Category: Second moment of area

The Moment of inertia is an important subject in statics, The posts include how to find the proofs of the values of inertia for different common shapes.

  • 9- A Solved problem-case-2-Mohr’s circle of inertia.

    9- A Solved problem-case-2-Mohr’s circle of inertia.

    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.

    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.

    The Value of the angle 2θp2 from the Tan equation.

    This is How to get the value of 2θp1 for the Solved problem for A Solved problem-case-2-Mohr’s circle of inertia.

    The value of angle 2θp1 from the x-axis to the u-axis case 2 mohr's circle.

    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.

    The value of Imax and I min from Mohr's circle.

    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 direction of principal axes U and 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 Imax from the general expression of Ix'.

    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.

    The value of Imin from the general expression

    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.

    oriented view of Ixy and U direction.

    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.

  • 8- Easy Approach to Mohr’s circle of inertia-second case

    8- Easy Approach to 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 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.

    The value of principal angles for Mohr 's circle of inertia-second case

    This is the case where Ix’ is minimal when 2θp2 will be located in the first quarter.

    The case where Ix' is minimum.

    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 U and V axes in the normal view.

    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.

    How to draw Mohr's circle of inertia for case 2?

    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.

    The value of the radius of Mohr's circle-case 2.

    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.

    Angle 2θp1 is the angle between the x-axis and the Major axis U-circle of inertia- first case.

    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.

    The mirror points A' and B'.

    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.

    The U and V directions 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.

    The oriented u and Ixy axes.

    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.

  • 7- A Solved problem-case-1-Mohr’s circle of inertia.

    7- A Solved problem-case-1-Mohr’s circle of inertia.

    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.

    This is the direction of U considering the x-axis as horizontal.

    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.

  • 6- Easy Approach to Mohr’s circle of inertia-First case.

    6- Easy Approach to 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.

    Mohr 's circle of inertia-First case.

    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.

    The value of principal angles for Mohr 's circle of inertia-first case

    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.

    The value of Ix' is equal to Ix in Mohr's circle of inertia- first case for point A.

    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 value of Ix' is equal to Iy in circle of inertia-first case for point B.

    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).

    The steps to draw Mohr's ciecle of inertia-first case.

    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 the Major axis U-circle of inertia- first case.

    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.

    The orientation of the x-axis In Mohr's circle-first case.

    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.

    The need of mirror points A' and B'.

    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.

    The direction of U- major axis in the normal view

    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’.

    The direction of minor axis in the normal view.

    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.

    The radius value for Mohr's circle iof inertia -first case.

    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.

    Orientation of x- axis and the U direction

    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.

  • 5- Easy introduction to Mohr’s Circle of inertia part-2.

    5- Easy introduction to Mohr’s Circle of inertia part-2.

    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.

    The point of minimum value of inertia.
    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.

  • 21- How to find Moment of Inertia Iy for Parallelogram?

    21- How to find Moment of Inertia Iy for Parallelogram?

    Moment of inertia Iy for parallelogram.

    Divide into areas and estimate inertia for each individual area about the y-axis.

    The post includes how to estimate the moment of inertia Iy for a parallelogram. The axis y is passing by the left corner of the parallelogram and intersects with the base of the Parallelogram at Point a.
    A parallelogram is a skewed rectangle with an angle=θ, between the base and the left side, when θ=90, the shape becomes a rectangle.

    The dimension of the parallelogram is b*a, whereas the side length and the height is=h, h can be considered as=a*sinθ. To get the expression for Iy, we will divide the parallelogram into two triangles and a rectangle.

    How to derive the expression for inertia Iy for parallelogram?

    We will use the previous data for the moment of inertia about the y-axis obtained for inertia for the right-angle triangle and rectangle.

    Iy for right-angle case-1, Iy for right-angle triangle case-2, and Iy for the rectangle. The sum of inertia for the two triangles will be deducted from the inertia of the big triangle.

    The inertia of the left triangle about the y-axis.

    The left triangle has an upper base of (a* cos θ) and height h, the inertia for the y-axis will be estimated about an axis passing by the left corner point of the base of the Parallelogram Can be considered as the sum of Iy about the CG plus the Inertia value from the product of(A* x bar^2).

    inertia Iy for parallelogram by diving into shapes.

    The expression for the inertia Iy for the left triangle can be simplified as shown in the next slide image.

    Iy for the left triangle.

    The inertia of the right triangle about the y-axis.

    As for the right triangle, it has a bottom base of (a* cos θ) and a height of h, the inertia for the y-axis will be estimated about an axis passing by the CG point then add the product of the area*x cg^2, where xcg is the distance from the Cg to the external Y-axis. Iy2 is the sum value, which is shown in the next slide image.

    Iy for the right angle portion.

    More simplification is done for Iy2 is shown in the next slide image. The detailed calculations for Iy for parallelogram is shown in the next slide image.

    Iy calculation for the right triangle.

    We will add the inertia of the two triangles to get their sum, which will be later subtracted from the inertia Iy of the big triangle.

    Iy calculation for the two triangles.

    The steps for inertia Iy for the parallelogram.

    The inertia of the big rectangle can be considered as the height* base^3/3, the left side of that rectangle coincides with Y-axis.

    Steps to estimate inertia Iy for parallelogram
    The calculation for Inertia Iy for Parallelogram.

    The final value of the inertia Iy for the parallelogram is to be obtained by deducting the inertia of the two triangles from the Iy of the rectangle. The steps of estimation are shown in the following slide images. The common items can be cleared.

    Rearrangement of terms for inertia.
    The calculations for Inertia Iy for Parallelogram

    The final expression inertia Iy for parallelogram is shown in a similar form to the expression shown in the NCEES Handbook. For the value of Iy for the rectangle, we have θ=90 degrees, when substituted in the equation of inertia Iy for the parallelogram, we get the same expression for the rectangle inertia about the y-axis.

    Final Iy value for inertia Iy for parallelogram.
    Inertia Iy for Parallelogram-case of rectangle checked.

    The value of the radius of gyration at the external corner for the parallelogram can be obtained by dividing Iy at the left edge by the area of the parallelogram. The expression for r^2y for the parallelogram is shown in the next slide image.

    The radius of gyration about y axis for the parallelogram
    Inertia Iy for Parallelogram-radius of gyration at Y

    The steps for Inertia Iy for Parallelogram at the Cg.

    The inertia Iy for the parallelogram at the Cg can be estimated by subtracting the product of the parallelogram Area by the Xcg^2. The Xcg is the horizontal distance from the parallelogram Cg to the external Y-axis from the inertia Iy for the parallelogram. The matching items are to be cleared.

    Iy value for inertia for parallelogram at CG.

    The steps of the calculations of the inertia Iy for the parallelogram at the Cg are shown in detail in the next slide images.

    Iyg calculation for the parallelogram.

    The final expression for the moment of inertia for the parallelogram about the Y-axis.

    The final expression for Iy g for the parallelogram.
    Inertia Iy for Parallelogram at the Cg

    The value of the radius of gyration at the Cg for the parallelogram can be obtained by dividing Iy at the Cg by the area of the parallelogram. The expression for r^2g for the parallelogram is shown in the next slide image.

    The radius of gyration for the parallelogram
    Inertia Iy for Parallelogram-radius of gyration.

    The slide image contains the values of the moment of inertia for the parallelogram, at the Y-axis Inertia Iy for the Parallelogram, and the radius of gyration and matches with the previous calculation as shown in this post.

    List of inertia for trapezoid and parallelogram

    This is the end of our post for the Moment of inertia Iy for the parallelogram.

    This is the pdf file used in the illustration of this post.

    This is the next post, Iy for the trapezium.

    For the use of a calculator for various shapes, please find Moments of Inertia – Reference Table.