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21st century skills activity · AS 9702 · 6.1 Deformation of solids

Young modulus: running Numbered Heads Together

A step-by-step guide to running Numbered Heads Together on a six-problem team set, with a full teacher answer key. The aim is that it can be run faithfully by any teacher, including a cover teacher.

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What it is, and why it works

Solve together, anyone explains

In Numbered Heads Together, each learner in a group of four takes a number. The teacher reveals one problem; the group works until every member can explain the answer, not just write it. The teacher then calls a random number, and that learner answers for the group with no further help. Because nobody knows in advance who will be called, every member has to understand the method and the units, not just the most confident one.

Numbered Heads Together: a group of four numbered learners agree one method and calculation, then a number is called and that member explains for the group.
Agree the method and the calculation, then any number may be called

It passes the PIES test:

Positive interdependence

The group must agree one method and calculation together, because any member may have to explain it for the group.

Individual accountability

A number is called at random, so each member is accountable for the working and the units, not just the value.

Equal participation

Every member numbers off and attempts the problem, so each has a view before the group agrees.

Simultaneous interaction

All groups work at once, so every learner is calculating and reasoning.

Running the structure

About eighteen minutes, six problems

Number off1 min

In each group of four, members take a number from 1 to 4.

Reveal one problem

Show one problem from the team set at a time. Keep the useful relationships on display.

Heads together~2 min

The group works until every member can explain the answer, with method and units, not just a value.

Call a number

Call a number from 1 to 4; that learner answers for the group with no further help. Award a point for a correct, clearly explained answer, then move on.

Sentence frames for the answer

"the equation I need is ..." "substituting, ... gives ..." "the unit is ... (or none, because strain is a ratio)" "for the group, number 3 says ..."

The teacher's role during the activity

Keep the rounds brisk. Circulate and listen for the key moves: the right equation, a clean substitution, and the correct unit. Choose a different number each round so accountability is real. Aim for about eighteen minutes for the six problems; skip Q6 if time is short and use it as a stretch.

Team problem set

Six problems, one set per group

Useful relationships: σ = F / A  ·  ε = x / L  ·  E = σ / ε  ·  A = π d2 / 4
Q1 · Define
State what is meant by tensile stress and by tensile strain, and give the SI unit of each.
Agreed answer: ____________
Q2 · Stress
A wire of cross-sectional area 2.0 × 10−7 m2 supports a load of 60 N. Calculate the tensile stress.
Agreed answer: ____________
Q3 · Strain
A wire of original length 1.50 m extends by 1.2 mm under load. Calculate the tensile strain.
Agreed answer: ____________
Q4 · Young modulus
A metal wire of length 1.80 m and diameter 0.40 mm extends by 1.8 mm under a force of 25 N. Calculate the Young modulus.
Agreed answer: ____________
Q5 · Graph
A stress-strain graph for a metal is a straight line from the origin to a point P, then curves. (i) Name P. (ii) State how the Young modulus is found from the graph. (iii) What does the region beyond the elastic limit represent?
Agreed answer: ____________
Q6 · Explain (stretch)
A learner claims: "A thick copper wire has a larger Young modulus than a thin copper wire of the same copper." Explain whether the learner is correct.
Agreed answer: ____________
Teacher answer keyclick to reveal
Q1. Define.

stress is the force per unit cross-sectional area, σ = F / A; unit pascal, Pa (N m−2). Strain is the extension per unit original length, ε = x / L; no unit (a ratio).

Q2. Stress.

σ = F / A = 60 / (2.0 × 10−7) = 3.0 × 108 Pa (300 MPa).

Q3. Strain.

ε = x / L = (1.2 × 10−3) / 1.50 = 8.0 × 10−4.

Q4. Young modulus.

A = π d2 / 4 = π (0.40 × 10−3)2 / 4 = 1.26 × 10−7 m2.
σ = F / A = 25 / (1.26 × 10−7) = 1.99 × 108 Pa.
ε = x / L = (1.8 × 10−3) / 1.80 = 1.0 × 10−3.
E = σ / ε = (1.99 × 108) / (1.0 × 10−3) = 2.0 × 1011 Pa (200 GPa).

Q5. Graph.

(i) P is the limit of proportionality. (ii) the Young modulus is the gradient of the straight-line region. (iii) the region beyond the elastic limit represents plastic (permanent) deformation: the wire does not return to its original length when unloaded.

Q6. Explain (stretch).

incorrect. The Young modulus is a property of the material, not of the dimensions of the sample. The thicker wire has a larger cross-sectional area, so for the same force it has a smaller stress and stretches less (it is stiffer as a specimen), but the ratio stress / strain is the same for both because the material is the same.

Troubleshooting and differentiation

When the room does not behave like the plan

One member answers for everyone: remind the group that any number may be called, so all four must agree and be able to explain.

A group gives a value with no working: ask for the equation, the substitution and the unit, not just the number.

An odd group of three: use numbers 1 to 3, or give one member two numbers.

It runs long: four problems are enough; keep Q6 as a stretch and protect the experiment and the exit ticket.

Original work by the TheLucidSTEM team. Designed for the lesson on this site; no past paper material is reproduced.
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