Design for Shear and Torsion

A member under combined bending, shear and torsion is designed in IS 456:2000 (cl. 41) by converting 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 factored $M_u$, $V_u$ and $T_u$ at the critical sections.
2. Section size: assume $b$, $D$, cover; find $d$.
3. Equivalent shear: $$V_e=V_u+1.6\frac{T_u}{b}$$ Equivalent nominal shear stress $\tau_{ve}=\dfrac{V_e}{bd}$.
4. Check section size: $\tau_{ve}\le\tau_{c,max}$ (Table 20: 2.8 for M20, 3.1 for M25). If not, increase the section.
5. Equivalent moments: $$M_t=T_u\,\frac{1+D/b}{1.7},\qquad M_{e1}=M_u+M_t$$ $M_{e1}$ is resisted by longitudinal steel on the tension face.
6. Compression face steel: if $M_t>M_u$, an extra moment $M_{e2}=M_t-M_u$ acts on the compression face and requires extra longitudinal steel there (acting as tension steel for $M_{e2}$, in the opposite direction).
7. Longitudinal steel from $M_{e1}$ (and $M_{e2}$) using flexure formulas: $$M_e=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)$$ If $M_{e1}>M_{u,lim}$, the section is doubly reinforced.
8. Transverse reinforcement: find $\tau_c$ from $p_t$ (Table 19). If $\tau_{ve}>\tau_c$, provide closed stirrups: $$\frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}\ \ \ge\ \frac{(\tau_{ve}-\tau_c)b}{0.87f_y}$$ $b_1$, $d_1$ = centre-to-centre distances of corner bars in width and depth.
9. Spacing limits: $s_v\le x_1$, $\le(x_1+y_1)/4$, $\le300$ mm, where $x_1$, $y_1$ are the short and long dimensions of the stirrup.
10. Detailing: stirrups must be closed with 135$^\circ$ hooks, and torsional steel must be carried a distance $d$ beyond the section where it is no longer needed. For $D>450$ mm provide side face reinforcement (0.1% of web area, 0.1%$b\,D$ divided on both faces).
11. Check deflection, cracking and development length.

Given: $b=300$ mm, $d=550$ mm, 4-20 mm bars (2 bent up at 45$^\circ$), $V_u=425$ kN at the support, $L=6$ m, M20, Fe415.

Step 1: Shear force. Reaction $=w_uL/2=425$ kN, so $w_u=\dfrac{2\times425}{6}=141.67$ kN/m. At $d$ from the face of support:

$$V_{u,d}=425-141.67\times0.55=347.1\ \text{kN}$$

Check: $\tau_v=\dfrac{425\times10^3}{300\times550}=2.576$ N/mm$^2<\tau_{c,max}=2.8$ N/mm$^2$, so the section is adequate.

Step 2: Concrete strength. Straight bars: 2-20, $A_{st}=628.3$ mm$^2$.

$$p_t=\frac{100\times628.3}{300\times550}=0.381\%,\quad\tau_c=0.423\ \text{N/mm}^2,\quad V_c=69.8\ \text{kN}$$

$V_{us}=V_{u,d}-V_c=347.1-69.8=277.3$ kN.

Step 3: Shear resistance of the bent-up bars ($A_{sb}=2\times314.16=628.3$ mm$^2$):

$$V_{us,bent}=0.87f_yA_{sb}\sin45^\circ=0.87\times415\times628.3\times0.7071=160.4\ \text{kN}$$

By IS 456 cl. 40.4 this should not exceed half of the total shear reinforcement load: $0.5\times277.3=138.7$ kN. Hence bent-up bars are credited with 138.7 kN.

Step 4: Additional stirrups for $V_{us}=277.3-138.7=138.7$ kN, using 10 mm 2-legged stirrups ($A_{sv}=157.1$ mm$^2$):

$$s_v=\frac{0.87\times415\times157.1\times550}{138.7\times10^3}=225\ \text{mm}$$

Limits: $0.75d=412$ mm, 300 mm, $\dfrac{0.87f_yA_{sv}}{0.4b}=473$ mm.

Answer: Shear resistance of bent-up bars $=160.4$ kN (limited to 138.7 kN by the code); additional stirrups: 10 mm 2-legged @ 220 mm c/c near the supports.

When a beam is loaded, shear force and bending moment together produce diagonal tension in the web. Concrete is weak in tension, so inclined cracks form at about 45$^\circ$ near the supports. After cracking, shear is carried by several mechanisms working together.

Shear transfer mechanisms (beam without stirrups)

1. Uncracked concrete compression zone: carries about 20-40% of the shear.
2. Aggregate interlock (friction) across the crack: rough crack faces transfer shear, about 35-50%.
3. Dowel action of longitudinal bars: bars crossing the crack resist transverse movement, about 15-25%.

          P
    ______|_______________
   |      |     inclined crack
   |  Vcz |  /   Vay (interlock)
   |______|_/____________
   o===o==/===o===o===o===   Vd (dowel)
  R

With shear reinforcement

Stirrups (and bent-up bars) cross the diagonal crack, carry tension, and hold the crack width small. This also improves aggregate interlock and dowel action. The behaviour is modelled as a truss (Ritter-Morsch analogy): concrete struts in compression and stirrups as vertical ties. Design shear strength:

$$V_u=V_c+V_{us},\qquad V_c=\tau_cbd,\qquad V_{us}=\frac{0.87f_yA_{sv}d}{s_v}$$

Failure conditions (depends on shear span ratio $a/d$)

Shear strength increases with concrete grade, percentage of tension steel $p_t$ (dowel action, smaller crack width) and axial compression, and decreases with depth (size effect). Since shear failures are brittle, beams are designed so flexural failure occurs first.

Given: $b=400$ mm, $d=550$ mm, 5-25 mm bars (two bent up at 45$^\circ$), $w_u=100$ kN/m on 6 m clear span, M20, Fe415.

Step 1: Design shear force

$$V_u=\frac{w_uL}{2}=\frac{100\times6}{2}=300.0\ \text{kN (at support face)}$$

Critical section at $d$ from the support face: $V_{u,d}=300.0-100\times0.55=245.0$ kN.

$\tau_v=\dfrac{245.0\times10^3}{400\times550}=1.114$ N/mm$^2<\tau_{c,max}=2.8$ N/mm$^2$, OK.

Step 2: Shear resistance of concrete. Bars continuing to the support: 3-25, $A_{st}=1472.6$ mm$^2$.

$$p_t=\frac{100\times1472.6}{400\times550}=0.669\%\ \Rightarrow\ \tau_c=0.534\ \text{N/mm}^2\ (\text{Table 19})$$ $$V_c=\tau_cbd=0.534\times400\times550=117.5\ \text{kN}$$

Shear to be carried by shear reinforcement: $V_{us}=245.0-117.5=127.5$ kN.

Step 3: Shear resistance of the bent-up bars

$$V_{us,bent}=0.87f_yA_{sb}\sin\alpha=0.87\times415\times981.7\times\sin45^\circ=250.6\ \text{kN}$$

IS 456 (cl. 40.4) allows bent-up bars to carry at most half of the shear reinforcement requirement, i.e. $0.5\times127.5=63.7$ kN. So the useful contribution of the bent bars is 63.7 kN.

Step 4: Additional stirrups. Balance shear $=127.5-63.7=63.7$ kN. For 8 mm 2-legged stirrups ($A_{sv}=100.5$ mm$^2$):

$$s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times415\times100.5\times550}{63.7\times10^3}=313\ \text{mm}$$

Limits: $\dfrac{0.87f_yA_{sv}}{0.4b}=227$ mm, $0.75d=412$ mm and 300 mm. Provide 8 mm 2-legged stirrups at 225 mm c/c near the support.

Answer: Shear resistance of bent-up bars $=250.6$ kN (limited to 63.7 kN by the 50% rule); additional stirrups 8 mm 2L @ 225 mm c/c.

Given: $b=400$ mm, $d=600$ mm, 5-25 mm bars (2 bent up at 45$^\circ$, 3 straight), $V_u=250$ kN at the support reducing linearly to 0 at mid-span (span 6 m), M20, Fe415.

Step 1: Shear force diagram. Slope $w_u=\dfrac{250}{3}=83.33$ kN/m. At $d$ from the support:

$$V_{u,d}=250-83.33\times0.6=200.0\ \text{kN},\qquad \tau_v=\frac{250\times10^3}{400\times600}=1.042\ \text{N/mm}^2\ (\text{at support})<2.8$$

Step 2: Concrete strength. Straight bars continuing to the support: 3-25, $A_{st}=1472.6$ mm$^2$.

$$p_t=\frac{100\times1472.6}{400\times600}=0.614\%,\quad\tau_c=0.516\ \text{N/mm}^2,\quad V_c=\tau_cbd=123.9\ \text{kN}$$ $$V_{us}=V_{u,d}-V_c=200.0-123.9=76.1\ \text{kN}$$

Step 3: Shear resistance of bent-up bars ($A_{sb}=2\times490.87=981.7$ mm$^2$):

$$V_{us,bent}=0.87f_yA_{sb}\sin45^\circ=0.87\times415\times981.7\times0.7071=250.6\ \text{kN}$$

This is more than the whole requirement $V_{us}=76.1$ kN, but IS 456 cl. 40.4 limits bent-up bars to 50% of the shear reinforcement, i.e. $0.5\times76.1=38.0$ kN.

Step 4: Additional stirrups to carry $V_{us}-V_{bent}=38.0$ kN, with 8 mm 2-legged stirrups ($A_{sv}=100.5$ mm$^2$):

$$s_v=\frac{0.87\times415\times100.5\times600}{38.0\times10^3}=573\ \text{mm}$$

Limits: $\dfrac{0.87f_yA_{sv}}{0.4b}=227$ mm, $0.75d=450$ mm, 300 mm. Provide 8 mm 2-legged stirrups at 225 mm c/c.

Step 5: Extent of shear reinforcement. Shear equals $V_c=123.9$ kN at $x=\dfrac{V_c}{w_u}=1.49$ m from mid-span, so shear reinforcement is needed over 1.51 m from each support; beyond it, nominal stirrups (minimum) are provided.

Answer: Shear resistance of the two bent-up bars $=250.6$ kN (usable 38.0 kN); additional stirrups 8 mm 2L @ 225 mm c/c.

In flanged beams (T- and L-beams), the web carries almost all the shear; the flange contributes little. IS 456 therefore uses the web width $b_w$ in all shear calculations.

Procedure

1. Factored shear force: find $V_u$ at the support; the critical section is taken at $d$ from the face of the support (when the support reaction compresses the end region).
2. Nominal shear stress: $$\tau_v=\frac{V_u}{b_wd}$$ Check $\tau_v\le\tau_{c,max}$ (Table 20); else enlarge the web.
3. Percentage of steel: $p_t=\dfrac{100A_{st}}{b_wd}$ (tension steel continuing to the support). Find design shear strength $\tau_c$ of concrete from Table 19.
4. Decide reinforcement:
   • If $\tau_v\le\tau_c$: provide minimum shear reinforcement, $\dfrac{A_{sv}}{b_ws_v}\ge\dfrac{0.4}{0.87f_y}$.
   • If $\tau_v>\tau_c$: design for $V_{us}=V_u-\tau_cb_wd$ and
   $$s_v=\frac{0.87f_yA_{sv}d}{V_{us}}$$
5. Spacing limits: $s_v\le0.75d$ and $\le300$ mm, and not more than $\dfrac{0.87f_yA_{sv}}{0.4b_w}$.
6. Bent-up bars (optional): $V_{us}=0.87f_yA_{sb}\sin\alpha$, at most half of the total shear reinforcement.
7. Detailing: stirrups are closed (two-legged vertical), carried into the flange top and anchored around the top bars. Vary spacing along the span: closer near supports, up to the maximum where $V_u\le\tau_cb_wd$.

  bf
 +------------------+
 |       flange     |  Df
 +-----+      +-----+
       |  bw  |   <- shear area = bw x d
       |      |
       | o  o |   stirrups around bars

Procedure for design of shear reinforcement (beam)

1. Compute factored shear $V_u$ at the critical section (distance $d$ from the support face).
2. Nominal shear stress $\tau_v=\dfrac{V_u}{bd}$; check $\tau_v\le\tau_{c,max}$.
3. Find $\tau_c$ from IS 456 Table 19 using $p_t=\dfrac{100A_{st}}{bd}$ and $f_{ck}$.
4. If $\tau_v\le\tau_c$, provide minimum stirrups: $\dfrac{A_{sv}}{b\,s_v}\ge\dfrac{0.4}{0.87f_y}$.
5. If $\tau_v>\tau_c$, shear carried by steel $V_{us}=V_u-\tau_cbd$, then $$s_v=\frac{0.87f_yA_{sv}d}{V_{us}}\le\min(0.75d,\ 300\ \text{mm})$$
6. Provide closer spacing near supports and wider spacing where shear is small.

Punching (two-way) shear in isolated footings

A column punches through the footing along a truncated pyramid. IS 456 (cl. 31.6) checks it at a section at $d/2$ from the face of the column (critical perimeter $b_0$), in addition to one-way shear at $d$ from the face.

   plan                         section
  +--------------+              col
  |  +--------+  |           ____|__|____
  |  | col    |  |   d/2         \  /  45
  |  +--------+  |   |<--->|____/__\____
  |  critical    |    critical perimeter
  +--------------+

Steps

1. Factored upward soil pressure $q_u=P_u/(\text{footing area})$ (net, excluding self weight).
2. Punching shear force: $V_u=q_u\left[BL-(a+d)(b+d)\right]$ for a column $a\times b$.
3. Critical perimeter: $b_0=2[(a+d)+(b+d)]$.
4. Nominal punching shear stress: $\tau_v=\dfrac{V_u}{b_0\,d}$.
5. Permissible punching shear stress: $\tau_c'=k_s\tau_c$, where $k_s=0.5+\beta_c\le1$ ($\beta_c$ = short side/long side of column) and $\tau_c=0.25\sqrt{f_{ck}}$.
6. Condition: $\tau_v\le\tau_c'$. If not satisfied, increase the depth $d$ of the footing (shear reinforcement is not normally used in footings).
7. Also check one-way (beam) shear at distance $d$ from the column face with $\tau_c$ from Table 19 for the footing steel percentage.

Bent-up bars are part of the main tension bars that are bent upwards at 45$^\circ$ (or 30$^\circ$ to 60$^\circ$) near the supports, where the bending moment is small but shear is large. They carry the diagonal tension across the inclined cracks, as well as serving as flexural steel at mid-span.

   crack
     \       bent-up bar (alpha = 45 deg)
 ====o\=====o=====
        \        \
         \_______/ stirrup

Contribution to shear

The tension $T = 0.87f_yA_{sb}$ in a bar inclined at angle $\alpha$ to the beam axis has a vertical component:

$$V_{us}=0.87f_yA_{sb}\sin\alpha$$

Here $A_{sb}$ is the total area of the bent-up bars in one plane. For $\alpha=45^\circ$, $V_{us}=0.615f_yA_{sb}$.

Code provisions (IS 456 cl. 40.4)

• Bent-up bars should not carry more than 50% of the total shear to be resisted by shear reinforcement; the rest is carried by vertical stirrups.
• For a single bar or a single group of parallel bars all bent at the same cross-section, $V_{us}$ is as above; for a series of bars bent at different sections, spacing is also considered.
• Bent-up bars must be properly anchored and bent at gradual radius; bends are placed where they are no longer needed for flexure.

Total shear resistance: $V_u=V_c+V_{us}(\text{stirrups})+V_{us}(\text{bent-up})$, with $V_c=\tau_cbd$.

Given: $b=400$ mm, $D=750$ mm, $T_u=150$ kN·m, $M_u=400$ kN·m (hogging), $V_u=110$ kN, M25, Fe500, mild exposure. Assume effective cover 50 mm: $d=700$ mm. Corner-bar centre distances: $b_1=400-2\times50=300$ mm, $d_1=750-2\times50=650$ mm.

Step 1: Equivalent shear and moment (IS 456 cl. 41.3)

$$\begin{aligned} V_e&=V_u+1.6\frac{T_u}{b}=110+1.6\times\frac{150}{0.4}=710.0\ \text{kN}\\ M_t&=T_u\,\frac{1+D/b}{1.7}=150\times\frac{1+750/400}{1.7}=253.7\ \text{kN·m}\\ M_{e1}&=M_u+M_t=400+253.7=653.7\ \text{kN·m (on the tension face, top)} \end{aligned}$$

Since $M_t=253.7$ kN·m is less than $M_u=400$ kN·m, $M_{e2}=M_t-M_u$ is negative, so no extra steel is needed on the compression face (provide nominal bars there).

Step 2: Check the section. $\tau_{ve}=\dfrac{V_e}{bd}=\dfrac{710.0\times10^3}{400\times700}=2.54$ N/mm$^2$ $<\tau_{c,max}=3.1$ N/mm$^2$ (M25). Section size is adequate.

$M_{u,lim}=0.133f_{ck}bd^2=654.67$ kN·m $\ge M_{e1}=653.7$ kN·m, so the section can be singly reinforced for $M_{e1}$.

Step 3: Longitudinal steel (tension face, top)

$$M_{e1}=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)\ \Rightarrow\ A_{st}=2647\ \text{mm}^2$$

Provide 6-25 mm dia bars ($2945$ mm$^2$) at the top, $p_t=1.05\%$.

Step 4: Transverse (stirrup) reinforcement (IS 456 cl. 41.4.3). $\tau_c=0.65$ N/mm$^2$ for $p_t=1.05\%$ (Table 19). Since $\tau_{ve}>\tau_c$, closed stirrups are needed:

$$\frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}=1.768+0.156=1.924\ \text{mm}^2/\text{mm}$$

Also $\dfrac{A_{sv}}{s_v}\ge\dfrac{(\tau_{ve}-\tau_c)\,b}{0.87f_y}=1.732$ mm$^2$/mm. The first governs.

Using 12 mm 2-legged closed stirrups ($A_{sv}=226.2$ mm$^2$): $s_v=118$ mm. Maximum spacing: $\le x_1=325$ mm, $\le\dfrac{x_1+y_1}{4}=250$ mm, $\le300$ mm.

Provide 12 mm closed 2-legged stirrups at 115 mm c/c.

Answer: Top steel 6-25 mm dia ($A_{st}=2647$ mm$^2$ required for $M_{e1}=653.7$ kN·m); bottom 2-16 mm hanger bars (nominal); closed stirrups 12 mm @ 115 mm c/c.

Given: $L=5$ m, $b=230$ mm, $D=450$ mm, central point load 30 kN (taken as service load), 4-16 mm bars with one bent up (3 bars, $A_{st}=603.2$ mm$^2$, continue to the support), M20, Fe415, 8 mm 2-legged stirrups. Assume effective cover 40 mm, $d=410$ mm.

Step 1: Design shear force

$$\begin{aligned} \text{Self weight}&=0.23\times0.45\times25=2.587\ \text{kN/m},\quad w_u=1.5\times2.587=3.881\ \text{kN/m}\\ P_u&=1.5\times30=45\ \text{kN}\\ V_u&=\frac{w_uL}{2}+\frac{P_u}{2}=32.20\ \text{kN (at support)} \end{aligned}$$

At the critical section $d$ from the support: $V_{u,d}=V_u-w_ud=32.20-3.881\times0.41=30.61$ kN.

Step 2: Nominal shear stress

$$\tau_v=\frac{V_{u,d}}{bd}=\frac{30.61\times10^3}{230\times410}=0.325\ \text{N/mm}^2<\tau_{c,max}=2.8\ \text{N/mm}^2$$

Step 3: Shear strength of concrete. $p_t=\dfrac{100A_{st}}{bd}=\dfrac{100\times603.2}{230\times410}=0.640\%$, so $\tau_c=0.525$ N/mm$^2$ (Table 19, M20). Since $\tau_v<\tau_c$, no stirrups are required by calculation, but minimum shear reinforcement is provided (cl. 26.5.1.6):

$$\frac{A_{sv}}{b\,s_v}\ge\frac{0.4}{0.87f_y}\ \Rightarrow\ s_v\le\frac{0.87\times415\times100.5}{0.4\times230}=395\ \text{mm}$$

Other limits: $s_v\le0.75d=308$ mm and $\le300$ mm.

Step 4: Contribution of the bent-up bar (available as extra capacity): $V_{us}=0.87f_yA_{sb}\sin45^\circ=0.87\times415\times201.1\times0.707=51.3$ kN. Total capacity $=\tau_cbd+V_{us,bent}=49.5+51.3$ kN $>V_u$.

Answer: Provide 8 mm 2-legged vertical stirrups at 300 mm c/c throughout (minimum shear reinforcement governs); the bent-up bar adds an extra 51.3 kN.

Given: $b=250$ mm, $d=500$ mm, $V_u=110$ kN, $T_u=20$ kN·m, 3-20 mm bars ($A_{st}=942.5$ mm$^2$), M20, Fe500. Assume overall depth $D=550$ mm and effective cover 50 mm, so $b_1=250-100=150$ mm, $d_1=550-100=450$ mm (centre to centre of corner bars).

Step 1: Equivalent shear (IS 456 cl. 41.3.1)

$$V_e=V_u+1.6\frac{T_u}{b}=110+1.6\times\frac{20}{0.25}=238\ \text{kN}$$ $$\tau_{ve}=\frac{V_e}{bd}=\frac{238\times10^3}{250\times500}=1.904\ \text{N/mm}^2<\tau_{c,max}=2.8\ \text{N/mm}^2\ \text{(OK)}$$

Step 2: Concrete strength. $p_t=\dfrac{100\times942.5}{250\times500}=0.754\%$, $\tau_c=0.561$ N/mm$^2$ (Table 19). $\tau_{ve}>\tau_c$, so transverse reinforcement is needed.

Step 3: Stirrup area (cl. 41.4.3)

$$\frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}=\frac{20\times10^6}{150\times450\times435}+\frac{110\times10^3}{2.5\times450\times435}=0.681+0.225=0.906\ \text{mm}^2/\text{mm}$$

Check with $\dfrac{(\tau_{ve}-\tau_c)b}{0.87f_y}=0.772$ mm$^2$/mm (smaller, so the above governs).

For 8 mm 2-legged closed stirrups ($A_{sv}=100.5$ mm$^2$): $s_v=\dfrac{100.5}{0.906}=111$ mm.

Step 4: Spacing limits: $s_v\le x_1=175$ mm, $\le\dfrac{x_1+y_1}{4}=162$ mm, $\le300$ mm.

Answer: Provide 8 mm dia closed 2-legged stirrups at 110 mm c/c.

Given: $b=425$ mm, $D=550$ mm, $M_u=100$ kN·m, $T_u=28$ kN·m, $V_u=100$ kN, M20, Fe415, mild exposure. Assume effective cover 50 mm: $d=500$ mm; $b_1=425-100=325$ mm, $d_1=550-100=450$ mm.

Step 1: Equivalent forces (IS 456 cl. 41.3)

$$\begin{aligned} V_e&=V_u+1.6\frac{T_u}{b}=100+1.6\times\frac{28}{0.425}=205.4\ \text{kN}\\ M_t&=T_u\frac{1+D/b}{1.7}=28\times\frac{1+550/425}{1.7}=37.8\ \text{kN·m}\\ M_{e1}&=M_u+M_t=100+37.8=137.8\ \text{kN·m} \end{aligned}$$

$M_t<M_u$, so $M_{e2}=M_t-M_u$ is negative and no compression-face steel is required.

Step 2: Section check. $\tau_{ve}=\dfrac{205.4\times10^3}{425\times500}=0.967$ N/mm$^2$ $<\tau_{c,max}=2.8$ N/mm$^2$, OK. $M_{u,lim}=0.138f_{ck}bd^2=293.17$ kN·m $>M_{e1}$, singly reinforced.

Step 3: Longitudinal steel

$$M_{e1}=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)\ \Rightarrow\ A_{st}=831\ \text{mm}^2$$

Provide 3-20 mm bars ($942$ mm$^2$, $p_t=0.44\%$). Minimum steel $0.85bd/f_y=435$ mm$^2$ OK. Provide 2-12 mm bars at the top as hangers.

Step 4: Stirrups. $\tau_c=0.45$ N/mm$^2$; since $\tau_{ve}>\tau_c$:

$$\frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}=0.530+0.246=0.776\ \text{mm}^2/\text{mm}$$

For 8 mm 2-legged closed stirrups ($A_{sv}=100.5$ mm$^2$): $s_v=129$ mm. Limits: $x_1=350$ mm, $(x_1+y_1)/4=206$ mm, 300 mm.

Answer: Longitudinal steel $A_{st}=831$ mm$^2$ (3-20 mm bars); closed 8 mm 2-legged stirrups at 125 mm c/c.

Given: $b=200$ mm, $D=400$ mm, $M_u=100$ kN·m, $V_u=80$ kN, $T_u=15$ kN·m, M25, Fe415. Assume effective cover 50 mm: $d=350$ mm. Centre-to-centre of corner bars: $b_1=200-100=100$ mm, $d_1=400-100=300$ mm.

Step 1: Equivalent shear and moment

$$\begin{aligned} V_e&=V_u+1.6\frac{T_u}{b}=80+1.6\times\frac{15}{0.2}=200.0\ \text{kN},\qquad \tau_{ve}=\frac{200.0\times10^3}{200\times350}=2.857\ \text{N/mm}^2\\ M_t&=T_u\frac{1+D/b}{1.7}=15\times\frac{1+400/200}{1.7}=26.47\ \text{kN·m}\\ M_{e1}&=M_u+M_t=100+26.47=126.47\ \text{kN·m} \end{aligned}$$

$\tau_{ve}<\tau_{c,max}=3.1$ N/mm$^2$ (M25), so the section is acceptable. $M_t<M_u$, so no $M_{e2}$.

Step 2: Longitudinal steel. $M_{u,lim}=0.138f_{ck}bd^2=0.138\times25\times200\times350^2=84.50$ kN·m $<M_{e1}$, so the section is doubly reinforced for $M_{e1}$.

Doubly reinforced design

$$M_{u2}=M_u-M_{u,lim}=126.47-84.50=41.97\ \text{kN·m}$$

Limiting depth $x_{u,max}=0.48d=168.0$ mm. Strain in compression steel (similar triangles):

$$\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{50}{168.0}\right)=0.00246$$

From the design stress-strain curve of Fe415 (IS 456 Fig. 23): $f_{sc}=344.0$ N/mm$^2$. Stress in concrete displaced by the steel: $f_{cc}=0.446f_{ck}=11.15$ N/mm$^2$.

$$A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{41.97\times10^6}{(344.0-11.15)(350-50)}=420\ \text{mm}^2$$ $$A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=838\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=387\ \text{mm}^2$$ $$A_{st}=A_{st1}+A_{st2}=1225\ \text{mm}^2$$

Provide tension 4-20 mm ($1257$ mm$^2$) and compression 3-16 mm ($603$ mm$^2$) bars.

Step 3: Transverse reinforcement. $p_t=\dfrac{100\times1257}{200\times350}=1.80\%$, $\tau_c=0.79$ N/mm$^2$; $\tau_{ve}>\tau_c$.

$$\frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}=1.385+0.295=1.680\ \text{mm}^2/\text{mm}$$

For 10 mm 2-legged closed stirrups ($A_{sv}=157.1$ mm$^2$): $s_v=93$ mm. Limits: $x_1=125$ mm, $(x_1+y_1)/4=112$ mm, 300 mm.

Answer: Tension steel $A_{st}=1225$ mm$^2$ (4-20 mm), compression steel $A_{sc}=420$ mm$^2$ (3-16 mm), closed 10 mm 2L stirrups at 90 mm c/c.

Given: $b=300$ mm, $d=400$ mm, 4-20 mm bars ($A_{st}=1256.6$ mm$^2$), M25, Fe415. Loads at service state: $V=130$ kN, $T=45$ kN·m. Assume $D=450$ mm and effective cover 50 mm.

Step 1: Factored loads (service $\times$ 1.5)

$$V_u=1.5\times130=195.0\ \text{kN},\qquad T_u=1.5\times45=67.5\ \text{kN·m}$$

Step 2: Equivalent shear and check of section (IS 456 cl. 41.3)

$$V_e=V_u+1.6\frac{T_u}{b}=195.0+1.6\times\frac{67.5}{0.3}=555.0\ \text{kN},\qquad \tau_{ve}=\frac{555.0\times10^3}{300\times400}=4.62\ \text{N/mm}^2$$

$\tau_{ve}>\tau_{c,max}=3.1$ N/mm$^2$ (M25). The 300 mm wide section is therefore not adequate; the size must be increased (IS 456 cl. 41.3.1). Increase the width to $b=400$ mm:

$$V_e=195.0+1.6\times\frac{67.5}{0.4}=465.0\ \text{kN},\qquad \tau_{ve}=\frac{465.0\times10^3}{400\times400}=2.91\ \text{N/mm}^2<3.1\ \text{(OK)}$$

Step 3: Stirrups for the revised section. $b_1=400-100=300$ mm, $d_1=450-100=350$ mm. $p_t=\dfrac{100\times1256.6}{400\times400}=0.79\%$, $\tau_c=0.58$ N/mm$^2$ (Table 19, M25); $\tau_{ve}>\tau_c$.

$$\frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}=1.781+0.617=2.398\ \text{mm}^2/\text{mm}$$

The code minimum $\dfrac{(\tau_{ve}-\tau_c)\,b}{0.87f_y}=2.577$ mm$^2$/mm is larger and governs, so $A_{sv}/s_v=2.577$ mm$^2$/mm.

For 12 mm 2-legged closed stirrups ($A_{sv}=226.2$ mm$^2$): $s_v=88$ mm. Limits: $x_1=325$ mm, $(x_1+y_1)/4=175$ mm, 300 mm. Provide 12 mm 2-legged closed stirrups at 85 mm c/c.

If the given loads are taken as already factored (so $V_u=130$ kN, $T_u=45$ kN·m on the original 300 mm section): $V_e=370.0$ kN, $\tau_{ve}=3.083$ N/mm$^2$ (just below 3.1, OK), $p_t=1.05\%$, $\tau_c=0.65$ N/mm$^2$, $b_1=200$ mm, $d_1=350$ mm:

$$\frac{A_{sv}}{s_v}=1.781+0.411=2.192\ \text{mm}^2/\text{mm}\ \Rightarrow\ s_v=103\ \text{mm for 12 mm 2L}$$

giving 12 mm 2-legged stirrups at 100 mm c/c.

Answer: With the loads treated as service loads, the 300 mm width fails the $\tau_{c,max}$ check; with $b=400$ mm provide 12 mm 2L closed stirrups at 85 mm c/c (longitudinal steel to be provided for $M_{e1}$ in addition).

Assumptions: the imposed load 35 kN/m and torsion 18 kN·m are service values, factored by 1.5. Section $b=300$ mm, $D=600$ mm, effective cover 50 mm, $d=550$ mm. M20, Fe415.

Step 1: Effective span (IS 456 cl. 22.2): least of $c/c=4.5+0.2=4.7$ m and clear $+d=4.5+0.55=5.05$ m, so $L_{eff}=4.70$ m.

Step 2: Loads

$$\begin{aligned} \text{Self weight}&=0.30\times0.60\times25=4.500\ \text{kN/m}\\ w_u&=1.5(35+4.500)=59.25\ \text{kN/m},\qquad T_u=1.5\times18=27.0\ \text{kN·m}\\ M_u&=\frac{w_uL^2}{8}=\frac{59.25\times4.70^2}{8}=163.60\ \text{kN·m}\\ V_u&=\frac{w_u\times4.5}{2}=133.31\ \text{kN (face)},\quad V_{u,d}=133.31-59.25\times0.55=100.72\ \text{kN} \end{aligned}$$

Step 3: Equivalent forces (cl. 41.3)

$$\begin{aligned} M_t&=T_u\frac{1+D/b}{1.7}=27.0\times\frac{1+600/300}{1.7}=47.65\ \text{kN·m}\\ M_{e1}&=M_u+M_t=163.60+47.65=211.25\ \text{kN·m}\\ V_e&=V_{u,d}+1.6\frac{T_u}{b}=100.72+1.6\times\frac{27.0}{0.3}=244.7\ \text{kN} \end{aligned}$$

$\tau_{ve}=\dfrac{244.7\times10^3}{300\times550}=1.483$ N/mm$^2<2.8$, OK. $M_t<M_u$, so no $M_{e2}$.

Step 4: Flexure. $M_{u,lim}=0.138\times20\times300\times550^2=250.40$ kN·m $>M_{e1}$, singly reinforced.

$$M_{e1}=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)\ \Rightarrow\ A_{st}=1265\ \text{mm}^2$$

Provide 5-20 mm bars ($1571$ mm$^2$), $p_t=0.95\%$. Minimum steel $0.85bd/f_y=338$ mm$^2$ OK.

Step 5: Stirrups. $\tau_c=0.61$ N/mm$^2$; $\tau_{ve}>\tau_c$. With $b_1=200$, $d_1=500$ mm:

$$\frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_{u}}{2.5d_1(0.87f_y)}=0.748+0.223=0.971\ \text{mm}^2/\text{mm}$$

The code minimum $\dfrac{(\tau_{ve}-\tau_c)b}{0.87f_y}=0.727$ mm$^2$/mm. For 8 mm 2-legged closed stirrups ($A_{sv}=100.5$ mm$^2$): $s_v=104$ mm; limits $x_1=225$, $(x_1+y_1)/4=188$, 300 mm.

Answer: $M_u=163.60$ kN·m; $A_{st}=1265$ mm$^2$ (5-20 mm) for $M_{e1}=211.25$ kN·m; closed 8 mm 2L stirrups at 100 mm c/c.