Determine the thickness of a 140 mm wide uniform plate for safe continuous operation
if the plate is to be subjected to a tensile load that has a maximum value of 150 KN and
a minimum value of 75 KN. The properties of the plate material are as follows:
Endurance limit stress = 225 MPa, and Yield point stress = 200 MPa. The factor of safety
based on yield point may be taken as 4.

To determine the required thickness of the uniform plate for safe continuous operation under the given conditions, we need to ensure that the stresses induced in the plate do not exceed the material's yield point stress when considering the given factor of safety. Additionally, the alternating stress should not exceed the endurance limit of the material.

### Given Data:
- Width of the plate, \( b = 140 \) mm
- Maximum tensile load, \( P_{\text{max}} = 150 \) kN = 150,000 N
- Minimum tensile load, \( P_{\text{min}} = 75 \) kN = 75,000 N
- Endurance limit stress, \( \sigma_e = 225 \) MPa
- Yield point stress, \( \sigma_y = 200 \) MPa
- Factor of safety, \( N_f = 4 \)

### Step 1: Calculate the Mean and Alternating Stresses
The mean load \( P_{\text{mean}} \) and the alternating load \( P_a \) can be calculated as:
\[
P_{\text{mean}} = \frac{P_{\text{max}} + P_{\text{min}}}{2}
\]
\[
P_a = \frac{P_{\text{max}} - P_{\text{min}}}{2}
\]
Substituting the values:
\[
P_{\text{mean}} = \frac{150,000 + 75,000}{2} = 112,500 \text{ N}
\]
\[
P_a = \frac{150,000 - 75,000}{2} = 37,500 \text{ N}
\]

### Step 2: Calculate the Mean and Alternating Stresses in the Plate
Let \( t \) be the thickness of the plate. The cross-sectional area \( A \) of the plate is given by:
\[
A = b \times t = 140 \times t \text{ mm}^2
\]

The mean stress \( \sigma_{\text{mean}} \) and the alternating stress \( \sigma_a \) are:
\[
\sigma_{\text{mean}} = \frac{P_{\text{mean}}}{A} = \frac{P_{\text{mean}}}{140 \times t}
\]
\[
\sigma_a = \frac{P_a}{A} = \frac{P_a}{140 \times t}
\]

Substituting the values:
\[
\sigma_{\text{mean}} = \frac{112,500}{140 \times t} \text{ MPa}
\]
\[
\sigma_a = \frac{37,500}{140 \times t} \text{ MPa}
\]

### Step 3: Ensure Safety Against Yielding
For safety against yielding, the maximum stress must be less than or equal to the yield stress divided by the factor of safety:
\[
\sigma_{\text{mean}} + \sigma_a \leq \frac{\sigma_y}{N_f}
\]

Substituting the known values:
\[
\frac{112,500}{140 \times t} + \frac{37,500}{140 \times t} \leq \frac{200}{4} = 50 \text{ MPa}
\]

Simplifying:
\[
\frac{150,000}{140 \times t} \leq 50 \text{ MPa}
\]
\[
t \geq \frac{150,000}{140 \times 50} = \frac{150,000}{7,000} = 21.43 \text{ mm}
\]

### Step 4: Ensure Safety Against Fatigue (Endurance Limit)
For safety against fatigue, the alternating stress must be less than or equal to the endurance limit:
\[
\sigma_a \leq \sigma_e
\]

Substituting the values:
\[
\frac{37,500}{140 \times t} \leq 225 \text{ MPa}
\]

Simplifying:
\[
t \geq \frac{37,500}{140 \times 225} = \frac{37,500}{31,500} = 1.19 \text{ mm}
\]

### Conclusion
The required thickness to satisfy both yielding and fatigue criteria is the larger of the two values calculated:
\[
t \geq 21.43 \text{ mm}
\]

Thus, the thickness of the plate should be at least **21.43 mm** to ensure safe continuous operation.

---

Would you like more details on any part of this solution, or do you have any questions?

### Related Questions:
1. How does the factor of safety influence the required thickness?
2. What would happen if the yield point stress was increased?
3. How would the solution change if the load fluctuated between 0 and 150 kN?
4. Can you explain the significance of the endurance limit in fatigue analysis?
5. How would the required thickness change if the plate width was different?

### Tip:
Always consider both static and fatigue failure criteria when designing components subjected to fluctuating loads.

Learn how to calculate the required thickness of a uniform plate to ensure safe continuous operation under varying tensile loads. This problem involves using the mean stress formula, alternating stress formula, and considering the factor of safety based on yield point. Suitable for mechanical engineering students and professionals.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation