Limit state design of beam, 45 kN/m

Given: $w=45$ kN/m (design load, i.e. factored, including self weight), clear span 6 m, M20, Fe415. The 45 kN/m is taken as the design (factored) load.

Assumptions: support width 300 mm; $b=300$ mm, $D=600$ mm, effective cover 50 mm, $d=550$ mm.

Step 1: Effective span: least of $c/c=6+0.3=6.3$ m and clear $+d=6+0.55=6.55$ m, so $L_{eff}=6.30$ m.

Step 2: Flexure

$$M_u=\frac{w_uL^2}{8}=\frac{45\times6.30^2}{8}=223.26\ \text{kN·m}$$ $$M_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times300\times550^2=250.40\ \text{kN·m}>M_u$$

so the section is singly reinforced.

$$M_u=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)\ \Rightarrow\ A_{st}=1355\ \text{mm}^2$$

Provide 5-20 mm dia bars ($1571$ mm$^2$); minimum steel $0.85bd/f_y=338$ mm$^2$ OK; $x_u=262.6$ mm $<0.48d=264$ mm OK.

Step 3: Shear

Support shear (at face) $V_u=\dfrac{w_u\times6}{2}=135.0$ kN. At the critical section $d$ from the face:

$$V_{u,d}=135.0-45\times0.55=110.2\ \text{kN},\qquad \tau_v=\frac{110.2\times10^3}{300\times550}=0.67\ \text{N/mm}^2<2.8$$

$p_t=\dfrac{100\times1571}{300\times550}=0.95\%$ gives $\tau_c=0.61$ N/mm$^2$ (IS 456 Table 19). As $\tau_v>0.61$:

$$V_{us}=V_{u,d}-\tau_cbd=110.2-100.4=9.9\ \text{kN}$$

Using 8 mm 2-legged stirrups ($A_{sv}=100.5$ mm$^2$):

$$s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times415\times100.5\times550}{9.9\times10^3}=2027\ \text{mm}$$

Limits: $\le0.75d=412$ mm, $\le300$ mm, $\le\dfrac{0.87f_yA_{sv}}{0.4b}=302$ mm. The spacing required for $V_{us}$ is much larger than these limits, so the maximum spacing (minimum shear reinforcement) governs. Provide 8 mm 2-legged stirrups at 300 mm c/c throughout the span.

Answer: Beam 300 x 600 mm, $A_{st}=1355$ mm$^2$ (5-20 mm), stirrups 8 mm 2L @ 300 mm c/c.