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

2. 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.

3. 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

4. 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

5. 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

6. 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

7. 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.

8. 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

9. 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

10. 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

11. 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

12. 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

13. 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

14. DESIGN
Analysis of the Torsional Resistance of the member considering SPACE TRUSS ANALOGY
According to ACI
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

15. 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 , 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

16. DESIGN
Basic Derivation for Space Truss Analogy

17. 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.

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

19. 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

20. 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

21. 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

22. DESIGN Basic Derivation for Space Truss Analogy V1 V2 V4 V3 16 17 Y0X0

23. 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

24. 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.

25. 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

26. DESIGN Required Longitudinal Reinforcement Minimum Longitudinal Steel24
Minimum Longitudinal Steel
When 25