Effect of finite size of component
IK Y a
where Y = configuration factor
The SIF derived earlier is for cracks in an infinite body. However the finite size, geometry of the component, loading conditions have effect on SIF. The shape or configuration effect is given by a factor Y such that
SIF for some of the standard geometry are shown below
What is Fracture Toughness?
Fracture toughness is a measure of the ability of a material to resist the growth of a pre-existing crack or flaw. The fracture toughness of a material is characterized by the energy per unit area which is required to create new crack surfaces, and thereby propagate a crack through the material. This value is known as the critical stress intensity factor.
The critical SIF at plane stress condition is denoted by KC ( )
In practice we may attain KC by increasing either or a. For a defect of fixed length a, Cis the critical value of the applied stress for fracture. Alternatively, for a constant applied stress , aC is the critical defect size for fracture.Note that KC is a macroscopic criterion for failure. It makes no assumptions about the precise
mechanism of fracture.
KIC is the critical SIF at plane strain condition
Typical Values of Fracture Toughness KIc
pure ductile metals 100--350 pressure vessel steel 170 high strength steel 50--150 titanium alloys 50--110 GFRP, fibreglass 20--60 aluminium alloys 20--45 cast iron 6--20 reinforced concrete 10--15
polystyrene 2 silicon nitride 4--5 magnesia 3 granite 3 wood 1 glass 0.5 ice 0.2 Units are . MPa m
Application of the Fracture Mechanics Approach
The material is Al alloy 2024, which has a yield stress of 200 MPa. The value of the fracture toughness is . We assume that the Y parameter is 1.12, corresponding to an edge crack. We assume that a safety factor of one half is built in to the design; the nominal stress must then not exceed one half of , i.e. 100 MPa. What is the size of the critical flaw, ? From the data given we have
We therefore conclude that a crack will propagate in an unstable fashion if the length exceeds 0.37 m. This is a reassuringly large magnitude. We conclude that the structure can tolerate small defects such as incipient cracks at rivet holes. Inspection of the wing is expected to allow cracks to be detected before they reach the critical size.
For civil airliners, the material is employed is commonly a high strength steel. The yield stress is 1200 MPa, MPa, i.e. a safety factor of 0.6 is employed. The fracture toughness is . Take Y=0.95 corresponding to a semi-elliptic crack. Therefore
We therefore conclude that a crack of length greater than 2.4 mm will cause catastrophic brittle failure. As a consequence, we need to inspect after every heavy landing and non-destructive testing methods need to be used frequently, typically every week. It is essential to detect cracks of length 1 mm.
Imagine that this is fabricated from pressure vessel steel, which has . The yield stress is 300 MPa and a safety factor of 0.66 is taken. We take Y=1 and hence
The failure of pressure vessels can be so catastrophic that special precautions are always undertaken. One concerns the need for the vessel to leak before breaking. Consider the above example in which acrit=115 mm. If the wall thickness is t = 250mm then acrit<t,
i.e. if the crack remains undetected and grows by fatigue, explosion of the vessel by catastrophic brittle fracture will occur. Alternatively, by appropriate choice of steel toughness, stress and thickness, we can redesign.Now if an undetected crack undergoes slow growth, it must penetrate the wall thickness before it can achieve a size sufficient for unstable fracture. But penetration of the wall will cause leakage which leads to a loss of pressure which in turn reduces the stress and lowers the driving force for failure, i.e. the
vessel leaks before it breaks.
I
3I C
3
K Y a
Since the plate dimension is large when compared
to crack size a, the plate can be considered as infinite
Hence Y=1.
K K 1.0 32.5X10 28.3
28.388.6 MPa
32.5X10Uniaxial stress at which yie
y
y
lding begins is 240MPa
240Factor of safety (FS) = 2.71
88.6Note : Fracture occurs before yielding
A steel ship deck is 30 mm thick, 12 m wide and 20 m long and hasa fracture toughness of KC = 28.3 MPa.m 1/2 . If a 65 mm long central transverse crack is discovered, calculate the nominal tensile stress that will cause catastrophic failure. Compare the stress found to the yield strength of . y 240MPa
2a=65mmb=12m
l=20m
A plate of width 1.4 m and length 2.8 m is required to support a tensile load of 4 MN (in the long direction). Inspection procedures are capable of detecting through-thickness edge cracks larger than 2.7 mm. The two titanium alloys are being considered for this application. (Alloy (A) KIC = 115 MPa.m1/2
and y=910 MPa; alloy (B) KIC = 55 MPa.m1/2 and y=1035
MPa. For a factor of safety of 1.3 against yielding and fracture, which one of the two alloys will give the lightest weight solution?
y y
y
Design based on yield
FS=(P / bt)
FS Pt
b
a=2.7mm
b=1.4m
l=2.8m
t
P=bt
6
y
y
6
y
y
For alloy (A)
FS P 1.3X4X10t 4.08mm ;
b 1400X910
910700MPa
FS 1.3For alloy (B)
FS P 1.3X4X10t 3.59mm ;
b 1400X1035
1035796MPa
FS 1.3
I
3
1/ 2
3I
C C
I
C
Design based on Fracture
K Y a
a a0.752 2.02 0.37 1 sin
2b a b 2bY tan
aa 2b cos2b
Y 1.1225
K 1.1225 2.7X10 0.103386
K KFS
K 0.103386
K
0.103
386FS
C
6
C
6
For alloy (A)(weak)
K 115855.65MPa
0.103386FS 0.103386X1.3
P 4X10t 3.34mm
b 1400X855.65For alloy (B)
K 55409.22MPa
0.103386FS 0.103386X1.3
P 4X10t 6.98mm
b 1400X409.22
Summary
For alloy (A)
Yielding : t=4.08mm; 700MPa
Fracture : t=3.34mm; 855.65MPa
For alloy (B)(strong)
Yielding : t=3.59mm; 796MPa
Fracture : t=6.98mm; 409.22MPa
Best design solutio
n: use alloy A with t=4.09 mm