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To solve the problem of calculating the static factor of safety (FOS) for both materials, we need to use the formula for FOS based on the applied stress and the material's yield or ultimate strength.

Given the diagram and information:

1. **Dimensions of the bar**:
   - Thickness of the bar = 6 mm
   - The applied load = 500 N
   - Geometry dimensions are provided for calculating the stress distribution.

2. **Static Factor of Safety Formula**:
   \[
   \text{FOS} = \frac{\text{Material Strength}}{\text{Applied Stress}}
   \]

We will go step by step:

### Step 1: Calculate the cross-sectional area

The stress will be calculated based on the load and the effective cross-sectional area. The load is applied as a bending moment at the end.

#### Cross-sectional area for stress calculation:
From the diagram, the area at the thinnest section (20 mm height × 6 mm thick):

\[
A = 20 \text{ mm} \times 6 \text{ mm} = 120 \text{ mm}^2 = 120 \times 10^{-6} \text{ m}^2
\]

### Step 2: Calculate the moment of inertia

The bar is subject to bending. The moment of inertia (I) for the rectangular cross-section is given by:

\[
I = \frac{1}{12} \times b \times h^3
\]
Where:
- \( b = 6 \) mm is the thickness,
- \( h = 20 \) mm is the height at the thinnest part.

\[
I = \frac{1}{12} \times 6 \times (20^3) = \frac{1}{12} \times 6 \times 8000 = 4000 \text{ mm}^4 = 4 \times 10^{-9} \text{ m}^4
\]

### Step 3: Calculate the bending moment

The applied force of 500 N acts at a distance of 175 mm (100 mm + 75 mm) from the weld point:

\[
M = F \times d = 500 \text{ N} \times 0.175 \text{ m} = 87.5 \text{ Nm}
\]

### Step 4: Calculate maximum bending stress

The maximum bending stress \( \sigma_{\text{max}} \) occurs at the farthest point from the neutral axis (y = 10 mm):

\[
\sigma_{\text{max}} = \frac{M \times y}{I} = \frac{87.5 \times 0.01}{4 \times 10^{-9}} = 218.75 \times 10^6 \text{ Pa} = 218.75 \text{ MPa}
\]

### Step 5: Material Strengths

For 1018HR steel, the yield strength (\( \sigma_{\text{yield}} \)) is approximately 370 MPa.
For ASTM20 gray cast iron, the ultimate tensile strength (\( \sigma_{\text{ultimate}} \)) is about 200 MPa.

### Step 6: Calculate the factor of safety

- **For 1018HR steel**:
  \[
  \text{FOS} = \frac{\sigma_{\text{yield}}}{\sigma_{\text{max}}} = \frac{370}{218.75} \approx 1.69
  \]

- **For ASTM20 gray cast iron**:
  \[
  \text{FOS} = \frac{\sigma_{\text{ultimate}}}{\sigma_{\text{max}}} = \frac{200}{218.75} \approx 0.91
  \]

### Final Answers:
- The **static factor of safety** for **1018HR steel** is **1.69**.
- The **static factor of safety** for **ASTM20 gray cast iron** is **0.91**.

Would you like further clarification or more details on any step?

Here are 5 related questions you might find useful:
1. How does the bending moment affect the stress distribution in different materials?
2. What is the significance of the factor of safety in design?
3. Why is the yield strength used for ductile materials like steel and ultimate strength for brittle materials like cast iron?
4. How does the moment of inertia influence the bending stress in structural elements?
5. What are the effects of different loading conditions (like shear or torsion) on the factor of safety?

**Tip:** Always use the correct strength value (yield or ultimate) based on whether the material is ductile or brittle when calculating the factor of safety.

What is the static factor of safety if the bar is made of 1018HR steel or ASTM20 gray cast iron based on the given dimensions and load?

Learn how to calculate the static factor of safety for a 6mm thick bar made of 1018HR steel and ASTM20 cast iron. This problem involves concepts like bending stress, moment of inertia, and material strength. Detailed calculations and results are provided for each material, including yield strength and ultimate tensile strength.

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