Theories of Failure From My SOM Book

T h e o r i e s o f Fa i l u r e T h e o r y a t a Gl G l a n c e ( f o r I ES ES, GA T E & PSU s ) 1. Introduction •

Failure: Every material has certain strength, expressed in terms of stress or strain, beyond

which it fractures or fails to carry the load.

•

Failure Criterion: A Criterion: A criterion used to hypothesize hypothesize the failure.

•

Failure Theory: A Theory behind a failure criterion.

Why Need Failure Theories? •

To design structural components and calculate margin of safety.

•

To guide in materials development.

•

To determine weak and strong directions.

Failure Mode •

Yielding: a process of global permanent plastic deformation. Change in the geometry of the object.

•

Low stiffness: excessive elastic deflection.

•

Fracture: a process in which cracks grow to the extent that the component breaks apart.

•

Buckling: the loss of stable equilibrium. Compressive loading can lead to bucking in columns.

•

Creep: a high-temperature effect. Load carrying capacity drops .

Failure Modes: Excessive elastic

Yielding

Fracture

deformation 1.

Stretch,

twist,

or

•

bending 2.

Buckling

3. Vibration

Plastic deformation at room

•

temperature

• •

Creep

Sudden fracture of brittle materials

at

elevated

•

Fatigue

(progressive

temperatures

fracture)

Yield stress is the important •

Stress rupture at elevated

design factor

temperatures

•

Ultimate

stress

is

the

important design factor

2. Maximum Principal Stress Theory (W. Rankin’s Theory- 1850) – Brittle Material The maximum principal stress criterion:

•

Rankin stated max principal stress theory as follows- a material fails by fracturing when the largest principal stress exceeds the ultimate strength the onset of fracture, |σ1| = σu OR |σ3| = σu

σu

in a simple tension test. That is, at

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S K Mondal’s Theories of Failure Page-3 • Crack will start at the most highly stressed point in a brittle material when the largest principal stress at that point reaches

•

σu

Criterion has good experimental verification, even though it assumes ultimate strength is same in compression and tension

Failure surface according to maximum principal stress theory

•

This theory of yielding has very poor agreement with experiment. However, the theory has been used successfully for brittle materials.

•

Used to describe fracture of brittle of brittle materials such as cast iron

•

Limitations o

Doesn’t distinguish between tension or compression

o

Doesn’t depend on orientation of principal planes so only applicable to isotropic materials

•

Generalization to 3-D stress case is easy:

3. Maximum Shear Stress or Stress difference theory (Guest’s or Tresca’s Theory-1868)- Ductile Material The Tresca Criterion:

•

Also known as the Maximum Shear Stress criterion.

•

Yielding will occur when the maximum shear stress reaches that which caused yielding in a simple tension test.

•

Recall that yielding of a material occurred by slippage between planes oriented at 45° to principal stresses. This should indicate to you that yielding of a material depends on the maximum shear stress in the material rather than the maximum normal stress. If σ 1 > σ 2 > σ 3 Then σ 1 − σ 3 = σ y

•

Failure by slip (yielding) occurs when the maximum shearing stress, τ max exceeds the yield stress τ f as determined in a uniaxial tension test.









S K Mondal’s Theories of Failure • This theory gives satisfactory result for ductile material.

Page-4

Failure surface according to maximum shear stress theory

4. Strain Energy Theory (Haigh’s Theory) The theory associated with Haigh This theory is based on the assumption that strains are recoverable up to the elastic limit, and the energy absorbed by the material at failure up to this point is a single valued function independent of the stress system causing it. The strain energy per unit volume causing failure is equal to the strain energy at the elastic limit in simple tension.

σ y 2 1 2 2 2 ⎡σ 1 + σ 2 + σ 3 − 2μ (σ 1σ 2 + σ 2σ 3 + σ 3σ 1 ) ⎤⎦ = U = 2E ⎣ 2E σ 12 + σ 22 + σ 32 − 2μ (σ 1σ 2 + σ 2σ 3 + σ 3σ 1 ) = σ y 2 σ 12 + σ 22 − 2μσ 1σ 2 = σ y 2

For 3D- stress

For 2D- stress

5. Shear Strain Energy Theory (Distortion (Distort ion Energy Theory or Mises-Henky Theory or Von-Misses Theory)-Ductile Material Von-Mises Criterion:

•

Also known as the Maximum Energy of Distortion criterion

•

Based on a more complex view of the role of the principal stress differences.

•

In simple terms, the von Mises criterion considers the diameters of all three Mohr’s circles as contributing to the characterization of yield onset in isotropic materials.

•

When the criterion is applied, its relationship to the uniaxial tensile yield strength is:

•

For a state of plane stress ( σ 3 =0)

σ 12 − σ 1σ 2 + σ 22 = σ y2 •

It is often convenient to express this as an equivalent stress, σ e:









S K Mondal’s

Theories of Failure σ

e

=

or σe •

Page-5

1/2 1 ⎡ 2 2 2⎤ ⎢⎣( σ1 − σ2 ) + (σ2 − σ3 ) + (σ3 − σ1 ) ⎥⎦ 2 1 ⎡ 2 2 2 2 2 2 ⎤1/2 ( = + + σ − σ ) + (σ − σ ) + (σ − σ ) + 6(τ τ τ y y z x z xy yz zx )⎥⎦ ⎢ x 2⎣

In formulating this failure theory we used generalized Hooke's law for an isotropic material so the theory given is only applicable to those materials but it can be generalized to anisotropic materials.

•

The von Mises theory is a little less conservative than the Tresca theory but in most cases there is little difference in their predictions of failure. Most experimental results tend to fall on or between these two theories.

•

It gives very good result in ductile material.

6. Maximum Principal Strain Theory (St. Venant Theory) According to this theory, yielding will occur when the maximum principal strain just exceeds the strain at the tensile yield point in either simple tension or compression. If and minimum principal strains corresponding to

σ1

ε1

and

ε2

are maximum

and σ2, in the limiting case

7. Mohr’s theory- Brittle Material Mohr’s Theory

•

Mohr’s theory is used to predict the fracture of a material having different properties in tension and compression. Criterion makes use of Mohr’s circle









S K Mondal’s Theories of Failure • In Mohr’s circle, we note that τ depends on σ, or represents states of stress on planes with same plane is the one with maximum

•

τ

σ

Page-6 τ

= f( σ ). Note the vertical line PC

but differing

τ

, which means the weakest

, point P .

Points on the outer circle are the weakest planes. On these planes the maximum and minimum principal stresses are sufficient to decide whether or not failure will occur.

•

Experiments are done on a given material to determine the states of stress that result in failure. Each state defines a Mohr’s circle. If the data are obtained from simple tension, simple compression, and pure shear, the three resulting circles are adequate to construct an envelope (AB & A’B’)

•

Mohr’s envelope thus represents the locus of all possible failure states.

Higher shear stresses are to the left of origin, since most brittle materials have higher strength in compression

8. Comparison A comparison among the different failure theories can be made by superposing the yield surfaces as shown in figure









S K Mondal’s

Theories of Failure

Page-7

O B J E C T I V E Q UESTIONS (GA (G A T E, I ES, ES, IA S) P r e v i o u s 2 0 -Y - Y e a r s GA G A T E Qu Qu e s t i o n s Maximum Shear stress or Stress Difference Theory GATE-1.

Match 4 correct pairs between list I and List II for the questions List-I List-II (a) Hooke's law 1. Planetary motion (b) St. Venant's law 2. Conservation Energy (c) Kepler's laws 3. Elasticity (d) Tresca's criterion 4. Plasticity (e) Coulomb's laws 5. Fracture (f) Griffith's law 6. Inertia GATE-1. Ans. (a) - 3, (c) -1, (d) -5, (e) -2 St. Venant's law: Maximum principal strain theory

[GATE-1994]

GATE-2.

Which theory of failure will you use for aluminium components under steady loading? [GATE-1999] (a) Principal stress theory (b) Principal strain theory (c) Strain energy theory (d) Maximum shear stress theory GATE-2. Ans. (d) Aluminium is a ductile material so use maximum shear stress theory

Shear Strain Energy Theory (Distortion energy theory) GATE-3.

According to Von-Mises' distortion energy theory, the distortion energy under three dimensional stress state is represented by [GATE-2006]

GATE-3. Ans. (c)

Vs =

1 2 2 2 (σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 ) 12G

{

GATE-4.

}

W here E = 2G(1 + μ ) si mplify and get result.

A small element at the critical section of a component is in a bi-axial state of stress with the two principal stresses being 360 MPa and 140 MPa. The maximum working stress according to Distortion Energy Theory is: [GATE-1997] (a) 220 MPa (b) 110 MPa (c) 314 MPa (d) 330 MPa GATE-4. Ans. (c) According to distortion energy theory if maximum stress ( σt) then or σ t2 = σ 12 + σ 22 − σ 1σ 2

or σ t2 = 3602 + 1402 − 360 × 140 or σ t = 314 MPa











S K Mondal’s

Theories of Failure

Page-8

P r e v i o u s 2 0 -Y -Y e a r s I E S Qu e s t i o n s Maximum Principal Stress Theory IES-1.

Match List-I (Theory of Failure) with List-II (Predicted Ratio of Shear Stress to Direct Stress at Yield Condition for Steel Specimen) and select the correct answer using the code given below the Lists: [IES-2006] List-I List-II A. Maximum shear stress theory 1. 1·0 B. Maximum distortion energy theory 2. 0·577 C. Maximum principal stress theory 3. 0·62 D. Maximum principal strain theory 4. 0·50 Codes: A B C D A B C D (a) 1 2 4 3 (b) 4 3 1 2 (c) 1 3 4 2 (d) 4 2 1 3 IES-1. Ans. (d) IES-2.

From a tension test, the yield strength of steel is found to be 200 N/mm2. Using a factor of safety of 2 and applying maximum principal stress theory of failure, the permissible stress in the steel shaft subjected to torque will be: [IES-2000] 2 2 2 (a) 50 N/mm (b) 57.7 N/mm (c) 86.6. N/mm (d) 100 N/mm 2 IES-2. Ans. (d) For pure shear τ = ±σ x IES-3.

A circular solid shaft is subjected to a bending moment of 400 kNm and a twisting moment of 300 kNm. On the basis of the maximum principal stress theory, the direct stress is σ and according to the maximum shear stress theory, the shear stress is τ . The ratio σ / τ is: [IES-2000]

(a )

1

(b)

5

IES-3. Ans. (c) σ =

3 9

16 M + M2 + T 2 3 πd

Therefore

(

σ τ

=

( c)

)

M + M2 + T 2 M2 + T 2

and τ =

=

16 π d3

(

4 + 4 2 + 32 4 2 + 32

9 5

M2 + T 2

=

( d)

11 6

)

9 5

IES-4.

A transmission shaft subjected to bending loads must be designed on the basis of [IES-1996] (a) Maximum normal stress theory (b) Maximum shear stress theory (c) Maximum normal stress and maximum shear stress theories (d) Fatigue strength IES-4. Ans. (a) IES-5.

Design of shafts made of brittle materials is based on [IES-1993] (a) Guest's theory (b) Rankine’s theory (c) St. Venant's theory (d) Von Mises theory IES-5. Ans. (b) Rankine's theory or maximum principle stress theory is most commonly used for brittle materials.

Maximum Shear stress or Stress Difference Theory IES-6.

Which one of the following figures represents the maximum shear stress theory or Tresca criterion? [IES-1999]









S K Mondal’s IES-7.

Theories of Failure

Page-9

According to the maximum shear stress theory of failure, permissible twisting moment in a circular shaft is 'T'. The permissible twisting moment will the same shaft as per the maximum principal stress theory of failure will be: [IES-1998: ISRO-2008] (a) T/2

(b) T

IES-7. Ans. (d) Given τ =

(c)

2T

(d) 2T

16T σ yt principal st stresses ffo or o on nly tth his sh shear st stress are = π d3 2

σ 1,2 = τ 2 = ±τ maximum principal stress theory of failuregives

max[σ 1,σ 2 ] = σ yt = IES-8.

16 ( 2T ) π d3

Permissible bending moment in a circular shaft under pure bending is M according to maximum principal stress theory of failure. According to maximum shear stress theory of failure, the permissible bending moment in the same shaft is: [IES-1995] (a) 1/2 M

(b) M

2M

(d) 2M

16 M2 + T 2 put T = 0 3 π d ⎛ 32M ⎞ 32M 16M′ σ yt ⎜⎝ π d3 ⎟⎠ 16M or σ yt = and τ = Therefore M′ = M = = = π d3 π d3 2 2 π d3

IES-8. Ans. (b) σ =

16 M + M2 + T 2 3 πd

(c)

(

)

(

and τ =

)

IES-9.

A rod having cross-sectional area 100 x 10- 6 m2 is subjected to a tensile load. Based on the Tresca failure criterion, if the uniaxial yield stress of the material is 200 MPa, the failure load is: [IES-2001] (a) 10 kN (b) 20 kN (c) 100 kN (d) 200 kN IES-9. Ans. (b) Tresca failure criterion is maximum shear stress theory.

W e know that,τ =

σ P sin 2θ P or τ max = = yt or P = σ yt × A A 2 2A 2

IES-10.

A cold roller steel shaft is designed on the basis of maximum shear stress theory. The principal stresses induced at its critical section are 60 MPa and - 60 MPa respectively. If the yield stress for the shaft material is 360 MPa, the factor of safety of the design is: [IES-2002] (a) 2 (b) 3 (c) 4 (d) 6 IES-10. Ans. (b) IES-11.

A shaft is subjected to a maximum bending stress of 80 N/mm2 and maximum shearing stress equal to 30 N/mm2 at a particular section. If the yield point in tension of the material is 280 N/mm 2, and the maximum shear stress theory of failure is used, then the factor of safety obtained will be: [IES-1994] (a) 2.5 (b) 2.8 (c) 3.0 (d) 3.5 2

IES-11. Ans. (b) Maximum shear stress =

⎛ 80 − 0 ⎞ 2 N/mm2 ⎜ 2 ⎟ + 30 = 50 N/ ⎝ ⎠

According to maximum shear stress theory, τ =

IES-12.

For a two-dimensional state stress ( σ 1

σ y 2

; ∴ F .S . =

280 2 × 50

> σ 2 , σ 1 > 0, σ 2 < 0 )

= 2. 8

the designed values

are most conservative if which one of the following failure theories were used? [IES-1998] (a) Maximum principal strain theory (b) Maximum distortion energy theory (c) Maximum shear stress theory (d) Maximum principal stress theory IES-12. Ans. (c)









S K Mondal’s

Theories of Failure

Page-10

Graphical comparison of different failure theories Above diagram shows that σ 1 > 0, σ 2 < 0 will occur at 4 th quadrant and most conservative design will be maximum shear stress theory.

Shear Strain Energy Theory (Distortion energy theory) IES-13.

Who postulated the maximum distortion energy theory? (a) Tresca (b) Rankine (c) St. Venant IES-13. Ans. (d)

[IES-2008] (d) Mises-Henky

IES-14.

[IES-2008] (d) Mises-Henky

Who postulated the maximum distortion energy theory? (a) Tresca (b) Rankine (c) St. Venant IES-14. Ans. (d) Maximum shear stress theory Tresca → Maximum principal stress theory Rankine → Maximum principal strain theory St. Venant → Maximum shear strain energy theory Mises - Henky → IES-15.

The maximum distortion energy theory of failure is suitable to predict the failure of which one of the following types of materials? [IES-2004] (a) Brittle materials (b) Ductile materials (c) Plastics (d) Composite materials IES-15. Ans. (b) IES-16. If σy is the yield strength of a particular material, then the distortion energy theory is expressed as [IES-1994] (a) (σ 1 − σ 2 )

(

2

(b) σ 1

2

2

2

+ (σ 2 − σ 3 ) + (σ 3 − σ 1 ) = 2σ y2

− σ 22 + σ 32 ) − 2μ (σ 1σ 2 + σ 2σ 3 + σ 3σ1 ) = σ y2

(c) (σ 1 − σ 2 )

2

2

2

+ (σ 2 − σ 3 ) + (σ 3 − σ 1 ) = 3σ y2

(d) (1 − 2μ ) (σ 1 + σ 2

2

+ σ 3 ) = 2 (1 + μ ) σ y2

IES-16. Ans. (a) IES-17.

If a shaft made from ductile material is subjected to combined bending and twisting moments, calculations based on which one of the following failure theories would give the most conservative value? [IES-1996] (a) Maximum principal stress theory (b) Maximum shear stress theory. (d Maximum strain energy theory (d) Maximum distortion distortio n energy theory. IES-17. Ans. (b)









S K Mondal’s

Theories of Failure

Page-11

Maximum Principal Strain Theory IES-18.

Match List-I (Failure theories) with List-II (Figures representing boundaries of these theories) and select the correct answer using the codes given below the Lists: [IES-1997] List-I List-II A. Maximum principal stress theory

B. Maximum shear stress theory

C. Maximum octahedral stress theory

D. Maximum shear energy theory

Code: (a) (c) IES-18. Ans. (d)

A 2 4

B 1 2

strain

C 3 3

D 4 1

(b) (d)

A 2 2

B 4 4

C 3 1

D 1 3

P r e v i o u s 2 0 -Y -Y e a r s I A S Qu e s t i o n s Maximum Principal Stress Theory IAS-1.

For σ 1

≠ σ 2

and

theory? (a) A rectangle

σ3

= 0, what is the physical boundary for Rankine failure

(b) An ellipse

(c) A square

[IAS-2004] (d) A parabola









S K Mondal’s

Theories of Failure

Page-12

IAS-1. Ans. (c) Rankine failure theory or Maximum principle stress theory.

Shear Strain Energy Theory (Distortion energy theory) IAS-2.

Consider the following statements: [IAS-2007] 1. Experiments have shown that the distortion-energy theory gives an accurate prediction about failure of a ductile component than any other theory of failure. 2. According to the distortion-energy theory, the yield strength in shear is less than the yield strength in tension. Which of the statements given above is/are correct? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2

IAS-2. Ans. (c) τ y

=

σ y 3

= 0.577σ y

IAS-3.

Consider the following statements: [IAS-2003] 1. Distortion-energy theory is in better agreement for predicting the failure of ductile materials. 2. Maximum normal stress theory gives good prediction for the failure of brittle materials. 3. Module of elasticity in tension and compression are assumed to be different stress analysis of curved beams. Which of these statements is/are correct? (a) 1, 2 and 3 (b) 1 and 2 (c) 3 only (d) 1 and 3 IAS-3. Ans. (b) IAS-4.

Which one of the following graphs represents Mises yield criterion? 1996]

[IAS-

IAS-4. Ans. (d)

Maximum Principal Strain Theory IAS-5.

Given that the principal stresses σ 1

> σ 2 > σ 3

and σe is the elastic limit stress in

simple tension; which one of the following must be satisfied such that the elastic failure does not occur in accordance with the maximum principal strain theory? [IAS-2004] (a)

σe

⎛σ <⎜ 1

μ

σ 2

μ

σ 3 ⎞

⎟

(b)

σe

⎛σ >⎜ 1

μ

σ 2

μ

σ 3 ⎞

⎟











S K Mondal’s

Theories of Failure

σe E

>

σ1 E

−μ

σ 2 E

− μ

σ 3 E

Page-13









S K Mondal’s

Theories of Failure

Page-14

P r e v i o u s Co Co n v e n t i o n a l Qu Qu e s t i o n s w i t h A n s w e r s Conventional Question ESE-2006 Question:

A mild steel shaft of 50 mm diameter is subjected to a beading moment of 1.5 kNm and torque T. If the yield point of steel in tension is 210 MPa, find the maximum value of the torque without causing yielding of the shaft material according to (i) Maximum principal stress theory (ii) Maximum shear stress theory.

Answer :

We know know that that,, Max Maximu imum m bendi bending ng str stres ess s (σb ) = and Maximum Maximum shear shear stre stress ss ( τ ) =

32M 3 πd

16T 3 πd

Principal stresses are given by:

⎛ σ b ⎟⎞2 16 ± ⎜⎜ ⎟ + τ 2 = 3 σ 1,2 = 2 ⎝⎜ 2 ⎟⎠ πd σ

b

⎡M ± M 2 +T 2 ⎤ ⎢⎣ ⎥⎦

(i ) According According to Maximum principal principal stress stress theory theory Maximum principal stress=Maximum stress at elastic limit or

16 3 πd

⎡M + M 2 + T 2 ⎤ = 210 ×106 ⎢⎣ ⎥⎦

16

or

3

π

(0.050)

⎡1500 + 15002 + T 2 ⎤ = 210 × 10 106 ⎢⎣ ⎥⎦

or T = 3332 Nm = 3.332 kNm (ii ) According According to Maximum Maximum shear stress stress theory theory

− σ2 σ y = 2 2 or , σ1 − σ2 = σ y τ

max

=

or , 2×

σ

1

16 3 πd

M2 +T 2 = 210 ×106

or , T = 2096 N m = 2.096 kNm Conventional Question ESE-2005

( ) σ

y









S K Mondal’s

Theories of Failure

Page-15

Conventional Question ESE-2004 Question: Answer :

State the Von- Mises's theory. Also give the naturally expression. According to this theory yielding would occur when total distortion energy absorbed per unit volume due to applied loads exceeds the distortion energy absorbed per unit volume at the tensile yield point. The failure criterion is 2

2

2

( 1 − 2 ) + ( 2 − 3 ) + ( 3 − 1) = 2 σ

σ

σ

σ

σ

σ

σ

2 y

[symbols has usual meaning] Conventional Question ESE-2002 Question:

Derive an expression for the distortion energy per unit volume for a body subjected to a uniform stress state, given by the σ 1 and σ 2 with the third principal stress

Answer :

σ

3

being zero.

According to this theory yielding would occur when total distortion energy absorbed per unit volume due to applied loads exceeds the distortion energy absorbed per unit volume at the tensile yield point. Total strain energy E T and strain energy for volume change E V can be given as

Substituting strains in terms of stresses the distortion energy can be given as

At the tensile yield point, σ1 = σy , σ2 = σ3 = 0 which gives

The failure criterion is thus obtained by equating E d and Edy , which gives









S K Mondal’s

Theories of Failure

(i) Principal ssttress σ

σ

2000 = 80 N/mm2 ; 5×5 5 00 = = 20 N/mm2; 5×5

z

=

σ

+ σy

x

2

= 60 ± ∴ a nd

, σ2, σ 3 :

1

1 00 0 = 40 N/mm2 5×5 80 0 τ = 32 N/mm 2 xy = 5×5

=

x

σ

σ

y

⎛ σx − σy ⎞⎟ 2 80 + 40 ⎟⎟ + τ xy = ± ⎜⎜⎜ ± 2 ⎝ 2 ⎠⎟ 2

2

σ

2

MPa = 22.96 N/mm2 or 22.96 MP

σ

3

=

z

⎛ 80 − 40 ⎟⎞2 2 ⎜⎜ + (32) ⎟ ⎟ ⎝⎜ 2 ⎠

(20) + (32) = 97.74, 22.26

N/mm2 , or 97.74 7.74 MPa MPa = 97.74 N/

σ

=

σ

1

σ

Page-16

= 20 N/mm2 or 22 MPa

Will thecu hecube yieldorno eldornott? (ii) Wil Acco Accord rdin ing g to Vo Vonn-Mi Mise ses s yield ield crit criter eria ia,, yield ieldin ing g will will occu occurr if 2

2

( 1 − 2) +( 2 − 3) +( σ

Now

σ

σ

σ

2

σ

2 − σ1 ) ≥ 2σyt

σ

− σ1 )

3

2

( 1 − 2) +( 2 − 3) +( σ

2

σ

σ

σ

2

3

2

2

2

= (97.74 − 22.96) + (22.96 − 20) + (20 − 97.74)

− − − (i )

= 11745.8 a nd ,

2

2 2σ yt = 2 ×(70) = 9800

Since 11745.8 > 9800 so yielding will occur.

− − − (ii )









S K Mondal’s

Theories of Failure

Page-17

Due to torque, shear stress,

Tr Tr T = = J 2πr 3t 2πr 3t

xy =

τ

J=

{(r + t ) 2

π

4

}

of t. − r 4 = 2πr 3t -n eglecting t2 higher power of

of stress in in th t he tu t ube is is, ∴ The ssttate of

σ

x

=

P T , σy = 0, τ xy = 2πrt 2πr 3t

(ii) Von Mises-Henky failure in tension for 2-dimensional stress is σ

2 0

σ

1

σ

2

= σ12 + σ22 − σ1σ2 =

=

σ

x

+ σy 2

σ

x

+ σy 2

In this case,

⎛ σx − σy ⎟⎞2 2 ⎟⎟ + τ xy + ⎜⎜⎜ ⎝ 2 ⎟⎠ ⎛ σx − σy ⎟⎞2 2 ⎟ + τ xy − ⎜⎜ ⎜⎝ 2 ⎟⎟⎠ σ

1

σ

∴ σ02

⎡σ = ⎢⎢ x + ⎢⎣ 2

=

2

σ

x

2

=

σ

+ x

2

−

2 x

σ

2 , a nd + τ xy

4

2 x

σ

4

⎤2 ⎡ σ + τ x2y ⎥⎥ + ⎢⎢ x − 4 ⎥⎦ ⎢⎣ 2

σ

(

2 + τ xy

2 x

σ

∵ σ y

2 x

4

+ τ x2y

= 0)

⎤2 ⎡ σ ⎥ −⎢ x + ⎥ ⎢2 ⎥⎦ ⎢⎣

2 x

σ

4

+ τ x2y

⎤ ⎡σ ⎥⎢ x − ⎥⎢ 2 ⎥⎦ ⎢⎣

⎡ σ2 σ2 2 ⎤ ⎡ σ 2 σ2 2 σ σ σ σ 2 2 ⎥ 2 x x x x x x x x ⎢ ⎢ =⎢ + + τ xy + 2. . + τ xy ⎥ + ⎢ + + τxy + 2. . + τ x2y 4 2 4 4 2 4 ⎢⎣ 4 ⎥⎦ ⎢⎣ 4 ⎡ σ x2 σ x 2 ⎤ 2 ⎥ − ⎢ − − τ xy ⎢4 ⎥ 4 ⎣ ⎦ 2 2 = σ x + 3τ xy σ

0

=

σ

2 x

2 x

σ

4

⎤ ⎥ ⎥ ⎥⎦

+

2 τ

xy

⎤2 ⎥ ⎥ ⎥⎦

2 + 3τ xy

Conventional Question GATE-1994 Question:

Answer :

Find the maximum principal stress developed in a cylindrical shaft. 8 cm in diameter and subjected to a bending moment of 2.5 kNm and a twisting moment of 4.2 kNm. If the yield stress of the shaft material is 300 MPa. Determine the factor of safety of the shaft according to the maximum shearing stress theory of failure. Given: d = 8 cm = 0.08 m; M = 2.5 kNm = 2500 Nm; T = 4.2 kNm = 4200 Nm σ

yield

( ) = 300 MPa = 300 MN/m σ

yt

2











Theories of Failure From My SOM Book

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