Horizontal shear stress is a critical factor in the design and analysis of building elements such as beams, slabs, and walls. For Building Technology professionals in Kenya, understanding how to compute horizontal shear stress distribution ensures structural safety and durability, especially under lateral loads like wind or seismic activity common in various regions. This chapter focuses on the mathematical principles and methods used to determine the distribution of horizontal shear stresses in structural members, enabling effective design and assessment.

5.1 Introduction to Shear Stress

Shear stress appears in structural elements when forces act parallel or tangential to a surface, causing layers of material to slide relative to each other. In building structures, horizontal shear stresses are often induced by lateral loads or load transfers between floors and walls. Accurate computation of these stresses is essential for preventing shear failure, which can be sudden and catastrophic.

5.1.1 Concept of Shear Stress

Shear stress, denoted by \(\tau\), is defined as the force per unit area acting parallel to the cross-sectional area of a material. It arises when forces cause deformation by sliding layers within the material.

The governing formula for shear stress is:

$$ \tau = \frac{VQ}{Ib} $$

Where:

• \(V\) = Shear force at the section (N)
• \(Q\) = First moment of area about the neutral axis (mm\(^3\))
• \(I\) = Moment of inertia of the entire cross-section about the neutral axis (mm\(^4\))
• \(b\) = Width of the section at the point where shear is calculated (mm)

Worked Examples

Example 1:
Given: A concrete beam with a shear force \(V = 50,000\) N, moment of inertia \(I = 8 \times 10^8\) mm\(^4\), width \(b = 200\) mm, and first moment of area \(Q = 1.2 \times 10^6\) mm\(^3\).
Calculate the shear stress \(\tau\).

$$ \tau = \frac{VQ}{Ib} $$

$$ \tau = \frac{50,000 \times 1.2 \times 10^6}{8 \times 10^8 \times 200} $$

$$ \tau = \frac{6 \times 10^{10}}{1.6 \times 10^{11}} = 0.375 \, \text{N/mm}^2 $$

Answer: 0.375 N/mm\(^2\)

Example 2:
Given: A steel beam with \(V = 120,000\) N, \(I = 1.5 \times 10^9\) mm\(^4\), \(b = 150\) mm, and \(Q = 2.5 \times 10^6\) mm\(^3\).
Find the shear stress \(\tau\).

$$ \tau = \frac{VQ}{Ib} $$

$$ \tau = \frac{120,000 \times 2.5 \times 10^6}{1.5 \times 10^9 \times 150} $$

$$ \tau = \frac{3 \times 10^{11}}{2.25 \times 10^{11}} = 1.333 \, \text{N/mm}^2 $$

Answer: 1.333 N/mm\(^2\)

Example 3:
Given: A rectangular timber beam with \(V = 30,000\) N, \(I = 6 \times 10^7\) mm\(^4\), width \(b = 100\) mm, and \(Q = 2 \times 10^5\) mm\(^3\).
Calculate \(\tau\).

$$ \tau = \frac{VQ}{Ib} $$

$$ \tau = \frac{30,000 \times 2 \times 10^5}{6 \times 10^7 \times 100} $$

$$ \tau = \frac{6 \times 10^9}{6 \times 10^9} = 1 \, \text{N/mm}^2 $$

Answer: 1 N/mm\(^2\)

Example 4:
Given: A reinforced concrete slab with \(V = 70,000\) N, \(I = 1.2 \times 10^9\) mm\(^4\), \(b = 250\) mm, and \(Q = 1.8 \times 10^6\) mm\(^3\).
Find the horizontal shear stress.

$$ \tau = \frac{VQ}{Ib} $$

$$ \tau = \frac{70,000 \times 1.8 \times 10^6}{1.2 \times 10^9 \times 250} $$

$$ \tau = \frac{1.26 \times 10^{11}}{3 \times 10^{11}} = 0.42 \, \text{N/mm}^2 $$

Answer: 0.42 N/mm\(^2\)

Example 5:
Given: A masonry wall element with \(V = 40,000\) N, \(I = 5 \times 10^8\) mm\(^4\), \(b = 300\) mm, and \(Q = 7.5 \times 10^5\) mm\(^3\).
Calculate the shear stress \(\tau\).

$$ \tau = \frac{VQ}{Ib} $$

$$ \tau = \frac{40,000 \times 7.5 \times 10^5}{5 \times 10^8 \times 300} $$

$$ \tau = \frac{3 \times 10^{10}}{1.5 \times 10^{11}} = 0.2 \, \text{N/mm}^2 $$

Answer: 0.2 N/mm\(^2\)

5.1.2 Calculation of First Moment of Area \(Q\)

The first moment of area \(Q\) is the measure of the area above (or below) the point where shear stress is being calculated, multiplied by the distance from the centroid of that area to the neutral axis.

$$ Q = \bar{y} A' $$

Where:

• \(A'\) = Area of the section above (or below) the point where shear is calculated (mm\(^2\))
• \(\bar{y}\) = Distance from the neutral axis to the centroid of \(A'\) (mm)

Worked Examples

Example 1:
Given: A rectangular beam of width 200 mm and total depth 400 mm. Calculate \(Q\) at a point 100 mm from the top.

Area above the point:

$$ A' = b \times h' = 200 \times 100 = 20,000 \, \text{mm}^2 $$

Centroid of \(A'\) from neutral axis (neutral axis at mid-depth 200 mm):

$$ \bar{y} = 200 - \frac{100}{2} = 150 \, \text{mm} $$

Calculate \(Q\):

$$ Q = \bar{y} A' = 150 \times 20,000 = 3,000,000 \, \text{mm}^3 $$

Answer: 3.0 \times 10^6 mm\(^3\)

Example 2:
Given: Same beam but find \(Q\) at 50 mm from the top.

$$ A' = 200 \times 50 = 10,000 \, \text{mm}^2 $$

$$ \bar{y} = 200 - \frac{50}{2} = 175 \, \text{mm} $$

$$ Q = 175 \times 10,000 = 1,750,000 \, \text{mm}^3 $$

Answer: 1.75 \times 10^6 mm\(^3\)

Example 3:
Given: A T-beam flange width \(b_f = 300\) mm, thickness \(t_f = 50\) mm, neutral axis at 180 mm from top. Find \(Q\) at the bottom of the flange.

Area of flange:

$$ A' = b_f \times t_f = 300 \times 50 = 15,000 \, \text{mm}^2 $$

Distance from neutral axis to centroid of flange:

$$ \bar{y} = 180 - \frac{50}{2} = 155 \, \text{mm} $$

Calculate \(Q\):

$$ Q = 155 \times 15,000 = 2,325,000 \, \text{mm}^3 $$

Answer: 2.325 \times 10^6 mm\(^3\)

Example 4:
Given: A rectangular section 250 mm wide and 500 mm deep. Calculate \(Q\) at 300 mm from the top.

Area above the point:

$$ A' = 250 \times 300 = 75,000 \, \text{mm}^2 $$

Neutral axis at mid-depth 250 mm:

$$ \bar{y} = 250 - \frac{300}{2} = 100 \, \text{mm} $$

Calculate \(Q\):

$$ Q = 100 \times 75,000 = 7,500,000 \, \text{mm}^3 $$

Answer: 7.5 \times 10^6 mm\(^3\)

Example 5:
Given: A rectangular beam 150 mm wide and 450 mm deep, neutral axis at 225 mm. Calculate \(Q\) at 100 mm from the bottom.

Area below point:

$$ A' = 150 \times 100 = 15,000 \, \text{mm}^2 $$

Distance from neutral axis to centroid of area below point:

$$ \bar{y} = 225 - \frac{100}{2} = 175 \, \text{mm} $$

Calculate \(Q\):

$$ Q = 175 \times 15,000 = 2,625,000 \, \text{mm}^3 $$

Answer: 2.625 \times 10^6 mm\(^3\)

5.1.3 Moment of Inertia \(I\) for Common Sections

Moment of inertia \(I\) quantifies a section’s resistance to bending and is vital in shear stress calculations. For common building elements, \(I\) is calculated about the neutral axis.

For a rectangular section:

$$ I = \frac{b h^{3}}{12} $$

Where:

• \(b\) = width of the section (mm)
• \(h\) = depth of the section (mm)

Worked Examples

Example 1:
Calculate \(I\) for a rectangular beam 200 mm wide and 400 mm deep.

$$ I = \frac{200 \times 400^{3}}{12} $$

$$ I = \frac{200 \times 64,000,000}{12} = \frac{12,800,000,000}{12} $$

$$ I = 1,066,666,667 \, \text{mm}^4 $$

Answer: 1.07 \times 10^9 mm\(^4\)

Example 2:
Calculate \(I\) for a rectangular timber beam 150 mm wide and 300 mm deep.

$$ I = \frac{150 \times 300^{3}}{12} $$

$$ I = \frac{150 \times 27,000,000}{12} = \frac{4,050,000,000}{12} $$

$$ I = 337,500,000 \, \text{mm}^4 $$

Answer: 3.38 \times 10^8 mm\(^4\)

Example 3:
Given a square column 250 mm by 250 mm, calculate \(I\).

$$ I = \frac{250 \times 250^{3}}{12} $$

$$ I = \frac{250 \times 15,625,000}{12} = \frac{3,906,250,000}{12} $$

$$ I = 325,520,833 \, \text{mm}^4 $$

Answer: 3.26 \times 10^8 mm\(^4\)

Example 4:
Calculate \(I\) for a rectangular slab 300 mm wide and 100 mm thick.

$$ I = \frac{300 \times 100^{3}}{12} $$

$$ I = \frac{300 \times 1,000,000}{12} = \frac{300,000,000}{12} $$

$$ I = 25,000,000 \, \text{mm}^4 $$

Answer: 2.5 \times 10^7 mm\(^4\)

Example 5:
Calculate \(I\) for a rectangular beam 100 mm wide and 450 mm deep.

$$ I = \frac{100 \times 450^{3}}{12} $$

$$ I = \frac{100 \times 91,125,000}{12} = \frac{9,112,500,000}{12} $$

$$ I = 759,375,000 \, \text{mm}^4 $$

Answer: 7.59 \times 10^8 mm\(^4\)

5.1.4 Width \(b\) at the Point of Shear

The width \(b\) in the shear stress formula represents the thickness of the material where the shear stress is being calculated. This value varies depending on the cross-sectional shape and the location along the depth.

For rectangular sections, \(b\) is constant and equals the width of the section. For I-beams and T-beams, \(b\) changes as the shear stress is calculated in the web or flange.

Worked Examples

Example 1:
Calculate shear stress at the mid-depth of a rectangular beam 200 mm wide (width is constant).

Width \(b = 200\) mm.

Answer: Width \(b\) = 200 mm

Example 2:
For an I-beam with flange width 300 mm and web thickness 10 mm, find \(b\) at the web.

Width \(b = 10\) mm.

Answer: Width \(b\) = 10 mm

Example 3:
For a T-beam with flange width 250 mm and web thickness 15 mm, calculate \(b\) at the flange-web junction.

Width \(b = 250\) mm at flange, \(b = 15\) mm at web.

Answer: Width \(b\) varies by location; at flange \(b=250\) mm, at web \(b=15\) mm

Example 4:
Calculate shear stress width at 100 mm from the top in a rectangular beam 150 mm wide.

Width \(b = 150\) mm (constant).

Answer: Width \(b = 150\) mm

Example 5:
For a slab 300 mm wide and 120 mm thick, calculate \(b\) at the mid-depth.

Width \(b = 300\) mm.

Answer: Width \(b = 300\) mm

Chapter Summary

This chapter introduces the concept of shear stress as an internal force that acts parallel to the cross-section of a structural element, crucial for understanding how materials resist sliding failure. It explains the relationship between shear force, bending moment, and shear stress, highlighting how variations in shear force along a beam influence the internal distribution of stresses. The chapter details the shear stress formula, demonstrating how to compute shear stress at any point in a beam’s cross-section using the shear force, moment of inertia, and the first moment of area. Additionally, it guides the plotting of shear force distribution across a beam’s section, enabling visualization of how shear forces vary along the length of structural members. Understanding these principles is essential for safe and efficient structural design, ensuring that beams can withstand applied loads without excessive deformation or failure. The integration of theory with calculation methods prepares students to analyze real-world structural elements accurately. Overall, the chapter builds a foundation for advanced analysis of internal forces in beams and other structural components.