Brittle Fracture Self Study Question

[oxon88]

Homework Statement

[/B]
A tensile load F is to be applied to a tie bar. The tie bar is 50 mm diameter. The surface of the bar has defects resulting from the manufacturing process, these have a maximum depth of 1.5 mm and have a shape factor of 1.4.

If the fracture toughness of the material used is 120 MPa m–0.5, determine the maximum load which can be applied if brittle fracture is not to occur. Use a factor of safety of 3.

Homework Equations

Fracture toughness, K_ic = σ_f√Mc = 120 MPa m-0.5

shape factor, Q = 1.4

surface flaw shape, M = 1.21π / Q = 2.7153. Attempt at an Answer

not really sure where to start here. can anyone advise?

 

[oxon88]

K_Ic=120 MPa m-0.5
Q = 1.4
M = 1.21π/1.4
120 MPa m-0.5 = σ_f √ ((1.21*π*1.5*10^-3) / 1.4)

120 MPa m-0.5 = σ_f * 0.06382σ_f = 120 MPa m-0.5 / 0.06382

σ_f = 1880.29 MPasafety factor = 3

σ_f = 1880.29 / 3 = 626.76 MPacross sectional area = π*r² = 3.14 * 25² = 1963.5 mm²

Force = Stress * Area

1963.5 * 626.76 = 1230643.26 pascals = 1.23Mpa1.23Mpa = 1200 Newtonswould this be correct?

 

[oxon88]

can anyone help with this please?

 

[oxon88]

anyone?

 

[darin khan]

K_Ic=120 MPa m-0.5
Q = 1.4
M = 1.21π/1.4
120 MPa m-0.5 = σ_f √ ((1.21*π*1.5*10^-3) / 1.4)

120 MPa m-0.5 = σ_f * 0.06382σ_f = 120 MPa m-0.5 / 0.06382

σ_f = 1880.29 MPasafety factor = 3

σ_f = 1880.29 / 3 = 626.76 MPacross sectional area = π*r² = 3.14 * 25² = 1963.5 mm²

Force = Stress * Area

1963.5 * 626.76 = 1230643.26 pascals = 1.23Mpa1.23Mpa = 1200 Newtonswould this be correct?

how did you get 1200 Newtons from 1.23 Mpa?