Brief decription for  post 13-A solved problem for Mohr's circle of inertia-fourth case.

13- A solved problem for Mohr’s circle of inertia case 4.

A solved problem for Mohr’s circle of inertia case 4.

In this post, we will introduce a solved problem for Mohr’s circle of inertia case 4. The moment of inertia about X-axis Ix is lesser than the moment of inertia about Y-axis Iy and the product of inertia Ixy is negative.


The given solved problem is for the unequal angle of dimension 3 inches by 2 inches and the thickness of the angle is (1/4) inch.

It is required to determine the principal moment of inertia and we will use Mohr’s circle of inertia to get the directions of the principal axes and later verify by the general equation to check the estimated values. the long leg 3″ is parallel to the x-axis, while the short leg I parallel to the Y-axis.

We will use table 1-7 for angles to get more information about Area and other dimensions

You can click on any picture to enlarge then press the small arrow at the right to review all the other images as a slide show.

A solved problem for Mohr's circle of inertia case 4.

The data of the solved problem is shown in the next slide image.

Area, Cg, and inertias for the angle in the solved problem for Mohr’s circle of inertia case 4.

The area of the angle-3″x2″x1/4″ equals 1.19 inch2, and the horizontal distance for the Cg from the right edge of the unequal angle is equal to 0.98″. The vertical distance for the Cg of the unequal angle to the upper edge equals 0.487 inches.

We can get the moment of inertia Ix for the given angle from table 1-7. We will reverse the values and let the Ix value from the table is the Iy in the given solved example since the angle is rotated. The long leg is parallel to the x-axis, unlike the data that is shown in table 1-,7 fo,r which is the long portion parallel to the y-direction.

Area, Cg, and inertias for the angle in the solved problem for Mohr's circle of inertia case 4

The Area, Cg distances are given from table as shown in the next slide image.

Inertia from table 1-7

The data of the solved problem is shown in the next slide image.

The product of inertia for the angle in the solved problem for Mohr’s circle of inertia case 4.

The x-axis and Y-axis are placed at the CG of the unequal angle, we want to estimate the product of inertia Ixy. The upper leg is divided into two areas, while the short leg is considered as one area.

For each area, we will estimate the product of inertia and then add the multiplication of each area by the product of x*y to the Cg axes.

the detailed estimations of the values of the product of inertia are shown in the next slide image. The final value of the product of inertia equals (-0.3798) inch4.

Estimation of Ixy at the Cg for the given angle.

The values of Ix, Iy, and Ixy are collected and it is found that Ix is less Iy and Ixy is negative. This case is case 4.

A solved problem for Mohr’s circle of inertia case 4-Drawing the circle.

For the solved problem for Mohr’s circle-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 could make 1 unit equal to 0.10 inch4.

We start by locating Point A, which has Ix, Ixy values as (0.39, -0.3798) inch4 and will be located below the axis of inertia by a value of negative Ixy equals 0.3978 inch4, and apart from the vertical axis Ixy by a distance of +0.39 inch4.

Similarly, we can draw point B, which has a coordinate of (Iy,+Ixy). In inch4 will be (1.39, +0.3798). 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),0) from the given data this value is equal to (0.39+1.09)*0.50=0.74 inch4.

The radius of the circle is estimated from the equation, R=sqrt ((Ix-Iy)^2+Ixy^2), this is applied since. We have Ix=0.39 inch4.Iy=1.09 inch4.Ixy=-0.3798 inch4.

The calculations for the center point and the radius value of the circle

The calculations for the center point and the radius value of the circle are shown in the next slide image.

A solved problem for Mohr’s circle of inertia case 4-The value of Imax and Imin.

The values of I max and Imin are estimated as follows for Imax=(Ix+Iy)*0.50+R=(0.39+1.09)*0.50+0.5165=1.2565 inch4. For the estimation of Imin it is equal to =(Ix+Iy)*0.50-R=(0.39+1.09)*0.50-0.5165=0.2235 inch4.

We need to have two mirror points, the first point is A’ which has the coordinate of (0.39,+0.0.3789) inch4 and the second point is point B’, which has a coordinate of (1.09,-0.3798)

The values of Imax and Imin and the directions of U and V.

To get The U direction Joint point O with point A’. For the V direction join point O with point B’.

Value of Angle 2φp2 for the Mohr’s circle of inertia case 4.

The direction of the principal axis V is making a clockwise direction angle 2φp2. the value of this angle can be estimated from the tan value as (2*Ixy)/(Iy-Ix), or (+*(-0.3798)/1.09-0.39)=-1.0851. The value of the angle is 47.338 degrees. The value of 2φp1 equal(180-47.338)=+132.662 degrees in the anti-clockwise direction.

The angle φp1 value and the value of the angle φp2.

The angle φp1 value is 66.831 degrees, while the value of the angle φp2 equals to-23.669 degrees.

To get U and V directions in the normal view, join Point C by A’.While for the V direction in the normal view, join point C by point B’.

The solved problem for Mohr’s circle -case-4-using general expression.

We can use the general expression for Ix’ to check the value of Imax, provided that (2φp1), is (132.662 degrees), Ix’ value will be equal to Imax. We plug in with Ix value= 0.39 inch4,Iy value=1.09 inch4 and Ixy=-0.3798 inch4.

The value of Ix’=1.2565 inch4 is exactly the same value as Imax estimated by the use of Mohr’s circle.

The value of Ix' is the same as Imax.

The complete details of Ix’s estimation by the general equation.

We can use the general expression for Ix’ to check the value of Imin, provided that (2φp1), is (132.662 degrees), Iy’ value will be equal to Imin. We plug in with Ix value= 0.39 inch4,Iy value=1.09 inch4 and Ixy=-0.3798 inch4.

The value of Iy’=0.2235 inch4 is exactly the same value as Imin estimated by the use of Mohr’s circle.

The equation is used to estimate the value.

The complete details of Iy’s estimation by the general equation.

This is a comparison between the estimated values for Imin and the value of Imin from the table of steel angles. Imin is the product of r^2*A=0.22106 inch4.

Comparison between Iv and imin value for the given angle.

There is a little difference between iv and the estimated imin by both Mohr’s circle of inertia and by the general equation.

The next post is a solved problem-inertia value for a given direction.

This is a link to a useful external resource. Calculator for Cross Section, Mass, Axial & Polar Area Moment of Inertia and Section Modulus.

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