Category: Second moment of area for plain shapes

The inertia for the different plain shape-rectangles-isosceles, Triangles, parallelograms, and trapezium.

  • 27-Two Practice problems for Ixy for trapezium.

    27-Two Practice problems for Ixy for trapezium.

    Two Practice problems for Ixy for trapezium.

    We will introduce two practice problems for Ixy for trapezium which are 8.43, 8.49. these problems are quoted from Engineering mechanics statics by BEDFORD.

    The first practice problem for inertia is 8.43. Find Ixy for a given trapezium.

    For the first practice problem of the two Practice problems for polar Ixy for trapezium, we have a trapezium of 7 ft base length, and the upper part length is 3 Ft and its height is 3Ft. It is required to determine the product of inertia Ixy.

    We will start to estimate Ixy in two parts, the first part is Ixy1 or the product of inertia for the triangular portion about the point of external of the two axes X and y, we call it Ixy1. For a small strip dA, the value of this area is dx*y. The area dA creates dIxy about the point of intersection of axes X and y.

    The product of inertia dIxy1=(dA*x*y/2)=(x*y^2*1/2*dx), but y has a relation of x as y=(h/(b-a)*x ,so the dIxy1can be rewritten as (h^2/(2*(b-a)^2)*(x^3*dx). We will integrate from 0 to (b-a) for the triangular portion.

    finally the expression for Ixy1 can be written as (h^2*(b-a)^2/8). We can substitute for the values of h, a, and b for h value it is equal to 3Ft and for a it is equal to 3feet and b equals 7 feet, we can substitute these values in the Ixy1 expression. The Ixy1 value is equal to 18Ft4. Please refer to the next slide image for more details.

    Find the value of Ixy1 for trapezium.

    Ixy2 for a given trapezium by integration using a vertical strip.

    The next step is to get the Ixy2 value for the rectangular portion, the dIxy2 equals (dx*h^2*x/2), we will integrate from b-a to b, and the final value of Ixy2 can be found as equal to 74.25 Ft4. The final product of inertia Ixy for the given Trapezium will be equal to 92.25 Ft4.

    The steps to find the value of Ixy.

    Derive a general expression for Ixy for trapezium.

    To derive an expression for Ixy we add the two values Ixy1 and Ixy2, there are common items that we can simplify. for a more detailed illustration please refer to the next slide image, the final Ixy expression is shown.

    Derive a general expression for Ixy.

    We can substitute for the values of b, a, and h then we can get the final value of Ixy as equal to 92.25 ft4 that the same value which we obtained earlier. for more details please refer to the next slide image.

    The final value of Ixy.

    The second practice problem for inertia is 8.48. Find Ixy for a given trapezium about the Cg.

    For the second practice problem 8.48 of the two practice problems for Ixy for trapezium it is required to find the Ixy value for the trapezium at the Cg. We have obtained the values of x bar and y bar from the previous posts and the area of the trapezium is 15Ft2. The product of inertia about the Cg is equal to Ixy minus the product of area by (xbar*ybar). The final value of Ixy at the Cg is 7.10 ft4.

    The final value of Ixy at the Cg.

    We have to an end after solving the two Practice problems for Ixy for trapezium.

    Thanks a lot, and see you in the next post.

    Please refer to post 19 for more information about Ix for a trapezium.

    Please refer to post 22 for more information about Iy for a trapezium.

    The previous post, post 26-Two, Practice problems for the polar of inertia for trapezium.

    A useful external resource for the area, Cg, and inertia for the parallelogram, please refer to this link.

    A new subject for maximum and minimum moment of inertia will be followed. Please refer to post 1; Easy illustration of the theory of pure bending.

  • 26-Two Practice problems for polar of inertia for trapezium.

    26-Two Practice problems for polar of inertia for trapezium.

    Two Practice problems for polar of inertia for trapezium.

    We will introduce two practice problems for the polar of inertia for trapezium which are 8.42, 8.48. these problems are quoted from Engineering mechanics statics by BEDFORD.

    Two Practice problems for polar of inertia for trapezium.

    The first practice problem for inertia is 8.42. Find Jo and ko for a given trapezium.

    For the first practice problem of the two Practice problems for polar of inertia, we have a trapezium of 7 ft base length, and the upper part length is 3 Ft and it is required to determine the moment of inertia ix and the radius of gyration Kx.

    We have already estimated the Ix value of the trapezium from the previous post, the value of Ix equals 36 ft4, please refer to the previous post for more information.

    Since Jo=Ix+Iy, we have the value of Ix, we need to estimate Iy or the moment of inertia about  the y-axis that passes by the left corner

    Iy for a given trapezium by integration using a vertical strip.

    As we can from the next slide image we can use a vertical strip of y height and of width equals dx. The inertia of this strip about the Y-axis is the sum of two inertias the first inertia is dIy1 plus dIy2.

    To get the value of dIy1, we multiply the area of the strip by the square of its distance x  From the y-axis dIy1=y*dx)*X^2.

    We can get the equation of the line of the triangle as y=3/4 x, and y for the rectangular part equals h which is 3 feet. dIy1 is equal to (3/4*X^3*dx) . Integrate from x=0 to x=4. We get the value of Iy1=48 Ft4.

    i

    How to get the value of Iy1 for the vertical strip?

    The value of Iy2 for the rectangular portion can be found by integrating the vertical strip from x=4 feet to x=7 feet. The inertia of the strip about the Y-axis for the rectangular portion is equal to the area of the strip multiplied by the square distance from the Y-axis.

    Then dIy2=(h*dx*( x+4)^, h value is 3 Ft , and the final value is Iy2 will be found to be equal to 279 Ft4.

    How to get the value of Iy2 for the vertical strip?

    The Total Ix value is 36 Ft4. The square of the radius of gyration equals Ix/A=36/15. Then kx equals 1.55 Ft.

    The second practice problem for inertia is 8.47. Find Jo and ko for a given trapezium about the Cg.

    For the second practice problem 8.47 of the two practice problems the polar of inertia is required to find the moment of inertia for the trapezium at the Cg and also to determine the Ko value. The same Trapezium of 7 ft base length and the upper part length is 3 Ft is dealt with.

    Determine the J0 for the Trapezium.

    Since the Jo is the sum of ix plus Iy, we have Ix=36Ft4, and the value of Iy final is the sum of Iy1 plus Iy2 which is (48+279)=327 ft4. Jo can be found as the sum of 36 ft4 plus 327 ft4, which is equal to 363 ft4. The next step is to get ko which is the square root of (Jo/A). the area of the trapezium is 15 Ft2. So K0 can be written as equal to sqrt( 363/15)=2.218 ft.

    Please refer to the details of the calculations in the next slide image.

    The final value of Jo and Ko.

    Derive a general expression for Iy value.

    If you wish to derive a general expression for Iy of the trapezium. The Iy is the sum of Iy1 and Iy2. For  Iy1 we can write its expression as equal to the integration from 0 to (b-a) of ((h/b-a)*x^3*dx)) the integration will give a value of (h/4)*(b-a)^3.

    While for the integration of Iy2, or the rectangular portion, is the integration from (b-a) to b of (h*dx*x^2), and its final value is ((b^3-(b-a)^3*(h/3)).

    Adding Iy1 to Iy2 and after substitution of the values of h, a, and b.Iy= equal to 327 ft4 which is the same as the previously estimated value.

    The general expression for Iy for trapezium.

    Determine the final expression for Jo for the Trapezium.

    Adding the expression for Ix which we have obtained from the previous post, to the general expression for Iy will give us the final expression for jo or the polar moment of inertia for the trapezium.

    The final value of jo can be found to be equal to 363 ft4.

    The final value of Jo.

    Determine the Jo and ko at the Cg for the Trapezium.

    We need to find the x bar of the Cg which is the horizontal distance between the Y axis and the Cg. The x bar can be found from the expression of A(X bar)A1*x1+A2*x2.

    Please refer to the next slide image for more detailed information. The x-bar is 4.3667 Ft.

    The value of x bar for the trapezium.

    Find the value of Iy at the Cg.

    We need to find Iy value at the Cg which is equal to Iy at the external maxis y minus the product of A*X BAR^2.

    Iy value is equal to 327 Ft4 and the product of area by the square of x bar is equal to (15*19.06)=285.9 ft4. Iy final value at teh Cg is equal to 40.98 ft4.

    To get the final expression for Jo at the Cg, take the square root of (51.63/15)=1.855 ft.

    We have to an end after solving the two Practice problems for polar of inertia for trapezium.

    The final value of ko at the Cg.

    Thanks a lot, and see you in the next post. There are two practice problems for Ixy for trapezium.

    Please refer to post 19 for more information about Ix for a trapezium.

    Please refer to post 22 for more information about Iy for a trapezium.

    The next post is post 27, Two Practice Problems for Ixy for trapezium.

    Please refer to this link for a useful external resource on the area, Cg, and inertia of the parallelogram.

  • 25-Two Practice problems for inertia for trapezium.

    25-Two Practice problems for inertia for trapezium.

    Two Practice problems for inertia for trapezium.

    We will introduce practice problems for trapezium; there are two practice problems for inertia for trapezium, which are 8.41 and 8.47. These problems are quoted from Engineering mechanics statics by BEDFORD.

    The first practice problem for inertia is 8.41. Find Ix and kx for a given trapezium.

    A quick estimate of Ix for a given trapezium.

    For the first practice problem of the two Practice problems for inertia, we have a trapezium of 7 ft base length, and the upper part length is 3 Ft. It is required to determine the moment of inertia Ix and the radius of gyration Kx.

    We have a quick answer for the moment of inertia, as we can see from the upper right corner of the next slide image, that the trapezium consists of a triangle of base (b-a) equals 4 feet with height h equals 3 feet and a rectangle of the base a equals 4 feet and a height h=3 feet.

    From our previous posts, we have the expression for triangle inertia about the base as Ix = base *height^3/12, which is equal to (4*3^3/12)=5 Ft4.

    At the same time, the inertia for a triangle equals the base *height ^3/3 which equals (3^4)/3=81/3=27 ft4. Then, the moment of inertia for the trapezium is 9+27=36 feet4.

    Ix for a given trapezium by integration using a vertical strip.

    As we can see from the next slide image, we can use a vertical strip of y height and of width equals dx. The inertia of this strip about the X-axis is the sum of two inertias. The first inertia is dIx1 plus d Ix2

    To find the value of y, we can determine the equation of the triangular line as y = 3/4 x, while for the rectangular section, y equals h, which is 3 feet. The expression for dIx1 is dIx1 = (dx * y^3 / 3). This expression can be rewritten as dIx1 = 1/3 * (3/4 x^3) * dx.

    The integration is from x=0 to x=4. The Ix1 value is equal to 9 ft4, which matches the previously estimated inertia for the triangular part.

    Two Practice problems for inertia for trapezium.

    The value of Ix2 for the rectangular portion can be found by integrating the vertical strip from x=4 feet to x=7 feet.

    The inertia of the strip about the X-axis is dIx2=(dx)*(h^3/3), the h value is 3 Ft, and the final value is Ix2, which will be found to be equal to 27Ft4.

    The Total Ix value is 36 Ft4. The square of the radius of gyration equals Ix/A=36/15. Then the radius of gyration kx equals 1.55 Ft.

    Detailed solution for ix and kx for a trapezium.

    The second practice problem for inertia is 8.47. Find Ix and kx for a given trapezium about the Cg.

    For the second practice problem, 8.47, of the two practice problems for inertia, it is required to find the moment of inertia for the trapezium at the Cg and determine the kx value.

    The same Trapezium of 7 ft base length and the upper part length is 3 Ft is dealt with.

    Determine the x-bar for the Trapezium.

    We need to find the value of the area and x bar or the horizontal distance from the Cg to the vertical axis y1 passing by the left point of the trapezium, later we will use it for the forthcoming practice problems.

    Use the expression of A(X bar)=(A1*X1+A2*X2), A1=6 ft2, while A2 equals 9 ft2. The X1 is the horizontal distance from the Cg of the triangular part to the axis y1. The value is (2/3)*4=8/3 Ft.

    While X2 is the horizontal distance from the Cg of the rectangular part to the axis y1.

    The x2 value equals (7-1.50)=5.50 ft. Using the expression A(X bar)=(A1*X1+A2*X2), and substitute for the known values we can get x bar as equal to 4.3667 Ft. Please refer to the details of calculations in the next slide image.

    Derive a general expression for Ix value.

    If you wish to derive a general expression for Ix of the trapezium. The Ix is the sum of Ix1 and Ix2. For  Ix1 we can write its expression as equal to the integration from 0 to (b-a) of x^3*dx*(h^3)/3*(b-a)^3.

    Proceeding with the integration will give us h^3/12*(b-a). While for the integration of Ix2, or the rectangular portion, is the integration from (b-a) to b of (dx*(h^3/3), and its final value is a*h^3/3.

    Adding Ix1 to Ix2 will give us (h^3/12)*(3a+b). Substitute to get the value of Ix equal to 36 Ft4.

    Determine the y-bar for the Trapezium.

    We need to find the value of the area and Y bar or the vertical distance from the Cg to the trapezium base. Use the expression of A(y bar)=(A1*y1+A2*y2), A1=6 ft2, while A2 equals 9 ft2.

    The y1 is the vertical distance from the triangular part’s Cg to the trapezium’s base. The value is (1/3)*3=1 Ft.

    The y2 is the vertical distance from the rectangular part’s Cg to the trapezium’s base. The value is (1/2)* 3=1.5 Ft. We can get the value of y bar as equal to 1.3 ft.

    The vertical distance from the Cg of the Trapezium to the base.

    Determine the final value for Ix at the Cg for the Trapezium.

    The final value of Ix at the cg is equal to the value of ix about the base minus the product of the trapezium area by the square of y bar distance.

    From the estimated data, we have Ix equaling 36 ft4, while the area of the trapezium equals 15 ft2. The y bar of the trapezium is equal to 1.3 ft. Apply the expression for Ixcg, and we can get the value of 10.65 ft4.

    The radius of gyration at the Cg is equal to the square root of Ixcg/area and can be found as equal to 0.843 ft.

    The value of Ix at the Cg for the trapezium.

    Derive a general expression for Ix at the Cg value.

    If you wish to derive a general expression for Ix at the Cg of the trapezium, we can use the general expression of Ix for the trapezium. As a reminder, the Ix value equals h^3/12*(3a+b) and deducts the product of area *y bar cg ^2.

    Please refer to the next slide image for more details. We can get the  final expression as Ixcg =(h^3/(36)*(a^2+4*a*bn+b^2)/(a+b).

    Derive a general expression for ix at the Cg.

    We can use the known values as a=3 feet, b=7 ft, and h equals 7 feet and substitute in the final expression for Ix cg. Finally, we get the value of Ixcg equals 10.65 ft4.

    The final value of ix at the Cg from the general expression for trapezium.

    Thanks a lot and see you in the next post-Two Practice problems for polar of inertia for trapezium.

    Please refer to post 22 for more information about Iy for a trapezium.

    Please refer to post 19 for more information about Ix for a trapezium.

    The next post is post 26-Two, Practice problems for the polar of inertia for trapezium.

    A useful external resource for the area, Cg, and inertia for the parallelogram, please refer to this link.

  • 24- Ix for the Trapezium second option

    24- Ix for the Trapezium second option

    Ix for the Trapezium second option.

    Derive the expression for k1 for the horizontal strip.

    For the Ix for the trapezium second option, we will use a horizontal strip of width dy.

    The strip is located at y distance from the Trapezium with a length of (k1+a+k2) where a is the top length of the trapezium.

    For a rectangle of base b, we want to get an expression for the value of k1, we consider the left triangle with a base equal to b1 and height equal to h. Using the relation of (k1/b1)=(h-y)/h to get the K1 value. K1 can be written as equals b1*(h-y)/h. Please refer to the next slide image for more details.

    Ix for the Trapezium second option.
    Ix for the Trapezium second option.

    Derive the expression for k2 for the horizontal strip.

    Similarly, we want the value of k2, we refer to the right triangle that has a base of b2 and height equals h. Using the relation of (k2/b2)=(h-y)/h to get the k2 value. K2 can be written as equals b2*(h-y)/h.  Please refer to the next slide image for more details.

    How to find the value of k2 for the horizontal strip?

    Integrate the horizontal strip.

    The moment of inertia Ix for the Trapezium second option can be obtained by integrating the horizontal strip from y=0 to y=h. The length of the horizontal strip can be written as equal to (a+b1+b2)=(a+b1*(h-y)/h+b2*(h-y)/h).

    The inertia of the strip  dIx=(a+b1*(h-y)/h+b2*(h-y)/h)*y^2*dy

    We can simplify the expression as consisting of two items Ix1 plus Ix2

    The expression for ix for the trapezium.

    We can simplify the expression as consisting of two items Ix1 plus Ix2. We can clear the common items and get the expression as shown in the next slide image. Ix1 can be written as (b1+b2)*h^3/12.

    The value of Ix1 for the trapezium.

    We can clear the common items and get the expression as shown in the next slide image. Ix2 can be written as (4*a*h^3)/12. please refer to the next image for more details.

    The value of Ix2 for the trapezium.

    Derive the final expression for Ix for the Trapezium second option.

    We can add the value of ix1 plus Ix2 to get the final expression for Ix for the Trapezium second option.

    The Ix value for the trapezium can be expressed as equal to h^3*(b+3a)/12. For the next post, we will solve two practice problems for trapezium. Thank you.

    The final expression for Ix of the trapezium.

    For the use of a calculator for various shapes, please find Moments of Inertia – Reference Table.
    This is a link to the complete details of how to get an x-bar and y-bar for a Trapezium.


    This is the next post-
    Two Practice problems for inertia for trapezium.

  • 23- Product of inertia Ixy for the parallelogram

    23- Product of inertia Ixy for the parallelogram

    Product of inertia Ixy for the parallelogram.

    For the product of inertia Ixy for the parallelogram, we will consider that is composed of one triangle marked I and a rectangle marked II, but we need to subtract the area of one triangle marked III.

    our x-axis is the horizontal axis passing by the lower base of the parallelogram, while the Y-axis is the vertical axis passing by the left corner.

    Product of inertia Ixy for triangle and rectangular part.

    We know already the product of inertia Ixy for the left triangle for which it is equal to the square of the base multiplied by the square of height divided by 8.

    The base equals a*cos θ, where θ is the enclosed angle between the inclined length and the horizontal length of the parallelogram, please refer to the first slide image. the height of the first triangle equals h.

    For the rectangle, the product of inertia about its Cg equals zero, because of symmetry.

    To estimate the product of inertia for the rectangle about the X and Y axes, we need to find the horizontal distance x bar and the vertical distance y bar from the Cg to the left corner.

    The x-bar is found to be equal to (b/2+a cos (θ), while the Y-bar distance equals (h/2). The product of inertia for the rectangle equals the product of area by x-bar by y- bar. The value is shown in the next slide image.

    Step 1 for the Product of inertia Ixy for the parallelogram
    Step 1 for the Product of inertia Ixy for the parallelogram

    Product of inertia Ixy for a right triangle.

    We will estimate the product of inertia Ixy for the third shape, the triangle, its product of inertia at its Cg is equal to the square of its base by the square of its height divided by 72.

    The base is equal to (a cos(θ), and the height equals h.

    The product of inertia Ixy at Cg=(a cos(θ))^2*h^2/72.

    We need to add the product of area by x-bar by y-bar from the Cg to the left corner to the product of area by the area of the triangle equals (1/2)(a cos(θ))*h. X-bar and y- bar.

    The final value for the product of inertia for the third triangle can be found in the next slide image.

    The product of inertia for the third shape Ixy.
    The product of inertia for the third shape Ixy.

    In the next slide, all the products of inertia for the left triangle, rectangle, and the third shape, triangle are listed.

    The product of the inertia of the Ixy-for the second triangle is to be subtracted from the sum of the product of the inertia of the triangle and the rectangle.

    I have made a simplification for the common terms to reach a value for the product of inertia Ixy for the parallelogram.

    The sum of product of inertias for the shapes.

    The final value for the Product of inertia Ixy for the parallelogram.

    The final expression for the product of inertia for the parallelogram can be found as equal to (h^2*b)*(3b+4*a*cos (θ)) divided by 12.

    Since we can substitute for the height h as equals (a* cos (θ), then the final value for the Product of inertia Ixy for the parallelogram can be rewritten as equals (a^2*b*sin^2(θ)*(3b+4*a*cos (θ)) divided by 12.

    For the square of the radius of gyration r^2xy, it can be obtained by diving Ixy /area, since the area of the parallelogram equals (b*a*sin(θ)).

    The final r^xy equals (a*b*sin(θ)*(3*b+4*a*cos (θ)) divided by 12. the detailed estimation is shown in the next slide image.

    The final value of ixy for the parallelogram and radius of gyration.

    Check the product of inertia for a rectangle as a special case of a parallelogram.

    We can check the value of the product of inertia for a rectangle with base b and a height a since the rectangle is a special case of a parallelogram but setting the angle (θ) as equals 90 degrees. first the h=a *sin (90)=a.

    The Ixy for the rectangle at the left corner can be found after substituting the equation of the product of inertia for the parallelogram and letting cos (θ)=0. the value will be equal (a^2)*b^2)/4, which we obtained earlier from post 5.

    Check the value of Ixy when the parallelogram becomes a rectangle.
    Check the value of Ixy when the parallelogram becomes a rectangle.

    The final value for the Product of inertia Ixy for the parallelogram at the Cg.

    The final value for the Product of inertia Ixy for the parallelogram at the Cg can be obtained by subtracting the product of the area of the parallelogram by (x-bar*y bar) from the final expression of the product of inertia at the y-axis.

    We know that the x-bar value equals ( b+a*cos(θ)), while y bar=h/2=a*sin(θ)/2.

    The final value for the Product of inertia Ixy for the parallelogram at the Cg can be found to be equal to (a^3*b*(sin^2(θ)*cos(θ)/12.

    The final value for the product of inertia Ixy for the parallelogram at the Cg.
    The final value for the product of inertia Ixy for the parallelogram at the Cg.

    Check the product of inertia for a rectangle as a special case of a parallelogram at the CG.

    We can check the value of the product of inertia, at the Cg, for a rectangle with base b and a height a since the rectangle is a special case of a parallelogram but setting the angle (θ) as equals 90 degrees.

    Let h=a *sin (90)=a. the expression for the product of inertia as proved is equal to (a^3*b*(sin^2(θ)*cos(θ)/12. substitute for θ=90 degrees, we will find out the product of inertia for the rectangle equals zero, which matches the well-known fact that the product of inertia for the rectangle at the Cg equals zero.

    Check the value of Ixy when the parallelogram becomes a rectangle at the Cg.
    Check the value of Ixy when the parallelogram becomes a rectangle at the Cg.

    For more information on the moment of inertia Ix for the parallelogram please refer to post 20.

    For the moment of inertia for the parallelogram about the Y-axis, please refer to post 21.

    For a useful external resource for the area, Cg, and inertia for the parallelogram, please refer to this link.


Moment of inertia Iy for the Trapezium.

For the moment of inertia Iy for the trapezium, we will start to estimate the Iy about Cg and later we will find Iy at the left corner.

First, we will consider a trapezium that is symmetric about a vertical axis passing by the Cg which we call y’.

The Trapezium with base b and the upper top side a and height h is divided into the following shapes:

1-Two rectangles of base a/2 and a height of h, Areas are A1, A2 each area =1/2*(a/2) h, it is CG is apart from the y’-axis by a distance x1= (a/4).

2-A left triangle of base C and a height of h, its area A3=1/2*C*h, it is CG is apart from the y’-axis by a distance x2=-(a/2+1/3C).

3-A right triangle of base C and a height of h, its area A4=1/2*C*h, it is CG is apart from the y-axis by a distance x3=(a/2+1/3C).

The division of the trapezium into four areas is shown in the next slide image, where Y’ is the axis of symmetry.

Moment of inertia Iy for the trapezium.

We will estimate the inertia Iy’ for only one rectangle and one triangle and later we will multiply the value by 2. The value of Iy’ for the rectangle is as shown in the next slide equal to (h*a^3/24). For the triangle, we will estimate the inertia about its Cg as equal to h*c^3/36 as C is the base. we will add the product of the triangle area by the square of the Cg distance to Y’ axis.

Moment of inertia Iy for the Trapezium. For the right rectangle and triangle.

The Cg distance is equal to (1/6*(3a+2c). The calculation of both of the inertias for rectangle and triangle are shown in the previous slide image.

Moment of inertia Iy’ for the Trapezium about the Cg vertical axis.

Adding the two values of inertia and then multiplying by 2 will give us the final value of inertia Iy for the trapezium. We have used c as the base distance later we will equate it to (b-a)/2.

The value of the moment of inertia Iy for the Trapezium.

For the terms inside the bracket, we will get their values in terms of b and a. We have the first term as 6h*c^3, and the second term is 9h*(a^2*c). The third term is 12*hC^2*a. the last term 3h*a^3. Iy for the trapezium is obtained after getting the values of the four terms which are shown in detail in the next slide image.

The four terms value with respect to a and b values.

Moment of inertia Iy for the Trapezium in terms of b &a and h.

Finally, the Iy’ for the trapezium can be obtained. This is the final value for Iy for the trapezium, where the y’-axis is passing by the CG.

The final value of Iy for the trapezium about the Cg.
The final value of Iy for the trapezium about the Cg.

Moment of inertia Iy for the Trapezium at an external axis.

The inertia Iy for the trapezium about the Y-axis that passes by the left point can be obtained by adding the value of Iy’ to the product of the area of the trapezium by the square value of the Cg distance to the y-axis. The final value of Iy is shown in the next slide image.

The final value of Iy for the trapezium.

The radius of gyration value for inertia Iy for the trapezium.

The radius of the gyration value is shown in detail in the next slide image.

The radius of gyration value for inertia Iy for the trapezium.
The radius of gyration value for inertia Iy for the trapezium.

The Polar of inertia value IP for the trapezium.

This is the polar inertia value for the trapezium shown in detail in the next slide image.


The Polar of inertia value IP for trapezium.
The Polar of inertia value IP for trapezium.

For an external resource for the values of inertia, please find this link to the e-funda.
This is a link to the complete details of how to get the x-bar and y-bar for a Trapezium.

This is a link to the post for the Ix value for the trapezium.

the next post will be about getting the product of inertia for a parallelogram.