Math Problem Statement

A wide flange section for a 4 m long column (hinged at both ends) has the following properties:

Cross-sectional area = 6,000 mm² Radius of gyration, rx = 150mm Radius of gyration, ry = 100mm Modulus of Elasticity, E = 200,000 MPa

Determine the value of Euler critical stress.

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Solution

Given:

  • A=6,000mm2A = 6,000 \, \text{mm}^2
  • rx=150mmr_x = 150 \, \text{mm}
  • ry=100mmr_y = 100 \, \text{mm}
  • E=200,000MPaE = 200,000 \, \text{MPa}
  • L=4m=4000mmL = 4 \, \text{m} = 4000 \, \text{mm}

Step-by-step solution:

  1. Calculate Effective Length (hinged at both ends):
    Leff=L=4000mmL_{\text{eff}} = L = 4000 \, \text{mm}

  2. Calculate Slenderness Ratio (we use the smaller ryr_y for conservative calculation):
    λ=Leffry=4000100=40\lambda = \frac{L_{\text{eff}}}{r_y} = \frac{4000}{100} = 40

  3. Euler’s Critical Stress Formula: σcr=π2Eλ2\sigma_{\text{cr}} = \frac{\pi^2 E}{\lambda^2}

  4. Substitute values: σcr=π2×200,000402\sigma_{\text{cr}} = \frac{\pi^2 \times 200,000}{40^2}

  5. Calculate σcr\sigma_{\text{cr}}: σcr=9.8696×200,0001600\sigma_{\text{cr}} = \frac{9.8696 \times 200,000}{1600} σcr=1,973,9201600\sigma_{\text{cr}} = \frac{1,973,920}{1600} σcr=1233.7MPa\sigma_{\text{cr}} = 1233.7 \, \text{MPa}

Answer: σcr1233.7MPa\sigma_{\text{cr}} \approx 1233.7 \, \text{MPa}

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Math Problem Analysis

Mathematical Concepts

Mechanics of Materials
Euler's Buckling Theory
Structural Engineering

Formulas

Effective length for hinged ends: Leff = L
Slenderness ratio: λ = Leff / ry
Euler's Critical Stress: σcr = (π^2 * E) / λ^2

Theorems

Euler's Buckling Theory

Suitable Grade Level

Undergraduate Engineering