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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