Materials 335

Mid-semester test 2005

Test 2 (Closed book, 40 minutes)

(This test accounts for 15% of the grade for the unit)

All intermediate calculations, assumptions, and related workings out should be provided in order to gain full credit.

The propeller and shaft of an aircraft engine are shown schematically in the figure below. The propeller has a mass, m, and generates a forward force, F, relative to the aircraft at a constant rotational speed of N revolutions per minute. The shaft has a diameter, d, where it joins the propeller and increases with a fillet of radius, r, to a diameter of D at a distance, L, from the propeller centre of mass. The shaft transmits a power, P, to the propeller.

Question continued on following page

Student Name: _______________________________ Student No.: ________________ Total Pages: ____

D:\Data\Curtin\Lecture courses\Materials 335\2005\Mid-semester test\September 23rd\Mid-semester_test_2_solutions_003.doc 9/23/2005

(i) Show that the effective mean stress, σm, and alternating stress, σa, at the fillet can be given by the following:

( )3

32a f b

mgL Kd

σπ

=

[4 marks]

( ) ( ) ( )

2 2

2 2 2 3

2 2 480m f a f a f t

F F PK K Kd d N d

σπ π π

⎛ ⎞ ⎛= + +⎜ ⎟ ⎜⎝ ⎠ ⎝

⎞⎟⎠

[8 marks] Where Kf(b), Kf(a), and Kf(t) are the fatigue stress intensity factors for bending, axial, and torsional loading, respectively. (ii) Estimate the factor of safety for the following conditions: d = 50 mm, D = 100 mm, r = 4 mm, L = 100 mm, F = 4000 N, m = 1000 kg, P = 50 kW, and N = 3000 rpm. Assume the shaft to be made from commercially polished steel with an ultimate tensile strength, Su, of 1200 MPa.

[16 marks]

[Total marks: 28]

Notes:

In order to gain full credit you will need to clearly indicate (for example, by drawing intersecting lines in the appropriate positions) how the fatigue data figures in the following pages were utilised. You are required to fill out the details on the front page of this question booklet and return this question booklet with your answer booklet.

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Fatigue property sheet

All symbols have their usual meaning.

106 cycle strength: Load type

Factor Bending Torsion Axial

CL 1.0 0.58 0.9*

CD1.0 for D ≤ 10 mm

0.9 for 10 ≤ D ≤ 50 mm 1.0

CS From figure shown below

Sn = CLCDCSSn’ * A lower value may be used to account for known or suspected undetermined bending due to load eccentricity. 103 cycle strength: Bending loads: 0.9 SuTorsional loads: 0.9 SusAxial loads: 0.75 Su

Reduction of endurance strength due to surface finish – steel parts (Adapted from R. C. Juvinall, p. 234 in Engineering considerations of stress, strain and strength, McGraw-Hill, Inc., New York, 1967).

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Fatigue property sheet (continued)

Theoretical stress concentration factors for a shaft with a fillet: (a) bending, (b) axial, and (c) torsional loads (R. E. Peterson, in Stress concentration design factors, John Wiley & Sons, Inc., 1953).

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Fatigue property sheet (continued)

Notch-sensitivity curves for use with theoretical factors, Kt (Adapted from R. E. Peterson, “Notch sensitivity”, in

Metal fatigue, G. Sines and J. L. Waisman, eds., McGraw-Hill Book Company, New York, 1959).

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Mid-semester test 2005 – Test 2 (solutions)

The system contains the following stresses acting at the fillet: (a) Alternating bending stress, σa(b), due to mg (b) Constant axial stress, σm(a), due to F (c) Constant torsional stress, τm, due to P (a) Alternating bending stress, σa(b), due to mg. From the supplied figure:

( ) 3

32a b

Md

σπ

= [1 mark]#

M = mg·L [1 mark] We need to take into account Kf for bending:

⇒ ( ) ( )3

32a b f b

mgL Kd

σπ

= [1 mark]

There is only one alternating stress:

∴ ( )3

32a f b

mgL Kd

σπ

= [1 mark]

(b) Constant axial stress, σm(a), due to F. From the supplied figure:

( ) 2

4m a

Pd

σπ

= (note different P from that given in figure) [1 mark]#

P = F We need to take into account Kf for axial:

⇒ ( ) ( )2

4m a f a

F Kd

σπ

= [1 mark]

# Equivalent expression also accepted.

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(c) Constant torsional stress, τm, due to P. From the supplied figure:

3

16m

Td

τπ

= [1 mark]#

60

2T

Nπ⎛ ⎞= ⎜ ⎟⎝ ⎠

P [1 mark]

We need to take into account Kf for torsion:

⇒ ( )2 3

480m f t

P KN d

τπ

= [1 mark]

We can obtain the equivalent mean stress, σm, by assuming failure due to the maximum distortion energy theory as follows: [1 mark]

2

2

2 2x y x y

m

σ σ σ σxyσ τ

+ −⎛ ⎞= + +⎜ ⎟

⎝ ⎠ [1 mark]

⇒ ( ) ( ) ( )

2 2

2 2 2 3

2 2 480m f a f a f t

F F PK K Kd d N d

σπ π π

⎛ ⎞ ⎛= + +⎜ ⎟ ⎜⎝ ⎠ ⎝

⎞⎟⎠

[1 mark]

# Equivalent expression also accepted.

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(ii) Estimate the factor of safety for the following conditions: d = 50 mm, D = 100 mm, r = 4 mm, L = 100 mm, F = 4000 N, m = 1000 kg, P = 50 kW, and N = 3000 rpm. Assume the shaft to be made from commercially polished steel with an ultimate tensile strength, Su, of 1200 MPa.

First of all we need to calculate the respective Kf values.

Theoretical stress concentration factors for a shaft with a fillet: (a) bending, (b) axial, and (c) torsional loads (R. E. Peterson, in Stress concentration design factors, John Wiley & Sons, Inc., 1953).

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r/d = 4/50 = 0.08 D/d = 100/50 = 2.0 From the previous figure: Kt(b) ≈ 1.82 Kt(a) ≈ 2.14 [1½ marks] Kt(t) ≈ 1.53 We also need to calculate the Peterson notch sensitivity factor, q:

Notch-sensitivity curves for use with theoretical factors, Kt (Adapted from R. E. Peterson, “Notch sensitivity”, in Metal fatigue, G. Sines and J. L. Waisman, eds., McGraw-Hill Book Company, New

York, 1959).

r = 4 mm Su = 1200 MPa ⇒ q(b) ≈ 0.93 q(a) ≈ 0.93 [1½ marks] q(t) ≈ 0.95

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We also need to know the surface finish:

Reduction of endurance strength due to surface finish – steel parts (Adapted from R. C. Juvinall, p. 234 in Engineering considerations of stress, strain and strength, McGraw-Hill, Inc., New York, 1967).

⇒ CS ≈ 0.88 [½ mark] We need to make use of the following equation:

( )1 1f t SK K qC= + − [1 mark]

∴ Kf(b) ≈ 1 + (1.82 – 1)(0.93)(0.88) ≈ 1.67(1)

∴ Kf(a) ≈ 1 + (2.14 – 1)(0.93)(0.88)

≈ 1.93(3) [1½ marks]

∴ Kf(t) ≈ 1 + (1.53 – 1)(0.95)(0.88) ≈ 1.44(3)

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( )3

32a f b

mgL Kd

σπ

=

( )

3

33

32 1000 9.81 100 10 1.67150 10

ax x x x

xσ

π

−

−=

⇒ σa = 133.6 MPa [½ mark]

( ) ( ) ( )

2 2

2 2 2 3

2 2 480m f a f a f t

F F PK K Kd d N d

σπ π π

⎛ ⎞ ⎛= + +⎜ ⎟ ⎜⎝ ⎠ ⎝

⎞⎟⎠

( ) ( ) ( )

2 24

2 23 3 2 3

2 4000 2 4000 480 5 101.933 1.933 1.44350 10 50 10 3000 50 10

mx x x x

x x x x xσ

π π π− − −

⎛ ⎞ ⎛⎜ ⎟ ⎜= + +⎜ ⎟ ⎜⎝ ⎠ ⎝

3

⎞⎟⎟⎠

)

( ) (2 26 61.969 10 1.969 10 9.357 10m x x xσ = + + 6

⇒ σm = 11.53 MPa [½ mark] Construct a Goodman line:

Load type Factor

Bending Torsion Axial CL 1.0 0.58 0.9*

CD1.0 for D ≤ 10 mm

0.9 for 10 ≤ D ≤ 50 mm 1.0

CS From figure shown below

CL = 1 [1 mark]#

CD = 0.9 Sn = CLCDCSSn’ [1 mark]*

Assume Sn’ = 0.5 Su [½ mark]

# The loading type is always assumed to be fully reversed bending (i.e., CL = 1) when constructing the Goodman line. * The specimen is assumed to be stress-free (i.e., Kf = 1) when constructing the Goodman line.

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∴ Sn = CLCDCS(0.5 x Su) Sn = (1)(0.9)(0.88)(0.5 x 1200) ⇒ Sn = 475.2 MPa

Sn (475.2 MPa)

σmO Su (1200 MPa)

σa

11.53 MPa

133.6 MPa

[2½ marks]#

We can obtain the following equations:

475.2475.21200a mσ σ= −

⇒ 475.2 0.396a mσ σ= − (i) [1 mark]

133.611.53a mσ σ=

⇒ 11.59a mσ σ= (ii) [1 mark] Solving these equations gives:

σm = 39.66 MPa [1 mark]

∴ Factor of safety 39.6611.53

= [1 mark]

⇒ Factor of safety = 3.44

# Marks of ½ each for: (i) σm on x-axis, (ii) σa on y-axis, (iii) labelling of Su or 1200 MPa on x-axis, (iv) labelling of Sn or 475.2 MPa (or similar), (v) drawing of line between Sn and Su.

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Materials 337

Mid-Semester Test 2008 (a) The fatigue surface finish factor, Cs, is not included in the equation for the one thousand

(103) cycle data point. Briefly explain the reason behind this omission? (1 Mark)

(b) Name three (3) characteristics or properties that are required in order to determine the Peterson notch sensitivity factor, q ?

(2 Marks) (c) A square steel plate of width, w, and thickness, t, contains a circular hole of radius, r, at its

centre (w >> r > t). The plate is subjected to axial and shear stresses as shown in Figure 1 overleaf.

(i) When analysing the possible failure of this component it is stated that the presence of

the hole will have negligible influence on the stress state for any region at a distance, X, from the hole centre where X > Nr and N is typically 5. What engineering principle may be invoked to justify this decision?

(1 Mark) (ii) Sketch a Mohr’s circle to represent the stress state of the plate at a position far away

from the hole, i.e., unaffected by the presence of the hole. (Note: A hand drawn Mohr’s circle is satisfactory)

(4 Marks) (iii) It is a requirement that the maximum strain at the edge of the hole should be

estimated using a strain gauge. Use the Mohr’s circle to estimate at what angle, θ, the strain gauge should be placed.

(4 Marks) (iv) Using approximate values for the mechanical properties of the plate, estimate the

value of the maximum strain measured by the strain gauge. Assume that the strain gauge is “small” and located exactly at the edge of the hole.

(5 Marks)

(v) Use the maximum distortion strain energy theory (von Mises theory) to estimate the factor of safety assuming an ultimate tensile strength, σu, of 1500 MPa.

(3 Marks)

(Total 20 Marks)

- 2 -

Figure 1: Schematic diagram of a square steel plate containing a hole together with a strain gauge (shown in red) attached at the hole edge

150 MPa

400 MPa

θ

50 MPa

- 3 -

Mid-Semester Test 2008 (Solutions) (a) The fatigue surface finish factor, Cs, is not included in the equation for the one thousand

(103) cycle data point. Briefly explain the reason behind this omission?

The stress required to produce failure after 103 cycles is sufficiently high (on the order of 0.9σu) that the surface is being damaged to such an extent (e.g., plastic deformation if the stress is above σy) that the initial state of the surface becomes unimportant.

(1 Mark)

(b) Name three (3) characteristics or properties that are required in order to determine the Peterson notch sensitivity factor, q ?

If we consider the information required to determine q from the standard figure then any 3

from: (i) notch radius, (ii) material type (steel or aluminium), (iii) loading type, and (iv) ultimate tensile strength, σu.

(2 Marks) (c) A square steel plate of width, w, and thickness, t, contains a circular hole of radius, r, at its

centre (w >> r > t). The plate is subjected to axial and shear stresses as shown in Figure 1 overleaf.

(i) When analysing the possible failure of this component it is stated that the presence of

the hole will have negligible influence on the stress state for any region at a distance, X, from the hole centre where X > Nr and N is typically 5. What engineering principle may be invoked to justify this decision?

Saint-Venant’s Principle

(1 Mark) (ii) Sketch a Mohr’s circle to represent the stress state of the plate at a position far away

from the hole, i.e., unaffected by the presence of the hole. (Note: A hand drawn Mohr’s circle is satisfactory)

(4 Marks)

(275, 0)

+ve

+ve -ve

(150, 50)

(400, -50)

-ve

σx1 (MPa)

τx1y1 (MPa)

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(iii) It is a requirement that the maximum strain at the edge of the hole should be estimated using a strain gauge. Use the Mohr’s circle to estimate at what angle, θ, the strain gauge should be placed.

Strictly speaking, the correct answer is that no angle will give the maximum

strain for the strain gauge configuration shown (as the strain gauge is measuring the radial strain, σr, whereas we have generally relied upon measuring the tangential strain, σθ (as σθ>>σr for most stress concentrations around a hole)).

(4 Marks) or

Alternatively, it would be sensible to put the strain gauge at the angle of maximum principal stress, σ1.

(2 Marks)

From the Mohr’s circle we can estimate that 2θ ≅ 158.2o

(1 Mark) ⇒ θ ≅ 79.1o

(1 Mark)

(iv) Using approximate values for the mechanical properties of the plate, estimate the value of the maximum strain measured by the strain gauge. Assume that the strain gauge is “small” and located exactly at the edge of the hole.

The stress state is biaxial so we need to use the following equation (from Lecture 2): (½ Mark)

For steel a typical value for E would be 200 GPa (±10%) and ν would be 0.25∼0.3.

(½ Mark)

σx1 (MPa) (275, 0)

+ve

+ve -ve

(150, 50)

(400, -50)

-ve

2θ

σ1

σ2

τx1y1 (MPa)

( )1x x yE

ε σ νσ= −

- 5 -

In this case: σx = σ1 and σy = σ2

(1 Mark)

From the Mohr’s circle we can estimate σ1 and σ2 as follows: σ1 ≅ 409.6 MPa σ2 ≅ 140.4 MPa

(½ Mark) For a square plate containing a hole with w >> r then (from Lecture 4): Kt → 3

(1 Mark)

Therefore the respective stresses at the edge of the hole would be: ⇒ σ1 ≅ 409.6 x 3 σ1 ≅ 1229 MPa σ2 ≅ 140.4 x 3 σ2 ≅ 421.1 MPa

(½ Mark)

Therefore, the estimated strain will be given by:

⇒ εx = 5.51 x 10-3

(1 Mark) Note that inclusion of σ1 only results in εx = 6.15 x 10-3

(v) Use the maximum distortion strain energy theory (von Mises theory) to estimate the factor of safety assuming an ultimate tensile strength, σu, of 1500 MPa.

We need to use the following equation: 21

22

21 σσσσσ −+=equiv

(1 Mark)

Substituting in the appropriate values: 2 21229 421.1 1229 421.1equiv xσ = + − ⇒ σequiv = 1082 MPa

( ) 69

1 1229 0.3 421.1 10200 10x x

xε = − ⋅

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(1 Mark)

Thus, our specimen contains the same amount of distortion strain energy as for a uniaxial tensile specimen at 1082 MPa.

The factor of safety can thus be calculated from:

Factor of safety = u

equiv

σσ

= 15001082

= 1.387

(1 Mark)