THEORIES OF ELASTIC FAILURE class notes for Mechanical Engineering

 

https://www.mesubjects.net/wp-admin/post.php?post=4215&action=edit         MCQ Factor of safety

https://www.mesubjects.net/wp-admin/post.php?post=4378&action=edit             MCQ theories of elastic failure

THEORIES OF ELASTIC FAILURE

class notes for Mechanical

Engineering

There are five well known theories of

elastic failures. These theories explain

material under simple or complex loaded

condition. It is elastic under five different

criterions. It does not meet a failure in four

of the five. Failure occurs in one of the

remaining criterions. The material is safe

and elastic in four remaining ways. For

example, cast iron fails under first theory

only. Mild steel fails under second theory

only.

  • Maximum Principal Theory (Rankine Theory)

             Failure due to maximum principal stress

  • Maximum Shear Stress Theory (Guest Theory)

              Failure due to maximum shear stress

  • Maximum principal strain theory (Saint Venant’s Theory)

             Failure due to maximum principal strain

  • Maximum Principal Strain Energy Theory (Haigh Theory)

           Failure due to maximum principal strain energy

  • Distortion Energy Theory (Maximum Shear Strain Energy Theory OR Von Mises-Hencky Theory)

          Failure due to maximum shear strain energy

THEORIES OF ELASTIC

FAILURE

 Assumed in all the theories

Firstly σ1  is tensile

Secondly σ2 is tensile

Thirdly σ3 is tensile

Fourthly σ1  > σ2 > σ3

FOR DUCTILE MATERIALS

σallowyp/FOS

FOR BRITTLE MATERIALS

σallowutm/FOS

ALLOWABLE VALUES σallow, σallow, 0

Maximum shear allowed from allowable

= (σallow— σallow)/2,

allow—o)/2,     (0–σallow)/2

We know     Stresses actually present σ1, σ2 and σ3=0

Actually present maximum shear stress

1– σ2)/2, (σ23)/2 and (σ3– σ1)/2

MAXIMUM PRINCIPAL STRESS THEORY (RANKINE THEORY)

MAXIMUM PRINCIPAL STRESS THEORY (RANKINE THEORY)

ALLOWABLE stress VALUES σallow, σallow, 0

Stresses actually present σ1, σ2 and σ3=0

Finally Maximum Principal Stress Actually present≤ σult/FOS

MAXIMUM SHEAR STRESS THEORY (Guest Theory or Tresca Theory)

We know Maximum shear stress from allowable stress

(i) (σallow— σallow)/2,

(ii) (σallow—o)/2,

(iii) (0–σallow)/2

Principal Stresses actually present σ1, σ2 and σ3=0

Actually present maximum shear stress

(i) (σ1– σ2)/2

(ii) (σ23)/2

(iii) (σ3– σ1)/2

Actually present maximum shear stress≤ allowable shear stress

MAXIMUM PRINCIPAL STRAIN THEORY

(SAINT VENANT’S THEORY)

ALLOWABLE stress VALUES σallow, σallow, 0

Allowable strains  (1/E) (σallow-µ σallow),

(1/E)(σallow–0) and     (1/E)(0–σallow)

Principal Stresses actually present σ1, σ2 and σ3=0

Actual strains present

(i) (1/E)(σ1– µσ2)

(ii) (1/E)(σ2–µ σ3)

(iii) (1/E)(σ3— µσ1)

If Actually present≤ Allowable strain, it is safe

MAXIMUM PRINCIPAL STRAIN ENERGY THEORY-HAIGH THEORY

ALLOWABLE stress VALUES σallow, σallow, 0

Allowable strain energy

u=(1/2E)( σ2allow+ σ2allow–2µ σallow σallow)

Stresses actually present σ1, σ2 and σ3=0

Strain energy ACTUALLY present

u=(1/2E)( σ12+ σ22–2µ σ1 σ2)

FINAL IS

( σ12+ σ22–2µ σ1 σ2)= σ2allow

MAXIMUM SHEAR STRAIN ENERGY THEORY (VON MISES HENCKY THEORY)

ALLOWABLE stress VALUES σallow, σallow, 0

Allowable shear strain energy

uSSE =(1/12G)[( σallow—σallow)2 +(σallow–0)2 + (0- σallow)2

uSSE =(1/6G) σ2allow

Stresses actually present σ1, σ2 and σ3=0

ACTUAL SHEAR STRAIN ENERGY PRESENT

uSSE =(1/12G)[( σ1— σ2)2 +(σ2–σ3)2 + (σ3– σ1)2]

Actually present≤ Allowable

FINAL SHEAR STRAIN ENERGY THEORY

[( σ1—σ2)2 +(σ2–0)2 + (0- σ1)2 = 2 σ2allow

GRAPHICAL REPRESENTATION OF VARIOUS THEORIES OF FAILURE

ULTIMATE FINAL RESULTS

  • Maximum principal theory

Largest of σ1, σ2 and σ3 =σyp/FOS

  • Maximum shear stress theory

Largest of 1σ2)/2, (σ2σ3)/2 and (σ3 σ1)/2 = σyp/2 FOS

  • Maximum principal strain theory

σ1µσ2 = σYp/FOS

  • Maximum Principal Strain energy theory

( σ12+ σ22–2µ σ1 σ2)= (σyp/FOS)2

  • MAXIMUM SHEAR STRAIN ENERGY THEORY

OR

MAXIMUM DISTORSION ENERGY THEORY

[( σ1—σ2)2 +(σ2–0)2 + (0- σ1)2  = 2 (σyp /FOS)2      

Conclusions

(i)  In any theory, if the word ‘Principal’ exists, that will be applicable to brittle materials only.

(ii)  In any theory, if the word ‘Shear’ exists, that will be applicable to ductile materials only.

(iii) A particular theory is applicable to few specific brittle material(s) or few specific ductile material(s).

(iv) Different theories won’t be applicable to the same particular brittle or same ductile material.

PROBLEM

A bolt is under an axial thrust of 9.6 k N together with a transverse force of 4.8 k N. Calculate the diameter according to:

  • Maximum principal stress theory

  • Maximum shear stress theory

Given factor of safety =3,

yield strength of the material=270 N/mm2

Poisson’s ratio=0.3

SOLUTION

Let A be the area of the bolt.

σ =9600/A

τ = 4800/A

We know      σ1,2 =σ/2 + (1/2)[(σ/2)2+4τ2]0.5

Firstly   σ1,2=9600/2A/2)±[(9600/2A)2+4x(4800/A)2]0.5

Secondly σ1=11588.2/A

Thirdly σ2 =–1988.2/A

Allowable stress =σyp/FOS=270/3=90

  • As per maximum principal theory

11588.2/A=90

A=128.76

(π/4)d2=128.76

d=12.8 mm ANS

  • As per maximum shear stress theory

[11588.2/A—(–1988.2/A)] =90

A = 150.8

(π/4)d2=150.8

= 13.8 mm ANS