User:Eas4200c.f08.team12.hepsworth/HW2

Lecture 6 - 9/8/08
Problem 1.1 Continued

Case 1: Method

We begin by assuming that $$\sigma_{max}$$ will reach the allowable stress first. This will allow the maximum bending moment allowable, Mallowable, to be computed.

Compute $$\sigma_{max}$$ for $$z = \frac{b}{2}$$ , the center of the cross section

$$\displaystyle \sigma = \frac{M z}{I}$$ , $$\displaystyle =>\sigma_{max}= \frac{M b}{2 I}$$ $$\displaystyle => M = \frac{2 I \sigma_{max}}{b} = 2\sigma_{allowable} (\frac{I}{b})$$

Looking at the above equation we see that M is directly proportional to $$\frac{I}{b}$$, therefore maximizing M is the same as maximizing $$\frac{I}{b}$$

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

The next step is to find the moment of inertia, $$I$$, in terms of b. To achieve this we must find the value of b that maximizes $$\frac{I}{b}$$ . Recall that $$a = \frac{L}{2} - b$$ . After finding the optimal cross section, $$\frac{b^1}{a^1}$$, then $$\frac{I}{b}$$ can be obtained.

$$ (\frac{I}{b})_{max} = \frac{I}{b} => M_{max} = 2\sigma_{all}(\frac{I}{b}) = T_{max}$$

Next we must check the torsional shear stress,

$$\tau_{max} = \frac{T_{max}}{2abt} = \frac{M_{max}}{2abt}$$

does not exceed what is allowable, $$\tau_{allowable} = \frac{\sigma_{allowable}}{2}$$

If $$\tau_{max} < \tau_{allowable}$$ then, $$\frac{b^1}{a^1}$$ is acceptable

If $$\tau_{max} > \tau_{allowable}$$ then, $$\frac{b^1}{a^1}$$ is not acceptable

Case 1: Answer

Using the parallel axis theorem and taking the sum of the moments of inertia of the four segments of the cross section we get:

$$I = \Sigma [\frac{b_i (h_i)^3}{12} + A_i d_i^2] = 2\frac{t b^3}{12} + 2[\frac{a t^3}{12} + at(\frac{b}{2})^2] = \frac{t b^2}{6}(3a + b)$$

therefore:

$$f(b) := \frac{I}{b} = \frac{t b}{6}(3a + b) = \frac{t b (3L - 4b)}{12} $$

The goal is to maximize f(b):

$$\frac{df (b^1)}{db} = 0 = \frac{t}{12} (3L - 8b^1) => b^1 = \frac{3L}{8}$$

Lecture 7 - 9/10/08
Problem 1.1 Continued

Case 1: Assume σ max = σ allowable

Recall: $$\displaystyle \sigma = \frac{M z}{I}$$ ,       $$z = \frac{b}{2}$$

Solving for Moment: $$M = \frac{2 I \displaystyle \sigma}{b}$$

$$M = 2\displaystyle \sigma ( \frac{I}{b} )$$   :(I/b) is a function of the length dimensions for the cross section (a,b)

Recall Assumptions:
 * 1) L = 2(a+b) = Constant
 * 2) M = T
 * 3) σ = 2τ

Solving for dimension a : a = (L / 2) - b. Solving the parameter for a will give us the opportunity to solve the moment equation as a function of only b.

Now maximize the moment equation:  $$M = 2\displaystyle \sigma ( \frac{I}{b} )$$ max. This equation shows that in order to maximize the Moment (M), the ratio of (I/b) must be maximized.

I = Second Moment of Inertia. I can be found by using the Parallel Axis Theorem.

I = I y + Ad 2

Iundefined is the area moment of inertia through the parallel axis to the center of mass

Iy is the area moment of inertia through the center of mass

A is the surface area

d is the distance the y axis to the cg axis

Calculation for beam in Problem 1.1: $$I= 2( \frac{b^3 t}{12} ) + 2( \frac{3a b^2 t}{12} )$$

Simplify: $$I= ( \frac{t b^2}{6} )(3a + b)$$

Maximizing $$( \frac{I}{b} ) = ( \frac{t b}{6} )(3a + b)$$, Remember $$a= ( \frac{L}{2} ) - 2$$

Now: $$( \frac{I}{b} ) = ( \frac{t b}{12} )(3L + 4b)$$

Solving the equation for b yields two distinct solutions: $$b = 0$$   and   $$b = \frac{3L}{4}$$

These two points are the zero crossings on the graph for a parabola. To find the desired solcution (b max) you need to find the maximum of that parabola. This can be done easily by realizing that the maximum must be half way between the two zero crossings. This gives the solution to the Case 1 as the following.

b 1 = $$ ( \frac{3L}{8} )$$

Now that we have solved for b, we can now evaluate the entire beam structure.

Plugging b into the parameter equation $$a = L - ( \frac{b}{2} )$$    yields a 1 = $$ ( \frac{L}{4} )$$

Thus comparing the Length of the two sides gives: $$ ( \frac{b^1}{a^1} ) = 3$$

The relationship between the second moment of inertia to the side be can also help solve for the Moment M.

$$M = \frac{2 I \displaystyle \sigma}{b}$$

Case 1 Momentum: $$M = ( \frac{3t L^2}{32} )\displaystyle \sigma$$

Lecture 8 - 9/12/08
Motivation: Study of a box beam (Problem 1.1) with a rectangular cross-section is a model for aerospace structures (fuselage, wings).

Problem 1.1 continued shear stress due to T max : $$\displaystyle \tau = ( \frac{T^1}{2a^1 b^1 t} )$$ This derivation is from slide 5-2 in the lecture notes.

Due to Assumption 2 T max = M max, and plugging that relationship and the calculated values for a and b into the shear stress yields:

$$\displaystyle \tau_{max} = ( \frac{M^1}{2 (\frac{L}{8}) (\frac{3L}{8}) t} )$$

Thus: $$\displaystyle \sigma = ( \frac{32}{3t L^2} ) M^1$$

This solution looks shows that: $$\tau_{max} = \displaystyle \sigma_{allow}$$

Recall Assumption 3) $$\displaystyle \sigma_{allow} = 2\tau_{allow}$$

This gives the result: $$2\displaystyle \tau_{allow} > \tau_{allow}$$ This relationship is not acceptable. The relationship would also mean that: $$\displaystyle \tau_{max} > \tau_{allow}$$. The relationship means failure for the beam.

Case 2: $$\tau_{max} = \displaystyle \tau_{allow}$$    and verify    $$\displaystyle \sigma_{max} < \sigma_{allow}$$

$$\displaystyle \tau = \frac{T}{2a b t} =  \tau_{max}  =  \tau_allow$$

Solving for the Torque (T) : $$T = 2t\displaystyle \tau_{allow}(ab)$$

$$2t\displaystyle \tau_{allow}$$ is a constant. So if you were trying to maximize the Torque that would mean that you would have to maximize the product of $$ab$$.

Lecture 9 - 9/15/08
Problem 1.1 Case 2 Continued: $$\tau_{max} = \displaystyle \tau_{allow}$$

$$\tau_{max} = \frac{T}{2 a b t}$$

replace τmax with τallow and move the denominator to the left hand side.

Now, we have $$T = ( 2 t \tau_{allow} ) * ( a b )$$ where we want to maximize the last factor...

Proof:

$$T_{max} = ( 2 t \tau_{allow} ) * ( a b )_{max}$$

Remember that $$L = 2 * ( a + b )$$ from which we get $$a = \frac{L}{2} - b$$

Multiply through by a factor of b to get $$( a b ) = \frac{L b}{2} - b^2$$

Taking the derivative of both sides of this equation gives: $$\frac{d( a b )}{db} = \frac{L}{2} - 2b$$

Setting this derivative equal to zero and rearranging the equation reveals: $$\frac{L}{2} = 2b$$

from which we get $$b = \frac{L}{4}$$

Plugging this answer back into $$L = 2 * ( a + b )$$ gives the answer $$a = \frac{L}{4}$$

From Lecture 8: $$a^{(2)} = b^{(2)} = \frac{L}{4}$$

Solving to the maximum Torque. $$T_{max} = (2t\displaystyle \tau_{allow})( \frac{L}{4} )^2$$

$$T_{max} = ( \frac{1}{8} ) t L^2 \displaystyle \tau_{allow} =  M_{max}^{(2)}$$     By Assumption 2.

$$M_{max}^{(2)} = \frac{t L^{2}}{16} * \sigma_{allow}$$     By Assumption 3.

$$\sigma_{allow} = \frac{16 M_{max}^{(2)}}{t L^{2}}$$

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

$$\frac{I}{b} = \frac{t b^{(2)}}{6}*(3a+b^{(2)})$$

Replacing $$b^{(2)}$$ with $$\frac{L}{4}$$ and rearranging we get

$$\frac{I}{b} = \frac{t L^2}{24}$$

Recall $$\sigma_{max}^{(2)} = \frac{M_{max}^{(2)} b^{(2)}}{I^{(2)} 2}$$

Replacing $$\frac{b}{I}$$ in the above equation and rearranging yields:

$$\sigma_{max} = \frac{3}{4}*\sigma_{allow}$$ which is less than the maximum allowable stress! This case is a success! Thus, $$b^{(2)}=\frac{L}{4}$$

Lectures 11 - 9/19/08
Q: Why are the walls of a stringer not completely "vertical"?

A:One of the reasons is a very simple one, storage. It is easier to store the stringers if the walls are not completely vertical.

General case of axial torque of a right cylindrical beam
$$T = \oint_{}^{}{\rho q ds} $$

$$T = \oint_{A}^{}{\int 2 q dA} = 2 q A$$

where q = shear flow = shear stress x thickness = \tau * t and A = a*b = average area of the right circular cylinder.

Hooke's Law
$$\displaystyle \tau = G \displaystyle \gamma$$

G = shear modulus

$$\displaystyle \gamma$$ = shear strain, counterpart to normal strain $$\displaystyle \epsilon$$ , but describes the change in the angle between two lines in the material due to deformation.

$$\displaystyle \gamma = \frac{dv}{dy} + \frac{du}{dx} = \frac{du_x}{dy} + \frac{du_y}{dx}$$

u = u x displacement along x direction v = u y displacement along y direction

$$\gamma = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial x} = \frac{\partial u_{x}}{\partial y} + \frac{\partial u_{y}}{\partial x}$$

Assigned Problems
Problem 1.7 Compare the load-carrying capabilities of two beams having the respective cross-sections shown in Fig. 1.19. Use bending rigidity as the criterion for comparison. It is given that R o =10cm, t=1cm, and the two cross-sections have the same area.

Bending rigidity is defined as the... and is determined by multiplying the Young's modulus by the moment of inertia of the cross section. In order to compare the two cross-sections, we must first determine the unknown dimension (R 1 ) of the second beam.



$$A^{(1)} = A^{(2)}$$

$$A^{(1)}=\displaystyle \pi R^2$$

Plugging in the value of R=10cm gives the Area for Case 1:

$$A^{(1)}=\displaystyle \pi (10)^2 = 100\pi    =   314.2 cm^2$$

$$A^{(2)}= 2\displaystyle \pi R_1^2 + 2R_o(t)   = A^{(1)}$$

$$A^{(2)}= 2\displaystyle \pi R_1^2 + 2(10cm)(1cm)   = 314.2 cm^2$$

Solving for R 1: $$R_1 = 6.84 cm^2$$

Next, we compute the moment of inertia for both beams:

$$ I_y = \int R_o^2\,dA                                    dA = \pi R dr $$

=$$\int R^2(\displaystyle \pi R)\,dr $$

= $$ \displaystyle \pi \int R_o^3\,dr $$

$$ I_y = \frac{\displaystyle \pi }{4} R_o^4 $$

$$ I_y = \frac{\displaystyle \pi }{4} (10cm)^4 $$

$$ I_y = 2500\displaystyle \pi  = 7854 cm^4 $$

Now Calculate I 2 :

$$I_2 = \frac{1}{12}t(2R_o)^3 + 2( \frac{1}{4} \displaystyle \pi R_1^4 + (\displaystyle \pi R_o^2 (R_o +R_1)^2))$$

$$I_2 = \frac{1}{12}(1cm)(2(10cm))^3 +2[ \frac{1}{4} \displaystyle \pi (6.84cm)^4 + (\displaystyle \pi (6.84cm)^2 (10cm + 6.84 cm)^2)]     =      I_2 = 87468.5 cm^4$$

Thus I 2 > I 1. Assuming that the sections are made of the same material, we can conclude that beam 2 has a higher bending rigidity due to the larger second moment of inertia.

Extra Student Contributions
Works Cited

Sun, C.T. Aerospace Stuctures, 2006

www.Merriam-Webster.com

2007-02-17 20:31 Keenan Pepper 525×315×1 (1262 bytes) Parallel axes rule for area moment of inertia. Image made by MH 21:51, July 9, 2005 (UTC)

=Contributing Team Members HW 2=

Steven Hepsworth Eas4200c.f08.team12.hepsworth 21:51, 23 September 2008 (UTC)

Brian Taylor Eas4200c.f08.team12.taylor 20:53, 23 September 2008 (UTC)

Victoria WatlingtonEAS4200C.Fall08.Team12.Watlington.VG 03:51, 24 September 2008 (UTC)

Jacob Papp Eas4200c.f08.team12.papp 05:04, 26 September 2008 (UTC)

Melisa Gaar Eml4500.f08.group.gaar 17:43, 26 September 2008 (UTC)