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Mechanics of Materials II

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Title: Mechanics of Materials II


1
Mechanics of Materials II
  • UET, Taxila
  • Lecture No. (45)

2
Mechanisms of material failure
  •  In order to understand the various approaches to
    modeling fracture, fatigue and failure, it is
    helpful to review briefly the features and
    mechanisms of failure in solids.
  •  

3
Failure under monotonic loading
  • If you test a sample of any material under
    uni-axial tension it will eventually fail.  The
    features of the failure depend on several
    factors, including
  •   

4
  • 1-The materials involved and their
    micro-sctructure
  •    2-The applied stress state
  •    3- Loading rate
  •    4-Temperature
  •    5-Ambient environment (water vapor or
    presence of corrosive environments)

5
Brittle and Ductile failures
  • Materials are normally classified loosely as
    either brittle  or ductile depending on the
    characteristic features of the failure. 

6
Brittle Materials
  • Examples of brittle materials include
  • glasses,
  • ceramics (Oxides, Carbides Nitrides)
  • and Cast Iron

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Features of a brittle material are
  • 1-Very little plastic flow occurs in the specimen
    prior to failure
  • 2-The two sides of the fracture surface fit
    together very well after failure. 
  •  

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  • 3- The fracture surface appears faceted you can
    make out individual grains and atomic planes.
  • 4- In many materials, fracture occurs along
    certain crystallographic planes.  In other
    materials, fracture occurs along grain boundaries

9
Ductile Materials
  • Examples of ductile
  • Tin and lead
  • FCC metals at all temperatures
  • BCC metals at high temperatures
  • polymers at relatively high temperature.   

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  • Features of a ductile fracture are
  • Extensive plastic flow occurs in the material
    prior to fracture.
  • There is usually evidence of considerable necking
    in the specimen

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  • Fracture surfaces dont fit together.
  • The fracture surface has a dimpled appearance
    you can see little holes

12
  • Complex Materials
  • Of course, some materials have such a complex
    microstructure (especially composites) that its
    hard to classify them as entirely brittle or
    entirely ductile.
  •  

13
  • How Brittle Fracture occurs
  • Brittle fracture occurs as a result of a single
    crack, propagating through the specimen.  Most
    materials contain pre-existing cracks, in which
    case fracture is initiated when a large crack in
    a region of high tensile stress starts to grow. 

14
How ductile fracture occurs
  • Ductile fracture occurs as a result of the
    nucleation, growth and coalescence of voids in
    the material
  • Failure is controlled by the rate of nucleation
    of the voids and their rate of growth.

15
Schematic Representation for Ductile Brittle
Fracture
16
  • Does material fail under stresses lower than
    yield strength?

17
  • The answer is yes

18
  • This chapter would not be complete, therefore,
    without reference to certain loading conditions
    under which materials can fail at stresses much
    less than the yield stress, namely creep and
    fatigue.

19
Creep and Fatigue
  • In the preceding sections it has been suggested
    that failure of materials occurs when the
    ultimate strengths have been exceeded.

20
Definition of creep
  • Creep is the gradual increase of plastic strain
    in a material with time at constant load.

21
Creep at high temperature
  • At elevated temperatures some materials (most
    metals) are susceptible to this phenomenon and
    even under the constant load mentioned, strains
    can increase continually until fracture.

22
Applications of creep
  • This form of fracture is particularly relevant to
    turbine blades, nuclear reactors, furnaces,
    rocket motors, etc.

23
Typical creep curve.
24
In previous figure
  • The general form of the strain versus time graph
    or creep curve is shown for two typical operating
    conditions.

25
Four features
  • In each case the curve can be considered to
    exhibit four principal features

26
  • (a) An initial strain, due to the initial
    application of load. In most cases this would be
    an elastic strain.
  • (b) A primary creep region, during which the
    creep rate (slope of the graph) diminishes.

27
  • (c) A secondary creep region, when the creep rate
    is sensibly low (Constant).
  • (d) A tertiary creep region, during which the
    creep rate accelerates to final fracture.

28
Design for creep
  • It is clearly vital that a material which is
    susceptible to creep effects should only be
    subjected to stresses which keep it in the
    secondary (straight line) region throughout its
    service life.

29
  • This enables the amount of creep extension to be
    estimated and allowed in design.

30
Fatigue
  • Definition of Fatigue
  • Fatigue is the failure of a material under
    fluctuating stresses each of which is believed to
    produce minute amounts of plastic strain.

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Applications of fatigue
  • Fatigue is particularly important in components
    subjected to repeated and often rapid load
    fluctuations, e.g. aircraft components, turbine
    blades, vehicle suspensions, etc.

33
Representation of Fatigue
  • Fatigue behaviour of materials is usually
    described by a
  • fatigue life or S-N curve
  • in which the number of stress cycles N to produce
    failure
  • is plotted against S.

34
  • A typical S-N curve for mild steel is shown in
    next figure

35
S-N Curve
36
Fatigue limit
  • The particularly relevant feature of this curve
    is the limiting stress Sn since it is assumed
    that stresses below this value will not produce
    fatigue failure however many cycles are applied,
    i.e. there is infinite life.

37
Design for fatigue
  • In the simplest design cases, therefore, there is
    an aim to keep all stresses below this limiting
    level.

38
Solved
  • Examples

39
  • Example 1
  • Determine the stress in each section of the bar
    shown in next Figure when subjected to an axial
    tensile load of 20 kN. The central section is 30
    mm square cross-section the other portions are
    of circular section, their diameters being
    indicated.
  • What will be the total extension of the bar? For
    the bar material E 210GN/m2.

40
Figure (1)
41
Solution
42
Stress in section (1)
43
Stress in section 2
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Stress in section 3
45
  • Now the extension of a bar can always be written
    in terms of the stress in the bar since

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Example 2
  • (a) A 25 mm diameter bar is subjected to an axial
    tensile load of 100 kN. Under the action of this
    load a 200mm gauge length is found to extend by
    the distance
  • 0.19 x 10-3 mm.
  • Determine the modulus of elasticity for the bar
    material.

52
Solution (a)
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  • (b) If, in order to reduce weight whilst keeping
    the external diameter constant, the bar is bored
    axially to produce a cylinder of uniform
    thickness, what is the maximum diameter of bore
    possible, given that the maximum allowable stress
    is 240MN/m2?
  • The load can be assumed to remain constant at 100
    kN.

56
Solution (b)
  • Let the required bore diameter be d mm the
    cross-sectional area of the bar will then be
    reduced to

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  • But this stress is restricted to a maximum
    allowable value of 240 MN/m2.

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  • The maximum bore possible is thus 9.72 mm.

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  • (c) What will be the change in the outside
    diameter of the bar under the limiting stress
    quoted in (b)?
  • Where E 210 GN/m2
  • and ? 0.3

65
Solution (C)
  • The change in the outside diameter of the bar
    will be obtained from the lateral strain,

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?d/d
s/E
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  • Example 3
  • The coupling shown in next Figure is constructed
    from steel of rectangular cross-section and is
    designed to transmit a tensile force of 50 kN. If
    the bolt is of 15 mm diameter calculate

73
  • (a) the shear stress in the bolt
  • (b) the direct stress in the plate
  • (c) the direct stress in the forked end of the
    coupling.

74
Double Shear
75
Solution
  • (a) The bolt is subjected to double shear,
    tending to shear it as shown in Figure. There is
    thus twice the area of the bolt resisting the
    shear and from equation

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v
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  • (c) The force in the coupling is shared by the
    forked end pieces, each being subjected to a
    direct stress

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Problems
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