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Assume that the section shown in question 2 is of a simply suppported beam spanning 10 metres.  If the beam was subjected to a moment of 100 kNm about its x axis, what normal compressive stress would be induced at a distance of 50mm above the neutral axis?  

*Express your answer to the nearest 0.1 MPa (absolute value), but do not show the units (eg: an answer of 19.38 MPa should be shown as 19.4)*

To solve the problem, we need to calculate the normal compressive stress induced at a distance of 50 mm above the neutral axis of the section due to a bending moment of 100 kNm about the x-axis.

### Step 1: Determine the Second Moment of Area (Iₓ)

From the image provided, we can see the section is a composite of rectangular areas. The second moment of area about the x-axis (\(I_x\)) can be calculated using the parallel axis theorem:

\[
I_x = I_1 + I_2 + I_3
\]

Where:
- \(I_1\), \(I_2\), \(I_3\) are the second moments of area of each rectangular segment about the x-axis.

### Step 2: Calculate Individual Moments of Inertia
For each rectangular segment, \(I = \frac{b \cdot h^3}{12} + A \cdot d^2\), where:
- \(b\) is the width.
- \(h\) is the height.
- \(A = b \cdot h\) is the area.
- \(d\) is the distance from the centroid of the rectangle to the neutral axis.

Let's break down the calculation.

**1. Top Flange (horizontal segment at the top):**
- \(b_1 = 100 \, \text{mm}\)
- \(h_1 = 10 \, \text{mm}\)
- Distance from the centroid of this section to the neutral axis (\(d_1\)) = \(380 \, \text{mm} - \frac{h_1}{2} = 380 - 5 = 375 \, \text{mm}\)

\[
I_{1x} = \frac{100 \times 10^3}{12} + 100 \times 10 \times (375)^2
\]

**2. Vertical Web:**
- \(b_2 = 10 \, \text{mm}\)
- \(h_2 = 380 \, \text{mm}\)
- Distance from the centroid of this section to the neutral axis (\(d_2\)) = \(0\) (since it is at the neutral axis).

\[
I_{2x} = \frac{10 \times 380^3}{12}
\]

**3. Bottom Flange (horizontal segment at the bottom):**
- \(b_3 = 120 \, \text{mm}\)
- \(h_3 = 20 \, \text{mm}\)
- Distance from the centroid of this section to the neutral axis (\(d_3\)) = \(\frac{h_2}{2} + \frac{h_3}{2} = 190 + 10 = 200 \, \text{mm}\)

\[
I_{3x} = \frac{120 \times 20^3}{12} + 120 \times 20 \times (200)^2
\]

### Step 3: Sum the Moments of Inertia
\[
I_x = I_{1x} + I_{2x} + I_{3x}
\]

### Step 4: Calculate the Bending Stress
Once we have \(I_x\), we can calculate the stress using:

\[
\sigma = \frac{M \cdot y}{I_x}
\]

Where:
- \(M = 100 \times 10^3 \, \text{N·m}\)
- \(y = 50 \, \text{mm} = 0.05 \, \text{m}\)

### Let's calculate

I'll compute these step-by-step.The normal compressive stress induced at a distance of 50 mm above the neutral axis due to a moment of 100 kNm about the x-axis is **17.7 MPa**.

---

**Relative Questions:**
1. How would the stress change if the moment was applied about the y-axis instead of the x-axis?
2. What is the significance of the second moment of area in structural analysis?
3. How does the distance from the neutral axis affect the induced stress in a beam?
4. What would be the effect on stress if the beam span was reduced to 5 meters?
5. How does the shape and size of the cross-section influence the beam's bending strength?

**Tip:** Always ensure to calculate the correct centroid and moment of inertia when dealing with composite sections to obtain accurate stress results.

Learn how to calculate the normal compressive stress induced at a distance of 50mm above the neutral axis in a simply supported beam subjected to a moment of 100 kNm. Detailed step-by-step solution involving mechanics of materials and bending stress calculations.

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