Question

A pipe is subjected to a tension force of P = 90 kN. The pipe outside diameter is 45 mm, the wall thickness is 5 mm, and the elastic modulus is E = 150 GPa. Determine the normal strain in the pipe. Express the strain in mm/mm.

Answer 1

The normal strain in the pipe subjected to a tension force is calculated using Hooke's Law, which relates stress to strain via the elastic modulus. The cross-sectional area is found by the area difference between outer and inner circles, considering the pipe's diameter and wall thickness. The final strain is the normal stress divided by the elastic modulus.

Step-by-step explanation:

To determine the normal strain in the pipe due to the tension force, we can use Hooke's Law, which relates stress and strain via the elastic modulus. Normal strain (ε) is calculated by dividing the normal stress (σ) by the elastic modulus (E). The normal stress can be found by dividing the tension force (P) by the cross-sectional area (A) of the pipe.

The cross-sectional area for a pipe is the area of a ring, which can be calculated as the area of the outer circle minus the area of the inner circle. Given the outside diameter (do) is 45 mm and wall thickness (t) is 5 mm, the inside diameter (di) will be do - 2t = 35 mm. So the area is calculated with A = π/4 • (do2 - di2).

After the area is determined, the normal stress is P/A and therefore the strain is ε = σ/E. Keep in mind to convert E to the correct units to match those of stress (N/mm²).

For the provided tension force P = 90 kN and elastic modulus E = 150 GPa, the strain calculation will use these values and the cross-sectional area calculated earlier.

Answer 2

The strain is
`1.79*10^(-3)`

Step-by-step explanation:

Given that,

Force = 90 kN

Outside diameter = 45 mm

Thickness = 5 mm

Elastic modulus = 150 GPa

We need to calculate the inner diameter

Using formula of inner diameter

`D_(in)=D_(o)-t`

Where, t = thickness

Put the value into the formula

`D_(in)=45-5`

`D_(in)=40\ mm`

We need to calculate the area of the pipe

Using formula of area

`A=(\pi)/(4)(D_(o)^2-D_(in)^2)`

Put the value into the formula

`A=(\pi)/(4)((45)^2-(40)^2)`

`A=333.79\ mm^2`

We need to calculate the stress

Using formula of stress

`\sigma=(P)/(A)`

Put the value into the formula

`\sigma=(90*10^(3))/(333.79)`

`\sigma=269.63\ N/mm^2`

We need to calculate the strain in the pipe

Using formula of elastic modulus

`E=(stress)/(strain)`

Put the value into the formula

`150*10^(3)=(269.63)/(strain)`

`strain=(269.63)/(150*10^(3))`

`strain=0.00179`

`strain=1.79*10^(-3)`

Hence, The strain is
`1.79*10^(-3)`