In case of a simply supported I-section beam of span L and loaded with a central load W, the length of elasto-plastic zone of the plastic hinge, is A. $$\frac{{\text{L}}}{2}$$ B. $$\frac{{\text{L}}}{3}$$ C. $$\frac{{\text{L}}}{4}$$ D. $$\frac{{\text{L}}}{5}$$

$$rac{{ ext{L}}}{2}$$
$$rac{{ ext{L}}}{3}$$
$$rac{{ ext{L}}}{4}$$
$$rac{{ ext{L}}}{5}$$

The correct answer is $\boxed{\frac{{\text{L}}}{2}}$.

A plastic hinge is a region in a beam where the material has yielded and is undergoing plastic deformation. The length of the plastic hinge is determined by the following equation:

$$L_p = \frac{M_p}{\sigma_y}$$

where $M_p$ is the plastic moment of the beam, $\sigma_y$ is the yield stress of the material, and $L_p$ is the length of the plastic hinge.

For a simply supported I-section beam of span $L$ and loaded with a central load $W$, the plastic moment is given by the following equation:

$$M_p = \frac{WL^2}{8}$$

The yield stress of steel is typically 250 MPa. Therefore, the length of the plastic hinge is given by the following equation:

$$L_p = \frac{\frac{WL^2}{8}}{{250 \text{ MPa}}} = \frac{{\text{L}}}{2}$$

The other options are incorrect because they do not represent the correct length of the plastic hinge.

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