\({{\rm{\sigma }}_{{\rm{n}}1}} = \left( {\frac{{{{\rm{\sigma }}_1} + {{\rm{\sigma }}_2}}}{2}} \right) + \left( {\frac{{{{\rm{\sigma }}_1} - {{\rm{\sigma }}_2}}}{2}} \right){\rm{cos}}2{\rm{\theta }}\)

\({{\rm{\sigma }}_{{\rm{n}}2}} = \left( {\frac{{{{\rm{\sigma }}_1} + {{\rm{\sigma }}_2}}}{2}} \right) + \left( {\frac{{{{\rm{\sigma }}_1} - {{\rm{\sigma }}_2}}}{2}} \right){\rm{cos}}2{\rm{\theta }}\)

when θ = 45° Both cos 2θ = cos 90° =0

So the normal stress will be,\(\left( {\frac{{{{\rm{\sigma }}_1} + {{\rm{\sigma }}_2}}}{2}} \right)\) = Half of sum of Normal stresses

Concept:

1) Maximum principal stress theory (Rankine theory/ Lame’s theory):-

As per this theory, for no failure maximum principal stress should be less than yield stress under uniaxial loading.

i.e. σ_maj < f_y

For design, \({\sigma _{maj}} < \frac{{{f_y}}}{{F.O.S}}\)

This theory is most suitable for brittle materials, not applicable to ductile materials, not applicable to pure shear case because as per this theory τ should be less then f_y

2) Maximum Principle strain theory (Saint-Venant theory):-

As per this theory, for no failure maximum principal strain should be less than strain under uniaxial loading when the stress is f_y.

\(i.e\;{\varepsilon _{maj}} < \frac{{{f_y}}}{E}\)

For Design, \({\varepsilon _{maj}} < \frac{{\left( {{f_y}/FOS} \right)}}{E}\)

This theory is satisfactory for brittle material, not suitable for hydrostatic stress conditions.

This theory is not suitable for the pure shear case.

3) Maximum shear stress Theory (Tresca theory/ Guest Theory/ Coulomb Theory):-

As per this theory, for no failure absolute maximum shear stress should be less than maximum shear stress under uniaxial loading, when the stress is f_y.

Maximum shear stress under the uniaxial condition when the stress is f_y is given as f_y/2

\(\therefore {\tau _{abs}}\max < \frac{{{f_y}}}{2}\)

For Design,

\({\tau _{abs}}\max < \frac{{\left( {{f_y}/FOS} \right)}}{2}\)

4) Maximum strain energy theory (Beltrami-Haigh Theory):-

As per this theory, for no failure maximum strain energy per unit volume should be less than strain energy per unit volume under uniaxial loading when the stress is f_y.

\(\frac{1}{{2E}}\left[ {\sigma _1^2 + \sigma _2^2 + \sigma _3^2 - 2\mu \left( {{\sigma _1}{\sigma _2} + {\sigma _2}{\sigma _3} + {\sigma _3}{\sigma _1}} \right)} \right] < \frac{{f_y^2}}{{2E}}\)

Where, σ₁, σ₂, σ₃ are principal stresses.

This theory is applicable for ductile material, not suitable for brittle material, and not suitable for the pure shear case.

5) Maximum shear strain energy theory (Von mises/ Distortion energy theory):-

As per this theory, for no failure, maximum shear strain energy per unit volume should be less than maximum shear strain energy per unit volume under uniaxial loading.

\(\frac{1}{{12G}}\left[ {{{\left( {{\sigma _1}-{\sigma _2}} \right)}^2} + {{\left( {{\sigma _2} - {\sigma _3}} \right)}^2} + {{\left( {{\sigma _3} - {\sigma _1}} \right)}^2}} \right] < \frac{{f_y^2}}{{6G}}\)

This theory is in perfect agreement with the result of the test in case of pure shear.

This theory is the most suitable theory of failure for a ductile material.

Conclusion:

• All theories of failure give similar results in the case of uniaxial loading.

Concept:

Maximum shear stress theory (Tresca theory):

According to this theory, yielding would occur when the maximum shear stress just exceeds the shear stress at the tensile yield point. At the tensile yield point σ2= σ3 = 0 and thus maximum shear stress is σ_y/2. This gives us six conditions for a three-dimensional stress situation:

Mathematically,

Maximum shear stress= (Difference between two principle stresses)/2

Factor of safety=Allowed shear stress/Maximum shear stress

Calculation:

Difference between two principle stresses = 120 MPa

Allowed shear stress = 300/2 = 150 MPa

Maximum Shear stress = 120/2 = 60 MPa

F.O.S = 150/60 = 2.5

Concept:

The normal stress on a plane at angle θ is given by:

\({\sigma _n} = \frac{{{\sigma _x} + {\sigma _y}}}{2} + \frac{{{\sigma _x} - \;{\sigma _y}}}{2}\cos 2\theta + {\tau _{xy}}\sin 2\theta \)

Calculation:

Here, σ_x = σ_y = 500 MPa and τ_xy = 0

\({\sigma _n} = \frac{{500 + 500}}{2} + \frac{{500 - 500}}{2}\cos 90^\circ + 0\)

σ_n = 500 MPa

Alternate Solution:

Here σ_x = σ_y

Concept:

Calculation:

σ₁ = 300 MPa, σ₂ = 100 MPa

Radius of Mohr’s circle \( = \frac{{{\sigma _1} - {\sigma _2}}}{2} = \frac{{300 - 100}}{2} = \frac{{200}}{2} = 100\)

CM = 100

MN is shear stress at same plane where shear stress is 260 MPa

∴ ΔCMN: (CM)² = (CN)² + (MN)²

(100)² = (60)² + MN²

∴ τ_s = 80 MPa

Concept:

Plane strain condition: If the strain associated with two opposite faces are zero, the strain condition is called plane strain condition.

Stains existing may be either ϵ_x and ϵ_y or ϵ_y and ϵ_z or ϵ_x and ϵ_z along with shear strain on one plane respectively.

In a plane strain situation ; if

γ_max = Maximum shear strain.

γ_xy = Shear strain to which the element is subjected.

e_x, e_y = Axial strains along x- & y-direction respectively. 

Then ; Radius of the Mohr circle of strain ; R is given as:

\(R = \frac{{{\gamma _{max}}}}{2} = \sqrt {{{\left( {\frac{{{e_x} - {e_y}}}{2}} \right)}^2} + {{\left( {\frac{{{\gamma _{xy}}}}{2}} \right)}^2}}\)

Given:

e_x = 4 × 10^-6, e_y = 0, γ_xy = 12 × 10_-6

Calculation:

∴ \(R = \sqrt {{{\left( {\frac{{4 \times {{10}^{ - 6}}}}{2}} \right)}^2} + {{\left( {\frac{{12 \times {{10}^{ - 6}}}}{2}} \right)}^2}} = 6.325 \times {10^{ - 6}}\)

Thus ; Diameter, D = 2R = 2 × 6.325 × 10^-6 ⇒ D = 12.65 × 10^-6

Mistake Point:

Note that the radius of the Mohr circle in the stress problem is equal to the maximum shear stress. Whereas in plane strain problem, the radius of the Mohr circle equals half the value of the Maximum shear strain \(\left( {i.e.\;R = \frac{{{\gamma _{max}}}}{2}} \right)\)

Principle stresses for the given problem,

\(\begin{array}{l} {\sigma _1}\ and\ {\sigma _2} = \frac{{{\sigma _x} + {\sigma _y}}}{2} \pm \sqrt {{{\left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right)}^2} + \tau _{xy}^2} = \frac{{120 + \sigma }}{2} \pm \sqrt {{{\left( {\frac{{120 - \sigma }}{2}} \right)}^2} + {{40}^2}} \\ \Rightarrow {\sigma _1} = \frac{{120 + \sigma }}{2} + x\ and\ {\sigma _2} = \frac{{120 + \sigma }}{2} - x \end{array}\)

Where \(x = \sqrt {{{\left( {\frac{{120 - \sigma }}{2}} \right)}^2} + {{40}^2}}\)

Now, from the shear strain energy, σ₃ = 0

\(\begin{array}{l} U = \frac{1}{{12G}}\left[ {{{\left( {{\sigma _1} - {\sigma _2}} \right)}^2} + {{\left( {{\sigma _2} + {\sigma _3}} \right)}^2} + {{\left( {{\sigma _3} - {\sigma _1}} \right)}^2}} \right] = \frac{1}{{12G}}\left[ {{{\left( {{\sigma _1} - {\sigma _2}} \right)}^2} + \sigma _2^2 + \sigma _1^2} \right]\\ = \frac{1}{{12G}}\left[ {{{\left\{ {\frac{{120 + \sigma }}{2} + x - \frac{{120 + \sigma }}{2} + x} \right\}}^2} + {{\left\{ {\frac{{120 + \sigma }}{2} - x} \right\}}^2} + {{\left\{ {\frac{{120 + \sigma }}{2} + x} \right\}}^2}} \right]\\ = \frac{1}{{12G}}\left[ {6{x^2} + 2{{\left( {120 + \sigma } \right)}^2}} \right] = \frac{1}{{12G}}\left[ {6\left\{ {{{\left( {\frac{{120 - \sigma }}{2}} \right)}^2} + {{40}^2}} \right\} + 2{{\left( {120 + \sigma } \right)}^2}} \right]\\ = \frac{1}{{6G}}\left[ {{\sigma ^2} - 120\sigma + 16000} \right] \end{array}\)

Concept:

Stress transformation equation:

Consider the state of stress as shown in the figure.

\(\sigma _x' = \frac{{{\sigma _x} + {\sigma _y}}}{2} + \frac{{{\sigma _x} - {\sigma _y}}}{2}\cos 2\theta + {\tau _{xy}}\sin 2\theta \)

\({\tau _{x'y'}} = \frac{{ - \left( {{\sigma _x} - {\sigma _y}} \right)}}{2}\sin 2\;\theta + {\tau _{xy}}\cos 2\theta \)

Calculation:

The state of stress at the point is obtained as a result of a combination of stress state I and II.

σ_x = 45 + σ_x2       ----(i)

σ_y = σ_y2               ---(ii)

From the state of stress in II, θ = -30°

\({\sigma _{x2}} = \frac{{60 + 0}}{2} + \frac{{60 - 0}}{2}\cos \left( {2 \times \left( { - 30} \right)} \right) + 0\) ⇒ σ_x2 = 45 MPa

Also,

\({\tau _{{x_2}{y_2}}} = \frac{{ - \left( {60 - 0} \right)}}{2}\sin \left( {2x - 30} \right) + 0\)

⇒ \({\tau _{{x_2}{y_2}}} = 15\sqrt 3 \;\;MPa\)

The summation of normal stresses on two mutually perpendicular planes is constant.

∵ σ_x2 + σ_y2 = 60 + 0 ⇒ σ_y2 = 60 – 45 = 15 MPa

∴ The state of stress at the point ,

∴ σ_x = 90 MPa, σ_y = 15 MPa & τ_xy = 15√3 MPa

∴ Major principal stress \(= \frac{{{\sigma _x} + {\sigma _y}}}{2} + \sqrt {{{\left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right)}^2} + {{\left( {{\tau _{xy}}} \right)}^2}} \)

\(= \frac{{90 + 15}}{2} + \sqrt {{{\left( {\frac{{90 - 15}}{2}} \right)}^2} + {{\left( {15\sqrt 3 } \right)}^2}} \) = 98.12 MPa

Concept:

Maximum shear stress Theory (Tresca theory/ Guest Theory/ Coulomb Theory):-

As per this theory, for no failure absolute maximum shear stress should be less than maximum shear stress under uniaxial loading, when the stress is f_y.

Maximum shear stress under the uniaxial condition when the stress is f_y is given as f_y/2

\(\therefore {\tau _{abs}}\max < \frac{{{f_y}}}{2}\)

For Design,

\({\tau _{abs\;max}} < \;\frac{{\left( {\frac{{{f_y}}}{{FOS}}} \right)}}{{\;2}}\)

From the maximum shear stress theory, we have

\({\tau _{abs\;max}} = Max\left\{ {\left( {\frac{{{\sigma _1} - {\sigma _2}}}{2}} \right),\frac{{{\sigma _1}}}{2},\frac{{{\sigma _2}}}{2}} \right\} \le \left[ {\frac{{{\sigma _y}}}{{2 \times \left( {FOS} \right)}}} \right]\)

Maximum shear strain energy theory (Von mises/ Distortion energy theory):-

As per this theory, for no failure, maximum shear strain energy per unit volume should be less than maximum shear strain energy per unit volume under uniaxial loading.

\(\frac{1}{{12G}}\left[ {{{\left( {{\sigma _1}-{\sigma _2}} \right)}^2} + {{\left( {{\sigma _2} - {\sigma _3}} \right)}^2} + {{\left( {{\sigma _3} - {\sigma _1}} \right)}^2}} \right] < \frac{{f_y^2}}{{6G}}\)

This theory is in perfect agreement with the result of the test in case of pure shear.

This theory is the most suitable theory of failure for a ductile material.

Calculation:

From distortion energy theory,

\(\frac{1}{2}\left[ {{{\left( {{\sigma _1} - {\sigma _2}} \right)}^2} + {{\left( {{\sigma _2} - {\sigma _3}} \right)}^2} + {{\left( {{\sigma _1} - {\sigma _3}} \right)}^2}} \right] \le \sigma _y^2\)

\(\sigma _1^2 + \sigma _2^2 - {\sigma _1}{\sigma _2} \le {\left( {\frac{{{\sigma _y}}}{{FOS}}} \right)^2}\)      (∵ σ3 = 0)

As per maximum shear stress theory,

Max \(\left[ {\frac{{{p_1} - {p_2}}}{2},\frac{{{p_1}}}{2},\frac{{{p_2}}}{2}} \right] \le \frac{{{\sigma _y}}}{{2 \times FOS}}\)       

∵ p₁ is major principal stress \(\therefore \;\frac{{{p_1}}}{2} \le \frac{{{\sigma _y}}}{{2 \times FOS}}\)

Considering limiting case:

\(FOS = \frac{{{\sigma _y}}}{{{p_1}}} \Rightarrow {p_1} = \frac{{{\sigma _y}}}{{FOS}}\)

As per distortion energy theory:

\(\left( {p_1^2 + p_2^2 - {p_1}{p_2}} \right) \le {\left( {\frac{{{\sigma _y}}}{{FOS}}} \right)^2} \Rightarrow p_1^2 + p_2^2 - {p_1}{p_2} \le p_1^2\)         {∵ \({p_1} = \frac{{{\sigma _y}}}{{FOS}}\)}

Considering limiting case,

\(p_1^2 + p_2^2 - {p_1}{p_2} = p_1^2\)

\(\Rightarrow p_2^2 - {p_1}{p_2} = 0\) ⇒ p₂ = 0  or p₁ = p₂

p₂ = 0 is not given in the options. So, p₁ = p₂ is the correct answer.

Important Point:

Concept:

For 45° strain rosette

ϵ_x = ϵ₁, ϵ_y = ϵ₃ and γ_xy = 2ε₂ – (ϵ₁ + ϵ₃)

For 60° strain rosette

\({\epsilon_x} = {\epsilon_1},\;{\epsilon_y} = \frac{1}{3}\left( {2 \times {\epsilon_2} + 2{\epsilon_3} - {\epsilon_1}} \right){\rm{\;and\;}}{\gamma _{xy}} = \frac{2}{{\sqrt 3 }}\left( {{\epsilon_3} - {\epsilon_2}} \right)\) 

Calculation:

For given 45° strain rosette

Shear strain (γ_xy) is given by

γ_xy = 2ϵ₂ – (ϵ₁ + ϵ₃)

γ_xy = 2 × 40 – (25 + 30)

γ_xy = 25μ