Principal Stress or Strain MCQ Quiz - Objective Question with Answer for Principal Stress or Strain - Download Free PDF
Last updated on Jun 10, 2025
Latest Principal Stress or Strain MCQ Objective Questions
Principal Stress or Strain Question 1:
At a point in a stressed body there are normal stresses of 1 N/mm² (tensile) on a vertical plane and 0.5 N/mm² (tensile) on a horizontal plane. The shearing stresses on these planes are zero. What will be the normal stress on a plane making an angle
Answer (Detailed Solution Below)
Principal Stress or Strain Question 1 Detailed Solution
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:
\( \sigma_n = 1 \times 0.413 + 0.5 \times 0.587 = 0.413 + 0.2935 = 0.7065 \, \text{N/mm}^2 \)
Principal Stress or Strain Question 2:
Figure shows the stresses acting on a element ABCD. What will be the normal stress on the plane BE which is inclined at 45° to the plane BC?
Answer (Detailed Solution Below)
Principal Stress or Strain Question 2 Detailed Solution
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:
\( \cos 2\theta = \cos 90^\circ = 0, \quad \sin 2\theta = \sin 90^\circ = 1 \)
\( \sigma_n = \frac{100 + 50}{2} + \frac{100 - 50}{2} \cdot 0 + 30 \cdot 1 \)
\( \sigma_n = 75 + 0 + 30 = 105~\text{N/mm}^2 \)
Principal Stress or Strain Question 3:
A piece of material is subjected to two perpendicular tensile stresses (σx = 100 MPa. and σy = 60 MPa). What will be inclination (θ) of the plane on which the resultant stress (R) has maximum obliquity (φ) with the normal?
Answer (Detailed Solution Below)
Principal Stress or Strain Question 3 Detailed Solution
Explanation:
Determining the Inclination of the Plane for Maximum Obliquity:
When a piece of material is subjected to two perpendicular tensile stresses, σx and σy, 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:
- σx = 100 MPa
- σy = 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 (σx / σy)
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:
\( \theta=\tan ^{-1} \sqrt{\frac{5}{3}}\)
Principal Stress or Strain Question 4:
A circular shaft subjected to twisting moment result in maximum shear stress of 90 MPa. Then the maximum compressive stress in the material is
Answer (Detailed Solution Below)
Principal Stress or Strain Question 4 Detailed Solution
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, τxy = τ and σx = σy = 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.
Principal Stress or Strain Question 5:
At a point in the two dimensional stress system, σx = 10 N/mm2 and σy = τxy = 40 N/mm2. What is the radius of Mohr circle drawn with a scale of 1 cm = 10 N/mm2?
Answer (Detailed Solution Below)
Principal Stress or Strain Question 5 Detailed Solution
Concept:
The Mohr circle is a graphical representation of the state of stress at a point. The radius of the Mohr circle for a 2-dimensional stress system is given by:
∴ Radius of Mohr Circle \( = \sqrt{{\left(\frac{{\sigma_x - \sigma_y}}{2}\right)^2 + \tau_{xy}^2}}\)
Calculation:
Given:
σx = 10 N/mm2, σy = 40 N/mm2, τxy = 40 N/mm2
Using the formula for the radius of the Mohr Circle,
Given that the scale is 1 cm = 10 N/mm2,
Radius of Mohr Circle = \(\sqrt{(\frac{10~-~40}{2})^2 ~+~40^2}\) = 50 N/mm2
⇒ Radius of Mohr Circle = 50/10 = 5 cm
Top Principal Stress or Strain MCQ Objective Questions
If the principal stresses in a plane stress problem, are σ1 = 100 MPa, σ2 = 40 MPa, the magnitude of the maximum shear stress (in MPa) will be
Answer (Detailed Solution Below)
Principal Stress or Strain Question 6 Detailed Solution
Download Solution PDFConcept:
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:
\(\tau_{max,abs}=\frac{{\sigma _{max}}\;-\;{\sigma _{min}}}{{2}}=\frac{\sigma_1}{2}\)
The angle between Major Principal plane and Minor Principal plane for a strained body is:
Answer (Detailed Solution Below)
Principal Stress or Strain Question 7 Detailed Solution
Download Solution PDFExplanation:
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)
For the state of stress shown in the below figure, normal stress acting on the plane of maximum shear stress is -
Answer (Detailed Solution Below)
Principal Stress or Strain Question 8 Detailed Solution
Download Solution PDFConcept:
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:
σx = 100 MPa, σy = - 50 MPa.
\({\left( {{σ _n}} \right)_{{\tau _{max}}}} = \frac{{{σ _x} \;+ \;{σ _y}}}{2}\)
\(\Rightarrow {\left( {{σ _n}} \right)_{{\tau _{max}}}} = \frac{{{100} \;+ \;(-50)\;}}{2}\)
∴ σn = 25 MPa.
Shear stress on mutually perpendicular planes are
Answer (Detailed Solution Below)
Principal Stress or Strain Question 9 Detailed Solution
Download Solution PDFExplanation:
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 angle between the principle plane and the plane of maximum shear is
Answer (Detailed Solution Below)
Principal Stress or Strain Question 10 Detailed Solution
Download Solution PDFThe 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.
The state of stress represented by Mohr’s circle shown in the figure is
Answer (Detailed Solution Below)
Principal Stress or Strain Question 11 Detailed Solution
Download Solution PDFConcept:
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
The components of pure shear strain in a sheared material are given in the matrix form:
\(ε = \begin{bmatrix} 1 & 1 \\\ 1 & -1 \end{bmatrix} \)
Here, Trace (ε) = 0. Given, P = Trace (ε8) and Q = Trace (ε11).
The numerical value of (P + Q) is ________. (in integer)
Answer (Detailed Solution Below) 32
Principal Stress or Strain Question 12 Detailed Solution
Download Solution PDFExplanation-
- 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
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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.
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Some important properties of eigenvalues are-
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If λ1, λ2…….λn are the eigenvalues of A, then \(\lambda _1^k,\lambda _2^k,......,\lambda _n^k\) are the eigenvalues of Ak
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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
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Eigenvalues of real symmetric matrices are real.
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Eigenvalues of real skew-symmetric matrices are either pure imaginary or zero.
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Eigenvalues of unitary and orthogonal matrices are of unit modulus |λ| = 1
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If λ1, λ2…….λn are the eigenvalues of A, then kλ1, kλ2…….kλn are eigenvalues of kA.
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If λ1, λ2…….λn are the eigenvalues of A, then 1/λ1, 1/λ2…….1/λn are eigenvalues of A-1
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Eigenvalues of A = Eigen Values of AT (Transpose)
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Product of Eigen Values = |A|
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Maximum number of distinct eigenvalues of A = Size of A
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If A and B are two matrices of the same order then, Eigenvalues of AB = Eigenvalues of BA
Principal stresses at a point in a plane stressed element are σx = σy = 500 N/mm2. Normal stress on the plane inclined at 45° to the x-axis will be
Answer (Detailed Solution Below)
Principal Stress or Strain Question 13 Detailed Solution
Download Solution PDFConcept:
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 σn = Normal stress at inclination θ, σx = Principal stress in x-direction, σy = Principal stress in Y direction,
θ = Angle of inclinitaion of Normal stress to x axis
Calculation:
Given:
σx = σy = 500 N/mm2, θ = 45° ,
\(σ _n{}= \dfrac{σ _{x}+σ _{y}}{2}+\dfrac{σ _{x}-σ _{y}}{2}\cos 2θ\)
σn \(= \dfrac{500+500}{2}+\dfrac{500-500}{2}\cos 90\)
σn = 500 N/mm2
∴ The value of normal stress at an angle of 45° is 500 N/mm2
If a bar of cross section area 'A' is subjected to a tensile force 'P', resultant shear stress on a oblique plane inclined at an angle θ to its axis is -
Answer (Detailed Solution Below)
Principal Stress or Strain Question 14 Detailed Solution
Download Solution PDFConcept:
Maximum allowable shear in a square section (τmax) = (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
If the principle stress at a point in a stressed body are 150 kN/m2 tensile and 50 kN/m2 compressive, then maximum shear stress at this point will be:
Answer (Detailed Solution Below)
Principal Stress or Strain Question 15 Detailed Solution
Download Solution PDFConcept:
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, \({\tau _{max}} = \frac{{σ_1 - σ_3 }}{2}\)
⇒ \({\tau _{max}} = \frac{{150 \ + \ 50 }}{2} = 100\ kN/m^2\)