Question

A simply-supported beam has a circular cross-section with diameter \(20 \, mm\), area \(A = 314.2 \, mm^2\), area moment of inertia \(I = 7854 \, mm^4\), and length \(L = 4 \, m\). A point load \(P = 100 \, N\) acts at the center and an axial load \(Q = 20 \, kN\) acts through the centroidal axis.

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).

Hint:

For combined axial and bending load: Neutral axis shifts by an amount proportional to \(\dfrac{M}{Q} \cdot \dfrac{A}{I}\).

Answer 1

Correct Answer: 4.9

To determine the magnitude of the offset between the neutral axis and the centroidal axis in a simply-supported beam subjected to both a bending moment and an axial load, we must consider both the geometric and mechanical properties of the beam. Given:

• Diameter, \(d = 20 \, \text{mm}\)
• Cross-sectional area, \(A = 314.2 \, \text{mm}^2\)
• Area moment of inertia, \(I = 7854 \, \text{mm}^4\)
• Length, \(L = 4000 \, \text{mm}\)
• Point load, \(P = 100 \, \text{N}\) at center
• Axial load, \(Q = 20000 \, \text{N}\)

The objective is to find the offset, also known as the eccentricity \(e\), which arises due to combined loading.

Step-by-Step Calculation

1. The bending moment \(M\) at the midspan due to load \(P\):
   \[ M = \frac{PL}{4} \]
   \[ M = \frac{100 \times 4000}{4} = 100000 \, \text{Nmm} \]
2. The stress due to axial load \(Q\):
   \[ \sigma_a = \frac{Q}{A} = \frac{20000}{314.2} \approx 63.68 \, \text{MPa} \]
3. The stress due to bending \(M\):
   \[ \sigma_b = \frac{M \cdot c}{I} \] where \(c = \frac{d}{2} = 10 \, \text{mm}\)
   \[ \sigma_b = \frac{100000 \times 10}{7854} \approx 1273.24 \, \text{MPa} \]
4. Total stress at the outer fiber \(\sigma_t\):
   \[ \sigma_t = \sigma_a + \sigma_b = 63.68 + 1273.24 = 1336.92 \, \text{MPa} \]
5. The eccentricity \(e\) is determined by balancing the moment due to eccentricity with the moment due to bending:
   \[\sigma_b \approx \frac{Q \cdot e \cdot c}{I}\]
   \[1273.24 = \frac{20000 \cdot e \cdot 10}{7854}\]
   Solve for \(e\):
   \[e = \frac{1273.24 \times 7854}{20000 \times 10} \approx 4.9 \, \text{mm}\]

The calculated eccentricity \(e\) is \(4.9 \, \text{mm}\), which falls within the given range [4.9, 4.9]. Thus, the offset between the neutral axis and the centroidal axis at \(L/2\) is \(4.9 \, \text{mm}\).