Brief decription for post 11-Mohr's circle of inertia-third case-solved problem

11- A Solved problem-case-3-Mohr’s circle of inertia.

A Solved problem-case-3-Mohr’s circle of inertia.

In this post, we will introduce a solved problem-case-3-Mohr’s circle of inertia, where Ix is bigger than Iy and the product of inertia Ixy is negative.

The given solved problem is for the unequal angle of dimension 8 inches by 6 inches and the thickness of the angle is one 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.

We will check the x bar for the unequal angle by dividing the area of the unequal angle into two rectangles from which their difference will give the area. We set two axes x and Y at the Cg.

The first rectangle is 8*6 inches and the second rectangle is 5×7 inches. The x bar or the distance from the Cg can be found as equal to 1.65 inches measured from the axis Y”.

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-case-3-Mohr’s circle of inertia-estimate Ix,Iy, and Ixy at the Cg.

A Solved problem-case-3-Mohr's circle of inertia.

The shown calculations for the x-bar can be checked from the next slide image.

We can proceed to estimate the y bar for the given unequal angles. The Y bar or the distance from the Cg can be found as equal to 1.65 inches measured from the axis Y”.

Y bar for the Solved problem-case-3-Mohr's circle of inertia.

The shown calculations for the y-bar can be checked from the next slide image.

We will estimate the moment of inertia Ix about the Cg for the Unequal angle by considering two rectangles, the first one is 1-inch width by 8 inches in height, and the second one is a rectangle of 5 inches in length by one inch thick.

We will add the product of the area for each rectangle by the square of the vertical distance between each rectangle Cg to the final Cg distance to get the Value of the moment of inertia about the x-axis passing by the Cg.

Estimate the value of Ix at the Cg for the Solved problem-case-3-Mohr's circle of inertia.

The shown calculations for the Ix Cg can be checked from the next slide image.

We will estimate the moment of inertia Iy about the Cg for the Unequal angle by considering two rectangles, the first one is 1inch width by 8 inches in height, and the second one is a rectangle of 5 inches in length by one inch thick. We will add the product of the area for each rectangle by the square of the horizontal distance between each rectangle Cg to the final Cg distance.

Estimate the value of Iy at the Cg for the Solved problem-case-3-Mohr's circle of inertia.

The shown calculations for the Iy Cg can be checked from the next slide image.

The last step is to estimate the product of inertia for the unequal angle, by estimating the product of inertia for each rectangle, which is zero. We will add the product of each area by its horizontal and vertical distance to Cg.

Estimate the value of Ixy at the Cg for the Solved problem-case-3-Mohr's circle of inertia.

The Value of the product of inertia is found to be equal to (-32.31)inch4 as shown in the next slide image.

based on the previous estimation, we have case-3-Mohr’s circle of inertia, where ix is bigger than Iy and Ixy is negative.

Solved problem-case-3-Mohr’s circle of inertia-drawing the circle.

For the Solved problem-case-3-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(80.8, -32.31) units and will be located below the axis of inertia by a value of negative Ixy equals 32.31 units, and apart from the vertical axis Ixy by a distance of (80.80) units.

Similarly, we can draw point B, which has a coordinate of (Iy,+Ixy). In units will be (38.8, +32.31). 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 (80.81+38.8)*0.50=59.80 units.

The radius of the circle is estimated from the equation, R=sqrt ((Ix-Iy)^2+Ixy^2), this is applied since. We have Ix=80.8 inch4.Iy=38.80 inch4.Ixy=-32.31 inch4.

For the minimum value of inertia, the value will be equal to the (0.50*(Ix+Iy)-R, where R is the radius value.

The steps are used to get the radius of Mohr's circle of inertia.

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 38.534 inch4. Please refer to the slide image for more details.

The minimum value of the moment of inertia is equal to 21.266 inch4, point C represents the point of minimum moment of inertia on 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 value of the maximum moment of inertia.

The maximum value of the moment of inertia is equal to (59.80+38.534)=32 units, the product of inertia is zero, this point is point E.

The directions for both two axes X and Y are represented by lines OA and OB.

Solved problem-case-3-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 (80.8,+32.31) units. we will join point O, with point A’ and we will get the direction of the major Axis U.

Value of Angle 2φp1 for the Solved problem-case-3-Mohr’s circle of inertia

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)

The equation is written as (-Ixy)/(Ix-Iy)/2). We have Ix=80.80 units, Iy=38.80 units, Ixy=-32.31 units, the tan value is (-2*(-32.31)/(80.80-38.80)=+1.53857. The plus sign indicates that this angle is anticlockwise. The value of the angle is 56.978, this is the value of 2φp1.

The value of 2φp1 and 2φp2.

The value of 2φp2 is equal to 180-56.978=123.021 degrees n the clockwise direction. Please refer to the next slide image for more information.

The directions of U and V for the normal view.

There is a separate slide to show the orientation of both u and v axes in the normal view, construct a line between points C and A’ will give the U direction in the normal view while drawing a line between points C and B’ will give the direction of V axis in the normal view.

The directions of U and V for the normal view.

The value of φp1 equals 28.489 degrees, while the value of φp2=-61.511degrees in the clockwise direction.

Solved problem-case-3-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 (56.978 degrees), Ix’ value will be equal to Imax. We plug in with Ix value= 80.80 inch4,Iy value=38.80 inch4 and Ixy=-32.31 inch4.

The value of Ix’=98.334 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 details of estimation of Ix’ when 2θ = (2φp1)

We can use the general expression for Iy’ to check the value of Imin, We can use the general expression for Ix’ to check the value of Imax, provided that when the angle 2θ = (2φp1), which is (56.978 degrees), Ix’ value will be equal to Imax. We plug in with Ix value= 80.80 inch4,Iy value=38.80 inch4 and Ixy=-32.31 inch4.

The value of iy' equals Imin.

The details of estimation of Iy’ when 2θ = (2φp1)

The value of Iy’=21.266 inch4 is exactly the same value 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=80.80+38.8=119.60 inch4. imin+Imax=98.334+21.266=119.60 inch4.

Check Ip polar value.

Check that IP is the same value when adding Ix+Iy as compared by the sum of Iu plus Iv.

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 with reference to two new datums which are shown in the next slide image. Line OC represents the Y-direction with reference to two new datums.

the oriented two axes fulfill the condition of Mohr’s circle of inertia-case 3. Ix is bigger than Iy and Ixy is negative. Please refer to point E in the next slide image.

page 13 post 11 case 3 solved

The normal view angles are shown in the next slide image.

This is an extract from the table of Unequal angles, for our unequal angle 6x8x1 inches, the axis VV is the axis of minimum moment of inertia, its value is given as the product of A*rmin^2.

The Table of unequal angle 8x6x1 inches.

The product gives the value of Imin as equals 21.299 inch4, which is close to our estimated value of 21.266 inch4. Also the direction of the alpha angle is close to our angle φp1.

In the next post, we will solve a problem that covers Mohr’s circle of inertia-fourth case.

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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