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Singly Reinforce Concrete
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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 .

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