To determine the maximum uniform load $w$ (in kN/m) that can be applied to the timber beam, we need to calculate the load corresponding to the allowable bending stress. The procedure involves:

1. Finding the Moment of Inertia for the Circular Cross-Section
   For a circular cross-section of diameter $d = 360 \, \text{mm}$, the moment of inertia $I$ is given by the formula: $I = \frac{\pi d^4}{64}$ where $d$ is the diameter of the circle.

2. Calculating the Section Modulus for a Circular Cross-Section
   The section modulus $S$ is related to the moment of inertia by: $S = \frac{I}{c}$ where $c$ is the distance from the neutral axis to the outermost fiber, which for a circle is half the diameter: $c = \frac{d}{2}$

3. Allowable Bending Stress and the Maximum Moment
   The maximum bending moment $M_{\text{max}}$ that the beam can withstand is determined by the allowable bending stress $F_b$ and the section modulus $S$: $M_{\text{max}} = F_b \times S$

4. Relating the Maximum Moment to the Uniform Load
   For a simply supported beam with a uniform load $w$ and span $L$, the maximum bending moment occurs at the center of the span and is given by: $M_{\text{max}} = \frac{w L^2}{8}$

5. Setting the Maximum Moment Equal to the Allowable Moment
   Equating the moment from the uniform load to the maximum allowable bending moment gives: $\frac{w L^2}{8} = M_{\text{max}}$ Solving for $w$, we get: $w = \frac{8 M_{\text{max}}}{L^2}$

Let's calculate each part step by step:

Step 1: Moment of Inertia for the Circular Cross-Section

Given $d = 360 \, \text{mm} = 0.36 \, \text{m}$: $I = \frac{\pi (0.36)^4}{64} = \frac{\pi \times 0.01679616}{64} = 8.283 \times 10^{-5} \, \text{m}^4$

Step 2: Section Modulus for the Circular Cross-Section

$S = \frac{I}{c} = \frac{8.283 \times 10^{-5}}{0.18} = 4.605 \times 10^{-4} \, \text{m}^3$

Step 3: Maximum Bending Moment

The allowable bending stress $F_b = 19.35 \, \text{MPa} = 19.35 \times 10^6 \, \text{Pa}$. Thus: $M_{\text{max}} = F_b \times S = 19.35 \times 10^6 \times 4.605 \times 10^{-4} = 8.91 \times 10^3 \, \text{Nm} = 8.91 \, \text{kNm}$

Step 4: Maximum Uniform Load

The span of the beam is $L = 4.7 \, \text{m}$. Substituting in the formula for $w$: $w = \frac{8 \times 8.91}{4.7^2} = \frac{71.28}{22.09} = 3.23 \, \text{kN/m}$

Thus, the maximum uniform load that can be applied to the beam without exceeding the allowable bending stress is 3.23 kN/m.

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Would you like further clarification on any of the steps, or would you like to explore additional related questions?

Here are some follow-up questions for further exploration:

1. How does the allowable deflection limit influence the calculation of the maximum load?
2. How would the calculation change if the timber had a different diameter, say 400 mm?
3. What happens if the beam is subjected to a point load instead of a uniform load?
4. How can we use the Modulus of Elasticity (E) to calculate the deflection and check compliance with the allowable deflection limit?
5. How can we compute the shear force in the beam at various points to ensure the allowable shearing stress is not exceeded?

Tip: Always double-check units and ensure that all the terms (like moment, stress, deflection, etc.) are in consistent units throughout the calculation.