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

`R_A=(2w_0l)/(\pi)`

`M_A=(w_0l^2)/(\pi)`

Step-by-step 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((\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((\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((\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((\pi)/(l)x\right)dx`

`\Rightarrow F=\int_(0)^(l)w_0\sin\left((\pi)/(l)x\right)dx`

`=-\left[(lw_0)/(\pi)\cos\left((\pi)/(l)x\right)\right]_0^l`

`=-(l)/(\pi)w_0(-1-1)`

`\Rightarrow F=(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=(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* \frac l 2`

`\Rightarrow M_A=(2w_0l)/(\pi)* \frac l 2` [ using (iv)]

`\Rightarrow M_A=(w_0l^2)/(\pi)`

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

`R_A=(2w_0l)/(\pi)`

`M_A=(w_0l^2)/(\pi)`