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Choose a route or read from the beginning. All teaching chapters stay open; the grouped index below is expandable.
Define the restraint
Chapters 1–2. Draw the rod, connector and conditions the model assumes.
Build compatibility
Chapters 3–5. Separate strains, derive the force and work the numbers.
Change the stiffness
Chapters 6–7. Check limiting supports and a larger member area.
Interpret carefully
Chapters 8–10. Follow cooling, stability limits and common questions.
Practise and continue
Chapters 11–12. Transfer the model and choose related engineering guides.
Full chapter index · Practice and worked checks · Structural Engineering subject root
Heating a restrained member can create force even when its movement is small. A perfectly fixed model gives one limiting answer, but a flexible support shares the thermal expansion between member deformation and support movement. You need both stiffnesses to calculate that intermediate state.
This guide derives a one-dimensional rod-and-support model and checks its free and fully restrained limits. Every numerical property is stipulated for teaching. The model explains compatibility; it does not establish a structural member’s capacity, a safe detail or a design approval.
Chapter index
Define the restraint · Chapters 1–2
1. Ask where the expansion can go
2. State the geometry and material assumptions
Build compatibility · Chapters 3–5
3. Separate thermal strain from mechanical strain
4. Add the support compatibility
5. Work through a flexible support
Change the stiffness · Chapters 6–7
6. Test the limiting cases
7. Change the member without changing the support
Interpret carefully · Chapters 8–10
8. Follow the sign when the rod cools
9. Check movement and force together
10. Questions readers often ask
Practise and continue · Chapters 11–12
11. Practise with a changed thermal input
12. Continue through the connected engineering guides
Picture a straight rod fixed at its left end. Its right end is attached to a linear elastic connector that is anchored to ground further to the right. Heating the rod makes it try to grow. Its right end moves right and deforms the connector, which pushes back.
The resulting force depends on how much movement the connector permits. Calling the right end simply supported or restrained without defining its axial behaviour leaves the central calculation unfinished.
We will track three quantities: the rod’s free thermal extension, its actual end displacement and its mechanical shortening under compression. Their relationship makes the system understandable before any numbers are entered.
CHAPTER 2 OF 12 · Define the restraint
2. State the geometry and material assumptions
Let the rod have initial length L, uniform area A, Young’s modulus E and thermal-expansion coefficient α. It begins stress-free, and its temperature changes uniformly by ΔT. Take small deformation, constant properties and linear elastic axial behaviour.
Let the connector stiffness be kₛ, measured as force per displacement. Its ground anchor is fixed and it has no thermal expansion in this model. Assume an attached, bilateral connector: it can transmit either sign of axial force. A loose bearing or compression-only contact needs a different contact condition.
Ignore self-weight, other axial loads and bending. Assume the straight configuration remains stable. Buckling, yielding, creep, temperature gradients and connection failure are separate checks that the following calculation does not perform.
CHAPTER 3 OF 12 · Build compatibility
3. Separate thermal strain from mechanical strain
The free thermal strain is αΔT. It is the fractional length change the rod would undergo without axial restraint. Its free extension is δₜ = αLΔT.
Take tensile stress and extension as positive. The total axial strain is ε = σ/E + αΔT. The first term is mechanical strain; the second is thermal strain.
For a positive compressive-force magnitude F during heating, σ = −F/A. The actual extension is therefore δ = αLΔT − FL/(EA). Define the rod’s axial stiffness kᵦ = EA/L, giving δ = δₜ − F/kᵦ.
The member can extend overall while still carrying compressive stress. Its positive thermal strain exceeds the negative mechanical contribution. Confusing total extension with mechanical strain would miss that force.
When the rod’s right end moves by δ, the connector deforms by that same amount under the stated geometry. During heating its opposing force magnitude is F = kₛδ.
Substitute δ = F/kₛ into the rod equation: δₜ = F/kᵦ + F/kₛ. The free extension is divided between elastic shortening of the rod and movement against the connector.
For positive finite stiffnesses, F = δₜ/(1/kᵦ + 1/kₛ). Equivalently, F = [kᵦkₛ/(kᵦ + kₛ)]δₜ. The end displacement is δ = [kᵦ/(kᵦ + kₛ)]δₜ.
This is a series-compliance relationship, even though the physical story begins with a heated rod rather than an externally applied force. Equilibrium supplies the common force; compatibility supplies the displacement relationship.
Use L = 2,000 mm, A = 100 mm², E = 200,000 N/mm², α = 12 × 10⁻⁶/K and ΔT = 40 K. These are stipulated constant properties, not a material specification.
The free extension is 12 × 10⁻⁶ × 2,000 × 40 = 0.960 mm. The rod stiffness is 200,000 × 100/2,000 = 10,000 N/mm. Let the connector stiffness be 5,000 N/mm.
Then F = 0.960/(1/10,000 + 1/5,000) = 3,200 N. Stress is −3,200/100 = −32 N/mm², or −32 MPa. The end moves 3,200/5,000 = 0.640 mm.
| Quantity | Calculation | Result |
|---|---|---|
| Free extension | αLΔT | 0.960 mm |
| Rod stiffness | EA/L | 10,000 N/mm |
| Compressive force magnitude | 0.960/(0.0001 + 0.0002) | 3,200 N |
| Mechanical shortening | F/kᵦ | 0.320 mm |
| Actual end displacement | 0.960 − 0.320 | 0.640 mm |
| Compressive stress | −F/A | −32 MPa |
Check both sides: connector deformation gives 5,000 × 0.640 = 3,200 N, and the rod’s 0.320 mm mechanical shortening plus its 0.640 mm actual extension equals the 0.960 mm free extension.
As kₛ approaches zero, the support offers vanishing axial resistance. Force approaches zero and the end displacement approaches the full free extension. Use this limit rather than inserting zero into a reciprocal-stiffness formula.
As kₛ becomes arbitrarily large, displacement approaches zero. The force approaches kᵦδₜ = EAαΔT. The signed stress approaches −EαΔT for positive heating.
| Connector model | Force magnitude | End displacement |
|---|---|---|
| Free limit | 0 N | 0.960 mm |
| kₛ = 5,000 N/mm | 3,200 N | 0.640 mm |
| kₛ = 10,000 N/mm | 4,800 N | 0.480 mm |
| Perfectly fixed limit | 9,600 N | 0 mm |
The fully fixed stress is −96 MPa in this example. It is a limiting model result, not a capacity or an allowable stress. Zero total movement coexists with nonzero thermal and mechanical strains that cancel.
CHAPTER 7 OF 12 · Change the stiffness
7. Change the member without changing the support
Double the rod area to 200 mm² while retaining its length, material, temperature change and the 5,000 N/mm connector. The rod stiffness doubles to 20,000 N/mm, while its free extension remains 0.960 mm.
The force becomes 0.960/(1/20,000 + 1/5,000) = 3,840 N. Force has increased, but stress magnitude is 3,840/200 = 19.2 MPa, which is smaller.
End displacement becomes 0.768 mm; rod mechanical shortening becomes 0.192 mm. Their sum still recovers the free extension.
This is a useful transfer check: more area does not necessarily mean more thermal stress under the same finite support stiffness. The answer depends on what is held fixed. Changing the support along with the member would define another comparison.
For a negative temperature change, free extension becomes negative: the rod wants to contract. With the attached bilateral connector assumed here, the force direction reverses and the rod carries tension.
The signed strain relation ε = σ/E + αΔT remains a dependable starting point. The positive compression-magnitude notation used for heating should not be carried into cooling without changing its interpretation.
If the right support is instead a compression-only contact, cooling may release the contact. The bilateral formula would then predict a tensile force that the real contact cannot transmit. Model the change in contact state before using the equation.
Temperature history also matters when material behaviour is inelastic or connections slip. Our initially stress-free, reversible model has no memory of those effects; a real assembly may.
A measured end displacement alone is not a stress measurement. To infer force using F = kₛδ, you need a suitable connector stiffness and evidence that the assumed contact and load path apply.
Likewise, knowing ΔT does not establish the restraint. Identify the stress-free reference state, material temperature rather than an unrelated air reading, geometry, actual connection behaviour and any other imposed movement.
The model’s compression may require a stability assessment in a real slender member. Nonuniform heating can also cause bending or self-equilibrating stresses outside this one-dimensional picture. An elastic force calculation is one part of an analysis, not evidence of overall structural adequacy.
For a learning notebook, retain three separate lines: material law, equilibrium and compatibility. Each answers a different question and makes an incorrect support assumption easier to find.
If the member does not move, is its strain zero?
Its total axial strain is zero in the fully fixed model. Its thermal strain and mechanical strain are individually nonzero and cancel. Stress follows the mechanical part.
Does heating always produce compression?
Not without the stated restraints. Free uniform expansion produces no axial stress in this model. Geometry, temperature gradients, other members and connection behaviour can change the result.
Can I use a spring stiffness quoted for another direction?
No. Axial, vertical, rotational and distributed support properties describe different relationships. The stiffness must match the modeled displacement and force.
Is releasing force always an improvement?
The model alone cannot decide that. Greater movement may affect connections, alignment, finishes or other members. Structural assessment must consider the complete system and applicable requirements.
Keep the original rod and 5,000 N/mm connector but halve ΔT to 20 K. Predict the result before recalculating.
Free extension halves to 0.480 mm. Because the stipulated system is linear, compressive force halves to 1,600 N, stress to −16 MPa and displacement to 0.320 mm. Mechanical shortening is 0.160 mm.
Now restore ΔT = 40 K and make kₛ equal to kᵦ = 10,000 N/mm. The free extension divides equally: 0.480 mm mechanical shortening and 0.480 mm end movement. Force is 4,800 N.
Explain why equal division occurs here but not with the original connector. The answer is equal compliance, not the idea that every restrained rod must lose half its expansion.
CHAPTER 12 OF 12 · Practise and continue
12. Continue through the connected engineering guides
Use Structural Engineering: load paths that stay trustworthy for the subject root and the wider relationship between supports, strength, serviceability and construction.
For applied continuity and movement questions, read rail welding and continuously welded rail and rail expansion devices and special joints. Their rail systems have additional distributed restraint and operational requirements; the single connector here is not a rail design model.
For the primary strain framework, read MIT OpenCourseWare, Structural Mechanics Unit 9: Effects of the Environment. It distinguishes thermal, mechanical and total strain. The finite-connector calculation and numerical examples here are original extensions of that framework.
Return to The Gold Standard of Education or the How X Works hub. The practical learning outcome is to ask what a support permits before trusting the force produced by a model.
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Sources checked 21 October 2026. Worked examples are educational models with the assumptions stated above.