If nominal shear stress $${\tau _{\text{v}}}$$ exceeds the design shear strength of concrete $${\tau _{\text{c}}}$$, the nominal shear reinforcement as per IS : 456-1978 shall be provided for carrying a shear stress equal to A. $${\tau _{\text{v}}}$$ B. $${\tau _{\text{c}}}$$ C. $${\tau _{\text{v}}} – {\tau _{\text{c}}}$$ D. $${\tau _{\text{v}}} + {\tau _{\text{c}}}$$

The correct answer is A. $\tau_{\text{v}}$.

The nominal shear stress $\tau_{\text{v}}$ is the shear stress calculated from the applied loads on the member. The design shear strength of concrete $\tau_{\text{c}}$ is the maximum shear stress that concrete can carry without failure. If $\tau_{\text{v}}$ exceeds $\tau_{\text{c}}$, then the member will fail in shear. To prevent this, nominal shear reinforcement is provided to carry the excess shear stress. The nominal shear reinforcement is designed to carry a shear stress equal to $\tau_{\text{v}}$.

Option B is incorrect because $\tau_{\text{c}}$ is the maximum shear stress that concrete can carry without failure. If $\tau_{\text{v}}$ exceeds $\tau_{\text{c}}$, then the member will fail in shear. Therefore, the nominal shear reinforcement cannot be designed to carry a shear stress equal to $\tau_{\text{c}}$.

Option C is incorrect because $\tau_{\text{v}} – \tau_{\text{c}}$ is the shear stress that is not carried by concrete. This shear stress must be carried by the nominal shear reinforcement. Therefore, the nominal shear reinforcement cannot be designed to carry a shear stress equal to $\tau_{\text{v}} – \tau_{\text{c}}$.

Option D is incorrect because $\tau_{\text{v}} + \tau_{\text{c}}$ is the total shear stress on the member. This shear stress is carried by both the concrete and the nominal shear reinforcement. Therefore, the nominal shear reinforcement cannot be designed to carry a shear stress equal to $\tau_{\text{v}} + \tau_{\text{c}}$.