Question

A PMMA plate with a 25 mm (width) x 6.5 mm (thickness) cross-section has a contained crack of length 2c = 0.5 mm in the center of the specimen. If the applied load (along the 2 length direction) is 1000 N, calculate the stress intensity factor. If the fracture toughness of PMMA, K1c = 1.15 MPa.m1/2, at what load (in N) does the crack propagate?

Answer 1

LAOD = 6669.86 N

Step-by-step explanation:

Given data:

width
`= 25 mm = 25* 10^(-3) m`

thickness
`= 6.5 mm = 6.5* 10^(-3) m`

crack length 2c = 0.5 mm at centre of specimen

`\sigma _(applied) =  1000 N/cross sectional area`

stress intensity factor = k will be

`\sigma_(applied) = (1000)/(25* 10^(-3)* 6.5* 10^(-3))`

`= 6.154* 10^(6) Pa`

we know that

`k =\sigma_(applied) (√(\pi C))`

`=6.154\sqrt{\pi (2.5* 10^(-04))}` [c =0.5/2 = 2.5*10^{-4}]

K = 0.1724 Mpa m^{1/2} for 1000 load

if
`K_C = 1.15 Mpa m^(1/2)` then load will be

`Kc = \sigma _(frac)(√(\pi C))`

`1.15 MPa = \sigma _(frac)* \sqrt{\pi (2.5* 10^(-04))}`

`\sigma _(frac) = 41.04 MPa`

`load = \sigma _(frac)* Area`

`load = 41.04 * 10^6 * 25* 10^(-3)* 6.5* 10^(-3) N`

LAOD = 6669.86 N