User:Eas4200c.f08.nine.s/Lecture 2

Homework 1

Homework 2

Homework 3

Homework 4

Homework 5

Homework 6

Group nine - Week 2

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Second Moment of Inertia

Moment of inertia of different stringers

"The second moment of area, also known as the area moment of inertia or second moment of inertia, is a property of a shape that is used to predict its resistance to bending and deflection which are directly proportional. This is why beams with higher area moments of inertia, such as I-beams, are so often seen in building construction as opposed to other beams with the same area."

A good example of second moment of inertia would be a persons attempt to prevent a force from turning a large lever. The ferther the person puts his or her hand from the pivot the easier it is to keep the lever from moveing. The persons hand can be thought of as the sum of all of the small parts makeing up the object. The position of the persons hand is proportional to the squar of the distance from the pivot. Each part of the object adds its own contribution depending on its shape, size and position. Each part can be cut into smaller parts and then be summed up until the infinitesimal size is reached and the result is accurate. The basic equation for the second moment of inertia is as follows.

I_x=$\int \$y²dA

The figure to the right shows different types of stringers. The moment of inertia will be calculated assuming the areas of a,b,c and d are the same.

Mohr's Circle

Mohr's Circle (FELIX)

Mohr’s circle is a graphical tool that can express a plane stresses as a function of the angle theta.

The creation of a Mohr’s circle takes few simple steps.

1. Create a coordinate axis in which the horizontal axis represents the normal stress and the vertical axis will represent the shear stress.
2. In an x-y coordinate system plot $\sigma _{y}$ and $\tau _{xy}$ a on the numerical position on the normal stress axis on the y-face of the object.
3. The next point is the x-y coordinate of $\sigma _{x}$ and $\tau _{xy}$ noting that tension and clockwise shear on the x-face.
4. Mark the midpoint of the line between the points described in steps 2 & 3. Refer to Figure 2 for a reference
5. Create a circle as shown in Figure 2.
6. With the circle now constructed each point on the circle represents a normal/shear stress combination. The line that was drawn in Figure 2 represents the principal axis of the stress. In Figure 3 the horizontal axis of the Mohr's circle represents the two principal stresses in which all he stress on an object represents a normal stress.

Figure 1

Figure 2

Figure 3

In reading a Mohr's circle the primary consideration comes from the angle $\theta$ which is measured off the principal axis. a rotation of 2 $\theta$ represents a rotation of $\theta$ in the element.

Brief History of Mohr's Circle

In 1885 Christian Otto Mohr, a German civil Engineer proposed a graphical representation of two and three-dimensional stresses. Mohr was born on October 8, 1835 and died October 2, 1918 in the town of Dresden.

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Problem 1.1 Case 1

Problem 1.1 Case 1 (RICARDO) Problem: Finding the optimum ${\frac {b}{a}}$ ratio. Problem 1.1 Equations and assumptions: $t\ll a$ $t\ll b$ $L=2(a+b)$ $M=T$ $\sigma _{all}=2\tau _{all}$ $\tau ={\frac {T}{2abt}}\,$ $\sigma ={\frac {Mz}{I}}\,$, where $z={\frac {b}{a}}$ CASE 1: $\left(\sigma _{max}=\sigma _{all}\right)$ _________________________________________________________________________________________________________________________________________ Solving for M from our equations we have: from equation $\sigma ={\frac {Mz}{I}}\,$ where $z={\frac {b}{a}}$ we can solve for M, $M=2\sigma _{all}\left({\frac {I}{b}}\right)\,$ We need to find our I: $I=\sum _{i=1}^{4}{\frac {b_{i}h_{i}^{3}}{12}}\ +A_{i}d_{i}^{2}$ $I=2[{\frac {at^{3}}{12}}+at({\frac {b}{2}})^{2}]+2({\frac {tb^{2}}{12}})$ $I={\frac {at}{12}}(t^{2}+3b^{2})+2({\frac {tb^{3}}{12}})$ since $\ t\ll b$ and $\ t\ll a$ we have: ${\frac {I}{b}}={\frac {tb}{6}}(3a+b)$ Substituting for a in terms of L and b, $(a={\frac {L}{2}}-b)$, we have: ${\frac {I}{b}}={\frac {tb(3L-4b)}{12}}$ Differentiating this with respect to b and setting it equal to zero in order to obtain the optimized b value, we find: $b={\frac {3L}{8}}$ Substituting into $\ L=2(a+b)$, our optimized a value is ${\frac {L}{8}}$. Dividing ${\frac {b}{a}}$ we find the optimized ratio to be 3. Now we check to see if this case is possible: $M_{max}=2\sigma _{all}\left({\frac {I}{b}}\right)_{max}$ substituting the known value fore I/b, we have: $M_{max}={\frac {3tL^{2}}{32}}\sigma _{all}$ now we have to write $\tau _{max}={\frac {T}{2abt}}$ From the assumption $T=M$, this equation can be rewritten and we can also plug in the known values for optimized $a$ and $b$: $\tau _{max}=\sigma _{max}$ and from the assumption that $\sigma _{max}=2\tau _{all}$ We can conclude that $\tau _{all}<\tau _{max}$ and so this case is not acceptable!

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Problem 1.1 Case 2

Problem 1.1 Case 2 (RICARDO) _________________________________________________________________________________________________________________________________________ Problem: Finding the optimum ${\frac {b}{a}}$ ratio. Problem 1.1 Equations and assumptions: $t\ll a$ $t\ll b$ $L=2(a+b)$ $M=T$ $\sigma _{all}=2\tau _{all}$ $\tau ={\frac {T}{2abt}}\,$ $\sigma ={\frac {Mz}{I}}\,$, where $z={\frac {b}{a}}$ CASE 2: $\left(\tau _{max}=\tau _{all}\right)$ _________________________________________________________________________________________________________________________________________ In order to maximize $a$ and $b$, in this case their lengths would have to be the same, so $a=b={\frac {L}{4}}$ from equation $\tau ={\frac {T}{2abt}}\,$ and assumption $M=T$, we can plug in the known values for a and b and solve for $T_{max}$ $T_{max}={\frac {2t\tau _{all}L^{2}}{16}}=M_{max}$ since we know that $\sigma _{all}=2\tau _{all}$ , this equation can be written in terms of sigma, $T_{max}={\frac {tL^{2}\sigma _{all}}{16}}=M_{max}$ solving for $\sigma _{all}$ we have: $\sigma _{all}={\frac {16M_{max}}{tL^{2}}}$ Using the ${\frac {I}{b}}$ funcion found on Case 1, which is the same for this case, ${\frac {I}{b}}={\frac {tb}{6}}(3a+b)$ and plugging in our known values for $a$ and $b$, we have: ${\frac {I}{b}}={\frac {tL^{2}}{24}}$ from these equations: $\sigma ={\frac {Mz}{I}}\,$ where $z={\frac {b}{a}}$ we can solve for $\sigma _{max}$ in terms of $M_{max}$ and plug in our known ${\frac {I}{b}}$ value: $\sigma _{max}={\frac {M_{max}b}{2I}}=M_{max}{\frac {12}{tL^{2}}}$ Using this previously derived equation, $\sigma _{all}={\frac {16M_{max}}{tL^{2}}}$ we can know solve for $\sigma _{max}$ in terms of $\sigma _{all}$ $\sigma _{max}={\frac {12}{16}}\sigma _{all}$ So we conclude that this case is acceptable!

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Problem 1.7

Problem 1.7 (RICARDO) Given: $R_{0}=10cm$ $t=0.1cm$ Find: $R_{1}$ such that $A_{1}=A_{2}$ and solve $I_{y}$ for both areas and compare. Problem 1.7 ___________________________________________________________________________________________________________________________________________________________ Solution: First we solve for $R_{1}$ by setting both areas equal, so $A_{1}=A_{2}$ , and we have the following formula: $(\pi R_{0}^{2})=(2\pi R_{1}^{2})+(2R_{0}t)$ So we plug in the values given for $R_{0}$ and $t$ and solve for $R_{1}$ , so $R_{1}=7.05cm$ Now we solve for $I_{y}^{(1)}$ which is ${\frac {\pi R_{0}^{4}}{4}}=7854cm^{4}$ Now we need to solve $I_{y}^{(2)}$, so we need to use PAT (parallel axis theorem). $I_{y}^{(2)}=2(I_{centroid_{circle}}+Ad^{2})+I_{rectangle}$ where $I_{centroid_{circle}}={\frac {\pi R_{1}^{4}}{4}}$ $A=\pi R_{1}^{2}$ $d=(R_{0}+R_{1})^{2}$ and $I_{rectangle}={\frac {t(2R_{0})^{3}}{12}}$ When we plug in the known values, $I_{y}^{(2)}=94730cm^{4}$ So by comparing both area moment of inertia, we can observe that $I_{y}^{(2)}=12.06I_{y}^{(1)}$

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Contributing Team Members

The following students contributed to this report:

David Phillips Eas4200C.f08.nine.d (talk) 18:34, 26 September 2008 (UTC)

Oliver Watmough Eas4200c.f08.nine.o 10:07, 26 September 2008 (UTC)

Stephen Featherman Eas4200c.f08.nine.s 11:48, 26 September 2008 (UTC)

Ricardo Albuquerque Eas4200c.f08.nine.r 4:30, 26 September 2008 (UTC)

Felix Izquierdo Eas4200c.f08.nine.F 4:34, 26 September 2008 (UTC)