Torsion in Girders B 2 A 2 Mu

Torsion in Girders B 2 A 2 Mu = wuln 2/24 A 3 Mu = wuln 2/10 Mu = wuln 2/11 B 3 The beams framing into girder A 2 -A 3 transfer a moment of wuln 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 B 2 -B 3 due to the difference between the end moments in the beams framing into the girder.

Girder A 2 -A 3 A 2 Torsion Diagram A 3

Strength of Concrete in Torsion where, Acp = smaller of bwh + k 1 h 2 f or bwh + k 2 hf(h - hf) Pcp = smaller of 2 h + 2(bw+ k 1 hf) or 2(h + bw) + 2 k 2(h - hf) Slab one side of web Slab both sides of web k 1 4 8 k 2 1 2 As for shear, ACI 318 allows flexural members be designed for the torque at a distance ‘d’ from the face of a deeper support.

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

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.

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

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

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

Design of Torsion Reinforcement for Previous Example p. 21 notes Design the torsion reinforcement for girder A 2 -A 3. 30 ft 24 ft 30 ft