Math Problem Statement

f a beam whose length is 6 m with the modulus of elasticity of E = 200 GPa and I = 99.0 ✕ 106 mm4. The beam is designed to carry a load of 14,000 N/m.What is the maximum deflection of the beam (in m)? (Enter the y-value where the beam is deflected the most.)

Solution

To find the maximum deflection of a beam subjected to a uniformly distributed load, we use the formula for deflection of a simply supported beam:

δmax=5qL4384EI\delta_{\text{max}} = \frac{5qL^4}{384EI}

Where:

  • q=14,000N/mq = 14,000 \, \text{N/m} (uniform load),
  • L=6m=6000mmL = 6 \, \text{m} = 6000 \, \text{mm} (beam length),
  • E=200GPa=200×109PaE = 200 \, \text{GPa} = 200 \times 10^9 \, \text{Pa} (modulus of elasticity),
  • I=99×106mm4I = 99 \times 10^6 \, \text{mm}^4 (moment of inertia).

Step-by-step solution:

  1. Convert the given values into consistent units:

    • Convert qq: 14,000N/m=14N/mm14,000 \, \text{N/m} = 14 \, \text{N/mm},
    • LL is already 6000mm6000 \, \text{mm},
    • EE: 200×109Pa200 \times 10^9 \, \text{Pa},
    • I=99×106mm4I = 99 \times 10^6 \, \text{mm}^4.
  2. Substitute the values into the formula: δmax=514(6000)4384(200×109)(99×106)\delta_{\text{max}} = \frac{5 \cdot 14 \cdot (6000)^4}{384 \cdot (200 \times 10^9) \cdot (99 \times 10^6)}

  3. Perform the calculations:

    • Calculate L4=(6000)4=1.296×1015mm4L^4 = (6000)^4 = 1.296 \times 10^{15} \, \text{mm}^4,
    • Calculate the numerator: 5141.296×1015=9.072×1016Nmm45 \cdot 14 \cdot 1.296 \times 10^{15} = 9.072 \times 10^{16} \, \text{N} \cdot \text{mm}^4,
    • Calculate the denominator: 38420099109106=7.6032×1018Nmm2384 \cdot 200 \cdot 99 \cdot 10^9 \cdot 10^6 = 7.6032 \times 10^{18} \, \text{N} \cdot \text{mm}^2,
    • Divide numerator by denominator: δmax=9.072×10167.6032×10180.01193mm.\delta_{\text{max}} = \frac{9.072 \times 10^{16}}{7.6032 \times 10^{18}} \approx 0.01193 \, \text{mm}.
  4. Convert to meters: δmax=0.01193mm=0.00001193m.\delta_{\text{max}} = 0.01193 \, \text{mm} = 0.00001193 \, \text{m}.

Final Answer:

The maximum deflection of the beam is approximately: δmax0.00001193m.\delta_{\text{max}} \approx 0.00001193 \, \text{m}.

Would you like additional details or assistance with related concepts?


5 Related Questions:

  1. How does the length of a beam affect its maximum deflection under a uniform load?
  2. What changes in deflection occur if the modulus of elasticity EE is halved?
  3. How does the moment of inertia II influence the beam's stiffness?
  4. Can this formula be used for a cantilever beam? Why or why not?
  5. What are the real-world factors that might affect the accuracy of this deflection calculation?

Tip: Always check units carefully when performing structural calculations—convert everything to a consistent system!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Structural Engineering
Mechanics of Materials
Beam Deflection Analysis

Formulas

Maximum deflection of a simply supported beam under uniform load: δ_max = (5qL^4) / (384EI)

Theorems

Euler-Bernoulli Beam Theory

Suitable Grade Level

Undergraduate (Engineering Level)