User:Eas4200c.f08.vqcrew.c/Homework 2

= Class Notes =

Problem 1.1
Given Figure 1, find the optimum ratio $$\frac{b}{a}$$  to maximize the load bearing capability of the thin-walled beam.

The assumptions for this problem are as follows:
 * 1) The cross-sectional perimeter is constant, i.e. $$L=2(a+b)=Constant$$
 * 2) The torque $$T$$ is equal to the bending moment $$M$$, i.e. $$T = M$$
 * 3) The maximum allowable bending normal stress is twice the maximum allowable shear stress, i.e. $$\sigma_{allow} = 2\tau_{allow}$$

The assumption that $$t$$ is very small compared to $$a$$ and $$b$$ is also valid.

Since the cross-section walls are very thin, the shear stress can be assumed to be uniform along the walls, as shown in the shear flow diagram in Figure 2. The shear distribution and resultant shear vector $$V$$ for a small piece of the wall is also shown. This distribution is also mechanically equivalent to the linear distribution shown in Figure 3, where $$\tau =\frac{V}{t}$$.



The total torque ($$T$$) can be described as the sum of the torques on each wall of the box beam as follows: $$T = T_{AB}+T_{BC}+T_{CD}+T_{DA}$$

$$T_{AB}=(\frac{b}{2})Va=\frac{1}{2}\tau tab$$

$$T_{BC}=(\frac{a}{2})Vb=\frac{1}{2}\tau tab$$

$$T_{CD}=(\frac{b}{2})Va=\frac{1}{2}\tau tab$$

$$T_{DA}=(\frac{a}{2})Vb=\frac{1}{2}\tau tab$$

Therefore, $$T=2t\tau ab$$, and $$\tau =\frac{T}{2abt}$$.

There are two cases that must be analyzed in order to solve the problem:
 * 1) Assume the bending normal stress (σ) reaches $$\sigma_{allow}$$ first, then verify that $$\tau \le \tau_{allow}$$.
 * 2) Assume the shear stress (τ) reaches $$\tau_{allow}$$ first, then verify that $$\sigma \le \sigma_{allow}$$.

Case 1
Case 1 begins by recalling the formula for bending normal stress: $$\sigma =\frac{Mz}{I}$$, where $$M$$ is the bending moment, $$z$$ is the ordinate of a point on the axis perpendicular to the bending neutral axis, and I is the second area moment of inertia defined by $$I=\iint_A z^2 \ dy\, dz$$. In this case, $$z=\frac{b}{2}$$. Figure 4 shows each item pictorially.



Rewriting this equation yields $$M=\frac{2I\sigma _{max}}{b}$$, where $$\sigma _{max}=\sigma _{allow}$$ from the Case 1 assumption. We proceed to find $$M_{max}$$ by maximizing the ratio of $$\frac{I}{b}$$.

The expression for finding the moment of inertia $$I$$ of the box beam is: $$I=\sum_{i=1}^{4}{\left[\frac{b_{i}h_{i}^3}{12} + A_{i}d_i^2\right]}$$, where $$i$$ represents each wall of the box beam, varying from one to four. This simplifies to $$I=2 \frac{tb^3}{12}+2\left[\frac{at^3}{12}+\frac{atb^2}{4} \right]$$.

The first sum is the result of the vertical bars, and the second sum is the result of the horizontal bars. Because $$t\ll b$$, the $$t^3$$ term will be very small compared to $$b$$ and can be neglected. After some algebra, $$I$$ can be simplified to: $$I=\frac{tb^2}{6}\left(3a+b \right) $$.

Let us now define a function $$f(b)\equiv \frac{I}{b}$$ and maximize it to find the maximum bending moment, $$M_{max}$$. By using Assumption 1, we can substitute $$a$$ with $$a=\frac{L}{2}-b$$. Replacing this in the expression for $$I$$ results in the concave-downward quadratic shown in Figure 5: $$f(b)=\frac{tb}{12}\left(3L-4b \right)$$. By setting $$f(b)=0$$, the roots are shown to be $$b=0$$ and $$b=\frac{3L}{4}$$. Since the function is a quadratic, the maximum value of $$f(b)$$ is located halfway between the two roots at $$b^{(1)}=\frac{3L}{8}$$. By Assumption 1, we find $$a^{(1)}=\frac{L}{8}$$.

The maximum bending moment $$M_{max}^{(1)}$$ can then be written as:

$$M_{max}^{(1)} = \left(2\sigma _{allow} \right)\left(\frac{I^{(1)}}{b^{(1)}} \right) = \left(2\sigma _{allow} \right)\left(\frac{tb^{(1)}}{6} \right)\left(3a^{(1)}+b^{(1)} \right)$$

Substituting in for $$a^{(1)}$$ and $$b^{(1)}$$ yields: $$M_{max}^{(1)} = \frac{3tL^2}{32}\sigma _{allow}$$

This is also equal to $$T_{max}^{(1)}$$ based on Assumption 2. We will now use this assumption and the shear stress equation to determine the maximum shear stress for Case 1.

$$\tau _{max}=\frac{T_{max}^{(1)}}{2a^{(1)}b^{(1)}t} = \frac{\frac{3tL^2}{32}\sigma _{allow}}{2\frac{L}{8}\frac{3L}{8}t}$$

This reduces to:

$$\tau _{max}= \sigma _{allow}$$

Substituting Assumption 3 for $$\tau _{max}$$ gives:

$$\tau _{max}= 2 \tau _{allow}$$.

Therefore, Case 1 is an unacceptable solution because $$\tau _{max} > \tau _{allow}$$.

Case 2
For this case, the assumption is made that $$ \tau_{max} = \tau_{allow}$$  (i.e.  $$ \tau_{max}$$  reaches  $$ \tau_{allow}$$  first).

$$ \tau = \frac{T}{2abt} = \tau_{max} = \tau_{allow} $$

Rearrange to solve for the torque, T.

$$ T = (2t\tau_{allow})(ab) $$

$$ T_{max} = (2t\tau_{allow})(ab)_{max} $$

From Assumption 1, the perimeter of the beam is constant, L = 2(a + b). This allows for a to be solved in terms of only L and b.

$$a = \frac{L}{2} - b$$

Now, substituting in for a the maximum torque becomes

$$ T_{max} = (2t\tau_{allow})(\frac{bL}{2} - b^{2})_{max} $$

The first term in the maximum torque equation $$(2t\tau_{allow}) $$  is a constant. In order to maximize the torque, the second term $$(\frac{bL}{2} - b^{2})$$ must be maximized. A function f(b) is defined as the second term to allow for the maximum to be found.

$$ f(b) = (\frac{bL}{2} - b^{2}) $$

The derivative of f(b) is taken and set to zero to find the maximum of the function.

$$ f'(b) = (\frac{L}{2} - 2b) = 0 \Rightarrow b^{(2)} = \frac{L}{4} $$

Next, $$ a^{(2)}$$ is found through the perimeter relation.

$$ a^{(2)} = (\frac{L}{2} - \frac{L}{4}) = \frac{L}{4}$$

The value of $$ T^{(2)}$$ can now be determined and equated to the maximum moment by Assumption 2, T = M.

$$ T^{(2)}_{max} = (2t\tau_{allow})(\frac{L}{4})^{2} = \frac{1}{8}tL^{2}\tau_{allow} = M_{max}^{(2)}$$

The allowable normal stress can now be substituted in to the equation per Assumption 3, $$\sigma_{allow} = 2\tau_{allow}$$

$$M_{max}^{(2)} = \frac{1}{8}tL^{2}\tau_{allow} = \frac{1}{16}tL^{2}\sigma_{allow}$$

Solving for $$\sigma_{allow}$$  ,

$$\sigma_{allow} = \frac{16M_{max}^{(2)}}{tL^{2}}$$ Recall: $$I=\frac{tb^2}{6}(3a+b)$$

For $$ a=b=\frac{L}{4} $$

$$I^{(2)} = \frac{t(\frac{L}{4})^2}{6}(3(\frac{L}{4})+(\frac{L}{4})) = \frac{tL^3}{96}$$

Solve for $$\sigma_{max}^{(2)}$$  and substitute in values for $$I^{(2)}$$ and $$b^{(2)}$$ ,

$$\sigma_{max}^{(2)} = \frac{M_{max}^{(2)}b^{(2)}}{2I^{(2)}} = \frac{M_{max}^{(2)}\frac{L}{4}}{2\frac{tL^3}{96}} = \frac{12M_{max}^{(2)}}{tL^2}$$

Substituting in the previously found expression for $$M_{max}^{(2)}$$ ,

$$\sigma_{max}^{(2)} = \frac{12(\frac{1}{16}tL^{2}\sigma_{allow})}{tL^2} = \frac{12}{16}\sigma_{allow}$$

Since $$ \frac{12}{16}\sigma_{allow}$$ is less than $$\sigma_{allow}$$ , this configuration of dimensions for the thin-walled beam are acceptable.

Stringers, Shear Panels, and Shear Strain
= Homework Problems =

2nd Area Moment of Inertia of a Circle
Given a circle of radius R, the second area moment of inertia can be derived from the definition:
 * $$I_{y} = \iint_A z^2 \,dA$$

With circles it is often convenient to work with polar coordinates rather than with the traditional Cartesian coordinate system. To convert the above equation, a transformation must be done. Noting that


 * $$ \begin{align}

\\y &= r\ cos(\theta) \\z &= r\ sin(\theta) \\dA &= r\ dr\ d\theta \end{align}$$

Where the $$r$$ in the $$dA$$ term is the Jacobian of the transformation which is found by


 * $$J = \det\frac{\partial(x,y)}{\partial(r,\theta)}

=\begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} =\begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} =r\cos^2\theta + r\sin^2\theta = r.$$

When substituting into the original definition, the following relation is obtained
 * $$I_{y} = \int_0^R \int_{0}^{2\pi} r^3\ sin^2(\theta)\ d\theta\ dr$$

and integrating
 * $$I_y = \frac{R^4}{4}\ \frac{2\pi}{2}$$

which yields
 * $$I_y = \frac{\pi R^4}{4}$$

Comparing a Circle to a Thin-Walled Three-Sided Box Beam of Equal Area
Given a circle of radius $$R$$ and a three-sided box beam of equal areas, the moments of inertia can be computed and compared. The moment of inertia for a circle was proven above and the moment of inertia of a rectangle is trivial and given as
 * $$I_y = \frac{b h^3}{12}$$

and applying the parallel axis theorem to the case of the three-sided beam would cause the resulting equation to be
 * $$I_y = \sum_{i=1}^3 \frac{b_i h_i^3}{12}+A_i d_i^2$$

the area of the circle will be
 * $$A_{circle} = \pi R^2$$

while the area of the three-sided box beam is
 * $$A_{box} = \sum_{i=1}^3 b_i h_i = 3 a t = \frac{3 a^2}{10}$$

Note: the relation $$t = \frac{a}{10}$$ was used in simplifying the area

Equating the two area terms
 * $$\pi R^2 = \frac{3 a^2}{10} \therefore a = R \sqrt{\frac{10 \pi}{3}}$$

Determining the $$I_y$$ for each case in terms of R yields for the circular case (case 1)
 * $$I_y^{(1)} = \frac{\pi R^4}{4} = .7854 R^4 $$

and for the three-sided box beam
 * $$I_y^{(2)} = 2 \frac{t a^3}{12} + (\frac{a t^3}{12} + (a t)(\frac{a}{2} + \frac{t}{2})^2)$$

plugging in $$t = \frac{a}{10}$$


 * $$I_y^{(2)} = \frac{2 a^4}{120} + (\frac{a^4}{12(10)^3} + \frac{a^2}{10}(\frac{a}{2} + \frac{a}{20})^2)$$


 * $$I_y^{(2)} = \frac{a^4}{60} + (\frac{a^4}{12000} + \frac{a^2}{10}(\frac{11 a}{20})^2)$$


 * $$I_y^{(2)} = \frac{a^4}{60} + \frac{a^4}{12000} + \frac{a^4(121)}{4000}$$


 * $$I_y^{(2)} = a^4(\frac{1}{60} +\frac{1}{12000} + \frac{121}{4000}) = .047 a^4 = .047(R\sqrt{\frac{10 \pi}{3}})^4 = 5.1541 R^4$$


 * $$\frac{I_y^{(2)}}{I_y^{(1)}} = \frac{5.1541 R^4}{.7854 R^4} = 6.5624$$

meaning that altering the cross section of a beam from circular to three-sided box while keeping the area the same will increase the moment of inertia by a factor of roughly 6.56.

Comparing a Circular Cross-Section to a Dumbbell Cross-Section of Equal Area
The area for the circular cross-section is
 * $$A_{circle} = \pi R_o^2$$

while the area for the dumbbell is
 * $$A_{dumbbell} = 2 \pi R_1^2 + 2 R_o t$$

equating the areas of each case and setting $$t = \frac{R_o}{10}$$ yields
 * $$\pi R_o^2 = 2 \pi R_1^2 + 2 R_o t \therefore R_1 = \sqrt{\frac{R_o^2}{2} - \frac{R_o^2}{10 \pi}} = .4682 R_o$$

The moment of inertia of a circle was derived above as
 * $$I_y^{(1)} = \frac{\pi R_o^4}{4} = .7854 R_o^4$$

while the moment of inertia of the dumbbell is
 * $$I_y^{(2)} = 2 (\frac{\pi R_o^4}{4} + \pi R_1^2(R_o + R_1)^2) + \frac{.1 R_o(2 R_0^3)}{12}$$


 * $$I_y^{(2)} = \frac{\pi}{2}(.04804) R_o^4 + \pi (.2192 R_o^2) (.2156 R_o^2) + .0666 R_o^4$$


 * $$I_y^{(2)} = 1.560 R_o^4 + .0666 R_o^4$$


 * $$I_y^{(2)} = 1.627 R_o^4$$

comparing the two moments of inertia
 * $$\frac{I_y^{(2)}}{I_y^{(1)}} = \frac{1.627 R_o^4}{.7854 R_o^4} = 2.0716$$

thus, altering the cross-section of a beam from circular to dumbbell-shaped while keeping the area the same serves to increase the moment of inertia by a factor of roughly 2.07

Proof that $$(a\ b)_{Max}$$ is a Square with Constant Perimeter
Given a rectangle of constant width, to maximize area it is intuitive that the optimal ratio of $$\frac{b}{a}$$ would be one, resulting in a square. Noting that L is constant and defining and substituting



F(b) = a\ b$$ $$ a = \frac{L}{2} - b$$ $$ F(b) = b\ (\frac{L}{2} - b) = \frac{L}{2}\ b - b^2$$

Recalling in calculus that setting the first derivative to zero and solving for the variable results in the critical number, which is a relative minimum or maximum of the function. Also noting that if the second derivative is negative at the critical number, that point is a relative maximum and alternatively if second derivative is positive at the critical number, that point is a relative minimum. Taking the first derivative and solving



\frac{dF(b)}{db} = \frac{L}{2} - 2b = 0$$ $$ b_{crit} = \frac{L}{4}$$

Meaning that at the point $$L/4$$, the function $$F(b)$$ is either a relative minimum or maximum. Taking the derivative again and plugging in the critical point



\frac{d^2F(b)}{db^2} = -2$$

Showing that when $$b = \tfrac{L}{4}$$ the function $$F(b)$$ is a maximum, showing that the area at that point is at its maximum. When $$b = \tfrac{L}{4}$$ and substituting to solve for a
 * $$a = \frac{L}{2} - \frac{L}{4} = \frac{L}{4}$$

This shows that $$a = b$$ meaning that the sides have equal length, resulting in the geometry of a square.

= Contributing Team Members = The following students contributed to this report:

John Saxon Eas4200c.f08.vqcrew.c 02:22, 29 September 2008 (UTC)

Adam Edstrand Eas4200c.f08.vqcrew.a 20:25, 26 September 2008 (UTC)

Javier Stober Eas4200c.f08.vqcrew.b 20:26, 26 September 2008 (UTC)

Darin Toscano Eas4200c.f08.vqcrew.d 20:08, 26 September 2008 (UTC)

Kevin Klauk Eas4200c.f08.vqcrew.E 16:22, 26 September 2008 (UTC)

=References=