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A simply supported beam of length 10 units has a point load of 6 units at its center. Calculate the reactions at the supports A and B.

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The image shows a simply supported beam with a point load at its center.

  • The point load at C is 6 units.
  • The distance from A to C is 5 units.
  • The distance from C to B is 5 units.
  • The total length of the beam is 5+5=105 + 5 = 10 units.

Your handwritten calculations show RA+RB=10R_A + R_B = 10. However, the diagram clearly indicates a single downward load of 6 units. For the beam to be in static equilibrium, the sum of upward reactions must equal the sum of downward loads. Therefore, RA+RBR_A + R_B should be equal to 6, not 10.

I will proceed by assuming the point load shown in the diagram (6 units) is correct. This means the sum of vertical forces must be equal to this load (RA+RB=6R_A + R_B = 6). This will result in different reaction forces for RAR_A than you calculated.

  1. Calculate the Reactions at Supports A and B

    Let RAR_A be the reaction force at support A and RBR_B be the reaction force at support B.

    • Sum of vertical forces (∑Fy=0\sum F_y = 0):

      RA+RB−6=0R_A + R_B - 6 = 0 RA+RB=6(∗)R_A + R_B = 6 \quad (*)

    • Sum of moments about point A (∑MA=0\sum M_A = 0): (Assume counter-clockwise moments are positive)

      (RB×10)−(6×5)=0(R_B \times 10) - (6 \times 5) = 0 10RB−30=010 R_B - 30 = 0 10RB=3010 R_B = 30 RB=3010R_B = \frac{30}{10} RB=3 unitsR_B = 3 \text{ units}

    • Substitute RBR_B into equation (∗)(*) to find RAR_A:

      RA+3=6R_A + 3 = 6 RA=6−3R_A = 6 - 3 RA=3 unitsR_A = 3 \text{ units}

    So, the correct reaction forces are RA=3R_A = 3 units and RB=3R_B = 3 units.

  2. Calculate the Shear Force (V(x)V(x))

    We will analyze the beam in two sections:

    1. From A to C (0≤x<50 \le x < 5)
    2. From C to B ($5 < x \le 10
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