Analytical Method of Analysis MCQ Quiz - Objective Question with Answer for Analytical Method of Analysis - 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:

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

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:

The maximum and minimum normal stress is given by

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

where σx and σy are direct stress and τxy is the shear stress

Calculation:

Given:

From the given condition we have,

σx = p, σy = 0, τxy = q

Now maximum normal stress is:

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

\(\sigma_{max}= \frac{p }{2} + \frac{1}{2}\sqrt {{p ^2} + 4{q ^2}} \)

Concept:

Principal stresses are given by:

\({σ _{1,2}} = \frac{{{σ _x}\;+\;{σ _y}}}{2} \pm \sqrt {{{\left( {\frac{{{σ _x}\;-\;{σ _y}}}{2}} \right)}^2} +\;{τ _{x{y^2}}}} \)

Maximum shear stress:

\(\begin{array}{l} {τ _{max}} = \frac{{{σ _1}\;-\;{σ _2}}}{2}\\ {τ _{max}} = \frac{{{σ _1}\;-\;{σ _2}}}{2} = \sqrt {{{\left( {\frac{{{σ _x}\;-\;{σ _y}}}{2}} \right)}^2} + τ _{xy}^2} \end{array}\)

Calculation:

Given:

σ_x = 50 MPa, σy = -10 MPa and τ_xy = 40 MPa

\({σ _{1,2}} = \frac{{{σ _x}\;+\;{σ _y}}}{2} \pm \frac{1}{2}\sqrt {{{\left( {{σ _x} - {σ _y}} \right)}^2} + 4τ _{xy}^2} \)

\(= \frac{{50\;+\;\left( { - 10} \right)}}{2} \pm \frac{1}{2}\sqrt {{{\left( {50 + 10} \right)}^2} + 4{{\left( {40} \right)}^2}}\)

⇒  20 ± 50

\(\therefore{σ _{1,2}} = 70~MPa, - 30~MPa\)

Concept

The coordinates of the centre of Mohr’s circle given by σAvg:

\(\sigma_{avg}=\left( {\frac{{{σ _x}\; + \;{σ _y}}}{2},0} \right)\)

Calculation

Given

σ_x = 200 MPa, σ_y = -100 MPa

Co-ordinates of the center of Mohr’s circle is

\(\sigma_{avg}=\left( {\frac{{{σ _x}\; + \;{σ _y}}}{2},0} \right)\)

\(\sigma_{avg}=\left(\frac{200\;+\;(-100)}{2},0\right)\)

\(\sigma_{avg}= \left( {\;\frac{{100}}{2},\;0} \right)=(50,0)\)

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:

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

Concept:

The maximum and minimum normal stress is given by

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

where σx and σy are direct stress and τxy is the shear stress

Calculation:

Given:

From the given condition we have,

σx = p, σy = 0, τxy = q

Now maximum normal stress is:

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

\(\sigma_{max}= \frac{p }{2} + \frac{1}{2}\sqrt {{p ^2} + 4{q ^2}} \)

Concept

The coordinates of the center of Mohr’s circle given by σAvg:

\(\sigma_{avg}=\left( {\frac{{{σ _x}\; + \;{σ _y}}}{2},0} \right)\)

Calculation

Given

σ_x = 200 MPa, σ_y = -100 MPa

Co-ordinates of the center of Mohr’s circle is

\(\sigma_{avg}=\left( {\frac{{{σ _x}\; + \;{σ _y}}}{2},0} \right)\)

\(\sigma_{avg}=\left(\frac{200\;+\;(-100)}{2},0\right)\)

\(\sigma_{avg}= \left( {\;\frac{{100}}{2},\;0} \right)=(50,0)\)

Concept:

Principal Stress:

\({\sigma _{1,2}} = \frac{{{\sigma _x} + {\sigma _y}}}{2} \pm \sqrt {{{\left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right)}^2} + τ _{xy}^2\;} \)

Where σx and σy are the normal stress in x and y-direction respectively and τ_xy is the shear stress in x-y plane. 

The minimum principal stress will be 

\({\sigma _2} = \frac{{{\sigma _x} + {\sigma _y}}}{2} - \sqrt {{{\left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right)}^2} + τ _{xy}^2\;} \)

Calculation:

Given:

σ_x = 100 kPa, τ_xy = 50 kPa, σ₂ = 10 kPa

Minimum Principal Stress:

\({\sigma _2} = \frac{{{\sigma _x} + {\sigma _y}}}{2} - \sqrt {{{\left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right)}^2} + τ _{xy}^2\;} \)

\(10 = \frac{{100 + {\sigma _y}}}{2} - \sqrt {{{\left( {\frac{{100 - {\sigma _y}}}{2}} \right)}^2} + {{50}^2}\;} \)

\(\sqrt {{{\left( {\frac{{100 - {\sigma _y}}}{2}} \right)}^2} + {{50}^2}\;} = \frac{{100 + {\sigma _y}}}{2} - 10\)

\({\left( {\frac{{100 - {\sigma _y}}}{2}} \right)^2} + 2500 = {\left( {\frac{{100 + {\sigma _y}}}{2} - 10} \right)^2} = {\left( {\frac{{100 + {\sigma _y}}}{2}} \right)^2} + {10^2} - 10\left( {100 + {\sigma _y}} \right)\)

\({\left( {100 - {\sigma _y}} \right)^2} + 10000 = {\left( {100 + {\sigma _y}} \right)^2} + 400 - 40\left( {100 + {\sigma _y}} \right)\)

\({100^2} + \sigma _y^2 - 200{\sigma _y} + 10000 = {100^2} + \sigma _y^2 + 200{\sigma _y} + 400 - 4000 - 40{\sigma _y}\)

\(13600 = 360\;{\sigma _y} \)

\(\Rightarrow {\sigma _y} = 37.78\;MPa\)

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:

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:

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

Concept:

The maximum and minimum normal stress is given by

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

where σx and σy are direct stress and τxy is the shear stress

Calculation:

Given:

From the given condition we have,

σx = p, σy = 0, τxy = q

Now maximum normal stress is:

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

\(\sigma_{max}= \frac{p }{2} + \frac{1}{2}\sqrt {{p ^2} + 4{q ^2}} \)

Concept

The coordinates of the center of Mohr’s circle given by σAvg:

\(\sigma_{avg}=\left( {\frac{{{σ _x}\; + \;{σ _y}}}{2},0} \right)\)

Calculation

Given

σ_x = 200 MPa, σ_y = -100 MPa

Co-ordinates of the center of Mohr’s circle is

\(\sigma_{avg}=\left( {\frac{{{σ _x}\; + \;{σ _y}}}{2},0} \right)\)

\(\sigma_{avg}=\left(\frac{200\;+\;(-100)}{2},0\right)\)

\(\sigma_{avg}= \left( {\;\frac{{100}}{2},\;0} \right)=(50,0)\)

Concept:

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

Calculation:

Given:

σ₁ = 2σ, σ₂ = 0

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

\(\therefore {\tau _{max\;}} = \frac{{2\sigma - 0}}{2}\)

\(\therefore {\tau _{max\;}} = \frac{{2\sigma }}{2}\)

\(\therefore {\tau _{max\;}} = \sigma\)

Concept:

Principal stresses are given by:

\({σ _{1,2}} = \frac{{{σ _x}\;+\;{σ _y}}}{2} \pm \sqrt {{{\left( {\frac{{{σ _x}\;-\;{σ _y}}}{2}} \right)}^2} +\;{τ _{x{y^2}}}} \)

Maximum shear stress:

\(\begin{array}{l} {τ _{max}} = \frac{{{σ _1}\;-\;{σ _2}}}{2}\\ {τ _{max}} = \frac{{{σ _1}\;-\;{σ _2}}}{2} = \sqrt {{{\left( {\frac{{{σ _x}\;-\;{σ _y}}}{2}} \right)}^2} + τ _{xy}^2} \end{array}\)

Calculation:

Given:

σ_x = 50 MPa, σy = -10 MPa and τ_xy = 40 MPa

\({σ _{1,2}} = \frac{{{σ _x}\;+\;{σ _y}}}{2} \pm \frac{1}{2}\sqrt {{{\left( {{σ _x} - {σ _y}} \right)}^2} + 4τ _{xy}^2} \)

\(= \frac{{50\;+\;\left( { - 10} \right)}}{2} \pm \frac{1}{2}\sqrt {{{\left( {50 + 10} \right)}^2} + 4{{\left( {40} \right)}^2}}\)

⇒  20 ± 50

\(\therefore{σ _{1,2}} = 70~MPa, - 30~MPa\)