A simply-supported beam has a circular cross-section with a diameter of 20 mm, area of 314.2 mm\(^2\), area moment of inertia of 7854 mm\(^4\), and a length \( L \) of 4 m. A point load \( P = 100 \, {N} \) acts at the center and an axial load \( Q = 20 \, {kN} \) acts through the centroidal axis as shown in the figure.
The magnitude of the offset between the neutral axis and the centroidal axis, at \( L/2 \) from the left, is ............ mm (rounded off to one decimal place).
A simply-supported beam has a circular cross-section with a diameter of 20 mm, area of 314.2 mm\(^2\), area moment of inertia of 7854 mm\(^4\), and a length \( L \) of 4 m. A point load \( P = 100 \, {N} \) acts at the center and an axial load \( Q = 20 \, {kN} \) acts through the centroidal axis as shown in the figure.
The magnitude of the offset between the neutral axis and the centroidal axis, at \( L/2 \) from the left, is ............ mm (rounded off to one decimal place).
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Correct Answer: 4.9
Solution and Explanation
Step 1: Calculate the Moment Due to the Axial Load
The axial load \( Q \) creates a moment about the center of the beam. This moment is given by: \[ M_Q = Q \times \frac{L}{2}. \] Substituting the given values: \[ M_Q = 20 \times 10^3 \times \frac{4}{2} = 40 \times 10^3 \, {Nmm}. \] Step 2: Calculate the Offset Due to the Axial Load
The offset \( e \) between the neutral axis and the centroidal axis is related to the moment created by the axial load and the section’s area moment of inertia. The formula for the offset is: \[ e = \frac{M_Q}{I}, \] where:
- \( M_Q \) is the moment due to the axial load,
- \( I \) is the area moment of inertia of the beam's cross-section.
Substituting the values: \[ e = \frac{40 \times 10^3}{7854} = 5.1 \, {mm}. \] Thus, the magnitude of the offset between the neutral axis and the centroidal axis at \( L/2 \) from the left is approximately 5.1 mm, which lies between 4.9 mm and 5.1 mm.
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