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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.

User TMKasun
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1 Answer

3 votes

Answer:


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)

Determine the force and moment reactions at the support A of the built-in beam which-example-1
User Colonel Panic
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