Category: Maximum value of Ix and Iy

How to get the maximum value of Ix and Iy for plane shapes?

9- A Solved problem-case-2-Mohr’s circle of inertia.
Table Of Contents
Solved problem-case-2- Mohr's circle of inertia.
Post content.
Solved problem-case-2-Mohr's circle of inertia-drawing the circle.
Solved problem-case-2-Mohr's circle of inertia-direction of principal axes U&V.
The directions of U and V for the normal view.
Solved problem-case-2-Mohr's circle of inertia using general expression.
Oriented datums to let x-direction as a horizontal axis.
Solved problem-case-2- Mohr’s circle of inertia.
Post content.
In this post, we will introduce a solved problem-case-2-Mohr’s circle of inertia, where Ix is lesser than Iy and the product of inertia Ixy is positive.

For a given Area the following data are presented as follows: the moment of inertia Ix is 14 inch4. the moment of inertia about y-axis Iy is 24 inch4. The product of inertia Ixy is 12 inch4.
It is required to determine the principal moment of inertia and the angle alpha that the x-axis makes with the maximum moment of inertia.
In the first slide, I have evaluated the coordinates of points A, and B, as well as the radius R which we will use to draw Mohr’s circle of inertia.
Solved problem-case-2-Mohr’s circle of inertia-drawing the circle.
For the Solved problem-case-2-Mohr’s circle of inertia, We start by drawing two intersecting axes. The horizontal axis represents the value of the moment of inertia. The vertical axis represents the value of the product of inertia, with a positive value pointing up. One unit represents 1 inch4.
We start by locating Point A, which has Ix, Ixy values as(14,12) units and will be located above the axis of inertia by a value of positive Ixy equals 12 units, and apart from the vertical axis Ixy by a distance of (14) units.
Similarly, we can draw point B, which has a coordinate of (Iy,-Ixy). In units will be (24,-12), for the negative value the point will be below the axis of inertia by 12 units and apart from the vertical axis by the positive value of 24 units. We will join both two points, The line AB will intersect the horizontal axis at point O, which will be the center of the circle.
We can draw the circle by getting the middle point O, which has a coordinate of 0.50*(Ix+Iy), from the given data this value is equal to (14+24)*0.50=19 units.
The radius of the circle is estimated from the equation, R=sqrt ((Iy-Ix)^2+Ixy^2), this is applied since Iy is bigger than Ix. We have Ix=14 inch4.Iy=24 inch4.Ixy=12 inch4. We can use the data to get the radius of Mohr’s circle of inertia, applying the known formula, we can get the radius value equal to 13 inch4. Please refer to the slide image for more details.
This is How to get the value of 2θp1 for the Solved problem for A Solved problem-case-2-Mohr’s circle of inertia.
To get the maximum value of inertia Imax we can add the value of the radius of the circle to the value of the center point, the maximum value of the moment of inertia is equal to (19+13)=32 units, and the product of inertia is zero, this point is point E.
For the minimum value of inertia, the value will be equal to the (0.50*(Ix+Iy)-R, where R is the radius value.
I minimum is equal to 6 units, the product of inertia is equal to zero, and the point of minimum value of inertia is point C.
Solved problem-case-2-Mohr’s circle of inertia-direction of principal axes U&V.
To draw the direction of the major axis U in Mohr’s circle of inertia, we will locate point A’, which is the mirror point of Point A. Point A’ has a coordinate of (10,-12) units. we will join point O, and with point A’ we will get the direction of the major Axis U.
To draw the direction of the major axis V in Mohr’s circle of inertia, we will locate point B’, which is the mirror point of Point B. Point B’ has a coordinate of (24,+12) units. we will join point O, and with point B’ we will get the direction of the major Axis V.
The directions of U and V for the normal view.
Draw a line from point C to A’ , this is the direction of the U-axis in the normal view. Draw a line from C to point B’ this is the direction of V -axis in the normal view.
Solved problem-case-2-Mohr’s circle of inertia using general expression.
We can use the general expression for Ix’ to check the value of Imax, provided that when the angle 2θ = (2φp1), which is (-112.62 degrees ), Ix’ value will be equal to Imax. We plug in with Ix value=14 inch4, Iy value=24 inch4 and Ixy=12 inch4.
The value of Ix’=32 inch4 is the same as Imax estimated by the use of Mohr’s circle.
We can use the general expression for Iy’ to check the value of Imin, provided that when the angle 2θ = (2φp1), which is (-112.62 degrees ), Iy’ value will be equal to Imin. We plug in with Ix value=14 inch4, Iy value=24 inch4, and Ixy=12 inch4.
The value of Iy’=6.0 inch4 is the same as Imin estimated by the use of Mohr’s circle.
It is important to check that the sum of Ix and Iy, will be the same value of the sum of Iu and Iv, where Iu is the maximum value of inertia, While Iv is the minimum value of inertia. Ix+Iy=14+24=38 inch4. imin+Imax=6+32=38 inch4.
Oriented datums to let x-direction as a horizontal axis.
To let the x-axis as horizontal Mohr’s circle of inertia, the x-axis is represented by line OE. The datums of Ix&Ixy are adjusted to represent the direction of U.
Line OE represents the x-direction concerning two new datums which are shown in the next slide image. Line OC represents the Y-direction concerning two new datums. The full data that covers our Solved problem case-1-Mohr’s circle of inertia is included in the slide image.
The following post will discuss Mohr’s circle of inertia-third case.
There is a video illustrating the content of this post-introduction video for Mohr’s Circle of Inertia.
This is a link to a useful external resource: a calculator for Cross Section, Mass, Axial and polar area moment of Inertia, and Section Modulus.
March 25, 2022
8- Easy Approach to Mohr’s circle of inertia-second case
Table Of Contents
Mohr's circle of inertia-second case.
Post content.
Mohr's circle of inertia- second case.
What is the orientation of the principal axes for Mohr's circle of inertia-second case?
The steps used to draw Moh's circle of inertia-second case.
Need for two mirror points A' and B' in Mohr's circle of inertia-second case.
The direction of U and V axes in Mohr's circle of inertia-second case.
Mohr’s circle of inertia-second case.
Post content.
In this post, we will be talking about Mohr’s circle of inertia-second case. Mohr’s circle of the inertia-second case is the case where the moment of inertia about the x-axis is lesser than the moment of inertia about axis Y, and the product of inertia Ixy is positive.
Mohr’s circle of the inertia-second case can be found as an example in two cases, the first case is where we have a rectangular shape where the height is lesser than the width of the rectangle. The value of the moment of inertia about the x-axis and Y-axis together with the value of the product of inertia about the left corner is included in the next slide image.
Mohr’s circle of inertia- second case.
Mohr’s circle of the inertia-second case is the case of inertia about the left corner of the unequal angle of dimension (b,h).
What is the orientation of the principal axes for Mohr’s circle of inertia-second case?
We have the general expression for the moment of inertia for any oriented axis, we call it Ix’. For Mohr’s circle of the inertia-second case, the value can be a maximum value when the angle 2θ has a cosine of a negative value and the sine value is also negative, this angle is called 2φp1 or ( 2θp1). This angle is measured from the x-axis.
The other orthogonal direction has an angle of 2φp2 or ( 2θp2) is also measured from the x-axis, but this angle will have a positive cosine value and a positive sine value. The angle 2θp1 will be located in the third quarter.
This is the case where Ix’ is minimal when 2θp2 will be located in the first quarter.
How to get the U and V direction for the normal view? by rotating the x-axis by an angle of 2θp1 in the clockwise direction?
The steps used to draw Moh’s circle of inertia-second case.
The first two steps are to draw the two orthogonal axes. The first axis represents the Moment of inertia values, Ix, Iy, and Imax, and minimum values, while the second axis represents the values of the product of Inertia Ixy.
The coordinate of point A(Ix, Ixy), +Ixy is given from the date of the case. Point B has a coordinate (Iy,-Ixy), and the value of Ixy is the positive value of the Ixy of point A. Locating Points A and B with their respective values of Ix and Iy, As shown in the next slide image.
Join Points A and B and get the middle point of line AB, this point is the point of the Circle center.
The radius value is the sqrt( (Iy-Ix/2)^2+Ixy^2).
Start from point O and draw the circle with the radius value just estimates. The circle will interest the line in two points C and E.
Angle 2θp1 is the angle between the X-axis and Major axis U, which is the angle enclosed between OA and Line OE. Point E is the point of the maximum value of inertia.
Angle 2θp2 is the angle between the x-axis and Minor axis V, which is the angle enclosed between OA and Line OC. Point C is the point of the minimum value of inertia.
The distance from the vertical axis Ixy to point C will give the Minimum value of inertia, while the distance from the same axis to point E will give the maximum value of inertia.
Need for two mirror points A’ and B’ in Mohr’s circle of inertia-second case.
In the normal view the x-axis is a horizontal axis, while in Mohr’s circle of the inertia-second case, the x-axis is oriented by an angle of 2φp1 from the U-direction. We need a mirror point A’ to let the x-direction in Mohr’s circle of the inertia-second case be a horizontal line.
Setting point A’, which is a mirror of point A and has a coordinate of (Ix,-Ixy) and gives the direction of the maximum moment axis U. This point A’ will enable us to rotate the x-axis or line OA a clockwise rotation by the value of 2θp1. The x-axis will be horizontal, the U line will also have a new direction represented by line OA’.
Point B’ is the mirror of point B and has a coordinate of (Iy,+Ixy) and is used to indicate the direction of the V-axis, which is the direction of the minimum moment of the inertia axis.
From the relation of tan 2φp, we have a positive value of tangent which means that the V axis will have an enclosed angle measured in the anti-clockwise direction.
The direction of U and V axes in Mohr’s circle of inertia-second case.
Join points OA’ to get the U-direction. Join the two points O and B’ to get the V- direction. The directions of both the U and V axes are shown in the next slide image.
The V axis is rotated by an angle of (θp2) anti-clockwise from axis X, While the U axis is rotated by an angle of ( φθ1) in the clockwise direction from axis X in the normal view.
For the orientation of two axes Ixy and Ix, so the x-axis is horizontal, we can find that point E is the point at which Ix’ = Ix at 2θ = zero and Ix’y’ =+Ixy since the line OE is the horizontal line, while point A’ has Ix’=Iy and Ix’y’=-Ixy.
The line OA’ is the direction of U, This line has an angle equal to ( -2θp1) measured from the horizontal line OE. The line OB’ is the direction of V. This line has an angle equal to (+2θp2) measured from the horizontal line OE as an anti-clock direction.
In the next post, we will introduce a solved problem for Mohr’s circle of inertia-second case.
This is a link to a useful external resource. Calculator for Cross Section, Mass, Axial and Polar Area Moment of Inertia, and Section Modulus.
March 25, 2022
7- A Solved problem-case-1-Mohr’s circle of inertia.
Table Of Contents
Solved problem-case-1-Mohr's circle of inertia.
Solved problem-case-1-Mohr's circle of inertia.
Solved problem-case-1-Mohr's circle of inertia-Imax and Imin.
Solved problem-case-1-Mohr's circle of inertia-direction of principal axis U.
Solved problem-case-1-Mohr's circle of inertia- Check the value of the sum of Iu &Iv.
Solved problem case-1- Mohr's circle of inertia- Check the value of Ix' from the general equation =Imax.
Solved problem case-1- Mohr's circle of inertia- Check the value of Iy' from the general equation =Imin.
Oriented datums to let x-direction as a horizontal axis.
Solved problem-case-1-Mohr’s circle of inertia.
In this post, we will introduce solved problem case 1: Mohr’s circle of inertia, where Ix is bigger than Iy, and the product of inertia Ixy is positive.
For a given triangle with a breadth of 150mm and a height of 200mm, Mohr’s circle of inertia must be used to get the maximum and minimum moments of inertia for the two major principal axes and their orientation concerning the left corner point of the triangle.
Solved problem-case-1-Mohr’s circle of inertia.
For the case of the right-angle triangle, we have already proved that for a corner point of a right-angle triangle, we have the value of the moment of inertia about the x-axis x equal to b*h^3/12. The value of the moment of inertia about the y-axis Iy is equal o h*b^3/12, while the product of inertia is equal to h^2*b^2/24.
Using the given data for b=150mm and h=200mm, we can get the value of the moment of inertia about the x-axis as equal to 100*10^6mm4. The moment of inertia about the Y-axis will be equal to 56.25*10^6 mm4.

The product of inertia Ixy will be equal to 37.50*10^6 mm4. Since Ix is bigger than Iy and Ixy is positive, this is case-1-Mohr’s circle of inertia.
Solved problem-case-1-Mohr’s circle of inertia-Imax and Imin.
For the Solved problem case-1-Mohr’s circle of inertia, We start by drawing two intersecting axes. The horizontal axis represents the value of the moment of inertia. The vertical axis represents the value of the product of inertia, with a positive value pointing up.
We start by locating Point A, which has Ix, Ixy values as(100, 37.50) units and will be located above the axis of inertia by a value of positive Ixy, and apart from the vertical axis Ixy by a distance of (100) units.
Similarly, we can draw point B, which has a coordinate of (Iy,-Ixy). In units will be (56.25,-37.50), for the negative value the point will be below the axis of inertia by 37.50 units and apart from the vertical axis by the positive value of 56.25 units. We will join both two points. Line AB will intersect the horizontal axis at point O, which will be the center of the circle.
We can draw the circle by getting the middle point O, which has a coordinate of 0.50*(Ix+Iy), from the given data this value is equal to (100+56.25)*0.50=78.125 units.
The radius of the circle is estimated from the equation, R=sqrt ((Ix-Iy)^2+Ixy^2). we have Ix=100*10^6mm4.Iy=56.25×10^6 mm4.Ixy=37.50×10^6mm4. We can use the data to get the radius of Mohr’s circle of inertia, applying the known formula, we can get the radius value is equal to 43..4138*10^6 mm4. Please refer to the slide image for more details.
To get the maximum value of inertia Imax we can add the value of the radius of the circle to the value of the center point, the maximum value of the moment of inertia is equal to (34.414+78.125)=121.539 units, the product of inertia is zero, this point is point E.
For the minimum value of inertia, the value will be equal to (0.50*(Ix+Iy)-R, where R is the radius value.
I minimum is equal to 34.711 units, the product of inertia is equal to zero, and the point of minimum value of inertia is point C.
Solved problem-case-1-Mohr’s circle of inertia-direction of principal axis U.
The direction of the principal axis U, can be obtained by estimating the tangent value of the angle from that principal axis and the horizontal axis X. The value will be(-Ixy)/(Ix-Iy)/2).
We have Ix=100 units, Iy=56.25 units, Ixy=37.50 units, the tan value is (2*(37.50)/(100-56.25)=-1.714. Since tan is negative, the value of the angle (2φp1 ) is minus (59.743).
To draw the direction of the major axis U in Mohr’s circle of inertia, we will locate point A’, which is the mirror point of Point A. Point A’ has a coordinate of (100,-37.50) units. We will join point O with point A’ to get the direction of the major Axis U.
The angle (- 59.743) is a central angle, by joining line CA’, we can get the direction of U but in the normal view, where x and Y have a 90-degree angle between them. Line CA’ is shown in Mohr’s circle of inertia, in the slide image.
Solved problem-case-1-Mohr’s circle of inertia- Check the value of the sum of Iu &Iv.
It is important to check that the sum of Ix and Iy, will be the same value as the sum of Iu and Iv, where Iu is the maximum value of inertia, While Iv is the minimum value of inertia. we have already proved that the two sums have the same values.
Solved problem case-1- Mohr’s circle of inertia- Check the value of Ix’ from the general equation =Imax.
We can use the general expression for ix’ to check the value of Imax, provided that when the angle 2θ = (2φp 1), which is (-59.743), Ix’s value will be equal to Imax. We plug with the value of (2φp 1) as=-59.743 and estimate the value of Ix’.
We will find that the value is the same as estimated from Mohr’s circle for I max. This means that we have the correct value of the principal angle.
Solved problem case-1- Mohr’s circle of inertia- Check the value of Iy’ from the general equation =Imin.
We can use the general expression for ix’ to check the value of Imin, provided that when the angle 2θ = (2φp 1), which is (-59.743), Iy’s value will be equal to Imin. If we plug with the value of (2φp 1) as=-59.743 and estimate the value of Iy’, we will find that the value is the same as estimated from Mohr’s circle for I min.
Oriented datums to let x-direction as a horizontal axis.
In the normal view, the x-axis is horizontal, while in Mohr’s circle of the inertia-first case, the x-axis is oriented by an angle of (2φp1) from the U-direction.
To let the x-axis be horizontal Mohr’s circle of inertia, the x-axis is represented by line OE. The datums of Ix&Ixy are adjusted to represent the direction of U we have already estimated, we will find that the directions of U and V are achieved.
Point B is at angle 180 from Line OE and still has the same coordinate of (56.25,-37.50). The full data that covers our Solved problem, case-1-Mohr’s circle of inertia, is included in the slide image.
In the next post, we will introduce case 2 in Mohr’s circle of inertia -the second case.
A video covers the topic of this Post.
This is a link to a useful external resource. Calculator for Cross Section, Mass, Axial and Polar Area Moment of Inertia, and Section Modulus.
March 24, 2022
6- Easy Approach to Mohr’s circle of inertia-First case.
Table Of Contents
Mohr's circle of inertia-First case.
What is the orientation of the principal axes for Mohr's circle of inertia-first case?
The steps used to draw Mohr's circle of inertia-first case.
Need for two mirror points A' and B' in Mohr's circle of inertia-First case.
The direction of U and V axes in Mohr's circle of inertia-first case.
Mohr’s circle of inertia-First case.
In this post, we will be talking about Mohr’s circle of inertia-first case. Mohr’s circle of the first case is the case where the moment of inertia about the x-axis is bigger than the moment of inertia about axis Y, and the product of inertia Ixy is positive.

Mohr’s circle of inertia-first case can be found as examples in two cases, the first case is where we have a rectangular shape where the height is bigger than the width of the rectangle. The value of the moment of inertia about the x-axis and Y-axis together with the value of the product of inertia about the left corner is included in the next slide image.
The second case for Mohr’s circle of inertia-first case is the case of inertia about the left corner of a right-angle triangle of dimension (b,h).
The Ix value is (bh^3/12, Iy value is hb^3/12, and the height of the triangle is bigger than the breadth. The product of inertia Ixy is (b^2*h^2/124), which is a positive value.
What is the orientation of the principal axes for Mohr’s circle of inertia-first case?
We have the general expression for the moment of inertia for any oriented axis, we call it Ix’. For Mohr’s circle of the inertia-first case, the value can be a maximum value when the angle 2θ has a cosine of a positive value and the sine value is also negative, this angle is called 2φp1 or ( 2θp1).

This angle is measured from the x-axis. The other orthogonal direction has an angle of 2φp2 or ( 2θp2) is also measured from the x-axis, but this angle will have a negative cosine value and a positive sign value.
The angle 2φp1 will be located in the 4th quarter, while angle 2φp2 will be located in the second quarter.
While drawing Mohr’s circle of inertia, we locate Point A, with 2θ equal to zero, which means that Point A is located at the X-axis. Substituting in the general equation, for the values of Ix’ & Iy’ and Ixy’. The value of Ix’ is equal to Ix while Iy’ value will be equal to Iy, Ixy’ value will be equal to Ixy.
While drawing Mohr’s circle of inertia, we locate Point BA, with 2θ equal to 90 degrees, which means that point B is located at the Y-axis. Substituting in the general equation, for the values of Ix’ & Iy’ and Ixy’. The value of Ix’ is equal to Iy while Iy’ value will be equal to Ix, Ixy’ value will be equal to -Ixy.
The steps used to draw Mohr’s circle of inertia-first case.
The first two steps are to draw the two orthogonal axes.
The first axis represents the Moment of inertia values, Ix, Iy, and Imax, and minimum values., while the second axis represents the values of the product of Inertia Ixy. Locating Points A and B with their respective values of Ix and Iy. Note that the Ixy value of point B is (-Ixy).
Join Points A and B and get the middle point of line AB, this point is the point of the Circle center.
The radius value is the sqrt( (Ix-Iy/2)^2+Ixy^2).
Start from point O and draw the circle with the radius value just estimates. The circle will interest the line in two points C and E.
Angle 2φp1 is the angle between the x-axis and Major axis U, which is the angle enclosed between OA and Line OE. Point E is the point of the maximum value of inertia.
Angle 2φp2 is the angle between the x-axis and Minor axis v, which is the angle enclosed between OA and Line OC. Point C is the point of the minimum value of inertia.

The distance from the vertical axis Ixy to point c will give the value of the Minimum value of inertia, while the distance from the same axis to point E will give the maximum value of inertia.
Need for two mirror points A’ and B’ in Mohr’s circle of inertia-First case.
In the normal view the x-axis is a horizontal axis, while in Mohr’s circle of the inertia-first case, the x-axis is oriented by an angle of 2φp1 from the U-direction. We need a mirror point A’ to let the x-direction in Mohr’s circle of the inertia-first case be a horizontal line.
Setting point A’, which is a mirror of point A and has a coordinate of (ix,-Ixy) will enable us to rotate the x-axis or line OA a clockwise rotation by the value of 2φp1. the x-axis will be horizontal, The U line will also have a new direction represented by line OA’.The other mirror line B’ that has a coordinate of (Iy and Ixy) will be used to get the direction of the minor axis of inertia V.
From the relation of tan 2φp, we have a negative value of tangent which means that the U axis will have an enclosed angle measured in the clockwise direction.
The direction of U and V axes in Mohr’s circle of inertia-first case.
Since the range is a central angle, we can get the direction of U, the major axis in the normal view by joining point C, the point of minimum value of inertia by point A’.
We can use the mirror point b’ to get the direction of minor axes in the normal view. Join Point C with Point B’ to get the V- direction in the normal view.
This is my understanding of the orientation of two axes Ixy and Ix, so the x-axis is horizontal, we can find that point E is the point at which Ix’ = Ix at 2θ = zero and Ix’y’ =Ixy since the line OE is the horizontal line, while point B has Ix’=Iy and Ix’y’=-Ixy.
The line OE is the direction of U, This line has an angle equal to ( -2θp1) measured from the horizontal line OE.
In the next post, we will solve a problem that covers Mohr’s circle of inertia-first case.
This is a link to a useful external resource. Calculator for Cross Section, Mass, Axial and Polar Area Moment of Inertia, and Section Modulus.
March 23, 2022
5- Easy introduction to Mohr’s Circle of inertia part-2.
Table Of Contents
Mohr's Circle of Inertia part-2.
Point of minimum value of inertia.
The expression of tan 2θp.
The expression of Imax using Mohr's circle.
The expression of Imin using Mohr's circle.
The expression of pole point.
The pole point in Mohr's circle for the normal view.
Mohr’s Circle of Inertia part-2.
Point of minimum value of inertia.
We will continue our discussion about Mohr’s circle of inertia part -2.
The point of the minimum value of inertia is point Z’, where the Ixy value equals zero. What about the direction of the V axis?
The direction is obtained from Mohr’s circle of inertia by an enclosed angle of (180-2θp) in the anti-clockwise or (2θp )from the Y-axis in the clockwise direction.
While in the Normal view, it has an angle of (90-θp) from the x-axis in the anti-clockwise direction.
Mohr’s Circle of inertia part-2.
The expression of tan 2θp.
In Mohr’s Circle of Inertia part-2, the value 2θp can be estimated from the third equation by setting Ix’y’ to zero. This will give the 2θp value equal to (-Ixy/(Ix-Iy/2)). The negative sign will indicate that the angle is in the clockwise direction.
The expression of Imax using Mohr’s circle.
The value of Imax or Iu can be estimated as equal to the distance between Ixy and the center of the circle, which equals (Ix+Iy)/2 plus the radius value. Checking the tan 2θp expression, we can draw an angle and get the radius value for Ixy, Ix, and Iy.
Please refer to the next slide image for deriving the value of the radius R.
The expression of Imin using Mohr’s circle.
From the next slide in Mohr’s Circle of Inertia part-2, the value of I minimum or Imin or Iv can be estimated as equal to shift which is (Ix+Iy)/2 minus the radius value.
Checking the tan 2θp expression, we can draw an angle, get the radius value in terms of Ixy, Ix, and Iy, and deduct the value of (Ix+Iy)/2.
The expression of pole point.
The pole point in Mohr’s circle of inertia is the point from which you can get direction, and the circle’s center is the pole point. When we join to point x, it will give the X-axis.
While joining with point y, you can get the Y direction. Similarly, a line from the center of the circle to point Z will give the principal maximum direction or the U direction.
The pole point in Mohr’s circle for the normal view.
From the last slide, we can get the pole of the standard view as point Z’ joining to the mirror point of x, which will give the U direction in the normal view, and joining to the mirror of point y will give the V direction.
Thanks a lot, and peace be upon you all.
A video covers the topic of the previous post and this Post.
The next post is Easy Approach to Mohr’s Circle of inertia-First Case.
This link to a useful external resource: a calculator for Cross-Section, Mass, Axial and polar area moment of Inertia, and Section Modulus.
August 29, 2021
4-Easy introduction to Mohr’s Circle of inertia part-1.
Table Of Contents
Mohr's Circle of inertia.
Introduction to Mohr's circle of inertia.
Brief content of the post.
How do you draw Mohr's circle of inertia?
How to prove that Mohr's circle of inertia is a shifted circle?
How to find the point of maximum value for inertia?
The expression of Imax using Mohr's circle.
Mohr’s Circle of inertia.
In this post, we will be starting our discussion about Mohr’s circle of inertia. Still, we will use the relation of x’ and y’ for an area about an inclined axis, with the x and y distance for the same area but about horizontal and vertical axes x, and y. The Inertia Ix is bigger than Iy and Ixy is positive.
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Introduction to Mohr’s circle of inertia.
Brief content of the post.
The post items that will be explained are summarized here.
When Ix is greater than Iy, we will begin to learn how to draw Mohr’s circle of inertia using the provided data for two points. As we can see, y’=ycos θ -x sin θ represents the perpendicular distances from the little area dA to x’ and the axes. In contrast, x’=xcos θ + y sin θ.
We have two theories regarding the composition of Mohr’s circle of inertia: is it a shifted circle or a centric circle with the intersection of Ix, Iy, and Ixy as its center point?
What is the value and direction of the primary angle 2θp at the third point?
What are the values of Iu and Iv, or the maximum and minimum moments of inertia, along with their directions, constitute the fourth point?
Where is the pole point in Mohr’s circle and the normal view is the fifth point?
How do you draw Mohr’s circle of inertia?
Firstly, how can Mohr’s circle of inertia be drawn when the Ix value exceeds the Iy value?
We begin by locating point X, which has the coordinates Ix, and Ixy as positive values. Point Y, on the other hand, has the coordinates (Iy,-Ixy). A circle’s center is where the intersection of these two locations with the vertically shifted axis occurs.
This circle’s radius is equal to the square root of Ixy^2+The difference between (Ix-Iy/2)^2. The axis that connects point X to the circle’s center is known as the X-axis.
On the other hand, the Y-axis is the axis that connects point Y to the circle’s center. Our axis, X’, is produced by connecting the point x’, which has the coordinates (Ix’, Ix’y’), to the center of Mohr’s circle of inertia. Point y’, which has a coordinate of (Iy’,- Ix’y’), is the opposite point.
The enclosed angle of 2θ, which is twice the value of theta from the normal view—from which we derived the three equations Ix’, Iy’, and Ix’y’—between axis X and axis x’.
Given that Mohr’s circle of inertia has both points X and X’, it follows that the square of the values of (Ix’-(Ix-Iy)/2) and the square of the value of Ix’y’ can likewise be used to determine the radius of that circle.
How to prove that Mohr’s circle of inertia is a shifted circle?
As seen in the slide, we have three equations for Ix’, Iy, and Ix’y’ from the previous post.
Two points on Mohr’s circle of inertia—a shifted circle with the values of the moment of inertia, Ix, and Iy—are represented by the three equations.
The sum of the squares of (Ix’-(Ix+Iy)/2 and Ix’y’ should equal the sum of the squares of (Ix-Iy)/2 and Ixy to demonstrate that the Mohr circle has been relocated.
The term an in equation 1 can be used to equal (Ix-Iy)/2, while the term b can be used to equal Ixy if we subtract the value of (Ix+Iy)/2 from both sides. Acos(2θ)-bsin(2θ) equals Ix’-1/2(Ix-Iy). Its value for Ix’y will be (a* sin(2θ)+b*sin(2θ).
Square (Ix’-1/2*(Ix+Iy) will give a^2*cos^2(2θ)+b^2 sin^2((2θ)-2a*b*sin((2θ)*cos((2θ).
While squaring Ix’y’ will give a^2*sin^2(2θ)+b^2 cos^2((2θ)+2a*b*sin(2θ)*cos(2θ).
If we have a look at the fourth slide. Adding together will give (a^2+b^2), which we will rewrite as (Ix-Iy)^2/4+Ixy^2.
This value represents the square value of the radius the square of the distance of Mohr’s circle of inertia and point X.
We’ll proceed to the following slide to see what the square of (Iy’ minus (Ix+Iy)/2 plus the square of Ixy will equal. We will move on to equation II and subtract the value of (Ix+Iy)/2 from both sides using the same terms, a and b.
The formula for the new right-hand side is (-a *cos(2 θ) +b sin(2θ).
Adding together the square (Iy’ minus (Ix+Iy)/2 plus the square of Ix’y’ will give (a^2+b^2), which we will rewrite as (Ix-Iy)^2/4+Ixy^2. The result matches with radius value from point X and the center of the circle.
How to find the point of maximum value for inertia?
The point Z, where the Ixy value equals zero, is the greatest value of inertia.
What about the primary axis’s direction? By utilizing an enclosed angle of 2θp in a clockwise direction from the x-axis, one can determine the direction from Mohr’s circle of inertia. In contrast, it has an angle of θp from the x-axis in a clockwise direction when viewed normally.
By utilizing an enclosed angle of 2θp in a clockwise direction from the x-axis, one can determine the direction from Mohr’s circle of inertia. In contrast, it has an angle of θp from the x-axis in a clockwise direction when viewed normally.
The expression of Imax using Mohr’s circle.
One can estimate the value of Imax or Iu by taking the shift, that is equal to (Ix+Iy)/2 plus the radius value. We can draw an angle and obtain the radius value in terms of Ixy, Ix, and Iy by checking the tan 2θp equation.
Thanks a lot and peace be upon you all.
The next post will be an Easy introduction to Mohr’s Circle of Inertia part 2.
This is a link to a useful external resource. Calculator for Cross Section, Mass, Axial and Polar Area Moment of Inertia, and Section Modulus.
July 25, 2020