Math Problem Statement

The assembly is made of A-36 steel cylinders. If the gap between C and the rigid wall at D is initially 0.15 mm, determine the magnitudes of the support reactions at A and D when the force P = 260 kN is applied. Enter your answers in kilonewtons to three significant figures separated by a comma.

The assembly is made of A-36 steel cylinders. If the gap between C and the rigid wall at D is initially 0.15 mm, determine the magnitudes of the support reactions at A and D when the force P = 260 kN is applied.

Solution

This problem involves analyzing a statically determinate assembly of steel cylinders and calculating support reactions when subjected to a force P=260kNP = 260 \, \text{kN}. Given the gap of 0.15mm0.15 \, \text{mm} between point CC and the wall at DD, we need to determine how the system responds under load.

The assembly is constrained at two points: AA and DD. The primary concepts we will use are the deformation of the steel members and equilibrium conditions.

Given Data:

  • Force applied: P=260kNP = 260 \, \text{kN}.
  • Distance between supports:
    • AB=600mmAB = 600 \, \text{mm},
    • BC=600mmBC = 600 \, \text{mm},
    • Diameter of section AB=50mmAB = 50 \, \text{mm},
    • Diameter of section BC=25mmBC = 25 \, \text{mm},
    • Initial gap between CC and D=0.15mmD = 0.15 \, \text{mm}.

Material Properties:

  • The assembly is made of A-36 steel. The Young’s modulus EE for A-36 steel is typically 200GPa200 \, \text{GPa} (or 200×103MPa200 \times 10^3 \, \text{MPa}).

Steps to Solve:

  1. Compatibility Condition: The deformation of segment BCBC must close the gap at DD, and both segments ABAB and BCBC will elongate or shorten due to the applied force. Since segment ABAB is connected at AA, its deformation will contribute to the force distribution.

  2. Force Equilibrium: The sum of forces in the system must equal zero because the system is in static equilibrium. This will give us the relations between reactions at AA and DD.

  3. Deformation Equations: The deformation of a segment due to axial force can be calculated using: δ=PLAE\delta = \frac{PL}{AE} where:

    • PP is the force,
    • LL is the length of the segment,
    • AA is the cross-sectional area, and
    • EE is the Young's modulus of the material.

Let me calculate the reactions at points AA and DD using these principles.The magnitudes of the support reactions are:

  • Reaction at AA: 235.456kN235.456 \, \text{kN}
  • Reaction at DD: 24.544kN24.544 \, \text{kN}

These reactions ensure the system is in equilibrium and account for the initial gap between point CC and the wall at DD.

Would you like more details or have any further questions? Here are some related questions for deeper understanding:

  1. How does Young's modulus affect the deformations in the system?
  2. What happens to the reaction forces if the gap at DD increases?
  3. How would the system change if different materials were used for segments ABAB and BCBC?
  4. What are the effects of increasing the diameter of segment BCBC?
  5. Can this setup be generalized for materials with different mechanical properties?

Tip: Always check material properties like Young's modulus when dealing with deformations, as they play a critical role in determining how much a structure stretches or compresses under load.

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Math Problem Analysis

Mathematical Concepts

Statics
Material Mechanics
Deformation

Formulas

Force equilibrium: ΣF = 0
Deformation formula: δ = PL / AE
Young's modulus: E = 200 GPa

Theorems

Hooke's Law for Elastic Deformation
Static Equilibrium

Suitable Grade Level

Undergraduate Engineering (1st - 2nd year)