Files 2-Lectures LEC 17 Coulomb-Mohr Theory

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ME ME 307 307 MACHINE DESIGN IMACHINE DESIGN I

Dr. A. Aziz Bazoune

King Fahd University of Petroleum & MineralsKing Fahd University of Petroleum & MineralsKing Fahd University of Petroleum & MineralsKing Fahd University of Petroleum & MineralsMechanical Engineering Department

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ME ME 307 307 MACHINE DESIGN IMACHINE DESIGN I2

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FAILURE CRITERIA

Failure Criteria Failure Criteria ((Static Load Static Load ++Ductile MaterialDuctile Material))

66--66-- Coulomb Mohr Theory for Ductile MaterialsCoulomb Mohr Theory for Ductile Materials

� Not all materials have their tensile strength equal to their compressive strength. Examples are

Magnesium

Cast Iron

0.5uc utS S=

( )3 4S S� ∼

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Cast Iron

� This section deals with materials

( )3 4uc ut

S S� ∼

uc utS S≠

syS

Practical difficulties to construct 3 Mohr circles for three tests.

Figure 6-20 Three Mohr circles, one for the uniaxial compression test, one for the test in pure shear, and one for

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pure shear, and one for the uniaxial tension test are used to define failure by the Mohr hypothesis.The compressive and tensile strengths, respectively; they can be used for yield or ultimate strength .

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Figure 6-20 Three Mohr circles, one for the uniaxial compression test, one for the test in pure shear, and one for the uniaxial tension test are used to define failure by the Mohr hypothesis. The compressive and tensile strengths, respectively; they can be used for yield or ultimate strength

Coulomb Mohr Theory for Ductile MaterialsCoulomb Mohr Theory for Ductile Materials

1 2 3σ σ σ≥ ≥

Failure will occur if Mohr’s circle (3-D) describing a state of

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describing a state of stress in a body grows until it becomes tangent to the failure envelope.

1 3

1 31 0, 0

t cS S

σ σσ σ− = > ≤

Coulomb Mohr Theory for Ductile MaterialsCoulomb Mohr Theory for Ductile Materials

Failure will occur if

1 3

1 31 0, 0

t cS S

σ σσ σ− = > ≤

Failure criteria according to Coulomb-Mohr Theory

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Factor of safety

1 31

t cS S n

σ σ− =

Special Cases: Application of CoulombSpecial Cases: Application of Coulomb--Mohr for Mohr for Plane Stress (Failure will occur ifPlane Stress (Failure will occur if☺☺☺☺☺☺☺☺

Case Case 11::

A tSσ ≥

⇒A B

σ σ≥

0A B

σ σ≥ ≥1 2 3

, , 0A B

σ σ σ σ σ= = =

Case Case 22:: ⇒0A B

σ σ≥ ≥1 2 3

, 0,A B

σ σ σ σ σ= = =

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⇒0A B

σ σ≥ ≥1 2 3

, 0,A B

σ σ σ σ σ= = =

1A B

t cS S

σ σ− ≥

Case Case 33::

B cSσ ≤ −

⇒0A B

σ σ≥ ≥1 2 3

0, ,A B

σ σ σ σ σ= = =

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Factor of safety

1 31

t cS S n

σ σ− =

� For Coulomb-Mohr Theory we do not need the torsional shear strength circle, we can deduce it from Eq.(6-22). For pure shear .

� The torsional yield strength occurs when1 3

,τ σ σ τ= − =

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� The torsional yield strength occurs when

� For yielding, Eq. (6-22) gives

y t y c

s y

y t y c

S SS

S S=

+

max sySτ =

Example Example 66--2 2 ((Textbook Page Textbook Page 268268))

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66--7 7 Failure of Ductile Materials SummaryFailure of Ductile Materials Summary

� This study is limited materials and parts that are known to fail in a ductile manner.

� Experimental data

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� Experimental data collected for several materials from many sources are shown in Figure 6-23.

Figure 6-23Experimental data superposed on failure theories


� Figure 6-23 shows that either the Maximum Shear Stress Theory (MSS) or the Distortion Energy Theory (DE) is acceptable for design and analysis of materials that would fail in a ductile manner.

� The selection of one or the other of these theories is something

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� The selection of one or the other of these theories is something that you, the engineer, MUST DECIDE.


� For Design Purposes:

Maximum Shear Stress (MSS) Theory is

� easy

� quick to use

� conservative

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� If the problem is to learn why a part failed, then the

Distortion Energy (DE) Theory may be best to use. Figure 6-23 shows that the locus of the DE passes closer to the central area of the data points, and thus is generally a better predictor of failure.


� For ductile materials with , the Mohr Theory is the best available. However, the theory requires the results from three separate modes of tests, graphical construction of the failure locus, and fitting the largest Mohr’s circle to the failure locus.

yt ycS S≠

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� Coulomb-Mohr Theory (CM) which requires only tensile and compressive yield strengths and represents an alternative to Mohr Theory. It is easily dealt with in equation form.

Example Example 66--33

A certain force F applied at D nearthe end of the 15-in lever shownin Fig. 6-24, which is quite similarto a socket wrench, results incertain stresses in thecantilevered bar OABC. The barOABC is of AISI 1035 steel, forgedand heat treated so that it has a

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and heat treated so that it has aminimum (ASTM) yield strength of81 Kpsi. We presume that thiscomponent would be of no valueafter yielding. Thus the force Frequired to initiate yielding canbe regarded as the strength of thecomponent part. Find this force.

Figure. 6-24

Example Example 66--3 3 (Cont.’ d)(Cont.’ d)

Assuming lever DC strong enough

( )

( )

( )

( )

( )

( )

( )

( )

34 3

34 3

2 32 1432142.6

64 1

2 16 151676.4

32

x

zx

M d FMc MF

I d d

T d FTr TF

J d d

σπ π π

τπ π π

= = = = =

= = = = =

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( ) ( )34 3

76.432 1

zx FJ d d

τπ π π

= = = = =

Using the DE theory, we find, from Eq. (6-15)

Equating the Von Mises stress to , we solve for and get

( ) ( ) ( )1 21 2 2 2' 2 2

3 142.6 3 76.4 194.5x zx

F F Fσ σ τ = + = + × =

yS

81,000416

194.5 194.5lbf

yS

F = = =

F

Example Example 66--3 3 (Cont.’ d)(Cont.’ d)

In this example the strength of the material point A is . The strengthof the assembly or component is .

Using the MMS Theory, for a point undergoing plane stress with only one non-zero normal stress and one shear stress, the two nonzero principal stresses and will have opposite signs and hence fit case 2 for the MSS Theory. From Eq. (4-13)

( )1 2

2

σ

81kpsiy

S =416 lbfF =

Aσ Bσ

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case 2 for the MSS Theory, Eq. (6-5) applies and hence

( )2

1 22 2 2

2 42

xA B zx x zx

σσ σ τ σ τ

− = + = +

( )

( ) ( )

1 22 2

1 22 2

4

142.6 4 76.4 209 81,000

388 lbf

x zx yS

F F F

F

σ τ+ =

+ = =

=


The cantilever tube shown in Fig. 6-25 is to be made of 2014 aluminum alloy treated to obtain a specified minimum yield strength of 276 MPa. We wish to select a stock-size tube from Table A-8 using a

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tube from Table A-8 using a design factor . The bending load is F=1.75 kN, the axial tension is P = 9.0 kN, and the torsion is T = 72 N.m. What is the realized factor of safety?

4dn =

Figure 6-25


( )( )120 1.75 29 9 105o ox

dP Mc d

A I A I A Iσ = + = + = +

SOLUTION

The normal stress for an element on the top surface of the tube at the origin is

(1)

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The torsional stress at the same point is

( ) ( )72 2 36o o

zx

d dTr

J J Jτ = = = (2)

Example Example 66--4 4 ((Cont.’dCont.’d))

0.276S

For accuracy, we choose the DE as the design basis. The Von Mises stress is

(3)( )1 2

2 2' 3

x zxσ σ τ= +

On the basis of the given design factor, the goal for is 'σ

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0.276' 0.0690

4GPa

y

d

S

nσ ≤ = =

Programming Using MATLAB or EXCEL , entering metric sizes from Table A-8 reveals that a 42-X 5- mm is satisfactory. The Von Mises stress is found to be

for this size. Thus the realized factor of safety is

(4)

' 0.06043GPaσ =

0.2764.57

' 0.06043

ySn

σ= = =

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Files 2-Lectures LEC 17 Coulomb-Mohr Theory

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