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