Concept:

Pure torsion will be the case of pure shear in which principal normal stresses occur 45° to the axis.

The maximum and minimum principal stresses are equal in magnitude and opposite in direction.

2θ₁ = 90° ⇒ θ1 = 45°

Concept:

Concept:

${\tau }_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{{\sigma }_1-{\sigma }_2}{2}$

Calculation:

Given:

σ₁ = 2σ, σ₂ = 0

$\therefore {\tau }_{max}=\frac{{\sigma }_1-{\sigma }_2}{2}$

$\therefore {\tau }_{max}=\frac{2\sigma -0}{2}$

$\therefore {\tau }_{max}=\frac{2\sigma }{2}$

$\therefore {\tau }_{max}=\sigma$

Graphical Method:

Stresses on X face and Y face:

A(σ_x1 ,-τ_xy), B(0, τ_xy)

$\begin{array}{l}{\tau }_{max}=R=\frac{D}{2}=\frac{AB}{2}\\ {\tau }_{max}=\frac{1}{2}\sqrt{{\left({\sigma }_x-0\right)}^2+\left(2{\tau }_{xy}^2\right)}=\frac{1}{2}\sqrt{{\sigma }_x^2+4{\tau }_{xy}^2}\end{array}$

Analytical Method:

$\begin{array}{l}{\sigma }_1/{\sigma }_2=\frac{{\sigma }_x+0}{2}\pm \sqrt{\left(\frac{{\sigma }_x-0}{2}\right)+{\tau }_{xy}^2}\\ =\frac{{\sigma }_x}{2}\pm \frac{1}{2}\sqrt{{\sigma }_x^2+4{\tau }_{xy}^2}\\ {\tau }_{max}=\frac{1}{2}\sqrt{{\sigma }_x^2+4{\tau }_{xy}^2}\end{array}$

Radius of Mohr’s circle = 1000 MPa

Radius $={\tau }_{max}=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}$

And centre of Mohr’s circle is at a distance,

${\sigma }_{avg}=\frac{{\sigma }_x+{\sigma }_y}{2}$ from the origin. Here it is in origin.

So, $\frac{{\sigma }_x+{\sigma }_y}{2}=0$

As, ${\sigma }_x+{\sigma }_{\mathrm{y}}=0$ both the normal stress may be zero which is a pure shear case or opposite in nature which don’t exist in any of the options. So it is the pure shear case, where the radius is

${\tau }_{max}=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}=1000MPa$

Concept:

Normal stress on the plane PR

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

Tangential stress on the plane PR

${\tau }_{\theta }=\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)\sin 2\theta -{\tau }_{xy}\cos 2\theta$

Calculation:

Given:-

σ_x = 60 MPa, σ_y = 0 MPa, τ_xy = 40 MPa

θ = 90 - 60 = 30°

Now,

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

${\sigma }_{\theta }=\left(\frac{60+0}{2}\right)+\left(\frac{60-0}{2}\right)\cos 2(30)+40\sin 2(30)$

σ_θ = 79.64 MPa

Now,

${\tau }_{\theta }=\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)\sin 2\theta -{\tau }_{xy}\cos 2\theta$

${\tau }_{\theta }=\left(\frac{60-0}{2}\right)\sin 2(30)-40\cos 2(30)$

τθ  = 5.98 MPa

Concept:

The principle stress given for a state of plane stress consists of biaxial tensile stress is

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

And maximum shear stress is given by

${\tau }_{max}=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}$

Where σ_x and σ_y are normal stress in x and y direction respectively.

And σ₁ and σ₂ are maximum and minimum principle stress.

Calculation:

Given uniaxial tensile stress so let us assume that it is acting in x-direction

then σ_x = 8 kPa and σ_y = 0 and largest principal stress σ₁ = 10 kPa       

${\sigma }_1=\frac{8}{2}+\sqrt{{\left(\frac{8}{2}\right)}^2+{\tau }_{xy}^2}$

$10=4+\sqrt{4^2+{\tau }_{xy}^2}$

$\Rightarrow {\left(10-4\right)}^2=16+{\tau }_{xy}^2$

$\Rightarrow {\tau }_{xy}^2=20$

∴  τ_xy = 4.47 kPa

Concept:

Maximum Principal Stress is given by:

${\sigma }_{max}=\frac{{\sigma }_x+{\sigma }_y}{2}+\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\left({\tau }_{xy}\right)}^2}$

Calculation:

Given:

σ_max = 30 MPa, σ_x = - 50 MPa

τ_xy = 40 MPa

Now,

$30=\frac{-50+{\sigma }_y}{2}+\sqrt{{\left(\frac{-50-{\sigma }_y}{2}\right)}^2+{\left(40\right)}^2}$

$\therefore \frac{60+50-{\sigma }_y}{2}=\sqrt{{\left(\frac{-50-{\sigma }_y}{2}\right)}^2+{\left(40\right)}^2}$

${\left(\frac{110-{\sigma }_y}{2}\right)}^2={\left(\frac{-50-{\sigma }_y}{2}\right)}^2+{\left(40\right)}^2$   

$\therefore {\left(110-{\sigma }_y\right)}^2-{\left(50+{\sigma }_y\right)}^2={\left(40\right)}^2\times 4$

Now, from the relation  

a² – b² = (a + b)(a - b)

∴ (110 - σ_y + 50 + σ_y)(110 - σ_y – (50 + σ_y)) = 1600 × 4

∴ (60 – 2σ_y)(160) = 1600 × 4

∴ σ_y = 10 MPa

Concept:

From Mohr’s circle

σ_x = 100 Mpa

σ_y = 0, τ = 50 MPa

Calculation:

Major Principle stress

${\sigma }_1=\frac{1}{2}[\left({\sigma }_x+{\sigma }_y\right)+\sqrt{{\left({\sigma }_x-{\sigma }_y\right)}^2+4{\tau }^2}$

$\therefore {\sigma }_1=\frac{1}{2}[\left(100+0\right)+\sqrt{{\left(100-0\right)}^2+4{\left(50\right)}^2}$

$\therefore {\sigma }_1=\frac{1}{2}\left[100+141.421\right]$

$\therefore {\sigma }_1=\frac{241.421}{2}$

∴ σ₁ = 120.71 MPa

 Similarly

Minor Principle stress

${\sigma }_2=\frac{1}{2}\left[\left({\sigma }_x+{\sigma }_y\right)-\sqrt{{\left({\sigma }_x-{\sigma }_y\right)}^2+4{\tau }^2}\right]$

$\therefore {\sigma }_2=\frac{1}{2}\left[100-141.421\right]$

$\therefore {\sigma }_2=\frac{-41.421}{2}$

∴ σ₂ = -20.71

Concept:

Principal stresses are given by:

${\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{x{y}^2}}$

Maximum shear stress:

$\begin{array}{l}{\tau }_{max}=\frac{{\sigma }_1-{\sigma }_2}{2}\\ {\tau }_{max}=\frac{{\sigma }_1-{\sigma }_2}{2}=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}\end{array}$

Calculation:

Given:

σ_x = 50 MPa, σy = -10 MPa, τ_xy = 40 MPa

${\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \frac{1}{2}\sqrt{{\left({\sigma }_x-{\sigma }_y\right)}^2+4{\tau }_{xy}^2}$

$=\frac{50+\left(-10\right)}{2}\pm \frac{1}{2}\sqrt{{\left(50+10\right)}^2+4{\left(40\right)}^2}$

= 20 ± 50

${\sigma }_{1,2}=70MPa,-30MPa$