---
title: "Similitude and scale model of a bracket"
subject: "Machine Design II"
chapter: 1
key: "semester_7/machine_design_ii:similitude-scale-model-cantilever"
source: https://www.bctengineer.com/q/semester_7/machine_design_ii/similitude-scale-model-cantilever
site: BCT Engineering
---

# Similitude and scale model of a bracket

Machine Design II · Chapter 1: Modeling and Simulation

- Asked in: year not given, 3+5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/machine_design_ii/similitude-scale-model-cantilever
- Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-1

## Question

(a) What is similitude? Explain geometric, kinematic and dynamic similarity. (b) A geometrically similar 1:4 scale model (model = 1/4 of prototype) of a steel cantilever bracket, made of the same steel, is loaded at its tip with 2 kN. It shows a maximum bending stress of 75 MPa and a tip deflection of 1.2 mm. Find the tip load on the prototype that gives the same maximum stress, the prototype tip deflection, and the stress in the prototype if it is loaded with the same 2 kN.

## Short answer

Similitude allows model-to-prototype prediction when geometric (length ratio λ), kinematic (similar motions) and dynamic (similar force ratios) similarity hold. For the 1:4 steel cantilever, using stress ∝ F/L² and deflection ∝ F/L: prototype load for equal stress is 32 kN, tip deflection 4.8 mm, stress under 2 kN is 4.69 MPa.

## Answer

#### (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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does geometric similarity require between model and prototype?
   A. Equal velocities at corresponding points
   B. Equal Reynolds numbers
   C. Equal forces at all points
   D. All linear dimensions in the same ratio and equal angles
   Answer: D. Geometric similarity means a constant length ratio lambda for all linear dimensions with equal angles. Equal velocities are kinematic and equal force ratios are dynamic similarity. (Key point: Geometric similarity)

2. For the same steel and equal maximum stress, how does tip load scale with the length ratio lambda = 4, and what is the prototype load?
   A. F is proportional to lambda, so 8 kN
   B. F is proportional to lambda^2, so 32 kN
   C. F is proportional to lambda^3, so 128 kN
   D. F is the same, so 2 kN
   Answer: B. Stress is proportional to F/L^2, so equal stress needs Fp/Fm = lambda^2 = 16, giving 32 kN. (Key point: Load for equal stress)

3. Deflection is proportional to F/L when E is constant. What is the prototype tip deflection at the equal-stress load?
   A. 1.2 mm
   B. 19.2 mm
   C. 4.8 mm
   D. 0.3 mm
   Answer: C. delta_p/delta_m = (F_p/F_m)/lambda = 16/4 = 4, so delta_p = 4 x 1.2 = 4.8 mm. (Key point: Prototype deflection)

4. If the prototype carries only the same 2 kN as the model, what happens to the maximum stress?
   A. It falls to 75/16 = 4.69 MPa because stress is proportional to F/L^2
   B. It stays at 75 MPa
   C. It rises to 1200 MPa
   D. It falls to 18.75 MPa
   Answer: A. The same force acts over a section that is 16 times larger in area-like terms (L^2), so stress is 75/16 = 4.69 MPa. (Key point: Stress at the same load)

## Related in Machine Design II

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- [Gear ratios of a four-speed car](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aautomobile-gear-ratio-selection) — Machine Design II, chapter 5
- [Simple band brake absorbing power](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aband-brake-power-absorbed) — Machine Design II, chapter 7
- [Design of a helical compression spring](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Acompression-spring-design-static) — Machine Design II, chapter 6
- [Screw jack with square thread](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Ascrew-jack-square-thread-numerical) — Machine Design II, chapter 8

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- [Cantilever 3 m: bending stress 25 mm below top](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acantilever-3m-udl-20kn-t-section) — Strength of Materials, chapter 7
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