![DESIGN OF
REINFORCED
CONCRETE
STRUCTURES
[CEM β 603]
(Additional Assignment)](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Design of torsion reinforcement
- 2. DESIGN OF REINFORCED CONCRETE STRUCTURES [CEM β 603] (Additional Assignment)
- 3. TOPIC: TORSION REINFORCEMENT IN R.C.C. STRUCTURES PRESENTED BY: MUSKAN DEURA B.TECH - III YEARβ CIVIL ROLL NO. - 1700655
- 4. Torsion β’ The torsional moments produced in a beam are of two types: (i) Primary torsion is the one which could be determined only by using static equilibrium condition (ii) Secondary or compatibility torsion is induced by rotation applied to one or more points along the length of the member through connected members. Twisting moment induced here is proportional to the torsional stiffness of member. (i) (ii)
- 5. Effects of Torsional Moment 1. For this longitudinal reinforcements are provided which helps in reducing the crack width through dowel action and stirrups crossing the cracks resist shear due to vertical loads and torsion. 2. Where the torsional resistance or stiffness of members is taken into account in the analysis , the members shall be designed for torsion. β’ Because of torsion, beam fails in diagonal tension forming spiral cracks around the beam. β’ Warping of the section does not allow a plane section to remain as plane after twisting.
- 6. Design Of Torsion Reinforcement: Clause 41 of IS 456:2000 provides the provisions for design of torsional reinforcements. β’ The design rules for torsion are based on equivalent moment. β’ Types of reinforcements required to resist load:
- 7. Steps In Designing :- 1. Critical Section 2. Shear and Torsion 3. Reinforcements In Members Subjected To Torsion oLongitudinal Reinforcement o Transverse Reinforcement 4. Final designing for placing of reinforcement in beam
- 8. 1. Critical Section (Clause 41.2) : β’ The details of this step is provided under the Clause 41.2 of IS 456:2000 β’ For design of torsion, section located at a distance less than βdβ from the face of support may be designed for the same torsion as computed at a distance βdβ (where βdβ is the effective depth).
- 9. 2. Shear and Torsion (Clause 41.3) : β’ The details of this step is provided under the Clause 41.3 of IS 456:2000 β’ The combined effect of torsion and shear is considered and is called equivalent shear, which is calculated as: π½ π = π½ π + π. π π» π π where, ππ= equivalent shear ππ’= shear ππ’= torsional moment π= breadth of beam β The equivalent nominal shear, π ππ = π½ π πβπ β The maximum shear stress, π πππ= from Table-20 under Clause 40 β π ππ should not be greater than π πππ β π π is found from Table-19 under Clause 40 with respect to value of πππ π¨ ππ πβπ and concrete grade.
- 10. βIf (π ππ < π π), minimum reinforcement should be provided as per criteria under Clause 26.5.1.6 π΄ π π£ π β π π£ β₯ 0.4 0.87 β ππ¦ where, π΄ π π£=total cross-sectional area of stirrup legs effective in shear π π£=stirrup spacing along the length of the member π=breadth of beam/web of flanged beam ππ¦=characteristic strength of stirrup reinforcement βIf (π ππ > π π), both longitudinal and transverse reinforcement should be provided in accordance with Clause 41.4 βIf (π ππ > π πππ), redesigning of the reinforcement should be done. Thus, this case should never happen.
- 11. 3. Reinforcements In Members Subjected To Torsion (Clause 41.4) β’ The details of this step is provided under the Clause 41.4.2 of IS 456:2000 β’ The longitudinal reinforcement shall be designed to resist an equivalent bending moment of π΄ ππ = π΄ π + π΄ π where, π π’= bending moment of cross section and, π΄ π = π» π π+ π« π π.π where, ππ’= torsional moment π·= overall depth of beam π= breadth of beam o LongitudinalReinforcement(Clause41.4.2)
- 12. In longitudinal reinforcement:- βIf (π΄ π > π΄ π) , Reinforcement shall be provided on the flexural compression face, to resist equivalent moment, given by π΄ ππ = π΄ π β π΄ π βIf (π΄ π < π΄ π) , No additional steel reinforcement is required on bending compression side.
- 13. β’ The details of this step is provided under the Clause 41.4.3 of IS 456:2000 β’ Two-legged closed hoops enclosing the corner longitudinal bars shall have an area of cross-section (stirrups), π΄ π π£ = ππ’ β π π£ {π1 β π1 β 0.87 β ππ¦ } + ππ’ β π π£ {2.5 β π1 β 0.87 β ππ¦ } β’ But the total reinforcement should be greater than: π΄ π π£ β₯ π π£π β π π β π β π π£ 0.87 β ππ¦ Where, π1 = center-to-center distance between corner bars in the direction of the width π1 = center-to-center distance between corner bars in the direction of the depth o TransverseReinforcement(Clause41.4.3)
- 14. Since, π1 = ππππ‘ππ β π‘π β ππππ‘ππ πππ π‘ππππ πππ‘π€πππ ππππππ ππππ ππ π‘βπ ππππππ‘πππ ππ π€πππ‘β π1 = π β πππππ πππ£ππ β 2 β ππππππ‘ππ ππ π π‘ππππ’ππ β 2 β ππππππ‘ππ ππππππππ‘π’πππππ πππ 2 And π1 = ππππ‘ππ β π‘π β ππππ‘ππ πππ π‘ππππ πππ‘π€πππ ππππππ ππππ ππ π‘βπ ππππππ‘πππ ππ ππππ‘β π1 = π· β πππππ πππ£ππ β 2 β ππππππ‘ππ ππ π π‘ππππ’ππ β 2 β ππππππ‘ππ ππππππππ‘π’πππππ πππ 2 From figure π₯1 = π βπππ‘ππ ππππππ πππ ππ π π‘ππππ’π π₯1 = π β 2 β ππππππ‘ππ ππ π π‘ππππ’π 2 And π¦1 = ππππππ ππππππ πππ ππ π π‘ππππ’π π¦1 = π· β 2 β ππππππ‘ππ ππ π π‘ππππ’π 2
- 15. βLongitudinal Reinforcement: βͺ Shall be placed as close as practicable to the corners of the cross section and in all cases there shall be at least one longitudinal bar at each corner of the ties. βͺ When the cross-sectional dimensions (either b or D) of the member exceeds 450 mm, additional longitudinal bars shall be provided along the two faces. The total area of such reinforcement shall not exceed 0.1% of the web area and shall be distributed equally on two faces at a spacing not exceeding 300 mm or web thickness (whichever is less). βͺ Minimum diameter of this bar shall not be less than 10 mm. 4. Final designing for placing of reinforcement in beam
- 16. βTransverse Reinforcement: βͺ This reinforcement for torsion shall be rectangular closed stirrups placed perpendicular to the axis of the member. βͺ The spacing of the stirrups shall not exceed the least of π π ππππππππππ , π π , π π+π π π , πππ ππ
- 18. QUESTION Determine the reinforcement required of a beam of b = 400 mm, d = 650 mm, D = 700 mm and subjected to factored bending moment (π π’) = 200 kN-m, factored torsional moment (ππ’) = 50 kN-m and factored shear (ππ’) = 100 kN. Use M-20 and Fe 415 for the design.
- 19. Given: β’ Factored bending moment (π π’) = 200 kN-m β’ Factored torsional moment (ππ’) = 50 kN-m β’ Factored shear (ππ’) = 100 kN β’ M- 20 grade concrete β’ Fe-415 steel hence, ππ¦=250 N/sq.mm β’ Breadth(b) = 400 mm β’ Depth(D) = 700 mm β’ Effective depth(d) = 650 mm
- 20. Solution: Equivalent shear, ππ = ππ’ + 1.6 ππ’ π = 100 + 1.6 50 0.4 = πππ kN The equivalent nominal shear, π π£π = ππ πβπ = 300 0.4β0.65 = π. πππ N/sq.mm The maximum shear stress(from table-20), π πππ₯ = π. π N/sq.mm Since, π ππ < π πππ Hence, the section is OK and does not require redesigning. 1. CheckforDepth
- 21. Assuming, 100 π΄ π π‘ πβπ = 50% π π(found from Table-19) = 0.48 N/sq.mm Since, π π < π ππ So, both longitudinal and transverse reinforcement should be provided. 2. Checkforshearreinforcement
- 22. β’ Equivalent Bending Moment, ππ1 = π π’ + ππ‘ Here, π΄ π = ππ’ 1+ π· π 1.7 = 50 1+ 0.7 0.4 1.7 = ππ. ππ kN-m Hence, ππ1 = π π’ + ππ‘ = 200 + 80.88 = πππ. ππkN-m Since (π΄ π < π΄ π) , No additional steel reinforcement is required on bending compression side. β’ From Table 2 of SP-16, For π π’ πβπ2 = π. ππN/sq.mm 3. LongitudinalReinforcement:
- 23. β’ By linear interpolation π΄ π π‘ πβπ = 0.5156 π΄ π π‘ = 0.5156β400β650 100 = ππππ. ππ ππ. ππ β’ If we provide 2-25T and 2-16T = 981 + 402 = 1383 sq.mm This gives π΄ π π‘ πβπ = 0.532 for which,π π= 0.488 N/sq.mm β’ Minimum percentage of tension reinforcement = 0.85 ππ¦ β 100 = 0.205 β’ Maximum percentage of tension reinforcement is 4.0. β’ So, 2-25T and 2-16T bars satisfy the requirements
- 24. According to dimensions, We have π« > πππππ Hence, side face reinforcement should be provided Let us provide 2-10 mm diameter bars at the mid- depth of the beam and one on each face Area of bars = 157 sq.mm The total area required = 0.1β400β300 100 = 120 sq.mm < 157 sq.mm Hence, OK
- 25. Using twoβlegged 10 mm - diameter bars for stirrups π΄ π π£ = 2 β πβ102 4 = πππ. ππ sq.mm And, π1 = π β πππππ πππ£ππ β 2 β ππππππ‘ππ ππ π π‘ππππ’ππ β α α 2 β ππππππ‘ππ ππππππππ‘π’πππππ πππ 2 = πππ mm π1 = π· β πππππ πππ£ππ β 2 β ππππππ‘ππ ππ π π‘ππππ’ππ β α α 2 β ππππππ‘ππ ππππππππ‘π’πππππ πππ 2 = πππ mm π₯1 = π β 2 β ππππππ‘ππ ππ π π‘ππππ’π 2 = πππmm π¦1 = π· β 2 β ππππππ‘ππ ππ π π‘ππππ’π 2 = πππmm 4. TransverseReinforcement:
- 26. Also, we know that π΄ π π£ = ππ’ β π π£ {π1 β π1 β 0.87 β ππ¦ } + ππ’ β π π£ {2.5 β π1 β 0.87 β ππ¦ } 157.08 = 50 β 106 β π π£ {305 β 600 β 0.87 β 250 } + 100 β 103 β π π£ {2.5 β 600 β 0.87 β 250 } π π£(πππππ’πππ‘ππ) = πππ ππ And, π π£ = least of π π£ πππππ’πππ‘ππ , π₯1 , π₯1+π¦1 4 , 300 ππ π π£ = least of 160, 340, 242, 300 ππ π π£ = 160 mm Area check, π΄ π π£ β 0.87βππ¦ π π£ β₯ π π£π β π π β π πππ. π > πππ. π Hence, OK
- 27. β’ Final Diagram will look like: