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# Flexural design of Beam...PRC-I

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Prepared by madam rafia firdous. She is a lecturer and instructor in subject of Plain and Reinforcement concrete at University of South Asia LAHORE,PAKISTAN.

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### Flexural design of Beam...PRC-I

1. 1. Plain & Reinforced Concrete-1 By Engr. Rafia Firdous Flexural Analysis and Design of Beams (Ultimate Strength Design of Beams) 1
2. 2. Plain & Reinforced Concrete-1 Ultimate Strength Design of Beams (Strength Design of Beams) Strength design method is based on the philosophy of dividing F.O.S. in such a way that Bigger part is applied on loads and smaller part is applied on material strength. fc’ 0.85fc’ Stress Strain Crushing Strength 0.003 favg favg = Area under curve/0.003 If fc’ ≤ 30 MPa favg = 0.72 fc’ β1 = Average Strength/Crushing Strength β1 = 0.72fc’ / 0.85 fc’ = 0.85 2
3. 3. Plain & Reinforced Concrete-1 Ultimate Strength Design of Beams (contd…) Cc T = Asfs la = d – a/2 N.A. εcu=0.003 Strain Diagram Actual Stress Diagram Internal Force Diagram In ultimate strength design method the section is always taken as cracked. c = Depth of N.A from the extreme compression face at ultimate stage a = Depth of equivalent rectangular stress diagram. εs h c d b 0.85fc fs 0.85fc a Equivalent Stress Diagram/ Whitney’s Stress Diagram a/2 fs 3
4. 4. Plain & Reinforced Concrete-1 Ultimate Strength Design of Beams (contd…) Actual Stress Diagram 0.85fc 0.85fc Equivalent Stress Diagram/ Whitney’s Stress Diagram c Cc Cca • The resultant of concrete compressive force Cc, acts at the centriod of parabolic stress diagram. • Equivalent stress diagram is made in such a way that it has the same area as that of actual stress diagram. Thus the Cc, will remain unchanged. a/2 ab'0.85fcbf cav ××=×× a'0.85fc'0.72f cc ×=× c '0.85f '0.72f a c c ×= cβa 1 ×= 4
5. 5. Plain & Reinforced Concrete-1 Ultimate Strength Design of Beams (contd…) Factor β1 β1 = 0.85 for fc’ ≤ 28 MPa Value of β1decreases by 0.05 for every 7 MPa increase in strength with a minimum of 0.65 0.65'0.00714f1.064β c1 ≥−= 85.0≤ 5
6. 6. Plain & Reinforced Concrete-1 Determination of N.A. Location at Ultimate Condition CASE-I: Tension Steel is Yielding at Ultimate Condition ys εε ≥ or ys ff ≥ CASE-II: Tension Steel is Not Yielding at Ultimate Condition yf yε sε yε<sε or ys ff < 0.0015 200,000 300 E f ε y y === 0.0021 200,000 420 E f ε y y === For 300 grade steel For 420 grade steel 6
7. 7. Plain & Reinforced Concrete-1 CASE-I: Tension Steel is Yielding at Ultimate Condition ysss fAfAT ×=×= ab'0.85fC cc ××= 2 a da −=l abffA cys ××=× '85.0 For longitudinal Equilibrium T = Cc bf fA a c ys × × = '85.0 1β a c =and Cc T = Asfs Internal Force Diagram a/2 la 7
8. 8. Plain & Reinforced Concrete-1 CASE-I: Tension Steel is Yielding at Ultimate Condition (contd…) Nominal Moment Capacity, Mndepending on steel = T x la       −××= 2 Mn a dfA ys Design Moment Capacity       −××= 2 M bnb a dfA ysφφ Nominal Moment Capacity, Mn depending on Concrete = Cc x la       −×××= 2 a dab0.85fc'Mn       −×××= 2 a dab0.85fc'M bnb φφ 8
9. 9. Plain & Reinforced Concrete-1 Minimum Depth for Deflection Control I 1 αΔ 3 (Depth) 1 αΔ For UDL 4 ωLαΔ ( ) 3 LωLαΔ Deflection Depends upon Span, end conditions, Loads and fy of steel. For high strength steel deflection is more and more depth is required. 9
10. 10. Plain & Reinforced Concrete-1 Minimum Depth for Deflection Control (Contd…) ACI 318, Table 9.5(a) Steel Grade Simply Supported One End Continuous Both End Continuous Cantilever 300 L/20 L/23 L/26 L/10 420 L/16 L/18.5 L/21 L/8 520 L/14 L/16 L/18 L/7 10
11. 11. Plain & Reinforced Concrete-1 Balanced Steel Ratio, ρb It is corresponding to that amount of steel which will cause yielding of steel at the same time when concrete crushes. At ultimate stage: ys εε = ys ff =and 0.003εcu = From the internal Force diagram bc ab'0.85fCc ××= ab = depth of equivalent rectangular stress block when balanced steel ratio is used. εcu=0.003 Strain Diagram εy cb d- cb 11
12. 12. Plain & Reinforced Concrete-1 Balanced Steel Ratio, ρb(contd…) ybys fd)b(ρfAT ×××== For the longitudinal equilibrium cCT = bcyb ab'0.85ffd)b(ρ ××=××× d a f 'f 0.85ρ b y c b = (1) 12
13. 13. Plain & Reinforced Concrete-1 Balanced Steel Ratio, ρb(contd…) From the strain diagram εcu= 0.003 Strain Diagram εy cb d- cb B C A E D ADE&ABCΔs b y b cd ε c 0.003 − = y b ε0.003 0.003d c + = s s s y b E E d E f 0.003 0.003 c ×× + = 13
14. 14. Plain & Reinforced Concrete-1 Balanced Steel Ratio, ρb(contd…) d f600 600 c y b × + = d f0.003E 0.003E c ys s b × + = b1b Cβa ×=As we know d f600 600 βa y 1b × + ×= (2) Put (2) in (1) y 1 y c b f600 600 β f 'f 0.85ρ + ××= 14
15. 15. Plain & Reinforced Concrete-1 Types of Cross Sections w.r.t. Flexure at Ultimate Load Level 1. Tension Controlled Section A section in which the net tensile strain in the extreme tension steel is greater than or equal to 0.005 when the corresponding concrete strain at the compression face is 0.003. εcu= 0.003 Strain Diagram εs=>0.005 c d- c cd 0.005 c 0.003 − ≥ d 0.008 0.003 c ≤ d 8 3 c ≤ d 8 3 βa 1≤and 15
16. 16. Plain & Reinforced Concrete-1 Types of Cross Sections w.r.t. Flexure at Ultimate Load Level (contd…) 2. Transition Section The section in which net tensile strain in the extreme tension steel is greater than εy but less than 0.005 when corresponding concrete strain is 0.003. εcu= 0.003 Strain Diagram εy<εs<0.005 c d- cd 8 3 βa 1> baa < 16
17. 17. Plain & Reinforced Concrete-1 Types of Cross Sections w.r.t. Flexure at Ultimate Load Level (contd…) 2. Transition Section (contd…) εcu= 0.003 Strain Diagram 0.004 c d- c To ensure under-reinforced behavior, ACI code establishes a minimum net tensile strain of 0.004 at the ultimate stage. cd 0.004 c 0.003 − = d 7 3 c = d 7 3 βa 1= 17
18. 18. Plain & Reinforced Concrete-1 Types of Cross Sections w.r.t. Flexure at Ultimate Load Level (contd…) Both the “Tension Controlled Section” and “Transition Section” are “Under-Reinforced Section” In Under-Reinforced Sections steel starts yielding before the crushing of concrete and: ρ < ρ b It is always desirable that the section is under-reinforced otherwise the failure will initiate by the crushing of concrete. As concrete is a brittle material so this type of failure will be sudden which is NOT DESIREABLE. 18
19. 19. Plain & Reinforced Concrete-1 Types of Cross Sections w.r.t. Flexure at Ultimate Load Level (contd…) 3. Compression Controlled Section (over-reinforced section) The section in which net steel strain in the extreme tension steel is lesser than εy when corresponding concrete strain is 0.003.  Capacity of steel remain unutilized.  It gives brittle failure without warning. baa > bCC > 19
20. 20. Plain & Reinforced Concrete-1 Strength Reduction Factor (Resistance Factor), Φ It is reciprocal of minor part of overall factor of safety that is applied on strength of member to obtain its design strength.  Tension Controlled Section, Φ = 0.9  Compression Controlled Section Member with lateral ties, Φ = 0.65 Members with spiral reinforcement, Φ = 0.75  Transition Section For transition section Φ is permitted to be linearly interpolated between 0.65 or 0.75 to 0.9. 20
21. 21. Plain & Reinforced Concrete-1 Strength Reduction Factor (Resistance Factor), Φ  Transition Section (contd…) ( )yt y εε ε0.005 0.25 0.65Φ − − += For members with ties For members with Spirals ( )yt y εε ε0.005 0.15 0.75Φ − − += 21
22. 22. Plain & Reinforced Concrete-1 Maximum Steel Ratio, ρmax For T = C ab'0.85ffA cys ××=× ab'0.85ffbdρ cy ××=×× d a f 'f 0.85ρ y c ×= For tension controlled section d 8 3 βa0.005ε 1s == So       ×= 1 y c max β 8 3 f 'f 0.85ρ 22
23. 23. Plain & Reinforced Concrete-1 Maximum Steel Ratio, ρmax(contd…) For transition section d 7 3 βa0.004ε 1s ==       ×= 1 y c max β 7 3 f 'f 0.85ρ 23
24. 24. Plain & Reinforced Concrete-1 Minimum Reinforcement of Flexural Members (ACI – 10.5.1)  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 slabs this formula gives a margin of 1.1 to 1.5.  This formula is not used for slabs. yy c min f 1.4 4f 'f ρ ≥= 24
25. 25. Plain & Reinforced Concrete-1 Under-Reinforced Failure Cc T Internal Force Diagram a/2 la Stage-II, Cracked Section When section cracks, N.A. moves towards compression face means “la” increases. “T” and “Cc” also increase. Stage-I, Un-cracked Section N.A. position is fixed, means “la” remains constant. Only “T” and “Cc” increase with the increase of load 25
26. 26. Plain & Reinforced Concrete-1 Under-Reinforced Failure (contd…) Stage-III, Yielding in Steel Occur T = Asfy remains constant and Cc also remains constant. “la” increases as the N.A. moves towards compression face because cracking continues. Failure initiates by the yielding of steel but final failure is still by crushing of concrete Cc T Internal Force Diagram a/2 la 26
27. 27. Plain & Reinforced Concrete-1 Under-Reinforced Failure (contd…) Derivation for ρ Design Moment Capacity abnb TM l×Φ=Φ       −×= 2 a dfAΦMΦ ysbnb For tension controlled section Φ = 0.9       −×= 2 a df0.9AMΦ ysnb And b'0.85f fA a c ys = (1) (2) 27
28. 28. Plain & Reinforced Concrete-1 Under-Reinforced Failure (contd…) Put value of “a” from (1) to (2)       × −= b'0.85f2 fA df0.9AMΦ c ys ysnb       × × −××= b'0.85f2 fρbd dfρbd0.9MΦ c y ynb For economical design unb MMΦ =       × × −××= '0.85f2 fρ 1fρbd0.9M c y y 2 u       ×−×= '0.85f f 2 ρ 1f0.9ρ bd M c y y2 u 28
29. 29. Plain & Reinforced Concrete-1 Under-Reinforced Failure (contd…) Let (MPa)R bd M 2 u = ω f 0.85fc' y =And Hence       −×= 2ω ρ 1f0.9ρR y       −= 2ω ρ 1ρ 0.9f R y 2ω ρ -ρ 0.9f R 2 y = 0 0.9f R2ω ρ2ω-ρ y 2 = × +× 0 fc' fc' 0.85 0.85 0.9f R2ω ρ2ω-ρ y 2 =×× × +× 0 '0.3825f Rω ρ2ω-ρ c 2 2 = × +× 2 0.3825fc' ωR 44ω2ωρ 2 2 × ×−±= 29
30. 30. Plain & Reinforced Concrete-1 Under-Reinforced Failure (contd…) By simplification         −±= 0.3825fc' R 11ωρ We have to use –ve sign for under reinforced sections. So         −−= fc' 2.614R 11ωρ Reason For under reinforced section ρ <ρb If we use positive sign ρ will become greater than ρb, leading to brittle failure. y 1 y c b f600 600 β f 'f 0.85ρ + ×= < 1.0 ω ωρb < 30
31. 31. Plain & Reinforced Concrete-1 Trial Method for the determination of “As” b'0.85f fA a c ys = (A)       −= 2 a df0.9AM ysu (B)       − = 2 a d0.9f M A y u s (C) Trial # 1, Assume some value of “a” e.g. d/3 or d/4 or any other reasonable value, and put in (C) to get “As” Trial # 2, Put the calculated value of “As” in (A) to get “a”. Put this “a” value in (C) to get “As” Keep on doing the trials unless “As” from a specific trial becomes equal to the “As” calculated from previous trial. THIS VALUE OF AS WILL BE THE FINAL ANSWER. 31
32. 32. Plain & Reinforced Concrete-1 Is The Section is Under-Reinforced or NOT ? 1. Calculate ρ and if it is less than ρmax, section is under reinforced 2. Using “a” and “d” calculate εt if it is ≥ 0.005, section is under-reinforced (tension controlled) 3. If section is over-reinforced than in the following equation –ve term will appear in the under-root.         −−= 'f 2.614R 11ωρ c 32
33. 33. Plain & Reinforced Concrete-1 Is The Section is Under-Reinforced or NOT ? (contd…) 1. For tension controlled section, εt = 0.005, d 8 3 βa 1= Using formula of Mn from concrete side acbnbu CΦMΦM l×==       −××= 2 a dba'0.85f9.0M cu             −×      ×= 2 d 8 3 0.85 dd 8 3 0.85b'0.85f0.9M cu 2 cu bd'0.205fM = b'0.205f M d c u min × = If we keep d > dmin the resulting section will be under- reinforced. d > dmin means that section is stronger in compression. 33
34. 34. Plain & Reinforced Concrete-1 Over-Reinforced Failure Cc T Internal Force Diagram a/2 la Stage-II, Cracked Section These two stages are same as in under-reinforced section. Stage-I, Un-cracked Section Stage-III, Concrete reaches strain of 0.003 but steel not yielding We never prefer to design a beam as over- reinforced (compression controlled) as it will show sudden failure. Φ = 0.65 εs < εy fs<fy 34
35. 35. Plain & Reinforced Concrete-1 Over-Reinforced Failure Stage-III, Concrete reaches strain of 0.003 but steel not yielding (contd…)       ××=Φ 2 a -dba'.85f00.65M cnb aCcMnb l×=Φ “a” is unknown as “fs” is not known b'0.85f fA a c ss = (i) (ii) 35
36. 36. Plain & Reinforced Concrete-1 Over-Reinforced Failure Stage-III, Concrete reaches strain of 0.003 but steel not yielding (contd…) εcu= 0.003 Strain Diagram εs c B C A E D d- c Comparing ΔABC & ΔADE c cd 0.003 εs − = 1 1s a d 0.003 ε β β a− =       − = a aβ 0.003ε 1 s ss εEf ×=       − ×= a aβ 0.003200,000f 1 s       − ×= a aβ 600f 1 s (iii) (iv) Eq # (iv) is applicable when εs < εy 36
37. 37. Plain & Reinforced Concrete-1 Over-Reinforced Failure Stage-III, Concrete reaches strain of 0.003 but steel not yielding (contd…) Putting value of “fs” from (iv) to (ii) b'0.85f a adβ 600A a c 1 s       − × = (v) Eq. # (v) is quadratic equation in term of “a”. Flexural Capacity       −Φ=      −Φ=Φ 2 '85.0 2 a dbaf a dCM cbcbnb       −=      −= 2 a dfAΦ 2 a dTΦMΦ ssbbnb Calculate “a” from (v) and “fs” from (iv) to calculate flexural capacity from these equations 37
38. 38. Plain & Reinforced Concrete-1 Extreme Tensile Steel Strain εt Type of X-section c/d a/d ρmax Φ < εy Compression Controlled 0.65 ≥ εy Transition Section (Under-Reinforced) 0.65 to 0.9 ≥ 0.004 Under- Reinforced (minimum strain for beams) 0.65 to 0.9 ≥ 0.005 Tension Controlled 0.9 ≥ 0.0075 Redistribution is allowed 0.9         + > yf600 600         + > y 1 f600 600 β         + > yy c 1 f600 600 f '0.85f β         + ≤ yf600 600         + ≤ y 1 f600 600 β         + ≤ yy c 1 f600 600 f '0.85f β 7 3 ≤ 7 3 β1≤ 7 3 f '0.85f β y c 1 ×≤ 8 3 ≤ 8 3 β1≤ 8 3 f '0.85f β y c 1 ×≤ 7 2 ≤ 7 2 β1≤ 7 2 f '0.85f β y c 1 ×≤ 38
39. 39. Plain & Reinforced Concrete-1 Capacity Analysis of Singly Reinforced Rectangular Beam by Strength Design method Data 1. Dimensions, b, h, d and L (span) 2. fc’, fy, Ec, Es 3. As Required 1. ΦbMn 2. Load Carrying Capacity 39
40. 40. Plain & Reinforced Concrete-1 Capacity Analysis of Singly Reinforced Rectangular Beam by Strength Design method (contd…) Solution Step # 1Calculte the depth of N.A assuming the section as under-reinforced ys ff = ys εε ≥and b'0.85f fA a c ys = 1β a c =and 40
41. 41. Plain & Reinforced Concrete-1 Capacity Analysis of Singly Reinforced Rectangular Beam by Strength Design method (contd…) Solution Step # 2 Calculate εs and check the assumption of step# 1 c cd 0.003εε ts − == For extreme point If εs ≥ εy, the assumption is correct If εs ≤ εy, the section is not under-reinforced. So “a” is to be calculated again by the formula of over reinforced section b'0.85f a adβ 600A a c 1 s       − × = 41
42. 42. Plain & Reinforced Concrete-1 Capacity Analysis of Singly Reinforced Rectangular Beam by Strength Design method (contd…) Solution Step # 3 Decide Φ factor For εs ≥ 0.005, Φ = 0.9 (Tension controlled section) For εs ≤ εy, Φ = 0.65 (Compression controlled section) For εy ≤ εs≤0.005, Interpolate value of Φ (Transition Section) Step # 4 Calculate ΦbMn       −Φ=Φ 2 a dfAM ysbnb       −×Φ=Φ 2 a dba'f85.0M cbnb For under-reinforced Section For over-reinforced Section 42
43. 43. Plain & Reinforced Concrete-1 Capacity Analysis of Singly Reinforced Rectangular Beam by Strength Design method (contd…) Alternate Method Step # 1 to step # 3 are for deciding whether the section is over reinforced or under-reinforced. Alternatively it can be done in the following manner. 1. Calculate ρ and ρmax if ρ < ρmax section is under- reinforced. 2. Calculate dmin, if d ≥ dmin, section is tension controlled 43
44. 44. Plain & Reinforced Concrete-1 Selection of Steel Bars for Beams 1. When different diameters are selected the maximum difference can be a gap of one size. 2. Minimum number of bars must be at least two, one in each corner. 3. Always Place the steel symmetrically. 4. Preferably steel may be placed in a single layer but it is allowed to use 2 to 3 layers. 5. Selected sizes should be easily available in market 6. Small diameter (as far as possible) bars are easy to cut and bend and place. 44
45. 45. Plain & Reinforced Concrete-1 Selection of Steel Bars for Beams (contd…) 7. ACI Code Requirements There must be a minimum clearance between bars (only exception is bundled bars).  Concrete must be able to flow through the reinforcement.  Bond strength between concrete and steel must be fully developed. Minimum spacing must be lesser of the following  Nominal diameter of bars  25mm in beams & 40mm in columns  1.33 times the maximum size of aggregate used. We can also give an additional margin of 5 mm. 45
46. 46. Plain & Reinforced Concrete-1 Selection of Steel Bars for Beams (contd…) 8. A minimum clear gap of 25 mm is to be provided between different layers of steel 9. The spacing between bars must not exceed a maximum value for crack control, usually applicable for slabs What is Detailing?  Deciding diameter of bars  Deciding no. of bars  Deciding location of bent-up and curtailment of bars  making sketches of reinforcements. 25mm Not O.K. 46
47. 47. Plain & Reinforced Concrete-1 Concrete Cover to Reinforcement Measured as clear thickness outside the outer most steel bar. Purpose  To prevent corrosion of steel  To improve the bond strength  To improve the fire rating of a building  It reduces the wear of steel and attack of chemicals specially in factories. 47
48. 48. Plain & Reinforced Concrete-1 Concrete Cover to Reinforcement (contd…) ACI Code Minimum Clear Cover Requirements 1. Concrete permanently exposed to earth, 75 mm 2. Concrete occasionally exposed to earth,  # 19 to # 57 bars 50 mm  # 16 and smaller bars 40 mm 1. Sheltered Concrete  Slabs and Walls 20 mm  Beams and Columns 40 mm 48
49. 49. Plain & Reinforced Concrete-1 Load Carried by the Beam Beam Supporting One-way Slab lx lx Exterior Beam Interior Beam Width of slab supported by interior beam = lx Width of slab supported by exterior beam = lx/2 + Cantilever width ly (ly/lx > 2)
50. 50. Plain & Reinforced Concrete-1 Load Carrie by the Beam Beam Supporting Two-way Slab (ly/lx≤ 2) lx lx ly ly Exterior Long Beam Interior Long Beam Exterior Short Beam Interior Short Beam 45o
51. 51. Plain & Reinforced Concrete-1 Load Carrie by the Beam Beam Supporting Two-way Slab (ly/lx≤ 2) contd… 45olx/2 lx/2 [ ] o 45cos 2/xl [ ] 2 o 45cos 2/x    = lArea of Square Shorter Beams For simplification this triangular load on both the sides is to be replaced by equivalent UDL, which gives same Mmax as for the actual triangular load. 2 2       = xl
52. 52. Plain & Reinforced Concrete-1 Load Carried by the Beam Beam Supporting Two-way Slab (ly/lx≤ 2) contd… 45o 45o Equivalent Rectangular Area 2 2 x 3 2 2 x 3 4 l l =×= Factor of 4/3 convert this VDL into UDL. Equivalent width supported by interior short beam lx x x 3 2 l l 2 = x 3 2 l= Equivalent width supported by exterior short beam += 3 xl Cantilever x 3 2 l
53. 53. Plain & Reinforced Concrete-1 Load Carried by the Beam Beam Supporting Two-way Slab (ly/lx≤ 2)contd… lx lx ly Exterior Long Beam ( ) 2 2 x xy 2 x       +−×= l ll l Supported Area lx/2 lx/2ly - lx 4 x 2 x 2 yx 22 llll +−= 4 x 2 yx 2 lll −=       −= 2 x yx 2 1 2 l ll
54. 54. Plain & Reinforced Concrete-1 Load Carried by the Beam Beam Supporting Two-way Slab (ly/lx≤ 2)contd… Exterior Long Beam 2 R1 3 R1 F 2 − − = y x R l l =where Factor F converts trapezoidal load into equivalent UDL for maximum B.M. at center of simply supported beam.For Square panel R = 1 and F = 4/3
55. 55. Plain & Reinforced Concrete-1 Load Carried by the Beam Beam Supporting Two-way Slab (ly/lx≤ 2)contd… Exterior Long Beam Equivalent width lx/2 lx/2ly - lx Equivalent width Length...Span F)Supported..Area( × = ly y 1 2R1 3R1 2 x yx 2 1 22 l l ll ×      − − ×      −=       − − ×      −= 2R1 3R1 2 x 1 2 x 2 lyll ( )    −= 3R1 2 x 2l + Cantilever (if present)
56. 56. Plain & Reinforced Concrete-1 Load Carried by the Beam Beam Supporting Two-way Slab (ly/lx≤ 2) contd… Interior Long Beam Equivalent width lx/2 lx/2ly - lx Equivalent width ly ( )3R1x 2 −l
57. 57. Plain & Reinforced Concrete-1 Wall Load (if present) on Beam tw (mm) H (m) 1000 81.9 1930 1000 tw Hm1 ××      ××= UDL on beam Htw019.0 ××= (kN/m) Htw019.0 ××
58. 58. Plain & Reinforced Concrete-1 Wall Load on the Lintel Equivalent UDL on lintel if height of slab above lintel is greater than 0.866L 0.866L 60o 60o L Ltw11.0UDL ××= kN/m tw = wall thickness in “mm” L = Opening size in “m” If the height of slab above lintel is less than 0.866L Total Wall Load + Load from slab in case of load bearing wall UDL = (Equivalent width of slab supported) x (Slab load per unit area) = m x kN/m2 = kN/m
59. 59. Plain & Reinforced Concrete-1 Slab Load per Unit Area Top Roof Slab Thickness = 125 mm Earth Filling = 100 mm Brick Tiles = 38 mm Dead Load 2 m/kg3002400 1000 125 =×=Self wt. of R.C. slab Earth Filling 2 m/kg1801800 1000 100 =×= Brick Tiles 2 m/kg741930 1000 38 =×= 554 kg/m2 Total Dead Load, Wd =
60. 60. Plain & Reinforced Concrete-1 Slab Load per Unit Area (contd…) Top Roof Live Load WL = 200 kg/m2 Ldu W6.1W2.1W += Total Factored Load, Wu ( ) 1000 81.9 2006.15542.1Wu ××+×= 2 u m/kN66.9W =
61. 61. Plain & Reinforced Concrete-1 Slab Load per Unit Area (contd…) Intermediate Floor Slab Thickness = 150 mm Screed (brick ballast + 25% sand) = 75 mm P.C.C. = 40 mm Terrazzo Floor = 20 mm Dead Load 2 m/kg3602400 1000 150 =×=Self wt. of R.C. slab Screed 2 m/kg1351800 1000 75 =×= Terrazzo + P.C.C 2 m/kg1382300 1000 )4020( =× + = 633 kg/m2 Total Dead Load, Wd =
62. 62. Plain & Reinforced Concrete-1 Slab Load per Unit Area (contd…) Intermediate Floor Live Load Occupancy Live Load = 250 kg/m2 Moveable Partition Load = 150 kg/m2 WL= 250 + 150 = 400kg/m2 Ldu W6.1W2.1W += Total Factored Load, Wu ( ) 1000 81.9 4006.16332.1Wu ××+×= 2 u m/kN73.13W =
63. 63. Plain & Reinforced Concrete-1 Slab Load per Unit Area (contd…) Self Weight of Beam Service Self Wight of Beam = b x h x 1m x 2400 2 L11.112400m1 18 L 12 L =×××= Kg/m Factored Self Wight of Beam 22 L131.0 1000 81.9 2.1L11.11 =××= kN/m Self weight of beam is required to be calculated in at the stage of analysis, when the beam sizes are not yet decided, so approximate self weight is computed using above formula.
64. 64. Plain & Reinforced Concrete-1 Design of Singly Reinforced Beam by Strength Method (for flexure only) Data:  Load, Span (SFD, BMD)  fc’, fy, Es  Architectural depth, if any Required:  Dimensions, b & h  Area of steel  Detailing (bar bending schedule)
65. 65. Plain & Reinforced Concrete-1 Design of Singly Reinforced Beam by Strength Method (contd…) Procedure: 1. Select reasonable steel ratio between ρmin and ρmax. Then find b, h and As. 2. Select reasonable values of b, h and then calculate ρ and As.
66. 66. Plain & Reinforced Concrete-1 Design of Singly Reinforced Beam by Strength Method (contd…) 2. Using Trial Dimensions I. Calculate loads acting on the beam. II. Calculate total factored loads and plot SFD and BMD. Determine Vumax and Mumax. III. Select suitable value of beam width ‘b’. Usually between L/20 to L/15. preferably a multiple of 75mm or 114 mm. IV. Calculate dmin. b'f205.0 M d c u min = hmin = dmin + 60 mm for single layer of steel hmin = dmin + 75mm for double layer of steel Round to upper 75 mm
67. 67. Plain & Reinforced Concrete-1 Design of Singly Reinforced Beam by Strength Method (contd…) V. Decide the final depth. minhh ≥ For strength minhh ≥ For deflection ahh ≈ Architectural depth 12 hh ≈ Preferably “h” should be multiple of 75mm. Recalculate “d” for the new value of “h”
68. 68. Plain & Reinforced Concrete-1 Design of Singly Reinforced Beam by Strength Method (contd…) VI. Calculate “ρ” and “As”.         −−= fc' 2.614R 11wρ Four methods y c f 'f 0.85w = 2 u bd M R = Design Table Design curves Using trial Method a) b) c) d)
69. 69. Plain & Reinforced Concrete-1 Design of Singly Reinforced Beam by Strength Method (contd…) VII. Check As ≥ Asmin. Asmin=ρmin bd (ρmin= 1.4/fy to fc’ ≤ 30 MPa) VIII. Carry out detailing IX. Prepare detailed sketches/drawings. X. Prepare bar bending schedule.
70. 70. Plain & Reinforced Concrete-1 Design of Singly Reinforced Beam by Strength Method (contd…) 1. Using Steel Ratio I. Step I and II are same as in previous method. III. Calculate ρmax and ρmin & select some suitable “ρ”. IV. Calculate bd2 from the formula of moment V. Select such values of “b” and “d” that “bd2 ” value is satisfied. VI. Calculate As. VII. Remaining steps are same as of previous method. ( )       −== '1.7f ρf 1fρbd0.9MΦM c y y 2 nbu