A simply supported beam of span '5 m' has a moment '100 kN-m' acting at mid span. The reaction at each support is:

Concept:

In a simply supported beam subjected to a pure moment, the reactions at the supports are equal and opposite, ensuring equilibrium. The beam span is given as 5 meters, with a moment of 100 kN-m applied at mid-span.

Vertical Force Equilibrium:

For a simply supported beam, the sum of vertical forces must equal zero because no vertical forces are applied:

\( R_P + R_Q = 0 \)

This implies that:

\( R_P = -R_Q \)

Moment Equilibrium:

Taking moments about point P, the equation becomes:

\( \mu - R_Q \cdot 5 = 0 \)

Substituting the applied moment:

\( 100 - R_Q \cdot 5 = 0 \)

Solving for \( R_Q \):

\( R_Q = \frac{100}{5} = 20 \, \text{kN} \)

Calculate Reaction at P:

Using the relationship \( R_P = -R_Q \), we get:

\( R_P = -20 \, \text{kN} \)

Conclusion:

The reactions at the supports are:

• \( R_P = -20 \, \text{kN} \) (acting downward)
• \( R_Q = 20 \, \text{kN} \) (acting upward)