Question

A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the tube if the stress is 120 MPa?

Answer 1

D =119.35 mm

Step-by-step explanation:

given data:

inside diameter = 100 mm

load = 400 kN

stress = 120MPa

we know that load is given as

`P = \sigma A`

where:

P=400kN = 400000N

`\sigma = 120MPa`

`A =((1)/(4) \pi D^2 - (1)/(4)\pi (100^2)`

`A=(1)/(4) \pi (D^2 - 10000)`

putting all value in the above equation to get the required diameter value

`400 =  120*(1)/(4) \pi (D^2 - 10000)`

solving for

D =119.35 mm

Answer 2

119.35 mm

Step-by-step explanation:

Given:

Inside diameter, d = 100 mm

Tensile load, P = 400 kN

Stress = 120 MPa

let the outside diameter be 'D'

Now,

Stress is given as:

stress = Load × Area

also,

Area of hollow pipe =
`(\pi)/(4)(D^2-d^2)`

or

Area of hollow pipe =
`(\pi)/(4)(D^2-100^2)`

thus,

400 × 10³ N = 120 ×
`(\pi)/(4)(D^2-100^2)`

or

D² = tex]\frac{400\times10^3+30\pi\times10^4}{30\pi}[/tex]

or

D = 119.35 mm