2005-10-22 - Division of Solid Mechanics

SOLUTIONS
to
Examination in Damage Mechanics and Life Analysis (TMHL61)
LiTH 2005-10-22
1. (1 point)
(a) What type of equation is this
Part 1
ε ij = 1 2 (u i, j + u j,i )
(b) Write the equation in full (without using tensor notations) for i = z and j = x.
Explain the meaning and the contents of the expression you give.
(c) What is the relationship between ε zx and ε xz ?
(d) What is the relationship between ε zx and γ zx ?
Solution and answer here:
(a) Equation of deformations (compatibility equation)
(b) With i = z and j = x one obtains the shear strain ε zx
ε zx = 1 ⎛ ∂u z
⎜
2 ⎝ ∂x + ∂u x⎞
⎟
∂z ⎠
where u x = u and u z = w are displacements in the x and z direction, respectively. The
displacements are differentiated with respect to x and z, respectively.
(c)
ε zx =ε xz
(d)
ε zx = 1 2 γ zx
2. (1 point)
When using the fracture criterion K I = K Ic three conditions must be fulfilled. Give, and
explain, these three conditions.
Solution and answer here:
Plate thickness (crack tip width) t, crack length a, and remaining length in front of the
crack, W−
a, should all be large enough. One has
⎧ t ⎫
⎪ ⎪
⎨ a ⎬
⎪ ⎪
⎩W − a⎭
≥ 2.5 ⎛ ⎜ ⎝
K Ic
σ Y
⎞ ⎟⎠
2
where K Ic is the fracture toughness of the material and σ Y is the yield limit of the
material.
Plate thickness (crack tip width) t ensures that plane deformation is at hand, crack
length a and remaining length in front of the crack W−
a ensure that linear elastic
fracture mechanics (LEFM) can be used.
7 2005-10-19/TD

Examination in Damage Mechanics and Life Analysis (TMHL61)
LiTH 2005-10-22
Part 1
3. (1 point)
Define the stress intensity range ∆K I (the range of the stress intensity factor) that we
use in Paris’ law.
Solution and answer here:
In a case where the loading is varying with time, let K Imax be the maximum value of the
stress intensity factor, and K Imin be the minimum value. Then the stress intensity range
∆K I is
∆K I = K Imax − K Imin if K Imin >0
∆K I = K Imax if K Imin

5. (3 points)
( ) 0
Examination in Damage Mechanics and Life Analysis (TMHL61)
LiTH 2005-10-22
y
2 a
2 W
P
x
Part 2
A large thin plate contains a through-thickness crack
with initial crack length 2a 0 = 15 mm. The plate is
loaded by a force P. Upon loading the plate,
symmetric stable crack growth was observed. At load
P = 300 kN the plate fractured. During the loading
the crack opening δ(0) was recorded, and just prior to
the fracture, the crack opening displacement was
found to be δ(0) = 0.12 mm.
Use the Dugdale model to determine the crack length
just before the plate fractured.
P
Numerical data: Modulus of elasticity E = 210 GPa, yield strength σ Y = 480 MPa, plate
width 2W = 0.4 m and plate thickness t = 0.002 m (i.e. the plate is so thin that the
conditions at the crack tip may be regarded as plane stress).
Solution:
The load P = 300 kN gives stress σ ∞ = P/2Wt = 375 MPa in the plate.
Since plane stress is at hand, the Dugdale model may be used. It gives
δ(0)= 8 σ Y a
E π
Enter the given numerical values and solve for the unknown crack (half-)length a = a crit .
One then obtains
a crit =δ(0) E π ⎛ ⎜⎝ ln 1 + sin(πσ − 1
∞ / 2σ Y ) ⎞
⎟
8 σ Y cos(πσ ∞ / 2σ Y ) ⎠
− 1
210 000 ⋅π⎛
1 + sin(π ⋅ 375 / 2 ⋅ 480) ⎞
= 0.00012 ⎜ln ⎟ = 0.01177 m
8 ⋅ 480 ⎝ cos(π ⋅ 375 / 2 ⋅ 480) ⎠
Thus, the total crack length at failure is 2a crit = 23.54 mm (= 1.57⋅2a 0 ).
ln⎧ ⎨
⎩
1 + sin(πσ ∞ / 2σ Y )
cos(πσ ∞ / 2σ Y )
⎫
⎬
⎭
9 2005-10-19/TD

6.
Examination in Damage Mechanics and Life Analysis (TMHL61)
LiTH 2005-10-22
Determine, by use of the energy release rate G, the
q 0 stress intensity factor K I for the split beam with a
long crack, see figure. The crack length is a, beam
2h
thickness is b and beam height is 2h, where a >> b
a
and a >> h. The material is linear elastic with
b modulus of elasticity E. The beam is loaded
x q 0
symmetrically with two opposite distributed forces
q 0 (N/m). Assume plane stress conditions at the
crack tip.
For a cantilever beam the deflection w(x) due to a
distributed force q 0 is
w(x)= q 0 a 4 ⎧ x 4
⎨
24EI ⎩ a − 4 x 3
4 a + 6 x 2 ⎫
⎬ 3 a 2 ⎭
Solution:
The potential energy Π due to the work performed by the load q 0 is
Π= ⎛ ⎜ ⎝
− P∆
2
⎞
⎟
⎠ = 2 ⎧ ⎨ − 1
⎩ 2
a
⌠ q0 w(x) dx ⎫ ⎬ =−
⌡ ⌠ a q 0 a 4
q0
0 ⎭ ⌡ 0 24EI
=− q 2 0 a 4 ⎧ a 5
⎨
24EI ⎩ 5a − 4 a 4
4 4a + 6 a 3 ⎫
⎬
3 3a =−q 0
2 ⎭
The energy release rate is
(This result can be checked using Problems 5/7(b) and 5/2 in the textbook.)
2 a 4
24EI
⎧
⎨
⎩
6a ⎫
⎬
5
⎧
⎨
⎩
x 4
a 4 − 4 x 3
2 a 5
=−q 0
⎭ 20EI
G =− dΠ
dA =−dΠ bda = 1 q 2 0 5 a 4
b 20EI = q 2 0 5 a 4 12
20b Ebh = 3 q 2 0 a 4
3 Eb 2 h 3
Finally, for plane stress, one has
E 3 q 0
2 a 4
K I = √⎯⎯⎯GE =
√⎯⎯ ⎯
Eb 2 h = √⎯3 q 0 a 2
3
bh√⎯h
a + 6 x 2 ⎫
⎬
3 a dx 2 ⎭
10 2005-10-19/TD

Examination in Damage Mechanics and Life Analysis (TMHL61)
LiTH 2005-10-22
7. (3 points) A loading sequence of a structure with an edge crack
0
(see figure) is composed of m cycles with pulsating
stress 0 to 100 MPa to 0 and one single cycle with
pulsating stress 0 to 175 MPa to 0, see the lower
figure.
Paris’ law for the material reads
a
da
W >> a
dN = 5.1 × 10 − 12 (∆ K I ) 4.17
m per cycle
where ∆K I has the unit MPa m 1/2 (and n = 4.17).
0
(a) Which is the minimum crack length
Stress [MPa]
a min required to give crack propagation at
200
the stress cycle to 100 MPa? The
material has threshold value ∆K th =8
MPa m 1/2
100
0
m cycles
one sequence
Solution:
(a) The stress intensity range is
time
(b) Assuming a > a min , how many cycles
m to stress level 100 MPa are required to
give the same crack growth as one single
cycle to stress level 175 MPa?
⎛ a ⎞
∆K I =∆σ√⎯⎯πa f 5⎜⎝ ⎟
W ⎠ =∆σ √⎯⎯πa 1.12
Crack propagation will occur when the stress intensity range ∆K I is larger than the
threshold value ∆K th . Stress range 100 MPa gives
∆K I = 100√⎯⎯πa 1.12 = 8 (=∆K th )
from which a = 0.0016 m is solved. Thus, a should be larger than 1.6 mm.
(b) Paris’ law gives, for a sequence of m cycles to stress level 100 MPa, the crack
growth per sequence
da
dN s
= 5.1 × 10 − 12 {m ⋅(100 √⎯⎯πa 1.12) 4.17 }
m per sequence
Paris’ law gives, for one cycle to stress 175 MPa, the crack growth per cycle
da
dN = 5.1 × 10 − 12 (175 √⎯⎯πa 1.12) 4.17
m per cycle
The same crack growth in the two cases gives
da
= da
dN s dN giving 5.1 × 10 − 12 m ⋅(100 √⎯⎯πa 1.12) 4.17 = 5.1 × 10 − 12 (175 √⎯⎯πa 1.12) 4.17
from which m = 175 4.17 /100 4.17 = 10.3 is solved.
Thus, 10.3 cycles to stress level 100 MPa give the same crack growth as one cycle to
stress level 175 MPa.
11 2005-10-19/TD

8.
Examination in Damage Mechanics and Life Analysis (TMHL61)
LiTH 2005-10-22
2b
2 a
A flat plate has a small elliptical hole, see figure.
The major axis 2b is perpendicular to the direction
of the stress. The minor axis is 2a, and b = 1.5a. The
plate is subjected to the stress σ ∞ = 100 ± 150 MPa.
Estimate the number of stress cycles to fatigue
failure by use of Neuber’s method. Assume that K f =
0.9K t .
Use the Ramberg-Osgood’s material relation for amplitudes (E = 200 GPa, K’ = 1344
MPa, n’ = 0.18)
ε a = σ a
E + ⎛ 1 / n’
σ a ⎞
⎜ ⎟ ⎝ K’ ⎠
with appropriate modifications of the formula for changes ∆σ of the stress.
For the fatigue life analysis, use the Coffin-Manson relation, with numerical values σ U
= 800 MPa, and Ψ = 0.65.
Solution:
The stress concentration factor K t is (1 + 2b / a) = 4. No information for calculation of
the notch sensitivity factor q is given here. Instead, K f = 0.9K t was given. Thus, use K f
= 0.9⋅4 = 3.6.
The maximum local stress σ max and the maximum local strain ε max at the notch are not
needed here, because in Coffin-Manson’s relation the stress mean value is not included.
Only the change of strain, i.e. the strain range ∆ε, is required. This strain range is
obtained as the point of intersection between the Neuber hyperbola and the material
stress-strain relation for load changes. One obtains, for a nominal change of stress ∆σ ∞
= 300 MPa (which is the change of stress far away from the stress concentration),
⎧
⎪∆σ ⋅ ∆ε = K 2 f (∆σ ∞ ) 2
= 3.62 ⋅ 300 2
E 200 000 = 5.832
⎨
⎪∆ε = ∆σ
⎩ E + 2 ⎛ 1/n’
∆σ ⎞ ∆σ
⎜ ⎟ = ⎝ 2K’ ⎠ 200 000 + 2 ⎛ 1/0.18
∆σ ⎞
⎜ ⎟ ⎝ 2 ⋅ 1344⎠
where ∆σ and ∆ε are changes of stress and strain at the stress concentration. From this
system of equations ∆ε = 0.007038378 and ∆σ = 828.6 MPa are solved.
Now Coffin-Manson’s rule gives
∆ε = 3.5 σ U
E ( N 800
)− 0.12
+ D 0.6 ( N ) − 0.6 = 3.5
200 000 ( N ) − 0.12 + 1.0498 0.6 ( N ) − 0.6
For ∆ε = 0.007038378 one obtains N = 19 375 cycles.
Thus, failure is expected after, approximately, 19 000 cycles.
12 2005-10-19/TD