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### 10.01.03.153

1. 1. A PRESENTATION ON RECTANGULAR BEAM SINGLY & DOUBLY BY USD METHOD
2. 2.  DEFINITION & ASSUMPTIONS  EVALUATION OF DESIGN PARAMETERS MOMENT FACTORS STRENGTH REDUCTION FACTOR  BALANCED REINFORCEMENT RATIO b  DESIGN PROCEDURE FOR SINGLY REINFORCED BEAM CHECK FOR CRACK WIDTH  DESIGN PROCEDURE FOR DOUBLY REINFORCED BEAM
3. 3. Precast beams made with Elematic machinery offers an extremely wide range of technical possibilities. Beams have a rectangular crosssection and are used mostly horizontally for the structure of multi-store buildings, and also for bridges. With Elematic beams can be either prestressed or reinforced. FIG- Rectangular FIG- Rectangular Beam
4. 4. Types of reinforced concrete beams: a)Singly reinforced beam b)Doubly reinforced beam c)Singly or Doubly reinforced flanged beams
5. 5. Plane sections before bending remain plane and perpendicular to the N.A. after bending Strain distribution is linear both in concrete & steel and is directly proportional to the distance from N.A. Strain in the steel & surrounding concrete is the same prior to cracking of concrete or yielding of steel Concrete in the tension zone is neglected in the flexural analysis & design computation b εc=0.003 c h 0.85fc’ a a/2 C d d-a/2 T εs = fy / Es TO SLIDE-5
6. 6. Concrete stress of 0.85fc’ is uniformly distributed over an equivalent compressive zone. fc’ = Specified compressive strength of concrete in psi. Maximum allowable strain of 0.003 is adopted as safe limiting value in concrete. The tensile strain for the balanced section is fy/Es Moment redistribution is limited to tensile strain of at least 0.0075 fs Actual fy Idealized Es 1 εy εs
7. 7. EVALUATION OF DESIGN PARAMETERS Total compressive force Total Tensile force C = 0.85fc’ ba (Refer stress diagram) T = As fy C=T 0.85fc’ ba = As fy a = As fy / (0.85fc’ b) = d fy / (0.85 fc’)   = As / bd Moment of Resistance, Mn = 0.85fc’ ba (d – a/2) or Mn = As fy (d – a/2) =  bd fy [ d – (dfyb / 1.7fc’) ] TO SLIDE-7  = Strength Reduction Factor
8. 8. Balaced Reinforcement Ratio ( b) From strain diagram, similar triangles cb / d = 0.003 / (0.003 + fy / Es) ; Es = 29x106 psi cb / d = 87,000 / (87,000+fy) Relationship b / n the depth `a’ of the equivalent rectangular stress block & depth `c’ of the N.A. is a = β1 c β1= 0.85 β1= 0.85 - 0.05(fc’ – 4000) / 1000 ; 4000 < fc’ 8000 β1= 0.65 b ; fc’ 4000 psi ; fc’> 8000 psi = Asb / bd = 0.85fc’ ab / (fy. d) = β1 ( 0.85 fc’ / fy) [ 87,000 / (87,000+fy)]
9. 9. In case of statically determinate structure ductile failure is essential for proper moment redistribution. Hence, for beams the ACI code limits the max. amount of steel to 75% of that required for balanced section. For practical purposes, however the reinforcement ratio ( = As / bd) should not normally exceed 50% to avoid congestion of reinforcement & proper placing of concrete.   0.75  b Min. reinforcement is greater of the following: Asmin = 3fc’ x bwd / fy or 200 bwd / fy min = 3fc’ / fy or 200 / fy For statically determinate member, when the flange is in tension, the bw is replaced with 2bw or bf whichever is smaller The above min steel requirement need not be applied, if at every section, Ast provided is at least 1/3 greater than the analysis
10. 10. DESIGN PROCEDURE FOR SINGLY REINFORCED BEAM       Determine the service loads Assume `h` as per the support conditions according to the ACI code Calculate d = h – Effective cover Assume the value of `b` by the rule of thumb. Estimate self weight Perform preliminary elastic analysis and derive B.M (M), Shear force (V) values  Compute min and b  Choose  between min and b  Revise & repeat the steps, if necessary  BACK
11. 11.    With the final values of , b, d determine the Total As required Design the steel reinforcement arrangement with appropriate cover and spacing stipulated in code. Bar size and corresponding no. of bars based on the bar size #n. Check crack widths as per codal provisions EXAMPLE 
12. 12. DESIGN PROCEDURE FOR DOUBLY REINFORCED BEAM  Moment of resistance of the section Mu = Mu1 + Mu2 Mu1 = M.R. of Singly reinforced section =  As1 fy (d – a/2) ; As1 = Mu1 / [  fy (d – a/2) ] Mu2 =  As2 fy (d – d’) ; As2 = Mu2 / [ fy (d – d’) ] Mu =  As1 fy (d – a/2) +  As2 fy (d – d’) If Compression steel yields, ε’  fy / Es I.e., 0.003 [ 1 – (0.85 fc’ β1 d’) / ((- ’) fyd) ]  fy / Es If compression steel does not yield, fs’ = Es x 0.003 [ 1 – (0.85 fc’ β1 d’) / ((- ’) fyd) ] Balanced section for doubly reinforced section is END b = b1 + ’ (fs / fy) b1 = Balanced reinforcement ratio for S.R. section
13. 13. FIG: DOUBLY REINFORCED BEAM
14. 14. DESIGN STRENGTH  Mu =  Mn The design strength of a member refers to the nominal strength calculated in accordance with the requirements stipulated in the code multiplied by a Strength Reduction Factor , which is always less than 1. Why  ? To allow for the probability of understrength members due to variation in material strengths and dimensions To allow for inaccuracies in the design equations To reflect the degree of ductility and required reliability of the member under the load effects being considered. To reflect the importance of the member in the structure RECOMMENDED VALUE Beams in Flexure………….……….. Beams in Shear & Torsion ………… 0.90 0.85  BACK
15. 15. AS PER ACI CODE Simply One End Both End Cantilever Supported Continuous Continuous L / 16 L / 18.5 L / 21 L/8 Values given shall be used directly for members with normal weight concrete (Wc = 145 lb/ft3) and Grade 60 reinforcement  For structural light weight concrete having unit wt. In range 90-120 lb/ft3 the values shall be multiplied by (1.65 – 0.005Wc) but not less than 1.09  For fy other than 60,000 psi the values shall be multiplied by (0.4 + fy/100,000)  `h` should be rounded to the nearest whole number  BACK
16. 16.  Reduced sustained load deflections.  Creep of concrete in compression zone  transfer load to compression steel  reduced stress in concrete  less creep  less sustained load deflection
17. 17. Spacing of Reinforcement and Concrete Cover The spacing of reinforcement and the concrete cover should be sufficient to make concreting more easier; consequently, the concrete surrounding the reinforcement can be efficiently vibrated, resulting in a dense concrete cover which provides suitable protection of the reinforcement against corrosion.  BACK
18. 18. FIGURE - \$- Spacing of steel bars (a) in one row or (b) in two rows Spacing of Reinforcement: Figure-\$, shows two reinforced concrete sections. The bars are placed such that the clear spacing s shall be at least equal to the maximum diameter of the bars, or 25 mm, or 1.50 times maximum size of aggregate, whichever is greatest, according to the Egyptian Code. Vertical clear spacing between bars, in more than one layer, shall not be less than 25 mm.
19. 19. BAR SIZE  #n = n/8 in. diameter for n 8. Ex. #1 = 1/8 in. …. #8 = 8/8 i.e., I in. Weight, Area and Perimeter of individual bars Bar No Wt.per Foot (lb) 3 4 5 6 7 8 9 10 11 14 18 0.376 0.668 1.043 1.502 2.044 2.670 3.400 4.303 5.313 7.650 13.600  BACK Stamdard Nominal Dimensions C/S Area, Perimeter Diameter db (in.) Ab (in2) inch mm 0.375 9 0.11 1.178 0.500 13 0.20 1.571 0.625 16 0.31 1.963 0.750 19 0.44 2.356 0.875 22 0.60 2.749 1.000 25 0.79 3.142 1.128 28 1.00 3.544 1.270 31 1.27 3.990 1.410 33 1.56 4.430 1.693 43 2.25 5.319 2.257 56 4.00 7.091
20. 20. CRACK WIDTH w = Where, w = = = fs = dc = A = Aeff = = N = 0.000091.fs.3(dc.A) Crack width 0.016 in. for an interior exposure condition 0.013 in. for an exterior exposure condition 0.6 fy, kips Distance from tension face to center of the row of bars closest to the outside surface Effective tension area of concrete divided by the number of reinforcing bars Aeff / N Product of web width and a height of web equal to twice the distance from the centroid of the steel and tension surface Total area of steel As / Area of larger bar  BACK