Graphical Method of Analysis MCQ Quiz - Objective Question with Answer for Graphical Method of Analysis - Download Free PDF

Concept:

Stress invariants in 3D stress plane:

\(J_1=σ_x+σ_y+σ_z\)

\(J_2=σ_xσ_y+σ_yσ_z+σ_zσ_x-\tau_{xy}^2-\tau_{yz}^2-\tau_{xz}^2\)

\(J_3=σ_xσ_yσ_z+2\tau_{xy}\tau_{yz}\tau_{xz}-σ_x\tau_{yz}^2-σ_y\tau_{xz}^2-σ_z\tau_{xy}^2\)

Calculation:

Given

σ_x = 10 MPa, σy = 8 MPa, σz = -5 MPa, \(\tau_{yz}=\tau_{zy}=5 \, MPa, \tau_{zx}=\tau_{xz}=-4 \, MPa, and \,\tau_{xy}=\tau_{yx}=-8 \, MPa.\)

J₁ = 10 + 8 + (-5) = 13 MPa

J₂ = 10 × 8 + 8 × -5 + (-5) × 10 - (5)² - (-4)² - (-8)² = - 115 MPa

J₃ = 10 × 8 × (-5) + 2 × 5 × (-4) × (-8) - 10 × 5² - 8 × (-4)2 - (-5) × (-8)2 = -138 MPa

J₁ = 13 MPa, J₂ = -115 MPa, and J₃ = -138 MPa.

Hence option (2) is correct.

Concept:

Mohr's Circle:

• Mohr's circle is a graphical method of finding normal, tangential, and resultant stresses on an oblique plane.

Mohr's circle will be drawn for the following cases :

1. A body subjected to two mutually perpendicular principal tensile stresses of unequal intensities.
2. A body subjected to two mutually perpendicular principal stresses which are unequal and unlike (i.e., one is tensile and other is compressive ).
3. A body subjected to two mutually perpendicular principal tensile stresses accompanied by a simple shear stress.

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Similarly,

In the given figure minor principal stress is represented by OB (σ₂) and major principal stress is represented by OA (σ₁). 

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 τmax.
• 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. σx = σy = σ and τxy = 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 σx and σy = 0, τxy = 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.

Hence,the correct answer is option 1 and 3.

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.
• Distance of the center of the Mohr's circle from origin of coordinates gives the average of normal stress ie (σ₁ + σ₃)/2
• 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 τmax.
• 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.

Explanation:

From the Mohr circle the Major, minor, and normal stress are drawn in the diagram below, and shear stress corresponding to the normal stress of 115 MPa is given by τ. Point X represent the Point with normal stress 115 MPa and shear stress of τ 

Given:

Major principal stress = σ₁ = 120 MPa and Minor principal stress = σ₃ = 70 MPa, Normal Stress, σ = 115 MPa and shear stress corresponding to the same plane is τ.

Radius of circle = AC = CX = (σ₁ - σ₃)/2 = (120 - 70)/2 = 25 MPa

BD in Mohr circle = Major principal stress - Normal stress = OB - OD = 120 - 115 = 5 Mpa = y

So, CD will be = Radius of mohr circle - BD = 25 - 5 = 20 MPa

From, the Mohr circle, we can say that

CX² = CD² + DX²

25² = 20² + DX²

DX² = 625 - 400 = 225

DX = τ = 15 MPa

Therefore the shear stress will be 15 MPa

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

Concepts:

Mohr's strength theory (1900): 

• Mohr’s failure theory of materials states that a material fails because of a critical combination of normal stress and shearing stress and not from either maximum normal or shear stress alone.
• Thus, the functional relationship between normal stress and shear stress on a failure plane can be expressed as, \({\rm{\;}}{{\rm{\tau }}_{\rm{f}}} = {\rm{s}} = {\rm{f}}\left( {\rm{\sigma }} \right)\) where \({{\rm{\tau }}_{\rm{f}}}\) is shear stress at failure, s is shear resistance of the material and \({\rm{f}}\left( {\rm{\sigma }} \right)\) function of normal stress. Mohr’s failure envelope is shown in the following diagram:

\({\tau _f} = c + {\sigma _f} \times \tan \phi \) (this is the most general form of mohr's equation)

Where, τf = shear stress on the failure plane, σf = normal stress on the failure plane, c = apparent cohesion, ϕ = angle of internal friction, which is constant till all the failure plane cuts the same failure envelop

Concept:

When the normal stresses on the two mutually perpendicular planes are equal and like then the radius of the Mohr circle will be zero.

The radius of Mohr's circle,

R = \(\sqrt { (\frac{σ_x-σ_y}{2})^2 + τ_xy^2}\)

Here,

σ_x = σ_y and τ_xy = 0

∴ R = 0 means Mohr's circle reduces to a point.

Concept:

General Mohr Circle for biaxial stress condition:

Principal stresses are given by

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

Radius of circle (Mohr’s circle) \(=\sqrt{{{\left( \frac{{{\sigma }_{x}}-{{\sigma }_{y}}}{2} \right)}^{2}}+\tau _{xy}^{2}}\)

Calculation:

Given state of stress is represented by

∴ σ_x = 10 MPa. σ_y = 10 MPa τ_xy = 0

For this the Mohr Circle will be a point:

∴ σ₁ = σ₂ = 10 MPa

Radius of circle (Mohr’s circle) will be zero.

The co-ordinate of the centre of Mohr’s circle on σ-axis at a distance of 10 units from the origin.

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

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

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

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

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

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

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

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

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

Concept:

Calculation:

σ₁ = 300 MPa, σ₂ = 100 MPa

Radius of Mohr’s circle \( = \frac{{{\sigma _1} - {\sigma _2}}}{2} = \frac{{300 - 100}}{2} = \frac{{200}}{2} = 100\)

ON represent the normal stress = 260 MPa

MN is the shear stress at the same plane where the normal stress is 260 MPa

CM represents the maximum shear stress on the plane where the normal stress is 260 MPa

CM = 100 MPa

CN = ON - OC = 260 - 200 = 60 MPa

For ΔCMN,

(CM)² = (CN)² + (MN)²

(100)² = (60)² + MN²

MN = 80

∴ The magnitude of shear stress τs = 80 MPa

Explanation:

Mohr's circle

• The transformation equations for plane stress can be represented in graphical form by a plot known as Mohr’s Circle.
• This graphical representation is extremely useful because it enables you to visualize the relationships between the normal and shear stresses acting on various inclined planes at a point in a stressed body
• Using Mohr’s Circle you can also calculate principal stresses, maximum shear stresses, and stresses on inclined planes.
• The angle between two planes is represented as 2θ 

By Mohr's circle

The radius of Mohr's circle = \(R = \sqrt{[{1\over2}(σ_x-σ_y)]^2+\tau_{xy}^2}\)

Center of Mohr's Circle = (σ_avg, 0) 

σavg = \({{σ_x+σ_y}\over2}\)

Principle Stress \(σ_{x_{1}} = {{σ_x+σ_y}\over2}+{{σ_x-σ_y}\over2}cos2θ+\tau_{xy}sin2θ \)

Shear Stress \(\tau_{{x_{1}}{y_{1}}} = -{{σ_x+σ_y}\over2}sin2θ+\tau_{xy}cos2θ\)

So the center of Mohr's Circle from the origin is (\({{σ_x+σ_y}\over2}\),0)

Concept:

The general case of a stress system when two tensile stress σ_x and σ_y acting on a body on two perpendicular axes along with shear stress τ_xy is given below.

∵ no tensile or compressive stress exists, the Mohr’s circle diagram is simply a circle of radius τ_xy and center at the intersection of the two axes i.e origin.

It can also be verified by the transformation of the plane formula.

\(Radius=\sqrt{\left ( \frac{σ_x\;-\;σ_y}{2} \right )^2+{τ_{xy}}^2}\)

∵ σ_x = 0 and σ_y = 0

∴ Radius = τ_xy

\(Centre=\left ( \frac{σ_x\;+\;σ_y}{2},0 \right )\)

∴ centre = (0, 0).

Explanation:

Graphical method for locating principal axes:

1. Dyadic circle

2. Mohr-circle

3. Circle of inertial

Additional Information

Ellipse of inertia is not the graphical method for locating principal axes. Momental ellipse or ellipse of inertia is a graphical method used to locate the neutral axis in case of unsymmetrical bending.

Concept:

When the normal stresses on the two mutually perpendicular planes are equal and alike then the radius of the Mohr circle will be zero.

\({\rm{Radius\;of\;Mohr's\;Circle}},R = \sqrt {{{\left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right)}^2} + {\tau _{xy}}^2} \)

Only in figure 3, we can see that 

 \({\sigma _x} = {\sigma _y}\;and\;{\tau _{xy}} = 0\)

∴ option 3 is the correct answer