User:Eas4200c.f08.aero6.villalba/shear1-4

(1.4)
Problem statement: Use AS4/3501-6 carbon/epoxy composite to make the I-beam as stated in Problem 1.3. Compare its weight with that of the aluminum beam.

So, repeating the same process as with problem 1.3, we now use the AS4/3501-6 carbon/epoxy composite. Now we use the Young's modulus of this material:


 * $$E_{composite}=140 \ GPa$$

From problem 1.3, we know the expression for the moment of inertia of this I-beam (EQN1 ). This gave us the moment of inertia of the steel beam:


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

From the problem statement $$E_{steel}\cdot I_{steel} = E_{composite}\cdot I_{composite}\!$$, we find $$I_{composite}\!$$ to be
 * $$I_{composite} = \frac{E_{steel} \cdot I_{steel}}{E_{composite}} = \frac{(200 \ GPa)(7.612 \times 10^{-6} \ m^4)}{140 \ GPa} = 1.087 \times 10^{-5} \ m^4$$

Now, rearranging EQN1 to solve for $$b\!$$ > we obtain the following:


 * $$b_{composite} = \frac{1}{2}\left(I_{composite}-\frac{t(h-2t)^3}{12}\right)\Bigg/\left(\frac{t^3}{12}+t\left(\frac{h-2t}{2}+\frac{t}{2}\right)^2\right) = 0.0843 \ m = 84.3 \ mm$$

This will yield the following cross-sectional area:


 * $$A_{composite} = 2 \cdot b_{composite} \cdot t+(h-2t) \cdot t = 1793 \ mm^2\!$$

Now, knowing $$\rho_{composite} = 1.55 \times 10^{-3} \ \frac{g}{mm^3}$$, we can obtain the mass-per-unit length and our weight-per-unit length


 * $$m_{composite}=\rho_{composite} \times A_{composite}\!$$
 * $$m_{composite}=\left(1.55 \times 10^{-3} \ \frac{g}{mm^3}\right) \left(1793 \ mm^2\right)\!$$
 * $$m_{composite}=2.78 \ \frac{g}{mm}\!$$

The weight-per-unit length of the composite beam is
 * $$W_{composite}=27.2 \ \frac {N}{m}$$

Therefore, the composite beam is more efficient weightwise than both the aluminum and the steel beams.