User:Eas4200c.f08.spars.rondon/Homework 2

=Professor's Notes=

Problem 1.1


"1.1 The beam of a rectangular thin-walled section (i.e. t is very small) is designed to carry both bending moment M and torque T. If the total wall contour length L = 2(a + b) (see Figure 1) is fixed, find the optimum b/a ratio to achieve the most efficient section if M = T and $\sigma$allowable = 2$\tau$allowable. (Sun p. 17)"

Assumptions
There are four assumptions made in the problem statement.

Assumption 1:   $$\mathbf{L} = 2\mathbf{(a + b)} $$

Assumption 2:      $$\mathbf{M} = \mathbf{T} $$

Assumption 3:     $$\ \sigma_{allowable} = 2\tau_{allowable} $$

Assumption 4:     $$\ \mathbf{t<<a},\mathbf{t<<b}$$

Shear stress



 * The first step is to determine the relationship between the shear stress, $$\tau$$, and the torque, T. In actuality the shear stress takes a parabolic shape as shown in figure (2), but in this case since the thickness is very thin (assumption 4) the shear force can be approximated by a uniformly distributed shear flow q.


 * q = $$\tau$$t (3.47)


 * Where the shear flow is related to torque by,


 * dT = ρq ds


 * where ρ is equal to the lever arm and ds is the differential length of the very thin bar. Integrating, this equation becomes


 * T = ρqS


 * The torque in this problem can be broken up into the sum of the torques in each side, T = TAB + TBC + TCD + TDA. By symmetry TAB = TCD and TBC = TDA which simplifies the equation.


 * TAB = $$\frac{b}{2} (q\cdot a) = \frac{\tau abt}{2} $$


 * TAB = $$\frac{a}{2} (q\cdot b) = \frac{\tau abt}{2} $$


 * Therefore


 * T = $$ \mathbf{2\tau a b t} $$


 * $$ \tau = \frac{T}{2abt} $$


 * This is identical to to equation provided in the problem statement with no explanation.

Solution

 * In order to solve this problem it is necessary to consider two cases.

Case One

 * Assume the bending normal stress, $$ \sigma $$, reaches $$ \sigma_{allowable} $$ before shear stress, $$ \tau $$, reaches $$ \tau_{allowable} $$.
 * Recalling that the bending normal stress is related to the bending moment by ordinate of a point on an axis perpendicular to the bending neutral axis over the second moment of inertia according to the equation


 * $$ \sigma = \frac{Mz}{I} $$


 * From which the following relationship is derived


 * $$ \sigma_{allowable} = \frac{M_{max}b}{2I} $$ → $$ M_{max} = \frac{2I\sigma_{allowable}}{b} = 2\sigma_{allowable} (\frac{I}{b})_{max} $$


 * The second moment of inertia is represented by


 * $$ I = \sum_{i=1}^{4}{[\frac{b_i(h_i)^3}{12} + A_id_i^2]} = 2[\frac{tb^3}{12} + \frac{at^3}{12} + \frac{atb^2}{4}] = 2[\frac{tb^3}{12} + \frac{at}{12}(t^2 + 3b^2)]$$


 * Assuming that t is very small compared to b this equation can be approximated by


 * $$ I = 2[\frac{tb^3}{12} + \frac{3atb^2}{12}] = \frac{tb^2}{6}[b + 3a] $$


 * Recalling assumption 1 ($$ L = 2(a + b) $$) the variable a can be stated as


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


 * Which can then be used to express the second moment of inertia in terms of a single unknown variable b


 * $$ \frac{tb^2}{6}[3(\frac{L}{2} - b) + b] = \frac{tb^2}{6}[\frac{3L}{2} - 2b] = \frac{tb^2(3L - 4b)}{12} = \frac{1}{12}[-4tb^3 + 3Ltb^2]$$


 * Returning to the equation for Mmax it is evident that $$ 2\sigma_{allowable}$$ is a constant and therefore Mmax is maximized when $$ (\frac{I}{b}) $$ is maximized.


 * $$ f(b) := \frac{I}{b} = \frac{1}{12}[-4tb^2 + 3Ltb]$$
 * $$ f'(b) = -\frac{8tb}{12} + \frac{3Lt}{12} $$
 * $$ f''(b) = -\frac{8t}{12} $$


 * The negative second derivative indicates the function $$ f(b) $$ represents a concave parabola which reaches a maximum when it's slope, or first derivative, is equal to zero.


 * $$ f'(b^{(1)}) = 0 = -\frac{8tb^{(1)}}{12} + \frac{3Lt}{12}$$ → $$\frac{8tb^{(1)}}{12} = \frac{3Lt}{12} $$ → 8b(1) = 3L  → $$ b^{(1)} = \frac{3L}{8} $$


 * $$ b^{(1)} = \frac{3Lt}{12}$$ → $$ a^{(1)} = \frac{L}{2} - b^{(1)} = \frac{L}{2} - {3L}{8} = {L}{8} $$


 * $$ f(b^{(1)}) = f(\frac{3L}{8}) = \frac{1}{12}[-4t(\frac{3L}{8})^2 + 3Lt(\frac{3L}{8})] = \frac{1}{12} [\frac{9tL^2}{8} - \frac{36tL^2}{64}] = \frac{1}{12}[\frac{36tL^2}{64}] = \frac{3tL^2}{64}$$


 * Therefore


 * $$ M_{max} = 2\sigma_{allowable} \frac{3tL^2}{64}$$


 * Recalling assumption 2 ($$ M = T $$) and the equation relating the shear force to the torque ($$ \tau = \frac{T}{2abt} $$)


 * $$ \tau^{(1)}_{max} = \frac{T^{(1)}_{max}}{2a^{(1)}b^{(1)}t} = \frac{M^{(1)}_{max}}{2a^{(1)}b^{(1)}t} = \frac{2\sigma_{allowable}{\frac{3tL^2}{64}}} {2\frac{L}{8}\frac{3L}{8}t} = \sigma_{allowable}$$


 * Recall from assumption 3 ($$ \sigma_{allowable} = 2\tau_{allowable}) $$


 * $$ \tau^{(1)}_{max} = \sigma_{allowable} = 2\tau_{allowable} $$


 * $$ \tau^{(1)}_{max} > \tau_{allowable} $$


 * The maximum shear stress exceeds the allowable shear stress and therefore this solution is unacceptable.

Case Two

 * Assume the shear stress, $$ \tau $$, reaches $$ \tau_{allowable} $$ before the bending normal stress, $$ \sigma $$, reaches $$ \sigma_{allowable} $$.


 * Recall the equation determined earlier relating shear stress to torque ($$ \tau = \frac{T}{2abt} $$)


 * $$ \tau^{(2)}_{max} = \frac{T}{2abt} $$ → T(2)max = $$ 2\tau_{allowable} a^{(2)} b^{(2)} t $$


 * It's clear that maximizing the torque is equivalent to maximizing the product a(2) b(2) which recalling assumption 1 ($$ L = 2(a + b) $$) that product becomes


 * $$ \frac{L}{2}b^{(2)} + (b^{(2)})^2 = maximum$$


 * Taking the derivative


 * $$ \frac{L}{2} - 2b^{(2)} = 0 $$ → $$ 2b^{(2)} = \frac{L}{2} $$ → $$ b^{(2)} = \frac{L}{4} $$


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


 * Therefore the torque is maximized when a(2) = b(2) = $$ \frac{L}{4}$$, therefore


 * T(2)max = $$ 2\tau_{allowable} (\frac{L}{4})^2 t = \frac{1}{8}tL^2\tau_{allowable}$$


 * Recalling assumption 2 ($$\mathbf{M} = \mathbf{T} $$)


 * $$M^{(2)}_{max} = \frac{1}{8}tL^2\tau_{allowable}$$


 * Recalling assumption 3 ($$\ \sigma_{allowable} = 2\tau_{allowable} $$)


 * $$M^{(2)}_{max} = \frac{1}{16}tL^2\sigma_{allowable}$$ → $$\sigma_{allowable} = \frac{16M^{(2)}_{max}}{tL^2}$$


 * Recall from case one


 * $$ f(b) := \frac{I}{b} = \frac{1}{12}[-4tb^2 + 3Ltb]$$ → $$ f(b^{(2)}) = \frac{1}{12}[\frac{tL}{4}[3L - L] = \frac{tL^2}{24} $$


 * Recall also from case one


 * $$ \sigma_{max} = \frac{M_{max}b}{2I} $$ → $$ \sigma^{(2)}_{max} = \frac{1}{16}tL^2\sigma_{allowable}(\frac{12}{tL^2}) = \frac{12}{16}\sigma_{allowable} $$


 * $$\sigma_{max} < \sigma_{allowable}$$


 * The maximum normal stress does not exceed the allowable normal stress and therefore this solution is acceptable. Furthermore, it proves the optimal cross section is a square such that


 * $$ \frac{b}{a} = 1 $$

=Book Summary= In Chapter 2, the course begins introducing the concepts related to elasticity. Topics include displacement, stress (normal, uniform, principal, shear), strain, equilibrium, reviewing stress transformations, stress strain relations, elastic strain energy to name a few. These topics are covered are related to composites as well as non-composites to determine the failure charateristics of the materials.

2.1
2.1 is entitled "Concept of Displacement" which simply describes the movement of a point in a material from one position to another upon deformation of the material. This displacement is described by a vector, u. This vector has components of u, v, and w in the x, y, and z directions. Therefore, when a material is deformed, the new position of any point, denoted with a "prime" symbol, e.g. x', can be described by the following: x' = x + u, where x is the original position of the point in the material. The following figure illustrates a sample cross section which undergoes a deformation:



We seek to find an equation for displacement u based on x and uniform axial strain in a member, for example, a long rod. We assume some initial displacement u0 = u (x0). Then u1 = u(x0 + Δx) for the displacement of another point x0 + Δx. The elongation of the material between these two points is Δu = u1 - u0. Then the axial strain becomes ε = du/dx, and the equation for u is u = ε0 (x - x0) + u0. Again, this holds for any situation in which the strain is uniform in the member, if not, the displacement in one section will be different from that in another, and one must seek a more accurate solution or find an acceptable value for average displacement.

=Homework Problems=

Moments of Inertia of Circular Cross Sections vs Channels


What are the moments of inertia, Iy, for both case 1 and case 2 using polar coordinates. Compare the results to see which case is more efficient. Know that t = a/10 and that a = b. For case 1: Iy of a solid circular cross section is                                   $$Iy = \int\int_A Z^2 \,dy dz$$

with respect to the y and z axes. Z2 is also considered dA which, when the integral is taken, is the area of the circle $$A = \pi r^2$$. The formula then becomes $$Iy = \pi r^4/4 \,$$

where r is the perpendicular distance from the center to the edge of the circle.

For case 2: We know, since the mass is redistributed into a channel, that A1=A2. Since the area of a circle is πr2, then we can equate that to the area of the channel, 2(bt)+t(a-2t):

$$\pi r^2=2 (b t)+t (a-2 t)\,$$

Knowing that t=a/10, and b=a, then we can rewrite this as:

$$\pi r^2=2*(2 a/10)+(a/10)*(a-2(a/10))\,$$ This in turn simplifies to:

$$7 a^2/25=\pi r^2\,$$

so we can say:

$$a=(25 \pi/7)^{(1/2)} r\,$$

With this knowledge, we can use the parallel axis theorem for rectangles, shown to the right, to solve for the moment of inertia for the channel along the axis that goes through y=a/2 if y is the vertical coordinate, and the channel were to be standing on y=0.

$$I=((1/12) t a^3) + ((1/12) (a-2 t) (t^3) + (a t) (a/2)^2)\,$$

Once again using the relationship t=a/10 we can rewrite the moment of inertia equation as:

$$I=a^4 /120 + ((1/12) (4 a/5) (a^3 /1000) + (a^2 /10) (a^2 /4))\,$$ $$I=a^4 /120 + a^4/1500 + a^4 /40\,$$ $$I=17 a^4 /500\,$$

Recalling: $$a=(25 \pi/7)^{(1/2)} r\,$$ We can say: $$I=17 ((25 \pi/7)^{(1/2)} r)^4/500 = 85 \pi^2 r^4 /196\,$$

Dividing the I from case 1, and the I from case 2, by πr4 to get both I terms in simple numerical terms with which we can compare their values, we find: For I1: $$I=1/4=.25\,$$ And for I2: $$I=85 \pi /196=1.36\,$$

Dividing I2 by I1 we can see that the moment of inertia about the center of the channel is almost 5.5 times larger than the moment of inertia about the center of the circular cross section.

Problem 1.7


Compare the load carrying capabilities of two beams having the respective cross-sections shown in the figure. Use bending rigidity as the criterion for comparison. It is given that a = 4cm, t = 0.2cm, and the two cross-sections have the same area. (Sun, 18)

Case 1 To solve this, we first need the area of the square. The area of a square is $$A = a^2$$. Since a = 4cm, we get that A is 16cm2.

Now that we have the area, we need the moment of inertia. The formula used in this case is $$Iy = 1/12bh^3$$, where b is the length and h is the height. Since both b and h are 4cm, Iy = 21.33cm4.

Case 2 To solve the area, we need to break this cross section into parts; a rectangle and two squares. The area of a rectangle is $$A = at$$ where a is the length and t is the width. Using this data, A = 0.8cm2. The area of the squares on top and bottom are $$A = b^2$$. Since there are two squares, the area of both combined is 2b2. The total area of this cross section is the area of the rectangle + the area of both squares, which gives us 0.8cm2 + 2b2. Since both areas are equal to one another, our new equation is $$0.8cm^2 + 2b^2 = 16cm^2$$. Using algebra, we solve for b and find that it is 2.76cm.

The moment of inertia for this case will also be calculated in parts. For the rectangle, Iy = 1/12ta3 where t is the base and a is the height. When the values are plugged in, we get Iy = 1/12(0.2cm)(4cm)3. This gives a total of 1.067cm4. For the squares, the Iy formula is the same; Iy = 1/12bh3 where b is the base and h is the height. Since both b and h are the same value, Iy = 1/12(2.76cm)(2.76cm)4. This gives a total of 4.84cm4 for each square. The total Iy of this cross section is Itotal = 1.067cm4 + 4.84cm4 + 4.84cm4 which is 10.74cm4.

Comparison Although both cross sections have the same area, their moments of inertia are quite different. Iy for case 2 is much less and allows for less of a bending moment when used in practical applications. This allows for maximum potential of the materials used with less weight. A key ingredient in aerospace structures.

Problem 1.7 Modified


Find R1 such that the area of case 2 is equal to the area of case 1. Also, find the moment of inertia, Iy, for both cases and compare. R0 = 10cm and t = (1/10)*R0.

Case 1

The area of acircle is $$A = \pi r^2$$. Plugging in 10cm for R0 gives us $$A = \pi (10cm)^2$$ which is 314.159cm2.

The moment of inertia of a circle is $$I = \pi r^4/4$$. When we plug in R0 for r, the moment of inertia for this circle is 7853.98cm4.

Case 2

The area of case 2 will be broken down into three parts; a rectangle and two circles. For the rectangle, A=lw, where l is the length and w is the width. This will give the result of $$A = 20cm * 1cm$$ which is 20cm2. The area of both circles combined will be $$A = 2 \pi R^2$$. The total area will be the sum of both the rectangular and two circular areas. Since the areas of both cases are to be equivalent, we get $$314.159cm^2 = 20cm^2 + 2\pi R^2$$. by subtracting 20cm2 from both sides and dividing both sides by $$2\pi$$, we get R2 = 46.8169. When we square root both sides, the value of R1 is found to be 6.84cm

The total moment of inertia will also be broken down into parts; the rectangle and the two circles. The moment of inertia for a rectangle can be expressed as $$I = 1/12bh^3$$. The moment of inertia for each circle will use $$I = \pi r^4/4$$. Since b = 1cm and h = 20cm, plugging those values in gives the total of 666.667cm4 for the rectangular moment of inertia. For each circle, plug in 6.84cm for r and the result is 1719.15cm4. When totaled, I for case 2 is 1719.15cm4 + 1719.15cm4 + 666.667cm4= 4104.97cm4.

Comparison

Although both cases had the same area, it is shown that their moments of inertia are quite different. The moment of inertia for case 2 is much less which allows for much better resistance to bending.

Do stringers always have a rectangular cross section?


Stringers do not always have a rectangular cross section. They are made of bars with open, thin walled cross sections. (See Figure 9) This method is used to resist the bending moments imposed from external pressures on      an aircraft structure. It increases the amount of moment of inertia, while keeping the material fixed.

Why use angled beam-walls for stringers?


Angled beam-walls are used for stringers because this design allows for better stockpiling, manufacturing, and construction. Looking at Figure 10, it is shown that by the stringers being angled, more can fit when stacked on top of each other. This allows for maximizing a large quantity of stringers in a small storage area. When it comes to manufacturing, stringers are usually fabricated from flat sheets of metal by a stamper. When the stamper releases, it is easier for it not to get caught on an angled wall rather than a vertical wall (Dashed line of Figure 10). Lastly, when it comes time to rivet the stringers into place, it is easier with angled walls than with vertical walls. The rivet gun will not have to squeeze into place and every rivet will be accessible rather than closed.

=Contributing Team Members= Eduardo Rondon Eas4200c.f08.spars.rondon 15:22, 25 September 2008 (UTC)

Michael Lee   Eas4200c.f08.spars.lee 16:23, 25 September 2008 (UTC)

Eas4200c.f08.spars. 18:40, 25 September 2008 (UTC)

--Eas4200c.f08.spars.prey 07:28, 26 September 2008 (UTC)

Thomas McGilvray Eas4200C.f08.spars.mcgilvray 19:02, 26 September 2008 (UTC)