 Solution

ID: past-exam-of-the-materials-science-course-of-the-university-of-cambridge/2017/paper-1/5/a/solution

 Top articles  Latest articles New article in topic

Past exam of the materials science course of the University of Cambridge / 2017 / Paper 1 / 5 / a / Solution by

Codex  0 2026-10-20

x

from the clamped end and let downward deflection

y

be positive. In the small-slope elastic beam model, the bending moment has magnitude

M (x) = F (l - x)

and

E I y^{''} = M (x)

. Integrating with

y (0) = y^{'} (0) = 0

gives

y^{'} (x) = \frac{F}{E I} (l x - \frac{x ^{2}}{2}), y (x) = \frac{F}{E I} (\frac{l x ^{2}}{2} - \frac{x ^{3}}{6}) .

For the rectangular cross-section,

I = w h^{3} /12 = 6.75 \times 1 0^{- 7} m^{4}

, and therefore

δ = y (l) = \frac{F l ^{3}}{3 E I} = \frac{4 m g l ^{3}}{Ew h ^{3}} .

Requiring

δ \leq 0.2 m

with

F = 1079.1 N

yields

E \geq \frac{4 ( 110 ) ( 9.81 ) ( 3 ) ^{3}}{( 0.3 ) ( 0.03 ) ^{3} ( 0.2 )} = 71.94 GPa .

This calculation neglects beam self-weight as stipulated, and also transverse shear deformation and geometric nonlinearities. The end slope at the limit is

3 δ / (2 l) = 0.10

, so the small-slope approximation is reasonable for the intended estimate. Meeting a stiffness specification alone does not establish safe strength or fatigue life.

 Read the full article

 New to topics? Read the docs here!