Similitude and scale model of a bracket

(a) Similitude

Similitude is the set of conditions under which results measured on a model can be used to predict the behaviour of the full-size prototype.

• Geometric similarity: all linear dimensions have the same ratio $\lambda = L_p/L_m$, and angles are equal.
• Kinematic similarity: motions are similar; velocities at corresponding points have the same ratio and direction.
• Dynamic similarity: forces at corresponding points have the same ratio (for example equal Reynolds, Froude or Mach number); this includes similar load and stress patterns in solids.

(b) Numerical

Scale ratio $\lambda = L_p/L_m = 4$. Same material, so $E$ is the same.

Bending stress at the root: $\sigma = \dfrac{M c}{I} \propto \dfrac{F L \cdot L}{L^4} = \dfrac{F}{L^2}$.

Tip deflection: $\delta = \dfrac{F L^3}{3EI} \propto \dfrac{F L^3}{L^4} = \dfrac{F}{L}$ (E constant).

1. Prototype load for equal stress

$$\begin{aligned} \frac{F_p}{F_m} &= \lambda^2 = 16 \\ F_p &= 16 \times 2 = 32\ \text{kN} \end{aligned}$$

2. Prototype deflection

$$\begin{aligned} \frac{\delta_p}{\delta_m} &= \frac{F_p/F_m}{\lambda} = \frac{16}{4} = 4 \\ \delta_p &= 4 \times 1.2 = 4.8\ \text{mm} \end{aligned}$$

(Deflection scales with $\lambda$, as in geometric similarity at equal stress.)

3. Stress when the prototype carries only 2 kN

$$\sigma_p = \sigma_m \times \frac{F_p}{F_m}\times\frac{1}{\lambda^2} = 75 \times \frac{1}{16} = 4.69\ \text{MPa}$$

Answer: prototype load = 32 kN, prototype tip deflection = 4.8 mm, stress under 2 kN = 4.69 MPa.