Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 13C / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii
- iii b
  - Solution iii
- iv b
  - Solution iv
- v b
  - Solution v

i

Solution

Here

F (x, y, y^{'}, y^{''}) = \frac{A}{2} (y^{''})^{2} + ρ g y .

Substitution into the higher-order Euler-Lagrange equation gives the fourth-order ordinary differential equation

A y^{''''} + ρ g = 0 .

ii

Solution

Integrating the ordinary differential equation four times gives a quartic polynomial. It is useful to use the linearity of the equation and split the solution into a gravity part and a force part:

y = y_{0} + y_{F},

where

y_{0} (x) = - \frac{ρ g}{24 A} x^{2} (6 L^{2} - 4 Lx + x^{2})

satisfies the clamped and torque-free boundary conditions with

F = 0

, while

y_{F} (x) = \frac{F}{6 A} x^{2} (3 L - x)

satisfies the homogeneous equation and contributes the endpoint force. Direct differentiation verifies

y (0) = y^{'} (0) = 0, y^{''} (L) = 0, - A y^{'''} (L) = F .

Thus

y (x) = - \frac{ρ g}{24 A} x^{2} (6 L^{2} - 4 Lx + x^{2}) + \frac{F}{6 A} x^{2} (3 L - x),

the clamped-free beam under uniform load and endpoint force.

iii

Solution

Evaluating the function at the endpoint gives the displacement

Δ = y (L) = - \frac{ρ g L ^{4}}{8 A} + \frac{F L ^{3}}{3 A} .

Therefore

h_{0} = - \frac{ρ g L ^{4}}{8 A}, h = \frac{F L ^{3}}{3 A}, Δ = h_{0} + h .

The force-induced displacement

h

is a linear map of

F

.

iv

Solution

Substitute

y = y_{0} + y_{F}

into the energy functional and integrate the resulting polynomial:

E = \int_{0}^{L} [\frac{A}{2} (y^{''})^{2} + ρ g y] d x = - \frac{ρ ^{2} g ^{2} L ^{5}}{40 A} + \frac{F ^{2} L ^{3}}{6 A} .

The term linear in

F

cancels. Hence

E_{0} = - \frac{ρ ^{2} g ^{2} L ^{5}}{40 A}, E = \frac{F ^{2} L ^{3}}{6 A}, E = E_{0} + E .

As required,

E_{0}

is independent of the force and

E

is a quadratic function of it.

v

Solution

Taking the derivative of the force-dependent energy gives

\frac{d E}{d F} = \frac{F L ^{3}}{3 A} = h .

Thus the derivative of the additional minimized internal energy with respect to the applied force equals the resulting endpoint displacement. This is the endpoint force derivative of clamped-free beam energy.

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