Question

Calculate the apparent circumferential stress, when Pmax=3.5MPa, D=140mm, and t=8mm?

(a) 12.3N/mm^2

(b) 14.4N/mm^2

(c) 15.3N/mm^2

(d) 16.1N/mm^2

The question was asked in semester exam.

Question is taken from IC Engine topic in section Design of IC Engine Components of Automotive Engine Design

Answer 1

To calculate the apparent circumferential stress, we can use the formula for the hoop stress or circumferential stress in a thin-walled cylinder:

σcirc=Pmax⋅D2⋅t\sigma_{\text{circ}} = \frac{P_{\text{max}} \cdot D}{2 \cdot t}σcirc​=2⋅tPmax​⋅D​

Where:

• σcirc\sigma_{\text{circ}}σcirc​ is the circumferential stress
• PmaxP_{\text{max}}Pmax​ is the maximum pressure (3.5 MPa)
• $D$ is the diameter (140 mm)
• $t$ is the wall thickness (8 mm)

Let's plug in the values:

σcirc=3.5×103 N/m2⋅140 mm2⋅8 mm\sigma_{\text{circ}} = \frac{3.5 \times 10^3 \, \text{N/m}^2 \cdot 140 \, \text{mm}}{2 \cdot 8 \, \text{mm}}σcirc​=2⋅8mm3.5×103N/m2⋅140mm​

First, convert units for consistency:

• Pmax=3.5 MPa=3.5×103 N/m2P_{\text{max}} = 3.5 \, \text{MPa} = 3.5 \times 10^3 \, \text{N/m}^2Pmax​=3.5MPa=3.5×103N/m2
• Diameter $D=140\text{mm}=0.14\text{m}$
• Wall thickness $t=8\text{mm}=0.008\text{m}$

Now, calculate the circumferential stress:

σcirc=3.5×103⋅0.142⋅0.008\sigma_{\text{circ}} = \frac{3.5 \times 10^3 \cdot 0.14}{2 \cdot 0.008}σcirc​=2⋅0.0083.5×103⋅0.14​ σcirc=4900.016\sigma_{\text{circ}} = \frac{490}{0.016}σcirc​=0.016490​ σcirc=30.625 N/mm2\sigma_{\text{circ}} = 30.625 \, \text{N/mm}^2σcirc​=30.625N/mm2

After calculating the above, we see that the correct formula or calculations might have been misunderstood, since the options listed seem inconsistent with the result. However, based on given options, the closest option is:

Answer: (d) 16.1 N/mm².