Question

A beam has the same section throughout its length with moment of inertia \(I = 1 \times 10^8 \, \text{mm}^4\). It is subjected to a uniform BM = 40 kN·m, \(E = 2 \times 10^5 \, \text{N/mm}^2\). What is the radius of curvature of the circle into which the beam will bend in the form of an arc of a circle?

• A. 1000 m
• B. 500 m
• C. 400 m
• D. 350 m

Hint:

The radius of curvature is inversely proportional to the bending moment. A higher bending moment results in a smaller radius of curvature.

Answer 1

The Correct Option is C

Step 1: Using the Formula for Radius of Curvature.
The formula for the radius of curvature \(R\) of a beam is given by: \[ R = \frac{EI}{M} \] Where: - \(E\) = Modulus of elasticity = \(2 \times 10^5 \, \text{N/mm}^2\) - \(I\) = Moment of inertia = \(1 \times 10^8 \, \text{mm}^4\) - \(M\) = Bending moment = 40 kN·m = \(40 \times 10^3 \, \text{N·m}\)
Step 2: Substituting the Values.
Substituting the given values into the formula: \[ R = \frac{(2 \times 10^5) \times (1 \times 10^8)}{40 \times 10^3} = \frac{2 \times 10^{13}}{40 \times 10^3} = 5 \times 10^5 \, \text{mm} = 500 \, \text{m} \]
Step 3: Conclusion.
Therefore, the radius of curvature is 500 m, making option (2) the correct answer.

Final Answer: \[ \boxed{500 \, \text{m}} \]