5.failures resulting from static loading

8/10/2019 5.Failures Resulting From Static Loading

1/56

MECopyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 5

FailureResulting from

Static Loading


2/56

ME Failure Resulting from Static Loading

- Strengthis aproperty or characteristic of a mechanical element. This property

results from the material identity, the treatment and processing incidental to

creating its geometry, and the loading, and it is at the controlling or critical

location. In addition to considering the strength of a single part, we must becognizant that the strengths of the mass-produced parts will all be somewhat

different from the others in the collection or ensemble because of variations in

dimensions, machining, forming, and composition.

- A static loadis astationary forceor couple applied to a member. To be

stationary, the force or couple must be unchanging in magnitude, point or points

of application, and direction. A static load can produce axial tension or

compression, a shear load, a bending load, a torsional load, or any combination of

these. To be considered static, the load cannot change in any manner.


3/56

ME Failure of machine components

- In this chapter we consider the relations between strength and static loading in

order to make the decisions concerning material and its treatment, fabrication,

and geometry for satisfying the requirements of functionality, safety, reliability,

competitiveness, usability, manufacturability, and marketability.

-Failurecan mean a part has separated into two or more pieces; has become

permanently distorted, thus ruining its geometry; has had its reliability

downgraded; or has had its function compromised, whatever the reason. A

designer speaking of failure can mean any or all of these possibilities.

- In this chapter our attention is focused on the predictability of permanent

distortion or separation.


4/56


Fail ure of a truck drive

shaf t spline due to

corrosion fatigue.

Impact fai lure of a lawn mower blade

driver hub. The blade impacted a

sur veying pipe marker.


5/56


Failure of an overhead-pulley retaining bolt on a weightl i f ting

machine. A manufactur ing error caused a gap that forced the bolt

to take the entire moment load.


6/56


Chain test f ixture that failed in one cycle. To alleviate complaints of

excessive wear, the manufactur er decided to case-harden the material.

(a) Two halves showing f racture; this is an excellent example of bri ttle

fracture ini tiated by stress concentration.

(b) Enlarged view of one portion to show cracks induced by stress

concentration at the support-pin holes.


7/56ME Failure of machine components

Valve-spring failur e caused by

spring surge in an oversped

engine. The fr actures exhibit the

classic 45shear failure.


8/56MEStatic Strength

- In designing any machine element, the engineer should have available the

results of a great many strength tests of the particular material chosen. These tests

should be made on specimens having the same heat treatment, surface finish, and

size as the element the engineer proposes to design; and the tests should be made

under exactly the same loading conditions as the part will experience in service.

- The cost of gathering such extensive data prior to design is justified if failure of

the part may endanger human life or if the part is manufactured in sufficiently

large quantities.

- More often than not it is necessary to design using only published values of

yield strength, ultimate strength, percentage reduction in area, and percentage

elongation.

- How can one use such poor data to design against both static and dynamicloads, two- and three-dimensional stress states, high and low temperatures, and

very large and very small parts?


9/56MEStress Concentration

- Stress concentrationis a highly localized

effect. In some instances it may be due to a

surface scratch. If the material is ductile and

the load static, the design load may cause

yielding in the critical location in the notch.This yielding can involve strain strengthening

of the material and an increase in yield

strength at the small critical notch location.

- Since the loads are static and the material isductile, that part can carry the loads

satisfactorily with no general yielding. In

these cases the designer sets the geometric

(theoretical)stress concentration factorKtto

unity.

An idealized stress-strain curve. The dashed

l ine depicts a strain-strengthening material.


10/56MEStress Concentration

- Designers do not applyKtinstatic loadingof a ductile materialloaded

elastically, instead settingKt= 1.

- When using this rule for ductile materials with static loads, be careful to assure

yourself that the material is not susceptible to brittle fracture in the environmentof use.

- The usual definition of geometric (theoretical)stress-concentration factorfor

normal stress Ktandshear stress Ktsis

nomtK max nomtsK max

- An exception to this rule is a brittle material that inherently contains

microdiscontinuity stress concentration, worse than the macrodiscontinuity that the

designer has in mind. Sand molding introduces sand particles, air, and water vapor

bubbles. The grain structure of cast iron contains graphite flakes, which are literallycracks introduced during the solidification process. The strength of a cast iron

reported in the literature includes this stress concentration. In such casesKtorKts

need not be applied.


11/56ME Failure Theories

- If thefailure mechanismis simple, then simple tests can give clues. Just what is

simple? The tension test is uniaxial (thats simple) and elongations are largest in

the axial direction, so strains can be measured and stresses inferred up to

failure. Just what is important: a critical stress, a critical strain, a criticalenergy?

- Next, failure theories will be shown that have helped answer some of these

questions. Unfortunately, there is no universal theory of failure for the general

case of material properties and stress state. Instead, over the years several

hypotheses have been formulated and tested, leading to todays accepted

practices. These practices will be charactersied as theories.


12/56ME Failure Theories

Structural metal behavior is typically classified as being ductile or brittle.

Ductile :f 0.05

have an identifiable yield strength (often Syt= Syc= Sy)

Brittle :f < 0.05do not exhibit an identifiable yield strength

classified by Sut, Suc

Ductile materials (yield criteria)

- Maximum shear stress (MSS)- Distortion energy (DE)

- Ductile Coulomb-Mohr (DCM)

Brittle materials (fracture criteria)

- Maximum normal stress (MNS)

- Brittle Coulomb-Mohr (BCM)

- Modified Mohr (MM)


13/56ME


14/56ME Maximum-Shear-Stress Theory (MSS) for ductile material

The maximum-shear-stresstheory predicts that yielding begins whenever the

maximum shear stress in any element equals or exceeds the maximum shear stress

in a tension test specimen of the same material when that specimen begins to yield.

TheMSS theoryis also referred to as the Trescaor Guesttheory.

As a strip of a ductile material is subjected to tension, slip lines (calledLder

lines) form at approximately 45 with the axis of the strip. These slip lines are the

beginning of yield, and when loaded to fracture, fracture lines are also seen at

angles approximately 45 with the axis of tension. Since the shear stress is

maximum at 45 from the axis of tension, it makes sense to think that this is the

mechanism of failure.

However, it turns out the MSS theory is an acceptable but conservative predictor of

failure; and since engineers are conservative by nature, it is quite often used.



45

ductile brittle

F

F F

F

sI



For simple tensile test, the maximum shear stress at yield is

The MMS theory predicts yielding when

This implies that

The design equation

2

31

max

22

31

max

yS yS 31

ysy S.S 50

For a general state of stress, three principal stresses can be ordered that 1 2 3

n

S

y

2max

n

S

y 31

2max

yS

or

or


17/56

ME Maximum-Shear-Stress Theory (MSS) for ductile material

The MSS theory for plane

stress, where Aand Bare the

two nonzero pri ncipal stresses.

For plane stress (one of the principal stresses

is zero and )BA

A 1Case 1:

yA S

0 BA 03

Case 2:

Case 3:

A 1BA 0 B 3

yBA S

01

BA 0 B 3

yB S

Shaft design problems typically fall into

case 2 where a normal stress exists from

bending and/or axial loading, and a shear

stress arises from torsion (sx, txy, sy= 0)


18/56

ME Distortion-Energy Theory (DE) for ductile material

The distortion-energy theorypredicts that yielding occurs when the distortion

strain energy per unit volume reaches or exceeds the distortion strain

energy per unit volume for yield in simple tension or compression of the

same material

Yielding of ductile materials was not a simple tensile or compressive

phenomenon at all, but, rather, that it was related somehow to the angular

distortion of the stressed element.

The stress tensor can be divided into 2 components:

a) Hydrostatic component

b) Distortional component


19/56


Due to the stress sav

acting in each of the

same principal direction.

3

321

av

The element undergoes

pure volume change, no

angular distortion.

This element subjected to

pure angular distortion, no

volume change.


20/56


The strain energy per unit volume for simple tension is

For the element shown in the last page the strain energy per unit volume is

u

2

1

332211

2

1u


21/56


Strain energy per unit volume for general state of stress can be found from

Strain energy for producing only volume uvchange can be obtained by substitute

1= 2= 3= av

Then the distortion energy is obtained by

For yielding in simple tension; 1= Sy, 2= 3= 0

133221

2

3

2

2

2

1 2

2

1

Eu

133221

2

3

2

2

2

1 222

6

21

E

uv

23

1 2

13

2

32

2

21

E

uuu vd

Note that the distortion energy is zero if 1= 2= 3.

2

3

1yd S

E

u


22/56


Yielding is predicted when

where the von Mises stress is

yS

2122 /BBAA

For plane stress (1= A, 2= B, 3= 0)

21

2

13

2

32

2

21

2

/

The left of this equation can be thought of as asingle, equivalent, or effective

stressfor the entire general state of stress given by 1, 2,and 3. This effective

stress is usually called the von Mises stress, ,or it can be written as

y

/

S

21

2

13

2

32

2

21

2


23/56


Using xyz components of 3D stress, the von Mises stress can be written as

21222 3 /xyyyxx

For plane stress

21222222 62

1 /

zxyzxyxzzyyx

The DE theory for plane stress states.

The failure surface for DE is a

circular cylinder with an axis

inclined at 45 from each principal

stress axis, whereas the surface for

MSS is a hexagon inscribed withinthe cylinder.


24/56


- Consider an isolated element, in which the normal stresses on each surface are

equal to the hydrostatic stress av. There are eight surfaces symmetric to the

principal directions that contain this stress. This forms an octahedron. The shear

stresses on these surfaces are equal and are called the octahedral shear stresses.

Through coordinate transformations the octahedral shear stress is given by

21213

2

32

2

21

3

1 /

oct

yoct S3

2

- By comparing two above equations, yielding is

predicted when

- For yielding in simple tension; 1= Sy, 2= 3= 0

y

/

S

21

2

13

2

32

2

21

2


25/56


For the case of pure shear xyin plane stress x= y= 0

Principal stresses are 1= xy, 2= - xy

Thus the shear yield strength predicted by DE theory is

The design equation is

y/

xy S 212

3 yy

xy S.S

5770

3

ysy S.S 5770

n

S

y

or

The DE theory predicts no failure under hydrostatic stress and agrees well with all

data for ductile behavior. Hence, it is the most widely used theory for ductile

materials and is recommended for design problems unless otherwise specified.

is about 15 percent greater than the 0.5 Sypredicted by

the MSS theory.


26/56


The model for theMSStheory ignores the contribution of the normal stresses onthe 45 surfaces of the tensile specimen. However, these stresses areP/2A, and

notthe hydrostatic stresses which areP/3A. Herein lies the difference between

theMSSandDEtheories.


27/56

ME


28/56

ME MMS and DE theory for ductile material


29/56



30/56



31/56


F i l i ld h i


32/56

ME Coulomb-Mohr (CM) Theory for ductile material

For some materials, yield strength in

tension and compression are not

identical, e.g., magnesium (Syc

0.5Syt), gray cast iron (Suc 3-4Syt)

Mohrs hypothesiswas to use theresults of tensile, compression

and torsion shear test to construct

the three Mohrs circles.

Line ABCDE is calledfailureenvelope. It needs not be straight

but may be circular or quadratic.

Tension

Compression

Pure shear

Failure envelope

The three Mohr circles describe the

stress state in a body growing during

loading until one of them becametangent to the failure envelope,

thereby definingfailure. (yielding

occurs)

Three Mohr cir cles (uniaxial compression test,

test in pure shear, and the uniaxial tension test)

are used to define failure by the Mohrhypothesis. The strengths Scand Stare the

compressive and tensile strengths, respectively;

they can be used for yield or ultimate strength.


33/56

SSS


34/56


Simplifying above equation gives

For pure shear t, s1= -s3= t. The torsional yield strength occurs when tmax=Ssy.It can be obtained by substituting 1= -3=Ssy

The design equation is

131 ct S

S

22

22

22

22

31

31

tc

tc

t

t

SS

SS

S

S

ycyt

ycyt

sySS

SSS

nS

S

ct

131


35/56


Plot of the Coulomb-Mohr theory

of failur e for plane stress states.

For plane stress ( ) the Coulomb-

Mohr theory provides the hexagon as shown

BA

0 BA

A 1Case 1:

tA S

0 BA 03

Case 2:

Case 3:

A

1BA 0

B

3

1

c

B

t

A

S

S

01

BA 0 B 3

cB S

Case 1

Case 2

Case 3


36/56

ME


37/56

ME Coulomb-Mohr Theory for ductile material


38/56

ME Failure of ductile material (summary)

- The selection of one or the other

of these two theories is something

that you, the engineer, must

decide.

- von Mises theory passes closer

to the central area of the data. It is

the most precise criteria to predict

yielding of ductile materials.

While the Tresca theory is thesimplest to be used. Exper imental data

superposed on failur e

theories.

- For ductile materials with unequal yield strengths, Sytand Syc, the Mohr

theory is the best available. However, the theory requires the results fromthree separate modes of tests. The alternative to this is to use the Coulomb-

Mohr theory, which requires only the tensile and compressive yield

strengths and is easily to be dealt with.


39/56

ME Failure of ductile material


40/56


plane stress:


41/56


Plane stress Eq. (3-14):


42/56



43/56


plane stress:


44/56



45/56



46/56

ME Maximum-Normal-Stress Theory for brittle materials

The Maximum-Normal-Stress (MNS)theory states thatfailure occurs

whenever one of the three principal stresses equals or exceeds the strength.

For a general stress state in the ordered form

1 2 3,the failure occurs when

Graph of maximum-normal-stress (MNS) theory of failure for plane

stress states. Stress states that plot inside the fai lure locus are safe.

utS 1 ucS 3

utA S

or

For plane stress with A B,the failure

occurs when

ucB S or

ultimate

tensile

strength

ultimate

compression

strength


47/56

ME Maximum-Normal-Stress Theory for brittle materials

The design equations:

B ittl C l b M h (BCM)


48/56

ME Modification of the Mohr Theory for brittle materials

*These theories is restricted to plane stress and be of design type incorporating the factor of safety.

Brittle-Coulomb-Mohr (BCM)

n

S

utA

nS

S

uc

B

ut

A 1

n

S ucB

0 BA

BA 0

BA 0

Biaxial fr acture data of gray cast ironcompared with var ious fai lu re cri ter ia.

Case 1

Case 2

Case 3

M difi d M h (MM)


49/56


*These theories is restricted to plane stress and be of design type incorporating the factor of safety.

Modified-Mohr (MM)

n

S

utA

nS

SS

SS

uc

B

utuc

Autuc 1

n

S

ucB

0 BA

BA 0

BA 0

Case 1

Case 2

Case 3

BA 0

and 1A

B

and 1A

B

Biaxial fr acture data of gray cast iron

compared with var ious fail ure cri ter ia.Case 4


50/56

ME

31When ss ucut

SSNModified Mohr theory Normal stress theory


51/56

ME

mucut Nss

131

ss

31

311

When)(

ss

sss

utuc

mm

SSN

31

1

When sss

utmms

N

Modified Mohr theory

Coulomb Mohr theory

Normal stress theory

for brittle material


52/56



53/56


Eq. (3-13):


54/56


- Ductile materials may develop a


55/56

ME Failure of brittle materail (Summary)

y p

brittle fracture or crack if used

below the transition temperature.

- In the 1stquadrant the data appear

on both sides and along the failure

curves of MNS, CM and MM. All

failure curves are the same and fit

well.

- In the 4th

quadrant the MM theoryrepresents the data the best.

- In the 3rdquadrant the points data

are too few to make any suggestion

concerning a fracture locus.

A plot of exper imental data points obtained from

biaxial tests on cast i ron. Shown also are the graphs

of three failure theor ies for br ittle mater ials.


56/56

5.failures resulting from static loading

Documents