Reinforced Concrete Design / Serviceability: cracking, deflection and durability

The cracked elastic section

The state a beam spends its life in — and a different calculation entirely from the ultimate one.

Lesson review: expert verification remains open

This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.

Corrections checked in this pass

Corrected the modular ratio in crack-strain calculations and the elastic strain/stress figure.

Checks needed for project use

A concrete-design reviewer must check the stated first-generation clauses, National Annex choices, detailing rules and factor sets. Numerical agreement alone does not establish code compliance.
Check restraint, tension-stiffening assumptions, effective tension area, actual exposure and service action combinations.

These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.

Concrete and steel calculations retain their stated first-generation Eurocode teaching basis. Use the standards and amendments specified for your project; do not mix generations without an appropriate design basis.

Read the lesson register and outstanding review notes

Why this matters

Module 4 analysed the section at failure: rectangular stress block, steel at yield, concrete at its crushing strain. That section describes an instant the beam will almost certainly never reach. For the fifty years in between, the beam is cracked and elastic, and if you want to know how much it will sag or how wide its cracks will be, you must analyse that state instead. It uses a different neutral axis, a different stress distribution, and different material properties — and confusing the two produces answers that are wrong by a factor of two or more.

By the end of this lesson you should be able to

Find the cracked elastic neutral axis by first moments of area
Compute the cracked second moment of area
Compare it with the gross value and quantify what cracking costs
Calculate service stresses in steel and concrete
Explain why creep drives the neutral axis downward

What you should already know

Flexural behaviour, states 1 to 4 (Module 4)
Transformed sections and the modular ratio (Module 1)
Second moment of area (Structural Analysis Fundamentals, Module 9)

Two analyses of the same beam

	Ultimate	Service
Loads	Factored	Unfactored
Concrete stress	Rectangular block at f_cd	Linear, to E_cm
Steel stress	At yield, f_yd	Wherever equilibrium puts it
Tension concrete	Ignored	Ignored (below the crack)
Neutral axis	From force equilibrium of the block	From first moments of area
Typical x/d	0.2 to 0.45	0.25 to 0.40, and it MOVES with time

The common ground is that tension concrete is neglected in both. Everything else differs.

The service analysis is the ordinary elastic transformed-section calculation from any mechanics course, applied to a section that has lost its tension zone. The steel is replaced by an equivalent area of concrete, α_e A_s, where α_e = E_s/E_c is the modular ratio. Then the neutral axis is where it always is for an elastic section: at the centroid of the transformed area.

Cracked elastic neutral axis

\frac{b\,x^2}{2} = \alpha_e A_s\,(d - x)

What it calculates: The depth to the neutral axis of a cracked, elastic, singly reinforced section.

b: Section width (mm)
x: Neutral-axis depth from the compression face (mm)
$\alpha_e$: Modular ratio E_s/E_c,eff (—)
A_s: Tension steel area (mm²)
d: Effective depth (mm)

This assumes

Both materials are modelled as linearly elastic within the applicable service stress limits
Plane sections remain plane
Concrete below the neutral axis carries no tension
Perfect bond, so the steel strain equals the strain in the concrete at that level

In plain terms: This is a statement about first moments of area, not forces: the area above the axis times its lever arm equals the transformed area below times its lever arm. That is the definition of a centroid. Solving the quadratic gives x directly, and unlike the ultimate analysis there is no iteration and no stress block to choose.

Cracked elastic section under an explicitly illustrative 100 kNm service moment. The strain profile is linear, with εc =0.000367129 at the top and εs =0.000841048 at the steel. Concrete stress is triangular over x =136.742 mm; no ultimate rectangular block is used.

Worked example

Cracked section properties of a 300 × 500 beam

Given

b = 300 mm, h = 500 mm, d = 450 mm
Tension steel A_s = 1470 mm² (3 H25)
C30/37 concrete, E_cm ≈ 32.8 GPa
Short-term loading, so no creep: φ = 0

Find

The neutral axis depth, the cracked second moment of area, and how they compare with the uncracked section.

Assumptions

Section fully cracked in the region considered
E_s = 200 GPa
Tension concrete neglected

The cracking moment: where the change of state happens

Before the beam cracks it has the full gross stiffness. The transition happens at the cracking moment, when the extreme tensile fibre reaches the concrete's tensile strength.

Cracking moment

M_{cr} = \frac{f_{ctm}\,I_g}{y_t}

What it calculates: The moment at which flexural cracking begins.

$f_{ctm}$: MEAN tensile strength — not the characteristic value (N/mm²)
I_g: Gross second moment of area (mm⁴)
y_t: Distance from the neutral axis to the extreme tension fibre (mm)

This assumes

Elastic behaviour up to cracking, which is a good approximation in tension
No pre-existing shrinkage or thermal cracking — which is optimistic

In plain terms: Note that the MEAN tensile strength is used, not a characteristic value. That is deliberate: this is not a safety check but an estimate of what will actually happen, and using a lower fractile would make the beam appear to crack earlier than it does. For the beam above, M_cr = 2.90 × 3.125 × 10⁹/250 = 36.2 kNm — a small fraction of its ultimate capacity. Almost every reinforced concrete beam in service is cracked, and that is normal, expected and designed for.

Predict first

The beam above is loaded for twenty years. Creep gives it a creep coefficient φ = 2, so its effective modulus falls to a third. What happens to the neutral axis depth x?

Note

Creep is a stiffness problem, not a strength problem

Concrete under sustained load goes on straining for years. A creep coefficient of φ = 2 means that after a long time the total strain is three times the initial elastic strain, at the same stress.

The concrete has not become weaker. Its strength is, if anything, slightly higher at twenty years than at twenty-eight days. But it behaves as though its modulus were a third of the short-term value, and every consequence of that is a stiffness consequence: deflections grow, the neutral axis migrates, forces redistribute toward stiffer members, and prestress is lost.

This is why the effective modulus method exists — use E_c,eff = E_cm/(1 + φ) to obtain an approximate sustained-load analysis, with load history, cracking and model applicability checked separately.

Practice

A 300 × 550 beam with d = 500 mm has A_s = 1470 mm² and α_e = 6.09. What is the cracked neutral axis depth x, in mm?

Practice

What is the cracking moment of a 300 × 500 mm C30/37 beam, in kNm? Take f_ctm = 2.90 N/mm².

Summary

Service analysis: unfactored loads, linear stresses, cracked section
For this linear pure-bending model, the transformed-area centroid satisfies the axial-force equilibrium condition
Cracking removes roughly two thirds of the flexural stiffness
M_cr uses the MEAN tensile strength — it is an estimate, not a safety check
Almost every concrete beam in service is cracked, by design
For the illustrated creep coefficient φ = 2, the effective modular ratio trebles and the neutral axis moves deeper
Creep is a stiffness effect: use E_c,eff = E_cm/(1 + φ) and carry on