Shear design with curtailed bars, 30x45 cm

Given: $b=300$ mm, $d=450$ mm, 5-20 mm mild steel bars (Fe250), with 2 bars curtailed at a section where the service shear is 100 kN, M20. The factored shear is $V_u=1.5\times100=150.0$ kN. Stirrups: 8 mm 2-legged mild steel ($f_y=250$ N/mm$^2$, $A_{sv}=100.5$ mm$^2$).

Step 1: Nominal shear stress

$$\tau_v=\frac{V_u}{bd}=\frac{150.0\times10^3}{300\times450}=1.111\ \text{N/mm}^2<\tau_{c,max}=2.8\ \text{N/mm}^2$$

Step 2: Concrete strength. After curtailment, 3-20 mm bars remain: $A_{st}=942.5$ mm$^2$.

$$p_t=\frac{100\times942.5}{300\times450}=0.698\%\ \Rightarrow\ \tau_c=0.543\ \text{N/mm}^2,\quad V_c=\tau_cbd=73.4\ \text{kN}$$

Step 3: Stirrups for shear

$$V_{us}=V_u-V_c=150.0-73.4=76.6\ \text{kN}$$ $$s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times250\times100.5\times450}{76.6\times10^3}=128\ \text{mm}$$

Spacing limits: $\dfrac{0.87f_yA_{sv}}{0.4b}=182$ mm, $0.75d=337$ mm, 300 mm. Provide $s_v=125$ mm for shear.

Step 4: Extra provisions where bars are curtailed (IS 456 cl. 26.2.3.2). Flexural bars may be stopped in a tension zone only if (a) the shear at the cut-off point is not more than $\tfrac23$ of the shear capacity of the section (including the shear reinforcement provided), or (b) excess stirrup area is provided along each terminated bar over a distance $\tfrac34d$ from the cut-off point, such that

$$A_{sv,excess}\ge\frac{0.4\,b\,s_v}{f_y},\qquad s_v\le\frac{d}{8\beta_b}$$

where $\beta_b=\dfrac{\text{area of bars cut off}}{\text{total area of bars at the section}}=\dfrac{2}{5}=0.4$, so $s_v\le\dfrac{450}{8\times0.4}=141$ mm.

Check of option (a): with stirrups at 125 mm, $V_{us}=\dfrac{0.87\times250\times100.5\times450}{125}=78.7$ kN, so $\tfrac23(73.4+78.7)=101.4$ kN $<V_u=150.0$ kN. Option (a) is not satisfied, so excess stirrups (option b) are needed.

Option (b): the stirrups must carry the shear plus the excess:

$$\frac{A_{sv}}{s_v}=\frac{V_{us}}{0.87f_yd}+\frac{0.4b}{f_y}=\frac{76.6\times10^3}{0.87\times250\times450}+\frac{0.4\times300}{250}=0.783+0.480=1.263\ \text{mm}^2/\text{mm}$$ $$s_v=\frac{100.5}{1.263}=80\ \text{mm}\ (<141\ \text{mm, OK})$$

Provide 8 mm 2-legged stirrups at 75 mm c/c in this zone.

Answer: 8 mm 2-legged mild steel stirrups at 125 mm c/c for shear near the cut-off section, reduced to 75 mm c/c over a length $\tfrac34d\approx340$ mm from the cut-off point along the curtailed bars.