Here's the problem. Steel is clean: one material, linear elastic, predictable. You write \(\sigma = E\varepsilon\), go home, sleep well.
Reinforced concrete hands you two materials — one that cracks under tension, one that doesn't; one with a parabolic stress-strain curve, one with a linear one. There is no tidy closed-form formula that falls out of first principles.
So engineers did what engineers do: they made a set of carefully chosen simplifications. Five of them. Backed by decades of physical tests. Conservative enough to keep you safe. Simple enough to actually use.
These aren't wild guesses — they're the foundation of every RC beam ever designed. Learn them deeply and the rest of the course clicks into place.
- A1 Internal forces balance external loads (always)
- A2 Steel and concrete strain together — no slip
- A3 Flat cross-sections stay flat after bending
- A4 Cracked concrete carries zero tension
- A5 Actual (nonlinear) material behavior governs
Equilibrium — Internal Forces Balance External Loads
Stand on a scale. The scale pushes up on you with exactly the force equal to your weight. If it pushed less, you'd sink through it. If it pushed more, you'd fly off. The forces must balance. Structures obey the same law — every single moment of every single day.
When you cut through a beam and draw a free body diagram, the bending moment \(M\), shear \(V\), and axial force \(P\) that appear on the cut face must exactly equal the external load effects on the other side. No exceptions. No special RC clause. Newton's laws don't care what material you're using.
The lecture notes call this an "assumption" but it's really a law — it's mentioned explicitly because students sometimes lose track of it as RC analysis gets complicated. Stress redistribution happens, cracks form, steel yields — and through all of it, equilibrium holds perfectly at every section.
Every shear force diagram, every bending moment diagram, every reaction calculation flows from equilibrium. Violate it and your design isn't just wrong — it's physically impossible. The structure cannot exist in the state you've drawn.
\(\sum F_x = 0,\quad \sum F_y = 0,\quad \sum M = 0\)
Cut the beam anywhere. The stresses on the cut face integrate to give \(N, V, M\) — those must balance everything on the other side of the cut.
- When drawing a free body diagram and cutting a section, students forget to include internal forces \((N, V, M)\) on the cut face. Without those, equilibrium equations are missing terms — and the numbers won't work.
- A cracked, yielding RC beam still satisfies equilibrium — stress just redistributes. "This section yielded" doesn't mean equilibrium was violated; it means the force diagram changed shape.
Equilibrium tells us that internal forces exist at every section. But how do those forces split between the steel and the surrounding concrete? That depends on Assumption 2.
Perfect Bond — Steel and Concrete Strain Together
Glue a stiff rubber band and a soft rubber band side by side. Stretch them together. Even though one resists much more than the other, they both stretch the same amount. Now imagine the glue fails. Each band stretches independently — they fight each other, the assembly buckles, and everything falls apart. That's what slip between steel and concrete would do to a beam.
Modern deformed (ribbed) bars provide mechanical interlocking with the surrounding concrete on top of natural surface adhesion. The result: the strain in the steel bar equals the strain in the concrete immediately surrounding it — at every point along the bar. This shared deformation is what makes them act as a composite rather than two separate materials fighting over the same space.
Without perfect bond, you can't predict how load distributes between the two materials. The entire "transformed section" concept — replacing steel with \(n\) times its area of equivalent concrete — only works because equal strain lets you use one modulus ratio to relate stresses. Bond failure causes sudden, brittle collapse.
The ribs on a deformed bar interlock mechanically with the concrete paste. When the beam bends, the concrete deforms — and drags the bar with it, pixel-for-pixel. The force each material exerts then depends on its own stiffness: \(\sigma = E\varepsilon\).
| Symbol | Meaning | Unit |
|---|---|---|
| εs, εc | Strain in steel / surrounding concrete — they're equal | dimensionless |
| n | Modular ratio — how many times stiffer steel is vs concrete | — |
| Es | Steel modulus ≈ 200,000 MPa (constant for all steel) | MPa |
| Ec | Concrete modulus — varies with \(f'_c\) | MPa |
Requires: deformed bars with adequate anchorage. Breaks down near bar cutoffs or under severe cyclic loading.
- Same strain ≠ same stress. Steel and concrete at the same height have equal strain but very different stress — because \(E_s \approx 200{,}000\) MPa while \(E_c \approx 25{,}000\) MPa. The stress ratio is exactly \(n \approx 8\).
- Bond failure can occur near bar ends without adequate development length, or under high cyclic loading. When it does, composite action is lost and the beam loses significant capacity suddenly.
Steel and concrete now move together. The next question: how does strain vary from the top of the beam to the bottom?
Plane Sections Remain Plane — Bernoulli's Hypothesis
Take a rubber eraser. Draw a perfectly vertical line across its face. Now bend it. Look at the line — it's still straight. It tilted, but it didn't bow or wave or develop a kink. That straight line staying straight is the entire content of Assumption 3, applied to a beam cross-section.
A cross-section that was flat before bending remains flat after bending — it just rotates. The consequence is powerful: strain varies linearly across the section height. The top fibre (farthest from the neutral axis in compression) gets the most strain. The neutral axis gets zero. The bottom fibre gets the most strain in the other direction.
Double the distance from the neutral axis? Double the strain. That clean proportionality is why the flexure formula \(\sigma = Mc/I\) works — and why RC design can predict stress at any point in the cross-section without solving a full elasticity problem.
Linear strain distribution is the geometric foundation of everything. Without it, there's no rational way to predict stress at any point in the cross-section. It's also why the extreme fibre governs — that's where strain (and therefore stress) is highest, so that's where failure initiates.
\[ \varepsilon(y) = \phi \cdot y \]
Strain at height \(y\) from the neutral axis equals curvature \(\phi\) times \(y\). Linear. The farthest fibre from the N.A. always has the highest strain.
| Symbol | Meaning | Unit |
|---|---|---|
| ε(y) | Strain at distance y from neutral axis | dimensionless |
| φ | Curvature of the beam at that section | rad/mm |
| y | Distance from neutral axis (+ toward compression) | mm |
| ρ | Radius of curvature = 1/φ | mm |
Valid far from supports or point loads, no torsion, moderate deflections. Slight deviation near ultimate — but errors are minor and well-tested.
- Plane sections → linear strain diagram. But in RC, the stress diagram is not linear — because concrete's stress-strain curve is nonlinear (Assumption 5). Strain stays linear; stress doesn't. This trips up almost every student at first.
- The rectangular stress block used in strength design is a simplified approximation of the curved stress distribution that results from nonlinear concrete behavior under a linear strain diagram.
We know strain is linear from top to bottom. But what happens in the tension zone below the neutral axis — where concrete is being pulled apart? That's where Assumption 4 makes a bold, practical decision.
Cracked Concrete Carries Zero Tension
Imagine trying to pull apart a dry sand castle. You can push on it all day, but the moment you try to pull it apart, the grains just separate. Concrete isn't quite that bad — but it's in that spirit. Its tensile strength is roughly 10% of its compressive strength. In a bending beam, the tension zone cracks at service loads. A crack is a gap. A gap can't transfer tension. So we stop counting on it.
This is the structural logic of reinforcement: concrete handles compression above the neutral axis; steel handles tension below it. The cracked concrete below the neutral axis is still physically there — it provides cover to the bars, helps with shear — but it contributes zero tensile resistance in our calculations.
It's a conservative assumption (concrete between cracks does resist some tension) but the simplification is enormous and the error is small. It's also why you put steel at the bottom of a simply-supported beam and at the top over a continuous support — wherever the tension zone is, that's where the bars go.
Forgetting this and trusting concrete in tension leads to sudden brittle failure at loads far below design capacity. Hairline cracks at service load are expected — they're not a failure.
As tension stress reaches the modulus of rupture (~3–5 MPa), a hairline crack forms and propagates upward. The crack faces can't transfer tension. The steel bar is the only thing keeping the faces from separating.
In the tension zone: \(\sigma_c = 0\). The full tensile resultant is carried by steel:
\[ T = A_s \cdot f_s \]
Below the neutral axis, the concrete is cracked and contributes nothing in tension. Only the steel carries it — which is precisely why it's there.
| Symbol | Meaning | Unit |
|---|---|---|
| T | Total tensile force resultant in the tension zone | N or kN |
| As | Cross-sectional area of steel reinforcement | mm² |
| fs | Stress in steel at the section of interest | MPa |
Concrete in tension zone = fully cracked, zero stress. Steel bars bonded (Assumption 2 applies).
Calculator — Tensile Force T = As · fs
- The cracked concrete in the tension zone still physically exists — it provides cover to the bars, contributes to shear resistance via the "compression strut" mechanism, and limits crack widths. We only set its tensile stress to zero. We don't pretend it isn't there.
- Service-load cracks are not structural failures. They are designed and expected. The code limits crack width to protect the steel from corrosion and maintain appearance — not because the crack itself is catastrophic.
We've handled tension. Now what about the compression zone above the neutral axis — specifically, how does the compressive stress actually vary, and why can't we just use \(\sigma = E\varepsilon\) there?
Nonlinear Stress-Strain — The Parabolic Truth
Steel has a beautiful stress-strain curve: a straight line up to yield, then a flat shelf. You could set your watch to it. Concrete? It peaks around a strain of 0.002, curves over gently, then crushes somewhere around 0.003. And the shape of that curve changes with every concrete mix. There's no simple formula — just the curve. Assumption 5 says: use the real curve, or a good engineering approximation of it.
In practice this creates two design philosophies:
- Working Stress Design (WSD) / Elastic: Pretend both materials are linear elastic and stay within their elastic ranges. Use the transformed section. Simple and conservative — but leaves strength on the table.
- Strength Design (ACI/USD): Accept the nonlinearity. At ultimate load, the parabolic compressive stress block is approximated as an equivalent rectangle of depth \(a = \beta_1 c\) and intensity \(0.85f'_c\). This is more accurate near failure and gives more economical designs. It's the modern standard.
You cannot apply \(M = \sigma I/c\) to an RC beam at ultimate load — the stress distribution is no longer triangular. The ACI rectangular block replaces the curved reality with a rectangle that has the same area (same resultant force \(C\)) placed at the same effective centroid — giving you accurate capacity calculations without integrating a parabola.
\[ C = 0.85\,f'_c \cdot a \cdot b \]
\[ a = \beta_1 \cdot c \]
\(\beta_1 = 0.85\) for \(f'_c \leq 28\) MPa, decreasing by 0.05 per 7 MPa above 28 MPa, min 0.65.
| Symbol | Meaning | Unit |
|---|---|---|
| C | Compressive resultant in concrete | N |
| 0.85 f'c | Equivalent uniform compressive stress intensity | MPa |
| a | Depth of equivalent rectangular block | mm |
| b | Width of compression zone | mm |
| β1 | Block depth ratio (0.85 for f'c ≤ 28 MPa) | — |
| c | Neutral axis depth from compression face | mm |
Valid at or near ultimate load. For service-level checks (WSD), use elastic linear stress-strain instead.
- β₁ is not always 0.85. For \(f'_c > 28\) MPa, it reduces by 0.05 per 7 MPa, with a floor of 0.65. Always verify before using the rectangular block for higher-strength concretes.
- The rectangular stress block is only valid near ultimate load. For service-load stress checks and deflection calculations, use elastic (linear) stress distribution — the two regimes require different tools.
The Three Core Stress Formulas
Any internal force in a structure is some combination of axial, bending, and shear. Each one creates a different stress pattern on the cross-section. These three formulas are the translators: they convert force diagrams into stress numbers you can check against material limits.
| Symbol | Meaning | Unit |
|---|---|---|
| P, A | Axial force; cross-sectional area | N; mm² |
| M, c, I | Moment; distance from N.A. to fibre; 2nd moment of area | N·mm; mm; mm⁴ |
| V, Q, t | Shear force; 1st moment of area above point; section width at point | N; mm³; mm |
All three assume homogeneous elastic material + plane sections. For RC, apply to the transformed cracked section — not the gross section.
Calculator — Bending Stress σ = Mc / I
Worked Example — Full Bending Stress Check
A rectangular RC beam: \(b = 300\) mm, effective depth \(d = 450\) mm. Neutral axis at \(\bar{y} = 180\) mm from top. Cracked transformed \(I_{\text{cr}} = 2.1 \times 10^9\) mm⁴. Service moment \(M = 120\) kN·m. Find concrete stress at top fibre and steel stress. \(n = 8\).
Concrete top fibre: \(c_{\text{top}} = \bar{y} = 180\) mm. Steel level: \(d - \bar{y} = 450 - 180 = 270\) mm below N.A.
The N.A. divides the section. Every stress calculation needs a distance measured from the N.A., not the top or bottom of the beam.
\( M = 120 \text{ kN·m} = 120 \times 10^6 \text{ N·mm} \)
I is in mm⁴, c is in mm — M must be in N·mm or units don't cancel to MPa. This unit conversion causes most numerical errors.
\[ \sigma_c = \frac{Mc}{I} = \frac{120 \times 10^6 \times 180}{2.1 \times 10^9} = \mathbf{10.3 \text{ MPa}} \]
This is the governing compression in the concrete. Compare to allowable ≈ 0.45f'c.
Concrete "stress" at steel level: \( \dfrac{120 \times 10^6 \times 270}{2.1 \times 10^9} = 15.43 \text{ MPa} \)
Actual steel stress: \( f_s = n \times 15.43 = 8 \times 15.43 = \mathbf{123 \text{ MPa}} \)
The transformed section pretends steel is n × more concrete. To recover the real steel stress, multiply the "concrete equivalent" stress back by n.
Quick Reference — When to Use Each Formula
| Formula | Use When | Key Condition |
|---|---|---|
| σ = P/A | Pure axial (column, tie member) | Uniform stress; no bending |
| σ = Mc/I | Bending — elastic range, WSD | Plane sections, linear material |
| τ = VQ/It | Shear stress at any point | Elastic, no torsion |
From Blank Paper to Built Structure — The Two Phases
Most students think structural design starts with a load and a beam size. It doesn't. It starts with a person asking: "I need a hospital that can serve 500 patients a day." Everything structural flows from that. Understanding what Phase 1 demands is what makes Phase 2 possible.
What the building does: Adequate working spaces and clearances. Ventilation and HVAC routing. Vertical transportation (elevators, stairs). Lighting and acoustic requirements. Aesthetics and code compliance. The architect and owner drive this — the structural engineer listens and adapts.
How the building stands: Column/beam layout from Phase 1 constraints. Load determination. Preliminary member sizing. Structural analysis (V, M diagrams). Section design (reinforcement sizing). Iteration until all checks pass. Final construction documents.
Phase 2 is an iterative loop — not a one-pass calculation. You estimate member sizes, run the analysis, check stresses and deflections, find something doesn't work, adjust, and re-run. The first trial section is a trial — don't over-optimize it before you've confirmed the layout works.
Phase 2 — The Iterative Loop
| Step | What You Do | Output |
|---|---|---|
| Layout | Column/beam grid from Phase 1 architecture | Structural configuration |
| Loads | Dead + live + lateral loads per code | Factored load combinations |
| Prelim. size | Rules of thumb for member dimensions | Trial cross-sections |
| Analysis | Shear, moment, axial diagrams | V(x), M(x), P(x) |
| Design | Size reinforcement; check capacity ≥ demand | Bar schedule and layout |
| Iterate | Adjust if checks fail; optimize if over-designed | Revised member sizes |
| Final | All members pass all code checks | Construction drawings |
- Phase 1 constraints can make Phase 2 very difficult. If the architect forces a column grid that requires 12 m spans with a shallow floor-to-floor height, you're in for a challenging Phase 2. The best structural engineers negotiate Phase 1 early — not after the layout is fixed.
- The iterative nature of design is not a sign that something is wrong. It's the nature of structural engineering. Build the iteration into your process rather than hoping the first trial works.
The Five Assumptions — The Full Picture
Cheat Sheet
| # | Assumption | Key Equation | When It Governs |
|---|---|---|---|
| A1 | Equilibrium — internal = external | \(\sum F = 0,\; \sum M = 0\) | Every FBD, always |
| A2 | Perfect bond — equal strain | \(\varepsilon_s = \varepsilon_c\); \(n = E_s/E_c\) | Transformed section, anchorage |
| A3 | Plane sections remain plane | \(\varepsilon = \phi \cdot y\) (linear) | Any bending calculation |
| A4 | Cracked concrete = zero tension | \(T = A_s f_s\) only | Below N.A. in tension zone |
| A5 | Nonlinear stress-strain governs | \(C = 0.85 f'_c \cdot a \cdot b\) | Strength design, ultimate load |