## 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, 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 and 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. While 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 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 get the value of y 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. dIx1 is equal to (dx*y^3/3) . Rewrite the expression as dIx1=1/3*(3/4x^3)*dx.

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

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 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 kx equals 1.55 Ft.

### 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 also to 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 base of the trapezium. 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 Cg of the triangular part to the base of the trapezium. The value is (1/3)*3=1 Ft.

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

#### 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 equals 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 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.

#### 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).

We can use the knowns val;ues 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.

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.

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