FRACTURE MECHANICS
Failure can occur due to various reasons: Uncertainties in loading or environment Defects in material Inadequacy in design Deficiency in construction or maintenance
The strength of materials can be increased significantly. But sometimes this comes at the cost of reduced ductility of the material. For example, strength of steel can be increased by manipulating the microstructure. But this increases the brittleness of the material. Thus, a member can fail catastrophically due to crack formation and propagation, with little or no warning. A crack in a member modifies the local stresses that sometimes elastic stress analysis is insufficient for design. A structure can fail even before reaching its capacity, if the crack …show more content…
As the crack propagates from 0 to a, only if the stress is increased. After reaching the critical crack length a=ac, the crack growth is spontaneous and catastrophic.
As the energy associated with crack at a=ac is maximum, we can compute the value of critical crack length by equating the first derivative of total energy (S+U) to zero.
(∂ (S+U))/∂a=2 γ - 〖σ_(f^.)〗^2/E πa=0
From the above equation, the stress can be written as: σf = √((2Eγ )/πa)
It is important to note that Griffith’s original work was true for brittle materials, specifically glass rods. When a ductile material is considered, an accurate fracture model cannot be provided considering only the surface energy. Later on, Irwin and Orowan independently suggested that in a ductile material, the released strain energy was absorbed by energy dissipation due to plastic flow near the crack tip. Accordingly, the Griffith’s equation was modified …show more content…
WESTERGAARD’S SOLUTION FOR SIF
The set of equations given by Westergaard for the stress field is: σxx = Re Z – yImZ ’ σyy = Re Z + yImZ ’ τxy = -yRe Z ’ where Z=σ_(∞^.)/√((1-〖(a/z)〗^2 )
Z'=(-σ_(∞^.) a^2)/(z^3 [〖(1-(a/z)^2]〗^(3/2) )
For y = 0, σyy = σ_(∞^.)/√((1-〖(a/x)〗^2 ) IRWIN’S NEAR CRACK TIP SOLUTION
Irwin further simplified the result obtained by Westergaard in the area immediately surrounding the crack tip. He included a new term r in the expression of z which is measured from the crack tip (i.e. r = 0 at x = a)
He expressed z as: z = a + reiθ Thus,
Z (z)=σ_(∞^.)/√((1-(a/(a+〖re〗^iθ ))^2 )
On solving the above equation, we get
Z (z)=σ_(∞^.)/√(((2a〖re〗^iθ+ r^2 e^i2θ)/(a^2+2a〖re〗^iθ+ r^2 e^i2θ ))^2 )
Irwin proposed that the area near the crack tip corresponds to very small value of r as compared to a, (i.e., a >> r). Therefore, a2 >> ar >> r2. Thus, the above equation can be simplified as:
Z (z)=σ_(∞^.)/√(((2a〖re〗^iθ)/a^2 )^2 ) which is further simplified as:
Z (z)= σ_(∞^.) √(a/2r ) e^(-i θ/2)
As we know, eiθ = cos θ + i sin θ , the above equation becomes:
Z (z)= σ_(∞^.) √(a/2r ) (cos〖θ/2〗-i sin〖θ/2〗 )
Similarly, we can compute Z’ as
Z(_^')(z) =σ_(∞^.)/2r √(a/2r) (cos〖3θ/2〗-i sin〖3θ/2〗
Failure can occur due to various reasons: Uncertainties in loading or environment Defects in material Inadequacy in design Deficiency in construction or maintenance
The strength of materials can be increased significantly. But sometimes this comes at the cost of reduced ductility of the material. For example, strength of steel can be increased by manipulating the microstructure. But this increases the brittleness of the material. Thus, a member can fail catastrophically due to crack formation and propagation, with little or no warning. A crack in a member modifies the local stresses that sometimes elastic stress analysis is insufficient for design. A structure can fail even before reaching its capacity, if the crack …show more content…
As the crack propagates from 0 to a, only if the stress is increased. After reaching the critical crack length a=ac, the crack growth is spontaneous and catastrophic.
As the energy associated with crack at a=ac is maximum, we can compute the value of critical crack length by equating the first derivative of total energy (S+U) to zero.
(∂ (S+U))/∂a=2 γ - 〖σ_(f^.)〗^2/E πa=0
From the above equation, the stress can be written as: σf = √((2Eγ )/πa)
It is important to note that Griffith’s original work was true for brittle materials, specifically glass rods. When a ductile material is considered, an accurate fracture model cannot be provided considering only the surface energy. Later on, Irwin and Orowan independently suggested that in a ductile material, the released strain energy was absorbed by energy dissipation due to plastic flow near the crack tip. Accordingly, the Griffith’s equation was modified …show more content…
WESTERGAARD’S SOLUTION FOR SIF
The set of equations given by Westergaard for the stress field is: σxx = Re Z – yImZ ’ σyy = Re Z + yImZ ’ τxy = -yRe Z ’ where Z=σ_(∞^.)/√((1-〖(a/z)〗^2 )
Z'=(-σ_(∞^.) a^2)/(z^3 [〖(1-(a/z)^2]〗^(3/2) )
For y = 0, σyy = σ_(∞^.)/√((1-〖(a/x)〗^2 ) IRWIN’S NEAR CRACK TIP SOLUTION
Irwin further simplified the result obtained by Westergaard in the area immediately surrounding the crack tip. He included a new term r in the expression of z which is measured from the crack tip (i.e. r = 0 at x = a)
He expressed z as: z = a + reiθ Thus,
Z (z)=σ_(∞^.)/√((1-(a/(a+〖re〗^iθ ))^2 )
On solving the above equation, we get
Z (z)=σ_(∞^.)/√(((2a〖re〗^iθ+ r^2 e^i2θ)/(a^2+2a〖re〗^iθ+ r^2 e^i2θ ))^2 )
Irwin proposed that the area near the crack tip corresponds to very small value of r as compared to a, (i.e., a >> r). Therefore, a2 >> ar >> r2. Thus, the above equation can be simplified as:
Z (z)=σ_(∞^.)/√(((2a〖re〗^iθ)/a^2 )^2 ) which is further simplified as:
Z (z)= σ_(∞^.) √(a/2r ) e^(-i θ/2)
As we know, eiθ = cos θ + i sin θ , the above equation becomes:
Z (z)= σ_(∞^.) √(a/2r ) (cos〖θ/2〗-i sin〖θ/2〗 )
Similarly, we can compute Z’ as
Z(_^')(z) =σ_(∞^.)/2r √(a/2r) (cos〖3θ/2〗-i sin〖3θ/2〗