Download presentation

Presentation is loading. Please wait.

Published byHope Melton Modified over 5 years ago

1
Torsion in Girders A2 A3 M u = w u l n 2 /24 M u = w u l n 2 /10M 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
Girder A2-A3 A2 A3 Torsion Diagram A2A3

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 ) k1k1 k2k2 Slab one side of web 41 Slab both sides of web 82

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

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

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google