Shear design of T-beam on 300x600 section

Assumptions: the 40 kN/m service load includes self weight; web $b_w=300$ mm, $D=600$ mm, effective cover 50 mm, $d=550$ mm. Of the four 20 mm bars, 2 are bent up at 45$^\circ$ near the support and 2 continue straight. M20, Fe415. For a T-beam, shear is calculated on the web width.

Step 1: Design shear

$$w_u=1.5\times40=60.0\ \text{kN/m},\quad V_u=\frac{60.0\times6}{2}=180.0\ \text{kN}$$

At $d$ from the face: $V_{u,d}=180.0-60.0\times0.55=147.0$ kN, so $\tau_v=\dfrac{147.0\times10^3}{300\times550}=0.891$ N/mm$^2<2.8$, OK.

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

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

$V_{us}=147.0-69.8=77.2$ kN is to be carried by shear reinforcement.

Step 3: Bent-up bars (2-20 at 45$^\circ$)

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

As bent bars may carry only half of the shear reinforcement load (cl. 40.4), the usable value is $0.5\times77.2=38.6$ kN.

Step 4: Stirrups for the balance $=77.2-38.6=38.6$ kN. With 8 mm 2-legged stirrups ($A_{sv}=100.5$ mm$^2$):

$$s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times415\times100.5\times550}{38.6\times10^3}=517\ \text{mm}$$

Maximum spacing limits: $\dfrac{0.87f_yA_{sv}}{0.4b}=302$ mm, $0.75d=412$ mm and 300 mm, so the limit of 300 mm governs.

Answer: Provide 8 mm 2-legged vertical stirrups at 300 mm c/c near the support, in addition to the 2 bars bent up at 45$^\circ$.