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For a given area, the moments of inertia with respect to two rectangular centridal \(x\) and \(y\) axes are \(\bar{I}_{x}=1200 \mathrm{in}^{4}\) and \(\bar{I}_{y}=300 \mathrm{in}^{4}\), respectively. Knowing that, after rotating the \(x\) and \(y\) axes about the centroid \(30^{\circ}\) counterclockwise, the moment of inertia relative to the rotated \(x\) axis is \(1450 \mathrm{in}^{4},\) use Mohr's circle to determine (a) the orientation of the principal axes, \((b)\) the principal centroidal moments of inertia.

Short Answer

Expert verified
The principal axes are oriented at approximately 36.17 degrees. The principal moments of inertia are 2150 in鈦 and -650 in鈦.

Step by step solution

01

Understand Mohr's Circle

Mohr's circle is a graphical representation used to determine the moments of inertia about rotated axes. It uses the information of the original moments of inertia and the given angle of rotation to find the principal moments of inertia and their directions.
02

Identify the Known Values

From the problem, we have \( \bar{I}_x = 1200 \text{ in}^4, \bar{I}_y = 300 \text{ in}^4 \), and \( I_{x'} = 1450 \text{ in}^4 \) after a \( 30^{\circ} \) rotation. The angle \( 2\theta \) is used in Mohr's circle, so for rotation by \( 30^{\circ} \), \( 2\theta = 60^{\circ} \).
03

Calculate the Average Moment of Inertia

The average moment of inertia \( I_{avg} \) is calculated as:\[ I_{avg} = \frac{\bar{I}_x + \bar{I}_y}{2} = \frac{1200 + 300}{2} = 750 \text{ in}^4 \]
04

Calculate the Radius of Mohr's Circle

The radius \( R \) of Mohr's circle is given by:\[ R = \sqrt{ \left( \frac{\bar{I}_x - \bar{I}_y}{2} \right)^2 + \bar{I}_{xy}^2 } \]But first, find \( \bar{I}_{xy} \) using:\[ I_{x'} = I_{avg} + R \cos(60^{\circ}) = 1450 \text{ in}^4 \]\( I_{avg} = 750 \text{ in}^4 \), so:\[ 1450 = 750 + R \times 0.5 \]\[ R = 1400 \text{ in}^4\]
05

Determine the Product of Inertia

Using the radius formula again:\[ 1400 = \sqrt{ \left( \frac{1200 - 300}{2} \right)^2 + \bar{I}_{xy}^2 } \]\[ 1400 = \sqrt{ 22500 + \bar{I}_{xy}^2 } \]\[ 1960000 = 22500 + \bar{I}_{xy}^2 \]\[ \bar{I}_{xy} = \sqrt{1937500} \approx 1392.85 \text{ in}^4 \]
06

Determine the Principal Axes Orientation

The angle \( \theta_p \) of the principal axes is given by:\[ \tan(2\theta_p) = \frac{2 \cdot \bar{I}_{xy}}{\bar{I}_x - \bar{I}_y} = \frac{2 \cdot 1392.85}{900} \approx 3.096 \]\[ 2\theta_p = \tan^{-1}(3.096) \approx 72.34^{\circ} \]Thus, \( \theta_p \approx 36.17^{\circ} \).
07

Calculate the Principal Moments of Inertia

The principal moments of inertia \( I_1, I_2 \) are given by:\[ I_1 = I_{avg} + R = 750 + 1400 = 2150 \text{ in}^4 \]\[ I_2 = I_{avg} - R = 750 - 1400 = -650 \text{ in}^4 \] (Note: values might be switched if negative, reflecting choice of axis in graphical construction).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moments of Inertia
Moments of inertia describe an object's resistance to rotational motion about a certain axis. These values are critical in understanding how an object will behave when subjected to rotational forces.For example, in the exercise, the moments of inertia about the centroidal axes are given as:
  • \( \bar{I}_x = 1200 \text{ in}^4 \)
  • \( \bar{I}_y = 300 \text{ in}^4 \)
This means that the area has a greater resistance to rotation about the x-axis compared to the y-axis.When we rotate these axes, the moments of inertia change, which leads us to find through Mohr's Circle how they are transformed.Using graphical methods helps visualize and calculate the new moments of inertia after rotation.
Principal Axes
Principal axes are special axes around which the moment of inertia is at a maximum or minimum.These axes are oriented such that the product of inertia becomes zero, simplifying many calculations.In the exercise, we determine the orientation of these principal axes using the rotation angle and Mohr's circle.Knowing the angle of rotation (here \(30^\circ\)), we find the orientation of principal axes:
  • The tangent of two times the principal angle \(2\theta_p\) is determined by:\[ \tan(2\theta_p) = \frac{2 \cdot \bar{I}_{xy}}{\bar{I}_x - \bar{I}_y} \]This is calculated as approximately \(3.096\).
  • The calculated principal angle \(\theta_p\) is thus approximately \(36.17^\circ\).
This shows the rotation relative to centroidal axes where the moments of inertia become principal moments.
Centroidal Axes
Centroidal axes pass through the centroid of a geometric shape or area.They are often chosen as reference axes because they reduce complexity in the mathematical calculations of inertia.In this problem, our given moments of inertia are with respect to these centroidal axes with coordinates \(\bar{I}_x\) and \(\bar{I}_y\).These centroidal inertias are essential for simplifying certain properties when rotating or determining stability conditions.By understanding the moments about these axes, we better analyze how the entire area will react to rotations or forces.
Product of Inertia
The product of inertia \(\bar{I}_{xy}\) is a measure of how the area is distributed with respect to two different axes.For rectangular centroidal axes, \(\bar{I}_{xy}\) represents an off-diagonal component when the inertia tensor is expressed in matrix form.In our exercise, using Mohr's circle helps determine this value when other known values (moments of inertia) change after an axis rotation:
  • Calculated against the rotation value to find \(\bar{I}_{xy} \approx 1392.85 \text{ in}^4\).
The product of inertia is significant in calculating the orientation of principal axes and understanding how an area will rotate or resist rotation.A zero product often marks the axes as principal axes where it simplifies the moment of inertia tensor, leading to calculations that are easier to handle.

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Most popular questions from this chapter

A piece of thin, uniform sheet metal is cut to form the machine component shown. Denoting the mass of the component by \(m,\) determine its mass moment of inertia with respect to \((a)\) the axis \(A A^{\prime},(b)\) the axis \(B B^{\prime},\) where the \(A A^{\prime}\) and \(B B^{\prime}\) axes are parallel to the \(x\) axis and lie in a plane parallel to and at a distance \(a\) above the \(x\) plane.

Show that the resultant of the hydrostatic forces acting on a sub- merged plane area \(A\) is a force \(P\) perpendicular to the area and of magnitude \(P=\gamma A \bar{y} \sin \theta=\bar{p} A,\) where \(\gamma\) is the specific weight of the liquid and \(\bar{p}\) is the pressure at the centroid \(C\) of the area. Show that \(P\) is applied at a point \(C_{P},\) called the center of pressure, whose coordinates are \(x_{p}=I_{y} / A \bar{y}\) and \(y_{p}=I_{x} / A \bar{y},\) where \(I_{y}=\int x y d A\) (see Sec. 9.3 ). Show also that the difference of ordinates \(y_{p}-\bar{y}\) is equal to \(\vec{k}_{x} / y\) and thus depends upon the depth at which the area is submerged.

The cover for a 0.5 -m-diameter access hole in a water storage tank is attached to the tank with four equally spaced bolts as shown. Determine the additional force on each bolt due to the water pressure when the center of the cover is located 1.4 m below the water surface.

Knowing that the thin hemispherical shell shown has a mass \(m\) and thickness \(t\), determine the mass moment of inertia and the radius of \(f\) gyration of the shell with respect to the \(x\) axis. (Hint: Consider the shell as formed by removing a hemisphere of radius \(r\) from a hemisphere of radius \(r+t ;\) then neglect the terms containing \(t^{2}\) and \(t^{3}\) and keep those terms containing \(t .)\)

Prove that the centroidal polar moment of inertia of a given area \(A\) cannot be smaller than \(A^{2} / 2 \pi\). (Hint: Compare the moment of inertia of the given area with the moment of inertia of a circle that has the same area and the same centroid.)

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