If the principal stresses in a plane stress problem, are σ_{1} = 100 MPa, σ_{2} = 40 MPa, the magnitude of the maximum shear stress (in MPa) will be

Option 3 : 30

**Concept:**

In the case of **3-D analysis**, maximum shear stress can be calculated as:

maximum of \([\frac{{{σ _1} - {σ _2}}}{2},\frac{{{σ _2} - {σ _3}}}{2},\frac{{{σ _1} - {σ _3}}}{2}]\)

And In the case of **plane stress analysis**, **maximum shear stress** can be calculated by,

\({{\bf{\tau }}_{{\bf{max}}}} = \frac{{{{\rm{\sigma }}_1} - {{\rm{\sigma }}_2}}}{2}\), where σ_{1} and σ_{2} are principal stresses in a plane.

**Calculation:**

**Given:**

σ1 = 100 MPa , σ_{2} = 40 MPa

Now, In plane stress analysis, we have

\({{\bf{\tau }}_{{\bf{max}}}} = \frac{{{{\rm{\sigma }}_1} - {{\rm{\sigma }}_2}}}{2}\)

\(\therefore {\tau _{max}} = \frac{{{\sigma _1} - {\sigma _2}}}{2} = \frac{{100 - 40\;}}{{2\;}} = 30\;MPa\)

Maximum In-Plane shear stress/Surface shear stress:

\(\tau_{max,inplane}=\frac{{\sigma _1}\;-\;{\sigma _2}}{{2}}\)

Maximum wall shear stress/Out plane shear stress/Absolute maximum shear stress:

\(\tau_{max,abs}=\frac{{\sigma _{max}}\;-\;{\sigma _{min}}}{{2}}=\frac{\sigma_1}{2}\)

Option 2 : 0 unit

**Concept:**

Mohr's circle is very useful in determining the relationships between normal and shear stresses acting on any inclined plane at a point in a stresses body. It is helpful in finding maximum and minimum principal stresses, maximum shear stress etc.

**Calculation:**

Since the fluid element will be subjected to hydrostatic loading (\(\sigma_1=\sigma_2=p\)) , when all principal stresses are equal and compressive.

Mohr circle will reduce into a point on σ - axis.

∴ Radius of Mohr circle = 0 unit

**For 2D hydrostatic state of stress:**

- All planes feel equal normal stress.
- There is no shear stress on any plane

Option 4 : None of these

The plane on which normal stress attains its maximum and minimum values are called principal planes. The shear stress on the principal plane is zero.

**The planes of maximum and minimum normal stresses are at an angle of 90° to each other.**

**Planes of maximum stress occur at 45° to the principal planes.**

To understand this concept in more detail and simple way, Click Here.

Option 4 : 400 MPa

__Concept: __

The maximum and minimum normal stress id given by

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

Where, σ_{x} and σ_{y} are direct stress and τ_{xy} is the shear stress

__Calculation:__

Given, Direct stress σ_{x} = 300 MPa, τ_{xy} = 200 MPa

Now maximum normal stress

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

\({\sigma _{max}} = \frac{{300}}{2} + \sqrt {{{\left( {\frac{{300}}{2}} \right)}^2} + {{200}^2}} = 400\;MPa\)

Option 4 : 90°

__Explanation:__

Principal Planes:

Those planes on which normal stresses acted alone (i.e. no shear stress) are called principal planes.

Principal Stresses:

Normal stresses acting on mutually perpendicular principal planes on which shear stress is zero are called principal stress.

The maximum value of normal stresses is called Major Principal stress and the minimum value is called Minimum Principal stress.

**Principle planes are perpendicular to each other** and the can be calculated by,

\(\tan 2θ = \frac{{2{τ _{xy}}}}{{{σ _x} - {σ _y}}}\)

where, τxy = shear stress, σx = normal stress in x-direction, σy = normal stress in y-direction,

θ_{2} = θ_{3} = θ + 90° (θ, θ_{2}, θ_{3} location of principle planes)

Option 3 : A point on the normal stress axis

**Explanation:**

**Mohr Circle:**

- It is a two-dimensional graphical representation (σ as x-axis and τ as y-axis) of the state of stress inside a body.
- The abscissa and ordinate of each point on the circle are the magnitudes of the normal stress and shear stress components, respectively.
- In other words, the Mohr circle is the locus of those points which represents the normal and shear stress on various planes passing through a point on a loaded body.

**Properties of Mohr Circle:**

- Centre of Mohr circle always lies on x-axis (σ-axis) i.e. Mohr circle is always symmetrical about the σ-axis.
- The co-ordinate of centre is \(\left ( \frac{σ_x\;+\;σ_y}{2} \right )\;or\;\left ( \frac{σ_1\;+\;σ_2}{2} \right )\) which represents normal stress on the plane of τ
_{max}. - Radius of Mohr circle represents maximum shear stress i.e. \(τ_{max}= \sqrt {{{\left({\frac{{{σ _{x}}\;-\;{σ _{y}}}}{2}} \right)}^2} + τ _{xy}^2}\;\Rightarrow\frac{σ_1\;-\;σ_2}{2}\)
- If two-point on the circumference of Mohr circle subtends an angle 2θ at centre, then the angle between those plane will be θ.

**Special case:**

**1) Hydrostatic loading / Hydrostatic stress:**

In the case of hydrostatic fluid, equal and alike normal stress acts on two mutually perpendicular planes without any shear i.e. σ_{x} = σ_{y} = σ and τ_{xy} = 0.

\(Centre =\left ( \frac{σ_x\;+\;σ_y}{2} \right )\;and\;Radius=τ_{max}\Rightarrow\sqrt {{{\left({\frac{{{σ _{x}}\;-\;{σ _{y}}}}{2}} \right)}^2} + τ _{xy}^2}\)

∴ centre is at (σ, 0) and radius = 0, **which represents a point on x-axis / σ-axis or normal stress axis**.

**2) Pure shear**

In pure shear σ_{x} and σ_{y} = 0, τ_{xy} = τ

\(Centre =\left ( \frac{σ_x\;+\;σ_y}{2} \right )\;and\;Radius=τ_{max}\Rightarrow\sqrt {{{\left({\frac{{{σ _{x}}\;-\;{σ _{y}}}}{2}} \right)}^2} + τ _{xy}^2}\)

∴ centre is at origin (0, 0) and radius = τ.

Option 3 :

450 MPa

__Concept:__

The maximum and minimum normal stress id given by

\({{\rm{σ }}_{{\rm{max,}}\;\;{\rm{min}}}} = \frac{{{{\rm{σ }}_{\rm{x}}}{\rm{ + }}{{\rm{σ }}_{\rm{y}}}}}{2}\; \pm \sqrt {{{\left( {\frac{{{{\rm{σ }}_{\rm{x}}}\;{\rm{ - }}\;{{\rm{σ }}_{\rm{y}}}}}{2}} \right)}^2} + {\rm{\tau }}_{{\rm{xy}}}^{\rm{2}}} \)

Where, σx and σy are direct stress and τxy is the shear stress

__Calculation:__

Given,

Direct stress σx = 400 MPa, τxy = 150 MPa

σ_{y} = 0 MPa

Now maximum normal stress

\({{\rm{σ }}_{{\rm{max,}}\;\;{\rm{min}}}} = \frac{{{{\rm{σ }}_{\rm{x}}}{\rm{ + }}{{\rm{σ }}_{\rm{y}}}}}{2}\; + \sqrt {{{\left( {\frac{{{{\rm{σ }}_{\rm{x}}}\;{\rm{ - }}\;{{\rm{σ }}_{\rm{y}}}}}{2}} \right)}^2} + {\rm{\tau }}_{{\rm{xy}}}^{\rm{2}}} \)

\({{\rm{σ }}_{{\rm{max}}}} = \frac{{400}}{2} + \sqrt {{{\left( {\frac{{400}}{2}} \right)}^2} + {{150}^2}} \)

**σ _{max} = 450 MPa**

**Hence the maximum normal stress is 450 MPa**

Option 3 : 3

**Explanation:**

**Principle planes:**

- These are the complementary oblique planes on which shear stress is zero but normal stress either maximum or minimum.
- They are also known as planes of zero shear or planes of pure complementary normal stresses.
- Principle planes are perpendicular to each other and the can be calculated by:

\(\tan 2θ = \frac{{2{τ _{xy}}}}{{{σ _x} - {σ _y}}}\)

where, τ_{xy} = shear stress, σ_{x} = normal stress in x-direction, σ_{y} = normal stress in y-direction,

θ_{2} = θ_{3} = θ + 90° (θ, θ_{2}, θ_{3} location of principle planes)

**At any point within a stressed body, no matter how complex the state of stress may be, there are always exist three mutually perpendicular planes on each of which the resultant stress is normal stress.**- The normal stresses (σ) and shear stresses (τ) are shown in the figure.

Option 3 : 50

**Concept:**

Maximum and minimum values of normal stresses occur on planes of **zero shearing stress**. The maximum and minimum normal stresses are called the principal stresses, and the planes on which they act are called the principal plane.

\({\sigma _{max,\;min}} = \frac{{{\sigma _{xx}} + {\sigma _{yy}}}}{2} \pm \sqrt {{{\left( {\frac{{{\sigma _{xx}} - {\sigma _{yy}}}}{2}} \right)}^2} + \tau _{xy}^2} \)

But the maximum shear stress planes may or may not contain normal stresses as the case may be.

\({\tau _{max}} = \frac{{{\sigma _{max}} - {\sigma _{min}}}}{2} = \sqrt {{{\left( {\frac{{{\sigma _{xx}} - {\sigma _{yy}}}}{2}} \right)}^2} + \tau _{xy}^2} \)

**Calculation:**

σ_{x} = 20 MPa, σ_{y} = 80 MPa and τ_{xy} = 40MPa

**\(\tau_{max}= \sqrt {{{\left( {\frac{{80 - 20}}{2}} \right)}^2} + {{40}^2}} =50\;MPa\)**

**\({\tau _{max}} = \max \left\{ {\frac{{{\sigma _1} - {\sigma _2}}}{2},\frac{{{\sigma _1}}}{2},\frac{{{\sigma _2}}}{2}} \right\}\)**

\({\sigma _1}/{\sigma _2} = \frac{{{\sigma _x} + {\sigma _y}}}{2} \pm \sqrt {{{\left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right)}^2} + \tau _{xy}^2} \)

\({\sigma _1}/{\sigma _2} = \frac{{20 + 80}}{2} \pm \sqrt {{{\left( {\frac{{20 - 80}}{2}} \right)}^2} + {{40}^2}} \)

\({\sigma _1}/{\sigma _2} = 50 \pm 50\)

\({\sigma _1} = 100\;MPa\;and\;{\sigma _2} = 0\;MPa\)

\({\tau _{max}} = \max \left\{ {\frac{{100 - 0}}{2},\frac{{100}}{2},0} \right\}\)

\({\tau _{max}} = 50\;MPa\)

For the state of stress shown in the below figure, normal stress acting on the plane of maximum shear stress is -

Option 1 : 25 MPa tension

**Concept:**

Principal stress and shear stress can be calculated for a system of stresses acting on a body is given by -

**\(σ_n=(\frac{σ_x\;+\;σ_y}{2})\;+\;(\frac{σ_x\;-\;σ_y}{2})cos\;2\theta\;+\;τ_{xy}sin\;2\theta\)**

**\(\tau_n=-(\frac{σ_x\;-\;σ_y}{2})sin\;2\theta\;+\;\tau_{xy}cos\;2\theta\)**

The plane of maximum shear stress occurs at 45° to the principal planes.

∴ normal stress at a plane of maximum shear stress is -

**\(∴ {\left( {{σ _n}} \right)_{{\tau _{max}}}} = \frac{{{σ _x}\; + \;{σ _y}}}{2}\)**

**Calculation:**

**Given:**

σ_{x} = 100 MPa, σ_{y} = - 50 MPa.

\({\left( {{σ _n}} \right)_{{\tau _{max}}}} = \frac{{{σ _x} \;+ \;{σ _y}}}{2}\)

\(\Rightarrow {\left( {{σ _n}} \right)_{{\tau _{max}}}} = \frac{{{100} \;+ \;(-50)\;}}{2}\)

**∴ σ _{n} = 25 MPa.**

__Concept:__

\(Strain=\frac{{{\rm{\Delta }}l}}{l} = \frac{{{{\epsilon_{xx}}} + {{\epsilon_{yy}}}}}{2} + \frac{{{{\epsilon_{xx}}} - {{\epsilon_{yy}}}}}{2}\cos 2\theta + \frac{{{\gamma _{xy}}}}{2}\sin 2\theta \)

__Calculation:__

__Given:__

ϵxx = 0.001, ϵyy = 0.002, γxy = 0.003, l = 5 units

tan θ = ¾ ⇒ θ = 36.86°

\(Strain=\frac{{{\rm{\Delta }}l}}{l} = \frac{{{{\epsilon_{xx}}} + {{\epsilon_{yy}}}}}{2} + \frac{{{{\epsilon_{xx}}} - {{\epsilon_{yy}}}}}{2}\cos 2\theta + \frac{{{\gamma _{xy}}}}{2}\sin 2\theta \)

\( = \frac{{0.001 + 0.002}}{2} + \frac{{0.001 - 0.002}}{2}\cos 2\theta + \frac{{0.003}}{2}\sin 2\theta \)

\(\frac{{{\rm{\Delta }}l}}{l} = 2.8 \times {10^{ - 3}}\)

Δl = 5 × 2.8 × 10^{-3} = 0.014

Deformed length = l + Δl = 5 + 0.014 = 5.014 units

Option 2 : 50 MPa

__Concept:__

Maximum and minimum normal stress is given by,

\({{\rm{σ }}_{{\rm{max}}}}{\rm{\;or\;}}{{\rm{σ }}_{{\rm{min}}}} = \frac{{{{\rm{σ }}_{\rm{x}}} + {{\rm{σ }}_{\rm{y}}}}}{2} \pm \sqrt {{{\left( {\frac{{{σ _x} - {σ _y}}}{2}} \right)}^2} + τ _{xy}^2\;}\)

Maximum shear stress is given by,

\({τ _{max}} = \sqrt {{{\left( {\frac{{{σ _x} - {σ _y}}}{2}} \right)}^2} + τ _{xy}^2\;} = \frac{{{σ _{max}} - {σ _{min\;}}}}{2}\)

**Calculation:**

**Given:**

σ_{max} = 130 MPa, σ_{min} = 30 MPa

Maximum shear stress is given by,

τ_{max} = (σ_{max} - σ_{min})/2 = (130 - 30)/2 = 50 MPa

The principal stresses at a point in a two-dimensional stress system are σ_{1} and σ_{2 }and corresponding principal strains are ϵ_{1} and ϵ_{2}. If E and μ denote Young’s modulus and Poisson’s ratio, then which one of the following is correct ?

Option 2 : \({\sigma _1} = \frac{E}{{1 - {\mu ^2}}}\;\left( {{\epsilon_1} + \mu {\epsilon_2}} \right)\)

__Concept:__

For a **two-dimensional stress system**,

We have,

\({\epsilon_1} = \frac{1}{E}\left( {{\sigma _1} - \mu {\sigma _2}} \right)\)

\({\epsilon_2} = \frac{1}{E}\left( {{\sigma _2} - \mu {\sigma _1}} \right)\)

**Principal stresses = σ _{1}, σ_{2}, Principal strains = ϵ_{1}, ϵ_{2}, Young’s modulus (E), Poisson’s ratio (μ)**

__Given:__

__Calculation:__

\({\epsilon_1} = \frac{1}{E}\left( {{\sigma _1} - \mu {\sigma _2}} \right)\) → (1)

\({\epsilon_2} = \frac{1}{E}\left( {{\sigma _2} - \mu {\sigma _1}} \right)\) → (2)

**From equation (1),**

ϵ_{1} × E = σ_{1} – μ × σ_{2 }

\({\sigma _2} = \frac{1}{\mu }\left( {{\sigma _1} - \;{\epsilon_1}E} \right)\)

Now, **substituting σ _{2} in equation (2),**

ϵ_{2} × E = σ_{2} – μ × σ_{1 }

\({\epsilon_2}E = \frac{{{\sigma _1}}}{\mu } - \frac{{{\epsilon_1}E}}{\mu } - \mu {\sigma _1}\)

\({\epsilon_2}E = \sigma \left( {\frac{1}{\mu } - \mu } \right) - \frac{{{\epsilon_1}E}}{\mu }\)

μ × ϵ_{2} × E = σ_{1}(1 – μ^{2}) - ϵ_{1}E

σ_{1} (1-μ^{2}) = E (ϵ_{1} + μ ϵ_{2})

\({\sigma _1} = \frac{E}{{1 - {\mu ^2}}}\;\left( {{\epsilon_1} + \mu {\epsilon_2}} \right)\)

Similarly,

\({\sigma _2} = \frac{E}{{1 - {\mu ^2}}}\;\left( {{\epsilon_2} + \mu {\epsilon_1}} \right)\)

Option 3 : 180.3

**Concept:**

Maximum and minimum values of normal stresses occur on planes of **zero shearing stress**. The maximum and minimum normal stresses are called the principal stresses, and the planes on which they act are called the principal plane.

\({\sigma _{max,\;min}} = \frac{{{\sigma _{xx}} + {\sigma _{yy}}}}{2} \pm \sqrt {{{\left( {\frac{{{\sigma _{xx}} - {\sigma _{yy}}}}{2}} \right)}^2} + \tau _{xy}^2} \)

But the maximum shear stress planes may or may not contain normal stresses as the case may be.

\({\tau _{max}} = \frac{{{\sigma _{max}} - {\sigma _{min}}}}{2} = \sqrt {{{\left( {\frac{{{\sigma _{xx}} - {\sigma _{yy}}}}{2}} \right)}^2} + \tau _{xy}^2} \)

which is equal to the radius of the Mohr's Circle.

**Calculation:**

**Given:**

σ_{x} = −200MPa, σ_{y} = 100MPa and τ_{xy} = 100MPa

Option 1 : Less than the maximum shear stress τ_{max}

__Explanation:__

**Stress at a point-Mohr circle of stress:**

**Important relationship obtained from the mohr circle:**

(i) Maximum shearing stress occurs on planes inclined at 45° to principal planes

\({\tau _{\max }} = \frac{{{\sigma _1} - {\sigma _3}}}{2}\)

(ii) The normal stresses on plane of maximum shear are equal to each other and they are given by

\({\sigma _{1,2}} = \frac{{{\sigma _1} + {\sigma _3}}}{2}\)

(iii) The sum of normal stresses on mutually perpendicular planes is a constant. i.e.,

σ_{1 }+ σ_{3} = σ_{n1} + σ_{n2} = constant

(iv) When the principal stresses are equal to each other, the radius of the mohr’s circle becomes zero, which means that shear stresses vanish on all planes. Such a point is called ISOTROPIC point.

(v) The resultant stress at any point is √(σ^{2} + τ^{2}) and the obliquity, β, equal to tan^{-1}(τ / σ)

(vi) the maximum angle of obliquity (β_{max}) is obtained from the failure envelope and is given by

θ_{c} = 45° + β_{max }/ 2

(vii) **the plane of maximum obliquity is most liable to failure and not the plane of maximum shear means the shear stress on the plane of maximum obliquity is less than the maximum shear stress.**

(viii) Failure becomes incipient when β_{max} approaches and equals the angle of internal friction.

The state of stress at a point when completely specified enables one to determine the

1. Maximum shearing stress at the point

2. Stress components on any arbitrary plane containing that point

Which of the above is/are correct?

Option 3 : Both 1 and 2

__Concept:__

**Normal stress at inclined plane is given by **

\({\sigma _n} = \frac{1}{2}\left[ {{\sigma _x} + {\sigma _y}} \right] + \frac{1}{2}\left[ {{\sigma _x} - {\sigma _y}} \right]\cos 2\theta + {\tau _{xy}}\sin 2\theta \)

**Shear stress at inclined plane is given by **

\({\tau _s} = - \frac{1}{2}\left[ {{\sigma _x} - {\sigma _y}} \right]\sin 2\theta + {\tau _{xy}}\cos 2\theta \)

**Sign Convention which is used in deriving the above formula is as follows:**

- Tensile stress is treated positive and compressive stress is treated as negative.
- Shear stress is treated positive when it is causing a couple in clockwise direction and when it is causing a couple in anticlockwise direction it is taken as negative.
- Shear stress in the inclined plane is treated as positive when it causes a couple in the anticlockwise direction and it is treated as negative when it is causing a couple in the clockwise direction.
- Inclination of the inclined plane is treated as positive when it is measured from y-axis in the clockwise direction and it is treated as negative when it is measured from the x-axis in the clockwise direction.

__Calculation:__

It is given that, σ_{x} = P and σ_{y} = -P, θ = 45° since no shear stress is acting to the given point ∴ τ_{xy} = 0

Therefore the normal stress on the plane inclined at 45°

\({\sigma _n} = \frac{1}{2}\left[ {{\sigma _x} + {\sigma _y}} \right] + \frac{1}{2}\left[ {{\sigma _x} - {\sigma _y}} \right]\cos 2\theta + {\tau _{xy}}\sin 2\theta \)

\({\sigma _n} = \frac{1}{2}\left[ {P + \left( { - P} \right)} \right] + \frac{1}{2}\left[ {P - \left( { - P} \right)} \right]\cos 90\)

\({\sigma _n} = 0\)

Maximum shear stress is given by

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

Option 3 : 150, 0

__Concept:__

Mohr’s Circle is represented by,

In mohr’s circle, X-axis represents normal stress and Y-axis represents shear stress.

Where σ_{x} = Maximum principle stress, σ_{y} = Minor principle stress, τ = Shear stress

__Calculation:__

∴ Two co-ordinates obtained are (200,0) and (100,0).

\({\rm{Radius\;of\;Mohr's\;circle\;}} = \frac{{200 - 100}}{2} = 50\)

∴ Coordinates of centre of the mohr’s circle = 100 + 50 = **150**

The tensile stresses at a point across two mutually perpendicular planes are 150 N/mm^{2} and 75 N/mm^{2}. What is the normal stress on the plane inclined at 35° to the axis of the major stresses?

Option 2 : 125.33 N/mm2

**Concept:**

For an element under the effect of bi-axial state of normal stress, the normal stresses on the plane inclined at θ° to the axis of the major stresses is given by

\({{\rm{σ }}_{{\rm{n}}1}} = \left( {\dfrac{{{{\rm{σ }}_1} + {{\rm{σ }}_2}}}{2}} \right) + \left( {\dfrac{{{{\rm{σ }}_1} - {{\rm{σ }}_2}}}{2}} \right){\rm{cos}} \ 2{\rm{θ }}\)

Where,

σ_{1} and σ_{2} are stresses at two mutually perpendicular planes.

**Calculation:**

Given,

σ_{1} = 150 N/mm^{2}

σ_{2} = 75 N/mm2

θ = 35°

\({{\rm{σ }}_{{\rm{n}}1}} = \left( {\dfrac{{{{\rm{σ }}_1} + {{\rm{σ }}_2}}}{2}} \right) + \left( {\dfrac{{{{\rm{σ }}_1} - {{\rm{σ }}_2}}}{2}} \right){\rm{cos}} \ 2{\rm{θ }}\)

\({σ _{{\rm{n}}1}} = \left( {\dfrac{{150 + 75}}{2}} \right) + \left( {\dfrac{{150 - 75}}{2}} \right){\rm{cos}}\ 2 \times 35 = 112.5 + 12.826 = 125.33\rm \ N/m{m^2}\)

What will be the magnitude of the shear stress on the principal plane?

Option 4 : Zero

__Concept:__

Maximum and minimum values of normal stresses occur on planes of zero shearing stress. The maximum and minimum normal stresses are called the principal stresses, and the planes on which they act are called the **principal plane**.

\({\sigma _{max,\;min}} = \frac{{{\sigma _{xx}} + {\sigma _{yy}}}}{2} \pm \sqrt {{{\left( {\frac{{{\sigma _{xx}} - {\sigma _{yy}}}}{2}} \right)}^2} + \tau _{xy}^2} \)

But the maximum shear stress planes may or may not contain normal stresses as the case may be.

\({\tau _{max}} = \frac{{{\sigma _{max}} - {\sigma _{min}}}}{2} = \sqrt {{{\left( {\frac{{{\sigma _{xx}} - {\sigma _{yy}}}}{2}} \right)}^2} + \tau _{xy}^2} \)

In a two-dimensional stress analysis, the state of stress at a point P is

\(\left[ \sigma \right] = \left[ {\begin{array}{*{20}{c}} {{\sigma _{xx}}}&{{\tau _{xy}}}\\ {{\tau _{xy}}}&{{\sigma _{yy}}} \end{array}} \right]\)

The necessary and sufficient condition for existence of the state of pure shear the point P, isOption 3 : σ_{xx} + σ_{yy} = 0

**Concept:**

For the state of pure shear, normal stresses on the plane must be equal to zero.

Also, the principal stresses are equal to shear stress.

Mohr’s circle for the state of pure shear is

**Calculation:**

**Given:**

\(\sigma = \left[ {\begin{array}{*{20}{c}} {{\sigma _{xx}}}&{{\sigma _{xy}}}\\ {{\tau _{xy}}}&{{\sigma _{yy}}} \end{array}} \right]\)

Principal stress \( = \frac{{{\sigma _{xx}} + {\sigma _{yy}}}}{2} \pm \sqrt {{{\left( {\frac{{{\sigma _{22}} - {\sigma _{yy}}}}{1}} \right)}^2} + \tau _{xy}^2}\)

Shear stress \( = \sqrt {{{\left( {\frac{{{\sigma _{xx}} - {\sigma _{yy}}}}{2}} \right)}^2} + {{\left( {{\tau _{xy}}} \right)}^2}}\)

For the **pure shear state** of stress –

**Principal stress = shear stress**

\(\Rightarrow \frac{{{\sigma _{xx}} + {\sigma _{yy}}}}{2} \pm \sqrt {{{\left( {\frac{{{\sigma _{xx}} - {\sigma _{yy}}}}{2}} \right)}^2} + \tau _{xy}^2} = \sqrt {{{\left( {\frac{{{\sigma _{xx}} - {\sigma _{yy}}}}{2}} \right)}^2} + \tau _{xy}^2}\)

⇒ σ_{xx} + σ_{yy} = 0

__Important Points__

**In the case of pure shear:**

- The magnitude of principal stresses are equal and they are opposite in nature.
- Mohr’s circle is centered at origin and radius = τ
- τ
_{max}in-plane (τ) which is also the absolute maximum shear stress.