lecture 8 stress concentration class

8/9/2019 Lecture 8 Stress Concentration Class

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Mechanical Response of Engineering

Materials: EMch 315

Stress Concentration Factors Lecture 8

Chapter 2.11 : Mechanical Response of Engineering Materials


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Stress Concentration Factors

Geometric discontinuities cause an object to

experience a local increase in the intensity of a

stress field. The examples of shapes that cause theseconcentrations are: cracks, sharp corners, holes and,

sudden changes in the cross-sectional area of the

object. High local stresses can cause the object to

fail more quickly than if they were not there.

Engineers must design the geometry to minimize

such stress concentrations.


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http://en.wikipedia.org/wiki/Stress_concentration

Stress Concentration FactorsA hole in a component)


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De Havilland Comet


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Two de Havilland Comet passenger jets broke up in mid-air and

crashed within a few months of each other in 1954.

As a result systematic tests were conducted on a fuselage immersed

and pressurized in a water tank. After the equivalent of 3,000 flightsinvestigators at the Royal Aircraft Establishment (RAE) were able

to conclude that the crash had been due to failure of the pressure

cabin at the forward Automatic Direction Finder window in the roof.

The failure was a result of metal fatigue caused by the repeated pressurization and de-pressurization of the aircraft cabin.

Another fact was that the supports around the windows were riveted,

not bonded, as the original specifications for the aircraft had called

for. The problem was exacerbated by the punch rivet construction

technique employed. Unlike drill riveting, the imperfect nature of the

hole created by punch riveting caused manufacturing defect cracks

(due to stress concentrations) which may have caused the start of

fatigue cracks around the rivet.


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Stress Concentrations

PP

W

t

Stress uniformly

distributed on

cross-section

Consider a bar loaded with auniaxial force P

s = =P P

A Wt


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If we were to reduce the cross-section of the bar by drilling a hole

through it,

Stress Concentration

PP

t

W

2r

We might calculate the stress in the bar to be: s = __________

The normal area is Wt and we reduced this area by 2rt.

Area lost by drilling hole


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This calculation assumes a uniform

stress distribution.


In reality, the stress distribution is not uniform -- high stress is present

at the edge of the hole.

hole

holex


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Calculations based on uniform distribution of stress members having stress

concentrators have led to disastrous failures for highly stressed components. The

error is that the stress at the edge of the hole is not the nominal stress it is a muchhigher stress. This higher stress is obtained with one of the following equations and

knowledge of the Elastic stress concentration factor K s.


Where sw is the far field stress and snom is the nominal stress.The stress concentration factor is a function of the geometry of the

stress concentrator.

(uses net cross-section)

(uses gross cross-section)

Specific values of K for various geometries are generally reported in

handbooks related to Stress analyses.


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Where to find Stress Concentration Factors (K)

Peterson’s Stress

Concentration

Factors, 2nd edition

Stress Concentration Factors, On-Line

http://www.knovel.com/web/portal/brow

se/display?_EXT_KNOVEL_DISPLAY _bookid=583


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Stress ConcentrationStress concentrators include:

shoulders

gooves

holes

keyways

threads

surface finish

inclusions

Groove in

a shaft

2nd phase

in a metal


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1. Stress concentrations occur at sections where the cross-

sectional area suddenly changes. The more severe the changes,

the larger the concentration

2. For safe design, it is only necessary to determine the max.

stress acting on the smallest cross-sectional area.

3. The Concentration factor K is independent of material

properties, it is function of the geometry only, and can be readout from an appropriate graph.


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Stress ConcentrationProblem

Given P = 20,000 lbs, d = 0.4 in,

w = 1 in, calculate the maximumstress in a bar similar to that

shown.PP

1”

1” 0.4”

The maximum stress occurs at the edges of the hole

Concentrated local stress =

K s (graph) =

so the stress at the edge of the hole =

maximum stress =

snom = =

K s (snom)


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P P

t

W2r

0.05

Also K s is 3 when W >> r


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Stress ConcentrationProblem

Given P = 20,000 lbs, d = 0.4 in,

w = 1 in, calculate the maximumstress in a bar similar to that

shown.PP

1”

1” 0.4”

The maximum stress occurs at the edges of the hole

Concentrated local stress =

K s (graph) =

so the stress at the edge of the hole =

maximum stress =

snom

= 20,000 =

(1-0.4)1

K s (snom)2.45 (r/w = 0.2)


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Consider an infinitely large plate with a cylindrical hole through it, inthis case W>>r. The plate has a working stress sw. Where sw = TS/4.The factor of four is a _______________.

Stress ConcentrationA more generalized case

2r

W

sw

t

Width an order

of magnitude or

greater

Where working

stress is tensile

strength (TS) of

material

by 4.


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The tangential stress at the hole is given by


As x , s tends to ___.At the edge of the hole, or x = r, s = ____. In this case, the stressconcentration factor, K s = ___.

2r

3sW

sW s x

3 1


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A more typical example of a stress concentration factor would be a

flat plate with an elliptical flaw in it. (more realistic situation) Defects in materials more typically have elliptical geometry

(Flaws, inclusions, voids etc.)


sw

2b

2a

r

r = radius of curvature ofthe ellipse edge (crack tip)


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The geometry of the ellipse is defined by its major and minor axislengths 2a and 2b or by one axis length and the radius of the

curvature r (measured at the sharp tip). For this case, the stressconcentration factor is given by

K s =

and when r


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ExampleEstimate the stress concentration factor for the smallest “sharp”

surface flaw that one could hope to observe nondestructively in aweld of a nuclear reactor vessel.

Assume the geometry of the flaw is an ellipse with sharp edge.

Current non-destructive evaluation procedure limits the size (amin

) of

the flaw to 0.001 inch or larger

Now if the flaw has been generated during welding of the vessel or by

in service by fatigue or corrosion, the sharpest possible flaw (r) will

be 2-3 interatomic spacing ~ 10 Angstrom.

r = 10 A = 10 x 10-8 cm = 4 x 10-8 inches

K s = = 316 (huge)

when r


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Stress Concentration: Brittle vs Ductile

materials

If a true brittle material fails in elastic range, large safety factors are

needed in the design of a components to prevent catastrophic failure

In case of ductile material the large stresses at notches or at sudden

changes in the geometry will cause permanent local deformations,

the resulting stresses will redistribute themselves (relax) and no

catastrophic flaw will occur. So safety factors will not be much

stringent.


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In general we have seen that

K s


Since K s the local stress increases as the crack becomeslonger and sharper. K as a and r

Therefore, if we were to double the radius of curvature, for any type of geometry, we

would reduce the stress concentration factor by 29%.

r0

r Condition 0Condition 1

r here isdouble r

0

~ 71% Of K 0

a

1

r

K 1 1/ 2 0 1

K 1 1/ 0 2= ; K 1 = 0.707==


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Prevention of stress concentrations

A counter-intuitive method of reducing one of the worst types of

stress concentrations, a crack, is to drill a large hole at the end of the

crack. The drilled hole, with its relatively large diameter, causes a

smaller stress concentration than the sharp end of a crack. This is

however, a temporary solution that must be corrected at the first

opportune time.

I l bl


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In-class problemYou are designing a member with a U shaped notch identical to the

one seen in the sketch below. Your member has the following

dimensions: D =2.25 in, d =1.5 in, h =0.25 in, r = 0.3. From theinformation presented on the graph determine:

U

h

Dd

ra. The stress concentration factor, K s.

b. The maximum tolerable bending moment, M b for the member if the

maximum stress smax is not to exceed 85 ksi.

D/d =

r/d =

K s = (from graph)

K s = smax / snom snom = smax / K s

= 6M b / hd2snom M b =


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Stress concentrations apply to all loading types.Generalized expressions are as follows:

Axial loadingTorsion

Bending

Polar moment of inertia of

cross-section

Moment of Inertia of

cross section about

neutral axis

J

Tc K

s

I

Mc K b s s

A

P K

s s


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Bar of circular

cross-section

with U-groove

loaded in tension.

Peterson’s Stress Concentration Factors:

Second Edition, W.D. Pilkey, editor.John Wiley & Sons, Inc., New York, NY, 1997, pg. 99.


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Bar of circular

cross-section

with U-groove

loaded in bending.




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Bar of circular

cross-section

with U-groove

loaded in torsion.




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Homework ProblemsReading Assignment:

Ch 2 sec. 2-11II.

II.

12

13

J

Tc K s

K s r

1Use

Answer (B): ~0.01387”

0.01”

Consider this an infinite plate.

Max. K s is 3 or from the graph2.85.

Answer: 58,333-61,403 lbs


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You will need it for HW 2 13

lecture 8 stress concentration class

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