Principal Stress or Strain MCQ Quiz - Objective Question with Answer for Principal Stress or Strain - Download Free PDF

Concept:

When a body is subjected to bi-axial stresses (direct stresses P₁ and P₂ ) along with a shear stress q , the maximum and minimum normal stresses can be determined using stress transformation equations. The maximum normal stress is particularly important for failure analysis.

Given:

• Direct stress in the first direction, \( P_1 \)
• Direct stress in the second direction, \( P_2 \)
• Shear stress, \( q \)

Step 1: Understand the Stress State

The given stress state consists of two normal stresses P₁ and P₂ acting in mutually perpendicular directions, along with a shear stress q .

Step 2: Use the Formula for Principal Stresses

The maximum and minimum normal stresses (principal stresses) are given by:

\[ \sigma_{\text{max}}, \sigma_{\text{min}} = \frac{P_1 + P_2}{2} \pm \frac{1}{2} \sqrt{(P_1 - P_2)^2 + 4q^2} \]

Step 3: Identify the Maximum Normal Stress

The maximum normal stress corresponds to the positive root of the expression:

\[ \sigma_{\text{max}} = \frac{P_1 + P_2}{2} + \frac{1}{2} \sqrt{(P_1 - P_2)^2 + 4q^2} \]

Concept:

Normal stress on a plane inclined at angle \( \theta \) is given by:

\( \sigma_n = \sigma_v \cos^2 \theta + \sigma_h \sin^2 \theta \)

Given:

• \( \sigma_v = 1 \, \text{N/mm}^2 \)
• \( \sigma_h = 0.5 \, \text{N/mm}^2 \)
• \( \cos^2(50^\circ) = 0.413 \Rightarrow \sin^2(50^\circ) = 0.587 \)

Calculation:

\( \sigma_n = 1 \times 0.413 + 0.5 \times 0.587 = 0.413 + 0.2935 = 0.7065 \, \text{N/mm}^2 \)

Concept:

To calculate the normal stress on an inclined plane at an angle \( \theta \), use the stress transformation equation:

\( \sigma_n = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2} \cos 2\theta + \tau_{xy} \sin 2\theta \)

Given:

\( \sigma_x = 100~\text{N/mm}^2, \quad \sigma_y = 50~\text{N/mm}^2, \quad \tau_{xy} = 30~\text{N/mm}^2, \quad \theta = 45^\circ \)

Calculation:

\( \cos 2\theta = \cos 90^\circ = 0, \quad \sin 2\theta = \sin 90^\circ = 1 \)

\( \sigma_n = \frac{100 + 50}{2} + \frac{100 - 50}{2} \cdot 0 + 30 \cdot 1 \)

\( \sigma_n = 75 + 0 + 30 = 105~\text{N/mm}^2 \)

Explanation:

Determining the Inclination of the Plane for Maximum Obliquity:

When a piece of material is subjected to two perpendicular tensile stresses, σ_x and σ_y, the inclination of the plane on which the resultant stress (R) has maximum obliquity (φ) with the normal can be determined using the principles of stress transformation in mechanics of materials.

In this problem, we are given:

• σ_x = 100 MPa
• σ_y = 60 MPa

The maximum obliquity of the resultant stress occurs when the shear stress (τ) on the plane is maximized. This typically happens at an angle θ where the plane is oriented such that the normal stress is balanced by the shear stress. Using Mohr's circle of stress, the inclination θ of the plane can be found.

The formula for the inclination θ of the plane on which the resultant stress R has maximum obliquity φ with the normal is given by:

θ = tan^-1 (τ / σ)

Here, τ is the shear stress and σ is the normal stress on the plane. For the given problem, we need to derive the correct expression for θ by considering the relationship between the normal and shear stresses.

From the theory of Mohr's circle, the inclination θ for the plane on which the resultant stress has maximum obliquity can be derived as:

θ = tan^-1 (σ_x / σ_y)

Plugging in the given values:

θ = tan^-1 (100 / 60)

Simplifying this expression, we get:

\( \theta=\tan ^{-1} \sqrt{\frac{5}{3}}\)

Therefore, the correct inclination θ of the plane on which the resultant stress has maximum obliquity is:

\( \theta=\tan ^{-1} \sqrt{\frac{5}{3}}\)

Concept:

Given:

The circular shaft is subjected to twisting moment only which results in maximum shear stress of 90MPa.

Maximum Shear stress is given by:

\({τ _{max}} = \;\frac{{16T}}{{\pi {d^3}}}\)

∴ It is a case of pure shear. And the Mohr circle for pure shear is represented by:

• Here, the compressive stress will be equal to the maximum shear stress.
• In the case of pure shear, τ_xy = τ and σ_x = σ_y = 0

σ₁ = τ , σ2 = -τ

Centre of Mohr circle = 0

Radius = \(\frac{{{\sigma _1} - {\sigma _2}}}{2}\) = \(\frac{{\tau - \left( { - \tau } \right)}}{2} = \;\tau \)

Hence, the maximum compressive stress in the material is 90MPa.

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 σ₁ and σ₂ are principal stresses in a plane.

Calculation:

Given:

σ1 = 100 MPa , σ₂ = 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:

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

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, 

θ₂ = θ₃ = θ + 90°  (θ, θ₂, θ₃ location of principle planes)   

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.

Explanation:

The shearing stresses in two mutually perpendicular planes are equal in magnitude to maintain the rotational equilibrium but not in same direction.

Consider the following figure:

ABCD is rectangular element let τ1 and τ2 shear stress acting on plane DC and BC respectively.

Let consider the unit length of this element.

For ABCD to be in rotational equilibrium

∑MA = 0 ⇒ (τ1 b) (a) + (τ2 a) (b) = 0

τ1 (ab) = τ2 (ab)

τ1  = τ2

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.

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.

Uniaxial tension: Tensile stress acts along one axis only.

Mohr’s circle:

Biaxial Tension of equal magnitude:'

Mohr’s circle: It is a point on the normal stress axis.

Hydrostatic stress:

Mohr’s circle: It is a point on the normal stress axis.

Mistake Points

The Mohr’s circle for Biaxial tension of equal magnitude and Hydrostatic stress represent a point on σ-axis but in case of hydrostatic stress, the point is on the negative side and on biaxial tension the point is on the positive side of the Mohr’s circle 

Explanation-

• The eigenvector of a matrix A is a vector represented by a matrix X such that when X is multiplied with matrix A, then the direction of the resultant matrix remains the same as vector X.
• AX = λX

  • where A is any arbitrary matrix, λ are eigen values and X is an eigen vector corresponding to each eigen value.​Here, we can see that AX is parallel to X.So X is an eigen vector.

Some important properties of eigenvalues are-​

• If λ₁, λ₂…….λ_n are the eigenvalues of A, then \(\lambda _1^k,\lambda _2^k,......,\lambda _n^k\) are the eigenvalues of A^k

• Sum of Eigen Values = Trace of A (Sum of diagonal elements of A)

Given data and Calculation-

Given matrix = \(ε = \begin{bmatrix} 1 & 1 \\\ 1 & -1 \end{bmatrix} \)

\(A - \lambda I = \left( {\begin{array}{*{20}{c}} {1 - \lambda }&1\\ 1&{ - 1 - \lambda } \end{array}} \right)\) = 0

\(\Rightarrow (1 - \lambda )( - 1 - \lambda ) - 1\) = 0

\( \Rightarrow {\lambda ^2} - 2 = 0\)

\( \Rightarrow {\lambda ^2} = 2\)

\( \Rightarrow \lambda = \pm \sqrt 2 \)

Eigenvalues of ε8 are \({\left( {\sqrt 2 } \right)^8} \) and \({\left( { - \sqrt 2 } \right)^8}\).

 Eigenvalues of ε¹¹ are \({\left( {\sqrt 2 } \right)^{11}}\) and \({\left( { - \sqrt 2 } \right)^{11}}\).

P = Trace (ε8) = 16 + 16 = 32

Q = Trace (ε11).= 0

P + Q = 32

Additional Information 

• Eigenvalues of real symmetric matrices are real.

• Eigenvalues of real skew-symmetric matrices are either pure imaginary or zero.

• Eigenvalues of unitary and orthogonal matrices are of unit modulus |λ| = 1

• If λ1, λ2…….λn are the eigenvalues of A, then kλ1, kλ2…….kλn are eigenvalues of kA.

• If λ1, λ2…….λn are the eigenvalues of A, then 1/λ1, 1/λ2…….1/λn are eigenvalues of A-1

• Eigenvalues of A = Eigen Values of AT (Transpose)

• Product of Eigen Values = |A|

• Maximum number of distinct eigenvalues of A = Size of A

• If A and B are two matrices of the same order then, Eigenvalues of AB = Eigenvalues of BA

Concept:

Principal Stress:

Principal stresses are the maximum and minimum value of normal stresses on a plane (when rotated through an angle) on which there is no shear stress.

Principal Plane:

It is that plane on which the principal stresses act and shear stress are zero.

Principal Angle:

The orientation of the principal plane with respect to the original axis is the principal angle.

Normal stress at any angle is calculated as

\(σ _n{}= \dfrac{σ _{x}+σ _{y}}{2}+\dfrac{σ _{x}-σ _{y}}{2}\cos 2θ\)

Where σ_n = Normal stress at inclination θ, σ_x = Principal stress in x-direction, σ_y = Principal  stress in Y direction,

θ = Angle of inclinitaion of Normal stress to x axis

Calculation:

Given:

σ_x = σ_y = 500 N/mm², θ = 45° , 

\(σ _n{}= \dfrac{σ _{x}+σ _{y}}{2}+\dfrac{σ _{x}-σ _{y}}{2}\cos 2θ\)

σ_n \(= \dfrac{500+500}{2}+\dfrac{500-500}{2}\cos 90\)

σn = 500 N/mm²

∴ The value of normal stress at an angle of 45° is 500 N/mm²

Concept:

Maximum allowable shear in a square section (τ_max) = (σ sin 2θ)/2

Where σ is the maximum tensile stress 

Let,

P = Maximum tensile force,

A = Area of cross-section 

∴ σ = P/A

Calculation:

Given,

P, A, θ 

σ = P/A

τmax = ((P/A) sin 2θ)/2 = (P/2A) sin 2θ

Concept:

Maximum normal stress, \({σ _{n,max}} = \frac{{σ_1 + σ_3 }}{2}\)

Maximum shear stress, \({\tau _{max}} = \frac{{σ_1 - σ_3 }}{2}\)

Where, σ1 = major principle stress and  σ3 = minor principle stress

calculation:

Given,

Major Principal stress (σ₁) = 150 kN/m² 

Minor Principal stress (σ₂) = -50 kN/m²

Maximum shear stress is given by, \({\tau _{max}} = \frac{{σ_1 - σ_3 }}{2}\)

⇒ \({\tau _{max}} = \frac{{150 \ + \ 50 }}{2} = 100\ kN/m^2\)