Fracture
Mechanics
Course No. MEE4110/ME419
Problems
#1
A large plate has a small elliptical hole in it. The ratio of the axes (major to minor
axis) of the elliptical hole is 2 to 1. Determine the stress concentration factor Kt
when the plate is loaded in parallel to the
(a) minor axis,
(b) major axis of the ellipse.
Figure (a)
Figure (b)
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Problems
Solution
The maximum stress at an elliptical hole is given by
where a and b are the half-axes of the ellipse.
(a) If the load is parallel to the minor axis, Figure (a), one obtains
(b) If the load is parallel to the major axis, Figure (b), one obtains
Stress concentration factor is Kt = 5 and 2, respectively.
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Problems
#2
A plate contains an elliptical hole as shown in the figure. The plate is supported at
two opposite sides; the support being such that all motion in the y-direction is
prevented whereas motion in the x-direction is possible. The plate is loaded with an
uni-axial stress σ∞ along the two other sides.
Under which condition (ratio b / a) will fracture start at point A and at point B,
respectively? The material is linearly elastic (with Young’s modulus E and
Poisson’s ratio ν) up to its brittle fracture at the ultimate strength σU.
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Problems
Solution
Due to the prevented contraction in the y direction
(strain εyy = 0), stress σyy will appear in that direction.
Let σxx = σ∞ . Hooke’s law gives
Knowing that σzz = 0, one obtains
• Stresses σxx (= σ∞) and σyy = νσxx give rise to stresses at the elliptical hole. It is
seen that if half-axis b is large (b >> a), then the largest stress at the hole will be
at point A.
• Contrary, if a >> b, then the largest stress will appear at point B. Determine the
ratio a / b that gives the same stress at point A as at point B.
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Problems
Solution
Using the elementary case for uni-axial stress:
one obtains (σyy = νσxx is used)
Setting σA = σB gives
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Problems
Solution
Simplifying (f) gives
Let b = βa. It gives
Knowing that β > 0, one obtains the solution β = ν.
Thus, when β = ν, i.e. when b = νa, one has σA = σB. If b > νa, then the stress at
point A is the largest, whereas b < νa gives that the stress at point B is the largest.
Ratio,
b / a > ν gives fracture (largest stress) at point A and
b / a < ν gives fracture at point B.
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Problems
#3
A through-thickness crack of length 2a has been found in a large plate. The plate is
subjected to a bending moment M0 (Nm/m) per unit length. Determine at which
moment M0max crack growth will occur.
Given data:
a = 0.02 m (crack length 2a),
plate thickness t = 0.03 m,
yield strength σY = 1300 MPa, and fracture toughness KIc = 110 MN/m3/2.
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Problems
Solution
We know for a crack of length 2a the SIF is given by
which is smaller than a, t and W - a. Thus LEFM can be used.
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Problems
Solution
The fracture criterion KI = KIc gives
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