User:Eas4200c.f08.aero6.villalba/modification 1-3

(1.3)
Problem statement: The dimensions of a steel (300M) I-beam are $$b = 50 \ mm,\!$$ $$t = 5 \ mm,\!$$ and $$h = 200 \ mm,\!$$ as shown in Figure 1. Assume that $$t\!$$ > and $$h\!$$ > are to be fixed for an aluminum (7075-T6) I-beam. Find the width $$b\!$$ > for the aluminum beam such that its bending stiffness $$EI\!$$ > is equal to that of the steel beam. Compare the weight-per-unit length of these two beams. Which is more efficient weightwise? Question quoted from Mechanics of Aircraft Structures by C.T. Sun



Young's Modulus for specified materials:
 * $$E_{steel} = 200 \ GPa\!$$
 * $$E_{aluminum} = 71 \ GPa\!$$

The Moment of Inertia for an I-beam as shown in Figure 1 is found by breaking up the cross-section into three rectangles and using the Parallel Axis Theorem to compensate for the centroids of each shape. The I-beam will be broken up into one $$t$$ by $$h-2t$$ rectangle and two $$b$$ by $$t$$ rectangles. Remembering the general form of the Parallel Axis Theorem and Moment of Inertia calculation, $$I$$ will be as follows:
 * $$I = \frac{t(h-2t)^3}{12} + 2\left(\frac{bt^3}{12} + bt\left(\frac{h-2t}{2}+\frac{t}{2}\right)^2\right)$$ EQN 1

Therefore
 * $$I_{steel} = 7.612 \times 10^{-6} \ m^4\!$$

The problem states that $$E_{steel}\cdot I_{steel} = E_{aluminum}\cdot I_{aluminum}\!$$

Therefore
 * $$I_{aluminum} = \frac{E_{steel} \cdot I_{steel}}{E_{aluminum}} = \frac{(200 \ GPa)(7.612 \times 10^{-6} \ m^4)}{71 \ GPa} = 2.144 \times 10^{-5} \ m^4$$

The next step is to solve EQN 1 for $$b\!$$ >, in terms of the known and fixed values specified in the problem statement, as shown below.
 * $$b = \frac{1}{2}\left(I_{aluminum}-\frac{t(h-2t)^3}{12}\right)\Bigg/\left(\frac{t^3}{12}+t\left(\frac{h-2t}{2}+\frac{t}{2}\right)^2\right) = 0.196 \ m = 196 \ mm$$

Now, to obtain the weight-per-unit length of the beams we need both the density of the materials being used and the cross-sectional areas for the beams.


 * For steel,
 * $$A_{steel} = 2 \cdot b_{steel} \cdot t+(h-2t) \cdot t = 1450 \ mm^2\!$$
 * $$\rho_{steel} = 7.8 \times 10^{-3} \ \frac{g}{mm^3}$$
 * so the mass-per-unit length for steel is:
 * $$m_{steel}=\rho_{steel} \times A_{steel}\!$$
 * $$m_{steel}=\left(7.8 \times 10^{-3} \ \frac{g}{mm^3}\right) \left(1450 \ mm^2\right)\!$$
 * $$m_{steel}=11.31 \ \frac{g}{mm}\!$$


 * For aluminum,
 * $$A_{aluminum} = 2 \cdot b_{aluminum} \cdot t+(h-2t) \cdot t = 2910 \ mm^2\!$$
 * $$\rho_{aluminum} = 2.78 \times 10^{-3} \ \frac{g}{mm^3}$$
 * so the mass-per-unit length for aluminum is:
 * $$m_{aluminum}=\rho_{aluminum} \times A_{aluminum}\!$$
 * $$m_{steel}=\left(2.78 \times 10^{-3} \ \frac{g}{mm^3}\right) \left(2910 \ mm^2\right)\!$$
 * $$m_{aluminum}=8.09 \ \frac{g}{mm}\!$$

The weight-per-unit length of each beams is then
 * $$W_{steel}=110.84 \ \frac {N}{m}$$
 * $$W_{aluminum}=79.28 \ \frac {N}{m}$$

Therefore, the aluminum beam is more efficient weightwise.