Singly Reinforce Concrete

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this slide will clear all the topics and problem related to singly reinforced beam by limit state method, things are explained with diagrams , easy to understand .

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Singly Reinforce Concrete

  1. 1. SINGLY REINFORCED BEAM By limit state method of design
  2. 2. DIFFERENT METHODS OF DESIGN OF CONCRETE <ul><li>Working Stress Method </li></ul><ul><li>Limit State Method </li></ul><ul><li>Ultimate Load Method </li></ul><ul><li>Probabilistic Method of Design </li></ul>
  3. 3. LIMIT STATE METHOD OF DESIGN <ul><li>The object of the design based on the limit state concept is to achieve an acceptable probability, that a structure will not become unsuitable in it’s lifetime for the use for which it is intended, </li></ul><ul><li>i.e. It will not reach a limit state </li></ul><ul><li>A structure with appropriate degree of reliability should be able to withstand safely. </li></ul><ul><li>All loads, that are reliable to act on it throughout it’s life and it should also satisfy the subs ability requirements, such as limitations on deflection and cracking. </li></ul>
  4. 4. <ul><li>It should also be able to maintain the required structural integrity, during and after accident, such as fires, explosion & local failure. </li></ul><ul><li>i.e. limit sate must be consider in design to ensure an adequate degree of safety and serviceability </li></ul><ul><li>The most important of these limit states, which must be examine in design are as follows </li></ul><ul><li>Limit state of collapse </li></ul><ul><li>- Flexure </li></ul><ul><li>- Compression </li></ul><ul><li>- Shear </li></ul><ul><li>- Torsion </li></ul><ul><li>This state corresponds to the maximum load carrying capacity. </li></ul>
  5. 5. TYPES OF REINFORCED CONCRETE BEAMS <ul><li>Singly reinforced beam </li></ul><ul><li>Doubly reinforced beam </li></ul><ul><li>Singly or Doubly reinforced flanged beams </li></ul>
  6. 6. SINGLY REINFORCED BEAM <ul><li>In singly reinforced simply supported beams or slabs reinforcing steel bars are placed near the bottom of the beam or slabs where they are most effective in resisting the tensile stresses . </li></ul>
  7. 7. <ul><li>Reinforcement in simply supported beam </li></ul><ul><li>COMPRESSION b </li></ul><ul><li>STEEL REINFORCEMENT D d </li></ul><ul><li>TENSION </li></ul><ul><li>SUPPORT SECTION A - A </li></ul><ul><li>CLEAR SPAN </li></ul>
  8. 8. <ul><li>Reinforcement in a cantilever beam </li></ul><ul><li>A </li></ul><ul><li>TENSION </li></ul><ul><li>D </li></ul><ul><li>d </li></ul><ul><li>COMPRESSION SECTION A - A </li></ul><ul><li>A </li></ul><ul><li>CLEAR COVER </li></ul>
  9. 9. <ul><li>STRESS – STRAIN CURVE FOR CONCRETE </li></ul><ul><li>f ck </li></ul><ul><li>STRESS </li></ul><ul><li>.20 % .35% </li></ul><ul><li>STRAIN </li></ul>
  10. 10. <ul><li>STRESS ― STARIN CURVE FOR STEEL </li></ul>
  11. 11. STRESS BLOCK PARAMETERS <ul><li>! </li></ul><ul><li>0.0035 0.446 f ck </li></ul><ul><li>X 2 X 2 a </li></ul><ul><li>X 1 X 1 </li></ul>
  12. 12. <ul><li>x = Depth of Neutral axis </li></ul><ul><li>b = breadth of section </li></ul><ul><li>d = effective depth of section </li></ul><ul><li>The depth of neutral axis can be obtained by considering the equilibrium of the normal forces , that is, </li></ul><ul><li>Resultant force of compression = average stress X area </li></ul><ul><li>= 0.36 f ck bx </li></ul><ul><li>Resultant force of tension = 0.87 f y A t </li></ul><ul><li>Force of compression should be equal to force of tension, </li></ul><ul><li>0.36 f ck bx = 0.87 f y A t </li></ul><ul><li>x = </li></ul><ul><li>Where A t = area of tension steel </li></ul>
  13. 14. <ul><li>Moment of resistance with respect to concrete = compressive force x lever arm </li></ul><ul><li>= 0.36 f ck b x z </li></ul><ul><li>Moment of resistance with respect to steel = tensile force x lever arm </li></ul><ul><li>= 0.87 f y A t z </li></ul>
  14. 15. MAXIMUM DEPTH OF NEUTRAL AXIS <ul><li>A compression failure is brittle failure. </li></ul><ul><li>The maximum depth of neutral axis is limited to ensure that tensile steel will reach its yield stress before concrete fails in compression, thus a brittle failure is avoided. </li></ul><ul><li>The limiting values of the depth of neutral axis x m for different grades of steel from strain diagram. </li></ul>
  15. 16. MAXIMUM DEPTH OF NEUTRAL AXIS f y N/mm 2 x m 250 0.53 d 415 0.48 d 500 0.46 d
  16. 17. LIMITING VALUE OF TENSION STEEL AND MOMENT OF RESISTANCE <ul><li>Since the maximum depth of neutral axis is limited, the maximum value of moment of resistance is also limited . </li></ul><ul><li>M lim with respect to concrete = 0.36 f ck b x z </li></ul><ul><li>= 0.36 f ck b x m (d – 0.42 x m ) </li></ul><ul><li>M lim with respect to steel = 0.87 f ck A t (d – 0.42 x m ) </li></ul>
  17. 18. LIMITING MOMENT OF RESISTANCE VALUES, N MM Grade of concrete Grade of steel Fe 250 steel Fe 450 steel Fe 500 steel General 0.148 f ck bd 0.138 f ck bd 0.133 f ck bd M20 2.96 bd 2.76 bd 2.66 bd M25 3.70 bd 3.45 bd 3.33 bd M30 4.44 bd 4.14 bd 3.99 bd
  18. 19. TYPES OF PROBLEM <ul><li>Analysis of a section </li></ul><ul><li>Design of a section </li></ul>
  19. 20. <ul><li>For under reinforced section, the value of x/d is less than x m /d value. </li></ul><ul><li>The moment of resistance is calculated by following equation: </li></ul><ul><li>M u = 0.87 f y A t d – </li></ul><ul><li>For balanced section, the moment of resistance is calculated by the following equation: </li></ul><ul><li>M u = 0.87 f y A t ( d – 0.42x m ) </li></ul><ul><li>For over reinforced section, the value of x/D is limited to xm/d and the moment of resistance is computed based on concrete: </li></ul><ul><li>Mu = 0.36 f ck b x m ( d – 0.42 x m ) </li></ul>
  20. 21. <ul><li>Analysis of section </li></ul>
  21. 22. <ul><li>Determine the moment of resistance for the section shown in figure. </li></ul><ul><li>(i) f ck = 20 N/mm , f y = 415 N/mm </li></ul><ul><li>Solution: </li></ul><ul><li>(i) f ck = 20 N/mm , f y = 415 N/mm </li></ul><ul><li>breadth (b) = 250 mm </li></ul><ul><li>effective depth (d) = 310 mm </li></ul><ul><li>effective cover = 40 mm </li></ul><ul><li>Force of compression = 0.36 f ck b x </li></ul><ul><li>= 0.36 X 20 X 250x </li></ul><ul><li>= 1800x N </li></ul>
  22. 23. <ul><li>Area of tension steel A t = 3 X 113 mm </li></ul><ul><li>Force of Tension = 0.87 f y A t </li></ul><ul><li>= 0.87 X 415 X 3 X 113 </li></ul><ul><li>= 122400 N </li></ul><ul><li>Force of Tension = Force of compression </li></ul><ul><li>122400 = 1800x </li></ul><ul><li>x = 68 mm </li></ul><ul><li>x m = 0.48d </li></ul><ul><li>= 0.48 X 310 </li></ul><ul><li>= 148.8 mm </li></ul><ul><li>148.8 mm > 68 mm </li></ul><ul><li>Therefore, </li></ul><ul><li>Depth of neutral axis = 68 mm </li></ul>f y x m 415 0.48d 500 0.46d
  23. 24. <ul><li>Lever arm z = d – 0.42x </li></ul><ul><li>= 310 – 0.42 X 68 </li></ul><ul><li>= 281 mm </li></ul><ul><li>As x < x m ( It is under reinforced ) </li></ul><ul><li>Since this is an under reinforced section, moment of resistance is governed by steel. </li></ul><ul><li>Moment of resistance w.r.t steel = tensile force X z </li></ul><ul><li>M u = 0.87 f y A t z </li></ul><ul><li>= 0.87 X 415 X 3 X 113 X 281 </li></ul><ul><li>Mu = 34.40kNm </li></ul>
  24. 25. <ul><li>Design of a section </li></ul>
  25. 26. <ul><li>Question : Design a rectangular beam to resist a bending moment equal to 45 kNm using (i) M15 mix and mild steel </li></ul><ul><li>Solution : </li></ul><ul><li>The beam will be designed so that under the applied moment both materials reach their maximum stresses. </li></ul><ul><li>Assume ratio of overall depth to breadth of the beam equal to 2. </li></ul><ul><li>Breadth of the beam = b </li></ul><ul><li>Overall depth of beam = D </li></ul><ul><li>therefore , D/b = 2 </li></ul><ul><li>For a balanced design, </li></ul><ul><li>Factored BM = moment of resistance with respect to concrete </li></ul><ul><li>= moment of resistance with respect to steel </li></ul><ul><li>= load factor X B.M </li></ul><ul><li>= 1.5 X 45 </li></ul><ul><li>= 67.5 kNm </li></ul>
  26. 27. <ul><li>For balanced section, </li></ul><ul><li>Moment of resistance M u = 0.36 f ck b x m (d - 0.42 x m ) </li></ul><ul><li>Grade for mild steel is Fe250 </li></ul><ul><li>For Fe250 steel, </li></ul><ul><li>x m = 0.53d </li></ul><ul><li>M u = 0.36 fck b (0.53 d) (1 – 0.42 X 0.53) d </li></ul><ul><li>= 2.22bd </li></ul><ul><li>Since D/b =2 or, d/b = 2 or, b= d/2 </li></ul><ul><li>Mu = 1.11 d </li></ul><ul><li>Mu = 67.5 X 10 Nmm </li></ul><ul><li>d=394 mm and b= 200mm </li></ul>f y x m 250 0.53d 415 0.48d
  27. 28. <ul><li>Adopt D = 450 mm </li></ul><ul><li>b = 250 mm </li></ul><ul><li>d = 415mm </li></ul><ul><li>Area of tensile steel At = </li></ul><ul><li>= </li></ul><ul><li>= 962 mm </li></ul><ul><li>= 9.62 cm </li></ul><ul><li>Minimum area of steel A o = 0.85 </li></ul>
  28. 29. <ul><li>= </li></ul><ul><li>= 353 mm </li></ul><ul><li>353 mm < 962 mm </li></ul><ul><li>In beams the diameter of main reinforced bars is usually selected between 12 mm and 25 mm. </li></ul><ul><li>Provide 2-20mm and 1-22mm bars giving total area </li></ul><ul><li>= 6.28 + 3.80 </li></ul><ul><li>= 10.08 cm > 9.62 cm </li></ul>
  29. 30. <ul><li>! </li></ul>
  30. 31. <ul><li>SLIDES BY : </li></ul><ul><li>HARSIMRAN SINGH TIWANA </li></ul><ul><li>ROLL NO :5059 </li></ul><ul><li>UNIVERSITY : 514010017 </li></ul><ul><li>GROUP MEMBERS : </li></ul><ul><li>DILRAJ SINGH D3/CIVIL/5051 </li></ul><ul><li>HARSIMRAN SINGH D3/CIVIL/5059 </li></ul><ul><li>AMNINDER SINGH D3/CIVIL/5060 </li></ul>

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