User:Eas4200c.f08.gator.edwards/Additional Questions

Circle Moment of Inertia
$$I_y=\iint\limits_A z^2\,dA=\frac\pi4 r^2$$

In polar coordinates, Iy for a circle is calculated below:

$$J_o = \int_A{\rho^2 dA} = \int_0^r{\rho^2 (2\pi\rho d\rho)} = 2\pi \int_0^r{\rho^3 d\rho} = \frac{1}{2}\pi r^4$$

$$J_o = I_z + I_y = 2I_y \Rightarrow I_y = {1 \over 4} \pi r^4$$

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Channel Cross-Section Moment of Inertia


To compute the moment of inertia of a channel cross section we first recall formula for the moment of inertia of a rectangle

$$I_{y}=\frac{1}{12}bh^{3}$$

and the parallel axis theorem

$$I_{y}=\bar{I}_{y'}+Ad^{2}$$

We also remember that the moment of inertia of a composite area is equal to the sum of the moments of inertia of its parts. Using the first equation we find the moments of inertia of the vertical sections:

$$I_{y,vert}=2(\frac{1}{12}tb^{3})=\frac{1}{6}tb^{3}$$

Next we find the moment of inertia of the horizontal piece

$$I_{y,horiz}=\frac{1}{12}(a-2t)b^{3}$$

and use the parallel axis theorem to combine the components:

$$I_{y}=I_{y,vert}+(I_{y,horiz}+Ad^{2})=\frac{1}{6}tb^{3}+[\frac{1}{12}(a-2t)t^{3}+(a-2t)(t)(\frac{b-t}{2})]$$

This procedure can be simplified with a few parameters: 1) a = b   2)  t = a / 10 3) A = 3at   4)  t << a so that a is the distance from the base to the middle of the horizontal section and the area of the horizontal section is equal to at

Applying these simplifications and reassembling the equation we get

$$I_{y}=I_{y,vert}+(I_{y,horiz}+Ad^{2})=\frac{1}{6}tb^{3}+\frac{1}{12}at^{3}+3at\left(\frac{b}{2} \right)^{2}$$

$$=\frac{1}{6}\left(\frac{a}{10} \right)a^{3}+\frac{1}{12}a\left(\frac{a}{10} \right)^{3}+3a\left(\frac{a}{10} \right)\left(\frac{a}{2} \right)^{2}=\frac{a^{4}}{60}+\frac{a^{4}}{120}+\frac{3a^{4}}{20}=\frac{7a^{4}}{40}$$

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Problem 1.7 with Modification


The objective of this problem is to compare the moments of inertia of a solid circle (case 0) and a composite section consisting of two circles connected with a rectangle (case 1, shown in figure at right). The radius of the solid circle is R0, the radius of the smaller circles on the composite section is R1, the length of the rectangular portion is 2R0, and the thickness of the rectangular portion is t.

R0 = 10 cm t = 1 cm

R1 is to be found so that the cross sectional areas are the same for both sections.

Starting with the areas

$$A_{0}=\pi R_{0}^{2}=(10 cm)^{2}\pi = 314 cm^{2}$$

$$A_{1}=2(\pi R_{1}^{2})+2R_{0}t=A_{0}=\pi R_{0}^{2}$$

Solve for R1

$$R_{1}^{2}=\frac{\pi R_{0}^{2}-2R_{0}t}{2\pi }$$

Plugging in the known values we get R1

$$R_{1}=\sqrt{\frac{\pi \cdot10^{2}-2\cdot 10\cdot 1}{2\pi }}= 6.84 cm^{2}$$

From the above section the moment of inertia of a circle and rectangle are

Circle: $$I_{y}=\frac{1}{4}\pi r^{4}$$

Rectangle: $$I_{y}=\frac{1}{12}bh^{3}$$

Using these the moment of inertia for Case 0 is

$$I_{y,0}=\frac{1}{4}\pi R_{0}^{4}=\frac{1}{4}\pi \cdot 10^{4}= 7854 cm^{4}$$

Combining with the parallel axis theorem stated above the moment of inertia for Case 1 is

$$I_{y,1}=\frac{1}{12}bh^{3}+\frac{1}{4}\pi R_{1}^{4}+(\pi R_{1}^{2})(R_{0}+R_{1})^{2}=\frac{1}{12}\cdot 1\cdot (2\cdot 10)^{3}+\frac{1}{4}\cdot \pi \cdot 6.84^{4}+(\pi \cdot 6.84^{2})(10+6.84)^{2}= 44,068 cm^{4}$$

From the results it is clear that the cross section of Case 1 yields a much higher moment of inertia and therefore can withstand higher forces than the section of Case 0.

Mohr's Circle




Mohr's circle is a graphical representation of any 2-D stress state proposed in 1892 by Christian Otto Mohr. It can be applied to many engineering quantities such as stresses, strains, and moments of area.

Mohr's circle may also be applied to three-dimensional stress. In this case, the diagram has three circles, two within a third.

You can use Mohr's circle to find the planes of maximum normal, principle and shear stresses, as well as the stresses on known weak planes. For example, if the material is brittle, the engineer might use Mohr's circle to find the maximum component of normal stress (tension or compression); and for ductile materials, the engineer might look for the maximum shear stress.

Looking at the stress figure on the right, we see that with only a force in the x direction, in the enlarged section the stress in the x-direction $$(\sigma _{xx})$$ is equal to $$\sigma$$. The stress in the y-direction $$(\sigma _{yy})$$ is equal to 0.

Now if they are plotted on a stress-strain graph as a Mohr's Circle, we see that by plotting $$(\sigma _{xx})$$ and $$(\sigma _{yy})$$ and connecting them with a circle, $$\tau _{max} = \frac{\sigma _{allow}}{2}$$.