Ultimate Moment Capacity of RC Beams
Analysis of flexural behaviour • When load is initially applied to a beam all sections are uncracked unless there has been a history of restrained shrinkage. When the load reaches its serviceability limit state and the peak moment regions are cracked, there are still extensive regions where the moment is not large enough for the cracks to form. P
Moment at midspan Bending Moment
Mcr
Mu My
P>Pcr Mcr
Deformation at mid-span
Bending Moment
Mcr My
Uncracked region RMIT University©
cracked region
Uncracked region
• Last week we looked at analysis of RC beams for Flexure in service limit state – Uncracked elastic section analysis of RC beams εo
σo C
dn d σst
εst Ast
σct Beam
Cross section
Strain
Stress in concrete
– Cracked elastic section analysis of RC beams • Using the Transformed section method of analysis
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Tc Ts
Stress Internal in steel Forces
Post-cracking, service load behaviour – Cracked elastic section analysis of RC beams εo
σo C
dn
M
d
z σst
εst Ast
Beam
Cross section
Strain
Tension stiffening between cracks
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Stress
Ts Internal Forces
Analysis of flexural behaviour • When load reaches its ultimate limit state and the peak moment regions are cracked and overloaded, the reinforcing steel has yielded, and the ultimate moment capacity of the beam at the section of maximum moment is reached. P>Pcr
Bending Moment
Mcr My
Uncracked region
cracked region
Uncracked region
Moment at midspan Mu My
P>Pcr Mcr
Deformation at mid-span Bending Moment
Mcr My cracked region
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Mu
Non-Linear Overload Behaviour • As the moment increases further above the working load level the concrete and steel start to yield and stresses become non-linear. This is known as overload behaviour up to the point where the beam develops its ultimate moment capacity, Mu. • The concrete stress increases less than proportionally with strain, so that the stress block above the crack becomes progressively more curvilinear. σo = f’c f’c
Changes in concrete stress block with increasing moment Stress f’c
Ec
Strain εcpeak Typical Stress-strain curve of Concrete RMIT University©
f’c
Failure • Beyond peak load : concrete crushing after steel yielding
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Condition at Ultimate Moment, Mu • At Ultimate Moment, M u, the stress and strain profile at the cross section is represented as: εu C dc
kud d
Note: dn= ku.d Mu
zu εst>εsy Strain
T
fst=fsy Stress
Forces
• In design we normally assume that the cross-section is under-reinforced”, ie the steel reinforcement will exceed its yield strength before the concrete in the extreme compressive fibre crushes. (and we limit the ultimate compressive strain εu = 0.003) Idealised Stressstrain curve of steel
Stress fsy
failure Yield Plateau εsy
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Strain
Condition at Ultimate Moment, Mu • The steel reinforcement has yielded ie it carries a stress equal to the yield strength of the steel, Therefore the internal tensile force is now a constant value (T = Ast .fsy) which will remain unchanged with increase in external load • The concrete compressive region above neutral axis depth is highly stressed and at point of crushing • The compressive C force and the tensile T force are equal in magnitude and for a resisting couple moment with a lever arm z= d-dc • For equilibrium we have T= C and Mu = Tz = Cz
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Rectangular Stress Block – AS3600-2009 • As shown previously the concrete compressive stress profile is non-linear at ultimate • AS3600 allows for the non-linear concrete stress profile to be simplified with the rectangular stress block given in Clause 8.1.3 and εu = 0.003
εu
α2f’c
εst>εsy Strain
γkud
kud
kud
σs=fsy
σs=fsy
d
dc C z T
Neutral axis
Note: dn= ku.d
Actual stress profile Rectangular Stress Block
• The parameter γ is used to define both the magnitude and location of C. • Values for γ given in AS3600 for normal strength concrete is:
γ = 1.05-0.007f 1.05-0.007f’’c
( 0.67 ≤ γ ≤ 0.85 )
α2 is used to determine the magnitude of the uniform compressive stress ie. rectangular stress block assumptions given in AS3600-2009 Clause 8.1.3
α2 = 0.85 RMIT University©
for f’c ≤ 50MPa
Rectangular Stress Block – AS3600 εu kud
α2f’c
εst>εsy Strain
γkud
kud σs= fsy
σs=fsy
d
dc C z T
Neutral axis
Note: dn= ku.d
Actual stress profile Rectangular Stress Block
• The resultant compressive force, C = stress x area = 0.85f’cbγdn
(Remember α2 = 0.85 for f’c ≤ 50MPa ) and acts at depth dc = 0.5γdn • The resultant tensile force, T = Astfsy • Equating C = T gives neutral axis depth dn, with dn we can determine the location of the compressive force dc this in turns allows the internal lever arm z=d-dc and the moment capacity Mu to be calculated
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Neutral Axis parameter ku • To calculate the ultimate moment capacity Mu, it is often convenient to use the dimensionless neutral axis depth parameter ku = dn/d • This term can be used in lieu of dn to specify the neutral axis depth and is also an indicator of ductility • To derive equations for Mu we equate T = C • The resultant compressive force C = 0.85f’cγkudb and acts at depth dc = 0.5γdn • The resultant tensile force T = Astfsy
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Moment Capacity of a singly reinforced Section • Equating C = T gives
ku =
1 f sy Ast 0.85γ f 'c bd
• The lever arm z = d-dc = d(1-0.5γku) • Which gives Mu= Tz = fsyAstd(1-0.5γku) • Substituting for ku :
Ast f sy ⎞ ⎛ ⎟⎟ M u = f sy Ast d ⎜⎜1 − 0.6 bdf 'c ⎠ ⎝
• And substituting for ρ = Ast/bd :
f sy ⎞ 2 ⎛ ⎟⎟bd M u = f sy ρ ⎜⎜1 − 0.6 ρ f 'c ⎠ ⎝
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Moment Capacity of a singly reinforced Section The moment capacity of the section shown below is to be calculated using the limiting strain criterion εu = 0.003 and the AS3600 rectangular stress block. The concrete strength f’c is 40MPa so that γ = 1.05-0.007(40) = 0.77 B=300
D=600mm
d
Ast = 3N28 Clear cover to main bars = 30mm
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Failure mechanism of R/C beam • There are three possible failure mechanisms which determine the ultimate moment capacity. They are: – Failure through steel reinforcement. That is, reinforcing bars reaching their yield capacity well before concrete starts to crush – Failure through concrete reaching its compressive strength prior to steel reinforcement yielding – Both materials reaching their capacities simultaneously (less probable in practice) • The mode of flexural failure of a reinforced concrete beam is dependent upon the quantity of the steel reinforcement present. It is for this reason that the code AS3600 specifies limits on both the minimum and maximum allowable reinforcement ratios.
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Failure • Beyond peak load : concrete crushing after steel yielding
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Ductility • • •
A region of a beam is ductile if it undergoes large plastic deformations prior to failure. Ductility is characterised by an extended flat plateau in the moment curvature relation. Good ductility achieved if the quantity of reinforcement is kept small. M / f'cbd2
ρ = 0.0375
ρ = 0.030 0.2
ρ = 0.0225
ρ = 0.015 0.1
ρ = 0.0075
Curvature 5
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10
Ductility and ku (neutral axis parameter) εu<0.003 N
kud
εu=0.003
A N
kubd
εu=0.003 kud
A
A
N
εst>εsy Figure 1
εst=0.0025
εst<εsy
Figure 2
Figure 3
• Concrete strain at failure, εu, namely the strain at which concrete crushes, is considered as 0.003 (0.3%) and also independent of the concrete strength. Steel reinforcement strain at yield, εst is considered as 0.0025 (0.25%) for N500 or L500 steel
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Under-reinforced, over-reinforced and balanced sections • We know that the failure through steel (ie yielding) is ductile in nature and therefore desirable. This means condition given in figure (1) should be encouraged where εst>εsy = 0.0025. Such as beam is called ‘underreinforced reinforced’’. • Condition shown in figure (2), where steel and concrete both reach their limiting values at the same time is called the “balanced failure failure’’ and regarded as the point where beam changes from being ductile to being brittle • We also know that failure through crushing concrete is sudden and brittle in nature and therefore undesirable and should be avoided as the steel strain is less than εst<εsy=0.0025. The beam is then called “over reinforced ”.Figure reinforced” (3) • Although “balanced” failure is also not desirable as it is close to being “brittle”, however in theory, this will allow us to demonstrate the margin which we can base our design recommendations • The neutral axis parameter kub (under balanced failure) is therefore a good parameter to use as a guide.
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How do we find the kub value? From Similar triangles we get
εu=0.003
kub
d =
kubd
A
N
0.003
0.0025+0.003
d
0.003 kub =
εst+εu = 0.0025+0.003
• Therefore, – When kub = 0.55 - balanced – ku ≤ 0.55 - under reinforced – ku ≥ 0.55 – over reinforced
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0.0055
= 0.55
Minimum ductility requirements in AS3600-2009
1. Clause 8.1.5 states that the neutral axis depth parameter ku must not be greater than 0.36 to ensure ductility
(When ku is very close to, yet less than 0.55, still there will be very little warning of failure. In other words member is not as ductile as we would like it to be. Leaving a safety margin, the design code AS3600 prescribes to limit ku ≤ 0.36 to regard the beam as ductile). –
Hence the designer must also calculate ku and satisfy the ductility requirement
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Minimum strength requirements in AS3600-2009 Clause 8.1.6 places a lower limit on the steel content. The intention is to avoid steel fracture and hence sudden collapse and cracking. Minimum required strength in bending is given by
M u ,min = 1.2 M cr = 1.2. f 'ct . f .Z Where f’ct.f = 0.6√f’c
Z is the elastic section modulus and for a rectangular section is Z=bD2/6 Minimum required tensile reinforcement is given by 2 Ast ⎛ D ⎞ f 'c t . f ≥ αb ⎜ ⎟ bw d ⎝ d ⎠ f sy
Where αb = 0.2 for rectangular sections
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Moment Capacity of a Doubly Reinforced Section b dsc
kud dst
εsc
0.85f’c
Cs Cc
dc
γkud
zc
zs
fst=fsy
εst>εsy
T = Ast.fsy
Stress
Strain
Note: dn= ku.d
dsc
εu
C zu
Mu T
Internal Forces
Cc= Ts – Cs Tentatively assuming both steels have yielded 1 dn = ( f sy Ast − f sy Asc ) 0 . 85 f 'c b γ With this value of dn and εu=0.003, the strain distribution in the section is fixed and the strains in the compressive and tensile steels can be determined
ε sc
d − d sc = εu n dn
εs = εu
d st − d n dn
If the steel strains are in fact at yield then all internal forces are known and M u is calculated as
M u = Ts d st − Cc d c − C s d sc If the compressive steel is not at yield a trial value for d n is chosen until Cc=TsCs RMIT University©
Example Moment Capacity of a Doubly Reinforced Section Determine the moment capacity Mu of the section shown below. The section contains both tensile and compressive reinforcement. Concrete strength f’c=30MPa, γ = 0.836 350
Asc = 2N28 = 1232mm2
60 550 Ast = 8N28 = 4926mm2
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Design Consideration for Ultimate Moment Capacity of Beam Section
• In designing the beam for strength in flexure, AS3600 clause 8.5.1 requires that M* ≤ φMu and ku ≤ 0.36 to ensure ductile failure ( assumes under reinforced section) φ = 0.8 for bending • In actual fact the beam section details such as reinforcement are not known at this stage of the design but what is known is the design moment M M** which is obtained from an analysis of the structure for various load combinations • The beam section then needs to be proportioned ie. determine the amount of reinforcement needed to ensure the ultimate moment capacity is not exceeded • Design tables for proportioning steel have been set-up for example by C&CA and in house spreadsheets.
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To proportion beam section M * < φM u
⎛ Ast f sy ⎞ 2 ⎜ ⎟ ÷ bd M < φf sy d ⎜1 − 0.6 ⎟ bd f 'c ⎠ ⎝ *
* M Ast ⎛ Ast f sy ⎞ ⎜ ⎟⎟ < φ f 1 − 0 . 6 sy 2 ⎜ bd bd ⎝ bd f 'c ⎠
The steel proportion p can be used as a useful non-dimensional measure of the amount of tensile steel reinforcement in the section
p=Ast/bd , submit this into the above equation to give * f sy ⎞ ⎛ M ⎜ ⎟⎟ < φ f p 1 − 0 . 6 p sy ⎜ 2 bd f 'c ⎠ ⎝
Alternatively *
M φM u < 2 bd bd 2 RMIT University©
