Cantilever, Supported Beam and Relation between Sagitta and Curvature

A horizontal beam whose one end is fixed and load is applied to the other end is called a cantilever.

Expression for depression on the lever(when the weight of beam is not taken into account)

Consider a horizontal beam AB of length 'l' fixed at the end 'A' and a load 'mg' be applied at the free end B so that the depression of neutral surface is 'y' at B.

Let, l-x part of the beam undergoes depression while the length 'x' of the beam remains undepressed. In this case, a force 'mg' equal to the applied force acts vertically upward at point P. Thus, the length l-x of the beam under the action of two equal but opposite parallel forces which forms couple and tries to rotate the beam.

Moment of the force mg on (l-x) part of the beam is given by

\begin{align*}moment=mg(l-x) \end{align*}

Also, the bending moment of the beam due to the applied force is given by\begin{align*}B=\frac{Y}{R}I_g \end{align*}where, Y=Young's modulus of elasticity of material of the beam

R= Radius of curvature of bending plane

\(I_g \)=Geometrical moment of inertia of cross-section.

At equilibrium, the moment of the beam should be balanced by the bending moment of the beam\begin{align*}\frac{Y}{R}I_g=mg(l-x) \end{align*}\begin{align*}or,\space \frac{1}{R}=\frac{mg}{YI_g}(l-x)\rightarrow 1 \end{align*}From figure, y=f(x), thus the radius of curvature is defined by the relation,\begin{align*}\frac{1}{R}=\frac{y"}{(1+(y')^2)^{3/2}} \end{align*}

In case of rigid body, \(\frac{dy}{dx}\) i.e. the slope of the tangent at a point P is very small. So we can neglect the higher powers of\(\frac{dy}{dx}\). Then, \begin{align*}\frac{1}{R}=y" \end{align*}Thus, from 1,\begin{align*}\frac{d^2y}{dy^2}=\frac{mg}{YI_g}(l-x) \end{align*}On integration,\begin{align*}\frac{dy}{dx}=\frac{mg}{YI_g}(lx-\frac{x^2}{2})+c\rightarrow 2 \end{align*},\begin{align*}At\space x=0,\frac{dy}{dx}=0\implies c=0 \end{align*}Then from 2,\begin{align*}\frac{dy}{dx}=\frac{mg}{YI_g}(lx-\frac{x^2}{2}) \end{align*}

Again, integrating, we get\begin{align*}y-\frac{mg}{YI_g}[\frac{lx^2}{2}-\frac{x^3}{6}]+c\rightarrow 3 \end{align*}Again at x=0, y=0\(\implies c=0\)

From equation 3,\begin{align*}y=\frac{mg}{YI_g}[\frac{lx^2}{2}-\frac{x^3}{6}] \end{align*}The depression of beam \(\delta \) at the end is of the beam, where x=l is\begin{align*}\delta =\frac{mg}{YI_g}[\frac{l^3}{2}-\frac{l^3}{6}] \end{align*}\begin{align*}\implies \delta =\frac{mgl^3}{3YI_g} \end{align*}

\(\therefore \) This is the required expression for the depression on cantilever.

Expression for the depression of the lever (when the weight of beam is taken into account)

Considering again the section PB of distance x from the fixed end, we have in addition to the load acting at B, a weight equal to the length of portion (L-x) of the beam also acting at its mid-point. So, that if w be the weight per unit length of the beam, we have an additional weight w(L-x) acting at a distance (L-x)/2 from P.
Therefore, total moment of external couple applied \begin{align*}=W(L-x)+w(L-x).(L-x)/2 \end{align*}\begin{align*}W(L-x)+\frac{w}{2}(L-x)^2 \end{align*}The beam being in equilibrium, this must be balanced by bending moment \(\frac{YI_g}{R}\).
\begin{align*}W(L-x)+w(L-x)^2=\frac{YI_g}{R}=YI_g(\frac{d^2y}{dx^2}) \end{align*}which on integration gives\begin{align*}YI_g\frac{dy}{dx}=W(Lx-\frac{x^2}{2})+\frac{w}{2}(L^2x-2L\frac{x^2}{2}+\frac{x^3}{3})+c \end{align*}where c is constant of integration.
Since at x=0,\(\frac{dy}{dx}=0\), we have c=0. So that,\begin{align*}YI_g\int_0^ydy=W\int_0^L(Lx-\frac{x^2}{2})dx+\frac{w}{2}\int_0^L(L^2x-2L\frac{x^2}{2}+\frac{x^3}{3})dx \end{align*}\begin{align*}=\frac{WL^3}{3}+\frac{wL^4}{8} \end{align*}Since,\(wL=W_0\), the weight of the beam, we have\begin{align*}YI_gy=\frac{WL^3}{3}+\frac{W_0L^3}{8} \end{align*}\begin{align*}hence,\space y=(W=\frac{3}{8}W_0)\frac{L^3}{3YI_g} \end{align*}Thus the beam now acts as though the load (W) at its free end (B) is increased by \(\frac{3}{8}\) of its own weight.

Supported Beam:

A beam whose two ends are supported by two rigid supports is called a supported beam. Let PQ be a supported beam of length L. A load 'mg' is applied at its middle. If we consider only half part of the beam, it is like cantilever. Thus, the depression produced at O is given by \begin{align*}\delta =\frac{(mg)'.{l'}^3}{3YI_g}\space (mg)'=\frac{mg}{2}\space and\space l'=\frac{L}{2}\end{align*}\begin{align*}Thus,\space \delta =\frac{mgL^3}{2.8.3YI_g} \end{align*}\begin{align*}\implies \delta = \frac{mgL^3}{48YI_g}\end{align*}

Sagitta:

The perpendicular bisector of the chord and arc. In fig, OX is the sagitta.

Curvature:

Reciprocal of radius of curvature of an arc. If R be the radius of curvature then \(curvature=\frac{1}{R}\)\begin{align*}curvature=\frac{2\times sagitta}{(semi-chord)^2}\end{align*}\begin{align*}\implies \frac{1}{R} =\frac{2\times (OX)}{(PO)^2}\end{align*}

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Feynman, Richard P. The Feynman Lectures on Physics Volume 1. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.

Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.

Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.