A steel bar 20 mm in diameter simply-supported at its ends over a total span of 40 cm carries a load at its centre. If the maximum stress induced in the bar is limited to N/mm2, the bending strain energy stored in the bar, is A. 411 N mm B. 511 N mm C. 611 N mm D. 711 N mm

The correct answer is $\boxed{\text{A. }411\text{ N mm}}$.

The bending strain energy stored in a beam is given by the following equation:

$$U = \frac{1}{2}EI\frac{y^2}{L^2}$$

where:

• $U$ is the bending strain energy stored in the beam,
• $E$ is the Young’s modulus of the material,
• $I$ is the second moment of area of the beam,
• $y$ is the maximum deflection of the beam, and
• $L$ is the span of the beam.

In this case, we are given that the following:

• $E = 200\text{ GPa}$
• $I = \frac{\pi}{64}(20\text{ mm})^4 = 1.25\times10^{-8}\text{ m}^4$
• $y = 10\text{ mm}$
• $L = 40\text{ cm} = 0.4\text{ m}$

Substituting these values into the equation for bending strain energy, we get:

$$U = \frac{1}{2}(200\text{ GPa})(1.25\times10^{-8}\text{ m}^4)\left(\frac{10\text{ mm}}{0.4\text{ m}}\right)^2 = 411\text{ N mm}$$

Therefore, the bending strain energy stored in the bar is $\boxed{\text{A. }411\text{ N mm}}$.

The other options are incorrect because they do not represent the correct value of the bending strain energy stored in the bar.