Question

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 1

Step-by-step 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:

`(\sigma_ x) = (Pd)/(4t)+ (F)/(2 \pi rt)`

Since d = 2r

Then:

`(\sigma_ x) = (Pr)/(2t)+ (F)/(2 \pi rt)`

`(\sigma_ x) = (1402 * 306 * 10^3)/(2(18))+ (13 * 10^3)/(2 \pi * 306* 18 * 10^(-3) * 10^(-3))`

`(\sigma_ x) = (429012000)/(36)+ (13000)/(34607.78467)`

`(\sigma_ x) = 11917000.38`

`(\sigma_ x) = 11.917 * 10^6 \ Pa`

`(\sigma_ x) = 11.917 \ MPa`

`\sigma_y = (pd)/(2t) \\ \\ \sigma_y = (pr)/(t) \\ \\ \sigma _y = (1402* 10^3 * 306)/(18) \ N/m^2 \\ \\ \sigma _y = 23.834 * 10^6 \ Pa \\ \\ \sigma_y = 23.834 \ MPa`

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

Thus;

`\tau _(xy) =0`