A hollow circular tube of inner radius 12.6 inches and a thickness
of 0.5 inches is subjected to a load of 48 kips. Determine the normal
stress.
strength of materials: normal stress of hollow circular tube
New forum topics
- shear and moment diagram
- Diameter of bolts
- How old is Ann?
- Area of a triangle
- Differential equations: Newton's Law of Coolin
- I need guidance in designing a beam supporting specified ultimate moment of 1100 kN.m (doubly reinforced beam)
- I need guidance in solving the ultimate moment capacity (doubly reinforced beam)
- I need guidance in solving the balance steel area
- Please help me solve this problem using WSD (Working Stress Design) Method
- Physics: Uniform Motion
Because it is not stated in the problem, let us assume that the 48 kips is an axial load.
Solution
Inner radius, $r = 12.6 ~ \text{inches}$
Outer radius, $R = 12.6 + 0.5 = 13.1 ~ \text{inches}$
Calculate the cross-sectional area of the tube:
$A = \pi (R^2 - r^2)$
Calculate the normal stress.
$\sigma = \dfrac{P}{A}$ where $P = 48^k$
Your answer will be in the unit ksi (kilopound per square inch).