**Given data:**
- Width (b) = 40 mm
- Depth (h) = 60 mm
- Radius of curvature (R) = 80 mm
- Moment (M) = 500 Nm

**To find:**
- Position of the neutral axis
- Bending stresses at the inner and outer faces

**Step-by-step solution:**

**1. Calculation of moment of inertia (I):**
The moment of inertia is required to determine the position of the neutral axis. For a rectangular cross-section, the moment of inertia can be calculated using the formula:

I = (b * h^3) / 12

Substituting the given values, we get:

I = (40 * 60^3) / 12
I = 7,200,000 mm^4

**2. Calculation of position of the neutral axis (y):**
The position of the neutral axis can be determined using the formula:

y = (I * e) / A

Where:
- y is the distance of the neutral axis from the centroidal axis
- I is the moment of inertia
- e is the distance from the centroidal axis to the extreme fiber (in this case, half the depth, h/2)
- A is the cross-sectional area

Substituting the given values, we get:

y = (7,200,000 * 30) / (40 * 60)
y = 180,000 mm^3 / 2400 mm^2
y = 75 mm

Therefore, the neutral axis is located at a distance of 75 mm from the centroidal axis.

**3. Calculation of bending stresses:**
The bending stress at any point on the cross-section can be calculated using the formula:

σ = (M * y) / I

Where:
- σ is the bending stress
- M is the applied moment
- y is the distance from the neutral axis
- I is the moment of inertia

For the inner face (y = -h/2 = -30 mm):

σ_inner = (500 * -30) / 7,200,000
σ_inner = -2.08 N/mm^2 (compressive)

For the outer face (y = h/2 = 30 mm):

σ_outer = (500 * 30) / 7,200,000
σ_outer = 2.08 N/mm^2 (tensile)

Therefore, the bending stress at the inner face is -2.08 N/mm^2 (compressive), and the bending stress at the outer face is 2.08 N/mm^2 (tensile).

**Summary:**
- The position of the neutral axis is located at a distance of 75 mm from the centroidal axis.
- The bending stress at the inner face is -2.08 N/mm^2 (compressive), and the bending stress at the outer face is 2.08 N/mm^2 (tensile).

Ans????