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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

  1. Analysis and loads: find the factored MuM_u, VuV_u and TuT_u at the critical sections.
  2. Section size: assume bb, DD and cover; find dd.
  3. Equivalent shear: Ve=Vu+1.6TubV_e=V_u+1.6\frac{T_u}{b} Equivalent nominal shear stress τve=Vebd\tau_{ve}=\dfrac{V_e}{bd}.
  4. Check section size: τve≤τc,max\tau_{ve}\le\tau_{c,max} (Table 20: 2.8 for M20, 3.1 for M25). If not, increase the section.
  5. Equivalent moments: Mt=Tu 1+D/b1.7,Me1=Mu+MtM_t=T_u\,\frac{1+D/b}{1.7},\qquad M_{e1}=M_u+M_t Me1M_{e1} is resisted by longitudinal steel on the flexural tension face.
  6. Compression face steel: if Mt>MuM_t>M_u, an extra moment Me2=Mt−MuM_{e2}=M_t-M_u acts in the opposite sense and needs extra longitudinal steel on the compression face (designed as tension steel for Me2M_{e2}).
  7. Longitudinal steel from Me1M_{e1} (and Me2M_{e2}) using the flexure formula: Me=0.87fyAstd(1−Astfybdfck)M_e=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right) If Me1>Mu,limM_{e1}>M_{u,lim}, the section is doubly reinforced.
  8. Transverse reinforcement: find τc\tau_c from ptp_t (Table 19). If τve>τc\tau_{ve}>\tau_c, provide closed stirrups: Asvsv=Tub1d1(0.87fy)+Vu2.5d1(0.87fy)  ≥ (τve−τc)b0.87fy\frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}\ \ \ge\ \frac{(\tau_{ve}-\tau_c)b}{0.87f_y} b1b_1, d1d_1 = centre-to-centre distances between the corner bars in the width and depth directions.
  9. Spacing limits: sv≤x1s_v\le x_1, ≤(x1+y1)/4\le(x_1+y_1)/4 and ≤300\le300 mm, where x1x_1 and y1y_1 are the short and long dimensions of the stirrup.
  10. Detailing (cl. 26.5.1.7): stirrups must be closed, with 135∘^\circ 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.
  11. Check deflection, cracking and development length.

✓ Answer checked by Claude Opus 5.5 (high effort)

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