Math Problem Statement

Determine the total vertical load acting on the beam and use moments to determine the two vertical support reactions RA and RB acting at positions A and B, given a UDL of 5 kN/m and point loads of 4 kN and 3 kN.

Solution

Let's solve the problem step by step.


(i) Total Vertical Load Acting on the Beam

  1. Uniformly Distributed Load (UDL):

    • UDL = 5kN/m5 \, \text{kN/m}
    • UDL length = 6m6 \, \text{m} (from the diagram: 2m+4m2 \, \text{m} + 4 \, \text{m})
    • Total UDL = 5×6=30kN5 \times 6 = 30 \, \text{kN}

    The UDL acts at its center, which is at the midpoint of the 6 m span. The center point is located: 2m+62=5m from A.2 \, \text{m} + \frac{6}{2} = 5 \, \text{m} \text{ from A}.

  2. Point Loads:

    • Point load 4kN4 \, \text{kN} is located 6m6 \, \text{m} from A.
    • Point load 3kN3 \, \text{kN} is located 7.4m7.4 \, \text{m} from A.

    Total vertical load: Total Load=UDL+Point Loads=30+4+3=37kN.\text{Total Load} = \text{UDL} + \text{Point Loads} = 30 + 4 + 3 = 37 \, \text{kN}.


(ii) Determining Support Reactions (RAR_A and RBR_B)

We will use the equilibrium equations:

  1. Fy=0\sum F_y = 0: The sum of vertical forces equals zero.
  2. MA=0\sum M_A = 0: The sum of moments about point A equals zero.

Step 1: Taking Moments About A

To calculate RBR_B, take moments about A. Clockwise moments are positive: MA=0\sum M_A = 0

Moments due to all loads:

  1. UDL:

    • Total UDL = 30kN30 \, \text{kN}
    • Acts at a distance of 5m5 \, \text{m} from A.
    • Moment = 30×5=150kN\cdotpm30 \times 5 = 150 \, \text{kN·m}.
  2. Point Load 4kN4 \, \text{kN}:

    • Acts at 6m6 \, \text{m} from A.
    • Moment = 4×6=24kN\cdotpm4 \times 6 = 24 \, \text{kN·m}.
  3. Point Load 3kN3 \, \text{kN}:

    • Acts at 7.4m7.4 \, \text{m} from A.
    • Moment = 3×7.4=22.2kN\cdotpm3 \times 7.4 = 22.2 \, \text{kN·m}.
  4. Reaction at B (RBR_B):

    • RBR_B acts at 8.6m8.6 \, \text{m} from A.
    • Moment = RB×8.6R_B \times 8.6.

Equilibrium equation for moments: RB×8.6=150+24+22.2R_B \times 8.6 = 150 + 24 + 22.2 RB×8.6=196.2R_B \times 8.6 = 196.2 RB=196.28.6=22.81kN.R_B = \frac{196.2}{8.6} = 22.81 \, \text{kN}.


Step 2: Using Fy=0\sum F_y = 0 to Find RAR_A

The sum of vertical forces is zero: RA+RB=Total LoadR_A + R_B = \text{Total Load} RA+22.81=37R_A + 22.81 = 37 RA=3722.81=14.19kN.R_A = 37 - 22.81 = 14.19 \, \text{kN}.


Final Results:

  1. Total Vertical Load: 37kN37 \, \text{kN}
  2. Reactions:
    • RA=14.19kNR_A = 14.19 \, \text{kN}
    • RB=22.81kNR_B = 22.81 \, \text{kN}

Do you want a detailed explanation of any step, or help with further concepts?


Related Questions:

  1. How do you calculate the location of the resultant force for a UDL?
  2. What is the significance of moments in equilibrium problems?
  3. How would the reactions change if the UDL intensity increased to 10kN/m10 \, \text{kN/m}?
  4. Can you explain how to verify equilibrium conditions for this problem?
  5. How are distributed loads handled differently from point loads in calculations?

Tip: When solving for reactions, always double-check the distances and ensure all forces and moments are accounted for correctly.

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Math Problem Analysis

Mathematical Concepts

Static equilibrium
Moments
Vertical forces
Uniformly distributed load (UDL)
Point loads

Formulas

Total vertical load = Sum of all loads
Moment about a point = Force × Perpendicular distance
Sum of vertical forces (ΣFy) = 0
Sum of moments about a point (ΣM) = 0

Theorems

Equilibrium of rigid bodies

Suitable Grade Level

Grades 10-12