LIMIT STATE METHOD OF DESIGN OF REINFORCED CONCRETE STRUCTURES
SCOPE (LIMIT STATE METHOD) 1.VARIOUS LIMIT STATES, DESIGN VALUES, SAFETY FACTORS. 2.LIMIT STATE OF COLLAPSE – ITS APPLICATIONS TO DIFFERENT STRUCTURAL MEMBERS. 3.LIMIT STATE OF SERVICEABILITY. 4.PROVISIONS OF IS 456. 5.DESIGN AIDS SP 16
MAIN MENU General Methods of Design Working Stress Method Ultimate Load Method Limit State Method
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GENERAL Civil engineering structures are a combinaton of structural elements – beams, columns, slabs, footings etc. Due to external loads on the structure, forces are generated in these elements which cause bending and shear. Axial forces are also generated in some elements. Design involves providing a member of a suitable size and material to withstand these forces. Main Menu
METHODS OF DESIGN Theoretical (a)
Working Stress Method
(b)
Ultimate Load Method
(c)
Limit State Method
Experimental Main Menu
WORKING STRESS METHOD • This method assumes that both concrete and steel act together and are elastic at all stages. • It adopts permissible stresses which are derived by applying a factor of safety to the ultimate strength of the materials. The factor of safety for concrete is 3 and for steel it is 1.8.
STRESS-STRAIN DIAGRAMS FOR CONCRETE AND STEEL B A
б
O
A C
Є Concrete
OC – Linear
B
б
C
B
Є Mild Steel (Fe 250)
CA – Elastic Range
C
б
Є HYSD Bars (Fe 415)
AB – Plastic Range
WS Method confines itself to the linear and elastic ranges in both concrete and steel.
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STRESS-STRAIN BLOCKS FOR WORKING STRESS METHOD OF DESIGN Єc
бcb
kd d
C=0.5бcb(kd)b jd
NA
Ast b Cross Sec
Єs
T= бst Ast
Strain Diag
Stress Diag
cb k m cb st k j 1 3 M 0.5 kj cb bd
2
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ULTIMATE LOAD (LOAD FACTOR) METHOD
A load factor is applied to the service loads.
Does not take into account the variations in the strength of materials.
It has sound experimental backing.
It uses the idealised stress-strain curve and takes into account the plastic range also. Design by this method results in an economical design.
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Design results in slender members which may lead to excessive deformations (deflections) and cracking. It does not take into consideration effects of creep, shrinkage etc. Єcu
kбcu=0.55 бcu
бcb
C
C d
a=0.43d
z
NA
Ast
b Cross Sec
Єsy Strain Diag
T Stress Diag
T Whitney’s stress block Main Menu
LIMIT STATE METHOD OF DESIGN Design Philosophy Characteristic Values Definitions Characteristic Strength Characteristic Load Partial Safety Factors Exit
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LIMIT STATE METHOD OF DESIGN Limit State of Collapse (Assumptions) Limit State of Collapse (Flexure) Balanced Section (Rectangular Beam) Under Reinforced Section (Rectangular Beam) Over Reinforced Section (Rectangular Beam) Doubly Reinforced Section (Rectangular Beam) Flanged Beam Exit
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Design Philosophy
Structure is designed to withstand safely all loads likely to act on it throughout the life of the structure without any danger to life, property and its function.
Limit State corresponds to each of the states in which it can become unfit for use such as: • Limit State of failure – strength, shear, bond, torsion etc. • Serviceability Limit State – deflection, cracking, vibration etc. • Working Stress Method deals with serviceability limit state.
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All design procedures are concerned about the safety of the structures, which depend on: • Inaccuracies in design assumptions or errors in calculations. • Inaccuracies in assessing the properties of the materials. • Degree of supervision, quality of materials, execution of work etc. • Possible deterioration of the structure with time. …
Probability calculations are done taking into account all the random factors which can lead to the limit state to arrive at the : • Magnitude of load causing the most unfavourable effects. • Strength of materials, their quality and variation. •
Deterioration with passage of time.
Code has introduced characteristic values depending on the dispersion of data relating to intensity of loading and the strength of the materials. …
Characteristic values (Qc) are expressed in the form of an equation in terms of the mean value (Qm), standard deviation (б) and a variable (k) depending on the probability accepted. Qc = Qm ± kб ) This leads to : Design load
= Amplification Factor γf x characteristic load
Design strength = characteristic load/ Reduction Factor γm These factors are called Partial Safety Factors and Overall Factor of safety = Product of Menu LS
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Amplification and
the
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DEFINITIONS Limit State: It is the acceptable limit for the safety and serviceability of a structure before failure. • The structure should be able to withstand all the loads liable to act on it throughout its life satisfying the limits of collapse as well as serviceability (deflection and cracking). • The aim is to achieve acceptable probabilities such that the structure will not become unfit for the use for which it is intended i.e. it will not reach its limit state. …
• The design is based on characteristic values of material strengths and the applied loads which take into account the variations in the strengths of the material and the loads which are applied on the structure.
• The characteristic values are based on statistical data which are obtained from frequency distribution curve called Gaussian Distribution Curve.
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Characteristic Strength: It is that value of the strength of the material below which not more than 5% of the test results are expected to fall.
fk= fm – ks where fk = characteristic strength
F r e q u e n c y
Mean Strength (fm) 1.64 s fk
fm = mean strength k = coefficient depending on the accepted probability of tests falling below charac strength (5% in this case) & is equal to 1.64 s = standard deviation
5% chances that results will fall below this value
Strength
s
2
n1
where Δ = deviation of individual test mean
strength from
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Characteristic Load is that value of the load which has a 95% probability of not exceeding during the life of the structure. F r e q u e n c y
fk = fm + ks
fk Mean Strength
1.64 s
5% chances that results will exceed this value
Load
Design Load = γf x Characteristic load γf = partial safety factor which is different for different types of loading and are given in IS 456:2000 Design strength = Characteristic strength / γm γm = partial safety factor 1.15 for steel 1.5 for concrete
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Partial Safety Factors for Loads •
For LS of Collapse
* *
DL IL EL
1.5(DL+LL or IL) + 1.0 WL or EL 1.5(DL + WL) or (0.5DL + 1.5WL or EL) Dead Load LL Imposed Load WL Earthquake Load
Live Load Wind Load
• It is assumed that WL and EL do not act together. • The structure is to be designed for the worst combination of loads. …
•
For LS of Serviceability * * *
•
1.0(DL + LL) 1.0(DL + LL) 1.0DL + 0.8(LL + WL)
Design Criteria
The collapse of a member may occur in flexure, shear, bond, axial compression/ tension, torsion or a combination of any of these. Therfore, in LS Method of design the structure is designed for LS of Collapse and checked for LS of Serviceability (i.e. deflection, cracking etc). Menu LS
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LIMIT STATE OF COLLAPSE
Assumptions • Plane sections normal to the plane of bending remain plane after bending. • Tensile stress in concrete is ignored. Tensile stress in concrete is ignored. • There is perfect bond between concrete and steel. • The strain in concrete at the outermost compression fibre reaches a specified value of 0.0035 in bending at the Limit State. Click here to view diagram
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• Max strain in tensile reinforcement (Fe 415) at failure shall not be less than 0.002 + fy/(1.15Es) where fy = characteristic strength of steel Es = Modulus of elasticity of steel (2x105 N/mm2) Click here to view diagram
•
Stress in reinforcement is derived from representative stress- strain diagram of the steel used with partial safety factor of 1.15. Click here to view diagram
• Distribution of compressive stress in concrete is defined by an idealised stress-strain curve with partial factor of safety of 1.5. Click here to view diagram
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RECTANGULAR BEAM
d’ Effective cover to compression steel
Asc Effective depth
d Ast
Effective cover to tensile steel dc
Asc
Area of compression steel
Ast
Area of tensile steel
b Width of beam
Cross Section
Case 1: Balanced Section
Case 3: Over Reinforced Section
Case 2: Under reinforced Section
Case 4: Doubly Reinforced Section
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BALANCED SECTION 0.0035 0.002
xu d
0.45fck 3xu/7 =0.43xu 4xu/7 =0.57xu
0.42xu C = 0.36fck xu b z = d - 0.42xu
Ast 0.002+0.87fy/Es b
Cross Section
Strain Diagram
T = 0.87fy Ast
Stress Diagram
STRAIN AND STRESS DIAGRAMS
Depth of Neutral Axis for Balanced Section (xu= xu,max) is derived from : xu ,max d
0.0035 Es 0.0055 0.87 f y
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Depth of Neutral Axis for a Balanced Section for different grades of steel
Grade
fy (N/mm2)
xu,max/d
Fe 250
250
0.53
Fe 415
415
0.48
Fe 500
500
0.46
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Limiting value of percentage of steel for different grades of concrete and steel: pt ,lim
xu ,max f ck 41.4 d fy
pt,lim for fy = fck
250
415
500
15
1.136
0.718
0.571
20
1.755
0.958
0.762
25
2.194
1.197
0.952
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Limiting value of Moment of Resistance of Balanced Section: M u ,lim 0.36 f ck xu ,max b d 0.42 xu ,max M u ,lim
xu ,max xu ,max 1 0.42 0.36 d d
f ck bd 2
For Fe 250 Mu,lim = 0.148 fckbd2 For Fe 415 Mu,lim = 0.138 fckbd2 For Fe 500 Mu,lim = 0.133 fckbd2
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UNDER (SINGLY) REINFORCED SECTION • Section is under reinforced if %steel (or area of steel) provided is less than that required for the balanced section i.e. pt < pt,lim or Ast < Ast,lim 0.45fck
0.0035 xu,max d
xu
0.42xu C = 0.36fck xu b z = d - 0.42xu
Ast 0.002+0.87fy/Es
T = 0.87fy Ast
b
Cross Section Strain Diagram Strain Diagram Stress Diagram Balanced Sec Under Reinf Sec
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• This will imply that the design moment or the moment acting at the section is less than the limiting moment of the section (moment of resistance of the balanced section) i.e. Mud < Mu,lim
• Consequently the depth of the neutral axis will be less than the max depth of the neutral axis for than section i.e. xu/d < xu,max/d
or
xu < xu,max
Moment of Resistance (under reinforced section):
f y Ast M u 0.87 f y Ast d 1 f ck bd
• If Ast is known/given, Mu of the section can be found.
Variations • If Mu is known/given, Ast can be obtained by solving the above quadratic equation. • Ast can also be found by: * calculating xu from the quadratic equation below: and … M u 0.36 f ck xu b d 0.42 xu
f ck bxu then obtaining Ast from: Ast 0.414 f y
OR *
Directly from the equation: 4.6 M u 1 f ck Ast bd 1 1 2 f y f ck bd 2
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OVER REINFORCED SECTION • Section is over reinforced if %steel (or area of steel) provided is more than that required for the balanced section i.e. pt > pt,lim or Ast > Ast,lim TO AVOID COMPRESSION FAILURE OF CONCRETE, WE NEVER PROVIDE AN OVER REINFORCED SECTION. •
IMPORTANT: IN THE ABOVE CIRCUMSTANCES, THE LOAD ON THE STRUCTURE HAS TO BE RESTRICTED SUCH THAT IT DOES NOT GENERATE A MOMENT MORE THAN THE MOMENT OF RESISTANCE OF THE Solve Menu LS BALANCED SECTION. Problem
DOUBLY REINFORCED SECTION • If the design moment acting on the section is more than the moment of resistance of the balanced section, the section is required to be designed as a doubly reinforced section. Mud > Mu,lim ;
Mu,lim = Mu1
• Tensile steel (Ast1) is provided for the balanced moment Mu1. • Additional tensile (Ast2) and compressive steel (Asc) are provided to resist the moment Mu2 i.e moment in excess of Mu,lim
Єsc 0.0035
d’ Asc d
xu
0.002
0.002+0.87fy/Es
Cross Section
3xu/7 =0.43xu 4xu/7 =0.57xu
0.42xu C = 0.36fck xu b z = d - 0.42xu
Ast
b
0.45fck
T = 0.87fy Ast
Strain Diagram Stress Diagram STRESS-STRAIN DIAGRAMS
xu ,max d Strain in compression steel: sc 0.0035 x u , max
Stress in compression steel: For Fe 250, fsc= 0.87fy For Fe 415 and Fe 500 it is obtained from Table ‘A’ of SP 16 for different values strains or from Table ‘F’ for different values of d/d’ To Flanged Beam
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Additional area of tensile steel:
Ast 2
Total tensile steel:
Ast Ast 1 Ast 2
Area of compression steel:
Asc
M u2 0.87 f y ( d d )
M u2 f sc ( d d ) OR
0.87 f y Ast 2 Asc f sc
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FLANGED BEAMS: T AND L BEAMS •
•
Intermediate beams ( T Beam)
Edge Beams ( L Beam)
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C=(Cweb+Cflange)
bf Єc=0.0035 Df
xu d
0.45fck C 0.43xu
0.002
NA 0.57xu
Ast Єs
T=0.87fckAst
bw Cross Sec
Strain Diag
Stress Diag
CASE 1:
xu < Df
i.e. Neutral Axis in the Flange.
CASE 2:
Df <= 3xu/7 i.e. 0.43xu
or xu >= 7Df/3
or Df/d <= 0.2
CASE 3:
Df> 3xu/7
or xu <
or Df/d > 0.2
i.e. 0.43xu
7Df/3
In all above cases xu < xu,max i.e. section is under reinforced CASE 4:
xu > xu,max i.e. doubly reinforced section
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CASE 1:
xu < Df i.e. Neutral Axis in the Flange. This is the condition when Design Moment is less than the Moment of Resistance of the flange (Muf) which is obtained by substituting xu with Df and b with bf in the formula for Rectangular Beam
M u 0.36 f ck xu b d 0.42 xu Area of steel can be obtained from the formula for Rectangular Beam given below or from direct formula.
M u 0.87 f y Ast d
f y Ast 1 f ck bd
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CASE 2:
Df <= 3xu/7 i.e. 0.43xu or xu >= 7Df/3 or Df/d <= 0.2 i.e entire flange is within the rectangular portion of the stress block To calculate Mu,lim for a section, substitute xu with xu,max in the formula given below. To calculate xu for a given Design Moment Mud, substitute Mu with Mud and solve the quadratic equation.
M u 0.36 f ck bw xu d 0.42 xu 0.45 f ck b f bw D f d 0.5 D f Contribution of web
Contribution of flange
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Area of steel is calculated by equating the tensile force (T) in steel to the compressive force in concrete due to flange (Cf) and web (Cw) T = Cw + Cf
0.36 f ck bw xu 0.45 f ck b f bw D f Ast 0.87 f y
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CASE 3:
Df> 3xu/7 i.e. 0.43xu or xu < 7Df/3 or Df/d > 0.2 i.e flange is in the parabolic portion of the stress block. To calculate Mu,lim for a section, substitute xu with xu,max in the formula given below. xu is calculated by trial and error by solving the quadratic equation and substituting the value of yf as given below. Alternatively, xu can be calculated by substituting Mu with Mud and expressing yf in terms of xu and solving the quadratic equation.
M u 0.36 f ck bw xu d 0.42 xu 0.45 f ck b f bw y f d 0.5 y f
where
y f Modified thickness of the flange 0.15 xu 0.65 D f D f
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Area of steel is calculated by equating the tensile force (T) in steel to the compressive force in concrete due to flange (Cf) and web (Cw) T = Cw + Cf
0.36 f ck bw xu 0.45 f ck b f bw y f Ast 0.87 f y
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Case 4:
Doubly reinforced section If Design Moment(Mud) > Mu,lim, the section is to be designed as a Doubly Reinforced section. Tensile steel (Ast1) is provided for the balanced moment (Mu1 = Mu,lim). Additional tensile (Ast2) and compressive steel (Asc) are provided to resist the moment Mu2 i.e moment in excess of Mu,lim
Mu2 = Mud – Mu1 &
Ast = Ast1 + Ast2
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Area of steel for Balanced Section
Additional Area of tensile steel
M u2 Ast 2 0.87 f y d d
Total tensile steel
Ast Ast1 Ast 2
Area of Compression steel
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0.36 f ck bw xu 0.45 f ck b f bw y f Ast1 0.87 f y
M u2 Asc fsc d d
where fsc is calculated as in Rectangular Beam
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