Saturday, 25 February 2017

MOMENT OF INERTIA OF RECTANGULAR SECTION ABOUT ITS BASE

We were discussing “The perpendicular axis theorem and its proof”, “The theorem of parallel axis about moment of inertia”, “Method to determine the area moment of inertia for the rectangular section” and “Area moment of inertia” in our previous posts.

Today we will see here the method to determine the area moment of inertia for the rectangular section about a line passing through the base of the rectangular section with the help of this post.

Let us consider one rectangular section ABCD as displayed in following figure. Let us assume that one line is passing through the base of the rectangular section and let us consider this line as line CD and we will determine the area moment of inertia for the rectangular section about this line CD.
B = Width of the rectangular section ABCD
D = Depth of the rectangular section ABCD
ICD = Moment of inertia of the rectangular section about the CD line

Now we will determine the value or expression for the moment of inertia of the rectangular section about the line CD

Let us consider one rectangular elementary strip with thickness dY and at a distance Y from the line CD as displayed in above figure.

Let us determine first the area and moment of inertia of the rectangular elementary strip about the line CD

Area of rectangular elementary strip, dA = dY.B
Moment of inertia of the rectangular elementary strip about the line CD = dA.Y2
Moment of inertia of the rectangular elementary strip about the line CD = B Y2 dY

Now we will determine the moment of inertia of entire area of rectangular section about the line CD and it could be easily done by integrating the above equation between limit 0 to D.

Therefore, moment of inertia of entire area of rectangular section about the line CD will be as displayed here in following figure
Therefore, moment of inertia of the rectangular section about the line CD
ICD =BD3/3
Do you have any suggestions? Please write in comment box

Reference:

Strength of material, By R. K. Bansal
Image Courtesy: Google
We will see another important topic i.e. Determination of the moment of inertia of a hollow rectangular section in the category of strength of material.

Also read

Continue Reading

AREA MOMENT OF INERTIA FOR RECTANGULAR SECTION

We were discussing “The perpendicular axis theorem and its proof”, “The theorem of parallel axis about moment of inertia” and “Area moment of inertia” in our previous posts.

Today we will see here the method to determine the area moment of inertia for the rectangular section with the help of this post.

Let us consider one rectangular section ABCD as displayed in following figure. Let us assume that centre of gravity of the given rectangular section is C.G and axis X-X passing through the centre of gravity of the rectangular section as displayed in following figure.
B = Width of the rectangular section ABCD
D = Depth of the rectangular section ABCD
IXX = Moment of inertia of the rectangular section about the X-X axis

Now we will determine the value or expression for the moment of inertia of the rectangular section about the X-X axis

Let us consider one rectangular elementary strip with thickness dY and at a distance Y from the X-X axis as displayed in above figure.

Let us determine first the area and moment of inertia of the rectangular elementary strip about the X-X axis

Area of rectangular elementary strip, dA = dY.B
Moment of inertia of the rectangular elementary strip about the X-X axis = dA.Y2
Moment of inertia of the rectangular elementary strip about the X-X axis = B Y2 dY

Now we will determine the moment of inertia of entire area of rectangular section about the X-X axis and it could be easily done by integrating the above equation between limit (-D/2) to (D/2).
Therefore, moment of inertia of the rectangular section about the X-X axis after calculation, we will have

IXX =BD3/12

Similarly, we will determine the moment of inertia of the rectangular section about the Y-Y axis and we will have

IYY =DB3/12

Do you have any suggestions? Please write in comment box

Reference:

Strength of material, By R. K. Bansal
Image Courtesy: Google

We will see another important topic i.e. Determination of the moment of inertia of the rectangular section about a line passing through the base of rectangular section in the category of strength of material.

Also read

Continue Reading

Wednesday, 22 February 2017

STATE AND PROVE THE THEOREM OF PARALLEL AXIS ABOUT MOMENT OF INERTIA


We were discussing the “Theorem of perpendicular axis” and “Radius of gyration” and also we have seen the “Difference between centroid and centre of gravity” with the help of our previous posts.

Now we are going further to start our discussion to understand the theorem of parallel axis with the help of this post.

Let us see now theorem of parallel axis

According to the theorem of parallel axis, the moment of inertia for a lamina about an axis parallel to the centroidal axis (axis passing through the center of gravity of lamina) will be equal to the sum of the moment of inertia of lamina about centroidal axis and product of area and square of distance between both axis.
IAB = IG + Ah2
Where,
IG = Moment of inertia of entire area A about the centroidal axis,
IAB = Moment of inertia of entire area A about the axis AB which is parallel to the centroidal axis
A = Entire area of the lamina
h= Perpendicular distance between both axis i.e. centroidal axis and Axis AB and it is displayed in following figure.
Let us see the above figure which indicates one lamina and area of the lamina is A. Let us assume that axis X-X is passing through centre of gravity of lamina and axis AB is parallel to the axis X-X or centroidal axis as displayed in above figure.

Let us assume one small strip, in given lamina, parallel to the centroidal axis and at a disthance Y from the centroidal axis. Let us assume that area of small strip is dA.

Moment of inertia of small strip area dA about the X-X axis = dA.Y2

Hence, moment of inertia of entire area A about the X-X axis, IXX will be determined as mentioned here.

IXX =Ʃ dA.Y2

Moment of inertia of entire area A about the X-X axis could also be written as moment of inertia of entire area about the centroidal axis because axis X-X is passing through the centre of gravity of the lamina i.e. we can write IG in above equation instead of IXX.

IG =Ʃ dA.Y2

Now we will determine the moment of inertia of small strip area dA about the AB axis.

IAB= dA. (h +Y) 2

Therefore moment of inertia of entire area of lamina about the AB axis will be concluded as

IAB= Ʃ dA. (h +Y) 2
IAB= Ʃ dA. (h2 +Y2+2hY)
IAB= Ʃ dA. h2 + Ʃ dA Y2 + Ʃ dA.2hY

We can easily observe here that Ʃ dA = A, because distance between axis i.e. h will be constant. Therefore we can write above equation as mentioned here

IAB= A. h2 + IG + 2h. ƩdA Y

As we can observe here that dA.Y represents the moment of area of small strip dA about the XX axis and Ʃ dA.Y represents the moments of entire area about the XX axis or centroidal axis.

 As we know that the moments of entire area about the XX axis will be equal to the product of entire area and distance between C.G of entire area and XX axis. But as axis XX passes through the C.G of entire area and therefore distance between C.G of entire area and XX axis will be zero and therefore Ʃ dA.Y will be zero.

Hence, we can write above equation as mentioned here
IAB= A. h2 + IG

Therefore we can brief here the theorem of parallel axis as the moment of inertia for a lamina about an axis parallel to the centroidal axis (axis passing through the center of gravity of lamina) will be equal to the sum of the moment of inertia of lamina about centroidal axis and product of area and square of distance between both axis.

Do you have any suggestions or any amendment required in this post? Please write in comment box.

Reference:

Strength of material, By R. K. Bansal
Image Courtesy: Google
We will see another important topic i.e., in the category of strength of material, in our next post.

Also read

Continue Reading

Tuesday, 21 February 2017

STATE THE PERPENDICULAR AXIS THEOREM WITH PROOF

We were discussing “Area moment of inertia” and “Radius of gyration”,” and also we have seen the “Basic principle of complementary shear stresses” with the help of our previous posts.

Now we are going further to start our discussion to understand the theorem of perpendicular axis with the help of this post.

Let us first see here theorem of perpendicular axis

According to the theorem of perpendicular axis, if IXX and IYY are the moments of inertia of an irregular lamina about tow mutually perpendicular axis X-X and Y-Y respectively in the plane of lamina, then there will be a moment of inertia IZZ of lamina around the Z-Z axis which will be perpendicular to the plane of lamina and passing through the intersection of X-X and Y-Y axis.

Moment of inertia IZZ of lamina around the Z-Z axis will be given by following equation as per the theorem of perpendicular axis.

IZZ = IXX + IYY

Let us see the following figure which indicates one irregular lamina in the plane of X-Y and area of the lamina is A. Let us assume one small elemental area dA in the plane of X-Y.
Where,
X= Distance of the C.G of the small elemental area dA from OY axis
Y= Distance of the C.G of the small elemental area dA from OX axis
R= Distance of the C.G of the small elemental area dA from OZ axis

We can easily note it here that distance of the C.G of the small elemental area dA from OZ axis could be written as mentioned here.
R2= X2+Y2
Moment of inertia of small elemental area dA about the OX axis = dA.Y2
Hence, moment of inertia of entire area A about the OX axis, IXX will be determined as mentioned here.
IXX =Ʃ dA.Y2

Moment of inertia of small elemental area dA about the OY axis = dA.X2
Hence, moment of inertia of entire area A about the OY axis, IYY will be determined as mentioned here.
IYY =Ʃ dA.X2

Moment of inertia of small elemental area dA about the OZ axis = dA.R2
Hence, moment of inertia of entire area A about the OZ axis, IZZ will be determined as mentioned here.

IZZ =Ʃ dA. R2
IZZ =Ʃ dA. [X2+ Y2]
IZZ =Ʃ dA.X2+ Ʃ dA.Y2
IZZ = IYY +IXX

IZZ = IXX + IYY

Therefore we can brief here the theorem of perpendicular axis as moment of inertia of an irregular lamina about an axis normal or perpendicular to it will be equal to the sum of the moment of inertia about any two mutually perpendicular axis in the plane of the lamina.

Do you have any suggestions or any amendment required in this post? Please write in comment box.

Reference:

Strength of material, By R. K. Bansal
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