Question

Derive an expression for the depression at the free end of a cantilever due to load.

Answer 1

Explanation:

Answer in the attachment

Answer 2

Explanation:

EXPRESSION FOR THE DEPRESSION OF THE LOADED END OF A

CANTILEVER

A cantilever is a beam fixed horizontally at one end and loaded at the other end.

AB represents the neutral axis of a cantilever of length.

The end A is fixed. The end B is loaded with a weight W.

The end B is displaced to the position B. The neutral axis of the cantilever shifts to the

position AB.

Let the depression BB of the free end be y.

Consider an element PQ, at a distance x from the end A and of radius of curvature R

Bending moment = $\frac{EI_g}{R}$

Deflecting couple = W ( l - x)

For equilibrium,

$\frac{EI_g}{R}$ = W ( l - x)

But, $\frac{1}{R} = \frac{d^2y}{dx^2}$

$\frac{d^2y}{dx^2} = \frac{W(l - x) }{El_g}$

Integrating

$\frac{dy}{dx} = \frac{W}{EI_g} (lx - \frac{x^2}{2} )$+ C₁

$\frac{dy}{dx} = 0$

Therefore

C₁ =0

$\frac{dy}{dx} = \frac{W}{EI_g} (lx - \frac{x^2}{2} )$

Integrating

$y = \frac{W}{EI_g} \frac{lx^2}{2} - \frac{x^3}{6}$+ C₂

When,

X = 0, y = 0

C₂ = 0

y =  $\frac{W}{EI_g} \frac{lx^2}{2} - \frac{x^3}{6}$

For the depression of the free end, x = l.

y =  $\frac{W}{EI_g} \frac{l^3}{2} - \frac{l^3}{6}$

y = $\frac{Wl^3}{3EI_g}$