Problems and Solutions in Fracture Mechanics

January 5, 2017 | Author: Kumar Saheb | Category: N/A
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Problems in Fracture Mechanics PROBLEM: 1 If the specific surface energy for Polmethyl acrylate is 0.0365 J / m2 and its corresponding modulus of elasticity is 2.38 GPa, compute the critical tensile stress required for unstable propagation of a central internal crack whose length is 30 mm. If the strength of the sound glass is 70 MPa, calculate the reduction in strength due to the presence of the crack.

E = 2.38 x 109 Pa

a = 0.015 m

s = 0.0365 J / m2

c= [2 x 2.38*109 x 0.0365 /  x 0.015]0.5 = 60,719 Pa = 60.7 MPa % Reduction in strength =

13.29%

PROBLEM: 2 A sheet of glass measuring 2 m by 200 mm by 2 mm contains a central slit parallel to the 200 mm side. The sheet is restrained at one end and loaded in tension with a mass of 500 kg. What is the maximum allowable length of slit before fracture occurs? Assume plane stress condition and the following material property values: E = 60 GPa, surface energy is 0.5 J/m2. E = 60 x 109 Pa s = 0.5 J/m2 P = 500 Kg

PROBLEM: 3 A thin sheet of maraging steel has a tensile strength of 1950 MPa. Calculate the percentage reduction in strength due to the presence of a central crack in the sheet, which is 4 mm long and orientated perpendicular to the stressed direction. For this steel, E can be taken as 200 GPa, the energy of fracture surface as 2 J/m2, and the work of plastic deformation of each crack tip is 2x104 J/m2. E = 200 x 109 Pa a = 2 * 10-3 m

p = 104 J / m2 s = 2 J / m2

PROBLEM: 4 A rectangular perspex plate 600 mm by 300 mm by 6 mm thick is scribed into two equal squares by a knife, leaving a uniform cut of depth 0.3 mm. What is the bending moment required to break the plate if the perspex has a work to fracture of 500 J/m2? Note that E = 2.5 GPa for perspex.

Problem description : Stage 1 - Scribing

Stage 2 : Bending the sheet as shown

Since the value for Poisson's ratio is not provided, it is reasonable to assume plane stress conditions here, even though the plate is fairly thick and perspex is moderately brittle at ambient temperatures. This problem has two stages to the solution, firstly to calculate the Griffith fracture stress and, secondly, to find the bending moment that corresponds to this. Note that this technique is often used in practice to fracture brittle and quasi-brittle materials, e.g. glass, tiles and polymers. Recalling Griffith's equation as:

and noting that this is an edge crack, i.e. a = 0.3 mm, we can substitute in the values to get:

We can find the required bending moment from the simple bend equation:

PROBLEM: 5 Determine the energy release rate using elementary Beam analysis for the following configuration (h < a)

General Beam Analysis

Energy approach =

= 2

=2 = 

=

=

=

=

=

Compliance approach = =

=

;

C=

;

(bottom arm)

C=

=

;

=

=

C=

=

: Mode of Fracture : Opening mode (Mode I)

= ;

PROBLEM: 6 Determine the energy release rate using elementary Beam analysis for the following configuration (h < a)

=

;=

= ;=

;=

;=

Energy approach =

=

=

=

= 

=

;

=

=

= =

Compliance approach = = =

;

C=

;

=

C=

(bottom arm)

=

;

=

=

: Mode of Fracture : Sliding mode (Mode II)

;=

;=

PROBLEM: 7 Determine the energy release rate using elementary Beam analysis for the following configuration (h < a)

M=P*a

PROBLEM: 8 Determine the energy release rate using elementary Beam analysis for the following configuration (h < a)

This problem is treated as constraint at the crack tip causing bending as shown:

PROBLEM: 9 Load on a 30 mm thick plate with an edge crack of 50 mm length was increased very slowly and the displacement of the load point was monitored. It was observed that at the load of 2100 N and displacement u = 4.1 mm, the crack started growing. The rate of crack growth was much faster than the rate of load increase and therefore the crack essentially was grown at the load of 2100 N. Through a rapid camera recording it was found that the crack grows up to 65 mm length with rapid increase in displacement to u = 7.5 mm. Determine the critical energy release rate.

= a = (65 x

– 50 x

;

=

) = 15 x

m

= 1.0794 x 104 =

=

;

= 7.93 kJ/m2

PROBLEM: 10 A large plate of 36 mm thickness with an edge crack a = 32 mm length is pulled very slowly under displacement control loading. At the displacement of 7.2 mm, when the recorded load is 2750 N, the crack starts growing. At a = 41.7 mm, the crack is arrested and the load was found to decrease to 1560 N. Determine the critical release rate.

=

= 12268 J/

=

= 12.27 kJ/m2

=

PROBLEM: 11 Determine the shape of the DCB specimen [i.e. relationship between depth (2h) and crack length (a)], if GI is to remain constant. Following data are given: Initial crack length (a0): 50 mm Total depth of specimen up to initial crack length (2h0): 24 mm Thickness of specimen (B): 30 mm Calculate the total depth of the cantilever when crack length (a) is 80 mm.

PROBLEM: 12 Consider a large thin sheet of Mild steel and determine minimum crack length at the center that may not require invoking Fracture mechanics solution. Material properties are E: 200 GPa, Gc (J/m2) and UTS = 650 MPa

PROBLEM: 13 If the fracture stress of a large sheet of maraging steel , which contains a central crack of length 40 mm, is 480 MPa, calculate the fracture stress of a similar sheet containing a crack of length 100 mm.

Fracture stress of sheet containing central crack of 100 mm is:

PROBLEM: 15 Grinding wheels are fabricated, typically, from alumina powder, which is compacted and sintered at high temperature and pressure. The powder is sieved before compacting to remove impurities which may later act as defects in the grinding wheel. Hence residual impurities are related in size to the sieve mesh dimension. One particular type of alumina wheel has a density of 3800 kg/m3, a bore diameter of 140 mm and an outer diameter of 1.0 m. It spins at 3000 rpm. The maximum stress in the wheel is given by:

Calculate the allowable size of the sieve mesh if the wheel is to have a factor of safety of two on critical defect size when operating at 3000 rpm. Note that, for alumina, the fracture toughness R = 0.10 kJ/m2 and E = 371 GPa. You may assume plane strain conditions. Ans : 1.1 mm PROBLEM: 16 Consider a plate with an edge crack (see figure). The plate thickness is such that a plane strain condition is present. Given: W = 1000 mm; stress intensity factor KI = C  a where C = 1.12

Answer the next questions for the three materials given in the table above: a) Does fracture occur at a stress 

ys and a crack length a = 1 mm?

b) What is the critical defect size at a stress 

ys?

c) What is the maximum stress for a crack length a = 1 mm without permanent consequences? PROBLEM: 16 A 3 mm thick tension panel 10 cm wide containing an edge crack of 1 mm yielded at a load of 150 kN. However at a load of 120 kN, another panel of the same material cracked into two pieces when the crack was 5 mm long. Estimate the yield stress and Fracture Toughness of the material. Assume SIF for an edge crack subjected to tension stress as KI = 1.12  a. PROBLEM: 17 Predict the failure mode and load of the cantilever beam of thickness 5 mm as indicated below: Assume the fracture toughness and yield stress as 30 MPam and 300 MPa. What is the failure mode if the yield stress is doubled? Assume nominal stress at the crack as M/Z where M is the bending moment and Z is the section modulus = 400 mm3. SIF for the edge crack is KI = 1.12  a.

Computation of Failure load assuming fracture to take place at crack location:

i) Computation of Failure load assuming yielding (ys = 300 MPa) to take place at the support:

Since yield load is lower than fracture load, failure mode will be by yielding ii) Computation of Failure load assuming yielding (ys = 600 MPa) to take place at the support

Since fracture load is lower than yield load, failure mode will be by Fracture

PROBLEM: 17 A cracked cantilever beam is deflected 8 mm by a 10kN load. At the same load the deflection is increased by 1 mm due to a crack extension of 0.5 mm. Calculate the initial stress intensity factor. Assume E = 200 GPa, Section thickness = 0.5 m Initial compliance = 8/10000 mm/N Final compliance = 9/10000 mm/N dC / da = 1/5000 N

PROBLEM: 18 A 3 mm thick tension panel 10 cm wide containing an edge crack of 1 mm yielded at a load of 150 kN. However, at a load of 120 kN, another panel of same material cracked into two pieces when the crack was 5 mm long. With this information, calculate the yield stress and fracture toughness of the material. SIF for the edge crack is KI = 1.12  a.

Yield analysis

Fracture analysis For edge crack, SIF is KI = 1.12  a

PROBLEM: 19 Apply Irwin’s plasticity correction for the case of 5 mm thick and 80 mm wide plate containing an edge crack of 20 mm and determine the plastic zone size and effective crack length and effective stress intensity factor. The plate is loaded in Mode I with a far field stress of 150 MPa and the yield strength of the material is 350 MPa. The stress intensity factor (SIF) is indicated in the sketch.

Solution Computation of Plastic zone size

Computation of effective crack length

Computation of effective SIF

PROBLEM: 20 Consider a 25 mm thick steel specimen with an edge crack. A test is performed in which the specimen is crack line loaded with 10,000 N at the edge of the crack. While artificially increasing the crack size, the crack opening displacement  at the edge is measured. The following relation between crack length ’a’ and  (both in mm) is found.

What is the maximum crack length for this steel assuming plane strain conditions? Following are the data for the material

Solution

PROBLEM: 20 Determine maximum central crack permissible for a 300 wide plate subject to tensile stress of 150 MPa. Assume the fracture toughness of the material to be 90 MPam. Take SIF as:

Solution



a in m

a/W

0.02

0.067

0.143

0.04

0.133

0.209

0.06

0.200

0.272

0.08

0.266

0.345

0.10

0.333

0.447

It can be seen that the value of a lies between 0.06 and 0.08. Carry out iterations between 0.06 and 0.08, till  is close to 0.339. Accordingly, the crack length is approximately 0.078 m (78 mm).

PROBLEM: 14 A cylindrical pressure vessel, with a diameter of 6.1 m and a wall thickness of 25.4 mm, underwent a failure when the internal pressure reached 17.5 MPa. The material properties are as under: E = 210 GPa,, ys = 2450 MPa and GC = 131 kJ/m2. i) Using von Mises criteria assess whether the failure is due to yielding ii) Based on Griffith's analysis assess the failure and determine the size of crack that might have caused this failure, stating assumptions that you have made. Solution For a pressure vessel, the three principal stresses (p is the internal pressure, r is the mean radius and t is the thickness) are : 1 = [17.5 x 3.05 / 0.0254] = 2100 MPa

:

2 = [17.5 x 3.05 / 2 x 0.0254] = 1050 MPa 3 = - 17.5MPa

Von Mises yield criterion

[(1 - 2)2 + (2 - 3)2 + (3 - 1)2] = 2 [ys]2 LHS = 10502 + 1067.52 + 2117.52

= 6.725 x 106

RHS = 12 x 106 Since LHS is less than RHS, failure is not due to yielding Fracture criterion A crack oriented along the axial direction (i.e. perpendicular to the hoop stress, which is the largest principal stress) would represent the worst case with respect to fracture. We can use Griffith analysis to calculate the size of the crack:

PROBLEM: 21 A thin –wall pressure vessel is made from Ti alloy with KIC of 57 MPa√m and yield strength of 900 MPa. The internal pressure produces a circumferential hoop stress of 360 MPa. The vessel has a semi-elliptical surface flaw with the major plane oriented perpendicular to the uniform tensile hoop stress. For this type of loading, the stress intensity factor is given by: KI = 1.21  √ ( a) / Q where ‘a’ is depth of flaw , Q is the flaw shape parameter ( = 2.35) and  is the applied stress Determine i) whether the vessel will fracture or leak when the crack reaches critical length. Assume the thickness of the wall to be 12 mm and the crack length is twice the depth. ii) If the thickness of the vessel is twice the given value, for the same stress re-assess whether the vessel will fracture or leak. Solution

Since the critical crack length is more than the thickness, the vessel will leak. This condition is called “leak – before – break” condition. If the thickness of the vessel is doubled, then the critical crack length is less than the thickness, the vessel will fracture. This condition is called “break– before – leak” condition.

A circumferentially cracked rotor shaft is transmitting a power of 0.88 MW. Inspection records indicate a current crack depth of 2 mm and the crack growth for this configuration is 1 mm over 5000 hr. What is the expected life of the rotor shaft. Assume the following ; Mode III KIC = 10 MPa√m Speed = 420.2 rpm ; Diameter of the shaft = 100 mm Solution

A critical component of a off-shore structure is subjected to fluctuating tensile stress as per the histogram shown below. During a routine check up, an edge crack of length 1.5 mm is detected. If the crack is not allowed to exceed 25 mm, determine the remaining life of the component. SIF for the edge crack is KI = 1.12  a

On integrating the crack growth equation with limits as 1.5 and 25, we get the Number of cycles.

On a large plate used as a critical component of a machine, amplitude of the fatigue load shifts several times in a sequence of every 1000 cycles as indicated below. The crack growth in the material follows the Paris Law with C = 2.2 x 10-12 (MPa)-3.4 (m)-0.7, m = 3.4. Determine how many sequences are needed to cause the failure if the initial crack of 2a = 7.2 mm is detected near the center of the plate and KIC = 80 MPam.

Determine KI for the problem as shown in the sketch wherein pressure distribution on the cracked face of an infinite plate is as indicated.

p0 p0 2a

Solution

p0 p0 Config. ( m )

p0

p0

=

p0 Config. ( g )

+

p0 Config. ( h )

KI (Configuration m) = KI (Configuration g) + KI (Configuration h) p0 a = KI (Configuration g) + KI (Configuration g) … Conf. g and Conf. h are same = 2 x KI (Configuration g) KI (Configuration g) = [ p0 / 2] a

Determine KI for the problem as shown in the sketch wherein pressure distribution on the cracked face of an infinite plate is as indicated.

0 0 2a Solution Config. ( h )

Config. ( g )

Config. ( m )

= 2a

+ 2a

2a

KI (Configuration m) = KI (Configuration g) + KI (Configuration h) 0 a = KI (Configuration g) + KI (Configuration g) … Conf. g and Conf. h are same = 2 x KI (Configuration g) KI (Configuration g) = [0 / 2] a

KI (Configuration z) = KI (Configuration g’) + KI (Configuration h’) = [0 / 2] [1/2]a + [0 / 2] [1/2]a = [0 ] [1/2]a

In a large plate, a crack length 2a is inclined with an angle  with x1 axis. The plate is loaded in x2 direction with a far field stress of , find the stress intensity factors if  = 80 MPa ; 2a = 20 mm ;  = 30

Solution

During water quenching of steel components with a section thickness of 30 mm, heat transfer calculations indicate that a peak stress of 130 MPa is generated in the section. Prior to heat treatment, the components were ultrasonically inspected to detect defects. The inspection technique has a minimum detection size of 0.5 mm. a) What type of defect will be most critical? b) Calculate the size of defect which would cause fracture of the component during the quenching operation, given that the aspect ratio of the crack is 2c/a = 10. c) Would this inspection procedure guarantee integrity of the component if the quenching stresses approached the proof stress of the steel? Note that the value of the plane strain fracture toughness K1C = 30 MPa m1/2 and the proof stress = 620 MPa. The stress intensity calibration for this component and crack geometry is given in the figure below.

Where, for surface flaws:

Solution Surface defects 15.4 mm Procedure does not guarantee integrity : ac = 0.54 mm

and for embedded flaws:

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