Flexural design

Contents
1. Strength calculations for original beam
2. Strength calculations for Damaged beam
3. Repair of damaged portions – options and suggestions
4. Strengthening of beam – options and suggestions
5. Strength calculations for rehabilitated structure (ductile failure)
6. Strength calculations for rehabilitated structure (
ittle failure)
7. Implications
8. Report to the owne
Part 1 - Strength Calculation for Original Beam
Given data:
Width of the web, b    = 200 mm
Effective depth, d = 500 – (25 + 25/2) = 462.5 mm
f’c = 30 MPa
fy = 400 MPa
As = 4 * (π / 4 ) *25 * 25 = 1962.5 mm2
Calculations:
α1 = 0.85 – 0.0015 f’c
    = 0.85 – 0.0015 * 30 = 0.805 > 0.67
β1 = 0.97 – 0.0025 f’c
    = 0.97 – 0.0025 * 30 = 0.905 > 0.67
φs = 0.85 – factor of reduction – material - steel
φc = 0.65 – factor of reduction – material - conrete
consider b = 200 mm
a = ( φs As fy ) / (φc b fc’)
    = (0.85 * 1962.5) * 400) / (0.65 * 0.805 * 30 * 200)
    =    212.56
c = (a / β1 )
    = 212.56 / 0.905
    = 234.9
For the failure to be ductile i.e. tensile reinforcement to reach yield (according to CSA A23.3-14 (10.5.2) )
(c/d ) should be less than or equal to (700 / 700 + fy )
    234.9 / 462.5 = 0.507 < 0.64 (OK)
This implies that the failure is ductile
jd = d – a/2 = 462.5 – 212.56/2
    =356.22
Therefore, j = 0.77    
Moment capacity of the given section CSA A23.3-14 (10.1.7)
Mf = As φs fy jd
     = 1962.5 * 0.85 * 400 * 356.22
    = 205.57 x 106 Nmm
    = 205.57 kNm
Load is assumed as UDL and support conditions as simply supported
Given, effective span = 7.6m
Factored load w =( M * 8 )/ (L * L)
            = (205.57 * 8 * 8 )/ (7.6 * 7.6)
            = 32.72 kN/m
@@@@@@@@@@@@@@new set of calculations
Location of max shear = dv =maximum (0.9d, 0.75h)
= maximum (0.9 * 462.5, 0.75 * 500)
= maximum 416.25mm, 375mm)
Location of max shear = 375 mm from the face of the support
Maximum shear force = wL / 2 = (32.72 * 7.6 / 2) = 124.4 kN
Shear strength provided by concrete = φc β √fc’ bw dv
                                CSA A23.3-14 (Eq. 11.5)
= 0.65 * 0.18 * √30 * 200 * 375
= 78.4 * 1000 N = 78.4 kN
The nominal shear strength required to be provided by shear reinforcement is
124.4- 78.4 = 46.025 kN
Ꝋ = 35 degrees CSA A23.3-14 (11.3.6.2)
(Av / s)req = (Vf-Vc) ) / ( φ *fyt * dv * cot Ꝋ)
CSA A23.3-14 (11.3.5.1)
    =    (46.25 * 1000) / ( 0.85 * 400 * 375 * 1.428)
    = 0.253 mm2 /mm
10mm – 2 legged – 300 mm c/c
Actual (Av / s) = (200 / 200) = 0.67 mm2 /mm
0.67 > 0.253 (OK)
(Av / s) min = 0.06 √fc’(bw / fyt )        CSA A23.3-14 (11.2.8.2)
        = (0.06 * √30 * 200) / 400
        = 0.165 mm2 /mm    OK
Check if the failure is ductile when b = 500 mm
α1 = 0.85 – 0.0015 f’c
    = 0.85 – 0.0015 * 30 = 0.805 > 0.67
β1 = 0.97 – 0.0025 f’c
    = 0.97 – 0.0025 * 30 = 0.905 > 0.67
Take b = 500 mm
φs = 0.85 – factor of reduction – material - steel
φc = 0.65 – factor of reduction – material - concrete
Reference:
https:
ooks.google.co.in
ooks?id=qdsufl6tSV8C&pg=PA89&lpg=PA89&dq=neutral+axis+in+flange+-+tbeam+-+moment+calculation+-+CSA+a23&source=bl&ots=DY5RL0u8Wd&sig=ACfU3U3f8TRR0KgeGcSdXLkvP6WVvS--xQ&hl=en&sa=X&ved=2ahUKEwj4objD9_XnAhW583MBHdH3DYEQ6AEwDnoECAsQAQ#v=onepage&q=neutral%20axis%20in%20flange%20-%20tbeam%20-%20moment%20calculation%20-%20CSA%20a23&=false
For the failure to be ductile i.e. tensile reinforcement to reach yield (according to CSA A23.3-14 (10.5.2) )
The value of c max = (700 / 700 + fy ) * d
                    = (700 / 700 + 400 ) * 462.5
                    = 294 mm
C max = (a max/ β1 )
Therefore a max = 294 * 0.905 = 266.07
jd = d – a/2
    = 462.5 – (266.07 / 2)
    = 329.465
a = ( φs As fy ) / (φc b )
    = (0.85 * 1962.5) * 400) / (0.65 * 0.805 * 30 * 500)
    =    85.024
c = (a / β1 )
    = 85.024/ 0.905
    = 94.05
Moment capacity of the given section
Mf = As φs fy (d – a /2 )
     = 1962.5 * 0.85 * 400 * (462.5 – 85.024 /2 )
    = 280.2 x 106 Nmm
    = 280.2 kNm
Load is assumed as UDL and support conditions as simply supported
Given, effective span = 7.6m
Factored load w =( M * 8 )/ (L * L)
            = (280.2 * 8 )/ (7.6 * 7.6)
            = 38.8 kN/m
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Part 2 - Strength Calculations for the Damaged Beam
Given data:
Width of the web, b    = 200 mm
Width of the flange = 500 mm
Effective depth, d = 500 – (25 + 25/2) = 462.5 mm
f’c = 30 MPa
fy = 400 MPa
area of each bar (pie * 25 * 25 ) / 4 = 490.625 mm2
Area of two bars = 981.25 mm2
Area of two outer bars = 981.25 mm2
Area of two inner bars = 981.25 mm2
30% of...