Module 10 · Lesson 10.2
Deriving τ = VQ/It
From longitudinal equilibrium of an element to the shear stress at any level.
Try it
Shear stress is not spread evenly — see where it goes
τ = VQ/It gives one number in a formula, but a shape over the depth. The left is the section; the right is the shear stress at every height. Change the section and watch the profile change with it.
Section
- Peak shear stress
- 79.2 N/mm²
- Average V/A
- 31.4 N/mm²
- Peak / average
- 2.52 ×
- Where the peak is
- at the neutral axis
- Shear the web carries
- 97.2 %
- Shear the flanges carry
- 2.8 %
- Web area / gross area
- 42 %
Things worth trying
- Start on the rectangle. The profile is a clean parabola — zero at the top and bottom faces, greatest at the neutral axis. The peak is 1.5 times the crude V/A, and the average alone would understate the real stress by a third.
- Why zero at the faces? Q is the first moment of the area BEYOND the cut. At the very top there is no area beyond, so Q is zero and so is τ. At the neutral axis the whole half-section is beyond, so Q — and τ — is greatest.
- Switch to the circle. Same story but gentler: the peak is only 4/3 the average, because the section is widest exactly where the shear is largest. Shape decides how peaked the distribution is.
- Now the I-section. The shape changes character completely: small in the flanges, then a sudden JUMP into a much larger, fairly flat stress in the web.
- The jump is the width. Q barely changes across the flange–web junction, but the width collapses from the flange to the thin web, and τ = VQ/It divides by that width — so the stress leaps up by the width ratio.
- Read the web-share rows. The web carries over 90% of the shear on well under half the area. The flanges are for bending; the web is for shear. That division is the whole logic of the I-shape.
- Thin the web down to 6 mm. The peak stress climbs steeply and the web's share rises further — a thin web is efficient but is the first thing to check, which is exactly why plate girders are governed by web shear.
- This is why steel design takes the 'shear area' as roughly the web area, not the gross area: the flanges are barely working in shear, so counting them would be optimistic.
Why this matters
This is the formula that sizes the welds in a plate girder and the shear studs in a composite deck. It is short, it is derived in half a page, and it explains the shape of every shear stress diagram you will meet.
By the end of this lesson you should be able to
- Derive the shear formula from longitudinal equilibrium
- Interpret Q physically
- Apply it to rectangular and I-sections
- Use shear flow for connector design
From first principles
The beam shear formula
We want to show: that the shear stress at any level in a beam is τ = VQ/It.
Cut a short length of beam, δx long. Then cut it again horizontally at the level you care about, and keep only the piece above that cut. Bending stresses push on the two vertical end faces of that piece. Because the bending moment differs by δM across the slice, those two pushes do not balance. The only remaining surface that can supply the missing force is the horizontal cut at the bottom, so a shear stress must act there. Work out the imbalance, divide by the area it acts over, and you have the formula.
Worked example
Shear stress distribution in a rectangular beam
Given
- Rectangular section 120 mm wide × 300 mm deep
- Shear force at the section V = 90 kN
Find
The maximum shear stress, and the stress one quarter of the depth from the top.
Assumptions
- Elastic behaviour
- Shear stress uniform across the width
Worked example
Why the web of an I-section carries the shear
Given
- Symmetric I-section: flanges 180 mm wide × 15 mm thick, overall depth 400 mm, web 10 mm thick
- Shear force V = 250 kN
Find
The maximum shear stress, and how it compares with the gross average.
Assumptions
- Elastic behaviour
- Shear stress uniform across the web thickness
Practice
A rectangular timber beam 100 mm wide and 250 mm deep carries a shear force of 30 kN. What is the maximum shear stress?
Practice
A solid circular shaft of diameter 120 mm carries a shear force of 60 kN. What is the maximum shear stress?
Practice
A built-up beam has I = 180 × 10⁶ mm⁴. At the glue line, Q = 900 × 10³ mm³. The beam carries a shear force of 45 kN. What is the shear flow the glue must resist, in N/mm?
Check yourself
Where is the shear stress zero in a rectangular beam?
Summary
- τ = VQ/It comes from balancing the unbalanced bending force on a slice
- Q is the first moment about the neutral axis of the area beyond the cut
- τ = 0 at a free surface and peaks at the neutral axis for a solid section
- Rectangle: 1.5 × average. Circle: 4/3 × average. I-section: the web takes nearly all of it
- Shear flow q = VQ/I sizes welds, bolts and glue lines
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint