Stress and strain are fundamental concepts in structural analysis that describe how materials respond to forces and deformations. Understanding and calculating these quantities enable building technologists to assess the safety and performance of structural elements in buildings. In Kenya, where construction materials and environmental conditions vary widely, accurate computation of stress and strain ensures reliability in design and construction practices. This chapter introduces these concepts and guides students through detailed calculations relevant to building technology.

1.1 Definition of Stress and Strain

Stress and strain describe the internal forces and deformations within materials subjected to external loads. Stress quantifies the intensity of internal forces per unit area, while strain measures the relative deformation experienced by the material. Both are essential for predicting structural behavior under various loading conditions in building elements such as beams, columns, and slabs.

1.1.1 Stress: Concept and Formula

Stress is the internal resistance offered by a material to an applied force, distributed over the cross-sectional area. It is expressed as force per unit area and is a measure of intensity rather than total force.

The formula for normal stress, \( \sigma \), is:

$$ \sigma = \frac{F}{A} $$

where

\( \sigma \) = normal stress (Pa or N/m²),
\( F \) = axial force applied (N),
\( A \) = cross-sectional area perpendicular to the force (m²).

Worked Examples

Example 1: A concrete column in a county government office building carries an axial load of 150 kN. The cross-sectional area of the column is 0.15 m². Calculate the normal stress in the column.

Given:
\( F = 150,000 \, \text{N} \),
\( A = 0.15 \, \text{m}^2 \)

$$ \sigma = \frac{F}{A} $$$$ \sigma = \frac{150,000}{0.15} $$$$ \sigma = 1,000,000 \, \text{Pa} $$ Answer: 1,000,000 Pa (1 MPa)

Example 2: A steel beam in a university building supports a tensile force of 200 kN. The beam's cross-sectional area is 4000 mm². Calculate the stress.

Given:
\( F = 200,000 \, \text{N} \),
\( A = 4000 \, \text{mm}^2 = 4000 \times 10^{-6} \, \text{m}^2 = 0.004 \, \text{m}^2 \)

$$ \sigma = \frac{F}{A} $$$$ \sigma = \frac{200,000}{0.004} $$$$ \sigma = 50,000,000 \, \text{Pa} $$ Answer: 50,000,000 Pa (50 MPa)

Example 3: A wooden strut in a retail business structure has a cross-sectional area of 0.01 m² and is subjected to a compressive load of 10 kN. Determine the compressive stress.

Given:
\( F = 10,000 \, \text{N} \),
\( A = 0.01 \, \text{m}^2 \)

$$ \sigma = \frac{F}{A} $$$$ \sigma = \frac{10,000}{0.01} $$$$ \sigma = 1,000,000 \, \text{Pa} $$ Answer: 1,000,000 Pa (1 MPa)

Example 4: A steel reinforcing bar in a hospital foundation has a diameter of 20 mm and carries a tensile load of 25 kN. Calculate the normal stress.

Given:
Diameter \( d = 20 \, \text{mm} = 0.02 \, \text{m} \),
\( F = 25,000 \, \text{N} \)

Cross-sectional area of circular bar:$$ A = \frac{\pi d^2}{4} = \frac{3.1416 \times (0.02)^2}{4} = 3.1416 \times 0.0001 = 0.00031416 \, \text{m}^2 $$

$$ \sigma = \frac{F}{A} $$$$ \sigma = \frac{25,000}{0.00031416} $$$$ \sigma = 79,577,471 \, \text{Pa} $$ Answer: 79,577,471 Pa (approximately 79.58 MPa)

Example 5: A steel tie rod in a hotel structure has a cross-sectional area of 3000 mm² and carries a tensile load of 45 kN. Calculate the stress.

Given:
\( F = 45,000 \, \text{N} \),
\( A = 3000 \, \text{mm}^2 = 0.003 \, \text{m}^2 \)

$$ \sigma = \frac{F}{A} $$$$ \sigma = \frac{45,000}{0.003} $$$$ \sigma = 15,000,000 \, \text{Pa} $$ Answer: 15,000,000 Pa (15 MPa)

1.1.2 Strain: Concept and Formula

Strain represents the deformation of a material relative to its original length due to applied stress. It is a dimensionless quantity expressing the ratio of change in length to the original length.

The formula for normal strain, \( \varepsilon \), is:

$$ \varepsilon = \frac{\Delta L}{L_0} $$

where
\( \varepsilon \) = strain (dimensionless),
\( \Delta L \) = change in length (m),
\( L_0 \) = original length (m).

Worked Examples

Example 1: A steel rod in a SACCO office building is originally 2 m long. Under tension, it elongates by 1.5 mm. Calculate the strain in the rod.

Given:
\( L_0 = 2 \, \text{m} \),
\( \Delta L = 1.5 \, \text{mm} = 0.0015 \, \text{m} \)

$$ \varepsilon = \frac{\Delta L}{L_0} $$$$ \varepsilon = \frac{0.0015}{2} $$$$ \varepsilon = 0.00075 $$ Answer: 0.00075 (dimensionless)

Example 2: A wooden beam in a school structure has an original length of 3 m. It shortens by 2 mm under compression. Determine the strain.

Given:
\( L_0 = 3 \, \text{m} \),
\( \Delta L = 2 \, \text{mm} = 0.002 \, \text{m} \)

$$ \varepsilon = \frac{\Delta L}{L_0} $$$$ \varepsilon = \frac{0.002}{3} $$$$ \varepsilon = 0.000667 $$ Answer: 0.000667 (dimensionless)

Example 3: A steel cable in a retail building is 5 m long and stretches by 5 mm when loaded. Find the strain.

Given:
\( L_0 = 5 \, \text{m} \),
\( \Delta L = 5 \, \text{mm} = 0.005 \, \text{m} \)

$$ \varepsilon = \frac{\Delta L}{L_0} $$$$ \varepsilon = \frac{0.005}{5} $$$$ \varepsilon = 0.001 $$ Answer: 0.001 (dimensionless)

Example 4: A steel reinforcement bar in a hospital foundation has an original length of 1.5 m and elongates by 0.75 mm under load. Calculate the strain.

Given:
\( L_0 = 1.5 \, \text{m} \),
\( \Delta L = 0.75 \, \text{mm} = 0.00075 \, \text{m} \)

$$ \varepsilon = \frac{\Delta L}{L_0} $$$$ \varepsilon = \frac{0.00075}{1.5} $$$$ \varepsilon = 0.0005 $$ Answer: 0.0005 (dimensionless)

Example 5: A timber strut in a cooperative farm building originally 4 m long shortens by 3 mm under load. Find the strain.

Given:
\( L_0 = 4 \, \text{m} \),
\( \Delta L = 3 \, \text{mm} = 0.003 \, \text{m} \)

$$ \varepsilon = \frac{\Delta L}{L_0} $$$$ \varepsilon = \frac{0.003}{4} $$$$ \varepsilon = 0.00075 $$ Answer: 0.00075 (dimensionless)

1.1.3 Relationship Between Stress and Strain (Hooke's Law)

Within the elastic limit of a material, stress and strain are proportional to each other. This relationship is governed by Hooke's Law, which allows calculation of one quantity when the other and the material's modulus of elasticity are known.

Hooke's Law is expressed as:

$$ \sigma = E \times \varepsilon $$

where
\( \sigma \) = stress (Pa),
\( E \) = modulus of elasticity (Pa),
\( \varepsilon \) = strain (dimensionless).

Worked Examples

Example 1: A steel bar with modulus of elasticity \( E = 200 \times 10^9 \, \text{Pa} \) experiences a strain of 0.0005 in a county government building. Calculate the stress.

Given:
\( E = 200 \times 10^9 \, \text{Pa} \),
\( \varepsilon = 0.0005 \)

$$ \sigma = E \times \varepsilon $$$$ \sigma = 200 \times 10^9 \times 0.0005 $$$$ \sigma = 100,000,000 \, \text{Pa} $$ Answer: 100,000,000 Pa (100 MPa)

Example 2: A concrete column in a hospital project has modulus of elasticity \( E = 25 \times 10^9 \, \text{Pa} \) and experiences a strain of 0.0012. Calculate the stress.

Given:
\( E = 25 \times 10^9 \, \text{Pa} \),
\( \varepsilon = 0.0012 \)

$$ \sigma = E \times \varepsilon $$$$ \sigma = 25 \times 10^9 \times 0.0012 $$$$ \sigma = 30,000,000 \, \text{Pa} $$ Answer: 30,000,000 Pa (30 MPa)

Example 3: A steel tie rod in a retail building has a modulus of elasticity \( E = 210 \times 10^9 \, \text{Pa} \) and is subjected to a stress of 63 MPa. Find the strain.

Given:
\( E = 210 \times 10^9 \, \text{Pa} \),
\( \sigma = 63,000,000 \, \text{Pa} \)

$$ \varepsilon = \frac{\sigma}{E} $$$$ \varepsilon = \frac{63,000,000}{210 \times 10^9} $$$$ \varepsilon = 0.0003 $$ Answer: 0.0003 (dimensionless)

Example 4: A wooden beam in a school has modulus of elasticity \( E = 12 \times 10^9 \, \text{Pa} \). If the stress is 6 MPa, calculate the strain.

Given:
\( E = 12 \times 10^9 \, \text{Pa} \),
\( \sigma = 6,000,000 \, \text{Pa} \)

$$ \varepsilon = \frac{\sigma}{E} $$$$ \varepsilon = \frac{6,000,000}{12 \times 10^9} $$$$ \varepsilon = 0.0005 $$ Answer: 0.0005 (dimensionless)

Example 5: A steel reinforcement bar in a SACCO building experiences a strain of 0.0012. Given modulus of elasticity \( E = 205 \times 10^9 \, \text{Pa} \), calculate the stress.

Given:
\( E = 205 \times 10^9 \, \text{Pa} \),
\( \varepsilon = 0.0012 \)

$$ \sigma = E \times \varepsilon $$$$ \sigma = 205 \times 10^9 \times 0.0012 $$$$ \sigma = 246,000,000 \, \text{Pa} $$ Answer: 246,000,000 Pa (246 MPa)

1.1.4 Types of Stress and Strain in Building Materials

In building technology, stress and strain occur in various forms depending on the nature of the load and deformation. The main types include tensile, compressive, shear, and volumetric stress and strain.

Types of Stress

• Tensile stress: Occurs when a material is subjected to forces trying to elongate it. It causes stretching and elongation.
• Compressive stress: Results from forces pushing or squashing a material, causing shortening.
• Shear stress: Arises when forces act parallel to the surface, causing sliding deformation.
• Volumetric stress: Involves uniform pressure acting in all directions, changing the volume without shape distortion.

Types of Strain

• Tensile strain: Elongation per unit length due to tensile stress.
• Compressive strain: Shortening per unit length due to compressive stress.
• Shear strain: Angular distortion caused by shear stress.
• Volumetric strain: Change in volume relative to original volume under volumetric stress.

Worked Examples

Example 1: A steel plate in a county government office experiences a shear force of 10 kN over an area of 0.02 m². Calculate the shear stress.

Given:
\( F = 10,000 \, \text{N} \),
\( A = 0.02 \, \text{m}^2 \)

$$ \tau = \frac{F}{A} $$$$ \tau = \frac{10,000}{0.02} $$$$ \tau = 500,000 \, \text{Pa} $$ Answer: 500,000 Pa (0.5 MPa)

Example 2: A concrete cube in a hospital foundation has a volume of 0.001 m³ and undergoes a volumetric change of 0.0001 m³ under pressure. Calculate the volumetric strain.

Given:
\( V_0 = 0.001 \, \text{m}^3 \),
\( \Delta V = 0.0001 \, \text{m}^3 \)

$$ \varepsilon_v = \frac{\Delta V}{V_0} $$$$ \varepsilon_v = \frac{0.0001}{0.001} $$$$ \varepsilon_v = 0.1 $$ Answer: 0.1 (dimensionless)

Example 3: A timber beam in a school subjected to a shear force causes an angular displacement of 0.002 radians. Calculate the shear strain.

Given:
Shear strain \( \gamma = 0.002 \, \text{radians} \)

Answer: 0.002 (dimensionless shear strain)

Example 4: A steel rod 1.2 m long undergoes elongation of 0.6 mm under tensile load. Calculate the tensile strain.

Given:
\( L_0 = 1.2 \, \text{m} \),
\( \Delta L = 0.6 \, \text{mm} = 0.0006 \, \text{m} \)

$$ \varepsilon = \frac{\Delta L}{L_0} $$$$ \varepsilon = \frac{0.0006}{1.2} $$$$ \varepsilon = 0.0005 $$ Answer: 0.0005 (dimensionless)

Example 5: A wooden post shortens by 1 mm under compressive load. Original length is 0.8 m. Calculate compressive strain.

Given:
\( L_0 = 0.8 \, \text{m} \),
\( \Delta L = 1 \, \text{mm} = 0.001 \, \text{m} \)

$$ \varepsilon = \frac{\Delta L}{L_0} $$$$ \varepsilon = \frac{0.001}{0.8} $$$$ \varepsilon = 0.00125 $$ Answer: 0.00125 (dimensionless)

Chapter Summary

This chapter introduced the fundamental concepts of stress and strain, defining stress as the internal force per unit area within materials and strain as the measure of deformation representing the change in length relative to the original length. It then detailed how to perform calculations for both stress and strain using relevant formulas, emphasizing the importance of accurate measurement of forces and dimensions in structural elements. The chapter also explored the stress-strain diagram, illustrating the relationship between stress and strain for different materials and highlighting key points such as the elastic limit, yield point, and ultimate strength. This graphical representation helps in understanding material behavior under load, including elastic and plastic deformation phases. Overall, the chapter provided the essential theoretical and computational tools necessary for analyzing material response in structural engineering contexts.