Answer :
Answer:
LAOD = 6669.86 N
Explanation:
Given data:
width[tex] = 25 mm = 25\times 10^{-3} m[/tex]
thickness [tex]= 6.5 mm = 6.5\times 10^{-3} m[/tex]
crack length 2c = 0.5 mm at centre of specimen
[tex]\sigma _{applied} = 1000 N/cross sectional area[/tex]
stress intensity factor = k will be
[tex]\sigma_{applied} = \frac{1000}{25\times 10^{-3}\times 6.5\times 10^{-3}}[/tex]
[tex]= 6.154\times 10^{6} Pa [/tex]
we know that
[tex]k =\sigma_{applied} (\sqrt{\pi C})[/tex]
[tex]=6.154\sqrt{\pi (2.5\times 10^{-04})}[/tex] [c =0.5/2 = 2.5*10^{-4}]
K = 0.1724 Mpa m^{1/2} for 1000 load
if[tex] K_C = 1.15 Mpa m^{1/2}[/tex] then load will be
[tex]Kc = \sigma _{frac}(\sqrt{\pi C})[/tex]
[tex]1.15 MPa = \sigma _{frac}\times \sqrt{\pi (2.5\times 10^{-04})}[/tex]
[tex]\sigma _{frac} = 41.04 MPa[/tex]
[tex]load = \sigma _{frac}\times Area[/tex]
[tex]load = 41.04 \times 10^6 \times 25\times 10^{-3}\times 6.5\times 10^{-3} N[/tex]
LAOD = 6669.86 N