Module 7 · Lesson 7.3
Shear strain and the Poisson effect
The second kind of stress, and why stretching one way shrinks the others.
Why this matters
Direct stress and strain describe pulling and pushing along one axis. Real members are also sheared, and they change shape sideways when you stretch them. Both effects have their own constants, both appear in almost every later topic — bending, torsion, Mohr's circle, composite sections — and neither is difficult once you have seen the picture.
By the end of this lesson you should be able to
- Define shear stress and shear strain, and say how they differ from the direct quantities
- Use Hooke's law in shear, τ = Gγ
- Define Poisson's ratio and calculate lateral strain
- Explain why a restrained member develops stress in more than one direction
What you should already know
- Direct stress and strain (this module, lesson 1)
- Hooke's law and Young's modulus (this module, lesson 2)
Direct stress acts perpendicular to a surface: it pulls the material apart or pushes it together. Shear stress acts along the surface — it tries to slide one part of the material past the neighbouring part.
The definition looks the same as for direct stress, but the area means something different:
What it calculates: the mean shear stress on a surface
- τ
- shear stress (N/mm²)
- V
- force acting along the surface (N)
- A
- area of the surface being sheared (mm²)
This assumes
- The shear is spread evenly over the area, which is an approximation
- Elastic behaviour
In plain terms: For direct stress the area is perpendicular to the force. For shear stress it is parallel to it. Getting the two mixed up is the commonest error in a bolt or weld check.
Where direct strain measures a change of length, shear strain measures a change of angle.
Take a small square of material and shear it. It becomes a parallelogram. The corners were right angles and are no longer. The shear strain γ is the amount by which a right angle has changed, measured in radians:
Because it is an angle change, shear strain is dimensionless — like direct strain, and for the same reason: it is a ratio of a sideways movement to the distance over which that movement occurs.
What it calculates: shear stress from shear strain
- τ
- shear stress (N/mm²)
- G
- shear modulus, sometimes called the modulus of rigidity (N/mm²)
- γ
- shear strain, the change in a right angle (radians)
This assumes
- Linearly elastic material
- Small strains
In plain terms: Exactly parallel to σ = Eε. G plays the part for shape change that E plays for length change. For structural steel G is about 79 000 N/mm², a little under two-fifths of E.
Predict first
A single bolt of 20 mm diameter connects two plates that pull apart in line with each other. Which area does the shear stress act on?
Now the second effect. Stretch a rubber band and it gets visibly thinner. Steel does the same thing; you just cannot see it.
When a material is stretched along one axis it contracts along the other two. The ratio of that contraction to the stretch causing it is Poisson's ratio:
What it calculates: how much a material contracts sideways when stretched
- ν
- Poisson's ratio (dimensionless)
- εlateral
- strain across the member (dimensionless)
- εaxial
- strain along the member (dimensionless)
This assumes
- Isotropic, linearly elastic material
- Uniaxial stress
In plain terms: The minus sign makes ν positive for ordinary materials, since the two strains have opposite signs. Steel is about 0.30, aluminium 0.33, concrete 0.20.
Try it
Stretch a bar
Stress depends on the load and the area. Extension also depends on the length and the material.
kN (tension)
mm
mm²
Material
- Stress σ = P/A
- 100.0 N/mm²
- Strain ε = σ/E
- 4.878e-4
- Extension δ = PL/AE
- 1.220 mm
- Member stiffness AE/L
- 98.4 kN/mm
A stiffer material gives a steeper line: the same strain needs more stress.
Worked example
A bolt in shear, and a bar that gets thinner
Given
- A single 20 mm diameter bolt carries a shear force of 45 kN across one plane
- Separately: a steel bar 60 mm wide and 20 mm thick carries 240 kN in tension
- E = 205 000 N/mm², ν = 0.30
Find
The shear stress in the bolt, and the change in width of the bar.
Practice
A single bolt 20 mm in diameter carries a shear force of 45.0 kN across one shear plane. What is the average shear stress, in N/mm²?
Practice
A steel bar 60.0 mm wide carries a tensile stress of 200 N/mm². With E = 205 000 N/mm² and ν = 0.30, what is the magnitude of the lateral strain?
Practice
A block of material is sheared so that a right angle changes by 0.0012 radians. With G = 79 000 N/mm², what shear stress is acting, in N/mm²?
Summary
- Direct stress acts perpendicular to a surface; shear stress acts along it
- Shear strain is a change of angle, not of length, and is measured in radians
- Hooke's law in shear: τ = Gγ, with G ≈ 79 000 N/mm² for steel
- Poisson's ratio ν = −εlateral/εaxial, about 0.30 for steel
- Lateral strain produces lateral stress only where the movement is restrained
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