Question

The fracture toughness of a stainless steel is 137 MPa*m12. What is the tensile impact load sustainable before fracture that a rod can withstand with a pre-existing surface crack of 2 mm, given a square cross-section of 4.5 mm on each side, in kiloNewtons

Answer 1

7.7 kN

Step-by-step explanation:

The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.

It can be expressed by using the formula:

`K = \sigma Y √(\pi a)`

where;

fracture toughness K = 137 MPa
`m^(1/2)`

geometry factor Y = 1

applied stress
`\sigma` = ???

crack length a = 2mm = 0.002

∴

`137 =\sigma * 1 √( \pi * 0.002 )`

`137 =\sigma * 0.07926`

`(137)/(0.07926) =\sigma`

`\sigma = 1728.489 MPa`

Now, the tensile impact obtained is:

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

P = A × σ

P = 1728.289 × 4.5

P = 7777.30 N

P = 7.7 kN