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Design of member for bending, shear, torsion
Short answer
Convert torsion to equivalent shear Ve = Vu + 1.6Tu/b and equivalent moments Me1 = Mu + Mt, Me2 = Mt - Mu (if Mt > Mu). Design longitudinal steel for Me1 (and Me2 on compression face). Provide closed stirrups for combined shear and torsion: Asv/sv = Tu/(b1d1·0.87fy) + Vu/(2.5d1·0.87fy) with spacing limits.
Explain how a RC structural member subjected to bending moment, shear force and torsion is designed by IS code method (write down the step-by-step procedure).
A member under combined bending, shear and torsion is designed in IS 456:2000 (cl. 41) by converting the torsion into an equivalent shear and an equivalent bending moment, and then designing for flexure and shear in the usual way.
Step-by-step procedure
- Analysis and loads: find the factored , and at the critical sections.
- Section size: assume , and cover; find .
- Equivalent shear: Equivalent nominal shear stress .
- Check section size: (Table 20: 2.8 for M20, 3.1 for M25). If not, increase the section.
- Equivalent moments: is resisted by longitudinal steel on the flexural tension face.
- Compression face steel: if , an extra moment acts in the opposite sense and needs extra longitudinal steel on the compression face (designed as tension steel for ).
- Longitudinal steel from (and ) using the flexure formula: If , the section is doubly reinforced.
- Transverse reinforcement: find from (Table 19). If , provide closed stirrups: , = centre-to-centre distances between the corner bars in the width and depth directions.
- Spacing limits: , and mm, where and are the short and long dimensions of the stirrup.
- Detailing (cl. 26.5.1.7): stirrups must be closed, with 135 hooks, and there must be at least one longitudinal bar in each corner of the stirrups. When a cross-sectional dimension exceeds 450 mm, provide additional longitudinal (side face) bars as in cl. 26.5.1.3: a total of 0.1% of the web area, divided equally between the two faces, at a spacing of not more than 300 mm or the web thickness.
- Check deflection, cracking and development length.
✓ Answer checked by Claude Opus 5.5 (high effort)
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