mechanics of materials ii n.
Download
Skip this Video
Download Presentation
Mechanics of Materials II

Loading in 2 Seconds...

play fullscreen
1 / 80

Mechanics of Materials II - PowerPoint PPT Presentation


  • 127 Views
  • Uploaded on

Mechanics of Materials II. UET, Taxila Lecture No. (4&5). 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. Failure under monotonic loading.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Mechanics of Materials II' - robert-page


Download Now An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
mechanics of materials ii

Mechanics of Materials II

UET, Taxila

Lecture No. (4&5)

mechanisms of material failure
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.

failure under monotonic loading
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:

slide4
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)

slide5
Materials are normally classified loosely as either `brittle’  or `ductile’ depending on the characteristic features of the failure. 

Brittle and Ductile failures

brittle materials
Brittle Materials

Examples of `brittle’ materials include:

glasses,

ceramics (Oxides, Carbides & Nitrides)

and Cast Iron

features of a brittle material are
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.

slide8
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

ductile materials
Ductile Materials

Examples of `ductile’

Tin and lead

FCC metals at all temperatures;

BCC metals at high temperatures;

polymers at relatively high temperature. 

slide10
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

slide11
Fracture surfaces don’t fit together.

The fracture surface has a dimpled appearance – you can see little holes

slide12
Complex Materials

Of course, some materials have such a complex microstructure (especially composites) that it’s hard to classify them as entirely brittle or entirely ductile.

slide13
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.

how ductile fracture occurs
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.

slide18
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.
creep and fatigue
Creep and Fatigue

In the preceding sections it has been suggested that failure of materials occurs when the ultimate strengths have been exceeded.

definition of creep
Definition of creep

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

creep at high temperature
Creep at high temperature

At elevated temperaturessome materials (most metals) are susceptible to this phenomenon and even under the constant load mentioned, strains can increase continually until fracture.

applications of creep
Applications of creep

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

in previous figure
In previous figure:

The general form of the strain versus time graph or creep curveis shown for two typical operating conditions.

four features
Four features

In each case the curve can be considered to exhibit four principal features:

slide26
(a) An initial strain, due to the initial application of load. In most cases this would be an elastic strain.

(b) A primary creepregion, during which the creep rate (slope of the graph) diminishes.

slide27
(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.

design for creep
Design for creep

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

fatigue
Fatigue

Definition of Fatigue

Fatigueis the failure of a material under fluctuating stresses each of which is believed to produce minute amounts of plastic strain.

applications of fatigue
Applications of fatigue

Fatigue is particularly important in components subjected to repeated and often rapid loadfluctuations, e.g. aircraft components, turbine blades, vehicle suspensions, etc.

representation of fatigue
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.

fatigue limit
Fatigue limit

The particularly relevant feature of this curve is the limiting stress Snsince it is assumed that stresses below this value will not produce fatigue failure however many cycles are applied, i.e. there is infinite life.

design for fatigue
Design for fatigue

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

solved
Solved

Examples

slide39
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.

example 2
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.

slide55
(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.

slide56
Let the required bore diameter be ‘d’ mm; the cross-sectional area of the bar will then be reduced to:

Solution (b)

slide64
(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

slide65
The change in the outside diameter of the bar will be obtained from the lateral strain,

Solution (C)

slide68

d/d

σ/E

slide72
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:

slide73
(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.

solution1
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:

slide79
(c) The force in the coupling is shared by the forked end pieces, each being subjected to a direct stress