Optimization of the cross-sectional geometry of a double - clamped beam

Consider the double-clamped beam with a distributed load. The cross-section is constructed with two parabolic curves in the form b(y) that depend on b_t, b_w and h. Distributed load has a parabolic shape p(x) that depend on the p_max, p_min and L. The Young's modulus of the material is E, the yield stress is σ_y, and the deflection is u(x).

Our goal is to optimize the cross-sectional geometry of a double-clamped beam. We can consider 10 tasks to get the solution for this goal.

Task 1. Derivation the area A(y), the first moment of area Q(y) and the second moment of area I(y)

As we can see, our cross-sectional area is symmetric. It means that we can integrate A(y) from 0 to h/2 but need to multiply by 2.

Define b(y): We have parabolic shapes. The main equation for a parabola is x =ay²+by+c, where a,b,c are constants that need to be defined. Using ‘boundary’ conditions:

If y=0 then x=b_w

If y=h/2 then x=b_f

If y=-h/2 then x=b_f

Substitute these conditions to a parabolic equation, and we can obtain that: a=4(b_f - b_w )/h² , b=0, c=b_w

The same consideration should be made for the second moment of inertia. We need to multiply our integral by 2 because our cross-sectional area is symmetric through the axis y.

Task 2. Governing ODE equation

The governing equation for the beam is

Derivation of the parabolic shape of p(x). The main equation for parabola is p =ax²+bx+c, where a,b,c are constants that need to be defined. Using ‘boundary’ conditions:

If x=L/2 then p=p_max

If x=0 then p=p_min

If x=L then p=p_min

Substitute these conditions to a parabolic equation we can obtain that: a=(-4(p_max - p_min))/L² , b=4(p_max - p_min)/L, c=p_min

Task 3. Derivation the general solution to the governing ODE

We solve the ODE in a symbolic equation with corresponding constants C1, C2, C3, and C4.

Task 4. State the boundary conditions

In this case, the beam has two fixed walls. It means that the displacement on the ends equals zero (fixed), and the slope equals zero.

Task 5. Calculate the integration constants

We can substitute boundary conditions in our equation and obtain the solution for displacement for a simply supported beam.

Task 6. Derivation of the moment, shear, and square of the von mises stress

We use the following equations:

We can substitute derivatives of displacement in our equation and obtain moment, shear force, and stresses.

Optimization criteria

The first optimization criteria mean we want to construct and design a minimum-mass supported beam. This is important from an economic point of view. The mass is proportional to the cross-sectional area because length and density are fixed in our case. m=ρV=ρAL.

The second optimization criteria mean there is a stress constraint. By varying the b_w and h in the beam we don’t want to exceed a maximum stress constraint (σ_y). But in this case, we have shear and normal stresses, that’s why we want to minimize Von-Mises stress (combination of σ_xx and σ_xy). Otherwise, we will face the problem of beam destruction.

Third optimization criteria mean there is a deflection constraint. This is one of the critical limitations. Minimizing the maximum deflection with a specific value in real design is very important. Otherwise, we can run into the problem of the destruction of the structure.

The forth and fifth optimization criteria mean that in the construction of a beam (manufacture), there are dimension restrictions, that width or height shouldn’t be greater than the certain values that are used in production during manufacture.

Task 7. Substitution of the constants in the above equation

It is evident from the symmetry of the problem that since the distributed load is symmetrical, the maximum deflection will be in the middle of the beam, i.e., x=L/2. To find the maximum stress, we can consider four points:

(x,y)=(0, h/2)

(x,y)=(L/2, -h/2)

(x,y)=(0, 0)

(x,y)=(L/4, h/4)

To find the maximum stress we need to understand where the Von-Mises stress is maximum. The greater contribution in Von-Mises stress is made by the normal stress, because: σ_vm²=σ_xx²+3σ_xy,max. Therefore we can simplify our problem by considering a rectangular distribution load. Then the maximum moment is equal to: M_ends,max=(wL²)/12. The maximum will be at points x=0 and x=L. At the same time, y should be equal to –h/2: y=-h/2 to obtain the maximum normal stress and the maximum Von-Misses stress. Substitute all the constants in the solution.

Task 8. Lambdify operation into Python function definition

Lambdified the expressions into Python function definitions.

Task 9. Generate the mesh grid of the test points

Meshgrided n² points and generated values for the mass, the displacement and the stress constraints.

Task 10. Plot the optimization objective a filled contour plot

Plot the optimization objective a filled contour plot

In the figure, two yellow lines are visible which restrictions for b_w and h are.

This figure shows an approximation for displacement isoline. Here I had to zoom the previous graph because displacement isoline is not visible. Displacement isoline is out of bounds for h and b_w.

At this point, we have the maximum moment. So, in this case, we will get the correct solution for an optimization problem. The red arrows show a feasible area. The area indicated by the orange arrow satisfies all the conditions and is in the feasible area. So we can choose optimizing dot. The red dot will be optimized for this case because we are looking at the area to the right of the purple line, and the contour for the mass has the smallest value.

[h=0.23, b_w=0.1]