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Mechanics of Materials · Chapter 6 of 10 · Intermediate

Shear Stress in Beams

Bending is not the only stress in a loaded beam. The shear force also creates shear stress, greatest at the neutral axis, and it sets how many nails or bolts hold a built-up beam together.

Readiness check

This chapter builds on bending and the shear force. Tick only what you can do closed-notes.

Find the shear force V from a diagram.
Compute a moment of inertia I.
Find the first moment of an area (A·ȳ).
Recall the neutral axis at the centroid.
Work in N, mm, and MPa.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsReview the moment of inertia in Chapter 4 and shear in Chapter 5.

3 or more weak itemsRevisit section properties first.

The core idea

The transverse shear force produces shear stress that flows horizontally between the beam's layers, given by τ = VQ/It and greatest at the neutral axis.

τ = VQ/ItQ = A′ȳ′q = VQ/I (shear flow)

Because the bending stress changes along the beam, layers want to slide over one another, and the resistance to that sliding is horizontal shear stress, equal by complementarity to the vertical shear stress. The shear formula τ = VQ/It uses the first moment Q of the area beyond the level of interest. The stress is zero at the top and bottom fibers and maximum at the neutral axis. Integrated over a joint, the shear flow q = VQ/I sets the spacing of the fasteners holding a built-up beam.

The skill works when: you take Q as the first moment of the area beyond the cut and use the width t at that level.

The skill breaks down when: shear stress is assumed uniform (V/A) or maximum at the top fiber instead of the neutral axis.

The concept. Shear stress in a rectangular beam is parabolic: zero at the outer fibers, maximum at the neutral axis. The first moment Q grows toward the center, so τ peaks there.

The skills, taught in order

Beam shear is one formula plus the geometry it needs. Five skills cover the origin, the shear formula, the first moment, the distribution, and shear flow.

6.1 Why beams carry shear stress

Along a beam the bending moment changes, so the bending stress on adjacent cross-sections differs. That difference would slide the layers past one another; horizontal shear stress resists the sliding. By complementarity, the horizontal and vertical shear stresses at a point are equal.

6.2 The shear formula

The transverse shear stress at a level is τ = VQ/It, where V is the shear force, I the section's moment of inertia, t the width at that level, and Q the first moment of the area beyond it. It comes from the change in bending stress over a slice, integrated.

6.3 The first moment Q

Q = A′ȳ′ is the area beyond the level of interest times the distance from the neutral axis to that area's centroid. Q is largest at the neutral axis (the whole half-section contributes) and zero at the extreme fiber (no area beyond it), which sets where the stress peaks.

6.4 The shear-stress distribution

For a rectangle the distribution is parabolic, with a maximum of 1.5·V/A at the neutral axis, half again the average.

Section	Maximum τ
Rectangle	1.5·V/A (at neutral axis)
Solid circle	(4/3)·V/A
I-beam (web)	≈ V/A_web (web carries nearly all shear)

6.5 Shear flow and built-up members

For a beam assembled from parts, the force per unit length crossing a joint is the shear flow q = VQ/I, with Q the first moment of the attached part about the neutral axis. Fasteners carrying a force F each are spaced at s = F/q, so higher shear means closer fasteners.

Engineering connection: short, deep beams and the webs of I-beams are shear-critical; shear flow sizes the welds, bolts, and nails in plate girders and glued-laminated timber.

Worked example 1: shear stress in a rectangular beam

A rectangular beam 60 mm wide and 120 mm deep carries a transverse shear force of 30 kN. Find the maximum shear stress and compare it with the average V/A.

Figure 1. The shear stress peaks at the neutral axis at 1.5 times the average. Treating it as the uniform V/A would underestimate the maximum by half again.

ProblemFind the maximum shear stress for the beam in Figure 1.
Given / findb = 60 mm, h = 120 mm, V = 30 kN. Find τ_max and compare with V/A.
AssumptionsTransverse shear formula applies; stress uniform across the width.
ModelCompute I and the first moment Q at the neutral axis, then τ = VQ/It.
EquationsI = bh³/12 Q_NA = (bh/2)(h/4) τ_max = VQ/(It)
SolveI = 60(120)³/12 = 8.64×10⁶ mm⁴. Q_NA = (60 × 60)(30) = 108 000 mm³. τ_max = (30 000 × 108 000)/(8.64×10⁶ × 60) = 6.25 MPa. The average V/A = 30 000/7200 = 4.17 MPa, so τ_max = 1.5·V/A.
CheckThe factor 1.5 is the known rectangular result, a good confirmation. The shear stress (6.25 MPa) is far below typical bending stresses, which is why long beams are bending-critical and only short, deep ones are shear-critical.
ConclusionShear stress is not uniform; it peaks at the neutral axis at 1.5 times the average for a rectangle. The shear formula τ = VQ/It gives the true value at any level.

Result. τ_max = 6.25 MPa at the neutral axis (1.5 times the 4.17 MPa average).

Worked example 2: nail spacing in a built-up beam

A beam is built up as a symmetric I from boards: two 200 mm × 40 mm flanges and a 40 mm × 200 mm web. It carries a shear of 8 kN, and nails at each flange-web joint hold 2.5 kN each. Find the shear flow and the nail spacing.

Figure 2. The shear flow at each flange-web joint must be carried by the nails. Their capacity divided by the shear flow gives the spacing; more shear means tighter nailing.

ProblemFind the shear flow at the joint and the nail spacing for the beam in Figure 2.
Given / findflanges 200×40, web 40×200, V = 8 kN, nail capacity F = 2.5 kN. Find q and s.
AssumptionsSymmetric section (NA at mid-height), nails at the flange-web joint share the flow, linear-elastic.
ModelCompute I of the whole section, Q of one flange about the NA, the shear flow q = VQ/I, then s = F/q.
Equationsq = VQ/I Q = A_flange·d s = F/q
SolveThe section I = 2.59×10⁸ mm⁴ and Q for one flange = (200 × 40)(120) = 960 000 mm³ (its centroid is 120 mm from the NA). q = (8000 × 960 000)/2.59×10⁸ = 29.6 N/mm. Spacing s = 2500/29.6 = 84 mm.
CheckThe units of q are N/mm, force per length, and s comes out a sensible joinery spacing. Doubling the shear would halve the spacing to about 42 mm, as q ∝ V.
ConclusionShear flow turns a stress problem into a fastener-spacing rule. Near the supports, where V is largest, nails must be closest; toward mid-span they can spread out.

Result. Shear flow q = 29.6 N/mm; nail spacing s = 84 mm.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Uniform shear stress	τ taken as V/A everywhere	"Is the distribution uniform or parabolic?"	It is parabolic; the maximum is 1.5·V/A for a rectangle.
Max shear at the top fiber	Stress placed where Q = 0	"Where is Q largest?"	Q (and τ) peaks at the neutral axis, not the outer fiber.
Wrong Q	First moment of the wrong area	"Is Q the area beyond the cut, A′ȳ′?"	Use the area beyond the level times its centroid distance.
Ignoring shear flow in built-up beams	Fasteners under-provided	"What force per length crosses the joint?"	Size fasteners from q = VQ/I and s = F/q.

Practice ladder

Level 1 · Direct skill

A rectangular beam 40 mm × 100 mm carries 12 kN of shear. Find the maximum shear stress.

Show answer

τ_max = 1.5·V/A = 1.5 × 12 000/(40 × 100) = 1.5 × 3.0 = 4.5 MPa, at the neutral axis.

Level 2 · Mixed concept

For the Worked Example 2 beam, what nail spacing is needed near a support where the shear is 16 kN?

Show answer

q ∝ V, so q doubles to 59.2 N/mm and s = 2500/59.2 = 42 mm. Twice the shear means half the spacing, which is why nailing tightens toward the supports.

Level 3 · Independent problem

Why does the web of an I-beam carry almost all the shear, while the flanges carry almost all the bending?

Show answer

Shear stress τ = VQ/It is largest near the neutral axis where Q is large and the width t (the thin web) is small, so the web dominates shear. Bending stress σ = My/I is largest at the extreme fibers (the flanges), far from the neutral axis. The shapes are optimised for each: flanges for bending, web for shear.

Level 4 · Transfer to real engineering

Find a built-up or layered beam (a plywood I-joist, a nailed box beam, a welded plate girder). Estimate the shear flow at a joint and the required fastener or weld spacing.

What good work looks like

Section I and the joint Q computed, shear flow q = VQ/I evaluated, and a fastener spacing s = F/q with a note that it tightens where V is high.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that my Q is the first moment of the area beyond the cut."

"Give me five sections; I will say where the shear stress peaks."

"Compute the shear stress." Building Q and applying τ = VQ/It is the skill.

"What nail spacing?" Finding the shear flow yourself is the point.

Portfolio task

Analyse one beam for shear: find τ_max at the neutral axis, and for a built-up section compute the shear flow and fastener spacing.

Must include: I and Q, τ = VQ/It at the NA, and a shear-flow fastener spacing.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the shear formula.

τ = VQ/It, with Q the first moment of the area beyond the level.

2. Where is the shear stress maximum and minimum?

Maximum at the neutral axis (Q largest), zero at the top and bottom fibers (Q = 0).

3. Give the maximum shear stress for a rectangle.

1.5·V/A at the neutral axis.

4. What is shear flow, and how does it set fastener spacing?

q = VQ/I (force per length at a joint); spacing s = F/q for fasteners of capacity F.

5. In an I-beam, what carries the shear?

The web (high Q, small width); the flanges carry the bending.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive τ_max and the shear flow from a blank page.

+3 daysOne shear-stress and one shear-flow problem.

+7 daysCombine bending and shear stresses at a point in Chapter 7.

+30 daysRecall shear flow when designing welded or bolted girders.

Textbook mapping

Item	Mapping
Primary source	Beer, Johnston, DeWolf and Mazurek, Mechanics of Materials (6th ed), Chapter 6 (Shearing Stresses in Beams)
Cross-reference	Hibbeler, Ch. 7 · Gere and Goodno, Ch. 5
Core topics	6.1 Origin of beam shear · 6.2 Shear formula · 6.3 First moment Q · 6.4 Distribution · 6.5 Shear flow and fasteners
Engineering connection	Shear-critical short beams, I-beam webs, and fasteners in built-up girders.
Read next	Chapter 7: Stress Transformation and Mohr's Circle.