Structural analysis in civil engineering relies heavily on understanding the geometric properties of sections, which influence the behavior of structural elements under loads. Determining these properties accurately ensures safe and efficient design of beams, columns, slabs, and other components in buildings, bridges, and infrastructure. This chapter covers fundamental geometric properties, focusing on moment of inertia and radius of gyration, essential for stability and strength calculations in Kenyan civil engineering projects.

4.1 Introduction to Geometric Properties of Sections

Geometric properties of sections describe the shape and size characteristics that affect the stress distribution and deformation of structural members. Kenyan civil engineers use these properties to predict how beams and columns will perform under loads, ensuring compliance with standards like the Kenya Building Code. This section introduces key definitions and examines common section shapes with their geometric properties.

4.1.1 Definitions

Geometric properties quantify the dimensions and shape of a structural cross-section influencing its mechanical response. The main properties include area, centroid, moment of inertia, and radius of gyration.

Area (A) = Cross-sectional size in m²

Centroid (x̄, ȳ) = Geometric centre coordinates

Moment of Inertia: Iₓ = ∫ y² dA, I_y = ∫ x² dA

Radius of Gyration: r = √(I ÷ A)

4.1.1.1 Area

The cross-sectional area (A) is the amount of surface enclosed by the outline of a member's cross-section, measured in mm² or m². It determines how much direct (axial) load a member can carry, since stress = force ÷ area, and it also gives the self-weight of a member per metre when multiplied by the density of the material. For composite shapes, the total area is found by dividing the section into simple rectangles, triangles or circles and adding their areas, subtracting any holes. Example: the engineer checking timber struts for a dairy shed at a farm in Ol Kalou calculated the cross-sectional areas of 75 × 100 mm and 100 × 100 mm members to compare their axial capacities.

Example 1: Calculate the area of a rectangular beam section 300 mm wide and 500 mm deep.

Given: Width b = 0.3 m, Depth h = 0.5 m

A = b × h

A = 0.3 × 0.5

A = 0.15 m²

Answer: 0.15 m²

Uses of Cross-Sectional Area

• Axial stress calculation: Direct stress in columns and ties equals the axial force divided by the area.
• Self-weight: Area × density × gravity gives the weight per metre of a member.
• Material quantities: Area × length gives the volume of concrete, steel or timber required.
• Shear stress estimation: Average shear stress in a section is the shear force divided by the relevant area.
• Basis for other properties: Area is needed to calculate the centroid and the radius of gyration.

4.1.1.2 Centroid

The centroid of a section is the geometric centre of its area, the point through which the whole area can be considered to act, given by its coordinates x̄ and ȳ from chosen reference axes. In a beam, the axis through the centroid is the neutral axis in bending, where the bending stress is zero, so its position must be known before bending stresses or moments of inertia can be calculated. Symmetrical sections have their centroid on the axis of symmetry, while for unsymmetrical shapes it is found by taking moments of the component areas about a reference axis. Example: the designer of precast T-beams for a footbridge serving a secondary school in Kakamega located the centroid of the T-section to set the neutral axis before checking the stresses.

Centroid of a composite section: ȳ = Σ(A × y) ÷ ΣA

Example 2: Find the centroid of a rectangular section 400 mm wide and 600 mm deep, with origin at the bottom-left corner.

Given: b = 0.4 m, h = 0.6 m

x̄ = b ÷ 2

x̄ = 0.4 ÷ 2 = 0.2 m

ȳ = h ÷ 2

ȳ = 0.6 ÷ 2 = 0.3 m

Answer: Centroid at (0.2 m, 0.3 m)

Steps for Locating the Centroid of a Composite Section

1. Choose reference axes: Select convenient axes, usually the bottom edge and the left edge or an axis of symmetry.
2. Divide the section into simple parts: Split the shape into rectangles, triangles or circles.
3. Find each area: Calculate the area of each part.
4. Find each part's centroid: Locate the centroid distance of each part from the reference axis.
5. Calculate first moments: Multiply each area by its centroid distance.
6. Divide by the total area: Divide the sum of the first moments by the total area to get the centroid position.

4.1.1.3 Moment of inertia

The second moment of area, commonly called the moment of inertia (I), measures how the area of a section is distributed about an axis, and it is calculated by adding the products of each small area and the square of its distance from the axis, giving units of mm⁴ or m⁴. It is the key property governing a beam's resistance to bending and deflection: the further the material is placed from the neutral axis, the greater I and the stiffer and stronger the beam. This is why beams are placed with their greater depth vertical and why I-sections concentrate material in their flanges. Example: by turning 50 × 150 mm timber joists on edge instead of flat, the contractor building a hostel floor for a TVET institute in Nyeri increased their moment of inertia nine times, greatly reducing floor deflection.

Example 3: Determine the moment of inertia about the x-axis for a rectangular beam 200 mm wide and 400 mm deep.

Given: b = 0.2 m, h = 0.4 m

Iₓ = b h³ ÷ 12

Iₓ = 0.2 × (0.4)³ ÷ 12

Iₓ = 0.2 × 0.064 ÷ 12 = 0.0128 ÷ 12

Iₓ = 0.00107 m⁴

Answer: 0.00107 m⁴

Parallel Axis Theorem

The moment of inertia of an area about any axis parallel to its own centroidal axis is:

I = I_c + A × d²

where I_c is the moment of inertia about the part's own centroidal axis, A is its area and d is the distance between the two axes. This theorem is used to find the moment of inertia of composite sections such as T-sections and I-sections.

Importance of Moment of Inertia

• Bending stress: Bending stress equals M × y ÷ I, so a larger I gives lower stress for the same moment.
• Deflection: Deflection is inversely proportional to I, so stiffer beams deflect less.
• Section modulus: The elastic section modulus Z = I ÷ y_max is used directly in beam design.
• Buckling of columns: The buckling load of a column depends on its least moment of inertia.
• Efficient sections: Placing material far from the neutral axis increases I without adding weight.

4.1.1.4 Radius of Gyration

Example 4: Calculate the radius of gyration for the rectangular section in Example 3.

Given: A = b × h = 0.2 × 0.4 = 0.08 m², Iₓ = 0.00107 m⁴

r = √(I ÷ A)

r = √(0.00107 ÷ 0.08)

r = √0.013375

r = 0.1156 m

Answer: 0.1156 m

4.1.2 Common shapes and their geometrical properties

Civil engineers frequently encounter standard shapes such as rectangles, circles, I-sections, and T-sections. Each shape has established formulae for area, centroid, and moments of inertia, facilitating quick calculations during design.

Basic geometric and structural shapes used in construction (Source: Untitled (20))

Properties of Common Sections Compared

4.1.2.1 Rectangular

The rectangular section is the most common shape in construction, used for reinforced concrete beams and columns, timber joists, lintels and masonry piers. Its properties are simple: the centroid is at the centre, and its moment of inertia about the horizontal axis (bh³ ÷ 12) depends on the cube of the depth, so doubling the depth makes the beam eight times stiffer, while doubling the width only doubles the stiffness. This explains why rectangular beams are always used with their greater dimension vertical. Example: the engineer for a rural health centre in Makueni chose 230 × 450 mm concrete beams rather than 450 × 230 mm, because the deeper orientation gave nearly four times the moment of inertia.

Properties of a Rectangular Section

• Area: A = b × h.
• Centroid: At mid-width and mid-depth, x̄ = b ÷ 2 and ȳ = h ÷ 2.
• Moment of inertia about the horizontal centroidal axis: Iₓ = b h³ ÷ 12.
• Moment of inertia about the vertical centroidal axis: I_y = h b³ ÷ 12.
• Elastic section modulus: Z = b h² ÷ 6.
• Radius of gyration: r = h ÷ √12, about 0.289 h.

Worked Example: Rectangular Concrete Beam

Find A, Iₓ and Z for a beam 230 mm wide and 450 mm deep.

A = b × h

A = 230 × 450

A = 103,500 mm²

Iₓ = b h³ ÷ 12

Iₓ = 230 × 450³ ÷ 12

Iₓ = 230 × 91,125,000 ÷ 12

Iₓ = 1,746,562,500 mm⁴ (about 1.75 × 10⁹ mm⁴)

Z = b h² ÷ 6

Z = 230 × 450² ÷ 6

Z = 230 × 202,500 ÷ 6

Z = 7,762,500 mm³ (about 7.76 × 10⁶ mm³)

4.1.2.2 I-beams

An I-beam (or universal beam) is a steel section with two horizontal flanges joined by a vertical web, shaped like the letter I. Most of its material is placed in the flanges, far from the neutral axis, where it is most effective in resisting bending, while the thin web carries shear and holds the flanges apart, giving a very high moment of inertia for little weight. I-beams are therefore used as floor beams, roof beams, bridge girders and columns in steel structures, and their properties are listed in steel section tables. Example: the steel frame of a new warehouse for a SACCO-owned agricultural produce store in Eldoret used I-beams for the floor beams of its mezzanine, selected from section tables by their moment of inertia and section modulus.

Parts of an I-Beam and Their Functions

• Top flange: Resists compression in sagging bending and must be restrained against buckling.
• Bottom flange: Resists tension in sagging bending.
• Web: Resists shear forces and connects the flanges.
• Root radius: The curved junction between flange and web reduces stress concentrations.
• Depth: The overall depth largely controls the moment of inertia and stiffness.

Worked Example: Moment of Inertia of an I-Section

A symmetrical I-section has flanges 150 mm wide and 12 mm thick, a web 8 mm thick and an overall depth of 300 mm. Find A, Iₓ and Z.

Clear web depth, d = D − 2 t_f

d = 300 − (2 × 12)

d = 276 mm

A = 2 B t_f + d t_w

A = (2 × 150 × 12) + (276 × 8)

A = 3,600 + 2,208

A = 5,808 mm²

Iₓ = (B D³ − (B − t_w) d³) ÷ 12

Iₓ = (150 × 300³ − 142 × 276³) ÷ 12

Iₓ = (4,050,000,000 − 2,985,489,792) ÷ 12

Iₓ = 1,064,510,208 ÷ 12

Iₓ = 88,709,184 mm⁴ (about 88.7 × 10⁶ mm⁴)

Z = Iₓ ÷ (D ÷ 2)

Z = 88,709,184 ÷ 150

Z = 591,395 mm³ (about 591 × 10³ mm³)

Advantages of I-Sections

• High strength-to-weight ratio: Material in the flanges gives high bending resistance for little weight.
• High stiffness: The large moment of inertia reduces deflection.
• Economy: Less steel is needed than for a solid section of similar strength.
• Easy connections: Flanges and web provide convenient surfaces for bolting and welding.
• Standard sizes: Sections are available in standard sizes with tabulated properties, simplifying design.

4.1.2.3 T-section

A T-section consists of a horizontal flange on top of a vertical web, and it occurs as steel tees, precast concrete T-beams and, very commonly, as the flanged beams formed when a reinforced concrete beam is cast together with the slab it supports. Because a T-section is symmetrical only about its vertical axis, its centroid is not at mid-depth but closer to the flange, so it must be calculated by the method of moments, and its moment of inertia is found using the parallel axis theorem. Example: in the floor of a hotel in Nanyuki, the engineer treated the beams and the slab above them as T-sections, taking advantage of the slab acting as a wide compression flange.

Example 2: Calculate the centroid of a T-section composed of a flange 150 mm wide and 30 mm thick, and a web 30 mm wide and 200 mm deep.

Given:

• Flange: b_f = 0.15 m, t_f = 0.03 m

• Web: b_w = 0.03 m, h_w = 0.2 m

Calculate areas:

A_f = b_f × t_f = 0.15 × 0.03 = 0.0045 m²

A_w = b_w × h_w = 0.03 × 0.2 = 0.006 m²

Set bottom of web as reference:

Centroid of flange from bottom:

y_f = h_w + t_f ÷ 2 = 0.2 + 0.015 = 0.215 m

Centroid of web from bottom:

y_w = h_w ÷ 2 = 0.1 m

Calculate overall centroid:

ȳ = (A_f y_f + A_w y_w) ÷ (A_f + A_w)

ȳ = (0.0045 × 0.215 + 0.006 × 0.1) ÷ (0.0045 + 0.006)

ȳ = (0.0009675 + 0.0006) ÷ 0.0105 = 0.0015675 ÷ 0.0105

ȳ = 0.1493 m

Answer: Centroid is 0.1493 m from bottom

Example 3: Calculate the moment of inertia about the centroidal x-axis for the T-section in Example 2.

Using parallel axis theorem:

Moment of inertia of flange about its own centroid:

I_f = b_f t_f³ ÷ 12 = 0.15 × (0.03)³ ÷ 12 = 0.15 × 0.000027 ÷ 12 = 0.0000003375 m⁴

Distance flange centroid to overall centroid:

d_f = 0.215 − 0.1493 = 0.0657 m

Moment of inertia flange about overall centroid:

I_f,total = I_f + A_f d_f² = 0.0000003375 + 0.0045 × (0.0657)²

I_f,total = 0.0000003375 + 0.0045 × 0.004315 = 0.0000003375 + 0.00001942 = 0.00001976 m⁴

Moment of inertia of web about its own centroid:

I_w = b_w h_w³ ÷ 12 = 0.03 × (0.2)³ ÷ 12 = 0.03 × 0.008 ÷ 12 = 0.00002 m⁴

Distance web centroid to overall centroid:

d_w = 0.1493 − 0.1 = 0.0493 m

Moment of inertia web about overall centroid:

I_w,total = I_w + A_w d_w² = 0.00002 + 0.006 × (0.0493)²

I_w,total = 0.00002 + 0.006 × 0.00243 = 0.00002 + 0.00001458 = 0.00003458 m⁴

Total moment of inertia:

Iₓ = I_f,total + I_w,total

Iₓ = 0.00001976 + 0.00003458

Iₓ = 0.00005434 m⁴

Answer: 0.00005434 m⁴

Example 4: Determine the radius of gyration for the T-section in Example 3.

Total area:

A = A_f + A_w = 0.0045 + 0.006 = 0.0105 m²

r = √(I ÷ A)

r = √(0.00005434 ÷ 0.0105) = √0.00518

r = 0.07196 m

Answer: 0.07196 m

4.1.2.4 Circular Section

Example 1: Compute the moment of inertia about the centroidal x-axis for a circular column section diameter 300 mm.

Given: Diameter d = 0.3 m

Iₓ = π d⁴ ÷ 64

Iₓ = 3.1416 × (0.3)⁴ ÷ 64

Iₓ = 3.1416 × 0.0081 ÷ 64 = 0.02545 ÷ 64

Iₓ = 0.000397 m⁴

Answer: 0.000397 m⁴

Chapter Summary

This chapter introduced the fundamental geometric properties of structural sections essential for structural analysis. It began with clear definitions of key concepts including area, centroid, and moment of inertia, establishing the basis for understanding section behavior. The chapter then examined common section shapes such as rectangular sections, I-beams, and T-sections, detailing their specific geometric properties. Methods for calculating the moment of inertia were presented, covering both basic shapes and composite sections formed by combining simpler shapes. The calculation of the radius of gyration was explained along with its significance in assessing the stability of structural elements. Emphasis was placed on how these properties influence the design and analysis of structural components under load. The chapter provided a comprehensive framework for determining section properties critical to structural performance and safety.