The correct answer is: B. Is safe with stirrups
The shear force at the critical section is $V = 12 \times 0.6 = 7.2$ kN. The shear stress is therefore $\tau = \frac{V}{bd} = \frac{7.2}{0.3 \times 0.55} = 3.4$ kg/cm$^2$. This is greater than the allowable shear stress of 5 kg/cm$^2$, so the beam is not safe in shear without stirrups.
Stirrups are steel bars that are placed perpendicular to the main reinforcement in a beam. They help to resist shear forces by providing a mechanism for the concrete to transfer the shear stress to the main reinforcement.
If stirrups are added, the shear stress in the beam will be reduced. The new shear stress can be calculated using the following equation:
$$\tau = \frac{V}{bd} + \frac{As_v f_y}{s}$$
where:
- $V$ is the shear force at the critical section
- $b$ is the width of the beam
- $d$ is the effective depth of the beam
- $As_v$ is the area of the stirrups
- $f_y$ is the yield stress of the stirrups
- $s$ is the spacing of the stirrups
In this case, the new shear stress is:
$$\tau = \frac{7.2}{0.3 \times 0.55} + \frac{0.005 \times 400}{0.1} = 2.75$$
This is less than the allowable shear stress of 5 kg/cm$^2$, so the beam is safe in shear with stirrups.