Lec 7-flexural analysis and design of beamns
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This document discusses the analysis and design of plain and reinforced concrete beams, emphasizing the importance of minimum reinforcement to prevent sudden failure under loads. It covers key concepts such as the determination of the 'k' value, reduced moment of inertia due to cracking, and long-term deflection caused by factors like creep and shrinkage. An example is provided to apply the discussed principles, calculating stresses on a specified beam under a given bending moment.
![Plain & Reinforced Concrete-1
Reduced Moment of Inertia Due to Cracking
(contd…)
Ie= [ Mcr/Ma ]3
Ig + [ 1 – {Mcr/Ma}3
]Icr < Ig
Where
Mcr= Cracking moment Mcr = fr Ig/yt
Ma = Maximum moment in the member at a load level where we want
to calculate deflection
Ig = Gross moment of inertia of un-cracked transformed section
Icr = Moment of inertia of cracked transformed section
fr = Modulus of rupture
yt= Extreme tension fiber distance from N.A.
Effective moment of inertia calculated by the above expression can be
used to calculate the deflection of beams after cracking](https://image.slidesharecdn.com/lec-7-flexuralanalysisanddesignofbeamns-181114184240/85/Lec-7-flexural-analysis-and-design-of-beamns-8-320.jpg)
Lec 7-flexural analysis and design of beamns
- 1. Plain & Reinforced Concrete-1 CE3601 Lecture # 7 23th Feb 2012 Flexural Analysis and Design of Beams
- 2. Plain & Reinforced Concrete-1 Minimum Reinforcement of Flexural Members (ACI – 10.5.1) Asmin = √fc’ / (4fy) x bwd ≥ 1.4/fy x bwd Critical if fc’ < 28 Mpa The minimum steel is always provided in structural members because when concrete is cracked then all load comes on steel, so there should be a minimum amount of steel to resist that load to avoid sudden failure. For a design to be safe M < Mr For an economical design M = Mr M = Asfs x jd As = M / (fs x jd)
- 3. Plain & Reinforced Concrete-1 Minimum Depth of Section To satisfy a cross section against concrete failure M = Mr M = Cc x jd M = (fc/2 x kd x b) x jd dmin = √ (2M / fc k j b)
- 4. Plain & Reinforced Concrete-1 Determination of “k” value k = √ (n2 ρ2 + 2nρ) –nρ Value of k can not be determined as ρ is not know. There are two different approaches to establish the value of k. 1. Simultaneous Occurring of Maximum Permissible Steel and Concrete Stresses. 2. Assuming some suitable steel ratio
- 5. Plain & Reinforced Concrete-1 Determination of “k” value (contd…) 1- Simultaneous Occurring of Maximum Permissible Steel and Concrete Stresses. εc Strain Diagram εs kd d- kd d B C A E D N.A . Consider Δ ABC & Δ ADE εs/(d-kd) = εc/(kd) (fs / Es) / (d-kd) = (fc/Ec) / (kd) fs x kd = (Es/Ec) x fc x (d-kd) fs x k = n x fc x (1-k) fs x k + nfck = nfc k = nfc / (fs + nfc)
- 6. Plain & Reinforced Concrete-1 Determination of “k” value (contd…) 2- Assuming some suitable steel ratio In this approach, some suitable value of steel ratio (less than ρ max) is selected at the start of calculations and used for the determination of “k” ρ max= 0.85 x 3/8 x fc’/fy(600/(600 + fy) Calculate ρ maxand select some value less than this.
- 7. Plain & Reinforced Concrete-1 Reduced Moment of Inertia Due to Cracking Elastic Deflection can be expressed in general form Δ = f (loads, spans, supports) / ( EI ) Deflection of a cracked section can be expressed as Δ = f (loads, spans, supports) /( EIe) As the number of cracks and their width increases the M.O.I of cross section reduces and deflection increases. Ie = Effective moment of inertia of cracked section
- 8. Plain & Reinforced Concrete-1 Reduced Moment of Inertia Due to Cracking (contd…) Ie= [ Mcr/Ma ]3 Ig + [ 1 – {Mcr/Ma}3 ]Icr < Ig Where Mcr= Cracking moment Mcr = fr Ig/yt Ma = Maximum moment in the member at a load level where we want to calculate deflection Ig = Gross moment of inertia of un-cracked transformed section Icr = Moment of inertia of cracked transformed section fr = Modulus of rupture yt= Extreme tension fiber distance from N.A. Effective moment of inertia calculated by the above expression can be used to calculate the deflection of beams after cracking
- 9. Plain & Reinforced Concrete-1 Long Term Deflection Long term deflection is caused by creep and shrinkage. Δt = λ Δi where Δt =Long term deflection Δi=Instantaneous Deflection λ = Multiplier for additional deflection due to long term effect Total Deflection = Δi+Δt Total Deflection = Δi+λ Δi Total Deflection = (1+λ) Δi
- 10. Plain & Reinforced Concrete-1 Long Term Deflection (contd…) λ = ξ / (1 + 50 ρ’) ρ’ = compression steel ratio = Area of compression steel / (b x d) = As’ / (b x d) ξ = Time dependent factor for sustained loads Elapsed Time ξ 5 years or more 2.0 12 months 1.4 6 months 1.2 3 months 1.0
- 11. Plain & Reinforced Concrete-1 Example (un-cracked transformed section) A rectangular beam of size 250mm x 650mm with effective depth of 590mm and is reinforced with 3 # 25 US customary bars. Concrete has specified compressive strength 28MPa. Yield strength of steel is 420 MPa. Modulus of rupture (tensile strength) is 3.5 MPa. Determine the stresses caused by the bending moment of 50kN-m. Data fc’ = 28 MPa fy = 420 MPa fr = 3.5 MPa M = 50 kN-m (positive moment) ftop = ? fbottom = ? b = 250mm d = 590mm
- 12. Concluded