Math Problem Statement

calculate the deflection at points B and C using Area Moment Method. Given that Reaction at A is 2.11kN and Reaction at D is 5.89kN

Calculate the deflection at points B and C using Area Moment Method. Given that Reaction at A is 2.11 kN and Reaction at D is 5.89 kN.

Solution

To calculate the deflection at points B and C using the Area Moment Method, we will break the problem into steps. Here's how we proceed:

Step 1: Analyze the beam

The beam has:

  • A moment of 5kN-m5 \, \text{kN-m} at point B.
  • A point load of 8kN8 \, \text{kN} at point C.
  • Supports at points A and D.

The distances are:

  • AB=3mAB = 3 \, \text{m}
  • BC=3mBC = 3 \, \text{m}
  • CD=3mCD = 3 \, \text{m}

Reactions at supports:

  • Reaction at A, RA=2.11kNR_A = 2.11 \, \text{kN}
  • Reaction at D, RD=5.89kNR_D = 5.89 \, \text{kN}

Step 2: Bending moment diagram

For this method, the bending moment diagram is essential. The bending moments can be calculated as follows:

Section AB (0 ≤ x ≤ 3):

  • The bending moment at point A (x = 0) is 00 since it's a pinned support.
  • The moment at point B (x = 3 m): MB=RA×35=2.11×35=6.335=1.33kN-mM_B = R_A \times 3 - 5 = 2.11 \times 3 - 5 = 6.33 - 5 = 1.33 \, \text{kN-m}

Section BC (3 ≤ x ≤ 6):

  • At point C, just before the 8 kN force, the moment is: MC=MB8×3=1.3324=22.67kN-mM_C = M_B - 8 \times 3 = 1.33 - 24 = -22.67 \, \text{kN-m}

Step 3: Calculate the areas under the bending moment diagram

Using the bending moment diagram, divide the diagram into two key areas:

Area between A and B:

  • This is a triangle with height MB=1.33kN-mM_B = 1.33 \, \text{kN-m} and base 3m3 \, \text{m}. A1=12×1.33×3=1.995kN-m2A_1 = \frac{1}{2} \times 1.33 \times 3 = 1.995 \, \text{kN-m}^2 The centroid of this triangle is 23×3=2m\frac{2}{3} \times 3 = 2 \, \text{m} from point A.

Area between B and C:

  • This is a trapezoid with heights MB=1.33kN-mM_B = 1.33 \, \text{kN-m} and MC=22.67kN-mM_C = -22.67 \, \text{kN-m}, and a base 3m3 \, \text{m}. A2=12×(1.33+(22.67))×3=12×(21.34)×3=32.01kN-m2A_2 = \frac{1}{2} \times (1.33 + (-22.67)) \times 3 = \frac{1}{2} \times (-21.34) \times 3 = -32.01 \, \text{kN-m}^2 The centroid of this trapezoid is calculated based on the shape, but generally lies slightly closer to the larger moment (closer to C).

Step 4: Deflection calculations using the Area Moment Method

To calculate the deflection at points B and C:

  • Deflection at B: Use the area A1A_1 and the centroidal distance for point B to calculate the deflection.

  • Deflection at C: Use areas A1A_1 and A2A_2 along with their respective centroidal distances to calculate the deflection at C.

We also need the modulus of elasticity EE and the moment of inertia II of the beam to finish the calculation.

Would you like to provide the values for EE and II, or should we assume standard values to continue the calculations?


Follow-up Questions:

  1. What is the modulus of elasticity EE and moment of inertia II for the beam?
  2. Would you like to see the detailed bending moment diagram?
  3. Do you need assistance with drawing the shear force diagram?
  4. Would you like the final deflection values once we have EE and II?
  5. Would you prefer this explanation in another method like the conjugate beam method?

Tip: In structural analysis, always confirm boundary conditions and load distributions before starting calculations.

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Math Problem Analysis

Mathematical Concepts

Structural Analysis
Beam Deflection
Area Moment Method

Formulas

M(x) = Reaction * distance - applied moments
Area of a triangle: A = (1/2) * base * height
Deflection: Δ = (Area * centroidal distance) / (E * I)

Theorems

Area Moment Theorem
Principle of Superposition in Structural Analysis

Suitable Grade Level

College/University Level - Civil or Mechanical Engineering