The compressed-air tank has an inner radius r and uniform wall thickness t. The gage pressure inside the tank is p and the centric axial load F is applied at the end cap. Use p = 1402 kPa, F= 13 kN, t 18 mm and r =306 mm.

Required:
Obtain the state of plane stress in the x - y coordinate system

Answer:

Explanation:

Given that:

The Inside pressure (p) = 1402 kPa

= 1.402 × 10³ Pa

Force (F) = 13 kN

= 13 × 10³ N

Thickness (t) = 18 mm

= 18 × 10⁻³ m

Radius (r) = 306 mm

= 306 × 10⁻³ m

Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)

Then;

the state of the plane stress can be expressed as follows:

[tex](\sigma_ x) = \dfrac{Pd}{4t}+ \dfrac{F}{2 \pi rt}[/tex]

Since d = 2r

Then:

[tex](\sigma_ x) = \dfrac{Pr}{2t}+ \dfrac{F}{2 \pi rt}[/tex]

[tex](\sigma_ x) = \dfrac{1402 \times 306 \times 10^3}{2(18)}+ \dfrac{13 \times 10^3}{2 \pi \times 306\times 18 \times 10^{-3} \times 10^{-3}}[/tex]

[tex](\sigma_ x) = \dfrac{429012000}{36}+ \dfrac{13000}{34607.78467}[/tex]

[tex](\sigma_ x) = 11917000.38[/tex]

[tex](\sigma_ x) = 11.917 \times 10^6 \ Pa[/tex]

[tex](\sigma_ x) = 11.917 \ MPa[/tex]

[tex]\sigma_y = \dfrac{pd}{2t} \\ \\ \sigma_y = \dfrac{pr}{t} \\ \\ \sigma _y = \dfrac{1402\times 10^3 \times 306}{18} \ N/m^2 \\ \\ \sigma _y = 23.834 \times 10^6 \ Pa \\ \\ \sigma_y = 23.834 \ MPa[/tex]

When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.

Thus;

[tex]\tau _{xy} =0[/tex]

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