Math Problem Statement

The rigid bar AB, attached to two vertical rods as shown in the figure, is horizontal before the load is applied. If the load P = 70 kN, determine its vertical movement.

Solution

The problem you have provided shows a rigid bar ABAB attached to two vertical rods, one made of steel and the other made of aluminum. The bar is initially horizontal before a load P=70kNP = 70 \, \text{kN} is applied vertically downward at the center of the bar. We are tasked with determining the vertical movement of the bar.

Let's summarize the key information for both materials:

  • Steel rod:

    • Length Ls=4mL_s = 4 \, \text{m}
    • Cross-sectional area As=300mm2A_s = 300 \, \text{mm}^2
    • Modulus of elasticity Es=200GPaE_s = 200 \, \text{GPa}
  • Aluminum rod:

    • Length La=5mL_a = 5 \, \text{m}
    • Cross-sectional area Aa=500mm2A_a = 500 \, \text{mm}^2
    • Modulus of elasticity Ea=70GPaE_a = 70 \, \text{GPa}

Approach:

  1. Assumption of compatibility: The rigid bar stays horizontal after the load is applied, so the vertical displacement of both rods at points A and B must be the same.

  2. Force distribution: The load PP will be distributed between the steel and aluminum rods. Let's denote the force in the steel rod as FsF_s and in the aluminum rod as FaF_a. Therefore: Fs+Fa=P=70kNF_s + F_a = P = 70 \, \text{kN}

  3. Deformation relationship: The deformation δ\delta in each rod is related to the force, the material properties, and the geometry by the formula: δ=FLAE\delta = \frac{F L}{A E} where FF is the force in the rod, LL is the length, AA is the cross-sectional area, and EE is the modulus of elasticity.

Since the bar remains horizontal, the deformations in both rods must be equal, i.e., δs=δa\delta_s = \delta_a. We can set up an equation using the deformation formulas for both rods and solve for the forces.

I'll calculate the forces and the vertical displacement of the bar next.The forces in the steel and aluminum rods are as follows:

  • Force in the steel rod: Fs=47.73kNF_s = 47.73 \, \text{kN}
  • Force in the aluminum rod: Fa=22.27kNF_a = 22.27 \, \text{kN}

The vertical displacement of the bar (which is the same for both rods) is: δ=0.00318m=3.18mm\delta = 0.00318 \, \text{m} = 3.18 \, \text{mm}

Thus, the vertical movement of the rigid bar is 3.18mm3.18 \, \text{mm} downward.

Do you need more details or have any questions?

Here are five related questions to expand on this:

  1. How would the displacement change if the cross-sectional area of the steel rod was increased?
  2. What happens to the distribution of forces if the modulus of elasticity of aluminum changes?
  3. How would the system behave if the lengths of both rods were the same?
  4. How do different materials' properties affect the overall stiffness of the system?
  5. Can we analyze this system if the bar is not initially horizontal?

Tip: Remember, the material's modulus of elasticity directly influences how much it stretches under a load. A higher modulus means less deformation for the same applied force.

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Math Problem Analysis

Mathematical Concepts

Statics
Deformation of materials
Stress-strain relationships

Formulas

Force balance: F_s + F_a = P
Deformation: δ = FL / (A * E)

Theorems

Hooke's Law for linear elastic materials
Compatibility of deformations

Suitable Grade Level

University Level (Engineering - Mechanics of Materials)