Plan and reinforce concrete by zahid ahmad sidiqi

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)
As min = √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 = √ (n2ρ2 + 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 max and 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