Types of Beams, design example and design sheet in excel file download.

Types of Beams, design example and design sheet in excel file download.


Types of Beams

1.    Rectangular Beam

Beams can often adopt a rectangular cross-section.

Applications: The relative ease with which they can be designed and constructed makes them particularly suitable for use as a base structure in buildings, bridges or other engineering projects.

Delights: Its standard dimensions make it an easy piece to calculate or fabricate.

Downsides: Supposing beams need more than usual strength (i.e., larger dimensions), the heaviness of result designs shall increase.

2.    Circular Beams (Circular, Cylindrical)

Shape: Cylindrical or circular cross-section.

Applications: Used in structures that require symmetry for aesthetic reasons; for examples, columns serving the dual purpose as beams or very special industrial applications.

Advantages: Every direction of resistance has equally effect induced upon it. (Isotropic in bending.) It can also be aesthetically pleasing.

Disadvantages: Less efficient than other shapes in regard to use of materials.


Design Example: Rectangular Beam

Problem: Design a simply supported rectangular beam of span L = 6 m, subjected to a uniformly distributed load of w = 10 kN/m. The allowable bending stress is f_b = 10 MPa.

Steps:

1.    Calculate Maximum Bending Moment (M_max): M_max = (wL²) / 8
Substituting the values:
M_max = (10 × 6²) / 8 = (10 × 36) / 8 = 45 kN·m = 45,000 N·m = 45,000,000 N·mm.

2.    Section Modulus (Z) Requirement: Z = M_max / f_b
Substituting the values:
Z = 45,000,000 / 10,000,000 = 4.5 m³ = 4,500,000 mm³.

3.    Select Dimensions of Beam: For a rectangular section, Z = (b d²) / 6.
Assume b = 250 mm (breadth).
Solve for d (depth): 4,500,000 = (250 × d²) / 6
d² = (4,500,000 × 6) / 250 = 27,000,000 / 250 = 108,000
d = √108,000 ≈ 328.63 mm.
Round d to a practical value:
d = 350 mm.

4.    Verify the Design: Recalculate Z = (b d²) / 6:
Z = (250 × 350²) / 6 = (250 × 122,500) / 6 = 5,102,083.33 mm³.
Check if Z > required Z:
5,102,083.33 mm³ > 4,500,000 mm³ (Safe!)

Final Beam Dimensions:
Breadth b = 250 mm, Depth d = 350 mm.


Design Example: Circular Beam

Problem: Design a simply supported circular beam of span L = 6 m, subjected to the same uniformly distributed load of w = 10 kN/m and the allowable bending stress f_b = 10 MPa.

Steps:

1.    Calculate Maximum Bending Moment (M_max): M_max = (wL²) / 8
Substituting the values:
M_max = (10 × 6²) / 8 = 45 kN·m = 45,000,000 N·mm.

2.    Section Modulus (Z) Requirement: Z = M_max / f_b
Substituting the values:
Z = 45,000,000 / 10,000,000 = 4.5 m³ = 4,500,000 mm³.

3.    For Circular Section: Section modulus Z = (π d³) / 32.
Solve for d (diameter): 4,500,000 = (π × d³) / 32
Rearranging:
d³ = (4,500,000 × 32) / π = 144,000,000 / 3.1416 = 45,873,572.8
d = 
45,873,572.8 ≈ 357.7 mm.
Round d to a practical value:
d = 375 mm.

4.    Verify the Design: Recalculate Z = (π d³) / 32:
Z = (π × 375³) / 32 = (3.1416 × 52,734,375) / 32 = 5,203,703.7 mm³.
Check if Z > required Z:
5,203,703.7 mm³ > 4,500,000 mm³ (Safe!)

Final Beam Diameter:
Diameter d = 375 mm.

 


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