Example for Design of beams

• Required design load: 10 kN
• Span of the beam: 4 meters
• Material: Steel (yield strength = 250 MPa)
• Allowable deflection: L/250 (where L is the span)

1. Determine the maximum bending moment (M) at the mid-span of the beam:
   • For a simply supported beam with a uniformly distributed load, the maximum bending moment occurs at the mid-span and is equal to W*L/4, where W is the total load.
   • In this case, W = 10 kN = 10,000 N, and L = 4 meters. Therefore, M = (10,000 * 4) / 4 = 10,000 Nm = 10 kNm.
2. Calculate the section modulus (S):
   • The section modulus is a geometric property of the cross-sectional shape of the beam that relates to its resistance to bending. It can be calculated using the formula S = (b * h^2) / 6, where b is the width of the beam and h is the height of the beam.
   • Since we don’t have specific dimensions, let’s assume a rectangular cross-section with a width (b) of 150 mm and a height (h) of 300 mm.
   • Plugging these values into the formula, we get S = (150 * 300^2) / 6 = 2,250,000 mm^3 = 2,250 cm^3.
3. Determine the required moment of inertia (I):
   • The moment of inertia is another geometric property of the cross-section that relates to the stiffness of the beam. It can be calculated using the formula I = (b * h^3) / 12.
   • Using the same dimensions as before, we have I = (150 * 300^3) / 12 = 13,500,000 mm^4 = 1350 cm^4.
4. Check the allowable stress:
   • The maximum stress in the beam should be less than or equal to the yield strength of the material.
   • The bending stress (σ) can be calculated as σ = M * y / I, where y is the distance from the neutral axis to the extreme fiber (half of the beam’s height).
   • Assuming a rectangular cross-section, y = h/2 = 150 mm.
   • Plugging in the values, σ = (10,000 * 150) / 1350 = 111.11 MPa (approximately).
   • Since this value is less than the yield strength of 250 MPa, the beam is within the allowable stress limits.
5. Check the deflection:
   • The deflection of the beam should be less than or equal to L/250.
   • The formula for deflection is δ = (5 * W * L^4) / (384 * E * I), where E is the modulus of elasticity of the material.
   • Assuming the modulus of elasticity for steel is 200 GPa (200,000 MPa), we have δ = (5 * 10,000 * 4^4) / (384 * 200,000 * 1350) = 0.034 mm.
   • Since this value is less than L/250 = 4/250 = 0.016 mm (approximately), the beam satisfies the deflection requirement.

Based on the calculations, a rectangular steel beam with a width of 150 mm and a height of 300 mm would be suitable for this design. However, please note that this is just an example,