DESIGN FOR TORSION Reinforced Concrete Structures (MSc Structural Engineering -2012) LECTURE 2 By Dr. Rashid

DESIGN Design before 1995 Design after 1995 Design for shear and torsion was combined Design after 1995 Design for shear, flexure and torsion are carried out separately and then the transverse steel for shear and torsion are added and also the longitudinal steel of flexure and torsion are combined at the time of detailing. Concrete shear and compressive strength For shear design, full shear strength of concrete is considered while no shear strength of concrete is considered in torsion design. However, comp. Strength of concrete is indirectly used as concrete compression diagonals in the space truss analogy.

DESIGN Types of Reinforcement Required to resist Torque Transverse Reinforcement in the form of stirrups (closed loops) Note: open stirrups are for shear not for torsion Longitudinal reinforcement in addition to steel required for flexure, specially in corners and around perimeter

DESIGN Concrete section for Design The design of a cracked reinforced concrete section is performed considering it as an equivalent hollow tubular section This is done because….. Experimental and theoratical evidence show that, after cracking, the concrete in the central portion of a solid member has little effect on the torsional strength of the member

DESIGN Concrete section for Design t (A solid cross-section of beam subjected to torsion is idealized as a thin walled tube with core concrete neglected) t Shear flow, q

DESIGN Analysis of the Torsional Resistance of the member considering SPACE TRUSS ANALOGY Shear stresses are considered constant over a finite thickness around the periphery of the member allowing the beam to be presented by an equivalent tube. t Shear flow, q Applied torque is resisted by the shear flow, q which is taken as constant around the perimeter

DESIGN Analysis of the Torsional Resistance of the member considering SPACE TRUSS ANALOGY After cracking the tube is idealized as a hollow truss consisting of closed stirrups, longitudinal bars in the corners, and compression diagonals. The diagonals are idealized as being between the cracks that are at angle , generally taken as 45 degrees for RC.

DESIGN Analysis of the Torsional Resistance of the member considering SPACE TRUSS ANALOGY Ao = Xo x Yo , Xo and Yo are measured from the center of the wall t Internal area is neglected and shear flow is calculated from average shear stress instead of maximum. Y0 Shear flow, q A0 X0

DESIGN Analysis of the Torsional Resistance of the member considering SPACE TRUSS ANALOGY Relationship between applied torque and shear flow for a tube section Y0 1 2 3 X0

DESIGN Analysis of the Torsional Resistance of the member considering SPACE TRUSS ANALOGY Relationship between applied torque and shear flow for a tube section 4 Y0 avg. shear stress due to torsion at any point along the perimeter of the tube gross area enclosed by the shear flow path. X0

DESIGN Analysis of the Torsional Resistance of the member considering SPACE TRUSS ANALOGY Let area enclosed by the outside perimeter of concrete section or gross area of the section outside perimeter of the concrete section As per ACI Code Prior to cracking, approximate values of thickness (t) of hollow section and Ao 5 6

DESIGN Analysis of the Torsional Resistance of the member considering SPACE TRUSS ANALOGY Cracking of Concrete Cracking of concrete occurs when the principal tensile stress of the concrete exceeds the tensile strength of the concrete Concrete Cracking Tensile strength of the concrete is appx. = ACI Code considers a conservative value of

DESIGN Analysis of the Torsional Resistance of the member considering SPACE TRUSS ANALOGY Cracking Torque The twisting moment at which cracking starts is cracking torque and is denoted by Tcr In pure torsion, the principal tensile stress is equal to the torsional shear stress, hence 7 8 9

DESIGN Analysis of the Torsional Resistance of the member considering SPACE TRUSS ANALOGY According to ACI 11.6.1 Torsional effects may be neglected when the factored torsional moment is less than one-fourth the cracking torque along with the Φ factor 10 or 11

DESIGN Critical Section for Design Torque In non-prestressed members, the design torque is calculated at d distance from the face of the support Critical Section d As per ACI 11.6.2.4, If a concentrated torque acts within d distance from the edge of support, the critical section for design must be taken at the face of the support

DESIGN Basic Derivation for Space Truss Analogy

DESIGN Basic Derivation for Space Truss Analogy A space truss consisting of longitudinal bars in the corners, closed stirrups and diagonal concrete compression member between the cracks. The height of the truss between centers of bars is taken equal to yo and the width between centers is taken equal to xo. The angle of crack is close to 45° but actually it varies and become lesser at high torques.

DESIGN Basic Derivation for Space Truss Analogy Face-1 Face-4 Face-2

DESIGN Basic Derivation for Space Truss Analogy The shear force force acting over the wall in cross-section for each face is denoted by Vi. The tensile force required in the longitudinal bars required per face may be denoted by Ni. The symbol used for diagonal compressive force required per face is Di and the corresponding diagonal truss is fcd

DESIGN Basic Derivation for Space Truss Analogy Y0 X0 Shear flow = 12 Shear flow is converted into total shear force per face by multiplying with the length of the face. Y0 12 13 X0

DESIGN Basic Derivation for Space Truss Analogy V1 V2 V4 V3 Y0 14 15 By taking the sum of moments produced by the shears on each face , the resisting torque (Tn) is found and is proved that it is equal to the applied torque. V1 V2 Y0 V4 14 V3 Values of V1, V2, V3 and V4 from equations 12 and 13 15 X0

DESIGN Basic Derivation for Space Truss Analogy V1 V2 V4 V3 16 17 Y0 X0

DESIGN Basic Derivation for Space Truss Analogy D2 V2 N2 Compressive Stress, fcd Considering face 2, the force V2 is resolved into two component; a diagonal compressive force D2 and axial tensile force N2. D2 V2 18 N2

DESIGN Basic Derivation for Space Truss Analogy Compressive Stress, fcd The force D2 acts over a width equal to and thickness equal to t and the resulting compressive stress then becomes; 19 20 Same stresses are developed in all four walls of the tube and, combined with the stress due to direct shear force, must not exceed the crushing strength of concrete.

DESIGN Required Transverse Reinforcement 21 22 23 Discussion about Ao Discussion about angle Minimum amount of transverse reinforcement combined with that of for the shear

DESIGN Required Longitudinal Reinforcement Minimum Longitudinal Steel 24 Minimum Longitudinal Steel When 25