Lightweight panel¶
The company Metallicious Aluminati Extrusions, proudly using the slogan Profiles so good, they're almost delicious, has recived a request for a lightweight panel. Being a specialist in aluminum extrusion, the material choice is easy. However, the requirements are challenging. Stated as constraints:
- The length is $L_0 = 2.4 \text{ m}$
- The width is $b_0 = 1.2 \text{ m}$
- Maxium thickness is $h_m = 40 \text{ mm}$
When loaded by a center force $F_0 = 10 \text{ kN}$ while being simply supported at both ends, the maximum deflection must be less than $w_0 = 50\text{ mm}$
The panel shall not fail in any modes when subjected to a center force $F_0 = 10 \text{ kN}$
Objective: As light as possible.
Initially, the problem can be approximated to that of a beam, where the deflection is
\begin{equation} w = \frac{FL^3}{48EI} \tag{1} \end{equation}The modulus is given (aluminum), $E = 70 \text{ GPa}$, while the cross section and therefore $I$ is a free variable.
The minimum magnitude of $I$ in order to satisfy the stiffness requirement is
\begin{equation} I_{min} = \frac{F_0L_0^3}{48Ew_0} \tag{2} \end{equation}L0, F0, E, w0 = 2400, 10000, 70000, 50
I_min = (F0*L0**3)/(48*E*w0)
print('Minium I = {:.1f}'.format(I_min))
A solid, homogeneous cross section having thickness $h_m$ as reference:
b0, hm = 1200, 40
I_ref = (b0*hm**3)/12
rho = 2700
mass_ref = L0*b0*hm*rho*1E-9
print('Solid homogeneous, I = {:.1f}'.format(I_ref))
print('Solid homogeneous, mass = {:.1f} kg'.format(mass_ref))
The stiffness of the solid homogeneous cross section is about eight times the required stiffness:
print(I_ref/I_min)
Thus, the thickness can be reduced:
\begin{equation} I = \frac{bh^3}{12} \Rightarrow h_{min} = \Big( \frac{12 I_{min}}{b_0} \Big)^{\frac{1}{3}} \tag{3} \end{equation}h_min = (12*I_min/b0)**(1/3)
print('Minium thickness, h = {:.1f} mm'.format(h_min))
mass_min = L0*b0*h_min*rho*1E-9
print('Solid homogeneous, mass = {:.1f} kg'.format(mass_min))
One of the rules for lightweight design is utlilize the design space, and the thickness $h$ is one parameter in the design space worth utilizing to the maximum. Flexural stiffness at a relatively low weight can be obtained using the principles of sandwich structures or beam shapes such as the I-profile.
Let the profile of the panel resemble a series of I-profiles as illustrated above. The number of unit-cell I-profiles is $n$ such that the dimension $d_1$ is
$$d_1 = \frac{b_0}{n}$$The parameters $t_1$ and $t_2$ are now free variables, and $I$ as well as the mass can be computed for any combinations:
hm=40
t1=0.78
t2=0.78
n = 30
d1 = b0/n
I_new = n*(d1*hm**3)/12 - n*((d1-t2)*(hm-2*t1)**3)/12
print('Profile, I = {:.1f}'.format(I_new))
print('I, relative to requirement = {:.5f}'.format(I_new/I_min))
mass_pro = n*(2*d1*t1 + t2*(hm-2*t1))*L0*rho*1E-9
print('Profile, mass = {:.3f} kg'.format(mass_pro))
print('Profile, relative mass compared to optimized solid homogeneous solution = {:.2f}'.format(mass_pro/mass_min))
Strength assessment¶
An approximation for the maximum/minium stress in the panel is
\begin{equation} \sigma = \pm \frac{M}{I}\frac{h}{2} = \pm \frac{FL}{4I} \frac{h}{2} \tag{4} \end{equation}F0, L0, hm = 10000, 2400, 40
sigma = (F0*L0*hm)/(8*I_new)
print('Maximum stress = {:.1f} MPa'.format(sigma))
The stress level is well within the capacity of relevant aluminum alloys.
However, the rather thin wall sections (0.78 mm) of the profile may be susceptible to buckling. Therefore, a more detailed finite element analysis will be useful.
Finite element analysis¶
