Horizontal shear stress is a critical factor in the design and analysis of beams in civil engineering structures such as bridges, buildings, and retaining walls commonly found in Kenya. Understanding how shear stress distributes across a beam’s cross-section ensures safety against shear failure and serviceability under applied loads. This chapter focuses on the principles governing horizontal shear stress in beams under bending and the relationship between vertical loading and horizontal shear, followed by detailed analysis of shear stress distribution in typical beam sections used in Kenyan construction.

5.1 Fundamentals of Horizontal Shear Stress

Horizontal shear stress occurs in beams subjected to vertical loads causing bending, where internal forces resist deformation. It is essential to quantify this shear stress to prevent failure modes such as web buckling or shear cracking, particularly in beams with varying cross-sectional shapes.

5.1.1 Horizontal shear stress in beams under bending

Horizontal shear stress, denoted as τ, arises due to the vertical shear force V acting on a beam cross-section. The fundamental formula for shear stress distribution in the beam’s cross-section is:

τ = V Q ÷ (I t)

where:

• V = shear force at the cross-section (N)
• Q = first moment of area about the neutral axis for the area above (or below) the point where shear stress is calculated (mm³)
• I = moment of inertia of the entire cross-section about the neutral axis (mm⁴)
• t = thickness of the section at the point of interest (mm)

When a beam bends, its layers try to slide over one another, just as the sheets in a stack of loose paper slide when the stack is bent. In a solid beam this sliding is prevented by horizontal shear stresses acting along the layers, and these are always accompanied by equal vertical shear stresses on the cross-section. Example: glued laminated timber beams for the roof of a church hall in Kericho were checked for horizontal shear at the glue lines near the supports, because that is where the layers tend to slide apart.

Characteristics of Horizontal Shear Stress Distribution

• Zero at the top and bottom surfaces: There is no area beyond the outer fibres, so Q is zero and so is the shear stress.
• Maximum at the neutral axis: For most sections, Q is greatest at the neutral axis, giving the highest shear stress there.
• Parabolic in rectangular sections: The shear stress varies as a parabola from zero at the edges to a maximum at the centre.
• Maximum in a rectangle is 1.5 times the average: τ_max = 1.5 V ÷ A for a rectangular section.
• Concentrated in the web of I-sections: The thin web carries most of the shear force, while the flanges carry very little.
• Jumps where the width changes: At the junction of flange and web, the stress increases suddenly because t becomes smaller.

Worked Examples

Example 1: A rectangular concrete beam 300 mm wide and 500 mm deep carries a shear force of 50 kN at a section. Calculate the horizontal shear stress at the neutral axis.

Given:
Width b = 300 mm
Depth h = 500 mm
Shear force V = 50,000 N

I = b h³ ÷ 12

I = 300 × 500³ ÷ 12 = 300 × 125,000,000 ÷ 12

I = 3.125 × 10⁹ mm⁴

Q = A′ × ȳ

Q = (b h ÷ 2) × (h ÷ 4) = (300 × 500 ÷ 2) × 125 = 75,000 × 125

Q = 9,375,000 mm³

At the neutral axis, thickness t = b = 300 mm.

Calculate τ:

τ = V Q ÷ (I t)

τ = 50,000 × 9,375,000 ÷ (3.125 × 10⁹ × 300)

τ = 4.6875 × 10¹¹ ÷ 9.375 × 10¹¹

τ = 0.5 N/mm² = 0.5 MPa

Answer: 0.5 MPa

Example 2: A steel I-beam with a web thickness of 10 mm carries a shear force of 120 kN. The moment of inertia I is 2.5 × 10⁸ mm⁴, and the first moment of area Q at the web is 8 × 10⁵ mm³. Find the shear stress in the web.

Given:
V = 120,000 N
I = 2.5 × 10⁸ mm⁴
Q = 8 × 10⁵ mm³
t = 10 mm

τ = V Q ÷ (I t)

τ = 120,000 × 8 × 10⁵ ÷ (2.5 × 10⁸ × 10)

τ = 9.6 × 10¹⁰ ÷ 2.5 × 10⁹

τ = 38.4 N/mm² = 38.4 MPa

Answer: 38.4 MPa

Example 3: A timber beam 150 mm wide and 250 mm deep supports a shear force of 30 kN. Calculate the horizontal shear stress at a point 50 mm above the neutral axis.

Given:
b = 150 mm
h = 250 mm
V = 30,000 N
Distance from neutral axis to point: 50 mm

Calculate moment of inertia:

I = b h³ ÷ 12

I = 150 × 250³ ÷ 12 = 150 × 15,625,000 ÷ 12

I = 1.953 × 10⁸ mm⁴

Calculate Q: Area above the point is A′ = b × (h ÷ 2 − 50) = 150 × (125 − 50) = 150 × 75 = 11,250 mm²

Centroid of this area from the neutral axis:

ȳ = 50 + (75 ÷ 2)

ȳ = 50 + 37.5

ȳ = 87.5 mm

Hence:

Q = A′ × ȳ

Q = 11,250 × 87.5

Q = 984,375 mm³

Thickness t = 150 mm.

Calculate τ:

τ = V Q ÷ (I t)

τ = 30,000 × 984,375 ÷ (1.953 × 10⁸ × 150)

τ = 2.953 × 10¹⁰ ÷ 2.930 × 10¹⁰

τ = 1.008 N/mm² = 1.008 MPa

Answer: 1.008 MPa

Check: the maximum shear stress at the neutral axis is 1.5 V ÷ A = 1.5 × 30,000 ÷ 37,500 = 1.2 MPa, so the stress 50 mm above the neutral axis is lower, as expected.

5.1.2 Relationship between vertical load and horizontal shear

The vertical shear force V applied on a beam section causes horizontal shear stresses that vary in magnitude across the cross-section. This relationship is governed by equilibrium and internal force transfer in the beam.

The shear force V at a section results in a horizontal shear stress distribution, which can be expressed as:

τ = V Q ÷ (I t)

where the variables are as defined previously. The magnitude of V directly influences the maximum horizontal shear stress, especially near supports where shear is greatest.

The link between vertical load and horizontal shear comes from the change of bending moment along the beam: the rate at which bending moment changes with distance equals the shear force (dM/dx = V), and because the compression and tension forces in the beam layers grow as the moment grows, horizontal shear must develop between the layers to balance them. The horizontal force transferred per unit length is called the shear flow, q = V Q ÷ I, and it governs the spacing of nails, bolts, welds or shear connectors that join the parts of built-up beams. Example: in a composite steel and concrete floor for a shopping centre in Kisumu, the engineer used the shear flow near the supports to space the shear studs that tie the concrete slab to the steel beams.

Practical Implications of the Relationship

• Highest shear near supports: Vertical shear is usually greatest at supports, so horizontal shear stresses are highest there.
• Shear reinforcement in concrete beams: Stirrups are placed closer together near supports, where shear is greatest.
• Connector spacing in built-up beams: Nails, bolts and welds are spaced more closely where shear flow is high.
• Short, heavily loaded beams: Deep, short-span beams are often governed by shear rather than bending.
• Timber beams: Timber is weak in shear along the grain, so horizontal shear must be checked near supports.

Worked Examples

Example 1: A simply supported beam with span 6 m carries a uniformly distributed load of 5 kN/m. Calculate the maximum horizontal shear stress at the neutral axis of a rectangular cross-section 200 mm wide and 400 mm deep.

Given:
Uniform load w = 5 kN/m
Span L = 6 m
Width b = 200 mm
Depth h = 400 mm

Reaction force at support:

R = w L ÷ 2

R = 5 × 6 ÷ 2

R = 15 kN = 15,000 N

Maximum shear force V = 15,000 N at the support.

Calculate I:

I = b h³ ÷ 12

I = 200 × 400³ ÷ 12 = 200 × 64,000,000 ÷ 12

I = 1.067 × 10⁹ mm⁴

Calculate Q at neutral axis:

Q = (b h ÷ 2) × (h ÷ 4)

Q = (200 × 200) × 100 = 40,000 × 100

Q = 4,000,000 mm³

Thickness t = b = 200 mm.

Calculate τ:

τ = V Q ÷ (I t)

τ = 15,000 × 4,000,000 ÷ (1.067 × 10⁹ × 200)

τ = 6.0 × 10¹⁰ ÷ 2.134 × 10¹¹

τ = 0.281 N/mm² = 0.281 MPa

Answer: 0.281 MPa

Example 2: A beam with a T-section carries a point load of 80 kN at mid-span, so the shear force near each support is 40 kN. The moment of inertia I = 4.5 × 10⁸ mm⁴, thickness of web t = 12 mm, and Q = 1.2 × 10⁶ mm³ at the web. Find the shear stress.

Given:
V = 40,000 N
I = 4.5 × 10⁸ mm⁴
Q = 1.2 × 10⁶ mm³
t = 12 mm

τ = V Q ÷ (I t)

τ = 40,000 × 1.2 × 10⁶ ÷ (4.5 × 10⁸ × 12)

τ = 4.8 × 10¹⁰ ÷ 5.4 × 10⁹

τ = 8.89 N/mm² = 8.89 MPa

Answer: 8.89 MPa

Example 3: A rectangular beam 250 mm wide and 600 mm deep supports a triangular load increasing from zero at the left support to 10 kN/m at the right support over an 8 m span. Find the horizontal shear stress at the neutral axis at a section 2 m from the left support.

Given:
Width b = 250 mm
Depth h = 600 mm
Span L = 8 m
Distance x = 2 m

Total load and left reaction:

W = ½ × L × w_max

W = 0.5 × 8 × 10

W = 40 kN, acting at 2L ÷ 3 = 5.333 m from the left support

R_A = W × (L − 5.333) ÷ L

R_A = 40 × 2.667 ÷ 8

R_A = 13.33 kN

Load intensity at x = 2 m:

w_x = w_max × x ÷ L

w_x = 10 × 2 ÷ 8

w_x = 2.5 kN/m

Shear force at x = 2 m:

V = R_A − (½ × x × w_x)

V = 13.33 − (0.5 × 2 × 2.5)

V = 13.33 − 2.5

V = 10.83 kN = 10,830 N

Calculate I:

I = b h³ ÷ 12

I = 250 × 600³ ÷ 12 = 250 × 216,000,000 ÷ 12

I = 4.5 × 10⁹ mm⁴

Calculate Q at neutral axis:

Q = (b h ÷ 2) × (h ÷ 4)

Q = (250 × 300) × 150 = 75,000 × 150

Q = 11,250,000 mm³

Thickness t = b = 250 mm.

Calculate τ:

τ = V Q ÷ (I t)

τ = 10,830 × 11,250,000 ÷ (4.5 × 10⁹ × 250)

τ = 1.218 × 10¹¹ ÷ 1.125 × 10¹²

τ = 0.108 N/mm² = 0.108 MPa

Answer: 0.108 MPa

Chapter Summary

This chapter introduced the fundamentals of horizontal shear stress in beams subjected to bending, explaining how internal shear forces arise and how they relate to applied vertical loads. It detailed the connection between vertical loading and the resulting horizontal shear stresses within beam cross-sections. The discussion then focused on the distribution of shear stress across different beam profiles, including rectangular, T-shaped, and I-shaped sections, highlighting how geometry influences stress patterns. Methods for analyzing and calculating shear stress distributions were presented to enable accurate assessment of structural behavior. Practical applications were explored by demonstrating how to plot and interpret shear stress distribution curves effectively. The chapter concluded with case examples illustrating the use of these concepts in real beam scenarios, emphasizing their importance for structural design and safety evaluation. Overall, the content provided a comprehensive understanding of horizontal shear stresses and their critical role in structural analysis.