Beam Design Optimisation

Problem Statement

We want to design a cantilever 1 meter-long beam that can support a vertical load of 10 N at its free end. If the cross-section is circular, what should be its diameter to minimize the deflection of the beam?

Mathematical Model

From structural mechanics, the maximum deflection of a cantilever beam under a point load at its free end is given by:

$\delta =\frac{F{L}^3}{3EI}$

Where:

• $F$ is the applied force (10 N in this problem)
• $L$ is the length of the beam (1 m)
• $E$ is the Young’s modulus of the material
• $I$ is the second moment of area of the cross-section

For a circular cross-section with diameter $D$, the second moment of area is:

$I=\frac{\pi D^4}{64}$

Optimisation Setup

Design Variable

• $D$ = diameter of the circular cross-section (scalar, 1-dimensional problem)

Objective Function

We want to minimize the deflection $\delta$:

$L(D)=\delta =\frac{F{L}^3}{3EI}=\frac{64F{L}^3}{3\pi E{D}^4}$

Constraints

In a practical setting, we might have additional constraints such as:

• Minimum diameter for manufacturing feasibility
• Stress constraints to prevent failure
• Weight or cost constraints

Solution Approach

To find the optimal diameter that minimizes deflection, we would:

1. Take the derivative of the objective function with respect to $D$
2. Set this derivative equal to zero: $\frac{dL}{dD}=0$
3. Solve for $D$

However, in this case, the deflection $L(D)$ is a monotonically decreasing function of $D$ (as $D$ increases, deflection decreases). This means there is no minimum within the feasible region - we want the largest possible diameter.

This is why additional constraints (such as weight or cost) are typically necessary in real beam design problems to obtain a meaningful optimum.

Including a Weight Constraint

Let’s extend the problem by adding a weight constraint. The weight of the beam is proportional to its cross-sectional area:

$W=\rho L\cdot \frac{\pi D^2}{4}$

Where $\rho$ is the density of the material.

If we have a maximum weight constraint $W_{\max }$, we can solve:

$\rho L\cdot \frac{\pi D^2}{4}\le W_{\max }$

This gives us a maximum feasible diameter:

$D\le \sqrt{\frac{4W_{\max }}{\rho L\pi }}$

Since the deflection decreases monotonically with increasing diameter, the optimal solution would be exactly at this bound:

$D^{\ast }=\sqrt{\frac{4W_{\max }}{\rho L\pi }}$

Multi-Objective Formulation

A more realistic approach would be to formulate this as a multi-objective optimization problem:

1. Minimize deflection
2. Minimize weight (or cost)

This leads to a Pareto frontier of solutions where we trade off between deflection and weight.

Key Insights

1. For beam design problems, often the design variables appear in the denominator of the objective function
2. Simple minimization of deflection alone often leads to impractical solutions (infinitely large beams)
3. Real engineering design requires balancing multiple objectives (deflection, weight, cost, etc.)
4. Constraints are crucial for making the optimization problem well-posed
5. This problem illustrates how engineering optimization often involves trade-offs between competing factors