Maximum Distortion Energy Theory MCQ Quiz - Objective Question with Answer for Maximum Distortion Energy Theory - Download Free PDF

Concept:

Maximum distortion energy theory states that the failure or yielding occurs at a point in member when the distortion strain energy per unit volume becomes equal to the limiting distortion strain energy per unit volume at the yielding point.

(σ₁ - σ₂)² + (σ2 - σ3)2 + (σ3 - σ1)2 ≤ $2{\left(\frac{S_y}{N}\right)}^2$

S_y = yield stress, N = factor of saftey

Calculation:

Given:

σx = 3P, σy = -2P and τxy = √2 P

Sy = 350 MPa, N = 1

Principal stress are calculated 

σ_1,2 = $\frac{1}{2}[({\sigma }_x-{\sigma }_y)\pm \sqrt{({\sigma }_x+{\sigma }_y{)}^2+(2{\tau }_{xy}{)}^2}]$

σ1,2 = $\frac{1}{2}[(3P-2P)\pm \sqrt{(3P+2P{)}^2+(2\times \sqrt{2}P{)}^2}]$

σ1,2  = $\frac{1}{2}[P\pm \sqrt{33}P]$

σ1 = 3.375 P, σ2 = - 2.375 P

According to Maximum distortion energy theory:

(σ1 - σ2)2 + (σ2 - σ3)2 + (σ3 - σ1)2 ≤ $2{\left(\frac{S_y}{N}\right)}^2$

here σ₃ = 0, on simplyfying we get 

σ₁² + σ22 - σ1σ2 = $\frac{S_y}{N}$

(3.375 P)² + ( - 2.375 P)² - (3.375 P)( - 2.375 P) = (350/1)²

11.4 P² + 5.64 P² + 8.015 P² = 350²

25.055 (P)² = (350)²

P = 350 / 5 = 70 MPa

Explanation:

Maximum shear strain energy / Distortion energy theory / Mises – Henky theory:

It states that inelastic action at any point in body, under any combination of stress begging, when the strain energy of distortion per unit volume absorbed at the point is equal to the strain energy of distortion absorbed per unit volume at any point in a bar stressed to the elastic limit under the state of uniaxial stress as occurs in a simple tension/compression test.

$\frac{1}{2}\left[{\left({\sigma }_1-{\sigma }_2\right)}^2+{\left({\sigma }_2-{\sigma }_3\right)}^2+{\left({\sigma }_3-{\sigma }_1\right)}^2\right]\le {\sigma }_{\mathrm{y}}^2$ for no failure

$\frac{1}{2}\left[{\left({\sigma }_1-{\sigma }_2\right)}^2+{\left({\sigma }_2-{\sigma }_3\right)}^2+{\left({\sigma }_3-{\sigma }_1\right)}^2\right]\le {\left(\frac{{\sigma }_{\mathrm{y}}}{\mathrm{F}\mathrm{O}\mathrm{S}}\right)}^2$ For design

• It is the best suitable theory for ductile material.
• It cannot be applied to the material under hydrostatic pressure.

​Additional Information

Maximum shear stress theory (Guest & Tresca’s Theory):

According to this theory, failure of the specimen subjected to any combination of a load when the maximum shearing stress at any point reaches the failure value equal to that developed at the yielding in an axial tensile or compressive test of the same material.

Graphical Representation:

${\tau }_{\mathrm{m}\mathrm{a}\mathrm{x}}\le \frac{{\sigma }_{\mathrm{y}}}{2}$ For no failure

${\sigma }_1-{\sigma }_2\le \left(\frac{{\sigma }_{\mathrm{y}}}{\mathrm{F}\mathrm{O}\mathrm{S}}\right)$ For design

σ1 and σ2 are maximum and minimum principal stress respectively.

Here, τmax = Maximum shear stress

σy = permissible stress

This theory is justified but a conservative theory for ductile materials. It is an uneconomical theory. 

Maximum principal stress theory (Rankine’s theory):

According to this theory, the permanent set takes place under a state of complex stress, when the value of maximum principal stress is equal to that of yield point stress as found in a simple tensile test.

For the design criterion, the maximum principal stress (σ1) must not exceed the working stress ‘σy’ for the material.

${\sigma }_{1,2}\le {\sigma }_{\mathrm{y}}$ for no failure

${\sigma }_{1,2}\le \frac{\sigma }{\mathrm{F}\mathrm{O}\mathrm{S}}$ for design

Note: For no shear failure τ ≤ 0.57 σy

Graphical representation:

For brittle material, which does not fail by yielding but fail by brittle fracture, this theory gives a satisfactory result.

The graph is always square even for different values of σ1 and σ2.

Maximum principal strain theory (ST. Venant’s theory):

According to this theory, a ductile material begins to yield when the maximum principal strain reaches the strain at which yielding occurs in simple tension.

${ϵ}_{1,2}\le \frac{{\sigma }_{\mathrm{y}}}{{\mathrm{E}}_1}$ For no failure in uniaxial loading.

$\frac{{\sigma }_1}{\mathrm{E}}-\mu \frac{{\sigma }_2}{\mathrm{E}}-\mu \frac{{\sigma }_3}{\mathrm{E}}\le \frac{{\sigma }_{\mathrm{y}}}{\mathrm{E}}$ For no failure in triaxial loading.

${\sigma }_1-\mu {\sigma }_2-\mu {\sigma }_3\le \left(\frac{{\sigma }_{\mathrm{y}}}{\mathrm{F}\mathrm{O}\mathrm{S}}\right)$ For design, Here, ϵ = Principal strain

σ1, σ2, and σ3 = Principal stresses   

Graphical Representation:

This theory overestimates the elastic strength of ductile material.

Maximum strain energy theory (Haigh’s theory):

According to this theory, a body complex stress fails when the total strain energy at the elastic limit in simple tension.

Graphical Representation:

$\left\{{\sigma }_1^2+{\sigma }_2^2+{\sigma }_3^2-2\mu \left({\sigma }_1{\sigma }_2+{\sigma }_2{\sigma }_3+{\sigma }_3{\sigma }_1\right)\right\}\le {\sigma }_{\mathrm{y}}^2$  for no failure

$\left\{{\sigma }_1^2+{\sigma }_2^2+{\sigma }_3^2-2\mu \left({\sigma }_1{\sigma }_2+{\sigma }_2{\sigma }_3+{\sigma }_3{\sigma }_1\right)\right\}\le {\left(\frac{{\sigma }_{\mathrm{y}}}{\mathrm{F}\mathrm{O}\mathrm{S}}\right)}^2$ for design

This theory does not apply to brittle material for which elastic limit stress in tension and in compression are quite different.

Important Points  

• For Brittle material:- Maximum  Principal Stress Theory (Rankine criteria) is used.
• Maximum Shear Stress Theory (Tresca theory), Total strain energy theory, Maximum Distortion Energy Theory (von Mises) useful for a ductile material.
• Tresca's theory fails in the hydrostatic state of stresses.
• All theories will give the same results if loading is uniaxial.

Explanation

Distortion Energy Theory OR The von Mises - Hencky Theory:

It has been observed that a solid under hydro-static external pressure (e.g. volume element subjected to three equal normal stresses) can withstand very large stresses. But when there is also the energy of distortion or shear to be stored, as in the tensile test, the stresses that may be imposed are limited.

As it was recognized that engineering materials could withstand enormous amounts of hydro-static pressures without damage, it was postulated that a given material has a definitely limited capacity to absorb the energy of distortion and that any attempt to subject the material to greater amounts of distortion energy results in yielding failure.

This theory states that no shearing stresses and shearing strains will be present anywhere in the block but only the volume changes.

Equation & Diagrammatic Representation:

For no failure

It cannot be applied for material under hydrostatic pressure.

Concept:

Von Mises yield criterion theory: 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 for the Von Mises theory is given by:

${\left({\mathbf{\sigma }}_1-{\mathbf{\sigma }}_2\right)}^2+{\left({\mathbf{\sigma }}_2-{\mathbf{\sigma }}_3\right)}^2+{\left({\mathbf{\sigma }}_3-{\mathbf{\sigma }}_1\right)}^2\le 2\times {\mathbf{\sigma }}_{\mathbf{y}}^2$

if σ₃ = 0, then, ${\sigma }_1^2+{\sigma }_2^2-{\sigma }_1{\sigma }_1\le {\sigma }_y^2$

where, ${\sigma }_1,{\sigma }_2,{\sigma }_3$ are the principal stresses, ${\sigma }_y$ = yield stress

Calculation:

Given,

σ_x = 40 MPa, σ_y = -20 MPa, τ_xy = 10 MPa

Major (σ1) and Minor (σ2) principal stresses are given by:

${\sigma }_1\mathrm{\setminus }{\sigma }_2=\frac{{\sigma }_{\mathrm{x}}+{\sigma }_{\mathrm{y}}}{2}\pm \sqrt{{\left(\frac{{\sigma }_{\mathrm{x}}+{\sigma }_{\mathrm{y}}}{2}\right)}^2+{\tau }_{\mathrm{x}\mathrm{y}}^2}$

${\sigma }_1=\frac{{\sigma }_{\mathrm{x}}+{\sigma }_{\mathrm{y}}}{2}+\sqrt{{\left(\frac{{\sigma }_{\mathrm{x}}-{\sigma }_{\mathrm{y}}}{2}\right)}^2+{\tau }_{\mathrm{x}\mathrm{y}}^2}$

${\sigma }_1=\frac{40-20}{2}+\sqrt{{\left(\frac{40-(-20)}{2}\right)}^2+{10}^2}$

σ₁ = 10 + 31.62 = 41.62 Mpa

σ₂ = 10 - 31.62 = -21.62 MPa

As per Von and mises criteria:

${\sigma }_1^2+{\sigma }_2^2-{\sigma }_1{\sigma }_1\le {\sigma }_y^2$

(41.62)² + (-21.62)² - (41.62) × (-21.62) ≤ (σ_y)²

 (σ_y)² = 3099.473

σ_y = 55.67

∴ yield strength of material, σy = 55.67 MPa

Concept:

We know that in a 3D element,

Distortion Energy = $\frac{1}{12G}\left[{\left({\sigma }_1-{\sigma }_2\right)}^2+{\left({\sigma }_2-{\sigma }_3\right)}^2+{\left({\sigma }_3-{\sigma }_1\right)}^2\right]$

For two dimensional, σ₃ = 0

∴ Distortion Energy = $\frac{1}{12G}\left[{\left({\sigma }_1-{\sigma }_2\right)}^2+{\sigma }_2^2+{\sigma }_1^2\right]$

= $\frac{1}{6G}\left[{\sigma }_1^2+{\sigma }_2^2-2{\sigma }_1{\sigma }_2\right]$

Calculation:

Given,

σ₁ = 90 N/mm², σ₂ = 0 and σ₃ = 0

∵ We know, Distortion energy = $\frac{1}{6G}\left[{\sigma }_1^2+{\sigma }_2^2-2{\sigma }_1{\sigma }_2\right]$

∴ Distortion Energy = $\frac{1}{6G}\times {90}^2=\frac{1350}{G}$

Concept:

von-Mises stress is given by

${({\sigma }_1-{\sigma }_2)}^2+{({\sigma }_2-{\sigma }_3)}^2+{({\sigma }_1-{\sigma }_3)}^2\le 2\cdot {\left({\sigma }_{per}\right)}^2$

σ_per = von-Mises stress

Calculation:

σ₁ = 300 N/mm²

σ₂ = 300 N/mm², σ₃ = 0

Then,

$2{\sigma }_{per}^2={(300-300)}^2+{(300-0)}^2+{(300-0)}^2$

$2{\sigma }_{per}^2=2\times {300}^2$

Explanation:

The Strain Energy of Distortion or Distortion energy per unit volume is given by:

Distortion energy per unit volume = (Total Strain energy) - (Energy of dilation)

Total Strain Energy is given by:

$\mathrm{U}=\frac{1}{2\mathrm{E}}\left\{{\sigma }_1^2+{\sigma }_2^2+{\sigma }_3^2-2\mu \left({\sigma }_1{\sigma }_2+{\sigma }_2{\sigma }_3+{\sigma }_3{\sigma }_1\right)\right\}$

Energy of dilation:

$U_v=\frac{(1-2\mu )}{6E}\times \frac{({\sigma }_1+{\sigma }_2+{\sigma }_3{)}^2}{3}$

Distortion energy per unit volume = (Total Strain energy) - (Energy of dilation)

Ud = $\frac{1+\mu }{6E}[({\sigma }_1-{\sigma }_2{)}^2+({\sigma }_2-{\sigma }_3{)}^2+({\sigma }_3-{\sigma }_1{)}^2]$..... eq (i)

In simple tension test:

U_d = $\left(\frac{1+\mu }{6E}\right)2{\sigma }_{ym}^2$ .... eq (ii)

Maximum distortion energy theory (Von mises theory)

• According to this theory, the failure or yielding occurs at a point in a member when the distortion strain energy per unit volume reaches the limiting distortion energy (i.e. distortion energy at yield point) per unit volume as determined from a simple tension test.
• von misses stress under triaxial condition is given by:

${\sigma }_{vm}=\sqrt{\frac{1}{2}\left\{{\left({\sigma }_1-{\sigma }_2\right)}^2+{\left({\sigma }_2-{\sigma }_3\right)}^2+{\left({\sigma }_3-{\sigma }_1\right)}^2\right\}}$

Now if we compare von misses stress and distortion energy per unit volume equation then,

Ud =$\frac{1+\mu }{3E}\times {\sigma }_{vm}^2$

U_d ∝ σ²_vm

σ_vm ∝  $\sqrt{U_d}$

The von Mises stress at a point in a body subjected to forces is proportional to the square root of the distortional strain energy per unit volume.

Explanation:

Maximum shear strain energy / Distortion energy theory / Mises – Henky theory:

It states that inelastic action at any point in body, under any combination of stress begging, when the strain energy of distortion per unit volume absorbed at the point is equal to the strain energy of distortion absorbed per unit volume at any point in a bar stressed to the elastic limit under the state of uniaxial stress as occurs in a simple tension/compression test.

$\frac{1}{2}\left[{\left({\sigma }_1-{\sigma }_2\right)}^2+{\left({\sigma }_2-{\sigma }_3\right)}^2+{\left({\sigma }_3-{\sigma }_1\right)}^2\right]\le {\sigma }_{\mathrm{y}}^2$ for no failure

$\frac{1}{2}\left[{\left({\sigma }_1-{\sigma }_2\right)}^2+{\left({\sigma }_2-{\sigma }_3\right)}^2+{\left({\sigma }_3-{\sigma }_1\right)}^2\right]\le {\left(\frac{{\sigma }_{\mathrm{y}}}{\mathrm{F}\mathrm{O}\mathrm{S}}\right)}^2$ For design

• It is the best suitable theory for ductile material.
• It cannot be applied to the material under hydrostatic pressure.

​Additional Information

Maximum shear stress theory (Guest & Tresca’s Theory):

According to this theory, failure of the specimen subjected to any combination of a load when the maximum shearing stress at any point reaches the failure value equal to that developed at the yielding in an axial tensile or compressive test of the same material.

Graphical Representation:

${\tau }_{\mathrm{m}\mathrm{a}\mathrm{x}}\le \frac{{\sigma }_{\mathrm{y}}}{2}$ For no failure

${\sigma }_1-{\sigma }_2\le \left(\frac{{\sigma }_{\mathrm{y}}}{\mathrm{F}\mathrm{O}\mathrm{S}}\right)$ For design

σ1 and σ2 are maximum and minimum principal stress respectively.

Here, τmax = Maximum shear stress

σy = permissible stress

This theory is justified but a conservative theory for ductile materials. It is an uneconomical theory. 

Maximum principal stress theory (Rankine’s theory):

According to this theory, the permanent set takes place under a state of complex stress, when the value of maximum principal stress is equal to that of yield point stress as found in a simple tensile test.

For the design criterion, the maximum principal stress (σ1) must not exceed the working stress ‘σy’ for the material.

${\sigma }_{1,2}\le {\sigma }_{\mathrm{y}}$ for no failure

${\sigma }_{1,2}\le \frac{\sigma }{\mathrm{F}\mathrm{O}\mathrm{S}}$ for design

Note: For no shear failure τ ≤ 0.57 σy

Graphical representation:

For brittle material, which does not fail by yielding but fail by brittle fracture, this theory gives a satisfactory result.

The graph is always square even for different values of σ1 and σ2.

Maximum principal strain theory (ST. Venant’s theory):

According to this theory, a ductile material begins to yield when the maximum principal strain reaches the strain at which yielding occurs in simple tension.

${ϵ}_{1,2}\le \frac{{\sigma }_{\mathrm{y}}}{{\mathrm{E}}_1}$ For no failure in uniaxial loading.

$\frac{{\sigma }_1}{\mathrm{E}}-\mu \frac{{\sigma }_2}{\mathrm{E}}-\mu \frac{{\sigma }_3}{\mathrm{E}}\le \frac{{\sigma }_{\mathrm{y}}}{\mathrm{E}}$ For no failure in triaxial loading.

${\sigma }_1-\mu {\sigma }_2-\mu {\sigma }_3\le \left(\frac{{\sigma }_{\mathrm{y}}}{\mathrm{F}\mathrm{O}\mathrm{S}}\right)$ For design, Here, ϵ = Principal strain

σ1, σ2, and σ3 = Principal stresses   

Graphical Representation:

This theory overestimates the elastic strength of ductile material.

Maximum strain energy theory (Haigh’s theory):

According to this theory, a body complex stress fails when the total strain energy at the elastic limit in simple tension.

Graphical Representation:

$\left\{{\sigma }_1^2+{\sigma }_2^2+{\sigma }_3^2-2\mu \left({\sigma }_1{\sigma }_2+{\sigma }_2{\sigma }_3+{\sigma }_3{\sigma }_1\right)\right\}\le {\sigma }_{\mathrm{y}}^2$  for no failure

$\left\{{\sigma }_1^2+{\sigma }_2^2+{\sigma }_3^2-2\mu \left({\sigma }_1{\sigma }_2+{\sigma }_2{\sigma }_3+{\sigma }_3{\sigma }_1\right)\right\}\le {\left(\frac{{\sigma }_{\mathrm{y}}}{\mathrm{F}\mathrm{O}\mathrm{S}}\right)}^2$ for design

This theory does not apply to brittle material for which elastic limit stress in tension and in compression are quite different.

Important Points  

• For Brittle material:- Maximum  Principal Stress Theory (Rankine criteria) is used.
• Maximum Shear Stress Theory (Tresca theory), Total strain energy theory, Maximum Distortion Energy Theory (von Mises) useful for a ductile material.
• Tresca's theory fails in the hydrostatic state of stresses.
• All theories will give the same results if loading is uniaxial.

Concept:

Maximum principal stress theory (Rankine’s theory)

According to this theory, the permanent set takes place under a state of complex stress, when the value of maximum principal stress is equal to that of yield point stress as found in a simple tensile test.

For the design criterion, the maximum principal stress (σ1) must not exceed the working stress ‘σy’ for the material.

${\sigma }_{1,2}\le {\sigma }_{\mathrm{y}}$ for no failure

${\sigma }_{1,2}\le \frac{\sigma }{\mathrm{F}\mathrm{O}\mathrm{S}}$ for design

Note: For no shear failure τ ≤ 0.57 σy

Graphical representation

For brittle material, which does not fail by yielding but fail by brittle fracture, this theory gives a satisfactory result.

The graph is always square even for different values of σ1 and σ2.

Maximum principal strain theory (ST. Venant’s theory)

According to this theory, a ductile material begins to yield when the maximum principal strain reaches the strain at which yielding occurs in simple tension.

${ϵ}_{1,2}\le \frac{{\sigma }_{\mathrm{y}}}{{\mathrm{E}}_1}$ For no failure in uniaxial loading.

$\frac{{\sigma }_1}{\mathrm{E}}-\mu \frac{{\sigma }_2}{\mathrm{E}}-\mu \frac{{\sigma }_3}{\mathrm{E}}\le \frac{{\sigma }_{\mathrm{y}}}{\mathrm{E}}$ For no failure in triaxial loading.

${\sigma }_1-\mu {\sigma }_2-\mu {\sigma }_3\le \left(\frac{{\sigma }_{\mathrm{y}}}{\mathrm{F}\mathrm{O}\mathrm{S}}\right)$ For design, Here, ϵ = Principal strain

σ1, σ2, and σ3 = Principal stresses   

Graphical Representation

This theory overestimates the elastic strength of ductile material.

Maximum shear stress theory

(Guest & Tresca’s Theory)

According to this theory, failure of the specimen subjected to any combination of a load when the maximum shearing stress at any point reaches the failure value equal to that developed at the yielding in an axial tensile or compressive test of the same material.

Graphical Representation

${\tau }_{\mathrm{m}\mathrm{a}\mathrm{x}}\le \frac{{\sigma }_{\mathrm{y}}}{2}$ For no failure

${\sigma }_1-{\sigma }_2\le \left(\frac{{\sigma }_{\mathrm{y}}}{\mathrm{F}\mathrm{O}\mathrm{S}}\right)$ For design

σ1 and σ2 are maximum and minimum principal stress respectively.

Here, τmax = Maximum shear stress

σy = permissible stress

This theory is well justified for ductile materials.

Maximum strain energy theory (Haigh’s theory)

According to this theory, a body complex stress fails when the total strain energy at the elastic limit in simple tension.

Graphical Representation.

$\left\{{\sigma }_1^2+{\sigma }_2^2+{\sigma }_3^2-2\mu \left({\sigma }_1{\sigma }_2+{\sigma }_2{\sigma }_3+{\sigma }_3{\sigma }_1\right)\right\}\le {\sigma }_{\mathrm{y}}^2$  for no failure

$\left\{{\sigma }_1^2+{\sigma }_2^2+{\sigma }_3^2-2\mu \left({\sigma }_1{\sigma }_2+{\sigma }_2{\sigma }_3+{\sigma }_3{\sigma }_1\right)\right\}\le {\left(\frac{{\sigma }_{\mathrm{y}}}{\mathrm{F}\mathrm{O}\mathrm{S}}\right)}^2$ for design

This theory does not apply to brittle material for which elastic limit stress in tension and in compression are quite different.

Maximum shear strain energy / Distortion energy theory / Mises – Henky theory.

It states that inelastic action at any point in body, under any combination of stress begging, when the strain energy of distortion per unit volume absorbed at the point is equal to the strain energy of distortion absorbed per unit volume at any point in a bar stressed to the elastic limit under the state of uniaxial stress as occurs in a simple tension/compression test.

$\frac{1}{2}\left[{\left({\sigma }_1-{\sigma }_2\right)}^2+{\left({\sigma }_2-{\sigma }_3\right)}^2+{\left({\sigma }_3-{\sigma }_1\right)}^2\right]\le {\sigma }_{\mathrm{y}}^2$ for no failure

$\frac{1}{2}\left[{\left({\sigma }_1-{\sigma }_2\right)}^2+{\left({\sigma }_2-{\sigma }_3\right)}^2+{\left({\sigma }_3-{\sigma }_1\right)}^2\right]\le {\left(\frac{{\sigma }_{\mathrm{y}}}{\mathrm{F}\mathrm{O}\mathrm{S}}\right)}^2$ For design

It cannot be applied for material under hydrostatic pressure.

All theories will give the same results if loading is uniaxial.

Conclusion:

For Brittle material:- Maximum  Principal Stress Theory (Rankine criteria) is used.

Maximum Shear Stress Theory (Tresca theory), Total strain energy theory, Maximum Distortion Energy Theory (von Mises) useful for ductile material.

Tresca theory fails in hydro static state of stresses.

Concept:

Maximum distortion energy theory (Von mises theory)

According to this theory, the failure or yielding occurs at a point in a member when the distortion strain energy per unit volume reaches the limiting distortion energy (i.e. distortion energy at yield point) per unit volume as determined from simple tension test.

Equivalent stress under triaxial condition is given by:

${\left({\sigma }_1-{\sigma }_2\right)}^2+{\left({\sigma }_2-{\sigma }_3\right)}^2+{\left({\sigma }_3-{\sigma }_1\right)}^2\le 2\left(\frac{S_Y}{FOS}\right)$ 

In shear σ₁ = τ, σ₂ = -τ, σ₃ = 0

Substituting all the values,

$\tau =\frac{S_Y}{\sqrt{3}}=0.577S_Y$ 

Concept:

According to Tresca Theory,

 $Max\left\{\left({\sigma }_1-{\sigma }_2\right),\left({\sigma }_2-{\sigma }_3\right),\left({\sigma }_3-{\sigma }_1\right)\right\}\le \frac{{\sigma }_{yt}}{n_T}\dots \left(1\right)$

According to Von–Mises

${\left({\sigma }_1-{\sigma }_2\right)}^2+{\left({\sigma }_2-{\sigma }_3\right)}^2+{\left({\sigma }_3-{\sigma }_1\right)}^2+6\left({\tau }_{xy}^2+{\tau }_{yz}^2+{\tau }_{zx}^2\right)\le 2{\left(\frac{{\sigma }_{yt}}{n_v}\right)}^2\dots \left(2\right)$

Where,

Maximum shear stress (Tresca) theory,  FOS = nT, Von-Mises (Maximum distortion energy) Theory, FOS = nV

Calculation:

Given:

σ₁ = 100 MPa, σ₂ = 100 MPa, σ₃ = 0, σ_yt = 200 MPa

By using equation (1),

$Max\left\{0,100,-100\right\}\le \frac{200}{n_T}\Rightarrow n_T=2$

By using equation (2),

${\left(0\right)}^2+{\left(100\right)}^2+{\left(0-100\right)}^2\le 2{\left(\frac{200}{n_V}\right)}^2$

$\Rightarrow 10000+10000\le 2{\left(\frac{200}{n_V}\right)}^2$

_{⇒ nv} = 2

Concept:

Von Mises yield criterion theory: 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 for the Von Mises theory is given by:

${\left({\mathbf{\sigma }}_1-{\mathbf{\sigma }}_2\right)}^2+{\left({\mathbf{\sigma }}_2-{\mathbf{\sigma }}_3\right)}^2+{\left({\mathbf{\sigma }}_3-{\mathbf{\sigma }}_1\right)}^2\le 2\times {\mathbf{\sigma }}_{\mathbf{y}}^2$

if σ₃ = 0, then, ${\sigma }_1^2+{\sigma }_2^2-{\sigma }_1{\sigma }_1\le {\sigma }_y^2$

where, ${\sigma }_1,{\sigma }_2,{\sigma }_3$ are the principal stresses, ${\sigma }_y$ = yield stress

Calculation:

Given,

σ_x = 40 MPa, σ_y = -20 MPa, τ_xy = 10 MPa

Major (σ1) and Minor (σ2) principal stresses are given by:

${\sigma }_1\mathrm{\setminus }{\sigma }_2=\frac{{\sigma }_{\mathrm{x}}+{\sigma }_{\mathrm{y}}}{2}\pm \sqrt{{\left(\frac{{\sigma }_{\mathrm{x}}+{\sigma }_{\mathrm{y}}}{2}\right)}^2+{\tau }_{\mathrm{x}\mathrm{y}}^2}$

${\sigma }_1=\frac{{\sigma }_{\mathrm{x}}+{\sigma }_{\mathrm{y}}}{2}+\sqrt{{\left(\frac{{\sigma }_{\mathrm{x}}-{\sigma }_{\mathrm{y}}}{2}\right)}^2+{\tau }_{\mathrm{x}\mathrm{y}}^2}$

${\sigma }_1=\frac{40-20}{2}+\sqrt{{\left(\frac{40-(-20)}{2}\right)}^2+{10}^2}$

σ₁ = 10 + 31.62 = 41.62 Mpa

σ₂ = 10 - 31.62 = -21.62 MPa

As per Von and mises criteria:

${\sigma }_1^2+{\sigma }_2^2-{\sigma }_1{\sigma }_1\le {\sigma }_y^2$

(41.62)² + (-21.62)² - (41.62) × (-21.62) ≤ (σ_y)²

 (σ_y)² = 3099.473

σ_y = 55.67

∴ yield strength of material, σy = 55.67 MPa

Explanation

Distortion Energy Theory OR The von Mises - Hencky Theory:

It has been observed that a solid under hydro-static external pressure (e.g. volume element subjected to three equal normal stresses) can withstand very large stresses. But when there is also energy of distortion or shear to be stored, as in the tensile test, the stresses that may be imposed are limited.

As, it was recognized that engineering materials could withstand enormous amounts of hydro-static pressures without damage, it was postulated that a given material has a definite limited capacity to absorb energy of distortion and that any attempt to subject the material to greater amounts of distortion energy result in yielding failure.

This theory states that no shearing stresses and shearing strains will be present anywhere in the block but only the volume changes.

Equation & Diagrammatic Representation:

For no failure

It cannot be applied for material under hydrostatic pressure.

Concept:

Maximum distortion energy theory states that the failure or yielding occurs at a point in member when the distortion strain energy per unit volume becomes equal to the limiting distortion strain energy per unit volume at the yielding point.

(σ₁ - σ₂)² + (σ2 - σ3)2 + (σ3 - σ1)2 ≤ $2{\left(\frac{S_y}{N}\right)}^2$

S_y = yield stress, N = factor of saftey

Calculation:

Given:

σx = 3P, σy = -2P and τxy = √2 P

Sy = 350 MPa, N = 1

Principal stress are calculated 

σ_1,2 = $\frac{1}{2}[({\sigma }_x-{\sigma }_y)\pm \sqrt{({\sigma }_x+{\sigma }_y{)}^2+(2{\tau }_{xy}{)}^2}]$

σ1,2 = $\frac{1}{2}[(3P-2P)\pm \sqrt{(3P+2P{)}^2+(2\times \sqrt{2}P{)}^2}]$

σ1,2  = $\frac{1}{2}[P\pm \sqrt{33}P]$

σ1 = 3.375 P, σ2 = - 2.375 P

According to Maximum distortion energy theory:

(σ1 - σ2)2 + (σ2 - σ3)2 + (σ3 - σ1)2 ≤ $2{\left(\frac{S_y}{N}\right)}^2$

here σ₃ = 0, on simplyfying we get 

σ₁² + σ22 - σ1σ2 = $\frac{S_y}{N}$

(3.375 P)² + ( - 2.375 P)² - (3.375 P)( - 2.375 P) = (350/1)²

11.4 P² + 5.64 P² + 8.015 P² = 350²

25.055 (P)² = (350)²

P = 350 / 5 = 70 MPa

Concept:

Von Mises yield criterion theory: 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 for the Von Mises theory is given by:

${\left({\mathbf{\sigma }}_1-{\mathbf{\sigma }}_2\right)}^2+{\left({\mathbf{\sigma }}_2-{\mathbf{\sigma }}_3\right)}^2+{\left({\mathbf{\sigma }}_3-{\mathbf{\sigma }}_1\right)}^2\le 2\times {\mathbf{\sigma }}_{\mathbf{y}}^2$

where, ${\sigma }_1,{\sigma }_2,{\sigma }_3$ are the principal stresses, ${\sigma }_y$ = yield stress

Calculation:

Given:

${\sigma }_1=\sigma ,{\sigma }_2=0,{\sigma }_3=-\sigma$, yield strength (${\sigma }_y$) = 700 MPa

${\left({\mathbf{\sigma }}_1-{\mathbf{\sigma }}_2\right)}^2+{\left({\mathbf{\sigma }}_2-{\mathbf{\sigma }}_3\right)}^2+{\left({\mathbf{\sigma }}_3-{\mathbf{\sigma }}_1\right)}^2\le 2\times {\mathbf{\sigma }}_{\mathbf{y}}^2$

${\left(\sigma -0\right)}^2+{\left(0-\left(-\sigma \right)\right)}^2+{\left(\left(-\sigma \right)-\sigma \right)}^2\le 2\times {\sigma }_{\mathrm{y}}^2$

${\sigma }^2+{\sigma }^2+4\times {\sigma }^2\le 2\times {\sigma }_{\mathrm{y}}^2$

${\sigma }^2\le \frac{{\sigma }_{\mathrm{y}}^2}{3}$

$\sigma =\frac{700}{\sqrt{3}}$

∴ σ  = 404.14 MPa

Additional Information

The types of theory of failures and their graphical representation is given below: