2003-01-18 - Division of Solid Mechanics

Examination in Damage Mechanics and Life Analysis (TMHL61)
LiTH 2003-01-18
7. (3 points) One type of test specimen to determine the
p
fracture toughness of a material is the double
cantilever beam, see figure. The height is 2h
= 30 mm (the width t is large).
At plane strain, the stress intensity factor can
2 h be written
p
a
where ν is the Poisson ratio, ν = 0.3, and p is
the loading per unit of the width of the test
specimen.
Paris’ law for the material reads da
dN = 5.1 × 10 − 12 (∆ K I ) 3
m per cycle
where ∆K I has the unit MPa m 1/2 .
The fracture toughness of the material is (in this example) determined in the following
way: First a crack of length a 0 = 20 mm is created by machining. Then, to obtain a
sharp crack tip, the test specimen is subjected to a loading p such that the load p is
cycled between 0 och 0.9 MN/m. This is done for N = 15 000 cycles. When this is done
the cyclic loading is stopped.
(a) Calculate the crack length a final obtained by the cyclic loading.
After this, the test specimen is loaded by a static, monotonically increasing load p until
fracture occurs. One obtains the final fracture of the specimen at load p = 1.8 MN/m.
(b) Determine the fracture toughness of the material.
The yield limit of the material is σ s = σ Y = 1000 MPa.
Solution: (a) First determine the crack length after the cyclic loading. Paris’ law gives
Which gives
da
dN = 5.1 × 10 ⎛ 3
− 12 2√⎺3 0.9 a ⎞
⎜ ⎟⎠ ⎝ 0.91 0.015 3 / 2
1 ⎧
− 2
from which is solved a final = 25.754 mm.
(b) At the final, monotonically increasing loading the critical crack length is a final =
25.754 mm. The fracture toughness is obtained from
K I = K Ic = 2 √⎺3 p failure a final
= 96.06 MPa√⎺m
1 − ν 2 h 3/2
Thus K Ic = 96 MPa m 1/2 .
Here linear elastic fracture mechanics (LEFM) was used. Was it allowed?
One has
2.5 ⎛ 2
K Ic ⎞
⎜ ⎟⎠ = 2.5 ⎛ 2
96 ⎞
⎜ ⎟ = 0.023 m < a ⎝ σ Y ⎝ 1000⎠
final = 0.02575 m
Thus, it was OK to use LEFM for the final failure.
⎨
a final
m per cycle
K I = 2 √⎺3
1 − ν 2 p a
h 3/2
1
− 1 ⎫
⎬
2 2
a = 0.033078 ⋅ N = 496.16 0 ⎭ (1/m2 )
10 2003-01-23/TD