Moment of Inertia and Centroid MCQ Quiz in मल्याळम - Objective Question with Answer for Moment of Inertia and Centroid - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

A radius of gyration is given by:

\(r = \sqrt {\frac{I}{A}} \)

I = Moment of inertia about C.G

A = Area of the x-section

\(I = \frac{{B{D^3}}}{{12}}\)

A = BD

\(r = \sqrt {\frac{{B{D^3}}}{{12 \times BD}}}\)

\(r = \frac{D}{{2\sqrt 3 }}\)

Explanation:

Parallel axis theorem: 

The moment of inertia of a body about an axis parallel to the body passing through its center is equal to the sum of moment of inertia of the body about the axis passing through the center in the same plane and product of the mass of the body times the square of the distance between the two axes.

I = Icom + Mx2, Where I is the total moment of inertia of the body, Icom is the moment of inertia about the centre of mass, M is mass, x is the perpendicular distance from an axis passing from centre of mass.

Additional Information

Perpendicular axis theorem: 

This theorem states that the moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.

Hence it can be expressed as \({I_z} = {I_y} + {I_x}\)

Here Ix, Iy and Iz are the moment of inertia of an object along X, Y and Z- axis respectively

Mass moment of inertia of a uniform thin rod of mass (m) and length (l) about its mid-point and perpendicular to its length is ml2 /12. 

Moment of Inertia of the triangular section about an axis through its CG and parallel to the base:

\({I_G} = {I_{x'x'}} = \frac{{b{h^3}}}{{36}}\)

Applying Parallel Axis Theorem:

\({I_{BC}} = {I_{x'x'}} + Area\;of\;triangle \times {\left( {\frac{h}{3}} \right)^2}\)

\({I_{BC}} = \frac{{b{h^3}}}{{36}} + \left( {\frac{1}{2} \times b \times h} \right) \times {\left( {\frac{h}{3}} \right)^2}\)

\({I_{BC}} = \frac{{b{h^3}}}{{36}} + \frac{{b{h^3}}}{{18}}\)

\({I_{BC}} = \frac{{b{h^3} + 2b{h^3}}}{{36}} = \frac{{3b{h^3}}}{{36}} = \frac{{b{h^3}}}{{12}}\)

Concept:

For Neutral axis

\(̅{y}=\frac{A_{1}× y_{1}+A_{2}× y_{2}}{A_{1}+A_{2}}\)

Taking bottom section as reference

where, A₁ = L × L/2

A₂ = L/2 × L/2 

y₁ = (L/2 + L/4) = 3L/4

y₂ = L/4

\(̅{y}=\frac{\left [ (L× \frac{L}{2})(\frac{3L}{4})+(\frac{L}{2}× \frac{L}{2})(\frac{L}{4}) \right ]}{\left ( L× \frac{L}{2}+\frac{L}{2}× \frac{L}{2} \right )}\)

y̅ = \({7L \over 12}\) from bottom 

So, neutral axis from top, y̅' = \(L\;-{7L \over 12}={5L\over 12}\)

Concept:

The centroid of the section shown in the question = 

Where, 

y1 is the distance centroid of the rectangle from base AB

y2 is the distance centroid of the circle from base AB

Explanation:

Given, 

Area of rectangle = 300 x 200 = 60000 mm²

The centroid of the rectangle from AB = 150 mm

Area of circle = \( \frac{\pi }{4}\)x1502 = 17671.46 mm2

The centroid of the circle from AB = 300-100 = 200 mm

Thus, Centroid = = \( \frac{60000*150-17671.46*200 }{60000-17671.46}\)=129.1 mm

Explanation:

MOI for a Triangle:

About centroidal axis: \(I_G=\frac{bh^3}{36}\)

About base: \(I_{BC}=\frac{bh^3}{12}\)

A square can be divided into two triangles about its base.

Diagonal of square = Base of the triangle (b) = \(a\sqrt{2}\)

Height of triangle = Half of square diagonal (h) = \(\frac{a}{\sqrt{2}}\)

MOI of the square about the diagonal:

\(I_{square}=2\times \left(I_{triangle}\right)_{base}\)

\(I_{square}=2\times \frac{bh^3}{12}\)

\(I_{square}=2\times \frac{(a\sqrt{2})\left(\frac{a}{\sqrt{2}}\right)^3}{12}\)

\(I_{square}=\frac{a^4}{12}\ \)

Concept:

Moment of inertia of Rectangle about each of its centroidal axis is:

\({{\bf{I}}_{{\bf{xx}}}} = \frac{{{\bf{b}}{{\bf{d}}^3}}}{{12}}\;{\bf{and}}\;{{\bf{I}}_{{\bf{yy}}}} = \frac{{{\bf{d}}{{\bf{b}}^3}}}{{12}}\)

where b = width of the rectangle, h = height of the rectangle

Moment of inertia of the I section can be calculated by calculating the Moment of inertia of the rectangular section shown below

Now the Moment of inertia of I section = Moment of inertia of rectangle section 1 - (2 × Moment of inertia of rectangle Section 2)

Calculation:

Given:

b₁ = 4t, d₁ = 6t, \(b_2 = (2t-\frac{t}{2})=1.5t\), d₂ = (6t - t - t ) = 4t

where b1 = width rectangle section 1, b2 = width rectangle section 2, d1 = height rectangle section 1, d1 = height rectangle section 2

\(Moment\ of\ inertia\ of\ I\ section= \frac{b_1d_1^3}{12}-(2\times\frac{b_2d_2^3}{12})\)

\(Moment\ of\ inertia\ of\ I\ section= \frac{4t×(6t)^3}{12}-(2\times\frac{1.5t×(4t)^3}{12})\)

\(Moment\ of\ inertia\ of\ I\ section= 72t^4-{16t^4}\)

\(Moment\ of\ inertia\ of\ I\ section= 56t^4\)

Concept:

Moment of inertia of a triangle with base b and height h

\({I_{base}} = \frac{{b{h^3}}}{{12}}\)

Parallel axis theorem

Parallel axis theorem is used to find a moment of inertia about an axis which is at some distance from the centroidal axis and parallel to centroidal.

Suppose an axis is at distance d from the centroidal axis and parallel to the centroidal axis.

according to the parallel axis theorem, the moment of inertia about that axis is given by,

\({I_{about~axis~parallel~to~centroida~laxis}} = {I_{about~centroidal~axis}} + A{d^2}\)

Moment of inertia about the centroidal axis of the triangle

Applying parallel axis theorem we get

\({I_{xx}} = {I_{cc}} + A{\left( {\frac{h}{3}} \right)^2}\)

\({I_{cc}} = {I_{xx}} - A{\left( {\frac{h}{3}} \right)^2}\)

\(I_{cc}=\frac{{b{h^3}}}{{12}} - \frac{{bh}}{2} \times \frac{{{h^2}}}{9} = \frac{{b{h^3}}}{{36}}\)

\({I_{centroid}} = \frac{{b{h^3}}}{{36}}\)

Concept:

Centre of gravity of plane figures (T-sections, I-sections, L-sections, etc.)

The plane geometrical figures (such as T-section, I-section, L-section etc.) have only areas but no mass. The centre of gravity of such figures is found out in the same way as that of solid bodies. The centre of area of such figures is known as centroid and coincides with the centre of gravity of the figure. It is a common practice to use the centre of gravity for centroid and vice versa.

Let \(\bar x\) and \(\bar y\) be the co-ordinates of the centre of gravity with respect to some axis of reference, then

\(\bar y={A_1y_1+A_2y_2+A_3y_3+.....+A_ny_n\over A_1+A_2+A_3+.....+A_n}={{\sum_{i=1}^{i=n}A_iy_i}\over \sum_{i=n}^{i=n}A_i}\)

\(\bar x={A_1 x_1+A_{2} x_2+A_3 x_{3}+.....+A_n x_n\over A_1+A_2+A_3+.....+A_n}={{\sum_{i=1}^{i=n}A_i x_i}\over \sum_{i=n}^{i=n}A_i}\)

where A₁, A₂, A₃........ etc., are the areas into which the whole figure is divided x₁, x₂, x₃ ..... etc., are the respective coordinates of the areas A₁, A₂, A₃....... on X-X axis with respect to same axis of reference.
y₁, y₂, y₃....... etc., are the respective coordinates of the areas  A1, A2, A3....... on the Y-Y axis with respect to the same axis of the reference.

Calculation:

Given

\(\bar y={A_1y_1+A_2y_2+A_3y_3\over A_1+A_2+A_3}\)

\(A_1 =800\, mm^2\, y_1=69\, mm\)

\(A_2=400 \, mm^2\, y_2=44\, mm\)

\(A_3=600 \, mm^2\, y_3=12\, mm\)

\(\bar y={800\times 69+400\times 44+600\times 12\over 800+400+600}\)

\(\bar y={80000\over 1800}=44.44\, mm\)

\(\bar y=44.44\, mm\)