Question

Determine the force and moment reactions at the support A of the built-in beam which is subjected to the sine-wave load distribution. The force reaction RA is positive if upward, negative if downward. The moment reaction MA is positive if counterclockwise, negative if clockwise.

Answer 1

Answer:

$R_A=\frac{2w_0l}{\pi}$

$M_A=\frac{w_0l^2}{\pi}$

Explanation:

The beam is subjected to the sine-wave load distribution as shown in the figure.

As the beam is in equilibrium condition, so net force and moment in any direction are zero.

Assuming the length, $l$, of the beam is along the x-axis and the loading direction is along the y-axis.

The load density, w, per unit length, at a distance of x from the point A, for the sine-wave load is

$w=w_0\sin\left(\frac{\pi}{l}x\right)$,

where $w_0$ is constant (maximum load density)

$R_A$ is positive if upward, so w is negative as it is acting in the downward direction.

A small force, dF, in the downward direction, due to load on a small element dx at a distance of x from the point A is

$dF=wdx$ in the downward direction

$\Rightarrow dF=-w_0\sin\left(\frac{\pi}{l}x\right)dx\cdots(i)$

The moment, dM about point A, due to small force, dF, is

$dM=(dF)x$

As the moment in the clockwise direction is negative, so

$dM=-(dF)x \cdots(ii)$

$\Rightarrow dM=w_0\sin\left(\frac{\pi}{l}x\right)xdx\cdots(i)$

At equilibrium state, net force along the y-direction will be zero, i.e

$\Sigma F_y=0$

$\Rightarrow R_A+\int_{0}^{l}dF=0$

$\Rightarrow R_A=-\int_{0}^{l}dF\cdots(iii)$

From equation (i)

$\int_{0}^{l}dF=\int_{0}^{l}w_0\sin\left(\frac{\pi}{l}x\right)dx$

$\Rightarrow F=\int_{0}^{l}w_0\sin\left(\frac{\pi}{l}x\right)dx$

$=-\left[\frac{lw_0}{\pi}\cos\left(\frac{\pi}{l}x\right)\right]_0^l$

$=-\frac{l}{\pi}w_0(-1-1)$

$\Rightarrow F=\frac{2w_0l}{\pi}\cdots(iv)$

The center of F is at the centroid of the sine-curve in the downward direction.

Putting this value in the equation (iii), we have

$R_A=\frac{2w_0l}{\pi}$

Again, at the equilibrium state, net force along the y-direction will be zero, i.e

$M_A+\int_{0}^{l}dM=0$

$\Rightarrow M_A-\int_{0}^{l}(dF)x=0$  [from (ii)]

$\Rightarrow M_A=\int_{0}^{l}\left\(dF)x$

$\Rightarrow M_A=F \bar{x}$

Where $\bar{x}$ is the x-coordinate of the centroid.

Due to symmetry, $\bar{x}=\frac l 2$

So, $M_A=F\times \frac l 2$

$\Rightarrow M_A=\frac{2w_0l}{\pi}\times \frac l 2$ [ using (iv)]

$\Rightarrow M_A=\frac{w_0l^2}{\pi}$

Hence, the reaction force and the moment at point A are

$R_A=\frac{2w_0l}{\pi}$

$M_A=\frac{w_0l^2}{\pi}$