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Construction structural design principles require a solid understanding of the types of stresses that materials and components experience under load. For architects in Kenya, applying these principles ensures buildings are safe, durable, and compliant with local building codes. This chapter focuses on identifying key stress types, combined stress, bending stress, and shear stress, that commonly occur in architectural structures. Each stress type will be analyzed with relevant formulas and worked examples to equip students with practical calculation skills.
Stress is a measure of internal force per unit area within materials subjected to external loads. In architectural structural components such as beams, columns, and slabs, stresses arise from various load conditions. Understanding how to identify and calculate these stresses is fundamental for safe design.
Combined stress occurs when a structural element is subjected simultaneously to more than one type of stress, typically axial stress combined with bending stress or shear stress. This is common in columns supporting eccentric loads or beams with axial forces.
The principal formula for combined normal stress is:
$$ \sigma = \sigma_{axial} \pm \sigma_{bending} $$
where
Example 1: A rectangular column with cross-section \(300\,mm \times 400\,mm\) carries an axial load of \(500\,kN\) and a bending moment of \(20\,kNm\). Calculate the maximum and minimum normal stresses at the extreme fibers.
Given:
\(P = 500,000\,N\),
\(M = 20,000\,Nm\),
\(b = 0.3\,m\),
\(h = 0.4\,m\),
\(y = \frac{h}{2} = 0.2\,m\),
\(A = b \times h = 0.3 \times 0.4 = 0.12\,m^2\),
\(I = \frac{b h^3}{12} = \frac{0.3 \times 0.4^3}{12} = 0.0016\,m^4\)
Step 1: Calculate axial stress$$ \sigma_{axial} = \frac{P}{A} = \frac{500,000}{0.12} = 4,166,667\,Pa = 4.167\,MPa $$
Step 2: Calculate bending stress$$ \sigma_{bending} = \frac{M y}{I} = \frac{20,000 \times 0.2}{0.0016} = 2,500,000\,Pa = 2.5\,MPa $$
Step 3: Calculate max and min stresses$$ \sigma_{max} = \sigma_{axial} + \sigma_{bending} = 4.167 + 2.5 = 6.667\,MPa $$$$ \sigma_{min} = \sigma_{axial} - \sigma_{bending} = 4.167 - 2.5 = 1.667\,MPa $$
Answer: Maximum stress = 6.667 MPa (tension), Minimum stress = 1.667 MPa (tension)
Example 2: A steel beam with circular cross-section diameter \(200\,mm\) supports an axial compressive load of \(300\,kN\) and a bending moment of \(15\,kNm\). Determine the maximum compressive and tensile stresses.
Given:
\(P = 300,000\,N\),
\(M = 15,000\,Nm\),
\(d = 0.2\,m\),
\(A = \frac{\pi d^2}{4} = \frac{3.1416 \times 0.2^2}{4} = 0.0314\,m^2\),
\(I = \frac{\pi d^4}{64} = \frac{3.1416 \times 0.2^4}{64} = 7.854 \times 10^{-5}\,m^4\),
\(y = \frac{d}{2} = 0.1\,m\)
Step 1: Axial stress$$ \sigma_{axial} = \frac{P}{A} = \frac{300,000}{0.0314} = 9,554,140\,Pa = 9.554\,MPa $$ (compression)
Step 2: Bending stress$$ \sigma_{bending} = \frac{M y}{I} = \frac{15,000 \times 0.1}{7.854 \times 10^{-5}} = 19,098,593\,Pa = 19.099\,MPa $$ (tension on one side, compression on other)
Step 3: Max compressive stress (axial + bending)$$ \sigma_{max} = 9.554 + 19.099 = 28.653\,MPa $$
Step 4: Min stress (axial, bending)$$ \sigma_{min} = 9.554 - 19.099 = -9.545\,MPa $$ (tension)
Answer: Maximum compressive stress = 28.653 MPa, Maximum tensile stress = 9.545 MPa
Example 3: A rectangular timber beam \(150\,mm \times 250\,mm\) is subjected to an axial tension of \(50\,kN\) and a bending moment of \(5\,kNm\). Calculate the stresses at the top and bottom fibers.
Given:
\(P = 50,000\,N\),
\(M = 5,000\,Nm\),
\(b = 0.15\,m\),
\(h = 0.25\,m\),
\(A = 0.15 \times 0.25 = 0.0375\,m^2\),
\(I = \frac{b h^3}{12} = \frac{0.15 \times 0.25^3}{12} = 1.953 \times 10^{-4}\,m^4\),
\(y = \frac{h}{2} = 0.125\,m\)
Step 1: Axial stress$$ \sigma_{axial} = \frac{P}{A} = \frac{50,000}{0.0375} = 1,333,333\,Pa = 1.333\,MPa $$ (tension)
Step 2: Bending stress$$ \sigma_{bending} = \frac{M y}{I} = \frac{5,000 \times 0.125}{1.953 \times 10^{-4}} = 3,200,000\,Pa = 3.2\,MPa $$
Step 3: Top fiber stress (compression side)$$ \sigma_{top} = \sigma_{axial} - \sigma_{bending} = 1.333 - 3.2 = -1.867\,MPa $$ (compression)
Step 4: Bottom fiber stress (tension side)$$ \sigma_{bottom} = \sigma_{axial} + \sigma_{bending} = 1.333 + 3.2 = 4.533\,MPa $$ (tension)
Answer: Top fiber stress = 1.867 MPa compression, Bottom fiber stress = 4.533 MPa tension
A steel column of cross-section \(250\,mm \times 250\,mm\) carries an axial load of \(400\,kN\) and a bending moment of \(25\,kNm\). Calculate the maximum and minimum stresses at the extreme fibers. (10 marks)
A concrete beam with circular cross-section diameter \(300\,mm\) is subjected to an axial load of \(100\,kN\) and a bending moment of \(30\,kNm\). Find the normal stresses at the top and bottom of the cross-section. (10 marks)
A timber beam \(200\,mm \times 350\,mm\) carries an axial tension of \(70\,kN\) and a bending moment of \(8\,kNm\). Determine the stresses at the extreme fibers. (10 marks)
Bending stress arises in structural elements when external moments cause the member to bend. In architectural beams, bending stress varies linearly from zero at the neutral axis to a maximum at the outermost fibers. The bending stress formula is:
$$ \sigma = \frac{M y}{I} $$
where \(M\) is the bending moment, \(y\) is the distance from the neutral axis, and \(I\) is the moment of inertia of the cross-section.
Example 1: A simply supported rectangular timber beam \(100\,mm \times 200\,mm\) carries a bending moment of \(12\,kNm\) at mid-span. Calculate the maximum bending stress.
Given:
\(M = 12,000\,Nm\),
\(b = 0.1\,m\),
\(h = 0.2\,m\),
\(y = \frac{h}{2} = 0.1\,m\),
\(I = \frac{b h^3}{12} = \frac{0.1 \times 0.2^3}{12} = 6.67 \times 10^{-5}\,m^4\)
Step 1: Apply formula$$ \sigma = \frac{M y}{I} = \frac{12,000 \times 0.1}{6.67 \times 10^{-5}} = 18,000,000\,Pa = 18\,MPa $$
Answer: Maximum bending stress = 18 MPa tension or compression
Example 2: A steel beam with an I-section has \(I = 8 \times 10^{-5}\,m^4\) and distance from neutral axis to extreme fiber \(y = 0.15\,m\). If the bending moment is \(25\,kNm\), find the bending stress.
Given:
\(M = 25,000\,Nm\),
\(I = 8 \times 10^{-5}\,m^4\),
\(y = 0.15\,m\)
Step 1: Calculate bending stress$$ \sigma = \frac{M y}{I} = \frac{25,000 \times 0.15}{8 \times 10^{-5}} = 46,875,000\,Pa = 46.875\,MPa $$
Answer: Maximum bending stress = 46.875 MPa
Example 3: A concrete beam \(300\,mm \times 500\,mm\) is subjected to a bending moment of \(40\,kNm\). Calculate the maximum bending stress.
Given:
\(M = 40,000\,Nm\),
\(b = 0.3\,m\),
\(h = 0.5\,m\),
\(y = \frac{h}{2} = 0.25\,m\),
\(I = \frac{b h^3}{12} = \frac{0.3 \times 0.5^3}{12} = 3.125 \times 10^{-3}\,m^4\)
Step 1: Calculate bending stress$$ \sigma = \frac{M y}{I} = \frac{40,000 \times 0.25}{3.125 \times 10^{-3}} = 3,200,000\,Pa = 3.2\,MPa $$
Answer: Maximum bending stress = 3.2 MPa
Calculate the maximum bending stress in a steel beam with \(I = 1.2 \times 10^{-4} m^4\), \(y = 0.2 m\) under a bending moment of \(50 kNm\). (8 marks)
A timber beam \(150 mm \times 300 mm\) carries a bending moment of \(15 kNm\). Find the bending stress at the extreme fibers. (8 marks)
A concrete beam \(250 mm \times 450 mm\) is subjected to a bending moment of \(35 kNm\). Calculate the bending stress. (8 marks)
Shear stress occurs when forces act parallel to the cross-sectional area, causing layers within the material to slide relative to each other. In beams, shear stresses are highest near supports and can significantly affect design. The shear stress formula is:
$$ \tau = \frac{V Q}{I t} $$
where
\(V\) = shear force,
\(Q\) = first moment of area about the neutral axis,
\(I\) = moment of inertia,
\(t\) = thickness of the section at the point of interest.
Example 1: A rectangular beam \(200\,mm \times 300\,mm\) carries a shear force of \(20\,kN\). Calculate the shear stress at the neutral axis.
Given:
\(V = 20,000\,N\),
\(b = 0.2\,m\),
\(h = 0.3\,m\),
\(I = \frac{b h^3}{12} = \frac{0.2 \times 0.3^3}{12} = 4.5 \times 10^{-4}\,m^4\),
\(t = b = 0.2\,m\),
Calculate \(Q\), the first moment of area of the area above the neutral axis:
Area above neutral axis \(A' = b \times \frac{h}{2} = 0.2 \times 0.15 = 0.03\,m^2\)
Distance from neutral axis to centroid of \(A'\): \(\bar{y} = \frac{h}{4} = 0.075\,m\)$$
Q = A' \times \bar{y} = 0.03 \times 0.075 = 0.00225\,m^3
$$
Step 1: Calculate shear stress$$ \tau = \frac{V Q}{I t} = \frac{20,000 \times 0.00225}{4.5 \times 10^{-4} \times 0.2} = 500,000\,Pa = 0.5\,MPa $$
Answer: Shear stress at neutral axis = 0.5 MPa
Example 2: A steel I-beam has a web thickness \(t = 10\,mm\), \(I = 8 \times 10^{-5} m^4\), and \(Q = 4 \times 10^{-4} m^3\). If the shear force is \(80\,kN\), calculate the shear stress in the web.
Given:
\(V = 80,000\,N\),
\(t = 0.01\,m\),
\(I = 8 \times 10^{-5}\,m^4\),
\(Q = 4 \times 10^{-4}\,m^3\)
Step 1: Calculate shear stress$$ \tau = \frac{V Q}{I t} = \frac{80,000 \times 4 \times 10^{-4}}{8 \times 10^{-5} \times 0.01} = 40,000,000\,Pa = 40\,MPa $$
Answer: Shear stress in the web = 40 MPa
Example 3: A concrete rectangular beam \(250\,mm \times 500\,mm\) experiences a shear force of \(50\,kN\). Calculate the shear stress at the neutral axis.
Given:
\(V = 50,000\,N\),
\(b = 0.25\,m\),
\(h = 0.5\,m\),
\(I = \frac{b h^3}{12} = \frac{0.25 \times 0.5^3}{12} = 2.604 \times 10^{-3}\,m^4\),
\(t = b = 0.25\,m\)
Calculate \(Q\):
Area above neutral axis \(A' = b \times \frac{h}{2} = 0.25 \times 0.25 = 0.0625\,m^2\)
Distance from neutral axis to centroid of \(A'\): \(\bar{y} = \frac{h}{4} = 0.125\,m\)$$
Q = A' \times \bar{y} = 0.0625 \times 0.125 = 0.0078125\,m^3
$$
Step 1: Calculate shear stress$$ \tau = \frac{V Q}{I t} = \frac{50,000 \times 0.0078125}{2.604 \times 10^{-3} \times 0.25} = 600,000\,Pa = 0.6\,MPa $$
Answer: Shear stress at neutral axis = 0.6 MPa
Calculate the shear stress at the neutral axis of a beam \(150 mm \times 250 mm\) carrying a shear force of \(15 kN\). (8 marks)
A steel beam web thickness is \(12 mm\) with \(I = 1.5 \times 10^{-4} m^4\), \(Q = 6 \times 10^{-4} m^3\), and shear force \(90 kN\). Find the shear stress in the web. (8 marks)
Find the shear stress at the neutral axis of a concrete beam \(300 mm \times 600 mm\) subjected to a shear force of \(60 kN\). (8 marks)
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Create a free accountThis chapter explored the fundamental principles of structural analysis with a focus on applying construction structural design principles. It began by identifying different types of stresses including combined stress, bending stress, and shear stress, explaining how these stresses affect structural elements. The calculation of compression stresses was then examined, highlighting key characteristics of structural elements such as strength, stiffness, durability, elasticity, and fire resistance. Various structural components including beams, columns, foundations, slabs, walls, roof trusses, and bracing systems were discussed in relation to their behavior under compression. The chapter also addressed the nature of compression members, emphasizing their positive or negative effects on structural performance. Overall, the content provided a comprehensive understanding of how to analyze and design structural elements to ensure safety and functionality in construction projects.
A rectangular concrete column in a school building has a cross-sectional area of \(0.3 \, m \times 0.5 \, m\). If the axial load applied is \(150 \, kN\), calculate the compression stress in the column. (2 marks)
A timber beam supporting a roof truss has a bending moment of \(12 \, kNm\) and a section modulus of \(0.002 \, m^3\). Calculate the bending stress in the beam. (2 marks)
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