Question

A wooden beam 6 m long, simply supported at its ends, is carrying appoint load of 7.8 KN at its centre. The cross-section of the beam is 150 mm wide and 240 mm deep. if E for the beam is 6 x 103 N/m2 ,Find the deflection at the centre.

Answer 1

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From the fig., the depression at the center of the beam is given by Pythagoras' Theorem as,

$\longrightarrow\sf{d=\sqrt{\left(\dfrac{L+\Delta L}{2}\right)^2-\left(\dfrac{L}{2}\right)^2}}$

Since $\sf{\Delta L}$ is so small,

$\longrightarrow\sf{d=\sqrt{\dfrac{L^2+2L\,\Delta L}{4}-\dfrac{L^2}{4}}}$

$\longrightarrow\sf{d=\sqrt{\dfrac{L\,\Delta L}{2}}}$

$\longrightarrow\sf{d^2=\dfrac{L\,\Delta L}{2}}$

$\longrightarrow\sf{\Delta L=\dfrac{2d^2}{L}\quad\quad\dots(1)}$

From the fig., we see that,

$\longrightarrow\sf{\cos\theta=\dfrac{2d}{L+\Delta L}}$

From (1),

$\longrightarrow\sf{\cos\theta=\dfrac{2d}{\left(L+\dfrac{2d^2}{L}\right)}}$

We consider $\sf{d}$ to be very small, so that $\sf{d^2}$ can be neglected.

$\longrightarrow\sf{\cos\theta=\dfrac{2d}{L}\quad\quad\dots(2)}$

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The area of cross section along the deforming force $\sf{F}$ is,

$\longrightarrow\sf{A'=2A\cos\theta}$

From (2),

$\longrightarrow\sf{A'=\dfrac{4Ad}{L}}$

The elasticity of the beam is given by,

$\longrightarrow\sf{E=\dfrac{FL}{A'\,\Delta L}}$

Substituting values for $\sf{A'}$ and $\sf{\Delta L,}$

$\longrightarrow\sf{E=\dfrac{FL}{\left(\dfrac{4Ad}{L}\cdot\dfrac{2d^2}{L}\right)}}$

$\longrightarrow\sf{E=\dfrac{FL^3}{8Ad^3}}$

Hence the depression at the center is given by,

$\longrightarrow\sf{d^3=\dfrac{FL^3}{8AE}}$

$\longrightarrow\underline{\underline{\sf{d=\dfrac{L}{2}\left[\dfrac{F}{AE}\right]^{\frac{1}{3}}}}}$

According to the question,

• $\sf{L=6\ m}$

• $\sf{F=7.8\times10^{3}\ N}$

• $\sf{E=6\times10^{3}\ N\,m^{-2}}$

Area of cross section of the beam is,

$\longrightarrow\sf{A=150\times10^{-3}\times240\times10^{-3}\ m^2}$

$\longrightarrow\sf{A=3.6\times10^{-2}\ m^2}$

Hence the depression is,

$\longrightarrow\sf{d=\dfrac{L}{2}\left[\dfrac{F}{AE}\right]^{\frac{1}{3}}}$

$\longrightarrow\sf{d=\dfrac{6}{2}\left[\dfrac{7.8\times10^{3}}{3.6\times10^{-2}\times6\times10^3}\right]^{\frac{1}{3}}}$

$\longrightarrow\underline{\underline{\sf{d=9.92\ m}}}$