By the end of this chapter, you will be able to:
Mastering these skills helps you understand how materials behave under load, which is essential for designing safe and reliable structures in your trade.
Structural analysis is crucial for quantity surveyors to accurately estimate material quantities and costs by understanding how forces affect building elements. This chapter focuses on computing stress and strain, key parameters that describe internal forces and deformations in structural components. Mastery of these calculations enables quantity surveyors to assess structural integrity and plan resource allocation effectively in Kenyan construction projects.
Stress and strain quantify the internal forces and deformations that structural materials experience under loads. For quantity surveyors, understanding different stress types such as combined, bending, and shear stress is essential to interpret structural behavior and estimate reinforcement and material quantities. This section covers the fundamental formulas and calculation methods for these stresses and strain in common structural elements.
Combined stress occurs when a structural element is subjected to more than one type of load simultaneously, resulting in a combination of stresses such as axial and bending stress acting together. This situation is typical in columns supporting beams or in beams subjected to axial loads and bending moments. The ability to compute combined stress helps quantity surveyors determine accurate load effects and material requirements.
The general formula for combined normal stress at a section is:
$$\sigma = \frac{P}{A} \pm \frac{M y}{I}$$
where:
Example 1: A rectangular concrete column has a cross-section of 300 mm by 400 mm and carries an axial load of 500 kN. If the bending moment at the section is 20 kNm, compute the combined stress at the extreme fiber 200 mm from the neutral axis.
Given:
\(P = 500 \times 10^3\, N\),
\(b = 0.3\, m\),
\(h = 0.4\, m\),
\(M = 20 \times 10^3\, Nm\),
\(y = 0.2\, m\).
Calculate cross-sectional area:$$A = b \times h = 0.3 \times 0.4 = 0.12\, m^2$$
Calculate moment of inertia for rectangle about neutral axis:$$I = \frac{b h^3}{12} = \frac{0.3 \times (0.4)^3}{12} = \frac{0.3 \times 0.064}{12} = 0.0016\, m^4$$
Calculate axial stress:$$\frac{P}{A} = \frac{500000}{0.12} = 4166666.67\, Pa = 4.17\, MPa$$
Calculate bending stress:$$\frac{M y}{I} = \frac{20000 \times 0.2}{0.0016} = \frac{4000}{0.0016} = 2,500,000\, Pa = 2.5\, MPa$$
Calculate combined stress at extreme fiber (compression side):$$\sigma = 4.17 + 2.5 = 6.67\, MPa$$
Answer: 6.67 MPa (compression)
Example 2: A steel beam with a circular cross-section diameter 100 mm carries an axial tension of 100 kN and a bending moment of 5 kNm. Find the combined stress at the outer fiber.
Given:
\(P = 100 \times 10^3\, N\),
\(d = 0.1\, m\),
\(M = 5 \times 10^3\, Nm\),
\(y = \frac{d}{2} = 0.05\, m\).
Calculate cross-sectional area:$$A = \frac{\pi d^2}{4} = \frac{3.1416 \times (0.1)^2}{4} = 0.00785\, m^2$$
Calculate moment of inertia for circle:$$I = \frac{\pi d^4}{64} = \frac{3.1416 \times (0.1)^4}{64} = 4.91 \times 10^{-6}\, m^4$$
Calculate axial stress:$$\frac{P}{A} = \frac{100000}{0.00785} = 12,738,853.5\, Pa = 12.74\, MPa$$
Calculate bending stress:$$\frac{M y}{I} = \frac{5000 \times 0.05}{4.91 \times 10^{-6}} = \frac{250}{4.91 \times 10^{-6}} = 50,918,367\, Pa = 50.92\, MPa$$
Calculate combined stress (tension side):$$\sigma = 12.74 + 50.92 = 63.66\, MPa$$
Answer: 63.66 MPa (tension)
Example 3: A rectangular timber beam 150 mm wide and 250 mm deep is subjected to an axial compression of 30 kN and a bending moment of 4 kNm. Find the combined stress at the top fiber.
Given:
\(P = 30 \times 10^3\, N\),
\(b = 0.15\, m\),
\(h = 0.25\, m\),
\(M = 4 \times 10^3\, Nm\),
\(y = 0.125\, m\).
Calculate cross-sectional area:$$A = 0.15 \times 0.25 = 0.0375\, m^2$$
Calculate moment of inertia:$$I = \frac{b h^3}{12} = \frac{0.15 \times (0.25)^3}{12} = \frac{0.15 \times 0.015625}{12} = 0.000195\, m^4$$
Calculate axial stress:$$\frac{P}{A} = \frac{30000}{0.0375} = 800,000\, Pa = 0.8\, MPa$$
Calculate bending stress:$$\frac{M y}{I} = \frac{4000 \times 0.125}{0.000195} = \frac{500}{0.000195} = 2,564,102.56\, Pa = 2.56\, MPa$$
Calculate combined stress (compression side):$$\sigma = 0.8 + 2.56 = 3.36\, MPa$$
Answer: 3.36 MPa (compression)
A steel column 200 mm by 300 mm carries an axial load of 600 kN and a bending moment of 50 kNm. Calculate the combined stress at the extreme fiber 150 mm from the neutral axis. (10 marks)
A circular concrete column diameter 400 mm is subjected to an axial tension of 80 kN and a bending moment of 10 kNm. Determine the combined stress at the outer fiber. (10 marks)
A rectangular timber beam 100 mm wide and 200 mm deep carries an axial compression load of 40 kN and a bending moment of 6 kNm. Compute the combined stress at the extreme fiber. (10 marks)
Bending stress arises when a beam or structural member is subjected to moments causing it to bend, resulting in tension on one side and compression on the other. Quantity surveyors need to understand bending stress to estimate reinforcement and material strength requirements. The calculation relies on the bending moment, section geometry, and material properties.
The bending stress formula is:
$$\sigma = \frac{M y}{I}$$
where:
Example 1: A simply supported concrete beam 250 mm wide and 500 mm deep carries a bending moment of 40 kNm. Calculate the bending stress at the extreme fiber.
Given:
\(b = 0.25\, m\),
\(h = 0.5\, m\),
\(M = 40 \times 10^3\, Nm\),
\(y = 0.25\, m\).
Calculate moment of inertia:$$I = \frac{b h^3}{12} = \frac{0.25 \times (0.5)^3}{12} = \frac{0.25 \times 0.125}{12} = 0.0026\, m^4$$
Calculate bending stress:$$\sigma = \frac{M y}{I} = \frac{40000 \times 0.25}{0.0026} = \frac{10000}{0.0026} = 3,846,153.85\, Pa = 3.85\, MPa$$
Answer: 3.85 MPa
Example 2: A steel I-beam section has \(I = 8 \times 10^{-6} m^4\) and the distance from neutral axis to extreme fiber is 75 mm. If the bending moment is 15 kNm, find the bending stress.
Given:
\(I = 8 \times 10^{-6} m^4\),
\(y = 0.075\, m\),
\(M = 15 \times 10^3\, Nm\).
Calculate bending stress:$$\sigma = \frac{M y}{I} = \frac{15000 \times 0.075}{8 \times 10^{-6}} = \frac{1125}{8 \times 10^{-6}} = 140,625,000\, Pa = 140.63\, MPa$$
Answer: 140.63 MPa
Example 3: A wooden beam 120 mm wide and 300 mm deep carries a bending moment of 8 kNm. Calculate the bending stress at the extreme fiber.
Given:
\(b = 0.12\, m\),
\(h = 0.3\, m\),
\(M = 8 \times 10^3\, Nm\),
\(y = 0.15\, m\).
Calculate moment of inertia:$$I = \frac{b h^3}{12} = \frac{0.12 \times (0.3)^3}{12} = \frac{0.12 \times 0.027}{12} = 0.00027\, m^4$$
Calculate bending stress:$$\sigma = \frac{M y}{I} = \frac{8000 \times 0.15}{0.00027} = \frac{1200}{0.00027} = 4,444,444.44\, Pa = 4.44\, MPa$$
Answer: 4.44 MPa
Calculate the bending stress at the extreme fiber of a steel beam with \(I = 1.2 \times 10^{-5} m^4\), \(y = 50 mm\), under a bending moment of 20 kNm. (10 marks)
A timber beam 150 mm wide and 350 mm deep carries a bending moment of 12 kNm. Find the bending stress at the extreme fiber. (10 marks)
A reinforced concrete beam with cross-section 300 mm by 500 mm is subjected to a bending moment of 60 kNm. Calculate the bending stress at the outer fiber. (10 marks)
Shear stress occurs when forces act parallel to the cross-section of a structural element, causing sliding failure along planes. Quantity surveyors must calculate shear stress to ensure adequate shear reinforcement and prevent shear failure. Shear stress is often critical near supports and load points.
The shear stress formula for rectangular sections is:
$$\tau = \frac{V Q}{I b}$$
where:
For rectangular sections, \(Q\) at the neutral axis is:
$$Q = \bar{y} A'$$
where:
Example 1: Calculate the maximum shear stress in a rectangular concrete beam 300 mm wide and 500 mm deep carrying a shear force of 100 kN.
Given:
\(b = 0.3\, m\),
\(h = 0.5\, m\),
\(V = 100 \times 10^3\, N\).
Calculate moment of inertia:$$I = \frac{b h^3}{12} = \frac{0.3 \times (0.5)^3}{12} = 0.0026\, m^4$$
Calculate \(Q\) at neutral axis:
Area above neutral axis:$$A' = b \times \frac{h}{2} = 0.3 \times 0.25 = 0.075\, m^2$$
Centroid of \(A'\) from neutral axis:$$\bar{y} = \frac{h}{4} = 0.125\, m$$
Calculate \(Q\):$$Q = \bar{y} \times A' = 0.125 \times 0.075 = 0.009375\, m^3$$
Calculate shear stress:$$\tau = \frac{V Q}{I b} = \frac{100000 \times 0.009375}{0.0026 \times 0.3} = \frac{937.5}{0.00078} = 1,201,923\, Pa = 1.20\, MPa$$
Answer: 1.20 MPa
Example 2: A steel beam 150 mm wide and 300 mm deep carries a shear force of 50 kN. Calculate the maximum shear stress.
Given:
\(b = 0.15\, m\),
\(h = 0.3\, m\),
\(V = 50 \times 10^3\, N\).
Calculate moment of inertia:$$I = \frac{b h^3}{12} = \frac{0.15 \times (0.3)^3}{12} = 0.00034\, m^4$$
Calculate \(Q\):$$A' = 0.15 \times 0.15 = 0.0225\, m^2$$$$\bar{y} = \frac{h}{4} = 0.075\, m$$$$Q = 0.075 \times 0.0225 = 0.0016875\, m^3$$
Calculate shear stress:$$\tau = \frac{50000 \times 0.0016875}{0.00034 \times 0.15} = \frac{84.375}{0.000051} = 1,653,676\, Pa = 1.65\, MPa$$
Answer: 1.65 MPa
Example 3: A rectangular timber beam 100 mm wide and 200 mm deep carries a shear force of 20 kN. Find the maximum shear stress.
Given:
\(b = 0.1\, m\),
\(h = 0.2\, m\),
\(V = 20 \times 10^3\, N\).
Calculate moment of inertia:$$I = \frac{b h^3}{12} = \frac{0.1 \times (0.2)^3}{12} = 0.000067\, m^4$$
Calculate \(Q\):$$A' = 0.1 \times 0.1 = 0.01\, m^2$$$$\bar{y} = \frac{h}{4} = 0.05\, m$$$$Q = 0.05 \times 0.01 = 0.0005\, m^3$$
Calculate shear stress:$$\tau = \frac{20000 \times 0.0005}{0.000067 \times 0.1} = \frac{10}{0.0000067} = 1,492,537\, Pa = 1.49\, MPa$$
Answer: 1.49 MPa
A concrete beam 250 mm wide and 400 mm deep carries a shear force of 80 kN. Calculate the maximum shear stress in the beam. (10 marks)
A steel beam 200 mm wide and 350 mm deep is subjected to a shear force of 60 kN. Compute the maximum shear stress. (10 marks)
Calculate the shear stress at the neutral axis of a timber beam 120 mm wide and 250 mm deep carrying a shear force of 25 kN. (10 marks)
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Create a free accountThis chapter focused on identifying various types of stress including combined stress, bending stress, and shear stress, essential for understanding how materials respond under different loading conditions. It then explored the calculation of compression stresses, emphasizing key characteristics of structural elements such as strength, stiffness, durability, elasticity, and fire resistance. The discussion extended to different structural components like beams, columns, foundations, slabs, walls, roof trusses, and bracing systems, highlighting their roles in supporting loads and maintaining stability. The nature of compression members was examined to understand their behavior under axial loads. Finally, the chapter demonstrated how to draw the stress-strain diagram, which is fundamental in visualizing material deformation and failure limits. Together, these topics provide a comprehensive foundation for analyzing and computing stress and strain in structural engineering contexts.
A steel rod with a cross-sectional area of \(50 \text{ mm}^2\) is subjected to an axial tensile force of \(10 \text{ kN}\). Calculate the normal stress in the rod. (2 marks)
A rectangular beam of width \(100 \text{ mm}\) and depth \(200 \text{ mm}\) carries a bending moment of \(5 \text{ kNm}\). Calculate the bending stress at the extreme fiber of the beam. (3 marks)
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