Question

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

Answer 1

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

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

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.

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

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