In the case of a simply supported beam with a concentrated load at the center, the reactions at the supports can be determined using the principles of static equilibrium. The conditions for static equilibrium are:

1. The sum of vertical forces is zero:
   $\Sigma F_y = 0$

2. The sum of moments about any point is zero:
   $\Sigma M = 0$

Given:

• Length of the beam $L = 3 \, \text{m}$
• Concentrated load at the center $P = 52 \, \text{kN}$
• Let the reactions at supports $A$ and $B$ be $R_A$ and $R_B$, respectively.

Step 1: Use equilibrium of moments to find reactions

Take moments about point $A$: $\Sigma M_A = 0 \implies R_B \times L - P \times \left(\frac{L}{2}\right) = 0$ Substitute the known values: $R_B \times 3 - 52 \times \left(\frac{3}{2}\right) = 0$ $R_B \times 3 = 52 \times 1.5$ $R_B = \frac{78}{3} = 26 \, \text{kN}$

Step 2: Use equilibrium of vertical forces to find $R_A$

From $\Sigma F_y = 0$: $R_A + R_B = P$ Substitute the known value of $R_B$: $R_A + 26 = 52$ $R_A = 52 - 26 = 26 \, \text{kN}$

Final Answer:

The upward reactions at both supports are:

• $R_A = 26 \, \text{kN}$
• $R_B = 26 \, \text{kN}$

Would you like a detailed explanation of the moment calculations or have any questions?

Here are five related questions to deepen understanding:

1. How would the reactions change if the load was applied off-center?
2. What would happen to the reactions if the beam had a uniform distributed load?
3. How can we calculate the deflection of the beam under this load?
4. What is the significance of the moment at the center of the beam?
5. How does the span length affect the reactions in a simply supported beam?

Tip: Always check for symmetry in beam loading conditions, as it simplifies calculations for reactions and moments.