Math Problem Statement

solve for the reactions at the supports in this structure

From the figure shown, solve for the reactions at the supports and check equilibrium.

Solution

Step 1: Analyzing the Structure

The given structure is a simply supported beam with various types of loads:

  1. Distributed Load: From AA to BB (left), with a magnitude of 3 kN/m over 3 meters.
  2. Point Loads: Located at different intervals:
    • 6 kN at 3 meters from AA.
    • 10 kN at 5 meters from AA.
    • 10 kN at 7 meters from AA.
  3. Triangular Load: From 9 to 12 meters with a maximum intensity of 5 kN/m.

The reaction forces at supports are:

  • RAR_A at AA (vertical).
  • RBxR_{Bx} and RBR_B at BB.

Step 2: Calculating the Reaction Forces

  1. Calculate the total load of the distributed load: w=3kN/m,Length=3mF1=w×L=3×3=9kNw = 3 \, \text{kN/m}, \quad \text{Length} = 3 \, \text{m} \quad \Rightarrow \quad F_1 = w \times L = 3 \times 3 = 9 \, \text{kN} This acts at the centroid, i.e., 1.5 meters from point AA.

  2. For the triangular load: The triangular load extends from 9 to 12 meters with a maximum value of 5 kN/m. The total equivalent load is: F2=12×Base×Height=12×3m×5kN/m=7.5kNF_2 = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 3 \, \text{m} \times 5 \, \text{kN/m} = 7.5 \, \text{kN} The location of this load is 13\frac{1}{3} of its base from the larger end, i.e., 1 meter from BB.

Step 3: Applying Equilibrium Conditions

The structure is in static equilibrium, so the sum of forces and moments should be zero.

  1. Sum of vertical forces: RA+RB=9+6+10+10+7.5=42.5kNR_A + R_B = 9 + 6 + 10 + 10 + 7.5 = 42.5 \, \text{kN}

  2. Sum of moments about AA: Taking moments about AA: RB×12=9×1.5+6×3+10×5+10×7+7.5×11R_B \times 12 = 9 \times 1.5 + 6 \times 3 + 10 \times 5 + 10 \times 7 + 7.5 \times 11

    12RB=13.5+18+50+70+82.5=234kNm12R_B = 13.5 + 18 + 50 + 70 + 82.5 = 234 \, \text{kN} \cdot \text{m} Solving for RBR_B: RB=23412=19.5kNR_B = \frac{234}{12} = 19.5 \, \text{kN}

  3. Solving for RAR_A: RA=42.519.5=23kNR_A = 42.5 - 19.5 = 23 \, \text{kN}

Step 4: Final Reaction Forces

  • Reaction at AA: RA=23kNR_A = 23 \, \text{kN}
  • Reaction at BB: RB=19.5kNR_B = 19.5 \, \text{kN}

Step 5: Checking Equilibrium

The moments and forces are balanced, confirming the structure is in equilibrium.


Do you need more details or have any questions?

Additional Questions:

  1. How would the reaction forces change if the triangular load had a steeper slope?
  2. What is the effect of changing the location of point loads on the reactions?
  3. How can we verify the results using shear and moment diagrams?
  4. What happens to the reactions if the beam length is increased?
  5. What would be the change in reactions if a moment is applied at the midpoint?

Tip:

When calculating moments, always ensure the correct distance is used relative to the chosen pivot point.

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Math Problem Analysis

Mathematical Concepts

Statics
Equilibrium of Forces
Moment of Forces
Distributed Loads
Point Loads
Triangular Loads

Formulas

Sum of Vertical Forces: ∑Fy = 0
Sum of Moments: ∑M = 0
Moment about a Point: M = F * d
Equivalent Load of Distributed Load: F = w * L
Equivalent Load of Triangular Load: F = 1/2 * base * height

Theorems

Equilibrium Conditions: For a structure in static equilibrium, the sum of forces and the sum of moments must both equal zero.

Suitable Grade Level

Undergraduate Civil Engineering