Principle Stress MCQ Quiz in मराठी - Objective Question with Answer for Principle Stress - मोफत PDF डाउनलोड करा

Concept:

τ1 = In-plane maximum shear stress

τ2 = Absolute shear stress (as given that it is the maximum shear stress at the given point)

In the bi-axial state of stress, σ3 = 0

Calculation:

Given:

σx = 100 MPa (Tensile) and σy = - 50 MPa (Compressive)

 Since it is a bi-axial state of stress ⇒ σ3 = 0 

Now, maximum shear stress is given by

\({\tau _2} = {\rm{max}}\left( {\frac{{{\sigma _x} - {\sigma _y}}}{2},{\rm{\;}}\frac{{{\sigma _x}}}{2},\;\frac{{{\sigma _y}}}{2}} \right)\)

\({\tau _2} = {\rm{max}}\left( {\frac{{100 + 50}}{2},{\rm{\;}}\frac{{100}}{2},\;\frac{{50}}{2}} \right)\)

\({\tau _2} = {\rm{max\;}}\left( {75,{\rm{\;}}50,\;25} \right)\)

∴ τ2 = 75 MPa.

Concept:

Normal stress at a plane  with an ∠θ from plane under the action of σ_x,

\({τ _{xx}} = \left( {\frac{{{σ _x} + {σ _y}}}{2}} \right) + \left( {\frac{{{σ _x} - {σ _y}}}{2}} \right)\cos 2θ + {σ _{xy}}\sin 2θ {{\;\;\;\;\;}} \ldots \left( 1 \right)\)

The shear stress at a plane with an ∠θ from plane under the action of σ_x,

\({τ _{xy}} = - \left( {\frac{{{σ _x} - {σ _y}}}{2}} \right)\sin 2θ + {σ _{xy}}\cos 2θ {{\;\;\;\;\;}} \ldots \left( 2 \right)\)

(θ = - 45° as the plane under +P is rotated by 45° anti-clockwise to get the plane under +τxx)

Also,

τxx + τyy = σx + σ_y      ...(3)

Calculation:

Given:

Since the given plane is the principal plane ⇒ σxy = 0, σx = P, σy = - P

By using above equation (1),

\({τ _{xx}} = \left( {\frac{{P - P}}{2}} \right) + \left( {\frac{{P - \left( { - P} \right)}}{2}} \right)\cos( -90^\circ) + 0 = 0\)

By using above equation (2),

\({τ _{xy}} = - \left[ {\frac{{P - \left( { - P} \right)}}{2}} \right]\sin \left( {-90^\circ } \right) - 0 = P\)

By using equation (3),

⇒ τyy = (σx + σy) - τxx = 0

∴ τ_xx = τ_yy = 0, τ_xy = P

Alternate method:

The given plane is the principal plane because σ_xy = 0

At 45° from the principal plane, the plane of maximum shear occurs on the plane of maximum shear.

\({τ _{xx}} = {τ _{yy}} = \frac{{{σ _1} + {σ _2}}}{2} = \frac{{P - P}}{2} = 0\)

Explanation:

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

Concept:

The equation of normal stress and shear  is given by:

\(σ_{n} = \frac{σ_{x}+σ_{y}}{2} + \frac{σ_{x}-σ_{y}}{2}\ cos\ 2θ + τ_{xy} sin\ 2θ \)

\(τ = -τ_{xy}\ cos\ 2 θ \)

In case of pure shear, σx = σy = 0

σ_n = τ_xy sin 2θ 

\(τ = -τ_{xy}\ cos\ 2 θ \)

In the principal plane, there is no shear stress i.e. τ = 0

​\(τ = -τ_{xy}\ cos\ 2 θ \)​ = 0 ⇒  cos 2θ = 0

\(\theta = \pm 45^{\circ}\)

Additional Information

By the Mohr circle method equation of normal stress:

\(σ_{1/2} = \frac{σ_{x}+σ_{y}}{2} \pm \sqrt{\left (\frac{σ_{x}-σ_{y}}{2}\right )^2 + (τ_{xy})^2}\)

σ_x = σy = 0

\(σ_{1/2} = \pm\ τ_{xy}\)

2θ1 = 90° ⇒ θ1 = 45°

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.

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​

Concept:

The stress tensor is given by:

\(σ = \left[ {\begin{array}{*{20}{c}} {{σ _x}}&{{\tau _{yx}}}&{{\tau _{zx}}}\\ {{\tau _{xy}}}&{{σ _y}}&{{\tau _{zy}}}\\ {{\tau _{xz}}}&{{\tau _{yz}}}&{{σ _z}} \end{array}} \right]\;\)

First stress invariant is given by:

I₁ = σ_x + σ_y + σ_z 

i,e. the trace of the stress tensor is known as the first stress tensor.

Calculation:

Given:

σx = 50, σy = 60,  σz  = 80

\(σ = \left[ {\begin{array}{*{20}{c}} {{σ _x}}&{{\tau _{yx}}}&{{\tau _{zx}}}\\ {{\tau _{xy}}}&{{σ _y}}&{{\tau _{zy}}}\\ {{\tau _{xz}}}&{{\tau _{yz}}}&{{σ _z}} \end{array}} \right]\;\)

First stress invariant is given by:

I1 = σx + σy + σz 

I1 = 50 + 60 + 80 ⇒ 190

Concept: 

The normal stress is given by

\({\sigma _n} = \frac{{{\sigma _x} + {\sigma _y}}}{2} + \frac{{{\sigma _x} - {\sigma _y}}}{2} \times \cos 2θ + {\tau _{xy}}\sin 2θ {{\;\;\;\;\;}} \ldots \left( 1 \right)\)

Calculation:

Given:

Sectional area (A) = 200 mm²

Axial Force (P) = 20 kN, σ y = 0 , τ xy = 0

The oblique plane angle is 45° with Y-Y in a clockwise sense.

⇒ θ = - 45° 

\(⇒ {\sigma _x} = \frac{P}{A} = \frac{{20\; \times \;1000\;N}}{{200\;m{m^2}}} = 100\ N/mm^{2}\)

Using equation (1),

\(⇒ {\sigma _n} = \frac{{100\; + \;0}}{2} + \frac{{100\; - \;0}}{2} \times \cos (-90) + 0\)

⇒σ_n = 50 N/mm²

Explanation:

Mohr's circle is a graphical method of representing stresses and strains. This enables the calculation of components of stresses and strains in an arbitrary oblique plane.

A circle is defined by the coordinate of the center point and its radius.

This can be examined in Mohr's circle for stresses as follows:

 The radius of Mohr's circle is found as:

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

This is also the value of maximum shear stress possible.

Coordinates of center:

The center of Mohr's circle is always on the x-axis, therefore coordinates are given by

\(OC = {\sigma_x +\sigma_y\over2}\)

Explanation:

Principal plane and principal stresses:

The planes, which have​​​​​​​​​​​​​​​​​​​​​​ no shear stress, are known as principal planes. Hence principal planes are the planes of zero shear stress. These planes carry only normal stresses.

The normal stresses, acting on principal plan​​​​​​​​​​​​​​​​​​​​​​e are known as principal stresses.

Stresses on an inclined plane:

\({{\rm{\sigma }}_{\rm{θ }}} = \frac{{{{\rm{\sigma }}_{\rm{x}}} + {{\rm{\sigma }}_{\rm{y}}}}}{2} + \frac{{{{\rm{\sigma }}_{\rm{x}}} - {{\rm{\sigma }}_{\rm{y}}}}}{2}\cos 2{\rm{θ }} + {{\rm{τ }}_{{\rm{xy}}}}\sin 2{\rm{θ }}\)

\({{\rm{τ }}_{\rm{θ }}} = \frac{{{{\rm{\sigma }}_{\rm{x}}} - {{\rm{\sigma }}_{\rm{y}}}}}{2}\sin 2{\rm{θ }} - {{\rm{τ }}_{{\rm{xy}}}}\cos 2{\rm{θ }}\)

The principal plane is at angle θ 

At principal plane τ_θ = 0

\(0 = \frac{{{{\rm{\sigma }}_{\rm{x}}} - {{\rm{\sigma }}_{\rm{y}}}}}{2}\sin 2{\rm{θ }} - {{\rm{τ }}_{{\rm{xy}}}}\cos 2{\rm{θ }}\)

∴ \(\tan 2{\rm{θ }} = \frac{{2{{\rm{\tau }}_{{\rm{xy}}}}}}{{{{\rm{\sigma }}_{\rm{x}}} - {{\rm{\sigma }}_{\rm{y}}}}}\)

θ = Angle of major principle plane

Principle stress equation is given by,

\({{\rm{σ }}_1},\;{{\rm{σ }}_2} = \frac{{{{\rm{σ }}_x} + \;{{\rm{σ }}_y}}}{2} \pm \sqrt {{{\left( {\frac{{{{\rm{σ }}_x} - \;{{\rm{σ }}_y}}}{2}} \right)}^2} + {{\left( {{τ _{xy}}} \right)}^2}} \)

where σ1 = Maximum principal stress, σ2 = Minimum principal stress, σx = Stress in X-direction, σy = Stress in Y-direction and

τxy = Shear stress in XY-plane.

Explanation:

Given Data

Major tensile stress(\(\sigma _x\)) = 150 MPa

Minor tensile stress (\(\sigma_y\)) = 50 MPa

Inclination angle (θ) = 30° (with the axis of major tensile stress)

Therefore, Inclination angle (θ) = 60° (with the vertical axis)

For an element under the effect of the bi-axial state of normal stress, the normal stress on the plane inclined at \(θ^0 \) to the axis of the major stress is given by

\(\sigma_n = { (\sigma_x + \sigma_y) \over 2} + { (\sigma_x - \sigma_y) \over 2} \times \cos2θ\)

\(\sigma_n = { (150 + 50) \over 2} + { (150 - 50) \over 2} \times \cos120\)

\(\sigma_n = 75\)

Please try once again through Mohr Circle, the answer will be the same as 75 MPa. The centre will be the point (100,0). From centre make an angle 60 in the anti-clockwise direction, the perpendicular line on X-axis mentioned as x in the image will be the normal stress.

Normal stress will be (50cos 60 + 50 =  75).

Additional Information An element under the effect of the bi-axial state with shear so the normal stress is given by 

\(\sigma_n = { (\sigma_x + \sigma_y) \over 2} + { (\sigma_x - \sigma_y) \over 2} \times \cos2θ +\tau_{xy} \sin2θ\)

where \(\tau_{xy}\) = shear stress