510580)'(
2
6
20
0'
===
bhRahARM
b
sscm .
Since m = 0,173 < R = 0,39 (see Table 3.2), required area of
tensile reinforcement is calculated using formula (3.26)
.mm2727
942355/)173.0211(65030017/)211(2
'0
==+=+= ssmbs ARbhRA
One can take 336 (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 20 (Rb = 11.5
MPa); reinforcement class 400 (Rs = Rsc= 355 MPa); area of tensile
reinforcement is As = 4826 mm2 (632), area of compressed
reinforcement is A 's = 339 mm
2 (312); moment of flection = 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 is calculated using formula (3.16):
7.4613005.11
)3394826(355' ===
bRARARx
b
sscss mm.
One can find R = 0.531 and R = 0.39 in Table 3.2. Since
,531.0733,0
6307.461
0
=>=== Rhx cross-section durability is checked using
condition (3.18):
,6302.606H102.606
)30630(3393556303005.1139.0)'(6
20
'20
=
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
(0.7 . 0.39 + 0.3 . 0.464)11.5 . 300 . 6302 + 355 . 339 . 600 =
636.6 .
106 N.mm = 636.6 kN. m > = 630 kN.m, i.e. durability is
ensured.
T-sections and I-sections
Example 7. Given: a cross section with dimensions 'fb = 1500 mm,
'fh = 50 mm, b = 200 mm, h = 400 mm; = 80 mm; concrete of class
25
(Rb = 14.5 MPa), reinforcement class 400 (Rs = 355 MPa); moment
of flection = 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 'sA = 0: Rbb 'f 'fh
(h0 0.5 'fh ) = 14.5
. 1500 . 50(320 0.5 . 50) = 320.8 . 106 N.mm = = 320.8 kN.m >
= 260 kN.m,
i.. compressed reinforcement zone is in the flange and
calculation is performed in the same way as for a rectangular
cross-section with width b =
'fb = 1500 mm in accordance with paragraph 3.21.
The value is calculated: 117,0
32015005.1410260
2
6
20
===
bhRM
bm < R = 0.39 (see Table 3.2),
i.. compressed reinforcement is in fact not required. Area of
tensile reinforcement is calculated using formula (3.22)
2446355/)117.0211(32015005.14/)211(0 === smbs RbhRA mm2. One can
take 428(As = 2463 mm2). Example 8. Given: a cross section with
dimensions 'fb = 400 mm,
120' =fh mm, b = 200 mm, h = 600 mm; = 65 mm; concrete of class
15 (Rb = 8.5 MPa); reinforcement class 400 (Rs = 355 MPa); moment
of flection = 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
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
Rbb 'f 'fh (h0 0.5 'fh ) = 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 24000120)200400()( ''
=== ffov hbbA mm2. Value m is calculated with 'sA = 0
Table3.2), (see 39.0356.05352005.8
)1205.0535(240005.810270)5,0(2
6
20
'0
=Rb '' ff hb =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
20000100)200400()( '' === ffov hbbA mm2 : 140
2005.14200005.141964355 =
==bR
ARARx
b
ovbss mm < Rh0 = 0.531 .530 = 281 mm (R 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> =
300 kN.m,
i.. cross-section durability is ensured.
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
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
Mx Rb[Awb(h0 x1/3) + Sov,x] + RscSsx, (3.35) where is a
component of flection moment in the plane of axis (two
mutual perpendicular axes crossing gravity center of tensile
reinforcement in parallel with cross-section sides are taken as
axes and ; for a cross section with a flange axis is taken in
parallel with the jack rib plane);
Awb = Ab Aov; (3.36) Ab area of concrete compressed zone, which
is equal to
b
sscssb R
ARARA'= ; (3.37)
b0
h oi
f
b
x
h o bov
y y
h'x 1
A'S
AS
2
A b
b0i
x1
b'ov
b'f
ASx
b0
1 A'Sb
2
y
1xh o
iAb
0ib
x
h o
y
a) b)
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; 1 size of
concrete compressed zone along the most compressed side
face of the cross-section calculated using the formula ctg221
webAttx ++= , (3.38)
where ;ctgctg5.1 00,,
+= hb
ASS
tweb
xovyov
-
REFERENCE MANUAL for CONCRETE AND REINFORCED CONCRETE structures
without REINFORCEMENT PRETENSIONING based on sp 52-101-2003
Sov,y,,Sov,x static moments of area Aov in relation to axes and
y;
bending angle of plane of flection moment to axis , i.. ctg =
Mx/My (My a component of flection moment in the plane of axis
);
b0 distance from gravity center of tensile reinforcement
cross-section 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 = 'fb .
If the condition
ov
web
bbA
x + 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 1 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
. 18541193025.1655.165tg2
. 5.165
36049011930
212600048606005.1ctgctg
5.1
221
00,,
=++=++==
=
+=
+=
web
web
xovyov
Attx
hbA
SSt
Condition (3.39) is checked: 5.79
75150119305.15.1 =+
=+ ovweb
bbA mm < x1 = 185 mm,
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;
;434.1
119302185
2tg
221 === webA
x
531.0562.0370434.1)7530(
185434.175tg)'(
tg'
00
1 =>=+++=++
+= Riovi
ovi hbb
xb (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 s, determined using formula (3.41) and correction of
values h0 and b0.
5.3353
3552)1562.0/8.0(7003
2)1/8.0(700 =+=+= sis R MPa = =0.945 Rs. Since all the rods are
of the same diameter new values Ab,b0 and h0
are equal to:
mm. 8.359945.02
30130400
mm; 1.91945.02
30945.01202 ;mm 183383
945.0218680
0
02
=+=
=++==+=
h
bAb
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;
mm. 1.17341158823.1813.181
mm; 3.1818,35941.9111588
212500048680005.1
21 =++=
=
+=
x
t
Let us check cross-section durability using condition (3.35)
assuming Ssx =0 and 1.80
4146.82
ctg1ctg
os22
=+
=+
== MMM x kN.m:
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,
-
MANUAL to CP 52-101-2003 concrete and REinforced concrete
structures without prior reinforcement stress
i.. cross-section durability is ensured. Example 11. It is
required to find tensile reinforcement area with
moment in the vertical plane = 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 and are: 52.15
4164
ctg1sin
22=
+=
+==
MMM y kNm;
Mx = My tg = 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 mx and my:
.072.0360905.14
1006.865,141052.15
;185.0360905.14
106.2125.14101.62
2
46
020
,
2
46
200
,
===
===
hbRSRM
hbRSRM
b
yovbymy
b
xovbxmx
Since mx> 0, calculation is continued for a T-section. Since
a point with coordinates mx = 0.185 and my = 0.072 in the
graph of Drawing. 3.7 is located at the right side from the
curve corresponding to the parameter 5.2
9075150
0
=+=+b
bb ov , and at the left side of the
curve corresponding to the parameter 83.090/75/ 0' ==bbov ,
calculation is continued with allowance for biaxial bending and
total design strength of reinforcement, i.. condition (3.40) is
fulfilled.
In the graph value s = 0.20 corresponds to coordinates mx =
0.185 and my = 0.072. Then in accordance with formula (3.42) area
of tensile reinforcement will be equal to
As = (sb0h0 + Aov)Rb/Rs = (0.2 . 90 . 360 + 6750)14.5/355 =
540.4 mm2. Rods are taken as 316 (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;
-
MANUAL to CP 52-101-2003 concrete and REinforced concrete
structures without prior reinforcement stress
- 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.
-
MANUAL to CP 52-101-2003 concrete and REinforced concrete
structures without prior reinforcement stress
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
-
MANUAL to CP 52-101-2003 concrete and REinforced concrete
structures without prior reinforcement stress
,c
MQ bb = (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
S
CC
maxQ
Q
wSS
maxQ
swR swA
qF
Q=Q -qC-Fmax
bQ
o
RswAsw
ARsw sw
0h
h'
b
Asw
b'f
f
ww
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
,w
swswsw s
ARq = (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
EQUALITY, I.E., ASSUME MB = 6H20 QSW.
PROJECTION Q
-
MANUAL to CP 52-101-2003 concrete and REinforced concrete
structures without prior reinforcement stress
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
sw
bq
Mc75,0
= BUT NO LESS THAN H0, IF THIS VALUE IS LESS THAN 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 IS TAKEN EQUAL TO 1q
M b , AND
CC
1
Q=
Q -
F
Q
2
F F1 2
2
1
1
21
1
Q
DRAWING 3.10. ARRANGEMENT OF CALCULATED INCLINED
SECTIONS FOR CONCENTRATED FORCES
1 INCLINED SECTION CHECKED FOR THE IMPACT OF TRANSVERSE FORCE
Q1; 2 THE SAME, FORCE
Q2
IF THEREBY 2or 5.01
2 01
>
I, i
iibtisw bRq
0)( 75.0
/5.1
= (3.51) WHERE 0I IS THE SMALLEST OF VALUES I 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 11 2 qMQ bb =
:
IF max01 /2 QhMQ bb ,
b
bsw M
QQq3
21
2max = ; (3.52)
IF QB1
-
MANUAL to CP 52-101-2003 concrete and REinforced concrete
structures without prior reinforcement stress
THEREBY, IF QB1 < RBTBH0, 0
100max5,1
35,0h
qhbhRQq btsw= , (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
0
max2
10max10max
5.15.18/
5.18/
++=
hQqhQqhQqsw (3.55)
AND TAKEN NO LESS THAN 5.33/ 10max qhQ .
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 QSW =
0.75 (QSW1 QSW2) AS FOLLOWS:
- IF QSW < Q1,
sw
swb
qcqQcqcMcl
++= 1max011 75,0/ , (3.58)
-
MANUAL to CP 52-101-2003 concrete and REinforced concrete
structures without prior reinforcement stress
s w1 s w1 s w2 s w2
F
l1
C0C
DRAWING 3.11. FOR CALCULATION OF INCLINED SECTIONS AFTER A
CHANGE OF STIRRUPS INTENSITY
WHEREsw
bqq
Mc = 1, BUT NO MORE THAN 3H0,
WHEREBY, IF 21
0
1 75.0 ,
5.01
2
sw
l
bt
swsw
l
qqMc
bRq
hqq
M+=
-
MANUAL to CP 52-101-2003 concrete and REinforced concrete
structures without prior reinforcement stress
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).
a)
C
h01 h0
1
q
C
01 h
h 0
qb)
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
2101 tg5,1)/(5.1+= bRqhc bt , (3.61)
WHEREBY, IF THIS VALUE IS LESS THAN bR
qhc
bt
sw
5.0)tg21(
tg2122
01
=
OR, IF QSW/(RBTB) > 2(1-2TG)2, THEN LEAST FAVORABLE VALUE OF
C IS EQUAL TO
.tg5,1)/()75,0(5.1
21
01 ++= bRqqhc btsw (3.62) THE TAKEN VALUE OF C SHALL NOT EXCEED
3H01/(1-
3TG), OR THE LENGTH OF THE BEAM SECTION WITH A CONSTANT VALUE
.
HERE: H01 IS THE EFFECTIVE DEPTH OF THE BEARING BEAM
SECTION;
-
MANUAL to CP 52-101-2003 concrete and REinforced concrete
structures without prior reinforcement stress
Q1 - SEE PARA 3.32; IS THE ANGLE BETWEEN THE COMPRESSED AND
TENSIONED BEAM FACES. THE EFFECTIVE DEPTH IS TAKEN EQUAL TO H0 =
H01+ C
TG . 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
tg21
2 101
+< lhc . 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/(1-2TG), THE INCLINED
SECTIONS LEADING TO THE SUPPORT ARE CHECKED.
-
MANUAL to CP 52-101-2003 concrete and REinforced concrete
structures without prior reinforcement stress
Q
Q
h 0 h01
F F F
c
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,INCSIN, (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 Q
-
MANUAL to CP 52-101-2003 concrete and REinforced concrete
structures without prior reinforcement stress
PROJECTION LENGTH EQUAL TO C, BUT NO MORE THAN 2H0 (DRAWING
3.14);
IS THE ANGLE OF OFFSET BENDS SLOPE TO THE LONGITUDINAL ELEMENT
AXIS.
FAs,inc
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 20 /Q.
-
MANUAL to CP 52-101-2003 concrete and REinforced concrete
structures without prior reinforcement stress
CC
C
C
s,inc1
Qmax1
1
A
3
2
4
02h
2h0
23
4
As,inc2 As,inc3
DRAWING 3.15. FOR DETERMINATION OF INCLINED SECTIONS IN AN
ELEMENT WITH OFFSET BENDS 1- 4 ARE THE CALCULATED INCLINED
SECTIONS
S
S
S1 2
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;
-
MANUAL to CP 52-101-2003 concrete and REinforced concrete
structures without prior reinforcement stress
B) cbhRQ bt
205.1 , (3.65)
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
61
bRq bt , (3.66) CONDITION (3.65) TAKES THE FORM QMAX<
0.5RBTBH0 + 3H0Q1 (3.67) (WHICH CORRESPONDS TO C = 3H0),
Q
C
1
1C
F
Q=
Q -
F
1
1
Q
2
1
F
2
2
1 2
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
PROJECTION Q
-
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 - 1
20max 6 qbhRQ bt
(WHICH CORRESPONDS TO 1
05.1q
bRhc bt= ).
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 VALUE CMAX IS TAKEN EQUAL TO
tg5,11
3 01max +=
hc , WHILE FOR
FLAT SLABS, MENTIONED IN PARA 3.41, tg2,114.2 01
max +=hc ,
WHERE H01 IS THE EFFECTIVE DEPTH AT THE BEARING SECTION;
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
,)5.1/(4/tg 1
201
bRqhc
bt+= (3.68)
BUT NO MORE THAN CMAX, WHERE Q1 SEE PARA 3.32.
-
Benefit to CP 52-101-2003 concrete and ferroconcrete
constructions without the preliminary stress of the
reinforcement
CALCULATION OF REINFORCED CONCRETE ELEMENTS ACCORDING FOR THE
IMPACT OF MOMENTS ALONG INCLINED
SECTIONS
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)
sw
Q
sw
o
C
l
S SZ h
x
w w
swsw
sb
swsws R A R AR A
N
R A
s s
DRAWING 3.18. DIAGRAM OF FORCES AT THE INCLINED SECTION FOR
ITS FLECTION MOMENT CALCULATION
-
Benefit to CP 52-101-2003 concrete and ferroconcrete
constructions without the preliminary stress of the
reinforcement
C
y
M =Qy- -Fa M
Q
Ns
a)q
iFia
M =ql (C+ ) +FC
iqy2 i
2
M
l12 i
C
Nb
Nb
)q
sN
1
Fi
0
0
1
l
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;
PROJECTION M PROJECTION
M
-
Benefit to CP 52-101-2003 concrete and ferroconcrete
constructions without the preliminary stress of the
reinforcement
ZS IS THE ARM OF THE INTERNAL COUPLE, DETERMINED USING THE
FORMULA
bRNhz
b
ss 20
= (WHERE B IS THE WIDTH OF THE COMPRESSED 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 CUT-OFFS) 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:
a
ssss l
lARN = , (3.73)
-
Benefit to CP 52-101-2003 concrete and ferroconcrete
constructions without the preliminary stress of the
reinforcement
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 = ANDS,
WHERE
bond
san R
R4
= (3.74) RBOND IS THE CALCULATED COHESION RESISTANCE OF
THE REINFORCEMENT WITH THE CONCRETE, EQUAL TO RBOND = 12RBT,
1 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;
2 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;
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 B/RB 0.75 - 0.75; IF
B/RB < 0.25 OR B/RB > 0.75 - 1.0,
HERE B = 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 COEFFICIENT IS DIVIDED BY VALUE
asAsw61+ (WHERE ASW AND S ARE THE SECTIONAL AREA OF THE
ENVELOPING STIRRUP AND ITS SPACING) AND IS TAKEN AT NO LESS THAN
0.7;
B) FOR THE FREE CANTILEVER ENDS 1.0.
-
Benefit to CP 52-101-2003 concrete and ferroconcrete
constructions without the preliminary stress of the
reinforcement
IN ANY CASE THE AN 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 AN 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 NWW 2wd RBT, (3.75) TAKEN AT NO MORE THAN 0.8 RS 2wd
NW. HERE: NW IS THE NUMBER OF WELDED-ON RODS FOR THE
LENGTH LS; W 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 = 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:
-
Benefit to CP 52-101-2003 concrete and ferroconcrete
constructions without the preliminary stress of the
reinforcement
TABLE 3.3
REIN-FORCEM
-ENT CLASS
COEF-FICIENT
RELATIVE LENGTH OF REINFORCEMENT ANCHORING AN = LAN/DS FOR
CONCRETE CLASSES
V10 V15 V20 V25 V30 V35 V40 V45 V50 V55 V60
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 AN SHOULD BE DIVIDED BY B1 = 0.9.
TABLE 3.4.
DW 6 8 10 12 14 W 200 150 120 100 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:
qq
Qcsw +
= max , (3.76) HERE QSW IS - SEE FORMULA (3.48).
-
Benefit to CP 52-101-2003 concrete and ferroconcrete
constructions without the preliminary stress of the
reinforcement
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
QSWL1 AND OF THE DENOMINATOR BY QSW, (WHERE L1 IS THE LENGTH OF THE
SECTION WITH INTENSITY QSW1, QSW1=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
2
0maxmax
tg4tg4tg ,tg
sw
sws
sw
sqq
hqNQcqq
NQc +=+
= , (3.77) ARE CHECKED, WHERE H0 IS THE EFFECTIVE DEPTH AT
THE BEARING SECTION; IS THE ANGLE OF SLOPE OF THE COMPRESSED
FACE
TO THE HORIZONTAL. WHEN THE TENSIONED FACE IS SLOPED AT AN
ANGLE
TOWARD THE HORIZONTAL, VALUE TG IN THESE FORMULAS IS REPLACED BY
SIN.
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
swqQc 1= , (WHERE
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 01 2hq
Qsw
> , C = L1 SHOULD BE TAKEN. IF THESE CANTILEVERS HAVE AN
INCLINED COMPRESSED FACE, VALUE Q1/QSW IS REPLACED BY (Q1-
NSTG)/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
)( qql
zARcswa
sss+= , (3.78)
BUT NO MORE THAN 2H0.
-
Benefit to CP 52-101-2003 concrete and ferroconcrete
constructions without the preliminary stress of the
reinforcement
IF C < LLAN, 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
2
M
w1
ult
DRAWING 3.20. TENSIONED RODS BREAKING-OFF WITHIN THE SPAN
1 - THEORETICAL BREAK-OFF POINT; 2 - PROJECTION M
ssw
dqQw 5
2+= ; (3.79)
-
Benefit to CP 52-101-2003 concrete and ferroconcrete
constructions without the preliminary stress of the
reinforcement
WHEREBY, IF sswsw
dQ
hqhwhqQ 512 ,
20
00 +
=> , (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
TG 0.2 VALUE W IS TAKEN EQUAL TO W = H0 + 5DS, (3.81)
THEREBY, IF >1, W = H0(2.2 1.2/) + 5DS, (3.82) WHERE ,
2tg
0hqNQ
sw
s = 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 TG BY SIN.
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).
-
Benefit to CP 52-101-2003 concrete and ferroconcrete
constructions without the preliminary stress of the
reinforcement
S
M
X 0,5h0
Z
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)
3.1431003.50285 ===
w
swswsw s
ARq N/MM.
SINCE 25.025.28575.03.143 >==bR
q
bt
sw , I.E., CONDITION (3.49) IS
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.
-
MANUAL to CP 52-101-2003 concrete and REinforced concrete
structures without prior reinforcement stress
LET US DETERMINE THE LENGTH OF THE LEAST FAVORABLE IN