1 / 9

Torsion in Girders

Torsion in Girders. 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. B2. A2. M u = w u l n 2 /24. M u = w u l n 2 /10. M u = w u l n 2 /11.

chogan
Download Presentation

Torsion in Girders

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Torsion in Girders The beams framing into girder A2-A3 transfer a moment of wuln2/24 into the girder. This moment acts about the longitudinal axis of the girder as torsion. B2 A2 Mu = wuln2/24 Mu = wuln2/10 Mu = wuln2/11 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. B3 A3

  2. Girder A2-A3 A2 A3 A2 A3 Torsion Diagram

  3. Strength of Concrete in Torsion where, Acp = smaller of bwh + k1h2f or bwh + k2hf(h - hf) Pcp = smaller of 2h + 2(bw+ k1hf) or 2(h + bw) + 2k2(h - hf) As for shear, ACI 318 allows flexural members be designed for the torque at a distance ‘d’ from the face of a deeper support.

  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 Tu 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 Tmax = 4Tc if other members are available for redistribution of forces.

  5. Torsion Reinforcement • Stresses induced by torque are resisted with closed stirrups and longitudinal reinforcement along the sides of the beam web. • The distribution of torque along a beam is usually the same as the shear distribution resulting in more closely spaced stirrups.

  6. Design of Stirrup Reinforcement Aoh= area enclosed by the centerline of the closed transverse torsional reinforcement At = cross sectional area of one leg of the closed ties used as torsional reinforcement st = spacing required for torsional reinforcment only sv = spacing required for shear reinforcement only S = stirrup spacing

  7. Design of Longitudinal Reinforcement Al = total cross sectional area of the additional longitudinal reinforcement required to resist torsion Ph = perimeter of Aoh max. bar spacing = 12” minimum bar diameter = st/24

  8. Cross Section Check To prevent compression failure due to combined shear and torsion:

  9. Design of Torsion Reinforcementfor Previous Example p. 21 notes Design the torsion reinforcement for girder A2-A3. 30 ft 30 ft 30 ft 30 ft 24 ft 24 ft 24 ft

More Related