Find the circumferential (hoop) stress in a cylindrical bar under axial load, we use the formula for hoop stress:
[tex]\sigma_h = \frac{F}{A}[/tex]
where:
- (sigma_h) is the circumferential (hoop) stress.
- (F) is the force applied.
- (A) is the cross-sectional area perpendicular to the force.
Given:
[tex] \text{ - Diameter of the bar (\(D\)): 10 inches}[/tex]
[tex] \text{ - Wall thickness of the bar (\(t\)): 1 inch}[/tex]
[tex] \text{ - Applied force (\(F\)): 100,000 lbs (each side, so the total force is \(100,000 \, \text{lbs} + 100,000 \, \text{lbs} = 200,000 \, \text{lbs}}[/tex]
Need to calculate the inner and outer radii of the bar:
[tex]{- \: \text{Inner radius} (\(r_i\)): \(\frac{D}{2} - t = \frac{10}{2} - 1 = 4 \, \text{in}}[/tex]
[tex]{- \: \text{Outer radius} (\(r_o\)): \(\frac{D}{2} = \frac{10}{2} = 5 \, \text{in}}[/tex]
Next, calculate the cross-sectional area (A) of the bar:
[tex]A = \pi (r_o^2 - r_i^2)[/tex]
[tex]{A = \pi (5^2 - 4^2) = \pi (25 - 16) = \pi \cdot 9 = 28.27433 \, \text{in}^2}[/tex]
Calculate the circumferential stress (sigma_h)):
[tex]
{\sigma_h = \frac{F}{A} = \frac{200,000 \, \text{lbs}}{28.27433 \, \text{in}^2} = \boxed{7071.07 \, \text{psi}}} [/tex]
Calculate The Circumferential Stress in the bar is 7071 psi
Explanation:
what is (sigma_h)?
[tex]\sigma_h [/tex]
