Math Problem Statement

A cantilever beam of rectangular cross-section is used to support a pulley. The tension in the wire rope is 5 KN. The beam is made of cast iron FG 200 and the factor of safety is 2.5. The ratio of depth to width of the cross-section is 2. Determine the dimensions of the cross-section of the beam

A cantilever beam of rectangular cross-section is used to support a pulley. The tension in the wire rope is 5 KN. The beam is made of cast iron FG 200 and the factor of safety is 2.5. The ratio of depth to width of the cross-section is 2. Determine the dimensions of the cross-section of the beam.

Solution

To solve this problem, let's analyze the situation and calculate the required dimensions of the cross-section of the beam. Here's the step-by-step procedure:


Given Data:

  1. Load on the beam due to tension in the rope: F=5kN=5000NF = 5 \, \text{kN} = 5000 \, \text{N}
  2. Length of the cantilever beam: L=1500mm+500mm=2000mmL = 1500 \, \text{mm} + 500 \, \text{mm} = 2000 \, \text{mm}
  3. Depth-to-width ratio: dw=2\frac{d}{w} = 2
  4. Material: Cast iron FG 200.
    • Ultimate tensile strength σut=200N/mm2\sigma_{\text{ut}} = 200 \, \text{N/mm}^2
  5. Factor of Safety: FOS=2.5\text{FOS} = 2.5

Step 1: Bending moment at the fixed end

The maximum bending moment occurs at the fixed end of the cantilever due to the applied load at the end: M=FL=50002000=10,000,000N\cdotpmmM = F \cdot L = 5000 \cdot 2000 = 10,000,000 \, \text{N·mm}


Step 2: Allowable stress

The allowable bending stress is given by: σallow=σutFOS=2002.5=80N/mm2\sigma_{\text{allow}} = \frac{\sigma_{\text{ut}}}{\text{FOS}} = \frac{200}{2.5} = 80 \, \text{N/mm}^2


Step 3: Section modulus for the rectangular section

For a rectangular cross-section, the section modulus ZZ is: Z=wd26Z = \frac{w \cdot d^2}{6}

Using the bending stress formula: σallow=MZorZ=Mσallow\sigma_{\text{allow}} = \frac{M}{Z} \quad \text{or} \quad Z = \frac{M}{\sigma_{\text{allow}}}

Substitute values: Z=10,000,00080=125,000mm3Z = \frac{10,000,000}{80} = 125,000 \, \text{mm}^3


Step 4: Dimensions of the cross-section

Using the section modulus formula Z=wd26Z = \frac{w \cdot d^2}{6}, and the depth-to-width ratio dw=2\frac{d}{w} = 2, let d=2wd = 2w. Substitute into ZZ: 125,000=w(2w)26125,000 = \frac{w \cdot (2w)^2}{6} 125,000=w4w26=4w36125,000 = \frac{w \cdot 4w^2}{6} = \frac{4w^3}{6} 125,000=2w33125,000 = \frac{2w^3}{3} w3=125,00032=187,500w^3 = \frac{125,000 \cdot 3}{2} = 187,500 w=187,500358.4mmw = \sqrt[3]{187,500} \approx 58.4 \, \text{mm}

Since d=2wd = 2w: d=258.4=116.8mmd = 2 \cdot 58.4 = 116.8 \, \text{mm}


Final Dimensions:

  • Width (ww): 58.4 mm
  • Depth (dd): 116.8 mm

Summary:

The dimensions of the cross-section of the beam are approximately:

  • Width: 58.4mm58.4 \, \text{mm}
  • Depth: 116.8mm116.8 \, \text{mm}

Would you like to see detailed explanations for any step or further clarifications? 😊


Related Questions:

  1. What is the importance of the factor of safety in design calculations?
  2. How is the bending stress distributed across a rectangular cross-section?
  3. Can the section modulus be improved without increasing the material volume?
  4. What is the significance of using FG 200 cast iron for this design?
  5. How would the design change if the depth-to-width ratio were altered?

Tip:

In beam design, maximizing the section modulus ZZ for a given material reduces stress and increases safety margins!

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Math Problem Analysis

Mathematical Concepts

Statics
Bending Moment
Allowable Stress
Section Modulus

Formulas

Maximum bending moment M = F × L
Allowable stress σ_allow = σ_ut / FOS
Section modulus Z = w × d^2 / 6
Z = M / σ_allow

Theorems

Bending Stress Formula
Section Modulus Relationship

Suitable Grade Level

Grades 11-12 / Engineering Foundation