13 torsion in concrete beams

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13 torsion in concrete beams

  1. 1. Torsion in Girders A2 A3 M u = w u l n 2 /24 M u = w u l n 2 /10 M u = w u l n 2 /11 B2 B3 The beams framing into girder A2-A3 transfer a moment of w u l n 2 /24 into the girder. This moment acts about the longitudinal axis of the girder as torsion. A torque will also be induced in girder B2-B3 due to the difference between the end moments in the beams framing into the girder.
  2. 2. Girder A2-A3 A2 A3 Torsion Diagram A2 A3
  3. 3. Strength of Concrete in Torsion As for shear, ACI 318 allows flexural members be designed for the torque at a distance ‘d’ from the face of a deeper support. where, A cp = smaller of b w h + k 1 h 2 f or b w h + k 2 h f (h - h f ) P cp = smaller of 2h + 2(b w + k 1 h f ) or 2(h + b w ) + 2k 2 (h - h f ) 2 8 Slab both sides of web 1 4 Slab one side of web k 2 k 1
  4. 4. Redistribution of Torque p. 44 notes-underlined paragraph When redistribution of forces and moments can occur in a statically indeterminate structure, the maximum torque for which a member must be designed is 4 times the T u for which the torque could have been ignored. When redistribution cannot occur, the full factored torque must be used for design. In other words, the design T max = 4T c if other members are available for redistribution of forces.
  5. 5. Torsion Reinforcement <ul><li>Stresses induced by torque are resisted with closed stirrups and longitudinal reinforcement along the sides of the beam web. </li></ul><ul><li>The distribution of torque along a beam is usually the same as the shear distribution resulting in more closely spaced stirrups. </li></ul>
  6. 6. Design of Stirrup Reinforcement A oh = area enclosed by the centerline of the closed transverse torsional reinforcement A t = cross sectional area of one leg of the closed ties used as torsional reinforcement s t = spacing required for torsional reinforcment only s v = spacing required for shear reinforcement only S = stirrup spacing
  7. 7. Design of Longitudinal Reinforcement max. bar spacing = 12” minimum bar diameter = s t / 24 A l = total cross sectional area of the additional longitudinal reinforcement required to resist torsion P h = perimeter of A oh
  8. 8. Cross Section Check <ul><li>To prevent compression failure due to combined shear and torsion: </li></ul>
  9. 9. Design of Torsion Reinforcement for Previous Example p. 21 notes <ul><li>Design the torsion reinforcement for girder A2-A3. </li></ul>30 ft 30 ft 30 ft 30 ft 24 ft 24 ft 24 ft

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