Static Loading _failure

8/27/2010 [email protected] 1

I/C: KALLURI VINAYAK


Introduction• The static load is a stationary force or couple

applied to a member. Stationary means, the load should be unchanging in…

– magnitude

– point or points of application

– direction

– Or in any other manner

• A stationary force may cause

– axial tension or compression, shear load or bending moment or a torsional load or any combination of these

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Introduction• Failure means a part has had:

– separted into two or more pieces

– become permanently distorted and thus ruined its geometry

– its reliability downgaded

– its function compromised

• Static strength data of various materials may be available in two forms

– Data obtained in tests conducted in the similar conditions as the actual service life

– Data collected in tests conducted in certain ideal conditions that may not necessary simulate the exact service conditions

• The first type of data is always desirable but difficult to produce

• When second type of data is used, then the engineer must be clever enough to apply suitable factors



IntroductionIn designing any machine element, the results of a

great many strength tests are very important for a

chosen material and tests should be made under

exactly the same conditions as the part will

experience in service.

Following four design categories:

1) Mass production or the product can endanger human life: You must generate the strength data in real service life conditions and use in the design.

2) Medium quantity of production and does not endanger lives: moderately accurate tests are ok

3) The part is made in very small quantities: no testing is required; use the published mean data in handbook

4) The part is already designed, manufactured but proved defective; needs analysis to find the problem



Stress Concentration• Basic stress equations for tension, compression,

bending, and torsion were developed with

assumption that no geometric irregularities

occurred in the member under consideration

• Difficult to design a machine without permitting

some changes in the cross sections

Ex 1: Rotating shafts must have shoulders to seat bearings

properly and to take thrust loads; and key slots for securing

pulleys and gears

Ex 2: A bolt has a head on one end and screw threads on the

other end



Stress Concentration• Any discontinuity in a machine part alters the stress

distribution in the neighborhood of the discontinuity so

that the elementary stress equations no longer describe

the state of stress in the part at these locations

• Such discontinuities are called stress raisers, and the

regions in which they occur are called areas of stress

concentration.

• A theoretical, or geometric, stress-concentration factor Kt

or Kts is used to relate the actual maximum stress at the

discontinuity to the nominal stress

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Where

areNominal normal and shear

stress respectively

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Stress Concentration

• Stress concentration is a highly localized effect.

• In some instances it may be due to a surface

scratch

• Stress-concentration factor depends for its value

only on the geometry of the part

• The particular material used has no effect on the

value

• If the load is static and material is ductile, then

Stress concentration factor is unity



Stress Concentration• Most stress-concentration factors are found by

using experimental techniques like photoelasticity,

grid methods, brittle-coating methods, and

electrical strain-gauge methods

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Ex: Thin plate in tension or

simple compression with a

transverse central hole. The net

tensile force is F = σwt,

(t = thickness of the plate)

The nominal stress is given by

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Stress Concentration

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�Stress-concentration factors for a variety of geometries

may be found in Table A–13

Rectangular filleted bar

in tension or simple

compression.

σ0 = F/A, where A = dt

and t is the thickness.

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Failure Theories• There is no general theory that is sacrosanct and unique;

hence we call them “failure theories”

• Rather it is a choice based on the design requirement and nature of material, whether ductile or brittle

• Ductile materials (εf ≥ 0.05 and identifiable yield

strength that is often the same in compression as in

tension i.e Syt = Syc = Sy ) are designed based on yield

criteria

�Maximum shear stress (MSS) theory

�Distortion energy (DE) theory

�Ductile Coulomb-Mohr (DCM) theory

• Brittle materials, (εf < 0.05 and typically classified by

ultimate tensile and compressive strengths i.e Sut and

Suc) are designed based on fracture criteria

�Maximum normal stress (MNS) theory

�Brittle Coulomb-Mohr (BCM) theory8/27/2010 [email protected]


Failure Theories• How do we know that the given material is ductile?

The judgement is based on the fracture strain.

• Ductile materials

Fracture strain, εf ≥ 0.05 and

identifiable yield strength is often the same in

compression as in tension i.e Syt = Syc = Sy

• Brittle materials

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ucut

f

SS

and

≠

< 05.0strain Fracture ε

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Maximum-Shear-Stress (MSS)

Theory• Also referred to as the Tresca or Guest theory.

• Theory 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”

• For a general state of stress, the maximum-

shear-stress theory predicts yielding when




Theory• For design purposes, Equation can be modified to

incorporate a factor of safety, n.

for plane stress (one of the principal stresses is zero)

and Assuming that σA ≥ σB:

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Case 1: σA ≥ σB ≥ 0.

For this case, σ1 = σA and σ3 = 0. Equation reduces to a yield condition of σA ≥ Sy

Case 2: σA ≥ 0 ≥ σB .

Here, σ1 = σA and σ3 = σB , and Equation becomes σA − σB ≥ Sy

Case 3: 0 ≥ σA ≥ σB .

For this case, σ1 = 0 and σ3 = σB and Equation gives σB ≤ −Sy

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Theory

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unmarked lines are cases for σB ≥ σA

yAy

BA

BA

SS ≥≥−

===

≥≥

σσσ

σσσσσ

σσ

,

0,,

,0

31

321

yBAy

BA

BA

SS ≥−≥−

===

≥≥

σσσσ

σσσσσ

σσ

,

,0,

,0

31

321

yB

y

B

A

BA

S

S

−≥−

≥−

=

=

=

≥≥

σ

σσ

σσ

σσ

σ

σσ

,

,

,0

,0

31

3

2

1

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Distortion-Energy (DE) Theory

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�Other names o f distortional energy criterion

• The von Mises or von Mises-Hencky

theory

• The shear energy theory

• The octahedral shear stress theory

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8/27/2010 16

ijσ pijδ'

ijij orS σ

ijijij Sp += δσ

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• The strain energy per unit volume for simple tension

• For tri-axial stress shown in fig (a)., the strain energy per

unit volume is

• Substituting for the principal strains gives



• The strain energy for producing only volume change is

• Substituting σav = (σ1 + σ2 + σ3)/3 gives

• Then the distortion energy is



Distortion-Energy (DE) Theory• Theory predicts 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”

• For a general state of stress, the Distortion-

Energy Theory predicts yielding when

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or σ′ ≥ Sy

Where “σ′ “is a single, equivalent, or effective stress called von Mises stress

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• For plane stress, let σA and σB be the two

nonzero principal stresses, then the von Mises

stress is

• The above equation is a rotated ellipse in the

σA, σB plane with σ′ = Sy




• Using xyz components of three-dimensional

stress, the von Mises stress can be written as

• and for plane stress





Octahedral shear stress:� Considering the principal directions as the coordinate

axes, a plane whose normal vector makes equal angles

with each of the principal axes (i.e. having direction

cosines equal to ) is called an octahedral plane

� The shear stress acting on the so-called ‘octahedral

planes’ which are the eight planes that are normal to the

equi-inclined directions in the eight quadrants formed by

the principal stresses.

3

1



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THE OCTAHEDRAL PLANES: Octahedral shear stress

theory

Failure occurs when the octahedral shear stress in the

given state of stress reaches the octahedral shear

stress in simple tension at yield.

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• According to DE (von Mises) criterion,

substituting the pure shear state of stress in the

2-D DE criterion, the two normal stresses being

zero,

ysy

y

y

xyyxy

SSyieldAt

SS

S

577.0,

577.03

3 2

=

=== ττ

ysy SS 5.0=

According to the MSS criterion,

DE criterion predicts the shear yield strength to be 15 percent more than

that predicted by the MSS criterion. Hence MSS is more conservative.

SHEAR YIELD STRENGH:




Ductile Coulomb-Mohr Theory or Internal friction

theory


To be applied when the material has unequal strength in tension and

compression,ycyt SS ≠

Examples of such materials are magnesium alloys, for whichytyc SS 5.0≅

DCM is a simplification of the old Mohr’s theory of failure.

OR ⇒

where either yield strength or ultimate strength can be used


Ductile Coulomb-Mohr Theory or Internal friction

theory


Maximum-Normal-Stress Theory• The maximum-normal-stress (MNS) theory

states that failure occurs whenever one of

the three principal stresses equals or

exceeds the strength.

• For the principal stresses for a general

stress state in the ordered form σ1 ≥ σ2 ≥

σ3; This theory then predicts that failure

occurs whenever

σ1 ≥ Sut or σ3 ≤ −Suc

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where Sut and Suc are the ultimate tensile and compressive strengths, respectively,

given as positive quantities.

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Maximum-Normal-Stress Theory• For plane stress, with σA ≥ σB,

σA ≥ Sut or σB ≤ −Suc



Brittle-Coulomb-Mohr Theory

• Design equations for a brittle material are



Modified Mohr Theory


Failure theory selection flowchart


Problem• The figure shows a crank loaded by a force F = 800 N which causes twisting and bending of the 20 mm-diameter shaft fixed to a support at the origin of the reference system. In actuality, the support may be an inertia which we wish to rotate, but for the purposes of a strength analysis we can consider this to be a staticsproblem. The material of the shaft AB is hot-rolled AISI 1018 steel (Table A–18). Using the maximum-shear-stress theory, find the factor of safety based on the stress at point A.

F

Static Loading _failure

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