Each of the four vertical links has an 8 X 36-mm uniform rectangular cross section, and each of the four pins has a 16-mm diameter. Determine the maximum value of the average normal stress in the links connecting (a) points B and D, (b) points C and E.
The Correct Answer and Explanation is :
To determine the maximum value of the average normal stress in the links connecting points B and D and C and E, we use the formula:
[
\sigma = \frac{P}{A}
]
where:
- ( \sigma ) = Average normal stress (MPa),
- ( P ) = Load carried by the link (N),
- ( A ) = Cross-sectional area of the link ((\text{mm}^2)).
Given Data:
- Rectangular link dimensions: ( 8 \times 36 ) mm
- Pin diameter: ( 16 ) mm
Step 1: Calculate the Cross-Sectional Area of Each Link
[
A = \text{width} \times \text{thickness} = 36 \times 8 = 288 \text{ mm}^2
]
Step 2: Load Distribution
Each link carries a portion of the total force ( P ). Assuming symmetric loading, each pair of links at B-D and C-E carries half of the applied load.
Thus, force per link:
[
P_{\text{link}} = \frac{P}{2}
]
Step 3: Compute Normal Stress
[
\sigma = \frac{P_{\text{link}}}{A} = \frac{P/2}{288}
]
To obtain the numerical value, we need the applied load ( P ). However, the formula provides a straightforward approach to solving for ( \sigma ) given any applied force.
Conclusion
- The normal stress depends on the applied force ( P ).
- Each link carries half of ( P ).
- Formula to compute normal stress in links BD and CE: [
\sigma = \frac{P}{576}
]
where ( P ) is in Newtons and ( \sigma ) will be in MPa.
Generated Image
I will now generate an image illustrating the structural system with forces, links, and stress distribution.
Here is the generated image illustrating the mechanical structure with vertical links, pins, force distribution, and stress annotations. Let me know if you need any modifications or additional details!
